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https://thatsmaths.com/2018/04/05/fouriers-wonderful-idea-ii/	[ "### Fourier’s Wonderful Idea – II\n\nSolving PDEs by a Roundabout Route", null, "Joseph Fourier (1768-1830)\n\nJoseph Fourier, born just 250 years ago, introduced a wonderful idea that revolutionized science and mathematics: any function or signal can be broken down into simple periodic sine-waves. Radio waves, micro-waves, infra-red radiation, visible light, ultraviolet light, X-rays and gamma rays are all forms of electromagnetic radiation, differing only in frequency [TM136 or search for “thatsmaths” at irishtimes.com].\n\nThe ability to break down signals into components of different frequencies is very useful. Some examples are for tuning a radio to a specific station, using ultrasound to examine a developing foetus or extracting a single phone conversation from a complex combination of inputs.\n\nSolving PDEs\n\nThe equations that govern many physical processes were formulated in the seventeenth century but, although many of them were linear – having the output proportional to the input – nobody knew how to solve them. These partial differential equations remained intractable for one hundred years or more. Around 1807, Joseph Fourier developed a powerful method of expressing the solutions as combinations of simple trigonometric functions and thereby solving the equations. His ideas had a profound effect, and now permeate all areas of mathematical physics and engineering.\n\nThe Essential Idea\n\nFourier showed that (almost) any periodic function can be expressed as a sum of sine and cosine functions. This allowed him to solve many of the equations of physics. Fourier analysis also turned out to be an ideal language for quantum mechanics. Fourier showed that functions that are not periodic, but that become small rapidly outside a bounded range, can be analyzed into periodic components using what is now called the Fourier transform. This transform is like a mathematical prism, breaking up inputs into constituents with different frequencies, just as a prism splits light into colours having different wavelengths.\n\nSolving Problems by a Roundabout Route\n\nPartial differential equations are solved by Fourier analysis via a “detour” from the physical space (or the time domain) to Fourier space (or the frequency domain), where the solution is simple, and then returning to physical space by an inverse transformation of the solution. We might compare this to solving the problem of calculating with Roman numerals: what is LXXXVI multiplied by XLI? Roman accountants and book-keepers had various tables to help them, but it was still a cumbersome process.\n\nWith the Hindu-Arabic numerals to hand, there is an easier way: we convert the two numbers to the “decimal domain”, LXXXVI = 86 and XLI = 41. Then we multiply to get 3526 and convert this back to the “Roman domain” to get MMMDXXVI.\n\nPartial differential equations are solved in an analogous manner: we transform from the time domain, where all the components are entangled, to the frequency domain where they are distinguishable, solve for each component and transform back to the time domain to get the solution.\n\nFaster Computation\n\nWith the development of computers it became possible to analyze complicated signals using Fourier analysis. But the level of calculation becomes rapidly greater with the signal length. The Fast Fourier Transform (FFT), developed around 1965, provided a solution to this problem. It dramatically reduced the way computation scales with problem size, making it practicable to solve many problems that could not otherwise be tackled.\n\nArticle on the Fast Fourier Transform: later." ]	[ null, "https://thatsmaths.files.wordpress.com/2018/04/fourier-3.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9287561,"math_prob":0.9587309,"size":3508,"snap":"2020-45-2020-50","text_gpt3_token_len":684,"char_repetition_ratio":0.1081621,"word_repetition_ratio":0.0036968577,"special_character_ratio":0.18557583,"punctuation_ratio":0.09421488,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98695606,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T02:01:59Z\",\"WARC-Record-ID\":\"<urn:uuid:48044d42-f7b1-447b-9e50-ef193c1ef0d8>\",\"Content-Length\":\"82802\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:641896bc-5c9b-47d5-a72a-9b07b49a1a34>\",\"WARC-Concurrent-To\":\"<urn:uuid:a73b419e-b863-43af-91db-16dd7055f424>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://thatsmaths.com/2018/04/05/fouriers-wonderful-idea-ii/\",\"WARC-Payload-Digest\":\"sha1:4UF4QBZCYHCKWFLCDITACVIN6C36EBUD\",\"WARC-Block-Digest\":\"sha1:HMFZKKEL4WJTKISLHESBSPVE2YFQJKTY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141733120.84_warc_CC-MAIN-20201204010410-20201204040410-00018.warc.gz\"}"}
https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.21/share/doc/Macaulay2/GKMVarieties/html/_make__K__Class_lp__G__K__M__Variety_cm__Flag__Matroid_rp.html	[ "# makeKClass(GKMVariety,FlagMatroid) -- the equivariant K-class of a flag matroid\n\n## Description\n\nA flag matroid of whose constituent matroids have ranks $r_1, \\ldots, r_k$ and ground set size $n$ defines a KClass on the (partial) flag variety $Fl(r_1,\\ldots, r_k;n)$. When the flag matroid arises from a matrix representing a point on the (partial) flag variety, this equivariant K-class coincides with that of the structure sheaf of its torus orbit closure. See [CDMS18] or [DES20].\n\n i1 : X = generalizedFlagVariety(\"A\",2,{1,2}) o1 = a \"GKM variety\" with an action of a 3-dimensional torus o1 : GKMVariety i2 : A = matrix{{1,2,3},{0,2,3}} o2 = \| 1 2 3 \| \| 0 2 3 \| 2 3 o2 : Matrix ZZ <--- ZZ i3 : FM = flagMatroid(A,{1,2}) o3 = a \"flag matroid\" with rank sequence {1, 2} on 3 elements o3 : FlagMatroid i4 : C1 = makeKClass(X,FM) o4 = an \"equivariant K-class\" on a GKM variety o4 : KClass i5 : C2 = orbitClosure(X,A) o5 = an \"equivariant K-class\" on a GKM variety o5 : KClass i6 : C1 === C2 o6 = true" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.70636153,"math_prob":0.99895746,"size":913,"snap":"2023-14-2023-23","text_gpt3_token_len":335,"char_repetition_ratio":0.11771177,"word_repetition_ratio":0.046511628,"special_character_ratio":0.37458926,"punctuation_ratio":0.16582915,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99376523,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T13:15:54Z\",\"WARC-Record-ID\":\"<urn:uuid:6bf14c24-7228-4687-acd7-a9507c461dbd>\",\"Content-Length\":\"7960\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:03c89289-899e-43d3-8c87-60f91755d004>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c2de02c-341e-4139-831f-7f3781b5c740>\",\"WARC-IP-Address\":\"128.174.199.46\",\"WARC-Target-URI\":\"https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.21/share/doc/Macaulay2/GKMVarieties/html/_make__K__Class_lp__G__K__M__Variety_cm__Flag__Matroid_rp.html\",\"WARC-Payload-Digest\":\"sha1:4T7UZNYGACXCMFH6LSJOCUGZISF6UJV5\",\"WARC-Block-Digest\":\"sha1:EFTOODGR4YBVWTAJROC7JKVEINPYVGWK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644855.6_warc_CC-MAIN-20230529105815-20230529135815-00592.warc.gz\"}"}
https://aws-amplify.github.io/aws-sdk-ios/docs/reference/AWSTextract/Classes/AWSTextractPoint.html	[ "# AWSTextractPoint\n\nObjective-C\n\n``@interface AWSTextractPoint``\n\nSwift\n\n``class AWSTextractPoint``\n\nThe X and Y coordinates of a point on a document page. The X and Y values that are returned are ratios of the overall document page size. For example, if the input document is 700 x 200 and the operation returns X=0.5 and Y=0.25, then the point is at the (350,50) pixel coordinate on the document page.\n\nAn array of `Point` objects, `Polygon`, is returned by DetectDocumentText. `Polygon` represents a fine-grained polygon around detected text. For more information, see Geometry in the Amazon Textract Developer Guide.\n\n• ``` X ```\n\nThe value of the X coordinate for a point on a `Polygon`.\n\n#### Declaration\n\nObjective-C\n\n``@property (nonatomic, strong) NSNumber _Nullable X;``\n\nSwift\n\n``var x: NSNumber? { get set }``\n• ``` Y ```\n\nThe value of the Y coordinate for a point on a `Polygon`.\n\n#### Declaration\n\nObjective-C\n\n``@property (nonatomic, strong) NSNumber _Nullable Y;``\n\nSwift\n\n``var y: NSNumber? { get set }``" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6519817,"math_prob":0.8258656,"size":939,"snap":"2021-04-2021-17","text_gpt3_token_len":229,"char_repetition_ratio":0.11336898,"word_repetition_ratio":0.16107382,"special_character_ratio":0.23109691,"punctuation_ratio":0.13407822,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9747817,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T13:44:33Z\",\"WARC-Record-ID\":\"<urn:uuid:d4aadcaa-221d-438d-86ec-f666520330ce>\",\"Content-Length\":\"15764\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09727d35-245c-48f5-a420-ad3685d1cf52>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb62cbb2-59de-4a6c-907b-1e8d3d6a40dc>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://aws-amplify.github.io/aws-sdk-ios/docs/reference/AWSTextract/Classes/AWSTextractPoint.html\",\"WARC-Payload-Digest\":\"sha1:V6BXU2C23TL5Y6GJZA4MF6VGGFJ747TH\",\"WARC-Block-Digest\":\"sha1:D3YTHBFMZSGIVPBDWXGORICOQV3WSJDO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703524858.74_warc_CC-MAIN-20210121132407-20210121162407-00522.warc.gz\"}"}
http://pike.lysator.liu.se/docs/ietf/rfc/13/rfc1320.xml	[ "Network Working Group\nObsoletes: RFC 1186\nR. Rivest\nMIT Laboratory for Computer Science\nand RSA Data Security, Inc.\nApril 1992\n\n# The MD4 Message-Digest Algorithm\n\n## Status of thie Memo\n\nThis memo provides information for the Internet community. It does not specify an Internet standard. Distribution of this memo is unlimited.\n\n## Acknowlegements\n\nWe would like to thank Don Coppersmith, Burt Kaliski, Ralph Merkle, and Noam Nisan for numerous helpful comments and suggestions.\n\n``` 1. Executive Summary 1\n2. Terminology and Notation 2\n3. MD4 Algorithm Description 2\n4. Summary 6\nReferences 6\nAPPENDIX A - Reference Implementation 6\nSecurity Considerations 20\n```\n\n## 1. Executive Summary\n\nThis document describes the MD4 message-digest algorithm . The algorithm takes as input a message of arbitrary length and produces as output a 128-bit \"fingerprint\" or \"message digest\" of the input. It is conjectured that it is computationally infeasible to produce two messages having the same message digest, or to produce any message having a given prespecified target message digest. The MD4 algorithm is intended for digital signature applications, where a large file must be \"compressed\" in a secure manner before being encrypted with a private (secret) key under a public-key cryptosystem such as RSA.\n\nThe MD4 algorithm is designed to be quite fast on 32-bit machines. In addition, the MD4 algorithm does not require any large substitution tables; the algorithm can be coded quite compactly.\n\nThe MD4 algorithm is being placed in the public domain for review and possible adoption as a standard.\n\nThis document replaces the October 1990 RFC 1186 . The main difference is that the reference implementation of MD4 in the appendix is more portable.\n\nFor OSI-based applications, MD4's object identifier is\n\nmd4 OBJECT IDENTIFIER ::=\n\n``` {iso(1) member-body(2) US(840) rsadsi(113549) digestAlgorithm(2) 4}\n```\n\nIn the X.509 type AlgorithmIdentifier , the parameters for MD4 should have type NULL.\n\n## 2. Terminology and Notation\n\nIn this document a \"word\" is a 32-bit quantity and a \"byte\" is an eight-bit quantity. A sequence of bits can be interpreted in a natural manner as a sequence of bytes, where each consecutive group of eight bits is interpreted as a byte with the high-order (most significant) bit of each byte listed first. Similarly, a sequence of bytes can be interpreted as a sequence of 32-bit words, where each consecutive group of four bytes is interpreted as a word with the low-order (least significant) byte given first.\n\nLet x_i denote \"x sub i\". If the subscript is an expression, we surround it in braces, as in x_{i+1}. Similarly, we use ^ for superscripts (exponentiation), so that x^i denotes x to the i-th power.\n\nLet the symbol \"+\" denote addition of words (i.e., modulo-2^32 addition). Let X <<< s denote the 32-bit value obtained by circularly shifting (rotating) X left by s bit positions. Let not(X) denote the bit-wise complement of X, and let X v Y denote the bit-wise OR of X and Y. Let X xor Y denote the bit-wise XOR of X and Y, and let XY denote the bit-wise AND of X and Y.\n\n## 3. MD4 Algorithm Description\n\nWe begin by supposing that we have a b-bit message as input, and that we wish to find its message digest. Here b is an arbitrary nonnegative integer; b may be zero, it need not be a multiple of eight, and it may be arbitrarily large. We imagine the bits of the message written down as follows:\n\n``` m_0 m_1 ... m_{b-1}\n```\n\nThe following five steps are performed to compute the message digest of the message.\n\n### 3.1 Step 1. Append Padding Bits\n\nThe message is \"padded\" (extended) so that its length (in bits) is congruent to 448, modulo 512. That is, the message is extended so that it is just 64 bits shy of being a multiple of 512 bits long. Padding is always performed, even if the length of the message is already congruent to 448, modulo 512.\n\nPadding is performed as follows: a single \"1\" bit is appended to the message, and then \"0\" bits are appended so that the length in bits of the padded message becomes congruent to 448, modulo 512. In all, at least one bit and at most 512 bits are appended.\n\n### 3.2 Step 2. Append Length\n\nA 64-bit representation of b (the length of the message before the padding bits were added) is appended to the result of the previous step. In the unlikely event that b is greater than 2^64, then only the low-order 64 bits of b are used. (These bits are appended as two 32-bit words and appended low-order word first in accordance with the previous conventions.)\n\nAt this point the resulting message (after padding with bits and with b) has a length that is an exact multiple of 512 bits. Equivalently, this message has a length that is an exact multiple of 16 (32-bit) words. Let M[0 ... N-1] denote the words of the resulting message, where N is a multiple of 16.\n\n### 3.3 Step 3. Initialize MD Buffer\n\nA four-word buffer (A,B,C,D) is used to compute the message digest. Here each of A, B, C, D is a 32-bit register. These registers are initialized to the following values in hexadecimal, low-order bytes first):\n\n``` word A: 01 23 45 67\nword B: 89 ab cd ef\nword C: fe dc ba 98\nword D: 76 54 32 10\n```\n\n### 3.4 Step 4. Process Message in 16-Word Blocks\n\nWe first define three auxiliary functions that each take as input three 32-bit words and produce as output one 32-bit word.\n\n``` F(X,Y,Z) = XY v not(X) Z\nG(X,Y,Z) = XY v XZ v YZ\nH(X,Y,Z) = X xor Y xor Z\n```\n\nIn each bit position F acts as a conditional: if X then Y else Z. The function F could have been defined using + instead of v since XY and not(X)Z will never have \"1\" bits in the same bit position.) In each bit position G acts as a majority function: if at least two of X, Y, Z are on, then G has a \"1\" bit in that bit position, else G has a \"0\" bit. It is interesting to note that if the bits of X, Y, and Z are independent and unbiased, the each bit of f(X,Y,Z) will be independent and unbiased, and similarly each bit of g(X,Y,Z) will be independent and unbiased. The function H is the bit-wise XOR or parity\" function; it has properties similar to those of F and G.\n\nDo the following:\n\n``` Process each 16-word block. /\nFor i = 0 to N/16-1 do\n\n/ Copy block i into X. /\nFor j = 0 to 15 do\nSet X[j] to M[i16+j].\nend /* of loop on j /\n\n/ Save A as AA, B as BB, C as CC, and D as DD. /\nAA = A\nBB = B\nCC = C\nDD = D\n\n/ Round 1. /\n/ Let [abcd k s] denote the operation\na = (a + F(b,c,d) + X[k]) <<< s. /\n/ Do the following 16 operations. /\n[ABCD 0 3] [DABC 1 7] [CDAB 2 11] [BCDA 3 19]\n[ABCD 4 3] [DABC 5 7] [CDAB 6 11] [BCDA 7 19]\n[ABCD 8 3] [DABC 9 7] [CDAB 10 11] [BCDA 11 19]\n[ABCD 12 3] [DABC 13 7] [CDAB 14 11] [BCDA 15 19]\n\n/ Round 2. /\n/ Let [abcd k s] denote the operation\na = (a + G(b,c,d) + X[k] + 5A827999) <<< s. /\n\n/ Do the following 16 operations. /\n[ABCD 0 3] [DABC 4 5] [CDAB 8 9] [BCDA 12 13]\n[ABCD 1 3] [DABC 5 5] [CDAB 9 9] [BCDA 13 13]\n[ABCD 2 3] [DABC 6 5] [CDAB 10 9] [BCDA 14 13]\n[ABCD 3 3] [DABC 7 5] [CDAB 11 9] [BCDA 15 13]\n\n/ Round 3. /\n/ Let [abcd k s] denote the operation\na = (a + H(b,c,d) + X[k] + 6ED9EBA1) <<< s. /\n/ Do the following 16 operations. /\n[ABCD 0 3] [DABC 8 9] [CDAB 4 11] [BCDA 12 15]\n[ABCD 2 3] [DABC 10 9] [CDAB 6 11] [BCDA 14 15]\n[ABCD 1 3] [DABC 9 9] [CDAB 5 11] [BCDA 13 15]\n[ABCD 3 3] [DABC 11 9] [CDAB 7 11] [BCDA 15 15]\n\n/ Then perform the following additions. (That is, increment each\nof the four registers by the value it had before this block\nwas started.) /\nA = A + AA\nB = B + BB\nC = C + CC\nD = D + DD\n\nend / of loop on i /\n```\n\nNote. The value 5A..99 is a hexadecimal 32-bit constant, written with the high-order digit first. This constant represents the square root of 2. The octal value of this constant is 013240474631.\n\nThe value 6E..A1 is a hexadecimal 32-bit constant, written with the high-order digit first. This constant represents the square root of 3. The octal value of this constant is 015666365641.\n\n``` See Knuth, The Art of Programming, Volume 2 (Seminumerical\nAlgorithms), Second Edition (1981), Addison-Wesley. Table 2, page\n\n660.\n```\n\n### 3.5 Step 5. Output\n\nThe message digest produced as output is A, B, C, D. That is, we begin with the low-order byte of A, and end with the high-order byte of D.\n\nThis completes the description of MD4. A reference implementation in C is given in the appendix.\n\n## 4. Summary\n\nThe MD4 message-digest algorithm is simple to implement, and provides a \"fingerprint\" or message digest of a message of arbitrary length. It is conjectured that the difficulty of coming up with two messages having the same message digest is on the order of 2^64 operations, and that the difficulty of coming up with any message having a given message digest is on the order of 2^128 operations. The MD4 algorithm has been carefully scrutinized for weaknesses. It is, however, a relatively new algorithm and further security analysis is of course justified, as is the case with any new proposal of this sort.\n\n## References\n\n``` Rivest, R., \"The MD4 message digest algorithm\", in A.J. Menezes\nand S.A. Vanstone, editors, Advances in Cryptology - CRYPTO '90\nProceedings, pages 303-311, Springer-Verlag, 1991.\n\n Rivest, R., \"The MD4 Message Digest Algorithm\", RFC 1186, MIT,\nOctober 1990.\n\n CCITT Recommendation X.509 (1988), \"The Directory -\nAuthentication Framework\".\n\n Rivest, R., \"The MD5 Message-Digest Algorithm\", RFC 1321, MIT and\nRSA Data Security, Inc, April 1992.\n```\n\n## APPENDIX A - Reference Implementation\n\nThis appendix contains the following files:\n\n``` global.h -- global header file\n\nmd4.h -- header file for MD4\n\nmd4c.c -- source code for MD4\n\nmddriver.c -- test driver for MD2, MD4 and MD5\n```\n\nThe driver compiles for MD5 by default but can compile for MD2 or MD4 if the symbol MD is defined on the C compiler command line as 2 or 4.\n\nThe implementation is portable and should work on many different plaforms. However, it is not difficult to optimize the implementation on particular platforms, an exercise left to the reader. For example, on \"little-endian\" platforms where the lowest-addressed byte in a 32- bit word is the least significant and there are no alignment restrictions, the call to Decode in MD4Transform can be replaced with a typecast.\n\n## /\n\n```/* PROTOTYPES should be set to one if and only if the compiler supports\nfunction argument prototyping.\nThe following makes PROTOTYPES default to 0 if it has not already\nbeen defined with C compiler flags.\n/\n#ifndef PROTOTYPES\n#define PROTOTYPES 0\n#endif\n```\n\n## typedef unsigned long int UINT4;\n\n```/ PROTO_LIST is defined depending on how PROTOTYPES is defined above.\n\nIf using PROTOTYPES, then PROTO_LIST returns the list, otherwise it\nreturns an empty list.\n/\n```\n\n#if PROTOTYPES\n#define PROTO_LIST(list) list\n#else\n#define PROTO_LIST(list) ()\n#endif\n\n## rights reserved.\n\nLicense to copy and use this software is granted provided that it is identified as the \"RSA Data Security, Inc. MD4 Message-Digest Algorithm\" in all material mentioning or referencing this software or this function.\n\nLicense is also granted to make and use derivative works provided that such works are identified as \"derived from the RSA Data Security, Inc. MD4 Message-Digest Algorithm\" in all material mentioning or referencing the derived work.\n\nRSA Data Security, Inc. makes no representations concerning either the merchantability of this software or the suitability of this software for any particular purpose. It is provided \"as is\" without express or implied warranty of any kind.\n\n``` These notices must be retained in any copies of any part of this\ndocumentation and/or software.\n/\n\n/* MD4 context. /\ntypedef struct {\nUINT4 state; / state (ABCD) /\nUINT4 count; / number of bits, modulo 2^64 (lsb first) /\nunsigned char buffer; / input buffer /\n} MD4_CTX;\n\nvoid MD4Init PROTO_LIST ((MD4_CTX ));\nvoid MD4Update PROTO_LIST\n((MD4_CTX , unsigned char , unsigned int));\nvoid MD4Final PROTO_LIST ((unsigned char , MD4_CTX ));\n```\n\n## /\n\nLicense to copy and use this software is granted provided that it is identified as the \"RSA Data Security, Inc. MD4 Message-Digest Algorithm\" in all material mentioning or referencing this software or this function.\n\nLicense is also granted to make and use derivative works provided that such works are identified as \"derived from the RSA Data Security, Inc. MD4 Message-Digest Algorithm\" in all material mentioning or referencing the derived work.\n\nRSA Data Security, Inc. makes no representations concerning either the merchantability of this software or the suitability of this software for any particular purpose. It is provided \"as is\" without express or implied warranty of any kind.\n\n``` These notices must be retained in any copies of any part of this\ndocumentation and/or software.\n/\n```\n\n## #include \"md4.h\"\n\n```/ Constants for MD4Transform routine.\n/\n#define S11 3\n#define S12 7\n#define S13 11\n#define S14 19\n#define S21 3\n#define S22 5\n#define S23 9\n#define S24 13\n#define S31 3\n#define S32 9\n#define S33 11\n#define S34 15\n\nstatic void MD4Transform PROTO_LIST ((UINT4 , unsigned char ));\nstatic void Encode PROTO_LIST\n((unsigned char , UINT4 , unsigned int));\nstatic void Decode PROTO_LIST\n((UINT4 , unsigned char , unsigned int));\nstatic void MD4_memcpy PROTO_LIST ((POINTER, POINTER, unsigned int));\nstatic void MD4_memset PROTO_LIST ((POINTER, int, unsigned int));\n\nstatic unsigned char PADDING = {\n0x80, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,\n0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0\n};\n\n/ F, G and H are basic MD4 functions.\n/\n#define F(x, y, z) (((x) & (y)) \| ((~x) & (z)))\n#define G(x, y, z) (((x) & (y)) \| ((x) & (z)) \| ((y) & (z)))\n#define H(x, y, z) ((x) ^ (y) ^ (z))\n\n/ ROTATE_LEFT rotates x left n bits.\n/\n#define ROTATE_LEFT(x, n) (((x) << (n)) \| ((x) >> (32-(n))))\n```\n\n## / Rotation is separate from addition to prevent recomputation /\n\n```#define FF(a, b, c, d, x, s) { \\\n(a) += F ((b), (c), (d)) + (x); \\\n(a) = ROTATE_LEFT ((a), (s)); \\\n}\n#define GG(a, b, c, d, x, s) { \\\n(a) += G ((b), (c), (d)) + (x) + (UINT4)0x5a827999; \\\n(a) = ROTATE_LEFT ((a), (s)); \\\n}\n#define HH(a, b, c, d, x, s) { \\\n(a) += H ((b), (c), (d)) + (x) + (UINT4)0x6ed9eba1; \\\n(a) = ROTATE_LEFT ((a), (s)); \\\n}\n\n/ MD4 initialization. Begins an MD4 operation, writing a new context.\n/\nvoid MD4Init (context)\nMD4_CTX context; /* context /\n{\ncontext->count = context->count = 0;\n\n/\ncontext->state = 0x67452301;\ncontext->state = 0xefcdab89;\ncontext->state = 0x10325476;\n}\n\n/* MD4 block update operation. Continues an MD4 message-digest\noperation, processing another message block, and updating the\ncontext.\n/\nvoid MD4Update (context, input, inputLen)\nMD4_CTX context; /* context /\nunsigned char input; /* input block /\nunsigned int inputLen; / length of input block /\n{\nunsigned int i, index, partLen;\n\n/ Compute number of bytes mod 64 /\nindex = (unsigned int)((context->count >> 3) & 0x3F);\n/ Update number of bits /\nif ((context->count += ((UINT4)inputLen << 3))\n< ((UINT4)inputLen << 3))\ncontext->count++;\ncontext->count += ((UINT4)inputLen >> 29);\n\npartLen = 64 - index;\n/ Transform as many times as possible.\n/\nif (inputLen >= partLen) {\nMD4_memcpy\n((POINTER)&context->buffer[index], (POINTER)input, partLen);\nMD4Transform (context->state, context->buffer);\n\nfor (i = partLen; i + 63 < inputLen; i += 64)\nMD4Transform (context->state, &input[i]);\n\nindex = 0;\n}\nelse\ni = 0;\n\n/ Buffer remaining input /\nMD4_memcpy\n((POINTER)&context->buffer[index], (POINTER)&input[i],\ninputLen-i);\n}\n\n/ MD4 finalization. Ends an MD4 message-digest operation, writing the\nthe message digest and zeroizing the context.\n/\nvoid MD4Final (digest, context)\nunsigned char digest; / message digest /\nMD4_CTX context; /* context /\n{\nunsigned char bits;\n\n/ Save number of bits /\nEncode (bits, context->count, 8);\n\n/ Pad out to 56 mod 64.\n/\nindex = (unsigned int)((context->count >> 3) & 0x3f);\npadLen = (index < 56) ? (56 - index) : (120 - index);\n\n/ Append length (before padding) /\nMD4Update (context, bits, 8);\n/ Store state in digest /\nEncode (digest, context->state, 16);\n\n/ Zeroize sensitive information.\n/\nMD4_memset ((POINTER)context, 0, sizeof (context));\n```\n\n## }\n\n```/* MD4 basic transformation. Transforms state based on block.\n/\nstatic void MD4Transform (state, block)\nUINT4 state;\nunsigned char block;\n{\nUINT4 a = state, b = state, c = state, d = state, x;\n\nDecode (x, block, 64);\n\n/ Round 1 /\nFF (a, b, c, d, x[ 0], S11); / 1 /\nFF (d, a, b, c, x[ 1], S12); / 2 /\nFF (c, d, a, b, x[ 2], S13); / 3 /\nFF (b, c, d, a, x[ 3], S14); / 4 /\nFF (a, b, c, d, x[ 4], S11); / 5 /\nFF (d, a, b, c, x[ 5], S12); / 6 /\nFF (c, d, a, b, x[ 6], S13); / 7 /\nFF (b, c, d, a, x[ 7], S14); / 8 /\nFF (a, b, c, d, x[ 8], S11); / 9 /\nFF (d, a, b, c, x[ 9], S12); / 10 /\nFF (c, d, a, b, x, S13); / 11 /\nFF (b, c, d, a, x, S14); / 12 /\nFF (a, b, c, d, x, S11); / 13 /\nFF (d, a, b, c, x, S12); / 14 /\nFF (c, d, a, b, x, S13); / 15 /\nFF (b, c, d, a, x, S14); / 16 /\n\n/ Round 2 /\nGG (a, b, c, d, x[ 0], S21); / 17 /\nGG (d, a, b, c, x[ 4], S22); / 18 /\nGG (c, d, a, b, x[ 8], S23); / 19 /\nGG (b, c, d, a, x, S24); / 20 /\nGG (a, b, c, d, x[ 1], S21); / 21 /\nGG (d, a, b, c, x[ 5], S22); / 22 /\nGG (c, d, a, b, x[ 9], S23); / 23 /\nGG (b, c, d, a, x, S24); / 24 /\nGG (a, b, c, d, x[ 2], S21); / 25 /\nGG (d, a, b, c, x[ 6], S22); / 26 /\nGG (c, d, a, b, x, S23); / 27 /\nGG (b, c, d, a, x, S24); / 28 /\nGG (a, b, c, d, x[ 3], S21); / 29 /\nGG (d, a, b, c, x[ 7], S22); / 30 /\nGG (c, d, a, b, x, S23); / 31 /\nGG (b, c, d, a, x, S24); / 32 /\n/ Round 3 /\nHH (a, b, c, d, x[ 0], S31); / 33 /\nHH (d, a, b, c, x[ 8], S32); / 34 /\nHH (c, d, a, b, x[ 4], S33); / 35 /\nHH (b, c, d, a, x, S34); / 36 /\nHH (a, b, c, d, x[ 2], S31); / 37 /\nHH (d, a, b, c, x, S32); / 38 /\nHH (c, d, a, b, x[ 6], S33); / 39 /\nHH (b, c, d, a, x, S34); / 40 /\nHH (a, b, c, d, x[ 1], S31); / 41 /\nHH (d, a, b, c, x[ 9], S32); / 42 /\nHH (c, d, a, b, x[ 5], S33); / 43 /\nHH (b, c, d, a, x, S34); / 44 /\nHH (a, b, c, d, x[ 3], S31); / 45 /\nHH (d, a, b, c, x, S32); / 46 /\nHH (c, d, a, b, x[ 7], S33); / 47 /\nHH (b, c, d, a, x, S34); / 48 /\n\nstate += a;\nstate += b;\nstate += c;\nstate += d;\n\n/ Zeroize sensitive information.\n/\nMD4_memset ((POINTER)x, 0, sizeof (x));\n}\n\n/ Encodes input (UINT4) into output (unsigned char). Assumes len is\na multiple of 4.\n/\nstatic void Encode (output, input, len)\nunsigned char output;\nUINT4 input;\nunsigned int len;\n{\nunsigned int i, j;\n\nfor (i = 0, j = 0; j < len; i++, j += 4) {\noutput[j] = (unsigned char)(input[i] & 0xff);\noutput[j+1] = (unsigned char)((input[i] >> 8) & 0xff);\noutput[j+2] = (unsigned char)((input[i] >> 16) & 0xff);\noutput[j+3] = (unsigned char)((input[i] >> 24) & 0xff);\n}\n}\n```\n\n## static void Decode (output, input, len)\n\n```UINT4 output;\nunsigned char input;\nunsigned int len;\n{\nunsigned int i, j;\n\nfor (i = 0, j = 0; j < len; i++, j += 4)\noutput[i] = ((UINT4)input[j]) \| (((UINT4)input[j+1]) << 8) \|\n(((UINT4)input[j+2]) << 16) \| (((UINT4)input[j+3]) << 24);\n}\n\n/ Note: Replace \"for loop\" with standard memcpy if possible.\n/\nstatic void MD4_memcpy (output, input, len)\nPOINTER output;\nPOINTER input;\nunsigned int len;\n{\nunsigned int i;\n\nfor (i = 0; i < len; i++)\noutput[i] = input[i];\n}\n\n/ Note: Replace \"for loop\" with standard memset if possible.\n/\nstatic void MD4_memset (output, value, len)\nPOINTER output;\nint value;\nunsigned int len;\n{\nunsigned int i;\n\nfor (i = 0; i < len; i++)\n((char )output)[i] = (char)value;\n}\n```\n\n## rights reserved.\n\nRSA Data Security, Inc. makes no representations concerning either the merchantability of this software or the suitability of this software for any particular purpose. It is provided \"as is\" without express or implied warranty of any kind.\n\nThese notices must be retained in any copies of any part of this documentation and/or software.\n\n``` /\n\n/ The following makes MD default to MD5 if it has not already been\ndefined with C compiler flags.\n/\n#ifndef MD\n#define MD MD5\n#endif\n```\n\n#include <stdio.h>\n#include <time.h>\n#include <string.h>\n#include \"global.h\"\n#if MD == 2\n#include \"md2.h\"\n#endif\n#if MD == 4\n#include \"md4.h\"\n#endif\n#if MD == 5\n#include \"md5.h\"\n#endif\n\n```/ Length of test block, number of test blocks.\n/\n#define TEST_BLOCK_LEN 1000\n#define TEST_BLOCK_COUNT 1000\n\nstatic void MDString PROTO_LIST ((char ));\nstatic void MDTimeTrial PROTO_LIST ((void));\nstatic void MDTestSuite PROTO_LIST ((void));\nstatic void MDFile PROTO_LIST ((char ));\nstatic void MDFilter PROTO_LIST ((void));\nstatic void MDPrint PROTO_LIST ((unsigned char ));\n```\n\n#if MD == 2\n#define MD_CTX MD2_CTX\n#define MDInit MD2Init\n#define MDUpdate MD2Update\n#define MDFinal MD2Final\n\n#endif\n#if MD == 4\n#define MD_CTX MD4_CTX\n#define MDInit MD4Init\n#define MDUpdate MD4Update\n#define MDFinal MD4Final\n#endif\n#if MD == 5\n#define MD_CTX MD5_CTX\n#define MDInit MD5Init\n#define MDUpdate MD5Update\n#define MDFinal MD5Final\n#endif\n\n## / Main driver.\n\n``` Arguments (may be any combination):\n-sstring - digests string\n-t - runs time trial\n-x - runs test script\nfilename - digests file\n(none) - digests standard input\n/\nint main (argc, argv)\nint argc;\nchar argv[];\n{\nint i;\n\nif (argc > 1)\nfor (i = 1; i < argc; i++)\nif (argv[i] == '-' && argv[i] == 's')\nMDString (argv[i] + 2);\nelse if (strcmp (argv[i], \"-t\") == 0)\nMDTimeTrial ();\nelse if (strcmp (argv[i], \"-x\") == 0)\nMDTestSuite ();\nelse\nMDFile (argv[i]);\nelse\nMDFilter ();\n```\n\n## }\n\n```/* Digests a string and prints the result.\n/\nstatic void MDString (string)\n```\n\nchar string;\n{\n\n``` MD_CTX context;\nunsigned char digest;\nunsigned int len = strlen (string);\n\nMDInit (&context);\nMDUpdate (&context, string, len);\nMDFinal (digest, &context);\n\nprintf (\"MD%d (\\\"%s\\\") = \", MD, string);\nMDPrint (digest);\nprintf (\"\\n\");\n}\n\n/* Measures the time to digest TEST_BLOCK_COUNT TEST_BLOCK_LEN-byte\nblocks.\n/\nstatic void MDTimeTrial ()\n{\nMD_CTX context;\ntime_t endTime, startTime;\nunsigned char block[TEST_BLOCK_LEN], digest;\nunsigned int i;\n```\n\nprintf\n\n(\"MD%d time trial. Digesting %d %d-byte blocks ...\", MD,\n\n``` TEST_BLOCK_LEN, TEST_BLOCK_COUNT);\n\n/ Initialize block /\nfor (i = 0; i < TEST_BLOCK_LEN; i++)\nblock[i] = (unsigned char)(i & 0xff);\n\n/ Start timer /\ntime (&startTime);\n\n/ Digest blocks /\nMDInit (&context);\nfor (i = 0; i < TEST_BLOCK_COUNT; i++)\nMDUpdate (&context, block, TEST_BLOCK_LEN);\nMDFinal (digest, &context);\n\n/ Stop timer /\ntime (&endTime);\n\nprintf (\" done\\n\");\nprintf (\"Digest = \");\nMDPrint (digest);\n\nprintf (\"\\nTime = %ld seconds\\n\", (long)(endTime-startTime));\nprintf\n(\"Speed = %ld bytes/second\\n\",\n(long)TEST_BLOCK_LEN (long)TEST_BLOCK_COUNT/(endTime-startTime));\n}\n\n/* Digests a reference suite of strings and prints the results.\n/\nstatic void MDTestSuite ()\n{\nprintf (\"MD%d test suite:\\n\", MD);\n\nMDString (\"\");\nMDString (\"a\");\nMDString (\"abc\");\nMDString (\"message digest\");\nMDString (\"abcdefghijklmnopqrstuvwxyz\");\nMDString\n(\"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789\");\nMDString\n\n(\"1234567890123456789012345678901234567890\\\n1234567890123456789012345678901234567890\");\n}\n\n/ Digests a file and prints the result.\n/\nstatic void MDFile (filename)\nchar filename;\n{\nFILE file;\nMD_CTX context;\nint len;\nunsigned char buffer, digest;\n\nif ((file = fopen (filename, \"rb\")) == NULL)\nprintf (\"%s can't be opened\\n\", filename);\n```\n\nelse {\n\n``` MDInit (&context);\nwhile (len = fread (buffer, 1, 1024, file))\nMDUpdate (&context, buffer, len);\nMDFinal (digest, &context);\n\nfclose (file);\n\nprintf (\"MD%d (%s) = \", MD, filename);\nMDPrint (digest);\n\nprintf (\"\\n\");\n}\n}\n\n/ Digests the standard input and prints the result.\n/\nstatic void MDFilter ()\n{\nMD_CTX context;\nint len;\nunsigned char buffer, digest;\n\nMDInit (&context);\nwhile (len = fread (buffer, 1, 16, stdin))\nMDUpdate (&context, buffer, len);\nMDFinal (digest, &context);\n\nMDPrint (digest);\nprintf (\"\\n\");\n}\n\n/ Prints a message digest in hexadecimal.\n*/\nstatic void MDPrint (digest)\nunsigned char digest;\n```\n\n## unsigned int i;\n\n``` for (i = 0; i < 16; i++)\nprintf (\"%02x\", digest[i]);\n}\n```\n\n### A.5 Test suite\n\nThe MD4 test suite (driver option \"-x\") should print the following results:\n\nMD4 test suite:\nMD4 (\"\") = 31d6cfe0d16ae931b73c59d7e0c089c0\nMD4 (\"a\") = bde52cb31de33e46245e05fbdbd6fb24\nMD4 (\"abc\") = a448017aaf21d8525fc10ae87aa6729d\nMD4 (\"message digest\") = d9130a8164549fe818874806e1c7014b MD4 (\"abcdefghijklmnopqrstuvwxyz\") = d79e1c308aa5bbcdeea8ed63df412da9 MD4 (\"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789\") = 043f8582f241db351ce627e153e7f0e4 MD4 (\"123456789012345678901234567890123456789012345678901234567890123456 78901234567890\") = e33b4ddc9c38f2199c3e7b164fcc0536\n\n## Security Considerations\n\nThe level of security discussed in this memo is considered to be sufficient for implementing moderate security hybrid digital- signature schemes based on MD4 and a public-key cryptosystem. We do not know of any reason that MD4 would not be sufficient for implementing very high security digital-signature schemes, but because MD4 was designed to be exceptionally fast, it is \"at the edge\" in terms of risking successful cryptanalytic attack. After further critical review, it may be appropriate to consider MD4 for very high security applications. For very high security applications before the completion of that review, the MD5 algorithm is recommended.\n\n``` Ronald L. 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https://rdrr.io/cran/timetk/man/slidify.html	[ "# slidify: Create a rolling (sliding) version of any function In timetk: A Tool Kit for Working with Time Series in R\n\n## Description\n\n`slidify` returns a rolling (sliding) version of the input function, with a rolling (sliding) `.period` specified by the user.\n\n## Usage\n\n ```1 2 3 4 5 6 7``` ```slidify( .f, .period = 1, .align = c(\"center\", \"left\", \"right\"), .partial = FALSE, .unlist = TRUE ) ```\n\n## Arguments\n\n `.f` A function, formula, or vector (not necessarily atomic). If a function, it is used as is. If a formula, e.g. `~ .x + 2`, it is converted to a function. There are three ways to refer to the arguments: For a single argument function, use `.` For a two argument function, use `.x` and `.y` For more arguments, use `..1`, `..2`, `..3` etc This syntax allows you to create very compact anonymous functions. If character vector, numeric vector, or list, it is converted to an extractor function. Character vectors index by name and numeric vectors index by position; use a list to index by position and name at different levels. If a component is not present, the value of `.default` will be returned. `.period` The period size to roll over `.align` One of \"center\", \"left\" or \"right\". `.partial` Should the moving window be allowed to return partial (incomplete) windows instead of `NA` values. Set to FALSE by default, but can be switched to TRUE to remove `NA`'s. `.unlist` If the function returns a single value each time it is called, use `.unlist = TRUE`. If the function returns more than one value, or a more complicated object (like a linear model), use `.unlist = FALSE` to create a list-column of the rolling results.\n\n## Details\n\nThe `slidify()` function is almost identical to `tibbletime::rollify()` with 3 improvements:\n\n1. Alignment (\"center\", \"left\", \"right\")\n\n2. Partial windows are allowed\n\n3. Uses `slider` under the hood, which improves speed and reliability by implementing code at C++ level\n\nMake any function a Sliding (Rolling) Function\n\n`slidify()` turns a function into a sliding version of itself for use inside of a call to `dplyr::mutate()`, however it works equally as well when called from `purrr::map()`.\n\nBecause of it's intended use with `dplyr::mutate()`, `slidify` creates a function that always returns output with the same length of the input\n\nAlignment\n\nRolling / Sliding functions generate `.period - 1` fewer values than the incoming vector. Thus, the vector needs to be aligned. Alignment of the vector follows 3 types:\n\n• center (default): `NA` or `.partial` values are divided and added to the beginning and end of the series to \"Center\" the moving average. This is common in Time Series applications (e.g. denoising).\n\n• left: `NA` or `.partial` values are added to the end to shift the series to the Left.\n\n• right: `NA` or `.partial` values are added to the beginning to shift the series to the Right. This is common in Financial Applications (e.g moving average cross-overs).\n\nAllowing Partial Windows\n\nA key improvement over `tibbletime::slidify()` is that `timetk::slidify()` implements `.partial` rolling windows. Just set `.partial = TRUE`.\n\n## References\n\n• The Tibbletime R Package by Davis Vaughan, which includes the original `rollify()` Function\n\nTransformation Functions:\n\n• `slidify_vec()` - A simple vectorized function for applying summary functions to rolling windows.\n\nAugmentation Functions (Add Rolling Multiple Columns):\n\n• `tk_augment_slidify()` - For easily adding multiple rolling windows to you data\n\nSlider R Package:\n\n• `slider::pslide()` - The workhorse function that powers `timetk::slidify()`\n\n## Examples\n\n ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106``` ```library(tidyverse) library(tidyquant) library(tidyr) library(timetk) FB <- FANG %>% filter(symbol == \"FB\") # --- ROLLING MEAN (SINGLE ARG EXAMPLE) --- # Turn the normal mean function into a rolling mean with a 5 row .period mean_roll_5 <- slidify(mean, .period = 5, .align = \"right\") FB %>% mutate(rolling_mean_5 = mean_roll_5(adjusted)) # Use `partial = TRUE` to allow partial windows (those with less than the full .period) mean_roll_5_partial <- slidify(mean, .period = 5, .align = \"right\", .partial = TRUE) FB %>% mutate(rolling_mean_5 = mean_roll_5_partial(adjusted)) # There's nothing stopping you from combining multiple rolling functions with # different .period sizes in the same mutate call mean_roll_10 <- slidify(mean, .period = 10, .align = \"right\") FB %>% select(symbol, date, adjusted) %>% mutate( rolling_mean_5 = mean_roll_5(adjusted), rolling_mean_10 = mean_roll_10(adjusted) ) # For summary operations like rolling means, we can accomplish large-scale # multi-rolls with tk_augment_slidify() FB %>% select(symbol, date, adjusted) %>% tk_augment_slidify( adjusted, .period = 5:10, .f = mean, .align = \"right\", .names = str_c(\"MA_\", 5:10) ) # --- GROUPS AND ROLLING ---- # One of the most powerful things about this is that it works with # groups since `mutate` is being used data(FANG) mean_roll_3 <- slidify(mean, .period = 3, .align = \"right\") FANG %>% group_by(symbol) %>% mutate(mean_roll = mean_roll_3(adjusted)) %>% slice(1:5) # --- ROLLING CORRELATION (MULTIPLE ARG EXAMPLE) --- # With 2 args, use the purrr syntax of ~ and .x, .y # Rolling correlation example cor_roll <- slidify(~cor(.x, .y), .period = 5, .align = \"right\") FB %>% mutate(running_cor = cor_roll(adjusted, open)) # With >2 args, create an anonymous function with >2 args or use # the purrr convention of ..1, ..2, ..3 to refer to the arguments avg_of_avgs <- slidify( function(x, y, z) (mean(x) + mean(y) + mean(z)) / 3, .period = 10, .align = \"right\" ) # Or avg_of_avgs <- slidify( ~(mean(..1) + mean(..2) + mean(..3)) / 3, .period = 10, .align = \"right\" ) FB %>% mutate(avg_of_avgs = avg_of_avgs(open, high, low)) # Optional arguments MUST be passed at the creation of the rolling function # Only data arguments that are \"rolled over\" are allowed when calling the # rolling version of the function FB\\$adjusted <- NA roll_mean_na_rm <- slidify(~mean(.x, na.rm = TRUE), .period = 5, .align = \"right\") FB %>% mutate(roll_mean = roll_mean_na_rm(adjusted)) # --- ROLLING REGRESSIONS ---- # Rolling regressions are easy to implement using `.unlist = FALSE` lm_roll <- slidify(~lm(.x ~ .y), .period = 90, .unlist = FALSE, .align = \"right\") FB %>% drop_na() %>% mutate(numeric_date = as.numeric(date)) %>% mutate(rolling_lm = lm_roll(adjusted, numeric_date)) %>% filter(!is.na(rolling_lm)) ```\n\ntimetk documentation built on Jan. 19, 2021, 1:06 a.m." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6569476,"math_prob":0.9108771,"size":5983,"snap":"2021-31-2021-39","text_gpt3_token_len":1748,"char_repetition_ratio":0.14216425,"word_repetition_ratio":0.054136876,"special_character_ratio":0.31890357,"punctuation_ratio":0.16306306,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98632455,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T00:36:00Z\",\"WARC-Record-ID\":\"<urn:uuid:66cae201-329b-4648-9677-a81e5cc45203>\",\"Content-Length\":\"75723\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:619f9945-8ff8-49f9-b5fe-62dd11b94e18>\",\"WARC-Concurrent-To\":\"<urn:uuid:3654ea8c-c93c-4727-bb7f-9516ec20c651>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/timetk/man/slidify.html\",\"WARC-Payload-Digest\":\"sha1:BEXNDXEDIP6JEFWDTX3MKA62UBKXCSER\",\"WARC-Block-Digest\":\"sha1:HFUUEFRA3FVOA4OIW2E6LSMQ4UWWNYQP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154408.7_warc_CC-MAIN-20210802234539-20210803024539-00715.warc.gz\"}"}
https://socratic.org/questions/if-the-slope-of-a-line-is-17-13-what-is-the-slope-of-a-perpendicular-line	[ "# If the slope of a line is 17/13, what is the slope of a perpendicular line?\n\n##### 1 Answer\nOct 21, 2015\n\n$- \\frac{13}{17}$\n\n#### Explanation:\n\nSuppose a line has equation $y = m x + c$\n\nThis is in slope intercept form with slope $m$ and intercept $c$\n\nIf we reflect this line in the line $y = x$ then that is equivalent to swapping $x$ and $y$ in the equation, resulting in a line with equation:\n\n$x = m y + c$\n\nIf we then reflect that line in the $x$ axis, that is equivalent to replacing $y$ with $- y$, so we get a line with equation:\n\n$x = - m y + c$\n\nSubtract $c$ from both sides to get:\n\n$x - c = - m y$\n\nDivide both sides by $- m$ to get:\n\n$y = - \\frac{1}{m} x + \\frac{c}{m}$\n\nThis is in slope intercept format.\n\nNotice that the geometric result of the two reflections is a rotation through a right angle (Try it yourself with a square of paper with an arrow on one side).\n\nNotice that the effect on the slope is to replace $m$ by $- \\frac{1}{m}$.\n\nAny parallel line will just have a different intercept value, so we have shown that a line perpendicular to a line of slope $m$ has slope $- \\frac{1}{m}$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89601225,"math_prob":0.9999857,"size":748,"snap":"2021-21-2021-25","text_gpt3_token_len":186,"char_repetition_ratio":0.13172042,"word_repetition_ratio":0.01369863,"special_character_ratio":0.24732621,"punctuation_ratio":0.0397351,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000069,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-25T06:04:07Z\",\"WARC-Record-ID\":\"<urn:uuid:ba9c5d58-8662-4fe3-a24d-a85e957fd569>\",\"Content-Length\":\"35977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90458bef-423e-48e8-b445-0dbeaa8a3b9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:5ebc607c-653e-454b-9e55-12f1d699a58d>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/if-the-slope-of-a-line-is-17-13-what-is-the-slope-of-a-perpendicular-line\",\"WARC-Payload-Digest\":\"sha1:PSDDR3WM72546HWUQGBOMHYW3HJWVHKK\",\"WARC-Block-Digest\":\"sha1:WICN3LVRSVOBCY2NOIXNVB7CH4PU2BBF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487622113.11_warc_CC-MAIN-20210625054501-20210625084501-00236.warc.gz\"}"}
https://answers.everydaycalculation.com/compare-fractions/24-8-and-1-20	[ "Solutions by everydaycalculation.com\n\n## Compare 24/8 and 1/20\n\n1st number: 3 0/8, 2nd number: 1/20\n\n24/8 is greater than 1/20\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 8 and 20 is 40\n\nNext, find the equivalent fraction of both fractional numbers with denominator 40\n2. For the 1st fraction, since 8 × 5 = 40,\n24/8 = 24 × 5/8 × 5 = 120/40\n3. Likewise, for the 2nd fraction, since 20 × 2 = 40,\n1/20 = 1 × 2/20 × 2 = 2/40\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 120/40 > 2/40 or 24/8 > 1/20\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8384621,"math_prob":0.9904581,"size":413,"snap":"2021-31-2021-39","text_gpt3_token_len":170,"char_repetition_ratio":0.2616137,"word_repetition_ratio":0.0,"special_character_ratio":0.45762712,"punctuation_ratio":0.067307696,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9941625,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T03:36:36Z\",\"WARC-Record-ID\":\"<urn:uuid:82c00afe-2513-48bf-8787-ad6a3482b071>\",\"Content-Length\":\"8007\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7f135ca-53f5-4264-b604-6e51cffb6455>\",\"WARC-Concurrent-To\":\"<urn:uuid:0b555ac3-3c83-490c-8edc-f807f622dcf8>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/compare-fractions/24-8-and-1-20\",\"WARC-Payload-Digest\":\"sha1:BOD3RDTEPT2G7JIR6LSGSVCBKHTYKMP4\",\"WARC-Block-Digest\":\"sha1:YDLJP24BZBXXGKZ3NXF6TVXCEWVMB6WX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154158.4_warc_CC-MAIN-20210801030158-20210801060158-00637.warc.gz\"}"}
https://openstax.org/books/college-physics/pages/18-section-summary	[ "College Physics\n\n# Section Summary\n\nCollege PhysicsSection Summary\n\n### 18.1Static Electricity and Charge: Conservation of Charge\n\n• There are only two types of charge, which we call positive and negative.\n• Like charges repel, unlike charges attract, and the force between charges decreases with the square of the distance.\n• The vast majority of positive charge in nature is carried by protons, while the vast majority of negative charge is carried by electrons.\n• The electric charge of one electron is equal in magnitude and opposite in sign to the charge of one proton.\n• An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.\n• The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal magnitude; the magnitude of this basic charge $∣ q e ∣ ∣ q e ∣ size 12{ lline q rSub { size 8{e} } rline} {}$ is\n$∣ q e ∣ = 1.60 × 10 − 19 C . ∣ q e ∣ = 1.60 × 10 − 19 C . size 12{ lline q rSub { size 8{e} } rline =1 \".\" \"60\" times \"10\" rSup { size 8{ - \"19\"} } C} {}$\n• Whenever charge is created or destroyed, equal amounts of positive and negative are involved.\n• Most often, existing charges are separated from neutral objects to obtain some net charge.\n• Both positive and negative charges exist in neutral objects and can be separated by rubbing one object with another. For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletion of electrons.\n• The law of conservation of charge ensures that whenever a charge is created, an equal charge of the opposite sign is created at the same time.\n\n### 18.2Conductors and Insulators\n\n• Polarization is the separation of positive and negative charges in a neutral object.\n• A conductor is a substance that allows charge to flow freely through its atomic structure.\n• An insulator holds charge within its atomic structure.\n• Objects with like charges repel each other, while those with unlike charges attract each other.\n• A conducting object is said to be grounded if it is connected to the Earth through a conductor. Grounding allows transfer of charge to and from the earth’s large reservoir.\n• Objects can be charged by contact with another charged object and obtain the same sign charge.\n• If an object is temporarily grounded, it can be charged by induction, and obtains the opposite sign charge.\n• Polarized objects have their positive and negative charges concentrated in different areas, giving them a non-symmetrical charge.\n• Polar molecules have an inherent separation of charge.\n\n### 18.3Coulomb’s Law\n\n• Frenchman Charles Coulomb was the first to publish the mathematical equation that describes the electrostatic force between two objects.\n• Coulomb’s law gives the magnitude of the force between point charges. It is\n$F=k\|q1q2\|r2,F=k\|q1q2\|r2, size 12{F=k { {q rSub { size 8{1} } q rSub { size 8{2} } } over {r rSup { size 8{2} } } } } {}$\n\nwhere $q1q1$ and $q2q2$ are two point charges separated by a distance $rr$, and $k≈8.99×109 N·m2/C2k≈8.99×109 N·m2/C2$\n\n• This Coulomb force is extremely basic, since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies most macroscopic forces.\n• The Coulomb force is extraordinarily strong compared with the gravitational force, another basic force—but unlike gravitational force it can cancel, since it can be either attractive or repulsive.\n• The electrostatic force between two subatomic particles is far greater than the gravitational force between the same two particles.\n\n### 18.4Electric Field: Concept of a Field Revisited\n\n• The electrostatic force field surrounding a charged object extends out into space in all directions.\n• The electrostatic force exerted by a point charge on a test charge at a distance $rr size 12{r} {}$ depends on the charge of both charges, as well as the distance between the two.\n• The electric field $EE size 12{E} {}$ is defined to be\n$E = F q , E = F q , size 12{E= { {F} over {q,} } } {}$\n\nwhere $FF size 12{F} {}$ is the Coulomb or electrostatic force exerted on a small positive test charge $qq size 12{q} {}$. $EE size 12{E} {}$ has units of N/C.\n\n• The magnitude of the electric field $EE size 12{E} {}$ created by a point charge $QQ size 12{Q} {}$ is\n$E = k \|Q\| r 2 . E = k \|Q\| r 2 . size 12{E=k { {Q} over {r rSup { size 8{2} } } } } {}$\n\nwhere $rr size 12{r} {}$ is the distance from $QQ size 12{Q} {}$. The electric field $EE size 12{E} {}$ is a vector and fields due to multiple charges add like vectors.\n\n### 18.5Electric Field Lines: Multiple Charges\n\n• Drawings of electric field lines are useful visual tools. The properties of electric field lines for any charge distribution are that:\n• Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.\n• The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.\n• The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines.\n• The direction of the electric field is tangent to the field line at any point in space.\n• Field lines can never cross.\n\n### 18.6Electric Forces in Biology\n\n• Many molecules in living organisms, such as DNA, carry a charge.\n• An uneven distribution of the positive and negative charges within a polar molecule produces a dipole.\n• The effect of a Coulomb field generated by a charged object may be reduced or blocked by other nearby charged objects.\n• Biological systems contain water, and because water molecules are polar, they have a strong effect on other molecules in living systems.\n\n### 18.7Conductors and Electric Fields in Static Equilibrium\n\n• A conductor allows free charges to move about within it.\n• The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached.\n• Any excess charge will collect along the surface of a conductor.\n• Conductors with sharp corners or points will collect more charge at those points.\n• A lightning rod is a conductor with sharply pointed ends that collect excess charge on the building caused by an electrical storm and allow it to dissipate back into the air.\n• Electrical storms result when the electrical field of Earth’s surface in certain locations becomes more strongly charged, due to changes in the insulating effect of the air.\n• A Faraday cage acts like a shield around an object, preventing electric charge from penetrating inside.\n\n### 18.8Applications of Electrostatics\n\n• Electrostatics is the study of electric fields in static equilibrium.\n• In addition to research using equipment such as a Van de Graaff generator, many practical applications of electrostatics exist, including photocopiers, laser printers, ink-jet printers and electrostatic air filters.\nOrder a print copy\n\nAs an Amazon Associate we earn from qualifying purchases." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90977126,"math_prob":0.98885065,"size":6162,"snap":"2023-14-2023-23","text_gpt3_token_len":1203,"char_repetition_ratio":0.1654758,"word_repetition_ratio":0.0,"special_character_ratio":0.1944174,"punctuation_ratio":0.08671587,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9844634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T23:40:02Z\",\"WARC-Record-ID\":\"<urn:uuid:55ee8670-4248-4867-af47-3d62cfe117e8>\",\"Content-Length\":\"487942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa6546c3-1976-4dae-bd71-1de022471064>\",\"WARC-Concurrent-To\":\"<urn:uuid:99a95660-9caa-4120-8d20-87221aaa8d97>\",\"WARC-IP-Address\":\"18.160.46.96\",\"WARC-Target-URI\":\"https://openstax.org/books/college-physics/pages/18-section-summary\",\"WARC-Payload-Digest\":\"sha1:HGAD2ZH6GUQAPBMPVLTIFU37G4V4NMQB\",\"WARC-Block-Digest\":\"sha1:36O4RUT4XFBNT6AER2Q6TFH34MWG42QQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648209.30_warc_CC-MAIN-20230601211701-20230602001701-00429.warc.gz\"}"}
https://b-ok.org/book/2464071/d862b2	[ ",\n\nMedical Statistics Made Easy 2nd edition continues to provide the easiest possible explanations of the key statistical techniques used throughout the medical literature.\n\nFeaturing a comprehensive updating of the 'Statistics at work' section, this new edition retains a consistent, concise, and user-friendly format. Each technique is graded for ease of use and frequency of appearance in the mainstream medical journals.\n\nMedical Statistics Made Easy 2nd edition is essential reading for anyone looking to understand:\n\n* confidence intervals and probability values\n* numbers needed to treat\n* t tests and other parametric tests\n* survival analysis\n\nIf you need to understand the medical literature, then you need to read this book.\n\nReviews:\n\"This book helps medical students understand the basic concepts of medical statistics starting in a 'step-by-step approach'. The authors have designed the book assuming that the reader has no prior knowledge. It focuses on the most common statistical concepts that are likely to be faced in medical literature.\nAll chapters are concise and simple to understand. Each chapter starts with an introduction which consists of “how important” that particular statistical concept is, using a 'star' system. A 'thumbs-up' system shows how easy the statistical concept is to understand. Both these systems indicate time-efficient learning allowing yourself to focus on areas you find most difficult. Following this, there are worked out examples with exam-tips at the end of some chapters.\nThe last chapter, 'Statistics at Work', shows how medical statistics is put into practice using worked out examples from renowned journals. This helps in assessing the reader’s own knowledge and gives them confidence in analysis of statistics of a journal.\nIn conclusion, we would recommend this book as an introduction into medical statistics before plunging into the deep 'statistical' waters! It gives confidence to the reader in taking up the challenge of understanding statistics and [being] able to apply knowledge in analysing medical literature.\"\nStefanie Zhao Lin Lip & Louise Murchison, Scottish Medical Journal, June 2010\n\n\"If ever there was a book that completely lived up to its title, this is it...Perhaps above everything, it is the chapter layout and design that makes this book stand out head and shoulders above the crowd. At the beginning of each chapter two questions are posed – how important is the subject in question and how difficult is it to understand? The first is answered on the basis of how often the subject is mentioned / used in papers published in mainstream medical journals. A star rating is then given from one to five with five stars implying use in the majority of papers published. The second question is answered by means of a ‘thumbs up’ grading system. The more thumbs, the easier the concept is to understand (maximum of five). This, of course, provides a route into statistics for even the most idle of uneducated individuals! Five stars and five thumbs must surely indicate time-efficient learning! At the end of each chapter exam tips (light bulb icon!) are given – I doubt anyone could ask for more!\nThe whole way in which the authors have written this book is commendable; the chapters are succinct, easy to follow and a pleasure to read...Is it value for money? – a definite yes even at twice the price. Of course I never exaggerate but if you breathe, you should own this book!\"\nIan Pearce, Urology News, June 2010\n\nYear: 2008\nEdition: 2nd\nLanguage: english\nPages: 116\nISBN 10: 1904842550\nISBN 13: 9781904842552\nFile: PDF, 2.21 MB\n\nYou may be interested in\n\nMedical Statistics at a Glance Workbook\n\nYear: 2013\nLanguage: english\nFile: PDF, 5.88 MB\n\nPost a Review", null, "You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them.\n1\n\n日本語500問中級\n\nYear: 2009\nLanguage: japanese\nFile: PDF, 33.63 MB\n2\n\nPHP for Absolute Beginners\n\nYear: 2014\nLanguage: english\nFile: PDF, 3.98 MB\n```\fMEDICAL\nSTATISTICS\n\n2\n\nNew Clinical Genetics\n978 1 904842 31 6\n“This book is a very valuable tool that will be\nused by future geneticists all over Europe and\nbeyond, both as a teaching material and as a\nsource of excellent knowledge.”\nEuropean Journal of Human Genetics\n\nPuzzles for Medical Students\nR. Howard\n978 1 904842 34 8\n“Unlike bland medical textbooks offering\nlearning by rote, this book challenges the\nmedical student to learn key medical facts\nthrough the solving of puzzles and the author\nshould be congratulated for this novel\napproach.”\nDr Andrew Catto\n\nClinical Skills for OSCEs\nN. Burton\n978 1 904842 59 0\n\n“An invaluable guide to clinical skills for OSCEs.\nA must have for all students!”\nInternational Journal of Clinical Skills\n\nMEDICAL\nSTATISTICS\nMichael Harris\nGeneral Practitioner and Senior Lecturer in\nMedical Education, Bristol, UK\nand\n\nGordon Taylor\nSenior Lecturer in Medical Statistics,\nUniversity of Bath, UK\n\n2\n\nSecond edition © Scion Publishing Ltd, 2008\nISBN 978 1 904842 55 2\nReprinted 2008, 2009\nFirst edition published in 2003 by Martin Dunitz (ISBN 1 85996 219 X)\nReprinted 2004, 2005, 2007\ntransmitted, in any form or by any means, without permission.\nA CIP catalogue record for this book is available from the British Library.\nScion Publishing Limited\nBloxham Mill, Barford Road, Bloxham, Oxfordshire OX15 4FF\nwww.scionpublishing.com\nImportant Note from the Publisher\nThe information contained within this book was obtained by Scion\nPublishing Limited from sources believed by us to be reliable. However,\nwhile every effort has been made to ensure its accuracy, no responsibility\nfor loss or injury whatsoever occasioned to any person acting or\nrefraining from action as a result of information contained herein can be\naccepted by the authors or publishers.\nAlthough every effort has been made to ensure that all owners of\ncopyright material have been acknowledged in this publication, we\nwould be pleased to acknowledge in subsequent reprints or editions any\nomissions brought to our attention.\n\nTypeset by Phoenix Photosetting, Chatham, Kent, UK\nPrinted by Gutenberg Press Ltd, Malta\n\nCONTENTS\n\nAbbreviations\n\nvii\n\nPreface\n\nix\n\nx\n\nForeword\n\nxi\n\nHow to use this book\n\n1\n\nHow this book is designed\n\n4\n\nStatistics which describe data\nPercentages\nMean\nMedian\nMode\nStandard deviation\n\n7\n9\n12\n14\n16\n\nStatistics which test confidence\nConfidence intervals\nP values\n\n20\n24\n\nStatistics which test differences\nt tests and other parametric tests\nMann–Whitney and other non-parametric tests\nChi-squared test\n\n28\n31\n34\n\nStatistics which compare risk\nRisk ratio\nOdds ratio\nRisk reduction and numbers needed to treat\n\n37\n40\n43\n\nvi\n\nContents\n\nStatistics which analyze relationships\nCorrelation\nRegression\n\n48\n53\n\nStatistics which analyze survival\nSurvival analysis: life tables and Kaplan–Meier\nplots\nThe Cox regression model\n\n57\n60\n\nStatistics which analyze clinical investigations and screening\nSensitivity, specificity and predictive value\nLevel of agreement and Kappa\n\n62\n67\n\nOther concepts\n\n69\n\nStatistics at work\n\n73\n\nStandard deviation, relative risk and numbers\nneeded to treat\nOdds ratios and confidence intervals\nCorrelation and regression\nSurvival analysis\nSensitivity, specificity and predictive values\n\n74\n78\n81\n85\n88\n\nGlossary\n\n93\n\nIndex\n\n113\n\nABBREVIATIONS\n\nARR\nBMI\nBP\nCI\ndf\nHR\nIQR\nLR\nNNH\nNNT\nNPV\nP\nPPV\nRRR\nSD\nSE\n\nabsolute risk reduction\nbody mass index\nblood pressure\nconfidence interval\ndegrees of freedom\nhazard ratio\ninter-quartile range\nlikelihood ratio\nnumber needed to harm\nnumber needed to treat\nnegative predictive value\nprobability\npositive predictive value\nrelative risk reduction\nstandard deviation\nside effect\n\nPREFACE\n\nThis book is designed for healthcare students and\nprofessionals who need a basic knowledge of when\ncommon statistical terms are used and what they\nmean.\nWhether you love or hate statistics, you need to have\nsome understanding of the subject if you want to\ncritically appraise a paper. To do this, you do not\nneed to know how to do a statistical analysis. What\nyou do need is to know why the test has been used\nand how to interpret the resulting figures.\nThis book does not assume that you have any\nprior statistical knowledge. However basic your\nmathematical or statistical knowledge, you will find\nthat everything is clearly explained.\nA few readers will find some of the sections\nridiculously simplistic, others will find some bafflingly\nout concepts that suit your level of understanding.\nthe most important concepts if you are short of time.\nThis book is also produced for those who may be\ntips” sections if you are in a hurry.\nYou can test your understanding of what you have\nlearnt by working through extracts from original\npapers in the “Statistics at work” section.\n\nDr Michael Harris MB BS FRCGP MMEd is a\nGeneral Practitioner and Senior Lecturer in Medical\nEducation in Bristol, UK. He teaches nurses, medical\nstudents and GP Registrars. Until recently he was an\nexaminer for the MRCGP.\nDr Gordon Taylor PhD MSc BSc (Hons) is a Senior\nLecturer in Medical Statistics at the University of\nBath, UK. His main role is in the teaching, support\nand supervision of health care professionals involved\nin non-commercial research.\n\nFOREWORD\n\nA love of statistics is, oddly, not what attracts most\nyoung people to a career in medicine and I suspect\nthat many clinicians, like me, have at best a sketchy\nand incomplete understanding of this difficult\nsubject.\nDelivering modern, high quality care to patients now\nrelies increasingly on routine reference to scientific\npapers and journals, rather than traditional textbook\nlearning. Acquiring the skills to appraise medical\nresearch papers is a daunting task. Realizing this,\nMichael Harris and Gordon Taylor have expertly\nconstructed a practical guide for the busy clinician.\nOne a practising NHS doctor, the other a medical\nstatistician with tremendous experience in clinical\nresearch, they have produced a unique handbook. It\nis short, readable and useful, without becoming\noverly bogged down in the mathematical detail that\nfrankly puts so many of us off the subject.\nI commend this book to all healthcare professionals,\ngeneral practitioners and hospital specialists. It\ncovers all the ground necessary to critically evaluate\nthe statistical elements of medical research papers, in\na friendly and approachable way. The scoring of each\ncomprehension will efficiently guide the busy\npractitioner through his or her reading. In particular\nit is almost unique in covering this part of the\nsyllabus for Royal College and other postgraduate\nexaminations. Certainly a candidate familiar with\nthe contents of this short book and taking note of its\n\nxii\n\nForeword\n\nnumerous helpful examination tips should have few\ndifficulties when answering the questions on\nstatistics in both the MCQ and Written modules of\nthe new MRCGP exam.\nNovember 2007\nBill Irish\nBSc MB BChir DCH DRCOG MMEd FRCGP\n(Head of GP School, Severn Deanery, UK and\nSenior Examiner for the MRCGP(UK)).\n\n> HOW TO USE THIS BOOK\nYou can use this book in a number of ways.\n\nIf you want a statistics course\n∑ Work through from start to finish for a complete\ncourse in commonly used medical statistics.\n\nIf you are in a hurry\n∑ Choose the sections with the most stars to learn\nabout the commonest statistical methods and\nterms.\npercentages (page 7), mean (page 9), standard\ndeviation (page 16), confidence intervals (page\n20) and P values (page 24).\n\nIf you are daunted by statistics\n∑ If you are bewildered every time someone tries to\nexplain a statistical method, then pick out the\nsections with the most thumbs up symbols to find\nthe easiest and most basic concepts.\nmean (page 9), median (page 12) and mode (page\n14), then move on to risk ratio (page 37),\nincidence and prevalence (page 70).\n\n2\n\nXA\n\nMT\nIP\n\nE\n\nIf you are taking an exam\n∑ The “Exam Tips” give you pointers to the topics\n∑ You will find these in the following sections: mean\n(page 9), standard deviation (page 16), confidence\nintervals (page 20), P values (page 24), risk\nreduction and NNT (page 43), sensitivity,\nspecificity and predictive value (page 62),\nincidence and prevalence (page 70).\n\n∑ See how statistical methods are used in five\nextracts from real-life papers in the “Statistics at\nwork” section (page 73).\n∑ Work out which statistical methods have been\nused, why, and what the results mean. Then check\n\nGlossary\n∑ Use the glossary (page 93) as a quick reference for\nstatistical words or phrases that you do not know.\n\n∑ Go through difficult sections when you are fresh\nand try not to cover too much at once.\n∑ You may need to read some sections a couple of\ntimes before the meaning sinks in. You will find\nprinciples.\n\nHow To Use This Book\n\n3\n\n∑ We have tried to cut down the jargon as much as\npossible. If there is a word that you do not\nunderstand, check it out in the glossary.\n\nTHIS BOOK IS\n> HOW\nDESIGNED\n\nEvery section uses the same series of headings to help\nyou understand the concepts.\n\n“How important is it?”\nWe noted how often statistical terms were used in\n200 quantitative papers in mainstream medical\njournals. All the papers selected were published\nduring the last year in the British Medical Journal,\nThe Lancet, the New England Journal of Medicine\nand the Journal of the American Medical\nAssociation.\nWe grouped the terms into concepts and graded them\nby how often they were used. This helped us to\ndevelop a star system for importance. We also took\ninto account usefulness to readers. For example,\n“numbers needed to treat” are not often quoted but\nare fairly easy to calculate and useful in making\ntreatment decisions.\n\n88888\n\nConcepts which are used in the majority of medical\npapers.\n\n8888\n\nImportant concepts which are used in at least a third\nof papers.\n\n888\n\nLess frequently used, but still of value in decisionmaking.\n\nHow This Book Is Designed\n\n88\n\nFound in at least 1 in 10 papers.\n\n8\n\nRarely used in medical journals.\n\n5\n\nHow easy is it to understand?\nWe have found that the ability of health care\nprofessionals to understand statistical concepts varies\nmore widely than their ability to understand anything\nelse related to medicine. This ranges from those that\nhave no difficulty learning how to understand\nregression to those that struggle with percentages.\nOne of the authors (not the statistician!) fell into the\nlatter category. He graded each section by how easy\nit is to understand the concept.\nEven the most statistic-phobic will have little\ndifficulty in understanding these sections.\nWith a little concentration, most readers should be\nSome readers will have difficulty following these.\nYou may need to go over these sections a few times to\nbe able to take them in.\nQuite difficult to understand. Only tackle these\nsections when you are fresh.\nStatistical concepts that are very difficult to grasp.\n\nWhen is it used?\nOne thing you need to do if critically appraising a\npaper is check that the right statistical technique has\nbeen used. This part explains which statistical\nmethod should be used for what scenario.\n\n6\n\nWhat does it mean?\nThis explains the bottom line – what the results mean\n\nExamples\nSometimes the best way to understand a statistical\ntechnique is to work through an example. Simple,\nfictitious examples are given to illustrate the\nprinciples and how to interpret them.\n\nWatch out for . . .\nThis includes more detailed explanation, tips and\ncommon pitfalls.\n\nXA\n\nMT\nIP\n\nE\n\nExam tips\nSome topics are particularly popular with examiners\nbecause they test understanding and involve simple\ncalculations. We have given tips on how to approach\nthese concepts.\n\n> PERCENTAGES\nHow important are they?\n\n88888\n\nAn understanding of percentages is probably the first\nand most important concept to understand in statistics!\n\nHow easy are they to understand?\nPercentages are easy to understand.\n\nWhen are they used?\nPercentages are mainly used in the tabulation of data\nin order to give the reader a scale on which to assess\nor compare the data.\n\nWhat do they mean?\n“Per cent” means per hundred, so a percentage\ndescribes a proportion of 100. For example 50% is\n50 out of 100, or as a fraction 1⁄2. Other common\npercentages are 25% (25 out of 100 or 1⁄4), 75% (75\nout of 100 or 3⁄4).\nTo calculate a percentage, divide the number of items\nor patients in the category by the total number in the\ngroup and multiply by 100.\n\n8\n\nEXAMPLE\nData were collected on 80 patients referred for heart transplantation. The\nresearcher wanted to compare their ages. The data for age were put in\n“decade bands” and are shown in Table 1.\nTable 1. Ages of 80 patients referred for heart transplantation\nYearsa\n\nFrequencyb\n\n0–9\n\n2\n\nPercentagec\n2.5\n\n10–19\n\n5\n\n6.25\n\n20–29\n\n6\n\n7.5\n\n30–39\n\n14\n\n17.5\n\n40–49\n\n21\n\n26.25\n\n50–59\n\n20\n\n25\n\n≥ 60\n\n12\n\n15\n\nTotal\n\n80\n\n100\n\na\n\nb\n\nFrequency = number of patients referred;\n\nc\n\nPercentage = percentage of patients in each decade band. For example, in the\n\n30–39 age band there were 14 patients and we know the ages of 80 patients,\n14\nso\n¥ 100 = 17.5%.\n80\n\nWatch out for . . .\nAuthors can use percentages to hide the true size of\nthe data. To say that 50% of a sample has a certain\ncondition when there are only four people in the\nsample is clearly not providing the same level of\ninformation as 50% of a sample based on 400\npeople. So, percentages should be used as an\nthe actual data.\n\n> MEAN\nOtherwise known as an arithmetic mean, or average.\n\nHow important is it?\n\n88888\n\nA mean appeared in 90% papers surveyed, so it is\nimportant to have an understanding of how it is\ncalculated.\n\nHow easy is it to understand?\nOne of the simplest statistical concepts to grasp.\nHowever, in most groups that we have taught there\nhas been at least one person who admits not knowing\nhow to calculate the mean, so we do not apologize\nfor including it here.\n\nWhen is it used?\nIt is used when the spread of the data is fairly similar\non each side of the mid point, for example when the\ndata are “normally distributed”.\nThe “normal distribution” is referred to a lot in\nstatistics. It’s the symmetrical, bell-shaped distribution of data shown in Fig. 1.\n\n10\n\nFig. 1. The normal distribution. The dotted line shows the mean of the data.\n\nWhat does it mean?\nThe mean is the sum of all the values, divided by the\nnumber of values.\n\nEXAMPLE\nFive women in a study on lipid-lowering agents are aged 52, 55, 56, 58\nand 59 years.\n52 + 55 + 56 + 58 + 59 = 280\nNow divide by the number of women:\n280 = 56\n5\nSo the mean age is 56 years.\n\nWatch out for...\nIf a value (or a number of values) is a lot smaller or\nlarger than the others, “skewing” the data, the mean\nwill then not give a good picture of the typical value.\n\nMean\n\n11\n\nXA\n\nMT\nIP\n\nE\n\nFor example, if there is a sixth patient aged 92 in the\nstudy then the mean age would be 62, even though\nonly one woman is over 60 years old. In this case, the\n“median” may be a more suitable mid-point to use\n(see page 12).\nA common multiple choice question is to ask the\ndifference between mean, median (see page 12) and\nmode (see page 14) – make sure that you do not get\nconfused between them.\n\n> MEDIAN\nSometimes known as the mid-point.\n\nHow important is it?\n\n8888\n\nIt is given in over a half of mainstream papers.\n\nHow easy is it to understand?\nEven easier than the mean!\n\nWhen is it used?\nIt is used to represent the average when the data\nare not symmetrical, for instance the “skewed”\ndistribution in Fig. 2. Compare the shape of the\ngraph with the normal distribution shown in Fig. 1.\n\nFig. 2. A skewed distribution. The dotted line shows the median.\n\nWhat does it mean?\nIt is the point which has half the values above, and\nhalf below.\n\nMedian\n\n13\n\nEXAMPLE\nUsing the first example from page 10 of five patients aged 52, 55, 56, 58\nand 59, the median age is 56, the same as the mean – half the women are\nolder, half are younger.\nHowever, in the second example with six patients aged 52, 55, 56, 58, 59\nand 92 years, there are two “middle” ages, 56 and 58. The median is halfway between these, i.e. 57 years. This gives a better idea of the mid-point\nof this skewed data than the mean of 62.\n\nWatch out for...\nThe median may be given with its inter-quartile range\n(IQR). The 1st quartile point has the 1⁄4 of the data below\nit, the 3rd quartile point has the 3⁄4 of the sample below\nit, so the IQR contains the middle 1⁄2 of the sample. This\ncan be shown in a “box and whisker” plot.\n\nEXAMPLE\nA dietician measured the energy intake over 24 hours of 50 patients on a\nvariety of wards. One ward had two patients that were “nil by mouth”. The\nmedian was 12.2 megajoules, IQR 9.9 to 13.6. The lowest intake was 0,\nthe highest was 16.7. This distribution is represented by the box and\nwhisker plot in Fig. 3.\n18\nHighest intake\n16\nEnergy intake (MJ)\n\n14\n\n3rd quartile point\n\n12\n\nMedian (2nd quartile point)\n\n10\n\n1st quartile point\n\n8\n6\n4\n\nLowest intake,\nexcluding extreme values\n\n2\n0\n\nFig. 3. Box and whisker plot of energy intake of 50 patients over 24 hours. The ends\nof the whiskers represent the maximum and minimum values, excluding extreme\nresults like those of the two “nil by mouth” patients.\n\n> MODE\nHow important is it?\n\n8\n\nRarely quoted in papers and of limited value.\n\nHow easy is it to understand?\nAn easy concept.\n\nWhen is it used?\nIt is used when we need a label for the most\nfrequently occurring event.\n\nWhat does it mean?\nThe mode is the most common of a set of events.\n\nEXAMPLE\nAn eye clinic sister noted the eye colour of 100 consecutive patients. The\nresults are shown in Fig. 4.\n80\nNumber of patients\n\n70\n60\n50\n40\n30\n20\n10\n0\nBrown\n\nBlue\n\nGrey\n\nGreen\n\nEye colour\n\nFig. 4. Graph of eye colour of patients attending an eye clinic.\n\nIn this case the mode is brown, the commonest eye colour.\n\nMode\n\n15\n\nYou may see reference to a “bi-modal distribution”.\nGenerally when this is mentioned in papers it is as a\nconcept rather than from calculating the actual\nvalues, e.g. “The data appear to follow a bi-modal\ndistribution”. See Fig. 5 for an example of where\nthere are two “peaks” to the data, i.e. a bi-modal\ndistribution.\n400\n\nNumber of patients\n\n300\n\n200\n\n80–89\n\n70–79\n\n60–69\n\n50–59\n\n40–49\n\n30–39\n\n20–29\n\n10–19\n\n0–9\n\n0\n\n≥90\n\n100\n\nAges of patients\nFig. 5. Graph of ages of patients with asthma in a practice.\n\nThe arrows point to the modes at ages 10–19 and\n60–69.\nBi-modal data may suggest that two populations are\npresent that are mixed together, so an average is not\na suitable measure for the distribution.\n\n> STANDARD DEVIATION\nHow important is it?\n\n88888\n\nQuoted in two-thirds of papers, it is used as the basis\nof a number of statistical calculations.\n\nHow easy is it to understand?\nIt is not an intuitive concept.\n\nWhen is it used?\nStandard deviation (SD) is used for data which are\n“normally distributed” (see page 9), to provide\ninformation on how much the data vary around their\nmean.\n\nWhat does it mean?\nSD indicates how much a set of values is spread\naround the average.\nA range of one SD above and below the mean\n(abbreviated to ± 1 SD) includes 68.2% of the values.\n± 2 SD includes 95.4% of the data.\n± 3 SD includes 99.7%.\n\nStandard deviation\n\n17\n\nEXAMPLE\nLet us say that a group of patients enrolling for a trial had a normal\ndistribution for weight. The mean weight of the patients was 80 kg. For\nthis group, the SD was calculated to be 5 kg.\n1 SD below the average is 80 – 5 = 75 kg.\n1 SD above the average is 80 + 5 = 85 kg.\n± 1 SD will include 68.2% of the subjects, so 68.2% of patients will weigh\nbetween 75 and 85 kg.\n95.4% will weigh between 70 and 90 kg (± 2 SD).\n99.7% of patients will weigh between 65 and 95 kg (± 3 SD).\nSee how this relates to the graph of the data in Fig. 6.\n14\n\nNumber of patients\n\n12\n10\n8\n6\n\n± 1 SD (68.2%)\n\n4\n2\n0\n60\n\n± 2 SD (95.4%)\n± 3 SD (99.7%)\n65\n\n70\n\n75\n\n80\n85\nWeight (kg)\n\n90\n\n95\n\n100\n\nFig. 6. Graph showing normal distribution of weights of patients enrolling in a trial\nwith mean 80 kg, SD 5 kg.\n\n18\n\nIf we have two sets of data with the same mean but different SDs, then the\ndata set with the larger SD has a wider spread than the data set with the\nsmaller SD.\nFor example, if another group of patients enrolling for the trial has the\nsame mean weight of 80 kg but an SD of only 3, ± 1 SD will include 68.2%\nof the subjects, so 68.2% of patients will weigh between 77 and 83 kg\n(Fig. 7). Compare this with the example above.\n14\n\nNumber of patients\n\n12\n10\n8\n6\n4\n2\n0\n60\n\n65\n\n70\n\n75\n\n80\n85\nWeight (kg)\n\n90\n\n95\n\n100\n\nFig. 7. Graph showing normal distribution of weights of patients enrolling in a trial\nwith mean 80 kg, SD 3 kg.\n\nWatch out for...\nSD should only be used when the data have a normal\ndistribution. However, means and SDs are often\nwrongly used for data which are not normally\ndistributed.\nA simple check for a normal distribution is to see if 2\nSDs away from the mean are still within the possible\n\nStandard deviation\n\n19\n\nrange for the variable. For example, if we have some\nlength of hospital stay data with a mean stay of 10\ndays and a SD of 8 days then:\nmean – 2 ¥ SD = 10 – 2 ¥ 8 = 10 – 16 = -6 days.\nThis is clearly an impossible value for length of stay,\nso the data cannot be normally distributed. The\nmean and SDs are therefore not appropriate\nmeasures to use.\nGood news – it is not necessary to know how to\ncalculate the SD.\nIt is worth learning the figures above off by heart, so\na reminder –\n± 1 SD includes 68.2% of the data\n± 2 SD includes 95.4%,\n± 3 SD includes 99.7%.\n\nXA\n\nMT\nIP\n\nE\n\nKeeping the “normal distribution” curve in Fig. 6 in\nmind may help.\nExaminers may ask what percentages of subjects are\nincluded in 1, 2 or 3 SDs from the mean. Again, try\nto memorize those percentages.\n\n> CONFIDENCE INTERVALS\nHow important are they?\n\n88888\n\nImportant – given in three-quarters of papers.\n\nHow easy are they to understand?\nA difficult concept, but one where a small amount of\nunderstanding will get you by without having to\n\nWhen is it used?\nConfidence intervals (CI) are typically used when,\ninstead of simply wanting the mean value of a\nsample, we want a range that is likely to contain the\ntrue population value.\nThis “true value” is another tough concept – it is the\nmean value that we would get if we had data for the\nwhole population.\n\nWhat does it mean?\nStatisticians can calculate a range (interval) in which\nwe can be fairly sure (confident) that the “true value”\nlies.\nFor example, we may be interested in blood pressure\n(BP) reduction with antihypertensive treatment.\nFrom a sample of treated patients we can work out\nthe mean change in BP.\n\nConfidence intervals\n\n21\n\nHowever, this will only be the mean for our particular\nsample. If we took another group of patients we would\nnot expect to get exactly the same value, because\nchance can also affect the change in BP.\nThe CI gives the range in which the true value (i.e.\nthe mean change in BP if we treated an infinite\nnumber of patients) is likely to be.\n\nEXAMPLES\nThe average systolic BP before treatment in study A, of a group of 100\nhypertensive patients, was 170 mmHg. After treatment with the new drug\nthe mean BP dropped by 20 mmHg.\nIf the 95% CI is 15–25, this means we can be 95% confident that the true\neffect of treatment is to lower the BP by 15–25 mmHg.\nIn study B 50 patients were treated with the same drug, also reducing\ntheir mean BP by 20 mmHg, but with a wider 95% CI of -5 to +45. This CI\nincludes zero (no change). This means there is more than a 5% chance\nthat there was no true change in BP, and that the drug was actually\nineffective.\n\nWatch out for...\nThe size of a CI is related to the sample size of the\nstudy. Larger studies usually have a narrower CI.\nWhere a few interventions, outcomes or studies are\ngiven it is difficult to visualize a long list of means\nand CIs. Some papers will show a chart to make it\neasier.\nFor example, “meta-analysis” is a technique for\nbringing together results from a number of similar\nstudies to give one overall estimate of effect. Many\nmeta-analyses compare the treatment effects from\n\n22\n\nthose studies by showing the mean changes and 95%\nCIs in a chart. An example is given in Fig. 8.\nStudy A\nStudy B\nStudy C\nStudy D\nStudy E\nCombined estimate\n–40\n\n–30\n\n–20\n\n–10\n\n0\n\n10\n\nChange in BP (mmHg)\nFig. 8. Plot of 5 studies of a new antihypertensive drug. See how the results of studies\nA and B above are shown by the top two lines, i.e. a 20 mmHg reduction in BP, 95% CI\n15–25 for study A and a 20 mmHg reduction, 95% CI -5 to +45 for study B.\n\nThe vertical axis does not have a scale. It is simply\nused to show the zero point on each CI line.\nThe statistician has combined the results of all five\nstudies and calculated that the overall mean reduction\nin BP is 14 mmHg, CI 12–16. This is shown by the\n“combined estimate” diamond. See how combining a\nnumber of studies reduces the CI, giving a more\naccurate estimate of the true treatment effect.\nThe chart shown in Fig. 8 is called a “Forest plot” or,\nmore colloquially, a “blobbogram”.\nStandard deviation and confidence intervals – what is\nthe difference? Standard deviation tells us about\nthe variability (spread) in a sample.\nThe CI tells us the range in which the true value (the\nmean if the sample were infinitely large) is likely to be.\n\nXA\n\nMT\nIP\n\nE\n\nConfidence intervals\n\n23\n\nAn exam question may give a chart similar to that in\nFig. 8 and ask you to summarize the findings.\nConsider:\n∑ Which study showed the greatest change?\n∑ Did all the studies show change in favour of the\nintervention?\n∑ Were the changes statistically significant?\nIn the example above, study D showed the greatest\nchange, with a mean BP drop of 25 mmHg.\nStudy C resulted in a mean increase in BP, though\nwith a wide CI. The wide CI could be due to a low\nnumber of patients in the study.\nThe combined estimate of a mean BP reduction of\n14 mmHg, 95% CI 12–16, is statistically significant.\n\n> P VALUES\nHow important is it?\n\n88888\n\nA really important concept, P values are given in\nmore than four out of five papers.\n\nHow easy is it to understand?\nNot easy, but worth persevering as it is used so\nfrequently.\nIt is not important to know how the P value is\nderived – just to be able to interpret the result.\n\nWhen is it used?\nThe P (probability) value is used when we wish to see\nhow likely it is that a hypothesis is true. The\nhypothesis is usually that there is no difference\nbetween two treatments, known as the “null\nhypothesis”.\n\nWhat does it mean?\nThe P value gives the probability of any observed\ndifference having happened by chance.\nP = 0.5 means that the probability of the difference\nhaving happened by chance is 0.5 in 1, or 50:50.\nP = 0.05 means that the probability of the difference\nhaving happened by chance is 0.05 in 1, i.e. 1 in 20.\n\nP values\n\n25\n\nIt is the figure frequently quoted as being\n“statistically significant”, i.e. unlikely to have\nhappened by chance and therefore important.\nHowever, this is an arbitrary figure.\nIf we look at 20 studies, even if none of the\ntreatments work, one of the studies is likely to have a\nP value of 0.05 and so appear significant!\nThe lower the P value, the less likely it is that the\ndifference happened by chance and so the higher the\nsignificance of the finding.\nP = 0.01 is often considered to be “highly\nsignificant”. It means that the difference will only\nhave happened by chance 1 in 100 times. This is\nunlikely, but still possible.\nP = 0.001 means the difference will have happened\nby chance 1 in 1000 times, even less likely, but still\njust possible. It is usually considered to be “very\nhighly significant”.\n\n26\n\nEXAMPLES\nOut of 50 new babies on average 25 will be girls, sometimes more,\nsometimes less.\nSay there is a new fertility treatment and we want to know whether it affects\nthe chance of having a boy or a girl. Therefore we set up a null hypothesis\n– that the treatment does not alter the chance of having a girl. Out of the\nfirst 50 babies resulting from the treatment, 15 are girls. We then need to\nknow the probability that this just happened by chance, i.e. did this happen\nby chance or has the treatment had an effect on the sex of the babies?\nThe P value gives the probability that the null hypothesis is true.\nThe P value in this example is 0.007. Do not worry about how it was\ncalculated, concentrate on what it means. It means the result would only\nhave happened by chance in 0.007 in 1 (or 1 in 140) times if the treatment\ndid not actually affect the sex of the baby. This is highly unlikely, so we\ncan reject our hypothesis and conclude that the treatment probably does\nalter the chance of having a girl.\nTry another example: Patients with minor illnesses were randomized to\nsee either Dr Smith or Dr Jones. Dr Smith ended up seeing 176 patients in\nthe study whereas Dr Jones saw 200 patients (Table 2).\nTable 2. Number of patients with minor illnesses seen by two GPs\nDr Jones\n(n=200)a\nPatients satisfied\n186 (93)\nwith consultation (%)\nMean (SD) consultation\nlength (minutes)\n\n16 (3.1)\n\nPatients getting a\nprescription (%)\n\n65 (33)\n\nMean (SD) number of\ndays off work\nPatients needing a\nfollow-up\nappointment (%)\na\n\n3.58 (1.3)\n68 (34)\n\nDr Smith\n(n=176)\n168 (95)\n6 (2.8)\n67 (38)\n\nP value i.e. could have\nhappened by chance\n0.38\n\n– possible\n\n<0.001 < One time in 1000\n– very unlikely\n0.28\n\n– possible\n\n3.61 (1.3) 0.82\n\n– probable\n\n78 (44)\n\n0.044 Only one time in 23\n– fairly unlikely\n\nn=200 means that the total number of patients seen by Dr Jones was 200.\n\nP values\n\n27\n\nWatch out for...\nThe “null hypothesis” is a concept that underlies this\nand other statistical tests.\nThe test method assumes (hypothesizes) that there is\nno (null) difference between the groups. The result of\nthe test either supports or rejects that hypothesis.\nThe null hypothesis is generally the opposite of what\nwe are actually interested in finding out. If we are\ninterested if there is a difference between two\ntreatments then the null hypothesis would be that\nthere is no difference and we would try to disprove\nthis.\n\nXA\n\nMT\nIP\n\nE\n\nTry not to confuse statistical significance with\nclinical relevance. If a study is too small, the results\nare unlikely to be statistically significant even if the\nintervention actually works. Conversely a large study\nmay find a statistically significant difference that is\ntoo small to have any clinical relevance.\nYou may be given a set of P values and asked to\ninterpret them. Remember that P = 0.05 is usually\nclassed as “significant”, P = 0.01 as “highly\nsignificant” and P = 0.001 as “very highly\nsignificant”.\nIn the example above, only two of the sets of data\nshowed a significant difference between the two\nGPs. Dr Smith’s consultations were very highly\nsignificantly shorter than those of Dr Jones. Dr\nSmith’s follow-up rate was significantly higher than\nthat of Dr Jones.\n\n>\n\nt TESTS AND OTHER\nPARAMETRIC TESTS\n\nHow important are they?\n\n8888\n\nUsed in one in three papers, they are an important\naspect of medical statistics.\n\nHow easy are they to understand?\nThe details of the tests themselves are difficult to\nunderstand.\nThankfully you do not need to know them. Just look\nfor the P value (see page 24) to see how significant\nthe result is. Remember, the smaller the P value, the\nsmaller the chance that the “null hypothesis” is true.\n\nWhen are they used?\nParametric statistics are used to compare samples of\n“normally distributed” data (see page 9). If the data\ndo not follow a normal distribution, these tests\nshould not be used.\n\nWhat do they mean?\nA parametric test is any test which requires the data\nto follow a specific distribution, usually a normal\ndistribution. Common parametric tests you will\ncome across are the t test and the c2 test .\n\nt tests and other parametric yests\n\n29\n\nAnalysis of variance (ANOVA). This is a group of\nstatistical techniques used to compare the means of\ntwo or more samples to see whether they come from\nthe same population – the “null hypothesis”. These\ntechniques can also allow for independent variables\nwhich may have an effect on the outcome.\nAgain, check out the P value.\nt test (also known as Student’s t). t tests are\ntypically used to compare just two samples. They\ntest the probability that the samples come from a\npopulation with the same mean value.\nχ2 test. A frequently used parametric test is the c2\ntest. It is covered separately (page 34).\n\nEXAMPLE\nTwo hundred adults seeing an asthma nurse specialist were randomly\nassigned to either a new type of bronchodilator or placebo.\nAfter 3 months the peak flow rates in the treatment group had increased\nby a mean of 96 l/min (SD 58), and in the placebo group by 70 l/min\n(SD 52). The null hypothesis is that there is no difference between\nthe bronchodilator and the placebo.\nThe t statistic is 11.14, resulting in a P value of 0.001. It is therefore very\nunlikely (1 in 1000 chance) that the null hypothesis is correct so we reject\nthe hypothesis and conclude that the new bronchodilator is significantly\nbetter than the placebo.\n\nWatch out for...\nParametric tests should only be used when the data\nfollow a “normal” distribution. You may find\nreference to the “Kolmogorov Smirnov” test. This\n\n30\n\ntests the hypothesis that the collected data are from a\nnormal distribution and therefore assesses whether\nparametric statistics can be used.\nSometimes authors will say that they have\n“transformed” data and then analyzed them with a\nparametric test. This is quite legitimate – it is not\ncheating! For example, a skewed distribution might\nbecome normally distributed if the logarithm of the\nvalues is used.\n\n>\n\nMANN–WHITNEY AND\nOTHER NON-PARAMETRIC\nTESTS\n\nHow important are they?\n\n88\n\nUsed in one in five papers.\n\nHow easy are they to understand?\nNon-parametric testing is difficult to understand.\nHowever, you do not need to know the details of the\ntests. Look out for the P value (see page 24) to see\nhow significant the results are. Remember, the\nsmaller the P value, the smaller the chance that the\n“null hypothesis” is true.\n\nWhen are they used?\nNon-parametric statistics are used when the data are\nnot normally distributed and so are not appropriate\nfor “parametric” tests.\n\nWhat do they mean?\nRather than comparing the values of the raw data,\nstatisticians “rank” the data and compare the ranks.\n\n32\n\nEXAMPLE\nMann–Whitney U test. A GP introduced a nurse triage system into her\npractice. She was interested in finding out whether the age of the patients\nattending for triage appointments was different to that of patients who\nmade emergency appointments with the GP.\nSix hundred and forty-six patients saw the triage nurse and 532 patients\nsaw the GP. The median age of the triaged patients was 50 years (1st\nquartile 40 years, 3rd quartile 54), for the GP it was 46 (22, 58). Note how\nthe quartiles show an uneven distribution around the median, so the data\ncannot be normally distributed and a non-parametric test is appropriate.\nThe graph in Fig. 9 shows the ages of the patients seen by the nurse and\nconfirms a skewed, rather than normal, distribution.\n\n200\n\nAges of patients\nFig. 9. Graph of ages of patients seen by triage nurse.\n\n≥70\n\n60–69\n\n50–59\n\n40–49\n\n30–39\n\n0\n\n20–29\n\n100\n\n10–19\n\nNumber of patients\n\n300\n\nMann–Whitney and other non-parametric tests\n\n33\n\nThe statistician used a “Mann–Whitney U test” to test the hypothesis that\nthere is no difference between the ages of the two groups. This gave a U\nvalue of 133 200 with a P value of < 0.001. Ignore the actual U value but\nconcentrate on the P value, which in this case suggests that the triage\nnurse’s patients were very highly significantly older than those who saw\nthe GP.\n\nWatch out for...\nThe “Wilcoxon signed rank test”, “Kruskal Wallis”\nand “Friedman” tests are other non-parametric tests.\nDo not be put off by the names – go straight to the P\nvalue.\n\n> CHI-SQUARED TEST\nUsually written as c2 (for the test) or C2 (for its\nvalue); Chi is pronounced as in sky without the s.\n\nHow important is it?\n\n8888\n\nA frequently used test of significance, given in a\nquarter of papers.\n\nHow easy is it to understand?\nDo not try to understand the C2 value, just look at\nwhether or not the result is significant.\n\nWhen is it used?\nIt is a measure of the difference between actual and\nexpected frequencies.\n\nWhat does it mean?\nThe “expected frequency” is that there is no\ndifference between the sets of results (the null\nhypothesis). In that case, the C2 value would be zero.\nThe larger the actual difference between the sets of\nresults, the greater the C2 value. However, it is\ndifficult to interpret the C2 value by itself as it\ndepends on the number of factors studied.\nStatisticians make it easier for you by giving the P\nvalue (see page 24), giving you the likelihood there is\nno real difference between the groups.\n\nChi-squared test\n\n35\n\nSo, do not worry about the actual value of C2 but\nlook at its P value.\nEXAMPLES\nA group of patients with bronchopneumonia were treated with either\namoxicillin or erythromycin. The results are shown in Table 3.\nTable 3. Comparison of effect of treatment of bronchopneumonia with amoxicillin\nor erythromycin\nType of antibiotic given\nAmoxicillin\nErythromycin\nImprovement at 5 days\nNo improvement at 5 days\nTotal\n\nTotal\n\n144\n\n(60%)\n\n160\n\n(67%)\n\n304\n\n96\n\n(40%)\n\n80\n\n(33%)\n\n176\n\n(63%)\n(37%)\n\n240\n\n(100%)\n\n240\n\n(100%)\n\n480\n\n(100%)\n\nC2 = 2.3; P = 0.13\nA table like this is known as a “contingency table” or “two-way table”.\n\nFirst, look at the table to get an idea of the differences between the\neffects of the two treatments.\nRemember, do not worry about the C2 value itself, but see whether it is\nsignificant. In this case P is 0.13, so the difference in treatments is not\nstatistically significant.\n\nWatch out for...\nSome papers will also give the “degrees of freedom”\n(df), for example C2 = 2.3; df 1; P = 0.13. See page 98\nfor an explanation. This is used with the C2 value to\nwork out the P value.\nOther tests you may find. Instead of the c2 test,\n“Fisher’s exact test” is sometimes used to analyze\ncontingency tables. Fisher’s test is the best choice as it\nalways gives the exact P value, particularly where the\nnumbers are small.\n\n36\n\nThe c2 test is simpler for statisticians to calculate but\ngives only an approximate P value and is\ninappropriate for small samples. Statisticians may\napply “Yates’ continuity correction” or other\nadjustments to the c2 test to improve the accuracy of\nthe P value.\nThe “Mantel Haenszel test” is an extension of the c2\ntest that is used to compare several two-way tables.\n\n> RISK RATIO\nOften referred to as relative risk.\n\nHow important is it?\n\n888\n\nUsed in one in six papers.\n\nHow easy is it to understand?\nRisk is a relatively intuitive concept that we encounter\nevery day, but interpretation of risk (especially low\nrisk) is often inconsistent. The risk of death while\ntravelling to the shops to buy a lottery ticket can be\nhigher than the risk of winning the jackpot!\n\nWhen is it used?\nRelative risk is used in “cohort studies”, prospective\nstudies that follow a group (cohort) over a period of\ntime and investigate the effect of a treatment or risk\nfactor.\n\nWhat does it mean?\nFirst, risk itself. Risk is the probability that an event\nwill happen. It is calculated by dividing the number\nof events by the number of people at risk.\nOne boy is born for every two births, so the\nprobability (risk) of giving birth to a boy is\n⁄2 = 0.5\n\n1\n\n38\n\nIf one in every 100 patients suffers a side-effect from\na treatment, the risk is\n⁄100 = 0.01\n\n1\n\nCompare this with odds (page 40).\nNow, risk ratios. These are calculated by dividing the\nrisk in the treated or exposed group by the risk in the\ncontrol or unexposed group.\nA risk ratio of one indicates no difference in risk\nbetween the groups.\nIf the risk ratio of an event is >1, the rate of that\nevent is increased compared to controls.\nIf <1, the rate of that event is reduced.\nRisk ratios are frequently given with their 95% CIs –\nif the CI for a risk ratio does not include one (no\ndifference in risk), it is statistically significant.\n\nRisk ratio\n\n39\n\nEXAMPLES\nA cohort of 1000 regular football players and 1000 non-footballers were\nfollowed to see if playing football was significant in the injuries that they\nAfter 1 year of follow-up there had been 12 broken legs in the football\nplayers and only four in the non-footballers.\nThe risk of a footballer breaking a leg was therefore 12/1000 or 0.012. The\nrisk of a non-footballer breaking a leg was 4/1000 or 0.004.\nThe risk ratio of breaking a leg was therefore 0.012/0.004 which equals\nthree. The 95% CI was calculated to be 0.97 to 9.41. As the CI includes the\nvalue 1 we cannot exclude the possibility that there was no difference in\nthe risk of footballers and non-footballers breaking a leg. However, given\nthese results further investigation would clearly be warranted.\n\n> ODDS RATIO\nHow important is it?\n\n8888\n\nUsed in a third of papers.\n\nHow easy is it to understand?\nOdds are difficult to understand. Just aim to\nunderstand what the ratio means.\n\nWhen is it used?\nUsed by epidemiologists in studies looking for factors\nwhich do harm, it is a way of comparing patients\nwho already have a certain condition (cases) with\npatients who do not (controls) – a “case–control\nstudy”.\n\nWhat does it mean?\nFirst, odds. Odds are calculated by dividing the\nnumber of times an event happens by the number of\ntimes it does not happen.\nOne boy is born for every two births, so the odds of\ngiving birth to a boy are 1:1 (or 50:50) = 1⁄1 = 1\nIf one in every 100 patients suffers a side-effect from\na treatment, the odds are\n1:99 = 1⁄99 = 0.0101\nCompare this with risk (page 37).\n\nOdds ratio\n\n41\n\nNext, odds ratios. They are calculated by dividing\nthe odds of having been exposed to a risk factor by\nthe odds in the control group.\nAn odds ratio of 1 indicates no difference in risk\nbetween the groups, i.e. the odds in each group are\nthe same.\nIf the odds ratio of an event is >1, the rate of that\nevent is increased in patients who have been exposed\nto the risk factor.\nIf <1, the rate of that event is reduced.\nOdds ratios are frequently given with their 95% CI –\nif the CI for an odds ratio does not include 1 (no\ndifference in odds), it is statistically significant.\n\nEXAMPLES\nA group of 100 patients with knee injuries, “cases”, was matched for age\nand sex to 100 patients who did not have injured knees, “controls”.\nIn the cases, 40 skied and 60 did not, giving the odds of being a skier for\nthis group of 40:60 or 0.66.\nIn the controls, 20 patients skied and 80 did not, giving the odds of being\na skier for the control group of 20:80 or 0.25.\nWe can therefore calculate the odds ratio as 0.66/0.25 = 2.64. The 95% CI\nis 1.41 to 5.02.\nIf you cannot follow the maths, do not worry! The odds ratio of 2.64\nmeans that the number of skiers in the cases is higher than the number of\nskiers in the controls, and as the CI does not include 1 (no difference in\nrisk) this is statistically significant. Therefore, we can conclude that skiers\nare more likely to get a knee injury than non-skiers.\n\n42\n\nWatch out for...\nAuthors may give the percentage change in the odds\nratio rather than the odds ratio itself. In the example\nabove, the odds ratio of 2.64 means the same as a\n164% increase in the odds of injured knees amongst\nskiers.\nOdds ratios are often interpreted by the reader in the\nsame way as risk ratios. This is reasonable when the\nodds are low, but for common events the odds and\nthe risks (and therefore their ratios) will give very\ndifferent values. For example, the odds of giving\nbirth to a boy are 1, whereas the risk is 0.5. However,\nin the side-effect example given above the odds are\n0.0101, a similar value to the risk of 0.01. For this\nreason, if you are looking at a case–control study,\ncheck that the authors have used odds ratios rather\nthan risk ratios.\n\n>\n\nRISK REDUCTION AND\nNUMBERS NEEDED TO\nTREAT\n\nHow important are they?\n\n888\n\nAlthough only quoted in less than 5% of papers, they\nare helpful in trying to work out how worthwhile a\ntreatment is in clinical practice.\n\nHow easy are they to understand?\n“Relative risk reduction” (RRR) and “absolute risk\nreduction” (ARR) need some concentration.\n“Numbers needed to treat” (NNT) are pretty intuitive,\nuseful and not too difficult to work out for yourself.\n\nWhen are they used?\nThey are used when an author wants to know how\noften a treatment works, rather than just whether it\nworks.\n\nWhat do they mean?\nARR is the difference between the event rate in the\nintervention group and that in the control group. It is\nalso the reciprocal of the NNT and is usually given as\n100\na percentage, i.e. ARR =\nNNT\nNNT is the number of patients who need to be\ntreated for one to get benefit.\n\n44\n\nRRR is the proportion by which the intervention\nreduces the event rate.\n\nEXAMPLES\nOne hundred women with vaginal candida were given an oral antifungal,\n100 were given placebo. They were reviewed 3 days later. The results are\ngiven in Table 4.\nTable 4. Results of placebo-controlled trial of oral antifungal agent\nGiven antifungal\n\nGiven placebo\n\nImproved\n\nNo improvement\n\nImproved\n\nNo improvement\n\n80\n\n20\n\n60\n\n40\n\nARR = improvement rate in the intervention group – improvement rate in\nthe control group = 80% – 60% = 20%\nNNT =\n\n100 100\n=\n=5\nARR 20\n\nSo five women have to be treated for one to get benefit.\nThe incidence of candidiasis was reduced from 40% with placebo to 20%\nwith treatment , i.e. by half.\nThus, the RRR is 50%.\nIn another trial young men were treated with an expensive lipid-lowering\nagent. Five years later the death rate from ischaemic heart disease (IHD)\nis recorded. See Table 5 for the results.\nTable 5. Results of placebo-controlled trial of Cleverstatin\nGiven Cleverstatin\n\nGiven placebo\n\nSurvived\n\nDied\n\nSurvived\n\nDied\n\n998 (99.8%)\n\n2 (0.2%)\n\n996 (99.6%)\n\n4 (0.4%)\n\nARR = improvement rate in the intervention group – improvement rate in\nthe control group = 99.8% – 99.6% = 0.2%\n\nRisk reduction and numbers needed to treat\n\nNNT =\n\n45\n\n100 100\n=\n= 500\nARR 0.2\n\nSo 500 men have to be treated for 5 years for one to survive who would\notherwise have died.\nThe incidence of death from IHD is reduced from 0.4% with placebo to\n0.2% with treatment – i.e. by half.\nThus, the RRR is 50%.\nThe RRR and NNT from the same study can have opposing effects on\nprescribing habits. The RRR of 50% in this example sounds fantastic.\nHowever, thinking of it in terms of an NNT of 500 might sound less\nattractive: for every life saved, 499 patients had unnecessary treatment\nfor 5 years.\n\nWatch out for...\nUsually the necessary percentages are given in the\nabstract of the paper. Calculating the ARR is easy:\nsubtract the percentage that improved without\ntreatment from the percentage that improved with\ntreatment.\nAgain, dividing that figure into 100 gives the NNT.\nWith an NNT you need to know:\n(a) What treatment?\n∑ What are the side-effects?\n∑ What is the cost?\n(b) For how long?\n\n46\n\n(c) To achieve what?\n∑ How serious is the event you are trying to\navoid?\n∑ How easy is it to treat if it happens?\nFor treatments, the lower the NNT the better – but\nlook at the context.\n(a) NNT of 10 for treating a sore throat with\nexpensive blundamycin\n∑ not attractive\n(b) NNT of 10 for prevention of death from\nleukaemia with a non-toxic chemotherapy agent\n∑ worthwhile\nExpect NNTs for prophylaxis to be much larger. For\nexample, an immunization may have an NNT in the\nthousands but still be well worthwhile.\nNumbers needed to harm (NNH) may also be\nimportant.\nNNH =\n\n100\n(% on treatment that had SEs) – (% not on treatment that had SEs)\nIn the example above, 6% of those on cleverstatin had\npeptic ulceration as opposed to 1% of those on placebo.\nNNH =\n\n100 100\n=\n= 20\n6–1\n5\n\ni.e. for every 20 patients treated, one peptic ulcer was\ncaused.\n\nRisk reduction and numbers needed to treat\n\n47\n\nXA\n\nMT\nIP\n\nE\n\nYou may see ARR and RRR given as a proportion\ninstead of a percentage. So, an ARR of 20% is the\nsame as an ARR of 0.2.\nBe prepared to calculate RRR, ARR and NNT from\na set of results. You may find that it helps to draw a\nsimple table like Table 5 and work from there.\n\n> CORRELATION\nHow important is it?\n\n88\n\nOnly used in 15% of medical papers.\n\nHow easy is it to understand?\n\nWhen is it used?\nWhere there is a linear relationship between two\nvariables there is said to be a correlation between\nthem. Examples are height and weight in children, or\nsocio-economic class and mortality.\nThe strength of that relationship is given by the\n“correlation coefficient”.\n\nWhat does it mean?\nThe correlation coefficient is usually denoted by the\nletter “r” for example r = 0.8.\nA positive correlation coefficient means that as one\nvariable is increasing the value for the other variable\nis also increasing – the line on the graph slopes up\nfrom left to right. Height and weight have a positive\ncorrelation: children get heavier as they grow taller.\nA negative correlation coefficient means that as the\nvalue of one variable goes up the value for the other\nvariable goes down – the graph slopes down from left\n\nCorrelation\n\n49\n\nto right. Higher socio-economic class is associated\nwith a lower mortality, giving a negative correlation\nbetween the two variables.\nIf there is a perfect relationship between the two\nvariables then r = 1 (if a positive correlation) or\nr = -1 (if a negative correlation).\nIf there is no correlation at all (the points on the\ngraph are completely randomly scattered) then r = 0.\nThe following is a good rule of thumb when\nconsidering the size of a correlation:\nr = 0–0.2\n\n: very low and probably meaningless.\n\nr = 0.2–0.4\n\n: a low correlation that might warrant\nfurther investigation.\n\nr = 0.4–0.6\n\n: a reasonable correlation.\n\nr = 0.6–0.8\n\n: a high correlation.\n\nr = 0.8–1.0\n\n: a very high correlation. Possibly too\nhigh! Check for errors or other\nreasons for such a high correlation.\n\nThis guide also applies to negative correlations.\n\nExamples\nA nurse wanted to be able to predict the laboratory HbA1c result (a\nmeasure of blood glucose control) from the fasting blood glucoses which\nshe measured in her clinic. On 12 consecutive diabetic patients she noted\nthe fasting glucose and simultaneously drew blood for HbA1c. She\ncompared the pairs of measurements and drew the graph in Fig. 10.\n\n50\n\n14\n12\n\nHbA1c %\n\n10\n8\n6\n4\n2\n0\n\n0\n\n5\n\n10\n15\nFasting blood glucose (mmol/l)\n\n20\n\n25\n\nFig. 10. Plot of fasting glucose and HbA1c in 12 patients with diabetes. For these\nresults r = 0.88, showing a very high correlation.\n\nA graph like this is known as a “scatter plot”.\nAn occupational therapist developed a scale for measuring physical\nactivity and wondered how much it correlated to Body Mass Index (BMI)\nin 12 of her adult patients. Fig. 11 shows how they related.\n\nActivity measure\n\n10\n8\n6\n4\n2\n0\n0\n\n10\n\n20\n30\nBody Mass Index\n\n40\n\n50\n\nFig. 11. BMI and activity measure in 12 adult patients.\n\nIn this example, r = -0.34, indicating a low correlation. The fact that the r\nvalue is negative shows that the correlation is negative, indicating that\npatients with a higher level of physical activity tended to have a lower BMI.\n\nCorrelation\n\n51\n\nWatch out for...\nCorrelation tells us how strong the association\nbetween the variables is, but does not tell us about\ncause and effect in that relationship.\nThe “Pearson correlation coefficient”, Pearson’s r, is\nused if the values are sampled from “normal”\npopulations (page 9). Otherwise the “Spearman rank\ncorrelation coefficient” is used. However, the\ninterpretation of the two is the same.\nWhere the author shows the graph, you can get a\ngood idea from the scatter as to how strong the\nrelationship is without needing to know the r value.\nAuthors often give P values with correlations;\nhowever, take care when interpreting them. Although\na correlation needs to be significant, we need also to\nconsider the size of the correlation. If a study is\nsufficiently large, even a small clinically unimportant\ncorrelation will be highly significant.\nR2 is sometimes given. As it is the square of the r\nvalue, and squares are always positive, you cannot\nuse it to tell whether the graph slopes up or down.\nWhat it does tell you is how much of the variation in\none value is caused by the other.\nIn Fig. 10, r = 0.88. R2 = 0.88 ¥ 0.88 = 0.77. This\nmeans that 77% of the variation in HbA1c is related\nto the variation in fasting glucose.\nAgain, the closer the R2 value is to 1, the higher the\ncorrelation.\n\n52\n\nIt is very easy for authors to compare a large number\nof variables using correlation and only present the\nones that happen to be significant. So, check to make\nsure there is a plausible explanation for any\nsignificant correlations.\nAlso bear in mind that a correlation only tells us\nabout linear (straight line) relationships between\nvariables. Two variables may be strongly related but\nnot in a straight line, giving a low correlation\ncoefficient.\n\n> REGRESSION\nHow important is it?\n\n888\n\nRegression analysis is used in a half of papers.\n\nHow easy is it to understand?\nThe idea of trying to fit a line through a set of points\nto make the line as representative as possible is\nrelatively straightforward. However, the mathematics\ninvolved in fitting regression models are more difficult\nto understand.\n\nWhen is it used?\nRegression analysis is used to find how one set of\ndata relates to another.\nThis can be particularly helpful where we want to use\none measure as a proxy for another – for example, a\nnear-patient test as a proxy for a lab test.\n\nWhat does it mean?\nA regression line is the “best fit” line through the\ndata points on a graph.\nThe regression coefficient gives the “slope” of the\ngraph, in that it gives the change in value of one\noutcome, per unit change in the other.\n\n54\n\nEXAMPLE\nConsider the graph shown in Fig. 10 (page 50). A statistician calculated the\nline that gave the “best fit” through the scatter of points, shown in Fig. 12.\n14\n12\n\nHbA1c %\n\n10\n8\n6\n4\n2\n0\n\n0\n\n5\n\n10\n15\nFasting blood glucose (mmol/l)\n\n20\n\n25\n\nFig. 12. Plot with linear regression line of fasting glucose and HbA1c in 12 patients\nwith diabetes.\n\nThe line is called a “regression line”.\nTo predict the HbA1c for a given blood glucose a nurse could simply plot it on\nthe graph, as here where a fasting glucose of 15 predicts an HbA1c of 9.95%.\nThis can also be done mathematically. The slope and position of the\nregression line can be represented by the “regression equation”:\nHbA1c = 3.2 + (0.45 ¥ blood glucose).\n\nThe 0.45 figure gives the slope of the graph and is called the “regression\ncoefficient”.\nThe “regression constant” that gives the position of the line on the graph\nis 3.2: it is the point where the line crosses the vertical axis.\nTry this with a glucose of 15:\nHbA1c = 3.2 + (0.45 x 15) = 3.2 + 6.75 = 9.95%\n\nRegression\n\n55\n\nThis regression equation can be applied to any\nregression line. It is represented by:\ny = a + bx\nTo predict the value y (value on the vertical axis of\nthe graph) from the value x (on the horizontal axis),\nb is the regression coefficient and a is the constant.\n\nOther values sometimes given with regression\nYou may see other values quoted. The regression\ncoefficient and constant can be given with their\n“standard errors”. These indicate the accuracy that\ncan be given to the calculations. Do not worry about\nthe actual value of these but look at their P values.\nThe lower the P value, the greater the significance.\nThe R2 value may also be given. This represents the\namount of the variation in the data that is explained\nby the regression. In our example the R2 value is\n0.77. This is stating that 77% of the variation in the\nHbA1c result is accounted for by variation in the\nblood glucose.\n\nOther types of regression\nThe example above is a “linear regression”, as the\nline that best fits the points is straight.\nOther forms of regression include:\nLogistic regression. This is used where each case in\nthe sample can only belong to one of two groups (e.g.\nhaving disease or not) with the outcome as the\nprobability that a case belongs to one group rather\nthan the other.\n\n56\n\nPoisson regression. Poisson regression is mainly\nused to study waiting times or time between rare\nevents.\nCox proportional hazards regression model. The\nCox regression model (page 60) is used in survival\nanalysis where the outcome is time until a certain\nevent.\n\nWatch out for...\nRegression should not be used to make predictions\noutside of the range of the original data. In the\nexample above, we can only make predictions from\nblood glucoses which are between 5 and 20.\n\nRegression or correlation?\nRegression and correlation are easily confused.\nCorrelation measures the strength of the association\nbetween variables.\nRegression quantifies the association. It should only\nbe used if one of the variables is thought to precede\nor cause the other.\n\n>\n\nSURVIVAL ANALYSIS:\nLIFE TABLES AND\nKAPLAN–MEIER PLOTS\n\nHow important are they?\n\n888\n\nSurvival analysis techniques are used in 20% of\npapers.\n\nHow easy are they to understand?\nLife tables are difficult to interpret. Luckily, most\npapers make it easy for you by showing the resulting\nplots – these graphs give a good visual feel of what\nhas happened to a population over time.\n\nWhen are they used?\nSurvival analysis techniques are concerned with\nrepresenting the time until a single event occurs. That\nevent is often death, but it could be any other single\nevent, for example time until discharge from hospital.\nSurvival analysis techniques are able to deal with\nsituations in which the end event has not happened in\nevery patient or when information on a case is only\nknown for a limited duration – known as “censored”\nobservations.\n\nWhat do they mean?\nLife table. A life table is a table of the proportion of\npatients surviving over time.\n\n58\n\nLife table methods look at the data at a number of\nfixed time points and calculate the survival rate at\nthose times. The most commonly used method is\nKaplan–Meier.\nKaplan–Meier\nThe Kaplan–Meier approach recalculates the\nsurvival rate when an end event (e.g. death) occurs in\nthe data set, i.e. when a change happens rather than\nat fixed intervals.\nThis is usually represented as a “survival plot”. Fig.\n13 shows a fictitious example.\n100\n\nCumulative survival (%)\n\n80\n\n60\n\n40\n\n20\n\n0\n0\n\n10\n\n20\n30\nSurvival (years)\n\n40\n\n50\n\nFig. 13. Kaplan–Meier survival plot of a cohort of patients with rheumatoid arthritis.\n\nThe dashed line shows that at 20 years, 36% of this\ngroup of patients were still alive.\n\nSurvival analysis: Life tables and Kaplan–Meier plots\n\n59\n\nWatch out for...\nLife tables and Kaplan–Meier survival estimates are\nalso used to compare survival between groups. The\nplots make any difference between survival in two\ngroups beautifully clear. Fig. 14 shows the same\ngroup of patients as above, but compares survival for\nmen and women.\n100\n\nWomen\nMen\n\nCumulative survival (%)\n\n80\n\n60\n\n40\n\n20\n\n0\n0\n\n10\n\n20\n30\nSurvival (years)\n\n40\n\n50\n\nFig. 14. Kaplan–Meier survival plot comparing men and women with rheumatoid\narthritis.\n\nIn this example 46% of women were still alive at 20\nyears but only 18% of men.\nThe test to compare the survival between these two\ngroups is called the “log rank test”. Its P value will\ntell you how significant the result of the test is.\n\nCOX REGRESSION\n> THE\nMODEL\n\nAlso known as the proportional hazards survival\nmodel.\n\nHow important is it?\n\n88\n\nIt appeared in a quarter of papers.\n\nHow easy is it to understand?\nJust aim to understand the end result – the “hazard\nratio” (HR).\n\nWhen is it used?\nThe Cox regression model is used to investigate the\nrelationship between an event (usually death) and\npossible explanatory variables, for instance smoking\nstatus or weight.\n\nWhat does it mean?\nThe Cox regression model provides us with estimates\nof the effect that different factors have on the time\nuntil the end event.\nAs well as considering the significance of the effect of\ndifferent factors (e.g. how much shorter male life\nexpectancy is compared to that of women), the\nmodel can give us an estimate of life expectancy for\nan individual.\n\nThe Cox regression model\n\n61\n\nThe “HR” is the ratio of the hazard (chance of\nsomething harmful happening) of an event in one\ngroup of observations divided by the hazard of an\nevent in another group. An HR of 1 means the risk is\n1 ¥ that of the second group, i.e. the same. An HR of\n2 implies twice the risk.\n\nEXAMPLE\nThe Cox regression model shows us the effect of being in one group\ncompared with another.\nUsing the rheumatoid arthritis cohort on page 58, we can calculate the\neffect that gender has on survival. Table 6 gives the results of a Cox model\nestimate of the effect.\nTable 6. Cox model estimate of the effect of sex on survival in a cohort of patients\nwith rheumatoid arthritis\nParameter\n\nHRa (95% CI)b\n\ndf c\n\nP valued\n\nSex (Male)\n\n1.91 (1.21 to 3.01)\n\n1\n\n<0.05\n\na\n\nThe HR of 1.91 means that the risk of death in any particular time period for men was 1.91\n\ntimes that for women.\nb\n\nThis CI means we can be 95% confident that the true HR is between 1.21 and 3.01.\n\nc\n\nDegrees of freedom – see glossary, page 98.\n\nd\n\nThe P value of <0.05 suggests that the result is significant.\n\nSPECIFICITY\n> SENSITIVITY,\nAND PREDICTIVE VALUE\n\nHow important are they?\n\n888\n\nThey are discussed in 40% of papers, so a working\nknowledge is important in interpreting papers that\nstudy screening.\n\nHow easy are they to understand?\nThe tables themselves are fairly easy to understand.\nHowever, there is a bewildering array of information\nthat can be derived from them.\nsection until you are feeling fresh. You may need to\ngo over it a few days running until it is clear in your\nmind.\n\nWhen are they used?\nThey are used to analyze the value of screening or\ntests.\n\nWhat do they mean?\nThink of any screening test for a disease. For each\npatient:\n∑ the disease itself may be present or absent;\n∑ the test result may be positive or negative.\n\nSensitivity, specificity and predictive value\n\n63\n\nWe need to know how useful the test is.\nThe results can be put in the “two-way table” shown\nin Table 7. Try working through it.\nTable 7. Two-way table\nDisease:\nTest result:\n\nPresent\n\nAbsent\n\nPositive\n\nA\n\nB (False positive)\n\nNegative\n\nC (False negative)\n\nD\n\nSensitivity. If a patient has the disease, we need to\nknow how often the test will be positive, i.e.\n“positive in disease”.\nThis is calculated from:\n\nA\nA+C\n\nThis is the rate of pick-up of the disease in a test, and\nis called the Sensitivity.\nSpecificity. If the patient is in fact healthy, we want\nto know how often the test will be negative, i.e.\n“negative in health”.\nThis is given by:\n\nD\nD+B\n\nThis is the rate at which a test can exclude the possibility\nof the disease , and is known as the Specificity.\nPositive Predictive Value. If the test result is\npositive, what is the likelihood that the patient will\nhave the condition?\nLook at:\n\nA\nA+B\n\nThis is known as the Positive Predictive Value (PPV).\n\n64\n\nNegative Predictive Value. If the test result is\nnegative, what is the likelihood that the patient will\nbe healthy?\nHere we use:\n\nD\nD+C\n\nThis is known as the Negative Predictive Value (NPV).\nIn a perfect test, the sensitivity, specificity, PPV and\nNPV would each have a value of 1. The lower the\nvalue (the nearer to zero), the less useful the test is in\nthat respect.\n\nEXAMPLES\nConfused? Try working through an example.\nImagine a blood test for gastric cancer, tried out on 100 patients admitted\nwith haematemesis. The actual presence or absence of gastric cancers\nwas diagnosed from endoscopic findings and biopsy. The results are\nshown in Table 8.\nTable 8. Two-way table for blood test for gastric cancer\nGastric cancer:\n\nBlood result:\n\nSensitivity =\n\nPresent\n\nAbsent\n\nPositive\n\n20\n\n30\n\nNegative\n\n5\n\n45\n\n20\n20\n=\n= 0.8\n20 + 5 25\n\nIf the gastric cancer is present, there is an 80% (0.8) chance of the test\npicking it up.\nSpecificity =\n\n45\n45\n=\n= 0.6\n30 + 45 75\n\nIf there is no gastric cancer there is a 60% (0.6) chance of the test being\nnegative – but 40% will have a false positive result.\n\nSensitivity, specificity and predictive value\n\nPPV =\n\n65\n\n20\n20\n=\n= 0.4\n20 + 30 50\n\nThere is a 40% (0.4) chance, if the test is positive, that the patient actually\nhas gastric cancer.\nNPV =\n\n45\n45\n=\n= 0.9\n45 + 5 50\n\nThere is a 90% (0.9) chance, if the test is negative, that the patient does\nnot have gastric cancer. However, there is still a 10% chance of a false\nnegative, i.e. that the patient does have gastric cancer.\n\nWatch out for...\nOne more test to know about.\nThe “Likelihood Ratio” (LR) gives an estimate of\nhow much a test result will change the odds of having\na condition.\nThe LR for a positive result (LR+) tells us how much\nthe odds of the condition increase when the test\nresult is positive.\nThe LR for a negative result (LR–) tells us how much\nthe odds of the condition decrease when the test\nresult is negative.\nTo calculate LR+, divide the sensitivity by (1 –\nspecificity). To calculate LR–, divide (1 – sensitivity)\nby the specificity.\nHead spinning again? Try using the example above\nto calculate the LR for a positive result.\nLR =\n\n0.8\n0.8\nsensitivity\n=\n=\n=2\n(1 – specificity) 1 - 0.6 0.4\n\nIn this example, LR+ for a positive result = 2. This\nmeans that if the test is positive in a patient, the odds\nof that patient having gastric cancer are doubled.\n\n66\n\nTip: Invent an imaginary screening or diagnostic test\nof your own, fill the boxes in and work out the\nvarious values. Then change the results to make the\ntest a lot less or more effective and see how it affects\nthe values.\nOne thing you may notice is that in a rare condition,\neven a diagnostic test with a very high sensitivity may\nresult in a low PPV.\nIf you are still feeling confused, you are in good\ncompany. Many colleagues far brighter than us\nadmit that they get confused over sensitivity, PPV etc.\nTry copying the following summary into your diary\nand refer to it when reading a paper:\n∑ Sensitivity: how often the test is positive if the\npatient has the disease.\n∑ Specificity: if the patient is healthy, how often the\ntest will be negative.\n∑ PPV: If the test is positive, the likelihood that the\npatient has the condition.\n∑ NPV: If the test is negative, the likelihood that the\npatient will be healthy.\n\nXA\n\nMT\nIP\n\nE\n\n∑ LR: If the test is positive, how much more likely\nthe patient is to have the disease than not have it.\nExaminers love to give a set of figures which you can\nturn into a two-way table and ask you to calculate\nsensitivity, PPV etc. from them. Keep practising until\nyou are comfortable at working with these values.\n\nOF AGREEMENT AND\n> LEVEL\nKAPPA\n\nKappa is often seen written as k.\n\nHow important is it?\n\n8\n\nNot often used.\n\nHow easy is it to understand?\n\nWhen is it used?\nIt is a comparison of how well people or tests agree\nand is used when data can be put into ordered\ncategories.\nTypically it is used to look at how accurately a test\ncan be repeated.\n\nWhat does it mean?\nThe kappa value can vary from zero to 1.\nA k of zero means that there is no significant\nagreement – no more than would have been expected\nby chance.\nA k of 0.5 or more is considered a good agreement, a\nvalue of 0.7 shows very good agreement.\nA k of 1 means that there is perfect agreement.\n\n68\n\nEXAMPLE\nIf the same cervical smear slides are examined by the cytology\ndepartments of two hospitals and k = 0.3, it suggests that there is little\nagreement between the two laboratories.\n\nWatch out for...\nKappa can be used to analyze cervical smear results\nbecause they can be ordered into separate categories,\ne.g. CIN 1, CIN 2, CIN 3 – so-called “ordinal data”.\nWhen the variable that is being considered is\ncontinuous, for example blood glucose readings, the\n“intra-class correlation coefficient” should be used\n(see glossary).\n\n> OTHER CONCEPTS\nImportance: 8\nEase of understanding:\nOne fundamental principle of statistics is that we\naccept there is a chance we will come to the wrong\nconclusion. If we reject a null hypothesis with a P\nvalue of 0.05, then there is still the 5% possibility\nthat we should not have rejected the hypothesis and\ntherefore a 5% chance that we have come to the\nwrong conclusion.\nIf we do lots of testing then this chance of making a\nmistake will be present each time we do a test and\ntherefore the more tests we do the greater the chances\nof drawing the wrong conclusion.\nadjust the P value to keep the overall chance of coming\nto the wrong conclusion at a certain level (usually 5%).\nThe most commonly used\n\nmethod\n\nis\n\nthe\n\n1- and 2-tailed tests\nImportance: 8\nEase of understanding:\nWhen trying to reject a “null hypothesis” (page 27),\nwe are generally interested in two possibilities: either\n\n70\n\nwe can reject it because the new treatment is better\nthan the current one, or because it is worse. By\nallowing the null hypothesis to be rejected from\neither direction we are performing a “two-tailed\ntest” – we are rejecting it when the result is in either\n“tail” of the test distribution.\nOccasionally there are situations where we are only\ninterested in rejecting a hypothesis if the new\ntreatment is worse that the current one but not if it is\nbetter. This would be better analyzed with a onetailed test. However, be very sceptical of one-tailed\ntests. A P value that is not quite significant on a twotailed test may become significant if a one-tailed test\nis used. Authors have been known to use this to their\n\nIncidence\nImportance: 8888\nEase of understanding:\nThe number of new cases of a condition over a given\ntime as a percentage of the population.\nExample: Each year 15 people in a practice of 1000\npatients develop Brett’s palsy.\n15\n¥ 100 = yearly incidence of 1.5%\n1000\n\nPrevalence (= Point Prevalence Rate)\nImportance 8888\nEase of understanding:\nThe existing number of cases of a condition at a\nsingle point in time as a percentage of the population.\n\nOther concepts\n\n71\n\nExample: At the time of a study 90 people in a\npractice of 1000 patients were suffering from Brett’s\npalsy (15 diagnosed in the last year plus 75\ndiagnosed in previous years).\n90\n¥ 100 = a prevalence of 9%\n1000\nWith chronic diseases like the fictitious Brett’s palsy,\nthe incidence will be lower than the prevalence – each\nyear’s new diagnoses swell the number of existing\ncases.\n\nXA\n\nMT\nIP\n\nE\n\nWith short-term illnesses the opposite may be true.\n75% of a population may have a cold each year\n(incidence), but at any moment only 2% are actually\nsuffering (prevalence).\nCheck that you can explain the difference between\nincidence and prevalence.\n\nPower\nImportance: 88\nEase of understanding:\nThe power of a study is the probability that it would\ndetect a statistically significant difference.\nIf the difference expected is 100% cure compared\nwith 0% cure with previous treatments, a very small\nstudy would have sufficient power to detect that.\nHowever if the expected difference is much smaller,\ne.g. 1%, then a small study would be unlikely to have\nenough power to produce a result with statistical\nsignificance.\n\n72\n\nBayesian statistics\nImportance: 8\nEase of understanding:\nBayesian analysis is not often used. It is a totally\ndifferent statistical approach to the classical,\n“frequentist” statistics explained in this book.\nIn Bayesian statistics, rather than considering the\nsample of data on its own, a “prior distribution” is\nset up using information that is already available. For\ninstance, the researcher may give a numerical value\nand weighting to previous opinion and experience as\nwell as previous research findings.\nOne consideration is that different researchers may\nput different weighting on the same previous\nfindings.\nThe new sample data are then used to adjust this\nprior information to form a “posterior distribution”.\nThus these resulting figures have taken both the\ndisparate old data and the new data into account.\nBayesian methodology is intuitively appealing\nbecause it reflects how we think. When we read\nabout a new study we do not consider its results in\nisolation, we factor it in to our pre-existing opinions,\nknowledge and experience of dealing with patients.\nIt is only recently that computer power has been\nsufficient to calculate the complex models that are\nneeded for Bayesian analysis.\n\n> STATISTICS AT WORK\nIn this section we have given five real-life examples of\nhow researchers use statistical techniques to describe\nand analyze their work.\nThe extracts have been taken from research papers\npublished in the British Medical Journal (reproduced\nwith permission of The BMJ Publishing Group), The\nLancet (reproduced with permission from Elsevier),\nand the New England Journal of Medicine\n(reproduced with permission from Massachusetts\nMedical Society). If you want to see the original\n∑ The BMJ website http://www.bmj.com/\n∑ The Lancet website http://www.thelancet.com/\n∑ The NEJM website http://content.nejm.org/\nIf you wish, you can use this part to test what you\nhave learnt:\n∑ First, go through the abstracts and results and note\ndown what statistical techniques have been used.\n∑ Then try to work out why the authors have used\nthose techniques.\n∑ Next, try to interpret the results.\n∑ Finally, check out your understanding\ncomparing it with our commentary.\n\nby\n\n74\n\nThe following extract is reproduced with permission\nfrom The BMJ Publishing Group.\n\nStandard deviation, relative risk and numbers needed to\ntreat\nAlho, O.-P., Koivunen, P., Penna, T., et al. (2007). Tonsillectomy versus watchful\nwaiting in recurrent streptococcal pharyngitis in adults: randomised controlled\ntrial. BMJ 334: 939. (Originally published online 8 Mar 2007;\ndoi:10.1136/bmj.39140.632604.55.)\n\nAbstract\nObjective: To determine the short-term efficacy\nand safety of tonsillectomy for recurrent\nDesign: Randomised controlled trial.\nSetting: Academic referral centre in Finland.\nParticipants: 70 adults with documented recurrent\nepisodes of streptococcal group A pharyngitis.\nIntervention: Instant tonsillectomy (n=36) or\nremaining on waiting list as control (n=34).\nMain outcome measures: Percentage change in the\nrisk of an episode of streptococcal pharyngitis at\n90 days. Rates of all episodes of pharyngitis and\ndays with symptoms and adverse effects.\nResults: The mean (SD) follow-up was 164 (63)\ndays in the control group and 170 (12) days in the\ntonsillectomy group. At 90 days, streptococcal\npharyngitis had recurred in 24% (8/34) in the\ncontrol group and 3% (1/36) in the tonsillectomy\ngroup (difference 21%; 95% confidence interval\n6% to 36%). The number needed to undergo\ntonsillectomy to prevent one recurrence was 5 (95%\nconfidence interval 3 to 16). During the whole\nfollow-up, the rates of other episodes of pharyngitis\nand days with throat pain and fever were\nsignificantly lower in the tonsillectomy group than\n\nStatistics at work\n\n75\n\nin the control group. The most common morbidity\nrelated to tonsillectomy was postoperative throat\npain (mean length 13 days, SD 4).\nConclusions: Adults with a history of documented\nrecurrent episodes of streptococcal pharyngitis\nwere less likely to have further streptococcal or\nother throat infections or days with throat pain if\n\nWhat statistical methods were used and why?\nWhile tonsillectomy has been used to prevent\nrecurrent streptococcal throat infections, a recent\nreview showed no evidence that it was effective in\nwith recurrent episodes of streptococcal pharyngitis\nsought to determine the effects of tonsillectomy.\nThe authors assumed that the length of follow-up\nwas normally distributed, so compared the two\ngroups by giving their mean length of follow-up.\nThey wanted to indicate how much the length of\nfollow-up was spread around the mean, so gave the\nstandard deviation of the number of days.\nAs there were different numbers of adults in each\ngroup, percentages were given as a scale on which to\ncompare the number of patients with a recurrence of\nstreptococcal pharyngitis (the incidence of pharyngitis).\nThis was a prospective study, following two cohorts of\nadults, and used the difference in incidences of\nrecurrence of streptococcal pharyngitis to compare the\neffect of the two different methods of management.\nThe authors wanted to give the range that was likely\nto contain the true difference, so gave it with its 95%\nconfidence interval.\n\n76\n\nThe null hypothesis was that there would be no\ndifference between the incidences of recurrence of\nstreptococcal pharyngitis in the two groups.\n\nWhat do the results mean?\nThe standard deviation of the mean duration of\npostoperative throat pain was 4 days.\nIn this group:\n1 SD below the average is 13 – 4 = 9 days.\n1 SD above the average is 13 + 4 = 17 days.\n± 1 SD will include 68.2% of the subjects, so 68.2%\nbetween 9 and 17 days.\n95.4% had pain for between 5 and 21 days (± 2 SD).\n99.7% of the patients would have had postoperative\nthroat pain for between 1 and 25 days (± 3 SD).\nrecurrence of streptococcal pharyngitis. As a\npercentage, this is:\n8\n¥ 100 = 24%\n34\nThis is the same as the risk, or probability, that a\nrecurrence would happen in this group.\nThe difference in incidence between recurrences over\n90 days was 21%. The confidence interval (CI) of\n6% to 36% doesn’t include 0% (no difference in\nrisk), so it is statistically significant.\n\nStatistics at work\n\n77\n\nThe risk ratio wasn’t given in the abstract, but can be\ncalculated by dividing the risk in the tonsillectomy\ngroup with that in the control group:\n3%\n= 0.125\n24%\nThe risk ratio of <1 shows that the rate of recurrence\nin the tonsillectomy group was lower than that in the\ncontrol group.\nFrom the results of the research, the clinician can\ncalculate absolute risk reduction (ARR), number\nneeded to treat (NNT) and relative risk reduction\n(RRR) from doing a tonsillectomy.\nARR = [risk in the control group] – [risk in the\ntonsillectomy group] = 24% – 3% = 21%\n100\n100\nNNT = ᎏᎏ = ᎏᎏ = 4.8\nARR\n21\nSo, for every 5 adults, tonsillectomy would prevent\none patient from experiencing a recurrence of\nstreptococcal pharyngitis in the first 90 days.\nThe risk of reactions was reduced from 24% to 3%\nby tonsillectomy, so the RRR is given by:\n21\n24% – 3%\nRRR = ᎏ ᎏ = ᎏᎏ = 0.88 = 88%\n24\n24%\nThe researchers concluded that adults with recurrent\nepisodes of streptococcal pharyngitis were less likely\nto have further streptococcal throat infections or\ndays with throat pain if they had their tonsils\nremoved.\nTonsillectomy significantly reduced rates of a\nrecurrence of streptococcal pharyngitis in the first 90\ndays, with an NNT of 5.\n\n78\n\nThe following extract is reproduced with permission\nfrom Elsevier.\n\nOdds ratios and confidence intervals\nYin, P., Jiang, C.Q., Cheng, K.K., et al. (2007). Passive smoking exposure and risk of\nCOPD among adults in China: the Guangzhou Biobank Cohort Study. Lancet 370:\n751–57.\n\nSummary\nBackground: Chronic obstructive pulmonary\ndisease (COPD) is a leading cause of mortality in\nChina, where the population is also exposed to high\nlevels of passive smoking, yet little information\nexists on the effects of such exposure on COPD. We\nexamined the relation between passive smoking\nand COPD and respiratory symptoms in an adult\nChinese population.\nMethods: We used baseline data from the\nGuangzhou Biobank Cohort Study. Of 20 430 men\nand women over the age of 50 recruited in\n2003–06, 15 379 never smokers (6497 with valid\nspirometry) were included in this cross-sectional\nanalysis. We measured passive smoking exposure at\nhome and work by two self-reported measures\n(density and duration of exposure). Diagnosis of\nCOPD was based on spirometry and defined\naccording to the GOLD guidelines.\nFindings: There was an association between risk of\nCOPD and self-reported exposure to passive\nsmoking at home and work (adjusted odds ratio\n1.48, 95% CI 1.18–1.85 for high level exposure;\nequivalent to 40 h a week for more than 5 years).\nThere were significant associations between\nreported respiratory symptoms and increasing\npassive smoking exposure (1.16, 1.07–1.25 for any\nsymptom).\n\nStatistics at work\n\n79\n\nWhat statistical methods were used and why?\nThe authors felt that the evidence of effects of passive\nsmoking exposure on lung function showed mixed\nresults.\nThey therefore analyzed a large cohort of patients to\nstudy the association between passive smoking and\nCOPD. They studied the odds of COPD in passive\nsmokers with those of patients who had not been\nexposed to tobacco, and compared the two groups\nwith an odds ratio.\nThe authors wanted to give the ranges that were\nlikely to contain the true odds ratios, so gave them\nwith their 95% confidence intervals (CI).\n\nWhat do the results mean?\nIn the Lancet paper, the authors gave the results and\nodds ratios for 8 outcomes of each of the trials. We\nhave extracted the results for COPD risk for those\nwith the highest passive exposure to tobacco (Table 9).\nTable 9. Relation between self-reported passive smoking exposure and COPD in never\nsmokers: total hours of adulthood home and work exposure.\nn (%) without\nCOPD\n\nn (%) with\nCOPD\n\nOR (95% CI)\n\n<2 yrs of 40 h per week\n\n2999 (94.0)\n\n191 (6.0)\n\n1\n\n2–5 yrs of 40 h per week\n\n1409 (94.5)\n\n82 (5.5)\n\n0.91 (0.70–1.19)\n\n>5 yrs of 40 h per week\n\n1660 (91.4)\n\n156 (8.6)\n\n1.48 (1.18–1.84)\nP = 0.001\n\nConsider the risk of COPD:\nOdds for COPD in patients with low passive\nexposure (<2 yrs) to tobacco =\nnumber of times the event (COPD) happens\nnumber of times it doesn’t happen\n\n80\n\n=\n\n191\n= 0.064\n2999\n\nOdds for COPD in patients with high passive\nexposure (>5 yrs) to tobacco\n= 156 = 0.094\n1660\nOdds ratio (OR) =\nOdds of COPD in patients with high tobacco exposure\nᎏᎏᎏᎏᎏᎏ\nOdds of COPD in patients with low tobacco exposure\n=\n\n0.094\n= 1.48\n0.064\n\nThe OR of >1 indicates that the rate of COPD is\nincreased in patients with high levels of passive\nsmoking compared to patients with low levels.\nThe 95% confidence interval was calculated to be\n1.18–1.85. As the CI for the odds ratio doesn’t\ninclude 1 (no difference in odds), the difference\nbetween the results in this study is statistically\nsignificant.\nThe OR between reported respiratory symptoms and\nincreasing passive smoking exposure was 1.16.\nAgain, the OR of >1 indicates an increased risk with\nhigh levels of passive smoking. The fact that the CI\nfor the OR is 1.07–1.25, and doesn’t include 1,\nindicates that this result is also significant.\nThe authors concluded that exposure to passive\nsmoking is associated with an increased prevalence\nof COPD and respiratory symptoms.\n\nStatistics at work\n\n81\n\nThe following extract is reproduced with permission\nfrom The BMJ Publishing Group.\n\nCorrelation and regression\nPriest, P., Yudkin, P., McNulty, C. and Mant, D. (2001). Antibacterial prescribing and\nantibacterial resistance in English general practice: cross sectional study. BMJ\n323: 1037–1041.\n\nAbstract\nObjective: To quantify the relation between\ncommunity based antibacterial prescribing and\nantibacterial resistance in community acquired\ndisease.\nDesign: Cross sectional study of antibacterial\nprescribing and antibacterial resistance of routine\nisolates within individual practices and primary care\ngroups.\nSetting: 405 general practices in south west and\nnorth west England.\nMain outcome measures: Correlation between\nantibacterial prescribing and resistance for urinary\ncoliforms.\nResults: Antibacterial resistance in urinary\ncoliform isolates is common but the correlation\nwith prescribing rates was relatively low for\nindividual practices (ampicillin and amoxicillin\nrs=0.20, P=0.001). The regression coefficient was\nalso low; a practice prescribing 20% less ampicillin\nand amoxicillin than average would have about\n1% fewer resistant isolates.\n\n% of urinary coliforms resistant\nto ampicillin/amoxicillin\n\n82\n\n80\n70\n60\n\ny = 0.019x + 39.6\n\n50\n40\n30\n20\n10\n0\n0\n\n100\n\n200\n\n300\n\n400\n\n500\n\n600\n\n700\n\n800\n\n900\n\nAmpicillin/amoxicillin: prescriptions/1000 patients/year\n\nFig. 15. Graph of regression of antibacterial resistance of urinary coliforms on\nprescribing at practice level\n\nWhat statistical methods were used and why?\nThe researchers wanted to know whether there\nwas a relationship between antibacterial\nprescribing and antibiotic resistance, so needed to\nmeasure the correlation between the two.\nAs the data were skewed, i.e. not normally\ndistributed, they needed to use the Spearman rank\ncorrelation coefficient as a measure of the\ncorrelation.\nThey wished to predict how much a 20%\nreduction in antibiotic prescribing would be likely\nto reduce antibacterial resistance. To do this they\nneeded to calculate the regression coefficient.\n\nStatistics at work\n\n83\n\nWhat do the results mean?\nThe Spearman rank correlation coefficient for\npenicillin prescribing and resistance for urinary\ncoliforms is given as rs = 0.20.\nThe r indicates that it is the correlation coefficient,\nand the small s behind it shows that it is a Spearman\nrank correlation coefficient.\nThe r value of 0.20 indicates a low correlation.\nP = 0.001 indicates that an r value of 0.20 would\nhave happened by chance 1 in 1000 times if there\nwere really no correlation.\nFigure 15 shows the authors’ regression graph\ncomparing antibiotic prescribing with antibiotic\nresistance. Each dot represents one practice’s results.\nThe line through the dots is the regression line.\nThe authors show the regression equation in the\ngraph: y = 0.019x + 39.6.\nIn this equation:\nx (the horizontal axis of the graph) indicates the\nnumber of penicillin prescriptions per 1000 patients\nper year.\ny (the vertical axis of the graph) indicates the\npercentage of urinary coliforms resistant to\npenicillin.\nThe regression constant of 39.6 is the point at which\nthe regression line would hit the vertical axis of the\ngraph (when number of prescriptions = 0).\n\n84\n\nThe regression coefficient gives the slope of the line –\nthe percentage of resistant bacteria reduces by 0.019\nfor every one less penicillin prescription per 1000\npatients per year.\nUsing this value, the authors calculated that a\npractice prescribing 20% less penicillin than average\nwould only have about 1% fewer resistant bacteria.\nThe authors therefore suggested that trying to reduce\nthe overall level of antibiotic prescribing in UK\ngeneral practice may not be the most effective\nstrategy for reducing antibiotic resistance in the\ncommunity.\n\nStatistics at work\n\n85\n\nThe following extract is reproduced with permission\nfrom the Massachusetts Medical Society, © 2007; all\nrights reserved.\n\nSurvival analysis\nSjöström, L., Narbro, K., Sjöström, C.D., et al. (2007). Effects of bariatric surgery\non mortality in Swedish obese subjects. N Engl J Med 357: 741–52.\n\nAbstract\nBackground: Obesity is associated with increased\nmortality. Weight loss improves cardiovascular risk\nfactors, but no prospective interventional studies\nhave reported whether weight loss decreases\noverall mortality. In fact, many observational\nstudies suggest that weight reduction is associated\nwith increased mortality.\nMethods: The prospective, controlled Swedish\nObese Subjects study involved 4047 obese subjects.\nOf these subjects, 2010 underwent bariatric\nsurgery (surgery group) and 2037 received\nconventional treatment (matched control group).\nWe report on overall mortality during an average\nof 10.9 years of follow-up.\nCox proportional-hazards models were used to\nevaluate time to death while adjusting for\npotentially significant risk factors.\nResults: There were 129 deaths in the control\ngroup and 101 deaths in the surgery group.\nThe overall hazard ratio was 0.76 in the surgery\ngroup (P = 0.04), as compared with the control\ngroup.\n\n86\n\n14\n12\nCumulative mortality (%)\n\nControl\n10\n8\nSurgery\n6\n4\nP = 0.04\n2\n0\n0\n\n2\n\n4\n\n6\n\n8\n\n10\n\n12\n\n14\n\n16\n\n1260\n1174\n\n760\n749\n\n422\n422\n\n169\n156\n\nYears\nNo. at risk\nSurgery\nControl\n\n2010\n2037\n\n2001\n2027\n\n1987\n2016\n\n1821\n1842\n\n1590\n1455\n\nFig. 16. Cumulative mortality. The hazard ratio for subjects who underwent\nbariatric surgery, as compared with control subjects, was 0.76 (95% CI, 0.59 to\n0.99; P=0.04), with 129 deaths in the control group and 101 in the surgery group.\n\nWhat statistical methods were used and why?\nThe authors wanted to find out whether surgical\nprocedures to treat severe obesity (bariatric\nsurgery) affect mortality.\nThe null hypothesis was that there was no\ndifference between the surgery group and those\nreceiving conventional treatment.\nThe cumulative mortality plot (Fig. 16) was used\nto give a visual representation of the mortality\nover time in each group. The survival between\nthese two groups was compared using the hazard\nratio, and the likelihood that there was no real\n\nStatistics at work\n\n87\n\ndifference between the groups was given by the P\nvalue.\nThe Cox regression model was used so that the effect\nof different variables could be studied.\n\nWhat do the results mean?\nThe ratio of the chance (hazard) of death in the\nsurgery group, divided by the chance of death in the\nconventional treatment group, was calculated to be\n0.76. A hazard ratio (HR) of 1 would mean the risk\nis 1¥ that of the second group, i.e. the same. The\nhazard ratio of 0.76 in this study means that there\nwas a lower risk in the surgery group.\nP=0.04, so the probability of the difference having\nhappened by chance is 0.04 in 1, i.e. 1 in 25.\nAs P<0.05, this is considered to be statistically\nsignificant.\nThis is stated another way by giving the 95%\nconfidence interval, given in Fig. 16. The CI of 0.59\nto 0.99 doesn’t include 1 (no difference in risk), so\nagain demonstrating statistical significance.\nThe conclusion was that bariatric surgery for severe\nobesity is associated with decreased overall\nmortality.\n\n88\n\nThe following extract is reproduced with permission\nfrom The BMJ Publishing Group.\n\nSensitivity, specificity and predictive values\nHopper, A.D., Cross, S.S., Hurlstone, D.P. et al. (2007). Pre-endoscopy serological\ntesting for coeliac disease: evaluation of a clinical decision tool. BMJ 334; 729.\n(Originally published online 23 Mar 2007; doi:10.1136/bmj.39133.668681.BE.)\n\nAbstract\nObjective: To determine an effective diagnostic\nmethod of detecting all cases of coeliac disease in\npatients referred for gastroscopy without\nperforming routine duodenal biopsy.\nDesign: An initial retrospective cohort of patients\nattending for gastroscopy was analysed to derive a\nclinical decision tool that could increase the\ndetection of coeliac disease without performing\nroutine duodenal biopsy. The tool incorporated\nserology and stratifying patients according to\ntheir referral symptoms (“high risk” or “low risk”\nof coeliac disease). The decision tool was then\ntested on a second cohort of patients attending for\ngastroscopy. In the second cohort all patients had\na routine duodenal biopsy and serology\nperformed.\nreferred for gastroscopy recruited prospectively.\nMain outcome measure: Evaluation of a clinical\ndecision tool using patients’ referral symptoms,\ntissue transglutaminase antibody results, and\nduodenal biopsy results.\nResults: No cases of coeliac disease were missed\nby the pre-endoscopy testing algorithm.\n\nStatistics at work\n\n89\n\nEvaluation of the clinical decision tool gave a\nsensitivity, specificity, positive predictive value,\nand negative predictive value of 100%, 60.8%,\n9.3%, and 100%, respectively.\nTable 10. Two-way table of number of patients categorised as having coeliac\ndisease by clinical decision tool and by gold standard (duodenal biopsy): values\nderived from data given in the paper.\nOutcome of duodenal\nbiopsy\nPositive\nResult from clinical decision tool\nTotal\n\nTotal\n\nNegative\n\nPositive\n\n77\n\n753\n\n830\n\nNegative\n\n0\n\n1170\n\n1170\n\n77\n\n1923\n\n2000\n\nWhat statistical methods were used and why?\nThe researchers wanted to know whether they\ncould use their clinical decision tool to replace\nduodenal biopsy in diagnosing coeliac disease.\nThey decided whether patients actually had\ncoeliac disease by using a “gold standard” test,\nduodenal biopsy.\nWe have used a two-way table to show the results,\nshown in Table 10.\nSensitivity, specificity, predictive values and\nlikelihood ratios were used to give the value of\ntheir clinical decision tool.\n\n90\n\nWhat do the results mean?\nIn a perfect test, the sensitivity, specificity and\npredictive values would each have a value of 1. The\nlower the value (the nearer to zero), the less useful\nthe test is in that respect.\nSensitivity shows the rate of pick-up of the disease –\nif a patient has coeliac disease, how often the tool\nwill be positive.\nThis is calculated from:\n\n77\n= 1.0, or 100%\n77\n\nSpecificity is the rate at which the tool can exclude a\ncoeliac disease – if the patient is in fact healthy, how\noften the tool will be negative.\nThis is given by:\n\n1170\n= 0.608, or 60.8%\n1923\n\nPositive predictive value is the likelihood that the\npatient has coeliac disease if the tool is positive.\n77\n= 0.093, or 9.3%\n830\nSo, in patients for whom the tool is positive, the\nincidence of coeliac disease is 9.3%. However, 91 in\n100 patients with a positive result won’t actually\nhave coeliac disease.\nNegative predictive value shows the likelihood that\nthe patient won’t have coeliac disease if the tool\nresult is negative.\n1170\n= 1.0, or 100%\n1170\nFollowing a negative test, no patient will turn out to\nhave the condition.\n\nStatistics at work\n\n91\n\nThe likelihood ratio (LR) gives an estimate of how\nmuch a test result will change the odds of having\ncoeliac disease.\nThe LR for a positive result (LR+) tells us how much\nthe odds of coeliac disease increase when the tool\nresult is positive.\nThe LR for a negative result (LR–) tells us how much\nthe odds of coeliac disease decrease when the tool\nresult is negative.\nLR =\n\n1\n1\nsensitivity\n=\n=\n= 2.55\n(1 – specificity) 1 – 0.608 0.392\n\nSo, if the tool is positive in a patient, the odds of\nhaving coeliac disease are increased by a factor of\n2.6.\nPre-endoscopy serological testing in combination\nwith biopsy of high risk cases had a 100% negative\npredictive value – no patients who were negative\nwere found to have coeliac disease. The authors\ntherefore suggested that patients with a negative\nresult might not need a duodenal biopsy.\n\n> GLOSSARY\nCross-references to other parts of the glossary are\ngiven in italics.\n\nAbsolute risk reduction, ARR\nThe difference between the event rate in the\nintervention group and that in the control group. It is\nalso the reciprocal of the NNT.\n\nAlpha, a\nThe alpha value is equivalent to the P value and\nshould be interpreted in the same way.\n\nANCOVA (ANalysis of COVAriance)\nAnalysis of covariance is an extension of analysis of\nvariance (ANOVA) to allow for the inclusion of\ncontinuous variables in the model. See page 29.\n\nANOVA (ANalysis Of VAriance)\nThis is a group of statistical techniques used to\ncompare the means of two of more samples to see\nwhether they come from the same population. See\npage 29.\n\nAssociation\nA word used to describe a relationship between two\nvariables.\n\n94\n\nBeta, b\nThe beta value is the probability of accepting a\nhypothesis that is actually false. 1 – b is known as the\npower of the study.\n\nBayesian statistics\nAn alternative way of analyzing data, it creates and\ncombines numerical values for prior belief, existing\ndata and new data. See page 72.\n\nBinary variable\nSee categorical variable below.\n\nBi-modal distribution\nWhere there are 2 modes in a set of data, it is said to\nbe bi-modal. See page 15.\n\nBinomial distribution\nWhen data can only take one of two values (for\ninstance male or female), it is said to follow a\nbinomial distribution.\n\nBonferroni\nA method that allows for the problems associated\nwith making multiple comparisons. See page 69.\n\nBox and whisker plot\nA graph showing the median, range and interquartile range of a set of values. See page 13.\n\nGlossary\n\n95\n\nCase–control study\nA retrospective study which investigates the\nrelationship between a risk factor and one or more\noutcomes. This is done by selecting patients who\nalready have the disease or outcome (cases), matching\nthem to patients who do not (controls) and then\ncomparing the effect of the risk factor on the two\ngroups. Compare this with Cohort study. See page 40.\n\nCases\nThis usually refers to patients but could refer to\nhospitals, wards, counties, blood samples etc.\n\nCategorical variable\nA variable whose values represent different\ncategories of the same feature. Examples include\ndifferent blood groups, different eye colours, and\ndifferent ethnic groups.\nWhen the variable has only two categories, it is\ntermed “binary” (e.g. gender). Where there is some\ninherent ordering (e.g. mild, moderate, severe), this is\ncalled an “ordinal” variable.\n\nCausation\nThe direct relationship of the cause to the effect\nthat it produces, usually established in experimental\nstudies.\n\nCensored\nA censored observation is one where we do not have\ninformation for all of the observation period. This is\nusually seen in survival analysis where patients are\n\n96\n\nfollowed for some time and then move away or\nwithdraw consent for inclusion in the study. We\ncannot include them in the analysis after this point as\nwe do not know what has happened to them. See\npage 57.\n\nCentral tendency\nThe “central” scores in a set of figures. Mean,\nmedian and mode are measures of central tendency.\n\nChi-squared test, c2\nThe chi-squared test is a test of association between\ntwo categorical variables. See page 34.\n\nCohort study\nA prospective, observational study that follows a\ngroup (cohort) over a period of time and investigates\nthe effect of a treatment or risk factor. Compare this\nwith case–control study. See page 37.\n\nConfidence interval, CI\nA range of values within which we are fairly\nconfident the true population value lies. For\nexample, a 95% CI means that we can be 95%\nconfident that the population value lies within those\nlimits. See page 20.\n\nConfounding\nA confounding factor is the effect of a covariate or\nfactor that cannot be separated out. For example, if\nwomen with a certain condition received a new\ntreatment and men received placebo, it would not be\npossible to separate the treatment effect from the\n\nGlossary\n\n97\n\neffect due to gender. Therefore gender would be a\nconfounding factor.\n\nContinuous variable\nA variable which can take any value within a given\nrange, for instance BP. Compare this with discrete\nvariable.\n\nCorrelation\nWhen there is a linear relationship between two\nvariables there is said to be a correlation between\nthem. Examples are height and weight in children, or\nsocio-economic class and mortality.\nMeasured on a scale from -1 (perfect negative\ncorrelation), through 0 (no relationship between the\nvariables at all), to +1 (perfect positive correlation).\nSee page 48.\n\nCorrelation coefficient\nA measure of the strength of the linear relationship\nbetween two variables. See page 48.\n\nCovariate\nA covariate is a continuous variable that is not of\nprimary interest but is measured because it may\naffect the outcome and may therefore need to be\nincluded in the analysis.\n\nCox regression model\nA method which explores the effects of different\nvariables on survival. See page 60.\n\n98\n\nDatabase\nA collection of records that is organized for ease and\nspeed of retrieval.\n\nDegrees of freedom, df\nThe number of degrees of freedom, often abbreviated\nto df, is the number of independent pieces of\ninformation available for the statistician to make the\ncalculations.\n\nDescriptive statistics\nDescriptive statistics are those which describe the\ndata in a sample. They include means, medians,\nstandard deviations, quartiles and histograms. They\nare designed to give the reader an understanding of\nthe data. Compare this with inferential statistics.\n\nDiscrete variable\nA variable where the data can only be certain values,\nusually whole numbers, for example the number of\nchildren in families. Compare this with continuous\nvariable.\n\nDistribution\nA distinct pattern of data may be considered as\nfollowing a distribution. Many patterns of data have\nbeen described, the most useful of which is the\nnormal distribution. See page 9.\n\nFisher’s exact test\nFisher’s exact test is an accurate test for association\nbetween categorical variables. See page 35.\n\nGlossary\n\n99\n\nFields\nSee variables below.\n\nHazard ratio, HR\nThe HR is the ratio of the hazard (chance of\nsomething harmful happening) of an event in one\ngroup of observations divided by the hazard of an\nevent in a different group. An HR of 1 implies no\ndifference in risk between the two groups, an HR of\n2 implies double the risk. The HR should be stated\nwith its confidence intervals. See page 61.\n\nHistogram\nA graph of continuous data with the data categorized\ninto a number of classes. See example on page 15.\n\nHypothesis\nA statement which can be tested that predicts t```" ]	[ null, "https://b-ok.org/book/2464071/css/jscomments/loader.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.937829,"math_prob":0.8440683,"size":105232,"snap":"2019-26-2019-30","text_gpt3_token_len":24367,"char_repetition_ratio":0.14932337,"word_repetition_ratio":0.08029077,"special_character_ratio":0.2375038,"punctuation_ratio":0.10551791,"nsfw_num_words":6,"has_unicode_error":false,"math_prob_llama3":0.9653755,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-25T10:43:34Z\",\"WARC-Record-ID\":\"<urn:uuid:c0a6e859-d104-40ae-ba5f-ed11d3df0579>\",\"Content-Length\":\"141810\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3748afcf-682f-4596-b2ab-4ce5b0bf16f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:8d68ad12-35d6-45b4-8d1e-e4637173b170>\",\"WARC-IP-Address\":\"179.43.147.124\",\"WARC-Target-URI\":\"https://b-ok.org/book/2464071/d862b2\",\"WARC-Payload-Digest\":\"sha1:MJOKQGNOHIHQABVNSFYKN6O4NLNHHEUO\",\"WARC-Block-Digest\":\"sha1:3KCPEXCP2YZKUK6BX3ISRQJS5BDX3FGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999817.30_warc_CC-MAIN-20190625092324-20190625114324-00057.warc.gz\"}"}
https://developer.valvesoftware.com/w/index.php?title=Character_Setup_Overview:jp&oldid=18121	[ "# Character Setup Overview:jp\n\n(diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff)\n\noriginally translated by N-neko(C-SEC), 2005/3/25\noriginal English version: Character Setup Overview\n\n=\n\n=\n\n[[]]\n\n=\n\n```\"eyeball_r.tga\"<\"eyeball_l.tga\"<\"dark_eyeball_r.tga\"<\"dark_eyeball_l.tga\"< ```\n\n``` [[]] ```\n\n```= ```\n\n```\"mouth.tga\"<\"fmouth.tga\"< ```\n\n```= ```\n\n``` ```\n\n```= ```\n\n``` ```\n\n```- ```\n\n``` ```\n\n``` ```\n\n```= ```\n\n``` ```\n\n``` ```\n\n```= ```\n\n```キャラクタ顔アニメーションシェイプキー一覧 ```\n\n```- ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` [[]] ```\n\n```= ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n```[[]] ```\n\n```= ```\n\n```= ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` = ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` ```\n\n``` = ```\n\n``` ```\n\n``` [[]] ```\n\n```[[]] ```\n\n```= ```\n\n``` ```\n\n``` ```\n\n``` [[]] (modelname_reference.smd)< ```\n\n``` = ```\n\n```[http://www.c-- ```\n\n```[http://www.c-- ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7882362,"math_prob":0.90392554,"size":448,"snap":"2021-31-2021-39","text_gpt3_token_len":165,"char_repetition_ratio":0.18468468,"word_repetition_ratio":0.0,"special_character_ratio":0.390625,"punctuation_ratio":0.1875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99252987,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T15:50:29Z\",\"WARC-Record-ID\":\"<urn:uuid:bebc00bf-635e-43db-8c97-6c31796cf466>\",\"Content-Length\":\"19966\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb476aca-8706-45b5-a02e-51fa37e44e27>\",\"WARC-Concurrent-To\":\"<urn:uuid:67b26e97-33fa-4f6d-ada8-e873bf822490>\",\"WARC-IP-Address\":\"104.18.23.32\",\"WARC-Target-URI\":\"https://developer.valvesoftware.com/w/index.php?title=Character_Setup_Overview:jp&oldid=18121\",\"WARC-Payload-Digest\":\"sha1:OWYBXLQI5V3NJXXTTGLUJUDT7XNPYWLR\",\"WARC-Block-Digest\":\"sha1:YB2QG4VGLN5ZW2JOKXVZLGJYNERGCXR2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055684.76_warc_CC-MAIN-20210917151054-20210917181054-00302.warc.gz\"}"}
https://www.johncanessa.com/2021/02/10/binary-tree-pruning/	[ "# Binary Tree Pruning\n\nHave an on-line call at 02:30 PM. I believe I am well prepared. Will see how it goes.\n\nEarlier today I selected LeetCode 814 Binary Tree Pruning problem.\n\n```We are given the head node root of a binary tree,\nwhere additionally every node's value is either a 0 or a 1.\n\nReturn the same tree where every subtree (of the given tree)\nnot containing a 1 has been removed.\n\n(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)\n\nNote:\n\no The binary tree will have at most 200 nodes.\no The value of each node will only be 0 or 1.\n```\n\nWe are given a binary tree whose nodes are holding 0 or 1 as value. We need to prune the BT as described in the requirements. If you are interested in this problem please navigate to the LeetCode web site and read the actual requirements. In addition LeetCode has a set of test cases.", null, "In this post we will solve the problem using the Java programming language on a Windows 10 computer using the VSCode IDE. You can use any of the languages supported by LeetCode and solve the problem on the LeetCode side with the IDE provided.\n\nSince I am going to solve the problem on my computer I will need to generate test scaffolding. Note that such code IS NOT PART OF THE SOLUTON.\n\n```/*\n Definition for a binary tree node.\n* public class TreeNode {\n* int val;\n* TreeNode left;\n* TreeNode right;\n* TreeNode() {}\n* TreeNode(int val) { this.val = val; }\n* TreeNode(int val, TreeNode left, TreeNode right) {\n* this.val = val;\n* this.left = left;\n* this.right = right;\n* }\n* }\n/\nclass Solution {\npublic TreeNode pruneTree(TreeNode root) {\n\n}\n}\n```\n\nThe signature for the method in question is as expected. In addition we will be using the TreeNode class to represent and manipulate the binary tree.\n\n```1,null,0,0,1\nmain <<< strArr: [1, null, 0, 0, 1]\nmain <<< arr: [1, null, 0, 0, 1]\nmain <<< bt levelOrder:\n1\n0\n0,1\nmain <<< pruned levelOrder:\n1\n0\n1\n\n1,0,1,0,0,0,1\nmain <<< strArr: [1, 0, 1, 0, 0, 0, 1]\nmain <<< arr: [1, 0, 1, 0, 0, 0, 1]\nmain <<< bt levelOrder:\n1\n0,1\n0,0,0,1\nmain <<< pruned levelOrder:\n1\n1\n1\n\n1,1,0,1,1,0,1,0\nmain <<< strArr: [1, 1, 0, 1, 1, 0, 1, 0]\nmain <<< arr: [1, 1, 0, 1, 1, 0, 1, 0]\nmain <<< bt levelOrder:\n1\n1,0\n1,1,0,1\n0\nmain <<< pruned levelOrder:\n1\n1,0\n1,1,1\n\n0,null,0,0,0\nmain <<< strArr: [0, null, 0, 0, 0]\nmain <<< arr: [0, null, 0, 0, 0]\nmain <<< bt levelOrder:\n0\n0\n0,0\nmain <<< pruned levelOrder:\n\n```\n\nThe input line per test contains a single line with the depth first traversal of the nodes in the binary tree. This technique is typically used by LeetCode to define binary trees.\n\nOur test scaffold seems to read and parse the input line. The values are placed in a String[] array. Using the String[] array we appear to create a Integer[] and populate it with the same values. The two arrays are displayed.\n\nWe then appear to generate and populate a binary tree with the values from the Integer[] array. The binary tree is displayed in level order. The display seems to match the specified values.\n\nWe then seem to make a call to the method in question to prune the binary tree. The resulting binary tree is then displayed.\n\nWhile looking at the test cases and associated diagrams I noticed the end case in which all the nodes in the binary tree may need to be pruned. I forgot the condition before I submitted the code. After addressing the issue the code was accepted by LeetCode.\n\n``` /\n Test scaffolding\n\n @throws IOException\n/\npublic static void main(String[] args) throws IOException {\n\n// * open buffered reader \n\n// read String[] with node values for binary tree \n\n// close buffered reader \nbr.close();\n\n// ???? ????\nSystem.out.println(\"main <<< strArr: \" + Arrays.toString(strArr));\n\n// create and populate Integer[] \nInteger[] arr = new Integer[strArr.length];\nfor (int i = 0; i < strArr.length; i++) {\nif (strArr[i].equals(\"null\"))\narr[i] = null;\nelse\narr[i] = Integer.parseInt(strArr[i]);\n}\n\n// ???? ????\nSystem.out.println(\"main <<< arr: \" + Arrays.toString(arr));\n\n// create and populate a original binary tree \nBST bt = new BST();\nbt.root = bt.populate(arr);\n\n// ???? ????\nSystem.out.println(\"main <<< bt levelOrder:\");\nSystem.out.println(bt.levelOrder());\n\n// prune the BT \nbt.root = pruneTree(bt.root);\n\n// ???? ????\nSystem.out.println(\"main <<< pruned levelOrder:\");\nSystem.out.println(bt.levelOrder());\n}\n```\n\nThe test scaffold seems to follow (what a surprise) our description while looking at the test cases. Note that we create the binary tree using the BST class. Such code is only used in the test scaffold. As usual all the code can be found in the associated GitHub repository (https://github.com/JohnCanessa/BinaryTreePrunning) for this post.\n\n``` /\n* Return the same tree where every subtree (of the given tree)\n* not containing a 1 has been removed.\n\n Runtime: 0 ms, faster than 100.00% of Java online submissions.\n* Memory Usage: 36.5 MB, less than 77.69% of Java online submissions.\n/\nstatic TreeNode pruneTree(TreeNode root) {\n\n// * sanity check \nif (root == null)\nreturn null;\n\n// recursive call \nprune(root);\n\n// delete root node (if needed) \nif (leafNode(root))\nroot = null;\n\n// returned pruned BT \nreturn root;\n}\n```\n\nThe idea is to perform a depth first search traversal on the binary tree. We will be checking and pruning nodes.\n\nAfter the sanity check we prune the binary tree. If we end with a root node with value 0 (end conditioned that I initially forgot to code) we set the root of the binary tree to null (pruned). The binary tree is then returned.\n\n``` /\n* DFS traversal deleting nodes.\n* Recursive call.\n/\nstatic void prune(TreeNode root) {\nif (root != null) {\n\n// * visit left sub tree \nprune(root.left);\n\n// delete left node (if needed) \nif (root.left != null && leafNode(root.left))\nroot.left = null;\n\n// visit right sub tree \nprune(root.right);\n\n// delete right node (if needed) \nif (root.right != null && leafNode(root.right))\nroot.right = null;\n}\n}\n```\n\nThe prune() method implements DFS in a recursive fashion. We traverse the left and then the right subtrees. After each traversal we prune the left or the right node as needed.\n\n``` /\n* Auxiliary method.\n/\nstatic boolean leafNode(TreeNode root) {\n\n// * sanity check \nif (root.val == 1)\nreturn false;\n\n// check if end node \nif (root.left == null && root.right == null)\nreturn true;\n\n// not a leaf node **\nreturn false;\n}\n```\n\nThis is an auxiliary method. We only prune a node with value 0 if it is a leaf node. That implies it has no children that meets the requirements.\n\nHope you enjoyed solving this problem as much as I did. The entire code for this project can be found in my GitHub repository.\n\nIf you have comments or questions regarding this, or any other post in this blog, or if you would like for me to help out with any phase in the SDLC (Software Development Life Cycle) of a project associated with a product or service, please do not hesitate and leave me a note below. If you prefer, send me a private e-mail message. I will reply as soon as possible.\n\nKeep on reading and experimenting. It is one of the best ways to learn, become proficient, refresh your knowledge and enhance your developer toolset.\n\nOne last thing, many thanks to all 6,488 subscribers to this blog!!!\n\nKeep safe during the COVID-19 pandemic and help restart the world economy. I believe we can all see the light at the end of the tunnel.\n\nRegards;\n\nJohn\n\njohn.canessa@gmail.com\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]	[ null, "https://www.johncanessa.com/wp-content/uploads/2021/02/binary_tree_pruning-300x80.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7330223,"math_prob":0.8505913,"size":7534,"snap":"2022-27-2022-33","text_gpt3_token_len":1919,"char_repetition_ratio":0.13532537,"word_repetition_ratio":0.066967644,"special_character_ratio":0.3068755,"punctuation_ratio":0.19107479,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95110035,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T09:26:43Z\",\"WARC-Record-ID\":\"<urn:uuid:f61eda83-9c72-4a84-89fc-e26ef744791b>\",\"Content-Length\":\"62241\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf9ce6ac-386f-4001-af56-5b22a55a90f6>\",\"WARC-Concurrent-To\":\"<urn:uuid:1dcd5896-0dea-4093-8940-001455bcae54>\",\"WARC-IP-Address\":\"208.113.168.135\",\"WARC-Target-URI\":\"https://www.johncanessa.com/2021/02/10/binary-tree-pruning/\",\"WARC-Payload-Digest\":\"sha1:RTX6BJ4OJUU6VDCOASUU455HUIA5Q627\",\"WARC-Block-Digest\":\"sha1:5JWBBWLAK5UWIRDWZHO774R7GUUGKC7X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103626162.35_warc_CC-MAIN-20220629084939-20220629114939-00542.warc.gz\"}"}
https://forum.allaboutcircuits.com/threads/explanation-for-time-dependent-distribution-of-electric-field-in-a-closed-circuit.121867/	[ "# Explanation for time dependent distribution of electric field in a closed circuit\n\n#### Bastians\n\nJoined Mar 13, 2016\n2\nupon closing a electric circuit DC or AC , the potential difference generates an electrical field and electrons start moving through the circuit based on the electric field force.\n\nThe distribution of this electric field is setup at a very high speed. A light bulb lights up almost instantaneously.\nWhere does the electric field starts first? at the terminal where electrons are pushed into the circuit, where electrons are pulled out from the circuit , or at both terminals simultaneously ? There must be some initial time frame to get the electrical field throughout the entire circuit.\ncan someone explain this ?\n\n#### crutschow\n\nJoined Mar 14, 2008\n33,346\nThe electric field travels at the speed of light and emanates from the source generating the voltage difference across the lines (or line and return).\n\n#### Bastians\n\nJoined Mar 13, 2016\n2\nThe electric field travels at the speed of light and emanates from the source generating the voltage difference across the lines (or line and return).\ndoes the electric field is distributed at both + and - terminal of the source at the same time ?\nor only at the terminal where electrons enters the circuit , almost at speed of light, but there must be a delay, heading towards the other terminal ?\ni am looking for a mathematical formulation of this phenomena.\n\n#### Pinkamena\n\nJoined Apr 20, 2012\n22\n\"The speed at which energy or signals travel down a cable is actually the speed of the electromagnetic wave, not the movement of electrons. Electromagnetic wave propagation is fast and depends on the dielectric constant of the material.\"\n\nSo if you assume you have a voltage source that is terminated to ground or some other constant potential, and the voltage source turns on at some time t, then the electric field will propagate along the wire at a speed determined by the dielectric constant of the material. For a PCB trace, this will mostly depend on the dielectric constant of the PCB material, I believe. It's still quite fast, about 66% of the speed of light for most pcb materials.\n\nHere is a blog post that goes a bit more in depth about the propagation speed for pcb traces: https://blogs.mentor.com/hyperblog/blog/tag/velocity-of-propagation/\n\n#### nsaspook\n\nJoined Aug 27, 2009\n12,286\ndoes the electric field is distributed at both + and - terminal of the source at the same time ?\nor only at the terminal where electrons enters the circuit , almost at speed of light, but there must be a delay, heading towards the other terminal ?\ni am looking for a mathematical formulation of this phenomena.\nIt's not Either/Or, the electrons in the wire and the electric field act as a system together to guide electrical energy from both terminals at the source to the load.", null, "https://en.wikipedia.org/wiki/Energy_current\nhttps://en.wikipedia.org/wiki/Poynting_vector\n\nLast edited:" ]	[ null, "https://forum.allaboutcircuits.com/proxy.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9316884,"math_prob":0.94228727,"size":1049,"snap":"2023-40-2023-50","text_gpt3_token_len":188,"char_repetition_ratio":0.17607656,"word_repetition_ratio":0.80701756,"special_character_ratio":0.18303145,"punctuation_ratio":0.09677419,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9603944,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T11:50:24Z\",\"WARC-Record-ID\":\"<urn:uuid:f84b0d45-fd4e-4bdb-81ad-af03ec67b16a>\",\"Content-Length\":\"110521\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38bc7ac4-b3a2-4c5b-8f25-5520ee8dd63e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cc876a5-f377-4a14-9f2b-dd286fb16a4d>\",\"WARC-IP-Address\":\"104.17.144.194\",\"WARC-Target-URI\":\"https://forum.allaboutcircuits.com/threads/explanation-for-time-dependent-distribution-of-electric-field-in-a-closed-circuit.121867/\",\"WARC-Payload-Digest\":\"sha1:JSK727CJ4BNKBD5HGBUFPEPDQDVP4NO7\",\"WARC-Block-Digest\":\"sha1:45W7DJCOFVETRRKPTOHHMHDCYWYM2OPV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100499.43_warc_CC-MAIN-20231203094028-20231203124028-00013.warc.gz\"}"}
https://ch.mathworks.com/matlabcentral/cody/problems/135-inner-product-of-two-vectors/solutions/506777	[ "Cody\n\n# Problem 135. Inner product of two vectors\n\nSolution 506777\n\nSubmitted on 30 Sep 2014 by Mike\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% x = 1:3; y= 3:-1:1; z_correct = 10; assert(isequal(your_fcn_name(x,y),z_correct))\n\n2 Pass\n%% x = 1:6; y= ones(1,6); z_correct = 21; assert(isequal(your_fcn_name(x,y),z_correct))" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5477463,"math_prob":0.9893253,"size":404,"snap":"2020-10-2020-16","text_gpt3_token_len":139,"char_repetition_ratio":0.145,"word_repetition_ratio":0.0,"special_character_ratio":0.37128714,"punctuation_ratio":0.18604651,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9554686,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T09:08:10Z\",\"WARC-Record-ID\":\"<urn:uuid:63f464e9-e6b5-42a3-baec-ed7a224102d1>\",\"Content-Length\":\"72677\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b92f2a4d-de11-40e9-8089-30600f85143c>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef5fa7c2-dbbb-4c33-ac56-f7dd8cf995fc>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://ch.mathworks.com/matlabcentral/cody/problems/135-inner-product-of-two-vectors/solutions/506777\",\"WARC-Payload-Digest\":\"sha1:PA6QGOHLTRMNRMBI2HI2NSOVHH2Q2YEC\",\"WARC-Block-Digest\":\"sha1:RB3JV77VJXL5MSMDTWSQRC36Z4FTJGKV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370505550.17_warc_CC-MAIN-20200401065031-20200401095031-00201.warc.gz\"}"}
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Keywords:\n\n•", null, "Calculation Of Circulating Load On Quarry CAESAR Mining calculation of circulating loads in kil sweden . Sweden. After the millennium shift it slowly became obvious .\n\n•", null, "### calculation of circulating loads in kil sweden\n\nblack galaxy granite quarry calculation of circulating loads of electrical equipment causes load losses. . grid structured circulating currents will\n\n•", null, "" ]	[ null, "https://technologica.be/images/case/1105.jpg", null, "https://technologica.be/images/case/769.jpg", null, "https://technologica.be/images/case/57.jpg", null, "https://technologica.be/images/case/956.jpg", null, "https://technologica.be/images/case/63.jpg", null, "https://technologica.be/images/case/1187.jpg", null, "https://technologica.be/images/case/92.jpg", null, "https://technologica.be/images/case/958.jpg", null, "https://technologica.be/images/case/308.jpg", null, "https://technologica.be/images/case/795.jpg", null, "https://technologica.be/images/case/217.jpg", null, "https://technologica.be/images/case/715.jpg", null, "https://technologica.be/images/case/74.jpg", null, "https://technologica.be/images/case/93.jpg", null, "https://technologica.be/images/case/295.jpg", null, "https://technologica.be/images/case/262.jpg", null, "https://technologica.be/images/case/1280.jpg", null, "https://technologica.be/images/case/358.jpg", null, "https://technologica.be/images/case/142.jpg", null, "https://technologica.be/images/case/1177.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7698522,"math_prob":0.993508,"size":3956,"snap":"2020-24-2020-29","text_gpt3_token_len":716,"char_repetition_ratio":0.3861336,"word_repetition_ratio":0.110912345,"special_character_ratio":0.15419616,"punctuation_ratio":0.06081081,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9880718,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T17:51:29Z\",\"WARC-Record-ID\":\"<urn:uuid:7d1d6e28-bc7f-49f3-a0d0-3ed7b517bb5f>\",\"Content-Length\":\"12872\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef4fed91-fcd6-4fb5-9f61-105469f36e5f>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b1e66b1-8f83-4525-8238-6f6f28a92c7d>\",\"WARC-IP-Address\":\"104.28.13.179\",\"WARC-Target-URI\":\"https://technologica.be/quarry/calculation-of-circulating-load-on-quarry/\",\"WARC-Payload-Digest\":\"sha1:SG5TP22EVYTG5YXKSDB3OG55NHQHBAMV\",\"WARC-Block-Digest\":\"sha1:TXXTSFCUNWV3ZYPJECCSSRDRQNHBC2XX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657146247.90_warc_CC-MAIN-20200713162746-20200713192746-00025.warc.gz\"}"}
https://de.mathworks.com/matlabcentral/cody/problems/848-calculate-a-modified-levenshtein-distance-between-two-strings/solutions/489921	[ "Cody\n\n# Problem 848. Calculate a modified Levenshtein distance between two strings\n\nSolution 489921\n\nSubmitted on 23 Aug 2014 by rifat\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% s1 = 'I do not like MATLAB'; s2 = 'I love MATLAB a lot'; d_correct = 4; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'i' 'do' 'not' 'like' 'matlab' c2 = 'i' 'love' 'matlab' 'a' 'lot'\n\n2 Pass\n%% s1 = 'Which words need to be edited?'; s2 = 'Can you tell which words need to be edited?'; d_correct = 3; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'which' 'words' 'need' 'to' 'be' 'edited' c2 = 'can' 'you' 'tell' 'which' 'words' 'need' 'to' 'be' 'edited'\n\n3 Pass\n%% s1 = 'Are these strings identical?'; s2 = 'These strings are not identical!'; d_correct = 3; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'are' 'these' 'strings' 'identical' c2 = 'these' 'strings' 'are' 'not' 'identical'\n\n4 Pass\n%% s1 = 'Settlers of Catan is my favorite game'; s2 = 'Tic-tac-toe is also one of my favorite games'; d_correct = 6; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'settlers' 'of' 'catan' 'is' 'my' 'favorite' 'game' c2 = 'tic-tac-toe' 'is' 'also' 'one' 'of' 'my' 'favorite' 'games'\n\n5 Pass\n%% s1 = 'This one should be simple, but maybe it isn''t'; s2 = 'This one should be simple, but maybe it isn''t'; d_correct = 0; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'this' 'one' 'should' 'be' 'simple' 'but' 'maybe' 'it' 'isn't' c2 = 'this' 'one' 'should' 'be' 'simple' 'but' 'maybe' 'it' 'isn't'\n\n6 Pass\n%% s1 = 'Testing, testing, one, two, three,...'; s2 = 'Testing, testing, one, two,...'; d_correct = 1; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = 'testing' 'testing' 'one' 'two' 'three' c2 = 'testing' 'testing' 'one' 'two'\n\n7 Pass\n%% s1 = 'How many edits do you think there are in this example? I don''t know!'; s2 = 'Well, it is hard to tell how many edits are required because there are big differences in the two strings.'; d_correct = 15; assert(isequal(modlevenshtein(s1,s2),d_correct))\n\nc1 = Columns 1 through 10 'how' 'many' 'edits' 'do' 'you' 'think' 'there' 'are' 'in' 'this' Columns 11 through 14 'example' 'i' 'don't' 'know' c2 = Columns 1 through 10 'well' 'it' 'is' 'hard' 'to' 'tell' 'how' 'many' 'edits' 'are' Columns 11 through 19 'required' 'because' 'there' 'are' 'big' 'differences' 'in' 'the' 'two' Column 20 'strings'" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5955507,"math_prob":0.6466995,"size":2392,"snap":"2019-51-2020-05","text_gpt3_token_len":881,"char_repetition_ratio":0.11515913,"word_repetition_ratio":0.101333335,"special_character_ratio":0.38419732,"punctuation_ratio":0.13729978,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99562836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T11:47:27Z\",\"WARC-Record-ID\":\"<urn:uuid:a898983c-025b-4596-97c1-24eb40384c8a>\",\"Content-Length\":\"78844\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f3ab839-bec0-45a5-a01e-93cabd7c1426>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c72aa28-32d1-450c-b10c-4339746732f8>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://de.mathworks.com/matlabcentral/cody/problems/848-calculate-a-modified-levenshtein-distance-between-two-strings/solutions/489921\",\"WARC-Payload-Digest\":\"sha1:VQYALYVKPVSPQ7LB7CFLWK7CLTGBOUQL\",\"WARC-Block-Digest\":\"sha1:IRFH2LCCO5CBIIETF4CKNN5ER2VEPZK5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250619323.41_warc_CC-MAIN-20200124100832-20200124125832-00058.warc.gz\"}"}
https://sheir.org/edu/statistics-quiz/	[ "# Statistics Quiz\n\nMathematicsStatistics → Statistics Quiz\n\n### Page: 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10\n\n11. In a module, quiz contributes 10%, assignment 30%, and final exam contributes 60% towards the final result. A student obtained 80% marks in quiz, 65% in assignment, and 75% in the final exam. What are average marks?\n(A) 64.5%\n(B) 68.5%\n(C) 72.5%\n(D) 76.5%\n\n12. In a university, average height of students is 165 cm. Now, consider the following Table,\n\nStudents1620242016\nHEIGHT160–162162–164164–166166–168168–170\nWhat type of distribution is this?\n(A) Normal\n(B) Uniform\n(C) Poisson\n(D) Binomial\n\n13. What is the average of 3%, 7%, 10%, and 16% ?\n(A) 8%\n(B) 9%\n(C) 10%\n(D) 11%\n\n11. (C) 72.5%\n12. (A) Normal\n13. (B) 9%\n\nSOLUTIONS: STATISTICS QUIZ\n\n11. By using formula for calculating weighted average, we have", null, "12.", null, "13.", null, "" ]	[ null, "https://sheir.org/edu/wp-content/uploads/2020/08/statistics-quiz.png", null, "https://sheir.org/edu/wp-content/uploads/2020/08/basic-statistics-mcqs-solution.png", null, "https://sheir.org/edu/wp-content/uploads/2019/12/statistics-mcqs-question.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7413269,"math_prob":0.95836294,"size":889,"snap":"2021-43-2021-49","text_gpt3_token_len":316,"char_repetition_ratio":0.09378531,"word_repetition_ratio":0.0,"special_character_ratio":0.44656917,"punctuation_ratio":0.17525773,"nsfw_num_words":3,"has_unicode_error":false,"math_prob_llama3":0.972156,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T04:11:35Z\",\"WARC-Record-ID\":\"<urn:uuid:9780f09d-6aca-4567-96f8-f8828ab8f873>\",\"Content-Length\":\"18454\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9ba7202-0346-41ca-88c3-329255eb3558>\",\"WARC-Concurrent-To\":\"<urn:uuid:77b3d039-8958-4652-9226-67bf513b69df>\",\"WARC-IP-Address\":\"43.255.154.97\",\"WARC-Target-URI\":\"https://sheir.org/edu/statistics-quiz/\",\"WARC-Payload-Digest\":\"sha1:F2ZST35XP7GYHUWLCNMUUNUD63HROPFD\",\"WARC-Block-Digest\":\"sha1:K5Z6D36WV6HG2MTZXXNMM2TIJMWMJRAK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363659.21_warc_CC-MAIN-20211209030858-20211209060858-00409.warc.gz\"}"}
http://andeekaplan.com/2015/01/29/ModelVis	[ "# Reading: Visualization of the model fitting process\n\nToggle Code Viewing: ON\n\nAs part of the course Stat 503 I am taking this semester, I will be posting a series of responses to assigned course readings. Mostly these will be my rambling thoughts as I skim papers.\n\nThis week we are reading a paper by Hadley Wickham, Dianne Cook, and Heike Hofmann entitled “Visualizing Statistical Models: Removing the Blindfold” which was submitted to the Annals of Applied Statistics. This paper presents multiple ways that visualization can be useful throughout the model fitting process, rather than solely after the models have been fit and a “best” model has been selected. This paper has been my favorite reading assignment thus far in the semester; I feel as though I’d been thinking of some of these concepts without knowing how to put words to them properly.\n\n# Strategies\n\nThere are three main strategies presented in the paper to enhance model fitting through the use of visualization and graphics:\n\n1. Display the model in the data space\n2. Look at all members in the collection\n3. Understand the process of model fitting\n\nThese three ideas on their own can be very powerful, but together I imagine the improvement to the model fitting would be more than the sum of the parts. To get an idea for all three of these suggestions, I will run through a case study of fitting linear models to the mtcars dataset.\n\n# Model in the data space\n\nA common way to diagnose the fit of a model is to display the data in the model space. For example, with linear regression we often look at plots of residuals versus fitted values to assess the assumptions of a linear model. This maps each observation (data) to a point in two dimensions that are summaries from the model (in the model space). The authors suggest flipping this paradigm and mapping the model in the data space instead. One way to accomplish this for a predictive model with continuous response is to visualize the response surface over a grid of predictors.\n\nFor a (very simple) regression example of fitting a linear model to the mtcars dataset, predicting miles per gallon (mpg) with engine displacement (disp) and horespower (hp), we can fit the model and then view the resulting prediction surface (a plane) in the dataspace with the data overlaid in red.", null, "This is commonly done for OLS regression where we have 2 dimensions. Once we get beyond 2 predictors (3 dimensions), this visualization requires much more creativity and some special tools (like GGobi).\n\n# A look at the members\n\nIn many modeling activities, some models are fit within the same family of models (think linear models with a main effect for example) and then a “best” combination of variables is chosen through some criteria (AIC, BIC, etc.). However, the only model that gets explored in depth is this best model. The less optimal models are thrown away in the selection process. The authors argue that by visualizing multiple models in the selection process can give “more insight into the data and the relative merits of different models”. To quote John Tukey,\n\nThe greatest value of a picture is when it forces us to notice what we never expected to see.\n\nThis is an important concept that can be applied to the model selection process, and is what the authors suggest when they say “look at members in a collection”. We can explore multiple models and hopefully find out something we never expected about the data and benefits of different models.\n\nTo get an idea for how this works, I will fit everypossible combination of predictors in the mtcars example for models with at least two explanatory variables. We can then look at the values for the stimates of the coefficients associated with each variable in a parallel coordinates plot. Note that a value of zero means that variable wasn’t included in the model. We can aso plot some (standardized) model fit diagnostics for each model and look for a relationship between complexity and fit for our many models.", null, "", null, "It would be interesting to explore these plots with interactive linked brushing, as was done in the paper, however there are still interesting things to note about the models. First, number of carburetors (carb), number of cylinders in the engine (cyl), and weight (wt) always negatively affect fuel efficiency when included in the model. Secondly it appears the number of forward gears (gear), 1/4 mile time (qsec) and V/S (vs) can either negatively or positively affect fuel efficiency. It would be interesting to investigate which models these are to explore collinearity in the model.\n\n# Understanding the process\n\nBy understanding how a model is fit, and by that I mean the algorithm that fits the model, we can more fully understand how specific data affect the resulting model. This can often be accomplished by visualizing the iterations of a model fitting algorithm to view each step in the process. For example, using the Newton-Raphson method, we could see the steps taken to arrive at a maximum in maximum likelihood fitting for a specific model.\n\n# Conclusion\n\nTo sum of this approach, why only look at one model when you can look at hundreds? By visualizing multiple models we can find interesting things about the data that we may not have known from just looking at one model. Additionally, viewing the model in the data space can provide additional insight as to why a certain model is fit. Finally, understanding the model that is being fit is always a good idea (let’s avoid the dreaded black box!). Visualization of models can help with all three of these ideas." ]	[ null, "http://andeekaplan.com/images/blog/2015-01-29-ModelVis/unnamed-chunk-1-1.png", null, "http://andeekaplan.com/images/blog/2015-01-29-ModelVis/unnamed-chunk-2-1.png", null, "http://andeekaplan.com/images/blog/2015-01-29-ModelVis/unnamed-chunk-2-2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9344613,"math_prob":0.94046414,"size":5552,"snap":"2021-43-2021-49","text_gpt3_token_len":1120,"char_repetition_ratio":0.14329489,"word_repetition_ratio":0.010593221,"special_character_ratio":0.19560519,"punctuation_ratio":0.07219512,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98095423,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-28T21:26:04Z\",\"WARC-Record-ID\":\"<urn:uuid:0141a3ed-3c6d-4756-84c8-c5ea4c3e02a4>\",\"Content-Length\":\"31975\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:000e371c-25e7-45f8-ab7f-6597922db37b>\",\"WARC-Concurrent-To\":\"<urn:uuid:894f82f3-b13b-4d63-aac2-ef0d09547ac1>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://andeekaplan.com/2015/01/29/ModelVis\",\"WARC-Payload-Digest\":\"sha1:54P4QIZT67ITPMKBHLIYKNGBFVNWNTWO\",\"WARC-Block-Digest\":\"sha1:OT7PO2BGV4LQEZHJ5CR5VFVQTA6OUDKG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358591.95_warc_CC-MAIN-20211128194436-20211128224436-00583.warc.gz\"}"}
https://www.ahyungo.com/distance-formula-physics-no-longer-a-mystery-3/	[ "# Distance Formula Physics: No Longer a Mystery\n\n## Distance Formula Physics Ideas\n\nA heavy body moving at a quick velocity is hard to stop. In case the distance is less than one wavelength, then the solution is unique. Its motion is known as projectile motion.\n\nHere’s the trajectory (path) between the 2 points. When it regards projectile motion, there are lots of equations to consider. This change in position is known as displacement.\n\nThere are various kinds of tests included in the majority of standards. As is typical in any matter, there are assumptions hidden in the way in which the dilemma is stated and we need to work out the way to care for it. Mass spectrometers have many designs, and a lot of use magnetic fields to measure mass.\n\nSeveral the words may be incorrectly translated or mistyped. Charge Q functions as a point charge to make an electric field. It is possible to add distances and you may add times, but you cannot add rates.\n\nThe work done by friction on the vehicle is linked to the initial kinetic energy of the vehicle. http://publishing.wsu.edu/copyright/fair_use/ In case the traffic ahead has stopped, that could mess up your day promptly. Imagine that you’re driving your vehicle on a normal street.\n\nSo if, for instance, you’re travelling at 50mph in your vehicle and the road is wet, you are going to want to double the conventional figures given for stopping distance. Speed only extends to you a number which lets you understand how fast you’re going. Stopping distance is the typical speed of the vehicle times the average stopping time.\n\nStandard assumption that collisions are binary effects in severe problems when seeking to take several interactions into consideration. It is possible to use any of them dependent on the nature of a certain problem. You always need to approach an issue first by thinking about the crucial interactions described, no matter what quantity you’re requested to find.\n\nHigher fidelity can be accomplished https://www.buyessay.net by calculating higher order moments. Therefore, we use that which we can see to tell us about that which we can’t. The velocity is changing over the term of time.\n\n## Introducing Distance Formula Physics\n\nThe key idea about angular momentum, much much like linear momentum, is the fact that it’s conserved. Among the most fundamental scientific disciplines, the most important purpose of physics is to comprehend the way the universe behaves. The idea of moment in physics comes from the mathematical notion of moments.\n\nAfter all the ideal thing about physics is the fact that it can be utilised to address real world difficulties. The primary goal of these subjects is to study and attempt to know the universe and everything within it. A comprehensive discussion of modulo is beyond the range of this piece, but here is a fantastic link on Khan Academy.\n\n## Finding the Best Distance Formula Physics\n\nProjectile motion has two key components. Therefore, the angular velocity will differ. Acceleration doesn’t occur by itself.\n\nYes, it becomes rather complicated, but all you want to understand is that the angular momentum is dependent on how fast the rotors spin. The equation that’s utilized to figure distance and velocity is provided below. You should also think of whether it’s possible to pick the initial speed for the ball and just figure out the angle at which it is thrown.\n\nThe Synchrotron Light source is going to be employed by researchers to study a huge selection of scientific questions. The electric field strength isn’t dependent upon the number of charge on the test charge. There are also a number of organic units that are employed in astronomy and space science.\n\n## Facts, Fiction and Distance Formula Physics\n\nIt’s complicated to memorize each arrangement of the 2 equations and I recommend that you practice creating new combinations from the original equations. These equations are called kinematic equations. The equations of motion could be utilised to figure out the qualities of the trajectory.\n\nYou will need a calculator for that, but nevertheless, it shouldn’t be essential for the theory test. This is among the most typical procedures of calculus. It is derived from the Pythagorean theorem.\n\nRotational kinetic energy has to be supplied to the blades to make them rotate faster, and enough energy cannot be supplied in time to prevent a crash. That means you can observe that flying a drone is really simple if you enable the computer do all of the work. The whole thrust force will stay equal to the weight, or so the drone will stay at the identical vertical level.\n\nThe very first step, naturally, is to establish what we mean when we discuss acceleration for a method of talking about force. Two charges would always be essential to encounter a force. It’s the huge force acting for a tiny interval of time.\n\n## What Distance Formula Physics Is – and What it Is Not\n\nIt should be put approximately ten centimeters under the base of the streamlined bob when it’s suspended by the magnet. This sort of bond is quite strong. When the time was determined, a horizontal equation can be utilised to ascertain the horizontal displacement of the pool ball.\n\nFor the triangle to be a proper triangle, D3 the biggest of the 3 distances have to be the amount of the hypotenuse and all 3 sides must satisfy Pythagorean theorem. This is the point where the lines crossed. Be ready for black ice on cold days, and look out for loose surfaces like gravel.\n\nA lot of men and women spend far too much time reading as a means to revise. Another sort of directed distance is that between two distinct particles or point masses at a particular time. In order to make sure that the stopping sight distance provided is adequate, we are in need of a more in-depth comprehension of the frictional force.\n\n## Whispered Distance Formula Physics Secrets\n\nDetailed mathematical solutions of practical problems typically don’t have closed-form solutions, and so need numerical strategies to handle. There are quite a lot of approaches that were designed for just such systems. See below to learn more." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9417359,"math_prob":0.9493041,"size":6029,"snap":"2022-27-2022-33","text_gpt3_token_len":1188,"char_repetition_ratio":0.105892114,"word_repetition_ratio":0.0,"special_character_ratio":0.1909106,"punctuation_ratio":0.086766,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98525995,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T10:27:18Z\",\"WARC-Record-ID\":\"<urn:uuid:3a89a7dc-f8c7-4e4e-8697-99dadcda8bf3>\",\"Content-Length\":\"34817\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e1bd34a-851b-404f-a3c1-e87798a7d5ae>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7361ec2-1087-4c29-90df-68acc2f869a9>\",\"WARC-IP-Address\":\"182.92.175.202\",\"WARC-Target-URI\":\"https://www.ahyungo.com/distance-formula-physics-no-longer-a-mystery-3/\",\"WARC-Payload-Digest\":\"sha1:3VBDWTKGCC4TV5WGAQL54BZYVFCHAFZH\",\"WARC-Block-Digest\":\"sha1:B2JTJVCAQXNVFLZBVZSMWLLYGG7SK3KD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572898.29_warc_CC-MAIN-20220817092402-20220817122402-00067.warc.gz\"}"}
https://numbermatics.com/n/10718221303/	[ "# 10718221303\n\n## 10,718,221,303 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 10718221303 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 10 divisors.\n\n10718221303 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of ten divisors.\n\n## Prime factorization of 10718221303:\n\n### 1014 × 103\n\n(101 × 101 × 101 × 101 × 103)\n\nSee below for interesting mathematical facts about the number 10718221303 from the Numbermatics database.\n\n### Names of 10718221303\n\n• Cardinal: 10718221303 can be written as Ten billion, seven hundred eighteen million, two hundred twenty-one thousand, three hundred three.\n\n### Scientific notation\n\n• Scientific notation: 1.0718221303 × 1010\n\n### Factors of 10718221303\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 5\n• Sum of prime factors: 204\n\n### Divisors of 10718221303\n\n• Number of divisors d(n): 10\n• Complete list of divisors:\n• Sum of all divisors σ(n): 10930504520\n• Sum of proper divisors (its aliquot sum) s(n): 212283217\n• 10718221303 is a deficient number, because the sum of its proper divisors (212283217) is less than itself. Its deficiency is 10505938086\n\n### Bases of 10718221303\n\n• Binary: 10011111101101101100010011111101112\n• Base-36: 4X9COHJ\n\n### Squares and roots of 10718221303\n\n• 10718221303 squared (107182213032) is 114880267900083017809\n• 10718221303 cubed (107182213033) is 1231312134701016876948952185127\n• The square root of 10718221303 is 103528.8428555059\n• The cube root of 10718221303 is 2204.8255854831\n\n### Scales and comparisons\n\nHow big is 10718221303?\n• 10,718,221,303 seconds is equal to 340 years, 41 weeks, 6 days, 11 hours, 41 minutes, 43 seconds.\n• To count from 1 to 10,718,221,303 would take you about six hundred eighty-one years!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 10718221303 cubic inches would be around 183.7 feet tall.\n\n### Recreational maths with 10718221303\n\n• 10718221303 backwards is 30312281701\n• The number of decimal digits it has is: 11\n• The sum of 10718221303's digits is 28\n• More coming soon!\n\n#### Copy this link to share with anyone:\n\nMLA style:\n\"Number 10718221303 - Facts about the integer\". Numbermatics.com. 2023. Web. 5 December 2023.\n\nAPA style:\nNumbermatics. (2023). Number 10718221303 - Facts about the integer. Retrieved 5 December 2023, from https://numbermatics.com/n/10718221303/\n\nChicago style:\nNumbermatics. 2023. \"Number 10718221303 - Facts about the integer\". https://numbermatics.com/n/10718221303/\n\nThe information we have on file for 10718221303 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 10718221303, math, Factors of 10718221303, curriculum, school, college, exams, university, Prime factorization of 10718221303, STEM, science, technology, engineering, physics, economics, calculator, ten billion, seven hundred eighteen million, two hundred twenty-one thousand, three hundred three.\n\nOh no. 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https://ktbssolutions.com/1st-puc-physics-question-bank-chapter-5/	[ "# 1st PUC Physics Question Bank Chapter 5 Laws of Motion\n\n## Karnataka 1st PUC Physics Question Bank Chapter 5 Laws of Motion\n\n### 1st PUC Physics Laws of Motion TextBook Questions and Answers\n\nQuestion 1.\nGive the magnitude and direction of the net force acting on\n\n1. a drop of rain falling down at a constant speed.\n2. a cork of mass 10 g floating on water\n3. a kite skillfully held stationary in the sky.\n4. a car moving with a constant velocity of 30 km/h on a rough road.\n5. a high-speed electron in space far from ail material objects, and free of electric and magnetic fields.\n\n1. In accordance with the first law of motion, there is no net force on the drop since it is moving with constant speed.\n2. The weight of the cork is balanced by upthrust which is equal to the weight of water displaced. Hence no net force on the cork.\n3. Since the kite is in the state of rest net force on it is zero.\n4. From the first law of motion, since the velocity of the car is a constant net force on it is zero.\n5. Since the electron is in free space no gravitational or electric or magnetic force is acting on it. Hence net force on it is zero.\n\nQuestion 2.\nA pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,\n\n1. during its upward motion.\n2. during its downward motion.\n3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.\n\n1. When the pebble is moving upward the force acting on it is gravitational force in downward direction. F = mg = 0.05 × 10 = 0.5 N\n2. Even in this case F = mg = 0.5 N in downward direction.\n3. Since there is no force other than gravitational force acting on pebble, during the whole process F = mg = 0.5 N. Note that pebble moves in opposite direction because of its initial velocity. The situation remains same for pebble thrown at an angle.\n\nQuestion 3.\nGive the magnitude and direction of the net force acting on a stone of mass 0.1 kg,\n\n1. Just after it is dropped from the window of a stationary train.\n2. Just after It is dropped from the window of a train running at a constant velocity of 36 km/h\n3. Just after it is dropped from the window of a train accelerating with 1 ms-2\n4. Lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.\n\n1. Gravitational force is acting on the stone in downward direction F = mg = 0.1 × 10 = 1 N\n2. Once the stone is dropped from the train, the only force acting on it is gravitation force =1 N.\n3. Since there is no contact between train and stone the force acting on it is again gravitational force.\n4. Since the stone is lying on the floor of train its acceleration in the same as that of the train. Hence the force excreted by train on the stone is F = ma = 0.1 × 1 = 0. 1 N in the direction of the train. The weight is balanced by the normal reaction of the floor of the train.\n\nQuestion 4.\nOne end of a string of length L is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:\n\n1. T.\n2. T – $$\\frac{m v^{2}}{L}$$\n3. T + $$\\frac{m v^{2}}{L}$$\n4. 0\n\nT is the tension in the string. [Choose the correct alternative].\n1. The centripetal force necessary for the particle to move in a circular path is provided by the tension in the string. Hence net force on the particle is nothing but tension T in the string.", null, "Question 5.\nA constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop?\nGiven F = – 50 N (retarding force)\nm = 20 kg\nu = 15 m/s. V = 0 m/s\nt = ?\nF = ma ⇒ a = $$\\frac{\\mathrm{F}}{\\mathrm{m}}$$ = $$\\frac{-50}{20}$$ = – 2.5m/s2\nbut we know that\nV = u + at\n0 = 15 + (- 2.5) t\n⇒ t = 6 s.\n\nQuestion 6.\nA constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s-1 to 3.5 m s-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?\nGiven m = 3 kg\nu = 2 m/s\nv = 3.5 m/s\nt = 25 s.\nF = ma\nbut we know that a = $$\\frac{v-u}{t}$$\n∴ F = m $$\\left(\\frac{v-u}{t}\\right)$$ = 3 $$\\left(\\frac{3.5-2}{25}\\right)$$\n= 0.18 N\nsince direction acceleration ‘a’ is positive force is acting in the direction of motion.\n\nQuestion 7.\nA body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.", null, "Given Fa = 8 N\nFb = 6 N\nm = 5 kg\nThe result ant force F is given by,\nF = $$\\sqrt{\\mathrm{Fa}^{2}+\\mathrm{Fb}^{2}}$$\n= $$\\sqrt{64+36}$$ = 10 N\nwe know from the figure that\ntan θ = $$\\frac{F_{b}}{F_{a}}=\\frac{6}{8}$$ = 0.75\n⇒ θ = tan-1(0.75) = 37° with 8 N force\n\nQuestion 8.\nThe driver of a three wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three- wheeler is 400 kg and the mass of the driver is 65 kg.\nGiven, u = 36 km/h\n36 × $$\\frac{1000}{3600}$$ m/s\n=10 m/s\nv = 0\nt = 4 s\nm = 400 + 65 = 465 kg\na = $$\\frac{v-u}{t}$$ = $$\\frac{-10}{4}$$ = – 2.5 m/s²\nF = ma = 465 (- 2. 5)\n= – 11625 N (retarding force)\n\nQuestion 9.\nA rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms-2. Calculate the initial thrust (force) of the blast.\nGiven m = 20000 kg\na = 5ms-2 (against gravity) since the rocket has to move upwards against gravity the total initial thrust of the blast is given by\nF = ma + mg\n= m (a + g) = 20000 (5 + 9.8)\n= 296 × 105 N.\n\nQuestion 10.\nA body of mass 0.40 kg moving initially with a constant speed of 10 m s-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t= 0, the position of the body at that time to be x= 0, and predict its position at t = – 5 s, 25 s, 100 s.\nGiven mass m = 0.4 kg\nRetarding force F = – 8 N\n∴ acceleration a = $$\\frac{F}{m}$$ = $$\\frac{-8}{0.4}$$ = – 20 m/s²\nat t = – 5 s\na = 0 for t < 0\n∴ s = u + $$\\frac{1}{2}$$ at²\n= 10 (-5) + $$\\frac{1}{2}$$ (0)(-5)²\nat t = 25 s\nS = ut+$$\\frac{1}{2}$$ at²\n= 10 (25) + $$\\frac{1}{2}$$ (- 20) (25)²\n= – 6000 m\nat t = 100 s\nsince there is a retarding force for 30 s\nS1 = ut + $$\\frac{1}{2}$$ at²\n= 10 (30) + $$\\frac{1}{2}$$ (-20) (30)²\n= – 8700 m .\nafter 30 s it move with a constant velocity.\nV = u + at\n= 10 – 20 (30)\n= – 590 m/s\nfor rest of 70 s.\nS2 = – 590 (70) + $$\\frac{1}{2}$$ (0) (70)² = -41300 m\n∴ Total distance = S1 + S2 = 50000 m.", null, "Question 11.\nA truck starts from rest and accelerates uniformly at 2.0 m s-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the\n\n1. velocity, and\n2. acceleration of the stone at t = 11s? (Neglect air resistance.)\n\nWe have, Vt = u + at\ni.e. Vt = 0 + 2 (10) = 20 ms-1\nDuring the nextone second stone is under the effect of gravity.", null, "Vg = u + gt\n= 0 + 9.8 (1) = 9.8 m/s\n∴ Net velocity of stone at t = 11 s\nv = $$\\sqrt{V_{t}^{2}+V_{g}^{2}}$$ = 22.27 m/s.\ntan θ = $$\\frac{V_{g}}{V_{t}}=\\frac{9.8}{20}$$\n⇒ θ = 26.1° with horizontal.\n\n(b) The moment stone is dropped from truck only gravitational force is acting on it.\n∴ acceleration = g = 9.8 m/s².\n\nQuestion 12.\nA bob of mass 0.1 kg hung from the celling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s-1. What is the trajectory of the bob if the string is cut when the bob is\n\n1. at one of its extreme positions\n2. at its mean position.\n\n1. When the bob is at one of its extreme positions its velocity is zero. Hence if the string is cut, it will fall straight down due to gravitational force.\n\n2. At the mean position the bob has a horizontal velocity of 1 m/s. If the string is cut, bob is acted by vertical gravitational force = a = 9.8 ms-2. Hence bob will behave like a projectile and follows a parabolic path.\n\nQuestion 13.\nA man of mass 70 kg stands on a weighing scale in a lift which is moving\n\n1. upwards with a uniform speed of 10 m s-1\n2. downwards with a uniform acceleration of 5 m s-2\n3. upwards with a uniform acceleration of 5 ms-2. What would be the readings on the scale in each case?\n4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?\n\nThe weighing machine measures the reaction R which is nothing but the apparent weight.\n1. when the lift is moving upwards with uniform speed.\nR = mg = 70 × 9.8 = 686 N.\n\n2. When lift moves downwards with an acceleration of 5m/s²\nR = m (g – a) = 70 (9.8 – 5) = 336 N.\n\n3. When lift moves upwards with with an acceleration of 5m/s²\nR = m (g + a) = 70 (9.8 + 5) = 1036 N.\n\n4. If the lift falls down freely under gravity\nR = m (g – g) = 0.\n\nQuestion 14.\nFigure shows the position-time graph of a particle of mass 4 kg. What is the\n\n1. force on the particle for t < 0, t > 4 s, 0 < t < 4 s?\n2. impulse at t = 0 and t = 4 s? (Consider onedimensional motion only)", null, "1. From the position time graph we can see that the particle is in rest during t < 0 and t > 4. Hence net force on it is zero t < 0, t > 4 s. During 0 < t < 4; the graph has a constant slope i.e, particle\nhas uniform velocity = 3/4 = 0.75 m/s.\nHence net force is zero.\n\n2. at t = 0 u = 0 v = 0.75\nimpulse = change in momentum\n= M (v – u) = 4 (0.75 – 0)\n= 3 kg m/s\nat t = 4, u = 0.75 v = 0\nimpulse = 4 (0 – 0.75) = – 3 kg m/s.\n\nQuestion 15.\nTwo bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string, a horizontal force F = 600 N is applied to\n\n1. A\n2. B along the direction of string. What is the tension in the string in each case?", null, "1. Acceleration of the whole system a = $$\\frac{F}{M_{1}+M_{2}}=\\frac{600}{10+20}$$ = 20ms-2\nThe net force, acting on A\n= 600 – T = m1 (a)\n∴ 600 – T = 10 × 20\n⇒ T = 400 N.\n\n2. similar to the above case.\nThe net force acting on B\n= 600 – T = M2 a\n∴ 600 – T = 20 × 20\n⇒ T = 200 N.", null, "Question 16.\nTwo masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.", null, "Let ‘a’ be the acceleration of the masses. Then\nfor block m1, T – m1g = m1a → (1)\nfor block m2, m2g – T = m2 a → (2)\n(1) + (2) (m2 – m1) g = (m1 + m2) a\n⇒ a = $$\\frac{12-8}{12+8}$$ g = 2m/s\nsubstituting in (1)\nT – 8 × 10 = 8 × 2\n⇒ T = 96 N\n\nQuestion 17.\nA nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.\nLet m1 & m2 be the masses of smaller nuclei and let $$\\vec{v}_{1}$$ & $$\\vec{v}_{2}$$ be their velocities.\nAccording to the law of conservation of momentum.\nInitial momentum = final momentum\n0 = m1$$\\vec{v}_{1}$$ + $$\\vec{v}_{2}$$\nOr $$\\vec{v}_{2}$$ = – $$-\\frac{m_{1}}{m_{2}} \\vec{v}_{1}$$\nHence v1 & v2 are in opposite direction.\n\nQuestion 18.\nTwo billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s-1 collide and rebound at the same speed. What is the impulse imparted to each bail due to the other?\nImpulse = change in momentum\nInitial momentum of each ball = 0.05 × 6\n= 0.3 kg m/s\nFinal momentum of each ball = 0.05 × (-6)\n= – 0.3 kg m/s\nImpulse = 0.6 kgm/s (in magnitude).\n\nQuestion 19.\nA shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1, what is the recoil speed of the gun?\nGiven\nm1 = 0.02 kg, m2 = 100 kg\nv1 = 80 m/s v2 = ?\nAccording law of conservation of momentum\nm1u1 + m2u2 = m1v1 + m2v2\n0.02 (0) + 100 (0) = 0.02 × 80 + 100 × v2\nv2 = – 1.6 × 10-2 m/s\n\nQuestion 20.\nA batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)", null, "given\nm = 0.15 kg\nu = 54 kmph\n= 54 × $$\\frac{1000}{3600}$$\n= 15m/s\nAlong x -axis\nInitial velocity = – u cos θ\n= 15 cos (22.5°)\nFinal velocity = u cos θ\n= 15 cos (22.5°)\n∴ Impulse = change in momentum\n= 0.15 [15 cos (22.5) – (-15 cos (22.5)°)]\n= 4.16 kgm/s.\nAlong y-axis\nInitial velocity = Final velocity = – u sin θ\n∴ No impulse along the y-axis.\n\nQuestion 21.\nA stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?\nGiven m = 1.5m\nr = 1.5 m\nw = 40 rpm = $$40 \\frac{\\times 2 \\pi}{60}$$ rad/s\n= $$\\frac{4}{3}$$π rad/s\nNow Tension T = mrw²\n= 0.25 × 1.5 × $$\\left(\\frac{4}{3} \\pi\\right)^{2}$$\n= 0.58 N\nTmax = 200 N\nTmax = $$\\frac{\\mathrm{m} \\mathrm{v}_{(\\mathrm{max})}^{2}}{\\mathrm{r}}$$\n⇒ Vmax = $$\\sqrt{\\frac{200 \\times 1.5}{0.25}}$$ = 34.6 m/s.\n\nQuestion 22.\nIf, In Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:\n\n1. the stone moves radially outwards,\n2. the stone flies off tangentially from the instant the string breaks,\n3. the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?\n\nThe answer is 2. When a particle moves in a circular path, at each point the velocity is directed along the tangent of the circular path. Hence when string, breaks, it moves along the tangent in accordance with Newton’s 1st law of motion.\n\nQuestion 23.\nExplain why\n\n1. a horse cannot pull a cart and run in empty space.\n2. passengers are thrown forward from their seats when a speeding bus stops suddenly.\n3. it is easier to pull a lawnmower than to push it.\n4. a cricketer moves his hands backward while holding a catch.\n\n1. In accordance with Newton’s 1st law of motion since there is no external agent the horse cannot pull cart.\n2. The passenger continues to move forward when a speeding bus breaks because of their inertia of motion. Hence they are thrown forward from their seats.\n3. A lawn mover is pulled or pushed by applying a force at an angle. When it is pushed, the normal force (N) must be more than its weight, for equilibrium in the vertical direction. This results in greater friction and hence greater applied force to move. It is just opposite while pulling.\n4. The ball will have a large momentum. If the player tries to stop it instantaneous, the time of contact is low which results in a large impulse which may hurt his hand. Hence he tries to move his hands backward which increases the time of contact hence reducing the impulse.\n\n### 1st PUC Physics Laws of Motion Additional Exercises Questions and Answers\n\nQuestion 24.\nThe figure shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?", null, "The graph could be representing a ball rebounding between two walls separated by 2 cm with a constant velocity in free space. After receiving the impulse ball changes its direction. Hence time between two impulses is 2 seconds.\nvelocity = $$\\frac{\\text { displacement }}{\\text { time }}$$ = $$\\frac{2 \\times 10^{-2}}{2}$$ = 0.01m/s\nInitial momentum,\nmu = 0.04 × 10-2kgm/s\nFinal momentum,\nmv = – 0.04 × 10-2kgm/s\n∴ Change in momentum = 0.08 × 10-2kgm/s\n\nQuestion 25.\nFigure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s-2. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt?\n(Mass of the man = 65 kg.)", null, "Given acceleration of conveyor belt a = 1 m s-2\nµs = 0.2\nmass of man m = 65 kg\nThen man experiences a pseudo force Fs = ma as he is in an accelerating frame as shown in the figure. Hence to maintain his equilibrium he exerts a force F = – Fs = ma = 65 × 1 = 65 N in direction of motion of belt.\n∴ Net force acting on man = 65 N The man continue to be stationary with respect to belt if force of friction equal to force acting on man i.e.", null, "µs N = mamax\nµs . m .g = mamax\na(max) = µs × g\n= 0.2 × 10\n= 2m s-2\n\nQuestion 26.\nA stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]", null, "T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.", null, "The net force acting on stone at the lowest point directed vertically downward = mg – T1 & and the highest point = mg + T2. Hence option (a) is correct answer.\n\nQuestion 27.\nA helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the\n\n1. force on the floor by the crew and passengers,\n2. action of the rotor of the helicopter on the surrounding air,\n3. force on the helicopter due to the surrounding air.\n\nmass of helicopter = mh = 1000 kg mass of crew = mc = 300 kg\nvertical acceleration, a =15 m/s².\n1. force on the floor by crew & passenger = apparent weight of crew & passenger\n= mc (g + a)\n= 300(10 + 15)\n= 7500 N.\n\n2. The action of motor of helicopter on surrounding air is vertical downwards. The helicopter rises on account of reaction to this force\n= (mn + mc) (g + a)\n= (1000 + 300) (10 + 15)\n= 32500 N.\n\n3. force on helicopter due to the surrounding air is nothing but the reaction to the action of rolor = 32500 N in vertically upward direction.", null, "Question 28.\nA stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m² and hits a vertical wall nearby. What is the force exerted on the wall by the Impact of water, assuming It does not rebound?\nThe volume of water hitting the wall per second\n= (Area × Velocity) of a stream of water\n= 10-2 × 15\n= 0.15 m3s-1\ndensity of water = 1000 kg/m3\n∴ mass of water hitting the wall per second\n= 0.15 × 1000= 150 kg/s\nInitial momentum of water hitting the wall per second\n= 150 × 15\n= 250 kg m/s² or 2250 N\nFinal momentum per second = 0\n∴ Force exerted on the wall: change in momentum per second\n= 2250 N.\n\nQuestion 29.\nTen one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of\n\n1. the force on the 7th coin (counted from the bottom) due to all the coins on its top,\n2. the force on the 7th coin by the 8th coin,\n3. the reaction of the 6th coin on the 7th coin.\n\n1. There are 3 coins above the 7th coin\nHence force = (3m) g\n= 3mg N\n\n2. The 8th coin has two coins above it. Hence force exerted by 8th coin on 7th is, it’s weight plus the weight of two coins\n= mg + 2 mg\n= 3 mg N.\n\n3. The 6th coin is under the weight of 4 coins above it\nReaction R = – F = – 4 mg N.\n\nQuestion 30.\nAn aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?\nυ = 720 km/hr = 720 × $$\\frac{1000}{3600}$$\n= 200 m/s\nθ = 15°", null, "From the relation\ntan θ = $$\\frac{v^{2}}{r g}$$\n⇒ r = $$\\frac{v^{2}}{\\tan \\theta \\times g}$$ = $$\\frac{200 \\times 200}{\\tan 15^{\\circ} \\times 10}$$\n= $$\\frac{200 \\times 200}{0.2679 \\times 10}$$\n= 14931 m\n\nQuestion 31.\nA train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?\nvelocity υ = 54 km/h = 54 × $$\\frac{1000}{3600}$$= 15 m/s\nmass m = 106 kg\nThe centripetal force F = $$\\frac{m v^{2}}{r}$$ is provided by the lateral frictional force between rails and wheels of train.\nThe angle of banking required to prevent the wearing out of rail\ntan θ = $$\\frac{v^{2}}{r g}$$ = $$\\frac{15 \\times 15}{30 \\times 10}$$ = 0.75\nθ = tan-1 (0.75) ≈ 37°.", null, "Question 32.\nA block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?", null, "In the first case man applies an upward force of 25 kg weight. Hence the action on the floor by man is\n= 50 kg weight + 25 kg weight = 75 kg weight\n= 75 × 10\n= 750 N.\n\nIn the second case man applies a downward force of 25 kg weight. Hence the action on the floor by man is\n= 50 kg weight – 25 kg weight = 25 × 10\n= 250 N.\n(other 500 N is applied on the ceiling) Hence man should adopt the second case.\n\nQuestion 33.\nA monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey\n\n1. climbs up with an acceleration of 6 m s-2\n2. climbs down with an acceleration of 4 m s-2\n3. climbs up with a uniform speed of 5 m s-1\n4. falls down the rope nearly freely under gravity? (Ignore the mass of the rope).", null, "1. When monkey climbs up with an acceleration ‘a’ then\nT – mg = ma\nOr T = m (g + a)\n= 40 (10 + 6)\n= 640 N\nwhich exceeds the maximum tension which rope can withstand (600 N), hence rope breaks.\n\n2. when monkey is climbing down with an acceleration a\nmg – T = ma\nor T = m (g – a)\n= 40 (10-4)\n= 240 N\nThe rope will not break.\n\n3. when the monkey climbs up with uniform speed then\nT = mg\n= 40 × 10\n= 400 N\nThe rope will not break.\n\n4. when the monkey is falling freely, it would be a state of weightlessness. So, there won’t be any tension in the rope hence it will not break.\n\nQuestion 34.\nTwo bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wail (Fig). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are\n\n1. the reaction of the partition\n2. the action-reaction forces between A and B?\n3. What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? ignore the difference between µs and µk.", null, "1. As the blocks are at rest against the rigid walls, reaction of the partition = – (force applied on A)\n= 200 N towards left.\n\n2. The action-reaction force between A & B are 200 N each.\n\n3. when the wall is removed, the pushing force gives acceleration to the system. On taking the coefficient of friction into account,\n200 – µ (m1 + m2) g = (m1 + m2) a\na = $$\\frac{200-0.15(5+10) \\times 10}{(5+10)}$$\n= 11.8 ms-2\nLet the force exerted by A on B be FBA. On considering the equilibrium of the only block\nA, 200 – fk1 = m1 a + FBA\nFBA = 200 – µ m1 g – m1 a\n= 200 – 7.5 – 59\n= 133.5 N towards left.\n\nQuestion 35.\nA block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as\nviewed by\n\n1. a stationary observer on the ground,\n2. an observer moving with the trolley.\n\n1. Force experienced by block\nF = ma = 15 × 0.5 = 7.5 N\nForce of friction, Ff = µ mg = 0.18 × 15 × 10 = 27 N\nSince the force experienced block is less than frictional force it wilt remain stationary with respect to trolley. For an observer on the ground block appears to move with same acceleration as trolley,\n\n2. For an observer moving with trolley the block appears to he stationary as there is no relative motion between him and trolley and the block.\n\nQuestion 36.\nThe rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fail off the truck? (ignore the size of the box).", null, "Force experienced by box F = ma = 40 × 2 = 80 N\nfrictional force Ffriction= µ mg = 0.15 × 40 × 10 = 60 N\n∴ Net force on the box = F – Ffriction = 80 – 60 = 20 N .\n∴ The backward acceleration experienced by box is given by,\na = $$\\frac{\\text { Net force }}{\\text { mass }}$$ = $$\\frac{20}{40}$$ = 0.5 m/s²\nLet‘t’ be the time taken by box to move through 5m backwards\nWe have, S = ut + $$\\frac{1}{2}$$ at²\n∴ 5 = 0 × t + $$\\frac{1}{2}$$ × 0.5 × t²\nt = $$\\sqrt{20} \\approx$$ 4.47 s\nThe distance travelled by truck in t = 4.47s is\ns = ut + $$\\frac{1}{2}$$ at² (a = 2m /s2)\ns = 0 × $$(\\sqrt{20})$$ + $$\\frac{1}{2}$$ × 2 $$(\\sqrt{20})^{2}$$\ns = 20 m\nThe box will off the truck after 20 m from starting point.\n\nQuestion 37.\nA disc revolves with a speed of $$33 \\frac{1}{3}$$ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?\nIf the coin is to revolve with the record then the force of friction must be enough to provide the necessary centripetal force.\ni.e. mr ω² ≤ µs mg or r ≤ $$\\frac{\\mu_{\\mathrm{s}} \\mathrm{g}}{\\omega^{2}}$$\nHere, ω = $$33 \\frac{1}{3}$$ rpm", null, "$$\\approx$$ 0.12 m\nThe coin placed within the radial distance of 0.12 m will revolve with the record. Hence coin at 4 cm will revolve with the record.\n\nQuestion 38.\nYou may have seen in a circus a motorcyclist driving in vertical loops Inside a ‘deathwell’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?\nWhen the motor cyclist is at the highest point of death well, the normal readction R on the motor cycle by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal acting on him.\n∴ R + mg = $$\\frac{m v^{2}}{r}$$ → (1)\nThe minimum speed required to perform vertical loop is given by equation (1) when\nR = 0\n∴ mg = $$\\frac{m v^{2}_{(\\min )}}{r}$$\nυmin = $$\\sqrt{\\mathrm{rg}}$$ = $$\\sqrt{25 \\times 10}$$\n≈ 15.8 m/s.", null, "Question 39.\nA 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/mln. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?", null, "The horizontal force N exerted by the wall on the man provides the necessary centrepetal force.\n∴ N = m ω² r\nThe static frictional force (vertically upwards) balances the weight of the man mg.\nThe man remains stuck to the wall after the floor is removed if mg 〈 µ N\ni.e., mg 〈 µ mRω²\n∴ Minimum angular speed of rotation is\n⇒ ωmin = $$\\sqrt{\\frac{g}{\\mu_{s}} r}$$ = $$\\sqrt{\\frac{10}{0.15 \\times 3}}$$ = 4.6 rad/s\nThus the minimum rotational speed of cylinder required to hold the man stuck to the wall is 4.6 rad/s\n\nQuestion 40.\nA thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω\n\n1. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ $$\\sqrt{g / R}$$\n2. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = $$\\sqrt{2 \\mathrm{g} / \\mathrm{R}}$$? Neglect friction.", null, "1. Let the radius vector joining the bead to the center of the wire make an angle θ with vertical downward direction. Let N be the normal reaction. From fig,\nmg = N cos θ → (1)\nmr ω² = N sin θ → (2)\nor m (R sin θ ) ω² = N sin θ\nmRω² = N\nOn substituting in (1)\nmg = (m Rω²) cos θ\n0r ω = $$\\sqrt{\\frac{9}{\\mathrm{R} \\cos \\theta}}$$\nFor the bead to remain in lower most position θ = 0\n⇒ cos θ = 1\n⇒ ω ≤ $$\\sqrt{\\frac{\\mathrm{g}}{\\mathrm{R}}}$$\n\n2. when ω = $$\\sqrt{\\frac{2 g}{R}}$$\ncos θ = $$\\sqrt{\\frac{g}{R \\omega^{2}}}$$ = $$\\frac{g}{R\\left(\\frac{2 g}{R}\\right)}$$\n⇒ θ =60°\n∴ apparent weight = m (g – a).\n\n### 1st PUC Physics Laws of Motion One Mark Questions and Answers\n\nQuestion 1.\nDefine force.\nForce is defined as that external agent acting on a body changes its state of rest or uniform motion along a straight line.\n\nQuestion 2.\nDefine inertia.\nThe tendency of a body to oppose any change in its state of rest or of uniform motion is called inertia.\n\nQuestion 3.\nState Newton’s first law of motion (or Law of inertia).\nEverybody continues to be in its state of rest or uniform motion along a straight line unless compelled to change that state by an external force.\n\nQuestion 4.\nDefine Linear momentum.\nThe momentum of a body to defined to be the product of mass and velocity and is denoted by P $$\\overrightarrow{\\mathrm{P}}$$ = $$m \\bar{v}$$\n\nQuestion 5.\nState Newton’s Second law of motion.\nThe rate of change of momentum of the body is directly proportional to the applied force and takes place in the direction of the force.\n\nQuestion 6.\nDefine newton.\nOne newton is defined as that force which acting on a body of mass 1kg produces an acceleration of 1 m/s2\n\nQuestion 7.\nGive the dominion formula for\na)Force\nb) momentum\nForce – MLT-2\nMomentum – MLT-1", null, "Question 8.\nWhat quantity is conserved during rocket propulsion?\nLinear Momentum.\n\nQuestion 9.\nAction and reaction forces do not cancel each other. Why?\nAction and reaction forces do not cancel each other because they act on different objects.\n\nQuestion 10.\nWhat is the apparent weight measured? When a person of mass ‘m’ is standing in a lift accelerating with an acceleration of ‘a’\n(i) Downwards\nThe weighting machine measures the reaction force given by the floor. So when lift is going\ni) downwards the weight measured is lesser\n\nQuestion 11.\nii) Upwards\nii) Upwards the weight measured is more\n∴ apparent weight = m (g + a)\n\nQuestion 12.\nState Newton’s third law of motion.\nFor every action, there is an equal and opposite reaction.\n\nQuestion 13.\nWhich is the weakest force in nature?\nGravitational force.\n\nQuestion 14.\nIs it possible for the weight of a body to be zero?\nYes, whenever a body is on a free fall its weight is zero, but its mass remains unaltered.\n\nQuestion 15.\nTwo masses are In the Nation 1:2. What is the ratio of their Inertia?\nInertia of a body is directly proportional to its mass. Therefore the ratio of their inertia is Their inertia is also in the ration 1:2.\n\nQuestion 16.\nPassengers in buses tend to fall back as it accelerates. Why?\nDue to inertia, the passengers tend to continue their state of rest, when the bus moves by accelerating.\n\nQuestion 17.\nA cricket player catches the ball by moving his hand along the direction of the motion of the ball. Why?\nBy moving his hand along the direction of motion of the ball, the player increases the time of contact, thus reduces the impulse felt.\n\nQuestion 18.\nA stone breaks the window glass, but a bullet make only a hole. Why?\nSince the velocity of the bullet is much greater than that of the stone, the bullet is in contact with the glass for a very short time that the glass can’t give enough resistance.\n\nQuestion 19.\nCan a moving body be in equilibrium?\nYes, If a body is in a state of uniform motion in a straight line (net force acting on it is zero), its a moving body in equilibrium.\n\nQuestion 20.\nState law of conservation of momentum?\nWhen the net external force on a system is zero, then there is no change in momentum of the system.", null, "Question 21.\nWhat is friction?\nThe property by virtue of which an opposing force is created between two bodies in contact, which opposes their relative motion is called friction.\n\nQuestion 22.\nWhat is frictional force?\nThe force, which opposes the motion of one body over the other in contact with it, is called frictional force\n\nQuestion 23.\nWhat is static friction?\nFrictional force, which balances the applied force when the body is in the state of rest is called static friction.\n\nQuestion 24.\nWhat is limiting friction?\nThe maximum static friction that a body can exert on the other body in contact with it is called limiting friction.\n\nQuestion 25.\nWhat is sliding friction?\nThe frictional force that opposes the relative motion between the surfaces when one body slides over the other body is called sliding friction.\n\nQuestion 26.\nWhat is rolling friction?\nRolling friction is defined as the force of friction acting when a body rolls over the other body.\n\nQuestion 27.\nDefine angle of friction.\nAngle made by the resultant of normal reaction and limiting friction with the normal reaction is called angle of friction.\n\nQuestion 28.\nDefine angle of repose.\nAngle of repose is defined as the angle that an inclined plane makes with the horizontal when a body placed on it just starts sliding.", null, "Question 29.\nDefine coefficient of static friction.\nThe coefficient of static friction is defined as the ratio of limiting friction to the normal reaction between the surfaces.\n\nQuestion 30.\nWhat are the units and dimensions of the coefficient of friction?\nCo-efficient of friction is a ratio, so it is unitless.\n\nQuestion 31.\nWhen a wheel is rolling, what is the direction of the friction?\nFriction is tangential to the wheel and in the direction opposite to motion.\n\nQuestion 32.\nWhich of the following is a scalar quantity? Force, momentum & Inertia.\nInertia.\n\nQuestion 33.\nIf the string rotating stone is cut. Which direction will the stone move?\nThe stone will move in the direction tangential from the point where it got cut.\n\nQuestion 34.\nDoes a stone moving in a uniform circular motion (constant speed) has no net external force on it?\nThe speed is uniform but direction is changing, so, there is a change in velocity (acceleration is non zero) Hence stone is under the influence of a net external force.\n\nQuestion 35.\nA man of mass 60 kg is on a lift which is moving up with uniform speed, [g = 10ms 2]. Find apparent weight?\nSince it is moving with uniform speed, there no additional force (a = 0) So, apparent weight = m(g + 0) = (60 kg) × (10 ms-2)\n= 600 N.\n\nQuestion 36.\nWhat happens to the coefficient of friction if the weight of a body is doubled?\nThe coefficient of friction remain constant.\n\nQuestion 37.\nWhat provides the centripetal force for a car taking a turn on a level road.\nFrictional force.\n\nQuestion 38.\nFind force on a body if change in momentum of a body is 20 kg ms-1 over 5 seconds.\nForce is the rate of change of momentum\nF = $$\\frac{\\Delta(\\text { momentum })}{t}$$ = $$\\frac{20 \\mathrm{kg} \\mathrm{ms}^{-1}}{5 \\mathrm{s}}$$ 4 N\n\nQuestion 39.\nA 50 kg mass is subjected to a force of 5 N. What is acceleration of the body.\nWe know that, F = ma\n⇒ α = $$\\frac{F}{m}$$\n= $$\\frac{5 \\mathrm{N}}{50 \\mathrm{kg}}$$ = 0.1 ms-1", null, "Question 40.\nA 25 g body is moving with uniform velocity of 5ms-1. What is the force acting on the body?\nSince there is no change in velocity the a = 0.\n⇒ F = ma= (25 × 10-3) × (0) = 0 N\n\n### 1st PUC Physics Laws of Motion Two Marks Questions and Answers\n\nQuestion 1.\nState the 4 basic forces of nature.\n\n1. Gravitational force\n2. Elector magnetic force\n3. Strong nuclear force\n4. Weak nuclear force.\n\nQuestion 2.\nWhich is the strongest & weakest force in nature?\nStrongest – strong nuclear force Weakest – gravitational force.\n\nQuestion 3.\nExplain why does a cyclist bends while riding a curved road?", null, "On bending the cyclist part of his normal force to acts as centrepetal force that helps him staying within the circular path.\nFa = N sin θ = $$\\frac{m v^{2}}{\\alpha}$$\n\nQuestion 4.\nShow that impulse of force is equal to the change in momentum of a body.\nLet a force F act on a body of mass ‘m’ for a short interval of time‘t’.\nThen impulse of the force = F t\n= mat = $$m\\left(\\frac{v-u}{t}\\right) t$$ = m (v – u). Therefore impulse of force is equal to the change in momentum.\n\nQuestion 5.\nDefine Impulse of a force and Impulsive force.\nThe product of the force & the time for which it acts on a body is called impulse of a force. The force acting on a body for short interval of time is called impulsive force.\n\nQuestion 6.\nA ball hits the ground with a momentum $$\\overrightarrow{\\mathbf{p}}$$ and bounce back with the same magnitude of momentum. Find the change in momentum.\nInitial momentum = $$\\overrightarrow{\\mathbf{p}}$$\nFinal momentum = – $$\\overrightarrow{\\mathbf{p}}$$\nchange in momentum =\nΔ p = – $$\\overrightarrow{\\mathbf{p}}$$ – $$\\overrightarrow{\\mathbf{p}}$$ =- 2$$\\overrightarrow{\\mathbf{p}}$$", null, "Question 7.\nWhile Jumping on a cement floor, we weigh less than what weighs on cement floor. Why?)\nWhen we are on the floor, exerts reaction on us. When we jump the reaction on us is zero. Therefore while jumping on a cement Floor we weigh less than on cement floor.\n\nQuestion 8.\nDistinguish between conservative and nonconservative force.\nConservative forces are those forces against which the work done doesn’t depend on the path followed but depends only on the initial and final positions. Non-conservative forces are those in which the work done depends on the path taken\n\nQuestion 9.\nCalculate the Impulse of a force of 50 N acting for 0:1s.\nF = 50N t = 0:1 s\nimpulse = 50 N × 0.1s= 5 Ns\n\nQuestion 10.\nWhat are the methods of reducing friction?\n\n1. Friction between two surfaces can be reduced by polishing them.\n2. Jets, aeroplanes, and cars are given streamlined shape to reduce friction due to air resistance.\n3. The use of lubricants like oil, grease, etc. reduces the friction in machines.\n4. By using ball bearings friction in wheels of a car or cycle can be minimised.\n\nQuestion 11.\nStatic friction is a self-adjusting force comment.\nThe magnitude of static friction depends? on the magnitude of the applied force. As the applied force increases the magnitude of the static friction also increases. Thus static frictional force is a self-adjusting force.\n\nQuestion 12.\n\n1. Brakes of the vehicles work due to friction.\n2. Friction helps in driving vehicles.\n3. A match stick is lighted because of friction.\n\nQuestion 13.\nIs earth an inertial frame of reference?\nNo. earth can not be considered as an inertial frame of reference, because the earth is rotating and revolving, which means it is accelerating.\n\nQuestion 14.\nDerive an expression for recoil velocity of gun.\nLet mg be mass of gun,\n$$\\vec{v}_{g}$$ = recoil velocity of gun,\nmb be mass of Bullet $$\\vec{v}_{b}$$\nInitial momentum = 0,\nFinal momentum = mg $$\\vec{v}_{g}$$ + mb $$\\vec{v}_{b}$$\nBy law of conservation of momentum\n0 = mg $$\\vec{v}_{g}$$ + mb $$\\vec{v}_{b}$$\n⇒ $$\\vec{v}_{b}$$ = – $$\\left(\\frac{m_{b} \\vec{v}_{b}}{\\vec{m}_{g}}\\right)$$\n\nQuestion 15.\nHow does lubricants help in reducing friction?\nThe lubricants spread over the irregularities on the surface that makes the contact. So, the contact between the lubricant and the moving objects reduces the friction.\n\nQuestion 16.\nA bubble generator is kept at the bottom of an aquarium which is on a free fall. Will the bubbles generated rise to the top?\nNo, the bubbles generated at the bottom will not rise to the surface, ‘this is because the water in the aquarium is in a state of weightlessness and does not give the bubbles a reactional upward force.", null, "Question 17.\nWhy is it easier to pull a roller than push it?", null, "When we pull, the force ‘F’ the normal force is reduced by a value F sin θ. So the friction experienced is lesser which makes it is easier to pull.\n\nQuestion 18.\nAn object of mass m collides with a another object of mass ‘2 m’. if the initial velocity of the object of mass ‘m’ is v1 and of mass ‘2m’ is ‘0’. Then find the final velocity assuming they get stick to each other.\nInitial momentum =\n(m) (v1) + (2m) × (0)\n= mv1\nFinal momentum= m(vx) + 2m (vx)\n= 3m vx.\nWhere Vx is the velocitycombined mass By Law of conservation of momentum\nmv1 = 3mvx\nvx = $$\\frac{m v_{1}}{3 m}$$\nvx = $$\\frac{v_{1}}{3}$$\n\nQuestion 19.\nIf a boats sail is blown by air produced by a fan on the boat, can the boat move forward?\nNo, the boat can not be moved by a fan on the boat, this is because when the fan pushes the sail blowing air, the air pushes the fan backwards with the same force. Since there is no external force on the system – the net change in momentum is zero.\n\nQuestion 20.\nA retarding force Is applied to a motor car. If speed Is doubled how much more distance will it travel.\nLet original force be F1, mass be m1 velocity be V and distance be S then,\nF = ma, & V² = 2as\n⇒ F = $$\\frac{m v^{2}}{2 s}$$ ⇒ S = $$\\frac{m v^{2}}{2 F}$$\nIf velocity is doubled, then the distance it will travel before coming to halt will be increased by 4 times.\n\n### 1st PUC Physics Laws of Motion Three Marks Questions and Answers\n\nQuestion 1.\nDistinguish between mass and weight.\n\n1. Mass is the amount of matter contained in a body while weight is the gravitational force acting on a body.\n2. Mass of body remains same while weight of body varies from place to place.\n3. Mass is a scalar but weight is a vector.\n4. Unit of mass is kilogram and that of weight is newton.\n5. Mass is measured using a physical balance and weight is measured using a spring balance.\n\nQuestion 2.\nDerive the equation F = ma.\nConsider a body of mass ‘m’ moving with a velocity ‘u’. Let a constant force ‘F’ applied on a body changes its velocity to V in ‘t’ seconds.\nInitial momentum of the body = mass × initial velocity = m u\nFinal momentum = mass × Final velocity = m v\nChange of momentum in ‘t’ seconds = mv – mu.\n= $$\\frac{m v-m u}{t}$$ = m $$\\left(\\frac{v-u}{t}\\right)$$\n∴ α = ma\n∵ $$\\frac{v-u}{t}$$ = a, acceleration\nAccording to Newton’s second law, the rate of change of momentum is directly proportional to the applied force or vice versa.\ni. e. Force a rate of change of momentum\nF α ma\nF = kma\nWhere ‘k’ is a proportionality constant. In SI system k=1.\n∴ F = ma\n\nQuestion 3.\nState and explain Newton’s third law of motion. Give illustrations for the same.\nNewton’s third law states that for every action, there is an equal and opposite reaction.\nLet F1 be the force exerted by the body A on body B, F1 is called action. Then force F2 exerted by B on A is called reaction. According to the third law F1 = – F2.\nIllustrations:\n\n1. When a book is placed on the table the weight of the book is acting vertically downwards (action). The table exerts an equal and opposite force vertically upwards (reaction).\n2. A swimmer pushes the water in the backward direction with a certain force (action) and the water pushes him in the forward direction with equal and opposite force (reaction).\n3. The sailing of a boat is due to the action of the boat on water and reaction from water on the boat.\n4. When an object is suspended from the string, the weight of the object acts vertically downwards. The reaction in the string called the tension acts vertically upwards.\n5. The earth attracts the moon with a force that constitutes action. In turn the moon attracts earth with equal and opposite force (reaction).\n\nQuestion 4.\nDerive a relation for the safe velocity of negotiating a curve by a body in a banked curve with fractional coefficient ‘µ’.\nThe net force along the x-direction inwards should provide the centripetal force\n∴ Ffriction cos θ + N sin θ = $$\\frac{m v^{2}}{2}$$ → (1)\n∵ there is is no motion in y-direction\nN cos θ Ffriction sin θ = mg → (2)\nwe know that\nFfriction = µ N.\ndivide (1) by (2)", null, "$$\\frac{\\mu \\mathrm{N} \\cos \\theta+\\mathrm{N} \\sin \\theta}{\\mathrm{N} \\cos \\theta-\\mu \\mathrm{N} \\sin \\theta}$$ = $$\\frac{m v^{2}}{r(m g)}$$\n⇒ $$\\frac{N(\\mu+\\tan \\theta)}{N(1-\\mu \\tan \\theta)}$$ = $$\\frac{\\mathrm{v}^{2}}{\\mathrm{rg}}$$\n⇒ v² = rg$$\\left(\\frac{\\mu+\\tan \\theta}{1-\\mu \\tan \\theta}\\right)$$\nThe max velocity to safely negotiate the turn is $$\\sqrt{r g\\left(\\frac{\\mu+\\tan \\theta}{1-\\mu \\tan \\theta}\\right)}$$", null, "Question 5.\nWrite the equation corresponding to the ones given, for rotational motion about a fixed axis.\n(i) x (t) = x (0) + v (0) t + 1a/2 t²\nθ (t) = θ (0) + ω(0) + $$\\frac{1}{2}$$ α t²\n\n(ii) v² (t) = v² (0) + 2a [x t) – x (0)]\nω² (t) = ω² (0) + 2 a α [θ (t) – θ (0)]\n\n(iii) $$\\overline{\\mathbf{v}}$$ = $$\\frac{v(t)-v(0)}{2}$$\n$$\\bar{\\omega}$$ = $$\\frac{\\omega(t)-\\omega(0)}{2}$$\n\n(iv) v(t) = v(0) + at\nω (t) = ω (0) + α t\n\nQuestion 6.\nTwo masses m1 and m2 are connected to ends of string passing over a pulley. Find tension and acceleration associated.", null, "Assuming mass mt moves down with an\nacceleration ‘a’\nm1g -T = m1a1 ………….. (1)\nT – m2 g = m2 a1 ………. (2)\n(1) + (2)\nm1g – m2g = (m1 + m2) a1\n⇒ a1 = $$\\left(\\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\\right) g$$\n& T = m1g – m1 $$\\left(\\left(\\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\\right) g\\right)$$\n⇒ T = m1g $$\\left[1-\\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\\right]$$\nT = $$\\frac{2 m_{1} m_{2} g}{m_{1}+m_{2}}$$\n\nQuestion 7.\nName a mass varying system. Derive an expression for the velocity of the rocket at any instant of time\nA rocket-propelled into space is a mass varying system as it losses the weight of the fuel burnt.\nLet the velocity of gas used for propelling be ‘vg’ & let the rate of decrease in mass of the body be $$\\frac{\\mathrm{d} m}{\\mathrm{dt}}$$\nThen, by law of conservation of momentum since initial momentum is zero, dp = 0\n⇒ d (mv) = 0\n⇒ (dm) v + m d v = 0\n⇒ vdm = – mdv ⇒ dm = – $$\\frac{\\mathrm{d} \\mathrm{v}}{\\mathrm{v}}$$\nIntegrating on both sides\nv = vg (Inm) + c\n⇒ v = – vg logc m + c\nwhere c is a constant.\n\nQuestion 8.\nIndicate the force acting on a block of mass ‘m’ at rest on an Inclined plane of angle θ.", null, "Ffriction = μ N, N = mg cos θ & mg sin θ = Ffriction\n\nQuestion 9.\nDistinguish between static friction, limiting friction & kinetic friction. How do they vary with applied force? Explain.\nThe static friction is a friction that acts on a body at rest.\nLimiting friction is the maximum value of static friction. It is the force that is required for the body to just start moving.\nKinetic friction is the frictional force that action a body which is in motion.\nOn increasing the applied force, static friction increase, until it reaches limiting friction which is fixed, and kinetic friction also remains constant.", null, "Question 10.\nDefine Impulse. What graphical methods can be used to calculate impulse in the following cases\n\n1. constant force\n2. variable force acting on a body\n\nImpulse is a force that acts a body for a very short duration of time. It is defined as the product of force and the time for which it acts.\n1.", null, "In case of a constant force, say F1, the impulse is simply product of force and duration\nImpulse = F1 × t1\n2.", null, "In case of a variable force, the impulse will be the integral of F2 one of the interval [0,t2]\nImpulse = $$\\int_{0}^{t_{2}} F_{2}(t) d t$$\n\nQuestion 11.\nAn object of mass ‘m’ is on a inclined plane (θ). Find\n\n1. The effective resistant force on the body if it’s moving downwards.\n2. The minimum force if it is being pushed upwards.\n\nAssume a friction with a coefficient of μ", null, "1. The net force along the slope is\n⇒ Feq = F1 – Mg sin θ\n= μ N – Mg sin θ\nBut, N = mg cos θ\n⇒ Feq = mg (μ cos θ – sin θ)", null, "2. Additional force required be Fa\nFa = mg sin θ + μ N\n= mg sin θ + μ mg cos θ\n⇒ Fa = mg (sin θ + μ cos θ)\n\n### 1st PUC Physics Laws of Motion Five Marks Questions and Answers\n\nQuestion 1.\nState and prove the law of conservation of momentum.\nIn a closed system, the total linear momentum of the system remains constant or conserved.\nProof:\nConsider two bodies A and B of masses m1 and m2 moving in the same direction with uniform velocities u1 and u2 respectively. After the collision let their uniform velocities be v1 and v2. Let‘t’ be the time of impact.\nChange in momentum of A\n= m1v1 – m1u1\nRate of change of momentum of A\n= $$\\frac{m_{1} v_{1}-m_{1} \\mu_{1}}{t}$$\nchange in momentum of B\n= m2v2 – m2u2\nRate of change of momentum of B\n= $$\\frac{m_{2} v_{2}-m_{2} u_{2}}{t}$$\nIf F1 is the force exerted by A on B then according to second law,\nF1 = $$\\frac{m_{2} v_{2}-m_{2} u_{2}}{t}$$ (action)\nIf F2 is the force exerted by B on A then\nF2 = $$\\frac{m_{1} v_{1}-m_{1} \\mu_{1}}{t}$$ (reaction)\nAccording to Newton’s third law, action and reaction are equal and opposite i.e.\nF1 = – F2\n$$\\left[\\frac{m_{2} v_{2}-m_{2} u_{2}}{t}\\right]$$ = – $$\\left[\\frac{m_{1} v_{1}-m_{1} \\mu_{1}}{t}\\right]$$\nm2v2 – m2u2 = – m1v1 + m1u1\nOR\nm1u1 + m2u2 = m1v1 + m2v2\ni.e., Total momentum before collision = Total momentum after collision. Hence the momentum is conserved.", null, "Question 2.\nA stone weighing 5kg. falls from the top of a tower 100m high and buries itself 1 m deep in the sand. What is the average resistance offered by sand?", null, "Mass of the stone, m = 5kg\nHeight of the tower h = s = 100m\nInitial velocity u =0\nFinal velocity v = ?\nFrom the relation, v² = u² + 2gs\nv² = 0 + 2 × 9.8 × 100\nv² = 1960\nV = $$\\sqrt{1960}$$\nv = 44.27 ms-1\nThen the stone penetrates through the sand with a initial velocity, u = 44.27 ms-1\nDistance travelled, S = 1 m\nFinal velocity, v = 0\nacceleration, a =?\nFrom the equation, V² = u²+2as\n0² = (44.27)² + 2 × a × 1\n– 1960 = 2a\na = -980ms-2\n∴ The average resistance offered by the sand is F = ma\n= 5 × 980\nF = 4900 N\n\nQuestion 3.\nState the Newton’s laws of motion. Write any two illustrations.\nNewton’s first law of motion states that, everybody continues to be in its state of rest or uniform motion along a straight line unless compelled to change its state by an external force. Newton’s second law of motion states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.\nNewton’s third law of motion states that for every action there is. an equal and opposite reaction.\nIllustrations for Newton’s third law of motion are\n\n• when a book is placed on the table the book exerts force on the table (action), in turn the table exerts an equal and opposite force (reaction) on the book in the upward direction.\n• The sailing of a boat is due to the action of the boat on water and the reaction from water on boat.\n\nQuestion 4.\nConsider a body of mass ‘m’ attached to a string of length ‘L’. If the ring is forming a vertical circle, derive an expression for velocity and tension at any point.\nAlso, find the velocity that is required for mass to just reach the peak point of the circle.", null, "Let ‘θ’ be the angle the string makes with the vertical at any instant of time. Let vx be the velocity of body at the lowermost point x. The distance traveled by the body from the given point ‘p’ to ‘X’ in ‘y’ direction is given by,\nXY = L – Lcos θ = L(1 – cos θ)", null, "T – mg cos θ = $$\\frac{m v^{2}}{L}$$\nUsing the equation v² = u² + 2as, we can write\nVx² = v² + 2g (XY)\ni.e. V² = Vx² – 2g L (1 – cos θ) ……….. (1)\nTension, T = mg cos θ + $$\\frac{m v^{2}}{L}$$ using (1)\nT = mg cos θ + $$\\frac{m}{L}$$ (Vx² – 2gL (1 – cos θ)\n= mg cos θ + $$\\frac{m v_{x}^{2}}{L}$$ – 2 mg (1 – cos θ)\nT = $$\\frac{m v_{x}^{2}}{L}$$ + mg (3 cos θ – 2)\nFor the vertical circle to be reached v = 0 & vx = ? & θ = 180°\n⇒ vx² = v² + 2g L (1 – cos θ)\nvx² = 0 + 2g L (1 – (- 1))\nvx = $$\\sqrt{4g L}$$\nvx = $$2 \\sqrt{g L}$$\n\nQuestion 5.\nIf the system Is on a frictionless surface. Find the ratio of tensions in the string.", null, "The acceleration of the system is given\nby a = $$\\frac{\\text { Force applied }}{\\text { Total mass }}$$\n= $$\\frac{120 \\mathrm{N}}{(10+20+30) \\mathrm{kg}}$$ = 2 ms-2\nFor the last block\n120 – T2 = ma\n⇒ T2 = 120- (30) × (2)\n= 60 N\nFor the middle block\nT2 – T1 = ma\n60 – T1 = (20) × (2)\nT1 = 60 – 40 = 20 N\nThe ratio of tensions is,\nT1 : T2 = 20 : 60 = 1 : 3\n\nQuestion 6.\nFor the figure shown, find acceleration produced and the force of contact between the blocks. What is the effect of this force if it is applied to other blocks.", null, "The acceleration of system is\na = $$\\frac{F}{\\left(m_{1}+m_{2}\\right)}$$\nWhen the force is applied on block m, we have,\nF – Fc = m1 a\nwhere Fc is force of contact\n⇒ Fc = F – m1 $$\\left(\\frac{F}{m_{1}+m_{2}}\\right)$$ = $$\\left(\\frac{F m_{2}}{m_{1}+m_{2}}\\right)$$\nIf F is applied to other block ,", null, "Fc = m1 a = $$\\left(\\frac{\\mathrm{m}_{1}}{\\mathrm{m}_{1}+\\mathrm{m}_{2}}\\right)$$ F\n\nQuestion 7.\nIn the system shown if μk (kinetic friction coefficient) is 0.04. Find acceleration of the trolley,\n[g = 10ms-2]", null, "Free body diagram of trolley", null, "We know that Ff = μ N\n= μk mg\n= 0.04 × 15 × 10\n= 6N\n⇒ T – 6N = ma\n= 15 × a\n⇒ T – 5 a = 6 …………. (1)\nFree body diagram of mass\n⇒ 2g – T = ma\n⇒ 20 – T = 20\n⇒ 2a + T = 20 ……….. (2)\n0n,(1) – (2)\n– 17 a = – 14\na = $$\\frac{14}{17}$$ = 0.82ms-2", null, "Question 8.\nWeights of 250 g & 200 g are connected by a string over a smooth pulley. I system is traveling 4.95 m in the first 3 second. Find the value of g.", null, "For the pulley system,\nT – (200 g × 10-3) g = (200 × 10-3kg) a\nT – (0.29) = 0.2a …………….(1)\nan (0.25g) – T = (0.25a) …………(2)\n⇒ From (1) & (2)\n0.45 a = (0.25 – 0.2) g\na = $$\\frac{0.05}{0.45}$$g = $$\\frac{g}{9}$$ ms-2\nNow, S = 4.95 m, t = 3s u = 0\nFrom S = ut + $$\\frac{1}{2}$$ at²\n4.95 = 0 + $$\\frac{1}{2}$$ × a × (3)²\n4.95 = $$\\frac{1}{2}$$ × $$\\frac{10}{9}$$ × g × 9²\ng = $$\\frac{4.95}{5}$$ = 9.9 ms-2\n\nQuestion 9.\nA force of 80N acting on a body at rest for 2 sec imparts it a velocity of 20ms-1 what is the mass of the body calculate the distance traveled by the body in 2 seconds?\nForce, F = 80N\nInitial velocity, u =0\nTime for which force acts on the body, t = 25s,\nFinal velocity, v = 20ms-1\nFrom the equation v = u + at\n20 = 0 + a × 2\na = 10ms-2\n∴ The mass of the body, from the equation F = ma\nm = $$\\frac{F}{a}$$ = $$\\frac{80}{10}$$ = 8kg\n∴ The distance travelled by the body in a time, t = 2s. From the equation\ns = ut + $$\\frac{1}{2}$$ at²\n= 0 × 2 + $$\\frac{1}{2}$$ × 10 × 2²\n= 0 + 20 = 20m.\n\nQuestion 10.\nA man of 60 kg is standing on a weighing machine placed on the floor of a lift which reads the force in newtons. Find the reading of the weighing machine when the lift is\n\n1. stationary\n2. moving upwards with uniform speed of 10 ms-1\n3. moving downwards with uniform acceleration of 5 ms-2\n4. moving upwards with uniform acceleration of 5 ms-2.\n5. What would be the reading of the weighing machine if the connecting rope of the lift suddenly breaks and lift begins to fall freely under gravity, g = 10 ms-2.\n\nWeight of the man is due to the reaction from the floor of the lift. It is given by,\nR = mg + ma\nwhere a is the acceleration of the lift.\n1. when the lift is at rest a* = 0\n∴ Reading of the weighing machine\nR = mg = 60 × 10 = 600 N.\n\n2. when the lift is moving up or down with uniform velocity, a* = 0\n∴ Reading of the weighing machine,\nR = 600 N.\n\n3. When the lift is moving downwards with an acceleration a.\nR = mg – ma = m(g – a)\n= 60 × (10 – 5) = 300 N.\n\n4. When the lift is moving upwards with acceleration a,\nR = mg + ma* = m(g + a)\n= 60 × (10 + 5)\n= 900 N.\n\n5. If the lift falls freely under gravity, a = g\n∴ R = m(g – a*) = 0.\n\nQuestion 11.\nA rubber ball of mass 0.1 Kg is dropped on the ground from a height of 2.5 m and it rises to a height of 0.4m. Assuming the time of contact with the ground to be 0.01 s, calculate the force exerted by the ground on the wall, g = 10ms-2\nMass of the ball m = 0.1 kg\nTime in contact with the ground, t = 0.01s\n1. When the ball is dropped on ground,\nu =0, s = 2.5m, g = 10ms-2\nFrom the equation, v² = u² + 2gs\nv² = 0 + 2 × 10 × 2.5\nv = 7.07ms-1, downwards.\n\n2. When the ball rises up from the ground,\nv = 0, s = 0.4m, g = 10ms-2\nFrom the equation v² = u² + 2gs\n0 = u² + 2(- 10)0.4\nu² = 8\nu =2.83ms-1, upwards.\nAssuming the upward velocity +ve & downward velocity -ve,\nchange of velocity = 2.83 – (- 7.07)\ni.e v – u = 9.9ms-1\n∴ Force exerted by the ground on the ball is,\nF = m $$\\left(\\frac{v-u}{t}\\right)$$ = 0.1 $$\\left(\\frac{9.9}{0.01}\\right)$$ = 99 N.\n\nQuestion 12.\nA Bullet flying with a velocity of 50ms-1 hits a block of wood and penetrates through a distance of 0.2 m before coming to rest. The Mass of the bullet is 0.03kg. Calculate the resistance offered by the block of wood.\nInitial velocity, u = 50ms-1\nDistance travelled, s = 0.2\nFinal velocity v = 0\nMass of the bullet, m = 0.03 kg\nFrom the equation, v² = u² + 2as\n0 = 50² + 2 × a × 0.2\na = – $$\\frac{2500}{0.4}$$ = -6250ms-2\n∴ The resistance offered by the wood is\nF = ma = 0.03 × 6250 = 187.5 N.\n\nQuestion 13.\nA machine gun fires 200 bullets per minute with a velocity of 60ms-1. If the mass of each bullet Is 0.02kg, calculate the power of the gun.\nNumber of bullets fired in one minute = 200\nwork done by the gun in one minute is,\nW = kinetic energy of 200 bullets\nW = 200 ($$\\frac{1}{2}$$mv²)\n= 200 × $$\\frac{1}{2}$$ × 0.02 × (60)²\n= 7200J\n∴ Power, P = $$\\frac{W}{t}$$, t = 60 seconds\nP = $$\\frac{7200}{60}$$ = 120 watt.\n∴ Power of the gun = 120 watts.\n\nQuestion 14.\nTwo metal balls of masses 10kg and 8kg are moving In the same direction with velocities 10m/s and 4m/s respectively. They stick together after collision. Find their common velocity after collision. If they are moving\n\n1. In the same direction,\n2. In opposite direction before collision.\n\nm1 = 10kg, m2 = 8kg\nu1 = 10m/s and u2= 4m/s.\n1. v1 = v2 = v, common velocity.\nm1u1 +m2u2 = m1v1 + m2v2\n10 × 10 + 8 × 4 = (l0 + 8)v\nor v = 7.33m/s\n\n2. m1u1 – m2u2 = (m1 + m2) v\n10 × 10 – 8 × 4 = (10 + 8)v\nv = 3.778m/s.\n\nQuestion 15.\nDerive the equation F = ma.\nConsider a body of mass ‘m’ moving with a velocity ‘u’. Let a constant force ‘F’ applied on a body changes its velocity to ‘v’ in ‘t’ seconds.\nInitial momentum of the body = mass × initial velocity = m u\nFinal momentum = mass × Final velocity = m v\nChange of momentum in ‘t’ seconds = mv – mu.\nRate of change of momentum\n= $$\\frac{m v-m u}{t}$$ = m$$\\left(\\frac{v-u}{t}\\right)$$ = ma\n∵ $$\\frac{v-u}{t}$$ = a, acceleration\nAccording to Newton’s second law, the rate of change of momentum is directly proportional to the applied force or vice versa.\ni.e. Force a rate of change of momentum\nF α ma\nF = kma\nWhere ‘k’ is a proportionality constant. In SI system k =1.\n∴ F = ma.", null, "Question 16.\nName the basic forces in nature.\nBasic forces in nature are,\n\n1. Gravitational force\n2. Electromagnetic force\n3. Nuclear force and\n4. Weak force\n\nQuestion 17.\nA body is moving on a frictionless curved path of radius of 1.8 km with a speed of 30 ms-1. Find the banking angle required.\nThe centripetal force required to keep the body in circular motion is $$\\frac{m v^{2}}{r}$$\nHere, v = 30 ms-1\nr = $$\\frac{1.8 \\times 10^{3} m}{2}$$ = 0.9 103 = 900m\nN cos θ = mg\nand $$\\frac{m v^{2}}{r}$$ = N sin θ\n⇒ $$\\frac{m v^{2}}{r}$$ = $$\\frac{m g}{\\cos \\theta}$$ sin θ\n⇒ tan θ = $$\\frac{v^{2}}{r g}$$\n⇒ θ = tan-1 $$\\left(\\frac{30^{2}}{900 \\times 10}\\right)$$\n⇒ θ = tan-1 (0.1) = 5.71°.\n\nQuestion 18.\nWhat is the acceleration of a body moving on a circular path of radius 400 m. If it has\n\n1. constant speed of 40 ms-1\n2. speed increases at 3 ms-2\n\nA body on a circular path has two kinds of accelerations: radial & linear\n1. If speed is constant linear acceleration is zero & radical acceleration is ar = $$\\frac{v^{2}}{r}$$\nv = 40ms-2\nr = 400 m ⇒ ar = $$\\frac{40 \\times 40}{400}$$ = 4ms-2\n\n2. If the speed increases at 3 m/s², it has a linear acceleration of 3ms-2\na = $$\\sqrt{a_{r}^{2}+a_{1}^{2}}$$\n= $$\\sqrt{3^{2}+4^{2}}$$\n= 5 ms-2.\n\nQuestion 19.\nAn aeroplane at 360 km hr-1 has its wing banked at an angle 20°. Find the radius of the circle traversed by the plane, [g = 10ms-2]\nSpeed of the plane = 360 km/hr\n= $$\\frac{360 \\times 10^{3}}{3600}$$ = 100 ms -1\nwe know that tan θ = $$\\frac{v^{2}}{r g}$$\n⇒ r = $$\\frac{v^{2}}{\\tan \\theta \\times g}$$ = $$\\frac{100 \\times 100}{\\tan \\left(20^{\\circ}\\right) \\times 10}$$\n= 2.747 km.\n\nQuestion 20.\nA uniform chain of length L is kept on a table of coefficient of static friction μ(limiting value). Find the maximum length of chain that can be outside the table, without it sliding away.", null, "Let x be the length of the chain that can be outside the table.\nLet ‘M’ be the total mass of the chain.\nMass on the table is $$\\frac{M}{L}$$(L – x)\nMass of the chain outside = $$\\frac{M}{L}$$ x\nFor, equilibrium,\nForce of friction = weight of the hanging part.\ni.e., μN = $$\\left(\\frac{M}{L} x\\right)$$g × α\ni.e., μ$$\\left(\\frac{M}{L}(L-x) g\\right)$$ = $$\\left(\\frac{M}{L} x\\right)$$g\nμ(L – x) = x or x = $$\\frac{\\mu L}{1+\\mu}$$\n\nQuestion 21.\nFor the system shown in the figure, the coefficient of Kinetic friction between the mass and plane is 0.25.\nGiven that M2 = 5kg & M3 = 7kg. Find M1, such that the body M1, is moving with uniform velocity. Sin37° = $$\\frac{3}{5}$$, cos 37° = $$\\frac{4}{5}$$", null, "For the mass m1 to have a uniform velocity the system should be in equilibrium.\n⇒ T1 = m1 g\nT1 = 10m1 …………… (1) (g =10 m/s²)\nT1 =T2 + F11 + m2 g sin θ\n= T2 + μ N + m2 g sin 37°\n= T2 + μm2g cos 37° + m2g sin 37°\n= T2 + m2g $$\\left(\\mu \\frac{4}{5}+\\frac{3}{5}\\right)$$ (g =10 m/s²)\nT1 = T2 + m2[8μ + 6] ………….. (2)\nT2 = F12\nT2 = μ N\n= μ m3 g\nT2 = (0.25) (7) (10)\nT2 = 17.5 N ………….. (3)\nSubstituting (1) & (3) in (2)\n10m1 = 17.5 + 5 [8(0.25) + 6]\n10m1 = 57.5 N\nm1 = 5.75 kg.\n\n1st PUC Physics Laws of Motion Numerical Problems Questions and Answers\n\nQuestion 1.\nTwo masses 4 kg & 2 kg are connected by a massless string and they are placed on a smooth surface. The 2 kg mass is pulled by a force of 12 N as shown\n\n1. Find the acceleration of the system.\n2. If the string is replaced by a spring then what change do you notice in the acceleration\n3. In the string, system find the Tension", null, "1. We know that from Newtons second Law,\nF = ma\nF Force on the system\n⇒ a = $$\\frac{F}{m}$$ = $$\\frac{\\text { Force on the system }}{\\text { Total mass }}$$\n= $$\\frac{12 \\mathrm{N}}{(4+2) \\mathrm{kg}}$$\na = 2 ms-2\n\n2. If the string is replaced by spring, there is no change in mass of system. So there is no change in the acceleration a = 2m s-2\nc) We know that, F = m a\n(12 – T) = (m a)\nT = 12 – (2 × 2)\nT = 8 N", null, "a = 2 ms-2\n\nQuestion 2.\nA force of 98 N acts on a body of mass 10 kg which is at rest. Calculate\n\n1. Velocity at the end of 5 seconds.\n2. Distance traveled by the body in 5 seconds.\n\nSolution:\n1. To find the velocity at the end of 5 seconds.\nWe have, F = ma\nGiven, F = 98 N and\nm = 10 kg\n∴acceleration a = $$\\frac{F}{m}$$ = $$\\frac{98}{10}$$\n= 9.8 ms-2\nvelocity v = u + at\nHere a = 0,\na = 9.8 ms-2 and\nt = 5 s\n∴ v = 0 + 9.8 × 5\n= 49.0 ms-1\n\n2. To find the distance travelled\nwe have, s = ut + $$\\frac{1}{2}$$ at²\nHere, u = 0,\na = 9.8 ms-2\nt = 5 seconds\n∴ s = 0 × 5 + $$\\frac{1}{2}$$ × 9.8 × (5)²\n= 122.5 m.\n\nQuestion 3.\nA truck of mass 3000 kg is moving with a velocity of 10 m/s is accelerated by a force of 600N.\n\n1. What is the rate at which its velocity increases?\n2. How far will it travel In 10s?\n\nSolution:\n1. To find the rate at which velocity is increasing\nForce F = 600 N\nmass m = 3000 kg\nRate of increase in speed,\na = $$\\frac{F}{m}$$\n= $$\\frac{600}{3000}$$\n= 0.2 ms-2\n\n2. To find the distance travelled in 10 s\nWe have, s = ut + $$\\frac{1}{2}$$ at²\nHere, u =10 ms-1\nt = 10 s\na = 0.2 ms-2\n∴ s = 10 × 10 + $$\\frac{1}{2}$$ × 0.2 × (10)²\n= 100 + 10\n= 110 m.\n\nQuestion 4.\nA certain force acting on a body of mass 10 kg at rest moves it through 125 m in 5 seconds. If the same force acts on a body of mass 15 kg, what is the acceleration produced?\nSolution:\nin the case of the first body,\nu = 0;\nt = 5 s;\ns = 125 m\nSubstituting these values in the equation,\ns = ut + $$\\frac{1}{2}$$ at\n125 = 0 × 5 + $$\\frac{1}{2}$$ × a × (5)²\n$$\\frac{1}{2}$$ × a × 25\n∴ a = $$\\frac{2 \\times 125}{25}$$ = 10 ms-2\nF = m × a = 10 × 10 = 100 N\nIf the same force acts on another body of mass 15 kg, the amount of acceleration produced is,\na = $$\\frac{F}{m}$$ = $$\\frac{100}{15}$$\n= 6.67 ms-2", null, "Question 5.\nA cricket ball of mass 0.15 kg is moving with a velocity of 12 ms-1 and is hit by a bat so that the ball is turned back with a velocity of 20 ms-1. If the force of blow acts for 0.01 s, find the average force exerted on the ball by the bat.\nSolution:\nInitial velocity u = 12 ms-1\nFinal velocity v = 20 ms-1\nChange in velocity =20 – (- 12)\n= 20 + 12\n= 32 ms-1\n(-ve sign is taken because initial and final velocities are in opposite direction)\nTime for which force is acting, t = 0.01 s\n∴ acceleration a = $$\\frac{\\text { change in velocity }}{\\text { time }}$$\n= $$\\frac{32}{0.01}$$\n= 3200 ms-2\nForce F = ma\n= 0.15 × 3200\n= 480 N.\n\nQuestion 6.\nA hammer of mass 1 kg moving with a speed of 6 ms-1 strikes a wall and comes to rest in 0.1 s. Find the\n\n1. Impulse\n2. Retarding force on the hammer\n3. Retardation\n\nSolution:\n1. The initial momantum of the hammer is,\nm × v = 1 kg × 6 m s-1\n= 6 kg m s-1\n= 6 Ns\nImpulse = F . t = Δ P\n= 0 – mv\n= – 6 Ns.\n\n2. The force on the hammer\nF = $$\\frac{\\text { Impulse }}{\\text { time }}$$ =$$\\frac{6 \\mathrm{Ns}}{0.1 \\mathrm{s}}$$\n60 N.\n\n3. Retardation = a =$$\\frac{F}{m}$$ = $$\\frac{60 \\mathrm{N}}{1 \\mathrm{kg}}$$\n= 60 m s-2\n\nQuestion 7.\nA disc of mass 200 g is kept floating horizontally by throwing 40 pebbles per second against it from below. If the mass of each pebble is 2g, calculate the velocity with which the pebbles are striking the disc. Assume the pebbles strike the disc normally and rebound with the same speed.\nSolution:\nMass of the disc M = 200 g\n= 0.2 kg\nTotal downward force\nF = Mg\n= 0.2 × 9.8 =1.96 N\nMass of one pebble m = 2 g = 2 × 10-3 kg Let v be the velocity with which the pebbles strike the disc. Momentum given by one pebble = mv The pebbles rebound downward and strike from below.\n∴ net momentum given to the disc in the upward direction\n= change in velocity of the pebble × m = (2v) m\nTotal momentum given in one second\n=40 × m × 2v\n= 80 mv\nThe disc remains horizontal if this is equal to the weight of the disc, Mg\n∴ 80 mv = Mg\n80 × 2 × 10-3 × v = 1.96\nv = 12.25 ms-1\n\nQuestion 8.\nWater ejects with a speed of 0.2 ms-1 through a pipe of area of cross-section 1 × 10-2 m². If the water strikes a wall normally, calculate the force on the wall in newtons, assuming the velocity of the water normal to the wall is zero after the collision.\nSolution:\nVolume of water striking the wall per second = 0.2 × 10-2 = 2 × 10-3 m3\nMass of the water striking the wall in one second = volume × density = 2 × 10-3 × 1000\n= 2 kg\nChange in velocity of water on striking the wall in one second = 0.2 – 0 = 0.2 ms-1\nForce acting on the wall\n= change in momentum\n=2 × 0.2\n= 0.4N.\n\nQuestion 9.\nA gun of mass 5 tons fires a bullet of mass 20g with a velocity of 110.2ms-1. Find the velocity of the gun.\nSolution:\nInitially, both the gun and the bullet are at rest.\n∴ The total initial momentum of the system is f zero.\nIf v1 and v2 are the final velocity of the gun and the bullet, final momentum is given by,\npf = m1v1 + m2v2\nAccording to the law of conservation of momentum, pi = pf\nm1v1 + m2v2 = 0\ni.e., v1 = – $$\\frac{m_{2} v_{2}}{m_{1}}$$\n= $$\\frac{-20 \\times 110.2}{5 \\times 1000}$$\n= – 0.44 ms-1\n∴ Recoil velocity of the gun is 0.44 ms-1.\n\nQuestion 10.\nA gun weighing 1000 kg recoils with a velocity of 3 × 10-2 m/s when a shell of mass 1 kg is shot from it. If the shell hits the target in 8 seconds, find the gun target distance.\nSolution:\nInitial momentum of the gun & that of shell is zero as they are at rest.\nRecoil velocity of the gun v1 = – 3 × 10-2ms-1\nMass of the gun m1 = 1000kg\nMass of the shell m2 = 1 kg\nvelocity of the shell v2 =?\nFrom the equation\nm1 u1 + m2 u2 = m1v1 + m2 v2\n0 = 1000(- 3 × 10-2) +1 .v2\n∴ v2 = 30ms-1\nThe gun target distance,\ns = v2 × t\n= 30 × 8\n= 240m.\n\nQuestion 11.\nA machine gun has a mass of 20 kg. The firing rate of 500 bullets per second and mass of each bullet Is 20 g. If the speed of the bullets 500 m s-1. Find the force required to keep the gun in its position.\nSolution:\nmgun = 20 kg, mb = 20 g\nvgun = ? vb = 500 ms-1\nFrom Law of conservation of momentum\nMgun Vgun + mb vb = 0\n⇒ Vgun = –$$\\frac{20 \\times 10^{-3} \\times 500}{20}$$\n= – 0.5 -1\n∴ Force required to hold its position\nF = m$$\\left(\\frac{v-u}{t}\\right)$$ = 20 × $$\\frac{(0.5-0)}{\\left(\\frac{1}{500}\\right) s}$$ = 5000 N\n\nQuestion 12.\nA body of mass 20 kg moving with a velocity of 10 ms-1 collides with another body of mass 40 kg moving In the same direction with a velocity of 5 ms-1. If both the bodies stick together after the collision, find the common velocity after collision.\nSolution:\nIf u1 and u2 are the initial velocities of the two bodies before the collision, the total momentum before the collision is\npi = m1u1 + m2u2\nLet v be the common velocity of the two bodies after collision. Then the final mo-mentum after collision is,\npf = (m1 + m2)v\nFrom the law of conservation of momentum\nPi = Pf\nm1u1 + m2u2 = ( m1 + m2) v", null, "= 6.67 ms-1.\n\nQuestion 13.\nA shell of mass 10 kg flying horizontally with a velocity of 36 kmph explodes in air into two fragments. The larger fragment has a velocity of 25ms-1 & is directed In the same direction as the initial velocity of the shell. The smaller fragment has a velocity of 12.5 ms-1 in the opposite direction. Find the masses of the fragments.\nSolution:\nLet the mass of larger fragment be m1 = x Then the mass of smaller fragment is m2 = 10 – x\nInitial velocity of larger fragment,\nu1 = 36kmph = 10 ms-1.\nFinal velocity of larger fragment,\nv1 = 25ms-1.\nInitial velocity of smaller fragment,\nu2 = 10ms-1.\nFinal velocity of smaller fragment,\nv2 = – 12.5ms-1.\nAccording to the law of conservation of momentum,\nmu = m1v1 + m2v2\n10 × 10 = x.25 + (10 – x) – 12.5\n100 = 25x – 125 + 12.5x\n225 = 37.5x\n∴ x = $$\\frac{225}{37.5}$$ = 6Kg\n∴ Mass of larger fragment, m1 =. 6kg\nMass of smaller fragment m2 = (10 – x) = 4kg.", null, "Question 14.\nA neutron (mass = 1.67 × 1o-27kg) at a speed of 108m s-1. Collides with detron and gets sticked to it Find the velocity of the composite particle.\nSolution:\nMass of neutron 1.67 × 10-27 kg\n= Mn,\nmass of detron = (mn) = 3.34 × 10-27 kg\n= md,\nvelocity of neutron = 108m s-1 = Vn\nvelocity of detron = 0 m s-1 = Vd\nOn collision,\nmass of composite particle = Mc = Mn + Md\n= (1.67 + 3.34) × 10-27 kg\n= 6.01 × 10-27 kg\nvelocity of composite particle = vc\nFrom Law of conservation of momentum\nMn + Vn + McVc = McVc\n1.67 × 10-27 × 10+8 + 0 = (5.01 × 10-27) Vc\n⇒ Vc = $$\\left(\\frac{1.67}{5.01}\\right)$$ × 10+8\nVc = 0.33 × 108m s-1\n\nQuestion 15.\nA projectile is fired a with velocity ‘V’ at an angle ‘θ’. If the projective breaks into 2 equal parts and one of them retraces the path then find the velocity of the other part\nSolution:", null, "At the highest point the projective will have only x-direction velocity and it is constant throughout the path.\nS0, Vx = Vi cos θ\nLet ‘M’ be the initial mass, $$\\frac{M}{2}$$ be mass of the halves. Velocity of 1 half changes from vx to – vx. Let the velocity of other half be V0. 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https://rdrr.io/github/trevorld/piecepack/man/aabb_piece.html	[ "# aabb_piece: Calculate axis-aligned bounding box for set of game pieces In trevorld/piecepack: Board Game Graphics\n\n## Description\n\nCalculate axis-aligned bounding box (AABB) for set of game pieces with and without an “oblique projection”.\n\n## Usage\n\n ```1 2 3 4 5 6 7 8``` ```aabb_piece( df, cfg = getOption(\"piecepackr.cfg\", pp_cfg()), envir = getOption(\"piecepackr.envir\"), op_scale = getOption(\"piecepackr.op_scale\", 0), op_angle = getOption(\"piecepackr.op_angle\", 45), ... ) ```\n\n## Arguments\n\n `df` A data frame of game piece information with (at least) the named columns “piece_side”, “x”, and “y”. `cfg` Piecepack configuration list or `pp_cfg` object, a list of `pp_cfg` objects, or a character vector referring to names in `envir` or a character vector referring to object names that can be retrieved by `base::dynGet()`. `envir` Environment (or named list) containing configuration list(s). `op_scale` How much to scale the depth of the piece in the oblique projection (viewed from the top of the board). `0` (the default) leads to an “orthographic” projection, `0.5` is the most common scale used in the “cabinet” projection, and `1.0` is the scale used in the “cavalier” projection. `op_angle` What is the angle of the oblique projection? Has no effect if `op_scale` is `0`. `...` Ignored\n\n## Details\n\nThe “oblique projection” of a set of (x,y,z) points onto the xy-plane is (x + λ * z * cos(α), y + λ * z * sin(α)) where λ is the scale factor and α is the angle.\n\n## Value\n\nA named list of ranges with five named elements `x`, `y`, and `z` for the axis-aligned bounding cube in xyz-space plus `x_op` and `y_op` for the axis-aligned bounding box of the “oblique projection” onto the xy plane.\n\n## Examples\n\n ``` 1 2 3 4 5 6 7 8 9 10``` ``` df_tiles <- data.frame(piece_side=\"tile_back\", x=0.5+c(3,1,3,1), y=0.5+c(3,3,1,1), suit=NA, angle=NA, z=NA, stringsAsFactors=FALSE) df_coins <- data.frame(piece_side=\"coin_back\", x=rep(4:1, 4), y=rep(4:1, each=4), suit=1:16%%2+rep(c(1,3), each=8), angle=rep(c(180,0), each=8), z=1/4+1/16, stringsAsFactors=FALSE) df <- rbind(df_tiles, df_coins) aabb_piece(df, op_scale = 0) aabb_piece(df, op_scale = 1, op_angle = 45) aabb_piece(df, op_scale = 1, op_angle = -90) ```\n\ntrevorld/piecepack documentation built on July 22, 2021, 3:26 a.m." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63167727,"math_prob":0.94440603,"size":1574,"snap":"2021-31-2021-39","text_gpt3_token_len":511,"char_repetition_ratio":0.10955414,"word_repetition_ratio":0.05511811,"special_character_ratio":0.32909784,"punctuation_ratio":0.16332377,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99249405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-29T02:26:27Z\",\"WARC-Record-ID\":\"<urn:uuid:647ac000-cb4d-4086-9ab5-03f3971e2d2f>\",\"Content-Length\":\"47200\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08309c1a-59ae-48a1-a447-a5bf76ac7519>\",\"WARC-Concurrent-To\":\"<urn:uuid:14cddd94-258e-47b1-a10a-5dc4017a1cf0>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/github/trevorld/piecepack/man/aabb_piece.html\",\"WARC-Payload-Digest\":\"sha1:VSCLKUFOJ3HC65KPJXBT36BZEWR2E5HD\",\"WARC-Block-Digest\":\"sha1:OITEU3T7N7DPHFIOS4B2TJEO3FCSBRVH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153814.37_warc_CC-MAIN-20210729011903-20210729041903-00007.warc.gz\"}"}
https://physics.stackexchange.com/questions/494771/what-is-the-exact-role-of-friction-in-rolling-without-slipping	[ "# What is the exact role of friction in rolling without slipping?\n\nWhen an object rolls along a plane, we say that for the object to roll without slipping, the velocity of the center of mass must be equal to angular velocity times radius, so that at the point of contact with the ground the two velocities cancel each other, and the point is instantaneously at rest.\n\nI understand this.\n\nBut then we say that if the ground is rough, there exists a friction on the bottom of the wheel. Which gives it torque. And this friction is responsible for the rotational motion of the wheel. Now if there is a external force acting on the center of mass of the wheel, then the only way the wheel will roll without slipping is if the tangential acceleration is greater than or equal to the acceleration caused by the external force at the center of mass. I don't understand why this is.\n\nFirst of all, if the point is momentarily at rest, how does friction know which direction to apply the force in? Secondly, if friction applies a torque, couldn't it be the case that the linear acceleration caused by the same is greater than the acceleration due to the external force? In that case also, shouldn't the wheel slip?\n\n• if your car is rolling down a hill, linear acceleration can be greater than the external force applied by the drive train to the center of the wheel, but the wheel still may not slip. Aug 2, 2019 at 2:47\n• Related question: Can we know when rolling occurs without slipping? Aug 2, 2019 at 4:34\n• If the tangential acceleration is lesser than the force at CM then it would be as if the cylinder/wheel is skidding. It’s intuitive to think of pushing a wheel rather than rolling it uphill or downhill would be difficult rather than rolling. Rolling friction is lesser than surface friction hence its easier roll rather than push an object across provided the object is circular in shape. Aug 2, 2019 at 5:54\n• Your first question “if the point is at rest how does friction know where to act?” is like asking where does friction know to act if an object is pushed along a rough plane. For the second question, friction itself does not act on the cm it acts at a distance causing a torque. This friction cannot cause linear acceleration because the net forces cancel out across opposite point s of the circle. In that case the wheel cannot slip unless a large enough linear force on the cm is applied. Aug 2, 2019 at 6:18\n\nThe part you understand is correct. Let's get to the rest of your question.\n\nBut then we say that if the ground is rough, there exists a friction on the bottom of the wheel.\n\nNot ideally, no. Just like how there is no static friction acting on a book resting on a table with no other horizontal forces acting on it, there is no static friction force acting on the wheel if there are no other forces/torques trying to change the wheel's velocity.\n\nWhich gives it torque. And this friction is responsible for the rotational motion of the wheel.\n\nThis is invalidated by the above discussion. If the wheel is rolling then it will keep on rolling. Friction is not responsible for this, just like how it is not responsible for keeping the book at rest on the table.\n\nNow if there is a external force acting on the center of mass of the wheel, then the only way the wheel will roll without slipping is if the tangential acceleration is greater than or equal to the acceleration caused by the external force at the center of mass.\n\nHere is where friction now comes into play, just like how static friction would be required to keep our book at rest of we started to push on it. Although I must say your terminology confuses me here. The simplest way to think about it is just that the static friction has some maximum value. If friction needs to be larger than this maximum value to prevent slipping, then slipping will occur. Once again, just think about the book.\n\nFirst of all, if the point is momentarily at rest, how does friction know which direction to apply the force in?\n\nFriction opposes relative motion. It \"knows\" which direction to act because that is the direction that opposes slipping. Once again, think about the book.\n\nSecondly, if friction applies a torque, couldn't it be the case that the linear acceleration caused by the same is greater than the acceleration due to the external force? In that case also, shouldn't the wheel slip?\n\nThis is confusing again. Just talk about forces. I see many people, including you now, getting things mixed up with comparing \"accelerations due to forces and torques\".\n\nYou can work out the problem in more detail using Newton's second law. What is interesting is that depending on where you apply your force and the type of wheel you have the friction force could act either in the direction of the applied force or opposite the direction of the applied force to prevent slipping. I discuss this here and here\n\nWhen a wheel is rolling without slipping at constant speed, friction has no role: The rotational and translational speeds are perfectly matched, and neither force nor torque from friction is needed nor available.\n\nTo put it another way, friction is about force and torque. Those in turn are related to angular and linear acceleration, not velocity. No acceleration, no force and/or torque, no friction.\n\nNow, what happens if some (external) force accelerates the axle?\n\nIn the absence of friction, the forward linear motion of the wheel increases due to the force on the axle, but there's nothing that increases the angular speed of the wheel: With that constant angular speed, the touch point will be moving forwards.\n\nIn the frictionless case, the wheel will start sliding forward on the surface. But that sliding is what friction at that surface can oppose with a fore that generates a torque to speed the wheel up.\n\nImagine a thrown bowling ball. At first it's sliding, and the friction acts to make it spin until it's rolling.\n\nWith a small acceleration, the wheel might not reach the point of sliding and kinetic friction. For small accelerations and large friction, it'll stay in the static friction regime. But that static friction will be generating a force in that same direction, causing the wheel to speed up it's rotation until it's rolling at the right speed to be rolling without friction.\n\n• I believe the question mainly concerns acceleration. Aug 2, 2019 at 3:16\n• \"In the absence of friction, the forward linear motion of the wheel increases due to the force on the axle, but there's nothing that increases the angular speed of the wheel: With that constant angular speed, the touch point will be moving forwards.\" So the wheels will slip? Aug 2, 2019 at 23:24\n• In the absence of friction, yes, in the linear motion increases by itself the wheel will slip. With friction, the friction force speeds up the rotation to match. Aug 2, 2019 at 23:26" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9501795,"math_prob":0.9453659,"size":2446,"snap":"2023-40-2023-50","text_gpt3_token_len":501,"char_repetition_ratio":0.14086814,"word_repetition_ratio":0.013824885,"special_character_ratio":0.20032707,"punctuation_ratio":0.08921162,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97354984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T14:24:36Z\",\"WARC-Record-ID\":\"<urn:uuid:6ba319ed-3436-447f-872b-d9eeec55c650>\",\"Content-Length\":\"183011\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec310346-83b2-4035-aa09-e1ec8742503e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b9f91ada-6d45-4d29-8785-8d01a9e137e7>\",\"WARC-IP-Address\":\"104.18.43.226\",\"WARC-Target-URI\":\"https://physics.stackexchange.com/questions/494771/what-is-the-exact-role-of-friction-in-rolling-without-slipping\",\"WARC-Payload-Digest\":\"sha1:HI33U425J4YON5TNUYWQR7ZXSSDTB3EZ\",\"WARC-Block-Digest\":\"sha1:JFB2JZB2AOLZO4TSJGPFLDKWAWGOMM6U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100287.49_warc_CC-MAIN-20231201120231-20231201150231-00534.warc.gz\"}"}
https://lists.nongnu.org/archive/html/auctex-devel/2021-12/msg00016.html	[ "auctex-devel\n[Top][All Lists]\n\n## a function that converts expression1/expression2 to \\frac{expression1}{e\n\n From: Uwe Brauer Subject: a function that converts expression1/expression2 to \\frac{expression1}{expression2} Date: Wed, 29 Dec 2021 08:22:47 +0100 User-agent: Gnus/5.13 (Gnus v5.13) Emacs/29.0.50 (gnu/linux)\n\nHi\n\nThis has annoyed me for years, honestly.\n\nI sometimes receive latex files, with constructions like this\n\nIn fact, if $6/5<\\gamma<2$, there exists a solution\nThen\n\\begin{equation}\n\\rho(t, x)=\\left\\{\n\\begin{array}{ll}\n{\\left[\\frac{K \\gamma}{4 \\pi\n\\kappa(\\gamma-1)}\\right]^{1/\\left(2-\\gamma^{\\prime}\\right)} u(\|x\|)^{1\n/(\\gamma-1)}} & \\text { for } 0 \\leqq\|x\| \\leqq R \\\\\n0 & \\text { for } R<\|x\|\n\\end{array}\n\\right.\n\\end{equation}\n\nWhat annoys me is the use of 6/5 instead of \\frac{5}{6} (I know\nsometimes it is even advised to use it).\n\nBut in general I want \\frac\n\nSo I have the following function\n\n(defun my-change-frac ()\n\"Changes stuff like 1/2 to \\frac{1}{2}.\"\n(interactive)\n(query-replace-regexp \"\\$$\\\\<[0-9]\\$$\\$$[\\\\/]\\$$\\$$[0-9]\\\\>\\$$\"\n\"\\\\\\\\frac{\\\\1}{\\\\3}\"))\n\nIt covers the 6/5 case but not, for example,\n1/\\left(2-\\gamma^{\\prime}\\right)\n\nDoes anybody have an idea?\nIn general shouldn't auctex have such a function?\n\nRegards\n\nUwe Brauer", null, "smime.p7s\nDescription: S/MIME cryptographic signature" ]	[ null, "https://lists.nongnu.org/icons/generic.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6752379,"math_prob":0.98312944,"size":1515,"snap":"2022-05-2022-21","text_gpt3_token_len":506,"char_repetition_ratio":0.13103905,"word_repetition_ratio":0.041237112,"special_character_ratio":0.3518152,"punctuation_ratio":0.09608541,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971875,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T07:38:13Z\",\"WARC-Record-ID\":\"<urn:uuid:9fe56b7f-7ab2-43c6-966a-fade1e332661>\",\"Content-Length\":\"5765\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3f3e223-89db-4ffd-979e-c3750f502c6f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8ea7b13-36a1-4781-8843-4d62fc84d807>\",\"WARC-IP-Address\":\"209.51.188.17\",\"WARC-Target-URI\":\"https://lists.nongnu.org/archive/html/auctex-devel/2021-12/msg00016.html\",\"WARC-Payload-Digest\":\"sha1:GL4GAF3AJZNU3S7XI63QBMMEPOXMPN7Q\",\"WARC-Block-Digest\":\"sha1:LTXXX64BGU5UW5AOSQP3HOCJU5L5O3LY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662521152.22_warc_CC-MAIN-20220518052503-20220518082503-00199.warc.gz\"}"}
https://epub.ub.uni-muenchen.de/88963/	[ "", null, "", null, "Mousset, Frank; Noever, Andreas; Panagiotou, Konstantinos; Samotij, Wojciech (2020): ON THE PROBABILITY OF NONEXISTENCE IN BINOMIAL SUBSETS. In: Annals of Probability, Vol. 48, No. 1: pp. 493-525\nFull text not available from 'Open Access LMU'.\n\n### Abstract\n\nGiven a hypergraph Gamma = (Omega, chi) and a sequence p = (p(omega))(omega is an element of)(Q) of values in (0, 1), let Omega(p) be the random subset of Omega obtained by keeping every vertex omega independently with probability p(omega). We investigate the general question of deriving fine (asymptotic) estimates for the probability that Omega(p) is an independent set in Gamma, which is an omnipresent problem in probabilistic combinatorics. Our main result provides a sequence of upper and lower bounds on this probability, each of which can be evaluated explicitly in terms of the joint cumulants of small sets of edge indicator random variables. Under certain natural conditions, these upper and lower bounds coincide asymptotically, thus giving the precise asymptotics of the probability in question. We demonstrate the applicability of our results with two concrete examples: subgraph containment in random (hyper)graphs and arithmetic progressions in random subsets of the integers.", null, "", null, "" ]	[ null, "https://epub.ub.uni-muenchen.de/images/epub_logo_de.png", null, "https://epub.ub.uni-muenchen.de/images/header_open-access-lmu_mobile_en.png", null, "https://epub.ub.uni-muenchen.de/images/autor-author.svg", null, "https://epub.ub.uni-muenchen.de/images/export.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83806926,"math_prob":0.9393992,"size":1248,"snap":"2022-27-2022-33","text_gpt3_token_len":288,"char_repetition_ratio":0.10932476,"word_repetition_ratio":0.0,"special_character_ratio":0.20913461,"punctuation_ratio":0.13513513,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99497634,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T22:55:47Z\",\"WARC-Record-ID\":\"<urn:uuid:365a3566-2333-4ee3-bab0-52d7d9f9d741>\",\"Content-Length\":\"24351\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:231bcb95-a752-4aee-874d-0759512fa0ea>\",\"WARC-Concurrent-To\":\"<urn:uuid:83e1529e-acfa-451c-9fdb-e4373385146d>\",\"WARC-IP-Address\":\"141.84.147.156\",\"WARC-Target-URI\":\"https://epub.ub.uni-muenchen.de/88963/\",\"WARC-Payload-Digest\":\"sha1:KAAS66HMQ5757PBUZCE74KAECM6GEGU5\",\"WARC-Block-Digest\":\"sha1:Z5JLJ5CQHDWP6VWR5FINTCBWO5HYBNRS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104678225.97_warc_CC-MAIN-20220706212428-20220707002428-00304.warc.gz\"}"}
https://www.geogebra.org/m/d2Sfwu5z	[ "# The ratio of the areas of two similar triangles\n\nProof: We are given two triangles ABC and PQR such that Δ ABC ~Δ PQR We need to prove that ar(ABC) / ar(PQR) = (AB/PQ)² = (BC/QR)²=(CA/RP)² For Finding the areas of the two triangles, we draw altiude AM and PN of the triangle Now, ar(ABC) = 1/2 BC x AM and ar(PQR) = 1/2 QR x PN Therefore , AM/PN = AB/PQ Also, Δ ABC ~ Δ PQR So, AB/PQ = BC/QR = CA/RP Therefore, ar(ABC)/ar(PQR) = AB/PQ x AM/PN = AB/PQ x AB/PQ = (AB/PQ)² Weget ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)² So, ar(ABC)/ar(PQR) = (1/2 x BC x AM) / 1/2 x QR x PN = (BC x AM) / (QR x PN) Now, in Δ ABM and Δ PQN, ∠B = ∠Q (As Δ ABC ~ Δ PQR) ∠M = ∠N (Each is of 90⁰) and Δ ABM ~Δ PQN (AA similarity criterion)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89673793,"math_prob":0.99977785,"size":799,"snap":"2023-14-2023-23","text_gpt3_token_len":329,"char_repetition_ratio":0.14842768,"word_repetition_ratio":0.0,"special_character_ratio":0.3729662,"punctuation_ratio":0.07070707,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99993896,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T02:51:11Z\",\"WARC-Record-ID\":\"<urn:uuid:25c33120-e55c-487e-948d-95cb761b87d1>\",\"Content-Length\":\"46395\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e957e752-3411-4aab-86bc-68a19477d992>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bf743b8-84a0-427e-ade6-21ab7e0a916a>\",\"WARC-IP-Address\":\"18.160.10.6\",\"WARC-Target-URI\":\"https://www.geogebra.org/m/d2Sfwu5z\",\"WARC-Payload-Digest\":\"sha1:CHZSDZA4ILQLXS4TKDCEYEJJY6SJ3UIQ\",\"WARC-Block-Digest\":\"sha1:JTRLAUBLM6BEWF2B72VDB5MNOQDX7MIL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943749.68_warc_CC-MAIN-20230322020215-20230322050215-00566.warc.gz\"}"}
http://conceptmap.cfapps.io/wikipage?lang=en&name=Linearized_gravity	[ "# Linearized gravity\n\nLinearized gravity is an approximation scheme in general relativity in which the nonlinear contributions from the spacetime metric are ignored, simplifying the study of many problems while still producing useful approximate results.\n\n## The method\n\nIn linearized gravity the metric tensor, $g$ , of spacetime is treated as a sum of an exact solution of Einstein's equations (often Minkowski spacetime) and a perturbation $h$ .\n\n$g\\,=\\eta +h$\n\nwhere $\\eta$ is the nondynamical background metric that is being perturbed about, and $h$ represents the deviation of the true metric ($g$ ) from flat spacetime.\n\nThe perturbation is treated using the methods of perturbation theory, \"linearized\" by ignoring all terms of order higher than one (quadratic in $h$ , cubic in $h$ etc...) in the perturbation.\n\n## Applications\n\nThe Einstein field equations (EFE), being nonlinear in the metric, are difficult to solve exactly and the above perturbation scheme allows linearized Einstein field equations to be obtained. These equations are linear in the metric, and the sum of two solutions of the linearized EFE is also a solution. The idea of 'ignoring the nonlinear part' is thus encapsulated in this linearization procedure.\n\nThe method is used to derive the Newtonian limit, including the first corrections, much like for a derivation of the existence of gravitational waves that led, after quantization, to gravitons. This is why the conceptual approach of linearized gravity is the canonical one in particle physics, string theory, and more generally quantum field theory where classical (bosonic) fields are expressed as coherent states of particles.\n\nThis approximation is also known as the weak-field approximation, as it is only valid if the perturbation h is very small.\n\n### Weak-field approximation\n\nIn a weak-field approximation, the gauge symmetry is associated with diffeomorphisms with small \"displacements\" (diffeomorphisms with large displacements obviously violate the weak field approximation), which has the exact form (for infinitesimal transformations)\n\n$\\delta _{\\vec {\\xi }}h=\\delta _{\\vec {\\xi }}g-\\delta _{\\vec {\\xi }}\\eta ={\\mathcal {L}}_{\\vec {\\xi }}g={\\mathcal {L}}_{\\vec {\\xi }}\\eta +{\\mathcal {L}}_{\\vec {\\xi }}h$\n$=\\left[\\xi _{\\nu ;\\mu }+\\xi _{\\mu ;\\nu }+\\xi ^{\\alpha }h_{\\mu \\nu ;\\alpha }+\\xi _{;\\mu }^{\\alpha }h_{\\alpha \\nu }+\\xi _{;\\nu }^{\\alpha }h_{\\mu \\alpha }\\right]$\n$dx^{\\mu }\\otimes dx^{\\nu }$\n\nWhere ${\\mathcal {L}}$ is the Lie derivative and we used the fact that η does not transform (by definition). Note that we are raising and lowering the indices with respect to η and not g and taking the covariant derivatives (Levi-Civita connection) with respect to η. This is the standard practice in linearized gravity. The way of thinking in linearized gravity is this: the background metric η is the metric and h is a field propagating over the spacetime with this metric.\n\nIn the weak field limit, this gauge transformation simplifies to\n\n$\\delta _{\\vec {\\xi }}h_{\\mu \\nu }\\approx \\left({\\mathcal {L}}_{\\vec {\\xi }}\\eta \\right)_{\\mu \\nu }=\\xi _{\\nu ;\\mu }+\\xi _{\\mu ;\\nu }$\n\nThe weak-field approximation is useful in finding the values of certain constants, for example in the Einstein field equations and in the Schwarzschild metric.\n\n## Linearized Einstein field equations\n\nThe linearized Einstein field equations (linearized EFE) are an approximation to Einstein's field equations that is valid for a weak gravitational field and is used to simplify many problems in general relativity and to discuss the phenomena of gravitational radiation. The approximation can also be used to derive Newtonian gravity as the weak-field approximation of Einsteinian gravity.\n\nThe equations are obtained by assuming the spacetime metric is only slightly different from some baseline metric (usually a Minkowski metric). Then the difference in the metrics can be considered as a field on the baseline metric, whose behaviour is approximated by a set of linear equations.\n\n### Derivation for the Minkowski metric\n\nStarting with the metric for a spacetime in the form\n\n$g_{ab}=\\eta _{ab}+h_{ab}$\n\nwhere $\\,\\eta _{ab}$ is the Minkowski metric and $\\,h_{ab}$ — sometimes written as $\\epsilon \\,\\gamma _{ab}$ — is the deviation of $\\,g_{ab}$ from it. $h$ must be negligible compared to $\\eta$ : $\\left\|h_{\\mu \\nu }\\right\|\\ll 1$ (and similarly for all derivatives of $h$ ). Then one ignores all products of $h$ (or its derivatives) with $h$ or its derivatives (equivalent to ignoring all terms of higher order than 1 in $\\epsilon$ ). It is further assumed in this approximation scheme that all indices of h and its derivatives are raised and lowered with $\\eta$ .\n\nThe metric h is clearly symmetric, since g and η are. The consistency condition $g_{ab}g^{bc}=\\delta _{a}{}^{c}$ shows that\n\n$g^{ab}\\,=\\eta ^{ab}-h^{ab}$\n\nThe Christoffel symbols can be calculated as\n\n$2\\Gamma _{bc}^{a}=(h^{a}{}_{b,c}+h^{a}{}_{c,b}-h_{bc,}{}^{a})$\n\nwhere $h_{bc,}{}^{a}\\ {\\stackrel {\\mathrm {def} }{=}}\\ \\eta ^{ar}h_{bc,r}$ , and this is used to calculate the Riemann tensor:\n\n$2R^{a}{}_{bcd}=2(\\Gamma _{bd,c}^{a}-\\Gamma _{bc,d}^{a})=\\eta ^{ae}(h_{eb,dc}+h_{ed,bc}-h_{bd,ec}-h_{eb,cd}-h_{ec,bd}+h_{bc,ed})=$\n$=\\eta ^{ae}(h_{ed,bc}-h_{bd,ec}-h_{ec,bd}+h_{bc,ed})=h_{d,bc}^{a}-h_{bd,}{}^{a}{}_{c}+h_{bc,}{}^{a}{}_{d}-h^{a}{}_{c,bd}$\n\nUsing $R_{bd}=\\delta ^{c}{}_{a}R^{a}{}_{bcd}$ gives\n\n$2R_{bd}=h_{d,br}^{r}+h_{b,dr}^{r}-h_{,bd}-h_{bd,rs}\\eta ^{rs}$\n\nFor the Ricci scalar we have:\n\n$R=R_{bd}\\eta ^{bd}=h_{,ab}^{ab}-\\square h$ .\n\nThen the linearized Einstein equations are\n\n$8\\pi T_{bd}\\,=R_{bd}-R_{ac}\\eta ^{ac}\\eta _{bd}/2$\n\nor\n\n$8\\pi T_{bd}=(h_{d,br}^{r}+h_{b,dr}^{r}-h_{,bd}-h_{bd,r}{}^{r}-h_{s,r}^{r}{}^{s}\\eta _{bd})/2+(h_{,a}{}^{a}\\eta _{bd}+h_{ac,r}{}^{r}\\eta ^{ac}\\eta _{bd})/4$\n\nOr, equivalently:\n\n$8\\pi (T_{bd}-T_{ac}\\eta ^{ac}\\eta _{bd}/2)\\,=R_{bd}$\n$16\\pi (T_{bd}-T_{ac}\\eta ^{ac}\\eta _{bd}/2)\\,=h_{d,br}^{r}+h_{b,dr}^{r}-h_{,bd}-h_{bd,rs}\\eta ^{rs}$\n\n## With a coordinate condition\n\nIf one uses the Lorentz invariant harmonic coordinate condition\n\n$h_{\\alpha \\beta ,\\gamma }\\eta ^{\\beta \\gamma }={\\frac {1}{2}}h_{\\beta \\gamma ,\\alpha }\\eta ^{\\beta \\gamma }\\,,$\n\nthen the last form above of the linearized Einstein equation simplifies to\n\n$16\\pi (T_{bd}-T_{ac}\\eta ^{ac}\\eta _{bd}/2)\\,=\\,-h_{bd,rs}\\eta ^{rs}\\,.$\n\nTo solve it, this can be rewritten as\n\n$\\Delta h_{bd}={-16\\pi G \\over c^{4}}(T_{bd}-T_{ac}\\eta ^{ac}\\eta _{bd}/2)+{\\frac {\\partial ^{2}h_{bd}}{c^{2}{\\partial t}^{2}}}\\,$\n\nwhere ∆ is the Laplacian on a spatial slice. If the stress-energy changes slowly (velocities are low compared to c), then this gives\n\n$h_{bd}(r)={\\frac {-1}{4\\pi }}\\int \\left({-16\\pi G \\over c^{4}}(T_{bd}(s)-T_{ac}(s)\\eta ^{ac}\\eta _{bd}/2)+{\\frac {\\partial ^{2}h_{bd}(s)}{c^{2}{\\partial t}^{2}}}\\right){\\frac {1}{\\vert r-s\\vert }}d^{3}s\\,$\n\nas a generalization of the Newtonian formula for gravitational potential. This is solved iteratively by first replacing the second time derivative by zero and then inserting the h so obtained repeatedly until convergence.\n\n## Applications\n\nThe linearized EFE are used primarily in the theory of gravitational radiation, where the gravitational field far from the source is approximated by these equations." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8953738,"math_prob":0.99959105,"size":5509,"snap":"2019-13-2019-22","text_gpt3_token_len":1144,"char_repetition_ratio":0.15894641,"word_repetition_ratio":0.0,"special_character_ratio":0.19077873,"punctuation_ratio":0.09334764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000006,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-18T16:10:43Z\",\"WARC-Record-ID\":\"<urn:uuid:5ad11695-94a7-4367-b188-4a9524fab64c>\",\"Content-Length\":\"124477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f72ce757-d0fe-470e-890e-15e26f074493>\",\"WARC-Concurrent-To\":\"<urn:uuid:7fc28fd4-2133-4ada-8138-71d8eaaee97f>\",\"WARC-IP-Address\":\"52.207.110.63\",\"WARC-Target-URI\":\"http://conceptmap.cfapps.io/wikipage?lang=en&name=Linearized_gravity\",\"WARC-Payload-Digest\":\"sha1:BJRHTF2TUWBQOWEMYIOKXBQWIA2QTWX4\",\"WARC-Block-Digest\":\"sha1:7MZHTDLU5TLUDEZ2DPEPNR3NE27DQKNE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912201455.20_warc_CC-MAIN-20190318152343-20190318174343-00385.warc.gz\"}"}
https://gamedev.stackexchange.com/questions/151056/how-can-i-move-a-3d-object-to-a-known-cartesian-point-on-the-surface-of-a-sphere	[ "How can I move a 3d object to a known Cartesian point on the surface of a sphere?\n\nI'm trying to figure out how to move an object from one Cartesian point to another located on the surface of a 3D sphere. So that the object will follow the spherical coordinate system (Theta and Phi)\n\nI've a function that converts a Cartesian to Spherical but how can I make it work on the surface of a sphere just like we are using the equation of a line y=mx+b when moving from one point to another on a flat surface?\n\nI'm going to use it to move an object from its point to a selected point somewhere else on the sphere so that the object will move in a straight line but following the curvature.\n\n• Can you give some example start/stopping points and what exactly you need as the output? It sounds like you currently can get both the cartesian coordinates and the spherical coordinates but you need 3D vectors describing the position? Nov 17 '17 at 21:04\n• Yes that's right I can but I don't know how to make the movement in between the starting and stopping point in a straight line over the curved surface. I know how to do it in a completely straight line through the surface but not over on the curved surface. I don't even know where to begin my thinking on this one. Nov 17 '17 at 21:09\n\nThe general idea between interpolation between two points on a sphere is to use a \"Spherical Lerp\" function or \"slerp\" for short.\n\nSome engines have it as a built in function, for others, either it's only available for Quaternions or missing entirely and you'll have to write it yourself. The page on wikipedia provides a good explanation of both the Vector and Quaternion versions of slerp, but to summarize the Vector3 part here:\n\nThe basic formula involves finding the spherical combination of the start and end points with a ratio of the sin of the subtended angle.\n\nslerp(a , b, t) =\ntheta = acos(dot(a, b))\n(sin((1-t)theta) / sin(theta)) * a+ (sin(t * theta)/sin(theta)) * b\n\nif you are animating in a loop, you can use a more efficient algorithm by iteratively reflecting each point after you've found the first intermediate point.\n\nc = dot(p0, p1) * 2\npk+1 = c pk − pk−1\n• That was all I needed here's the final function in C# public Vector3 slerpIt(Vector3 a, Vector3 b, float t) { float theta = (float)Math.Acos(Vector3.Dot(Vector3.Normalize(a), Vector3.Normalize(b))); return (float)(Math.Sin((1 - t) * theta) / Math.Sin(theta)) * a + (float)(Math.Sin(t * theta) / Math.Sin(theta)) * b; } Nov 18 '17 at 12:36\n\nIt's very simple. First of all, you need to get the line between the starting and ending point (marked with purple on the following image):", null, "(The green point is the center of the sphere, a is the vector from the center to the start point and b is the vector from the center to the end point, the red line is the path on the surface of the sphere)\n\nNow to make the object follow the surface of the sphere you simply need to interpolate between the start and endpoint, get the vector between that point and the center (if c is the center and p is the interpolated position, then the vector is simply p - c), normalize it, multiply it by the radius of the sphere, then add the result to the center of the sphere to get the object's position.\n\nPseudo code:\n\nvec3 getPositionOnSphere(vec3 startPos, vec3 endPos, vec3 center, float radius, float time) {\nvec3 pos = (endPos - startPos) * time + startPos\nreturn normalize(pos - center) * radius + center\n}\n\nThe problem with this approach is that the end result won't be uniform. You should generate the line first, then move the object on that.\n\n• I've up voted your answer because I hade use of your explanation, it's very informative. My reputation is too low to make it seen here, so I wanted to tell you. Nov 18 '17 at 12:47" ]	[ null, "https://i.stack.imgur.com/PIFGt.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9363036,"math_prob":0.97477645,"size":598,"snap":"2022-05-2022-21","text_gpt3_token_len":130,"char_repetition_ratio":0.13131313,"word_repetition_ratio":0.035714287,"special_character_ratio":0.21070234,"punctuation_ratio":0.024793388,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99847883,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T11:51:02Z\",\"WARC-Record-ID\":\"<urn:uuid:24091221-5577-4860-bbbd-b3aa3932bc2b>\",\"Content-Length\":\"136295\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:084bd98a-2b3d-450f-b1c4-001fd20b9507>\",\"WARC-Concurrent-To\":\"<urn:uuid:11ef64e7-c543-4a78-aeac-fbc9a527156a>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://gamedev.stackexchange.com/questions/151056/how-can-i-move-a-3d-object-to-a-known-cartesian-point-on-the-surface-of-a-sphere\",\"WARC-Payload-Digest\":\"sha1:V4F3NSTQSO26O2SVZWPHW3NCB45DMZFU\",\"WARC-Block-Digest\":\"sha1:TZUQT6JX4NYPWKMKKBTJXHZ5TILJJS5W\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304528.78_warc_CC-MAIN-20220124094120-20220124124120-00384.warc.gz\"}"}
https://data-mining-tutorials.blogspot.com/2012/11/	[ "Monday, November 5, 2012\n\nLinear Discriminant Analysis - Tools comparison\n\nLinear discriminant analysis is a popular method in domains of statistics, machine learning and pattern recognition. Indeed, it has interesting properties: it is rather fast on large bases; it can handle naturally multi-class problems (target attribute with more than 2 values); it generates a linear classifier linear, easy to interpret; it is robust and fairly stable, even applied on small databases; it has an embedded variable selection mechanism. Personally, I appreciate linear discriminant analysis because we can have multiple interpretations (probabilistic, geometric), and thus highlights various aspects of supervised learning.\n\nIn this tutorial, we highlight the similarities and the differences between the outputs of Tanagra, R (MASS and klaR packages), SAS, and SPSS software. The main conclusion is that, if the presentation is not always the same, ultimately we have exactly the same results. This is the most important.\n\nKeywords: linear discriminant analysis, predictive discriminant analysis, canonical discriminant analysis, variable selection, feature selection, sas, stepdisc, candisc, R software, xlsx package, MASS package, lda, klaR package, greedy.wilks, confusion matrix, resubstitution error rate\nComponents: LINEAR DISCRIMINANT ANALYSIS, CANONICAL DISCRIMINANT ANALYSIS, STEPDISC\nTutorial: en_Tanagra_LDA_Comparisons.pdf\nDataset: alcohol\nReferences :\nWikipedia - \"Linear Discriminant Analysis\"" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81935644,"math_prob":0.78513086,"size":1422,"snap":"2019-26-2019-30","text_gpt3_token_len":293,"char_repetition_ratio":0.116361074,"word_repetition_ratio":0.0,"special_character_ratio":0.17651196,"punctuation_ratio":0.20083682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9508003,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-24T17:24:56Z\",\"WARC-Record-ID\":\"<urn:uuid:80ebf1a7-9c0a-4579-9f8e-9d0130fec3a7>\",\"Content-Length\":\"85191\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e92812ac-17a2-4c16-8ee8-8cdb1b82f433>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcdbb4b9-3748-4f4f-8843-beea64242ab3>\",\"WARC-IP-Address\":\"172.217.8.1\",\"WARC-Target-URI\":\"https://data-mining-tutorials.blogspot.com/2012/11/\",\"WARC-Payload-Digest\":\"sha1:GJM3S3K624NJHAOLADDJ3IVX75N7NC4L\",\"WARC-Block-Digest\":\"sha1:JQ5QSM3VRVJPTS6FMQLSCYHA6A7XDPE6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999620.99_warc_CC-MAIN-20190624171058-20190624193058-00500.warc.gz\"}"}
https://mathemerize.com/tag/types-of-functions-in-maths/	[ "## Types of Functions in Maths – Domain and Range\n\nHere you will learn types of functions in maths i.e polynomial function, logarithmic function etc and their domain and range. Let’s begin – Types of Functions in Maths (a) Polynomial function If a function is defined by f(x) = $$a_0x^n$$ + $$a_1x^{n-1}$$ + $$a_2x^{n-2}$$ + ….. + $$a_{n-1}x$$ + $$a_n$$ where n is a non …" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7629729,"math_prob":1.000001,"size":368,"snap":"2022-40-2023-06","text_gpt3_token_len":113,"char_repetition_ratio":0.18956044,"word_repetition_ratio":0.032786883,"special_character_ratio":0.3451087,"punctuation_ratio":0.07042254,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999976,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T19:30:38Z\",\"WARC-Record-ID\":\"<urn:uuid:65628842-5bef-414c-bcce-01ea4fc4ef36>\",\"Content-Length\":\"199669\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66b812ef-6c8b-4a17-9fd6-f689fd944c92>\",\"WARC-Concurrent-To\":\"<urn:uuid:eeafc03c-2de8-4f51-a273-a6aeb177cee2>\",\"WARC-IP-Address\":\"3.234.104.255\",\"WARC-Target-URI\":\"https://mathemerize.com/tag/types-of-functions-in-maths/\",\"WARC-Payload-Digest\":\"sha1:JDPWRFQMUNC6G54RY3PZDAUDFVMJYBYF\",\"WARC-Block-Digest\":\"sha1:553ZP5SCN4TFD74NUXTT2SM5B3MGH5D7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335276.85_warc_CC-MAIN-20220928180732-20220928210732-00162.warc.gz\"}"}
https://testbook.com/question-answer/the-characteristic-equation-of-a-control-system-is--5fd34ee1c59dd0a44872d503	[ "# The characteristic equation of a control system is given as:$$1+\\frac{K(s + 1)}{s(s+4)(s^2+2s+2)} =0$$For a large value of s, the root loci for K ≥ 0 are asymptotic to asymptotes, where do the asymptotes intersect on the real axis?\n\nThis question was previously asked in\nESE Electronics 2011 Paper 2: Official Paper\nView all UPSC IES Papers >\n1. 5/3\n2. 2/3\n3. 6/3\n4. 4/3\n\nOption 1 : 5/3\nFree\nCT 3: Building Materials\n2668\n10 Questions 20 Marks 12 Mins\n\n## Detailed Solution\n\nConcept:\n\nAll the asymptotes meet at a common point on the real axis known as centroid, it is given by,\n\nCentroid:\n\n$$\\frac{{\\sum real\\;part\\;of\\;pole - \\sum real\\;part\\;of\\;zero}}{{No.\\;\\;of\\;poles - No.\\;\\;of\\;zeros}}$$\n\nAnalysis:\n\nfrom characteristic equation open-loop transfer function have,\n\nZeros = -1\n\nPoles = 0, -4, -1 + i, -1 - i.\n\nTherefore centroid = -5 / 3.", null, "Important Points\n\nIn root locus when P ≠ Z then some of the root locus branches will tend to infinity along the direction of asymptotes.\n\nThe angle of Asymptotes = $$\\frac{{\\left( {2k \\pm 1} \\right)180}}{{P - Z}}$$ K = 0, 1, 2, 3, …\n\nRemember: Centroid always lies on a real axis not compulsory to be on root locus whereas saddle point can lie anywhere but it should present on the root locus." ]	[ null, "https://cdn.testbook.com/resources/lms_creative_elements/important-point-image.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72738487,"math_prob":0.9918089,"size":987,"snap":"2021-31-2021-39","text_gpt3_token_len":307,"char_repetition_ratio":0.12716176,"word_repetition_ratio":0.0,"special_character_ratio":0.31914893,"punctuation_ratio":0.16744186,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99094343,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T16:39:47Z\",\"WARC-Record-ID\":\"<urn:uuid:8e6e48c0-fda5-4112-a4fb-29c7a67c5017>\",\"Content-Length\":\"114938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ad6365a-a41d-4fbe-9117-6145e27854c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd5ed56b-f433-4a4e-b315-4e83dfaf329a>\",\"WARC-IP-Address\":\"104.22.44.238\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/the-characteristic-equation-of-a-control-system-is--5fd34ee1c59dd0a44872d503\",\"WARC-Payload-Digest\":\"sha1:5IU3JPDVNEEYQIX4QLROOLQB3YH25Q7E\",\"WARC-Block-Digest\":\"sha1:2ISP6KHSS2VJNRQ74BVE7O2UT4UPZ5DC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060877.21_warc_CC-MAIN-20210928153533-20210928183533-00184.warc.gz\"}"}
https://thebestuknow.com/maths/multiplication/maths-table-30/	[ "# Table of 30\n\nMultiplication table of 30 is in the tabular form. So, you can get it easily.\n\nIn the table of 30 given below, you can see numerals on the left side and how to learn or read is on the right side. So, this is the right way to understand and learn the table.\n\nAs this table is of 30: in the part of numerals, we have the number 30 and the results by multiplying factors from 1 to 20. So, we get the results from 1 to 20.\n\nNow let’s see how to learn-\n\n## Table of 30 in Tabular Form\n\n Numerals Read as 1. 30 x 1 = 30 Thirty ones are thirty 2. 30 x 2 = 60 Thirty twos are sixty 3. 30 x 3 = 90 Thirty threes are ninety 4. 30 x 4 = 120 Thirty fours are hundred and twenty 5. 30 x 5 = 150 Thirty fives are hundred and fifty 6. 30 x 6 = 180 Thirty sixes are hundred and eighty 7. 30 x 7 = 210 Thirty sevens are two hundred and ten 8. 30 x 8 = 240 Thirty eights are two hundred and forty 9. 30 x 9 = 270 Thirty nines are two hundred and seventy 10. 30 x 10 = 300 Thirty tens are three hundred 11. 30 x 11 = 330 Thirty elevens are three hundred and thirty 12. 30 x 12 = 360 Thirty twelves are three hundred and sixty 13. 30 x 13 = 390 Thirty thirteens are three hundred and ninety 14. 30 x 14 = 420 Thirty fourteens are four hundred and twenty 15. 30 x 15 = 450 Thirty fifteens are four hundred and fifty 16. 30 x 16 = 480 Thirty sixteens are four hundred and eighty 17. 30 x 17 = 510 Thirty seventeens are five hundred and ten 18. 30 x 18 = 540 Thirty eighteens are five hundred and forty 19. 30 x 19 = 570 Thirty nineteens are five hundred and seventy 20. 30 x 20 = 600 Thirty twenties are six hundred\n\n### Get more tables other than a table of 30-\n\n Table of 1 Table of 2 Table of 3 Table of 4 Table of 5 Table of 6 Table of 7 Table of 8 Table of 9 Table of 10 Table of 11 Table of 12 Table of 13 Table of 14 Table of 15 Table of 16 Table of 17 Table of 18 Table of 19 Table of 20 Table of 21 Table of 22 Table of 23 Table of 24 Table of 25 Table of 26 Table of 27 Table of 28 Table of 29 Table of 30\n\nYou may also like to learn the terms given below-\n\nMultiplication Table of 30 to List of Chapters in Mathematics" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8808053,"math_prob":0.9889413,"size":1800,"snap":"2021-43-2021-49","text_gpt3_token_len":556,"char_repetition_ratio":0.266147,"word_repetition_ratio":0.0,"special_character_ratio":0.37333333,"punctuation_ratio":0.08226221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99685997,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T09:02:33Z\",\"WARC-Record-ID\":\"<urn:uuid:92dd2bb3-d3ce-4131-8414-9d01a138512f>\",\"Content-Length\":\"36498\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e3482d8-936c-49a1-a7e8-a57c895c99af>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c965775-dafc-49e7-9443-33f9ea362a29>\",\"WARC-IP-Address\":\"204.93.216.87\",\"WARC-Target-URI\":\"https://thebestuknow.com/maths/multiplication/maths-table-30/\",\"WARC-Payload-Digest\":\"sha1:IL4OS6TEKIBJTQZOXMB2CYDJKD6M5OSW\",\"WARC-Block-Digest\":\"sha1:R6ZM7PGZQDXMF7HTDSXVO5ZNJDBM3E2R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359976.94_warc_CC-MAIN-20211201083001-20211201113001-00365.warc.gz\"}"}
https://uk.mathworks.com/matlabcentral/answers/489726-parfeval-doesn-t-call-function?s_tid=prof_contriblnk	[ "# parfeval() doesn't call function\n\n10 views (last 30 days)\nTimo Schmid on 7 Nov 2019\nEdited: Edric Ellis on 8 Nov 2019\nI am trying to process Data from an UDP Server close to realtime.\nFor that I wrote this MatLab Code to fill a Buffer with the UDP Datagrams and Process the Data (split the strings, etc.) once the Buffer from my MatLab function is full (myBuffer).\nDuring Processing the data (which takes about 0.9s) I need to go on receiving Data and store them in the (now) emptied Buffer.\nI found the parfeval-Function of the \"Parallel Computing Toolbox\" which might suit my needs as I need my function \"ProcessData\" to run in the background.\nThe Problem I encountered is that I can't make it run as the parfeval function doesn't enter my function ProcessData. I tested it with setting a breakpoint in ProcessData() but the program never stops. Did I do anything wrong with the function parameters?\nThat's what MatLab help says: F = parfeval(p,fcn,numout,in1,in2,...) requests asynchronous execution of the function fcn on a worker contained in the parallel pool p, expecting numout output arguments and supplying as input arguments in1,in2,....\nHope you guys can help me with this problem! Thanks in advance.\n%% Specify a Server (host name or IP address) with Port 8080\nu = udp('192.168.0.164', 8080); %UDP Object Zuhause\n%u = udp('169.254.38.221', 8080); %UDP Object Pilotfabrik\n% Buffer in the enclosing function\nmyBuffer = {}; %Initialisierung\nMAXBUFFLEN = 100; %Maximale Anzahl an Eintraegen in Buffer (1 Eintrage = 1 Datagram)\nu.InputBufferSize = 4060;\nu.ErrorFcn = @ErrorFcn;\nu.DatagramTerminateMode =\nu.Terminator = '!';\n%% Initialize Parallel pool\npool = gcp();\n%% Oeffnen der Verbindung\nfopen(u);\nif (~strcmp(u.Status,'open'))\nNetworkError(u,'Connection failed!');\nend\n%% Start Data transmission by trigger\nfprintf(u, 'Requesting Data')\n%% Callback Funktion\ndatagram = fscanf(u);\nmyBuffer{end+1} = datagram; %Appends datagram to buffer\n[~, bufflen] = size(myBuffer);\nif bufflen < MAXBUFFLEN\nreturn;\nelse\nf = parfeval(pool, @ProcessData, 1, myBuffer);\nmyBuffer = {}; %empty Buffer\nend\nend\nfunction ErrorFcn(u,~)\ndisp(\"An Error occured\");\nend\nend\nfunction datagram_values = ProcessData(myBuffer)\nstringvalues = split(myBuffer, \";\"); %Split Strings\ndoublevalues = str2double(stringvalues) %Convert Strings do Doubles\ndim_doublevalues = size(doublevalues); %Dimension of Double Output Array\ni_max = dim_doublevalues(2) %Anzahl der Datenpakete\nj_max = (dim_doublevalues(3))-1 %Anzahl der Werte pro Datenpaket; -1 wegen leerem Wert nach \";\" am Ende\nk_max = i_max*j_max %Gesamtanzahl der Werte in Buffer\nk=1;\nwhile k<=k_max\nfor i = 1:i_max\nfor j = 1:j_max\ndatagram_values(k,1)=doublevalues(1,i,j);\nk=k+1;\nend\nend\nend\ndisp(datagram_values);\nend\n\nEdric Ellis on 8 Nov 2019\nEdited: Edric Ellis on 8 Nov 2019\nUnfortunately, the MATLAB debugger can't stop in code running on the workers - only code running at the client.\nIn this case, you should try looking at the diary output of the future f, like this:\nf = parfeval(..);\nwait(f); % wait for the worker to complete\ndisp(f.Diary); % display the output\nIf you don't wish to block the client, you could use afterEach to invoke the call to disp, like this:\nf = parfeval(..);\nafterEach(f, @(f) disp(f.Diary), 0, 'PassFuture', true);" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6587103,"math_prob":0.7896169,"size":2199,"snap":"2020-24-2020-29","text_gpt3_token_len":590,"char_repetition_ratio":0.13986333,"word_repetition_ratio":0.0,"special_character_ratio":0.26466575,"punctuation_ratio":0.19952494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95653754,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-06T00:34:20Z\",\"WARC-Record-ID\":\"<urn:uuid:c84c4c43-deb9-4b4a-acce-22149670d49d>\",\"Content-Length\":\"117959\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cb3549e-3d21-4ed8-ad20-ba70199ca1cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:97a318af-2e1d-4421-8545-4cf8674bab74>\",\"WARC-IP-Address\":\"23.223.252.57\",\"WARC-Target-URI\":\"https://uk.mathworks.com/matlabcentral/answers/489726-parfeval-doesn-t-call-function?s_tid=prof_contriblnk\",\"WARC-Payload-Digest\":\"sha1:3KG46A4KYSIGVJ2WVQ3D76OVSW6ZQTAO\",\"WARC-Block-Digest\":\"sha1:3I6UFAGHLA2BLTUXZQAQSB2IBW3XRCAP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655889877.72_warc_CC-MAIN-20200705215728-20200706005728-00171.warc.gz\"}"}
https://www.folkstalk.com/2022/10/python-how-to-copy-a-2d-array-leaving-out-last-column-with-code-examples.html	[ "# Python How To Copy A 2D Array Leaving Out Last Column With Code Examples\n\nPython How To Copy A 2D Array Leaving Out Last Column With Code Examples\n\nIn this session, we will try our hand at solving the Python How To Copy A 2D Array Leaving Out Last Column puzzle by using the computer language. The following piece of code will demonstrate this point.\n\n```In : import numpy as np\n\nIn : H = np.meshgrid(np.arange(5), np.arange(5))\n\nIn : H\nOut:\narray([[0, 1, 2, 3, 4],\n[0, 1, 2, 3, 4],\n[0, 1, 2, 3, 4],\n[0, 1, 2, 3, 4],\n[0, 1, 2, 3, 4]])\n\nIn : Hsub = H[1:-1,1:-1]\n\nIn : Hsub\nOut:\narray([[1, 2, 3],\n[1, 2, 3],\n[1, 2, 3]])\n```\n\nThere are a lot of real-world examples that show how to fix the Python How To Copy A 2D Array Leaving Out Last Column issue.\n\n## How do you pull a column out of an array in Python?\n\nUse the syntax array[:, [i, j]] to extract the i and j indexed columns from array . Like lists, NumPy arrays use zero-based indexes. Use array[:, i:j+1] to extract the i through j indexed columns from array .\n\n## How do you remove the last element of a NumPy array in Python?\n\nApproach\n\n• Import numpy library and create numpy array.\n• Using the len() method to get the length of the given array.\n• Now use slicing to remove the last element by setting the start of slicing=0 and end = lastIndex.\n• lastIndex is calculated by decrementing the length of array by one.\n\n## How do I find the last column of an array?\n\nIf you want the values in the last column as a simple one-dimensional array, use the syntax ar[:, -1] . If you want the resulting values in a column vector (for example, with shape (n, 1) where n is the total number of rows), use the syntax ar[: [-1]] .\n\n## How do you extract a matrix element in Python?\n\nGiven a Matrix, Extract all the elements that are of string data type. Input : test_list = [[5, 6, 3], [\"Gfg\", 3], [9, \"best\", 4]] Output : ['Gfg', 'best'] Explanation : All strings are extracted.04-Aug-2022\n\n## How do you extract a column from a multidimensional array in Python?\n\nUse a list comprehension to extract a column from an array. Use the syntax [row[i] for row in array] to extract the i – indexed column from array . Further reading: A list comprehension is often useful for extracting elements from a list. You can read more about list comprehensions here.\n\n## How do I extract a column from a data frame?\n\nExtracting Multiple columns from dataframe\n\n• Syntax : variable_name = dataframe_name [ row(s) , column(s) ]\n• Example 1: a=df[ c(1,2) , c(1,2) ]\n• Explanation : if we want to extract multiple rows and columns we can use c() with row names and column names as parameters.\n• Example 2 : b=df [ c(1,2) , c(“id”,”name”) ]\n\n## How do I remove the last element from a list?\n\npop() function. The simplest approach is to use the list's pop([i]) function, which removes an element present at the specified position in the list. If we don't specify any index, pop() removes and returns the last element in the list.\n\n## How do I remove the last two elements from a list in Python?\n\nUsing len() + list slicing to remove last K elements of list. List slicing can perform this particular task in which we just slice the first len(list) – K elements to be in the list and hence remove the last K elements.14-Sept-2022\n\n## How do you pop the last element of a list in Python?\n\nPython list pop() is an inbuilt function in Python that removes and returns the last value from the List or the given index value.25-Aug-2022\n\n## How do you pick up the last value in a column?\n\nCOLUMNS Function – Example 3 If we wish to get only the first column number, we can use the MIN function to extract just the first column number, which will be the lowest number in the array. Once we get the first column, we can just add the total columns in the range and subtract 1, to get the last column number.14-Jun-2022" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.68599755,"math_prob":0.909925,"size":3714,"snap":"2023-40-2023-50","text_gpt3_token_len":998,"char_repetition_ratio":0.15956873,"word_repetition_ratio":0.083333336,"special_character_ratio":0.29348412,"punctuation_ratio":0.15238096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9877586,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T13:29:07Z\",\"WARC-Record-ID\":\"<urn:uuid:a6362513-47fc-40fc-b2e4-19be02a8d04a>\",\"Content-Length\":\"72206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d5927d9-40f7-4e34-ad6a-3f6643de98b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0991d43-b37a-4c5f-bc11-c101ba40480f>\",\"WARC-IP-Address\":\"104.21.49.226\",\"WARC-Target-URI\":\"https://www.folkstalk.com/2022/10/python-how-to-copy-a-2d-array-leaving-out-last-column-with-code-examples.html\",\"WARC-Payload-Digest\":\"sha1:BG44NFAGEISSPSDOTQ2QR4LHTXJQXCHK\",\"WARC-Block-Digest\":\"sha1:3NM4YZYSEKUFKZ5GTQZY2N5SWVMD2SLQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506481.17_warc_CC-MAIN-20230923130827-20230923160827-00361.warc.gz\"}"}
http://mathcentral.uregina.ca/QQ/database/QQ.09.14/h/victoria1.html	[ "", null, "", null, "", null, "SEARCH HOME", null, "Math Central Quandaries & Queries", null, "", null, "Question from Victoria, a student: find the area of a regular pentagon inscribed in a circle with radius 3 units", null, "Hi Victoria,\n\nIf you join each of the vertices of the pentagon to the center $C$ of the circle you will see that the pentagon is partitioned into five congruent isosceles triangles. I have labeled one of these triangles $ABC.$ Thus the area of the pentagon is 5 times the area of the triangle $ABC.$", null, "S\\Also you can see that the measure of the angle $BCA$ is $\\frac{360^o}{5} = 72^{o}.$ You can use the technique in my response to a question by Sela to find the area of triangle $ABC.$\n\nPenny", null, "", null, "", null, "", null, "", null, "", null, "Math Central is supported by the University of Regina and the Imperial Oil Foundation." ]	[ null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/search.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/QQ/database/QQ.09.14/h/victoria1.1.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null, "http://mathcentral.uregina.ca/lid/images/qqsponsors.gif", null, "http://mathcentral.uregina.ca/lid/images/mciconnotext.gif", null, "http://mathcentral.uregina.ca/lid/images/pixels/transparent.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91102237,"math_prob":0.9740514,"size":588,"snap":"2023-14-2023-23","text_gpt3_token_len":154,"char_repetition_ratio":0.16267124,"word_repetition_ratio":0.0,"special_character_ratio":0.26020408,"punctuation_ratio":0.060344826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99802303,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,2,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T21:13:42Z\",\"WARC-Record-ID\":\"<urn:uuid:fb7b9280-9b87-43a1-ae36-253f9452d7cf>\",\"Content-Length\":\"7147\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a186b9a-6c22-4ff5-a7d3-c73d02e69804>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d7b3f89-2949-4c6f-9563-98f214130b3b>\",\"WARC-IP-Address\":\"142.3.156.40\",\"WARC-Target-URI\":\"http://mathcentral.uregina.ca/QQ/database/QQ.09.14/h/victoria1.html\",\"WARC-Payload-Digest\":\"sha1:BMFX3X3XMR6NI3URI6CYE4TWN4GJIAY6\",\"WARC-Block-Digest\":\"sha1:POPKNQ6OFE5DO43RQ7RSHAGYBC55PHFC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655143.72_warc_CC-MAIN-20230608204017-20230608234017-00368.warc.gz\"}"}
https://answers.everydaycalculation.com/lcm/16-27	[ "Solutions by everydaycalculation.com\n\n## What is the LCM of 16 and 27?\n\nThe lcm of 16 and 27 is 432.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 16\n16 = 2 × 2 × 2 × 2\n2. Find the prime factorization of 27\n27 = 3 × 3 × 3\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 2 × 2 × 2 × 2 × 3 × 3 × 3\n4. LCM = 432\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:" ]	[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7436596,"math_prob":0.9993404,"size":456,"snap":"2020-10-2020-16","text_gpt3_token_len":147,"char_repetition_ratio":0.121681415,"word_repetition_ratio":0.0,"special_character_ratio":0.3991228,"punctuation_ratio":0.086021505,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99651504,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-17T07:39:19Z\",\"WARC-Record-ID\":\"<urn:uuid:c4baaaff-e5b3-415e-be13-12f3cbd41c5f>\",\"Content-Length\":\"5756\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c6e016b-e585-4722-a75c-ae06980381b0>\",\"WARC-Concurrent-To\":\"<urn:uuid:10e7e951-2695-424c-9b32-fa46e0cbe6a2>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/lcm/16-27\",\"WARC-Payload-Digest\":\"sha1:GRUTVFN2RIN5DQ3PF3M3NJMLN5YHTCFB\",\"WARC-Block-Digest\":\"sha1:MKHFTR52AIJPDHT37ZX3KVVPQCUCIBIU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875141749.3_warc_CC-MAIN-20200217055517-20200217085517-00039.warc.gz\"}"}
http://nonlinearsolve.sciml.ai/dev/basics/NonlinearProblem/	[ "# Nonlinear Problems\n\nSciMLBase.NonlinearProblemType\n\nDefines a nonlinear system problem. Documentation Page: https://nonlinearsolve.sciml.ai/dev/basics/NonlinearProblem/\n\nMathematical Specification of a Nonlinear Problem\n\nTo define a Nonlinear Problem, you simply need to give the function $f$ which defines the nonlinear system:\n\n$$$f(u,p) = 0$$$\n\nand an initial guess $u₀$ of where f(u,p)=0. f should be specified as f(u,p) (or in-place as f(du,u,p)), and u₀ should be an AbstractArray (or number) whose geometry matches the desired geometry of u. Note that we are not limited to numbers or vectors for u₀; one is allowed to provide u₀ as arbitrary matrices / higher-dimension tensors as well.\n\nProblem Type\n\nConstructors\n\nNonlinearProblem(f::NonlinearFunction,u0,p=NullParameters();kwargs...)\nNonlinearProblem{isinplace}(f,u0,p=NullParameters();kwargs...)\n\nisinplace optionally sets whether the function is in-place or not. This is determined automatically, but not inferred.\n\nParameters are optional, and if not given, then a NullParameters() singleton will be used, which will throw nice errors if you try to index non-existent parameters. Any extra keyword arguments are passed on to the solvers. For example, if you set a callback in the problem, then that callback will be added in every solve call.\n\nFor specifying Jacobians and mass matrices, see the NonlinearFunctions page.\n\nFields\n\n• f: The function in the problem.\n• u0: The initial guess for the steady state.\n• p: The parameters for the problem. Defaults to NullParameters.\n• kwargs: The keyword arguments passed on to the solvers." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6268191,"math_prob":0.89835334,"size":1575,"snap":"2022-40-2023-06","text_gpt3_token_len":374,"char_repetition_ratio":0.13812858,"word_repetition_ratio":0.00921659,"special_character_ratio":0.20761904,"punctuation_ratio":0.17704917,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9729779,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T07:16:04Z\",\"WARC-Record-ID\":\"<urn:uuid:753a6d93-609b-403a-8fa7-aa9a508f42ad>\",\"Content-Length\":\"9274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:454dd2c6-8d28-4b65-9cd3-53d22425c20c>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cb3161c-c8e3-4c68-8b17-637721ac9642>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"http://nonlinearsolve.sciml.ai/dev/basics/NonlinearProblem/\",\"WARC-Payload-Digest\":\"sha1:QYOOW776KHWWBJR7OULG6LRH3LZ5C7S2\",\"WARC-Block-Digest\":\"sha1:EDT4344UUKZYC6OTV7OGPVTJNTOB3TE3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334992.20_warc_CC-MAIN-20220927064738-20220927094738-00307.warc.gz\"}"}
https://zbmath.org/?q=an%3A1269.65063	[ "# zbMATH — the first resource for mathematics\n\nW-methods in optimal control. (English) Zbl 1269.65063\nSummary: This paper addresses the consistency and stability of W-methods up to order three for nonlinear ordinary differential equation-constrained control problems with possible restrictions on the control. The analysis is based on the transformed adjoint system and the control uniqueness property. These methods can also be applied to large-scale partial differential equation-constrained optimization, since they offer an efficient way to compute gradients of the discrete objective function.\n\n##### MSC:\n 65K10 Numerical optimization and variational techniques 49J15 Existence theories for optimal control problems involving ordinary differential equations 49M25 Discrete approximations in optimal control\n##### Software:\nDONLP2; CG_DESCENT; RODAS\nFull Text:\n##### References:\n Bonnans, JF; Laurent-Varin, J, Computation of order conditions for symplectic partitioned Runge-Kutta schemes with application to optimal control, Numer. Math., 103, 1-10, (2006) · Zbl 1112.65063 Büttner, M; Schmitt, BA; Weiner, R, W-methods with automatic partitioning by Krylov techniques for large stiff systems, SIAM J. Numer. Anal., 32, 260-284, (1995) · Zbl 0820.65043 Dhamo, V; Tröltzsch, F, Some aspects of reachability for parabolic boundary control problems with control constraints, Comput. Optim. Appl., 50, 75-110, (2011) · Zbl 1245.93016 Hager, WW, Runge-Kutta methods in optimal control and the transformed adjoint system, Numer. Math., 87, 247-282, (2000) · Zbl 0991.49020 Hager, WW; Zhang, H, A new active set algorithm for box constrained optimization, SIAM J. Optim., 17, 526-557, (2006) · Zbl 1165.90570 Hager, W.W., Zhang, H.: Algorithm 851: CG_DESCENT, a conjugate gradient method with guaranteed descent. ACM Trans. Math. Softw. 32, 113-137 (2006) · Zbl 1346.90816 Hairer, E., Wanner, G.: Solving Ordinary Differential Equations II, Stiff and Differential-Algebraic Equations, 2nd revised edn. Springer, Berlin (1996) · Zbl 1192.65097 Jacobson, D.H., Mayne, D.Q.: Differential Dynamic Programming. American Elsevier Publishing, New York (1970) · Zbl 0223.49022 Kammann, E.: Modellreduktion und Fehlerabschätzung bei parabolischen Optimalsteuerungsproblemen. Diploma thesis, Department of Mathematics, Technische Universität Berlin (2010) · Zbl 0889.65083 Kaps, P; Rentrop, P, Generalized Runge-Kutta methods of order four with stepsize control for stiff ordinary differential equations, Numer. Math., 33, 55-68, (1979) · Zbl 0436.65047 Lang, J.: Adaptive Multilevel Solution of Nonlinear Parabolic PDE Systems. Theory, Algorithm and Applications. Lecture Notes in Computational Science and Engineering, vol. 16. Springer, Berlin (2000) · Zbl 0964.90062 Murua, A, On order conditions for partitioned symplectic methods, SIAM J. Numer. Anal., 34, 2204-2211, (1997) · Zbl 0889.65083 Pulova, N.V.: Runge-Kutta Schemes in Control Constrained Optimal Control. Lecture Notes in Computer Science, LNCS, vol. 4818, pp. 358-365 (2008) · Zbl 1229.65123 Rosenbrock, HH, Some general implicit processes for the numerical solution of differential equations, Comput. J., 5, 329-331, (1963) · Zbl 0112.07805 Sandu, A.: On the Properties of Runge-Kutta Discrete Adjoints. Lecture Notes in Computer Science, LNCS, vol. 3394, pp. 550-557 (2006) · Zbl 1157.65421 Sandu, A.: On consistency properties of discrete adjoint linear multistep methods. Report TR-07-40, Computer Science Department, Virginia Polytechnical Institute and State University (2007) · Zbl 0900.65238 Schwartz, A; Polak, E, Consistent approximations for optimal control problems based on Runge-Kutta integration, SIAM J. Control Optim., 34, 1235-1269, (1996) · Zbl 0861.49002 Spellucci, P.: Donlp2-intv-dyn users guide. Version November 18, 2009 · Zbl 0820.65043 Spellucci, P, A new technique for inconsistent QP problems in the SQP method, Math. Methods Oper. Res., 47, 355-400, (1998) · Zbl 0964.90062 Spellucci, P, An SQP method for general nonlinear programs using only equality constrained subproblems, Math. Program., 82, 413-448, (1998) · Zbl 0930.90082 Steihaug, T; Wolfbrandt, A, An attempt to avoid exact Jacobian and nonlinear equations in the numerical solution of stiff ordinary differential equations, Math. Comput., 33, 521-534, (1979) · Zbl 0451.65055 Strehmel, K., Weiner, R.: Linear-implizite Runge-Kutta-Methoden und ihre Anwendungen. Teubner-Texte zur Mathematik, Bd. 127, Teubner (1992) · Zbl 0759.65047 Verwer, JG; Spee, EJ; Blom, JG; Hundsdorfer, WH, A second order rosenbrock method applied to photochemical dispersion problems, SIAM J. Sci. Comput., 20, 1456-1480, (1999) · Zbl 0928.65116 Walther, A, Automatic differentiation of explicit Runge-Kutta methods for optimal control, Comput. Optim. Appl., 36, 83-108, (2007) · Zbl 1278.49037\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6300074,"math_prob":0.5721491,"size":5525,"snap":"2021-31-2021-39","text_gpt3_token_len":1630,"char_repetition_ratio":0.12280384,"word_repetition_ratio":0.010403121,"special_character_ratio":0.3279638,"punctuation_ratio":0.26512456,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9502533,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T08:59:53Z\",\"WARC-Record-ID\":\"<urn:uuid:bd9dfaee-008e-4c47-ad06-0d80a74ce28d>\",\"Content-Length\":\"55948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e632b761-9799-4475-be4a-0fbe40830d3f>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9499936-8128-4754-a0a3-eb65c010c3ac>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A1269.65063\",\"WARC-Payload-Digest\":\"sha1:JIVZTUWW3OGJNTH5VSTF7IHJWYMDTFV7\",\"WARC-Block-Digest\":\"sha1:ZTNWG65X7MEGP2HDJYNS2CQSE7N72SRV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154432.2_warc_CC-MAIN-20210803061431-20210803091431-00216.warc.gz\"}"}
https://www.intechopen.com/books/thermodynamics-physical-chemistry-of-aqueous-systems/thermodynamics-and-the-glass-forming-ability-of-alloys	[ "Open access peer-reviewed chapter\n\n# Thermodynamics and the Glass Forming Ability of Alloys\n\nBy Chengying Tang and Huaiying Zhou\n\nSubmitted: November 10th 2010Reviewed: May 30th 2011Published: September 15th 2011\n\nDOI: 10.5772/20803\n\n## 1. Introduction\n\nBulk metallic glasses (BMGs) have received a great deal of attention due to scientific and technological interest ever since the first successful synthesis of an amorphous phase in the Au–Si system in 1960 (Klement et al., 1960). There has been a lot of interest to identify parameters to assess the glass forming ability (GFA) of various alloy systems and compositions. A great deal of scientific efforts for quantification of GFA of alloys has been devoted to investigation of the GFA of alloys. There have been a lot of parameters to assess the glass forming ability (GFA) of various alloy systems and compositions. As a result, many criteria, including the confusion rule and the deep eutectic rule, for evaluating the glass forming ability (GFA) of an amorphous alloy have been proposed. Among them, the criteria used usually are the supercooled liquid region △Tx(=TxTg, where Tgand Txare the glass transition temperature and the crystallization temperature, respectively) (Inoue et al., 1993), the reduced glass transition temperature Trg(=Tg/Tl, where Tlis the liquidus temperature) (Turnbull, 1969) and the recently defined parameters γ(=Tx/(Tg+Tl)) (Lu & Liu, 2002), δ(=Τ x/(T l-T g)) (Chen, et al., 2005), β[=T xT g/(T l+T x)2] (Yuan, et al., 2008), ϕ(=△Trg(Tx/Tg)0.143) (Fan, et al. 2007), ω[=T l(T l+T x)/(T x(T l–T x))] (Ji & Pan, 2009), γc[=(3Tx–2Tg)/Tl) (Guo, 2010), and so on. These criteria have generally proved useful parameters for evaluating the GFA of an amorphous alloy. In order to guide the design of alloy compositions with high GFA, Inoue et al. (Inoue et al., 1998) and Johnson (Johnson, 1999) have proposed the following empirical rules: (I) multicomponent systems, (II) significant atomic size ratios above 12%, (III) negative heat of mixing and (IV) deep eutectic rule based on the Trg criterion. However, Al-based metallic glasses with rare earth metal additions (Guo et al., 2000), rare earth (RE) based glasses and some binary BMGs such as Zr-Cu, Ni-Nb binary alloy (Xia et al., 2006), provide important exception from this generality, because most of above mentioned GFA parameters and rules capable of searching metallic glasses with high GFA are not applicable to these Al–based and RE-based amorphous systems. Furthermore, all the above parameters need the alloy to be first prepared in glassy form to be able to measure the crystallization temperature T x, the liquidus temperature Tl, and/or the glass transition temperature T g.Hence, the above parameters are not predictive in nature, as they cannot predict a good glass forming composition without actually making that alloy and rapidly solidifying it into the glassy state. It is well known that crystallization is the only event that prevents the formation of an amorphous phase. Metallic glass formation is always a competing process between the undercooled melt and the resulting crystalline phases. The GFA of a melt is thus virtually determined by the stability of the undercooled melt and the competing crystalline phases. Thermodynamic analysis could be useful in evaluating the stability of the undercooled melt and the formation enthalpies of crystalline phases. So far, several attempts have been made successfully to investigate the GFA and predict glass forming range (GFR) in several binary and ternary amorphous alloy systems, using a pure thermodynamic approach or a combined thermodynamics and kinetics approach.\n\nFrom a thermodynamic point of view, there are generally following methods for calculating the GFA and predicting glass forming range (GFR) of an alloy system. The first approach is based on the T0 curve, which has been used to predict the GFR on several binary and some ternary systems. The quality of these predictions depends critically on the accuracy of the thermodynamic description. The second method is based on the semi-empirical Miedema’s model, which has been successfully applied to calculate and predict the glass forming range of some binary or ternary systems. The third consideration is directly employed on the calculation of the driving forces of crystalline phases (minimum driving force criterion) in a supercooled melt using calculation of phase diagram (CALPHAD) database. By employing driving force criterion with the obtained thermodynamic description for the investigated system, the GFA and predicted GFR of an alloy system were determined by comparing the driving force of crystalline phases precipitated from an undercooled melt. This evaluation has been successfully used to evaluate the GFA of several binary or ternary systems. Especially, it can be used to analyze the GFA of some alloy systems with unique glass forming ability, such as Al-based system. The other thermodynamic considerations, such as suppression of the formation of intermetallic phases, have been introduced.\n\nFrom a combined thermodynamics and kinetics approach, the GFA of the alloys were evaluated by introducing thermodynamic quantities obtained from CALPHAD method into Davies–Uhlmann kinetic formulations. In this evaluation, by assuming homogeneous nucleation without pre-existing nuclei and following the simplest treatment based on Johnson-Mehl-Avrami’s isothermal transformation kinetics, the time–temperature-transformation (TTT) curves were obtained, which are a measure of the time tfor formation of the phase Φ with a minimum detectable mass of crystal as a function of temperature. The critical cooling rates (Rc) for the glass formation calculated on the basis of the TTT curves was used to evaluate the glass-forming ability of this binary or ternary alloy. The calculated GFA results show good agreement with the experimental data in the compositional glass formation range of the investigated systems.\n\nThis chapter is intended to present systematically the methods and progress on the glass forming ability investigated by a thermodynamic approach or a combined thermodynamics and kinetics approach.\n\n## 2. Calculation of GFA based on thermodynamics analysis\n\nUsually, it is regarded the formation of metallic glasses is controlled by two factors, i.e., the cooling rate and the composition of the alloy. The critical cooling, which is the most effective gauge for GFA of the alloys, is hard to be measured experimentally. Hence, a great deal of efforts has devoted to the investigation on the correlation between the GFA and the composition of glass forming alloys. Inoue et al. (Inoue et al., 1998) and Johnson (Johnson, 1999) proposed the empirical rules to predict the element selection and compositional range of glass forming alloy. These rules have played an important role as a guideline for synthesis of BMGs for the last decade. However, recent experimental results have shown that the “confuse principal” and “deep eutectic rule” cannot be applicable to the Cu-Zr, Ni-Nb binary system (Xia et al., 2006) and Al-based ternary system (Guo et al., 2000). From a thermodynamic point of view, it is well known that crystallization is the only event that prevents the formation of an amorphous phase. During a melt–quenching process for metallic glass formation, the glass formation is exposed to crystallization competition of other crystalline phases from the undercooled melt between liquidus temperature Tl and glass transition Tg. The GFA of a melt is thus virtually determined by the stability of the undercooled melt and the competing crystalline phases, which can be analyzed by thermodynamic analysis. In this section, several GFA calculation based on thermodynamics analysis were introduced.\n\n### 2.1. Calculation of the GFA of alloys based on T0 curve\n\n#### 2.1.1. Method\n\nGenerally, a glass can be formed during cooling when crystallization is avoided up to the occurrence of the glass transition. Thus, in order to predict the tendency to glass formation in a system and the composition regions where it is most probable, nucleation of crystals in the undercooled melt must be considered. The GFR will be the region of composition where nucleation of crystalline phases is less likely. Various models have been developed to analyze the GFA of alloys in the literature, as will be discussed in the following section, with different levels of approximation. T0 curve is one of the approaches used to estimate the GFA of the alloys.\n\nA T0 curve is the locus of the compositions and temperatures where the free energies of two phases are equal. Thus, T0 curves can be calculated provided that their Gibbs free energy is known, i.e. an assessment of the system is available. The T0 curve between the liquid and a solid phase determines the minimum undercooling of the liquid for the partitionless formation of a crystalline solid with the same composition (Boettinger & Perepezko, 1993). Fig. 1 showed one example for a simple eutectic system. Alloys with T0 curves plunge steadily at low temperatures (dashed line in Fig. 1a), there will be no driving force for partitionless transformation in the composition region between them. If the equilibrium crystalline phases are not prone to nucleation, the glass can thus form. On the contrary, if T0 curves that are only slightly depressed below the stable liquidus curves are good candidate for partionless transformation of crystalline phases in the entire composition range (dashed line in Fig. 1b).", null, "Figure 1.Hypothetical T0 curves for a binary eutectic A–B system. (a) T0 curves drop to low temperature: glass formation is possible. (b) T0 curves intersect at low temperature: partitionless crystalline phase formation occurs (redrawn fromBoettinger & Perepezko, 1993).\n\n#### 2.1.2. Application of T0 curve\n\nPredictions of GFR based on T0 curves have been performed on several binary and some ternary systems. The construction of T0 curves for alloy glass needs a precise knowledge of thermodynamic properties of the supercooled liquid alloy and the introduction of the transition to the glassy state (Kim, et al., 1998). The quality of these predictions depends critically on the accuracy of the thermodynamic description and the introduction of the excess specific heat contribution is expected to improve the quality of results (Palumbo & Battezzati, 2008). However, as pointed out by Schwarz and co-workers (Schwarz et al., 1987), some discrepancies have been observed between the prediction and experimental results. For example, even when using the most recent thermodynamic assessment (Kumar, 1996) to calculate T0 curves in the Cu–Ti system, the results are not agreement with the reported experimental GFR. In fact, T0 curves for terminal solid solutions do not plunge at low temperatures as expected for glass forming systems (Kumar et al. 1996). Battezzati and co-workers (Battezzati, et al., 1990) have shown that in the Cu–Ti system the contribution of the excess specific heat is essential for describing the glass forming ability. An excess specific heat contribution has also been considered in the Al–Ti system (Cocco, et al., 1990) and the Fe–B system (Palumbo, et al., 2001).\n\n### 2.2. Calculation of the GFA of alloys based on Miedema’s model\n\n#### 2.2.1. Method\n\nMiedema’s model is an empirical theory for calculating heat of mixing in various binary systems both for the solid state (Miedema et al., 1975) and liquid (Boom et al., 1976). This model involves the calculations of the formation enthalpy of metallic glasses (amorphous phase) (ΔHamor), solid solutions (ΔHSS), and intermetallic compounds (ΔHinter) according to the following equations (Bakker 1988; Boer et al. 1988).\n\nΔHamor=ΔHchem(amor)+ΔHtopoE1\nΔHss=ΔHchem(SS)+ΔHelastic+ΔHstructureE2\nand\nΔHinter=ΔHchem(inter)E3\n\nwhere ΔHchem(amor) is the chemical mixing enthalpy of the amorphous state, ΔHtopo is the topology enthalpy of a glass, ΔHchem(SS) is the chemical mixing enthalpy of a solid solution, ΔHelastic is the elastic enthalpy of the solid solution calculated based on the continuous elastic model proposed by Friedel (Friedel, 1954) and Eshelby (Eshelby, 1954& 1956), ΔHstructure is the structure enthalpy induced by the structural changes, and ΔHchem(inter) is the chemical mixing enthalpy of an intermetallic compound. The formation enthalpy ΔHinter of a composition between two adjacent intermetallic compounds can be calculated using the level principle.\n\nThe chemical contribution of enthalpy of mixing of solid solution can be written as\n\nΔHchem=xAxB[xAΔHBinASS+xBΔHAinBSS]E4\n\nwhere x Aand x Brepresent the mole fraction of A and B atoms and ΔHSS is the enthalpy of solution of one element in another at infinite dilution. The data have been taken from Niessen et al. (Niessen, et al., 1983).\n\nThe elastic term in the enthalpy of formation originates from the atomic size mismatch, which can be expressed as\n\nΔHelastic=xAxB[xAΔHBinAelastic+xBΔHAinBelastic]E5\n\nTheΔHiinjelastichas been obtained by using the formalism by Simozar and Alonso(Simozar & Alonso, 1984) as\n\nΔHiinjelastic=2μj(ViVj)2Vj(3+4μjKi)E6\n\nwhere μj is the shear modulus of the solvent, Vi and Vj are the molar volumes of the solute and the solvent, respectively and K iis the compressibility of the solute.\n\nThe structural contribution of enthalpy for solid solution originates from the valence and the crystal structure of the solute and the solvent atom. It is found to have a very minor contribution and it is difficult to calculate. Hence, the structural contribution to enthalpy has been usually neglected (Basu, et al., 2008). In the case of the elastic and structural contributions are absent, thus the formation enthalpy of glasses can be calculated as\n\nΔHamor=ΔHchem(amor)+3.51nxiTm,iE7\n\nWhere x irepresents the mole fraction of component iatom, Tm,iis the melting temperature of the component i.\n\nAccording to the Miedema’s model, an amorphous phase can be formed if the enthalpy of formation of the amorphous phase is less than that of the solid solution phase. The heat of formation in alloys generally arises from the interactions among the constituent atoms where the interfacial energy plays a major role. The interfacial energy mainly comes from the atomic size difference. It has also been postulated that the number of intermetallic phases appearing in an alloy system is a strong function of the heat of mixing. The number of intermetallic phase in an alloy system increases with the increase in the heat of mixing. This model can be directly used to determine the glass forming range in binary alloy systems and can be extended to ternary systems by neglecting the ternary interactions.\n\n#### 2.2.2. Calculation of the GFA for the binary alloy systems\n\nSince the metallic glass formation process is controlled by thermodynamic factors, Miedema’s model was firstly used to predict the composition range of amorphous binary transition metal alloys (Kolk et al., 1988; Coehoorn et al., 1988; Murty, et al., 1992; Basu, et al., 2008). It is found that the predicted glass forming composition ranges are in good agreement with the experimental results. In the work of Takeuchi and Inoue (Takeuchi & Inoue, 2000), this approach has been used to calculate the mixing enthalpy and mismatch entropy of a number of bulk metallic glass alloy systems. It has been observed that the mixing enthalpy and normalised mismatch entropy for glass forming alloys vary within a certain range.", null, "Figure 2.Enthalpy–composition curves for binary Ti–Ni, Zr–Ni, Hf–Ni, Ti-Cu alloy systems (a–d). The curve with (•) and the curve with (Δ) represent amorphous and solid solution phase, respectively. The enthalpy values are in J/mol (FromBasu, et al., 2008).\n\nAs shown in Fig. 2, in the work of Basu et al. (Basu, et al., 2008), glass forming range (GFR) has been determined for different binary (Ti–Ni, Zr–Ni, Hf–Ni, Ti–Cu, Zr–Cu, Hf–Cu) in (Zr, Ti, Hf)–(Cu, Ni) alloys based on the mixing enthalpy and mismatch entropy calculations. Though copper and nickel appear next to each other in the periodic table, the glass forming ability of the copper and nickel bearing alloys is different. Thermodynamic analysis reveals that the glass forming behaviour of Zr and Hf is similar, whereas it is different from that of Ti. The smaller atomic size of Ti and the difference in the heat of mixing of Ti, Zr, Hf with Cu and Ni leads to the observed changes in the glass forming behaviour. Enthalpy contour plots can be used to distinguish the glass forming compositions on the basis of the increasing negative enthalpy of the composition. This method reveals the high glass forming ability of binary Zr–Cu, Hf–Cu, Hf–Ni systems over a narrow composition.\n\nIn the recent work performed by Xia (Xia, et al., 2006), the GFA of an alloy is considered that the formation of the meta-stable amorphous state should include two aspects: (1) the driving force for the glass formation, i.e., −ΔHamor, and (2) the resistance of glass formation against crystallization, i.e. the difference between the driving force for glass phase and for the intermetallic compound formation ΔHamor−ΔHinter. When two glass forming alloys have the same −ΔHamor but different ΔHamor−ΔHinter, their GFA can then be dominated by ΔHamor−ΔHinter. The lower the value of ΔHamor−ΔHinter, the higher the GFA of the alloy. On the other hand, when two glass forming alloys have the same ΔHamor−ΔHinter but different ΔHamor, their GFA is dominated by −ΔHamor. The higher the value of −ΔHamor, the better the GFA. Since the contribution from entropies is much smaller as compared with that from the formation enthalpy of solid compounds (Delamare, et al., 1994), the GFA is expressed in terms of formation enthalpy alone. Based on this thermodynamic consideration, a new parameter γ* to evaluate GFA for glass formation was proposed by Xia et al. (Xia, et al., 2006) and expressed as\n\nγ=GFAΔHamorΔHinterΔHamorE8\n\nwhere ΔHamor and ΔHinter are the enthalpies for glass and intermetallic formation, respectively. Both ΔHamor and ΔHinter are calculated by Miedema’s macroscopic atom model.\n\nThis parameter has been successfully used to predict the GFR and the best GFA alloy compositions in Zr-Cu and Ni-Nb system by comparing the value of γ of various alloy systems, respectively (Xia, et al., 2006).", null, "Figure 3.Calculated dependence of the parameter γ* on Zr and Ni concentration in Cu-Zr (a) and Ni-Nb (b) binary alloys, respectively (fromXia, et al., 2006).\n\nFig. 3 shows the calculated dependence of the parameter γ* on Zr and Ni concentration in Cu-Zr (a) and Ni-Nb (b) binary alloys, respectively, suggesting that the alloys Cu64Zr36 and Cu50Zr50 in Cu-Zr system, and Ni61.5Nb38.5 in Ni-Nb system are the best glass former, respectively. These predicted results are in good agreement with the experimentally reported Cu64.5Zr35.5 and Cu50Zr50, and Ni62Ni38 that could be made into bulk metallic glass rods with 2 mm in diameter, indicating that γ* is an effective parameter in identifying the best glass former in the Zr-Cu and Ni-Nb binary system.\n\nSimilarly, considering both the stability of liquid employing ΔHliq/ΔHinter, and the competition of glass and crystal using ΔHamor/ΔHinter, Ji et al. (Ji, et al., 2009) proposed a new parameter γ’ of GFA as\n\nγ'=GFAΔHliqΔHamor(ΔHinter)2E9\n\nAs Ji et al. described, this parameter γ’ is not only verified in five different binary bulk metallic glasses (Cu–Hf, Ni–Nb, Cu–Zr, Ca–Al, Pd–Si) but also showed wider application range comparing with the former model, but also have a better GFA estimation on the different composition than the parameter γ* because it is including ΔHliq in the evaluation expression. The predicated results are in good agreement with the experiments in all five different kinds of binary BMG systems and the biggest deviation of the peak of γ’ from the best current GFA composition is only about 6 at.% in Ca–Al alloy. Comparing with former GFA parameter γ, γ’ takes account of liquid stability and shows more universal for evaluation GFA in different kinds of binary alloys (Ji, et al., 2009). Recently, Wang et al. also made a modification to Xia’s proposal and it works more convenient to describe the GFA of transition metal systems (Wang, et al., 2009).\n\n#### 2.2.3. Calculation of the GFA for the multicomponent alloy systems\n\nMiedema’s approach has been extensively used by Nagarajan and Ranganathan (Nagarajan & Ranganathan, 1994), Takeuchi and Inoue (Takeuchi & Inoue, 2001&2004) and other researchers (Murty, et al., 1992; Rao, et al.; 2007; Basu, et al, 2008; Wang & Liu, 2009; Sun, et al., 2010) to determine the glass forming composition range (GFR) in a number of ternary and multicomponent systems. In the work performed by Takeuchi and Inoue (Takeuchi & Inoue, 2001), the amorphous-forming composition range (GFR) was calculated for 338 ternary amorphous alloy systems on the basis of the database given by Miedema's model in order to examine the applicability of the model, to analyze the stability of the amorphous phase, and to determine the dominant factors influencing the ability to form an amorphous phase. The mixing enthalpies of amorphous and solid solution phases were expressed as a function of alloy compositions on the basis of chemical enthalpy. The GFR was calculated for 335 systems except for the Al-Cu-Fe, Al-Mo-Si and Au-Ge-Si systems. The calculated results are in agreement with the experimental data for Cu-Ni- and Al-Ti-based systems. For typical amorphous alloy systems exemplified by the Zr-, La-, Fe- and Mg-based systems, it was recognized that the calculated GFR had been overestimated as a result of the model being simplified. It is found that the elastic enthalpy term arising in a solid solution phase stabilizes the amorphous phase, and the stabilization mechanism is particularly notable in Mg-based amorphous alloy systems. Short-range order plays an important role in the formation of Al-, Fe- and Pd-metalloid based systems (Takeuchi & Inoue, 2001).\n\nBased on Miedema’s model and Alonso’s method, the glass forming ability/range (GFA/ GFR) of the Fe–Zr–Cu system was studied by thermodynamic calculation. It is found that when the atomic concentration of Zr is between 34% and 56%, no matter what the atomic concentrations of Fe and Cu are, amorphous phase could be obtained, thus the atomic mismatch playing a dominating role in inﬂuencing the GFA. While the atomic concentration of Zr is out of the above range, the GFA is highly inﬂuenced by the immiscibility between Fe and Cu (Wang & Liu, 2009).\n\nGlass forming composition range for ternary Zr–Ti–Ni, Zr–Hf–Ni, Ti–Hf–Ni, Zr–Ti–Cu, Zr–Hf–Cu and Ti–Hf–Cu systems has been determined by extending the Miedema’s model to ternary alloy systems and by neglecting the ternary interaction parameter (Basu, et al., 2008). In their calculations, solid pure metals have been chosen to be the standard state and their enthalpy has been assigned to be zero. It is seen that the glass forming composition range for Ni bearing alloys is higher than that of the Cu bearing alloys, as heat of mixing of Ni is higher than that of Cu with Ti, Zr and Hf. In these ternary (Zr, Ti, Hf)–(Cu, Ni) alloys mixing enthalpy and mismatch entropy varies between (−13) and (−42) kJ/mol and 0.13 and 0.25, which is within the range predicted for glass formation (Basu, et al., 2008).\n\nIn the work of Oliveira et al., the γ parameter proposed by Xia et al. was extended to the ternary Al–Ni–Y system. The calculated γ* isocontours in the ternary diagram are compared with experimental results of glass formation in that system. Despite some misﬁtting, the best glass formers are found quite close to the highest γ* values, leading to the conclusion that this thermodynamic approach can be extended to ternary systems, serving as a useful tool for the development of new glass-forming compositions (Oliveira et al., 2008).\n\nRao et al. (Rao et al. 2007) identified the composition with highest glass forming ability in Zr-Ti-Ni-Cu-Al quinary systems with the Gibbs-energy change between the amorphous and solid solution phases as the thermodynamic parameter by calculating the Gibbs-energy change with the help of Miedema, Miracle, mismatch entropy and configurational entropy models. ΔG shows the strong correlation with the reduced glass transition temperature (Tg/Tl) in Zr-based metallic glasses. Thus, ΔG can be used as a predictive GFA parameter to identify compositions with the highest GFA. The compositions with the highest GFA have been identified in a number of quinary systems by iso-free energy contour maps by representing quandary systems as quasiternary systems (Rao et al. 2007). The best glass forming composition has been identified by drawing iso-Gibbs-energy change contours by representing quinary systems as pseudo-ternary ones. Attempts have been made to correlate the Gibbs-energy change with different existing glass forming criteria and it is found that the present thermodynamic parameter has good correlation with the reduced glass transition temperature. Further, encouraging correlations have been obtained between the energy required for amorphization during mechanical alloying to the Gibbs-energy change between the amorphous and solid solutions.\n\n### 2.3. Calculation of the GFA of alloys based on driving force criterion\n\n#### 2.3.1. Method\n\nThe basic underlying concept to predict the compositions of alloys having high GFA using the thermodynamic approach is that the compositions exhibiting the local melting minimum points favour amorphous phase formation. Thermodynamic approach of driving force criterion is based on a different concept. During a melt–quenching process for metallic glass formation, the glass formation is exposed to crystallization competition of other crystalline phases from the undercooled melt between liquidus temperature Tl and glass transition Tg. It is well known that the crystallization is the only event that prevents the formation of amorphous phase. Considering that crystallization is usually through the nucleation and growth process, the high GFA can be inversely predicted by searching a condition where the nucleation and growth of crystalline phases can be retarded. There are three dominating factors for the kinetics, (i) chemical driving force, (ii) interfacial energy, as an energy barrier, between the amorphous phase and the crystalline phases, (iii) the atomic mobility for rearrangement or transport of the partitioning atoms. According to the classical nucleation theory, the driving force of formation of the crystalline phases and interfacial energy, among other things, affects the nucleation rate of product phases. The interfacial energy between liquid and crystalline phases is known to be small compared to surface energy or grain boundary energy (Porter & Easterling, 1992), and therefore the role of interfacial energy in the nucleation kinetics of crystalline phases would be small. Then, the driving force of formation becomes the major factor that affect the nucleation kinetics of crystalline phases from amorphous alloy melts. It is believed that alloys with lower driving force for the formation of crystalline phases under the supercooled liquid state suggest higher GFA in the glass forming range. Therefore, Kim and co-workers proposed the minimum driving force criterion as a new thermodynamic calculation scheme to evaluate the composition dependence of the GFA (Kim et al., 2004). The driving force for the crystalline phases can be calculated using the critical assessed thermodynamic parameters by the CALPHAD method (Kaufman & Bernstein, 1970). In the CALPAHD method, the Gibbs energies of individual phases are described using thermodynamic models. Then, the model parameters are optimized considering relevant experimental information on phase equilibria or the other thermodynamic properties. The calculation of phase equilibrium is performed based on the minimum Gibbs energy criterion.\n\n#### 2.3.2. Application of the driving force criterion\n\nDriving force criterion has been successfully used to explain the composition dependence of GFA in several glass forming alloys with unique GFA, such as Cu-Zr-Ti (Kim, et al., 2004), Mg-Cu-Y (Kim, et al., 2005), Al-Ce-Ni (Tang, et al. 2010), and Al-Cu-Zr (Bo, et al, 2010) systems, by calculating the driving force of formation of crystalline phases under metastable supercooled liquid states and by searching the local minima of the driving forces for crystallization. The calculated results are in good agreement with the experimental results. It has been indicated that the driving force criterion can be used as a new thermodynamic scheme to estimate the composition dependence of GFA in multicompoent alloy systems for the development of bulk amorphous alloys.", null, "Figure 4.Calculated driving forces of crystalline phases for Cu55Zr45–xTix alloys, versus Ti content at (a) 1073 K, (b) 973 K, and (c) 873 K (fromKim, et al., 2004).\n\nFor the Cu-Zr-Ti system, among a series of ternary alloys Cu60Zr40–xTix (x = 10, 20, 30), the alloy with the highest GFA should be the alloy Cu60Zr20Ti20 according to the maximum Trg criterion (Turnbull, 1969), while experiments (Inoue, et al., 2001) show it is Cu60Zr30Ti10. Although the other alloys, Cu55Ti35Zr10 (Lin & Johnson, 1995) and Cu47Ti33Zr11Ni8Si1 (Choi, et al., 1998) based on the Cu–Ti–Zr ternary system but at different region, have been published as alloys with high GFA, there is no empirical rule or factor that can explain why the high GFA is obtained at certain compositions in the Cu–Ti–Zr system (roughly Zr: Ti=3:1 and Zr: Ti=1:3). As already the thermodynamic parameters for all phases in this system obtained by the CALPHAD method, the GFA is estimated by calculating the driving forces of all crystalline phases under the undercooled liquid state. Fig. 4 shows the calculated driving forces of individual crystalline phases as a function of Ti content in a temperature range (600–800 oC) where the alloys correspond to supercooled liquids state. As shown in this figure, along the composition line Cu55Zr45–xTix with varying Ti content, the driving forces of crystalline phases show two local minimums, one at Zr-rich region (x = 7–10) and the other at Ti-rich region (x= 28–29). According to the driving force criterion, the two local minimum points in Fig. 4 are the compositions where the GFA is expected to be higher than other compositional region. In a sense that the Zr:Ti ratios in the two local minimum points are toughly close to 3:1 and 1:3, it can be said that the former is close to the Inoue’s composition and the latter is close to Johnson’s composition (Kim et al., 2004). This finding indicates that the composition dependency of the GFA in the Cu-Zr-Ti ternary alloy system can be explained by calculating the driving forces of formation of crystalline phases under metastable supercooled liquid states and by searching the local minima of the driving forces for crystallization (Kim, et al. 2004).\n\nSimilarly, Al–based amorphous, which was discovered in 1988 (He, et al., 1988& Inoue, et al., 1988), is also of particular interest because of its low density, good bending ductility and high tensile strength. It was found that, however, most of above mentioned parameters and rules capable of searching metallic glasses with high GFA are not applicable to Al–based amorphous (Guo, et al., 2000, Hackenberg, et al., 2002, Gao, et al., 2003, Zhu, et al. 2004).Al–Ce–Ni system is an unique Al–based system, which can be synthesized into a strong, flexible metallic glass with the widest GFR covering 2–15 at.% Ce and 1–30 at.% Ni (Inoue, 1998, Kawazoe et al., 1997). The alloys with high GFA are situated away from the eutectic point. Experimental results of the Al–Ce–Ni bulk amorphous alloys prepared with copper mold casting indicate that the amorphous sheets with 5 mm width and 0.2 mm thickness are obtained in Al86Ce4Ni10 and Al88Ce6Ni6 alloys without appreciable glass transition. On contrary, alloys Al82Ce8Ni10 and Al80Ce6Ni14 with △Tx values of 20 and 21K consist mainly of crystalline phases (Inoue, 1998). After a thermodynamic assessment of the Al-Ce-Ni system in the Al-rich corner was performed, a set of consistent thermodynamic parameters were obtained, and the thermodynamic properties of the Al-Ce-Ni amorphous alloys were calculated. The calculated results indicated that the alloys with high GFA in the Al–Ce–Ni system are far from the eutectic point, and the heats of mixing are from –15 to –49 kJ/mol of atom for the observed amorphous alloys (Tang et al., 2010).\n\nAs shown in Fig. 5, the relatively smaller nucleation driving forces for the formation of crystalline phases for the Al–10Ce based alloys (Fig. 5a) are generally indicative of their higher GFA with a reportedly wider GFR (1–30at.% Ni) (Kawazoe et al., 1997). In contrast, the relatively larger driving forces in the Al–10Ni based alloys (Fig. 7b) are associated with their poorer GFA and narrower GFR (2–10 at.% Ce) (Kawazoe et al., 1997). This finding is further confirmed by the melt spinning (Tang, et al., 2010) and the copper mold casting experimental results (Inoue, 1998).\n\nBased on the experimental enthalpies of mixing of ternary liquid and undercooled liquid alloys as well as the evaluated isothermal sections, the Al–Cu–Zr ternary system has been assessed using the CALPHAD method. Most of the calculated results show good agreement with the experimental thermodynamic data and the reported phase diagrams. By employing the driving force criterion with the present thermodynamic description, the observed glass-forming ability in the Al–Cu–Zr system can be accounted for satisfactorily (Bo, et al., 2010).", null, "Figure 5.Calculated normalized nucleation driving force (per mole of atoms) for crystalline phases from undercooled (a) Al–10Ce–Ni, (b) Al–Ce–10Ni and (c) Al-6Ce-Ni metastable liquid at 800 oC (fromTang, et al., 2010).\n\n### 2.4. Calculation of the GFA of alloys based on the other thermodynamic approach\n\nBy treating the glass transition as a second-order phase transformation from liquid phase (Palumbo, et al., 2001, Shao et al., 2005), which can give good predictions of all important GFA indicators such as the reduced glass transition temperature and the thermodynamic stability of the amorphous phase, Shao et al. established a full thermodynamic database for glass forming ability (GFA). The resultant thermodynamic database can be used to produce all major temperature-related GFA indicators such as Tg/Tl, Tg/Tm and Tx/(Tg+Tl). It is indicated that together with phase diagram prediction, such an extensive CALPHAD approach is a powerful tool for designing alloys with large GFA (Shao et al., 2005).\n\nBy using the computational thermodynamic approach exhibiting low-lying liquidus surfaces coupled with the reduced glass transition temperature criterion of Turnbull, regions of alloy composition suitable for experimental tests for glass formation of Zr–Ti–Ni–Cu–Al system were identify rapidly by Cao et al. (Cao, et al., 2006). The glass forming ability of the alloys we studied can be understood in terms of the relative liquidus temperature in a thermodynamically calculated temperature vs. composition section through a multicomponent phase diagram. It does not follow several other proposed thermodynamic or topological criteria.\n\nA thermodynamic parameter (ΔHchem ×Sσ/kB) in the configuration entropy (Sconfig/R) range of 0.8-1.0 has been developed to identify excellent BMG composition using enthalpy of chemical mixing (ΔHchem), the mismatch entropy normalized by Boltzmann’s constant (Sσ/kB) and the configurational entropy (Sconfig/R) by Bhatt et al. and it has been demonstrated for the Zr-Cu-Al based ternary system. It is found that this approach can be used to predict the best BMG composition more closely than the earlier models (Bhatt, et al., 2007).\n\nBased on the undercooling theory resulting from the existence of multicomponent chemical short-range order (CSRO) domains, the glass forming range (GFR) in Zr-Ni-Ti alloy system was predicted by thermodynamic calculation. The GFR predicted by the thermodynamic calculation is consistent with the experiment results (Liu, et al. 2008).\n\nOne of the ways to predict the possible bulk glass formation composition is the phase diagram calculation with suppression of the formation of intermetallic phases. The formation of stoichiometric intermetallic compounds which have the ordered structure of atoms into specific lattice sites can take a time for the rearrangement of atoms from liquid state. Thus, the formation of intermetallic compounds can be suppressed during the fast solidification process normally applied to the bulk glass production. Combining the obtained the thermodynamic database and the above concept, the amorphous formation diagram of the Cu–Zr–Ag system with the suppression of all binary and ternary intermetallic phases has been proposed by Kang and Jung (Kang & Jung, 2010)\n\n## 3. Calculation of the GFA of alloys based on a combined thermodynamics and kinetics approach\n\nAs discussed above, the thermodynamic approach is useful since the thermodynamic parameters can be used to calculate the GFR in binary alloys and can also be used to predict the GFR in ternary systems based on the constituent binaries. One of the limitations of a purely thermodynamic approach is that it does not give the critical cooling rates for the glass formation. A combined thermodynamic and kinetic treatment, based on time-temperature-transformation curves (TTT) in the manner of Uhlmann and Davies has been presented (Saunders & Miodownik, 1986& 1988). This combined approach takes the thermodynamic parameters obtained from the phase diagram calculations and derives values for the free energy barrier for nucleation, free energy driving forces, and melting points used in kinetic equations. The combined approach has been successfully used to calculate the glass forming ability (GFA) of a wide range of binary and ternary alloy systems (Saunders & Miodownik, 1988). The calculated glass forming ranges for a wide number of binary and ternary alloy systems are in good agreement with experiment. A significant advantage of the combined approach is that data from binary alloy systems, often with little or no ternary modification, can be used to calculate the necessary thermodynamic input for the kinetic equations in higher order systems. This section outlines briefly the combined thermodynamic and kinetics method used for the calculation of the GFA for alloy systems.\n\n### 3.1. Method\n\nCritical cooling rates for glass formation can be obtained by Johnson-Mehl-Avrami isothermal transformation kinetics using the equation\n\nX=1exp[(π/3)IvUc3t4]E10\n\nwhere X is the volume fraction of material transformed, Iv is the nucleation frequency, Uc is the crystal growth rate, and tis the time taken to transform X. In the early stages of transformation the value of X approximates to\n\nXπIvUc3t4E11\n\nFor homogeneous nucleation without pre-existing nuclei, the nucleation frequencyIvhis given by\n\nIvh=DnNva02exp(ΔG/kT)E12\n\nwhere Dn is the diffusion coefficient necessary for crystallisation, Nv is the number of atoms per unit volume. a0 is an atomic diameter. kis Boltzmann’s constant. T is the transformation temperature, and ΔGis the free energy barrier for nucleation of a spherical nucleus given by the expression\n\nΔG=16π3(σ3/Gv2)E13\n\nwhere σ is the liquid/crystal interfacial energy and Gv is the change in free energy per unit volume on solidification. An equation for Uc can be written as\n\nUc=fDga0[1exp(ΔGm/RT)]E14\n\nwhere Dg is the diffusion coefficient for the atomic motion necessary for liquid to crystal growth, ΔGmis the molar free energy driving force for liquid to crystal growth, and Ris the universal gas constant, and fis a structural constant denoted the fraction of sites on the interface where atoms may preferentially be added or removed, and is given by the following expression (Uhlmann, 1972)\n\nf=0.2(TmT)/TmE15\n\nwhere Tm is the liquidus temperature. By assuming that Dn=Dg=the bulk liquid diffusivity and invoking the Stokes-Einstein relationship between diffusivity and viscosity η, equation (12) and (14) can be derived to give the time tneeded to form a volume fraction X of transformed crystalline phase in an undercooled liquid as following\n\nt9.3ηkT{a09Xf3Nvexp(ΔG/kT)[1exp(ΔGm/RT)]3}1/4E16\n\nwhere tis the time taken to transformation volume fraction X of crystalline solid. η is the viscosity of liquid, a0 is an atomic diameter, fis a structural constant, Nvis the number of atoms per unit volume, ΔGis the Gibbs energy barrier to nucleation and ΔGmis the Gibbs energy driving force for the liquid-crystal transformation. The constants have been typically taken as X=10-6, a0=0.28×10-9m, f=0.1 and Nv=5×1028 atoms/m3. In order to apply this equation to a real alloy system, it is necessary to derive or estimate the parameters η,ΔG, and ΔGm(Saunders & Miodownik, 1988).\n\n### 3.2. Estimation of η,ΔG, and ΔGm\n\nSince it is very difficult to measure experimentally the viscosity of supercooled liquid, there have been few measurements of it. In this case, the viscosity can be generally described as being between the liquidus temperature Tm and the glass transition Tg using a Doolittle-type expression involving the relative free volume fT (Ramachandrarao, et al., 1977) as\n\nη=Aexp(B/fT)E17\n\nwhere\n\nfT=Cexp(EH/RT)E18\n\nEH is the hole formation energy and A, B, and C are constants. Because of the lack of experimental data and the EH value was estimated by means of a direct relationship from Tg (Ramachandrarao, et al., 1977). Assuming B is unity and fT and η are 0.03 and 1012Ns/m2, respectively at Tg, A and C have been approximated at 3.33×10-3 and 10.1, respectively. If Tg values are not available, crystallisation temperatures Tx are used as a first approximation.", null, "Figure 6.The construction used in calculating the driving force, ΔGm, for the crystallization of compound AB2 from a liquid of compositionx1 in the A-B system.\n\nFor the crystallization of compounds AB2 from a liquid of composition x1 in the A-B system (Fig. 6), the molar free energy driving force for liquid to crystal growth, ΔGm, represents the Gibbs energy required to form one mole of crystalline phase from the liquid of composition x1, which can be obtained from thermodynamic phase diagram calculations to give explicitly molar heats of fusion Hmfand driving force that can be used toΔG. Values of Hmfand ΔGmare calculated from partial molar Gibbs energies of elements A and B and free energy values. Therefore, ΔGmis expressed by the following equation\n\nΔGm=xAG¯AL+xBG¯BLGcrystE19\n\nwhere xA and xB are the mole fractions of elements A and B in the precipitating crystalline phase, respectively. G¯ALand G¯BLare the partial molar Gibbs energies of elements A and B in the liquid phase, respectively, and Gcryst is the integral free energy of the precipitating crystalline phase. The Gibbs energy functions in Eq. (19) can be obtained from the thermodynamic model parameters evaluated in the literature. In an A-B alloy system, a liquid composition x1 becomes unstable with respect to the compounds AB2 at the liquidus temperature Tm. At a given temperature T1, there is a driving force for the precipitation of the compound AB2 given by G1 (Fig. 6), where G1 is defined as the driving force to form one mole of compound AB2 in a liquid of composition x1. In all cases here ΔGm is equal to ΔG1. By using heats of formation in place of free energy values, Hmfcan be similarly evaluated.\n\nThe Gibbs energy barrier to nucleation of a spherical nucleus ΔG* can be described as\n\nΔG=16π3N(σm3/ΔGm2)E20\n\nwhere Nis Avogadro’s number and the σm the molar liquid/crystal interfacial energy. σm is directly related to the molar enthalpy of fusion Hmfand expressed as\n\nσm=αHmfE21\n\nwhere αis a proportional constant. Hmfcan be obtained in a similar way to evaluate ΔGmbased on bond energy values across the interface (Turnbull, 1950). Saunders and Miodownik empirically evaluated the constant αto be 0.41 (Saunders & Miodownik, 1988).\n\n### 3.3. Calculation of critical cooling rates below T0 of disordered solid phases\n\nThe expression for tin equation (16) is derived assuming that the kinetics of the liquid to crystal transformation are limited by the bulk diffusivity, which is appropriate when the crystal composition differs from that of the liquid, or at compound compositions where substantial diffusion is necessary before the correct spatial relationships that define the ordered structure of the compound are achieved. However, at the temperatures below the T0 temperature of a disordered solid solution phase, the liquid becomes unstable with respect to a molecularly simple phase of the same composition. Consequently, no long range diffusion is necessary for the liquid to crystal and the kinetics are governed by atom motions of less than one atom in diameter. Then, the transformation is considered to be extremely difficult to suppress and this forms the T0 criteria for GFA. In such cases, it has been suggested that the rate limiting step for crystal growth is proportional to the rate at which atoms collide at the liquid/crystal interface, and an expression for the crystal growth rate is then given (Boettinger et al., 1984) by\n\nUc=fV0[1exp(ΔGm/RT)]E22\n\nwhere V0 is the velocity of sound in the liquid metal. This is the same form as equation (14), but with V0 replacing the Dg/a0. Replacing the Dn/a0 in equation (12) with V0 and rearranging equation (12) and (14), an expression for tis derived as\n\nt1V0{Xa0πf3Nvexp(ΔG/kT)[1exp(ΔGm/RT)]3}1/4E23\n\nThe value for V0 has been taken as 1000 m/s by Saunders and Miodownik (Saunders & Miodownik, 1988), close to the a value used by Boettinger et al. (Boettinger et al., 1984) and no transformation is considered to occur below Tg.\n\nFrom equations (10) to (23), the time-temperature-transformation (TTT) curve can be obtained. The critical cooling rate Rc necessary for amorphous phase formation with a melt quenching method can be evaluated from TTT curve calculated and approximated as follows\n\nRc=TmTn5tnE24\n\nwhere Τmand tn are the temperature and time at the nose of the TTT curve, respectively, since the cooling rate calculated directly from the isothermal transformation curve is somewhat overestimated compared with that from the CCT (continuous cooling transformation) curve, the right side of Equation (24) has divided by a factor of 5 to emulate continuous cooling. In the composition range with Rc˂1×10-7 K/s, which has been generally known to be a maximum available cooling rate for melting quenching, the amorphous phase formation may be possible.\n\n### 3.4. Evaluation of glass forming ranges in alloy systems\n\nThe combined thermodynamic and kinetic approach has been undertaken to evaluate the GFA of a wide number of binary and ternary alloy systems since the pioneering work performed by Sanders and Miodownik (Saunders & Miodownik, 1988; Shim et al., 1999; Clavaguera-Mora, 1995; Tokunaga, et al., 2004; Abe, et al., 2006; Ge, et al., 2008; Palumbo, & Battezzati, 2008; Mishra & Dubey, 2009). They calculated the free energy driving forces from the thermodynamic databases, free energy barrier for nucleation and melting points, and employed this data to kinetic calculation. There is in good agreement between the predicted glass forming ranges and those experimentally observed. It is indicated that the approach has the potential to predict glass forming ability in multicomponent alloys using mainly binary input data.\n\nThe first attempts to couple kinetic models with reliable thermodynamic data using the CALPHAD methodology was performed by Saunders and Miodownik (Saunders & Miodownik, 1988). In his work, the combined thermodynamics and kinetics approach was presented in detail and undertaken to evaluate the GFA of a wide range of binary (Au-Si, Pd-Si, Ti-Be, Zr-Be, Hf-Be, Cu-Ti, Co-Zr, Ni-Zr, Cu-Zr, Ni-P, Pd-P) and ternary (Ni-Pd-P, Cu-Pd-P, Co-Ti-Zr, Zr-Be-Hf, Ti-Be-Hf) alloy systems (Saunders & Miodownik, 1988). The TTT curves and the critical cooling rate for glass formation Rc were estimated. There is excellent agreement between the predicted and observed GFRs of binary systems, apart from the discrepancies in the Ti-Be and Cu-Ti systems. The approach was then extended to give predications for critical cooling rates in ternary and multicomponent alloys using mainly binary information. The results would appear to indicate that the combined approach takes into account a number of the major effects that govern glass formation and has the potential to predict GFA in multicomponent systems (Saunders & Miodownik, 1988).\n\nIn the work performed by Ge et al., (Ge, et al., 2008), the glass forming ability (GFA) of nine compositions of Cu-Zr and thirteen of Cu-Zr-Ti alloys in terms of critical cooling rate and fragility were evaluated by combining CALPHAD technique with kinetic approach. The driving forces for crystallization from the undercooled liquid alloys were calculated by using Turnbull and Thompson-Spaepen (TS) Gibbs free energy approximate equations, respectively. As shown in Fig. 7, time-temperature-transformation (TTT) curves of these alloys were obtained with Davies-Uhlmann kinetic equations based on classical nucleation theory. With Turnbull and TS equations, the critical cooling rates are calculated to be in the range of 9.78 ×103-8.23×105 K/s and 4.32 ×102-3.63×104 K/s, respectively, for Cu-Zr alloys, and 1.38×102-7.34×105 K/s and 0.64-1.36×104 K/s, respectively, for Cu-Zr-Ti alloys (Ge, et al., 2008).", null, "Figure 7.Calculated TTT curve of Cu-Zr (left) and Cu-Zr-Ti (right) alloys by (a) Turnbull model and (b) TS model (from Ref.Ge, et al., 2008).\n\nBased on topological, kinetic and thermodynamic considerations, Yang et al. (Yang, et al., 2010) have discussed the existence of the multiple maxima in GFA in a single eutectic system in Al–Zr–Ni system. It is apparent that, when taken alone, none of the factors seemed to be able to fully explain this phenomenon we have observed. It is suggested that glass formation is an intricate balance of kinetic, thermodynamic and also topological factors. Perhaps in good glass formers, all factors could come to a consensus at one composition or one compositional zone, where the best glass former(s) are located. However, for marginal glass formers like Al-based alloys, each of these factors could point to a different alloy composition, where conditions are best suited for glass formation.\n\nRecently, Considering chemical short-range ordering and metastability of undercooled melts, Zhu and co-workers have applied a simplified quasi-kinetic approach in order to predict the GFR in binary Al-rare earth (Zhu, et al., 2004) and Al-based Al-Gd-Ni(Fe) ternary system (Zhu, et al., 2004), using CALPHAD databases. They derived the derived an expression for the reduced time t’ = t/tmin for the formation of a minimal quantity of crystalline solid, where trepresents the composition-dependent time needed for the transformation and tmin the minimum transformation time at a certain optimum composition:\n\nt'(exp(ΔG*/kT)[1exp(ΔGm/RT)]3)1/4E25\n\nThis formula is in fact equivalent to Eqs. (10) to (23), except that the effect of parameters related to atomic transport is neglected. The calculated reduced times for various solid crystalline phases are then used to predict the GFR, i.e. the region of the composition space where these times are higher. A qualitative satisfactory agreement can be observed (Zhu, et al., 2004). The ability to predict the GFR of candidate metallic glass systems indicates a simple but effective approach for reducing reliance on extensive experimental trial and error in the search for new metallic glass systems (Zhu, et al., 2004).\n\n## 4. Conclusion\n\nSearch for new bulk metallic glasses (BMGs) system or composition by predicting the GFA of an alloy system is of interesting and theoretical and practical significance. In this chapter, the progress on the calculation or predication the glass forming ability by thermodynamics approach or a combined thermodynamics and kinetics approach have been reviewed. It is found that a good agreement between the predicated glass forming ability and those experimentally observed has been obtained. It is indicated that the thermodynamic approach developed in the literature has proved useful to predict the glass forming ability of a number of alloys system. It has revealed that the combined thermodynamics and kinetics approach has the advantage to predict the glass forming ability of the multicomponent alloys using the reliable database of binary system assessed by CALPHAD method. It has been accepted that the thermodynamic approach and/or the combined thermodynamic and kinetic approach are effective ways for the prediction of the GFA of metallic glass alloys.\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nChengying Tang and Huaiying Zhou (September 15th 2011). Thermodynamics and the Glass Forming Ability of Alloys, Thermodynamics - Physical Chemistry of Aqueous Systems, Juan Carlos Moreno-Piraján, IntechOpen, DOI: 10.5772/20803. Available from:\n\n### chapter statistics\n\n3Crossref citations\n\n### Related Content\n\nNext chapter\n\nBy Bohdan Hejna\n\nFirst chapter\n\n#### Thermodynamics of Ligand-Protein Interactions: Implications for Molecular Design\n\nBy Agnieszka K. Bronowska\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities." ]	[ null, "https://www.intechopen.com/media/chapter/20155/media/image2.jpeg", null, "https://www.intechopen.com/media/chapter/20155/media/image11.png", null, "https://www.intechopen.com/media/chapter/20155/media/image13.png", null, "https://www.intechopen.com/media/chapter/20155/media/image15.jpeg", null, "https://www.intechopen.com/media/chapter/20155/media/image16.png", null, "https://www.intechopen.com/media/chapter/20155/media/image35.jpeg", null, "https://www.intechopen.com/media/chapter/20155/media/image53.jpeg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91870093,"math_prob":0.9300013,"size":36831,"snap":"2021-21-2021-25","text_gpt3_token_len":8461,"char_repetition_ratio":0.18258344,"word_repetition_ratio":0.062239945,"special_character_ratio":0.21180527,"punctuation_ratio":0.10906682,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9630499,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T20:07:43Z\",\"WARC-Record-ID\":\"<urn:uuid:17a652f5-4011-4a54-8528-3246614a5b9b>\",\"Content-Length\":\"708498\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1257a45-c549-4f81-bd2d-dc1167cb144c>\",\"WARC-Concurrent-To\":\"<urn:uuid:556e1144-34e0-4ad9-9452-d218a593ee55>\",\"WARC-IP-Address\":\"35.171.73.43\",\"WARC-Target-URI\":\"https://www.intechopen.com/books/thermodynamics-physical-chemistry-of-aqueous-systems/thermodynamics-and-the-glass-forming-ability-of-alloys\",\"WARC-Payload-Digest\":\"sha1:IYNE6TBI6IC4TOGOPB52LIRINABV5YH2\",\"WARC-Block-Digest\":\"sha1:I5JF5UXYQ5RPE2ON5ZU4IXUMNNBQXWA4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243992440.69_warc_CC-MAIN-20210517180757-20210517210757-00598.warc.gz\"}"}
https://byjus.com/poisson-distribution-formula/	[ "", null, "# Poisson Distribution Formula\n\nPoisson distribution is actually another probability distribution formula. As per binomial distribution, we won’t be given the number of trials or the probability of success on a certain trail. The average number of successes will be given in a certain time interval. The average number of successes is called “Lambda” and denoted by the symbol “λ”.\n\nThe formula for Poisson Distribution formula is given below:\n\n$\\large P\\left(X=x\\right)=\\frac{e^{-\\lambda}\\:\\lambda^{x}}{x!}$\n\nHere,\n\n$$\\begin{array}{l}\\lambda\\end{array}$$\nis the average number\nx is a Poisson random variable.\ne is the base of logarithm and e = 2.71828 (approx).\n\n### Solved Example\n\nQuestion: As only 3 students came to attend the class today, find the probability for exactly 4 students to attend the classes tomorrow.\n\nSolution:\n\nGiven,\nAverage rate of value(\n\n$$\\begin{array}{l}\\lambda\\end{array}$$\n) = 3\nPoisson random variable(x) = 4\n\nPoisson distribution = P(X = x) =\n\n$$\\begin{array}{l}\\frac{e^{-\\lambda} \\lambda^{x}}{x!}\\end{array}$$\n\n$$\\begin{array}{l}\\begin{array}{c}P(X = 4)=\\frac{e^{-3} \\cdot 3^{4}}{4 !} \\\\ \\\\P(X = 4)=0.16803135574154\\end{array}\\end{array}$$" ]	[ null, "https://www.facebook.com/tr", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8100217,"math_prob":0.9997458,"size":862,"snap":"2023-40-2023-50","text_gpt3_token_len":208,"char_repetition_ratio":0.13519813,"word_repetition_ratio":0.015625,"special_character_ratio":0.23201856,"punctuation_ratio":0.10759494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000081,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T08:04:41Z\",\"WARC-Record-ID\":\"<urn:uuid:39c63a3d-5d01-48e7-859a-0f241c4e237e>\",\"Content-Length\":\"552085\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d435dc38-9b79-460a-aeef-636f71abd036>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0021662-c425-4ed2-b1d3-bec3eb1ceab9>\",\"WARC-IP-Address\":\"34.36.4.163\",\"WARC-Target-URI\":\"https://byjus.com/poisson-distribution-formula/\",\"WARC-Payload-Digest\":\"sha1:WRCI6MLKAVZX5YIZ3LHHOS2KQKYYR7NI\",\"WARC-Block-Digest\":\"sha1:6RMP7V7QLAN74KH4IBYULIZT3XXI7ZIT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679101282.74_warc_CC-MAIN-20231210060949-20231210090949-00735.warc.gz\"}"}
https://zbmath.org/?q=an%3A0907.62017	[ "# zbMATH — the first resource for mathematics\n\nRecord statistics. (English) Zbl 0907.62017\nCommack, NY: Nova Science Publishers, Inc. 227 p. (1995).\nContents (Chapter headings): 1. Record statistics; 2. Exponential distribution; 3. Generalized extreme value distributions; 4. Generalized Pareto distribution; 5. Power function distribution; 6. Geometric distribution; 7. Some selected distributions; 8. Additional topics; Appendix; References.\n\n##### MSC:\n 62E15 Exact distribution theory in statistics 62G30 Order statistics; empirical distribution functions 62-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to statistics 60G70 Extreme value theory; extremal stochastic processes 62-02 Research exposition (monographs, survey articles) pertaining to statistics" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.63503313,"math_prob":0.50011665,"size":917,"snap":"2021-31-2021-39","text_gpt3_token_len":231,"char_repetition_ratio":0.16648412,"word_repetition_ratio":0.036697246,"special_character_ratio":0.25627044,"punctuation_ratio":0.2760736,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9612058,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T00:58:22Z\",\"WARC-Record-ID\":\"<urn:uuid:d04e9092-fc90-4ae8-8957-f3fd6c1c87b3>\",\"Content-Length\":\"45938\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f55b1e77-a1b2-4593-a788-bf4bf5ca0aaf>\",\"WARC-Concurrent-To\":\"<urn:uuid:f162eb26-ff0c-4a3f-9222-22c640daacde>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A0907.62017\",\"WARC-Payload-Digest\":\"sha1:TSCV6HMZWRMSDWZXCGMP7ERJMTU7NJ65\",\"WARC-Block-Digest\":\"sha1:AJVZJGKCRI75645QNZGPNZH7DM5WFBJJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154486.47_warc_CC-MAIN-20210803222541-20210804012541-00080.warc.gz\"}"}