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:<"?> m nil OBBB^BK *ra#&$# <!4c<-<X-^.<S:ccC' MATHEMATICS FOR ENGINEERS PART I The Directly-Useful Technical Series See Detailed Prospectus Mathematics for Engineers Part II By W. N. ROSE, B.Sc. Eng. (Lond.) 440 pages. Demy 8vo. 13/6 net. E two volumes of Mathematics for Engineers form a most comprehensive and practical treatise on the subject. Great care has been taken to show the direct bearing of all principles to engineering practice, and the complete book will prove a valuable reference work embracing all the mathematics needed by engineers in their practice, and by students in all branches of engineering science. The second part is on similar lines to the present, and contains exhaustive chapters on the following INTRODUCTION TO DIFFERENTIATION DIFFERENTIATION OF FUNCTIONS APPLICATIONS OF DIFFERENTIATION METHODS OF INTEGRATION MEAN VALUES, R.M.S. VALUES, CENTROIDS, MOMENT OF INERTIA, ETC. POLAR CO-ORDINATES DIFFERENTIAL EQUATIONS APPLICATIONS OF CALCULUS HARMONIC ANALYSIS SPHERICAL TRIGONOMETRY MATHEMATICAL PROBABILITY, ETC. Particulars of other books in this Series are given on page 515. The Directly-Useful D.U. Technical Series Founded by the late WILFRID J. LINEHAM, B.Sc., M.Inst.C.E. Mathematics for Engineers PART I INCLUDING ELEMENTARY AND HIGHER ALGEBRA, MENSURATION AND GRAPHS, AND PLANE TRIGONOMETRY BY W. N. ROSE B.Sc. ENG. (LOND.) Late Lecturer in Engineering Mathematics at the University of London, Goldsmiths' College ; Teacher of Mathematics at the Borough Polytechnic Institute THIRD EDITION - (p LONDON CHAPMAN & HALL, LTD. II HENRIETTA STREET, W.C* 2 1922 FEINTED IN GRF.AT BRITAIN BY RICHARD CLAY & SONS, LIMITED, BUNGAY, SUFFOLK. EDITORIAL NOTE" THE DIRECTLY USEFUL TECHNICAL SERIES requires a few words by way of introduction. Technical books of the past have arranged themselves largely under two sections : the Theoretical and the Practical. Theoretical books have been written more for the train- ing of college students than for the supply of information to men in practice, and have been greatly filled with problems of an academic character. Practical books have often sought the other extreme, omitting the scientific basis upon which all good practice is built, whether discernible or not. The present series is intended to occupy a midway position. The information, the problems, and the exercises are to be of a directly useful character, but must at the same time be wedded to that proper amount of scientific ex- planation which alone will satisfy the inquiring mind. We shall thus appeal to all technical people throughout the land, either students or those in actual practice. . THE EDITOR. AUTHOR'S PREFACE AN endeavour has here been made to produce a treatise so thorough and complete that it shall embrace all the mathematical work needed by engineers in their practice, and by students in all branches of engineering science. It is also hoped that it will prove of special value for private study, and as a work of reference. Owing to the vast amount of ground to be covered, it has been found impossible to include everything in one volume : and accord- ingly the subject-matter has been divided into two portions, with the first of which the present volume deals. Stated briefly, Part I treats fully of the fundamental rules and processes of Algebra, Plane Trigonometry, Mensuration, and Graphs, the work being carefully graded from an elementary to an advanced stage ; while Part II is devoted to the Calculus and its applications, Harmonic Analysis, Spherical Trigonometry, etc. It is felt that the majority of books on Practical Mathematics, in the endeavour to depart from a theoretical treatment of the subject, neglect many essential algebraic operations, and, in addi- tion, limit the usefulness of the rules given by the omission of the proofs thereof. Throughout the book great attention has been paid to the systematic development of the subject, and, wherever possible, proofs of rules are given. Practical applications are added in the greater number of instances, the majority of the exercises, both worked and set, having a direct bearing on engineer- ing practice, thus fulfilling the main purpose of the book : and strictly academic examples are only introduced to emphasise mathematical processes needful in the development of the higher stages. In order to make the work of the greatest use to the engineer as a means of reference, many practical features have been intro- viii AUTHOR'S PREFACE duced, including : Calculation of Weights, Calculation of Earthwork Volumes, Land Surveying problems, and the Construction of PV and T diagrams or other general Practical Charts; and great care has been exercised in order that the best possible use may be made of mechanical calculators, such as the slide rule and the planimeter. Chap. I deals with methods of calculation. The method of approximating for a numerical result, introduced by Mr. W. J. Lineham, has been found to be very effective and easily grasped; and it is felt that the device here described for investigation for units could be more universally employed, because of its simplicity and directness. Simple, simultaneous, quadratic, cubic and all equations solvable by simple algebraic processes are treated in Chap. II : also fac- torisation by the simple methods, including the use of the Factor theorem, and the simplification of algebraic fractions. Great stress is laid on the importance of facility in transposing both terms and factors from one side of an equation to the other; and in this respect numerous literal equations are considered. The various rules of the mensuration of the simple areas and solids are stated in Chap. Ill ; the conic sections being included in view of their importance in connection with the theory of structures and strength of materials. The chapter concludes with the appli- cation of the rules for the calculation of weights ; a variety of types of machine parts being treated. All the elementary graph work is included in Chap. IV, in which attention is specially directed to the derivation of one curve from another, necessary, for example, in the case of efficiency curves. The usefulness of graphical solutions for the problems on arith- metical and geometrical progression is emphasised in Chap. V, in which also methods of allowing for depreciation of plant are intro- duced as illustrations of the commercial use of series. Here also are numerous examples on evaluation of formulae containing frac- tional and negative powers; and in these examples the absolute necessity of analysing compound expressions into their elements is made clear. In Chap. VI a departure is made from the old convention of the measurement of angles from a horizontal line, calling them positive if measured in a counter-clockwise direction. Plane Trigonometry has its widest application in land surveying, in which angles are measured by a right-handed rotation from the AUTHOR'S PREFACE ix North direction. Hence the north and south line is here taken as the standard line of reference and all angles are referred to it. Also by doing this the mathematical work is simplified, since the trigonometrical ratio of an angle does not alter; the angle of any magnitude being converted to the " equivalent acute angle," viz. the acute angle made with the north and south line. The calcula- tion of co-ordinates in land surveying is introduced as a good instance of the solution of right-angled triangles. Many rules for the solution of triangles are stated, but two only, viz. the " sine " and the " cosine " rules, are recommended for general use. This chapter contains much of importance to the electrical engineer, in the way of hyperbolic functions and complex quantities. The mode of utilising the planimeter for all possible cases is shown in Chap. VII, including the case in which the area to be measured is larger than the zero circle area. Graphic integration is introduced, in addition to the rules usually given for the measure- ment of irregular curved areas. Chap. VIII should prove of great value to railway engineers and to surveyors, since in it are collected the various types of earthwork problems likely to be encountered. Chap. IX deals with the plotting of difficult curve equations, and in this chapter it is demonstrated how a curve representing a rather complex equation may be obtained from a simple curve by suitable change of scales. Thus all sine curves have the same form, and accordingly the curve representing the equation y =* 72 sin(ioo7r/ - io6) can be obtained directly from the simple sine curve y sin x. The work on the construction and use of PV and T< diagrams should commend itself to students of thermo- dynamics. In Chap. X emphasis is laid on the advantage of making suit- able substitutions when transforming any equation into the linear form for the purpose of determining the law correlating two variables. Chap. XI provides a novel feature in its presentation of some methods used in the construction of charts applicable to drawing office practice. Alignment charts are here explained in the fullest detail, and it is hoped that the explanation given will further the more universal employment of these charts. Chap. XII embraces the more difficult algebra, necessary chiefly in the study of the Calculus; and in addition, the application of continued fractions to dividing-head problems. For extremely valuable advice, helpful criticism and assistance x AUTHOR'S PREFACE at all stages of the progress of the book the Author desires to tender his sincere thanks to MESSRS. W. J. LINEHAM, B.Sc., M.I.C.E., J. L. BALE, C. B. CLAPHAM, B.Sc., and G. T. WHITE, B.Sc. While it is hoped that the book is free from errors, it is possible that some may have been overlooked; and notification of such will be esteemed a great favour. W. N. ROSE. Goldsmiths' College, New Cross, S.E., January, 19 iS. NOTE TO SECOND EDITION THE early demand for a second edition of this volume has afforded an opportunity for making a number of corrections both in the text and in the illustrations; whilst a few exercises have been added. To those who have contributed to the improvement thus made, whether by notifying errors or by offering valuable suggestions, the Author's sincere thanks are proffered. November, NOTE TO THIRD EDITION THE favourable reception accorded the first and second editions inspires the hope of similar appreciation of the third edition. In this edition the need for the inclusion of some explanation of the determinant mode of expression employed in treatises on aerodynamics has been recognised by the addition of a section dealing with determinants. The work has been subjected to thorough revision, corrections being made where necessary, and further exercises have been added. December, 1921. CONTENTS PACE INTRODUCTORY .... i Previous knowledge Definitions and abbreviations Tables of weights and measures Useful constants. CHAPTER I AIDS TO CALCULATION . 6 Methods of approximation Indices Logarithms: i. Reading from tables. 2. Determination of characteristic Antilogarithms Ap- plications of logarithms Investigation for units. CHAPTER II EQUATIONS 31 Solution of simple equations Solution of simultaneous equations : i. With two unknowns. 2. With three unknowns Methods of factorisation The remainder and factor theorems Simplification of algebraic fractions Solution of quadratic equations : i. By factorisation. 2. By completion of the square. 3. By use of a formula Solution of cubic equations Solution of simultaneous quadratic equations Solution of surd equations. CHAPTER III MENSURATION 79 Area of rectangle and triangle Area of parallelogram and rhombus Areas of irregular quadrilaterals and irregular polygons Areas of regular polygons Circle: i. Circumference and area. 2. Area of annulus. 3. Length of chord and maximum height of arc. 4. Length of arc by true and by approximate rules. 5. Area of sector. 6. Area of segment by true and by approximate rules Area and perimeter of the ellipse The parabola The hyperbola Surface area and volume of prism and cylinder Surface area and volume of pyramid and cone Frusta of cones and pyramids The sphere : i. Surface area. 2. Volume. 3. Volume and surface area of a zone. 4. Volume of a segment. Relations between sides, surface areas, and volumes of similar figures Guldinus' rules for surface areas and volumes of solids of revolution Positions of centroids of simple figures Calculation of weights Tables of areas and circumferences of plane figures Tables of volumes and surface areas of solids. xii CONTENTS CHAPTER IV PAGE INTRODUCTION TO GRAPHS 148 Object and use of graphs Rules for plotting graphs Interpolation The plotting of co-ordinates Representation of simple equations by straight-line graphs Determination of the equation of a straight line Plotting of graphs to represent equations of the second degree Solution of quadratic equations : i. By use of a graph. 2. On the drawing board Plotting of graphs to represent equations of degree higher than the second Graphs applied to the solution of maximum and minimum problems. CHAPTER V FURTHER ALGEBRA 193 Variation Arithmetic and geometric progression, treated both algebraically and graphically Practical applications of geometric progression The laws of series applied to the calculation of allow- ance for depreciation of plant The value of " e " Napierian logarithms : i. Reading from tables. 2. Calculation from common logarithms Use of logarithms in the evaluation of formulae con- taining fractional and negative powers Logarithmic equations. CHAPTER VI PLANE TRIGONOMETRY 232 Definitions of the trigonometric ratios Reading the values of the trigonometric ratios from the tables Solution of right-angled triangles Reading the values of the trigonometric ratios from the slide rule Calculation of co-ordinates in land surveying Meaning of the terms " reduced bearing " and " whole-circle " bearing in surveying Rules for the determination of the trigonometric ratios of angles of any magnitude Table of signs of the trigonometric ratios Rules for the solution of triangles, for any given conditions The " ambiguous " case in the solution of triangles Proof of the " s " rule for the area of a triangle Use of tables of the logarithms of the trigonometric ratios The expansion of sin (A + B), A sin (A B), etc. Ratios of 2 A, , 3 A and 4 A in terms of the ratios of A Rules for the change of a sum or difference of two trigonometric ratios to a product, and vice versa Solution of trigo- nometric equations Hyperbolic functions Complex quantities Rule for addition of vector quantities Inverse trigonometrig functions. CONTENTS xiii CHAPTER VII PACE AREAS OF IRREGULAR CURVED FIGURES 300 Areas of irregular curved figures by various methods : I. By the use of the Amsler planimeter and the Coffin averager and plani- meter : the use of the Amsler planimeter for large areas being fully explained. 2. By averaging boundaries. 3. By counting squares. 4. By the use of the computing scale. 5. By the trape- zoidal rule. 6. By the mid-ordinate rule. 7. By Simpson's rule. 8. By graphic integration. CHAPTER VIII CALCULATION OF EARTHWORK VOLUMES 319 Volumes of prismoidal solids Volume of a wedge-shaped excava- tion Area of section of a cutting or embankment Volume of a cutting having symmetrical sides Volume of a cutting having unequal sides Net volume of earth removed in making a road by both cutting and embankment Volume of a cutting with unequal sides, in varying ground Surface areas for cuttings and embank- ments Volumes of reservoirs. CHAPTER IX THE PLOTTING OF DIFFICULT CURVE EQUATIONS . . . 336 Curves representing equations of the type y = ax" Use of the log-log scale on the slide rule Expansion curves for gases Special construction for drawing curves of the type pv n = C Equations to the ellipse, parabola and hyperbola The ellipse of stress Curves representing exponential functions The catenary Graphs of sine functions Use of the sine curve " template " Simple har- monic motion Graph of tan x Compound periodic oscillations Equation of time Curve of logarithmic decrement Graphic solu- tion of equations insolvable or not easily solvable by other methods Construction of PV and r<j> diagrams : i. Drawing PV and TQ diagrams. 2. Drawing primary adiabatics and constant-volume lines. 3. Drawing secondary adiabatics. 4. Plotting the Rankine cycle for two drynesses. 5. Plotting the common steam-engine diagram for an engine jacketed and non-jacketed. 6. Plotting quality curves. 7. Calculating exponents for adiabatic expansions. 8. Plotting constant heat lines PV and T$' diagrams for the Stirling, Joule and Ericsson engines. CHAPTER X THE DETERMINATION OF LAWS 396 b Laws of the type : i. y = a +-; y = a + bx*, etc. 2. y = ax" ; the usefulness of the slide rule for log plotting being demonstrated. 3. y ^ ae bx . 4. y = a+bx+cx*. 5. y = #+b.x n ; y y == {? + fo nx ; y = ws m , xiv CONTENTS CHAPTER XI PAGE THE CONSTRUCTION OF PRACTICAL CHARTS . . . .419 Correlation charts, including log plotting Ordinary intercept charts of various types Alignment charts, their principle and use Alignment charts involving powers of the variable Alignment chart for four variables. CHAPTER XII VARIOUS ALGEBRAIC PROCESSES, MOSTLY INTRODUCTORY TO PART II 448 Continued fractions Application of continued fractions to dividing- head problems Resolution of a fraction into two or more partial fractions Determination of limiting values of expressions Per- mutations and combinations The binomial theorem : i. Rule for the expansion of a binomial expression. 2. Rule for the calculation of any particular term in the expansion Use of the binomial theorem for approximations The exponential and logarithmic series Calculation of natural logs - Determinants. ANSWERS TO EXERCISES 479 TABLES : Trigonometric ratios 491 Logarithms 492 Antilogarithms 494 Napierian logarithms . . . . 496 Natural sines 498 Natural cosines 500 Natural tangents 502 Logarithmic sines 504 Logarithmic cosines 506 Logarithmic tangents 508 Exponential and hyperbolic functions . . . . . . .510 INDEX 511 MATHEMATICS FOR ENGINEERS INTRODUCTORY Previous Knowledge. While this work is intended to supply all the mathematical rules and processes used by the engineer, certain elementary branches of the subject have necessarily been omitted. It is assumed that the reader has a sound working knowledge of arithmetic, and also is acquainted with the four simple rules of algebra, viz. addition, subtraction, multiplication and division. Thus the meaning of the following algebraic processes should be known x 45 a? = axaxa; (* 2 ) 9 = # 18 ; =- = x*; x s O% _ \) i2c) = 30*2 426+72^; = '5# i-25y; 4 (40-76) (0 Again, the use of the lo-inch slide rule is not explained in detail as regards multiplying, dividing, involution and evolution ; but the special application of the slide rule is dealt with as occasion arises. Definitions and Abbreviations. An expression is any mathematical statement containing numbers, letters and signs. Terms of an expression are connected one with another by + or signs. The factors of an expression are those quantities, numerical or literal, which when multiplied together give the expression. Thus considering the expression , 2ga?xb 2 and io8ay 6 are terms; and each of these terms can be broken up into a number of factors; e. g. = i$xaxaxb. Again (ga 46) (5^+76) = 45a 2 +43& 286 2 ; and (9^46) and +7^) are the/ac/ofs of 45 B 2 MATHEMATICS FOR ENGINEERS When an expression depends for its value on that given to one of the quantities occurring in it, the expression is said to be a function of that quantity. Thus gx 3 7# 2 +5 is a function of x; and this relation would be written in the shorter form 9 * 3 7*2+5 =/(*). If a letter or number is raised to a power, the figure which denotes the magnitude of that power is called the exponent. An obtuse angle is one which is greater than a right angle. An acute angle is one which is less than a right angle. A scalene triangle has three unequal sides. The locus of a point is the path traced by the point when its position is ordered according to some law. The abbreviations detailed below will be adopted throughout. = stands for " equals " or " is equal to." + ,, " plus." ,, ,, " minus." x ,, " multiplied by." T- ,, " divided by." .*. ,, ?, " therefore." ,, ,, " plus or minus." > ,, ,, " greater than." < ,, " less than." ,, ,, " circle." Qce ,, ,, " circumference." oc ,, ,, " varies as." co ,, ,, " infinity." /_ ,, ,, " angle." A ,, ,, " triangle " or " area of triangle." li or 4! ,, ,, " factorial four "; the value being that of the product 1.2.3.4 or 2 4- "P, " the number of permutations of n things taken two at a time." "C, ,, ,, " the number of combinations of n things taken two at a time." n t ,, n (n i) (n 2). 7) ,, ,, " efficiency." a ,, ,, " angle in degrees." 6 ,, ,, " angle in radians." I.H.P. ,, ,, " indicated horse-power." B.H.P. ,, ,, " brake horse-power." m.p.h. ,, ,, " miles per hour." r.p.m. ,, ,, " revolutions per minute." r.p.s. ,, ,, " revolutions per second." I.V. " independent variable." INTRODUCTORY 3 F stands for " degrees Fahrenheit." C ,, " degrees Centigrade." L.C.D. " lowest common denominator." E.M.F. ,, ,, " electro-motive force." 1 ,, " moment of inertia." E ,, ,, " Young's modulus of elasticity." S n ,, " the sum to n terms." S OT ,, ,, " the sum to infinity (of terms)." 2 ,, ,, " sum of." B.T.U. " Board of Trade unit." B.Th.U. " British thermal unit." T ,, ,, " absolute temperature." p. ,, ,, " coefficient of friction." sin -1 x ,, ,, " the angle whose sine is x." e ,, ,, " the base of Napierian logarithms." g ,, " the acceleration due to the force of gravity." cms. ,, ,, " centimetres." grins. ,, ,, " grammes." Ly. " limit to which y approaches as x approaches *-* a the value a." Tables of Weights and Measures. BRITISH TABLE OF LENGTH 12 inches (ins.) = I foot 3 feet (ft.) = i yard 5j yards (yds.) = i pole 40 poles (po.) =i furlong 8 furlongs (fur.) = i mile. I nautical mile = 6080 feet i knot = i nautical mile per hour i fathom = 6 feet. SQUARE MEASURE 144 square inches (sq. ins.) = i square foot 9 square feet (sq. ft.) = i square yard 30^ square yards (sq. yds.) = i square pole 40 square poles = i rood 4 roods or 4840 sq. yds. i acre 640 acres = i square mile. 4 MATHEMATICS FOR ENGINEERS CUBIC MEASURE 1728 cubic inches (cu. ins.) = I cubic foot 27 cubic feet (cu. ft.) = i cubic yard. Weight of i gallon of water = 10 Ibs. Weight of i cu. ft. of water = 62-4 Ibs. i cu. ft. = 6-24 gallons. METRIC TABLE OF LENGTH I kilometre (Km.) = 1000 metres . I hectometre (Hm.) = 100 I dekametre (Dm.) = 10 metre (m.) = 39-37" i decimetre (dm.) = -i metre i centimetre (cm.) = -01 (2-54 cms. = i".) i millimetre (mm.) = -ooi LAND MEASURE 100 links = i chain i chain = 66 feet 10 chains = i furlong 80 chains = i mile 10 square chains = i acre. Useful Constants. <? = 271828 TT = 3-14159 log<io = 2-3026 Iog 10 e = -4343 log M N = log.N X -4343 logeN = log 10 N X 2-303 g = 32-18 ft. per sec. per sec. i horse-power = 33000 foot Ibs. per min. = 746 watts. Absolute temperature r = /C.+273 or tF.-\-^6i. i radian = 57-3 degrees. pressure of one atmosphere = 14-7 Ibs. per sq. in. i inch == 2-54 centimetres. i sq. in. = 6-45 sq. cms. i kilometre = -6213 mile. i kilogram = 2-205 Ibs. i Ib. = 453-6 grms. The following are the statements of the propositions in Euclid, to which reference is made in the text Euc. I. 47. In any right-angled triangle, the square which is INTRODUCTORY 5 described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. Euc. III. 35. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them shall be equal to the rectangle contained by the segments of the other. Euc. III. 36. Corollary. If from any point without a circle there be drawn two straight lines cutting it, the rectangles contained by the whole lines and the parts of them without the circle equal one another. Euc. VI. 4. The sides about the equal angles of triangles which are equiangular to one another are proportionals. Euc. VI. 19. Similar triangles are to one another in the duplicate ratio of their homologous (i. e., corresponding) sides. Euc. VI. 20. Similar polygons have to one another the dupli- cate ratio of that which their homologous sides have. [From this statement it follows that corresponding areas or surfaces are pro- portional to the squares of their linear dimensions.] CHAPTER I AIDS TO CALCULATION Approximation for Products and Quotients. Whatever may be the calculations in which the engineer is involved, it is always desirable, and even necessary, to obtain some approximate result to serve as a check on that obtained by the use of the slide rule or logarithms ; only in this way is confidence in one's working assured. Speed in approximation is as important as reasonable accuracy, and the following method, it is hoped, will greatly assist in such acceleration, especially in the cases of products and quotients. The great trouble in the evaluation of such an expression as 47-83 x 3-142 X 9-41 X -0076 ..,<-. is the fixing of the position of the 33000 decimal point. The rules usually given in handbooks on the manipulation of the slide rule may enable this to be done, but they certainly give no ideas as to the actual figures to be expected. The method suggested for approximation may be thus stated Reduce each number to a simple integer, i. e., one of the whole numbers I, 2, 3, etc., if possible choosing the numbers so that cancelling may be performed ; this reduction involving the omission of multiples or sub-multiples of 10. To allow for this, for every " multiplying 10 " omitted place one stroke in the corresponding line of a fraction spoken of as a point fraction, and for every " dividing 10 " place one stroke in the other line of this fraction. Thus two fractions are obtained, the number fraction, giving a rough idea of the actual figures in the result, and the point fraction from which the position of the decimal point in the result is fixed. Accordingly, by combining these two fractions the required approximate result is obtained. To illustrate the application of the method consider the following AIDS TO CALCULATION 7 *o ^ Example I. Find an approximate value of the quotient .r- The whole fraction may be written approximately as | (the number fraction) and IT (the point fraction) ; that is, we state 4-81 as 5 (working to the nearest integer). By so doing we are not multiplying or dividing by any power of ten, so that there would be nothing to write in the point fraction due to this change : but by writing 5 in place of -05, we are omitting two " dividing tens " ; therefore, since 5 is in the numerator of the number fraction, two strokes appear in the denominator of the point fraction. The number fraction reduces to i ; and the point fraction indicates that the result of the number fraction is to be divided by 100, since two strokes, corresponding to two tens multiplied together, appear in the denomi- nator. Hence, a combination of the two fractions gives the approximate result as i -f- 100 or -01. It may be easier to effect the combination cf the two fractions according to the following plan The result of the number fraction being !; shift the decimal point two places to the left, because of the presence of the two strokes in the denominator of the point fraction, thus 01 Example 2. Determine the approximate value of ^|-i - 28-4 x -00074 To apply the method to this example State 9764 as 10,000, i. e., write i in the numerator of the number fraction and four strokes in the numerator of the point fraction. For -0213 write 2 in the numerator of the number fraction and two strokes in the denominator of the point fraction. The strokes are placed in the denominator because in substituting -02 for 2 we are multiplying by 100, and therefore, to preserve the balance, we must divide the result by 100. For 28-4 we should write 3 with i stroke in the denominator, and for -00074 we should write 7 with 4 strokes in the numerator. Thus Number fraction. Point fraction. 1X2 1111 1\\\ 3><7 \\ \ . e., -i and - by cancelling. Hence the approximate result is -i x io 5 , i. e., 10,000 ; or, alter- natively, the shifting of the decimal point would be effected thus 10000 8 MATHEMATICS FOR ENGINEERS It will be seen from this method that it is often an advantage to express a very large or very small number as an equivalent simpler number multiplied by some power of ten. Not only is a saving of time obtained, but the method tends to greater accuracy. Thus 2,000,000 may be written as 2 X io 6 , a very compact form ; also it is far more likely that an error of a nought may be made in the extended than in the shorter form. " Young's modulus " for steel is often written as 29 X io 6 Ibs. per sq. in., rather than 29,000,000 Ibs. per sq. in. Example 3. Find the approximation for 47'83 x 3-142 x 9-41 x -0076 33000 The method will be understood from the explanation given in the previous examples ; and for clearness the strokes are separated in the point fraction. The approximation is 5x3x1x8 \ \ 3 111 which reduces to 40 im '. e., 40 -r- io 5 = -0004. The change in the position of the decimal point would be -00040' Further examples on approximation will be found on pp. 18 to 21. Approximations for Squares and Square Roots. An ex- tension of this method can be made to apply to cases of squares and square roots, cubes and cube roots. As regards squaring and cubing, these may be considered as cases of multiplication, so that nothing further need be added. To find, say, a square root approxi- mately, we must remember that the square root of 100 or io 2 is io, the square root of io 4 is io 2 , and so on ; the approximation, therefore, must be so arranged that an even number of tens are omitted or added. Hence the rule for this approximation may be expressed Reduce the number whose square root is to be found to some number between i and 100, multiplied or divided by some even power of ten ; then the approximate square root of this number, combined with half the number of strokes in the point fraction, gives the approximate square root of the number. AIDS TO CALCULATION 9 In the case of cube roots, the number must be reduced to some number between i and 100 multiplied or divided by 3, 6 or 9 ... tens; then the approximate cube root of this number must be combined with one-third of the strokes in the point fraction. Example 4. Find an approximation for ^498-4. In place of 498-4 write 500, which can be written as 5 x io 2 , 11 or as 5 Then the approximate square root is 2-2 or 22. If the number had been 4984 the number would read 5 o ^ and the square root 7 -, 22; Example 5. Find approximately the cube root of -000182. If for 182 we write 200, then -000182 is replaced by 111111 so that the cube root of -000182 is that of 200 divided by io a , since two strokes (viz. ^ of 6) appear in the denominator of the point fraction. Thus, the cube root is s-s n or -058. Example 6. Evaluate approximately I/ - 154 x 2409 Disregarding the square root sign for the moment, the approxima- tion gives 2X I J \_ 1-5x2 i iu ,'.*, -67 n For the application of the method of this paragraph this result would be written 67 of which the square root is 8-2 or the approximate square root is -082. io MATHEMATICS FOR ENGINEERS Exercises 1. On Approximations. Determine the approximate answers for Exercises i to 20. 1. 49-57 x -0243 2. -00517 x -1724 3. ? 2 3 4 4. 8-965 x 72-49 x '094 5. -1167 x -0004 x 98-1 x 2710 4-176 X 25400 _ ~ 87235 7 - 11540 x -3276 x 3-142 x -0078 - 4-176 X 25400 _ 154 X -00905 6> ~ 87235 7 - "-847 8 - "00346 x -0209 Indices. The approximation being made, the actual figures can be determined either by logarithms or by the slide rule. Napier, working in Scotland, and Briggs in England, during the period 1614-17 evolved a system which made possible the evalua- tion of expressions previously left severely alone. Without the aid of their system much of the experimental work of modern times would lose its application, in that the conclusions to be drawn could not be put into the most beneficial forms ; and failing loga- rithms, arithmetic, with its cumbersome and exacting rules, would dull our faculties and prevent any advance. The great virtue of the system of logarithms is its simplicity : rules with which we have long been acquainted are put into a more practical form and a new name given to them. Many are familiar with the simpler rules of indices, such as a 3 X a 4 = 3+4 = a 7 ; a8_i_02 = a 8 - 2 = a 6 ; (a 3 ) 4 = 3X4 = a 12 , etc. Following along these lines we can find meanings for *, a, and a~ 3 , i. e., we can establish rules that will apply to all cases of positive, negative, fractional or integral indices. Thus, to find a meaning for a fractional power, consider the simplest case, viz. that in which the index is J. When multiplying a 3 X a 4 we add the indices ; this can be done whatever the indices may be, hence = a = a i. e., a* is that quantity which multiplied by itself is equal to a, or in other words, a Ms the square root of a. AIDS TO CALCULATION n In like manner, since a& x a$ x a& = a 1 = a, a? may be written as -ty 'a. For example, 27^ = ^27 = 3. To carry this argument a step further we may consider a numerical example, e. g., 64$, and from the meaning of this, derive a meaning for a* Thus 64^ might be written as 64^ X 64^, which again may be put into the form ^64 X ^64, i. e., (^64)* or ^S^ 2 . Hence the actual numerical value = ^64 x 64 = 16. We see that the denominator of the exponent indicates the p root, and the numerator the power; thus a = ^a r t To find a meaning for a a m x a = m+0 = m . Dividing through by a m , a = i, i. g., any number or letter raised to the zero power equals i : e. g., 465 = i; -2384 = i; 4*0 = 4 x i = 4. Assuming this result for a, we can show how to deal with negative powers, for a m xa~ m = a m ~ m = a = i. Hence, dividing through by a" 1 , -m _ T a" 1 Accordingly, in changing a factor (such as a" 1 ) from the top to the bottom of a fraction or vice versa, we must change the sign before its index. Thus 2 b~ 7 Example 7. Simplify (- 6 6 2 c 8 ) 2 x Va 3 &- 4 c 6 . The expression = a~ 10 &Vxa^&~^ . . Removing brackets. = o~ 10+i 6 4 - 2 c 6 +3 . . . Collecting like letters. -17,29 6V 6V = a -y-b c =. or - Vfl 17 Example 8. Simplify (64*"')* 2( 5 AT 2 ) 3 The expression 2X5 3 * 6 2XI25X* 25QX~ r ~ 12 12 MATHEMATICS FOR ENGINEERS Exercises 2. On Indices. 1. Express with positive indices 2. Find the numerical values of 32!; 6 4 -f; (j)~*; 6x512$; fox #6^*} + {i 5 x 3. Simplify (sa 2 6c- 3 ) x (^a 9 b- 5 c)-^. 4. Simplify "343^-*^ -r 8ix~ l y 5. Simplify na 2 6crf 8 x (a~*b 6 c 3 )-^ TJ^ n WI} n ~ ^ 6. Find the value of -- .-^- in terms of v, when = 1-37. 4 U 7. Find the value of vnCv~ n i in terms of p when n =1-41 and pv n = C. _ r i / ^Tj 2 8. Simplify j V/ i f , a formula referring to the flow of a gas through an orifice, a being the ratio of the outlet pressure to the pressure in the vessel. 9. Simplify 8 (**) x ()** -i- (e 2 ')* r , and find its value when e = 2-718. 10. Simplify the expression 11. The work done in the adiabatic expansion of a gas from volume Q t/i to volume w, may be written W = ^(Vt 1 -" v^ 1 -"). If p t v t n = piUi 1 = C, by substituting for C, find a simpler expression for W. Logarithms. It is necessary to deal with indices at this stage, because logarithms and indices are intimately connected. For example, 100 = io 2 , and the logarithm of 100 to the base 10 = 2 (written Iog 10 100 = 2). Here are two different ways of stating the same fact, for 2 is the index of the power to which the base io has to be raised to equal the number 100; but it is also called the logarithm of 100 to the base io, i. e., the index viewed from a slightly different standpoint is termed the logarithm. Hence the rules of logs (as they are called) must be the same as those connecting indices. In general : The logarithm of a number to a certain base is the index of the power to which the base must be raised to equal the number. * It is not necessary to understand the theory of logs to be able to use them for ordinary calculations, but the knowledge of the principles involved is of very great assistance. AIDS TO CALCULATION 13 Consider the three statements 64 = 2 6 ; 64 = 4 3 ; 64 = 8 2 . These could be written in the alternative forms Iog 2 64 = 6; Iog 4 64 = 3; Iog 8 64 = 2; where the numbers 2, 4 and 8 are called bases. It will be noticed that the same number has different logs in the three cases, i. e. t if we alter the " base " or " datum " from which measurements or calculations are made, we alter the log; in consequence, as many tables of logs can be constructed as there are numbers. This shows the need for a standard base, and accordingly logs are calculated either to the base of 10 (such being called Common or Briggian Logarithms) or to the base of e, a letter written to represent a series of vast importance, the approximate value of which is 2-718. Logarithms calculated to the base of e are called Natural, Napierian or Hyperbolic Logarithms. At present we shall confine our attention to the Common logs; in the later parts of the work we shall find the importance and usefulness of the natural logs. From the foregoing definition of a logarithm the logs of simple powers of ten can be readily written down ; thus, Iog 10 iooo = 3, since 1000 = io 3 , Iog 10 ioooooo = 6, etc. ; Iog 10 iooo = 3 is usually written in the shorter form log 1000 = 3, the base io being under- stood when the small base figure is omitted. For a number, such as 526-3, lying between 100 and 1000, *. e., between io 2 and io 3 , the log must lie between 2 and 3, and must therefore be 2 -f- some fraction. To determine this fraction recourse must be made to a table of logs. To read logs from the tables. The tables appearing at the end of this book are known as four-figure tables, and are quite full enough for ordinary calculations, but for particularly accurate work, as, for example, in Surveying, five- and even seven-figure tables are used. One soon becomes familiar with the method of using these tables, the few difficulties arising being dealt with in the following pages. To return to the number 526-3 : the fractional or decimal part of its logarithm is to be found after the following manner : Look down the first column of the table headed " logarithms " (Table II at the end of the book) till 52 is reached, then along this line until under the column headed " 6 " at the top the figure 7210 is found; this is the decimal part of the log of 526, so that the 3 is at present unaccounted for. At the end of the line in which 7210 occurs are what are known as " difference " columns. Under that headed 14 MATHEMATICS FOR ENGINEERS " 3 " and in the same line as the 7210, the fourth figure of our number, the figure 2, occurs; this, added to 7210, making 7212, completes the decimal portion of the log of 526-3. The figure from the tables is thus 7212, and since this is to be the fractional portion the decimal point is placed immediately before the first figure. The log of 526-3 is therefore 2-7212, or, in other words, 526-3 = 10 raised to the power 2-7212; similarly the log of 52630 must be 4-7212, because 52630 is the same proportion of a power of 10 above 10,000 as 526-3 is above 100, and also it lies between io 4 and io 5 , so that its logarithm must be 4 + some fraction. The log thus consists of two distinct parts, the decimal part, which is always obtained from the tables and is called the mantissa, and the integral or whole-number part, settled by the position of the decimal point in the number, and called the characteristic or distinguisher. The logs of 526-3 and 52630 are alike as regards the decimal part, but must be distinguished from one another by the addition of the relative characteristic. When the number was 526-3, i. e., having 3 figures before the decimal point, the characteristic was 2, i. e., 3 I ; when the number was 52630, i. e., having 5 figures before the decimal point, the characteristic was 4, *'. e., 5 i. This method could be applied for numbers down to i, i. e., 10, but for numbers of less value we are dealing with negative powers, and accordingly we must investigate afresh. So far, then, we can formulate the rule : " When the number is greater than 1, the characteristic of its log is positive and is one less than the number of figures before the decimal point." E.g., if the number is 2507640, the characteristic of its log is 6, because there are seven figures in the number before the decimal point. Referring to the figures 5263 already mentioned, place the decimal point immediately before the first figure, giving -5263. The number now lies between -i and i. Now i = = io -1 and i = 10 io so that the log of -5263 lies between i and o, being greater than i and less than o, and therefore is i + a fraction. The mantissa is as before, viz. 7212, hence log -5263 = i + -7212, or, as it is usually written, 1-7212, the minus sign being placed over the i to signify the fact that it applies only to the i and not to the 7212, which latter is a positive quantity and must be kept as such. 1-7212 actually means, then, I + -7212, or, in fact, -2788. AIDS TO CALCULATION 15 The figures taken from the tables are always positive, and accordingly the form 1-7212 is adhered to throughout. From similar reasoning, log -005263 = 3-7212 (i) and log -00005263 = 5-7212 .... (2) We can conclude, then, that : When dealing with the log of a decimal fraction the mantissa is found from the tables in just the same way as for a number larger than I, or, in other words, no regard is paid, when using the tables to find the mantissa, to the position of the decimal point in the number whose log is required. The characteristic of the log, however, is negative, and one more than the number of zeros before the first significant figure. In (i) there are 2 noughts before the first significant figure; therefore the characteristic is 3. In (2) the characteristic is 5, because there are 4 noughts before the first significant figure. For emphasis, the rules for the determination of the charac- teristic of the log of any number are repeated If the number is greater than 1, the characteristic is positive and one less than the number of figures before the decimal point : if the number is less than i, the characteristic is negative and one more than the number of noughts before the first significant figure. It will be observed that in the earlier part of the table of loga- rithms at the end of the book there are, for each number in the first column, two lines in the " difference " column. This arrange- ment (the copyright of Messrs. Macmillan & Co., Ltd.) gives greater accuracy as regards the fourth figure of the log, since the differences in this portion of the table are large. The log is looked out as explained in the previous case, care being taken to read the " difference " figure in the same line as the third significant figure of the number whose log is being determined. Thus the log of 1437 is 3-1553 + a difference of 21 = 3-1574, while the log of 1487 is 3-1703 + a difference of 20 = 3-1723. We are now in a position to write down the value of the log of any number, and a few examples are given Number. 40760 2359 70-08 0009 500000 Log. 4-6102 1-3728 1-8456 4-9542 5 -~ (The mantissa for the log of 9, 90, 900, and 9000 is -9542.) i6 MATHEMATICS FOR ENGINEERS Values of log 1 and log 0. If a be any number, then, as proved earlier, a = i ; or in the log form, log a I = o. Thus the log of i to any base = o. The log of o to any base is minus infinity ; or if a be any number, log a o = <x. For, by writing this statement in the alternative form a' = o where x is the required logarithm, we see that x must be an infinitely small quantity ; in fact, the smallest quantity conceivable. Antilogarithms. Suppose the question is presented to us in the reverse way : " Find the number whose logarithm is 2-9053." The table of antilogarithms (Table III) will be found more con- venient for this, although the log tables can be used in the reverse way. Just as the mantissa alone was found from the log tables when finding the logarithm, so this alone is used to determine the actual arrangement of the figures in the number. In the case under consideration the mantissa is "9053, hence look down the first column until -90 is reached, then along this line until in the column headed " 5 " 8035 is read off : to this must be added 6, the number found in the " difference " column headed " 3," so that the actual figuring of the number is 8035 + 6 = 8041. The characteristic 2 must now be considered so as to fix the position of the decimal point. Referring to our rule, we see that the characteristic is one less than the number of figures before the decimal point (since 2 is positive), hence, conversely, the number of figures before the decimal point must be one more than the characteristic; in this case there must be 3 figures before the decimal point, i. e., the required number is 804-1. If we had been asked to find the antilog of 2-0905, the line through -09 would havebeen followed and not that through -90, and the antilog is found to be 123-1. Many errors occur if this distinction is not appreciated; and the actual mantissa must be dealt with in its entirety, no noughts being disregarded wherever they may occur. Examples Log. No. 8-1164 130700000 or i 307 x io 8 2549 i -0062 1-799 10-14 3-8609 007259 AIDS TO CALCULATION Failing a table of logs, the log scale on the slide rule can be used in the following manner. Reverse and invert the slide so that the S scale is now adjacent to the.D scale: place the ends of the D and S scales level: then using the D scale as that of numbers, the corresponding logarithms are read off on the L scale it being remembered that although the scale is inverted the numbers increase towards the right. The mantissa alone is found in this way, whilst the characteristic is settled according to the ordinary rules. Fig. i shows the scales of an ordinary 10" Slide Rule lettered as they will be referred to throughout this book. On the front are the scales A, B, C and D, the B and C scales being on the " slide." If the slide is taken out and reversed the S, L and T scales will be noticed (see right-hand end of figure). Any special markings referred to through- out the text are also indicated, and it is to this sketch that the reader should refer, no other sketch of the slide rule being inserted. The slide rule is referred to from time to time, wherever its use is required, and a word or two is then said about the method of usage, but no special chapter is devoted to its use. For a full explanation of the method of using the slide rule reference should be made to Arithmetic for Engineers* Applications of Logarithms. It will be granted that 2+4 = 6 or log 100 + log 10,000 = log 1,000,000 from definition. But 1,000,000 = 100 x 10,000 /.log(ioo x 10,000)= log 100 + log 10,000. Simple powers of ten have been taken in this example, for convenience, but the rule demon- strated is perfectly general, holding for all numbers. In general, log (AxB) = log A+ log B, where A and B are any numbers. Thus, the log of a product = the sum of the logs of the factors. This and the succeeding rules hold for bases other than 10; in fact, they are general in all respects. * Arithmetic for^Engineers, by Charles B. Clapham, B.Sc. Chapman and Hall, Ltd., ?s.'6d. net. C HEUlJH to i8 MATHEMATICS FOR ENGINEERS In like manner it can be shown that log( = ) = log A log B \o/ i. e., the log of a quotient =- the difference of the logs of numerator and denominator. Again, 3x2 = 6 .'. 3Xlog 100 = log 1,000,000 = log (ioo) 3 or in general, log (A) 1 * = n log A, this holding whatever value be given to n. E.g., (a) ^4^76 =(42 -76)* 6 = J log 42-76. (b) log (-05I7)- 4 -* = 4-2 x log -0517. Stated in words this rule becomes : The log of a number raised to a power is equal to the log of that number multiplied by that power. Summarising, we see that multiplication and division can be performed by suitable addition and subtraction, whilst the trouble- some process of finding a power or root resolves itself into a simple multiplication or division. (The application of logarithms to more difficult calculations is taken up in Chap. V.) In any numerical example care should be taken to set the work out in a reasonable fashion ; especially in questions involving the use of logs. Example 9. Find the value of 48-21 x 7-429. Actual Working Approximation 50x7 = 350. Let x = 48-21 X 7-429 then log x = log 48 -2 1 + log 7-429= 1-6831 + -8709 = 2-5540 = log 358-1 from the antilog tables. /. x = 358-1. V Example 10. If C=^, a formula relating to electric currents, find the value of C, a current, when the voltage V is 2-41 and the resistance R is 28-7. Substituting the values of V and R .'. log C = log 2-41 log 28-7 = 3820- 1-4579 A pproximation. 2J4 3 ! i. e., -8 -7- 10 or -08 = 2-9241, since 2 subtracted from = = log -08397 /. C = -08397- AIDS TO CALCULATION Example n. If F, the centrifugal force on a rotating body, = , find its value when W = 28, v = 4-75, g=32-2, r= 1-875. Substituting the numerical values in place of the letters F _ 28 x (4-75) a 32-2 x 1-875 Taking logs throughout log F = (log 28 + 2 log 4-75) - (log 32-2 + log i -8 75) = I> 447 2 _ i '579 1-3534 2-8006 = 2-8006 = 1-0197 = log 10-47 ' F- 10-47. 1-7809 1-7809 Approximation.. 3x5x5 \ 3x2 \ i. e., 12-5. Explanation. log 475= '6767 .\2xlog 4-75 = 1-3534- j? Example 12. If/=~r, an equation giving the acceleration pro- duced by a force P acting on a weight W, find / when g = 32-2, P = 5-934, and m = 487. Substituting the numerical values , 32-2 x 5-934 487 Taking logs log/= (log 32 -2 + log 5-934) -log 4' 8 7 = (i'5079 +7734) -2-6875 = 2-2813 2-6875 = T-5938 = log -3924 /. / = -3924. Approximation. 3x6 \_ 5 \1 i. e., 3-6 -^ 10 or -36. Example 13. Find the value of -* 001872 Let _ -05229 ~~ -001872 then log x = log -05229 log -001872 = 2-7184 3-2723 = 1-4461 = log 27-94 .'. ^=27-94. A pproximation. 5 Ul 2 \\ i. e., 2-5 X 10 or 25. Note. In the subtraction the minus 3 becomes plus 3 (changing the bottom sign and adding algebraically) ; and this, combined with minus 2, gives plus i. 20 MATHEMATICS FOR ENGINEERS Example 14. Find the value of the expression s = QII54 47-61 x -0000753 Taking logs log s = log -01154 (log 47-61 + log -0000753) = 2 -0622 (l -6777 + 5-8768) = 2-0622-3-5545 = -5077 = log 3-219 /. s = 3-219. A pproximation. 4-8 x 7-5 \\\ i. e., -033 x 100 or 3.3. Note. In this subtraction the i borrowed for the 5 from o should be repaid by subtracting it from the 2, making it 3 : this, combined with + 3 (the sign being changed for subtraction), gives o as a result. Alternatively, the 3 must be increased by i to repay the borrowing, so that it becomes 2 ; and 2 subtracted from 2 gives o. Example 15. The formula V = -X3-i42xr 3 gives the volume of a sphere of radius r. Find the volume when the radius r is -56. Substituting the numerical values = X 3-142 x Taking logs log V = (log 4 + log 3-142 + 3 log -56) -log 3 = (-6021 + -4972 + 1-2446) - -4771 = '3439 -'4771 = 1-8668 = log -7358 .'. V = -7358. Approximation. 4x3x6x6x6 3 Hi i. e., 864 -f- 1000 or -864. Explanation. log -56 = 1-7482 3 x log -56 = 1-2446 i. e., there is + 2 to carry from the multiplication of the mantissa and this, to- gether with 3 which is ob- tained when T is multiplied by 3, gives T. Example 16. Find the fifth root of -009185. Let x= V -009185 = (-009185)^ then log x = \ log -009185 = \ x 3^9630 * = Jx {5 + 2 -9630} = 1-5926 = log -3913 .-. *= -3913. * We must not divide 5 into 3-9630 because the 3 is minus, whilst the -9630 is plus ; but the addition of 2 to the whole number and of + 2 to the mantissa will permit the division of each part separately, while not affecting the value of the quantity as a whole. AIDS TO CALCULATION 21 Example 17. Evaluate (2 1 64)* x (-001762)4 x (49'i8)f Let the whole fraction = x. Then log x = {3 log -2164 + \ log 745-4} (J log -001762 + f log 49-18}. Explanation. log -2164 = 1-3353 3 x log -2164 = 2-0059 lo 745 '4 = 2 -8724 jxlog 745-4=1-4362 log -001762 = 3-2460 \ x log -001762 = 1-5410 log 49-18 =1-6918 2 X log 49-18 =3-3836 .-. ^=-1677. .\fxlog49-i8 = -6767. = (2-0059+1-4362) - (1-5410+ -6767) = 1-4421 --2177 = 1-2244 = log -1677 Example 18. If x= \/ -- - - 5 find the value of x. v 9004 x -0050 Taking logs log x = {(log 29-17 + log -1245) - (log 9004 + log -0856)} = & (I-4649+ 1-0951) ~ 03-9544 + 2-9325)} = (-5600 - 2-8869) = t(3-673i) = i(8 + 5-673I). = 1-7091 = log -5118 .-. x = -5118. The following examples are worked by the slide rule. Example 19. Find the buckling stress P for a column of length /; - -- from the formula P = when k z = -575, c = and / = 180. Substituting these numerical values 48000 30000 -575, The second term of the denominator must be worked apart from the rest Thus, to evaluate 4X 180x180 eed as f oUows _ 30000 x -5 75 Actual figuring, found from the slide rule, Approximation. is 751, so that, in accordance with the 4x2x2 \\\\1 approximation the value of this term is 3x6 \\\\ i. e., -88 x 10 or 8-8. 22 MATHEMATICS FOR ENGINEERS Example 20. Find the value of E, Young's modulus for steel, W/ ( I 2 z~\ from the formula E = ~~c \(Tr + jr\ which expresses the result of a bending test on a girder. Given that A = -924 / = 60 W = 5000 D = -07 I = 11-15. Substituting values F 5 000 x 60 / 60 x 60 5 \ 8 x -07 1 6 x 11-15 -924! = 536000 {53-75 + 5-41} = 31-7 x io 8 . Exercises 3. On the Use of Logarithms and Evaluation of Formulae. Evaluate, using logarithms or the slide rule, Exs. i to 32; using approximations wherever possible. 1. 85-23x6-917 2. 876-4 x -1194x2-356 3. 75-42 x-ooo2835 4. _ 4 5. -005376 x -1009 6 9543 f] 2-896x347-2 g i2-o8x-02H2 08176 81-48 -01299 -0005 1Q 4843^29-85 1 154 x -07648 ' -007503 75132 " -009914x36-42 -9867 x -4693 13 36-87x2-57 '* -0863 x -1842 ' -085 x 13-77 x -05 .. 24-23 x -7529 x -00814 ._ -572 x -0086 3000 x-o 115x45 -2 7" -4539 x -0037 x -059 16. V94-03 17. (-0517)3 18. 19. "IX (-00 1 769) 3 20. v (-1182)3 21. 22 ' ^fx^ 23 ' ( ' 253)3X *^ (-0648) 2 xV2 -753 2g ^94-72x853^9" ('275) s ' (-2347) x 5x10* OR (9i'56) 2 x(3-i8 4 )l (4-72) 3 x V26-43 (2- 3 ) 2 xV8^ 28. V 7 ' 008 ?^;, 0372 29. 42-3x1-05 /-O5Oo6\ 3 _. V -6463 x (-086) ' (27-63) 2 x log 10 3-476 * V('4349) B x 5-007x1 33. The formula V = nr z l gives the volume of a cylinder. If r = = 3-142, / = 12-76 find V, AIDS TO CALCULATION 23 34. Given that L = . Find L when =11-7, ^B = 175-5. l z a 35. If R = 7-+ find R when / = 5-1 and a = 0-87. oa 2 J ' 36. The velocity ratio of a differential pulley block is found from the formula 2.d VR = -s L- (where d lt d t and d a are the diameters of the pulleys) "a~~ a a Find VR when ^!= 14-57, ^ 2 = 5'72, ^3 = 4-83. 37. If v = u+ft and 5 = ^+i// 2 , find values of v and 5 when M = 350, / = 27, and t = 4-8. 38. Find a velocity, v, from when g= 32-2, <Z= 0-84, A = 30, /= 5000. 39. If p= (-7854 x *~) + d, find its value when f s = $, t = o-J5 ft=6, d = i -04 ; p is the pitch of rivets, of diameter d, joining plates of thickness /. 40. Find the weight of a roof principal from Merriman's formula W = alii-i ) when a=io and /=8o. 41. To compare the cost of lighting by gas and electricity the b ~~ f following rule is often used, a = -5 where a = price of i Board of Trade (B.O.T.) unit in pence ; b = price per 1000 cu. ft. of gas in pence; d= watts per candle power (C.P.) ; e = candles per cu. ft. of gas per hour; c = cost in pence of lamp renewals per 1000 candle hours. Find the equivalent cost per electric unit when lamps take 2-5 watts per C.P., e = 2, c = i and gas is 25. zd. per 1000 cu. ft. 42. The length / of a trolley wire for a span L when the sag is d 8d 2 is given by the formula /= L+ -^-. Find /, when L = 500, d= 12, 3^ 43. The input of an electric motor, in H.P., is measured by the product of the amperes and the volts divided by 746. What is the input in the case where 8-72 amps, are supplied at a pressure of 112-5 volts ? If the efficiency of the motor at this load is 45 %, what is its output ? (Output = efficiency x input.) 44. 2-4 Ibs. of iron are heated from 60 F. to 1200 F. The specific heat of iron being -13, find the number of British Thermal Units (B.Th.U.) required for this, given B.Th.U. = weight x rise in temp, x specific heat. 45. The following rules for the rating of motor-cars have been stated at various times. (a) By Messrs. Rolls Royce, Ltd. where d = diameter of cylinder in inches, N = no. of cylinders S = stroke in inches. MATHEMATICS FOR ENGINEERS (6) By the Royal Automobile Club H.P. = -i97d(d- i)(r+ 2)N where N and d have the same meanings as before, and r = ratio of stroke to diameter. Find the rating of a 4-cylinder engine, whose cylinders are of 4" diameter, and stroke 8-6" ; by the use of each of the rules. 46. 130 grms. of copper (W) at 95 C. (T) are mixed with 160 grms. (w) of water at ioC. (t), the final temperature (t^ being iGC. Calculate the specific heat (s) of copper from <*<- ~ 47. The volume t; of a gas at a temperature of oC., or 273 C. absolute, and at a pressure corresponding to 760 mms. of mercury is 17-83 cu. ins. Find its volume at temperature t C. and pressure H where t = 83-7 and H = 797 from the formula V - 48. If P= find its value when 6 = /=!. =8, L=I2, y = 80000. L = length of a railway spring on each side of the buckle, n = number of leaves, / = thickness of leaves, /= working stress, P = load applied and b = width of leaves. 49. The increase in length of a steel girder due to rise of temperature can be found from the formula, new length = old length (i + at) when t rise in temperature, and a = coefficient of linear expansion. Find the increase in length of a girder of 80 ft. span due to change of temperature of 150 F. when a= -000006. 50. If c = *fo V(i ip + id) (p + id), find its value when p = 3", d = } *. The meaning of c will be understood by reference to the riveted joint shown in Fig. 2. 51. Find the thickness (tj of a butt strap from the B.O.T. rule p-d\. when p = 4f *, / = !", d = \%'. 52. Find the thickness (t) of a pipe for pressure p Ibs. per sq. in., when internal dia. = d, from d=2, =45- 53. Taking p = which gives the principal (or maximum) stress p due to a normal stress /and a shearing stress s : determine p when /= 3800, s= 2600. 54. If P= Fig. 2. Riveted Joint. (Gordon's formula for the buckling load 1500 on struts), find P when F = 38, d = 15, T = AIDS TO CALCULATION 25 55. If p= i ,s 2 (Rankine's formula for the buckling load on struts), find p when / = 48000, / = 14 \ x 12, c = -- , A 2 = 30-7. 56. The deflection d of a helical spring can be obtained from , _ 6^wnr 3 '' CD* ' Find the deflection for the case in which w= 48, D = J n= 12-73, r=i-5. C= 12 X IO 6 . 57. If the deflection d of a beam of radius a and length J, due to a load of W is measured, Young's Modulus for the material of which the 4\V7 3 beam is composed, can be found from E= , .. If in a certain case 37ra 4 the deflection was 4-2; and W, /, a and IT had the values 14-8, 17-56, 39 and 3-142 respectively, find the value of E. 58. For oval furnaces, if A = difference between the half axes before straining. 8 = . , after p = pressure in Ibs. per sq. in. E = Young's Modulus. D = diameter of furnace. AX32EI - - - ==rf - 32EI Find the value of 8 when A = -5, D = 40, p= 100, E = 30 x io and I = -0104. 59. The modulus of rigidity C of a wire of length / and diameter d may be found by attaching weights of m { and w 2 respectively at the end of the wire and noting the times, t l and t z respectively, taken for a complete swing. The formula used in the calculation is p _ I28irl(m 1 W 2 ) 2 Find C when w^g-S, m t = !$, t 1 = 2-i, /, = 1*6, =32, d= -126, I = 4-83, a = -97 and ?r = 3-142. 60. The weight W in tons of a flywheel is given by _ R 3 N 3 Find the weight when R = - , r = -2, n = 30, N = 120, H = 70. 61. The approximate diameter of wire (in inches) to carry a given current C with 6 rise in temperature can be obtained from Find D when J = -0935, & = 35. P = ^ c = 55. m = >O02 5 ir = 3-142. 62. Find the dimensions for the flanged cast-iron pipe shown in Fig. 3 (in each case to the nearest ^th of an inch), when P = 85, PD . d= , = F . There are n bolts and n = -6D+?, 26 MATHEMATICS FOR ENGINEERS Investigation for Units. A train covers a distance of 150 miles in 5 hours ; what is its average speed ? ^, . , ... 150 miles . 30 miles Obviously it is , i. e., , - or 30 miles per hour. 5 hours ' i hour It could with equal truth be expressed by *-* * , J 5 X 60 X 60 sees. i. e. t ', or 44 ft. per second. The figures in the results differ ' i sec. because they are measured in terms of different quantities, and it is essential that the units in which results are expressed should be clearly stated. Here we have another form of investigation to be performed before the actual numerical working is attempted. To find the units in which the result is to be expressed, these units, with their proper powers attached, are put down in the form of a fraction, all figures and constants being disregarded, and are treated for can- celling purposes as though they were pure algebraic symbols. Suppose a force of 100 Ibs. weight is exerted through a distance of 15 ft., then the work done by this force is 100 X 15 or 1500 units : these units will be "foot Ibs." since the result is obtained by multi- plication of Ibs. by feet. This statement might be written in the form, Work = Ibs. X feet = foot Ibs. If now we are told that the time taken over the movement was 12 minutes we can determine the average rate at which the work was done. The work done in i minute is evidently obtained by dividing the total work done in 12 minutes by the number of minutes : thus, rate of working = - = 125. This figure gives the number of foot Ibs. of work done in one minute, and the result would be expressed as, average rate of working = 125 foot Ibs. per minute. It will be seen from this and from the previous illustration that the word per implies division. To obtain a velocity in miles per hour, the distance covered, in miles, must be divided by the number of hours taken, or . number of miles , ,, miles velocity (miles per hour) - number of hours or, more shortly, j^. An acceleration = rate of change of speed = feet per second added every second (say) feet feet QJ* sees. X sees. (sees.) 2 Hence, wherever an acceleration occurs it must be written as distance . ,, . ,. ,. f ., -T-. TJ- in the investigation for units. AIDS TO CALCULATION 27 The " g " so frequently met with in engineering formulae is an acceleration, being 32-2 ft. per sec. per sec. or 32-2 -. '-^, and (sees.) 4 therefore must be treated as such wherever it occurs. Example 21. The steam pressure, as recorded by a gauge, is 65 Ibs. per sq. in. ; the area of the piston on which the steam is acting is 87 sq. ins. What is the total pressure on the piston ? Total pressure = area x intensity of pressure and is Jprgfy. Ibs *. e., is in Ibs. The true pressure is 65 + 14-7 Ibs. per sq. in., because the gauge records the excess over the atmospheric pressure .*. total pressure = 79-7 x 87 Ibs. = 60 Ibs. = 6930 Ibs. Example 22. Find the force necessary to accelerate a mass of 10 tons by 12 ft. per sec. in a minute. The formula connecting these w/ quantities is P = - where W = weight, / = acceleration and g has its usual meaning. Dealing merely with the units given, and forming our investigation for units _ (sees.) 8 feet ~~ X sees, x mins. It will be seen that no cancelling can be done until the minutes are brought to seconds ; then we have P=Tons x To find the force, therefore, the minutes must be multiplied by 60; t. e., the denominator must be multiplied by 60. Hence P= 10 x x |^ = -0621 ton or 139-2 Ibs. 32-2 oo ^ - ^^ Example 23. The modulus of rigidity C of a wire can be found by noting the time of a complete swing of the pendulum shown in Fig. 4 and then calculating from the formula, C = * , where / is O the length of the wire, d is its diameter, I is the moment of inertia 28 MATHEMATICS FOR ENGINEERS of the brass rod about the axis of suspension and t is the time of one swing. If / and d are measured in inches, / in seconds, and I in Ibs. x (feet) 2 [I being of the nature of mass x (distance) 2 ], in what units will C be expressed ? Investigating for units I287T / I C = constant x ins. x Ibs. x feet 2 x feet ins. 4 sec. _ constant x Ibs. x feet ins. 3 If the numerator is multiplied by 12, then _ constant x Ibs. x ins. _ Ibs. ins 2 . ~~ ins.* or the result would be expressed in Ibs. per sq. in. provided that the numerator was multiplied by 12. Example 24. The head lost in a pipe due to friction is given by the I v 2 formula h = -03 . -r . . Find its value if the pipe is 3" dia., 56 yards long, and the velocity of flow is 28 yards per min. The meanings of the various letters will be better understood by reference to Fig. 5. Dealing only with the units given, and disregarding the constants i yards 2 sees. 8 Head lost = yards x -. - X . - 9 x -7 ins. rmns. 2 feet This is not in a form convenient for cancelling; accordingly, bring all distances to feet and all times to seconds. i feet 2 sees. 2 feet sees. 2 feet Then the head = feet. Fig. 5. Flow of water through a pipe. Substituting the numerical values in place of the symbols- Head lost = *= ^3 x 56 x_ 3 x X2 x 28 x 3 x 28 x 3 = -612 foot. 3 x 60 x 60 X 64-4 AIDS TO CALCULATION 29 Example 25. Find the maximum deflection of a beam 24 ft. long, simply supported at its ends and loaded with 7 tons at the centre. The moment of inertia I of the section is 87-2 ins. 4 units, and Young's Modulus E for the material is 30 x io 6 Ibs. per sq. in. W/ 3 The maximum deflection = orT -, 40! > The investigation for units, as given, reads : i I W / 3 T E Deflection = tons x feet 3 x = -. x TC - ins. 4 IDS. No cancelling can be attempted until the tons are brought to Ibs. and the feet to inches or vice versa ; assuming the former, then i ins ^ Deflection = Ibs. x ins. 3 x -. . x ^^-^ = ins. ms. 4 Ibs. So that, since 7 tons = 7 x 2240 Ibs. and 24 ft. = 288 ins. 7 x 2240 x 288 3 Deflection = 5* ~ ^ ins. 48 x 87-2 x 30 x io 6 Calculation log d = (log 7+log 2240 + 3 log 288) (log 48+ log 87-2+ log 30,000,000) = (-845 1 + 3-3502 + 7-3782) - (1-6812 + 1-9405 + 7'477 J ) = 11.5735- 11-0988 Explanation- '4747 = log 2-984 log 288 = 2-4594 .*. deflection = 2-984 ins. A pproximation 7x2x3x3x3 m 5x9x3 UUVWU or 2-8. 3 x log 288 = 7-3782. Exercises 4. On the Finding of Units. 6E 1. In what units will / be expressed if / = -^ and 8 is in inches, E in Ibs. per sq. in. and D in ins. ? 2. If a H.P. = 33000 foot Ibs. of work per minute, find the H.P. necessary to raise 300 cwts. of water through a vertical height of i6 yards in half an hour. f2 3. If H=-^-g: find H in yards when / = 18 tons per sq. in.; E = 13000 per sq. in. ; p = 480 Ibs. per cu. ft. 4. Determine the stress / in a boiler plate in tons per sq. in. from bd f = ~ when / = -63 in., d = 8 feet, p = 160 Ibs. per sq. in. t is the thickness of plate, p is the pressure inside the boiler, and d is the diameter of the boiler. 30 MATHEMATICS FOR ENGINEERS 5. The jump H of the wheels of a gun is given by _ = Find the jump in inches when A = 40 ins., A=io ft., h=i yd., P=475 cwts., R= 1-15 tons, M= 1-2 tons, W = g cwts. 6. The tension in a belt due to centrifugal action can be calculated from T= . If w = wi. per foot run of belt in Ibs., u = veloc. in 8 ft. per sec., and g has its usual value, in what units will T be expressed ? 7. If, in the previous example, w = -43 Ib. per foot length of belt per sq. in. of surface, find a simple relation between the stress (in Ibs. per sq. in.) and the velocity (ft./sec.). 8. The I.H.P. of an engine is determined from the formula IHp=2 PLAN 33000 where P = mean effective pressure in Ibs. per sq. in., L = stroke in feet, A = area of piston in sq. ins., and N = revolutions per minute. If / is the stroke in ins. and A = '7854D 2 show that this equation may be written I.H.P. = - approximately. 1,000,000 J tw^ 9. Given that / = - , where w = weight in Ibs. per cu. in., v = veloc. in feet per sec. (a formula relating to tensile stress in revolving bodies). Arrange the formula so that / is given in Ibs. per sq. in. 10. Investigate for units answer in the following formula for the Horse Power transmitted by a shaft. H.P. = - - where R is inches, N is Revolutions per minute, 33000 IT is a constant, and p is in Ibs. per sq. in. If these are not found to be H.P. units, viz. foot Ibs. per minute, state what correction should be made. 11. The formula Q = a t a, \f -f^ *-f{ gives the quantity of water P\ a i ~ a t i passing through a Venturi Meter. In what units will Q be expressed if a t and a, are in sq. ft. ; p^ and p 2 in Ibs. per sq. ft. ; g in feet per sec. per sec. ; p in Ibs. per cu. ft. ? 12. Given that i Ib. = 454 grms., i*= 2-54 cms. i erg. = work done when i dyne acts through i cm. i grm. weight = 981 dynes. and i watt = io 7 ergs per sec.; find the number of watts per H.P. CHAPTER II EQUATIONS Simple Equations. A simple equation consists of a statement connecting an unknown quantity with others that are known; and the process of " Solving the equation " is that of finding the particular value of the unknown that satisfies the statement. To many, this chapter, on the methods of solving equations and of transposing formulae, must be as important and useful as any in the book, for it is impossible to proceed very far without a working knowledge of the ready manipulation of formulae. The methods of procedure always followed is the isolation of the unknown, involving the transposition of the known quantities, which may be either letters or numbers, from one side of the equation to the other. The transposition may be of either (a) terms or (b) factors ; and the rule for each change will now be developed. To deal first with the transposition of terms : When turning the spindle shown in Fig. 6 it was necessary to calculate the length of the " plain turned " portion, or the length marked I in the diagram. The conditions here are that the required length, together with the radius *375" must add to 1-5". A statement of conditions may thus be made, "in the form *+ '375 = i-5- The truth of this statement will be unaltered if the same quantity, viz. -375, is subtracted from each side, so that I + '375 - '375 = i'5 - '375 or / = 1-5 - -375 = 1-125. Thus, in changing the -375 from one side of the equation to the other, the sign before it has been changed ; + -375 on the one side becoming -375 when transferred to the other side. Again, suppose the excess of the pressure within a cylinder over that of the atmosphere (taken as 147 Ibs. per sq. in.) is 86-2 32 MATHEMATICS FOR ENGINEERS Ibs. per sq. in., and we require to determine the absolute pressure in the cylinder. Let p represent the absolute pressure, i. e., the excess over zero pressure. Then p 14-7 = 86-2. To each side add 14-7; then P = 86-2 -f- I4'7 = iQQ'9 Ibs. per sq. in. Thus, 14-7 on the left-hand side becomes + *4'7 when trans- ferred to the right-hand side of the equation. Accordingly, we may say that : When transferring a TERM from one side of an equation to the other, the sign before the term must be changed, plus becoming minus, and vice versa. To deal with the transposition of factors : Suppose we are told that 3 tons of pig iron are bought for 7 los. : we should say at once that the price per ton was \ of 7 ios., or 2 IDS. We might, however, use this case to illustrate one of the most vital rules in connection with transpositions, by expressing the statement in the form of an equation and then solving the equation. The unknown in this case is the price per ton, which may be called p shillings. Our equation then becomes 3x^ = 150 (i) Divide both sides by 3, which is legitimate, since the equation is not changed if exactly the same operation is performed on either side. :. P = ^ = so (2) or the cost is 505. per ton. Again, had we been told that | a ton could be bought for 255. we could express this in the form \P = ^ (3) If we multiply both sides by 2 we find that = 25x2 = 50 . . (4) which, of course, agrees with the above. It will be seen that, to isolate p and so find its absolute value, we transfer the multiplier in equation (i) or the divider in equation (3) to the other side, when its effect is exactly reversed : thus the multiplier 3 hi equation (i) becomes a divider when transferred to the other side of the equation, as in (2) ; and the dividing 2 in equation (3) becomes the multiplying 2 in equation (4). EQUATIONS 33 The motion of a swinging pendulum furnishes an illustration of the transposition of a factor which is preceded by a minus sign. The acceleration of the pendulum towards the centre of the move- ment increases proportionately with the displacement away from the centre. Taking a numerical case, suppose that we wish to find the displacement s when the acceleration / is 4-6 units and the relation between/ and s is/ = 255. Substituting the numerical value for/ 4-6 = 255. To isolate s we must divide both sides of the equation by 25, and then -s " -25 or s = '184 unit. The rule for the transposition of factors can now be stated, viz. To change a FACTOR (i. e., a multiplier or a divisor) from one side of an equation to the other, change also its position regarding the fractional dividing line, viz., let a denominator become a numerator and conversely ; and let the sign of the factor be kept unchanged. We have thus established the elementary rules of term and factor changing in simple equations. The following examples, as illustrations of these fundamental laws, should be most carefully studied, every step being thoroughly grasped before proceeding to another. *\X 7 Example i. Solve for x, in the equation, *-" r*g *4. I *O Transferring the 5 and 4 so that x is by itself, the 5 must change from the top to the bottom and the 4 from the bottom to the top, since 5 and 4 are factors. Then- * = -? 5 x4 = 3-ii. 10 5 'Example 2. Solve for a, in the equation, 40+17 = 2-509. Transposing, to get the unknowns together on one side 4-2-50 = 9-17. Here the change is that of terms, hence the change of signs. Grouping, or collecting the terms i-^a = 26 -26 34 MATHEMATICS FOR ENGINEERS Example 3. The weight of steam required per hour for an engine was a constant 60 Ibs., together with a variable 25 Ibs. for each H.P. developed. If, in a certain case, 210 Ibs. of steam were supplied in an hour, what was the H.P. developed ? Let h represent the unknown H.P. Then 25^ represents the amount of steam for this H.P., apart from the constant, and the equation including the whole of the statement of conditions is 25/1 + 60 = 210. Transferring the term + 60 to the other side, where it becomes 60, 25& = 210 60 = 150. Dividing throughout by 25 h = ^2. = 5 or, the H.P. developed was 6. Example 4. To convert degrees Fahrenheit to degrees Centigrade use is made of the following relation F-3, = |C. Find the number of degrees C., corresponding to 457 F. Substituting for F its numerical value 457-32 = |C ' 5= Transposing factors 5 and 9, 236-1 = C i.e., 236 C. correspond to 457 F. It might happen that in an engine or boiler trial only ther- mometers reading in Centigrade degrees were available, whereas for purposes of calculation it might be necessary to have the tem- peratures expressed in degrees Fahrenheit. This would mean that a number of equations would have to be solved ; but the work involved could be shortened by a suitable transposition of the formula given above. F- 3 2 = C ........... (i) F = c+ 3 2 ......... (2) Equation (2) is far more suitable for our purpose than equation (i), although the change is so slight. EQUATIONS Example 5. Convert 80, 15, 120, and 48 C. to degrees F. When C = 80, F = (% x 80) + 32 = 176, and so on. Or, we might tabulate, for the four readings given, thus : 35 C lC+33 F 80 144 + 32 176 15 27+32 59 I2O 216+ 32 248 48 86-4 + 32 118-4 Example 6. Ohm's law states that the drop in electrical pressure E when a current C flows through a resistance R, is given by the formula E = CR. Transpose this for R and C. To find C, Transposing the factor R, In like manner * f* = CR E R' E C' Brackets occurring in equations must be removed before applying the rules of transposition, and the same remark applies to fractions, which may always be regarded as brackets written in a different form. Example 7. Solve for w in the equation Removing brackets 3W-4-I Dissociating knowns and unknowns w^yjo = 9'3 _ i. 3 w = 27-4 Note that the sign of 1-3 is kept unchanged. Example 8. When finding the latent heat L of steam, the following equation was used Transpose this for L, i. e., find an expression for L in terms of the other letters, which must be regarded as representing known quantities. 36 MATHEMATICS FOR ENGINEERS Here L is the unknown, since the values of all the other letters are supposed to be known. Clearing of brackets, wTwt = qL+^T Transposing terms, wT wt ^-pT = qL r~ wTwtti + T Transposing the factor q, - = L Example 9. Solve the equation 4* 7* 8-1 _ i-gx , 7-21 5~2 + T : ~5 3~' The L.C.M. of 5, 2, 4 and 3 is 60, and multiplication throughout by this figure will remove the denominators. (4* x 12) (7* x 30) + (8-1 x 15) = (1-9* x 12) + (7-21 x 20) 48* 2io#+ 121-5 = 22-8^+144-2 48* 2IOX 22-8* = 144-2121-5 or 184-8* = 22-7 Example 10. The electro-motive force E of a cell was found on open circuit, and also the drop in potential V when a resistance of R was placed in the circuit. The internal resistance of the cell may be V calculated from the equation (E V) = -~ x R where R< is the internal resistance. Find the internal resistance for the case for which E = 1-34, V = -8965 and R = 5. V It being required to find R< we transpose the ^- and treat the bracketed letter as one quantity for the time being ; then *(E-V) = R, which completes the transposition. Substituting the numerical values R<= ^siV 1 ' 34 "' 896 ^ = = 2-47 ohms. Example n. Solve the equation , 5 _ ~ 8 4 16 8 Before proceeding to find the L.C.M. it will be found the safest plan to place brackets round the numerators of the fractions. This EQUATIONS 37 emphasises the fact that the whole of each numerator is to be treated as one quantity. Thus (3y 5) (7^+9) , (Sy+ig) 69 _ rz r Q 1 cf " 4 lo o o Failing this step, mistakes are almost certain to arise, especially with signs, e. g., the minus before the second fraction applies equally to the 9 and to the 7y. This fact would probably be overlooked if the bracket were not inserted. Multiplying throughout by 16, the L.C.M. of 4, 16 and 8 4(3y-5)-(7y-l-9)+2(8y+i9) + (2X 69) = o i.e., I2y 20 7y 9+i6y+38+i38 = o .*. I2y 7y+i6y = 20+938138 2iy = -147 v - - T 47 y ~ ~^i Example 12. If p is the intensity of pressure over an annular plate of outside diameter D and inside diameter d, then the total pressure on the plate is given by Assuming that p, P and D are known, transpose this equation into a form convenient for the calculation of the value of d. Treating the -785^ as one quantity, and transposing it - -r 7854^ Transferring D 8 to the right-hand side P Changing signs throughout -- - 7854 p Taking the square root of both sides Example 13. If / = 2n- /-, giving the time in seconds of I swing (periodic, or to and fro) of a simple pendulum of length / feet ; find^ an expression for /. It will be easiest in this case to square both sides (i. e. t to remove the square root sign which is merely one form of bracket). Then t z = ^ 2 - S ef* or, transposing the factors, 4, n- 2 and g, - a = *. 38 MATHEMATICS FOR ENGINEERS Example 14. Transpose for g, the dryness fraction of steam found by the Barrus test for superheated steam, in the equation 4 8(T A -T B -w) = (i-j)L+- 4 8(T s -T). T A , TB, Tg and T are temperatures, L is the latent heat of the steam, and n = loss of temperature of the superheated steam when the supply of moist steam is cut off. Treating -48^3 T) as a term, it may be transferred to the other side with change of sign before it 4 8(T A -T B -w)-. 4 8(T s -T) = (i-g)L or, since -48 multiplies each bracket, we can take it outside one large bracket 4 8{T A -T B -w-T 8 + T} = (i-ff)L. Dividing both sides by L (i-q) = ^{T A -TB-W-T S +T} q = I- Example 15. The equation - ^- = W(H+e) refers to the stress produced in a bar by a weight W falling through a height H on to the bar. Transpose this equation for / and also for e. To find/: 2E Transposing factors, /* = W x ^r (H+ e) Extracting the square root of both sides of the equation. / 2 EW(H+e) J ** V AL To find e 2EW / 2AL 2EW Example 16. One hundred electric glow lamps, each of 150 ohms resistance and each requiring -75 ampere, are connected in parallel. How many cells, each of -0052 ohm resistance and giving 2-08 volts, will be required to light these lamps ? (Cells to be in series.) Total resistance = Internal resistance + external resistance. External resistance = -- = i-<5 ohms (because lamps in parallel 100 offer less resistance, i. e., an easier path is made for the current). EQUATIONS 39 Suppose x cells are required Total E.M.F. = #x 2-08 Total internal resistance = x x -0052 Total resistance = -0052*+ 1-5 Current = * M *' ' Resistance and 100 x -75 = - 0052* + 1-5* Multiplying across, i. e., multiplying throughout by the common denominator -0052*+ 1-5. 75 (-0052*+ 1-5) = z-o8x *39# + 112-5 = 2-08* 112-5 = 2-o8x '39,v = 1-69,1; .-. x = 66-6 Or 67 cells would suffice. Exercises 5. On Simple Equations and Transpositions. Solve the equations in Exs. i to 6 1- 5*+7(*~2) = 3-4(*+ 6 ) o I , 2 2fl 2. g+-- 3 = 5-- 3 4'2ft = 9-58 ' 7-45 4-69 . . 52 -y 3^5 37-5 = 1-08 2-95 "*" 9-n 6. 8-2^ -475(3 -2x) + 2-14(5^ +7) = 17 -(i- -8^) + 5'43 7. Transpose for c in the equation - = 7 5c 8. If H = ws(T t), find an expression for T. 9. If P = CTAE, find E when A = 19-25, C = -000006, T = 442, P = 1,532,000. 10. Transpose for L, the latent heat of steam, in the equation w i(ti~ T+L) = w(T t), and hence find its value when Wi = -*, / x = 212, w = i J, T = 145, and t = 70. 11. A formula occurring in connection with Tacheometric Surveying fS is D =-+/+(?. Determine the value of S to satisfy this when D = 3600, /= 12, d 6 and S = 36. 12. Using the equation in Exercise n, find the value of /to satisfy it when = 310-7, S = 4'63, 8 =-015, and ^=-5. 13. If w = ^^-V find S when w = 8 ' 15 ' l = 5 ' W ** 83 ' 5> d = 4> and c = 1400. w is the weight of a girder in tons to carry an external load 40 MATHEMATICS FOR ENGINEERS W tons, d is the effective depth of the girder in feet, s is the shearing stress in tons per sq. in., and c is a coefficient depending on the type of girder. 14. If TM i-\ ) = 7^, find E in terms of C for the case when m = 4. HA m/ L/ In other words, find the relation between Young's modulus and the Rigidity modulus when " Poisson's ratio " is 4. V 15. Find the value of R,- from E V= ~ xRj when E= 136-4, V = 97-9, R = 5- The letters have the same meanings as in Example 10, page 36. 16. Given that A= - -- 1^ -- -, transpose for R and hence find its value when A = 35, D = 6-5, d= 4-7, a= -25. 17. The equation - ^ ^ = (i8*+2x6^x^)X5X4& occurred when finding the thickness of the flange of the section of a girder for an overhead railway. Find the value of t to satisfy this. 18. Transpose for q in the equation W(A, A x ) = w(qL+h h t ). [q is the dryness fraction of a sample of steam.] 19. How many electric cells, each having an internal resistance of i '8 ohms, and each giving 2 volts, must be connected up in series so that a current of -686 amperes may be passed through an external resistance of 12-2 ohms? SC 20. If D= +K, find n when D = 500, S = 12, C = 950, and K= 1-5. 21. The tractive pull P that a two-cylinder locomotive can exert is given by 8pd*L ~D~ where p = steam pressure in Ibs. per sq. in., d = diameter of cylinders in ins., L = stroke in ins., and D = diameter of driving wheels in inches. Find the diameter of the cylinders of the engine for which the pull is 19,000 Ibs., the steam pressure 200 Ibs. per sq. in., the stroke 2 x -3", and driving wheels are 4'-6" in diameter. 22. To determine the diameter of a crank the following rule is used P/ n Put this equation in a form convenient for the calculation of the value of d. 23. Lloyd's rule for the strength of girders supporting the top of chH the combustion chamber of a boiler is P = 7vfr T\TVF where P = (W pjDL, working pressure in Ibs. per sq. in.; t= thickness of girder at the centre ; L = width between tube plates ; p = pitch of stays ; h = depth of girder at the centre ; and D = distance from centre to centre of the girders. Find the value of p when c= 825, W= 27, L= 2j, D= 7^, / = i J, h= 61, and P= 160. EQUATIONS 41 24. Find the thickness of metal t (ins.) for Morrison's furnace tube from each of the given formulae (a) Board of Trade rule p _ 14000* D (6) Lloyd's rule p _ 1259(161-2) D where P= pressure in Ibs. per sq. in., and D = diameter (ins.) outside corrugations. Given that P= 160 and D = 43*. 25. (a) Transpose the given equation for A where 8 = proof strain of iron, a = area of section of bar of length / on to which a weight W is dropped from a height h inches ; A being the extension produced. (b) Find the value of /, which equals -?, when =30x10*; a= i'2, 3 = -ooi, h 132, and W= 40. 26. If / = \ \ i T ' find the value of L when W = 7000, A = 8000, and t= 1-62. 27. Find the pitch p of the rivets in a single-riveted lap joint from 7 8 54 d 2 /, = (p - d)tf t where d = t + fa t= J, /, = 23, and f t = 28. 28. Calculate the value of p to satisfy the equation B= CVpA. when C= -02, A= 200, B= 2-53. 29. The stress / in the material of a cylinder for a steam-engine may be found from t~, + \ where p = steam pressure = 80 Ibs. per sq. in., d = diameter = 14", and t= thickness of metal = ". Find /for this case. 30. Determine the value of p to satisfy the equation (pd)tf t = i'57irf 2 /,, relating to riveted joints, when/, = 23, ft =28, d= i, and /= f. 31. The diameter of shaft to transmit a torque T when the stress allowable is / is found from T = ^/^ 3 - Find the diameter of shaft to transmit a torque of 22,000 Ibs. ft., if the maximum permissible stress in the material is 5000 Ibs. per sq. in. (IT = 3-142). 32. The formula d V/ , / r~\ occurs in reinforced concrete v xbc(i-\x) design. Find M (a bending moment) when b g,c= 600, x -36, ^=15-3. 33. D = dY-- is Lamp's formula for thick cylinders of outside JP diameter D and inside diameter d. Calculate the value of p when D= 9-5*, d= 6", and /= 6 tons per sq. in. M E 34. An important formula in structural work is y = R where M is the bending moment applied to a beam, I is the moment of inertia of the section of the beam, E is Young's modulus for the beam, and 42 MATHEMATICS FOR ENGINEERS R is the radius of curvature of the bent beam. If M= 5600 Ibs. ft., I =-7854 in.* units, E =28x10* Ibs. per sq. in.; find the value of R, stating clearly the units in which it is expressed. 35. Compare the deflection d m of a beam due to bending moment with that d, due to shear, for the following cases (a) length = 10 x depth, i. e. t I = lod. b) length = 3 X depth. You are given that , W/ 3 i- 5 W/ .. d* , _ _ d m = Q ~. , ,, d, = \-x-, ft , andE = 2'5U. 48EAA 2 4AC 12 oo rr mc k 2tr ,. 36. If = - r and c = -r-, find an expression for r in terms of m I I n R and k : hence find its value when = -36, m = 15. 37. If E = 3K(i--) and E = 2C(i+i), find the relation between K, the bulk modulus, and C, the rigidity modulus. Find also an expression for E, Young's modulus, in terms of K and C only. 38. Find an expression for x from the equation i6W* _ i6W(r-x) 2W ird 3 Trd 3 + ird* 39. Find the internal pressure p for a thick cylinder from Lamp's formula where D= 12-74*. d = 9* /= 2IO Ibs. /D". State the units in which P is expressed. n T T 40. Given that W= - pv ~ *, a formula occurring in Ther- ** X J. i modynamics, and also that \V= 33000, T, is fTj, T!= 2190, v= 12-4, and p =2160, find the value of n. 41. A takes 2 hours longer than B to travel 60 miles ; but if he trebles his pace he takes 2 hours less than B. Find their rates of walking. 42. If H=4=^r and /= 0-01(1 + -^,), find v when H = 22-i, d= i, 2gd I2d' 1= 380, and g= 32. (H is the head lost when water flows through a length / of pipe of diameter d, and / is the coefficient of resistance.) 43. If M = moment of a magnet, H = strength of the earth's field, p = time of a complete oscillation of the magnet, and I = moment M d 3 T of inertia of the magnet, then Q = (expressing the result of a deflection experiment, d being the distance between the centre of the magnet and that of the needle, and T being a measure of the deflection) 47T 2 I and also MH = 3L (expressing the result of an oscillation experiment). Find the values of M and H when ^=20, I = 169, p = 13-3, TT = 3-142, and T= -325. 44. In finding the swing radius k (ins.) of a connecting rod, the following measurements were made : EQUATIONS 43 / = lime of a complete oscillation = 2-03 sees. p = distance of centre of gravity from the centre of suspension = 3I-43"- If h = distance of centre of percussion from centre of suspension t=27r*/-; and also k z = ph. Find k in inches {n- = 3-142, g= 32-2 f.p. sec. 2 }. 45. The maximum stress in a connecting rod can be found from the , D*p , vW equation /= i -05-^ + -00429 ^ If /= 4700, D = diameter of cylinder = 14, d= diameter of rod = 2-5, p = steam pressure at mid stroke = 65, v = velocity of crank, r = crank radius = 8, and / = length of connecting rod = 60, find the value of v. 46. It is required to find the diameter D of one pipe of length L, equivalent to pipes of length /j and l z and diameters d^ and d t respec- tively, from I, I, J, D 5 ~ d? + J? Put this equation in a form suitable for this calculation. /' \" / i 47. If - - = I ^j find an expression for y. _ y o^ 48. Transpose the equation -, - = , 2 , -; occurring in n y 2/t -\~2Cn C structural design, to give an expression for y. Simultaneous Equations. So long as only one of the quanti- ties with which we are dealing is unknown, one equation, or one statement of equality, is sufficient to determine its value. Cases often present themselves in which two, and in rarer cases three or even more, quantities are unknown ; then the equations formed from the conditions are termed simultaneous equations. Taking the more common case of two unknowns, one equation would not determine absolutely the value of either, but would simply connect the two, i. e., would give the value of one in terms of the other. For two unknowns we must have two sets of con- ditions or two equations. This rule holds throughout, that for complete solution there must be as many equations as there are unknowns. The treatment of such equations will be best understood by the aid of worked examples. Example 17. What two numbers add up to 5-4 and differ by 2-6? For shortness, take x and y to represent the numbers ; substituting these to form an equation to satisfy the first condition 5'4 ............ ( J ) 44 MATHEMATICS FOR ENGINEERS Here, by taking various values of y we could calculate corresponding values of x, and there would be no limit to the number of " solutions." The first statement in the question is, however, qualified by the second, from which we form equation (2), viz. x y=2-6 (2) If equations (i) and (2) are added 2X = 8-0 .'. *=4; or, in other words, y has been eliminated, i. e., the number of unknowns has been reduced by one. Our plan must therefore be to " eliminate,'' by some means, one unknown at a time until all become " knowns." This method will be followed in all cases. Reverting to our example, x is found, but y is still unknown. To find y, substitute the value found for x in either equation (i) or equation (2). In (i) 4+y = 5-4 and y = 5'4~4 1*4 x = 4-0 \ y = i'4J and we have completely solved our problem. Example 18. Determine values of a and 6 to satisfy the equations 40 + 36 =43 (i) 3 26 =11 (2) If equations (i) and (2), as they stand, were either added or sub- tracted, both a and b would remain, so that we should be no nearer a solution. To eliminate a, say, we must make the coefficients of a the same in both lines. E. g., if equation (i) be multiplied by 3 and equation (2) be multiplied by 4, each line would contain I2a, so that the subtraction of the equations would cause a to vanish. Thus 120+96 = 129 I2 86 = 44 Subtracting 176 = 85 whence 6=5. Substituting this value for b in equation (2) 3a 10 = ii 3 = 21 a = 7 Grouping the results Note. If it were desired to eliminate 6, equation (i) would have to be multiplied by 2 and equation (2) by 3, and the resulting equations added, since there would then be +66 in the top line and 66 below, which on addition would cancel one another. EQUATIONS 45 Example 19. The effort E, to raise a weight W, by means of a screw jack, is given by the general formula, E=aW+6. If =2-5 when W= 5; and if =5-5 when W= 20, find the values of a and b, and thence the particular equation connecting E and W. Substituting the numerical values for E and W 2-5= 5 a +*> ........... (i) In this case it is easier to subtract straight away ; thus eliminating b. Thus 3= i 5 a or Substituting in equation (i), 2-5 = i + b so that E= -2W+I-5. Example 20. Keeping the length of an electric arc constant and varying the resistance of the circuit, the values of the volts V and amperes A were taken. These are connected by the general equation _ n Find the value of m and n for the following case V = 54-5 when A = 4 ~\ V = 48-8 when A = ioj Substituting the numerical values, in the general equation , n 54-5 = w+- 48-8 = m+^ Changing the fractions into decimals to simplify the calculation 54-5 =w+*25n (i) 48-8 = m+'in (2) Subtracting 5-7 = -15^ M = f^ = 38 Substituting this value in equation (2) 48-8 = w+3-8 m = 45 or V Example 21. Karmarsch's rule states that the total strength Pof a wire in Ibs. is given by P= ad+bd z , where d is the diameter in inches. For copper (unannealed) P=42i when d= P= 55212 when d= i Find the actual law connecting P and d. 46 MATHEMATICS FOR ENGINEERS By substitution of the numerical values 55212= (ax 1-2)+ (6 xi -44) (i) 421= (ax -i)+(6x -01) (2) To eliminate a multiply equation (2) by 12 and subtract. Thus 55212 = i-2a+i'44& 5052 = i-2a+ -126. Subtracting 50160 = 1-326 6 = 5i6o 1-32 Substituting in equation (2) 421 = -ia+38o ia = 41 /. a 410. P= 410^ + 380000? i. e., for a diameter of -5", the total strength is (410 x -5) + (38000 x -25) = 9705 Ibs. Solution of Equations involving three unknowns. These may also be solved by the process of elimination, the method being similar to that employed when there are two unknowns only. Three equations are necessary and these may be taken together in pairs, the same quantity being eliminated from each pair, whence the question resolves itself into a problem having two equations and two unknowns. Example 22. Find the values of a, b and c to satisfy the equations 4~5 6 +7 c = -14 (i) 90+2&+3C = 47 (2) a-b-y = ii (3) The unknowns must be eliminated one at a time. Suppose we decide to commence with the elimination of c. This may be done by taking equation (i) and equation (2) together, multiplying equation (i) by 3 and equation (2) by 7, and then subtracting; an equation con- taining a and 6 only being thus obtained. For complete solution one other equation must be found to combine with this ; if equation (2) and equation (3) are taken together, equation (2) must be multiplied by 5 and equation (3) by 3 and the resulting equations then added. Hence, considering equations (i) and (2), and multiplying according to our scheme I2a I5&+2IC = 42 63a+i4&+2ic = 329. Subtracting 5ia-29& = 371 (4) EQUATIONS 47 Combining equations (2) and (3), multiplying equation (2) by 5 and equation (3) by 3 45a+io& + i5C = 235 3a- 36-15^ = 33 Adding 480+ jb =268 .......... ( 5 ) Equations (4) and (5) may now be combined and either a or b eliminated. To eliminate a, multiply equation (4) by 16 and equation (5) by 17 and add. Then- Adding 8i6a 4646 = 5936 8i6a+ii9& = 4556 -345& = -1380 6= 4 Substitute this value of 6 in equation (5) and the value for a is found i. e., 480 + 28 = 268 or 480 = 240 For a write 5, and for b write 4, in equation (2). Then 45 + 8+35 = 47 or Collecting the results Example 23. A law is required, in the form E = a+6T+cT 2 , for the calibration of a thermo-electric couple. The corresponding values of E and T are T (C.) IOO 600 1000 E (micro-volts) 45 3900 5600 In other words, we wish to find the values of the three unknowns, a, b, and c. The three equations formed from the given values are 5600= a+iooo&+ioooooos (i) 3900 = a+ 6oo&+ 3600005 (2) 450 = a+ 1006+ looooc (3) Grouping equations (i) and (2) and subtracting, a is eliminated; and similarly for equations (2) and (3). 48 MATHEMATICS FOR ENGINEERS Thus 5600 = a + 10006 + looooooc 3900 = a+ 6006 -f 3600006 .*. 1700= 4006+ 6400006 ........ (4) Also 3900=0+ 6006+ 3600006 450 = a+ 1006+ 100006 345= 5006+ 3500006 ........ ( 5 ) To eliminate b, multiply equation (4) by 5 and equation (5) by 4, and subtract. Then 8500 = 20006 +32000006 13800 =20006+14000006 5300 = 18000006 5300 c = -o^- - = -00294 1800000 Substituting in equation (4) 1700 = 40061884 or 4006 = 3584 b = 8-96. Substituting for b and c in equation (3) 450 = a+8g6 29 a = -417. Hence the law of calibration is E = 4I7+8-96T--OQ294T*. Exercises 6. On Solution of Simultaneous Equations. Solve the equations in Exercises i to 9. 1. ?x+3V = i 2 * 2a ~ 9& = 3 2 35* 6y = i 3a+io6 = i 3. sm6n = 6-6 4. 48* 27^ = 48 nn 25 = 2m y$ix = 51 5. y+i-37=4* 6 - '\x-\V = &*+-&& gx-ijy =-49-87 19^+27 = 268 4X-yy_ , 8. 2a+ 3 6+ 5C=-4- 5 ' ' 3C 7a+ 156 = 62-7 9. 2p- 55+4^ = -33 10. If E= a+6*+c/ 2 , and also P '- for 3 P-I2* + 2S = 8 9 values we have E t 4-6 10 -4-5 find the values of a, b, and c. 11. If P= ad-}- bd z and P= 17830 when d= -5) P= 2992 when d= -2) find the values of a and 6. (P and d have the same meanings as in Example 21, page 45.) EQUATIONS 49 12. You are given the following corresponding values of the effort E necessary to raise a load W on a machine. Find the connection between E and W in the form E=aW+&, given that E=7 when W 20 ; and E = 14-2 when W = 80. 13. Corresponding values of the volts and amperes (obtained in a test on an electric arc) are V = 48-75 when A = 4 ; and V = 75-75 when A = -8. Find the law connecting V and A in the form V= 14. The I.H.P. (I) of an engine was found to be 3-19 when the B.H.P. (B) was 2, and 6-05 when the B.H.P. was 5. Find the I.H.P. when the B.H.P. is 3-7. {I = aB + 6.} 15. The law connecting the extension of a specimen with the gauge length may be expressed in the form, e = a + feL, where L = length and e extension on that length. The extension on 6" was found to be 2-062", and that on 8" was 2-444*. Find the values of the constants a and b. 16. The electrical resistance R< of a conductor at temperature t may be found from R t == R (i + at) where R = resistance at o, and a = temperature coefficient, If the resistance at 20 is 5-38 ohms and at 90 is 7-71 ohms, find the resistance at o and also the temperature coefficient. 17. Find a simple law connecting the latent heat L with the tem- perature t when you are given that L 975 800 t 200 450 Find also the latent heat at 212. 18. Unwin's law for the connection between the length, the area, and the extension of a specimen is ,. c Varea . , Percentage elongation e = -. T. + 6. If the area a is -75 and e= 30-11 when the length 1=5", and if e= 25-6 when /== 8", find the law for this case (Mild steel specimen). 19. Repeat as for No. 18, when a= 2-12, and /= 3" when e= 59-2 and 10* when e =24-5 (Rolled brass specimen). 20. The difference in potential E between the hot and cold junction of a thermal couple for a difference of temperatures T is given by E = a + 6T + cT 2 . Find the law connecting E and T for the values T 5 IOO 3OO E 2O2-2 570-1 2058 21. /is the tenacity (in tons per sq. in) of copper at t F. /and t are connected by an equation of the form/= a b(t 6o) 2 . Find this equation, given that /= 14-8 at 60 F. and /= 13-2 at 400 F. E 50 MATHEMATICS FOR ENGINEERS 22. Repeat as for No. 21, the values of /and / (for cast phosphor- bronze) being / 16-06 13-1 1 t IOO 400 23. Given that W= Find the law connecting W and p if W= 21-11 when p= 80; and also W= 16-56 when p= 126. W is the weight of water used by a steam engine per H.P. hour, and p is the absolute pressure. 24. If w = steam per H.P. hour and I = H.P., then If 12000 Ibs. of steam were used per hour when the H.P. was 1000 and 3554 Ibs. when the H.P. was 180, find the law connecting w and I. 25. 500 cu. ins. of cast iron together with 240 cu. ins. of copper weigh 206-8 Ibs., whilst 13 cu. ins. of copper weigh as much as 16 cu. ins. of cast iron. Find the number of cubic inches per ton of each of these metals. 26. Measurements to find the constants of a telescope with stadia wires resulted in the following. At i chain distance from the instru- ment the difference between the readings on the staff for the top and bottom wires was -65 ft.; and at 2 chains the difference was 1-311 ft. Find the constants, C and K from CS-f-K=D where S = difference of staff readings and D = distance. (i chain = 22 yds.) 27. Three wires A, B, and C are successively looped together and the resistance of each loop measured. The resistance of A and B is found to be 260 ohms, of A and C is 280 ohms, and of B and C is 300 ohms. Determine the individual resistances of A, B, and C. 28. The following equations occurred when finding the fixing couples of a built-in girder =762-5 200 , 400 *i+- m s = 7186. o 3 Solve these equations for m^ and m t . 29. The " dead weight " tonnage of a ship is 700 tons, whilst the cubic capacity of its hold is 42000 cu. ft. To ensure the most profitable voyage, a mixed cargo of heavy and lighter goods must be carried, and the complete capacity of the hold must be utilised. Prove the truth of the following rule : " To obtain the weight of the lighter cargo, multiply the specific volume (i. e., the number of cu. ft. per ton) of the heavy cargo by the dead weight tonnage. Subtract this result from the total cubic capacity and divide the difference by the difference between the specific volumes of the heavy and light goods." If, in a certain case, the densities of the heavy and light goods are 35 cu. ft. per ton (saltpetre), and 80 cu. ft. per ton (ginger in bags) respectively, determine the weight of saltpetre carried and also the weight of the ginger. EQUATIONS 5I Methods of Factorisation. Reference has already been made to the word " factor " as denoting a number or symbol that multi- plies or divides some other numbers or symbols in an expression. Thus 3x5 = 15, and 3 and 5 are called factors of 15, i.e., when multiplied together their product is 15. Again 260? = 2Xi3XaXaXa. Here the quantity has been broken up into 5 factors. The process of breaking up a number or expression into the simple quantities, which, when multiplied together, reproduce the original, is known as factorisation. Little is said about this in works on Arithmetic, but the process is used none the less for that. To illustrate by a numerical example Find the L.C.M. of 18, 24, 15, and 28. These numbers could be factorised and written as follows 2x3x3, 2x2x2x3, 3x5, 2x2x7. The L.C.M. must contain each of these; it must, therefore, contain the first, any factor in the second not already included, and so on for the four. 2x3x3 2x2 5 7 *. e., L.C.M. = , x ^-^ x ^^ X ^^ = 2520. ist 2nd 3rd 4th The necessity for the presence of the two 2's in the second group should be realised. There must be as many 2's as factors in the result as there are 2's in the number having the greatest quantity of 2's in its factors : i. e., there must here be three 2's as factors in the result. It is, however, in Algebra that this process finds its widest application. Rather difficult equations can often be put into simpler forms from which the solution can be readily obtained, and by its use much arithmetical labour can be saved. Generally speaking, the factorised form of an expression demonstrates its nature and properties rather more clearly than does its original form. For practical purposes the following methods of factorisation will be found sufficient. Rule i. Often every term of an expression contains a common factor : this factor can be taken out beforehand and put outside a bracket. The multiplication is then done once instead of many times. 35+6055 is, we know = 40 But 35+60-55 = (5X7) + (5Xi2)-(5Xii) and the factor 5 is common to each term. If this factor is taken outside a bracket, the arrangement then becomes 5(7+1211), 52 MATHEMATICS FOR ENGINEERS or 5 x 8 = 40, which agrees with the previous result. The final arrangement is to be preferred, because the numbers with which we have to deal are much simpler. Hence for this numerical case we see that the common factor must be taken outside a bracket, ^vhilst the terms inside are the quotients of this factor derived from the original terms. Numbers have been taken for clearness of demonstration, but the method holds equally well for symbols of all kinds. Example 24. Factorise the expression, 7a 4 fc 2 28a 3 6c 2 + 42a 6 6 3 c 4 . In this expression, 7 is common to each, term, a 3 is the highest power of a common to each term, 6 the highest power of b, whilst no c occurs in the first term, and c is, therefore, not a factor common to all terms. Then, the factor to be taken outside a bracket = ja?b. Hence the expression = ya 3 b(ab~ 4C 2 +6a 3 & 2 c 4 ), or we have broken it up into two factors. Example 25. Find the volume of a hollow cylindrical column, 12 ft. long, i ft. external radius, and 9 ins. internal radius, from the formula Volume of a cylinder = irr z l (r= 3-142) In this case the net volume will be the difference between the volumes of the outside and inside cylinders .*. V= (ffXi 2 xi2) (7rx(J) 2 xi2) ......... working in feet. = i27r{i 2 (|) 2 } because i2?r is a factor common to both terms. = 16*48 cu. ft. Rule 2. The expression may be of a form similar to one whose factors are known, and the factors may be written down from inspection. If (A+B) be multiplied by (A B) the resulting product is A 2 -B 2 . Conversely, then, the factors of A 2 B 2 are (A B) and (A+B), or A 2 -B 2 = (A-B)(A+B), i.e., to factorise the difference of two squares, multiply the sum of the quantities by their difference. This rule is of wide application. Example 26. Write down the value of 9154* Squaring each and subtracting the results is far longer than making use of the rule just given Thus 9i54 2 -9i5 l2 = (9154+ 9i5 I )(9i54- 9I5 1 ) = 18305 x 3 = 54915- EQUATIONS 53 Example 27. Find the factors of 8i 8 166*. 8ia 8 166* = (9 4 ) 2 (4& 2 ) 2 , which is the difference of two squares, and therefore = (9 4 4& 2 )(9 4 -f 4& 2 ) In this example the rule is applied twice. Two other standard forms are here added, although their use is by no means so frequent as that of the above. A 3 -B 3 = (A-B)(A 2 +AB+B 2 ) A 3 +B 3 = (A+B)(A 2 -AB+B 2 ). Example 28. Find the factors of 2ja*b 3 + i25 3 c 9 . Let E denote the expression, then E= a 3 (27a 3 6 3 + 1256') by Rule i. Rule 3. In many cases of trinomial, i. e., three-term expressions, the factors must be found by trial, at any rate to a very large extent. There are certain rules applying to the signs, which can best be followed by first considering the following products : = * 2 +ii*+3o .......... (i) (*-5)(* 6) = * 2 -n*+3o .......... (2) (x-5)(x+6) = x*+x-3o ........... (4) In (i) and (2) there are like signs in the brackets and a plus sign precedes the third term in the expansion, which must be written in the order of ascending or descending powers of x or its equivalent. In (3) and (4) there are unlike signs in the brackets and a mintts sign comes before the 30. Hence the first rule of signs may be stated : So arrange the signs that the one before the first term is plus, an adjustment of signs throughout being made if necessary. Look to the sign before the third term of the expression ; if this is a plus then we conclude that the signs in the brackets will be like, and if this sign is a minus then the signs in the brackets will be unlike. If they are to be like, they must be either both plus or both minus, and the sign before the second term in the given expression indicates which of these is accepted. Thus, a plus sign before the second term indicates that the signs in the brackets are both plus. 54 MATHEMATICS FOR ENGINEERS If, however, the signs in the brackets are to be unlike, one product must be the greater and the sign before the second term indicates whether it is the product obtained by using the plus or the minus sign. E. g., in (3) we have 30 as the third term ; accordingly the signs in the brackets will be unlike : also the second term is x so that the minus product is to be the greater; hence the minus sign in the brackets must be before the 6. The actual numbers in the brackets must be found by trial. They must in each of the four instances multiply together to give 30 ; also, in (i) and (2) they must add together to give n, and in (3) and (4) their difference must be i. Example 29. Find the factors of # 2 + ijx no. In the given expression the third term is no, so that there must be unlike signs in the brackets. Also, the + product must be the greater, since +17* is the second term. Since the signs in the brackets are to be unlike, two numbers must be found which when multiplied together give no, and which differ by 17. These numbers are 5 and 22 ; and the signs placed before these must be so chosen that + ijx results when the brackets are removed. Thus the plus sign must be placed before the 22, and hence x z + ijx no = (x -f 22) (x 5). Example 30. Factorise the expression 2X 2 z8x go. Applying Rule i The expression = 2(# 2 + 14*+ 45). (Note the adjustment of signs, to ensure + before the first term.) Dealing with the part of the expression in brackets : plus signs throughout denote -f in brackets ; hence two numbers are required that multiplied give 45, and added give 14; these being 9 and 5. /. The factors = z(x+ 9)(*+ 5). Example 31. Find the factors of 6m 2 + nm 35. This expression could be reduced to the form of the previous examples by dividing by 6, but the fractions so obtained would render the further working rather involved. It is better, therefore, to proceed as follows : There will be unlike signs in the brackets, since the sign before the third term is minus, and the factors of 6 have to be combined with those of 35 to give a difference of products of +n. The varying of the factors at either end may result in many arrangements being tried EQUATIONS 55 before the correct one is found. After a little practice, however, the student disregards absurd arrangements and so reduces his work. The correct arrangement in this case is (yn 5)(2nt+j). The first terms when multiplied together give 6m 2 , the last ones give 35, the extreme terms give +2im, and the middle terms lorn, i. e., the last two combine to give +nw. The arrangement is more clearly shown if written down as (3xx5) The end terms are easily settled, but for the middle term the multi- plication must be performed as indicated by the arrows, and the results must be added or subtracted as the case may demand. When the correct arrangement of the figures has been found, the letters must be inserted. Hence, the expression has for its factors (3m 5)(2m+7). Example 32. Factorise the expression 72a 2 + i8a6 776*. In the first place disregard the letters; dealing only with the numbers. The factors of 72 are to be combined with those of 77 to give a difference of 18. 72 has many factors, but 77 = 7x11 or 77x1. The trial arrangements would be of this nature ox^T For the middle term, the difference = 43. 135- faXX H ., = '8. The last is the arrangement desired. To allocate the signs: the net result of the products is to be +18 : 7x12 gives the greater product, hence the + must be placed before the 7. .'. The expression = (6a + 7b)(i2a lib). The Remainder and Factor Theorems. Suppose we have to deal with an expression such as If this expression be divided by (xa), the remainder will be 56 MATHEMATICS FOR ENGINEERS which could have been more simply obtained by substituting a for x in the original expression. If (x a) is to be a factor of the original expression then the remainder after division by (x a] must be zero. Hence we obtain a rule enabling us to find factors of rather complicated expressions Find the value of the main quantity (usually the x) which makes the suggested factor zero ; substitute this value in place of the x in the expression, and if the result is zero one factor has been found. E. g~, if it be conjectured that (#-(-3) is a factor of an expression, its value would be found when x had the value 3. Example 33. Find the factors of X s + x z 14* 24. Let us try if (#4) is a factor; we will substitute, therefore, + 4 for x in the expression, which becomes ( 4 )3 + (4)2-14(4) -24 = 64+16-56-24 = o .*. (x 4) is a factor. Another likely factor would be (#+3), for 3x4 is part of 24, and there must be a plus sign to combine with the minus in (#4) to give 24. Substitute 3 for x, and the expression becomes (-3) s + (-3) 2 -i4(-3)-24 = -27+9+42-24 = o .*. ( x + 3) is a factor. The other factor may be found to be (#+2) /. x 3 + x z - 14*- 24 = (x + 2}(x + 3)(x 4). Multiplication and Division of Algebraic Fractions. The simplification of algebraic fractions furnishes useful examples on the appli cation of the rules of indices and of factorisation. When a number of fractions are to be multiplied together, cancelling can be performed as in the case of arithmetic fractions, always provided that the complete factors are cancelled and not portions thereof. E. ., -r^ is in its lowest terms ; we cannot cancel 2X into ' 4*+3 4* or strike out the 3's, because (2^+3) must be treated as one quantity, as also must (4*4-3). .. c . ,., Example 34. Simplify ~ -.-r^-. x p , . * ja*b z c 5 i8a 3 c 4 48a 3 fcc 2 3 The fraction = - Tr*-. x - alb EQUATIONS 57 Example 3 5 .-Simplify *+*+ 15 x *- I 4 7_ y 2* 2 + 3*- 35 A 20*2 + 28*- 96 No cancelling must be made until numerators and denominators are expressed in terms of their factors. Thus the fraction _ = (*+3)(*+5) x 3 (2* -7) (2 x + 7) ( 2 *-7)(*+ 5 ) 4(*+ 3 j ( 5 * -8) and in this fraction cancels with cancels with cancels with giving the answer j!|J^rf j in which no further cancelling can occur. Example 3 6.-Simplify 4* 2 +*~i4 x - - - 6xy-i 4 y A' 2 - 4 4*- 7 ' 3 #2-*-i 4 The numerators and denominators are first factorised giving the fraction in the form ( 4 *-7)(*+g) 4*2 (x-2) (3* - which by cancelling reduces to - 2 v3 . * T v __ /t^2 [_ *jf\ Example 37. Simplify the fraction ~ ~^-L- ^x + nx 20 The factors for the denominator are the more easily found ; they are (#+5) and (3* 4). The first of these is a possible factor of the numerator also ; applying the remainder theorem, the value of the numerator when x 5 is 2( 125) 4i( 5) (5)2 + 70, i.e., o; hence (#+5) is a factor. In like manner it would be found that (x 2) was also a factor; and by division of the numerator by the product of these, viz. by x*+ 3* 10, the remaining factor is found to be (2X 7). Hence the fraction^ (*-M+5K**-7) = <*-*)(**-7) ~4/ \3 X ~4) Addition and Subtraction of Algebraic Fractions. The same rules are adopted as for arithmetical fractions. The L.C.M. of the denominators (L.C.D.) must first be found by factor- ising the separate denominators according to the plan detailed on page 51. Example 38. Simplify the fraction _5_ ,_ioo_ 51 407 120 21 20fl 35 5 8 MATHEMATICS FOR ENGINEERS This becomes (after factorisation of the denominators) (4-7) + 3(4-7) ~~ 5(4 -7) and the L.C.D. = ($a 7) x 3 x 5 = 15(40 7) whence the expression Example 39. Simplify ey . ft + 15 12 This becomes (after factorisation of the denominators) # 15 12 C*+3)(#+ 2 ) (#+7) (#+2) (#+7) (# + 3) and the L.C.D. is (x + 3) (x + 2) (* + 7) . Dealing with the first term only and multiplying both numerator and denominator by this L.C.D. which after cancelling reduces to . + -/ + \(x+ 7)' In like manner the second and third terms reduce to i5(*+3) and " d 7) + I 5( ;r + 3) I2(#+ 2) Hence the fraction = -- g+ 3 &+ 2)(+ 7) 45- "* - 24 IPX + 21 (*+3)(*+7) ., a c ., a-{-b c 4- d Example 40. Show that if r= j. then ^ = . - and _ a b ~ c d From ^ = ^, by adding i to each side o a a c , b +I= d +1 Taking the L.C.D. of each side a+ b _ c+ d EQUATIONS 59 In like manner by subtracting i from each side of the original e< l uation - ^ = -^r ........... W Hence, dividing (i) by (2) q + b _ c+ d a-b ~ T^~d ........... 3 These results are of importance. Exercises 7. On Factors, and on Multiplication and Addition of Algebraic Fractions. Factorise the expressions in Examples i to 20. 1. x 2 + iSx- 88 2. x 2 - igx + 88 3. x z -26x+ 105 4. 8a 3 -i25& 6 5. 24** -#-44 6. (2fl+6) 2 - (30-46)2 7. a 2 + 4 a&-4 5 & 2 8. I2# 2 73#y + 105^2 9. 88 3*2 13* lOi 2om+ 2 on- 5 8m 11. <^!_^ 3 + 5J^_ 4 24 384 16 12. |n-R 3 $irr 3 , giving the volume of a hollow sphere of outside radius R, and internal radius r. 13. 94^ 2 +39^-963. wlx 3 wx* wl 3 x i^EI ~~ 2~EI ~ 2~El' an ex P resslon occurring in connection with the deflection of beams. 15. 540*6 30oo 2 6c 2 42a 3 6c 16. 4# 2 i6c 2 1206 + 9& 2 17. 64C 5 + _ 27 18. V v (giving the volume of the frustum of a cone; R and r being the radii of the ends of the frustum and h its thickness) where T , wR 2 /r , .\ Trr z k rh V = (h + A), v = - and k = ^ 3 3 K - y 19. 2# 3 + 7# 2 44*+ 35. {Hint. Try (x + 7) as a factor.] 20. 6p 3 + 2$p z -\-6p- 35. [(p i) is one factor.] 21. Find, by the methods of this chapter, the value of (199 x 46) + (398x69) -(199x92). 22. Find the value of n-R 2 / irr z l, which gives the volume of a hollow cylinder, when IT = 3-142, R= 12-72, ^=9-58, =64-3. 23. Find the L.C.M. of # 2 -#-6, 3^-21^ + 36, and 4*2 - 8* -32. 24. Simplify _ c . ,., 8^ 2 2AX 80 25. Simplify 2ox z + i^xno 4X 2 2xgo 26. Simplify + -^ ^ ^ ^ y X42X+4 x z 2X8 27. Simplify ^ + ^ + ^ + Y&^W^ 6o MATHEMATICS FOR ENGINEERS 28. Simplify _ 5* __ 8*+7* __ 14 8* 18** 100 + 30* 24** 4* 280 fo# 2 + 175- 155* 29. Solve the equation _ = 30. Solve the equation -I- _-3__ = __9__ [Hint. Multiply through by the L.C.D.] 31. A unit pole is attracted by a magnetic pole of strength m with a force ^. t and repelled by a force of What is the resultant attractive force ? Find the value of this force if / is very small compared with d. 32. Find the factors of (a) 3**+ 6x* 189* ; (b) 24 + 37* 72** ; M (3* + 7X)' - (2* - 37)*- 33. Find the factors of (** -j- 7*+ 6)(** + 7*+ 12) - 280. 34. M, a bending moment, is given by M = Pfl ( 2S +3) _ Pa(i 85*4-355 + 9) 8(5+2) " 24(65*+ 2 +135) Find a more simple expression for M. 35. The expression pjt>i - (ptVPivJ p t v t relates to the 7* *~~ X work done in the expansion of a gas. State this in a more simple form. 36. The depth of the centre of pressure of a rectangular plate, of width h, immersed vertically in a liquid, the top being a and the bottom f (&'-*') b units below the level of the surface of the liquid, is given by -4- - ^(6* -a*) Express thU in a simpler form. Quadratic Equations. Any equation in which the square, but no higher power, of the unknown, occurs, is termed a quadratic equation. The simplest type, or pure quadratic, is <f* = 25 ; to solve which, take the square root of both sides. Then d = either +5 or 5, because (-f5) 2 = 25 and also ( 5) 2 = 25. This result would be written hi the shorter form d = +5. It is essential that the two solutions should be stated, although in most practical cases the nature of the problem shows that the positive solution is the one required. The solution of the pure quadratic is elementary; but in the case of an equation of the type Jt 2 +7*+i2 = o (spoken of as an adfccted quadratic, i. e., one in which both the first and the second power of the unknown occur) new rules must be developed or stated. Three rules or methods of procedure are suggested for the solution of adfected quadratics, viz. EQUATIONS 61 Method 1. Solution of a Quadratic by Factorisation. Group all the terms to the left-hand side and factorise the expression so obtained. Next, let each of these factors in turn = : thus two solutions are determined. For all quadratics there must be two solutions or " roots " ; in some cases they may be equal, and in rare cases " imaginary." Applying this method to the example under notice : Example 41. Solve the equation x*+ jx+ 12 = 0. By factorisation of the left-hand side Then either * -f 3 = o, in which case x = 3, or x -f 4 = o, in which case x = 4, because, if one factor is zero, the product of the two factors must also be zero ; e. g., if *= 3, (x+ 3)(x+ 4) = ox i = o. Hence x= 3 or 4. Example 42. Solve the equation 24a a -f 170 = 20. Collecting terms, 240* + i ja 20 = o. Factorising, (8a 5) (30 -f- 4) = o. .'. either 83 5 = 0, i.e.,a= |1 or 30 -f 4 = o, i. e., a = | J If no factors can be readily seen we may proceed to Method 2. Solution of a Quadratic by completion of the Square. All the terms containing the unknown must be grouped to one side of the equation and the knowns or constants to the other side. The left-hand side, viz. that on which the unknown is placed, is next made into a perfect square by a suitable addition, the same amount being added also to the right-hand side, and then the square root of both sides is taken. The solution of the two simple equations thus obtained gives the " roots " of the original equation. Before proceeding further with this method a little preliminary work is necessary, the principle of which must be grasped if the reason of the method of solution is to be understood. Suppose that the first two terms of the right-hand side are 62 MATHEMATICS FOR ENGINEERS given, and it is desired to add the necessary quantity to make it into a complete square. /A8\ 2 a 2 +48a+576 might be written as (a) 2 + [2 X (24) X (flVH \ 2 / so that if 2 +48a is given, the term to be added is ( ) , *' , is \ 2 / the square of half the coefficient of a. /7\ 2 Similarly, # 2 -f 7* could be expressed as a perfect square if (-] (7\ #+' ) 2/ Returning to the method; a numerical example will best illustrate the processes. Example 43. Solve the equation x 2 + i$x 4-9 = 0. Grouping terms x 1 + i$x = 9. Adding the square of half the coefficient of x, viz. ( ) 'to each side, or Extracting the square root of both si #4-^= 6-88 .*. x= 7-5 6-88 = - 7'5 + 6-88 or - 7-5 - 6-83 = -62 or I4'38. The change from X s + 15*4- ^- J to fjr+ ) often presents diffi- culty : the reason for the omission of the i$x does not seem clear. It must be remembered that it is represented in the second form, for -~ 2 = (i st ) 2 + (2 nd ) 2 +2 (product) = If the coefficient of x 2 is not unity it must be made so by division throughout by its coefficient. Example 44. Find a value of B (the breadth of a flange) to satisfy the equation 3-64B 4 5I-8B 2 900 = o. This equation, though not a quadratic, may be treated as a quadratic and solved first for B 2 ; i. e., if for B 2 we write A the equation becomes 3-64A 2 5I-8A 900 = o. EQUATIONS 63 Dividing through by 3-64 (the coefficient of A 2 ) and transferring the constant term to the right-hand side A 2 I4-24A = 247. The coefficient of A is 14-24; half of this is 7-12, hence add 7-12*, i.e., 50-8 to each side i.e., A 2 I4-24A+ (7-I2) 2 = 247+50-8 = 297-8 or (A 7-i2) 2 = 297-8. Extracting the square root throughout A 7-12 = db 17-26 i.e., A = 7-12 17-26 whence A = 24-38 or 10-14. Now A =-B a , so that B 2 = 24-38 or 10-14. Of these values the former only is taken, since we cannot extract the square root of a negative quantity. Thus B 2 = 24-38 or B = 4-94, but evidently the negative solution has no meaning in this case. Example 45. If 4a 2 1506+ 2& 2 = o, find the values of a to satisfy this equation. Dividing through by 4 and transferring the constant term to the T - 2 right-hand side, a 2 -- -ab = -- 4 2 The coefficient of a is b; half of this is -b : hence we must add 4 (-a-) to each side. Thus a* - * + - . + = 4 \ 8 / 2 04 04 Extracting the square root = 3-616 or -146. Method 3. Solution of a Quadratic by the use of a Formula. It will be evident from the foregoing examples that all quadratics reduce to the general form Ax 2 + BX + C = 0. If Method 2 is applied to the solution of this, the result is a 64 MATHEMATICS FOR ENGINEERS formula giving the roots of any quadratic, provided that the par- ticular values of A, B and C are substituted in it. Thus A* 2 + Ex + C = o Dividing through by A and transposing the constant term x* + B * = - / B\ 2 To each side add the square of half the coefficient of x, viz. . B , /B\ 2 C . /B\ 2 IY& _L v _L I __ + A + \2\) ' A + Uv = -C+B A ' 4/ B \ 2 B 2 - 4AC QJ- I V _J \ _ _C ^ _B 2 -4AC A *1 i \ 2 Extracting the square root of both sides 2A 2A , B VB 2 - 4AC whence x = - , 2A Thus the " roots " of the general quadratic A* 2 + BA; + C = o - B + VB 2 - 4AC , - B - B 2 - 4AC are - and 2A 2A Example 46. Solve the equation $x z Sx = 12. Collecting all the terms to one side, $x z 8x 12 = 0. Then for this to be identical with the standard form A = 5, B = -8, C = -12 + 8 V6 4 + 240 = 8 V3Q4 ^ 8 17-4 10 10 = 2-54 or - -94. Great care must be exercised to avoid errors of sign. To obtain the value of 4AC, first find AC= 5 x ( 12) or 60; 4AC then = 240 and 4AC = + 240. EQUATIONS 65 Example 47. Solve for y, in -^y z i'$y -32 = o. It is always advisable to have the first term positive, so change all the signs before applying the formula. Then 4>/ 2 +1-5?+ -32 = o. Here A =-4, 6=1-5, C=32. -i'5 V2-25- -512 8 - 1-5 ^1738 = -23 or 3-52. Example 48. The stresses on the section of a beam due to the loading are a normal stress / and a shearing stress q. These produce an entirely normal stress / on a plane known as the plane of principal stress. Find an expression for / from the equation /(//) = q*. Removing the bracket and grouping the terms to one side / 2 -//n-<? 2 =0. Here A=i, B= -/, C=-q* f _ +fn V/n 2 +4g 2 The next example is instructive as showing the advantage of resolving large or small numbers into integers multiplied by powers of ten. Example 49. Solve the equation L# 2 + ~Rx + T? = ( an equation occurring in electrical work) when L= -00:1:5, R = 4OO, K='45XiO" 6 . Substituting the numerical values for L, R, and K ' - = o. The last term may be written in the more convenient form 2-22 x io 8 since ^- = 2-22 and -^4= io 8 - Thus -oo 1 5# 2 4- 400^+ (2-22 x io 6 ) = o Comparing with the standard form A = (1-5 x io- 3 ), B= (4 x io 2 ), C= (2-22 X io) 66 MATHEMATICS FOR ENGINEERS Hence x = ~ (4 x I{)2 ) V(r6 x *o 4 ) (6 x IP' 3 X 2-22 x io 6 ) 3 x io 3 - (4 x io 2 ) V(i6 x io 4 ) (1-33 x io*) 3 x io~ 3 the second term under the radical sign being written in this form so that io* is a factor common to both terms ; and the square root of io 4 is readily found. [6 X IO- 3 X 2-22 X IO 8 = 13-32 X IO 3 = 1-332 X IO 4 .] The square root of io 4 is io 2 ; this may be placed outside the radical sign, and then - (4 X io 2 ) io 2 Vi6- 1-33 3 x io- 3 _ io 2 (-4 VI4-6;) 3 x io- = (io 6 x 2-61) or io 5 x -053 = 261000 or 5600. Example 50. A formula given by Prony (in connection with the flow of water through channels) connecting the hydraulic gradient i with the velocity v and the hydraulic mean depth m was of the form mi = av + bv*. Under certain conditions a = -000044, & = -000094. Show that this is in close agreement with the formula given by Chezy, viz. v = 103 Vmi. mi = av + bv z or bv* + av mi = o a Va z 2b Inserting the numerical values for a and b 4 mib = 4 x -oooo94u Also, a 2 is very small, even in comparison with -000376, and can therefore be neglected. 000044 Hence v = -- 2 x -000094 2 x -000094 = -234 104-3 Vmt. Taking the + sign and neglecting the first term, v = 104-3 Vmt, which agrees well with the v = 103 Vmi given by Chezy. EQUATIONS 67 Quadratics with "imaginary" Roots. The question may have presented itself : What is done when (B 2 4AC) in the formula for the solution of the quadratic becomes negative ? ' How can the square root of a negative quantity be extracted ? The square root of a negative quantity is known as an imaginary quantity, and all imaginaries are reduced to terms of the square root of i, which is denoted byj. At present no meaning can be stated for this, but it is referred to again in a later chapter. Thus ;=V r I, j 2 =i, j 3 = V i, etc. E.g., V^so = Vsox i = Vsox V^ = 5-47.7. Example 51. Solve the equation 2x z 3* + 15 = o, employing Method 3. x 4 + 3 ViTi X V^ Expressions of the type a bj, where a and b may have any values, occur in Electrical theory and in the theory of Vibrations ; such being referred to in Chapter VI. Cubic Equations. Cubic Equations, i. e., equations contain- ing the cube of the unknown as its highest power, may be solved graphically, in a manner to be demonstrated in a later chapter, or use may be made of what is known as Cardan's solution. The three roots of a cubic equation may be either, one real and two imaginary, or, three real. Cardan's solution applies only to the former of these cases and gives the real root only. If x 3 -f- ax -f- b = be taken as the standard type of cubic equation, then the real solution is given by Cardan as / b, /^T^Uo-f b /*V X== \-2 + V 27+TJ + \-2~V 27+ The proof of this result is too difficult to be inserted here, but it is outlined in A Treatise on Algebra, by C. Smith (Macmillan and Co., Ltd., 75. 6d.). 3 I\2 If + be negative, the three roots are all real, but Cardan's 27 4 solution cannot be applied. 68 MATHEMATICS FOR ENGINEERS Example 52. Solve the equation x 3 I2X + 65 = o. (Imaginary roots are not required.) Here a= 12, b = 65, in comparison with the standard form. = {-32-5 + 31-5}* + {-32-5 -31-5}* = (-!) + (-4)= -5- If the equation is not of the form, x 3 -f- ax + 6 = o, it can be made so in the following manner. Example 53. Find a solution of the equation t; 2 + 1441; 1944 = ' v being a velocity. For this to be reduced to the standard form, the term containing v* must be eliminated. By writing (V+a) for v and suitably choosing a, this can be done, for (V + a) 8 + 2 4 (V + a) + I44(V+ a) - 1944 = o '.., V'+3V 2 a+3a 2 V+a 3 +24V 2 +24a 2 +48aV + I44V+ 1440 - 1944 =o ............ (i) Equating the coefficients of V 2 to zero (since the term containing V 2 is to be made to vanish), 30 + 24 = o, *. e., a = 8 ; so that v = V-8. Equation (i) can now be written ( 8 being substituted for a) V 3 24V 2 +I92V- 512 + 24V 2 +1536 384V +I44V 1152 1944 = o or V 8 48V 2072 = o Comparing with our standard formula a= 48, 6= 2072. Therefore, by Cardan V- = (2070)*+ (2)* = I2-75+I-26 = 14. Hence v = V 8 = 6. Equations of degree higher than the third (if not reducible to any of the forms already given) are best solved graphically. (Com- pare with Chapter IX.) EQUATIONS 69 Exercises 8. On Quadratic and Cubic Equations. Solve the equations in Exercises i to 10. 1. x z + $x + 4 = o 2. 2X Z 7* + 15 = 4* 3. -3* 2 +9*+i4 = o 4. 8p*-7p = 4p* + 5 p+i6 5. -ooia 2 -234a -764 = -417(1 -3250? 6. 9^ + 5^+2 = 7 *+4 ^29-5* g 2* 17 _5*_ ~ ' * *+2 #+3 z 10. 1700 + -oi26F = -000003F 2 r , V a 2X230X/ J 2 fi , , , V=33"O 11. If = - - z - , find the values of I when J I 2g 2x36 #=32-2) 02 i ^2 12. If r= -T , find h when f = 15, a = 5-5. This equation gives the radius of a circle when the height of arc A and length of chord 2a are known. 13. Solve for F the equation 3F 2 3F -- S - -- - = 100. 30 x io 6 7200 14. We are told that W( + A) = FA (a formula relating to the jr strength of bodies under impact), and also A = - , W=45, and h = 2-4. Find values of F to satisfy these conditions. ph z 15. The equation 6 a ab a? = - relates to masonry dams, where 6 = width in feet of base of a dam a feet wide at the top, and h feet deep; w being the weight of i cu. ft. of masonry, and p being the weight of i cu. ft. of water. Find b for the case when a = 5, h = 30, w = 144, and p = 62-4. 16. Find expressions for /, from 17. If |L solve (a) for u and (6) for v. 2U* + 3U 2 18. To find n (the depth from the compression edge to the neutral axis of a reinforced concrete beam of breadth 6) it was necessary to solve the equation bn z + 2A T ww 2wA T d= o. Determine the value of n to satisfy the conditions when ^=15, A T = i-56, 6 = 5, and d= io. 19. Solve for C the equation 75 x ioC 2 - io 10 C + 12 x io 10 = o. /P _ /wx _ /) 20. Find the values of t when L = s p Ap ' X -oogC, and P=i20, C=3375,T=67, 1^=1765. 21. Find the ratio, length of arc of approach a pitch p wheels) from '^- = + -, where n = number of teeth in the follower a np 4 wheel == 24. 70 MATHEMATICS FOR ENGINEERS 22. The equation mi = av + bv z relates to the flow of water in channels. If a =-000024 an( i &= -000014, P ut this in the form v = cVmi, making any justifiable approximation. (Compare Example 50, p. 66.) 23. Solve the equation 2# a 53 + $x *54 24. Find values of / to satisfy the equation W=$al(i-\ J (Merriman's formula for the weight of roof principals), given that W = 5400 and a = 10. 25. x is the distance of the point of contraflexure of a fixed beam of length / from one end. If x and / are connected by the equation I T H 72" = fi 11 ^ the positions of the points of contraflexure. i if 26. To find the position of a mechanism so that the angular velocities of two links should be the same it was necessary to solve the equation / 3 I9'5/ 2 + 42 5/ + 546 = o. Find values of / to satisfy this. 27. To find d, the depth of flow through a channel under certain conditions of slope, etc., it was necessary to solve the equation d 3 i '305*2 1-305 = o. Find the value of d to satisfy this. 28. The values of the maximum and minimum stresses in the metal of a rivet due to a shearing stress q and a tensile stress / due to con- traction in cooling are given by the roots of the equation /(/-/n)=<? 2 - If q = 4^ tons per sq. in. and / = 3 tons per sq. in., find the two values of /. 29. The length L of a wire or cable hanging in the form of a parabola is given by L - S 3 where S = span and D = droop or sag. Find the span if the sag is 3'-9* and the length of cable is 100-4023 ft. Simultaneous Quadratics. Consider the two equations = 1-8 (i) y and 5'6(5*6 y) = x 2 (2) Values of x and y are to be found to satisfy both equations at the same time (hence the term " simultaneous ") : also the second equation is of the second degree as regards x, and is therefore a quadratic. In most practical examples (the above being part of the investi- gation dealing with compound stresses) one equation is somewhat more complicated than the other, and therefore, for purposes of elimination, we substitute from the simpler form into the more difficult. EQUATIONS 7I In this example equation (i) is the simpler, and from it, by transposition i-8y X = - 2 = -ay. 2 ^ Substitute for x, wherever it occurs in equation (2), its value in terms of y. Then 5-6(5-6 -y) = (.gy) = -8iy2 31-36- 5 -6y = -8iya or '8iy 2 +5-6y 31-36 = o. = ~ 5 ' 6 V ^'3^ + 4 x -81 x 31-36 -5-6 i Hence, y = 1-62 *62 To find * * = -gy and, substituting in turn the two values found for y, x= -9 x 10-56 or -9 x 3-67 = 9-51 or 3-30 x= -9-51 or 3-30 \ and y= 10-56 or 3-67 / Example 54. In a workshop calculation for the thickness * of a packing strip or distance piece in a lathe the following equations occurred (9-9)2 = (i-7 5 + y) + * ........... (2) Solve these equations for x and y. The packing strip was required for a check for a gauge, and great accuracy was necessary in the calculation. By removal of the brackets the equations become 98-01 = 2-25 + y 2 +3y+-64 + * 2 +i-6* .... (3) 98-01 = 3-o625 + y 2 +3-5y + * 2 ........ (4) By subtracting equation (4) from equation (3) an equation is obtained giving y in terms of x, thus o= -8125 -5y+ -64+ i-6x or *5y = i'6x '1725 whence y = 3-2^ -345. Substituting for y in equation (4) 98-01 = 3-0625 + (3-2* --345) 2 + 3-5(3-2* --345)+** = 3-0625+ 10-24**+ -1200-2-208*+ 1 1 -2X I -2075 + x*. 72 MATHEMATICS FOR ENGINEERS Collecting terms, n -24^+ 8-992.* 96-035 = o , 8-992 V / (8-992) 2 + (4 X 96-035 x 11-24) whence #= ^ .a 8-992 66-3219 22-48 _ 57-3299 nr -75-3I39 ' 22-48 22-48 = 2-5503 or -3-3503- i.e., the thickness of the strip was 2-5503 (inches); the negative solution being disregarded. The two values of y would be obtained by substituting the two values found for x in the equation y = 3-2* -345. Thus y =-- (3-2 x 2-5503) - -345 or y = (3-2 x - 3'353) - '345 y = 7-816 or 11-066. The positive solutions alone have meaning in this example, so that x = 2-5503 and y = 7-816. Example 55. Solve the equations 5# 2 + y 2 +2# 77 4 = c* (i) 7*+ 3? =9 (2) From equation (2) yy = 97* or y = 9 ~ 7 - Substituting in equation (i) .-. 5*' + - + - 7 ^ = 95 Multiplying through by 9 45# 2 + 81 + 49# 2 126*+ iSx 189+ 147* = 855. Collecting terms 94* 2 + 39* -963 = Factorising (9 4 #+32i)(*-3) = o 321 i.e., x= - or 3. 94 Now, EQUATIONS 73 Substituting the two values for x 64 orv 9 21 3 3 1031 or = -4 94 * = 3 or 1 94 y = -4 1031 or 94 ] The method of substitution indicated in the previous three examples is to be recommended in preference to the " symmetrical " methods usually given, but which only apply to special cases. Occasionally one meets with an equation or pair of equations to which this method is not applicable. Thus if the equations are homogeneous and of the second degree, i, e., all the terms contain- ing the unknowns are of the second degree in those unknowns, proceed as in the following example ; the method being in reality an extension of the preceding. Example 56. Solve the equations Divide equation (i) by equation (2). Then ~ = Multiplying across 2i6y a . Collecting terms 115** + zgixy 2i6y z = o. Factorising 72 3 whence x = y or x = =y, Substitute each of these values for x in turn in equation (2). 72 Thus, taking x = *^y 74 MATHEMATICS FOR ENGINEERS ..2^*15*23 y ' 20 and # = -y = - .e., y = 36 when # = -f- and y = + 36 when x = ~ Taking x = -y -y z + ^y z = 1 15 23 and y = 5 and x = 3 Grouping results .'. ^ = 3 or ^ 36 y = 5 or Surds and Surd Equations. One often meets with such quantities as -y/3, 3/7 or $ a : such are known as surds or irrational quantities, since their exact values cannot be found. The value of V$ can be found to as many places of decimals as one pleases, but for ordinary calculations two, or at the most three, figures after the decimal point are quite sufficient : thus \/3 = 173 approximately, or 1-732 more nearly. It is both easier and more accurate to multiply by a surd than to divide by it, and therefore, if at all possible, one must rid the denominator of the surds by suitable multiplication. The process is known as rationalising the denominator. Example 57. Rationalise the denominator of ~~ To do this, multiply both numerator and denominator by VJl since V^x ^3 =3. Then -5* = -^ -^ = - -, or if the result is required in v 3 V 3 x V 3 _3_ decimals it is 2-89. Example 58. Find the value of ' ._ 4 V5 Multiply numerator and denominator by 4+ V^. EQUATIONS 75 Then the fraction -- 7(4 + V5) __ = 7(4+ ~ = 7(4 + 2-236) = 7 x 6-236 l6- 5 H = 3-968. Surds occurring in equations must be eliminated as early as possible by squaring or cubing as the case may demand. Example 59. Solve the equation 3/p 2 = 7. Cubing both sides p 2 = 343 P = 345.- Example 60. Find x from the equation 4 - 2#+3 = 5. Transposing so that the surd is on one side by itself Vzx +3 = 1. Squaring both sides 2X + 3 = i whence x = i. Note. The solutions of all surd equations should be tested. Reverting to the original equation and substituting i for x 4-V2#+3 = 4-V-2 + 3 = 4-1 = 3 and not 5, so that i is not a solution of this equation, but it would be of the somewhat similar equation 4+ \2#+3 = 5. When squaring, either ( \/2#+3) 2 or ( + V2# +3)* = 2^+3, so that the solution obtained may be that of either the one equation or the other. In this case, then, there is no solution to the equation as given. Example 61. Wohler's law for repeated stresses can be expressed where f l = original breaking stress, S = stress variation in terms of / if and f t = new breaking stress. Find f.j, when S = -537/1, /i = 52, x = 2. Substituting the numerical values y 2== ;53_7/2 + \/52 2 -(2 x -537/2X52) *. e., f t = -2685/ t + A/2704 - 76 MATHEMATICS FOR ENGINEERS Arranging so that the surd is isolated /, - -2685/3 = V270 4 - 55 -8/, or -7315/2 = V270 4 - 55- 8/, Squaring '535/2 2 = 2704 - 55-8/5, or '535/ 2 +55-8/ - 2704 = o. Solving for /, by means of the formula f - 55' 8 V3"Q + 579Q /a - ^o7~~ - 55'8 94'5 = 38-7 Qr _ 50^3 1-07 1-07 1-07 = 36-2 or 140-5. The negative solution has no meaning in the cases to which this formula is applied, hence the positive solution alone is taken. Example 62. Find the value or values of x to satisfy the equation V^x 7 + 3 Vzx + 17 = 18. It is best to separate the surds thus 3 Vzx +17 = 18 V^x 7. Square both sides and then 9(2* + 17) = (18)* +( V^x 7)* 2 x 18 x V4# 7 * i. e. t i8x + 153 = 324 + 4* 7 36 V<ix 7 or, isolating the surd 36 V4* 7 = 164 14* or i8V4# 7 = 82 7*. Squaring again 324(4* - 7) = 6724 + 4Q* 2 - 1148* whence 49# 2 2444* + 8992 = o. Factorising (49* 2248) (# 4) = o. 2248 * = W r4 ' To test whether these values satisfy the original equation When x = 4, left-hand side = Vi6 7 + 3 V8 +17 = 3+15 = 1 8, which is the value of the right-hand side x = 4 satisfies. * Always remember that when squaring a "two-term" expression three terms result ; of the character ; (ist squared) + (and squared) (twice product of ist and 2nd). . e., (A + B) 2 '= A 2 + B 2 + 2AB (A - B) 2 = A 2 + B 2 - 2AB. EQUATIONS 7? When * = , left-hand side = > - 7 + 3 49 _ 93 , 2ig 7 ' 7 312 = -r- = 44$ which does not = 18. Hence # = - is not a solution of the given equation ; it however satisfies the equation 3 Vzx + 17 V^x 7 =a 18. Exercises 9. On the Solution of Simultaneous Quadratic Equations and Surd Equations. 1. Find values of x and y to satisfy the equations *-3V= 16 * 2 + 3V Z 2*+ 4y = 50 2. Solve the equations a 2 = 8 + 4y 2 2a + 2y = 7 3. Solve for p and <y the equations 3p*-pq-7i 2 = 5 4. Solve the equations 5# 2 9^+ i2Ary zy z = 132 7^+8y = 54 5. Determine the values of a and b to satisfy the equations a 2 2ab +3 = 2a + 6 = 4 6. Solve for m the equation \/3w a qm = 5. 7. The following formula is used to calculate the length of hob required to cut a worm wheel, for throat radius r and depth of tooth d f= 2 Vd(zr d) Find the depth of tooth when the hob is 3* long and the throat radius is 2". 8. Find a from the equation V8a+ g 3 = 4. 9. Solve for H the equation 25*6H 346 VH= 10000. g _ 10. The formula /, = h V/i 2 xSfi is that given by Unwin, and refers to variation of stress. Bauschinger's experiments in a certain case gave S= '4 1 /*' /i = 22-8, and #=1-5. Find the value of/,. 11. Using the same formula as in the previous example (No. 10) find / 2 when S = / 2 , / x = 30, and x = 2. 12. Solve for a the equation Va +2+ Va = 77= V Ge ~\~ 2 13. Find a value or values of x to satisfy the equation Vi + gx = Vx+ i V^x + i 78 MATHEMATICS FOR ENGINEERS 14. The length x of a strut in a roof truss was required from the equation Vx z 36 + Vx* 4 = 16. Find this length. 15. Find the value of k to satisfy the equation referring to the discharge of water from a tank 12 " ^7 Given that /,= -45x60, ^=2-6x60, A=i5-6, a = - ^, =32-2, 144 A 1= =36, and A, = 25. 16. The following equations occurred when calculating a slope of a form gauge^- w 2 = (*o6 + n}n vn_ _ -0175 n 03 ~ -035 Find the values of m and n to satisfy these conditions. 17. If / = span of an- arch d = rise A = height of roadway above the lowest point of the arch c = highest l z then c+d = h and also c = J^+o Find the values of d when h = 23 ft., and / = 24 ft. 18. The dimensions for cast iron pipes for waterworks are related by the equation where H = head of water in feet t = thickness of metal in inches d = internal dia. of pipe in inches If H = 300 and t = .5 find d. CHAPTER III MENSURATION Introduction. Mensuration is that part of practical mathe- matics which deals with the measurement of lengths, areas, and volumes. A sound knowledge of it is necessary in all branches of practical work, for the draughtsman in his design, the works' manager in his preparation of estimates, and the surveyor in his plans, all make use of its rules. Our first ideas of mensuration, apart from the tables of weights and measures, are usually connected with the areas of rectangles. How much floor space will be required for a planer 4 ft. wide and 12 ft. long ? Here we have the simplest of the rules of mensuration, viz. the multiplication together of the two dimensions. Thus, in this case, the actual space covered is 4 X 12 =48 sq. ft. Area of Rectangle and Triangle. If the rectangle is bisected diagonally, two equal triangles result, the area of each being one-half that of the original rectangle, or we might state it, \ (length X breadth), or as it is more generally expressed, \ base X height or \ height X base. (Note that the \ is used but once; thus we do not multiply \ base by \ height.) This rule for the area of the triangle will always hold, viz. that the area of the triangle is one-half that of the corresponding rectangle, i, e., the rectangle on the same base and of the same height. Thus in Fig. 7, the triangles ABC, AB'C, and AB"C are all equal in area, this area being one-half of the rectangle ACB'D, i. e., %bh. It is the most widely used of the rules for the area of the triangle, because if sufficient data are supplied to enable one to construct the triangle, one side can be considered as the base, and the height (i. e., the perpendicular from the opposite angular point on to this side or this side produced) can be readily measured, whence one-half the product of these two is obtained. A special case occurs when one of the angles is a right angle; then the rule for the area becomes : Area (to be denoted by A) equals one-half the product of the sides including the right angle. 8o MATHEMATICS FOR ENGINEERS One further point in connection with the right-angled triangle must be noted, viz. the relation between the sides. The square on the hypotenuse (the side opposite the right angle, *'. e., the longest side) is equal to the sum of the squares on the other sides. (Euclid, I. 47.) D B B ( A Fig. 7- Fig. 8. In Fig. 8, AB is the hypotenuse, because the right angle is at C, and .(AB) 2 = (AC) 2 -f(BC) or c 2 = b z +a*. A word about the lettering of triangles will not be out of place here. It is the convention to place the large letters A, B and C at the angular points of the triangle, to keep these letters to repre- sent the angles, e. g., the angle ABC is denoted by B, and to letter the sides opposite to the angles by the corresponding small letters. Thus the side BC is denoted by a, because it is the side opposite the angle A. Rule for Area of Triangle when the three sides are given. As previously indicated the rule %bh can here be applied if the triangle is drawn to scale and a height measured. (The triangle can be constructed so long as any two sides are together greater than the third.) If, however, instruments are not handy proceed along the following lines : Add together the three sides a, b, and c, and call half their sum s ; s= Then the area is given by A = Vs(s - a)(s - b)(s - c) This rule will be referred to as the " s" rule, and the proof of it will be found in Chapter VI. Logarithms or the slide rule can be employed directly when using this formula, since products and a root alone are concerned. MENSURATION 81 Example i. One end of a lock gate, 7 ft. broad, is 2 ft. further along the stream than the other when the gates are shut: find the width of the stream. Let 2x = width of stream. (See Fig. 9.) Then f = x z +2* ~ JC - x * = 72 - 2 *= (7-2)(7 + 2) or x = 6-7 so that the width of the stream = 2x = 13-4 ft. Fig. 9. Example 2. Find the area of the triangle ABC, Fig. 10. Area = base x height = $x 14-6 x n -4 = 83-2 sq. ins. C. (Note that 11-4 is the perpendicular on to AB produced.) Fig- 13- Example 3. The pressure on a triangular plate immersed in a liquid is 4-5 Ibs. per sq. ft. The sides of the plate measure 18-1", 25-3", and 17*4" respectively : find the total pressure on the plate. Let a = i8'i, b = 25-3, c = 17-4. Using these figures, the area will be in sq. ins. s = 18-1 25-3+17-4 60-8 " = - - = 30-4 2 2 Then A = V3O- 4 (3O- 4 - i8-i)(3O- 4 - 25-3X30-4- 17-4) = A/30-4 x 12-3 x 5-1 x 13 Taking logs throughout log A = i{log 30-4 + log 12-3 + log 5-1 + log 13} 'I-4829 1 1-0899 7076 1-1139 = 2-1972 U'3943' = log 157-5 A = 157-5 sq. ins. Then total pressure = ^^ X 4-5 Ibs. [feet" x 144 L. Ibs. feet 1 Ibs i 82 MATHEMATICS FOR ENGINEERS A further rule for the area of a triangle will be found in Chapter VI. A rule likely to prove of service is Area of an equilateral triangle = '433 x (side) 2 . Thus if the sides of a triangle are each 8 units long its area is '433 X 8 2 , i. e., 277 sq. units. Exercises 10. On Triangles and Rectangles. 1. A boat sails due E. for 4 hours at 13-7 knots and then due N. for 7 hours at 10-4 knots. How far is it at the end of the n hours from its starling-point ? 2. Find the diagonal pitch of 4 boiler stays placed at the corners of a square, the horizontal and vertical pitch being 16*. 3. If a right-angled triangle be drawn with sides about the right angle to represent the electrical resistance (R), and reactance (2nfL), respectively, then the hypotenuse represents the impedance. Find the impedance when / = 50, L,= 'i^g, R = 5O, and ?r = 3-142. 4. It is required to set out a right angle on the field, a chain or tape measure only being available. Indicate how this might be done, giving figures to illustrate your answer. 5. A floor is 29'-$" long and n'-io" broad. What is the distance from one corner to that opposite ? 6. At a certain point on a mountain railway track the level is 215 ft.; 500 yds. further along the track the level is 227 ft. Express the gradient as (a) i in x (x being measured along the track). (b) i in x (x being measured along the horizontal). Fig. Tl. 7. For the Warren Girder shown at (a), Fig. n, find the length of the member AB. 8. A roof truss is shown at (fc), Fig. n. Find the lengths of the members AB, BC and AC. MENSURATION 83 9. A field is 24 J chains long and 650 yds. wide. What is its area in acres ? (Surveyors' Measure is given on p. 87.) 10. Find how many " pieces " of paper are required for the walls of a room 15 ft. long, i2'-6" wide and 8 ft. high, allowing 8 % of the space for window and fireplace (a " piece " of paper being 21" wide and 9 ft. long). 11. A courtyard 15 yds. by 12 yds. is to be paved with grey stones measuring 2 ft. x 2 ft. each, and a border is to be formed, 2 ft. wide, of red stones measuring i ft. x i ft. How many stones of each kind are required ? 12. A room 15 ft. by 12 ft. is to be floored with boards 4^* wide. How many foot run will be required ? 13. Before fracture the width of a mild steel specimen was 2-014" and its thickness '387". At fracture the corresponding dimensions were i'524" and -250". Find the percentage reduction in area. 14. A rectangular plot of land J mile long and 400 ft. wide is to be cut up into building plots each having 40 ft. frontage and 200 ft. depth. How many such plots can be obtained ? 15. The top of a tallboy is in the form of a cone ; the diameter of the base is 4", and the vertical height is ij". Find the slant height. 16. A bar of iron is at the same time subjected to a direct pull of 5000 Ibs., and a pull of 3500 Ibs. at right angles to the first. Find the resultant force due to these. 17. At a certain speed the balls of a governor are 5" distant from the governor shaft ; the length of the arms is 10". Find the " height " of the governor h and hence the number of revs, per sec. . -816 n from h = -. n* 18. A load on a bearing causes a stress of 520 Ibs. /a*. If the stress is reckoned on the " projected area " of the bearing, the diameter of which is 4" and the length 5%', find the load applied. 19. The sides of a triangle are 17-4", 8-4" and 15-7" respectively. Find its area, by (a) Drawing to scale and use of i base x height rule. (6) Use of " s " rule. (See p. 80.) 20. Find the rent of a field in the form of a triangle having sides 720, 484 and 654 links respectively, at the rate of 2 los. per acre. (See note to Ex. 9.) 21. Find the area of the joist section shown in Fig. ua. (Thickness of flange is 0-2*.) 22. Neglecting the radii at the corners, calculate the areas of the 8 4 MATHEMATICS FOR ENGINEERS sections in Fig. 12 : viz. (6) channel section, (c) unequal angle, and (d) tee section. h 4 "-H Ay/'y/jT A Fig. 12. Mild Steel Rolled Sections. Area of Parallelogram and Rhombus. From the three- sided figure one progresses to that having four sides, such being spoken of generally as a quadrilateral. Of the regular quadrilaterals reference has already been made to the simplest, viz. the rectangle (the square being a particular example), for which the area = length X breadth. Since the area of a triangle is given by the product, \ base X height, it follows that (a) Triangles on the same base and having the same height are equal in area, and (b) Triangles on equal bases and having the same height are equal in area. MENSURATION Thus, if in Fig. 13 the length FC is made equal to the length ED, the triangles AED and BFC will be equal in area, since the bases are equal and the height is the same. Also it will be seen that the sides AD and BC are parallel, so that the figure ABCD is a parallelogram. Then The area of the parallelogram ABCD = area of figure ABFD + area of triangle BFC = area of figure ABFD + area of triangle AED = area of rectangle ABFE = AB x BF. This result could be expressed in the general rule, " Area of a parallelogram = length of one side x the perpendicular distance from that side to the side parallel to it." In the case of the rhombus (a quadrilateral having its sides equal but its angles not right angles) one other rule can be added. Its diagonals intersect at right angles, and hence its area can be expressed as equal to one-half the product of its diagonals; i. e., Area = X BD X AC, the reference being to the rhombus in Fig. 14. This rule should be proved as an example on the J base X height rule for the triangle. Area of Trapezoid. A trapezoid is a quadrilateral having one pair of sides parallel. Its area = mean width X perpendicular height. = (sum of parallel sides) X perpendicular distance between them. Fig. 15- * 30' H Fig. 1 6 Cross Section of a Cutting. In Fig. 15, AB and CD are the parallel sides, and the area 86 MATHEMATICS FOR ENGINEERS Example 4. Calculate the area of the cross section of a cutting, having dimensions as shown in Fig. 16. Area = {70 + 30} x 16 sq. ft. = 800 sq. ft. Example 5. The kathode, or deposit plate, of a copper voltameter has the form shown in Fig. 17. Calculate, approximately, the area and hence the current density (i. e., amperes per sq. in. of surface) if 1-42 amperes are passing. Fig. 17. Fig. 1 8. We may divide the surface of the plate into three parts, A, B, and C. Area of A = 2-6 x 2-65 = 6-9 sq. ins. Area of B = ('7+ 2 ' 6 5\ x .3. = x \ 2 / Area of C = -6 x -7 = -42 Total area of one side = 8-74 This is the area of one side ; but the deposit would be on both sides total area = 2 X 8-74 = 17-48 sq. ins. and current density = Jj g = '0812 amp, per sq. in. or i amp. for 12-3 sq. ins. of surface. Example 6. Find the area of the rhombus, one side of which measures 5" and one diagonal 8". Let 2X= length of other diagonal in inches (Fig. 18). Then, by the right-angled triangle rule, ** = 5 2 -4 2 = 9 x = 3 and 2X = 6. Area = i (product of diagonals) = J x 8 X 6 = 24 sq. ins. MENSURATION 87 This example could also have been worked as an exercise on the " s" rule, the sides of the triangle being 5, 5 and 8 respectively. Areas of Irregular Quadrilateral and Irregular Poly- gons. Having dealt with the regular and the " semi " regular quadrilaterals, attention must now be directed to the irregular ones. No simple rule can be given that will apply to all cases of irregular quadrilaterals : the figure must be divided up into two triangles and the areas of these triangles found in the ordinary way. This method applies also to irregular polygons (many-sided figures) having more than four sides; but these figures split into more than two triangles. Example 7. Find the area of the quadrilateral ABCD, Fig. 19, in which AD = 17 ft., DC= 15 ft., BC= 19 ft., ^ AC= 26 ft., and the angle ABC is a right angle. It will be necessary to find the length of AB. By the rule for the right-angled triangle, (AB) 2 = 26 2 -ig 2 = 7 x 45 = 3'5 AB = 17-76 ft. The quadrilateral ABCD = Triangle ADC + Triangle ABC. Dealing first with the triangle ADC, its area must be found by the " s" rule. s = 2 A = A/29 X 12X14X3 = 120-9 sq. ft. Area of triangle ABC = x 17-76 x 19 = 169 sq. ft. /. Area of quadrilateral ABCD =121+169 = 290 sq. ft. Example 8. Find the area of the plot of land represented in Fig. 20 (being the result of a chain survey). Some of the dimensions are given in chains : it is worth while to remind ourselves of the magnitude of a chain. SURVEYORS' MEASURE i chain = 22 yards = 66 feet. i chain = 100 links (i link = 7-92 .) 10 chains = i furlong. 80 chains = i mile. i sq. chain - 22 2 = 44 sq. yards = T V of an acre. or 10 sq. chains = i acre. 10 sq. chains = 100,000 sq. links. i acre 100,000 sq. links, 88 MATHEMATICS FOR ENGINEERS The given figure is divided by the " offsets " into triangles and trapezoids ; the offsets being at right angles to the main chain lines. It will be convenient to work in feet. Dealing with the separate portions. Area AC J = \ xig8 X2&4 = 26136 sq. ft. ACB =ixig8x24 = 2376 192 H5 2 M II2O 4 80 650 28OO 225 CDB = CDEF = FEGH = (20+8)x8o = HGJ =x8xi2o = JKL =ix 100x13 = LKMN= ^(13 + 15) x 200 = NMA = 35131 sq.ft. .'. Total area = 35131 sq. ft. = 8-07 sq. chns. = -807 acre. Fig. 20. Areas of Regular Polygons. Regular polygons should be divided up into equal isosceles triangles ; and there will be as many of these as the figure has sides. The areas of the triangles are best found (at this stage) by drawing to scale, and as an aid to this the following rule should be borne in mind. The angle of a regular polygon of n sides X 90 degrees. Thus, for a regular pentagon -(a five-sided figure) n = 5 and the angle = [ (2X5) ~ 4 1x9o = 108. Alternatively, the following construction may be used. Suppose that the area of a regular heptagon, i. e., a seven-sided figure, is required, the length of side being i|"; and we wish to find its area by drawing to scale. Set out on any base line (Fig. 21) a semicircle with A as centre and radius ij" (the length of side). Divide the semi-circumference into seven (the number of sides) equal parts, giving the points a, B, c, d, e, f, G (this division to be done by trial). Through the second of these divisions, viz. B, draw the line AB; drawing also lines Ac, Ad, etc., radiating from A. With centre B and radius i" strike an arc cutting Ac in C ; then BC is a side of the heptagon. This process can be repeated until the figure ABCDEFG is completed, MENSURATION 89 Bisect AG and GF at right D To find the area of ABCDEFG. angles and note the point of intersection O of these bisectors ; this being the geometrical centre of the figure. Measure OH (it is found to be 1-56"). Then area of AOG = \ AG X OH = i x 1-5 X 1-56 = 1*17 sq. ins. /. Area of ABCDEFG = 7XAAOG = 7x1-17 = 8-19 sq. ins. Fig. 21. Area of Polygon. Exercises 11. On Areas of Quadrilaterals and Polygons. 1. The central horizontal section of a hook is in the form of a trapezoid 2$" deep, the inner width being 2" and the outer width $*. Find the area of the section. 2. The diagonals of a rhombus are I9'74" and 5-28" respectively. Find the length of side and the area. 3. Find the area of the quadrilateral ABCD shown at (a), Fig. 22. What is the height of a triangle of area equal to that of ABCD, the base being 5* long ? 4. A field in the form of a quadrilateral ABCD has the following dimensions in yards : CD = 38, DA = 29, AC = 54, BE (perpendicular from B on to AC) = 23. Find its area in acres. 5. Reproduce (6), Fig. 22, to scale, and hence calculate the area of ABCDEF. 6. Find the area, in acres, of the field shown at (c), Fig. 22. Fig. 22. 90 MATHEMATICS FOR ENGINEERS 7. A retaining wall has a width of 4 ft. at base and 2'-6* at top. The face of the wall has a batter of i in 12, and the back of wall is vertical. Find the area of section and also the length along the face. 8. The side slopes of a canal (for ordinary soil) are ij horizontal to i vertical. If the width of the base is 20 ft. and the depth of water is 5 ft. find the " area of flow " when the canal is full. 9. Find the hydraulic mean depth ( i. e., - Area of flow } for the \ Wetted perimeter/ canal section for which the dimensions are given in Question 8. 10. The end of a bunker is in the form of a trapezoid. Find its area if the parallel sides are 9'-$", and 15'-! i" respectively, the slant side being 24-8", while the other side is perpendicular to the parallel sides. 11. A regular octagon circumscribes a circle of 2* radius. Find its area. 12. Find the area of a regular hexagon whose side is 4-28". 13. The " end fixing moment " for the end A of the built-in girder, Fig. 220, is found by making the area ABEF equal to the area ABCD. Find this moment, i. e., find the length AF. 14. A plate having the shape of a regular hexagon of side 2.\" is to be plated with a layer of copper on each of its faces. Find the current required for this, allowing 1-6 amperes per 100 sq. ins. 15. An irregular pentagon of area 59-08 sq. ins. is made up of an equilateral triangle with a square on one of its sides. Find the length of side. 16. Neglecting the radii at the corners, find the approxi- mate area of the rail section shown at (a), Fig. 12. 22a Circumference and Area of Circle. When n, the number of sides of a polygon, is increased without limit, the sides merge into one outline and the polygon becomes a circle. A circle is a plane figure bounded by one line, called the circum- ference and is such that all lines, called radii, drawn to meet the circumference from a fixed point within it, termed the centre, are equal to one another. The meanings of the terms applied to parts of the circle will best be understood by reference to Fig. 23 and Fig. 24. If a piece of thread be wrapped tightly round a cylinder for, say, five turns and the length then measured and divided by 5, the length of the circumference may be found. Comparing this with the diameter, as measured by callipers, it would be found to be about 3, times as long. Repeating for cylinders of various sizes, the same ratio of these lengths would be found. The Greek symbol TT (pi) always denotes MENSURATION this ratio of circumference to diameter, which is invariable ; but its exact value cannot be found. It has been calculated to a large number of decimal places, of which only the first four are of use to the practical man. For considerable exactness TT can be taken as 3'i4i6 : however, \ 2 - or 3-1428 is quite good enough for general use, the error only being about 12 in 30,000 or about '04%. Even - 2 . 12 - need not be remembered if a slide rule be handy, for a marking will be found to represent TT (see Fig. i, p. 17). Fig. 24. Then circumference = TT x diameter = ird or 2irr where d = diam. and r = radius. Also Area = -n-r 2 or ^d 2 4 - = 7854 : a marking on the slide rule indicating this. 4 (The mark M on the slide rule is - J It is sometimes necessary to convert from the circumference to the area; thus ATT ATT ATT [ stands for circle and 0ce for circumference.] Example 9. Find the diameter of the driving wheel of a locomotive which in a distance of 3 miles makes 1010 revolutions (assuming no slipping). In one revolution, the distance covered = 0ce. total distance 3 x 5280 f . i* )CC ; ~~ " ' ""^ " A* number of revolutions 1010 and Aiim W = 92 MATHEMATICS FOR ENGINEERS Example 10. Find the area of the cross section of shafting, 3^" diam. Area = - x 3-5* = 9-62 sq. ins. 4 Notice that - or '7854 is in the neighbourhood of '75 or f ; therefore, 4 for approximation purposes, square the diameter (to the nearest round figure) and take f of the number so obtained. In this case, (3'5) 2 = 12 approximately, and f of 12 = 9. Areas of circles can most readily be obtained by the use of the slide rule, the method being as follows Set one of the C's (marked on the C scale) (refer Fig. I, p. 17) level with the diameter on the D scale, place the cursor over i on the B scale, then the area is read off on the A scale ; the approxima- tion being as before. This method is of the greatest utility, and several examples should be worked by means of it for the sake of practice. Examples Dia. Area Approximation 4-8 18-1 fx 25=18 79'5 4965 | x 6400 = 4800 65 332 | x -50 = -36 If the C's are not indicated on the C scale of the slide rule, markings should be made for them at 1-128 and 3-569 respectively. The reason for these markings may be explained as follows The area of a circle = -^ 2 , or, as it might be written, d 2 -^* 4 TT Now 2 = 1-286, of which the square root is 1-128. A marking is 7T thus placed at 1-128, so that when this mark is set level with the diameter on the D scale, the reading on the D scale opposite the index of the C scale gives the value of d -i- */4. By reading the figure on the A scale level with the index on the B scale, the square of d 4- A /3 or d 2 -f- - is found; this being the area of a V 7T circle of diameter d. For convenience in handling the rule a marking is made at MENSURATION 93 3-569 on the C scale also ; this figure being obtained by extracting the square root of 12-86 instead of that of 1-286. Some slide rules are supplied with a three-line cursor. If the centre line is placed over the dia. on the D scale then the left hand line is over the area on the A scale. Example n. Find the connection between circumferential pitch and diametral pitch (as applied to toothed wheels). The circumferential pitch . _ Qce of pitch circle _ ird number of teeth N i -4- v, .A number of teeth N The diametral pitch p d = - - - = diam. of pitch circle d I 7T Hence, p c ir X = N "P* ~d circumferential pitch = diametr " al pitch E. g., if pc is ~ , then = -375 or p d = ~ = 8-37". To find the area of an Annulus, i.e., the area between two concentric circles. It is evident that the area will be : Area of outer area of inner circle = TrR 2 - T' 2 - (Fig- 25.) r///y . This can be put into a form rather \///\ -^^.. more convenient for computation, thus Area of annulus = 7r(R 2 -f 2 ) or^(D 2 -rf 2 ) VAmTuTus^ X^VVX///' ' or, in a form more easily applied Fig. 25. Area of annulus = 7r(R-rXR+r) or -(D-d)(D+d). This rule can be written in a form useful when dealing with tubes, thus . iR 4- zr r\ Area = -n-(R - r}(R + r) = 2^(R - r)(- ) = 2v X t X average radius = TT X average diameter X / where t is the thickness of the metal of the tube. 94 MATHEMATICS FOR ENGINEERS Example 12. What is the area of a piece of packing in the form of a circular ring, of outside diameter gj" and width 3 J" ? Here, R = 4-75", r = 4'75 ~ 3'25 = i'5" Hence the area = (R- r)(R + r) = ^(4*75 + i'5)(4'75 ~ 1 '5) = x 6-25 x 3-25 = 63-7 sq. ins. Example 13. A hollow shaft, 5* internal diam., is to have the same sectional area as a solid shaft of n* diam. Find its external diam. Area of solid shaft =-xn a = -x 121. 4 4 Let D = outside diam. required. Area of hollow shaft = -(D 2 - 25) 4 The two areas are to be the same ; equating the expressions found for these (D 2 25) = x 121 4 4 whence D 2 25 = 121 and D 2 = 146 .'. D = 12-07". Products, etc., of TT. Certain relations occurring frequently are here stated for reference purposes. TT = 3-142 - = -318 = ff * = 9-87 4^2 = 39'4 8 sa Y 39'5 4* 1 = 4 -I 9 o TT TT 47T , (often taken i = .^236 = -7854 - = 1-256 v 6 J 4 10 as f). log TT = -4972 Exercises 12. On Circumference and Area of Circles. 1. Find the circumference of a circle whose diam. is 7-13*. 2. The semi-circumference of a circle is 91-4 ft. What is its radius ? 3. Find the area of a circle of diam. i^'-^". 4. The following figures give the girth of a tree at various points along its length. Find the corresponding areas of cross sections : 4-28, 5-19, 6-47, 2-10, -87. {Suggestion : area = - ; first find value of constant multiplier 4T (approx. -08). Keep the index of the slide-rule B scale fixed at this ; 4* MENSURATION 95 place cursor over Qce on the C scale and read off result on A scale ; the squaring and the multiplying are thus performed automatically.} 5. If the circumferential pitch of a wheel is i J*. find the diametral pitch. (See Example n, p. 93.) 6. A packing ring, for a cylinder 12* diam., before being cut is 12- 5" diam. How much must be taken out of its circumference so that it will just fit the cylinder ? 7. A circular grate burning 10 Ibs. of coal per sq. ft. of grate per hour burns 66 Ibs. of coal in an hour. Find the diameter of the grate. 8. Assuming that cast-iron pulleys should never run at a greater circumferential speed than i mile per min., what will be the largest diam. of pulley to run at 1120 revs, per min. ? 9. The wheel of a turbine is 30* diam. and runs at 10600 R.P.M. What is the velocity of a point on its circumference ? Note. The rule used in questions such as this is v = 271-fN, where v = velocity in feet per min., r = radius of wheel in feet, and N = number of revolutions per minute. 10. A piece, 4" long, is cut out of an elastic packing ring, fitted to a cylinder of 30" diam., so that the ends are now J* apart. Find the diam. of the ring before being cut. 11. Find the diameter of an armature punching, round the circum- ference of which are 40 slots and the same number of teeth. The width of the teeth and also of the slots (at circumference) is '35*. 12. While the load on a screw jack was raised a distance equal to the pitch of the screw ("), the effort was exerted through an amount corresponding to i turn of a wheel 10-51" in diam. Find the VelocRy- , . f .... distance moved by effort"! Ratio of the machine \ V.R. = ,-r-r- L distance moved by load ) 13. The stress / in a flywheel rim due to centrifugal action is given by / = , where w = weight of rim in Ibs. per cu. in., v = circum- ferential speed of rim in ft. per sec., and g = 32-2. Find the revs, per min. if /= 12 x 2240 when w = -28 and diam. = 10 ft. 14. Find the bending stress in a locomotive connecting rod revolving at revs, per sec. from the equation , p ATr 2 n z yrl z . 480 stress =| x * ^ where y = J, p = ^g r = 12, / = 120, k 2 = 3, and g = 32-2. The driving wheels are 6 ft. in diam., and the locomotive travels at 40 miles per hour. 15. Find the area of the section of a column, the circumference of which is 18-47". 16. Calculate the diameter of a circular plate whose weight would be the same as that of a rectangular plate measuring 2'-6* by 3'-2*, both plates being of the same thickness and material. 17. Find the area of the section of a rod of -498" diam. 18. If there is a stress of 48000 Ibs. per sq. in. on a rod of -566* diam., what is the load causing it ? 19. Find the total pressure on a piston 9* diam., when the other 9 6 MATHEMATICS FOR ENGINEERS side of the piston is under a back pressure of 3 Ibs. per sq. in. above a vacuum. Gauge pressure (pressure above atmosphere) = 64 Ibs. per sq. in. i atmosphere = 14-7 Ibs. per sq. in. 20. The driving wheel of a locomotive, 5 ft. in diameter, made 10000 revolutions in a journey of 26 miles. What distance was lost owing to slipping on the rails ? 21. The total pressure on a piston was 8462 Ibs. If the gauge registered 51 Ibs. per sq. in. (i. e., pressure above atmosphere) and there was no back pressure, what was the diameter of the piston ? 22. Find the area of section of a hollow shaft of external diam. 5^* and internal diam. 3". 23. A circular plot of land is to be surrounded by a fencing, the distance between the edge of the plot and the fencing being the same all round, viz. 6 ft. The length of the fencing required is 187 ft. Find the area of the space between the plot and the fencing. 24. Find the resistance of 60-5 cms. length of copper wire of diam. 068 cm. from R-* a where a = area in sq. cms., / = length in cms., and k = resistivity = 0000018 ohm per centimetre cube. 25. The buckling load P on a circular rod is given by Af c where A = area of section diameter P = (_L\ VK/ and K = Find the diameter when P = 188500, f c = 67000, c 26. A pair of spur wheels with pitch of teeth ij" is to be used to transmit power from a shaft running at 120 R.P.M. to a counter shaft run- ning at 220 R.P.M. The distance between the centres of the shafts is to be 24" as nearly as possible. If the diameters of the pitch circles are inversely as the R.P.M., find the true distance between the centres and the number of teeth on each wheel. 27. Calculate the area of the zero circle (the circle of no registration of the wheel), -the radius of which is BD, for the planimeter shown in outline in Fig. 26. 28. The resistance of i mile of copper wire is found from 5600 , and L = 50. Tracioq Po'iot Fig. 26. Amsler Planimeter. R 04232 area in sq. ins. Find the resistance of I mile of wire of No. 22 B.W.G. (diam. 03*). MENSURATION ^ Length of Chord and Maximum Height of Arc In Fig. 27 let h = maximum height of the arc, 2a = length of the chord, and r = radius. Then, by the right-angled triangle rule, applied to the triangle AGO Transposing terms a* = 2rhh 2 whence a = V2rhh? or length of chord = 2\/2r/i-/i 2 a rule giving the length of chord when the radius and maximum height of arc are given. B If h is expressed as a fraction of the radius, say h = fr, the rule for the length of chord becomes length of chord = 2r\/2/f 2 . From equation (i) 2rh = a 2 -}-/; 2 r = 2/1 a rule giving the radius when the chord and the maximum height of arc are known. From (i) also, h 2 2rh-\-a z = o, and from this, by solution of the quadratic , 2r v 2 2 or Vr*-i giving two values for h (i. e., for the arc less than, and the arc greater than, the semi-circumference) when the radius and length of the chord are known. If two chords intersect, either within or without the circle, the rectangles formed with their segments as sides are equal in area, Euc. Ill, 35 and 36. Thus, in Fig. 28, at both (a) and (&) AP.PB = CP.PD MATHEMATICS FOR ENGINEERS If C and D coincide as at (c), Fig. 28, then (PI) 2 = AP ._PB Fig. 28. Example 14. The hardness number of a specimen, according to load Brinell's test, is given by curved area ofdepression' Express this as a formula. The curved area (of segment of sphere) = 2nrh (see p. 120). r is radius of the ball making the indentation. D = diameter of depression. Then corresponds to a in the foregoing formulae, I * D 2 = r./r* V 4 so that h and hardness number = Length of Arc. Exact Rule. The length of the arc depends on the angle it subtends at the centre of the circle : the total angle at the centre is 360, this being subtended by the circumference. An angle of 36 would be opposite an arc equal to one-tenth of the circumference, whilst if the arc was = of Oce, the angle at the centre would be 120. arc angle in degrees In general - = ^ Oce 360 2-trr X angle in degrees __ angle in degrees X radius or, cure 7= 360 57-3 If the arc is exactly equal in length to the radius, the angle then subtended ought to serve as a useful unit of measurement, MENSURATION 9g for one always expresses the circumference in terms of the radius. This angle is known as a radian. If the chord were equal to the radius, the central angle would then be 60, so that when the arc is involved in the same way the angle must be slightly less than 60. Actually, the radius is contained 2ir times in the Oce, hence 2?r radians = 360, i. e., i radian = - = 57-3. 2ir Therefore, to convert from degrees to radians divide by 57'3. Thus 273 = -21 = 4.76 radians. Radian or circular measure is the most natural system of angular measurement ; ah 1 angles being expressed in radians in the higher branches of the subject. A simple rule for the length of an arc can now be established. ,, , 2-n-rx angle (degrees) Length of arc = - & ^ 2irrx angle (radian) /since 360 \ 27r \ = 27r radians/ = rx angle subtended by the arc, expressed in radians. Now it is usual to represent the measurement of an angle in radians by 6, and when in degrees by a. Thus the angle AOB in Fig. 27, subtended at the centre of the circle by the arc ADB would be expressed as 0, if in radians ; or a, if in degrees. Hence, length of arc = g or re Example 15. A belt passing over a pulley 10" diam. has an angle of lap of 115 : find the length of belt in contact with the pulley. In this case r = 5 and a = 115 /. length in contact = length of arc = Example 16. What angle is subtended at the centre of a circle of 14-8 ft. diam. by an arc of 37-4 ft. ? Arc = rd . a _ arc _ 37'4 v<2 _ .. . 6 - -r-^8 x Thus the angle required is 5-05 radians or 290 degrees. TOO MATHEMATICS FOR ENGINEERS It may be found of advantage to scratch a mark on the C scale of the slide rule at 57-3, so that the conversions from degrees to radians can be performed without any further tax on the memory. Example 17. It is required to find the diameter of the broken eccentric strap shown in the sketch (Fig. 29). Here a = 2* and h = i-z". Then r = a z + h z 2h 4+ i-44 2-4 " 2-4 = 2-265. diam. = 4-53* (probably Fig. 29. An approximate rule for the length of arc is that known as Huyghens' ; viz. Length of arc = *~ * o where c 2 and c x represent the chord of half the arc and the chord of the arc respectively (*'. e., c^ = 2a}. (See Fig. 27.) To find the Height of an Arc from any Point in the Chord. It is required to find the height EF (see Fig. 27) of the arc ADB, the length of chord AB, the maximum height CD of the arc ADB and the distance CF being given. If O is the centre of the circle, OE is a radius and its length can be found from r = 2h Then (OE) 2 = (EG) 2 + (GO) 2 = (EG) 2 + (CF) 2 and of these lengths, OE and CF are known ; therefore EG is found. But FG is known, since FG = OC = r h. :. the height EF, which = EG FG, is known also. A numerical example will demonstrate this more clearly. Example 18. A circular arc of radius 15* stands on a base of 24". Find its maximum height, and also its height at a point along the base 5* from its ex- tremity. (Deal only with the arc less than a semi- circle.) (See Fig. 30.) Fig. 30. MENSURATION I0 i To find h. We know that r 15*, and a = 12* hence h = r Vr z a* = 15 V225-I 44 = 15 9 = 6* or 24*. According to the condition stated in brackets h must be taken as 6*, i. e., the maximum height of the arc is 6*. Then I5 2 = EG 2 +7 EG 2 = is 2 -; 8 = 22 x 8 = 176 or EG = 13-26* CO = r-h = 15-6 = 9* .*. EF = 13-269 = 4-26* or the height of the arc at the 5* mark is 4-26*. Area of Sector. A sector is a portion of a circle bounded by two radii and the arc joining their extremities; it is thus a form of triangle, with a curved base (see Fig. 24). Its area is given by a rule similar to that for the area of a triangle, viz., | base* X height, but in this case the base is the arc and the height is the radius (the radius being always perpendicular to the circumference). Hence Area of sector = J arc x radius, or, in terms of the radius, and angle at the centre (in radians). Area = ^r 2 ^, since for the arc we may write rd. The area of the sector bears the same relation to the area of the circle as the length of arc does to the o ce, i. e. Area of sector angle (in degrees) area of O 360 .*. Area of sector = ^7; x Ti-r 2 BQU Area of Segment. The area of the segment, being the area between the chord and the arc (see Fig. 24), can be obtained by subtracting the area of the triangle from that of the sector. Thus, in Fig. 24 Area of segment ADB = area of sector OADB area of triangle OAB. In place of this exact rule, we may use an approximate one, viz. 2h f 1 chord + 3 arc\ Area of segment = y {- ^ t where h = maximum height of segment. 102 MATHEMATICS FOR ENGINEERS When the arc is very flat the chord and arc become sensibly the same, so that 2hfio chordl Area of segment = ! | 2 = - X h x chord 2 = X maximum height X width. The area of a segment may also be found from the approximate rule 4 Id Area of segment = -^h z *J -r '608 o v a where d = diam. of circle, and h = maximum height of segment. Example 19. The Hydraulic Mean Depth (H.M.D.) a factor of great importance in connection with the flow of liquids through pipes or channels is equal to the section of flow divided by the wetted perimeter. Find this for the case illustrated in Fig. 31. Here, section of flow = area of segment ACB = area of sector OACB area of tri- angle OAB = -~ X7rx6 2 4x6x6 360 = 977 18 = 10-3 sq. ins. Wetted perimeter _ arc ACB = ^ x 2 X n- X 6 = ^JT = 9-42" .*. H.M.D. (usually denoted by tri) e IQ'3 = 1-094 Note that for a pipe running full bore the H.M.D. MENSURATION 103 Exercises 13. On Arcs, Chords, Sectors and Segments of Circles. In Exercises i to 5, the letters have the following meanings as in Fig. 27, v = radius, c^ = chord of arc, c 2 = chord of half arc, h = maxi- mum height of arc, and a = angle subtended at the centre of the circle by the arc. 1. r = 8*, c t = 2-4"; find c l and h. 2. c t = 80", r = 50"; find c a and h. 3. Ci = 49*, c s = 25*; find r and h. 4. Ct = 6", r = 9*; find arc and area of segment. 5. c t = io*, h = 1-34*; find area of segment and a. 6. A circular arc is of 10 ft. base and 2 ft. maximum height Find the height at a point on the base i'-6* distant from the end, and also the distance of the point on the base from the centre at which the height is i ft. 7. A circular arc has a base of 3* and maximum height 4*. Find (a) radius, (&) length of arc, (c) area of segment, (d) height of arc at a point on the base i* distant from its end. 8. A crank is revolving at 125 R.P.M. Find its angular velocity (t. e., number of radians per sec.). 9. If the angular velocity of a flywheel of i2'-6* diam. is 4-5, find the speed of a point on the rim in feet per minute. 10. Find the area of a sector of a circle of 9-7* diam., the arc of the sector being 12-3" long. 11. One nautical mile subtends an angle of i minute at the centre of the earth; assuming a mean radius of 20,890,000 feet, find the number of feet in i nautical mile. 12. Find the difference between the apparent and true levels (i e CE), if AC = 1500 yards and R = 3958 miles. [See (a), Fig. 32.] Fig. 32. 13. (6), Fig. 32 (which is not drawn to scale), shows a portion of a curve on a tramway track. If R = radius of quickest curve allowable (in feet), T = width of groove in rail (in inches), and B = greatest permissible wheel base (in feet) for this curve, find an expression for B in terms of R and T. 14. A circle of 2-4" diam. rolls without slipping on the circum- ference of another circle of 6-14" diam. What angle at the centre is swept out in i complete revolution of the rolling circle ? 104 MATHEMATICS FOR ENGINEERS 15. A railway curve of J mile radius is to be set out by " i chain " steps. Find the " deflection angle " for this, i. e., the angle to which the theodolite must be set to fix the position of the end of the chain. The deflection angle is the angle between the tangent and the chord. 16. Fig. 33 shows a hob used for cutting serrations on a gauge. It was found that the depth of tooth cut when the cutting edge was along AB was not sufficiently great. Find how far back the cutter must be ground so that the depth of serration is increased from -025* to '027*, i. e., find x when AB = -025" and CD = -027". C25 Fig. 33. Gauge Hob. The Ellipse. The ellipse is the locus of a point which moves in such a way that the sum of its distances from two fixed points, called the foci is constant. This constant length is the length of the longer or major axis. In Fig. 34, if P is any point on the ellipse, PF -f- PF 1 = constant = AA 1 , F and F 1 being the foci. Fig. 34. The Ellipse. Let major axis = 2, and minor axis = 2&. Then from the definition, FB = F X B = a. ; In the triangle FOB, (FB) 2 = (FO) 2 + (OB) 2 or a 2 = (FO) 2 + (b) z FO = Vo 2 ^ 2 so that if the lengths of the axes are given the foci are located. Area = nab. (Compare with the circle, where area = -n-rr.) The perimeter of the ellipse can only be found very approximately as the expression for its absolute value involves the sum of an MENSURATION 105 infinite series. Various approximate rules have been given, and of these the most common are, perimeter = 7r(a+6), or the second of which might be written in the more convenient form 2 + b 2 . These rules, however, do not give good results when the ellipse is flat. A rule which appears to give uniformly good results is that of Boussinesq, viz. perimeter = 7r{l-5(a -f b) Vab} . The height of the arc above the major axis at any point can most easily be found by multiplying the corresponding height of the semi-circle described on the major axis as diameter by -, a e. g., referring to Fig. 34, QN = -XMN. Example 20. The axes of an ellipse are 4-8" and 7-4". Find its perimeter and its area. According to our notation, viz. as in Fig. 34, ia = 7-4, a = 3-7 26 = 4-8, 6 = 2-4. Then the perimeter = ir(a + b) = 77(6-1) = 19-15" 77 A/2(a 2 +& 2 ~) = TT V2(i9- 45 ) = 19-58" 7r{i-5(a + 6)- Vab} = 7r{i-5(6-i)- V^y x 2-4} or or Area Fig. 35. The Parabola. 106 MATHEMATICS FOR ENGINEERS The Parabola. The parabola is the locus of a point which moves in such a way that its distance from a fixed straight line, called the directrix, is always equal to its distance from a fixed point called the focus. Referring to Fig. 35, PZ = PF, where F is the focus, and P is any point on the curve. The distance BA, which is equal to AF, is always denoted by a. The chord LL 1 through the focus, perpendicular to the axis, is called the latus rectum, and from the definition it will be seen to be equal to 40. The latus rectum, in fact, determines the propor- tions of the parabola just as the diameter does the size of the circle. If PQ = y and AQ = x, then FQ = AQ AF = (x a) and PF = PZ = BQ = x+a. Then in the triangle FPQ, (FP) 2 = (PQ) 2 +(FQ) 2 or x z +a z -\-2ax = y 2 +# 2 +0 2 2ax whence y 2 = 4ax or (| width) 2 = latus rectum X distance along axis from vertex, Fig. 36. e. g., (MR) 2 = 4 x AR in Fig. 36. If a semi-circle be drawn with F as centre and with FP as radius, to cut the axis of the parabola in T and N, PT is the tangent at P and PN is the normal. (Fig. 35.) The distance along the axis, under the normal, *. e., QN in Fig. 35, is spoken of as the sub-normal. For the parabola, the length of the sub-normal is constant, being equal to 2a, i. e., % latus rectum. Use is made of this property in the design of governors. If the balls are guided into a parabolic path, the speed will be the same for all heights, for it is found that the speed depends on the sub-normal of the parabola, and as this is constant so also must the speed be constant. 2 The area of a parabolic segment = = of surrounding rectangle, . e., B area of P X AP (Fig. 36) = |xPP x xAQ. Length of parabolic 8 D 2 arc = S-f-s- -g- approximately, where S = span and D = droop or sag, as indicated in Fig. 36. Circular and other arcs are often treated as parabolic when the question of the areas of segments arises; and if the arcs are very flat no serious error is made by so doing. The rule for the MENSURATION 107 area of a parabolic segment is so simple and so easily remembered that one is tempted to use it in place of the more accurate but more complicated ones which may be more applicable. Take, for example, the case of the ordinary stress-strain diagram, as in Fig. 37. To find the work done on the specimen up to frac- ture it is necessary to measure the area ABCDE. Replacing the irregular curve (that obtained during the plastic stage) by a portion of a parabola BF, and neglecting the small area ABG, we can say that area ABCDE = rectangle AGHE -f- parabolic segment BFH A Extension E Fig- 37- Stress-strain Diagram. L . Fig. 38. The Hyperbola. If the ratio ^ is denoted by r, then the result may be written eMY . L\ Me, . . area ABCDE = (2+^) = y ( 2 +')> which is Kennedy's rule. So, also, in questions on calculations of weights, circular seg- ments are often treated as parabolic. Example 21. The bending moment diagram for a beam 28 feet long, simply supported at its ends, is in the form of a parabola, the maximum bending moment, that at the centre being 49 tons feet. Find the area of the bending moment diagram, and find also the bending moment at 6 feet from one end (this being given by the height of the arc at D, Fig. 39). Area of parabolic segment ACB = | x 49 X 28 = 915 units. These units are tons feet X feet or tons feet*. Fig. 39. io8 MATHEMATICS FOR ENGINEERS Now it can be shown that the moment of one-half the bending moment area (viz. AMC), taken round AG determines the deflection at A and also at B. Actually, the maximum deflection (at A or B) = x area of AMC x L where E = Young's modulus for the material of the beam and I = second moment of its section. Since E would be expressed in tons per sq. foot and I in (feet) 1 the deflection would be expressed in feet 2 x feet 2 tons x feet . . , 4 ? -ri t. e., in teet. tons x feet* To find the height ED (MB) 2 = 40 x MC from definition (MB) 2 14' jl / _* f ' ~ A MC 49 EF* = 4 a x CF 4 4 DE = MC - CF = 49 - 16 = 33. .*. Bending moment 6 feet from end = 33 foot tons. Example 22. Find the length of the sub-normal of the parabola y* 6y i6x 23 = o. The equation might be written (y* 6y + 9) i6x 32 = o or (y s) 2 = i6(# + 2). This is of the form Y 2 = 4X where Y = y 3, X = # + 2, a = 4 .*. Length of subnormal = 2 a = 8 units ; and is a constant. The Hyperbola. The hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points, called the foci, is constant. There are two branches to this curve, which is drawn in Fig. 38. If P 1 is any point on the curve, then PT I^F 1 = AA 1 = za. AA 1 = transverse axis, and BB 1 = conjugate axis = 26. DOD 1 and EOE 1 are called asymptotes, i. e., the curve approaches these, but does not meet them produced : they are, as it were, its boundaries. PM and PQ are parallel to EOE 1 and DOD 1 respectively : then a most important property of this curve is that the product PM X PQ is constant for all positions of P. If BB 1 = AA 1 , the asymptotes are at right angles and the MENSURATION 109 hyperbola is rectangular : e.g., the curve representing Boyle's law (the case of isothermal expansion) is a rectangular hyperbola, the constant product being denoted by C in the formula, PV = C. Exercises 14. On the Ellipse and the Parabola. 1. A parabolic arc (as in Fig. 36) stands on a base of 12*. The latus rectum of the parabola being 8", find (a) Maximum height of arc; (6) area of segment; (c) width at point midway between the base and the vertex. 2. A parabola of latus rectum 5* stands on a base of 6", find (a) Maximum height of arc ; (6) height at a point on the base 2* from the centre of the base ; (c) area of segment ; (d) position of focus. 3. A parabolic segment of area 24 sq. ins. stands on a base of 12*. Find the height of the arc at a point 2j* from the centre of the base and also the latus rectum. 4. The axes of an ellipse are 10* and 6* respectively. Find (a) The area ; (6) distance between foci ; (c) height of arc at a point on the major axis 4" from the centre; (d) perimeter by the 3 rules. 5. The lengths of the axes of an ellipse can be found from a* = 30, b 2 = 15, where a and 6 have their usual meanings (see Fig. 34). Find (a) Area of ellipse; (6) distance of foci from centre; (c) peri- meter by the three given rules. 6. A manhole is in the form of an ellipse, 21* by 13*. Find, approxi- mately, the area of plate required to cover it, allowing a margin of 2* all the way round and assuming that the outer curve is an ellipse. 7. A cantilever is loaded with a uniform load of 15 cwts. per foot run. The bending-moment diagram is a parabola having its vertex at wl* the free end, and its maximum ordinate (at the fixed end) is , where ' 2 w = load per foot run, and / is the span which is 18 ft. Find the bending moment at the centre, and at a point 3 ft. from the free end. 8. It is required to lay out a plot of land in the form of an ellipse. The area is to be 6 acres and the ratio of the axes 3 ; 2. Find the amount of fencing required for this plot. 9. There are 60 teeth in an elliptical gear wheel, for which the pitch is -235*. If the major axis of the pitch periphery is twice its minor axis, find the lengths of these axes. 10. Find the number of feet per ton of oval electrical conduit tubing, the internal dimensions being f|* x f* and the thickness being -042* (No. 19 B.W.G.). Weight of material = -296 Ib. per The Prism and Cylinder. A straight line moving parallel to itself, its extremities travelling round the outlines of plane figures generates the solid known as the prism. If the line is always at right angles to the plane figures at its extremities the prism is known as a right prism. If the plane figures are circles the prism becomes a cylinder. no MATHEMATICS FOR ENGINEERS A particular case of the prism is the cuboid, in which all the faces are rectangular, i. e., the plane figures at the extremities of the revolving line are rectangles. For all prisms, right or oblique Volume = area of base x perpendicular height. The lateral or side surface of a right prism = perimeter of base x height. Total surface = lateral surface + areas of ends. Applying to the Cuboid. Volume = area of base X height = acxb = axbxc. (Fig. 40.) Lateral surface = 2ab-{-2bc Total surface = zab+zbc+zac = 2(ab-\-bc+ca). Fig. 40. If a = b = c, the cuboid becomes a cube, and then vol. = a 3 and total surface = 2( 2 + 2 + 2 ) = 6a 2 If the diagonal of a cuboid is required it can be found from, diagonal = V 2 +6 2 +c 2 ; whilst for the cube, diagonal = aVs. Example 23. An open tank, made of material y thick, is 2'-6* long, 10" wide and 15* deep (these being the outside dimensions). Find the amount of sheet metal required in its construction if the plates are prepared for acetylene welding, and find also the capacity of the tank. If the plates are to be joined by acetylene welding no allowance must be made for lap ; the plates would be left as shown in the sketch at A, Fig. 41. MENSURATION in Total Surface = 2 x (15 - J)[io (2 x J)] H- 2 x Us - i)[3o - (2 x J)] + [30 - (2 x i)][io - (2 x i)] = 280 + 870 + 280 sq. ins. = 1430 sq. ins. = 9-94 sq. ft. Volume = (30 J) x (10 \) x (15 J) = 2 9-5 x 9-5 x 14-75 f . 1728 x 9-5 x 14-75 x 6-25 Capacity = 1728 14-9 gallons. ions or -jd 2 h, where r = radius If the weight of water contained is required Weight = 14-9 x 10 = 149 Ibs. Note. I cu. ft. of fresh water weighs 62-4 Ibs. I cu. ft. of salt water weighs 64 Ibs. i gallon of fresh water weighs 10 Ibs. 6 gallons occupy i cu. ft. i cu. cm. of water weighs i grm. Applying the foregoing rules to the cylinder. Vol. = area of base X height i. e., Volume of cylinder = irr z xh = ir of base, d = diam. of base, h = height or length. Lateral surface = 2-rh. Total surface = 2-jrrh-\-2Trr z = 2irr(h+r). Volumes of cylinders can most readily be obtained by the use of the slide rule, adopting an extension of the rule mentioned on p. 92. It is repeated here with the necessary extension : Place one of the C's, on the C scale of the rule, opposite the diameter on the D scale : then place the cursor over the length on the B scale, and the volume is read off on the A scale. Rough approximation, Vol. = f d 2 /!.* E. g., Diam. = 4-63" Length = 1875". Vol. (by approximation) = fx 20x20 = 300 cu. ins. Vol. (by slide rule) = 316 cu. ins. Exercises Dia. Length. Vol. 23 47'3 300 2-8 12-45 4945 ii2 MATHEMATICS FOR ENGINEERS Example 24. Find the weight, in Ibs. and grms., per metre of copper wire of diam. -045 cm. (Copper weighs -32 Ib. per cu. in.) Note 2-54 cms. = 1 in. 453'6 grms. = 1 Ib. Then I cu. cm. = , I >, cu. in. 1 2 '54J .'. Weight of i cu. cm. of copper = . t 32 . 3 Ib. Vol. of i metre of wire = -x (-O45) 2 xioo cu. cms. 4 = -159 cu. cm. .*. Weight = >:[ 59X -32 = OQ (2-54)3 2 or weight = -00311 x 453-6 = 1-409 grms. Example 25. A boiler contains 480 tubes, each 6 ft. long and 2| ins. external diameter. Find the heating surface due to these. The heating surface will be the surface in contact with the water, i. e. t the outside surface of the tubes. Lateral surface = ird x length x no. of tubes = jr x -Q x 6 x 480 4 s = 2070 sq. ft. Exercises 15. On Prisms and Cylinders. Prisms 1. A room 22 ft. long by IS'-IQ" wide is g'-^" high. Find the volume of oxygen in it, if air contains 21 % of oxygen and 79 % of nitrogen by volume. 2. A block of wrought iron is'xg'xf weighs 14-2 Ibs. Find the density of W.I. (Ibs. per cu. in.) and also its specific gravity if i cu. ft. of water weighs 62-4 Ibs. 3. The weight of a brass plate of uniform thickness, of length 6'-5" and breadth n" was found to be 79-4 Ibs. If brass weighs -3 Ib. per cu. in., find the thickness of this plate. 4. The sectional area of a ship at its water-line is 5040 sq. ft. ; how many tons of coal would be needed to sink her i f t ? (35 cu. ft. of sea water weigh i ton.) 5. The coefficient of displacement of a ship volume of immersed hull of ship volume of rectangular block of same dimensions If the displacement is 4000 tons and the hull can be considered to have the dimensions 32o'x35'xi5' find the coefficient of displace- ment. MENSURATION II3 6. The ends of a right prism 8'-4" long are triangles having sides 19", 27-2" and 11-4" respectively. Find the volume of this prism. 7. Water is flowing along a channel at the rate of 6-5 ft. per sec The depth of the channel is 9", the width at base 14", and the side slopes are i horizontal to 3 vertical. Find the discharge (a) In cu. ft. per sec. ; (b) in Ibs. per min. 8. A tightly-stretched telephone cable, 76 ft. long, connects up two buildings on opposite sides of the road. The points of attachment of the ends are 38 and 64 ft. above the ground respectively, one being 37 ft. further along the road than the other, and the buildings each standing 10 ft. back from the roadway. Find the width of the road. 9. The section of an underground airway is as shown in Fig. 42. Air is passing along the airway at 10-5 ft. per sec. ; find the number of cu. ft. of air passing per minute. Fig. 42. Fig. 420. 10. Find the volume of stone in a pillar 20 ft. high, the cross-section being based on an equilateral triangle of i foot side, having three circular segments described from the angular points as centres, and meeting at the mid points of the sides. Find also its weight at 140 Ibs. per cu. foot. (Fig. 42.) Cylinders 11. The diameter of a cylinder is 38-7", and its length is 28'3". Find its curved surface, its total surface and its volume. 12. Find the ratio of total heating surface to grate area in the case of a Caledonian Railway locomotive. The heating surface in the firebox is 119 sq. ft., the grate area is 20-63 sc l- ft and there are 275 tubes, of if external diameter, the length between the tube plates being io'-j". 13. A current of -6 ampere at 100 volts was passed through the two field coils of a motor. If the diam. of the coils was 4* and the length 4^', find the number of watts per sq. in. of surface. (Curved surface only is required.) 14. Find the weight of 5 miles of copper wire of -02* diam., when copper weighs -32 Ib. per cu. in. 15. Find the weight of a hollow steel pillar, 10 ft. long, whose external diam. is 5* and internal diam. is 4 (i cu. ft. of steel weighs 499 Ibs.). (See Area of Annulus, p. 93.) 16. Water flows at the rate of 288 lb.s. per min. through a pipe of ij" diam. Find the velocity of flow in feet per sec. 17. Find the heating surface of a locomotive due to 177 tubes of 1 1" diam., the length between the tube plates being io'-6*, I 114 MATHEMATICS FOR ENGINEERS 18. A piston is moving under the action of a mean effective pressure of 38-2 Ibs. per sq. in. at a speed of 400 ft. per min. If the horse-power developed is 70, find the diam. of the piston. r_j p _ Feet per min. x total pressure in lbs.~| 33000 19. In a ten-coupled locomotive there were 404 tubes of 2* diam. and the heating surface due to these was 3280 sq. ft. Find the length of each tube. 20. The diameter of a hydraulic accumulator is 12" and the stroke is 6 ft. Find the work stored per stroke if a constant pressure of 750 Ibs. per sq. in. be assumed. 21. In calculating the indicated horse-power of an engine at various loads it was found that a saving of time was effected if an " engine constant " was found. Vol. of cylinder If the engine constant = 12 x 33000 find this, if diam. = 5-5* and stroke = 10*. 22. The weight of a casting is to be made up from 4-14 Ibs. to 4-16 Ibs. by drilling a ^" diam. hole and plugging with lead. To what depth must the hole be drilled if the weights of lead and cast iron are -41 and -26 Ib. per cu. in. respectively ? 23. The conductivity of copper wire can be expressed by its resist- ance per gramme metre. Find the " conductivity " of a wire 5 metres long and of -762 cm. diam. (No. i S.W.G.) if the Resistance is given by -00000017 x 6 - ; the units being cms. and the weight of i cu. cm. of copper being 8-91 grms. 24. Find the weight, in Ibs. per 100 feet, of electrical conduit tubing of external diam. 2" and internal diam. I-872", the weight of the material being -296 Ib. per cu. in. 25. A 10* length of i" diam. steel rod is to be forged to give a bar ij" wide and " thick. Assuming no loss in the forging, find the length of this bar. Pyramid and Cone. If a straight line of variable length moves in such a way that one extremity is always on the boundary of a plane figure, called the base, whilst the other is at a fixed point, called the vertex, the solid generated is termed a pyramid. If the line joining the vertex to the geometrical centre of the base is at right angles to the base, then the pyramid is spoken of as a right pyramid. When the base is circular the figure is termed a cone ; right circular cones being those most frequently met with. These are cones for which all sections at right angles to the axis are circles. The lateral surface of a right pyramid will evidently be the sum of the areas of the triangular faces. Consider the case of a " square " pyramid, i. e., where the base is square [see (), Fig. 43]. The triangular faces are equal in area. MENSURATION II5 Area of each = base X height = i X AB x VL [see (a), Fig. 43] where VL is spoken of as the slant height of the pyramid ; its value being found from VL = WO 2 + OL 2 [see (b), Fig. 43] LO being \ side of base. Total lateral surface = 4 x AB x VL = 2 . AB x VL or lateral surface of pyramid = J perimeter of base x slant height. This rule will hold for all cases in which the base is regular. [Note that if the base is rectangular, there will be two distinct slant heights.] V B L o Fig. 43. Square Pyramid. Length of edge of pyramid = VB = VVO 2 -f- OB 2 [see (b), Fig. 43], where OB = diagonal of base. The three lengths or heights should be clearly distinguished. VO = perpendicular height, or more shortly, the height VL = slant height, and VB = length of edge. Volume of pyramid is one-third of that of the corresponding prism (i. e., the prism on the same base and of same vertical height). .'. Vol. of pyramid = J x area of base x perpendicular height. Example 26. A flagstaff, 15 ft. high, is kept in position by four equal ropes, one end of each being attached to the top of the staff, whilst the other ends are fastened to the corners of a square of 6 ft. side. Find the length of each rope. n6 MATHEMATICS FOR ENGINEERS Diagonal of base = 6\/2 (the diagonal of a square always = \/2~ x side). The length required is the length of the edges, viz. VB [see (6) Fig. 43]. Now VO = 15, OB = 3 Vz, hence VB = \/ (3 V (i 5)* = ^18+225 = ^243 VB = length of each rope = 15-6 ft. Applying to the Cone. If the lateral surface of the cone is developed, i. e., laid out into one plane, a sector of a circle results, the radius being the slant height /, and the arc being the circum- ference of the base of the cone or 2irr (see Fig. 44). Now area of sector of circle = arc X radius = = i. e., area of curved surface of a cone = vrl. Notice that this agrees with the result obtained from the rule for the pyra- mid, viz. perimeter of base X slant height. If the development of the cone were actually required it would be necessary to find the angle a (Fig. 44). Now Janldeufit I a 360 arc Oce 360? I 2irl Lateral surface, then, = Total surface = Fig. 44. = irr(l+r). As the cone is a special form of pyramid its volume will be one-third that of the cylinder on the same base and of the same height. 1 T Vol. of cone or ~d 2 h or -2618</ 2 /i \Lt d being the diameter of the base and h the perpendicular height. The approximation for the volume is |x(diam.) 2 x height. Example 27. A projectile is cylindrical with a conical point (see Fig. 45). Find its volume. MENSURATION 117 As the cone is on the same base as the cylinder its volume can be accounted for by adding J of its length to that of the cylinder, and treating the whole as one cylinder. Fig- 45- Hence, net length = 4 / "+(Jxi-8") = 4-6" .*. total vol. = -x(i-6) 2 X4-6 4 = 9-26 cu. ins. Frusta. If the pyramid or cone be cut by a plane parallel to its base the portion of the solid between that plane and the base is known as a frustum of the pyramid or cone. The lateral surface and the volume can be found by subtracting that of the top cone from that of the whole cone or by the following rules, which give the results of this procedure in a more advanced form. Lateral surface of frustum of pyramid or cone = {sum of perimeters of ends} X slant thickness. Vol. of frustum of pyramid or cone = where A and B are the areas of the ends, and h is the perpendicular height or thickness of the frustum. (The proofs of these rules are given on p. 123.) For the frustum of a cone these rules may be expressed in rather simpler fashion Lateral surface of frustum of cone = 7r/(R+r) / being the slant height of the frustum. Volume of frustum of cone = ^~-{R 2 -}- r 2 + Rr} where R and r are radii of ends, and h is the thickness of the frustum. Example 28. A friction clutch is in the form of the frustum of a cone, the diameters of the end being 6$" and 4!*, and length 3$". Find its bearing surface and its volume (see Fig. 46). n8 MATHEMATICS FOR ENGINEERS The slant height must first be found .-. / = 3-68*. Now R = 3*25, and r 2-13. .*. Lateral surface = 77 X 3-68(3-25 + 2-13) = 7T x 3-68 x 5-38 = 62-2 sq. ins. Also Volume = {R 2 + y 2 + Rr} = 3l 5 {10-54 + 4-53 + 6-92} TT X 3-5 X 21-99 o -2? 80-5 cu. ms. 3 - - - Fig. 46. Friction Clutch. Exercises 16. On Pyramids, Cones and Frusta. 1. The sides of the base of a square pyramid are each 13-7" and the height of the pyramid is 9-5*. Find (a) the volume, (b) the lateral surface, (c) the length of the slant edge. 2. The volume of a pyramid, whose base is an equilateral triangle of 5-2" side, is 79-6 cu. ins. Find its height. 3. Find the total area of slating on the roof shown at (a) Fig. 47. Plan 27-^ Fig- 47- 4. Find the volume of a hexagonal pyramid, of height 5-12*, the base being a regular hexagon of 1-74* side. 5. A square pyramid of height 5 ft., the sides of the base being each 2 ft., is immersed in a tank in such a way that the base of the MENSURATION 119 pyramid is along the surface of the water. Find the total pressure on the faces of the pyramid if the average intensity of pressure is the intensity at a depth of i'-3* below the surface ; the weight of i cu. ft. of water being 62-4 Ibs. 6. A turret is in the form of a hexagonal pyramid, the height being 25 ft. and the distance across the corners of the hexagon being 15 ft. Find the true length of the hip (i. e., the length of a slant edge), and also the lateral surface. Cones. 7. The curved surface of a right circular cone when developed was the sector of a circle of 11-42" radius, the angle of the sector being 127. Find the radius of the base of the cone, and also its height. (Refer p. 116.) 8. A piece in the form of a sector (angle at centre 66) is cut away from a circular sheet of metal of 9" diam., and the remainder is made into a funnel. Find the capacity of this funnel. 9. A right circular cone is generated by the revolution of a right- angled triangle about one of its sides. If the length of this side is 32-4 ft. and that of the hypotenuse is 55-9 ft., find the total surface and the volume of the cone. 10. A vessel is in the form of a right circular cone, the circum- ference of the top being 19-74 ft- anc ^ the ^ uu depth of the vessel being 12 ft. Find the capacity in gallons. Find also the weight of water contained when the vessel is filled to one-half its height. 11. A conical cap is to be fitted to the top of a chimney. The cap is to be of 7" height and the diam. of the base is 12". Find the amount of sheet metal required for this. If this surface be developed, forming a sector of a circle, what will be the angle of the sector ? Frusta of Pyramids and Cones. 12. A pier is in the form of a frustum of a square pyramid. Its ends are squares, of side 3 ft. and 8'-6" respectively, and its height is 6 ft. Find its volume and its weight at 140 Ibs. per cu. ft. 13. A circular brick chimney is too ft. high and has an internal diam. of 5 ft. throughout. The external diam. at base is n ft. and at the top 7 ft., the thickness being uniformly reduced from bottom to top. Find its weight at 120 Ibs. per cu. ft. 14. Find the lift h of the valve shown in Fig. 48, given that BC = i f * and AD = if*. It is necessary that the area of the lateral surface of ABCD should be 1-3750". 15. One of a set of weights had the form of a frustum of a cone, the thickness being 4^", the diam. at the Fig. 48. too being 10", and the diam. at the tStom bling 2*'. Find its volume and its weight at -26 Ib. per cu m 16. A square pyramid of height g" and side of base 15 is ' cut into two parts by a plane parallel to the base and distant 4* from it. Find the volume of the frustum so formed, and also its lateral surface. I2O MATHEMATICS FOR ENGINEERS 17. A cone 12" high is cut at 8" from the vertex to form a frustum of a cone of volume 190 cu. ins. Find the radius of the base of the cone. 18. The parallel faces of a frustum of a pyramid are squares on sides of 3* and 5" respectively, and its volume is 32! cu. ins. Find its altitude and the height and lateral edge of the pyramid from which it is cut. 19. A conical lamp-shade is 2 \" diam. at the top and 8J" diam. at the bottom. The shortest distance between these ends is 5*. Find the area of material required for this, allowing 4 % extra for lapping. By drawing to scale, find the area of the rectangular piece from which the shade would be cut. 20. A pyramid, having a square base of side 18*, and a height of 34*, is cut by a plane distant n" from the base and parallel to it. Find the total surface of the frustum so formed, and also its volume. The Sphere. If a semi-circle revolves about its diameter as axis it sweeps out the solid known as the sphere. The portion of the sphere between two parallel cutting planes is . known as a zone : thus CDFE in Fig. 49 is a zone. F The portion included be- B tween two planes meeting along a diameter is known as a lune. A plane section through the centre is called a great circle : any other planes will cut the sphere in small circles. Thus, the section on AB (Fig. 49) would be a great circle, and the sections on The portion CMD is a segment. Fig. 49 . CD or EF would be small circles. Let the radius of the sphere = r, and diam. = d. Then the surface of the sphere = 4 X area of a great = 4 X Try 2 = 4rrr 2 OF Trd 2 Vol. of sphere = ^n-r 3 = - . d 3 = or '5236d 3 Surface of a zone = curved surface of circumscribing cylinder (h being the distance between the parallel planes). Vol. of zone = ^{3 (/y 5 + r 2 2 ) -f h 2 } [The proof of these two rules will be found in Vol. II of Mathematics for Engineers.] MENSURATION 121 The zone may be regarded as a form of frustum, r v and r z being the radii of the ends and h being the thickness. If r 1 = o, the zone becomes a segment, and then Vol. of segment = ^{Sr-j 2 + A 2 } k being the height of the segment. A relation that exists between the volumes of the cone, sphere and cylinder should be noted. Consider a sphere, of radius r; its circumscribing cylinder (i. e., a cylinder with diam. of base = zr and height = 2r], and the cone on the same base and of the same height. Then. Vol. of the sphere = -Trr 3 = -7ry 3 X2 3 3 2 Vol. of the cylinder = Try 2 x zr = - Try 3 x 3 o TtY^ 2 Vol. of the cone = X 2r = - -n-r 3 x i. 3 3 Hence the respective volumes of the cone, sphere and cylinder of equal heights and diameters are in the proportion 1:2:3. Example 29. A disc of lead 14* diam. and -8" thick is melted down and cast into shot of (a) " diam., (6) J* diam. How many shot can be made in each case, supposing no loss ? Case (a). VoL Qf digc = ^ x I4 2 x . 8 cu ^ 4 = 39'27r cu. ins. Vol. of i shot = I x (|) 3 = - r ^ 39-27T x 6 x 512 .* No. of shot = ^ 7T = 120,300. Case (6). The diam. is twice that of Case (a); therefore the vol. of i shot is 2 3 , i. e., 8 times as great. A No. of shot -gggg- 15.038. Example 30. Find an expression for the weight in Ibs. of a sphere of any material, having given that the weight of a cu. in. of copper is 318 Ib. (approx.). Weight of a copper sphere of diam. D = volume x density = ^D 3 x -318 122 Hence the weight of a sphere of any material, its diameter being D _ D 3 x specific gravity of solid "~ 6 x specific gravity of copper Example 31. Find the total surface of a hemispherical dome, of inside diam. 5^" and outside diam. 7-4*. Outside surface = J X 4?r x (3'7) 2 = 85-6 sq. ins. Inside surface = \ x 477 x (2'75) 2 = 47-5 ,, Area of base = n-(3'7 2 275 2 ) = 19-2 /. Total surface area = 1523 sq. ins. Similar Figures. Similar figures are those^ having the same shape : thus a field and its representation on a drawing-board are similar figures. Triangles, whose angles are equal, each to each, are similar figures. On every hand one comes across instances of the application of similar figures ; and in connection with these, three rules should be remembered. (1) Corresponding lines or sides of similar figures are proportional. (Euclid, VI. 4.) (2) Corresponding areas or surfaces are proportional to the squares of their linear dimensions. (Euclid, VI. 20.) (3) Volumes or weights of similar solids are proportional to the cubes of their linear dimensions. E. g., consider two exactly similar cones, the height of one being three times that of the other. Then (i) the radius and hence the circumference of the base of the first are three times the radius and circumference of the second respectively. (2) The curved surface of the first = 3 2 X that of the second. (3) The volume or weight of the first = 3 3 X volume or weight of the second. To generalise, using the symbols L, S, and V for side, surfaces and volumes respectively If the ratio of the linear dimensions of two similar figures is . , , L! Sj /LA 2 represented by j-, then ^ = ( ~ ] (i) and ^Hri) MENSURATION If it is desired to connect up volumes with surfaces / C ^ 3 / T \H By cubing equation (i) [*) = ( tl) W2/ VLg By squaring equation (2) (*] = /~i 123 Hence or (3) Example 32. A conical lamp-shade has the dimensions shown in Fig. 50. Find the height of the cone of which it is a part. Let x inches be the height of the top triangle, viz. ABC. Then ABC and ADE are similar triangles, hence the ratio -v - is OclSG the same for both. x i. e., ^ for the small triangle must = x + 4 for the large triangle. Then, by multiplying across iox = 6x + 24 4* = 24 x = 6" Total height of cone = 6 + 4 = 10*. Fig. 50. It is convenient at this stage to insert the proofs of the rules for the lateral surface and the volume of a frustum, given on p. 117. In Fig. 50 let the height or thickness FG of the frustum BCED be denoted by h; let A be the area of the end DE and let B be the area of the end BC. [Note. The figure is taken in these proofs to be the elevation of a Pyramid, so that the proofs may be perfectly general.] Then, from the similarity of the triangles ABC and ADE perimeter of end DE AD AB + BD = :_^_i _ _ J ~Df~* A "D A "D ' BD perimeter of end BC AB , p. ofDE BD whence - , p/ ^ i = -7^5 p. of BC AB p. of DE- p. of BC _ BD AB ' AB (p. being written to denote perimeter) or p. ofBC AB 124 MATHEMATICS FOR ENGINEERS Lateral surface of frustum BCED = lateral surface of pyramid ADE lateral surface of pyramid ABC = J(p. of DExAD)-(p. of BCxAB) . = i[(p. of DExAB) + (p. of DE ^ xBD)-(p. of BCxAB)] = |[AB(p. of DE-p. of BC) + (p. of DExBD)] Substituting from equation (i) = |[(p. of BCxBD) + (p. of DExBD)] = | X BD X sum of perimeters of ends = | sum of perimeters of ends X slant thickness. Again, since ABC and ADE are similar solids, the areas of their respective bases are proportional to the squares of their respective heights _ ~~ B ~~ (AF) By transposition B = A X / AG ( 2 ......... (2) (3) Also by extraction of the square root VA_AG VB~AF Volume of frustum BCED = vol. of pyramid ADE vol. of pyramid ABC = JxAxAG-JxBxAF By substitution from equation (2) (AF^ 2 jfzzJL v A TT / A ,/- \ A /\ Xi.1/ _ r(AG)3-(AF)31 - aii-l / A ^\o Factorising the numerator (see p. 53) _ JA[(AG- AF)] [(AG) 2 + (AG X AF) + (AF) 2 ] (AC) 2 FAG AF-M i^rAxjAG) 2 AxAGxAF , Ax(AF) 2 ] [AG-AF-A] = ^LlAGF- + (AG) + -@ST. MENSURATION Substituting from equation (3) 125 = p[A+ VAB+B]. Example 33. A surveyor's chain line is to be continued across a river. Describe a method by which the line may be prolonged and show how the required distance may be deduced. Suppose C is a point on the line : select some station A on the opposite bank (Fig. 51) and put A, B and C in line. Set off BD and CE as offsets at right angles, so that E, D and A are in a straight line. AB _ DF BC BD Then *'. e., AB = FE ~ CE - BD BC x BD CE - BD or AB is found. Fig. 51- Example 34. The actual area of a field is 5 acres : on the plan it is represented by an area of 50 sq. ins. To what scale is the plan drawn ? We are told that 50 sq. ins. represent 5 acres or 50 sq. chains. Hence i sq. in. represents i sq. chain or i* represents i chain. So that the scale is i* to a chain, or the representative fraction _ __ 22 X 36 ~ 7Q2 126 MATHEMATICS FOR ENGINEERS Example 35. The heating surfaces of two exactly similar boilers are 850 and 996 sq. ft. respectively. The' capacity of the second being 750 gallons, what is the capacity of the first ? It is not necessary to determine the ratio of the linear dimensions, for statement (3) on p. 123 can be used, since the capacities are pro- portional to the volumes. Now S x = 850, S a = 996, V 2 = 750, and V x is required. V, \S, or V,- 7S .x(||) log Vj = log 750 + i -5 (log 850 - log 996) = 2-8751 + 1-5(2-9294 - 2-9983) = 2-8751 1-5 x -0689 = 2-7717 .*. V\ = 591 gallons. An application of similar figures is found in the engraving machine and in the reducing gear used in connection with indicators. In Fig. 52 such a gear is represented. The movement of the cross- head is reduced, the ratio of reduction being movement of crosshead OC DC __ ..i.- _ OT* _ movement of pencil OP AP The performance of large ships can be investigated by comparing with that of small models. Here, again, the laws of similarity are of great importance. Suppose the model is built to a scale of , i. e., any length on the ship is fifty times the corresponding length on the model. Then its wetted surface is - of that of the ship; while its 2500 displacement is - - U. e., gj of the ship's displacement. Also the resistance to motion of the ship would be 5o 3 times that of the model. An instance of the use of the rules for similar figures is seen in the following : If the circumference of a circle of 3" diam. is 9-426" and its area is 7-069 sq. ins., then the circumference of a circle of 30" diam. will be 9-426 X 10, i. e., 94-26", and its area = 7-069 X I0 2 = 706-9 sq. ins. MENSURATION 127 Hence one can form a most useful table, to be used for all sizes of circles. Diana. Circumference. Area. I 3-142 785 2 6-283 3-142 3 9-426 7-069 4 I2- 5 66 12-566 5 15-708 I9-635 6 18-850 28-274 7 2I-99I 38-485 8 25-I33 50-265 9 28-274 63-617 Suppose the circumference of a circle of -375" is required, ce of circle of -3" diam. = ^ of ce of Q of 3" diam. = -9426 ce of circle of -07" diam. = ^ of ce of of 7" diam. = -2199 ce of circle of -005" diam. = y^ of ce of of 5" diam. = -0157 .'. ce (-375" diam.) = 1-1782 Again, the area of a circle of -8" diam. = g X area of circle of 8" diam. = -503 sq. in. Exercises 17. On Spheres. 1. Find the surface and volume of a sphere of 7-14* diam. 2. A sphere of 8" diam. is weighed in air and its weight is found to be 80 Ibs. Its weight in water is 70-35 Ibs. If Specific Gravity weight of solid . , , , = ^rr -r - 7 - and loss of weight = weight of water weight of equal vol. of water displaced, find the specific gravity of the material of which this sphere is composed and the weight of i cu. ft. of it. 3. Find the volume of a spherical shell whose external diam. is 4-92", the thickness of the metal being ". 4. A storage tank, in the form of a cylinder with hemispherical ends, is 23^ ft. long over all and 4 ft. in diam. (these being the internal measurements). Calculate the weight of water contained when the tank is half full. 5. A sphere of diameter 22 cms. is charged with 157 coulombs of electricity. Find the surface density (coulombs per sq. cm.), which is , quantity in coulombs given by -. . area in sq. cms. 6. The volume of a sphere is 84-2 cu. cms. : find its diam. 7. Find the surface and volume of the zone of a sphere of radius 8* if the thickness of the zone is 2" and the radius of its larger end is 6*. 128 MATHEMATICS FOR ENGINEERS 8. The weight of a hollow sphere of gun-metal of external diam. 6* was found to be 22-3 Ibs. Find the internal diam., if the gun-metal weighs -3 Ib. per cu. in. 9. In a Brinell hardness test a steel ball of diam. 10 mm. was pressed on to a plate, and the diam. of the impression was measured to be 3-15 mm. Find the hardness number for the material of the plate if the load applied was 5000 kgrms. and hardness number = -5 T-J = . (Compare Example 14, p. 98.) curved area of depression On Similar Figures. 10. Find the area of section of the masonry dam shown at (b), Fig. 47- 11. The symmetrical template shown at (c), Fig. 47, was cut too short along the bottom edge ; the length dimensioned as 2-06" should be 2-22". Find the amount x to be cut off in order to bring the edge to the required length. 12. A plan is drawn to a scale of $$. The area on the paper is 4280*. What is the actual area of the plot represented ? 13. Find the diam. of the small end of the conical roller for a bearing shown in Fig. 53. 14. The wetted surface of a ship of 6500 tons displacement is 260003'. What will be the wetted surface of a similar vessel whose displacement is 3000 tons ? 15. One side of a triangle is 12*. Where must a point be taken in it so that a parallel to the base through it will be cut off a triangle whose area is that of the original triangle ? 16. The parallel sides of a trapezoid are 10* and 16*, and the other sides are 5" and 7*. Find the area of the total triangle obtained by producing the non-parallel sides. 17. The surface of one sphere is 6 times that of another. What is the ratio of their volumes ? Find also the ratio of their diameters. 18. The area of a field was calculated, from actual measurements taken, to be 52-7 acres. The .chain with which the lines were measured was tested immediately after the survey and found to be 100-8 links long. Find the true area of the field (i chain = 100 links and 10 sq. chains = i acre). 19. A plank of uniform thickness is in the form of a trapezoid where one end is perpendicular to the parallel sides and is 12 ft. long. The parallel sides are 12" and g" respectively. At what distance from the narrower end must the plank be cut (the cut being parallel to the 12" and 9* sides) so that the weights of the two portions shall be the same ? 20. A trapezoid has its parallel sides 24* and 14* and the other sides each 8". Find the areas of the 4 triangles formed by the diagonals. 21. The length of a model of a ship was 10-75 ft., whilst that of the ship itself was 430 ft. If the displacement of the ship was 11600 tons, what was the displacement of the model ? 22. To ascertain the height of a tower a post is fixed upright 27 ft. from the base of the tower, with its top 12 ft. above the ground. The MENSURATION 129 observer's eye is $'-4* above the ground and at 3 ft. from the post when the tops of the tower and post are in line with the eye. Find the height of the tower. 23. What should be the diameter of a pipe to receive the discharge of three pipes each J* diam. ? The Rules of Guldinus. These deal with surfaces and volumes of solids of revolution. A solid of revolution is a solid generated by the revolution of a plane figure about some axis ; e. g., a right-angled triangle revolving about one of its perpendicular sides traces out a right circular cone ; and a hyperbola rotating about either of its axes generates a hyperboloid of revolution. For the cases with which we deal here the axis must not cut the revolving section, and all sections perpendicular to the axis of revolution must be circular. The rules are Surface of solid of revolution = perimeter of revolving figure X path of its centroid. Volume of solid of revolution = Area of revolving figure x path of its centroid. The centroid of a plane figure is the centre of gravity of an extremely thin plate of the same shape as the figure. The motion of the centroid may be taken to be the mean of the motions of all the little elements of the curve or area. These rules are of great value in dealing with awkward solids ; e. g., suppose the volume of the nose of a projectile is required, it being generated by the re- volution of a curved area round the axis of the projectile (see Fig. 54). The area of ABCD and the position of its centroid G can be found by rules to be detailed later, and then Vol. of nose = area of revolving figure X path of its centroid = (ABCD)x(2irXOG) A simpler example is that of a flywheel rim. Example 36. Find the weight of the rim of a cast-iron flywheel of 5 ft. outside diam. ; the rim being rectangular, 8' across the face and 4' thick radially. (C.I. weighs -26 Ib. per cu. in.) K 130 MATHEMATICS FOR ENGINEERS Here, area of revolving figure =8x4 also the mean diam. = 56* whence path of centroid = TT x 56 and vol. of rim = x 56 x 32 cu. ins. .*. Weight of rim = x 56 x 32 X -26 Ib. = 1460 Ibs. The positions of the centroids (G) for a few of the simple figures is here given (Fig. 55). Triangular area (i) ..... OG = \h /BD is the median,"^ GD = BD I i. e., AD = DC j 2f Semicircular arc (2) ..... OG = = -637 r TT Semicircular area (2) . OG, = = -424 r 3* 2r Semicircular perimeter (2) . . OG 2 = = -389 r (i. e., arc + diameter). Parabolic segment (3) .... OG = f A Semi-parabolic segment (4) . . OQ = f h, QG = f b Area over parabolic curve (5) . OG = '^h; GP =- 4 Area of quadrant of circle (6) . OG = GP = -424 r Area over circular arc (quadrant) or Fillet (7). OG = GP = -223 r Trapezoid (8). Bisect AB at E and DC at F. Join EF. Set off BM = DC and DN = AB. Intersection of MN and EF is at G, or, by calculation, OG = - ( Quadrilateral (9). Bisect AC at F and BD at E. Make OP = OE and OQ = |OF Through Q draw a parallel to BD and through P, a parallel to AC. The intersection of these gives G, the centroid of ABCD. Exercises 18. On Guldinus* Rules. 1. An isosceles triangle, each of whose equal sides is 4 ft. and whose altitude is 3 ft., revolves about an axis through its vertex parallel to its base. Find the surface and volume of the solid generated. 2. Find the surface and volume of the anchor-ring described by a circle of 3* diam. revolving round a line 4* from the nearest point on the circle. 3. Find the surface and volume described by the revolution of a semicircle of 4" diam. about an axis parallel to its base and 5* distant from it. 4. An equilateral triangle of 5* side revolves about its base as axis. Find the surface and volume of the double cone thus generated. MENSURATION Fig. 55 ._Positions of Centroids (G) for Simple Figures. 132 MATHEMATICS FOR ENGINEERS 5. A parabola revolves about its axis. Compare the volume of the paraboloid thus generated with that of the circumscribing cylinder. 6. At (a). Fig. 56, T T -10 L Fig. 56. is shown in section jy|~J" the winding of the secondary wire of an induction coil. Find the volume of the winding. 7. Calculate the weight, in mild steel weighing -287 Ib. per cu. in., of the spindle weight for a spring compressor shown at (b), Fig. 56. [Hints. Area of a fillet, as at A, = '215r 2 where r is the radius of the circular arc. For the position of the centroid of a fillet refer to (7), in Fig. 55, and also to p. 130.] Application to Calculation of Weights. When calculating weights two rules should be borne in mind in addition to the foregoing. (a) The solid should be broken up into simple parts, i. e., those whose volumes can be found by the rules already given; and (b) suitable approximations should be made wherever possible. Circular segments may be replaced by parabolic segments if the rules for the latter are easier, the rounding of corners may be neglected, unless very large, mean widths may be estimated, etc. For purposes of reference the table of weights of materials and other useful data are inserted here ; but the values given must be considered as average values. WEIGHTS AND DENSITIES OF METALS. METAL. Weight in Ibs. per cu. in. Weight in Ibs. per cu. ft. Specific Gravity (gnus, per cu. cm.). Cast iron ... Wrought iron Steel ..... 26 28 2Q 450 485 <?OO 7-21 7-76 8-04 Brass or Gun-metal Copper (Cu) ... Lead (Pb) ... Tin (Sn) 3 .32 HI 27 518 553 710 /i6^ 8-31 8-87 n-34 7M& Aluminium (Al) Zinc (Zn) 0932 26 161 ACQ 2- 5 8 7*21 MENSURATION WEIGHTS AND DENSITIES OF EARTH, SOIL, ETC. 133 MATERIAL. Slate. Granite. Sandstone. Chalk. Clay. Gravel. Mud. WEIGHT \ (cwt. per cu. yd.) J ' 43 42 39 36 31 30 2 5 Useful Data. Wrought-iron plate weighs about 10 Ibs., and steel 10-4 Ibs. per sq. ft. of area per J" of thickness, i. e., 8 sq. ft. of W.I. plate |" thick would weigh 10 X 8 x 3 = 240 Ibs. Wrought-iron bar or rod weighs about 10 Ibs., and steel 10-4 Ibs. per yard for every sq. in. of section. Wrought-iron bar or rod, i"diam., weighs 8 Ibs. and steel 8-2 Ibs. per yard : also the weight is proportional to the diameter squared; thus, a yard of steel bar 2" in diam. would weigh 2 2 X 8-2 or 32-8 Ibs. Four hundred cu. ins. of wrought iron, 430 cu. ins. of cast iron, 390 cu. ins. of steel, each weigh about i cwt. A few examples are here worked out to give some idea of the method of treatment. Example 37. Calculate the weight, in cast iron, of the D slide valve shown in Fig. 57. e*- IT -]] .11 Fig. 57. D Slide Valve. In many cases where the solid is partially hollowed it is best to treat first as a solid and then subtract the volume cut away, 134 MATHEMATICS FOR ENGINEERS First, considering as a solid Vol. above AB = 15-5 x 8-5 x 3 cu. ins. = 395 cu. ins. Vol. below AB = 16-25 x J 3 x I>2 5 = 26 4 " .*. Total vol. (as a solid) = 659 To be subtracted Vol. of cavity = 14 X 7 x 3-5 = 343 /. Net vol. = 3i6 and weight = 316 X -26 = 82-1 Ibs. Example 38. Find the weight of a plate for a cast-iron tank. The plate (see Fig. 58) is 24* square and f thick; there are 20 ribs, each \" x ij" x ij", and 24 bolt-holes in the flanges, each f* square; also the flanges are 23!" x J* x ij". r -j-j- i" Z S (. T ; ! 5"^ JJ :: i- a 1 ::: ~li 8^ a ! :.; *;;j i IV* - ;: * a .. *"""! A 1 1 ? .c LZ- .0 ^" *"'1| B^ i- U II U U D >]f I; a El Fig. 58. Plate for Tank. J Dealing with the separate portions : Flat Plate (A). Vol. = 24 x 24 x | . = 216 cu. ins. Flanges (D). Length = (2 x 22|") + (2 X 23!") = 93" .*. Vol. =93*fx = 5 8 ' 1 Ribs (B). Area of face of one = \ x X f /. Vol. of 20 each \" thick = \ x f X f X \ X 20 = 7-8 Gross vol. = 281-9 >, Subtract for 24 bolt-holes (C) ; 24 x f x f x J = 4-7 Net vol. = 277-2 .*. Weight = 277-2 X -26 = 72 Ibs. MENSURATION I35 Example 39 Find the weight of the wrought-iron stampings for a dynamo armature as shown in Fig. 59) 14* diam. and IO * long io/ of the length being taken off by ventilation and insulation. There are 3 ventilating ducts, each 6* internal diam. and i* thick, the gaps between these being i%" long; and also 60 slots, each |" by f*. The shaft is 3* diam. Note. The stampings are only thin and are separ- ated one from the other by some insulator; also there would be a small gap for ventilation purposes, and hence the actual length of the stampings is less than 10 is to be taken as 90 % of 10", i. e., g". Area of face of stamping = - x i4 2 = *54 sq. ins. To be subtracted Area of 60 slots = 60 x J x f Mean length of ventilating ducts = (IT x 7) (3 x = 17-5* /. Area = 17-5 x i Fi S- 59- Stamping for Dynamo Armature, in this case it = 197 Area of hole for shaft = - x 3 2 .. 4 Thus the total area to be subtracted = 7-1 = 44'3 .. or the net area of the stamping = 109-7 Then the volume = 109-7 x 9 cu - i ns - and the weight = 109-7 x 9 x -28 Ibs. = 277 Ibs. Example 40. Find the weight of 150 yards of steel chain, the links of which have the form shown in Fig. 60. The effective length A of a link is the inside length, provided that a number of yards of chain are being considered. (For small lengths this is not quite correct.) In this case the effective length of a link = i^*, so that in i yard 36 of the chain there are i. e., 24 Fig. 60. Chain Link. & links, or in 150 yards of the chain there are 3600 links. The mean length of i link = 0ce of circle of i J* diam. + (2 x f") ~~" 3 V3 v "^ *) TV 136 MATHEMATICS FOR ENGINEERS Now i* diam. steel rod weighs 8-2 Ibs. per yard (see p. 133) ; therefore \" diam. steel rod weighs -^, i. e., 2-05 Ibs. per yard. Hence, weight of i link = 5_^> x 2-05 Ibs. and weight of 3600 links = 5 '43 * 2-05 x 3600 lbs = III5lbs Example 41. Two straight cast-iron pipes, making an angle of 135 with one another, have the centres of their ends 2 ft. apart (in a straight line). They are to be joined by a curved pipe (as in Fig. 6r), 4* external and 3* internal diam., with flanges 8" diam. and \" thick. Find the weight of the curved pipe if the flanges each have five bolt- holes, of \" diam. Fig. 61. Curved Cast-iron Pipe. This is a useful example on the application of Guldinus' rule. Path of centroid = arc of circle, which is -^- or 5 of the circum- 360 8 ference. By drawing to scale (or by Trigonometry), the radius is found to be 2-6 ft. .*. Path of centroid = Jx 77X5-2 = 2-04 ft. and length of the path of the centroid between the flanges-^ % =2-0 4 ft.-(2X") 5= ;-9oft, - 23-5*, MENSURATION Area of revolving section = f X4 2 J ( X3 2 ) = 5-5 sq. ins. hence the volume of the solid between the flanges = 23-5x5-5 cu. ins. = 129 cu. ins. Vol. of 2 flanges, each J* thick, 8* external and 3* internal diam. = 43-2 cu. ins. .*. Gross vol. of bend = 172-2 cu. ins. To be subtracted Vol. of ten |" diam. holes : Diam. Length. Vol. 625 5" i'5 /. Net vol. of bend = 170-7 cu. ins. and weight = I7O-7X-26 = 44-4 Ibs. Example 42. Find the weight of the wrought-iron crank shown in Fig. 62, allowing for the horns at the junctions of the web and bosses. Dealing with the three parts separately : Vol. of the upper boss is the difference of the volumes of two cylinders Diam. Length. Vol. 12* 8* 908 6* 8* 227 net volume = 681 cu. ins. Similarly, vol. of the lower boss Diam. Length. Vol. *5', 9 7-25 7- 2 5 1282 462 - 8 i 04 net volume = 820 cu. ins. Fig. 62. Wrought-iron Crank. The horns can be allowed for by adding J of the height of each to the length of the web (i. e., we replace the circular segment by a parabolic segment, because the rule for the area is simpler). To find the height h^ of the top horn A, {a x = 5, r z = 6}. hi = fi- Vrf-aJ = 6- -^36-25 = 6-3-32 = 2-68*. Hence add J of 2-68", i, f ,, >f tq the length of the web, 138 MATHEMATICS FOR ENGINEERS For the lower horn B, a 2 = 6", r t = 7-5* 3" Hence add on i* to the length of the web. Thus the effective length of the web its mean width so that its vol. = 18-4* = n* = i8-4X 11X4-5 = 9 IQ Total vol. of crank = 2411 cu. ins. Weight = 241 1 x -28 = 675 Ibs. Example 43. Determine the number of i* diam. rivets, as at (a) Fig. 63 (i. e., with snap or spherical heads) to weigh i cwt. (Given that d = /+ t V and length = 2t.) 1 17- Fig. 63. If d = i* then t = &" and length = i *. For the heads, a rough approximation is that the two together are one-half the volume of a sphere of diameter i-8d, this being the diameter of the sphere of which the heads are segments ; but the result will be somewhat more accurate if -52 is taken in place of -5. (This figure is arrived at by the use of the rule given on p. 121 for the segment of a sphere.) Then vol. of heads = ^x^n-x-g 8 = 1-58 cu. ins. vol. of body {Diam. = i", length = 1-125"}= -88 or vol. of i rivet = 2-46 Number of i* rivets per cwt. = 2 -46 x -29 Example 4^. Find the weight of the cast-iron hanger bearing shown in Fig. 64. This example illustrates well the method of breaking a solid up into its component parts ; the different parts being dealt with according to the letters on the diagram. 139 cub. ins. MENSURATION Treating first as a solid throughout A. Cuboid, length = 12", breadth = 6-75", thickness = -75". Volume = 12x6-75x75 = 60-75 B. 4 cylinders, of diam. 1-625* an d total length = 5* Volume (obtained from the slide rule) = 10-35 C. Area of section = semicircle + rectangle (5-5x2-5) = 10-86 + 1375 = 25-61 Volume = 25-61x275 D. Cylinder, diam. = 4*, length = 4* Volume E. Cylinder, diam. = 4-5*, length = -75 Volume F. 4 cylinders, diam. = 2", total length = i" Volume , = 70-48 = 50-30 = 11-92 = 3-14 Gross Volume = 206-94 Fig. 64. Cast-iron Hanger Bearing. To be subtracted cub. ins. = 28-20 G. Cylinder, diam. = 3*, length = 4* Volume H. Cylinder, diam. = 2%", length = 3^* Volume = I7 >:1 5 J. 4 cylinders, diam. = -75", total length = 9" Volume = 3 '97 Total volume to be subtracted = 49'3 2 Net volume = I57' 62 Hence, weight = 157-6 x -26 = 41 Ibs. 140 MATHEMATICS FOR ENGINEERS Exercises 19. On Calculation of Weights. 1. Find the weight of the cast-iron Vee-block shown at (6), Fig. 63. 2. Find the weight in steel of the crank axle shown in Fig. 65. Centre Lir\e of Engine. ' P36-9"-- Fig. 65. Steel Crank Axle. 3. Find the weight of sheet iron in a rectangular measuring tank ; the metal being i" thick. Inside dimensions of the tank are 4'-6* by 3 '-6* by 7 / -o"deep. Cut from the sides are openings to accommodate fittings as follows : One rectangular hole 4 / -o* by 2", two elliptical holes 4" X 2", two circular holes 4" diam. and eight f-diam. bolt holes. 4. Determine the weight of a wrought-iron boiler end plate, 8 ft. diameter and ^" thick. There are two flue holes, each 2 ft. diam. and an elliptical manhole i8*xi2*. 5. Find the weight of 22 yards of iron chain. The links are elliptical and are made of elliptical metal i"x%", the greatest width of section being at right-angles to the plane of the link. The mean lengths of the axes of the link are 4* and 2^". 6. How many f-diam. snap-headed rivets weigh i cwt. ? (Compare with Example 43, p. 138.) 7. Find the weight in cast iron of the flywheel of a steam engine having a rectangular rim, 7" wide by 4* radial thickness ; six straight arms of elliptical section, the axes of the ellipse being 4^* and 2" ; a boss 7** wide, 9* diam. and 4^" bore. The outer diameter of the wheel is 7-9*. 8. Required the weight of the cast-iron anchor plate shown in Fig. 66. --L, ajfef- ii_ \ / / "< -8' M t eb i M 1-6" Fig. 66. Anchor Plate, T ._. ?af^ L- IK Fig. 67. Planer Tool Holder. MENSURATION 141 9. Calculate the weight in cast iron of the tool holder for a planer shown in Fig. 67. 10. Find the weight of the cast-iron roll for a rubber mill as in Fig. 68. (Use the slide rule throughout.) Fig. 68. Roll for Rubber Mill. 11. A mild steel sleeve coupling for 3* shaft is shown at (a), Fig. 69. Find its weight. 12. The steelwork for Hobson's flooring has the sectional form shown at (b), Fig. 69. There are 20 such plates for each span of the bridge, each ^" thick and 22 ft. long. Find the total weight of the steelwork, neglecting the angle and T-bar. Mild Steel Sleeve Coupling. Section of Hobson's Flooring. Fig. 69. 13. Find the weight in cast iron of the simple plummer block shown in Fig. 70. Fig. 70. Plummer Block. 142 MATHEMATICS FOR ENGINEERS 14. Fig. 71 shows the worm shaft for a motor-car rear axle. It is made of nickel steel, weighing -291 Ib. per cu. in. Find its weight. L-3T- Fig. 71. Worm Shaft. 15. Calculate the weight in cast iron of the half coupling shown in Fig. 72. Fig. 72. Wrought-iron Coupling. 16. Find the weight in cast iron of the cylinder cover shown in Fig. 73- - 73- C.I. Cylinder Cover. MENSURATION 143 17. Fig. 74 shows the brasses for the crankshaft of a 61"x6 T launch engine. Find the weight of one of these in gun metal. ! -r- Hf -r-* - ^ \ 1 1 1 1 1 T | ft -t- MI ! 1 't 1 i ^i r ' "^ Fig. 74. Crank Shaft Brasses. 18. The brasses for a thrust block are shown in Fig. 75. Calculate the weight of one of these in gun metal. Fig. 75. Brasses for a Thrust Block. 19. An air vessel is shown in Fig. 76. Find its weight in cast iron. Oval 4^x3' ty, 3*3 T. Vj.2 VKO ..i?....S?-. li' -6' Fig. 76. Air Vessel for Pump. 144 MATHEMATICS FOR ENGINEERS TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES. 145 TABLE OF AREAS AND CIRCUMFERENCES OF PLANE FIGURES (continued). TUle. Hollow circle (annulus) . Hollow circle (eccentric) Sector of circle . . Sector of hol low circle Fillet . . Segment of circle . . Figure. Ellipse Irregular figures ircumference or Perimeter. 57'3 Area. or ir x mean dia. x thick- ness or 7r(R* - r 2 ) approx irnr* Ir ~?6o~ or - 360 2I5K 2 or approx. Area = sector triangle Various approx. for- mulae on p. 102. Vab] more nearly Step round curved por- tions in small steps, with dividers ; add in any straight pieces. Divide into narrow strips ; measure thei mid-ordinates. Then-- Area = aver, mid-ordi nate x length / 146 MATHEMATICS FOR ENGINEERS TABLE OF VOLUMES AND SURFACE AREAS OF SOLIDS. Title. Figure. Volume. Surface Area. Any prism . Rectangular prism or cuboid . . Cube . . . Square prism Hexagonal prism . Octagona! prism . Cylinder . . Hollow cylin- der . Elliptical prism , Sphere . . Hollow sphere . . Area of base X height llh S 3 S 2 / 2-6S 2 / or -866/ 12 / Circumference of base X height or -Sag/ 2 / or 7854^ r(R 2 - r z )h rabh or -523&D 3 Whole area/ = 2(lb+Ui+bh) Whole area = 6S* Lateral surface = 4S/ Ends = 2S a Whole 1 ^c/^7 i o\ surfacej= 2S ( 2/+S ) Lateral = 6S/ or y^bfl (For ends see Table on p. 1 44.) Lateral = 8S/ or 3-32/1 Lateral = 2irrh Two ends = 2irr* Whole area = 2jrr(& + r) Outer lateral) surface / Inner lateral) surface Lateral / = 2-irrh or Tr(a+b)h (less accurate) MENSURATION 147 TABLES OF VOLUMES AND SURFACE AREAS OF SOLIDS (continued). Title. Figure. Volume. Surface Area. Segment of sphere . Zone of sphere . Any pyramid Square pyra- mid . . . Cone . . . Frustum o: any pyra- mid . . Frustum o square pyramid Frustum o: cone . Anchor ring urved surface = 2irRA or 5236/i(3' 2 +A 2 ) area of base X height ^=height of frus- tum A=area of large end B=area of small end -(A + B + VAB) here R=rad. of sphere ateral = \ circum. of base X slant height Lateral = 2S/ Lateral = *y/ Lateral=| mean circum X slant height Lateral = 2/(S +'s) (I = slant height) Lateral = ir/(R + f (1 = slant height) Round section Square section DS rDS These four tables are reproduced from Arithmetic for Engineers by kind permission of the author, Mr. Charles B. Clapham. CHAPTER IV INTRODUCTION TO GRAPHS Object and Use of Graphs. A graph is a pictorial statement of a series of values all drawn to scale. Such a diagram will often greatly facilitate the understanding of a problem ; for the meaning is more readily transmitted to the brain by the eye than by descrip- tion or formulae. When reading a description, one has often to form a mental picture of the scenes before one can grasp and fully appreciate the ideas or facts involved. If, however, the scenes are presented vividly to us, much strain is removed from the brain. A few pages of statistics would have to be studied carefully before their meaning could be seen in all its bearings, whereas if a " graph " or picture were drawn to represent these figures, the variations of their values could be read off at a glance. To take another example : a set of experiments are carried out with pulley blocks; the results will not be perfect, some readings may be too high, others too low : and to average them from the tabulated list of values would be extremely laborious; whereas the drawing of a graph is itself in the nature of an averaging. Or, again, a graph shows not only a change in a quantity, but the rate at which that change is taking place, this latter being often the more important. On a boiler trial a graph is often drawn to denote the consumption of coal : from which is shown during what period the consumption is uniform, or when the demand has been greater or less than the average, and so on. A graph, then, is a picture representing some happenings, and is so designed as to bring out all points of significance in connection with those happenings. The full importance and usefulness of graphs can only be appreciated after many applications have been considered. To commence the study of this branch of our work let us consider an example based on some laboratory experiments. INTRODUCTION TO GRAPHS T49 Example i. In some experiments on the flow of water over notches the following figures were actually obtained. RIGHT-ANGLED V-NoxcH Head (ft.) H . . 1888 2365 2617 2878 3065 336i Quantity flowing (Ibs. per min.) Q J4I-5 249-8 323-5 411-4 483-6 608 The flow, in subsequent experiments, was to be gauged by the " head " of water at the notch, so that a good " calibration " curve was desired. The figures were plotted as shown in Fig. 77, H along a horizontal axis and Q parallel to a vertical axis. In such plotting as this the following points of detail should be observed. Select two lines at right angles for the main axes and thicken them in : these lines should be as far over to the left and as low down, respectively, as will permit of the scales being written to the outside of each. Look to the values to be plotted, noting the " range " in either direction, the scales for the plotting being selected so that the whole of the available space is utilised : but care must be taken to select a sensible scale. Generally a decimal scale is to be preferred, e. g., in the present case we take \" to represent -02 ft. of head, horizontally and |" to represent 100 Ibs. per min. vertically. Write figures along the axes to indicate the scales adopted, and also indicate clearly which quantity is plotted along the horizontal axis and which along the vertical axis ; for attention to such details greatly enhances the value of the graph. To plot : We wish to illustrate the fact that for each value of H there is a value of Q ; which we can do by selecting some value of H, running up the vertical through the marking denoting that value until we meet the horizontal through the given corresponding value of Q, and then making a small mark, e. g., the point denoting that H = -2878 when Q = 411-4, as shown on the diagram by the point P. The use of paper ruled in squares will ease matters, although in a good many instances a series of horizontal and vertical lines through points specified in a table of given values will suffice. When all the points have been plotted, the best average or 150 MATHEMATICS FOR ENGINEERS smooth curve must be drawn through them : the points above the line should about balance those below it, and any obviously in- accurate values must be disregarded. For good results the curve should be drawn with the aid of either a spline or a French curve. The curve is now what is called a calibration curve for the notch, i. e., for any head within the range for which experiments were carried out, the quantity flowing can be read off. 600 100 18 20 24- -26 -28 30 Values of H(feel-) - 77- Calibration Curve for V-notch. (Full size.) 32 34 This process of reading off intermediate values is spoken of as "interpolation." Without the graph, for any values not given in the table one would have either to estimate or to repeat the experiment if intermediate values were required. Also one further point should be noticed : even the figures in the table may not be quite the best, and better approximations can be obtained from the curve. Ex. To find the quantity when the head is -24 ft. : erect the perpendicular SQ through -24 on the scale of head, meeting the curve at Q. Draw QR horizontally to cut the axis of quantity at Q = 260. Then for a head of -24 feet, 260 Ibs. per min. are flowing. Ex. Find the head when Q = 480. From the diagram, H = -30.5 ft. INTRODUCTION TO GRAPHS 151 Example 2. The following figures were obtained in some trials on a gas engine. Draw the efficiency curve, i. e., the curve in which the efficiency is plotted against the output. I.H.P. (Input) i'54 3-09 4-58 5-67 6-50 B.H.P. (Output) o 1-62 3'33 4-71 5-8! The efficiency (to be denoted throughout this book by rj, the Greek , . output B.H.P. letter eta) = ^ t or - and could be calculated by taking corresponding values of B and I from the table. Fig. 78. Test on Gas Engine. It is better, however, to first plot B.H.P. against I.H.P. and average these points by a straight line, which can be drawn with more certainty than a curve (see Fig. 78). The efficiencies at various loads can now T> be calculated from this " curve " by taking the ratios of y for con- venient values of B; e. g., when B = i, I = 2-43 and / = '4 12 - Plotting the values of the efficiency so obtained to a base of output, a well-defined smooth curve is obtained, as in Fig. 78. The efficiencies worked from the experimental figures are B.H.P. . . O 1-62 3'33 4-71 5-81 T) ... 525 726 831 895 If now these values are plotted to a base of B.H.P. the points lie fairly equally about the efficiency curve already drawn. 152 MATHEMATICS FOR ENGINEERS The efficiency-output and the input-output curves now agree, whereas they would not do so in all probability if plotted quite separately. This derivation of one curve from another is of wide application. To illustrate by another example : Example 3. A test on a Morris-Bastert pulley block gave the following results : Load lited\ (Ibs.) W. j 27-5 47'5 67-5 87-5 107-5 127-5 147-5 167-5 187-5 Effort re-] quired (Ibs.) 2-07 2'5 3-15 4-05 4-52 5'2 5-85 6-4 7-1 P. j The velocity ratio (V.R.) of the machine was 48. Draw the efficiency curve to a base of loads. W Theoretical effort to raise a weight W = P = T?~ Actual effort = P! and efficiency = ^5- Fig. 79. Test on Pulley Block. First plot the given values, W horizontally and P vertically, and draw the straight line which best fits the points (see Fig. 79). To calculate values of P corresponding to the values of W set 48 on the C scale of the slide rule level with i on the D scale. Then the readings on the C scale will correspond to values of W and those on the D scale level with these to values of P; e. g., place the cursor over 27-5 on the C scale and -572 is read off on the D scale, so that the value of P when W = 27-5, is '572. All values of P can thus be read off with one, or, at the most, two settings of the rule. INTRODUCTION TO GRAPHS 153 The values of P are -572, -99, 1-41, 1-82, 2-24, 2-66, 3-07, 3-49, 3-9. Plotting these to the same scale as chosen for P l the lower line in Fig. 79 is obtained. By division of corresponding ordinates of these lines the efficiency can be calculated for any load, e. g., when W = 80, P = 1-63, P! = 3-65 and r) = - = -447. A scale must now be chosen for efficiencies, and the curve can then be put in ; this will be a smooth curve, because it is obtained from two straight lines.. Example 4. In some experimental work, only gramme weights were available, whilst for calculation purposes the weights were re- quired in pounds. To save the constant division by 453-6 (the number of grms. equivalent to i Ib.) a straight line could be drawn from which the required interpolations could be made. To construct such a chart : 6 4 JT 3 - 5-77 r ~r / / J 7 4-6 / jr - co- -o / ' o / / / U?-2-32 2-10 Y\ i-Oi > / \ / i 207* 44 i^~ " y/2QO *7<5 (O5O 985 ' Sec lie of 2t iOO 400 600 12OO iQOO OOO E4OO 28OO Fig. 80. Chart to convert grammes to Ibs. Suppose that the readings in grms. were 200, 476, 985, 1050, 2072, 2600. Plotting grms. along the horizontal as in Fig. 80, a scale must be chosen to admit of 2600 being shown. Draw a vertical through 453 -6 to meet a horizontal through i on the " Ib. " scale. The line joining this to the origin (i. e., the zero point for both scales) is the conversion 154 MATHEMATICS FOR ENGINEERS line. The required values can now be quickly read off as in the following table : grms. 200 476 985 1050 2072 2600 Ibs. '44 1-05 2-16 2-32 4-6 577 One axis might take the place of the two in the above diagram. Along this on one side the graduation would be in Ibs. and on the other side, in grms. ; thus amounting to putting a scale of Ibs. alongside one of grms. Tables of logarithms might be, and in fact are (in Farmer's Log Tables), replaced by a number of lines, graduated in numbers and also in logarithms. For great accuracy a great number of lines are required so that two pages do not suffice as in the case of the tables, this being rather a disadvantage : never- theless there is much to be said for this method of table construc- tion. There are no differences to add, nor is it necessary to remem- ber when differences have to be subtracted, since for any definite value in the one set of units the corresponding value in the other is read off directly. Exercises 20. On Simple Plotting. 1. In a test on Hobson's flooring the following figures were obtained. Total load (tons) 35 4 5 I 6o 70 80 90 IOO no Deflection (ins.) * A A| 1 i* ii i A 2 Plot a graph to give the deflection for any load between 35 and no tons ; and read off the deflection for a load of 55 tons and also the load causing a deflection of i*. 2. Plot a curve to show the decrease in the tenacity of copper with increase of heat, from the following table : Temperature F. . . . 212 350 380 400 500 530 580 620 720 Tenacity (Ibs. per sq. in.) 32000 30000 29500 29000 26500 255 23500 21500 2OOOO Read off from your graph : (a) the tenacity at 302 F. ; (b) the temperature at v.-hich the tenacity is 21000 Ibs. per sq. in. ; (c) the tenacity at 545 F. INTRODUCTION TO GRAPHS 155 3. Draw the calibration curve for a rectangular notch, given Head (foot) .... 0871 1115 1588 1838 2124 Quantity (Ibs. per min.) I39H 199 323'3 406-2 502-8 Find the discharge when the head is -19 ft. 4. The following figures are given for the working stress allowable on studs and bolts : Diam. of stud (ins.) . I 3 I I* ii 1 2 Stress (Ibs.persq. in.) 2OOO 3OOO 3900 4700 5500 6300 7OOO Find the stress allowable on a stud of |" diam. and also the stud to be used if the stress is 5100 Ibs. per sq. in. 5. Cast-iron pulleys should never run at a greater circumferential speed than i mile per minute. In the table the maximum revolutions per minute (R.P.M.) allowable are given for various diameters. Find the R.P.M. for a pulley of 14". diam. Check this figure by the ordinary rule of mensuration. Diam. (ins.) 5 6 8 10 12 15 18 2O 25 30 R.P.M. . . 4034 336i 2524 2017 1681 1345 1120 I008 807 673. 6. Plot a curve to give the diameter of a shaft for any twisting moment from -7 ton per sq. in. to 360 tons per sq. in. Equivalent twisting ~| moment (tons per > sq. in. } 701 2-367 5-611 10-95 18-94 30-07 44'9 63'9 87-7 152 359 Diam. of shaft (ins.) i i'5 2 2'5 3 3'5 4 4'5 . 5 6 8 7. The table gives the " time constant " of the coils of an electro- magnet for gaps of various lengths. Represent this variation by a graph. Distance apart (cms.) . 125 5 75 i i-5 2 2-5 3 Time constant (sees.) . 2-5 1-7 i-4 i-4 i-i I-I 9 9 8. The relation between pressure p and temperature t of steam shown in the table was found experimentally. Plot a curve to represent this, finding the value of t when p is 105, and the value of p when t is 300. p Ibs. per sq. in. . . 5 10 15 20-5 27 3i 36 44 50 60 70 80 90 100 IIO 120 <(F.) 235 243 251 260 270 276 282 290 296 306 3'4 322 329 336 342 34 9. and 10. Plot curves of Magnetic Induction for (i) Iron, and (2) Cobalt, from the figures given in the tables following. 156 MATHEMATICS FOR ENGINEERS (9. Iron.) H (magnetising! force) t o 5 10 i? 25 30 38 45 52 Co 65 B (mag. induc-i tion density) / 2400 4500 6000 7100 7800 8300 8500 8600 8600 8700 (10. Cobalt.) H. o 1-55 3'io 4-65 6.2O 775 12-40 I5-5 23-25 3i 3875 46-5 B. 99 268 j 642 1128 1298 2405 2995 4070 4860 5390 5810 11. Plot a curve to show the variation in the ratio Q /weight of armament and protection) \ load displacement / as given for a speed of 21 knots, from the following table : Load displace-\ mentP(tons)/ 18000 22OOO 24000 26000 30000 34000 38000 4000O Ratio Q . . 383 401 409 416 428 438 446 45 Find the weight of armament and protection when the displacement is 28000 tons. 12. Plot a curve, as for Question n, but the figures belonging to a speed of 27 knots. p . . 18000 2OOOO 24OOO 26000 28000 30000 32OOO 36000 40OOO Q 236 252 275 286 295 303 3 IO 324 336 Find the value of Q when P = 34000. 13. The temperature of the field coils of a motor was measured at various times during the passage of a strong current, with the following results : Time (mins.) .... 2 5 10 15 20 25 30 35 40 45 50 55 60 Temperature (C.) . . 14 16 23 32-4 39 43'4 47 50-5 52-5 55 56-8 58-6 59-3 59'4 Find the time that elapses before radiation losses, etc., balance the heating effect of the current, viz. when there is no further sensible rise of temperature ; and find also the maximum rise of temperature. 14. Repeat as for Question 13, taking the following results : Time (mins.) .... o 5 10 15 20 25 30 35 40 45 50 55 60 65 Temperature (C.) . . . 20 26 32-5 4i 46 49 52-5 54'5 56-5 58 59-5 61 617 62 15. The following figures were obtained by reading spring balances at the ends of a beam on which a weight of 7 Ibs. was hung. Plot INTRODUCTION TO GRAPHS 157 curves to give the values of the reactions for any position of the weight. Note their point of intersection. Distance (ins.) of weight \ from R.H. end / 2 4 6 8 10 12 14 16 18 24 28 30 32 Left-hand reaction (Ibs.) '4 8 1-25 1-7 2*1 2'55 3 3'45 3'9 5'2 6-05 6-5 7 R.H. reaction (Ibs.) . . 7 6-5 6*05 5'6 5'2 4-8 4'3 4-5 3H5 3 1-8 8 *4 o In Questions 16 to 19 draw to a base of loads (W) curves whose ordinates gives (a) Actual effort P x ; (6) theoretical effort P ; (c) efficiency 17. 16. Test on a 6 to i pulley block, i. e., V.R. = 6. W 28 48 68 88 108 128 148 1 68 188 208 P! 9-75 1475 20-25 2 575 30-75 3575 40-25 45-25 49-25 55-25 17. Test on a Single Purchase Crab (V.R. = 27). W . . 50-1 92-1 137 1 80 224 266 310 354 394 PI . . 3-6 5-35 7'9 9-9 n-8 13-9 14-7 16-9 19-5 18. Test on a Screw Jack (V.R. = 60-5). W . 34 54 74 94 114 134 T 54 174 194 214 234 PI . i-73 2-85 3'93 5-17 6-19 7-70 8-95 10 n-3 12 I2'9 19. Test on a Weston Pulley Block, when raising (V.R. = 24). W . . . 25 45 65 85 105 125 145 165 185 205 P, . . . 4 6-75 8-75 8-75 10 13-5 15 18-75 21 22-5 20. The table gives the current absorbed by a carbon brush at various pressures. Plot, to a base of amperes of current, curves giving resistance and voltage. ! Resistance = ' I amperesj The resistance curve should be obtained from that for voltage. Volts. . . . '35 65 9 88 i 18-75 1-3 21-5 1-45 I'S 1-65 i'75 37'5 i-77 1-8 1-825 1-85 Amps. . . . 4 13-5 24'5 27-5 32-5 40-5 42 45'5 47'5 21. To a base of frequency plot curves giving (a) voltage, (b) current taking the following figures : Frequency 40 43-5 47 50 52 54 56 60 64 75 80 88 Current 5-39 8-75 I4'35 18-67 1473 11-66 9-33 6-83 5-19 3-05 2-64 2-14 Voltage . 52 32 195 15 19 24 30 41 54 93 1 06 131 158 MATHEMATICS FOR ENGINEERS 22. The following figures were obtained in a tensile test on a sample of 25% nickel steel. Stress (Ibs. per sq. in.) . 4000 12000 20000 28000 36000 48000 52000 56000 60000 Extension (inches per j inch length) 00015 00047 0007 001 1 1 00145 00195 -00213 00235 00264 64000 68000 72000 76OOO 80000 84000 88000 92000 96000 IOOOOO 104000 00365 0065 02 1 035 052 068 0853 1025 134 171 201 Plot the " stress-strain " diagram, the stresses being vertical and extensions along the horizontal; also determine the stress at the " yield point," where the sudden change occurs. 23. The voltage supplied to a 4-volt lamp was varied, and the candle-power (C.P.) then measured for various values of the voltage, the results being as follows : C.P. . o 5 I-O !'5 2 2'5 3 Volts . 3-03 3'57 3-96 4- 2 5 4-44 4'75 Amps. . 1-16 1-29 1-36 1-48 1-63 1-71 If watts = volts x amps, plot to a base of C.P. curves whose ordinates represent (a) volts; (6) amps, and by a combination of corresponding ordinates of these (c) watts per C.P. 24. The drop in potential due to a standard resistance of -3 ohm was measured by a potentiometer, for various currents. The current was also measured on an ammeter. volts If current = , calculate the true currents flowing. Also resistance plot a curve of true current against registered current, and hence find the percentage error of the ammeter. Ammeter reading) (Registered current)/ I I< 5 2 2'5 3 3'5 4 4'5 475 Volts .... ^093 458 6149 7629 92 1-0487 1-204 I-37 1 1*437 25. From the following figures (taken from a test on a 10 H.P. Diesel engine) plot curves, to a base of B.H.P., to show (a) I.H.P., from which deduce (b) mechanical efficiency : (c) oil per hour, and hence (d) oil per B.H.P. hour. B.H.P. . . . o O.TO 6-71 8-i; 9-Q4. I.H.P. . . . 4'5 7^7 10-66 11-69 12-95 Oil per hour (Ibs.) i'5 2-37 3-63 4'35 5'45 INTRODUCTION TO GRAPHS 159 26. From the given figures plot to a base of I.H.P., curves with ordinates to represent (a) steam per hour and thence (b) steam per I.H.P. hour. Steam per hour (Ibs.) 513 452 436 403 37 327 182 I.H.P 13-12 iQ'54 9-83 8-85 8-15 6-57 1-84 27. Results of an efficiency test on a small motor gave the follow- ing :-r- Output (watts) . . . 6-46 24-2 33-8 37'5 40 55-3 138 61-5 | 64-9 77'i 9* "7 Input (watts) . . . 57-6 82-4 IO2 104-2 107-2 142-2 141-1 162-4 187-5 228 To a base of output plot curves giving (a) input and thence (b) efficiency. (Efficiency = ^J 28. The voltage of an accumulator, when discharging, fell according to the following : At 2 o'clock voltage = 2-15, at 2.30 o'clock and also at 3.30 voltage = 2-06, at 6.30 voltage = 1-87 and at 9 o'clock voltage = 1-72. Another cell was charged at a uniform rate from 2 o'clock to 7 o'clock, the voltage rising from 1-75 to 2-38. Assuming that the discharge was uniform, find the time at which the cells had the same voltage. Co-ordinates. So far, in these graph problems, we have been concerned with positive quantities only; the question now is, How to deal with negative quantities ? If the plotting " movement " has been in a certain direction for the positive, then clearly for a negative the motion must be reversed. The convention adopted is that to the right and upwards are positive directions for the horizontal and vertical axes respectively; and therefore to the left and downwards will be the corresponding negative directions. These are indicated in the diagram (Fig. 81). To admit of all arrangements of signs the paper must be divided into four parts or quadrants as shown, the point of intersection of the axes being termed the origin, viz. the point O. The points A! A 2 A 3 and A 4 are all distant 4 units from the vertical axis and 3 units from the horizontal, so that to distinguish between them we must make some mention of the quadrant in which each is placed by affixing the correct signs. The distances from the axes together are spoken of as co-ordinates, that along the horizontal being usually called the abscissa, while vertical distances are called ordinates. In representing a point by its co-ordinates the abscissa is always stated first. i Go MATHEMATICS FOR ENGINEERS Point A! is thus + 4 and + 3 or more shortly (4, 3) A 2 is 4 and + 3 or more shortly ( 4, 3) A 3 is 4 and 3 or more shortly (4, 3) A 4 is +4 and 3 or more shortly (4, - 3). Note that ( -4, - 3) does not imply -7, but a movement of 4 units to the left of the vertical axis and then 3 units down from the horizontal axis. E. g., Point B is (1-5, i) Point C is ( 3-2, o) Point Dis ( 1-4, 2-3). D.|... {-1-4, 2-3) I -^ -4 -3 -2 -1 (I-5.-I) -2 oAa (-4,-: -3 (4,-3: Fig. 81. Co-ordinates of Points. To fix the position of a point in space it would be necessary to state the three co-ordinates, viz. the distances from three axes mutually at right angles. For example, a gas light in a room would be referred to two walls and the floor to give its position in the air. Representation of an Equation by a Graph. If two quantities x and y depend in a perfectly definite way, the one upon the other, the relation between them may be illustrated by a graph which will take the form of a straight line or a smooth curve. From this curve much information can be gleaned to assist in the study of the function as it is called. [Explanation. If y = 2X + 5, INTRODUCTION TO GRAPHS 161 y is said to be a function of x, for y depends for its value on that given to x; if y = 4z 2 +jz 3 8 log z, y is a function of z or, as it would be expressed more shortly, y = /(), meaning that y has a definite value for every value ascribed to z : e.g., in the case first considered, y = /(*) = 2*-f 5, then /(a) would indicate the value of y when 3 was written in place of x, i. e.,/(3) = (2x3) +5 = u.] Dealing first with the simplest type of graph, viz. the straight line, whenever the equation giving the connection between the variables is of the first degree as regards the variables, . e., it contains the first power only of the variables, a straight line will result when the equation is plotted. Example 5. Plot a graph to represent the equation y = 5* 9. In all cases of calculation for plotting purposes it is best to tabulate in the first instance; for any error can thus be readily detected, and in any case some system must be adopted to reduce the mental labour and the time involved. The general plan in these plotting questions is to select various values for one of the variables, which we can speak of as the " in- dependent variable " (I.V.), and then to calculate the corresponding values of the other, which may be spoken of as the " dependent variable" In questions where x and y are involved it is customary to make x the I.V., and to plot its values along the horizontal axis. We may take whatever values for x we please, since nothing is said in the question about the range. Let us suppose that x varies from 4 to +4. The table, showing values of y corresponding to values of x would be as follows : X 5* - 9 y *-4 20 - 9 - 29 -3 - 15 - 9 - 24 2 10 - 9 - 19 I - 5 - 9 - 14 o - 9 - 9 I 5 - 9 - 4 2 10 - 9 i 3 15 - 9 6 4 20 - 9 II * i.e., 5* = 5* (-4) = - 20 ' M Fig. 82. Curve of y = 52 9- 162 MATHEMATICS FOR ENGINEERS When we come to the plotting we see that it is advisable to select different scales for x and y, since the range of x is 8 and that of y is 40. On plotting the above values a straight line passes through them all (Fig. 82). A straight line would be definitely fixed if one knew its slope or inclination and some point through which it passes. As regards the slope, a line sloping upwards towards the right has a positive slope, because the increase in the value of x is accompanied by an increase in the value of y, and the slope is measured by -r change of x In measuring the slope of a line, the denominator is first decided upon, a round number of units, say 2 or 10, being chosen, and the numerator corresponding to this change is read off in terms of the vertical units from the diagram. In the case of the line representing y = 5^9 the slope is 2 1 ) seen to be = 5, i. e., the slope is the coefficient of x in the original *J equation. The fixed point, a knowledge of which is necessary before the line can be located, is taken on the y axis through x = o, *'. e., the point of intersection of the line with the vertical axis through x = o must be known. In the case shown in Fig. 82 the line intersects at the point for which x = o, y = 9 : also 9 is noted to be the value of the constant term in the equation from which the graph is plotted. In general, if the equation to a straight line is written, y = ax+b ; a is the slope of the line and b is the intercept on the vertical axis through the zero of the horizontal scale. All equations of the first degree can be put into this standard form, and hence will all be represented by straight lines. Example 6. Consider the three equations = 8 ........... (i) = o ........... (2) A similarity is at once noticed between the equations ; a short investigation will show the full interpretation of that similarity when regarded from the graphical standpoint. Whenever an equation is to be plotted it is always the best plan to find an expression for one variable in terms of the other; and it is usual to find y in terms of x in these simpler forms. INTRODUCTION TO GRAPHS Q ,*. From (i) 5)/ = 84* = -12-4*, From (2) From (3) y 5 -- 1-6 -8* y = 2-4 -8x 163 - (4) - (5) (6) Fig. 83. Straight Lines and their Equations. Evidently all three equations, viz. (4), (5) and (6), are of the form y = ax + b, the value of a being constant throughout, viz. -8, whilst the value of 6 varies. From our previous work, then, we conclude that the three lines representing these equations have the same slope and are therefore parallel, being separated a distance vertically represented by the different values of b. To plot, first calculate from the equations (i) y = 1-68*. (2) y = 8*. (3) y = -2-48*. and tabulate the numerical values : (i) * i-6 -8* y -4 1-6+3-2 4-8 2 1-6+ 1-6 3-2 O i-6 o 1-6 2 i-6 1-6 i o 4 1-6-3-2 -1-6 * 2-4 -8x y -4 -2-4+3-2 8 2 -2-4 + 1-6 - -8 o 2-4 -2-4 2 2-4 i-6 -4-0 4 -2-4-3-2 -5-6 These lines are parallel (see Fig. 83) and cross the y axis, (i) at 1-6, (2) at o, and (3) at 2-4, or the values of b in the three cases are 1-6, o and 2 -4 respectively. 164 MATHEMATICS FOR ENGINEERS Solution of Simultaneous Equations by a Graphic Method. Knowing that a first-degree equation can be represented by a straight line, our attention must now be directed to some useful application of this property. One of the greatest advantages of graphs is that they can be utilised to solve equations of practically every description. As a first illustration we shall solve a pair of simultaneous equations by the graphic method. Example 7. To solve, by the graphic method, the equations 5*+3y =19 (i) gx -zy =12 (2) Each of these equations can be represented by a straight line ; and these lines will either be parallel or meet at a point, and at that point only. Such a point represents by its co-ordinates a value of x and a value of y ; and since this point is common to the two lines, these values must be the solutions of the given equations. Fig. 84. Solution of Simultaneous Equations. [If the given equations were 5#+3J> = 19 and 5#+3y = 9 it would be found on plotting that the lines were parallel ; there could thus be no values of x and y satisfying the two equations at the same time, or, in other words, the equations are not consistent.] For the example given, the lines are not parallel. Two points are sufficient to determine a line, and therefore two values only of y need be calculated, but for certainty three are here taken, because if two only were taken, and an error made in one, the line would be entirely wrong. INTRODUCTION TO GRAPHS Equation (i) 5* + yy = 19 from which $y = 19 $x or y = 6-33 1-67*. Table of values reads : 165 X 6-33- i-6jx y - 3 6-33 + 5 n-33 o 6-33- 6-33 4 6-33- 6-68 - '35 Equation (2) whence gx 2y = i2 2y = 12 -9* Table of values reads : 2y = 9^ 12 y = 4'5# 6. X 4*5^ ~~" y 2 - 9 - 6 - 15 o - 6 - 6 4 18 - 6 12 These two lines must be plotted (see Fig. 84) to the same scales and on the same diagram and their point of intersection noted, viz. (2, 3). x = 2, y = 3 are the solutions of the given equations. [The scales chosen must be such that the point of intersection will be shown ; to ensure that this shall be the case a rough mental picture of the diagram should be formed. This is not a difficult matter, as one soon becomes accustomed to reading a table from its graphical aspect. E. g., one can see at a glance in which direc- tion the line is sloping, and a little further consideration decides the rate of its rising or falling.] Exercises 21. On plotting Co-ordinates, and plotting of Straight Lines representing Linear Equations. 1. On the same diagram plot the points (2, -5) ; (3, 4) ; (-9, 3) .' (o, n); and (1-2, o). Indicate each point clearly. 2. Join up the four points (-10, 10) ; (5, 10) ; (15, -2-5) ; and (10, 2-5) in the order given, and find the area in sq. units of the figure so formed. 3. On the same diagram plot the points (1-4, 2500); (-75, 374) ' (-1-82, 1140); (-32, 4816). Indicate clearly the scales chosen. 4. Plot the straight line $x-8y .= 19 from x = -4 to x = +5. What is the slope of this line, and what is its intercept on the vertical axis through o on the horizontal ? 5. Plot a straight line to show the change of x consequent on change of y between -10 and +15; the connection between y and x being 'i6y = 4-28 4-06*. 166 MATHEMATICS FOR ENGINEERS 6. The illumination I (foot candles) of a single arc lamp placed 22 ft. above the ground, at d feet from the foot of the lamp is given by I = i -4 -oid. Plot a graph to show the illumination for distances o to 12 ft. from the foot of the lamp. 7. Unwin's law states that the velocity of water in ft. per sec. in town supply pipes is v = 1-45^+ 2, where d is the diam. of pipe in ft. Plot a graph to give the diam. of pipe for any velocity from o to 13 ft. per sec. 8. The law connecting the ratio (-A i- e-, ^-^ - of a journal with the speed (N, R.P.M.) is -3 = -oo^N + i. Plot a graph to show values of this ratio for values of N from 20 to 180. If the diam. is 4-5* what should the length be at 95 R.P.M. ? 9. Plot a conversion chart to give the number of radians correspond- ing to angles between o and 360. (i radian = 57-3.) 10. The law connecting the latent heat L with the absolute tem- perature T, for steam is L = 1437 - "Jr. Plot a graph to give the latent heat at any temperature between 460 and 1000 F. absolute. 11. Plot a graph giving the resistance R of an incandescent lamp at any voltage V between 40 and no. You are given that R = 2-5 V + 75. What is the slope of the resulting graph ? Solve graphically the equations in Exs. 12 to 16 : 12. yn 6n = 6-6 13. 48* 27^ = 48 nn 25 = am. y$ix = 51. 14. y+i-37=4* 15. 7#+3y=io gx-ijy =-49-87. 35*-6y = i. 16. y = 1-4* 3 2-6# y = 13. 17. The co-ordinates of two points A and B are : A. Latitude (vertically) N 400 links ; Departure W (horizontally) 700 links B. Latitude S 160 links; Departure W 1500 links. Plot the points A and B and find the acute angle which the line AB makes with the N and S line. Determination of Laws. The straight line as the representa- tion of an equation finds its most direct and important application in the determination of laws embodying the results of experiments. An experiment has been made with some machine and a number of readings of the variable quantities taken; and it is desirable to express the connection between these quantities in a simple yet conclusive manner. If this is done the law of the machine is known for the range dealt with. INTRODUCTION TO GRAPHS 167 Example 8. A test is carried out on a steam engine, and trials are made with the engine running at various loads. The amount of steam used per hour (W) and the Indicated Horse Power (I.H.P.) are calculated from the readings taken at each load, and the corresponding values are as follows : I (I.H.P.) 4 5 7 IO 12 W (Ibs. of steam per hour) 7i 103 121 153 197 234 Find a simple relation connecting W and I. It is reasonable to assume that to just start the engine a certain amount of steam would be required, which would in a sense be wasted, and that after once starting, the steam used would be practically 240 200 teo 120 QO o W - 3 <x 5 J z Values c O2. 46 8 10 / Fig. 85. Test on Steam Engine. proportional to the power developed : accordingly we should expect a formula of the type W = b + al where a and b are constants to be determined. This we see is of the standard type y = ax + b, or putting it in a more general form (Vertical) = a (Horizontal) + b, where (Vertical) stands for the quantity plotted along the vertical ; therefore, a straight line should result when W is plotted against I. On plotting (see Fig. 85) we see that a straight line fits the points very nearly, being above some and below others, i. e., averaging the results. The values of a and b may be found by either of two methods. The first is that used in the laboratory and is to be recommended when the slope of the line is more important than the intercept : it can be used on all occasions when the quantities given admit i68 MATHEMATICS FOR ENGINEERS of the vertical axis through the zero of the horizontal being drawn without diminishing the scale. This method is very quick, measure- ments on the paper being scaled off and a quotient easily found. The second method is the more general, but involves rather more calculation; both methods should, however, be studied. First Method- W = a I + b where a the slope of the line and b = intercept on the vertical axis. To find the slope, select some convenient starting-point, say, where the line passes through the corner of a square, and measure a round number of units along the horizontal, in this case (Fig. 85) 5 being taken. (Note. Distances are measured in terms of units, and not in inches.) The vertical from the end of the 5 to meet the sloping line measures 79 units ; , increase in W 79 hence slope = - ; ^- = = 15-8, .'. a = 15-8. increase in I 5 Intercept on axis of W through I = o is 40 units, .'. b = 40. Thus the equation is W = 15-81+40- Second Method, or Simultaneous Equation Method Select two convenient points on the line, not too close together e. g., W = 167-5 \ and w = 8 7'5 \ when I = 8 / when I = 3 J Substituting these corresponding values in the equation W = a I + b two equations are formed, the solutions of which are the required values of a and b. Thus 167-5 = 8a + b (i) 87-5 = 3 + b (2) Subtracting 80 = $a whence a = 16. Substituting in equation (2) b = 87-5 48 = 39-5 .*. as by first method (very closely) W _= 1 6 1 + 3 9 ' 5 This particular line connecting the weight of steam per hour with the indicated horse-power is known as a Willans' line (named after Mr. Willans, who first put the results of steam-engine tests into this form). To take a further example Example g. In a test on a crane the following values were found for the effort P! required to raise a weight W. Find the law of the crane. W (Ibs.) . IO 20 30 40 50 60 70 80 90 IOO P! (Ibs.) . I i-6 3 2-13 2-63 3-25 3-75 4-25 5 5'5 6 INTRODUCTION TO GRAPHS 169 To find the equation in the form P t = aW + 6 plot W along the horizontal (Fig. 86). First Method Slope = = -0564, .*. a = -0564 u Also the intercept on the axis through o of W = -41, /. b = -41 PI= -Q564W+ -41. Second Method when Subtracting Substituting in (2) and Pj when W 3-8 = 6oa + b 7= 5+& (i) a = -0564 6 = -7 - -282 = -418 ! = -0564 W + -418. or 2-82 5O Values of W, 20 50 40 50 GO 70 Fig. 86. Test on a Crane. SO 30 100 This result suggests that -41 Ib. is required to just start the machine, i. e., to overcome the initial friction, and that after that point for every pound lifted only -0564 Ib. of effort is required. If we are told, in addition, that the velocity ratio of the machine is 39, we can calculate the efficiency of the machine for any load. distance moved by effort Velocity ratio = b>Twei g ht and work done by effort = work done by weight ; hence, theoretically, i Ib. of effort should just lift 39 Ibs. of weight ; i. e., the connection between P and W (theoretically) is P = W. 170 MATHEMATICS FOR ENGINEERS Theoretical effort P Then the efficiency at any load = Actual effort -LW 39 _ . PI 0256W _ ____ + -418 -0564 W + -418 2-2 16-35 W e. g., if W = 50, efficiency = rj 2-2 + 396. 16-35 5 Example 10. The following are the results of a test on a 6-ton Hydraulic Jack (V.R. = 106). Load (Ibs.) 1 600 IO2O 1445 I88 5 2320 2740 3210 3625 4010 Effort (Ibs.) ii 17 22 27-9 327 37'5 43'4 45'2 49 It is required to find an expression for the efficiency at any load, and also the maximum efficiency. O . 5OO /OOO 2OOO 3OOO Fig. 87. Test on Hydraulic Jack. 4000 To a base of W (load) we plot the values of PI (practical effort) and average the results by a straight line, as in Fig. 87. Theoretically, each pound of effort applied should lift 106 Ibs. of load, hence a straight line can be drawn giving the theoretical effort (P) for all loads within the range dealt with, p Now, efficiency ? = p- ; and therefore for any load find the quotient INTRODUCTION TO GRAPHS 171 p-, which will be the efficiency at that load. A new scale must be chosen for efficiency, and the curve, a smooth one, because obtained from two straight lines, is plotted. [e.g., If W = 2000, P = 18-9, Pj = 28, , = ^ = -675} To find the maximum efficiency, i. e., the efficiency at 6 tons load. Also P = V PI = 5'6+,^4-W - P P, 2oo 5-6+-OH2W . . (See Fig. 87) i 5'6 -OII2W " - 7 + -oo9 44 W or efficiency at any load = 593 W + 1-19 Then for the efficiency at 6 tons load we must write 6 x 2240 for W, hence maximum efficiency = * -^3_ + i-i 9 I>23 = -814. 13440 Exercises 22. On the Determination of Laws. P 1. Find the average value of ^ (coefficient of traction) from the following figures (i. e., find the slope of the resulting straight line). W (Ibs.) . 3 5'5 7'5 9-5 "5 13-5 !5'5 17-5 P (Ibs.) . 1-25 2-25 2'75 375 4'25 5-25 6-25 7-25 This was for the case of wood on wood. 2. Recalculate but for cast iron on cast iron (dry). W (Ibs.) . 33 53-3 63-2 72-9 93'2 "3 P (Ibs.) . n-3 19 22 2 5 28 37'5 In Exs. 3 and 4 the slope of the line gives the value of the Young's Modulus E for the material. Find E in each case, stating the units. (Note that the stress is to be plotted vertically.) 3. For i" round, crucible cast steel. Stress (Ibs. per sq. in.) 2000 4000 6000 8000 IOOOO I2OOO 14000 16000 Extension (inch per) inch length) . . . / 00008 oooiS OOO2I 00028 00034 00041 00048 00053 i;2 MATHEMATICS FOR ENGINEERS 4. For i" round, hard-rolled phosphor-bronze. Stress (Ibs. per sq. in.) 2000 4000 6000 8000 IOOOO 12000 Extension (inch per\ inch length) . . J oooi 00022 00034 00044 00055 00067 5. Find the simple law connecting the Indicated Horse Power I with the Brake Horse Power B, given the following values of I and B; B o 3-33 6-71 8-35 9'94 I 4'5 7-27 10-66 11-69 12-95 {I = aB + b} 6. The diameter under the thread for various diameters of bolts is given in the table for the Whitworth standard thread. Find the law connecting the smaller diameter, d lt with the larger, d. d 0625 09375 125 15625 1875 25 375 '5 625 '75 d, 0411 067 0929 1162 1641 1859 2949 3932 5085 6219 7. Recalculate as for Ex. 6, but taking the figures for the British standard fine thread. d i i 4 f i t .1 ii d, 199 3 ii 420 534 643 759 872 1-108 Find the law connecting T and 6 in the following cases (Exs. 8 and 9). T = ad + 6. (T = twisting moment and = angle of twist.) T O 30,60 90 1 2O 150 1 80 2IO 240 270 300 330 360 o 4-9 1-56 2-1 2-7 3'4 4 4'5 5'i 5-3 6-25 6-82 9. T o 1 200 2400 3600 4800 6OOO 7200 e o 34 67 I'O2 1-36 I-7I 2-O6 Express the results of the tests on incandescent lamps given in Exs. 10, ii and 12 in the form R = aV + b. (R = resistance and V = voltage.) 10. Test on a metallic filament lamp. V 75 78 | 80 82 84 86 88 90 92 94 96 98 100 102 R 144 147 148 149 151 153 155 157 158 159 1 60 161 162-5 164-5 INTRODUCTION TO GRAPHS 11. Test on two metallic filament lamps in parallel. V 54 60 65 70 75 80 8 5 90 95 I0 5 A 5 55 '57 59 61 63 65 67 69 72 (Values of resistance R must first be calculated from R = , where A = ampere.) 12. Test on a metallic filament lamp. V 86 80 70 60 5 40 30 R 277 267 259 231 208 174 I5<> The following two examples refer to tests on the variation of the resistance of a conductor with variation of temperature. Find the values of R,, (resistance at o) and a (temperature coefficient) in each case. [R, is intercept on the vertical axis through o of the temperature Slope of line T scale, and a ~ -J -tv<> 13. Equation is of form R, = R u (i + at) where R t = resistance at temperature t. Temperature (/) . 10 2 5 35 50 80 90 IOO Resistance (R t ) 1-039 i-i 1-141 1-198 1-32 I- 357 1-402 14. t 17-1 25'4 30-3 36-2 41 49'4 61-3 67 74-3 8o'i i'532 93-8 R, 1-214 1-259 1-285 1-317 I-34I 1-369 144 1-473 1-505 1-622 15. The following results were obtained from the testing plant of the Pennsylvania Railroad Co. : x (B.Th.U. across heat- \ ing surface per min.) / 207200 24/500 295900 331000 367500 393500 44 3000 448500 481300 I (I.H.P.) 3657 4547 587-6 650 779-3 803-3 95l'4 975-1 1036 Find the law connecting I and x in the form I = ax + b. 16. From the following figures find the value of g. (g = 3-29 x slope of line : obtained by plotting t 2 horizontally and / vertically.) / 34'5 30 28 25 21 16 12 t 1-87 1-76 1-67 1-6 1-49 1-26 I'll [t = periodic time in seconds of a pendulum swing, / = length of simple pendulum in inches, and g = acceleration due to gravity, in feet per sec. per sec.] i 7 4 MATHEMATICS FOR ENGINEERS 17. Find the value of Young's Modulus E for the material of a beam, from the following : Load (W Ibs.) . 36-5 56-5 96-5 136-5 176-5 216-5 256-5 296-5 3I6-5 Deflection (d in.) 12 198 34 5i 63 '79 925 1-07 1-17 W/ 3 Also d = O FT . I = 40", and I = -0127. ( Hint. Slope of line = ^.-J Graphs representing Expressions of th'e Second Degree. Consideration must now be paid to the graphs of such equations as y = 5* 2 + 7* 9, or x = ay z -f- by + c. As mentioned be- fore, the curves representing these equations will be smooth and of standard forms. The preliminary calculation must be performed in a manner similar to that already employed for the straight-line graphs. The only trouble likely to be experienced is with the signs : it must be remembered that 3 or + 3 squared each gives 9, so that if x = 3 and x 2 is required, the value is (9), i. e., 9 ; also 6x z would be 6 X ( 3) 2 = 6 x 9 = 54. Since we are no longer dealing with straight lines, two points are not sufficient to determine the curve, so a number of values must be taken. Example n. Plot, from x=$ to # = +4, the graph repre- senting the equation y = 5 X * + 7* ~ 9- Arranging the calculation in tabular form : X x* 5 * 2 + 7* - 9 y - 5 25 I 2 5 - 35 - 9 81 - 4 16 80 28 9 43 - 3 9 45-21-9 15 2 4 20 14 9 - 3 I i 5-7-9 ii O o o +0-9 - 9 I i 5+7-9 3 2 4 20+14 9 25 3 9 45 + 21-9 57 4 16 80 + 28 9 99 The scale for x must admit of a range of 9 units, whilst that for y requires a range of no units : and as the greater part of the curve is to be on the positive side of the x axis, this axis should be drawn fairly INTRODUCTION TO GRAPHS 175 .90 80 .60 5o low down on the paper and not in the centre (see Fig. 88). After plotting the points from the table of values, a smooth curve should be sketched in, passing through all the points; and if any one point is not well on the curve, the portion of the table in which the calculation for that point occurs must be referred to. The curve is a form of parabola, whose axis is vertical, and whose ver- tex is at the bottom of the curve : indeed, in all equations of the type y = ax 2 + bx + c the curve will be of the form shown if a is posi- tive ; while if a is negative the axis will still be vertical, but the vertex will be at the top of the curve. As an illustration of the latter type Example 12. Plot the curve 4 y = - 3* 2 ~ **+ 2 '44- from x = - 6 to x = +3- 2 1 20 10 Fig. 88. Curve of y = 5* 2 + 7* - 9- Division by 4 gives, y = Table of values : = _ -75AT 2 2X+'6l. X x* '75* 2 2#+-6i y - 6 ~ 5 36 25 16 -27 + 12 + -61 - 18-75 + 10 + -61 -12 + 8 + -61 - 14-39 - 8-14 - 3-39 4 - 3 2 I 9 4 I o - 6-75 + 6 + -61 _ 3 + 4 + -61 - 75 + 2 + ' 6l o + o + -61 - -14 + 1-61 + 1-86 + 61 I 2 3 i 4 9 - '75 - 2 + ' 6l -3 -4+ -6i 6-75 6 + -61 -2-14 - 0-39 12-14 Here the greater part of the curve is negative; hence the axis o 176 MATHEMATICS FOR ENGINEERS x must be higher than the centre of the paper. The plotting is shown in Fig. 89. i 1 1 1 ,^^ 2 - Solution of Quadratic Equations. The equations 5* 2 +7*-9 = o and 75* 2 2*+-6i = o, or, in fact, any quadratic equa- tion, can be solved by the aid of graphs. For the equa- tions y = 5* 2 +7# 9 and 5* 2 +7# 9 = o to be alike, y must equal o. Now y is = o anywhere along the x axis : if, then, we wish to arrange that the y value or ordinate of the curve is to be o, we must select the value or values of x that make it so; or, in other words, we must find those values of x at the points where the curve crosses the x axis. These values of x are the solu- tions or roots of the equation 5^2+7^9 = o. From the dia- gram (Fig. 88) we see that the curve crosses the x axis when x = -82 and also when x = 2-22 : Fig. 89. Curve of 4? = - 3* 2 - 8 * + 2 ' therefore x = -82 or 2-22 gives the two solutions of 5* 2 +7# 9 = o. In like manner the roots of 75** 2#+-6i are 2-95 and -28. (See Fig. 89.) Solution of Quadratic Equations on the Drawing Board. Whilst on the question of the graphical solution of quadratic equations, mention may be made of a method that is simple and requires the use of set squares and compasses, but not squared paper The general quadratic equation is ax 2 -f- bx -f- c = o. To solve this equation by the method of this paragraph : Set off a length OA (see a, Fig. 90) along a horizontal line, working from left to right, to represent a units to some scale. Through A draw AB perpendicular to OA ; if & is positive a length to represent & must be measured, giving AB, so that the arrows continue in a right-hand direction. If c is positive draw BC perpendicular to AB, making BC to represent c units to the same scale as before, the INTRODUCTION TO GRAPHS arrows still continuing to indicate right-hand movement about O. (If c were negative BC would be measured to the other side of AB.) Join OC. On OC as diameter describe a circle to meet AB in the points D and E. Then the roots of the equation are 7^-7- and W?. OA OA Proof of the construction. Let F be the centre of the circle ODC (a, Fig. 90). Draw FG parallel to OA to cut AB in G and join C to H, the point at which the circle cuts OA. Then, from the property of intersecting chords OA X AH = EA X AD Fig. 90. Solution of Quadratic Equations. Dividing both sides by (OA) 2 OA AH OA X OA or AH OA EA OA EA AD X OA AD OA X OA Now the angle OHC is a right angle since it is the angle in a semicircle and since angle OAB is a right angle also, CH and AB are parallel and AH = BC = c. Also, since FG and OA are parallel and F bisects the line OC, then GA = GB. Then EA+DA = ED+DA-f-DA = 2GD+2DA = 2GA = BA = -b EA DA_ _6_ _& / OA + OA ~ OA a DA or Let ^ OA = 178 MATHEMATICS FOR ENGINEERS AH or BC = ap b OA = a/3 or aft = - a Then from equation (i) and from equation (2) Vtr The original equation ax z + bx + c = o might be written Or X* (a -f ft)x + a/3 = O which after factorisation becomes (x a)(x ft) = o whence % = a or (3. D\ EA In other w r ords, a and /3 or ^ and ^K are the ro ts of the original equation. Example 13. Solve the equation ^x* + yx 9 == o by this method. Starting from the point O (&, Fig. 90) set out OA to represent 5 units to some scale. Draw AB downwards from A, since 7, the coefficient of x, is positive, and make it 7 units long. From B draw BC 9 units long, to the left of the positive direction of AB (since the constant term is negative). Join OC and on it describe the circle cutting AB in D and also in E. Then DA = + 4-04 units, EA = n-i units and OA = 5 units DA or the roots are and EA OA' t. e., i. e. 5 ii-i or -81 or 2-22. Example 14. Solve by the same means the equation 1'5X* + 4* -f 1-22 = O. First, change the signs throughout to make the co- efficient of x z positive, i. e., the equation becomes I'5# 2 4* 1-22 = O. Set out, in Fig. 91, OA = 1-5 units, AB (upwards, for b is negative) = 4 units, and BC (to the right, to reverse the direction of movement about O, for c is negative) = i -22 units. The circle on OC as diameter cuts AB in D and E. DA = -42 (for this would give left-hand rotation about O) ; EA = + 4- 45 , OA =1-5. Fig. 91. Solution ot Quadratic Equation. INTRODUCTION TO GRAPHS = -28 179 Then the roots are 7=r-r- = OA and EA 1-5 Graphs representing Equations of Higher Degree than the Second. This work will best be understood by some examples. Example 15. Plot a curve to show the cubes of all numbers between o and 6. Use this curve to find the cube roots of 30 and 200. _ Fig. 92. Curve of y = x a . If x represents the numbers and y the cubes then the equation of the curve will be y = # s . i8o MATHEMATICS FOR ENGINEERS A few values of x may be taken, and the corresponding values of y calculated, the curve being plotted to pass through these points. All intermediate values can be interpolated from the curve. The table of values reads : * o 2 3 4 5 6 y = x* o I 8 27 64 I2 5 216 The points all lie on a smooth curve (see Fig. 92), which is known as a " cubic " parabola. To read cubes, we must work from the horizontal scale to the curve and thence to the vertical scale; thus the cube of 4-8 = in while for the determination of cube roots the process is reversed ; thus v'so = 3-1 and -^200 = 5-85. Example 16. Represent the equation y = x 3 8x z + 3* + 15, by a graph (x to range from 4 to +4). Fig. 93. Curve of y = x 3 8x 2 + $x + 15. The table is arranged thus : X x* x 3 8x* + 3* + 15 y - 4 16 64 128 12 + 15 - 189 - 3 9 - 27 - 72 -9+15 - 93 2 4 - 8 - 32 -6 + 15 -3i I I - i -8-3 + 15 + 3 O o o 0+0 + 15 + 15 I i i -8+3 + 15 + ii 2 4 8 - 32 +6+15 - 3 3 9 27 - 72 +9+15 21 4 16 64 128 +12+15 -37 INTRODUCTION TO GRAPHS 181 The greater part of this curve is negative, hence the axis of x is taken well up to the top of the paper (Fig. 93). A warning is again given concerning the evaluation of 8x 2 ; e. g., when x = 4. First find x 2 , i. e., ( 4)* or + 16, then find 8# 2 , j. e>> + 128, and finally Sx 2 = 128. Example 17. Solve, graphically, the equation 2# 3 gx 2 2X + 24 = o. Fig. 94. Curve of y = 2x 3 gx* 2x + 24. We shall first plot the curve y = 2* 3 - gx 2 -2^+24 and then deter- mine the values for x at the intersections of the curve with the x axis. Let x range from 3 to +5 ; and arrange the table as indicated : X x 2 X 3 2# 3 gx 2 - 2* + 24 y 3 9 2 7 54 81 +6+24 - 105 2 4 - 8 16 36 +4 + 24 -24 _. _ T i i -2 9+2 + 24 15 O o o o + 24 24 I i i 2 -92 + 24 15 2 4 8 16 - 36 - 4 + 24 o 3 4 9 16 27 64 54-81 6+ 24 128 - 144 8 + 24 -9 o 5 25 125 250 225 10 + 24 39 On plotting the values of y against those of x the curve in Fig. 94 is obtained. 182 MATHEMATICS FOR ENGINEERS We observe that the curve is of a different character from the " square " parabola, in that it bends twice whereas the latter bends but once ; there is thus one bend for a second-degree equation, two bends for a third-degree equation and so on. One can form some idea of the form of the curve from the equation by bearing in mind this fact. The curve crosses the x axis at three points and three points only; and the three values of x satisfying the given equation are found from these points of intersection. Thus in Fig. 94 x = 1-5, 2, and 4. A cubic equation has three roots, although in some cases only one may be evident, the others being imaginary : if the curve were drawn to represent an equation, two of the roots of which were imaginary, it would cross the x axis at one point only, the bends being either both above or both below it. Example 18. A cantilever, 30 ft. long, carries a uniformly-dis- tributed load of w tons per foot run. The deflection y at distance x from the fixed end is given by the formula where I = moment of inertia of section of cantilever E = Young's Modulus of material. / = span. If w = 5, I = 200, and E = 12500, show by a graph the deflected form of the cantilever. Span. 10 12 14 16 18 2O 22 24 26 28 30 20 Fig. 95. Deflection of Cantilever. Substituting values (5400** + x*) 24 X 12500 X 200 v = '833 X IQ- 7 (54OO* 2 I2O# S + X 4 ) = -833 x 10 7 x Y (Y is substituted in place of the expression 54OO* 2 i2O# 3 + x*.) Since the powers of x combined with their respective coefficients give large numbers, it is found to be better to express all these large INTRODUCTION TO GRAPHS 183 numbers as simple numbers multiplied by a power of ten Thus the product 5400*2 when x = 5, which has the value 135000, is written 1-35 x io 5 , and similarly the other products are written in this abbrevi- ated form. One has thus to deal with the addition and subtraction of small numbers, performing the multiplication or division by io- at the end once instead of three times. To find values of y from those of Y we must divide by io' and multiply by -833, and according to our scheme we find it convenient to note the values of Y x io- 5 (shown in the sixth column) and then multiply these by -833, dividing by io 2 . By arranging the work in columns one setting of the slide rule suffices for the multiplication by each particular constant, i. e., in evaluating the values of 5400*2, 54 on the D scale would be set level with i on the C scale ; and the figures in the second column would be taken on the C scale, while the figures on the D scale level with these would be the products of 5400 and # 2 . Tabulation : X X* X 3 X* 5400# a I20X 3 + X* Y-fio 5 y o O o + o o 5 25 125 625 i-35Xio 5 - -i5xio s + -o63Xio 5 1-26 0105 10 IOO 1000 I0 5-4 X io 5 1-2 Xio 5 -f -i xio 5 4'3 0358 15 225 3375 50600 I2-I4XI0 5 4'05Xio 5 + -5o6xio 5 8-6 0716 20 400 8000 160000 21-6 xio 5 9-6 xio 5 +i-6 xio 5 13-6 '"33 25 625 15630 390000 33-7 Xio 5 -i8'75Xio 5 +3-9 xio 5 18-85 -I 57 30 900 27000 810000 48-6 xio 5 32-4 xio s +8-i xio 5 24'3 2025 The deflected form is shown in Fig. 95, the scale for deflections being magnified in comparison with the linear scale. Turning-points of Curves : Maximum and Minimum Values. A quadratic curve has one bend, and a cubic has two : there must therefore be some one point on each of these bends which is either higher or lower than all other points in its immediate neighbourhood, for the curves are perfectly smooth and continuous. Such points are known as turning-points of the curve, and it is with these that we must now deal. If the curve is an ordinary parabola, let us say that representing the equation y = $x 2 + jx q (see Fig. 88), there can only be one turning-point, and that is lower than all points on the curve round about it. Referring now to the ordinate at that particular point we note that it is less, algebraically (i. e., taking account of sign), than any other ordinate near to it; it is therefore spoken of as a minimum value of the function. What is usually required is the value of the " inde- pendent variable " that makes the function a maximum or mini- mum : hence the highest or lowest point on the curve must be 184 MATHEMATICS FOR ENGINEERS found, by sliding a straight edge parallel to the x axis until it just touches the curve, the abscissa of this point being noted. Thus the function 5# 2 -f 7* 9 has its minimum value when x = 7. The curve y = *75# 2 2x -\- -61 would have no minimum value (" minimum ", being understood to imply " less than any other value in the immediate vicinity "), but would have its ordinate a maximum when x = 1-33 (see Fig. 89). It is possible for a minimum value of an ordinate to be greater than a maximum. Many instances occur in practice in which greatest or least values have to be found, or, more generally, values of some variables which cause some function to have maximum or minimum values. Questions of economy of material or time, best dimensions for certain conditions, etc., all arise, and may be classed under the heading of " maximum and minimum " problems. Before dealing with any of these, an ordinary theoretical example will be treated as a clear demonstration of the principles involved. Example 19. Find the value or values of x that make the function x 9 + 2X Z 4* + 7 a maximum or minimum. State clearly the nature of the turning-points. First plot the curve y = x 3 + 2X* 4^+7. is as follows : For this, the tabulation X x z x 3 + 2x z 4* + 7 y - 4 16 -64 + 32 + 16+7 - 9 - 3 9 -27+18+12 + 7 10 2 4 -8 +8 +8+7 15 I i -i +2 +4+7 12 O o o +o o + 7 7 I i i +2 -4+7 6 2 4 8 +8 -8 + 7 15 3 9 27+1812 + 7 40 4 16 64+32-16+7 87 A rough plotting is made (in Fig. 96) from the figures in this table ; and for greater accuracy the portion between x = 3 and x = i and that between o and 1*5 are drawn to a larger scale and more values of x are taken. One should always adopt such refinements as this ; and especially does this apply when solving equations, viz. disregard the portion of the curve that is of no immediate use and deal with the useful portion in greater detail. Apparently one turning-point is in the neighbourhood of 2 and another in the neighbourhood of i, therefore take as additional INTRODUCTION TO GRAPHS 185 values for x, 2-5, 1-5, -5 and 1-5. Thus the subsidiary table reads : X x z x 3 + 2X 2 4* + 7 y -2-5 6-25 - 15-63 + 12-5 + 10 + 7 13-87 - i'5 2-25 - 3-38 +4'5 +6+7 14-12 '5 25 13 + '5 -2 + 7 5-63 i'5 2-25 3-38 +4-5 -6+7 8-88 Y- -2. 40 30. -Jc -2. / JJ Fig. 96. Drawing only these portions of the curve (see (a) and (6), Fig. 97) we find that the trend is horizontal when x = 2 and also when x = -67. Therefore, y = x 3 + 2# 2 4* + 7 is a maximum when * = 2, and a minimum when x = -67. Example 20. We require to find for what external resistance R the power supplied from a battery of internal resistance r and electro- motive force E is a maximum. We are told that E = 8-4 and r = -57. The power = (Current) 8 X external resistance E.M.F. ) x external resistance Uotal resistance/ RE 2 R X (8- 4 ) 8 - (R + r) a (R + -57) 2 Since (8- 4 ) 2 is a constant it can be disregarded throughout as it i86 MATHEMATICS FOR ENGINEERS does not affect the resistance for which the power is a maximum, but only the magnitude of the power. Let W = 7^ r ; then we require a value of R that makes W (R + -57) 2 a maximum, and R must be treated as the I.V., i. e. t R is plotted along the horizontal. No negative values need be taken for R, but otherwise we have no idea as to its magnitude ; a preliminary tabulation, and if necessary a preliminary graph, must consequently be first made The table reads : R (R + -57) (R + -57) 2 R - w (R+-57) 2 o-o 57 325 o-ooo 5 1-07 1-14 438 I'O i'57 2-46 406 i'5 2-07 4-27 352 2'O 2'57 6-6 303 Apparently the curve rises fairly rapidly from R = o to R = -5 and then falls again : hence we conclude that the maximum value of W will be obtained when R = -5 or thereabouts. (If this reasoning cannot be followed from mere inspection of the table, a rough graph should be drawn to represent it.) Accordingly, let us take values between R = -2 and i -o. R R+-57 (R + -57) 2 R W (R + -57) 2 2 77 592 338 4 97 94 1 425 5 1-07 I-I45 4367 6 1-17 1-37 4383 8 i'37 1-88 4263 i-o i'57 2-46 406 INTRODUCTION TO GRAPHS 187 Plotting the portion between R = -4 and -8 (as in Fig. 98) we find that W has its maximum value when R = -57, i. e ., the external resistance is equal to the internal resistance. 44 43 -42 "T*^ j / ( s \ / \ S / \ / \ I i u Cp 5 ' 7 4 -5 '6 'J ^8 Fig. 98. Curve of Power from an Electric Battery. Example 21. The horse power transmitted by a belt passing round a pulley and running at v feet per sec. is given by H.P. = g where T = maximum stress permissible in belt = 350 Ibs./D* w mass of i foot length of belt = -4 Ib. g = 32-2. {The belt is 4* wide and i* thick.} Find the speed at which the greatest horse-power is transmitted under these conditions : find also the maximum horse-power trans- mitted. Substituting the values of T, w and g, the equation becomes / 3 N H.P. = Usoy-^- IIOO\ 32*2> IIOO (35 oy - The factor may be disregarded in the curve plotting, as it is simply a constant, and does not affect the value of v without similarly affecting H.P. i88 MATHEMATICS FOR ENGINEERS Hence we plot the curve, Hj_ = 3501; -oi24t; 3 ; and taking values of v from o to 160 we obtain the following table : V V 3 3501* -OI24?; 3 H, o o 20 8000 7000 99 6100 40 64000 14000 794 13206 60 216000 21000 2680 18320 80 512000 28000 6250 21750 IOO I0 6 35000 12400 22600 I2O 1-728 x io 6 42OOO 2 1 IOO 20900 140 2-352 x io 6 49OOO 28700 20300 1 60 4-096 x io 6 56000 50OOO 6000 H! is evidently a maximum somewhere in the neighbourhood of v = 100 ; accordingly, taking some intermediate values, the subsidiary table reads : 90 729000 31500 - 9040 22460 95 855000 33200 10600 22600 105 1-16 x io 6 36800 14400 22400 97 913000 33930 - 11310 22620 224-OO 9Q 197-6 100 Values af V Fig. 99- Curve of H.P. transmitted by Belt. Plotting the portion of the graph from v = 90 to v = 105 (Fig. 99), we find that H x is a maximum when v 97-6. Maximum value of H! is 22615, i.e., the maximum H.P. = y $ = 20-6. Hence I IOO we con- clude that the greatest H.P. is transmitted at a speed of 97-6 ft. per sec. and that the greatest H.P. transmitted is 20-6. INTRODUCTION TO GRAPHS 189 Exercises 23. On the plotting of Graphs of Quadratic and Cubic Expressions : and on Maximum and Minimum Values. 1. Plot from x = 5 to x +3 the curve y = $x z 5* + 13. 2. Plot from x = 3 to x = +6 the curve y = 4-15* -23*2 + 1.94. 3. The centrifugal force on a pulley rim running at v ft. per sec. is found from T - . If w = 3-36 and g = 32-2, plot a curve to give o values of T for values of v ranging from 70 to 200. 4. Plot a curve giving the H.P. transmitted by a belt running at velocity v from H.P. = 4< - when t = 400 and v is to range from o to 165. 5. Indicate by a graph the changes in B consequent on the variation of T from 10 to 50 when 6. If w = Ibs. of water evaporated per Ib. of fuel, and / = Ibs. of fuel stoked per hour per sq. ft. of grate w = ^ + 8-5. Plot a curve to give values of w as / ranges from 12 to 40. 7. The weight per foot W of certain railroad bridges for electrical traffic can be calculated from W = 50 + 5/, where I = span in feet. Plot a graph to give the total weight of bridges, the span varying from 12 to 90 ft. 8. Johnson's parabolic formula for the buckling stress (Ibs. per sq. in.) of struts is (for W.I. columns having pin ends) p = 34000 - -67 Jj Plot a curve to give values of p for values of -rj from o to 150. 9. Plot as for Ex. 8, but for C.I. columns, for which the relation 25 / I \ z is expressed by the formula p = 60000 -- - ( -r ) ; the range of the ratio / 4 \/ v being from o to 55. K 10. For Yorke's notched weir or orifice for the measurement of the flow of water, the quantity flowing being proportional to the head, Shape op Opeaiaq iaa Plcfte _^__,. h ^_^^x Lirae of No Head i Fig. 100. The Yorke Weir. igo MATHEMATICS FOR ENGINEERS the half width w (see Fig. 100) at head h is given by w = ^jj*- Show the complete weir for a depth of 6*, taking the range of h from -095* to 6-095*. 11. The length of hob /to cut a worm wheel with teeth of i* circular pitch, N being the number of teeth, is found from /" = .8 74 2N - -1373. Plot a curve to show values of /for N ranging from 10 to 120. 12. The resistance R, in Ibs. per ton for the case of electric traction, o(V + 12) V 2 at a speed V miles per hour is given by R = - v - -\ If V ranges from o to 40, show the variation of R by a graph. 13. The following equation occurs in connection with the reinforce- ment of rectangular beams k Vzrm + r z m z rm. Plot a curve to give values of k for values of r ranging from -005 to 02, taking the value of w as 15. Solve, graphically, the equations in Exs. 14 to 17. 14. x* 5* 6 = o. 15. 6x* 56 = $x. 16. -14** + -87* - 1-54 = o. 17. (3 x io 6 * 2 ) + (2-8 x io*x) + (31 x io 4 ) = o. 18. Find a value of x which makes M maximum or minimum, it being given that M = 3-42* -ix 2 . 19. The following values were given for the B.H.P. and I.H.P. for different values of the valve cut-off. Find the cut-off when the engine uses least steam, (a) per I.H.P. hour; (b) per B.H.P. hour. Cut-off .... 9f 6* 44* 3" B.H.P in H5 "5 no I.H P 118 I2S 127 114 Steam per hour 2160 2116 2080 2020 20. If 40 sq. ft. of metal are to be used in the construction of an open tank with square base, the dimensions being chosen in such a way that the capacity of the tank is to be a maximum for the metal used : Let x ft. be the length of the side of the base ; then the volume By taking values of x from o to 7, find that value Hence find also the height , ., , / Volume \ of the tank I 1 ). 21. The table gives figures dealing with gas-engine tests. is IQX cu. ft. 4 which gives the greatest volume of water. P-vHn nf ait > 11-7 10-43 9-13 774 5-38 4-40 3-60 3-14 gas J Gas per I.H.P.) hour J 3i'9 22 20-8 19 21-6 24-8 29-8 34'5 INTRODUCTION TO GRAPHS lgi What are the best proportions of the mixture for least consumption of gas per I.H.P. hour? 22. In a non-condensing engine running at 400 revs, per min. the following results were obtained ; Ratio of expansion r 4 4'4 4-8 5'2 5-6 6-0 8 Ibs. of steam per 1 I.H.P. hour / 2075 20-48 20-35 20-16 20 20-32 23-14 Find the most economic ratio of expansion. 23. The work done by a series electric motor in time / is given by w = g(E - e)t where e = back E.M.F., E = supply pressure, R = resistance of armature. The electrical efficiency is =.. Find the efficiency when the motor so runs that the greatest rate of doing useful work is reached. {R = -035, E = no, / = 20.} 24. The total cost C, in pounds sterling, of a ship's voyage of 3000 nautical miles is given by P _ 3Ooo/ v 3 \ V \ 22OO/ where v is the speed in knots. Find the speed at which the cost has its minimum value and state the cost at this speed. 25. To find the best angle of thread for a worm gear with steel worm and brass worm gear, calculations were made with the following results : Angle (degrees) o 10 20 30 45 60 75 Efficiency . o 466 61 -671 68 605 28 Find the best angle and the maximum efficiency. 26. If W = 4C a + ^, find a value of C between o and 5 that Vx makes W a maximum or minimum. 27. The efficiency t; of a Pelton wheel is given by ? = ^, ' Find the value of u in terms of v which makes 17 a maximum. Find also the maximum efficiency. 28. If r, = 2U ( V ~ U ^ and v = 25, find the value of u for the maximum i/ 8 value of 77 ; find also the maximum value of 17. 29. The ratio of horse-power to weight of a petrol motor is -j^j- where D = diam. of cylinder in inches. Find the value of D which makes this ratio a maximum. 30. Sixteen electric cells are connected up, in -- rows of x cells per row. The current from them is 16 + 4 Find the arrangement for maximum current. 31. Find a value of V between o and 10 that makes R a maximum, when _ 3 (V-i2) t V+I2 54 32. Plot from x = 4 to #= + 4 the curve 33. Plot from x = 2to x = + 6 the curve zy = -56*2 1-07* 1-48*3 + -88. Solve, graphically, the equations in Exs. 34 to 37. 34. 2* 3 - x 2 - 7* + 6 = o. 35. 2o# 3 + n# 2 + 27 = 138*. 36. x 3 + 5# 2 -o8x 8-82 = o. 37. $op 3 + 4 = 2$p $p*. 38. Find the turning-points of the function 2x 3 + 3# 2 36* + 15, stating their nature. 39. If x is the distance of the point of contraflexure from the end of a built-in girder whose length is /, find x in terms of / by the solution fax fax^ of the equation i j- + -, a - = o. 40. To find d, the depth of flow through a channel under certain conditions of slope, etc., it was necessary to solve the equation d 3 - i -305*2- 1-305 = o. Find the value of d to satisfy this equation. 41. From tests with model planes Thurston calculated the following figures : Inclination of plane to horizontal (degrees). -2 i i 2 3 4 5 6 8 10 it Weight supported perH.P. 16-8 3i'i SI'S 9<>'5 157 203 230 256-5 259-2 233 196 128-5 Plot these values, to a base of angles, and find for what inclination the greatest weight is lifted per H.P. developed. CHAPTER V FURTHER ALGEBRA Variation. If speed is constant during a journey, the time taken is proportional to the distance, i. e., the bigger the distance the longer is the time taken, or, to extend this statement, twice the time would be required for twice the distance. This is expressed by saying that the time varies as the distance, or more shortly t <x d where the sign oc stands for varies as. We cannot say that t = d, but the statement of the variation is well expressed by the equation - =-J or r-=^ 2 , where ^and d : are 1 2 2 <*1 **2 the values of the time and distance in one instance, and t 2 and d 2 are corresponding values in some other. If, in the second arrangement, k is the number to which each fraction is equal, it will be seen that - 2 = k ' t = kd or, in general, t = kd. Hence the sign of variation may be replaced by the sign of equality together with a constant factor. e. g., suppose the time for a journey of 300 miles is 15 hours, then 15 = k X 30 or k = ^ i. e., the constant factor is ^ so long as the units are miles and hours, and the speed is uniform. Variation such as this is known as direct variation, since / varies directly as d. Suppose now that the length of journey is fixed, then the bigger the speed the less will be the time taken ; halve o i 9 4 MATHEMATICS FOR ENGINEERS the speed and the journey takes double the time. Here the time varies inversely as the speed when the distance is constant; or t cc - v i. e., t = lx- = - v v where / is some constant. If both speed and distance vary, the time will vary directly as the distance and inversely as the speed ; or t cc d and also t cc - v . d i. e., t cc - v md or / = - (i) v This variation is known as joint variation. A proof of statement (i) is here given, as the reason for it is not self-evident. Suppose the original values of time, distance, and speed are /!, d v and v r respectively. Change the distance to d z , keeping the speed constant : the time will now be t, the value of which is determined from the equation t d z , \ fj flj Now make another change; keep the distance constant at d 2 but let the speed become v z , then the time will change to t 2 and ' (3) '2 Multiplying equations (2) and (3) together or = T^ or -M- = -~ constant = m, say. md md, = ~ or ^ = . md or, in general, t=- v Questions on variation should be worked in the manner out- lined in the following examples. FURTHER ALGEBRA I95 Example i The loss of head of water flowing through a pipe is proportional to the length and inversely proportional to the diameter. If in a length of 10 ft. of \ H diam. pipe the head lost is 4-6 ft., what will it be for 52 ft. of 3^" diam. pipe ? Taking the first letters to represent the words h oc / when d is constant ; and h oc -i when / is constant. Then, when both I and d vary h oc ^ or h = -^, where k is a constant. We must first find the value of k. In the first case ,. k x 10 4-6 - Substituting this value in the second case A='23X^ = '^f =3-68 ft. Example 2. The weight of shafting varies directly as its length and also as its cross section. If i yard of wrought-iron shafting of i" diam. weighs 8 Ibs., what is the weight of 50 ft. of W.I. shafting of J* diam. ? If for weight, length and area, W, / and a respectively are written, then W oc / and also W oc a; and when both I and a vary W oc la. Also we know that the area of a circle depends on the diameter squared; hence a oc d 2 and W oc Id* or W = kid* In the first case 8 = & x 3 X i 2 k=- and W = -ld* 3 3 Substituting this value in the second case Example 3. The diam. d of a shaft necessary to transmit a certain horse-power H is proportional to the cube root of the horse-power. If a shaft of 1-5* diam. transmits 5 H.P., what H.P. will a 4" diam. shaft transmit ? Here d oc vH or d = AH* 196 MATHEMATICS FOR ENGINEERS Substituting the first set of values i- 5 = k x 5* ,-s When d = 4 4 "^ Transposing H* = 4 5 ' _ , . T 3 Cubing- An application of this branch of the subject occurs in con- nection with the whirling of shafts. It is known that the deflection d of a shaft, as for a beam, is proportional to the cube of its length /, and also that the critical speed of rotation c is inversely propor- tional to the square root of the deflection. In mathematical language d oc I s ............ (i) i and c oc ~ ............ (2) \ d We desire to connect c with /. From equation (i) d = kP Substituting in the modified form of equation (2), viz. c = - vd c== where /> is some constant, *. e., the critical speed is inversely pro- portional to the f power of the length. Thus if the equivalent lengths of the shaft under different modes of vibration (i. e., for the higher critical speeds) are I, -, -, 2 3 etc., the critical speeds are in the ratio i, 2-82, 5-2, etc.; for comparing the first and third /! = I / 2 = \ Cj = I C 2 = ? but tl = 1 a *. e., P = c l /^ also p = c 2 1 FURTHER ALGEBRA Ig7 Thus- c 2 if = Cl if = 5-2 Hence c -* = $2 Ci i Example 4. The energy E stored in a flywheel varies as the fifth power of the diameter d and also as the square of the speed n. Find the energy stored in a flywheel of 6 ft. diam., whilst it changes its speed from 160 to 164 revs, per min., if the energy stored at 100 R.P.M. is 25000 ft. Ibs. E <x d*n* E = kd s n*. When n = 100, d = 6, E = 25000, so that 25000 = k x 6 s x ioo 8 k = f 5000^ 6 s x io 4 Thus E at n = 164 = k x 6 5 x 164* and E at n = 160 = k x 6 5 x 160* Difference = k x 6 5 (i64 2 i6o 2 ) = 25000 x 6 S ( 4 )( 3 24) 6 6 x 10* = 3240 ft. Ibs. Example 5. A direct-acting pump having a ram of io* diam. is supplied from an accumulator working under a pressure p of 750 Ibs. per sq. in. When no load is on, the ram moves through a distance of 80 ft. in i min. at a uniform speed v. Estimate the value of the coefficient of hydraulic resistance or the coefficient of friction, viz. the friction force when the ram moves at a velocity of i ft. per sec. ; the total friction force varying as the square of the speed. Find also the time the ram would take to move through 80 ft. when under a load of 15 tons. If the whole system is running light, the full pressure is used to overcome the friction, i. e., p oc v 2 , since total friction force varies as (velocity) 2 . Thus p = kv z where k is the coefficient of hydraulic resistance ; also v = g- = i -33 ft. per sec., and p = 750 then 750 = k x (i-33) a or *-ri^~ 4M . e., the coefficient of hydraulic resistance is 422 if the units of pressure and velocity are Ibs. per sq. in. and ft. per sec. respectively. I 9 8 MATHEMATICS FOR ENGINEERS The intensity of pressure due to a load of 15 tons i"5 x 2240 = = 428 IDS. per sq. m. If a - X I0 2 4 Then, to find the velocity in the second case Total pressure = pressure to overcome the friction + pressure to move the load, i. e., p = kv^ + pi where v l is the new velocity, and p^ = 428. Then 750 = kvf + 428 = 422 1^ 2 + 428 or v * - 75 ~ 42S - -76^2 Vi ~ 422 and P! = -8736. Hence the time required for 80 ft. of the motion 80 i x j~ = 1-526 mms. -8736 Example 6. The linear dimensions of a ship are X times those of a model. If the velocity of the ship = V, find the speed of the model at which the resistance is p times that of the ship, given that the fluid resistance varies as the area of surface S and also as the square of the velocity. Let R = resistance of ship ; then from hypothesis R oc S, and also R oc V 2 . Then R = KSV 2 and r = resistance of model = Ksv*. Now s i /for surfaces of similar solids are proportional to the\ *{ S X 2 \ squares of corresponding linear dimensions J and we are told that Hence i.e., or *'. e., v and V (which is v V\) are spoken of as " corresponding speeds." FURTHER ALGEBRA IQ9 Exercises 24. On Variation. 1. The weight of a sphere is proportional to the cube of the radius A sphere of radius 3-4" weighs 47-8 Ibs. ; what will be the weight of a sphere of the same material, of which the radius is 4-17* ? 2. The candle-power (C.P.) of a lamp is proportional to the square of its distance from a photometer. A lamp of 16 C.P. placed at 58 cms. from a screen produced the same effect as a second lamp placed 94 cms. from this screen. If this second lamp was absorbing 100 watts, find its efficiency, where 17 = watts per C.P. 3. The velocity of sound in air is proportional to the square root of the temperature r (centigrade absolute, i.e., t C. + 273). If the velocity is 1132 ft. per sec. at temperature 18 C., find the law con- necting v and T ; find also the velocity at 52 C. 4. The force of the earth's attraction varies inversely as the square of the distance of the body from the earth's centre. Assuming that the diameter of the earth is 8000 miles, find the weight a mass of 12 tons would have if it could be placed 200 miles above the earth's surface. 5. The total pressure on the horizontal end of a cylindrical drum immersed in a liquid is proportional to the depth of the end below the surface and to the square of the radius of the end. If the pressure is 1200 Ibs. when the depth is 14 ft. and the radius is i yard, find the pressure at a depth of 6 yards when the radius is 8ft. 6. The loss of head due to pipe friction is directly proportional to the length, to the square of the velocity and inversely proportional to the diameter. If 2-235 ft- of head are lost in 50 ft. of 2* pipe, the velocity of flow being 4 ft. per sec., find the diameter of pipe along which 447 ft. of head are lost, the length of the pipe being i mile and the velocity of flow 8-7 ft./sec. 7. The electrical resistance of a piece of wire depends directly on its length and inversely on its diam. squared. The resistance of 85 cms. of wire of diam. -045 cm. was found to be 2-14 ohms. Find the diam. of the wire of which 128 cms. had a resistance of 8-33 ohms. 8. The power in an electric circuit depends on the square of the current and also on the resistance. The power is 15-34 kilowatts when 23 amps, are flowing through a resistance of 29 ohms. If a current of 9 amps, flows through a resistance of 17 ohms for 50 mins., what would be the charge at 2d. per unit ? (i unit = i kilowatt-hour.) 9. The electrical resistance of a conductor varies directly as the length and inversely as the area of cross section. The resistance of 70 cms. of platinoid wire of diam. -046 cm. was found to be 1-845 ohms. Find the resistance of 1-94 metres of platinoid wire of diam. -028 cm. 10. The number of teeth T necessary for strength in a cast-iron wheel varies directly as the H.P. transmitted, inversely as the speed and inversely as the cube of the pitch p of the teeth. If T = 10 when p = 2" and ratio of H.P. to speed (in R.P.M.) = -101, find the H.P. transmitted when there are 30 teeth, the pitch of the teeth being 6", and the speed being 30 revs, per min. 11. The coefficient of friction between the bearing and shaft varies directly as the square root of the speed of the shaft and inversely as the pressure. The coefficient was -0205 when the speed was 10 and 200 the pressure was 30 ; find the pressure when the coefficient is -0163 and the speed is 45. 12. The I.H.P. of a ship varies as the displacement D, as the cube of the speed v, and inversely as the length L. If I.H.P. = 2880 when D = 8000 tons, v = 12 knots, and L = 400 ft., find the speed for which I.H.P. = 30600, the displacement being 20000 tons and the length being 580 ft. 13. The pressure of a gas varies inversely as the volume and directly as the absolute temperature r (see proof in Question 18). The pressure is i kgrm. per sq. in. when the volume is 6-90 and the absolute tem- perature is 468; find the absolute temperature when the pressure is 8-92 kgrms. per sq. in. and the volume is 1-39. 14. In some experiments on anti-rolling tank models, the number of oscillations per min. of a model of length 10-75 ft- was 2 7- I* the number of oscillations per min. is inversely proportional to the square root of the ratio of the linear dimensions, find the number of oscillations of a similar ship 430 ft. long. 15. Assuming the same relations between volume, pressure and absolute temperature as in Question 13; if the pressure is 108 Ibs. per sq. in. when the volume is 130-4 cu. ins. and the absolute temperature is 641, find the absolute temperature when the pressure is 41-3 Ibs. per sq. in. and the volume is 283 cu. ins. 16. The time of vibration of a loaded beam is inversely propor- tional to the square root of the deflection caused by the loading. When the deflection was '0424* the time was -228 sec. ; find the deflection when the time was -45 sec. 17. If the cost per foot of a beam of rectangular section of breadth b and depth h varies as the area of section, and the moment of resist- ance of the beam is proportional to the breadth and also to the square of the depth, find the connection between the cost per foot and the moment of resistance. 18. Boyle's law states that the pressure of a gas varies inversely as its volume, the temperature being constant ; Charles's law states that the pressure is proportional to the absolute temperature, the PV volume being kept constant. Prove rigidly that = constant. Series. A succession of numbers or letters the terms of which are formed according to some definite law is called a series. Thus 6, 9, 12 is a series for which the law is that each term is greater by 3 than that immediately preceding it. Again, 40,, i6ab, 6^ab z is a series in which any term is obtained by multiplying the next before it by 46. In these particular series, taken as illustrations, the terms are said to be in progression, the former in Arithmetical Progression, written A. P., and the latter in Geometrical Progression, written G.P. Other series with which the engineer has to deal are those known as the Exponential and the Logarithmic ; and in the expan- sion or working out of certain binomial or multinomial functions or expressions a "series" results. FURTHER ALGEBRA 201 Arithmetical Progression. Consider the series of numbers 2, 9, 16, 23 ... etc. The 2nd term is obtained from the ist by adding 7. 3rd 2nd 7. 4 ln 3rd 7. i. e., each term differs by the same amount from that imme- diately preceding it. The numbers in such a series are said to be in Arithmetical Progression ; and since the terms increase, this is an increasing series. Again, I, 4, 9, 14 ...... is an A.P., the common difference in this case being 5. This is a decreasing series. In general, an A.P. can be denoted by a, (a + d), (a + zd) where a is the ist term and d is the common difference. Now the 2nd term = a+d = a+(2 i)d and the 3rd term = a+2d = a+(3i)d So that the 20th term = a+igd i. e., the general term, or the /1 th term = a + (n l)d. Thus the I5th term is obtained by adding 14 differences to the ist term, or I5th term = a+md. If three numbers are in A.P., the second is said to be the arithmetic mean between the other two ; e.g., 95, 85, 75 are three numbers in A.P., where 85 is the A.M. between 95 and 75 and ^ = 95 + 75 . or t k e arithmetic mean of two numbers is one-half 3 2 their sum. To find the sum of n terms of an A. P., which is denoted by S n S n = a + (a + d) + (a + 2d) + ...... [a + (n-i}d} Also, by writing the terms in the reverse order S n = {a + (n - i)d} + {+(- 2}d} + {a + (n- 3)^} + ..... a Adding the two lines 2 S n = {2a +(n- i)d} + (2a + (n - i)rf} + {za -f- (n -L)d} ...... to n terms or 2S n = n {za +(n- i)d} If we call the last term /, then / = +(- i)d, and the formula for the sum can be written s, = ?(,+/} or 202 MATHEMATICS FOR ENGINEERS i. e., the sum can most easily be found by multiplying the average term, i. e., - , by the number of terms n. Many problems on A. P. can be worked by means of a graph. If ordinates represent terms, and abscissae the numbers of the terms, an A. P. will be represented by a sloping straight line for which the " slope " is the common difference d and the ordinate on the axis through i of the horizontal scale is the first term. The sum will be the area under the line, with one-half the sum of the first and last terms added. Term VW Fig. 101. Arithmetical Progression. For the area under the line, viz. ABCD (Fig. 101) IS but J(AD -f BC) X AB = (n-i) ^areaunderline Example 7. Find the sum of 12 terms of the series, 4, 2, o find also its loth term. -18 Fig. 102 Sum of a Series. -2 -12 I -14 -16 -18 FURTHER ALGEBRA 203 case, n = 12, a = 4 , d = 4 subtracted from 2 = - 2 . Si, - ^{(2X 4 ) + (12 - I) X - 2} = 6 {8 - 22} = -84. Also the lothterm =a + gd = 4 ~g X2 or graphically, area under the line"" = i (4 x 2} -$ {9 x - 18}. (Fig. 102 ) = 4 - 81 = - 77 and J(a + l) = l^LlI = _ 7 S = - 77 - 7 = - 84 and ordinate AB represents the loth term and = 14. Example 8. Insert 4 arithmetic means between z-6 and 0-4 i.e., insert 4 terms between 1-6 and 9-4 equally spaced so that together with the terms given they form an A.P. 9-4 2345 Te.rmJV Fig. 103. Arithmetic Means. 6 The total number of terms must be 6 (two end terms together with the 4 intermediate), so that ist term = 1-6 and 6th term = 9-4 but the 6th term = a + 5^ and a = 1-6 1-6 + sd = 9-4 and $d = 7-8 or d = i -56. Hence the means are 3-16, 4-72, 6-28, and 7-84. The graphical construction would be quicker in this instance. Referring to Fig. 103, draw a vertical through i on the horizontal scale to represent 1-6, and a vertical through 6 to represent 9-4; join 204 MATHEMATICS FOR ENGINEERS the tops of the ordinates by a straight line and read off the ordinates through 2, 3, 4 and 5. Example 9. In calculating the deflection of a Warren girder due to the strain in the members of the lower flange, if U = the force in a member caused by a unit load at the centre of the girder, F = the force in the bar due to the external loads, a = area of section of member, d = length of one bay, h = height of the girder, and n == number of P bays, then deflection = ^ x sum of all the separate values of the product U x length of member. p If d = 20 ft., h = i2'-6", n = 8, = 4 tons per sq. in., and E = 12500 tons per sq. in., find the deflection. I Fig. 104. Deflection of a Warren Girder. Dealing with the first bay (see Fig. 104) and taking moments round the point A UiXAD or Jx-^UjX* d u 1 = - A i.d For the second bay, by taking moments round E, U a = , ', while cd 4" for the third bay U 8 = ^r, and so on. 4* j _ j Hence the sum of the separate values of U = ? + ~ + . . . .to 4* 4* d n terms, i. e., it is the sum of an A. P. of which the first term is * and d * h the common difference is =- 2* Hence- S. = i {(a x .*) + (. - ,) The sum of the products of U x length of member is this total x d, since all the members have the same length, viz. d. Then the deflection F ~ X and for this particular case, by substituting the numerical values, 4 x 64 x 400 ,, deflection = - - ft. 12500 x 4 x 12-5 FURTHER ALGEBRA 20 5 7 Q . 8 "' and the Geometrical Progression.-The numbers are part of a series in which each term is one by the use of a common S^ as a Geometrical Progression, or a G.P. conLn muldp^er or ratio is - l^' ^ ISt term Generally a G.P. may be expressed by a ' ar>ar * ...... (' being the common ratio). The 2nd term = ar 1 = ar z ~^ the 3rd term = ar z = a^a-i the n lh term = e. g., the 5ist term = If three terms are in G.P., the middle one is said to be the geometric mean of the other two : it is equal to the square root of their product, for if a, m and b be in G.P. - = and m 2 = ab 2 or m = Fig. 105 Geometrical Progression. [If the true weight of a body is required, but the weighing _ ( balance has unequal arms, weigh in each pan, and call -2 the balancing weights W x and W 2 respectively : then the true weight W is the geometrical mean between W x and W 2 , or W= VW x W a .] To find the sum to n terms, written S n S = a -f- ar + ar 2 + ar n ~ 2 -f- ar n ~ l and r S,, = ar + ar 2 + ar n ~ z + ar"~ l -f ar" then S,,(i r] = a ar n (Subtracting) -(l-r-J or (r--l) 1 r r 1 and the first form being used when the ratio is less than i. Referring once again to the series 4, 2, i the numerical value of the terms, plus or minus, soon becomes so small that the sum, say, of 60 terms is practically the same as that of 50, and the series is said to be rapidly converging. This fact is well illustrated by the graph of term values plotted to a base of 206 MATHEMATICS FOR ENGINEERS term numbers as in Fig. 105 ; the area between the curve and the horizontal axis being extremely small after even the fifth term of the series has been reached. Hence the sum of the entire series, called the sum to infinity (of terms) and written S^, can be expressed definitely. -j-. c a(i r") r rom the rule S B = i r if r is very small and n is very great, r" will be very small indeed compared with I, and may be neglected. a O . 3 on 1-r Example 10. Find the sum of 5 terms of the series 2, -002, 000002 and compare with the sum to infinity. In this case a = 2 and r = -ooi. Then- S - a - - ' 5 r i -ooi = {i -(ix io~ 15 )} 999 ' d 22 whereas Soo = = = , and therefore the i r i -ooi -999 two are essentially the same. Example n. The 5th term of a G.P. is 243 and the 2nd term is 9 ; find the law of the series, viz. find the values of r and a. 5th term = ar* = 243 (i) 2nd term = ar l = 9 (2) Dividing equation (i) by equation (2) ar* 243 = -3* = 27 ar g r 3 = 27 and r = 3. Substituting in equation (2), a X 3 = 9 and = 3. Hence the series is 3, 9, 27, etc. It is of interest to note that the logarithms of numbers in G.P. will themselves be in A.P. Thus, if the numbers are 28-4, 284, 2840 (i. e., in a G.P. having the common ratio = io), then their logs are i'4533, 2-4533, 3-4533 (i. e., are in A.P. with common difference i). FURTHER ALGEBRA 207 Use may be made of this property when a number of geometric means are required to be inserted between two numbers. Suppose that five geometric means are required between 2 and 89. Mark off on a strip of paper a length to represent the distance between 2 and 89 on the A or B scale of the slide rule. Divide this distance into 5 + i, *. e., six equal divisions : place the paper alongside the scale with its ends level with 2 and 89 respectively : then the readings opposite the intermediate markings will be the required means to as great a degree of accuracy as is required in practice. The means are 376. 7-1, 13-3, 25-1, and 47-3. To check this by calculation a = 2, and ar* = 89. Hence, by division r 6 = = 44-5. Taking logs 6 log r = i -6484 .*. log r = -2747- Now log av = log 2 + log r = -3010 + -2747 = -5757 .'. ar = 3-763. Also log ar z = log ar + log r = -5757 + -2747 = -8504 /. ar 2 = 7-084. Similarly, the other means are found to be I3'34, 2 5' Il > an d 47-26. It has already been demonstrated that the plotting of the values of the terms in an A.P. to a base of " term numbers " gives a straight line. Consequently it will be seen that if the logs of the values of the terms in a G.P. are plotted to a base of " term numbers," a straight line will pass through the points so obtained, since the logs of numbers in G.P. are themselves in A.P. Conse- quently many problems on G.P. can be solved by means of a straight- line plotting. Example 12. The values of the resistances of an electric motor starter should be in G.P. Thus if r t = resistance of armature and rheostat on the first step, and r 2 , r 3 , r t , etc., are the corresponding values on the subsequent steps, then = = - 8 , etc., and the value *2 *3 M of this ratio is ^, where C, = starting current and C = full load working ^ current. Find the separate resistances of the 9 steps in a motor starting switch for a 220 volt motor, if the maximum (i.e. starting) current 208 MATHEMATICS FOR ENGINEERS must not exceed the full load working current of 80 amps, by more than 40%, and the armature resistance is '133 ohm. Q Here we are told that -^ = 1-4 or the common ratio of the G.P. \^ is : while the value of r can be calculated by Ohm's law, viz. ' voltage 2; ft. = 1-964 ohms. starting current 1-4 x 80 The problem now is to insert 7 geometric means between 1-964 and -133 ; and this can be done in the following simple manner. Along a horizontal line indicate term numbers as in Fig. 106, and erect verticals through the points i, 2 9. Set the index of the A scale of the slide rule level with the point i, and mark the point P at 1-964 (at the right-hand end of the rule) : similarly the point Q should be indicated, the distance gQ representing 133 (at the left-hand end of the rule). Join PQ. I 234-56789 Term N? Fig. 106. Resistances of an Electric Motor Starter. Then the ordinates to this line through the points 2, 3 .... 8, read off according to the log scale (i. e., by the use of the A scale of the slide rule, the index being placed at the horizontal in each case), give the required means, which are 1-403, 1-002, -716, -511, -365, '261, and -186. Compound Interest furnishes an example of geometrical progression. If the original principle be P and the rate of interest be r Then the interest at the end of ist year = Pr and the amount = A x = P-f Pr interest at the end of 2nd year = rA, = r(P+Pr) and amount == A^Ia FURTHER ALGEBRA 209 i.e., I 2 = Pf(i+r) and A 2 = I 3 = Pr(i+r) 2 and A 3 = /. I n = Pr(i+r)-i and A = The consecutive interests are thus Pr, Pr(i-fr), Pr(i+r) z ...... . e., they are in a G.P. of ist term Pr and common ratio (i + r). Hence total interest for n years or the amount at the end of n years = P+Interest Further Applications of G.P. If an electric condenser be discharged through a ballistic galvanometer, and the lengths of the consecutive swings of the needle are measured, it will be found that they form a G.P. ; the ratio, of course, being less than I, because the amplitude of the swing decreases. If ! = ist swing and a % = 2nd swing, then a = ka^ and On = k n ~ l a r The logarithm of the ratio (- i. e., log (^ J according to our notation, is called the logarithmic decrement of the galvanometer. Thus if the respective swings were, in divisions on a scale, 36, 31*4, 2175, etc., the ratio k = 3 ^ and the logar- ithmic decrement of the galvanometer ^6 = log -^ = -1594- To find the practical mechanical advantage (-^-j for the pulley-block aw shown in Fig. 107. The pull P on one Fig I07 ._puii e y Block, side of the pulley becomes cP after passing round the pulley (due to friction, and bending of the rope) : after passing round the second pulley, the pull is now c 2 P, and so on. 2io MATHEMATICS FOR ENGINEERS Hence W = cP(i+c+c 2 +c 3 +c 4 +c 5 ) if there are 6 strings i c for the case of 6 strings from the lower block. This result may be put into a more general form by writing n in place of 6 ; n being the velocity-ratio of the blocks. W c Thus- -J- = ^(i-' n ) In an actual experiment with a i : I block, the value of c was found to be -837. Taking this value, the result given above may then be written By the use of this formula the maximum efficiency of any pulley- block can be determined. Thus for a 4 : i block n = 4 W and p- = 5'i3[i-('837) 4 ] = 5'i3(i 497) = 2-613. W Theoretically, p- = 4, and hence the maximum efficiency = = 653 Series may occur which, whilst not actually in arithmetical or geometrical progression, may be so arranged that the rules of the respective series may be applied. Example 13. Find the sum of n terms of the series, the rth term of which is (i) 3' + 1 1 (2) 5 x 3 r - (i) rth term = 37+1 Hence the ist term = (3 x i) + I the 2nd term = (3x2) + ! and S n = (3X i) + (3X 2) + (3 x 3)+ .... +(1 + 1+1 . . . . to n terms) = 3 (sum of natural numbers to n terms) + n = 3 x "{2 +(n-i)i}+ FURTHER ALGEBRA 211 (2) rth term = 5 x 3 r Hence the ist term = 5 x 3 1 2nd term = 5 x 3* and the nth term = 5x3" also S = (5 x 3) + (5 x 3 2 ) + (5 x 3 3 ) + or 7.5(3' -i). Methods of allocating Allowance for Depreciation. The principles of the previous paragraph may be applied to deal with the various systems of allowance for the depreciation of machinery, etc., which may be calculated by one of three methods. First Method, involving arithmetical progression, and sometimes spoken of as the " straight-line method." According to this scheme, the annual contribution to the depreciation fund is constant, and no interest is reckoned. Let P = the original price of the machine, R = its residual value at the end of its life, n years, and let D = the annual con- tribution to the depreciation fund. E.g., if a machine costs 500, has a scrap value of 80, and its life is 21 years, the annual contribution = *^ r = 20. 21 Then Value at end of ist year = P R D (i. e., neglecting its value as scrap). Value at end of 2nd year = P R 2D and Value at end of wth year = P R D whilst the contributions to the depreciation fund would total nD. Its value as a working machine would be o at the end of the period, i. e., P R nD = o, or nD = P R; hence its value as p p> scrap would be taken into account in fixing D, for D = , which is not so great as - n Taking the figures suggested above Value of the machine at end of ist year = 500-80-20 = 400, . e., its value as a working machine, and the depreciation fund would then stand at 20. At end of 2nd year, value = 500-80-40 = 380 and depreciation fund = 40. Thus the value + depreciation fund always = 420 = " work- ing " value of machine, which is as it should be. 212 MATHEMATICS FOR ENGINEERS Second Method. According to this method of reckoning, the same amount is added to the depreciation fund yearly, but interest is reckoned thereon. Let the rate of interest = r per annum per i. At end of ist year, depreciation fund = D 2nd t , = D-f-fD+D (since rD is the interest on the first contribution). 3rd = (2D+rD)+f( 2 D+fD)+D We wish to find a general expression giving the magnitude of the depreciation fund at the end of any year; to do this, the ex- pression last obtained must be slightly transposed. 3D+yD+rD = D ( a ^ ^multiplying and\ r ^ ' \ dividing by r ) This is the value of the fund at the end of the 3rd year. In like fashion, the value of the fund at the end of the 4th year so that at the end of the wth year, the depreciation fund stands at This must be equal to the working value or P R, . e., ?{( I+r )._i} = P-R p E. g., if the original value = 500 the scrap value = 80 no. of years = 21 and rate of interest =3%, *' -, r = -03 FURTHER ALGEBRA 2I3 Then - D = /"3(Soo-8o) (i-03) 21 i x 420 Explanation. Let x = (i-o 3 ) 2 i log X =21 log 1-03 73^ = 21 x -0128 = -2688 x = 1-857 There is one disadvantage in connection with the second method : the depreciation fund does not grow rapidly enough in the early years. Keeping to the same figures as before, depreciation fund at end of ist year = 14-7 2nd = (2 X 147) + (-03 X 147) = 29-84 3rd = (3 x 147) -f- (-09 X 14-7) + (-0009 x 14-7) = 45-44. If the value of the machine decreases each year by 20, the depreciation fund would not be sufficiently large to ensure no loss in the event of the loss of machine in the first few years of its life : on the other hand, provided nothing untoward happens, only about three-quarters of the depreciation has to be allowed for yearly, i. e., 14-7 as against 20. Third Method. The disadvantage of the second method may be eliminated by setting aside each year a constant percentage of the value of the preceding year. Let this constant percentage be K : then at the end of the first year KP will be assigned to the depreciation fund. At the end of the 2nd year the fund will stand at KP + per- centage of value at end of ist year = KP+(P-KP) X K = P(2K-K 2 ) = P{i-(i-2K+K 2 )} At the end of the 3rd year depreciation fund = KP+K(P-KP)+K(P-KP)(i-K) = P{K+K-K 2 -f-K-2K 2 +K 8 } = P{ 3 K- 3 K 2 +K 8 } = P{i-(i-3K+3K 2 -K)} 214 MATHEMATICS FOR ENGINEERS Hence at the end of nth year depreciation fund = P{i (i K) n } This must = P R so that P-R = P-P(i-K) or P(i-K)* = R (!-*) = ? Taking the nth root of each side or R To compare with the results by the other method, take the figures as before, viz. P = 500, R = 80, and n = 21. Explanation. = i -9164 = -0836 Then depreciation fund at end of ist year Ditto end of 2nd year = 80 Ditto end of 3rd year = 116 /. e., the yearly allowance is greater at commencement . Let 21/8~ = Vo =-{9031-1-699} = 7959 21 = " '0379 = 1-9621 x = -9164 Exercises 25. On Arithmetic and Geometric Progressions. 1. Find the 7th term and also the 2gth term of the series 16, 18, 20 .... 2. Which term of the series 81, 75, 69 ... is equal to 33? 3. The 3rd term of an A. P. is 34 and the i7th term is 8; find the sum of the first 30 terms. 4. Insert 8 arithmetic means between 2-8 and 10-9. 5. Three numbers are in A. P. ; the product of the first and last is 216, and 4 times the second together with twice the first is 84. Find the numbers. 6. How many terms of the series 1-8, 1-4, i . . . must be taken so that the sum of them is 67-2 ? 7. In boring a well 400 ft. deep the cost is 2S. 3^. for the first foot and an additional penny for each subsequent foot ; what is the cost of boring the last foot and also of boring the entire well ? FURTHER ALGEBRA 215 8. A manufacturer finds that his expenses, which in a certain year are 4000, are increasing at the rate of 28 per annum. He, however, sells 4 more machines each year than during the preceding, and after 1 6 years his total profit amounts to 14240. Find the selling price of each machine and the total number sold over this period if his profit the first year was 800. 9. A tank is being filled at the rate of 2 tons the first hour, 3 tons the second hour, 4 tons the third hour, and so on. It is completely filled with water in 10 hours. If the base measures 10 ft. by 15 ft. find the depth of the tank. 10. A body falls 16 ft. in the ist second of its motion, 48 ft. in the 2nd, 80 ft. in the 3rd, and so on. How far does it fall during the igth second and how long will it take to fall 4096 ft. ? 11. A slow train starts at 12 o'clock and travels for the first hour at an average speed of 15 m.p.h., increasing its speed during the second hour to one of 17 m.p.h. for the hour, and during the third hour to 19 m.p.h., and so on. A fast train, starting at 1-30 from the same place travels in the same direction at a constant speed of 32 m.p.h. At what time does this train overtake the first ? 12. Find the 5th term of the series i, 1-2, 1-44 ... 13. Find the sum to infinity of the series, 40, 10, 2-5 ... 14. Insert 3 geometric means between ij and 6|. 15. Calculate the sum of 15 terms of the series 5, 6-5, 8-45 . . . 16. In levelling with the barometer it is found that as the heights increase in A. P., the readings decrease in G.P. At a height of 100 ft. the reading was 100; at a height of 300 ft. the reading was 80; at 500 ft. the reading was 64. What was the reading at a height of 2700 ft. ? 17. Find the sum of the series 15, 12, 9-6 ... to 7 terms and the sum to infinity of the series -8, -02, -005 . . . 18. When a belt passes round a pulley it i? known that the tensions at equal angular intervals form a G.P. If the tension for a lap of 15 is 21-08 Ibs. and that for 90 is 27-38 Ibs., find the least tension in the belt, i. e., at o (the angular intervals are each 15). 19. The sum of the first 6 terms of a G.P. is 1020 and the common ratio is 2-4; find the series. 20. Find the 2oth term of the series 3, 12, 33, 72, 135 . . . [the nth term is of the form n(a + bn + en 2 )]. 21. A contractor agrees to do a piece of work in a certain time and puts 150 men on to the work. After the first day four men drop off daily and the work in consequence takes 8 days longer than was anticipated. Find the total number of days which the work actually takes. 22. Find the deflection of the Warren girder shown in Fig. 104 due to the strain in the members of the upper flange. [H int. By taking moments about the point B, the value of U AB = ~ v etc.] 23 A lathe has a constant countershaft speed, four steps on the cone and one double back-gear. There are thus 12 possible speeds for the spindle; the greatest being 150 ^ vs. per sec. and the least being 3 revs, per sec. If the spindle speeds are in 6.P, find the respective speeds. 216 MATHEMATICS FOR ENGINEERS Napierian Logarithms. Suppose that i is lent out at 2% compound interest per annum. Then the amount at the end of ist year = (i + -02) ; and this is the principal for 2nd year. I 2 = (i + -02) X -02 and A 2 = (i + '02) 2 If, however, the interest is to be reckoned and added on each month the amount at the end of the first year will be greater, for *O2 the interest = (*'. e., per month), 12 and, amount at end of ist month = ( i -\ --- ) \ 12 / (O2\ 2 I + j / 'O2\ 12 ist year = (* + -^) ....... (i) Assume now that the interest is added day by day, i. e., practi- cally continuously, then at the end of ist year (02 \ 365 i + -ffi) ........ (2) If the interest is calculated and added on each second, that being as near continuity as we need approach Amount at end of year ->-? \31536000 1 + 3i536ooo/ ........ By means of laborious calculation the actual values of these amounts could be found, and it would be observed that the amount in (2) was greater than that in (i), and the amount in (3) was greater than that in (2) ; the difference in the values being very slight, and not perceptible unless a great number of decimal places were taken. It would appear at first sight that by increasing the number of additions of interest to the earlier amounts, the final amount could be made indefinitely large : this, however, is not the case, for the amount approaches a figure beyond which it does not rise, but to which it approximates more nearly the larger the value of the exponent (*. e., 12, 365, etc.). This final amount is 2718 for a principal of i; in other words, when the interest, added continuously, is proportional to the previous amount, the final amount will reach a limiting value, being 2718. The symbol FURTHER ALGEBRA 2I7 " e " is given to the figure 2718 . . . from Euler, the discoverer of the series. Later work will show that e can be expressed as a sum of a series, viz. 1 ^ _ ___ 1x2 1x2x31x2x3x4 and from the foregoing reasoning it will be seen that it is a natural number: it occurs as a vital factor in the statement of many natural phenomena. E.g., a chain hanging freely due to its own weight lies in a curve whose equation may be written or more simply Y = \ {e*- + e~*} i. e., it hangs in its natural curve (known as the " catenary "), and this curve, depending for its form entirely on e, can only have this one form if e is a constant, and, further, a particular constant. Again, if an electric condenser discharges through a large resist- ance, the rate at which the voltage (i. e. t the difference in potential between the coatings of the condenser) is diminishing is proportional to the voltage. The equation which gives the voltage at any time t t, is v = ae where K, R and a are constants settled from the given conditions. Then e is a constant, but one determined quite apart from any particular set of conditions. Actually the most natural way to calculate logarithms is to work from e as base, such logs being called natural, Napierian, or hyperbolic logs ; the common logs, *. e., those to base 10, which are far more convenient for ordinary use, being obtained from the Napierian logs. In higher branches of mathematics all the logs are those to base e, for if natural laws are being followed, then any logs that may be necessary must, of course, be natural logs. It is, therefore, desirable to understand how to change from logs of one base to logs of another. The rule can be expressed in this form log e N = log 10 Nxlog e io ........ (i) To remember this, omit the logs and write the law in the fractional form N_ N 10 ~i~ io e which is equally correct, as proved by cancelling. 218 MATHEMATICS FOR ENGINEERS Again log 10 N = log e N X Iog 10 5 N N e or - = - x - 10 e 10 Proof of statement (i) Let log e N = x, log 10 N = y and log e io = z then N = e*, N= 10" and 10 = e* and e* = io y = (f?) y = e y * or x = yxz i. e., log e N = log 10 Nxlog e io Taking e as 2-718 (its actual value, like that of TT, is not com- mensurate) Iog 10 = Iog 10 2718 = -4343 and logeio is the reciprocal of this, viz. 2-3026 or 2-303 approximately. Hence, from the rules given above log e N = 2-303 log 10 N logio N = '4343 logN. To avoid confusion with these multipliers it should be borne in mind that e is a smaller base than 10, and therefore it must be raised to a higher power to equal the same number. Hence the log to base e of any number must be greater than its log to base 10. If tables of Napierian logs are to hand, the foregoing rules become unnecessary ; but a few hints as to the use of such tables will not be out of place, for reading from tables of Napierian logs is somewhat more involved than that from tables of common logs. Examples are here added to demonstrate the determination of natural logs by the two processes. Example 14. Using the tables of natural logs (Table IV at the end of the book), find log* 48-72, log,. and log e -00234. log, 48-72 = log, (4-872 x 10) = log e 4-872 + log* 10 = 1 '5 8 35 + 2-3026 = 3-8861 log ' 4Ti = log ' Je? = loge 5 ' 7 ~ loge 4 ' 61 = 1-6233 ~~ 1-5282 = -0951 FURTHER ALGEBRA 2I9 log, -00234 = log, ^24 = log, 2-34 - log, 1000 = log, 2-34 - 3 log, 10 = -8502 3 x 2-3026 -= ^8502 6-9078 = 9424 It will be observed that for each power of ten in the number 2-3026 has to be added or subtracted as the case may be. Example 15. Without tables of natural logs, find values of log, 9-63, log, -?, and log, -2357. log, 9-63 = Iogj 9-63 x 2-303 = -9836 x 2-303 = 2-266 x 2-303 . : ' = 2-303 {Iog 10 i7i 7 - Iog lo453 } = 2-303{3-2347 - 2-6561} = 2-303 x -5786 log,-23 5 7 = logi -2357 x 2-303 = 1-3724 x 2-303 Separating the two distinct parts = (f x 2-303) + (-3724 x 2-303) = - 2-303 + -8576 = 2:5546 {the subtraction being performed so that the mantissa is kept positive}. Application of Logarithms to harder Computations. In the first chapter the method of applying logs for purposes of evaluation of simple expressions was shown. Such values were found as (2I-25) 5 , ^-03, etc., *. e. t numbers raised to positive powers only. The rules there used are applicable to all cases, whatever the powers may be. A negative power may be made into a positive power by changing the whole expression from top to bottom of the fraction or vice versa (for a~ n = ) so that the \ a n / evaluation is obtained on the lines already detailed ; or it may be obtained directly as here indicated. N.B. Great care must be observed in connection with the signs : when- ever distinct parts (e.g., a positive and a negative) occur in a logarithm, these should be treated separately. 220 MATHEMATICS FOR ENGINEERS Example 16. Evaluate ( Let the expression = x then x = (-005 134)' 184 and by taking logs, log x = -134 x log -005134 = -134 x 37 I0 4 = ('134 X 3) + (-134 x -7104) = -402 + -0952 = 1-6932 = log -4934 x = -4934 Notice that -402 is subtracted from -0952 although the former is the greater; this being done so that the mantissa of the log shall be positive. Example 17. Find the value of (-1473) ~ l-1 Let- *= Then log x = 2*1 x log -1473 = 2-1 x 1-1682 = ( 2-1 x I) + ( 2-1 x -1682) = + 2-1 - -3532 = 1-7468 = log 55-82 x = 55-82 Example 18. Evaluate {^3-187} -*-* Let x = {log* 3 -i87}- 2 _ y-'ost where y = log 3-187. The value of y must first be found. From the tables log* 3-187 = 1-1591 Hence y = 1-1591 and x = (1-1591) OM Now log x = -024 x log 1-159 = -024 x -0641 = -001538 = 1-998462 or 1-9985 = log -9965 x = -9965 Example 19. Evaluate -||- Let x = this fraction. Then- log x ={J log 42-17 - 2 log -0145} {2 log 8-91 - -116 log 58-27} =ii log 42-17 + -116 log 58-27} {2 log 8-91 + 2 log -0145} FURTHER ALGEBRA 22 j = (-5417 + -2046) - (1-8998 + 4-3228) B -7463 2'2226 = 2-5237 = log 333-9 = 333-9 Explanation. log 42-17 = 1-6250 X log 42-17= -5417 log 58-27 =1-7654 116 x log 58-27= -2046 log 8-91= -9499 2 x log 8-91 = 1-8998 log -0145 =2 -1614 2 X log -0145 =4-3228 When substituting figures for the letters in formulae and thence evaluating the formulae, the importance of the preceding rules will be recognised. Empirical formulae and also the direct results of rigid proofs are of no value at all if one cannot use them efficiently. It is necessary for this purpose that one or two rules, in addition to, or in extension of, those already given should be rigidly observed, viz. Work one step at a time : keep all terms quite distinct until their separate values have been found : and remember that statements including + and cannot be directly changed into log forms. e. g., x = 45 + (29) 1 ' 2 would not read, when logs were taken throughout, log x = log 45 + 1-2 log 29 which is wrong. To evaluate this equation, (29) 1>a would be found separately and its value afterwards added to 45. In cases in which a number of separate terms have to be evaluated it is advisable to keep the separate workings for these to one side of the paper and quite distinct from the body of the sum. Example 20. A gas is expanding according to the law ptf* = C. Find the value of the constant C when p = 85, v = 2-93 and n = 1-3. Substituting values C = 85 x (2-93) 1 -* In the log form log C = log 85+1-3 log 2-93 = 1-9294+ (1-3 x -4669) = 1-9294 + -607 = 2-5364 C = 343-9 Example 21. The insulation resistance of a length / inches of fibre-covered wire, of outside radius r,, and inside radius r l ; the specific resistance of the insulator being S, is given by the formula Find the resistance of the insulation of 50 ft. of wire, of outside diam. -25 cm. and inside diam. I2 cm., when S = 3000 megohms. 222 R = -366 x MATHEMATICS FOR ENGINEERS Explanation. J ** le. o 3000 50 X 12 125 ^06" 366 x 5 x 734 = 1-343 megohms. is a ratio, and .*. r t and ?! may be in any units so long as both are in the same log* 06 1-25 = logi-25 -log, -6 = -2231 1-4892 - 7339 Example 22. For the flow of water over a rectangular notch, the quantity Q = aLH 1 - 6 + 6H 2 ' 5 Find Q when a = -27, L = 11-5, 6 = 28, and H = -517. Making substitutions for the separate parts Let Q = x + y. Then the values of x and y must be first found quite separately and then added. It is preferable in this example to treat the determination of the values of x and y as the main portion, . ., to work these in the centre of the page. x = aLH 1 - 6 = -27 x 11-5 X (-5I7) 1 ' 5 Then log x = log -27 + log 11-5 + 1-5 log -517 = 1-4314 + 1-0607 + (1-5 X I7I35) = 1-4314 + 1-0607 1-5 + 1-0703 = -0624 x = 1-154 Also y = bW'* = 28 x (-5I7) 2 ' 5 Then log y = log 28 + 2-5 log '5 1 ? = 1-4472 + (2-5 x 17135) = 1-4472 - 2-5 + 17838 Alternative Method of Setting Out. V = 5-383 Q = x + y + 5'383 = 6-537 No Log 517 ii'5 27 i'7 I 35 i'5 35675 7135 1-0703 + i'5 - i -5703 1-0607 I-43I4 I-I54 0624 FURTHER ALGEBRA 223 Example 23. The dryness fraction q of a sample of steam, ex- panding adiabatically, viz. without loss or gain of heat, can be found from where r lt q^ and LJ are the original conditions of absolute temperature, dryness and latent heat respectively; and T, q and L are the final conditions of absolute temperature, dryness and latent heat. One Ib. of dry steam at 115-1 Ibs. per sq. in. absolute pressure, expands adiabatically to a pressure of 20-8 Ibs. per sq. in. absolute : find its final dryness. From the steam tables, r t (corresponding to pressure 115-1 Ibs. per sq. in.) = 799 F. absolute temperature, and r (for pressure = 20-8 Ibs. per sq. in.) = 691 F. absolute temperature. Also the respective latent heats are L t = 879 and L = 954. Then if q-i = i, since the steam is originally dry -725{i-i + -145} -725 x 1-245 -903- Explanation. 2-303(2-9025 - 2-8395) 2-303 x -0630 145 Example 24. For an air-lift pump for slimes (a mixture of water and very fine portions of crushed ore, of specific gravity = 1-1013) the formula for the horse-power per cu. ft. of free air can be reduced to H.P. = -015042 {P*(p-|) "- p i} Find H.P. when P! = 12-5, P, = 15- Substituting values H.P. = .oi50 4 2{i 5 (^ 5 ) ?1 - 12-5} = -015042 {15 x -8786 - 12-5} = -015042 (13-18 12-5) = -01023 Explanation. /I2-5Y 71 Let AT = (^P) log x = -71 (log 12-5 - log 15) = -71(1-0969 - 1-1761) = -71 x -0792 = - -0562 = I-9438 ... x = -8786. 224 MATHEMATICS FOR ENGINEERS Logarithmic Equations. Whenever it is required to solve equations containing awkward powers it is nearly always the best plan, and in many cases the only one, to use logarithms. Little explanation should be necessary after the previous work, and a few examples will suffice. Example 25. Find v from the equation, pv n = C, when C = 146, n = 1-37, and p = 22. Substituting values 22 x v 1 - 37 =146 Then log 22 + 1-37 log v = log 146 x '37 log v log J 46 log 22 log 146 log 22 log v = - * * _ 2-1644 1-3424 _ -822 _ 1-37 ~ i'37~~ v = 3-981 Example 26. If h = -^ , giving the head h lost in length / of pipe of diam. d, the velocity of flow of the water being v, find d when h = -87, v = 4-7 and I = 12. Transposing for d OOO4U 1 ' 87 / * = -j- n Taking logs of both sides 1-4 \ogd = log -0004 + 1-87 log v + log / log h = log -0004 + 1-87 log 4-7 + log 12 log -87 f4-6o2i J 1-2568 1-9395 Explanation. 1-0792 log 4-7 = -6721 12-9381 2-9986 1-87 x log 4-7 = 1-2568 Then- logrf = ^6 _ - 2 + -9986 = - 1-0014 1-4 1-4 1-4 = - 7 1 53 = d = Example 27. It is required to express the clearance in the cylinder of a gas engine as a fraction of the stroke. We are told that the tem- perature at the end of compression is 1061 F. abs. and at the end of expansion is 661 F. abs. ; and that expansion is according to the law pv 1 '* = C. Also = K. (This example is important and should be carefully studied.) FURTHER ALGEBRA 225 Let p c , r e and v e be the pressure, absolute temperature and volume respectively at the end of compression; and let p e , r t and v t be the corresponding quantities at the end of expansion. Then we can say p c v c 1 ' 3 = C = p e v e 1 ' 3 (Pc\ _ V 8 '' w'v 7 " 3 and also- = K = tS. Hence from equations (i) and (2) or, dividing through by -' -2 . (3) T t What is required is - - ; and this can be found if is known. v e v e v c For.simplicity let x = - Then from (3) g In the log form '3 log * = log 1061 log 661 log 1061 log 661 3-0257 2-8202 or log x = - -^=685 .. x = 4-842 which is thus the value of ^ Hence v e = 4-8421;,. and v e v e = 4-842^ iv e = 3 '842^ v c v e 3 ' 260^ Exercises 26. On Evaluation of Difficult Formulae and on Logarithmic Equations. 1. Find the natural logs of 21-42 ; 3-18; -164. i 1871 2. Find the values of log, -007254; log* 72-54; log,. -^ 3. Tabulate the values of log -^p when r = 461, 500, 560, 613, 800, and 1000 respectively. 4. Evaluate [log, 226 MATHEMATICS FOR ENGINEERS 5. Find the value of pv log e r when p = 120, v = 4-71 and r = 5-13. Evaluate Exs. 6 to 14. 6. (24-91)- ' 7l 7. (-1183)" 8. (-0054)' 1 ' 9. (3-418)" x (-4006)- 3-4 10 . S*2Lj3*5J>:* 11. (-04I05)- 2 ' 3 12. (-3724)- 2 ' 43 (lo ge i-62)3 X (log 10 325-6)- 247 14. 1-163 x (-0005) 7 ' 76 -i- V(log 10 2i-67)- 1 15. The heat (B.Th.U) generated per hour in a bearing = dlv 1 ' 33 where d = diameter of bearing in inches, / = length of bearing in inches, v = surface velocity of shaft in feet per sec. Find the number of B.Th.U. generated per hour by a shaft of 5" diam., rotating in a bearing 2 ft. long with surface velocity of 50 ft. per sec. 16. Find the value of a velocity v from when c = -97, K = -63, g = 32-2 and h = 49-5. 17. The collapsing pressure P Ibs. per sq. in. for furnace tubes with longitudinal lap-joints may be calculated from Fairbairn's formula p= 7-363 x i where t = thickness in inches, / = length in inches and d = diameter in inches. Find P when t = -043", / = 38", and d = 4". 18. Similarly for tubes with longitudinal and cross joints. Calculate P if / = -12", I = 60", and d = 5^* from fU P = 15547000^-^^, 19. The theoretical mean effective pressure (m.e.p.) in a cylinder is calculated from ft. , Eiii'og.") _ Ps where P = boiler pressure, Pj = back pressure, and r = ratio of expansion. The actual m.e.p. = p m x diagram factor. Find the actual m.e.p. in the case when P = 95, P& = 15, cut off is at -3 of stroke (. e., r = J and diagram factor = -8. 20. The H.P. required to compress adiabatically a given volume of free air, to a pressure of R atmospheres, is given by H.P. = -oi5P(R' 29 i) when the compression is accomplished in one stage and H.P. = -O3P(R' 145 i) when the compression is accomplished in two stages. Find H.P. in each case if P = 14-7 and R = 4-6. 21. Find H, a hardness number, from i6PD-* ' n(2d) Given that D = 24, d = 5, P = 58, n = 2-35. FURTHER ALGEBRA 227 22. Mallard and Le Chatelier give the following rule for the deter ruination of the specific heat at constant volume (K,,) of CO, (carbon dioxide) / } V367 44K,, = 4-33 ^ j where t = C. Find K v when t = 326. 23. Find H.P. from H.P. = oi 5 o 4 |p 2 (?- 1 )' n -P 1 | when Pj = 12-5, and P 2 = 22; the letters having the same meanings as in worked Example 24. 24. Find the efficiency i; of a gas engine from /j\n~l T; = I ~V~J when n = i4 and r 5 25. The H.P. lost in friction when a disc of diameter D ft. revolves at N revs, per min. in an atmosphere of steam of pressure p Ibs. per sq. in. abs., is given by H.P. = io- 13 D 5 N 3 Find the H.P. lost when the diameter is 5 ft., N = 500, and p = i. 26. If p = P( | }* =l and n= 1-41 find \i + nj 27. Calculate the entropy of water <j> w , and that of steam <t>, at absolute temperature T from and ^log The value of r is 682. 28. In the case of curved beams, as for a crane hook where R = radius of inside of crane hook in ins. = 1-5, D = diam. of cross section in ins. = 2-1, P = safe load of hook in Ibs., and 5 = maxi- mum allowable tensile stress = 17000 Ibs. per sq. in. Find the value of P. 29. A sample of steam of dryness -83 at 380 F. expands adiabatic- ally to 58 F. ; calculate its dryness at the latter temperature from is the initial temperature and L = 1115 '\ r = F. abrol. V i.e., t+ 46iJ 30. Steam 20% wet at 90 Ibs. per sq. in. ^S^f adiabatically to 25 Ibs. per sq. in. absolute. Find its second pressure. Note that : p = go ,t= 3 20F.; p = 2 5 ,t = wF.; [Note also the difference between examples 29 and 30 as given data.] 228 MATHEMATICS FOR ENGINEERS 31. The efficiency rj of a perfectly-jacketed engine is given by + 6(r 1 -r 1 ) e -l + a + &T! TJ where a = 1437, b = -7; r t and r, being the extreme temperatures (F. abs.). Find the efficiency of a jacketed engine working between 66 F. and 363 F. 32. Calculate the efficiency of an engine working on the Rankine cycle between 60 F. and 363 F., using the formula LI + T! T, T X and T Z are absolute temperatures and L= 1437 "jr. 33. Calculate the flow Q over a triangular notch from the formula Q = 1 tan | V2i- H* A where g = 32-2, H = -28, tan - = -577. 34. Find the number of heat units Hf supplied for the jacket to an engine working between 60 F. and 363 F. from the formula H,- = 1437 !og 7^ - ( r i - TI) where r t and T X are absolute temperatures, initial and final respectively. 35. Francis' formula for the discharge of water over a rectangular notch is Q (cu. ft. per sec.) = 3-33 (L If the breadth L = 5-4, the head H = -4, and n = 2, find Q. 36. If * = ^| , find * when m = 2-16, v = 1-65. 37. The volume of i Ib. of steam may be calculated from Callendar's equation where o> = -017, c = 1*2, R = 154 V = vol. in cu. ft., p = pressure in Ibs. per sq. foot, T = temperature in centigrade degrees absolute (i. e., t C. + 273). Find V when p = 10 Ibs. per sq. in. and t = 89-6 C. 38. Recalculate, when P = 7200 Ibs. per sq. foot and t = 138-2 C. 39. Similarly, when p = 100 Ibs. per sq. in. and temperature is 437 C. absolute. 40. In calculating the tensions of ropes on grooved pulleys we have the formula ~. -** where 6 is the angle of lap in radians, /x is the coefficient of friction, r is a coefficient depending on the angle of the groove, and T and t are the greatest and least tensions respectively. Calculate the value of T if the angle of lap is 66, /x = -22, t = 45 and r => 1-84. FURTHER ALGEBRA 22g 41. The efficiency of an ideal or perfect engine (working on the Diesel principle) is given by where d = volume at cut -ff = maximum volume volume of clearance' volume of clearance Find the efficiency when d = 1-56, r = 14-3 and n = 1-4. 42. Find the tensions T and * in a belt transmitting 20 H P the belt lapping 120 round the pulley, which is of 3 ft. diam. and runs at 1 80 R.P.M. The coefficient of friction /x between the belt and pulley is -3. Given that - = (P 9 and 6 = angle of lap in radians ; and 7rND(T-fl XT ""33060" ' = revs- P 61 min ' and D ft - = diam - of pulley. 43. The pressure of a gas is 165 Ibs. per sq. in. when its volume is 2-257 cu. ft. and the pressure is -98 Ib. per sq. in. when the volume is 286 cu. ft. If the law connecting pressure and volume has the form pv n = constant, find the values of n and this constant. 44. Find y from 4 2y = 58-7. 45. Solve for x in the equation x l ' n = i^x' n 46. When e Sc = 41 -aS 2 ' 9 , find the value of c. 47. If (# 2 )*' 8 = gx : solve this equation for x. 48. Given that /^pt'J =/,*p,-t, and also that Pl = -283, /i = 28, and /, = 19-5 : find p a . 49. In the law connecting pressures and temperatures of a perfect gas, find p, from the equation having given that = 1-37, p^ = 2160, T, = 1460 and T^ =2190. 50. For a gas engine, P i; 1 ' 83 = p(v + s) 1 ' 83 where P = compression pressure, p suction pressure, v = clearance volume and s = total volume swept out by the piston. If P = 8-91 p and s = -138, find v. 51. If v = aH n , and H = 3 when v = 387 : and H = 80 when v = 2000 ; find the values of a and n. 52. If the pressure be removed from an inductive electric circuit, the current dies away according to the law where C is the current at any time / sees, after removal of the voltage, R and L are the resistance and self-inductance of the circuit respec- tively, and V is the voltage. If R = 350, L = 5-5 and V = 40000, find the time that elapses before the current has the value 80 amperes. 53. 120 was lent out at r% per annum compound interest, the interest being added yearly; and in 5 years the amount became ^150. Find the rate per cent. [Amount = Principal (i + j I 54. If P,VV = P<jV 33 ; = -206; and P<, = 44000; find P,. 230 MATHEMATICS FOR ENGINEERS 55. The insulation resistance R of a piece of submarine cable is being measured ; it has been charged, and the voltage is diminishing according to the law _ t v = be KR where b is some constant, and t = time in sees, and K = -8 x io~ 6 . If v = 30, and at 15 sees, after it is noted to be 26-43, find the value of R. 56. Calculate the efficiency of a Diesel Engine from the formula ^CH where n = 1-41, r c = compression ratio = 13-8 and r e = expansion ratio = 7-4. 57. Determine the ratio of the maximum tension to the minimum tension in a belt lapping an angle 6 radians round a pulley, the co- efficient of friction being \L, from T ma *. ~ A min. 3 The coefficient of friction is -18 and the angle of lap is 154. 58. The work done in the expansion of a gas from volume i/ t to volume v x is given by i n Find this work when v^ = 10, v t = i, and n = 1-13. 59. If T = te* 9 (the letters having the same meanings as in Example 40) : 6 = 2-88 radians, p. = -15 and t = 40, find the value of T. T 60. Similarly if 6 = 165, and j = 1-78, find /i. 61. In the expansion of a gas it is given that pv n = c, and that p = 107-3 when v 3 : and p = 40-5 when v = 6 : find the law connecting p and v in this case. 62. In a " repeated load " test on a rotating beam of fa" rolled Bessemer steel, the connection between the stress F in Ibs. per sq. in. and the number of revolutions N to fracture was found to be F _ 214300 - N '147 Find the value of N when F = 40700. 63. In a similar test on a specimen of \" bright drawn mild steel p _ 733QO - N '04 Determine the value of F which makes N = 48300. 64. The total magnetic force at a point in a magnetic field ~ (r* + * 2 )i Find this force when C = -4, n = 10, r = 4 and x = 5-9. 65. From the results on a test on the measurement of the flow of water over a rectangular notch, complete the following table; it FURTHER ALGEBRA 23I being given that coeff. of discharge = j^^L$!^Lge_ . . theoretical discharge' retical discharge = 40-15 bh? (Ibs. per min.). ft k Actual Discharge (Ibs. per minute). Theoretical Discharge. c. i-75 829 35 i'75 I- 4 I 79 i'75 1-81 112-6 66. Also calculate as in the preceding Example, but for a submerged rectangular orifice, for which the theoretical discharge *2 *i 6 Actual Discharge. Theoretical Discharge. c. 2-325 1-075 I-2 5 88-8 3'34 2-09 1-25 109-6 4'4I5 3-165 I-2 5 133 6-1 1 4-86 I-2 5 156-6 67. The skin resistance per sq. ft. of a ship model is proportional to some power of the speed. If the resistance is -0821 at velocity 5, and -612 at velocity 14, find the law connecting resistance and velocity. 68. The loss of head due to pipe friction is proportional to some power of the velocity. If loss of head was 14-13 when velocity was 10-23, and loss was 6-31 when velocity was 6-76, find the law connecting loss of head h, and velocity v. 69. Relating to the flow of water through pipes it is required to find a value of d (the diameter of the pipe) to satisfy the two equations OOO45V 1 ' 95 , it ,, = -- and -*v-X If * (hydraulic gradient) = ^ , find this value. 2040 70. When a disc revolves in air the H.P. lost in air friction varies as the 5-5 power of the diameter of the disc and the 3-5 power of the revolutions. If H.P. lost is -i when diameter is 4 and disc makes 500 R.P.M. find diameter when 10 H.P. is lost, the disc revolving at 580 R.P.M. 71. When a disc revolves in a fluid it is found that the friction F per sq. foot of surface is proportional to some power of the velocity V. For a brass surface F per sq ft. . 22 1-26 V ft./sec. . . 10 25 Find a formula connecting F and V. CHAPTER VI PLANE TRIGONOMETRY Trigonometric Ratios. If the ordinary 30 : 60 set-square be examined it will be found that for all sizes the ratios of corre- sponding sides are equal. If one of the angles is selected and the sides named according to their position with regard to that angle, the ratios of pairs of sides may be termed the trigonometrical ratios of the angle considered. The word trigonometry implies measure- ment of angles ; the measurement of the angles being made in terms of lengths of lines. For example, let the sides of the set-square be as shown in Fig. 108 : then the angle 30 can be denned as that angle in a right- angled triangle for which the side opposite to it is 2", whilst the hypotenuse is 4", i. e., the ratio of , opposite side 2 hypotenuse 4 Again, the side 3-46" long is that " lying next " or adjacent to the angle 30, so that the angle 30 could thus be alternatively denned by the ratio of its adjacent side to the hypotenuse, or by the ratio of the adjacent side to the opposite side. To these ratios special names are given. The ratio g pposite side is called the " sine " of the angle considered, hypotenuse Tne ra tio ad]acent Slde is called the "cosine" of the angle considered, hypotenuse The ratio ffl oate s "* e is called the "tangent "of the angle considered, adjacent side These three are the most important ; if they are inverted PLANE TRIGONOMETRY three other ratios are obtained, viz. the cosecant or ( J-' and 233 secant ' or As a general rule these ratios, which, as denned, only apply to right-angled triangles, are written in the abbreviated form : sin, cos, tan, cosec, sec and cot. In the triangle ABC, Fig. 109 sin A = PP site to A _ a hypotenuse c whilst sin B = PP site to B _ b hypotenuse c cos A tan A cosec A sec A cot A _ adjacent to A b cos B tan B cosec B sec B cot B a hypotenuse _ opposite to A c' a c b adjacent to A hypotenuse ~ b' c ~ a' c a c ~ b c ~ a opposite hypotenuse adjacent adjacent ~ b' b opposite ~ a' ~b The angles A and B together add up to 90 ; each being called the complement of the other ; and it may be noticed that any ratio of one of the angles is equal to the co-ratio of its complement. Hence the syllable " co " in cosine, cosec and cotan, indicates the complement of the sine, sec and tan respectively. Thus sine A = co-sine of its complement B. tan B = co-tan of its complement A. For any angle the ratios could be found by careful drawing to scale and measurement of sides; this is not very accurate, however, and is certainly very tedious, and therefore tables are provided, in which the ratios of all angles from o to 90 are ex- pressed. The changes in the values of the sine and cosine as the angle increases from o to 90 are illustrated by Fig. no, in which the quadrant is that of a circle of unit radius "i, e., OA =s QC - op~ i'. 234 MATHEMATICS FOR ENGINEERS Now sin i L. EOC = EC *** * J-J.il. -j- j . _ . _ nT = -^n- = BA, and in like manner sin OA i and sin L FOD = FD. Also cos L BOA = OB, cos L EOC = OE and cos L FOD = OF. Thus the sine of the angle depends on the horizontal distance from the line ON of the end of the revolving line, while the cosine depends on the vertical distance from OP. When the angle is very small, A is very near to ON and conse- quently the sine is small ; and as O*A approaches ON more closely, the value of the sine decreases until, when the angle is o the sine is o, because the revolving line lies along ON. When the angle is 90, the revolving line '^^ Curve N %. / \f Curve of Cosines the lies along OP and the horizontal distance of its end from ON has its greatest value, viz. i. Thus the value of the sine increases from o to i as the angle increases from o to 90. Along OA, produced, set off ABj = AB = sin L BOA; and in like manner obtain the points Ej, Fj and O v Draw a smooth curve through the points M, B 1( Ej, 1 and O 1 : then this is a curve of sine values, since the intercepts between the quadrant perimeter and this curve give the values of the sine, thus sin L MOR = RR r Similarly the curve of cosine values can be drawn, and it is seen that it is of the same form as the curve of sine values, but it is reversed in direction. To read Table I at the end of the book. In this table one page suffices for the various ratios, these being stated for each degree only from o to 90. This table is compact and has educa- tional advantages, for it demonstrates clearly that as the angle increases the sine increases whilst the cosine decreases; and that a ratio of an angle is equal to the co-ratio of its complement, and so on. Down the first column and up the last are the angles expressed PLANE TRIGONOMETRY 235 in degrees, whilst in the adjacent columns the corresponding values in circular measure (radians) are given. Thus 31 = -5411 radian, and 73 = 1-2741 radians. The values of the sines appear in the 4th column from the beginning and the 4th from the end, as do also the cosine values; but for cosines the tables must be read in the reverse direction. No difficulty should be experienced in this connection if it be remembered that one must always work away from the title ol the column. Thus for cosines read down the 7th column and up the 4th column. Values of tangents and cotangents appear in the 5th and 6th columns; again working away from the title E.g., sine 17= -2924, sine 61 = -8746 cos 23 = -9205, cos 49 = -6561 tan 42 = -9004, tan 88 = 28-6363 cot 5 = 11-4301, cot 59 = -6009. To read Table V at the end of the book, which should be used when greater subdivision of angles is required. Suppose that sin 43 22' is required : if Table I is followed, sin 43 must be found, 22 viz. -6820, and sin 44, viz. -6947, and g- of their difference must be added to -6820. ' - ^ 22 Thus sin 43 22' = -6820+ (-6947 6820) = -6867 This process is rather tedious : accordingly, referring to Table V, look down the ist column until 43 is reached, then along the line until under 18', the figure is -6858 ; 4' have now to be accounted for; for this, use the difference columns, in which under 4', 8 is found .'. sine 43 22' = -68584-0008 = -6866. The tangent tables, Table VII, would be applied in the same manner, but here the value of the ratio gets very large when in the neighbourhood of 90 so that the difference columns cannot be given with accuracy. When the angle = 45, the tangent = I and the tangent continues to increase as the angle increases, there- fore it happens occasionally that the integral part of the value has to be altered in the middle of a line. To signify this a bar ( ) is written over the _first figure: e.g., tan 63 = 1-9626, whilst tan 63 30' is written 0057, and this means 2-0057, the bar indicating that the integer at the commencement of the line must be increased by i. 236 MATHEMATICS FOR ENGINEERS When using the cosine table, viz. Table VI, it must be remem- bered that an increase of the angle coincides with a decrease of the cosine, so that differences must be subtracted: e.g., if the value of cos 52 55' is required. cos 52 54' = -6032 ; diff . for i' = 2 /. cos 52 55' = -6032 0002 = -6030. Values of cosecants and secants can be found by inverting the values of sines and cosines respectively. Example i. The angle of advance 6 of an eccentric in a steam engine mechanism can be found from sin 6 = j^- - =- . Find 6 when the lap is -72", the lead is -12" and the travel is 3-6*. Substituting the numerical values, sin 6 = - = I "O I*O = -4667. We have now to find the angle whose sine is -4667. Turning to the table of natural sines we find -4664 (the sine of 27 48') to be the nearest figure under -4667; this leaves -0003 to be accounted for. In the difference columns in the same line we see that a difference of 3 in the sine corresponds to a difference of I min. in the angle; hence i' must be added to 27 48' to give the angle whose sine is -4667. Hence 6 = 2 7 49'. -- a Example 2. If cos A = - T - ' a = 4-2, 6 = 7-8 and c = 6 ; find A. Substituting the numerical values _ 7 -8'+6 2 -4-2 2 _ 60-84+36-17-64 '2x7-8x6' 93-6 = .3 6I . 93- From the table of natural cosines we find that the angle having the ratio the nearest above -8461 is 32 12'; for this the cosine is -8462, and therefore the difference of -oooi has to be allowed for. In the difference columns we see that a difference of -0002 corresponds to i'; and thus -oooi corresponds to 30*. Hence A = 32 12' 30*. Exercises 27. On the Use of the Tables of Trigonometric Ratios. 1. Read from the tables the values of: sin 61; tan 19; cos 87; tan -2269 radian. 2. Find the values of sin 77$; cos 15 24'; tan 58 13'; cos 1-283 radians. 3. Evaluate ' PLANE TRIGONOMETRY 237 4. In a magnetic field, if H = horizontal component and T = the total force due to the earth, then H = T cos d. Find T when H = -18 and d = 63. 5. The tangent of the angle of lag of an electric current = reactance resistance and reactance = 2n- x frequency x inductance. If frequency is 40, inductance -0021 and resistance 1-7, find the angle of lag. 6. The mean rate of working in watts = amperes x volts x cos (angle of lag). Find the mean rate when A = 2-43, V = no and lag = I9i. What is the mean rate of working if the current lags 90 behind the voltage ? rise 7. The pitch of a roof = -- = 4 tan A, where A is the angle of the span roof. Find the angle of the roof for which the span is 36 ft. and the rise is 12 ft. 8. If an axle of radius r runs on a pair of antifriction wheels of radius R, and 6 is the angle between the lines joining the respective centres, then 2 where F = force required to overcome the friction on a plane axle and F! = force required to overcome the friction when using the antifriction wheels. Find F if 6 = 47^, r = 3", 'R = 10* and F x = 47. 9. If D = pitch diameter of spiral toothed gear, N = number of teeth in gear, P = normal diametral pitch, and a = tooth angle of gear, then PCOS a If D = 5-108, N = 24 and P = 5, find a. 10. In calculating principal or maximum stresses, if tan 26 = -j t s = 2852 and /= 3819, find d. 11. The number of teeth in the cutter for spiral gears no. of teeth in the gear cos 8 (angle of spiral) Find the number of teeth in the cutter when the angle of the spiral is 50 and there are 48 teeth in the gear. (#.B. cos 3 A means the cube of the cosine of A ; but cos A 8 is the cos of A 8 .) 12. In connection with the design of water turbines the equation " = tan 6 occurs, where w = tangential velocity of the water at inlet, = radial velocity of water at inlet, V = velocity of the blade at inlet, and & is the inclination of the blade at inlet, per sec., V = 47-7 ft. per sec. and 6 = 6oJ, find w. 13. In the formula giving the value of the horizontal p ress ure on a retaining wall of height h, the earth surface being ^***" . , _ _ *. ~. earth and 4, i s the angle of repose c .he e artn, Find the value of p when w = 130, * = 2 3 J and h - 24 ft. 238 MATHEMATICS FOR ENGINEERS 14. Calculate the value of M, the moment of friction of a collar bearing, from __ asinaCRi 2 - R, 2 ) when R! = 4-5, R z = 3-75, W = 2000, p. = -17 and a = 12. 15. The total pressure P on the rudder of a ship is given by P= 4 -6KAV 2 - * in . "39 + '61 sin a where V = speed of ship in knots, A = area of rudder, K = -7, and a = angle of rudder with fore and aft plane. Calculate P, given that V = 16, A = 8 and a = 15%. 16. The force Pj applied horizontally to move a weight W up a rough plane inclined at an angle a to the horizontal, is given by p = WQi + tan a) i p. tan a Find P a if W = 3000, a = 8, and /*, the coefficient of friction, = -12. 17. The total extension d of a helical spring is given by Wa 2 /. d = --FQ-(I 2 sin 2 a) If a radius of coil = 4", G = 12 x io 6 Ibs. per D*, J = -15, / = length = 29*, W = 12 Ibs. and a = 14, find the total extension. V 2 sin 2A 18. The range of a projectile is given by - - , where V = o velocity of projection, A = elevation of gun and g = 32-2. Find the range, if the projectile is fired at an elevation of 29 15' with a velocity of 1520 ft. per sec. 19. p n = intensity of the normal pressure of wind on a surface inclined at 6 to the direction of wind, and p = intensity of pressure on the surface perpendicular to its direction 2 sin 6 **~*'i + *a*e If P = 35 and 6 = 22 J, find p n . TT "p i 20. The maximum power-factor of a motor = cos <f> = ' ' ~] If H.P. is 4-78 find <f>, the angle of lag of the current. 21. If P = effort on crosshead of a steam engine, T = crank-pin connecting rod effort, 6 crank angle, n = -- - ; and if P = 450 Ibs., = u Sm 2 * and 6 = 1-5 radians ; find T, from T = P/sin 6 + . I Vw 2 sin 2 6} {Hint. Sin 171-9 = sin 8-1} 22. Calculate the value of y = Re~ K ' sin (wt + d) when R = 3-5, K = -4, t = -02, w = 5, 6 = -16 ; the angle being expressed in radians. 23. The electrical induction B in an air gap is given by Csin-(i+-)Rxio 9 B= - 2 An x io 7 Find B when A = 3-515, n = 20, X = -0867, C = 42-05, R = 10382 and tan 20 = -1052. PLANE TRIGONOMETRY 239 24. Find a value of 6 to satisfy the equation 4. an A $d(l 2x) ~!*~ where d = 5, / = 30 and x = 4-5. This equation refers to stiffened suspension badges, where 6 is the angle of inclination of the cable to the horizontal at a horizontal distance x from one end of the bridge, / is the span of the bridge and d is the sag. Application of Trigonometric Ratios. We will first deal with a very simple case. Example 3. The angle of elevation of the top of a chimney at a point on the ground 120 ft. from the foot of the chimney is 25. Find the height of the chimney. Before proceeding to the actual working of the example, the term angle of elevation must be explained. The zero of the theodolite (an angle measuring instrument) would be observed when the telescope was directed along the horizontal : the telescope would then be moved in a vertical plane until the top of the chimney was seen and the angle then noted. This angle is called the angle of elevation and is the angle between the horizontal and the line joining the eye to the objeet. If the instrument be placed on the chimney top, the same angle would be read, but it would now be called the angle of depression because the object (the earth) is below the level of the eye. In the example before us, let h ft. = height of chimney (Fig. in) TT " _ PP!./< rt r or\ "T and therefore = tan 25. Now, from the tables tan 25 = -4 66 3 .'. = -4663 120 and h = 120 x -4663 = 55 '96, say 56 ft. Example 4. Two coils are connected in series over a 220 volt alternating-current main, and the drop across each coil is 126 volts. If the diagram illustrating the relation between the voltage drops j as in Fig. 112, find the difference in phase between the voltages in tl two coils, i. e., find the angle a. Since the sides AB and BC are equal, the perpendicular from B on to AC bisects AC. or DC = no; and also /.DBC =| 240 MATHEMATICS FOR ENGINEERS Then- and whence sin - = ^ == -8730 2 126 /3 = sin 60 49' =-*> a = I2I ( 49 Example 5. In a test on the Halpin thermal storage system, as fitted to a Babcock and Wilcox boiler, the volume of water taken from the storage tank to the boiler is to be deter- mined by the difference in water level between start and finish. The tank being a cylinder of 57*81* diam. and 2^1* length, with its axis horizontal, see Fig. 113, the water level is 52-96" from the tank bottom at the start and 14-86* at the finish. Find the volume of water abstracted in cu. ft. We have to find the area of ABCD, viz. the difference between the area AEBCD (at the start) and the area AEB (at the finish), and then multiply by the length of the tank. To find the area of the segment AEB and OF = OE - EF OF :OS a = pr-f- = -Q OA 28-91 a = 60 56' or H 28-91 14-86 = 14-05 = -4859 = cos 60 56' 60-93 57'3 1-063 radians. Fig. 113. Halpin Thermal Storage System. We can now use the rule previously given for the area of a segment, 9* viz. area = (8 sin S) where 6 is the central angle in radians, for r = 28-91, 6 = u AOB = 2a = 2-126, and sin 6 = sin 121 52' = sin(i8o 121 52^ = sin 58 8' = -8493. PLANE TRIGONOMETRY 241 proof of the rule sin A = sin (180- A) is given later in the book.] Thus- area of AEB = Mi ( 2 . I26 _ = 533*8 sq. ins. To find the area of the segment DHC OG = EG-OE = 52-96 - 28-91 = 24-05 and ft = 33 43' or -588 radian sin 2,3 = sin 67 26' = -9234. -'83.8 -cos 33' 43" Hence area of DHC = - -923) = 105-9 sq. ins. Area of the whole circle = - x 57-8i 2 = 2625 sq. ins. area of ABCD = 2625533-8105-9 = 1985 sq. ins. and volume = -Jlj cu ft> = 2 88- 4 cu. ft. Outcrop Example 6. A seam dips at an angle of 62 to the horizontal for a distance of 900 ft. measured along the seam and then continues dipping at an angle of 40 to the horizontal. A shaft is started to cut the seam at a distance of 1200 ft. horizontally from the outcrop; at what depth will it cut the seam ? This example introduces the solution of two right-angled tri- angles : the lengths AD and AC (Fig. 114) are given and we require to find DF. AB Fig. 114. Problem on a Coal Seam. In the triangle ABC, = sin 28 = -4695 Also Hence In the triangle CEF, AB = 900x^4695 = 422-6' = sin 62 = -8829 900 BC = 900 x -8829 = 79^4' CE = BD = I200-AB = 777-4' ^g = tan 40 = -8391 EF = 777-4 x -8391 = 652-4 DF = DE+EF = BC+EF = 794'4+ 6 5 2 '4 = I 44P'8ft. 242 MATHEMATICS FOR ENGINEERS Trigonometric Ratios from the Slide Rule. The sine and tangent scales of the slide rule may be usefully employed in trigonometry questions; the multiplication of the side of the triangle by the trigonometric ratio being performed without the actual value of the ratio being read off. To read values of trigonometric ratios : Reverse the slide so that the S scale is adjacent to the A scale and the T scale to the D scale. The sines of angles on the S scale will then be read off directly on the A scale. If the number is on the left-hand end of the rule, then -o must be prefixed to the reading, but if on the right- hand end of the rule, then a decimal point only. e. g., to find sin 4 : place the cursor over 4 on S scale, and on A scale read off 698; this being on the left-hand end of the rule sin 4 = -0698. Again, sin 67 = -921 for 921 is read off on the A scale above 67 on the S scale and is on the right-hand end of the rule. As the angle approaches 90 the sine does not increase very rapidly and therefore the markings for the angles on the S scale in this neighbourhood are very close together. From 70 the usual markings are for 72, 74, 76, 78, 80, 85 and 90, the longer mark being at 80. To use the S scale for a scale of cosines, first subtract the angle from 90, *. e., find its complement, and then find the sine of this. e. g., cos 37 = sin 53 = -799. To combine multiplication with the reading of ratios, use the S scale just as the ordinary slide or B scale, multiplying, as it were, by the angles instead of by mere numbers. e.g., suppose the value of the product 18-5 x sin 72 is required. The right hand of the S scale is set level with 185 on the A scale, the cursor is placed over 72 on the S scale, and the product 17-6 is read off on the A scale. The tangents of angles from o to 45 will be read in a similar fashion, the T and D scales being used. Tan 45 = i, and after this the tangent increases rapidly, being infinitely large at 90. For an angle greater than 45 : subtract the angle from 90 and divide unity by the tangent of the resulting angle. e. g., suppose tan 58 is required. i Actually tan 58 = , tan 32 Hence : set 32 on the T scale level with i on the D scale ; then PLANE TRIGONOMETRY 243 the reading on the D scale opposite 45, ,'. ., the end of the T scale is the required value and is 1-6. A further example. Find the value of - 7 tan 64 talV = 87Xtan 26 = 42 '4' [The setting being : 45 on the T scale against 87 on the D scale the cursor over 26 on the T scale; then 42-4 on the D scale.] Example 7. A boat towed along a canal is 12 ft. from the near bank and the length of rope is 64 ft. The horse pulls with a force of 500 Ibs. : find the effective pull on the boat, and that tending to pull the boat to the side of the canal. Fig. 115. Forces on Boat towed along a Canal. The " space " diagram is first set out and from this x is calculated, viz. x = -v/64 2 -i2 2 = 62-8 (a, Fig. 115). If a triangle be drawn (see b, Fig. 115) with sides parallel to those of the triangle ABC so that EF represents 500 Ibs. to some scale, then EG and GF represent the pulls required to the same scales. Or by calculation = cos E = cos A = 2--, *' GE = 500 cos E 500 64 i. e., the pull towards the bank = 93-8 Ibs. Also = sin E = sin A = -^-, *. e., GF = 500 sin E 500 64 , 49I i. e., the effective pull in the direction of the boat's motion = 49* Ibs. In general the components of a force R in two directions at right angles to one another (see Fig. 116) are R cos a, and R s'.n a 244 MATHEMATICS FOR ENGINEERS where a is the angle between R and the first of the components. As a further example of resolution into components, if T (Fig. 117) is the total magnetic force on a unit pole at some place and d is the angle of dip, H the horizontal component of the force is = T cos d, and V the vertical component = T sin d. H Fig. 1 1 6. Components of Forces. Fig. 117. Calculation of Co-ordinates in Land Surveying. When plotting the notes of a traverse survey, in which the sides of a polygon and the " included " or internal angles are measured in the field, it is necessary to first transform the dimensions of the lines and angles so as to give the co-ordinates of the comers as measured from the north and south line (or meridian) on the one hand, and from some chosen east and west line on the other hand. The survey is then plotted from the co-ordinates, with the object of introducing an accuracy of drawing which is impossible if the field-book dimensions are directly set out. In the latter case the angular error is cumulative, and, further, the plotting of angles at all times is more productive of error than the plotting of lines (e. g., co-ordinates). Quadrant bearings. The co-ordinate axes being chosen as just stated, viz. North-South and East-West, every line of the traverse is referred to the meridian in terms of the smallest angle between it and the meridian, with the further statement of the " quadrant " (N.E., S.E., S.W., or N.W.) in which it is placed. Such angles are termed quadrant or reduced bearings. Thus in Fig. 118 The reduced bearing of the line A is 27 N.E., that of the line B is 36 S.E., that of the line C is 66 S.W., and that of the line D is 11 N.W. PLANE TRIGONOMETRY 245 handed direction. This is better than the quadranf method requiring but one simple numerical statement N.W. QUADRANT* N N.E. QUADRANT S.W. QUADRANT c. S.E. QUADRANT Fig. 118. Reduced Bearings. Fig. 119. Whole-circle Bearings. For example, in Fig. 119, the whole-circle bearings of the lines A, B, C and D are respectively 27, 144, 246 and 349, all measured from the north line ON. Example 8. Measurements on a triangular plot of land ABC, Fig. 120, resulted in the following : AB = 7073 links, BC = 7736 links, CA = 5462 links, A = 75, B = 43 and C = 62. The reduced bearing (R.B.) of AB is 9 N.E. and the point A is taken as the origin for the co-ordinates. Find the reduced bearings of BC and CA, the co-ordinates of the points B and C, and also the area of ABC. Right-hand order should be adhered to throughout, as indicated by the letters ABC. To find the R.B. of BC. [It should be grasped that the bearing of C to B is not the same as the bearing B to C.] Mark on the diagram all the known angles, and then by combination with 90 or 180 all the required bearings can be found. Thus R.B. of BC = 43 9 = 34 S.E. , since 34 is the acute angle made by BC with the N. and S. line : the quadrant must also be stated, to definitely fix the direction of movement. Similarly, the R.B. of CA = i8o-62 -3 4 = 84 S.W. To calculate the co-ordinates of B also = sin 9 and therefore BD = AB sin 9 t5 AD = AB cos 9. 246 MATHEMATICS FOR ENGINEERS Thus the departure of B (i. e., its distance E. or W. from A) = AB x sin (R.B. of AB) and the latitude of B (i. e., its distance N. or S. from A) = AB x cos (R.B. of AB) Then In the log form BD = 7073 x sin 9 log BD = log 7073 + log sin 9 = 3-8496 + I-I943 = 3-0439 BD = 1106 links, which is the departure of B east of A JSL Fig. 1 20. Plot of Land. Again AD = AB cos 9 = 7073 x cos 9 In the log form log AD = log 7073 + log cos 9 = 3-8496 + T-994 6 = 3-844 2 .". AD = 6985 links, which is the latitude B north of A Hence the co-ordinates of the point B are 1106, 6985. For the point C In the log form BE = 7736 sin 34 log BE = log 7736 + log sin 34 = 3-888 5 + 1-7476 = 3-6361 BE = 4326 links, which is the departure of C east of B. PLANE TRIGONOMETRY 2 4? Again CE = 7736 cos 34 In the log form log CE = log 7736 + log cos 34 = 3-8885 + 1-9186 = 3-8071 = 6413 links, which is the difference of latitude between B and C. Thus the co-ordinates of Care (1106 + 4326) and (6985 - 6413) or (5432, ,572). The figure may now be accurately plotted by means of the co- ordinates. To calculate the area A ABC = ADEF-ABD-BEC-ACF = (5432 x 6985)- (x 6985x1106) = (37-95 x io)-(3-87X io)-(i3-88x io 6 )-(i-5 53 x io = 18647000 sq. links Dividing by ioo 2 , = 1864-7 S( l- chns. Dividing by 10, = 186-47 acres. For greater precision tables of log sines and log cosines (viz. Tables VIII and IX at the end of the book) have been utilised in the working of this example. For general work the accuracy of the slide rule is sufficient, but in all cases these tables, which are used in the same way as the tables of natural sines and natural cosines, are convenient. As shown earlier in the chapter the value of the sine or cosine of an angle varies between o and I, and accordingly the values of the logs of these ratios vary between oo (;'. e., the smallest quantity possible) and o, since log o = oo (refer Chapter I) and log i = o. Except for small angles, therefore, the log sine will be of the nature of 7- ..... or ~2' ..... whilst the value of the log cosine will be 7- ..... or 2- ..... unless the angle is large. e. g., sin 27 = -454 and Io g sin 2 7 = lo g '454 = sin o33' = -0096 and log sin o33' = log -0096 = 3-9823 cos 87 = -0523 and log cos 87 = log -0523 = 2-7185 Example g. From the following co-ordinates compute the true length, the bearing, and the angle with the horizontal of the line AB. Station. Feet. Feet. Feet above Sea Level. A B Northing 4501-2 Southing 20-1 Westing 56-1 Easting 4788-1 Reduced level 249-2 Reduced level 329-2 248 MATHEMATICS FOR ENGINEERS By plotting the points A and B from the co-ordinates given, their actual positions are represented. Complete the triangle ABC by drawing AC vertically and BC horizontally (see Fig. 121). Then AC = 4501-2 + 20-1 = 4521-3 and BC = 4788-1 + 56-1 = 4844-2 56-1 To express the results with the same accuracy as that with which the figures have been measured we must use five figure log tables. To find the angle CAB tan CAB =484412 4521-3 i.e., log tan CAB = log 4844-2 - log 452 1 -3 = 3-68523 - 3-655 2 7 = -01996 = log tan 46 19'. CAB = 46 19' The whole circle bearing of AB is thus 180 46 19' = 133 41^. To find the length (on the plan) of AB -pjs = sin 46 19' AB ' Bearing angle reauired I inclined length CB In the log form sin 46 19' = 4844-2 sin 46 19' log AB = log 4844-2 log sin 46 19' = 3-68523 - 1-85924 = 3-82599 = log 6698-7 AB= 6698-7 This length found is that of AB on the plan ; the true length will be slightly greater than this, since it is the hypotenuse of the triangle of which the base is 6698-7; and the height is 80. In the triangle ABB 1 : tan B'AB = - 6698-7 and log tan B 1 AB = log 80 log 6698-7 = 1-90309 3-82599 = 2-07710 = log tan 41' 4* .*. the inclination of AB to the horizontal is 41 '4* and the true length of AB (or AB 1 ) = V (66987) ~^8o* = 6699-3- PLANE TRIGONOMETRY 249 Exercises 28. On the Solution of Right-Angled Triangbs, and the Calculation of Co-ordinates. In the following Examples i to 7, ABC is a triangle right-angled at C. (In each case the figure should be drawn to scale.) 1. c = 45", A = 15, find a and b. 2. a = 12", B = 36, find 6 and c. 3. c = 65", A = 48, find a and b. 4. b = 34", B = 27, find a and c. 5. c = 27-37", A = 54, find a and 6. 6. 6 = 72 -5", A - 3 8i, find a and c. 7. c = 23-4", B = 27i, find a and b. 8. A bomb dropped from an aeroplane strikes a building which is known to be one mile away from an observing station, at which the elevation of the aeroplane is seen to be 29. Find the " range," i. e., the distance of the aeroplane from the observer, and also its height. 9. A mountain railway at its steepest rise has a gradient of i in 7. What is the inclination to the horizontal of this gradient ? [Note that the gradient is always the *-; hypotenuse J 10. From the top of a house, 37 ft. high, a bench mark (Government height above sea-level) is sighted, and the angle of depression is 48. Find the horizontal distance from the house of the B.M., which is placed at a point 3 ft. above the ground. 11. The crank and connecting rod of a reciprocating engine are at right angles to one another. If the value of the ratio connecting rod length length of crank is 4-7, find the angle which the crank makes with the line of stroke. 12. The rise of a roof is n ft. and the span is 84 ft. : find the angle of the roof. 13. The tangent of the angle of a screw is given by the pitch divided by the circumference of the screw. If the diameter is 5* and the pitch angle is 7 15', find the pitch. 14. If the screw in Ex. 13 becomes (a) double- or (b) treble- threaded, what are now the angles of the thread ? 665 | FT' Fig. 122. " Set-over " of Lathe Tailstock. Fig. 123. Brown and Sharpe Worm- thread. 15. Calculate the " set-over " of the tailstock of a lathe for turning a taper (the angle being 9 and the length of job 15-2). See Fig. 122. 16. Find the angle of thread 6 for the Brown and Sharpe worm- thread shown in Fig. 123. 250 MATHEMATICS FOR ENGINEERS 17. When using the Weldon Range Finder, one determines a length AB AB by comparison with a base AD. Find ratio of ^~. for the case illustrated (Fig. 124). B B 90 Base Fig. 124. D Fig. 125. 18. Determine the co-ordinates of the points A, B, C and D (Fig. 125) with references to the axes marked. Find the area of ABCD; and state also the " reduced bearings " of BC, CD and DA. The bearing of ABis 50-5 N.E. 19. In Fig. 126 calculate the co-ordinates of the points B and C, the reduced bearings of BC and CA, and the area of ABC, if the bearing of AB is 60 S.E. N Fig. 126. 20. In finding the length of a line CB, a line CA was set out by means of the optical square at right angles to CB and the distance CA was chained and found to be 1-14 chains. The angle CAB was then ob- served by a box sextant and found to be 71 54'. Calculate the length ofCB. 21. The co-ordinates of two stations A and B are A. Latitude N 400 links ; Departure W 700 links B. Latitude S 160 links; Departure W 1500 links Find the whole circle-bearing of AB. PLANE TRIGONOMETRY 251 22. You are 220 ft. horizontally away from the headgear of s mine. From a point on the same level as its base you finf that the headgear subtends a vertical angle of !8 3 o'. Find the height 23 A ball fitting down to the taper sides was used to test the correctness of the cup-shaped check shown in Fig. I2 6a. The test ra made by measurement of the distance AB. Calculate' this - to 10000 Toper of Sides 5 IP 8 A measured on diarn. Hole to be I drilled Fig. I26a. Test for Gauge. 95"dia " Fig. 1266. Block for Jig. 24. Determine the diameter of the largest drill that could be used for the hole in the jig block shown in Fig. 1266, when you are told that the drilled hole, which is made first to clear away part of the metal, must cut the taper hole at the level AA. Angles of any magnitude. Up to this point our work has been confined to angles of 90 and under, whose trigonometrical ratios can easily be found from tables or by the use of the slide rule. Angles greater than 90 must be reduced to those less than 90 by com- bination with 180 or 360, i. e., they must be reduced to the equivalent acute angle made with some standard line, which in all this work will be taken as the N and S line. N Fig. 127. Fig. 128. Fig. 129. If the N. and S. line and the E. and W. line be drawn, they divide the space into four " quadrants," and the position of an angle can always be stated by reference to the quadrant in which it lies. Angles are measured in a right-hand direction from the N. and S. line, and the quadrants are numbered as shown in Fig. 127. A minus sign before an angle indicates a movement from the north in a left-hand direction. 252 MATHEMATICS FOR ENGINEERS e. g., referring to Figs. 128 and 129 154 is in the 2nd quadrant ; and its equivalent acute angle = 180 154 = 26 258 is in the 3rd quadrant ; and its equivalent acute angle = 258-i8o = 78 76 is in the 4th quadrant ; and its equivalent acute angle = 76 472 is in the 3rd quadrant ; and its equivalent acute angle = 68 To sum up, it will be seen that the equivalent acute angle (written e.a. angle) is always the angle made with the N. and S. line ; *'. e., it is obtained by compounding with 180 or 360. It is now necessary to find the algebraic signs to be prefixed to the trigonometric ratios of any angle. Thus although the sine of 472 is numerically equal to the sine of +68, since 68 is the e.a. angle for 472 (see Fig. 129), it would not necessarily be correct to state that sin 472 = sin 68, because we have not yet examined for the algebraic sign. As a matter of fact, sin 472 = sin 68. Suppose that a line of unit length rotates in a right-hand direc- tion, starting from the north, thus sweeping out the various angles. Its " sense " will always be considered positive, whilst the usual convention will fix the signs for horizontal and vertical distances. [Note. In all that follows, be sure to measure every angle from the north point : thus in Fig. 130, the angle (180 A) is the angle aod, and the angle (360 A) is the angle aoh measured in a right-hand direction.] Let Laoc (Fig. 130) represent the magnitude of the e.a. angle in all the four quadrants : *. e., L aoc = L eod = L eof = t_ aoh = A, say. In the ist quadrant + ac + ac sin A = cos A = tan A = + oc + oa + oc -f- ac ~4- oa i oa = ac = + oa ac oa In the 2nd quadrant sin (180 - A) = ^ = ed : but ed = ac PLANE TRIGONOMETRY 253 so that sm (i8o-A) = sin A. Hence the reason for compounding with 180 to find the e.a. angle is seen. Again cos (180 A) = ~ oe = oe = oa - oe indicating that oe is a negative length, because measured downwards tan (i8o-A) = + ed e ac oe In the 3rd quadrant sin (i8o+A) = - oe oa = cf = ac cos (i8o+A) = oe = oe = oa tan (i8o+A) = e J- = oe In the 4th quadrant / r A \ Q-h sin (360 A) = cos (360 A) = - oe oa / r A \ tan (360 A) = oa oa ah oa oa i. e., summarising for the equivalent acute angles in all four quadrants, the algebraic signs vary as follows Quadrant. ISt 2nd 3rd. 4 th. sine and cosec . cos and sec . tan and cot . + + + + + + i OS i joLL I i + + 1 i + on sin '+ - - + @ SINE AND COSINEAND TANGENTAND COSECANT. SECANT. COTANGENT. Fig. 131. Variation in Sign of Ratios. This variation in sign may be better or more plainly denoted by the diagrams (a), (b), (c) and (<*), Fig. 131. Fig. (a) 131 may need an additional word of explanation. In each quadrant is written 254 MATHEMATICS FOR ENGINEERS the word to indicate which ratio or ratios is or are positive in that quadrant. Thus in the 3rd quadrant, the tangent alone is positive, and in the 4th quadrant the cosine alone. Fig. (b), (c) and (d) 131 are merely a representation of the table just given. Hence, to find the trigonometric ratio of an angle of any magni- tude : find first its e.a. angle and the quadrant in which the angle occurs, and then apply the sign of the quadrant for the ratio re- quired. (Numerically, the ratio of any angle is that of its e.a. angle.) In all cases it will be found that a diagram simplifies matters. Example 10. Find the value of sin 172. sin 172 = sin (180172) = sin 8, for 8 is the e.a. angle = +-I392 since 172 is in the 2nd quadrant, and the sine there is +. Example n. Find cos 994. 994 =[(2 X 360) + 2 74] 2 x 360 brings us back to the starting line, and so we deal only with the 274. Now 274 is in the 4th quadrant, and thus its cos is + ; also the e.a. angle = 360 274 86. cos 994 = + cos 86 = +-0698. Example 12. Find tan 327. The angle 327 is in the ist quadrant, and hence its tan is + ; also the e.a. angle = 33. tan -327 = + tan 33 = +-6494. Example 13. Find the sin, cos and tan of 115. What connection is there between them ? sine is +1 ;os is J- :an is } The angle is in the 2nd quadrant, hence sine cos tan is . also the e.a. angle 180 115 = 65 sin 115 = +sin 65 = + "9063 cos 115 cos 65 = -4226 tan 115 = tan 65 = 2-1445 sin 115 Now = tan 115. PLANE TRIGONOMETRY 255 This most important relation always holds, viz. that tanA = sin - A cos A The "reduced bearing/' in surveying, may be regarded as identical with the " equivalent acute angle " here used. In the general solution of triangles only angles up to 180 occur e we are concerned mainly with the 1st and 2nd quadrants. Exercises 29. On the Trigonometric Ratios of Angles of any Magnitude. Find from the tables, the values of the sin, cos and tan of the following angles (Exs. i to 5). 1. 116; 322; 218. 2. -82; -398; 1562. 3.199-2; 34i5'; 9842 3 '. 4. 4 ; u-62; -85; 1-16 radians. 5. 1194; 2-45 radians; 787 n'. 6. Find values of cot 126; COSCCTT; sec (-52). [Note. The angle TT radians is that subtended at the centre by the half-circumference and is thus 180.] 7. Find a value of A between o and 180 if cos A - b * + cZ ~ a * and b = 9-8' \*\Jj X*. _ _ _ zbc c = 6-4* a = 14-45* 2 V2 _|_ 7/2 -zoT-T 8. The equation cot 6 = - ^y - relates to the design of water turbines. If V = 53-4, u = 10, H = 100, e = 32-2, find 6 (between o and i So ). 9. As for the preceding question, but taking V = , n = V8o, g = 32-2 and H = 80. 10. If a = angle of the crank of a steam engine from the dead centre, m = ratio of connecting rod length to length of crank and / = -833 : find values of a to satisfy the equation cos a = m Vm 2 +i 2m/ when m = 4. Solution of Triangles. The " solution " of a triangle consists in the determination of -the magnitudes of the six parts, viz. the three sides and the three angles. In many cases sufficiently accurate results can be obtained by careful drawing to scale, but for great precision the values of the parts of the triangle must be calculated. In such calculation extremely exact tables, giving the relations between the sides and angles, are employed, and the results obtained are superior to those given by even skilled draughtsmanship. Again, it sometimes happens that the triangle is difficult to construct: thus if in Fig. 136 the base AC was very small compared with the sides AB and BC, the intersection of AB and CB would not easily 256 MATHEMATICS FOR ENGINEERS be detei mined, and, therefore, the lengths of the sides as measured would only be approximate. The angle at B would under these circumstances be termed " badly conditioned." There are a number of rules developed for the general solution of triangles, but of these the following will be found to be of the greatest service, while even this list may be reduced to the first two rules. Adopting the usual notation for the triangle, viz. A, B and C for the angles, and a, b and c for the sides opposite these angles respectively, the rules for the solution of all triangles are ft h C (1) -7 r- = =r = - ^r, usually referred to as the sine rule. ' sin A sin B sin C (2) a 2 = b 2 + c 2 2 be cos A, usually referred to as the cosine rule. / \ / \ -A / (s b)(s c) (3) () sin- =</* g- /IA a) (b) - A /(s - b)(s - c) (c) tan - = A / v - j-^ ^ - 2v $(s a) B C fb c\ A A (4) tan - - = ( j. - } cot - 2 \b + c) 2 These may be employed under the following conditions I. Given two sides and included angle: use either rule (2) to find the third side and then either rule (i) or rule (3) to find another angle ; or use rule (4) to find the remaining angles together with rule (2) for the third side. e. g., suppose b, c and A are given. Then from rule (2) the value of a can be found, , sin B sin A r , , , ., also - r = [from rule (i)l b a L B is found and C = i8o (A-f-B), since A+B+C = 180; or alternatively B-C fb c\ . A tan - - = ( j - ) cot - 2 \b + cj 2 T> _ /-> .*. the angle - is found, and hence also (B C). But (B+C), i. e., 180 A is known, and therefore B and C are found by solving the simultaneous equations. Also a can be found from rule (2). PLANE TRIGONOMETRY 257 II. Given two angles and a side, say a, A and B Then C = i8o-(A+B) From rule (i) c a b a and sin C sin A' sin B sin A and therefore all the sides are found. III. Given two sides and an angle not included by them, say b, c and B , , , sinC sinB From rule (i) = j C /. C is found, and also A. {For A = 180 (B+C)} a b . , , and since - T- = - a is found, sin A sm B IV. Given the three sides : it is more convenient in this case to use rule (3) to find one of the angles ; because logarithms can be applied. From Rule (3) c tan - tan - s(s a) Then use Rule (i) to find B. Otherwise a? = & 2 + c 2 2bc cos A .*. cos A = -L i" e -> A is found and thence by the sine rule B may be found. Thus if rules (i) and (2) are remembered, any triangle may be solved. Proof of the "Sine" Rule. Consider Figs. 132 and 133. In both figures j = sin B t p = c sin B also in Fig. 132 = & sin C and in Fig. 133 p = bsm (i8o-C) = Hence c sin B = & sin C c & " sinC ~ sin B s 258 MATHEMATICS FOR ENGINEERS Similarly it could be proved that b c a B D a Fig. 132. c B a c Fig. 133- Or, the sides of a triangle are proportional to the sines of the opposite angles. Proof of the Cosine Rule. In Figs. 135 and 136 let BD be perpendicular to AC. In Fig. 135 in the triangle ADB = p 2 +b*+n*-2bn ....... (i) and in the triangle BDC />+n 2 = a 2 .............. (2) B 6 Fig. 134- Fig- 135. Hence, by substitution from (2) into (i) Fig. 136. Again in Fig. 136, in the triangle ADB C 2 =^) 2 = 2 and in the triangle BDC (3) (4) (5) PLANE TRIGONOMETRY 259 Hence, by substitution from (5) into (4) c 2 = a 2 +b 2 +2bn (6) Now in Fig. 135 - = cos C or = a ccs C a so that, writing a cos C in place of n in (3) Also in Fig. 136 Tii - = cos L BCD = cos (180 C) = cos C or n = a cos C. Substituting this value for n in (6) c 2 = a z +b z 2ab cosC. We have thus proved that the rule holds for the case in which C is an acute angle, and also for the case in which C is obtuse. When C is a right angle, as in Fig. 134, its cosine is zero and accord- ingly it is correct to write _ = a 2 -\-b 2 2ab cosC. Hence the rule is perfectly general. The two other forms of the cosine rule can be written down by writing the letters one on in the sequence a, b, c, a. i, e , t a 2 = b z +c z 2bc cos A and 6 2 = c 2 + 2 2ca cos B. By transposition c sC= -- : S :f cosB = 2ca J,2_ cos A = 2bc the forms in which the rule must be used if the three sides are given and the angles are required. In every case of a solution of a triangle the figure should be drawn to scale, for this serves as the best check on the results obtained by calculation. The following examples should be carefully studied- Examples on the use of the Sine Rule. Example 14. Solve the A ABC completely when c - 1916 ft.. 6 = 1748 ft., and C = 59. [This triangle is drawn to scale in Fig. 137.] 260 MATHEMATICS FOR ENGINEERS sin B sin C To find B u c b sinC 1748 x sin 59 and hence sm B = - = - J 1016 Taking logs throughout log sin B log 1748 + log sin 59 log 1916 = 3-2425 + 1-9331 - 3-2823 = ^'8933 = log sin 51 28' B = 5 i28' Then A = 180 - (59 + 51 280 Fig. 137- Fig. 138. To find sin A sin B b sin A a = -. =- sinB In the log form log a = log 1748 + log sin 69 32' log sin 51 28' = 3-2425 + 1-9717 ~ 1-8933 = 3'3 2 9 a = 2093* Example 15. Solve the A ABC completely when a = 12-6*. b = 17-8", A = 40. (This is similar to the last Example up to a certain point.) To draw this to scale (see Fig. 138). Make the angle 40 with a horizontal line and along AC mark off a length to represent 17-8*; this is the side b. With centre C and radius = 12-6* (to scale) strike an arc to cut the horizontal ; and two points of section being found, call them B and B'. Both the A ABC and the A AB'C satisfy the given conditions, because AC = b = 17-8, CB = CB' = a = 12-6 and A = 40, so that in this case there are too solutions. This case is known as the " ambiguous " case in the solution of triangles. Since CB = CB', L CBB' = L CB'B L CB'A = 1 80- L CBA or B'=i8o-B PLANE TRIGONOMETRY 261 and the two values of the angle B, which are indicated on the figure, are supplementary, i. e., together they add to 180. AB and AB' are the two different lengths for c for the different cases, while ACB and ACB' give the two values for the angle at C. To solve by calculation. Two sides and one opposite angle are given, and therefore the sine rule is to be used. Taking the same diagram To find B sin B = &sinA = 1 7'8 sin 40 a 12-6 In the log form log sin B = log 17-8 + log sin 40 log sin 12-6 = 1-2504 + i -8081 1-1004 = 1-9581 = log sin 65 13' B = 6 5 i3' The value of B', which is alternative to B must be 180 B = 114 47'. The mode of calculation would be unchanged, for sin 114 47' = sin 65 13'. To find C In the first case C = i8o-(4O+ 65 130 = 74 47' In the second case C = i8o-(4O+ 114 47') = 25 13' To find c. This is the base, which is AB' or AB. Either the sine or the cosine rule can be here used, but the sine rule is more adapted for logarithmic computation. _ a sin C sin A In the first case log c = log 12-6 +_log sin 74 47' - log sin 40 = 1-1004 + 1-9845 - i -8081 = 1-2768 c = 18-91* In the second case log c - log 12-6 + log sin 25 13' - log sin 40 = 1-1004 + 1-6295 i -8081 = -9218 c = 8-352 Grouping the results B = 6.si3' or 114 47'. C=7447' or 2SiV. c=i8-gi" or 8-352" all respectively. The sine scale on the slide rule could be used with advantage in this example. To multiply or divide by sines of angles, multiply 262 MATHEMATICS FOR ENGINEERS or divide by the angles, as marked on the scale, in the ordinary way. E. g. _ 12-6 x sin 74 4 7* sin 40 Set the cursor over 12-6 on the A scale, move the sine scale until 40 is level with the cursor; then place the cursor over 74 47' on the S scale. The value of c is read off on the A scale, and = 18-9. A little confusion may arise regarding the graduations on the S and T scales. The markings usually shown are not for decimals of a degree, but for minutes. As regards the S scale : up to 10, a line is shown at every 5', *. e., there are 12 divisions for each degree. From 10 to 20 every 10' is shown, from 20 to 40 every 30', from 40 to 70 each degree, and thence 70, 72, 74, 76, 78, Fig. 139. Solutions of Triangles. 80, 85 and 90. On the T scale, up to 20, markings are at each 5' and then at every 10'. Whenever two sides and an opposite angle are given, we must consider the possibility of the two solutions. Ths drawing to scale is an excellent test, for the arc B'B in Fig. 138 must either cut or touch the base if the triangle is to be possible. The various cases that arise are illustrated in Fig. 139 : in which the sides a and b, and the angle A are given. Drawing a horizontal line of unlimited length to serve as a base, the angle A can be set out and the point C fixed, since the length of AC is given. Then an arc of radius equal to b is described from the centre C. If b is very small, the arc does not cut the base and case (i) arises ; there being no triangle to satisfy the conditions. If the radius of the circle, i. e., the length of the side b, is increased, we arrive at case (2), in which the arc just touches the base and so gives one PLANE TRIGONOMETRY 263 triangle only, viz. the right-angled triangle ACB 2 . By further increasing the length of b cases (3), (4) and (5) are found, in which there are two, one, and one, solutions respectively. It will thus be seen that there may be two solutions if two sides of a triangle and an angle opposite the shorter of these is given. In all cases, however, the triangle should be drawn to scale before any trigonometrical rules are applied. Example 16. A mill chimney stands on the even slope of a hill, which has a gradient of 4 (Fig. 140). Two points are chosen on the same side of the hill and in the same vertical plane as that including the chimney. These points are 75 ft. apart measured //3/i up the slope, and, viewed from the points, the chimney subtends angles of 48 and 59 from the horizontal. Find the height of the chimney above the ground on which it stands. It should be noted that the angles of elevation are measured from the horizontal, since the scale of the theodolite vertical circle reads zero when the tele- scope is horizontal. Hence L ABC = 4 8-4 = 44 and L ACD -55 Thus /.ACB=i8o-55 = i25 , ABAC=n /.ADC =94, A CAD = 31 Here we have two triangles, viz. ACB and ACD, one containing the known length and one containing the unknown length ; and these must be connected up through a side common to b >th. viz. AU Let the required height AD = h Then, in the A ACB sn 44 sum In the A ACD _ 75 sm 44_ sin 11 C AC 55 sin 94 sin 86 AC x sin 5j since sin 86 = sin 94 264 MATHEMATICS FOR ENGINEERS Substituting for AC its value = 75 x sin 44 x sin 55^ sin 11 x sin 86 = 225 ft. (from the slide rule). Example 17. The elevation of the top P of a mountain (see Fig. 141) at a point A on the ground is 32. The surveying instrument is directed to another station B, also on the ground, and 4600 ft. distant from A, the angle PAB being found to be 48 ; also L PBA is 77. Find the height of the mountain. The sloping triangle PAB is shown laid flat on the ground in Fig. 142. From this ground plan PA 4600 sin 77 ~ sin 55 pA = 4600 x sin 77 sin 55 P<z/?spec//W vieuf "^ t i or survey lints. A In the right-angled A PAQ and PQ = AP x sin 32 Substituting for AP _ 4600 x sin 77 x sin 32 sin 55 Height of mountain = 2900 ft. Example 18. It is required to lay out a circular arc to connect the two straight roads AB and CD (Fig. 143) : the radius r of the arc is known, but the meeting point E of AB and CD is inaccessible. Select two convenient stations F and G, and by directing a theodolite first along Fe and then along FG the angle EFG is measured. Similarly measure 265 Let the sum of L EFG and L EFG Then /_AED = i8o- and L AEO = L. EOS = a za = go-a Now ES -Qg = tan a, since L ESO is a right angle ES = OS tana = r tana ( x ) and also ET = r tan a, since ET = ES . . . . (i) To find FS and GT, EF and EG must first be found. In the A EFG _EF sin FG or EF = FG sin EGF sin FEG ~ sin FEG EF is known (2) Also, in the same way EG = FG i in _ EFG . sirTFEG EG is known (3) Finally, FS is found from (i) and (2) since FS = ES EF, and GT is also found from (i) and (3) since GT = EG - ET. Thus the points F and G having been taken at random, S and T can now be plotted therefrom, which show the starting-points of the curved road. Examples on the use of Cosine Rule. Example 19. In the triangle ABC (Fig. 144) find L C when a = 4-45', b = 7-85', and c = 11-94'. C= 11-94- Fig. 144. The longest side is always opposite the largest angle ; and therefore C is the largest angle. 266 MATHEMATICS FOR ENGINEERS C - V^ Now I- fe 2 ~ c 2 = (4-45) 2 +(7-85) 2 -(n-94) 2 2ab 2 X 4-45 X 7-85 19-8 + 61-5 142-5 69-8 61-2 ^ = "o'y^ OQ'o = cos 28 43' = cos (180 28 43') = cos 151 17' /. C = 151 i7 / . It will be seen that a negative value for the cosine implies that the angle is obtuse. To avoid remembering too many rules the reader is advised to work entirely with the sine or cosine rules : this example, how- ever, is worked out, in addition to the above, by another rule, to demonstrate its usefulness and ease of application. a = 4-45, 6 = 7-85, c = 1 1 -94 S = 12-12 C _ /(s-b)(s~a) _ / 4 -2 7 x 7-67 ^ 2 ~ V S (S - C) ~ V 12-12 X -18 .-. log tan - = J ^{(-6304 +-8848) 4-27 + log 7-67) - (log 12-12 + log -18)} (1-0835 + 1-2553)} = -5882 = log tan 75 32' and C = 151 4' =75 32' i. e., an error of 13' was made when using the slide rule. [Note that if this rule is used and the angle is required correct to the nearest minute, we must work throughout correct to a half-minute since the rule gives as the direct result the value of a half-angle.] Example 20. If in the triangle ABC find the side b (Fig. 145). Using the cosine rule b 2 = a 2 +c 2 2ac cos B = (5'93) 2 +(2-94) 2 (2 x 5 -93 x 2 -94 x cos65) = 35-1+8-62 -(2X5-93X2-94X-4226) = 5-39' = 5-93". c = 2-94", B = 65, a. = 5- 93" Fig. MS- PLANE TRIGONOMETRY 267 Example 21. Two forces, of 47-2 Ibs. and 98-4 Ibs. respectively, making an angle of 63 with one another, act on a small body at A'. Find the magnitude of their resultant, or single equivalent force. If AB and AD in Fig. 146 represent the given forces, AC represents their resultant, as shown in Mechanics. Then XI 7 Fig. 146. CD = 47-2, AD = 98-4, t-ADC = i8o-6 3 < (AC) 2 = (AD) 2 +(DC) 2 -(2xADxDCxcos 117) = (98-4)*+ (47'2) a - (2 x 98-4 x 47'2 x 4540) = 9670+2230+4220 = 16120 AC = 127 Ibs. and this is the resultant. Area of a Triangle. The following rule gives the area when two sides and the included angle are given ; it is simply an extension of the \ base X height rule, for AD = AB sin B or AC sin C (Fig. 147) = c sin B or b sin C .*. Area = \ X base X height = |XXcsinB or xaX&sinC = \ac sin B or \ ab sin C or, generally, area of triangle = -J- product of two sides x sine of included angle. AA Fig. 147- Fig. 148. This gives a rule for the area of a parallelogram Area of ABCD = 2 {area AOB + area AOD} (Fig. 148) = 2{ AO.OB sin L AOB + \ AO.OD sin L AOD} 268 MATHEMATICS FOR ENGINEERS = sin L AOB {AO.OB + AO.OD} since sin L AOB = sin (180 AOB) = sin L AOD = sin L AOB x AO {OB + OD} = AO.BD sin L AOB = \ AC.BD sin L AOB = \ product of diagonals X sine of angle included between them. Example 22. Find the area of AABC in which a = 5-93*, c = 2-94" and B = 65. Area = ac sin B = \ x 5-93 x 2-94 x sin 65 = i x 5-93 x 2-94 x -9063 = 7*91 sq. ins. This result should agree with that found by the " s" rule given in Chapter III ; it being possible to apply this rule since the three sides are known and are 5-93, 5-39 and 2-94 respectively (compare Example 20). Thus- 5= 5-93 + 5-39 + 2-94 = ^ and area = ^7-13 x 1-20 x 1-74 x 4-19 = 7-91 sq. ins. Proof of the "s" Rule for the Area of a Triangle. It has been demonstrated in the previous paragraph that Area of triangle = ab sin C. Now for any angle it is true that (sine) 2 + (cosine) 2 = i : hence sin 2 C + cos 2 C = i, or sin C = i cos 2 C a 2 +6 2 -c 2 Also cos C = - j 2ab 2r (a 2 then cos 2 C: zr and i cos 2 C = J '- [Factorising difference of two squares] _ (2ab a 2 6 2 +c 2 )(2a6+a 2 +ft 2 c 2 ) PLANE TRIGONOMETRY 269 [Factorising difference of two squares] = (c-<*+b)(c+a-b)(a+b-c)(a+b+c) = 2(s a) x 2(s b) x 2(s-c) x 2s i. e., sin C = ~y' S (s a)(s-b)(s c) :. area of triangle ABC = $ab x -, X Vs(s a)(s b)(s c) = Vs(sa)(sb)(sc) In all of these worked examples the results have been given to as great a degree of accuracy as four-figure log tables or the slide rule allow. When extremely careful observations have been made it is advisable to employ five- or even seven-figure log tables in any necessary calculations ; but it should be remembered that the results must not be given to a greater degree of accuracy than the observa- tions or measurements warrant. Thus it would be useless to express a length " correct " to eight figures when the least possible error in measurement was %. The rules used in such cases are those stated in this chapter, except that the cosine rule is put into a form more adapted for logarithmic computation by means of the following artifice a z = 6 2 +c 2 2bc cos A In place of this rule we may write a = (b+c} cose ......... (i) provided that is found from zVbc A , v sm * = T+^ COS 2 ......... ( Both (i) and (2) can be solved by the aid of logs ; and the angle 6 thus introduced is known as a subsidiary angle. Let us illustrate this by taking the figures of Example 20. Given a = 5-93, c 2-94, B = 65. To find b From the above b = (c+a) cos 6 zVca B and sin 6 = - cos - j. e., sin 2^2-94 X 5 '93 cos ,2 1 = -- . 5 s 3 2 i 270 MATHEMATICS FOR ENGINEERS In the log form log sin 6 = log 2 + {log 2-94 + log 5-93} + log cos 32 \- log 8-87 = 1-8998 = log sin 52 33'' /. = 52 33' Then b = 8-87 x cos 52 33' In the log form log b = log 8-87 + log cos 52 33' = -7318 Exercises 30. On the Solution of Triangles. In Exs. i to 14 solve the triangle ABC completely, being given that 1. a=3*,6 = 5-2", B= 7 8i. 2. a=79-5",C = 5 i 32',B = 4 7 3 6'. 3. C = 26 50', 6 = 8-86", c = 5-68". 4. 6 = 5-97", C = 6 4 18', A = 75. 5. c = 9-2, a = 10-31, = 46. 6. 6 = 6-ift,,c = 9-3ft.,A = 73i6'. 7. a =124-4, 6 = 93-7, c = 99-3. 8. a = 13-7*. 6= 10-5*, 0=130. 9. a = 4-27", A = 29, b = 5-86*. 10. c = 688o, B = 30, b = 5141. \D cross-section o/'u/'V'e Fig. 149. Solution of Triangles. 11. A = 50 50', 6 = 922-4, c = 1003-8. 12, B = 353o', 6 = 38-6, c= 4 3-57. 13. = 21-8,6= 15-7,0 = 47 32'. 14. c = 32-7, 6 = 39-4, B = 55 30'. Find also the area. 15. The area of a triangle is 120 sq. ft. and the angles are 75, 60 and 45. Find the longest side. 16. The connecting rod of an engine is 8 ft. in length and the crank i '-6*. Find the inclination of the connecting rod to the line of stroke when the crank has moved 52 from its inner dead centre position. 17. The sides of a " triangle of forces " represent the forces 3-7 tons, 2-275 tons an d 3' 02 5 tons respectively. Find the angles of this triangle. 18. Forces of 21-6 and 19-7 Ibs., making an angle of 126 with one another act at a point. Find the magnitude of their resultant and its inclination to the larger force. 19. In setting out a railway curve to connect the lines AO and OD ((a) Fig. 149), a line CB was measured and found to be 1*4 74 chains. PLANE TRIGONOMETRY 2?I points are E and F find the lengths of BE and CF 20. The diagram (6) Fig. 149, is necessary for the calculation of lag by the 3-voltmeter method. If <p is the angle of lag, find Ss value for the case illustrated. {V 8 = 107, V x = 90, V, = 48 } 21. The jib of a crane is inclined at 57 to the horizontal the kes ' 23. The tangents to a curve meet at 120. On the bisector of this angle is a point 100 ft. distant from the point of meeting of the tangents and through which the curve must pass. Find the radius of the' required curve and also the tangent distances. 24. It is required to find the height of a house on the opposite bank of a river. The elevation of the top of the house is read at a certain point as 17; approaching 86 ft. nearer to the bank, towards the house, the elevation is found to be 31. Find the height of the house. 25. A theodolite is set up at two stations A and B at the water's edge of a lake which is 1240 ft. above sea-level. A staff on a hill at C is sighted from each station. From A the elevation of C is 15 14' and the horizontal angles CAB and CBA are 59 10' and 71 48' respectively. If AB = 820 yds., find the height of C above sea-level. 26. From a station C on a hill, two stations A and B, on opposite sides of the hill are observed. The horizontal projection of L ACB is 43 23', the horizontal projection of CA is 3633 links and of CB is 4275 links. The angle of elevation of C at A is 44 37' and at B is 33 24'. Determine the horizontal distance between A and B and the difference of level between them. 27. It is required to set out a curve of mile radius between two straight portions of a railway, AB and DC, which intersect in an in- accessible point E. Rods are set up at points B and C on the two straight portions and the angles ABC and BCD are measured and found to be 110 20' and 120 30' respectively. If BC = 830 links, determine the distances of the tangent points G and H from B and C respectively. 28. In a theodolite survey to find the positions of two visible but inaccessible points B and C, the following measurements were made AD = 517-75 links, L BAC = 70 44' 10", L BAD = 108 9', L ADB = 36 i8'3o", and /_ADC= ioiiS'3o". Find the lengths of AB, DC, AC and BC in order. 29. When setting out the centre line for a tunnel between the two ends A and B, an observatory station C is chosen on the top of a hill from which both A and B are visible, but it is not on the centre line of the tunnel. Let D be a point on a vertical through C. The horizontal projection of L ACB = 45 58', the vertical angle ACD = 4945' and the vertical angle BCD = 5 7 42'. The horizontal projection of CA is 750 yards, and of CB is 800 yards. Find the horizontal distance between A and B and the difference of level. 30. A light railway is to be carried round the shoulder of a hill, and 272 MATHEMATICS FOR ENGINEERS its centre line is to be tangential to each of the three lines AB, BC and CD as follows Line. Bearing. Length. AB E. 30 N. BC E 600 feet CD S - Calculate the radius of the curve and the lengths required for setting out the tangent points. [Note. E. 30 N. means 30 north of east.] 31. In taking soundings from a boat the position is fixed by observations taken to three stations A, B and C on the shore. The lines AB and BC have been measured by the following traverse : A to B, 542 ft., bearing 70 14'; B to C, 714 ft., bearing iio33'. From the boat in a certain position P, the angles APB and BPC were read as 32 16' and 44 21' respectively. Calculate the distances AP, BP and CP. 32. The speed of the blades of a turbine is 600 ft. per sec., the velocity of the steam at entrance to the wheel is 1780 ft. per sec., and the nozzle is inclined at 20 to the blades. Find the relative velocity of the steam at discharge, and the inclination of the direction of this velocity to the line of motion of the blades. 33. Find the diameter of the wire, whose section is shown in (c) Fig. 149, in terms of the pitch p of the V-threaded screw. This wire is used as a gauge to test the accuracy of the form of the thread. Further Mensuration Examples. 34. A circular arch has a rise of 20 ft. and a span of 80 ft. Find the angle at the centre of the circle which is subtended by this arc, and also the length of the curved portion of the arch. 35. A wooden core, having as section an equilateral triangle, is placed in the tubes (internal diameter f *) of a surface condenser. Find the ratio of the tube surface to the water-carrying section. 36. A roof is in the form of the surface of a segment of a sphere of 6 ft. radius. The tangents at the eaves make 48 with the hori- zontal. Find the area of the roof surface, and the weight of sheet lead required to cover it at 7 Ibs. per sq. ft. 37. Find the diagonals of a rhombus in which one side is 6-5* and one angle is 70. 38. A quadrilateral has two adjacent sides equal and containing a right angle. The other pair of sides are equal and contain 60. The area is i sq. ft. Find the lengths of the sides. 39. A quadrilateral has two adjacent angles each 120. The side between them is 24 ft., and the perpendiculars on this side from the other angular points are 7 ft. and 10 ft. respectively. Find the area of the quadrilateral. 40. A trapezoid has its parallel sides 82" and 38* and two of its angles each 60. Find its area and the area of the triangle obtained by producing the non -parallel sides. 41. A quadrilateral with two opposite angles right angles and one of the remaining angles 60 is described about a circle of 2" radius. Find its area. 1 tan A tan B tan A tan B PLANE TRIGONOMETRY 273 The Addition Formulae. It is sometimes necessary, more particularly in electrical work, to express the ratio of a compound angle in terms of the ratios of the simpler angles, or vice versa; e. g., it might be easier to state tan (A+B) in terms of tan A and tan B, and then evaluate, than to evaluate directly. The following rules must be committed to memory for this purpose sin (A + B) = sin A cos B + cos A sin B sin (A B) = sin A cos B cos A sin B cos (A + B) = cos A cos B sin A sin B cos (A B) = cos A cos B + sin A sin B tan (A + B) tan (A B) . 1 + tan A tan B Considering sin (A+B) one might be tempted at a first glance to apply the ordinary rules of brackets, and write the expansion as sin A+sin B. That this is not correct may be readily seen by referring to any angles. e. g., suppose A = 46, and B = 15, then (A+B) = 61 *. e., sin (A+B) = sin 61 = -8746 whereas sin A+sin B = sin 46+sin 15 = 7193 +-2588 = -9781 and -9781 does not equal -8746. It will be observed, however, that the above rule holds, at any rate for these particular values of A and B. Thus sin 46 cos 15 + cos 46 sin 15 = (7193 X -9659) + ( >6 947 X -2588) = -8745 = sin 61 = sin (46+i5). A more general proof is necessary to establish the truth of these rules for all angles ; and the proofs are here given for the simplest cases only. To prove that sin (A+B) = sin A cosB + cos A sinB. Taking the simplest case, when A and B are both acute In Fig. 150 let L. PQR = A, and L RQM = L PQM = (A + B) ; also let QR be perpendicular to PM. Then APQM = APQR + AQRM .'. iPQ QM sin (A + B) = iPQ . QR . sin A + *QR . QM . sin B. T 274 MATHEMATICS FOR ENGINEERS Dividing through by PQ . QM sin (A+B) = ^ sin A + |^ sin B = cos B sin A + cos A sin B or sin A cos B -f- cos A sin B. Fig. 150. Fig. 151. To prove that cos (A + B) cos A cos B sin A sin B. In Fig. 151 let L MOQ = A, and L QOP = B _ PQO = right angle. Drop QN perpendicular to RP, RP being perpendicular to OM. Then L OQN = A, L NQP = 90 A, and therefore L QPN = A. TVT /A 1 TN TD^T, OR OM MR Now cos (A+B) = cos L ROP OP OP \JXi IX \J r . XT .~ uTT-n = T&-r [ smce N Q = MR 1 OM_NQ OP OP OM OQ_NQ QP = OQ'OP QP'OP = cos A cos B sin A sin B. To prove that tan ton ^ + ton B - i tan A tanB Assume the rules for sin (A+B) and cos (A+B). Then- tan(A+B)= sin i A +g 1 cos (A+B) [Dividing both numerator and denominator by cos A cos B.] sin A cos B + cos A sin B cos A cos B sin A sin B sin A , sinB cos A cos B sin A sin B _ cos A ' cos B _ tan A + tan B ~ i tan A tanB PLANE TRIGONOMETRY 275 23 .-Verify the rules for sin (A - B), cos (A + B) and tan (A - B) for the case when A = 164 and B = 29. sin (A - B) = sin (164 - 29) = sin 135 = sin 45 = -7<>7- Also sin A cos B - cos A sin B = sin 164 cos 29 - cos 164 sin 29 [cos 164 = - cos 16] = sin 16 cos 29 + cos 16 sin 29 = (-2756 x -8746) + (-9613 x -4848) = -241 + -465 = -706. For brevity we shall denote the side containing (A + B) by L.H.S. (left-hand side) ; the other by R.H.S. (right-hand side). /. L.H.S. = R.H.S. For cos (A + B) L.H.S. = cos (A + B) = cos (164 + 29) = cos 193 = - cos 13 = - -9744 R.H.S. = cos A cos B sin A sin B = cos 164 cos 29 sin 164 sin 29 = cos 16 cos 29 sin 16 sin 29 = (- -9613 x -8746) - (-2756 x -4848) = 841133 = 974 .*. L.H.S. = R.H.S. For tan (A-B) L.H.S. = tan (164- 29) = tan 135 = -tan 45 = -i R TT s tan 164 tan 29 _ - tan 16 tan 29 i + tan 164 tan 29 ~ i - tan 16 tan 29 2867 --5543 " i- (-2867 x -5543) = ^4L = 841 = t 1159 -841 /. L.H.S. = R.H.S. Example 24. Find the value of cos (A + B) when sin A = -5, cos B = -23. (Tables are not to be used.) Before proceeding with this example, a little preliminary investigation is necessary. In the right-angled triangle ABC (Fig. 152) b* , fl 2 Cos 2 A + sin 2 A = 1, since cos A = - and sin A = - 276 MATHEMATICS FOR ENGINEERS This is a most important relation between these ratios ; and it holds for every value given to the angle A. Two other rules obtained by similar methods are sec 2 A = 1 + tan 2 A cosec 2 A = 1 + cot 2 A. JCL Fig. 152. Fig- 153. Returning to Example 24 : cos (A + B) = cos A cos B sin A sin B. Values must first be found for cos A and sin B. Now cos* A + sin 2 A = i from which cos 2 A = i sin 2 A, or sin 2 A = i cos 2 A or cos A = Vi sin 2 A, sin A = Vi cos 2 A Then cos A = Vi - (-s) 2 = V^j$ = -866 and sin B = Vi (-23) 2 = ^-947 = -973 .*. cos (A + B) = cos A cos B sin A sin B = (-866 x -23)- (-5 x -973) = -1991 4865 2874. It is often necessary to change the binomial or two-term expres- sion a sin qt-\-b cos qt into an expression of the form M sin (qt + c) , where c is an angle. We must therefore find the values of M and c in terms of a and b, so that M sin (qt-\-c) = a sin qt + b cos qt (i) Take the addition formula, viz. sin (A+B) = sin A cos B + cos A sin B. Replacing A by qt and B by c, this statement becomes sin (qt-\-c) = sin qt cos c + cos qt sin c. Multiplying through by M M sin (qt+c) = M sin qt cos c + M cos qt sin c . . . . (2) PLANE TRIGONOMETRY 2?7 Since the right-hand sides of (i) and (2) are equal in total value, they can be made equal term for term by choosing suitable values for the coefficients. Thus M sin qt cos c = a sin qt and M cos qt sin c = b cos qt .'. M cos c = a and M sin c = b a, b i.e., cosc = ^ andsinc = v? If, now, a triangle be drawn (Fig. 153) with sides a, b and hypotenuse M, it will be seen that the angle opposite the side b is the angle c, for its adjacent side is a, and therefore its cosine = ^ M Hence c is found, for tan c = - a Knowing the values of b and a, the value of c is read off from the table of tangents and is usually expressed in radians. Also M 2 = 2 -f-6 a M = V a + b*, so that M is found. This investigation is valuable in cases of harmonic motion. Example 25. The voltage necessary to produce an alternating current C, after any particular period of time /, and in a circuit of resistance 2 ohms in which the current varies, being given by C = 100 sin 6oot, can be expressed as V = 200 sin 6oot + 300 cos foot Find a simpler expression for V. Let 200 sin 6oo/ + 300 cos 6oo/ = M sin (6oo/ + c). Then by the previous work M = V2OO 2 + 300* = 360-6 and tan c = |^ = 1-5 = tan 56-3 c 56-3 = ^ radians = -983 radian / J V = 360-6 sin (6ool + '983) or, as it might be written, V = 360-6 sin 600 (/ + -00164). Note. If the current were continuous, then voltage = current x resistance = 200 sin 6oo/. When the current is interrupted, "inertia" or "induction" effects set up another current to oppose that due to the impressed voltage, and therefore the amperes are not a maximum when the voltage is, 278 MATHEMATICS FOR ENGINEERS i. e., the current lags behind the E.M.F. ; in this case to the extent of -00164 second. The coefficient 600 in the formulae = zirf where /= frequency: thus in this case the frequency = =95*6 cycles per second. As a further example of transformation consider the following case : Example 26 Let S w = displacement of the main steam valve of an engine from"\ its central position S = displacement of the expansion plate from its central J position J for the case of an engine with Meyer valve gear. Then Sm = r m cos (6 + a x ), and S e = r e cos (6 + a,). To find a simple expression for the displacement of the expansion plate relative to the main valve. This relative displacement = S m S e = r m cos (d + a t ) r e cos (6 + a,). Then Sm-S e = r m cos (6 + a x ) - r e cos (6 + a s ) = r TO cos tfcosaj r m sin Osma^ r e cos 6 cos a t + r e sin tfsina, = cos 6(r m cos G! r e cos a,) + sin 6(r e sin a, r m sin aj) = A cos 6 + B sin 6 = VA* + B 2 sin (6 + c) as before proved = VA + B* cos (|-(0 + c)\ VA + B* cos (6 + p) where A = r m cos Oi r e cos a,, J5 = r e sin a, r m sin aj and #> = c~ = tan- 1 g | -{tan- 1 ^ is the awg/e whose tan is ^ j We have thus reduced the expression for the relative displacement to a form of a simple character which shows that this displacement is equivalent to that caused by an imaginary eccentric of radius VA 2 + B 8 and of angular advance p. Exercises 31. On the Addition Formulae in Trigonometry. 1. If sin A = -45, find cos A and tan A (without reference to the tables). 2. If sin B = *i6, cos A = -29, find the value of sin A cos B cos A sin B. 3. Find the values of cos (A + B), and sin (A B), when sin A = -65, sin B = -394. PLANE TRIGONOMETRY 279 4. Tan A == 1-62, tan B = -58; find the values of tan (A + B) and tan (A B). 5. The horizontal force P necessary to just move a weight W down a rough plane inclined at a to the horizontal, the coefficient of friction between the plane and the weight being /*, can be obtained from the formula P ^ = tan (* - a) P If tan <(> = ft, find an expression for ^ in terms of p. and tan a. Hence find the value of P when W = 48, /i = -21, and a = 8. 6. The effort P required to raise a load W by means of a screw, of pitch p and radius r, is given by P = W tan (<t> + a) where a =. angle of screw and tan <f> p = coefficient of friction. Find an expression for P in terms of W, p, r and /*. 7. If ^Tr = , and tan A = u, find a simple expression for P. W cos <t> 8. Given that tan (A B) = -537 and tan B = -388, find tan A. 9. Express 4*2 cos 5^ + 2-7 sin 5^ in the form M sin (5* + c). 10. Express 200 sin 50* 130 cos 50^ in the form M sin (50* + c). 11. If a bullet be projected from a point on ground sloping at an angle A to the horizontal, the elevation being 6 to the incline, the range R is given by the formula R = ut \gt z sin A. Find a simpler 2V sin Q expression for R, if u = V cos 6 and t = g A 12. The efficiency of a screw jack = j^-^ + * where 6 is the angle of the screw and <f> is the angle of friction. In a certain experi- ment the efficiency was found to be -3, and by measurement of the pitch and the mean circumference of the screw tan 6 was calculated as -083. Find tan<f>, which is the coefficient of friction between the screw and nut, and thence find $>. 13. If the E.M.F. in an inductive circuit is given by E = RI Sin 2irft + 2r/LI COS 2irft find a simpler expression for E, i. e., one having the form M sin (a*/* + c), when R = 4-6, / = 60, L = -02 and I = 13-8. Formula for the Ratios of the Multiple and Sub-multiple Angles. In the addition formulae let B be replaced by A; by so doing, expressions may be found for the ratios of 2A. Thus sin (A+B) = sin A cos B + cos A sin B sin (A+ A) = sin A cos A + cos A sin A or sin 2A = 2 sin A cos A. Also cos (A+B) = cos A cos B - sin A sin B cos (A+ A) = cos A cos A - sin A sm A or cos 2A = cos 2 A - sin 2 A. 280 MATHEMATICS FOR ENGINEERS If for cos 2 A we write i sin 2 A, which is permissible since cos 2 A -f- sin 2 A = i, then cos 2 A = i sin 2 A sin 2 A = i 2 sin 2 A. Also cos 2A = cos 2 A (i cos 2 A) = 2 cos 2 A i. ,. tan A + tanB Again- tan (A+B) == I _ . . . . tan A + tan A tan (A+A) == ^ 2 tan A i tan 2 A Grouping the results sin 2 A = 2 sin A cos A cos 2A = cos 2 A sin 2 Al = 2cos 2 A- 1 = l-2sin 2 A J 2 tan A tan 2A = 1 - tan 2 A If the ratios of the half-angles are required they can be obtained " from the foregoing by dividing all the angles by 2. E. g., cos A = cos 2 sin 2 2 A = 2 COS 2 1 2 2 A = i 2 sin 2 - 2 A - A A sin A = 2 sin cos 2 2 A 2 tan tan A = i-tan 2 - 2 Similarly, by multiplying all the angles by 2, expressions can be found for the ratios of the angle 4A e. g., sin 4A = 2 sin 2A cos 2A and this expansion can be further developed if necessary. Formulae for ratios of 3A can be obtained by writing 2A in place of B in the (A+B) formulae, and using the rules for the ratios of 2A PLANE TRIGONOMETRY 28l E. g., sin 3A = sin (aA+A) = sin 2A cos A + cos 2 A sin A = 2 sin A cos 2 A + (i_ 2 sin 2 A) sin A = 2 sin A cos 2 A + sin A - 2 sin 3 A = 2 sin A (i- S i n 2 A) + sin A - 2 sin* A = 3 sm A 4 sin 8 A. In like manner cos 3A = 4 cos 3 A - 3 cos A tan 3 A = 3tanA-tanA i -3 tan 2 A for For cos 2A L.H.S. = cos 2A = cos 48 = -669 R.H.S. = cos a A - sin 2 A = cos* 24 sin* 24 = (-9I35) 2 ~ (-4067)' = '835 ~ -165 = -670 .'. L.H.S. = R.H.S. For tan 2A L.H.S. = tan 2A = tan 48 = 1-1106 R H S = 2tan A - 2 tan 24 2 x -4452 i tan 2 A i- tan 2 24 i- (-4452) 2 89 = .855 - ri08 /. L.H.S. = R.H.S. (the small differences being due to slide-rule working). For sin 3A L.H.S. = sin 3 A = sin 72 = '9511. R.H.S. = 3 sin A 4 sin 3 A = 3 sin 24 4 sin* 24 = 3 X -4067 - = i -220 -269 A A A Example 28. If sin A = -85, find sin , cos and tan , without 22 2 the use of the tables. (Examples are set in this manner so that the reader may become familiar with the formulae and the method of using them ; but in practice the tables would be used.) 282 MATHEMATICS FOR ENGINEERS A A To find sin . The formula that contains sin , only, of the ratios 2 2 of the half-angles is . A cos A = i 2 sin 2 2 and, therefore, to use this, cos A must first be found. cos A = Vi sin 2 A = Vi (-85) 2 = -526. Then -526 = 12 sin 2 2sin 2 - = 1526 = -474 or sin 8 - = -237 .% sin = '487 {the positive root only being taken}. To find cos 2 . f or alternatively ~\ cos A = 2 cos 2 i 2 /. -526 = 2 cos 2 i 2 COS 2 = 1-526 cos 2 = -763 2 cos - = -875. To find tan - 2 . A .sin A 2 -48' tan T- 2 A cos 2 Example 29. Find the value of tan 2A if cos A = *g6. sin A = Vi cos 2 A = Vi - (-96) a = -2795 . sin A -2795 Then tan A = -r- = ^f 2 = -292 cos A '96 2 tan A 2 x -292 -584 and tan 2A = r 5-*- = ~ 7 ^rs = -^ - = '638. i tan 2 A i (-292) 2 -915 - > PLANE TRIGONOMETRY 283 Example 30. It was required to find, to an accuracy of -oooi', the dimension marked c in Fig. 154; the figure representing part of a gauge for the shape of a boring tool. There is a radius of -5* at the top of the sloping side, which is tangential to an arc of 3-4* radius at the bottom ; and other dimensions are as shown. B 329-KB Fig. 154. Gauge for Boring Tool. Introduce the three unknowns x, y and z as indicated on the figure ; y being the distance along the slant side from the point of contact with the arc to the base. Let the angle ACB = o, then L APT = | CB = 1-704 + x - 1-375 = '329 + In the triangle ACB tan a = ^j O -* * In the triangle APT tan | = ^ = *z (i) (2) 3.4 A From the properties of tana = ....... (3)! the circle. y 2 = (6-8 + *)* . . . . (4) ' (*- EDC * a ri g ht an 6 le ) also and Connecting (i) with (3) 329 + x y i. e., y* 1-36 (-329 + *) 1-8496 (-329 + Hence from (4) whence (6-8 + x]x = 1-8496 (-329 + *) 8496*' - 5-583* + '20021 - O 284 MATHEMATICS FOR ENGINEERS + 5*583 V3I-I699 -6804 so that x = J J - T? 1-6992 or the required value of x = - = -0361*. 1-6992 Now- y -329 + x so that y = -4965 also tan a = ^>- = 6-8482. 4965 It would be unwise to use the tables to find a from the previous equation, for in the neighbourhood of the required value the change in the value of the tangent is extremely rapid ; hence it is a good plan to make use of the rule for tan aA or its modification. 2 tan - 2 Thus tan a = - i-tan z - 2 i. e., 6-8482 = i for tan - = 2* from (2). i 42* 2 This is a quadratic in terms of z, and the solution applicable to this case is z = '4323. .*. C = 2-75 2 X '4323 = 1-8854*. Further Transpositions of the Addition Formulas sin (A+B) = sin A cosB+cosA sinB sin (A B) = sin A cos B cos A sin B Hence, by addition sin (A+B)+sin (A B) = 2 sin A cos B and by subtraction sin (A+B) sin (A B) = 2 cos A sinB Also cos (A+B) = cos A cos B sin A sin B cos (A B) = cos A cos B+sin A sin B .'. cos (A+B)+cos (A B) = 2 cos A cosB and cos (A B) cos (A+B) = 2 sin A sin B. [Note the change in the order on the left-hand side in this last formula.] Now A = ' ' '*\ L i.e.,= % sum of the two angles, and B = ( A +B)-(A-B) { e . > = $ difference of the two angles. 2 PLANE TRIGONOMETRY 28s Hence, the first of these formulae could be written Sine (one angle) + sine (another angle) = 2 sine (i their sum) x cos ( J their difference) A substitution is very often of great service ; thus, let (A+B) = C and (A-B) = D ' Then sin C + sin D = 2 sin S cos =? etc 2 2 and we have the summary If the change is to be made from a sum or difference to a product, use the (C+D) formulae sin C+sin D = 2 sin cos sin C-sin D = 2 cos sin . . ( 2 ) a 2 cos C+cos D = 2 cos -~- cos ^^ ...... (3) C+D C D cos D cos C = 2 sin ~ sin ^ ...... (4) If, however, the change to be made is from a product to a sum or difference, use the A and B formulae, which follow sinA cosB = {sin(A+B)+sin(A-B)} ..... (5) eosA sinB = J{sin(A+B)-sin(A-B)j ..... (6) cosA cosB = {cos(A+B)+eos(A-B)} ..... (7) sinA sinB = {cos (A-B) -cos (A+B)} ..... (8) In later work it will be found that certain operations can be performed on a sum or difference of two trigonometric ratios that cannot be done with products ; hence the great importance of this last set of formulae. It may appear to the reader that his memory will be severely taxed by the above long list of formulae, but a second thought will convince him that all are derived from the original (A + B) and (A B) formulae, which must be committed to memory to serve as the first principles from which all the later formulae are developed. Example 31. Express 17 sin 56 sin 148 as a sum or difference. sin 56 sin 148 = i {cos (148 - 56) - cos (148 + 56)} . . from (8) {A = 148, B = 56} /. 1 7 sin 56 sin 148 = 8-j, {cos 92 - cos 204). 286 MATHEMATICS FOR ENGINEERS To check by the use of tables L.H.S. = 1 7 sin 56 sin 1 48 = 17 sin 56 sin 32 = 17 x -8290 x '5299 = 7-47- R.H.S. = 8-5 {cos 92 cos 204} = 8-5 { - cos 88 + cos 24} = 8-5 {- -0349 + -9135} = 8-5 x -8786 = 7-47. /. L.H.S. = R.H.S. Example 32. The voltage V in an A.C. circuit, after a time /, is given by V = 200 sin 360*. and the current by C = 3-5 sin(36o/ + c). Find an expression for the watts at any time, expressing it as a sum or difference. Watts = amps x volts = 3'5 sin (360* + c) x 200 sin 360* = 700 sin (360^ + c) sin 360* = 222 {cos c cos (720* + c)} from (8) = 350 {cos c cos (7201 + c)}. Example 33. Express (4 sin 5*) (5 cos 3*) as a sum or difference. (4 sin 5/) (5 cos 3<) = 20 sin 5* cos 3* = 10 {sin 8t + sin 2t} . . . from (5) Exercises 32. On Transpositions of the Addition Formulae. 1. If sin aA = -824, find cos 2A and tan 2A. 2. If sin A = $, find sin 2A and cos 2A. 3. Express cos 2 14 in terms of cos 28. 4. Find an expression for sin 2B in terms of cos B alone. Hence find the value of sin 2B when cos B = -918. A A 5. If sin A = '317, find sin , cos and sin 3 A. * " 22 3 ^ 6. If sin 2 A = -438, find cos 4 A and tan 7. Change 5 sin 2 2t into a form containing the first power only of the trigonometric function. 8. Express 15-7 cos 160 sin 29 as a sum or difference. 9. Simplify sin 15* + sin 3* + cos lit cos yt. /k 10. Sin 2 A = "504. Find sin A, tan A and cos j i 2 11. A rise of level is given by 100 sin a cos a x s where 5 = difference between the readings of the top and bottom hairs of a tacheometric telescope. Express this statement in a more convenient form. If the angle of elevation a is 11 37' 30*, and the staff readings are 5-72 and 8-41, find the rise. PLANE TRIGONOMETRY 287 12. Express as products, and in forms convenient for computation (a) sin 48 - sin 17 ; (b) cos 99 + cos 176 ; (c) 12 cos 365 - 12 cos 985'. 13. When using a tacheometer and a staff it is found that if C and K are the constants of the instrument, 6 is the angle of depres- sion, s the difference of the staff readings, then depth of point below level of station = sin 20 + K sin 6 - E + Q, and distance of point from station = CS cos 2 6 + K cos 6. Find the depth and the distance when C = 98-87, sin 6 = -2753, K = -75, S = -69, E = 4-88 and Q = 9-55. 14. If tan a = - - and tan ^ = 22, find values of z to satisfy 1 o75 ~ * 2 the equations. [Refer to Fig. 154 and the worked Example 30.] 15. If V = 94 sin27r/* and A = -2 sin (zirft '117), express the product AV as a sum or difference. Trigonometric Equations. Occasionally one meets with an equation involving some trigonometric ratios; if only these ratios occur, *. e., if no algebraic terms are present in addition, the equations may be solved by the methods here to be detailed. The relations between the ratios themselves, already given, must be borne in mind, so that the whole expression can be put into terms of one unknown quantity, and the equation solved in terms of that quantity. For emphasis, the relations between the ratios are here repeated tan A =^-, cotA = -r, sec A = , - T r cos A tan A cos A sin A sin 2 A + cos 2 A = i, whence sin 2 A = i cos 2 A or cos 2 A = i sin 2 A sec 2 A = i+tan 2 A, cosec 2 A = i+cot 2 A. The idea in the solution of these trigonometric equations is to eliminate all the unknowns except one, by the use of the above relations, and then to apply the ordinary rules of equations to determine the value of that unknown. Example 34. Solve the equation 4 sin Q = 3*5. 4 sin 6 = 3-5 and sin 6 = 5 = -875. 4 Hence one value of 6, viz. the simplest, is 61 3', since sin 61 3' = -875 but sin (180 - 61 30, *' . sin n857', also - -875. so that a possible solution is n857'. 288 MATHEMATICS FOR ENGINEERS Again, 360 + 61 3' or 360 -f 118 57' would also satisfy, and so an infinite number of solutions could be found ; but whilst these could all be included in one formula, it is not at all necessary from the engineer's standpoint that they should be, for, at the most, the angles of a circle, viz. o to 360, are all that occur in his problems. Hence, throughout this part of the work the range of angles will be understood to be o to 360. .'. The solutions in this example are 61 3" and 118 57'. Example 35. If tan 6 = 5 sin 6, determine values of 6 to satisfy the equation. Apparently, in this one equation two unknowns occur, or the data are insufficient, but in reality two equations are given, for tan 6 Thus '- * = 5 sin 6. cos 6 J Dividing through by sin 6 [and in doing this we must put sin Q = o as a possible solution, since -* = o and 5x0 = 0] COS v Then ^ = 5 cos 6 J COS = '2 = COS 78 28' /. 6 = o; or 180; or 78 28'; or 360 - 78 28', i.e., 281 32'. Example 36. Solve the equation sin 6 + tan 6 = 3 cos 6 sin 6. sin 6 + tan 6 = 3 cos & sin 6 By substituting for tan 6 its value sin 6 A A sin 6 H 2i = 3 cos ^ sin COS C7 or '^^ {cos 6 + 1} = 3 cos 6 sin 0. COS t/ Dividing through by sin 6 {sin 6 = o thus being one solution} and multiplying through by cos 6 cos 6 + i = 3 cos 2 or 3 cos 2 6 cos 6 i = o. It may appear easier to solve this equation if X is written for cos 6 . e., 3 X 2 - X - 1=0 + I Vl + 12 whence X = 3 '606 6 4-606 2 '606 " 6 f -6 = -7677 or - -4343 cos 6 = -7677 or cos 6 = '4343- PLANE TRIGONOMETRY 2 8 9 Now for the cosine to be positive, the angle lies in the first and fourth quadrants ; and, since the smallest angle having its cosine =-7677 is 39 51'. the values of 6 are 39 51' or 36o - 3 9 5I ', i. e ., 320 9'. ' For the cosine to be negative, the angle lies in the second and third quadrants. Now cos 64 15' = -4343, and therefore the values of 6 are i8o-64i5', i.e., H545', or i8o+6 4 i 5 ', i.e., 24 4 i 5 '. Hence the solutions are = o; 180; 395i'; 320 g'\ 11.5 4.S' or 2 4 4is'. Example 37. The velocity of the piston of a reciprocating engine is given by the formula / . n . Y sin 20\ v = 2nnr sin 6 -\ -- = ) \ 21 I where r = crank radius, n = R.P.M., 6 crank angle from dead centre position, and / = length of connecting-rod. Y The velocity is a maximum when cos0+j-cos20 = o; find the crank angles for the maximum velocity when / = 8r. We require to solve the equation ., . cos 20 cos 6 H -- o = - o To change into terms of cos 6 write 2 cos 2 6 i in place of cos ^6. A , 2 COS 2 6 Theii cos 6 -\ -- ~ 8 cos 6 + 2 cos 2 6i = o or 2X 2 + 8X 1=0 where X = cos 6. The solutions of this equation are given by x _ - 8 V6 4 + 8 4 _- 8 8-485 4 = ~ 16-485 or -485 4 4 = 4-1212 or -12 12, which are the values of cos 6. But cos 6 cannot = - 4-1212, since cos 6 is never greater than i, hence the first root is disregarded. cos & = -12 12, which gives the required solutions, i.'e.. = 83 2' or 276 58'. If a skeleton diagram is drawn it will be observed that when 6 has these values the crank and connecting-rod are very nearly at nghi angles to one another. U 290 MATHEMATICS FOR ENGINEERS Exercises 33. On the Solution of Trigonometric Equations. Solve the equations (for angles between o and 360). 1. Sin 2 A + 2 sin A = 2 cos 2 A. 3. Cos0 6cos 2 - = 5. 2 sin* 6 5 cos 6 = 4. 7. Tan x tan 2X = i. 9. Cos 2 A + 2 sin 2 A 2-5 sin A = o. 2. 2 sin 2 6 + 4 cos 2 6 = 3. 4. Cot 14 tan 6 = 5. 6. 15 cos 2 $ + 9 sin = 12-6. 8. Tan A + 3 cot A = 4. 11. 3 tan 1 B 2 tan B i 13. 3 tan 2 Q + i = 4 tan 6. 15. Cos 2X + sin 2* = i. rj 17. Cos x sin # = -^ 16 10. Cos x tan # = -5842. 11. 3 tan 1 13 2 tan 13 i = o. 12. Tan # + cot x = 2. 14. Cos* # = 3 sin 2 x. 16. Cos # + \/3 sin # = i. 18. 2-35 sin # 1-72 cos x = -64. 19. The velocity of a valve actuated by a particular Joy valve gear is maximum when i'2p* cos pt + i-8p* sin pt = o where p = angular velocity of the crank shaft. Find the values of the angle pt for maximum velocity. 20. To find the maximum bending moment on a circular arch it is necessary to solve the equation ;R 2 sin 6 cos 6 + -9340^* sin 6 = o. Find values of 6 to satisfy this equation. 21. The following equation occurred when taking soundings from a boat, the position of the boat being fixed by reference to three points on the shore. [Compare Exercise 31, p. 272.] sin (70 14" + *) = 1-195 sin (4 8 5 6/ + *) Find the value of x to satisfy this equation, x being an acute angle. Hyperbolic Functions. Consider the circle of unit radius (Fig. 155) and the rectangular hyperbola whose half-axes are also unity (Fig. 156), '. e., OA in either case = i. Fig. 155- Fig. 156. Draw any angle EOA in each diagram, and let the " circle angle " EOA = Q, and let the " hyperbola angle " EOA = ft. PLANE TRIGONOMETRY 291 radkns T- * ****** ***** " circular " or "hyperbolic" length of circular or hyperbolic arc ""mean length of radius vector ' or by 2 x area of sector OAE EX Now in Fig. 155 _ =E X=sin0, and the corresponding ratio, viz. EX, of the hyperbolic angle is termed sinh p.* Similarly OX = cos in Fig. 155 and OX = cosh ft in Fig. 156 At = tan in Fig. 155 and A* = tanh p in Fig. 156. In Fig. 155 (EX)* + (OX) 2 = i i.e., sin 2 + cos 2 = i ......... (!) In Fig. 156 (OX) 2 -(EX) 2 = i, since the equation of the rectangular hyperbola is x*-y* = i if the semi-axes are each equal to unity and the centre is taken as the origin. Hence cosh 2 p sinh 2 p = i or cosh 2 p + ( i x sinh 2 )8) = i *. e., cosh 2 p + (V~i X sinh ) 2 = i or cosh 2 p + (j sinh /3) 2 = i .where / is written to indicate V^-i. Comparing the last equation with equation (i), we see that we may change from circular to hyperbolic functions if we write / sinh ft for sin 0, and cosh p for cos 6, and hence / tanh P for tan 6, If these substitutions are made, the ordinary rules for circular functions follow. E. g., sin (x -f- y) = sin % cos y -f cos x sin y and the corresponding expansion with hyperbolic functions is / sinh (X+ Y) = / sinh X cosh Y + cosh X . ; sinh Y or sinh (X+ Y) = sinh X cosh Y + cosh X sinh Y or again, cos 2x = sin 2 x -f- cos 2 x ........ see p. 280 t. e., cosh 2X = - (/ sinh X) 2 + (cosh X) 2 = sinh 2 X + cosh 2 X, since ; 2 = (V 11 !) 8 = i. It can be shown that these hyperbolic functions can be expressed in terms of the exponentials in the forms e - x = cosh x sinh x ? = cosh x -f- sinh x e e i.e., cosh*= *T5 + i. 2.3.4"* and - , .2.3.4.5 . * To avoid confusing with the circular functions, sinh is usually pronounced " shine," and tanh " tank." 292 MATHEMATICS FOR ENGINEERS The corresponding relations for the circular functions are e ix = cos x -f- j sin x e~ 5x = cos x ; sin x y* cos* = - _ = j 1.2 1.2.3.4 / O .A- . sm A; = = = x 2; 1.2.3-1.2.3.4.5 Hyperbolic functions occur frequently in engineering theory; e. g., in connection with the whirling of shafts the equation y = A cos mx + B sin mx + C cosh mx + D sinh mx plays a most important part : the equation of the catenary is y = cosh x ; and so on. It is in electrical work that these functions occur most fre- quently; thus, for a long telegraph line having a uniform linear leakage to earth by way of the poles the diminishing of the voltage is represented by a curve of the form y = cosh x, whilst the curve y = sinh x represents the current. Example 38. A cable weighing 3 Ibs. per foot hangs from two points on the same level and 60 feet apart ; and it is strained by a horizontal pull of 300 Ibs. The form taken by the cable is a catenary. Find the length of the cable from the formula Length = 20 sinh zc , , horizontal tension where L = span and c = J-T-T- 7 7 r 7 TT- weight of i foot of cable Here we have L = 60 and c = - = 100 ; hence = = -3 3 2C 200 Thus length of cable = 2X 100 sinh -3 Table XI at the end of the book may be utilised to find the value of sinh -3, in the following manner : Look down the first column until 3 is seen as the value for x : follow the line in which this value occurs until the column headed sinh x is reached. The value there shown is that of sinh -3 and is -3045. Hence length of cable = 200 x '3045 = 60*9 ft. This rule gives the exact length of the cable, but in practice the form of the cable is assumed to be parabolic, and the approximate length is given by Length = span + 8 (sag) ' 3 span PLANE TRIGONOMETRY 293 the sag also being calculated on the assumption of the parabolic form of the cable. In this case the sag is found to be 4-5 ft. and hence _ length of cable = 60 + = 6o . g In this instance the result obtained by the true and approximate methods agree exactly: and in the majority of cases met with in practice the approximate rule gives results sufficiently accurate. Example 39. The resistance of the conductor of a certain telegraph line is 8-3 ohms per kilometre and the insulation resistance is 600 meg- ohms per km. The difference in potential E between the line and earth at distance L kms. from the sending end is found from the formula E = A cosh Vrl. L + B sinh Vrl. L where A and B are constants, r = resistance of unit length of the conductor and I = conductance of unit length of the path between the line and earth. If the total length of the line is 100 kms., the voltage at the sending end is no, and at the receiving end is 85, find the values of A and B. We have two unknowns and we must therefore form two equations. At the sending end L = o and then 1 10 = A cosh Vrl . o + B sinh Vrl . o = A cosh o + B sinh o = Ax i = A Hence A= no. NnwrZ = - = i-iSix io~ 8 and Vrl= -0001176; also at a 600 x io 6 distance of 100 kms. from the sending end the value of E is to be 85. Substituting these numerical values in the original equation 85 = no cosh (-0001176 x 100) + Bsinh (-0001176 x 100) = no cosh -01176+ Bsinh -01176 ......... (i) In order to solve this equation for B the values of cosh -01176 and sinh -01176 must first be found; and as the given tables of values of cosh and sinh are not convenient for this purpose we proceed according to the following plan and sinh, = ^; . = .""; and to evaluate we must take logs. Let y = e' 01176 and then log y = -01176 x log e = -01176 x -4343 = -0051 so that y =1-012. Thus *- 1176 = 1-012 and e-' 01 "' which is the reciprocal of ' 01 is -9883. 294 Then and MATHEMATICS FOR ENGINEERS 2 + -988 cosh -01176 = sinh -01176 = 2 1-012 = I 012 Substituting these values in equation (i) 85 = (no x i) + (B x -012) whence -012 B = 25 or B = 2083 Hence E = no cosh -0001176 L 2083 sinh -0001176 L. Complex Quantities. Algebraic quantities generally may be divided into two classes, real and imaginary, and the former of these Fig. 157. Complex and Vector Quantities. may be further subdivided into rational and irrational or surd quantities. Thus, Vs and also 7 are real, whilst V 15 is imaginary; indeed, all quantities involving the square root of a negative quantity are classed as imaginary. An expression that is partly real and partly imaginary is spoken of as a complex quantity ; thus 4 + 7^9, and Vzx-}-i6V zy are complex quantities. The first expression might be written as 4 -f- (jxVq X V i), . e., 4+21; where / stands for V i. The general form for PLANE TRIGONOMETRY 395 these complex expressions is usually taken as a -f- jb, where a and b may have any real values. According to the ordinary convention of signs, if OA ((a) Fig. 157) represents + a units, then OA' would stand for a if the length of OA' were made equal to that of OA ; in other words, to multiply by I, revolution has been made through two right angles. Now X V I X Vi = a, so that the multiplication by V^i must involve a revolution one-half of that required for the multiplica- tion by i; or OA* must represent ja. Accordingly a meaning has been found for the imaginary quantity /, and that is : If -fa is measured to the right and a is measured to the left, from a given origin, then ja must be measured upward, and differs from the other quantities only in direction, which is 90 from either -fa or a. To represent a -f jb on a diagram, therefore, we must set out a distance OA to represent a, erect a perpendicular AB making AB equal to b, choosing the same scale as that used for the hori- zontal measurement, and then join OB ; then OB = a -f jb. For OB' would represent a + b and AB = / . AB', so that OB must be the result of the addition of a to jb. The addition is not the simple addition with which we have been familiar, but is spoken of as vector addition, i. e., addition in which attention is paid to the direction in which the quantity is measured as well as to its magnitude. It is often necessary to change from the form a + jb to the form r (cos 6 -f / sin 0} ; and this can be done in the following way If r(cos0 + / sin0) is to be identically the same as a+/6 then the real parts of each must be equal, and also the imaginary. i. e. t r cos 6 = a (i) rj sin = jb whence r sin = b (2) By division of (2) by (i) tan 6 = - and by squaring both (i) and (2) and adding f 2 cos 2 + f* sin 2 = a 2 +&* or r z a 2-f &a since cos 2 -f sin 2 = i Thus, at (b) Fig. 157 OB = r and L BOA = Example 40. To change 3 - yjj into the form r (cos 6 + / sin 6). 296 MATHEMATICS FOR ENGINEERS From the above, since a = 3, and b = 5-7 r* = 3 2 + (~5'7) 2 = 9 + 32-4 = 4iH r = 6-44 and tan = ^2 =* i-g or 6 = 62 14$'. This case is illustrated at (6) Fig. 157, in which OB represents r. and the angle BOA is the angle 6. Vector quantities, such as forces, velocities, electrical currents and pressures, may be combined by either graphic or algebraic methods; in the algebraic addition, for example, the components along two directions at right angles are added to give the com- ponents along these axes of the resultant. Thus if the vector 2+1-5; were added to the vector 4 + -6;', the resultant vector would be 2 4+i"5/+-6;, i.e., 2+2-1;. The addition is really simpler to perform by the graphic method, thus : OB at (c) Fig. 157 represents the vector quantity 2+1-5; and OD represents 4+-6/. Through B draw BE parallel and equal to OD and join OE; then OE is the resultant of OB and OD. It will be seen that OE is the vector 2+2-1; since OF = 2 units measured in a negative direction and FE = 2-1 units. To multiply complex quantities. Let OA ((d) Fig. 157) represent a + jb, i. e., r (cos 6 + / sin 0) and let OB represent a 1 + jb lt i. e., ^(cos 0j + / sin X ) Then : OA x OB = (r cos 6 + rj sin 0)(r 1 cos X + rjj sin X ) = TTj cos cos 6j + rr^j sin Oj cos + rrrf sin cos X + rr^' 2 sin sin X = rr-L (cos cos 0j sin sin X ) + rrtf (sin cos 0! + cos sin 0J = rr^cos (0 + ej + / sin (0 + X ) j ^ ,. ., .... T , a + jb r (cos +;sin0) To divide complex quantities, Let - f- v- = 4 '. . .. \ a i + 1i fi(cos X + ; sin 0j) Rationalise the denominator by multiplying by (cos 0j ;' sin X ) ~^~ r cos 9 ~^" sm ecos e ~ sm 9 Th (cos 2 X + sin 2 which can be expressed in the form A + ;'B if desired. These results might have been arrived at by expressing r (cos + ;' sin 0) as rei and r (cos X + ;' sin X ) as r-^ l . Thus (a + jb)(cii + jb^ = rei* x r^ = rr^'+W = rr^cos (0 + Oj) + / sin (0 + 0j)} PLANE TRIGONOMETRY 297 Example 41. The electric current C in a star-connected lighting system was measured by the product Potential P x admittance y. If P = (-068 - -0015;) (28 + 30;) and y = -9 + -18; find C. P = (-068 - -0015;) (28 + 30;) = 1-949 + 1-998; (by actual multi- plication). = 2-791 (cos 45 43' + ; sin 45 43') y = -9 -f- -i8j = -9179 (cos 11 19' + j sin 11 19') .'. C = Py = 2-563 (cos 57 2'4-y sin 57 2') or, alternatively, t-395 + 2-1497'. Inverse Trigonometric Functions. If sin * = y, then x is the angle whose sine is y, and this statement may be expressed in the abbreviated form x = sin~ 1 y. (Note that sin~ 1 y does not mean -^ ^ but the I indicates a converse statement, y being the value of the sine and not the angle.) sin" 1 y is called an inverse circular function. Similarly sinh -1 y is called an inverse hyperbolic function Angles are sometimes expressed in this way instead of in degrees ; e. g., when referring to the angle of friction for two surfaces : if the coefficient of friction between the surfaces is given, that is the value of the tangent of the angle of friction, and the angle of friction = tan~V*, where /t is the coefficient of friction. Example 42. Given sin" 1 * = y, find the values of cos y and tan y. sin" 1 x = y i. e., sin y = x and tan y = cos y = Vi sin'y = Vi X* sin y x cosy Vi - x* Example 43. The transformation from the hyperbolic to the logarithmic form occurs when concerned with a certain integration. If cosh y = x, show that cosh" 1 * = log ,(* + V* 2 - i). cosh y = or whence / : (Solving the quadratic) * + Vx*-i MATHEMATICS FOR ENGINEERS since x Vx 2 i = - - ._ x + Vx* - i as is seen if we multiply across log, (x + Vx* - i) = y, or log* i. e., y = log* (x + Vx z i) or if only the positive root is taken y = loge (x + V* 2 i) cosh" 1 * = loge (x + Vx* i) In like manner it can be proved that . , , x , x + Vx z + a 2 smh.- 1 - = log, a a , x . x -f Vx z a 3 - 1 - = log e - Exercises 34. On Hyperbolic and Inverse Trigonometric functions. 1. Read from the tables the values of cosh -7 and sinhi'5. 2. Evaluate 5 cosh -015 + ! sinh-oi5. 3. Find the true length of a cable weighing 1-8 Ibs. per foot, the ends being 120 ft. apart horizontally and the straining force being 90 Ibs. weight. [Refer to worked Example 38, p, 292.] 4. Calculate the sag of the cable in Question 3, from the rule sag = c ( cosh -- i ) \ 2C I straining force in Ibs. wt. where c = - -. ^ 77 T TT - wt. of i ft. of cable Hence find the approximate length of the cable, from the rule length = span +**J& 3 X span 5. The E.M.F. required at the transmission end of a track circuit used for signalling can be found from E, = Put this expression into a simpler form, viz. one involving hyper- bolic functions. 6. If the " angle of friction " for iron on iron is tan~ 1 -i9, find this angle. 7. The lag in time between the pressure and the current in an ...... , period , 2n-L , alternating current circuit is given by * ^ x tan -1 p where n = number of cycles per second, L = self-induction of circuit and R = resistance of circuit, the angle being expressed in degrees. If the frequency is 60 cycles per second, L = -025 and R = 1-2, find the. lag in seconds. PLANE TRIGONOMETRY 299 8. If cosh y = 1-4645, find the positive value of y. 9. A block is subjected to principal stresses of 255 Ibs. and 171 Ibs., both tension. The inclination of the resultant stress on a plane in- clined at 27 to the plane of the greater stress is tan^K? tan A where Vi / /! and / 2 are the greater and lesser stresses respectively and 6 is the inclination of the plane. Find the inclination of the resultant stress for this case. 10. The solution of a certain equation by two different methods gave as results s = sin ( 7* + tan~ x ) and s = sin ( 7* 2 tan" 1 2 53 V/ 45 7 53 V/ 2 respectively. By finding the 'numerical values of the angles tan" 1 and tan -1 -, show that the two results agree. 11. The following equation occurred in connection with alternator regulation a = 6 + <(> lid = sin" 1 ^4r- anc * cos" 1 -55, find sin a. OIoO 12. The speed V knots of waves over the bottom in shallow water is calculated from V 2 = i-8 .L, where d = depth in feet L = wave length in feet. If d = 40 ft., and L = 315 ft., calculate the value of V. 13. By calculating the values of the angles (in radians) prove the truth of the following relations : tan- 1 \ + tan- 1 = 4 4 tan- 1 tan- 1 ?fa = - 4 tan- 1 tan- 1 ^ + tan- 1 ^ = ~ Illustrate the first of these by a diagram. 14. An equation occurring in the calculation of the arrival current in a telegraph cable contained the following : N _ 9* _ 3_ cos 2a gink 2,b cos -za cosh 2b. 10 2 Find N when a = 4-5 and b = 2. 15 If C = 5-4 (cos 62 + j sin 62) and y = x-8 (cos 12 + ; sin 12 ) find P (in the form a + /&). The letters have the same meanings Example 41, p. 297. CHAPTER VII AREAS OF IRREGULAR CURVED FIGURES Areas of Irregular Curved Figures. Rules have already been given (see Chapter III) for finding the areas of irregular figures bounded by straight sides; if, however, the boundaries are not straight lines, such rules only apply to a limited extent. The mean pressure of a fluid such as steam or gas on a piston is found from the area of the " indicator diagram," the figure automatically drawn by an engine " indicator," correlating pressure and volume. By far the quickest and most accurate method of determining the area of this diagram is (a) to use an instrument called the planimeter or integrator. Other methods are (b) the averaging of boundaries, (c) the counting of squares, (d) the use of the computing scale, (e) the trapezoidal rule, (/) the mid-ordinate rule, (g) Simpson's rule and (h) graphic integration. To deal with these methods in turn : (a) The Planimeter. The Amsler planimeter is the instrument most frequently employed, on account of its combination of sim- plicity and accuracy. It consists essentially of two arms, at the end of one of which is a pivot (see Fig. 158), whilst at the end of the other is the tracing-point P. By unclamping the screw B the length of the arm AP can be varied, fine adjustment being made by the adjusting screw C : and this length AP determines the scale to which the area is read. The rim of the wheel W rotates or partially glides over the paper as the point P is guided round the outline of the figure whose area is being measured; the pivot O being kept stationary by means of a weight. The motion of the wheel W is measured on the wheel N in integers, and on the wheel D in decimals, further accuracy being ensured by the use of the vernier V. To use in the ordinary manner, the pivot O being outside the figure. By rough trial find a position for the pivot so that the figure can be completely traversed in a comfortable manner. Mark some convenient starting-point on the boundary of the figure and s: AS: : 8 r A ^^ ^ - .- case the area would be 844 sq. units. Along the arm AP are marks for adjustment to different scales f A is set at one of these marks the area will be in sq. ins., at Fig. 158. The Amsler Planimeter. another, in sq. cms., etc.; but if there be any doubt about the scale, a rectangle, say 3* X 2* should be drawn, and the tracer guided round its boundary. Whatever reading is thus obtained must represent 6 sq. ins. so that the reading for i sq. in. can be calculated therefrom. If, in the tracing for which the figures are given above, the zero mark A had been set at the line at which " -01 sq. in." is found on the long bar, then the area would be 8-44 sq. ins., since the divisions on the vernier scale represent -01 sq. in. each. For large areas it may be found necessary to place the pivot inside the area ; and in such cases the difference between the first and last readings will at first occasion surprise, for it may give an 302 MATHEMATICS FOR ENGINEERS area obviously much less than the true one. This is accounted for by the fact that under certain conditions, illustrated in Fig. 159, the tracer P traces out a circle, called the zero circle, for which the area as registered by the instrument is zero, since the wheel does not revolve at all. For a large area, then, the reading of the instru- ment may be either the excess of the required area over that of the zero circle, or the amount by which it falls short of the zero circle area. These areas are shown respectively at EEE and III in Fig. 159, while the zero circle is shown dotted. [Noie. For the ordinary Amsler planimeter the area of the zero circle is about 220 sq. ins., but it is indicated for other units by figures stamped on the bar AP.] Fig. 159. Zero Circle of Planimeter. To use in the special manner. By means of a set square, adjust the instrument so that the axis of the tracing arm is perpendicular to a line joining the fixed centre O in Fig. 159 to the point of con- tact of the wheel and the paper. Measure the radius from the fixed centre O to the tracing-point P, and draw a circle with this radius on a sheet of tracing-paper. Place this over the plot whose area is being measured, and endeavour to estimate whether the figure is larger or smaller than the zero circle. If this is at once apparent trace round the figure in the ordinary way and add the area of the zero circle to the reading, or subtract the reading from the zero 303 circle area as the case may demand. If not apparent, proceed thus Set the planimeter to some convenient reading, say 2000 and trace the area in a right-handed direction. Then, if the final reading is greater than 2000, the area is greater than that of the zero circle and vice versa. Then to obtain the area (1) If the area is greater than the zero circle, trace in a right- handed direction and add the excess of the last reading over the first to the area of the zero circle; i. e., if x is the excess of the final reading over the initial reading, the true area = x -f zero circle area. (2) If the area is less than the area of the zero circle, trace in a left-handed direction and subtract the difference of the first and last readings from the area of the zero circle. For x = excess of the last reading over the first, if the tracing is in a right-handed direction, and this becomes -f- x if the tracing is in a left-handed direction. The true area = x + zero circle area, and the tracing is performed in a left-handed direction in order to get a positive value for x. If the instrument is to be used as an averager, as would be the case if the mean height of an indicator diagram was required, LL in Fig. 158 must be set to the width of the diagram and the outline must be traced as before. Then the difference of the readings gives the mean height of the diagram. Further reference to the planimeter is made in Volume II of Mathematics for Engineers. The Coffin Averager and Planimeter (Fig. 160) is somewhat simpler in construction as regards the instrument itself, but there are in addition some attachments. It is, in fact, the Amsler instru- ment with the arm AO (Fig. 158) made infinitely long so that A, or its equivalent, moves in a straight line and not along an arc of a circle. Referring to Fig. 160 it will be seen that the pointer is con- strained to move along the slot GH. To use the instrument to find the mean height of a diagram : Trace the diagram on paper and draw a horizontal line and two perpendiculars to this base line to touch the extreme points on the boundary of the figure. Place the paper in such a way that the base line is parallel to the edge of the clip B and set the clips AE and CD along the perpendiculars already drawn. Then start from F, the reading of the instrument being noted, and trace the outline of the figure until F is again reached. Next move the tracing- point along the vertical through F, *'. e., keep the tracer P against 304 MATHEMATICS FOR ENGINEERS the clip, until it arrives at M at which stage the instrument records the initial reading. FM is then the mean height of the diagram. Fig. 1 60 The Coffin Averager. If the area of the figure is required, the reading of the instru- ment must be made when the tracing-point is at F. The outline 305 of the figure is then traced until F is again reached and the reading is again noted. Then the difference between the two readings is the area of the figure. (b) Method of averaging Boundaries. The area of a figure of the shape bounded by the wavy line in Fig. 161 being required, proceed as follows : Draw the polygon ABCRED so that it shall occupy the same area as the original figure, viz. the portions added are to be equal to those subtracted, as nearly as can be estimated. Then, by joining BD, BE, CE, etc., the polygon is divided into a number of triangles and the area of each is : X base X height. Therefore draw the necessary per- pendiculars, scale off the lengths of the bases and the heights, and tabulate as follows : Fig. 161. Area by Averaging Boundaries. Triangle. Base. Height. Sura of Heights. Area of the two triangles = J base X sum of heights. ABD BED BD BD AG EF | AG + EF BD(AG+ EF) CBE CRE CE CE BM RN | BM+RN iCE (BM+RN) The triangles are thus grouped in pairs and the area of the figure is the sum of the quantities shown in the last column. (c) Method of counting Squares. Draw the figure, whose area is required, on squared paper, choosing some convenient scales. Then count the squares, taking all portions of a square greater than one-half as one, and neglecting all portions smaller than a half-, square. If i linear inch represents x units horizontally and y units vertically and the paper is divided into n squares to the linear inch : each square is -^ sq. ins., and I sq. in. on the paper represents xy xy sq. units of area, so that each square represents -^ sq. units. x 306 MATHEMATICS FOR ENGINEERS If the total number of squares = N Area of figure = : a - sq. units (d) The Computing Scale is often employed in the drawing office to find the areas of plots of land. It consists of two main parts, viz. a slider A, Fig. 162, and a fixed scale C. The slider can Tracing Fig. 162. Area by the Computing Scale. be moved along the slot B by means of the handles D; and it carries a vertical wire, which is kept tightly in position by means of screws. Along the fixed scales are graduations for acres and roods according to a linear scale of 4 chains to i", and a scale of square poles, 40 of which make up i rood, is indicated on the slider. To use the instrument. Rule a sheet of tracing-paper with a number of parallel lines exactly J" apart, i. e., i chain apart according to the particular scale chosen. Place the tracing-paper AREAS OF IRREGULAR CURVED FIGURES 307 over the plot in such a way that the whole width in any one directio is contained between two of these parallel lines. Place the slider at the zero mark, and move the whole instrument bodily until the wire at a cuts off as much from the area as it adds to it. Next move the slider from left to right until b is reached Remove the instrument and without altering the position of the wire, place the scale so that the wire is in the position c : then run the slider along the slot until the wire arrives at d, and so on. Take the final reading of the instrument, and this is the total area of the plot. If the slider reaches the end of the top scale before the area has been completed, the movement can be reversed, *'. e., it becomes from right to left and the lower scale must be used'. It will be observed that by the movement of the slider the mean widths of the various strips are added. Now the strips are each I chain deep, so that if the mean lengths of the strips measured in chains are multiplied by i chain, the total area of the plot is found in square chains. But 10 sq. chains = i acre, and the scale to which the plan is drawn is i" = 4 chains. Hence 2j* = 10 chains, and the scale must be so divided that 2\" = i acre, since the strip depth is i chain. If the plot is drawn to a scale other than the one for which the scale is graduated the method of procedure is not altered in any way, but a certain calculation must be introduced. Thus if the figure is drawn to a scale of 3 chains to the inch and the computing scale is graduated according to the scale i" = 4 chains, then the true area = (f) 2 or T \ of the registered area. (e) The Trapezoidal Rule. When using this rule divide the base of the figure into a number of equal parts and erect ordinates through the points of division. The strips into which the figure is thus divided are approximately trapezoids. For a figure with a very irregular outline the ordinates should be drawn much closer than for one with a smooth outline. Then the area of the figure is the sum of the areas of the trapezoids, *. e., in Fig. 163, Area = %!+y 2 ) + % 2 +y 3 ) + ....... i%io+yii) yn) + y 2 + y 3 + Or, the area is equal to the length of one division multiplied by the sum of half the first and last ordinates, together with all the remaining ordinates. Example i. Find the area of the figure ABCD in Fig. 163. which is drawn to the scale of half full size. 308 MATHEMATICS FOR ENGINEERS The base is divided into 10 equal parts and the ordinates are measured. Then the calculation for the area is set down thus y\ = 2-5 y n = 2-0 sum of first and last = 2-25 y 2 = 4-40 ys = 5 >I0 y\ = 5-34 y& = 4-13 y^ = 3-83 y s = 3-80 Vg = 3'63 i(yi + yu) = 2-25 Sum = 39-88 Width of one division of the base = h = i" Area = i x 39-88 sq. ins. At Fig. 163. Area by Trapezoidal Rule. (/) The Mid - ordinate rule is very frequently used and is similar to the trapezoidal rule. The base of the figure is divided into a number of equal parts or strips, and ordinates are erected at the middle points of these strips; such ordinates being called mid-ordinates as distinct from the extreme ordinates through the actual points of section. The average of the mid-ordinates multi-" plied by the length of the base is the area of the figure. Example 2. Find the area of the figure ABCD in Fig. 164, which is an exact copy of Fig. 163, and is drawn to the scale of half full size. The lengths of the mid-ordinates are 3-66, 4-90, 5-24, 5-24, 4*4, 3-92, 3'8, 3'73, 3'3 and 2-13 ins. respectively, and the average = - = 4-032. AREAS OF IRREGULAR CURVED FIGURES 309 Hence the area = 40-32 sq. ins., as against the previous result of 39-88 sq. ins., showing a difference of i%. The mid-ordinate rule is much in vogue on account of its sim- plicity. As a modification of this method we may ascertain the total area by the addition of the separate strip areas. It is not necessary to divide the base into equal portions : but the divisions may be chosen according to the nature of the bounding curve. If 3s HO I Al _ Fig. 164. Area by Mid-ordinate Rule. the latter is pretty regular for a large width of base, the division may be correspondingly wide; but sudden changes in curvature demand narrower widths. Assuming that the area has been divided into strips in the manner suggested, find the lengths of the mid-ordinates and the widths of the separate strips and tabulate as in the following example. Example 3. Calculate the area of the figure ABMP (Fig. 165). Strip. Width (inches). Length of Mid-ordinate (ins.). Area of Strip (sq. ins.) Sum of Areas of Strip (sq. ins.) AB 1-8 3'3 5'94 5'94 BC 4 5-02 2-OI 7'95 CD '3 4'35 I-3I 9-26 DE 5 1-89 11-15 EF 4 3-85 i'54 12-69 FG GH i'3 6 3'54 2'53 4-60 1-52 17-29 18-81 HJ J K KL i'3 6 8 3-6 4'54 3'4 4-68 2-72 2-72 23H9 26-21 28-93 LM I-O 2-31 2-31 31-24 Thus the total area = 31-24 sq. MATHEMATICS FOR ENGINEERS (g) Simpson's Rule is the most accurate of the strip methods and is scarcely more difficult to remember or more complicated in its application than the trapezoidal rule. In this rule, the base must be divided into an even number of equal divisions ; the ordinates through the points of section being added in a particular way, viz. first ordinate + last ordinate + 4 (sum of even ordinates) + 2 (sum of odd ordinates, excluding the first and last). J If the portions of the curve joining pairs of ordinates are straight or parabolic, i. e., if the equations to these portions are of the form y = a-}-bx-\-cx 2 , the ordinates being vertical, the rule gives rn Area = width of one division of base 3 i I i 4 Scale of In heights and widfhs. 1 Fig. 165. Modification of Mid-ordinate Rule. 3 ins. perfectly correct results ; and the strip width should be chosen to approximately satisfy these conditions. Taking an example Example 4. Find the area of the indicator diagram shown in Fig. 1 66. A convenient horizontal line is selected to serve as a base and, in this instance, is divided into 10 equal parts. The ordinates are numbered y\, y z> y t , etc., and their heights are measured, being those between the boundaries of the figure, and not down to the base line. AREAS OF IRREGULAR CURVED FIGURES The working is set out thus Width of i division = i foot. Ordinates. Even. Odd. y, = 80 y 3 = 66 ?4 = 55'5 y&= 48 y 6 = 43'5 Vi= 38-5 Vs = 34'5 y= 30-5 Vw= 24 ist = y l = o last = y u = o Sum =o Sum = 237-5 and (Even) 4Xsum = 950 2xsum = 366 (Odd) .". Area = f[o + 950 + 366} = 439 sq. units, which in this case would represent to some scale the work done per stroke on the piston. KO ho Fig. 1 66. Area by Simpson's Rule. Notice that this rule agrees with our notion of " average height X base " for Number of ordinates considered = i+i+4(5)+ 2 (4) = 3<> and if A = sum of ordinates according to the particular scheme A h Area = base X average ordinate = ioAx = -XA If the area is of such a character that two divisions of the base are sufficient Area = |{ist + last + (4Xmid.)} O = k^ ( Ist + last + (4Xmid)} since length = 2* 312 MATHEMATICS FOR ENGINEERS (h) Graphic Integration is a means of summing an area with the aid of tee- and set-square, by a combination of the principles of the " addition of strips " and " similar figures." An area in Fig. 167 is bounded by a curve a'b'z' ', a base line az and two vertical ordinates aa' and ztf . The base is first divided as in method (/), where the widths of the strips are taken to suit the changes of curvature between a' and z' , and are therefore not necessarily equal ; and mid-ordinates (shown dotted) are erected for every division. Next the tops of the mid-ordinates are projected horizontally on Fig. 167. Graphic Integration. to a vertical line, as BB'. A pole P is now chosen to the left of that vertical; its distance from it, called the polar distance p being a round number of horizontal units. The pole is next joined to each of the projections in turn and parallels are drawn across the corresponding strips so that a continuous curve results, known as the Sum Curve. Thus am parallel to PB' is drawn from a, across the first strip; mn parallel to PC' is drawn from m across the second strip, and so on. The ordinate to the sum curve through any point in the base gives the area under the original or primitive curve from a up to the point considered. AREAS OF IRREGULAR CURVED FIGURES 313 Referring to Fig. 167 Area of strip abb' a' = ab x AB but, by similar figures B'a or BA _ bm P ~ ab whence AB x ab = p x bm area of strip i" -, . or area of strip = pxbm i. e., bm measures the area of the first strip to a particular scale, which depends entirely on the value of p. T , v / area of second strip In the same way nm' = P and by the construction nm' and bm are added, so that cn _ area of ist and 2nd strips P or area of ist and 2nd strips = p x cn Thus, summing for the whole area Area of aa'z*z = pXzL Thus the scale of area is the old vertical scale multiplied by the polar distance ; and accordingly the polar distance should be selected in terms of a number convenient for multiplication. e. g., if the original scales are i" = 40 units vertically and i" = 25 units horizontally and the polar distance is taken as 2", i.e., 50 horizontal units; then the new vertical scale = old vertical scale X polar distance = 40 X 50 = 2000 units per inch. If the original scales are given and a particular scale is desired for the sum curve, then the polar distance must be calculated as follows . . ., new vertical scale Polar distance in horizontal units = old yerticaHclOe" e. g., if the primitive curve is a " velocity-time " curve plotted to the scales, i" = 5 ft. per sec. (vertically) and i" = = -i sec. (hori- zontally) and the scale of the sum curve, which is a " displacement- time " curve, is required to be i" = 2-5 ft., then 2*5 Polar distance (in horizontal units) = -e = -5 314 MATHEMATICS FOR ENGINEERS = ! unit along the horizontal, the polar distance and since I* = must be made 5". The great advantages of graphic integration are (a) Its ease of application and its accuracy. (b) The whole or part of the area is determined without separate calculation ; the growth being indicated by the change in the sum curve. Thus, if the load curve on a beam is known, the sum curve indicates the shear values, because the shear at any section is the sum of the loads to the right or left of that section. Example 5. Draw the sum curve for the curve of acceleration given in Fig. 168. Find the velocity gained in 20 seconds from rest, and also in 35 seconds : find also the average acceleration. Polo, Scale .of time (sees) Fig. 1 68. Construction of Sum Curve. Method of procedure. Project be horizontally to meet the vertical AB in c. Draw AM parallel to PC to cut the second ordinate in M. Project de horizontally and draw MN parallel to Pe. Continue the construction till Z is reached on the last ordinate. The polar distance was chosen as 3", or 15 horizontal units, so that, whilst the old vertical scale was i" = 2 units of acceleration, the sum curve vertical scale (in this case a scale of velocity) will be i" = 2 X 15 = 30 units. This new scale is indicated on the extreme right by the title " scale of velocity." Note that Z is at the point * 3 tt^tttt^ m ^2**~ this being the average height of the Graphic integration can only be immediately applied when the base is a straight Hne. If it is otherwise, the figure m duced to one wrth a straight base by stepping oQ the re jjualion :--u -oxVkx+oc-H J Fig. 169. Comparison of Sum Curves. with the dividers. Therefore, if the full area is required, as in the case of an indicator diagram, the additional complication would neutralize any other gain ; but if separate portions of the area are wanted the method is the most efficient. It is of interest to note that if the original curve is a horizontal line The first sum curve is a sloping straight line, The second sum curve is a parabola of the second degree, or a " square " parabola. The third sum curve is a parabola of the third degree, or a " cubic " parabola. 316 MATHEMATICS FOR ENGINEERS These cases are illustrated in Fig. 169, the poles being chosen to bring the curves to about equal scales for comparison. Graphic Integration will be again referred to when dealing with the Calculus generally. Calculation of Volumes. All the rules for finding areas can be extended to the calculation of volumes. The area of the figure should then represent the volume : e. g.,if the cross-section at various distances through an irregular solid be noted or estimated, and ordinates be erected to represent these cross-sections at the proper distances along the base of the diagram ; the area of the figure on the paper will represent the volume of the solid. Thus If i" represents x feet of length and i" represents y sq. ft. of cross-section, then i sq. in. of area represents xy cu. ft. of volume. Example 6. Find the capacity of a conical tub of oval cross -section, the axes of the upper oval being 28" and 20", those of the base being 21" and 15", and the height being 12". In this case the rule for the three sections may be applied; the axes of the mid-section are 24^" and 17^" and the areas of the three sections are A = TT X 14 X 10 = 14077 sq. ins. B = 77 x 10-5 x 7-5 = 7 8 '75 7r .. M = 77 x 12-25 x 8*75 = 107-277 ,, Volume = -{A + B + 4 M} = ^{140 + 78-75 + 428-8} = 277 x 647-6 = 4070 cu. ins. .*. Capacity = 4 ' gallons = 14-7 gallons. Other worked Examples on the calculation of volumes will be found in Chapter VIII. Exercises 35. On the Areas of Irregular Curved Figures. 1. A gas expands according to the law pv = 150, from volume 3 to volume 25. Find the work done in this expansion. 2. An indicator card for a steam cylinder is divided into 10 equal parts by 9 vertical ordinates which have the respective values of 100, 84, 63, 50, 42, 36, 32, 28 and 26 Ibs. per sq. in. ; and the extreme ordinates are 100 and 25 Ibs. per sq. in. respectively. Find the mean pressure of the steam. 3. The end areas of a prismoid are 62-8 and 20-5 sq. ft., the section mid-way between is 36-7 sq. ft. and the length of the prismoid is 15 ft. Find the average cross-section and the volume. AREAS OF IRREGULAR CURVED FIGURES 317 4. The mid-ship section of a vessel is given, the height from keel to deck being 19$ ft. ; and the horizontal widths, at intervals of 3-21 ft are respectively 46-8, 46-2, 45-4, 43, 36-2, 26-2 and 14-4 ft., the first being measured at deck level and the last at the keel. Calculate the total area of the section. 5. To measure the area of the horizontal water plane, at load line of a ship, the axial length of the ship was divided into nine abscissa whose half-ordinates from bow to stern were -6, 2-85, 9-1, 15-54, 18, 18-7, 18-45, I 7' 6 I 5'i3 an d 6-7 ft. respectively; while the length of the ship at load line was 270 ft. Find the area of the water plane. 6. The velocity of a moving body at various times is as given in the table Time (sees.) . o i'5 2-8 3'6 5 6-2 77 8-9 10-3 12 Velocity (ft."l per sec.) . J 37'3 31-5 27-5 25H 22 -4 20'3 18-2 16-9 15-8 15 Find the total distance covered in the period of 12 seconds ('. e., find the area under the velocity curve plotted to a time base.) 7. To find the cross-section of a river 90 ft. in breadth, the following depths, marked y, in feet, were taken across the river ; x, in feet, being -the respective horizontal distances from one bank. X o 10 20 30 40 50 60 70 80 90 y 3 4'5 5-6 6 5'7 4-8 4'7 4'5 4 3 Find the area of the cross-section. If the average velocity of the water normal to the cross-section is 5-1 ft. per sec., find the flow in cu. ft. per sec. 8. A series of offsets was measured from a straight line to a river bank. Find, by Simpson's rule, the area between the line and the river bank. Offsets (ft.) . 7 9 8 5 2 3 7 9 ii '.5 20 13 3 Dist. along \ line (ft.) / o IOO 200 300 400 450 500 600 700 725 750 775 800 900 1000 9. The mean spherical candle-power (M.S.C.P.) of a lamp can be determined by calculating the mean height of a Rousseau diagram (candle power plotted to any linear base). Find the M.S.C.P. for the arc lamp for which the Rousseau diagram is constructed from the following figures : Dist. from one end\ of base (ins.) J i V 1-8 2'5 3'5 4-2 4'9 5'4 5-8 6 Candle power . o|n5 i 350 650 I IOO 1350 1500 1 200 400 10. Reproduce (a) Fig. 12 to scale and then determine its area. MATHEMATICS FOR ENGINEERS 11. Fig. 170 is a reproduction of an indicator card taken during a test on a 10 H.P. Diesel engine. Calculate the mean pressure for this case, i.e., find the mean height of the diagram. O O O O N O I I I CHAPTER VIII CALCULATION OF EARTHWORK VOLUMES IN this chapter a series of examples will be worked out to illus- trate the method of calculating volumes of earthwork, such as railway cuttings, embankments, and other excavation work, mostly for the purpose of estimating the cost of earth removal. Definitions of Terms introduced in these Examples. The formation surface is the surface at the top of an embank- ment or at the bottom of a cutting, and in all the cases here con- sidered it will be regarded as horizontal. The line in which the formation surface intersects the transverse section of the cutting or embankment is spoken of as the formation width. The natural surface of the ground is the surface existing before the cutting or embankment is commenced. The sides of a cutting or an embankment slope at an angle which is less than that of sliding for the particular earth ; and the slope is usually expressed as x horizontal to one vertical. A few typical values of the slope are given for various soils : SOIL. Compact Earth. Gravel. Dry Sand Vege- table Earth. Damp Sand. Wet Clay. ANGLE WITH HORIZONTAL 50 40 38 28 22 16 SLOPE (i. e., x horizontal to i vertical) 8 39 I I-IQ2 1-28 1-88 2-475 3-487 The unit of volume usually adopted in questions of earth removal is one cubic yard, and accordingly the weights in the following table are expressed in terms of that unit : MATERIAL. Slate. Granite. Sand- stone. Chalk. Clay. Gravel. Mud. WEIGHT (cwts. per cu. yd.) 43 42 39 36 31 30 25 Volumes of Prismoidal Solids. To find the volume of any irregular solid having two parallel faces or ends, find the average 320 MATHEMATICS FOR ENGINEERS cross-section parallel to these faces and multiply by the axial distance between them. + 4M} Then- volume = {A and average section = - {A + B + 4M} where L = axial length; and A, B, and M are the end sections and the middle section respectively. Example i. A solid with vertical sides. Let the base be horizontal and all the sides be vertical as in excavating foundations for a house. Referring to Fig. 171 A = (I2 + IO) X 20 = 220 sq. ft. _ (4 + 8) _ (7 + I0 ) x 20 = 170 sq. ft. L= 50 /. Volume = ^{220 + 120 + ( 4 x 170)} = 8500 cu. ft. = 314-8 cu. yds. Fig. 171. Fig. 172. Example 2. Calculate the weight of clay removed in making the simple wedge-shaped excavation shown in Fig. 172. In this case A = \ x 12 x 84 = 504 sq. ft. B = ^ x 20 x 140 = 1400 sq. ft. M = x i6x 112 = 896 sq. ft. and L = 60 ft. 60 then volume = -^-{504 + 1400 + (4 x 896)} = 54880 cu. ft. = 2032-6 cu. yds. and weight of clay removed = 2 32 ' 6 X 3I tons 20 = 3151 tons. CALCULATION OF EARTHWORK VOLUMES 321 Example 3. A more difficult wedge-shaped excavation, which is shown in Fig. 173. To calculate the volume of earth removed. The earth removed is represented by a wedge figure ADEF and a tri- angular pyramid AFBC. The volume of the pyra- mid can be found if the area of the base is first obtained. Axis of Sloping Ground whence AD = 62-65 sin L BAD = . 62-65 but sin L BAC = sin ( 1 80 -BAD) = sin L BAD and hence sin ,1 BAC =7^ 62-65 Also AC = 250 62-65 = I8 7'35- Fig. 173. Wedge-shaped Excavation. Then area of triangle ABC = BA . AC sin L BAC = \ x 78 x 187-4 x 60 = 7000 sq. ft. 62-65 Height of pyramid = 30 ft. /. Volume of pyramid = \ x 30 x 7000 = 70000 cu. ft. = 2592 cu. yds. For the volume of the prismoidal solid ADEF, using the general rule A = \ x 20 x 60 = 600 sq. ft. B = \ x 30 x 78 1170 sq. ft. M = \ x 25 x 69 = 862-5 an d L = 60 ft. .*. Volume = -^-{600 + 1170 + 3450} = 52200 cu. ft. = 1935 cu - y ds - .'. Total volume removed = 4527 cu. yds. Sections of Cuttings. It will be convenient at this stage to demonstrate the mode of calculation of the areas of simple sections. In Fig. 174 we have the first case, of a cutting whose sides are sloped and whose natural surface of ground DC is horizontal. Let AB be the base or " formation width " and let its value be 2. Y 322 MATHEMATICS FOR ENGINEERS GH = height from centre of base to the natural surface = h. 6 = inclination to the horizontal of the sloping sides. EC = horizontal projection of slope. Then cot is usually denoted by s; or, in other words, the slope of the sides is s horizontal to i vertical. EC FB = cot 6 = s and FC = FB x s = hs GC = half width of surface = a-\-hs. Area ABCD (/. e., the area of the section of the cutting) = -(DC+AB) xh = -(za+2a+2hs} = h(2a-\-hs) )M- JO. _ _ Fig. 174. Fig- 175- Fig. 175 shows the cutting section when the natural surface of the ground takes a slope DC. Let a. = inclination to the horizontal of the natural surface, and let cot a = r. CM and DN, though not equal, are called the " half-widths " of the section ; let these be represented by m and n respectively. To find m and n Also From (2) MG i , .. m - = tan a = --, so that MG = - m r r = tan = -, so that HK = - s s MK i m - = tan e = -, whence MK = - GK = GH+HK = h+- S (i) (2) (3) From (i) and (3) GK = MK-MG = - - - CALCULATION OF EARTHWORK VOLUMES 323 Hence- fcf.5-'.. 2_2 s s r and Similarly n = -^-(a+hs). To find the extreme heights CE and DF BE = HE-HB = m-a BE = cot = s, whence CE x s = BE .*. CExs = ma or CE = s 72 n and similarly DF = To find the area of the section Area ABCD = CDFE DFA-CBE FF /CE+FD\ ! .IBECE \ 2 J 2 2 if(m-\-n)(ma-\-na) (na}(na) 21 S S _(m a)(m a)} s 2s\ -\-2an m 2 a? -\-2amj nina z fm' 2 -\-n 2 -\-2mn2am2ann 2 a z Example 4. A cutting is to be made through ground having a transverse slope of 5 horizontal to i vertical, and the sides are to slope at ij horizontal to i vertical. If the formation width is Co ft. and the height of the cutting (at centre) is 12 ft., find the half-widths, the extreme heights and the area of the section. Adopting the notation as applied to Fig. 175 2a = 60, h = 12, s =' ij, and r = 5 Then- m = (a+hs) = -^-[30 + (12 x ij)] = 60 ft. : ^ -- (30 + 15) 36 ft. 324 MATHEMATICS FOR ENGINEERS CE =. =. ^^ = 6 " 3 = = 24 ft. Area = 5 I-25 n-a _ 36- 30 = g f 5 1-25 S mna 2 (60 x 36) 1-25 = looSsq. ft. Example 5. Volume of a cutting having symmetrical sides, the dimensions being as in Fig. 176. Calculate the volume of earth removed, if the cutting enters a hill normally to the slope of the latter and emerges at a vertical wall or cliff. Fig. 176. Cutting on a Hill. The volume is found by application of the general rule. Volume = ^{A+B+ 4 M} To find A h = 24, s = i, 2a = 40 /. Area = h(za + hs) = 24(40 + 36) = 1824 sq. ft. In the case of the other end section, h = o and thus B = o. For M h = 12, s = i, 20 = 40 Area = 12(40 + 18) = 696 sq. ft. also L = 170 ft. 170 Hence Volume = -^-{1824 + o + 2784} = 130600 cu. ft. or 4837 cu. yds. Example 6. To find the volume of a cutting having unequal sides. In this case, shown at (a), Fig. 177, the cutting enters the hill in an oblique direction, although the outcrop is vertical as before. The CALCULATION OF EARTHWORK VOLUMES 325 sides of the cutting slope at i horizontal to i vertical, while the natural surface of the ground slopes upward at 4^ horizontal to i vertical. The solid can be split up into a prismoidal solid SRFE, together with the two pyramids SE and RF. To deal first with the prismoidal solid SRFE : its volume can be found from the general rule Volume = ^{A + B + 4M}, and in order to find the values of A and B the lengths of BS and AR must first be found. Fig. 177. Cutting with Unequal Sides. Referring to (b), Fig. 177 CQ = m a, and m = (a+hs) Y 5 Also so that Hence Now Also Again and Hence r = s = h = 30, and 2a = 40 m = ^-(20+45) = 97'5 ft - CQ = 97-5- 20 = 77'5 f t- S = tan a = 4'5 SO = = 5 4'5 4'5 = I7 . 22 f t . BQ = 22^5 = 51 . 66 ft< * i-5 3 BS = BQ - SQ = 51-66 - 17-22 = 34-44 ft. v di n = ~-(a+hs) = ^(20+45) = 48-75 ft. f~T~S D PD = 48-75 - 20 = 28-75 ft - ^ = , whence PR = = 6-39 PD 4-5 4*5 AP = PD tan 6 = *^p- 19-17 ft - AR = AP + PR = 19-17 + 6 '39 = 25-56 ft- 326 MATHEMATICS FOR ENGINEERS We can now proceed to find the volume of the solid SRFE. A = xBSxBE = 1x34-44x150 = 2583 sq. ft. B = xARxAF = -1x25-56x100 = 1278 sq. ft. M = X3OXI25 = 1875 sq. ft. and L = 40 .'. Volume = ^{2583+1278+7500} = 75750 cu. ft. = 2805 cu. yds. To find the volume of the pyramid SE The base is the triangle CSB, of which the area = JxBSxCQ = *x 34-44x77-5 = 1335 sq. ft. The height = 150 ft., and hence Volume = - x -*~ x 1335 cu. yds. = 2471 cu. yds. To find the volume of the pyramid RF The base is the triangle ARD; and its area = xARxPD = 2X25-56x28-75 = 367-5 s q- ft - The height = 100 ft., and hence Volume = -x X 367-5 cu. yds. = 453-6 cu. yds. 3 / .*. Total volume = 2805 + 2471 + 453-6 = 5730 cu. yds. Cutting and Embankment continuously combined ; the Sides being Symmetrical. If a road or a railway track has to be constructed through undulating ground, both cuttings and embankments may be necessary. The cost of the road-making depends to a large extent on the " net " weight of earth removed, seeing that the earth may be transferred from the cutting to the embankment. The calculation of the net volume removed will be dealt with according to two methods : First Method. Example 7. A cutting is to be made through the hill AC (Fig. 1 78) 24^' and an embankment in the j^6QL-94O--H i050-^-772-V584'f. <* valley BC so as to give ^^~ JL^^^ a straight horizontal road t\S ? g *"-"JC ic el JR fromAtoB. The formation * "V. ~ ^ 7 ^ V '"* s jb__JJ<f width is to be 40 ft., and the >i i* sides of the cutting and the Fig. 178. embankment slope if hori- zontal to i vertical. Calculate the net weight of vegetable earth removed (25 cwt. per cu. yd.). The volume of the cutting will be found by considering it made up of three prismoidal solids, and the volume of the embankment will be CALCULATION OF EARTHWORK VOLUMES 327 found in the same way. Then the net volume is the difference of these separate volumes. Dealing with the portion between A and C, i. e., with the cutting : For the portion Aae A = o B = i8[40 + (18 x 1 1)] = 1287, since h = 18 M = 9[40 + (9 X 1 1)] = 501-75, since h = 9 L = 560 /. Volume = =^{o + 1287 + (4 x 501-75)} = 307400 cu. ft. For the portion abfe A = 1287 B = 15^0 + (15 X if)] = 994. since h = X 5 M= i6-5[40+ (16-5 X if)] = 1136, since h = 16-5 L = 940 .'. Volume = ^{1287 + 994 + (4 X 1136)} = 1,069,000 cu. ft. For the portion fbC A = 994 B = o M= 7 - 5 [ 4 o+ (7-5 x 1 1)] = 398 L = 1050 /. Volume = 6 5 {994 + o + (4 X 398)} = 453oo cu. ft. Thus the total volume removed to make the cutting = 307400 + 1,069,000 + 453000 = 1,829,400 cu. ft. Dealing with the embankment portion, viz. that from B to C : . For the solid Cch A = o B = 17^0 + (17 x i|)l = Il8 5 M = 8- 5 [ 4 o + (8-5 x if)] = 466 L = 772 .. Volume = 2|?{o + 1185 + (4 x 466)} = 392000 cu. ft. For the solid chid A =1185 6=1185 L = 584 M=n85 .-. Volume = 1185 x 584 = 692000 cu. ft. For the solid dl~B A =1185 B = o M=8-5[40 + (8-5 X if)] =466 L = 246 /. Volume = ^{1185 + o + (4 x 466)} = 125000 cu. ft. 328 MATHEMATICS FOR ENGINEERS Hence the total volume required for the embankment _ ( 3g2 _|_ 692 + 125) X I0 3 CU. ft. = 1,209,000 CU. ft. Then net volume removed = (1-829 1-209) x io 6 = 620000 cu. ft. or 22960 cu. yds. 22960 x 25 , and the net weight removed = - tons = 28700 tons Second Method. Example 8. Fig. 1 79 shows the longitudinal section of some rough ground through which the road AC is to be cut. The sides of the cutting and of the embankment slope at ij horizontal to i vertical, Lonqiludinol Section Fig. 179. Volume of Earth removed in making Road. and the road is to be 50 ft. wide. Calculate the net volume of earth removed in the making of the road. Divide the length AC into ten equal distances and erect mid- ordinates as shown. Scale off the lengths of these, which are the heights of the various sections. The areas of the sections at a, b, c, d, etc., can be found by cal- culation as before, or, if very great accuracy is not desired, the various sections may be drawn to scale and the areas thus determined. To illustrate the latter method : Draw DE = 50 ft., and also the lines DF and EG, having the required slope, viz. ij to i. Through R, the middle point of DE, erect a perpendicular RS, and along it mark distances like RM, RN, etc., to represent the respective heights of the sections : thus RM = 24, and RN = 48. Then to find the area of the section at a, which is really the figure DPQE, add the length of PQ to that of DE and multiply half the sum by RM. The area of the section at 6 is % (TV + DE) x RN, and so on. The areas of the CALCULATION OF EARTHWORK VOLUMES 329 respective sections are 2064, 5856, 4746, 1826, and 650 sq. ft., these being reckoned as positive ; and 650, 3744, 3444, 1386, and 496 sq. ft., these being regarded as negative. The average of all these sections, added according to sign, is 5422 sq. ft. or 602-4 sq. yds. Then the net volume of earth removed = 602-4 x 1000 = 602400 cu. yds. Cutting with Unequal Sides, in Varying Ground. First Method. To find the average cross-section of ground with twisted surface (Fig. 180) ; an end view being shown in Fig. 181. The surface slopes downwards to the left at A and to the right at B. Fig. 180. End View. Fig. 181. Let WL n lt h lt and r t be the half-widths, etc., for A; and w 2 , 2 , h. z , and r 2 the corresponding values for B. Then- Area of A = For the mid-section M m = , and 3 = 33 - and the area of M = - "-| /. Average cross-section A+B+4M 6 _ m 1 n l 6s 6a 2 6s 6s 330 MATHEMATICS FOR ENGINEERS Example 9. Find the average cross-section of ground with twisted surface, when the formation width is 20 ft. and the side-slopes are i horizontal to i vertical. At the one end of the embankment the height is 12 ft. and the natural surface of the ground slopes at 20 horizontal to i vertical downwards to the right; while at the other end the height is 6 ft. and the slope of the ground is 10 to i down- wards to the left. Adhering to the notation employed in the general description ; For the section A {10 + (12 X 1-5)} = 30-3 ft. {10 + 18} = 26-05 ft. 20 1-5 2O For the section B 10 + (6 x 1-5)} = 16-5 ft. - 10 - i * = , I o+ I . 5 ( I + 9) = 22-4 ft. Hence the average cross-section _ (30-3 x 26) + (16-5 x 22-4) + (46-8 x 48-4) 600 9 = 313-6 sq. yds. Second Method. Example 10. Calculate the volume of earth removed in making a cutting of which AE is a longitudinal centre section (Fig. 182). The formation width is 20 ft., the length of the cutting is 4 chains, the Fig. 182. 2O 1 Section oT B. Fig. 183. sections are equally spaced, and the slope of the sides is 2 horizontal to i vertical. All the sections slope downwards to the left, as indicated in Fig. 183. The heights of the sections, in feet above datum level, are : Section. Left. Centre. Right. A O O O B 2-6 4-6 6-6 C 4'i 6-4 8-7 D 4-0 6-2 8-4 E o o o The areas of the sections may be found by drawing to scale and CALCULATION OF EARTHWORK VOLUMES 331 then using the planimeter; and Simpson's rule can afterwards be employed, since there are an odd number of sections. The results in this case are as follows : Section. M Area A 10 10 O B 32 13-7 169-5 C 42-2 15-7 280 D 39-9 15-55 260 E 10 10 O Then the volume = {o + o + 4(169-5 + 260) + 2(280)} = 50100 cu. ft. or 1855 cu. yds. Surface Areas for Cuttings and Embankments. The area of land required for a cutting or an embankment can be determined when the half-widths of the various transverse sections are known ; the method of procedure being detailed in the following example : Example n. Fig. 184 represents the horizontal projection of the cutting dealt with in Example 10. Find the area of land required for this cutting if a space of 5 ft. between the outcrops and a fence be allowed. The width RM is the extreme width of the section, i. e., its value = m + n; accord- ingly, allowing 5 ft. on each side, the widths to be considered are of the form m + n + 10. Taking the values of m and n as in the previous example, the widths are as in the table : Fig. 184. Surface Area for Cutting Section. A B c D E m + n + 10 30 557 67-8 65H5 30 332 MATHEMATICS FOR ENGINEERS Applying Simpson's rule Area of land required = {30 + 3 + 4 (55 '7 + 6 5'45) + (2 x 67-8)} = 14960 sq. ft. or 1663 sq. yds. Volumes of Reservoirs. Example 12. Find the volume of water in the reservoir formed as shown in Fig. 185, when the water stands at a level of 45 ft. above datum level, the bottom of the reservoir being at the level 22 ft. Fig. 185. Volume of Reservoir. In the diagram the land is shown contoured, i. e., the line marked 40, for example, joins all points having the level 40 ft. above datum. The problem, then, is to find the volume of an irregular solid, and this may be done in either of two ways, viz. (a) By taking vertical sections. According to this method, we should find the extreme length of the reservoir, which is about 320 ft., and then draw the cross-sections at intervals of, say, 40 ft. The area of each cross-section would then be found, preferably by the planimeter, and the volume calculated by adding the areas according to Simpson's rule. This process is somewhat tedious, as each section must be plotted separately; and consequently it is better to proceed as in method (6). (b) By taking horizontal sections, i. e., sections at heights of 45, 40, 35, etc., ft. respectively. To find the area of the section at the height 45 ft., determine the area of the figure ABCD by means of the planimeter. This area is found to be 5-083 sq. ins. Now the linear scale is i" = 80 ft., and therefore each square inch of area on the paper represents 80 x 80 or CALCULATION OF EARTHWORK VOLUMES 333 6400 sq. ft. Thus the area of the section at the level of 45 ft. = 5' 8 3 x 6400 = 32500 sq. ft. ; and in the same way the areas at the levels 40, 35, 30, 25, and 22 ft. are 21550, 10560, 3780, 577, and o sq. ft. respectively. The length of the irregular solid is 23 ft., i. e., 45 22, and we may plot the various areas to a base of length'. 24000 - I6ooo - 8000 as indicated in Fig. 186. The area of the figure EFG, which is found to be 1-633, gives the volume of water in the reservoir, to some scale. In the actual drawing i" = 10 ft. (horizontally), and i" = 16000 sq. ft. (vertically), so that i sq. in. on the paper represents 10 x 16000 or 160000 cu. ft. Hence volume of the reservoir = 160000 x 1-633 = 261300 cu. ft. or its capacity = 1630000 gallons. Exercises 36. On the Calculation of Volumes and Weights of Earthwork. 1. Calculate the volume of the solid with vertical sides shown in Fig. 187. Fig. 187. 334 MATHEMATICS FOR ENGINEERS 2. Fig. 188 shows the plan of a wedge-shaped excavation, where the encircled figures indicate heights. Calculate the weight of clay removed in making the excavation. 3. Fig. 189 is the longitudinal section of some rough ground through which a straight horizontal road is to be cut, the width of the road being 64 ft. The soil is vegetable earth (25 cwts. per cu. yd.), and Af ROAD - ZOOO-yds. Fig. 189. the sides of the cutting and embankment slope at 2 horizontal to I vertical. Calculate the weight of earth removed in making the road, if the natural surface of the ground is horizontal. 4. Determine the area of land required for making the cutting from A to B in Fig. 189. The side-slopes are 2 horizontal to I vertical, the formation width is 64 ft., and a fence is to be built round the working at a distance of 6 ft. from the outcrops. 5. Calculate the capacity of a reservoir for which the horizontal sections at various heights have the values in the following table : Height above sea level (ft.) 180 170 1 60 155 IS 147 Area of section in sq. ft. 47200 31000 21700 19000 11300 o 6. The depth of a cutting at a point on the centre line is 20 ft., the width of the base being 30 ft. The slope of the bank is i hori- zontal to i vertical, and the sidelong slope of the ground is 12 horizontal to i vertical. Find the horizontal distances from the vertical centre plane to the top of each slope. 7. Find the volume of earth removed from a cutting, if the forma- tion width is 20 ft., the side-slopes are i to i, and the slope of the surface is 10 to i. The depth of the cutting at the first point is 25 ft. ; at the end of the cutting (200 ft. long) it is 30 ft. ; and half-way between it is 26 ft. 8. The base of a railway cutting is 32 ft. in width, the depth of the formation is 34 ft. below the centre line of the railway, the side-slopes are ij to i, and the surface of the ground falls i in 8. Calculate the half-breadths for the cutting. At a distance of i chain along the centre line the depth of forma- tion level is 28 ft., and at a distance of 2 chains it is 20 ft. Find the volume of earth to be removed. 9. On the centre line of a railway running due N. the difference in level between the natural ground and the formation level of the em- bankment is 5-6 ft., 8-4 ft. and 6 ft. at the 23rd, 24th, and 25th chain pegs respectively. The width of the formation level is 20 ft., and the sides of the embankment slope at 2 to i. CALCULATION OF EARTHWORK VOLUMES 335 The natural ground slopes down across the railway from E. to W. at i in 10. Determine at each chain peg the distances of the toes of the embankment from the centre line and the area of the cross-section ; determine also the volume of the embankment between the 23rd and 25th chain pegs. 10. A cutting runs due E. and W. through ground sloping N. and S. The formation level is 15 ft. below the surface centre line and is 20 ft. wide. The ground slopes upwards on the north side of the centre line i vertical to 6 horizontal, and on the south side the ground slopes downwards i vertical to 10 horizontal. The sides of the cutting slope I vertical to i horizontal. Calculate the positions of the outcrops. CHAPTER IX THE PLOTTING OF DIFFICULT CURVE EQUATIONS Plotting of Curves of the Type y = ax". The plotting in Chapter IV was of a rather elementary character in that integral powers only of the quantities concerned were introduced. All calculations could there be performed on the ordinary slide rule; e. g., such curves as that representing y = 5# 2 + 7^ possible. If, now, a formula occurs in were which one, say, of the quantities is raised to a fractional or negative power, and a curve is required to represent the connection between the two quantities for all values within a given range, the necessary cal- culations must be made by the aid of logs. Suitable substitutions will in some cases make these calcu- lations simpler, but unless great care is exercised over the arrangement of the calculations and the selection of suitable values for the quantities, Fig. 190. Curve of y = 2-2* 1 - 78 . . - 1 a great deal of time will be wasted. In fact, the method of tabulating values is of more importance than is the actual plotting. Example i. To plot the curve y = 2-2A 1 ' 75 , values of x ranging from o to 4. y 2*2#^* 7 ^ /. log y = log 2-2 + 1-75 log x. Arrange a table according to the following plan : In the first THE PLOTTING OF DIFFICULT CURVE EQUATIONS 337 column write the selected values of x; in the second column write the values of log x. With one setting of the slide rule the values of I- 75 log x can b 6 read off; and these must be written in the third column. In the fourth column we must write the values of log y. which are obtained by addition ; then the antilogs of the figures in column 4 will be the values of y in column 5. The advantage of working with columns rather than with lines is seen ; thus we write down all the values of log x before any figure is written in the third column, and this saves needless turning over of pages, etc. Table : X log AT i-75 log * + log 2-2 logy y 00 00 + -3424 00 5 1-699 = '3 01 - -527 + -3424 1-8154 654 i-o o o + -3424 3424 2-2 1 '5 1761 308 + -3424 6504 4-47 2-O 3010 527 + -3424 8694 7-40 2'5 3979 696 + -3424 1-0384 10-92 3-0 4771 835 + -3424 1-1774 15-04 3'5 '544 1 952 + -3424 1-2944 19-7 4-0 0021 1-054 + "3424 1-3964 24-91 The plotting is shown in Fig. 190. Use of the Log Log Scale on the Slide Rule. The use of the log log scale now placed on some slide rules would obviate a great amount of the calculation in this and similar examples. e. g., taking y = 2-2* 1 ' 75 and disregarding the factor 2-2 until the end logy = i -75 log x log (log y) = log 1-75 + log (log x) i.e., log Y = log i -75 + log X where Y = log y, and X = log x. Therefore if a length on the ordinary log scale, say the C scale, be added to a length on the log log scale, which is usually the extreme scale, the result on the log log scale will be that required. jf y = ^1-75 ; an d supposing the value of y is required when x = 2-5. Set the index of the C scale level with 2-5 on the log log scale ; move the cursor until over the power, 1-75- on the C scale : then the reading on the log log scale (4-96) is the value of 2-5"". Multip icatwn by 2-2 for y = 2-2*, can be done with one setting of the rule afte 338 MATHEMATICS FOR ENGINEERS X % vn y = 2-2* 1 - 75 2'5 4-96 10-92 all the powers have been found. The tabulation would in this case reduce to as an example. The log log scale is most useful for finding roots. E. g., to find >/432. Set 5 on the C scale level with 432 on the log log scale ; then the reading on the log log scale opposite the index of the C scale is 3-37, i. e., the 5th root of 432. Expansion Curves for Gases. The formula PV" = C, for the expansion or compression curves of gases, is of the same type as that in the last example. 600 10., , p ~ r 2O Values of u. Fig. 191. Expansion Curves for Gases. 25 30 In this formula p is the pressure in Ibs. per sq. in. or per sq. ft. and v is the " specific volume," i. e., volume in cu. ft. of i Ib. whilst n and C are constants varying with the conditions THE PLOTTING OF DIFFICULT CURVE EQUATIONS 339 Thus for air expanding adiabatically, . <?., without loss or gain of heat, n = 1-41 : for the gas in the cylinder of a gas engine n = J> 37 : f r isothermal expansion, i. e., expansion at constant temperature, n = i. It is instructive to plot two or three expan- sion curves on the same diagram, n alone varying, and thus to note the effect of this change. Example 2. Plot, on the same diagram and to the same scales, from v = 4 to v = 30, the curves representing the equations : (a) PV*-" = 2500, (6) pv = 2500, (c) pv = 2500. The plotting is shown in Fig. 191. Each equation is of the form pv n = C log p + n log v = log C or log p = log C n log v. Dealing with the separate cases (a) Adiabatic expansion of air; n = 1-41 log p = log 2500 1-41 log v. The arrangement of the table is as follows : 9 logy log 2500 1-41 log v log/> P 4 602 3.398- -847 2-55 1 356 7 845 1-191 2-207 161 10 o 1-41 1-988 97-3 14 146 1-615 1-783 60-7 18 255 1-770 1-628 42-5 20 301 1-834 1-564 36-6 24 380 1-945 1-453 28-4 27 43i 2 -O2 1-378 23-9 30 477 2-O82 1-316 20-7 (b) Expansion of superheated steam; n = 1-3. Values of v and log v are as above ; and the table is completed as shown : log 2500 1-3 logu log/) P 3-398 - 783 2-615 412 1-097 2-301 200 3 2-098 125 49 1-908 80-9 632 1-766 58-3 691 1-707 5 '9 794 1-604 40-2 860 1-538 34'5 920 1-478 30-1 340 MATHEMATICS FOR ENGINEERS (c) Isothermal expansion ; n = i . V 4 7 10 M 18 20 24 27 30 - 2 5 625 357 250 179 139 125 IO4 92-6 83-3 P V It will be seen that the bigger the value of n, the steeper is the curve, or, in other words, the slope of the curve depends on n. All these curves are hyperbolas, that for (c) being the special case of the rectangular hyperbola. A Construction for drawing Curves of the Type pv" = C. In this construction the position of one point on the curve must be known. LetP, (piVj, be the given point. Fig. 192. Choose any angle o, say 30 : set it off as shown, and also an angle ft, calculated from the equation tan ft = i (i tan a)r. Draw the horizon- tal PM to meet OA in M, and the vertical PR to meet OR in R. Draw MN making 45 with ON and RS making 45 with OR. A horizontal through N meets a vertical through S in Q ; then Q is another point on Fig. 192. Construction for Curves of the pv n = C type. the curve : also the construction for the point L on the other side of P is indicated. ProoJ oj the Method MT TN ^ ^- ^- tan a = OT = OT = tan - SW RW Pi v 9 v but OW ' " OW " tan/3 = I (i tan a) THE PLOTTING OF DIFFICULT CURVE EQUATIONS 341 =* *>2 W Pi or p^f = p z vf, i. e., pv* = Constant. Example 3. If n = -9 and a = 30, calculate the value of /3. tan /3 = i (i - tan 30)^ = i-(i .5774)1-111 = I-(-4226) l ' ul Let x = (-4226) 1 ' 111 then log* = i -i 1 1 x log -4226 = i-in x 1-6259 = i -i 1 1 + -696 = 1-585 whence x = -3846 Then tan j3 = i - -3846 = -6154 = tan 31 36' or = 31 36'. If n = i, then tan ft = tan a i. e., ft = a. Note. 30 is rather a large angle for a if the range to be covered is small. Accordingly, the value of /? is stated here, for a = 10 and n = 1-37. tan/3 = i (i tan io)^7 = i (i 1763)' 73 = i -871 = -129 .'. ft = 7 8 ai'. Example 4. A tube 3* internal and 8* external diameter is sub- jected to a collapsing pressure of 5 tons per sq. in. : show by curves the radial and circular stresses everywhere, it being given that at a point r ins. from the axis of the cylinder T? T^ The radial stress p = A + -, and the circular stress q = A , Note that p = 5 tons per sq. in. when r = 4"; and p = o when r = i -5* ; and the object is to first find the values of the constants A and B from the data given. From the given conditions A+ B 2-25 Subtracting 5 = B (^-5 = B (-0625 --4444) 5 = 382B or B -=-'3-x. 342 MATHEMATICS FOR ENGINEERS Also Hence 5 = A + (-0625 x - 13-1) = A --818 A = 5-818. p = ">-8i8 5. a = 5-818 -f- i f */ y ' 3 J yZ [Note that (p + q) = 11-636 = constant. The material is sub- jected to crushing stresses p and q in two directions at right angles to one another and in the plane of the paper : therefore dimensions at z: 1-5 3-5 Fig- 193- Curves of Radial and Hoop Stresses. right angles to the paper must elongate by an amount proportional to (P + q)- If the cross-section is to remain plane this elongation must be constant ; hence (p + q) must also be constant.] To calculate values of p and q the table would be set out as follows : r " 13-1 I3 .! -Sifi I3>1 * r* J r* ~ g 5 OI r i P 1-5 2-25 5-818 11-636 2-O 4 3-275 9-093 2'543 2-5 6-25 2-095 7-9I3 3'O 9 1-455 7-273 4-363 3'5 4-0 12-25 16 1-068 818 6-886 6-636 4-750 5 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 343 It is customary, how- Circular v/ on Hoop Stress The curves are shown plotted in Fig. 193. ever, to plot the curves of radial and hoop stress in the manner shown in Fig. 194, where curve (i) gives the radial stress at any point between a and b, and curve (2) gives the circular or hoop stress at any point between a l and b t . Example 5. According to a certain scheme (refer to p. 212), the depreciation fund in connection with a machine can be expressed by- Fig. 194. Curves of Radial and Koop where Stresses. D = amount contributed yearly to the sinking fund, and ioor = percentage rate of interest allowed on same. For a machine whose initial value is 500 and scrap value is 80, D is found to be 14 145., if 3% interest per annum be allowed. If the life of the machine is 21 years, plot a curve to show the state of the sinking fund at any time, i. e., plot the curve A = -2 {i-o3 i}, n varying from o to 21. wo A Si 350 / / 280 <> y <& < >< & - ^210 &] <n u A -S A r ~JC\ / ~ A ~ O ^ . . i 01 2 3 4- 5 6 7 8 9 10 II 12 13 14- 15 16 17 18 '9 2O 21 Values of n Fig. 195. Curve of Depreciation Fund for Machine. 344 MATHEMATICS FOR ENGINEERS It will be advisable to work out i -03" separately. Let 1-03" = x; then log* = wlog 1-03 = -oi28 also 14-7 ^ = 490. 03 Taking a few values only for n, between o and 21, the tabulation will be as follows : n oi28n = log x X X I !*(,_,).,. A 03 o I o 4 0512 1-126 126 61-7 8 1024 1-266 266 130 JO 128 1-343 343 1 68 12 1536 1-424 424 207-5 16 2048 1-603 603 296 21 2688 1-857 857 420 and the plotting is shown in Fig. 195. Equations to the Conic Sections. A knowledge of the form of the curve that represents some particular type of equation may ensure a great saving of time and thought. Values for the variables need not then be chosen at random and beyond the range of the curves. The equations to the conic sections are here given because many of the curves occurring in practice are of one of these forms. The Ellipse. If the origin be taken at the centre of the ellipse, the equation is where a and 6 are the half-major and half-minor axes respectively, or the maximum values of x and y. If the equation is given in a slightly different form, it should be put into the standard form before any values are selected. Example 6. Plot the curve representing the equation 3# 2 + $y z = 60 3* 2 + 5y* = 60 is the equation of an ellipse, and can be written 60 60 "~ i. e., the equation is divided throughout by 60, so that the right-hand side becomes unity. Thus 2O so that 12 and and a z = 20, and a = 4-472 fc 2 = 12, and b = 3-464. Hence the range of x is from 4-472 to +4-472, and no lower or higher values respectively should be taken. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 345 If the values of y are to be calculated, we have 5y 2 = 60 3# 2 y* = 12 -6x* y= Vi2 -6# a Dealing only with one-half of the ellipse, the table of values reads X X 2 12 -6# 2 y2 y o 12 O 12 3-464 I I 12 '6 n-4 3-38 2 4 12 2-4 9-6 3-10 3 9 12 - 5'4 6-6 2-55 4 16 12 9-6 2-4 J -547 4H72 20 12 12 o o . The other half can be obtained by projection, and Fig. 196 is plotted. If the graphic method of drawing an ellipse is known this Fig. 196. Curve of Equation to Ellipse. calculation is unnecessary : all that is required from the equation being the lengths of the axes. An application of the ellipse is found in the Ellipse of Stress, in the subject of Strengths of Materials. It is required to deter- mine the magnitude and direction of the resultant stress on a 346 MATHEMATICS FOR ENGINEERS plane BD, due to the stresses /j and / 2 acting as indicated in Fig- 197- It is found that the resultant stress / = \//! 2 cos 2 + / 2 2 sin 2 0, and if a is the angle made with/! tan a = Y tan # /i If an ellipse be constructed with axes to represent the original stresses, the resultant stress can very easily be read from it. Along OQ in Fig. 198 and perpendicular to BD, mark off a length OQ to represent /j, and a length OR to represent / 2 . Draw Fig. 197- Ellipse of Stress. a horizontal QM to meet a vertical PR in P; then OP represents /and ^.MOP = a. To show that P lies on an ellipse, we must prove that the equa- yv2 /y2 tion governing P's position is of the nature -^ -f- j- z =i. OM = OQ cos =f 1 cos 6 MP = RN = OR sin 6 =/ 2 sin :. (PO) 2 = (OM) 2 + (MP) 2 = /! 2 cos 2 +/ 2 sin 2 = f i. e., OP =/ If the origin is at O, and x and y are the co-ordinates of P then x = MP =/ 2 sin 0, and y = OM = / x cos e X V y- = sin 6, f-ss cos 0. /2 /I but sin 2 6 + cos 2 = i for all values of x 2 v 2 4- < T f 2 ~ / 2 * h i or P lies on an ellipse the lengths of whose axes are 2/ 2 and 2/ x . THE PLOTTING OF DIFFICULT CURVE EQUATIONS 347 The circle may be regarded as a special case of the ellipse, x z v 2 where a = b, i. e., -^ + ^ = * or x z + y 2 = a 2 , a being the radius of the circle. e. g., 5*.+ 5 y 2 = 45 can be written x* -f y z = g, which represents a circle of radius 3 units. The Parabola. If the axis is horizontal, and the vertex at the origin, then the equation is y 2 = 40*. If the axis is vertical, the equation is x z = qay, where 40 = length of the " latus rectum," the chord through the focus perpendicular to the axis. To make the investigation more general, let x be changed to x-\- c = say, x-\-j: and y to y+c x = say, y+n-45; also let 4 = -2. Then the case will be that of the parabola having a latus rectum of -2, and the axis will be vertical, with the vertex at the point 7, -11-45- The equation is (*+7) 2 = 5* 2 +7*+2-45-H'45 = y or y = 5* 2 +7*~ 9- (This curve is shown plotted in Fig. 88.) Conversely, the equation y = $x z + jx 9 might be put into the standard form, thus y = 5(*+i-4*-i-8) = 5(*+7) 2 -'45 = (*+7) 2 which equation is of the form 4#Y = X a where 4* = T> Y = y-HMS. and X = *+7- This analysis is useful if the position of the vertex, say, is desired and the curve itself is not needed. (Compare maximum and minimum values.) For the parabolas occurring in practical problems the simpler forms are sufficient. MATHEMATICS FOR ENGINEERS The Hyperbola. If the centre of the hyperbola is at the origin, the equation is ^ 2 _y 2 _ a 2 b 2 ~ where 20, = the length of the transverse axis (along the x axis) 2b = the length of the conjugate axis (along the y axis). No values should be taken for x between a and -\-a, for there is no part of the curve there. Example 7. Plot the curve representing the equation 2# 2 - 5V 2 = 4 8 - (Fig- I99-) By dividing throughout by 48 the equation may be written so that a = ^24 = 4-9 and & = Vg-6 = 3-1 If a rectangle be constructed by verticals through x = 4-9 and + 4-9, and horizontals through y = 3-1 and + 3-1, the diagonals of this rectangle will be the " asymptotes " of the hyperbola, i. e., the boundaries of the curves are known. To calculate values - 5 y 2 = 48-2*2 5 y z = 2*2-48 y 2 = -4# 2 -9-6 y = V'4# 2 9-6 i. e., an expression is found for y in terms of x, The table of values reads : X x z 4*2 9-6 y z y 4'9 24 9-6 9-6 o o 5'0 25 10 9-6 4 -632 5'5 30-3 I2'I 9-6 2-5 1-58 6-0 36 14-4 - 9-6 4-8 2-19 This is the calculation for one branch of the curve only ; and the other branch may be obtained by writing x for + x throughout, e. g., when x = 5, y = -632 ; therefore project across. _. _ T a 2 a 2 If a = b, then r a 2 or * 2 -y 2 = a 2 . For this case the asymptotes are at right angles, and the hyperbola is rectangular. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 349 To find the equation oj the hyperbola when rejerred to the asymptotes as axes (see Fig. 199.) From- P a point on the curve, draw PN parallel to OF and PM parallel to OE. Let PM = p, / \ \ \ y \ i X 1 - <* A \ Fig. 199. The Hyperbola. and PN = q ; then the co-ordinates, when the asymptotes are axes, are (p, q}. Note that PN and PM are parallel to the asymptotes, and not perpendicular to them. Let L EOA = a ; then tan a = - i.e., also COS a = and sin a = OM = NP = q. QM = OM = ML = q (From equality of angles.) PL = PM-ML = p-q PQ = PM+MQ = p+q PR h . V & PL = = also P-9 = \ or QS . i.e., Va*+b* (2) 350 MATHEMATICS FOR ENGINEERS By subtracting (i) from (2) = a?+b z ..... since -= - 2 = I a* o* But a and b are constants, therefore the product of the co- ordinates p and q is constant : this is a most important relation. If the hyperbola is rectangular, b = a (the asymptotes being at right angles) a 8 and -pa = 2 Compare the equation PV = C, for the isothermal expansion of a gas. Example 8. Find the equation of the hyperbola x 3 $y* = 3, referred to its asymptotes. Answer : Pq = i. Exercises 37. On the plotting of Equations of the Type y=ax" + b. 1. Plot, for values of x ranging from i to 9, the curve y = 5-j6x 1 ' 29 . 2. Plot the curve zy = -od^x"" from x = o to x = 2. 3. Plot on the same axes the curves y 1 =^-2x 1 ' 63 and y t = ^i* 3 ' 47 and by adding corresponding ordinates obtain the curve y = 4-2* 1 ' 83 + '3i.* 3 ' 47 . {x to range from -2 to 3-5.} 4. Plot, from y = -5 to y = +'5, a curve to give values of C, when C = 1-69 (log, 3 - -j ) 5. Formulae given for High Dams are as follows : where x = depth in feet of a given point from the top y horizontal distance in feet from such point to flank of dam z = horizontal distance in feet from such a point to face of dam P = safe pressure in tons per sq. ft. on the masonry Draw the section of a dam 30 ft. deep, allowing P = 4-5. 6. For a steam engine, if x = mean pressure (absolute) expressed as a percentage of the initial pressure (absolute), and y = cut-off expressed as a percentage of the stroke, then * = y(5-6o5-iogy)- Plot a curve giving values of x for values of y between o and 70. 7. If a number of observations have been made, say, for a length THE PLOTTING OF DIFFICULT CURVE EQUATIONS 351 of a chain line in a survey, then the probable error e of the mean of the observations can be calculated from where r difference between any observation and the mean observa- tion and n = number of observations. If 2 f 2 _. j. 2> p^t a curve to give values of e for values of n between 2 and 30. ///^7>v^A Worm 8. Plot on the same axes the curves (a) pv 1 ' 13 = 4000 and (6) pv' 8 = 2540, v ranging from 4 to 32. 9. In Fig. 200, T^T-rA |Worm -Wheel. D = the outside diameter of a worm wheel = 2A( i cos -j + d. If d = 4 and A = -75, show by a graph the Fig- 200. variation in D due to a variation in a from 20 to 60. 10. The calculated efficiency jj of worm gearing is found from _ tan a(i n tan a) p + tan a where /x = coefficient of friction and a = angle of the worm. If p. = -15, plot a curve to show efficiencies for angles from o to 50. 11. The ideal efficiency 17 of a gas engine is given by 77 = i f-J If n = 1-41, and r = compression ratio, plot a curve giving the efficiency for any compression ratio between 3 and 18. 12. A machine costs 500 ; its value as scrap is 80. Plot curves to show the state of the depreciation fund as reckoned by the two methods (a) Equal amounts put away each year. (b) A constant percentage of the value of the preceding year set aside each year. The fund at the end of n years = 500 [i(i 0836)**], and the life of the machine is 21 years. 13. The capacity K per foot of a single telegraph wire far removed from the earth is K = ^4-^ microfarads. Plot a curve to 2log---6i8 give the capacity for wires for which the ratio - increases from 500 to 20000. 14. Mutton's formula for wind pressure on a plane inclined to the actual direction of the wind is where P = pressure on a plane at right angles to the direction of the wind, p = pressure on a surface inclined at 6 to the direction of the wind. If P = 20 Ibs. per sq. ft., plot a curve giving values of p for any angle up to 90. 352 MATHEMATICS FOR ENGINEERS 15. Plot a curve showing the H.P. transmitted by a belt lapping 1 80 round a pulley for values of the velocity v from o to 140, the coefficient of friction p being -2. T = 350, w = -4, g = 32-2, 6 = angle of lap in radians. 16. Aspinall gives as a rule for determining the resistance to motion of trains T 65-82 where R = resistance in Ibs. per ton, V = velocity in miles per hour. Plot a curve to give values of R for all velocities up to 55 m.p.h. 17. Find the value of r (the ratio of expansion), which makes W (brake energy per Ib. of steam) a maximum. r --27 00833 , ^+-000903 18. The efficiency j of a three-stage air compressor with spray injection is given by where n = 1-2 and r = ratio of compression. Plot a curve giving the efficiency for any compression ratio between 2 and 12. 19. Determine the length of the latus rectum and also the co- ordinates of the vertex of the parabola $y = 2X z nx2y. 20. A rectangular block is subjected to a tensile stress of 5 tons per sq. in. and a compressive stress of 3 tons per sq. in. Draw the ellipse of stress and read off the magnitude and direction of the resultant stress on the plane whose normal is inclined at 40 to the first stress. [Hint. Refer to p. 346.] Curves representing Exponential Functions. To plot the curve y = e x , where e has its usual value, one may work directly from the tables, or a preliminary transformation of the formulae may be necessary. If tables of powers of e are to hand, the values of y corresponding to certain values of x are read off at a glance ; and in such a case the values of x selected are those appearing in these tables. Example 9. Plot the curves y = e x and y = e-* from x 4 to +4. From Table XI at the end of the book the figures are found thus : X - 4 - 3 2 I O I 2 3 4 y = e x 0183 0498 1353 3679 I 2-7183 7-3891 20-08 54'6 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 353 When x = - 4, - is required, and this is found in the 3 rd column. x = 3, e 3 is required, and this is found in the 2nd column. The plotting for these figures is shown in Fig. 201, by the curve (i). If tables of powers of e are not available, proceed as follows : y = e*, and therefore log y = x log e = -4343* and the table is arranged thus X 4343* = log x y 2 8686 7-389 Having drawn the curve y e?, draw the tangent to it at some point and measure its slope; and it will be found that the value of the slope is also the value of the ordinate to the point of contact of the tangent and the curve. Thus, the tangent is drawn to touch the curve at the point for which x = 3-5 : its slope is measured and found to be 33, and this is seen to be the value of y when x = 3-5. If for x we write x, i. e., we plot the curve y = e~*, we find that this gives a curve exactly similar to the last, but on the other side of the y axis : such would be expected, since x must now be measured as positive towards the left instead of to the right. The curve y = e~* is shown plotted in Fig. 201, and is curve (2). All equations of this type will be represented by exactly the same form of curve, drawn to different scales. A A 354 MATHEMATICS FOR ENGINEERS Example 10. To plot the curve y = e 3x . Write this as Y = e*, where Y = y and X = 3*. Plot the curve Y = e x exactly as before, and then alter the hori- zontal scale in such a way that i on it now reads J, and so on. For X = 3* i. e., construction scale = 3 x required scale construction scale or required scale = Example u. Plot the curve 5^ = 4K This can be written i. e., where Y = 4 Y = , X = -x. 4 7 Hence, plot Y = e x from the tables, and alter both scales in such a way that the New scale for y = - x construction scale > ** 7 ^ > > so that where the vertical construction scale reads 5, 4 must be written ; and 7 must be written in place of i along the horizontal. Example 12. If the E.M.F. is suddenly removed from a circuit containing resistance R, and self-induction (coefficient of self -inductance L), the current C at any time t after removal of the E.M.F. is given by the equation B< C = C e~i Plot a curve to show the dying away of the current for the case when C = 50 amps, R = -32 ohm, and L = -004 henry. Substituting the numerical values ^32* C = 50<T-<x> 4 = so*- 80 ' It will be sufficient to plot values of C for values of / between t = o and -05 sec. C = so*- 80 ' C = e-" 1 [ C is spoken of as C bar] f* where C = - and T = 8o/ 50 If the maximum value of t is -05, the maximum value of T must be 80 x -05, i. e., 4. Hence from the tables : T . . . 5 i 2 3 4 Ci.e.. e~ T i 6065 3679 1353 0498 0183 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 355 These values are shown plotted in Fig. 202, and then the scales are altered so that i on the vertical becomes 50, and i on the hori- zontal becomes Q . oo 5CL 45 40 ^ c I $ Q C*" So 15 *> 5 tt v A A i\ * \ ^ \ C-SOe" 60 * ? V \ > 3 V & X \ X ^_ /OI25 ^ [ l^ Consl-r-? Seal* * O 77-uo Sco/e 025 -O375 "OS Fig. 202. " Dying-away " of Current m an Electric Circuit. The saving of time and thought in the calculation of values more than compensates for the somewhat awkward scales that may result, and even this difficulty may be avoided by choosing the original or construction scales suitably. If it is found that the necessary values of the x or t cannot readily be used, i. e., if values are necessary for x for which no values of <?, etc., are given in the table, recourse must be made to calculation. In this case the work would be arranged thus : log C = log 50 - Sot log e = 1-699 80 y. -4343' = 1-699- 34 -74/ t 1-699- 3474' logC C 0025 005 etc. 05 i -699 o 1-699- 1-737 1-699 1-962 50 9162 356 MATHEMATICS FOR ENGINEERS Example 13. If a pull t is applied at one end of a belt passing over a pulley and lapping an angle 6 (radians), the pull T at the other end is greatly increased owing to the friction between the belt and the pulley. If ft, = coefficient of friction between belt and pulley T = lw* Plot a curve to show values of T as 6 increases from o to 180, taking t = 40, and /i = -3. The angle 6 ranges from o to 3-14. (n radians = 180.) 100 4o 5 1 i-S ~ -5^ 3 Values of ft Fig. 203. Pull on a Belt. It will be rather more convenient in this case to calculate Substituting values T = qoe' 39 Then log T = log 40 + -3^ log e = 1-6021 + -3 x -43430 = 1-6021 + -13030 6 1-6021 + -1303$ logT T o 1-6021 + o 1-6021 40 5 + -0652 I -6673 46-5 i-o + -1303 1-7324 54 i'5 + '*955 1-7976 62-8 2-O + -2606 1-8627 72-9 2'5 + -3258 1-9279 84-7 3-0 + -3909 1-9930 98-4 3-14 + -4180 2-O2OI 104-7 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 357 i. e., when the belt is in contact for half the circumference of the pulley the tension is increased in the proportion of 2-6 to i. In practice a ratio of 2 to i is very often adopted. The plotting for this example is shown in Fig. 203. Example 14. If an electric condenser of capacity K has its coats connected by a wire of resistance R, the relation between the charge q at any time t sees, and the initial charge q at zero sec. is given by 3. = e~Es. ft Find the time that elapses before the charge falls to a value = - X initial charge, and indicate the form of the curve which repre- 6 sents the discharge. If * = RK, then q = q^ = & = ^q i. e., the charge falls to - 5 f its initial value in time RK sees. 2*710 This time is termed the " time constant " of the condenser circuit. The curve representing this discharge would be similar to that plotted for Example 12, viz. in Fig. 202. The Catenary. Referring to the curves y = e* and y = e~ x if the " mean " curve of these is drawn it will represent the equation e* + e-* y = - - i. e.. y = cosh x. ' 2 This curve is known as the " catenary " ; and it is the curve taken by a cable or wire hanging freely under its own weight. The catenary when inverted is the theoretically correct shape for an arch carrying a uniform load per foot curve of the arch. If the cable is strained to a horizontal tension of H Ibs., and the weight per foot run of the cable is w Ibs., then the equation becomes * _ y __fl_+_l c 2 H where c = - The proof of this rule is rather difficult, and is given in Volume II of Mathematics for Engineers. From what has already been mentioned it should be seen that the catenary is the curve y = cosh x with the scales in both direc- tions multiplied by c, since its equation can be written 358 MATHEMATICS FOR ENGINEERS Provided Y = y - and c X=- then U = Therefore, to plot any catenary one can select values of x, read off corresponding values of cosh x from the tables and plot one against the other, afterwards multiplying both scales by c. If a definite span is suggested, the range of values for X must be selected in the manner indicated in the following example. Cor\st"rvr Scale. Fig. 204. The Catenary. Example 15. A cable weighing 3*5 Ibs. per ft. has a span of 50 ft., and is strained to a tension of 40 Ibs. Draw the curve representing the form of the cable. Find the sag, and the tension at 10 ft. from the centre. Here C = = H-42. 3'5 Also the span is to be 50 ft., i. e., on the " new " or " final " scale 25 ft. must be represented on either side of the centre line. But, new scale = c x construction scale .*. 25 ft. on new scale 11-42 x X on construction scale or X = - = 2-19 11-42 so that no values of X need be taken beyond, say, 2-2. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 359 Taking values of X from o to 2-2, the values of cosh X are found from Table XI, thus : X '5 i-o i'5 2-O 2-2 cosh X = Y i 1-128 1-543 2'352 3-762 4-568 The curve is now plotted, as in Fig. 204, and then for unity on the construction scales 11-42 must be written, and the 25 ft. is marked off on either side of the centre line. The sag is read off as 39-3 ft., using the final scale. The tension in the cable at any point is measured by the ordinate to the curve multiplied by w; e.g., the tension at 10 ft. from the centre = 3-5 x 15-9 = 55-6 Ibs. Exercises 38. On the plotting of Curves representing Exponential Functions. 1. Plot, for values of x from -8 to 2-9, the curve y = 2e~*. Find its slope when x = 1-6. 2. Plot the curve y = -250-$* from x = o to x = 15. 3. Plot, from # = 5 to #=+3, the curve y = -021 x 1-62*. 4. If C = C e~ at , C = 14-6, a = 410, and t ranges from o to -023, represent by a graph the change in C (the dying-away of a current). 5. Plot a curve to give the tension T at one end of a belt for various coefficients of friction p,; the angle of lap (6 radians) being constant. Given that T = lev*, 6 = 165, and / = 50 ; p. ranges from -i to -35. 6. A cable weighing 2-18 Ibs. per ft. and strained to a tension of 56 Ibs. hangs freely. Depict the form taken by the cable when the span is 30 ft., and find the tension in it 12 ft. from the centre. Rt 7. C = 48-7(1 e~i*). If R = -56, L = -008, plot a current-time (C /) curve for values of t from o to -062. 8. Trace a graph to show the drop in electric potential down a uniform conductor, if the potential at the receiving end is 200 volts, the resistance per kilometre r of the conductor is 10 ohms, the leakage g of the insulation is -5 x io~ 6 megohms per kilometre, and the dis- tance from the " home " end to the receiving end is 500 kilometres. If e = the potential at distance x from the receiving end e = 200 cosh Vgr . x. Graphs of Sine Functions. Consider the equation y = sin x. We have already seen in Chapter VI that as the angle x increases from o to 90, the sine y increases from o to i; and as x increases from 90 to 180, y decreases from i to o. Continuing into the 3rd and 4th quadrants : for x increasing from 180 to 270, y decreases from o to i ; and for x increasing further to 360, y increases from i to o. 360 MATHEMATICS FOR ENGINEERS After 360 has been reached the cycle of changes is repeated, i. e., 360 is what is called the period, for the function y = sin x. Because y and x are connected by a law, we conclude that the changes will not be abrupt or disjointed, or in other words, the curve representing y = sin x will be a smooth one. The sine curve is perhaps the most familiar of ah 1 curves, there being so many instances of periodic variation in nature. Thus, if a curve be plotted showing the variation in the magnetic declination of a place over a number of years, its form will be that of a sine curve : so also for a curve showing the mean tem- perature, considered over a number of years, for each week of the year. -2 -1-0 l/=SLM.JG Radionsi Angle 300" 360" Fig. 205. Sine Curve. Sine curves occur frequently in engineering theory and practice ; in fact, a sine curve results whenever uniform circular motion is represented to a straight line base. All sine curves are of the same nature, and therefore it is neces- sary to carefully study one case, and that the simplest, to serve as a basis. To plot y = sin x : select values of x between o and 90, thus : x degs. o 25 45 60 80 90 y o 423 707 866 985 i Choose suitable scales so as to admit the full period to be plotted and plot for these values, as in Fig. 205. No further recourse to the tables is necessary, this portion of the curve being simply drawn out three times. For sin 100 = sin (180 100) = sin 80 and therefore for 10 to the right of 90 the value of y is the same as that for 10 to the left of it, i. e., the curve already drawn can THE PLOTTING OF DIFFICULT CURVE EQUATIONS 361 be traced and pricked through to give the portion of the curve between x = 90 and x = 180. Again, sin 205 = sin (180 + 25) = sin 25 and sin 240 = sin (180 + 60) = sin 60 *. e., the 3rd portion of the curve is the ist portion " folded over " the horizontal axis. Similarly, the 4th will correspond to the 2nd " folded over " ; and accordingly we need only concern ourselves with calculations for the ist quarter of the curve. The maximum value of y, viz. i, is spoken of as the amplitude of the function. Thus in the case of a swinging pendulum, the greatest distance on either side of its centre position is the amplitude of its motion. If y = 5 sin x, then the amplitude is 5, and the curve could be obtained from y = sin x by multiplying the vertical scale by 5. Example 16. Plot the curve y = -5 sin 4*. v Writing this as f- = sin 4* o or Y = sin X [where Y = - = 2y, and X = 4*] we see that the simple sine function is obtained. Accordingly we plot the curve Y = sin X (making use of the table on p. 360), and then alter both scales so that x = and y = 360 Dealing with the last example, we see that the period is or 90 ; i. e.,ii x is multiplied by 4, the period must be divided by 4. Similarly for the curve representing y = sin \x, the period would be 360 -r r ~ 1800. We thus obtain the important rule : " To obtain the period for a ' sine ' function, divide 360 by the coefficient of x or t (whichever letter is adopted for the base or ' independent variable ' ") or briefly 360 Period in degrees = eoeffieient ot ]^T, Since 27r radians = 360, wherever we have written 360 above we should write 2ir, if the angle is to be expressed in radians, >'. e., the period in radians or seconds (of time) 27T = coefficient of the x or t Thus if y = 4 sin 6x Period = ->- = - 6 3 and amplitude = 4 Period = = 1 or 6o< 362 MATHEMATICS FOR ENGINEERS Example 17. The current in an electric circuit at any time / sees. is given by the expression C = 4-5 sin loowt. Plot a curve to show the change in the current for a complete period. The general formula is C = C sin 2nft, where / = number of cycles per second = frequency. In this case 2irf = IOOTT, .'. / = 50. If / = 50, the time for one cycle, or the period, must = ^ = -02 sec. Thus the periodic time = -02 sec. Notice that the period is given in terms of seconds (of time) in this case, and not in degrees. The same periodic time would have been obtained if our previous rule had been applied, for ,-. . , 2ir 2ir Period = ~ T-. = - = -02 sec. ~ T coeff. of IOOTT Fig. 206. Change in Current in Circuit. Either of two methods can be used for the calculation of values (a) Plotting from the simple sine function. According to this scheme write the equation in the form C = sin looirt 4 '5 C = sin T , and T = ioont 4'5 or where Hence to plot the curve (Fig. 206) C = sin T, select values of T between o and |- (o and 1-571), and thus read off values for C so that the first quarter of the curve can be plotted, remembering always that the base must be numbered in radians. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 363 Values for this portion would be of this character : T o 2 4 7854 96 i-i i'4 I-57I C o 198 385 707 819 891 985 I To obtain the scales so that the given equation is represented, multiply the vertical scale by 4-5, i. e., i on the original scale now reads 4-5 ; and divide the horizontal scale by IOOTT, so that - now reads -005 The curve is shown in Fig. 206. \2OO7T (6) According to the second method, the simple sine curve is not used and no alterations are necessary. Having found the period, 02 sec., it is known that values of t need only be taken for one quarter of this, i. e., between o and -005. The tabulation would be arranged as follows : t lOOirt O (radians) r iS.ooot (degrees) sin IOOTT* C = 4-5 sin looirt o O o o OOI 3H 18 309 1-39 OO2 628 36 588 2-645 003 9 4 2 5< 809 3-64 004 1-256 **l 951 4-275 005 I-57 1 90 i 4'5 Note that the 2nd column is not really necessary; it is only in- serted here to make clear the reason for the 3rd column. Example 18. A crank i'-6* long rotates uniformly in a right-hand direction, starting from the inner dead centre position, and making 30 revs, per minute. Construct a curve to show the height of the end of the crank above the line of stroke at any time, assuming pure harmonic motion. Time for i revolution = 60 30 2 sees. or, in 2 sees., 2ir radians is the angular distance travelled. ' In i sec. TT radians is the angular distance travelled, or the " angular velocity," usually denoted by a, = it radians per sec. Construction. Draw, to some scale, a circle of radius i'-6*, to the left of the paper. (Fig. 207.) Divide its circumference into a number of equal parts, say 10 or 12 ; in this case 10 is chosen, lines making 36 with one another being drawn. These division lines will correspond to the positions of the crank at time divisions of a tenth of a period, . e., -2 sec. apart. Number these divisions of the circumference o, i, 2, 3 . . . 364 MATHEMATICS FOR ENGINEERS Produce the horizontal through O and along it mark off to some scale a distance to represent 2 sees., and divide this into 10 equal parts. When the crank is in the position Oi, i. e., at time -2 sec. after start, its projection on the vertical axis is OA : hence produce lA to meet Fig. 207. the vertical through -2 at i x ; and this will be a point on the curve required. Proceeding similarly for the other positions of the crank, the full curve is obtained, and from its form we conclude that it is a sine curve. To prove that it is a sine curve Suppose that in time / sees, the crank moves to the position OC (Fig. 208). Fig. 208. Fig. 209. In I sec. the angle moved = ir (in this case) .'. In t sec. the angle moved = itt (in this case) where a>t is the angle in radians. or a> (in general) or <ot (in general) OA = CB = CO sin L COB = r sin f where r = radius of crank circle. Therefore the curve obtained by the construction is that repre- senting the equation y = r sin at. Hence a graphic means of drawing sine curves can be employed in place of that by calculation. Great care must, however, be taken in connection with the magnitudes involved. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 365 e. g., to plot C = 4-5 sin iooirt by this means. Radius of circle = 4-5, the amplitude of the function and tat = looirt or <> = ioor . e., 100 TT radians must be swept out per sec. 27T radians are swept out in -02 sec. Therefore, if the circle were divided into 10 equal parts, the dis- tances along the time base corresponding to the angular displacements would be -002 sec. each. Simple Harmonic Motion. If the crank in Fig. 209, which is supposed to revolve uniformly, were viewed from the right or left, it would appear to oscillate up and down the line OA. Such motion is known as simple harmonic motion, or more shortly S.H.M. Looking, also, from the top, the motion as observed would be an oscillation along OB, and this again would be S.H.M. ; there- fore, if the connecting-rod were extremely long compared with the crank the motion of the piston would be approximately S.H. In the case of the valve rod it would be more nearly true that the movement of the valve was S.H., for the valve rod would be very long compared with the valve travel. At a later stage of the work it will be shown that the accelera- tion along OB, say, is proportional to the displacement from O; and this is often taken as a basis for a definition of S.H.M. S.H.M., then, is the simplest form of oscillatory motion, and can be illustrated by a sine curve. Suppose that the crank does not start from the inner dead ' centre position, but from some position below the horizontal, what modification of the equation and of the curve results? If at time t sees, after starting, the crank is at OC (Fig. 209) (Oo is the initial position of crank} then L COo = <>* and L COB = ut c where c = L. BOo y = r sin L COB = r sin (<at c). Similarly, if the crank is inclined at an angle c above the hori- zontal at the start, y = r sin (wt+c). A moment's thought will show that the curve will be shifted along the horizontal axis one way or the other, but that its shape will be unaltered. Example 19. Plot a curve to represent the equation C = 4-5 sin (loont i-i) for a complete period. 366 MATHEMATICS FOR ENGINEERS Let us reduce the equation to a form with which we have already dealt; thus C = 4-5 sin (looirt i -i) C I, i-i \ = sm loon-! t 4-5 \ I007T/ = sin ioo?r(/ -0035) j. e., C = sin loon-Ti = sin 1 C where T l = t -0035, T = loou-T, and C = - 4 5 *$ r/vWw? 3-6 , I ^ AX ' 5 ^ \, ^0 Mi s > / \ 1-8 & 5.1 L / \ Vu C=4-5 sin dooTfl- 1-1} O -o | i/ \ ?2 '/ 00* 36 O8 .-Ol < 12 \< 14- O 16 10 < 2 Scale for I -OO35 -V 9 /, 2 ^ /O 1-57 ' 1 >co/e jQ T' \ \ . / / OO5 ' \ / 27 / / 1 g \ / ?'/ 1 I \ / 4-5 11 V ^ ' Fig. 210. We have already seen, viz. in Example 17, how to plot this curve, the period being -02 sec. Then, having altered the two scales accord- ing to previous instructions, the vertical axis must be shifted a dis- tance of -0035 sec. to the left (see Fig. 210), because < = Tj+-oo35. Hence the scale for t, which is the final scale, must be measured from an axis -0035 unit to the left of that used in the construction, i. e., the horizontal scale must again be altered, not in magnitude but in position. The changes in the scales may appear rather confusing, but on the other hand calculations have only to be made for the one fundamental curve (the table for this being given on p. 360), and all the others are derived from it. Therefore, when once the table of values for the simple sine curve has been set out, it will serve for all sine and cosine curves, i. e., it is a " template." It serves for the cosine curve because this curve is merely the sine curve shifted along the axis a distance equal to one quarter of the period. Thus y = sin t and y = cos t = sin (90 /) will be the same curve measured from different vertical axes. Graph of tan x. The graph representing y = tan x is not of the same type as the sine and cosine curves. As x increases from o to 45, y increases from o to i, but after x has the value 45 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 367 y increases very much more rapidly; while at 90 the value of y is infinitely large. After 90 the tangent is negative, for the angle is in the 2nd quadrant. Supposing some form of continuity in 3 Degrees 90 Radians Period /ao* tT rae/iano Fig. 211. Graph of tan x. the curve, it must now approach from infinity from the negative side and come up to cross the axis at 180. After this the curve is repeated, so that the period for the simple tangent function is 180 or TT. 368 MATHEMATICS FOR ENGINEERS Selecting values for x, those for y can be read off from the tables : x y = tan x o o 25 466 45 i-o 60 1-732 80 5-671 85 n-43 90- + 00 X y = tan x 90 + 00 95 - n-43 100 - 5-67I 120 - 1-732 135 I 155 - -466 180 o Note that 90 indicates that a value of x is supposed to be taken just less than 90, but practically differing nothing from 90 ; thus 90 would be of the nature 89-99. Similarly 90+ would indicate 90-01, say. The curve on either side is asymptotic to the vertical through 90, as will be seen from the curve plotted in Fig. 211. All other simple tangent curves can be obtained from this fundamental curve by suitable change of scales. Example 20. Plot the curve representing the equation y = 8 tan 40O/. Rewrite the equation as Y = tan T where Y = ^ and T = 400*. Then plot Y = tan T from o to 180, t. e., for a complete period, and afterwards alter the scales so that i on the vertical scale becomes 8, and i on the horizontal scale becomes 400 sin x cos*' and therefore, if we had drawn the- curves y x = sin x (i) and y a = cos x (2), we should obtain the value of the ordinate of the curve y = tan x by dividing any ordinate of curve (i) by the corresponding ordinate of curve (2). Example 21. The efficiency of a screw-jack is given by tan 6 ' "~ tan (0 + <t>) where 6 is the angle of the developed screw, and <f> is the angle of fric- tion. If 6 varies from o to 12, plot a curve to give the value of the efficiency; p, the coefficient of friction, being -1465. The angle of friction is such that its tangent is equal to the co- efficient of friction, i. e., <f> = tan -1 u. THE PLOTTING OF DIFFICULT CURVE EQUATIONS tan 6 369 Thus tan <j> = -1465 and <f> = 8 20'; also ; = The tabulation of values is as follows : tan (6 + 8 20') 6 6+ <*> tan 6 tan (6 + <t>) n o 8 20' o H^S 2 10 20' 0349 1823 191 4 12 2O' 0699 2186 320 6 14 20' 1051 2555 4i3 8 l620' 1405 2931 480 10 l820' 1763 33H 533 12 20 20' 2126 3706 574 and the plotting is shown in Fig. 212. Efficiency 7J O .1 ro (>i A ch <i r*. .. _ E - S ^ - X ? - A / - T f 1 1 1 i 1 \ 32 A- 6 8 IO 15 0, Aragle of Thread Fig. 212. Efficiency of Screw-jack. Compound Periodic Oscillations. In engineering practice one often meets with curves which are quite periodic, but are not of the sine or tangent type. Many of these can be broken up or " analysed " into a number of sine curves. The process is spoken of as harmonic analysis, and reference to this is made in Volume of Mathematics for Engineers. At this stage, however, it is well to consider the work from the reverse or the synthetic point of view, in which the resultant curve is constructed from its components by the addition of ordinates. B B 370 MATHEMATICS FOR ENGINEERS An example of importance to surveyors concerns the " equation of time," which is the difference between the " apparent " and the " mean " time of day. The apparent time is the actual time as recorded by a sun-dial, whilst the mean time is calculated from its average over a year. Two causes contribute to the difference between the two times, viz. (a) The earth in its journey round the sun moves in an ellipse , , . ., /distance between focf\ . i having an eccentricity ( -~ sequence of the laws of gravity its speed is greater when nearer to the sun than when more remote. (b) The earth's orbit is inclined to the plane of the equator. i\ , - -1 of ^-,and in Con- Set Watch FAST over J this Period by Amount"! given bv Ond" Set I Watch Period b Curve giving Variation duel ho J ^'Curve giving .- Variation due to IE EarftA Orbital Spee Curve Of Equation of Time Fig. 213. Curves for " Equation of Time." The corrections due to these two causes are found separately, and are represented by the respective curves (a) and (b) in Fig. 213. For curve (a) the period is one year, and the period of (b) is half a year. These, when combined by adding corresponding ordinates, due attention being paid to the algebraic sign, give curve (c), for which the period is one year. By the use of this curve the correction to be added to or subtracted from the observed " sun time " can be obtained. Thus to determine the longitude, i. e., the distance in degrees east or west of Greenwich, of, say, a village in Ireland, it would be first necessary to find the meridian of the place by observation of the pole star. Next the time of the crossing of the meridian by the sun i. e., the local time, would be noted, and this would be corrected by adding or subtracting the equation of time for the particular day. Then the difference between the corrected THE PLOTTING OF DIFFICULT CURVE EQUATIONS 371 local time and Greenwich mean time as given by a chronometer would give the longitude, since one hour corresponds to fifteen degrees. Example 22. Plot the curve y = 4 sin / + -5 sin 2t sufficiently far to show a complete period. Let y\ = 4 sin * (i), and y, = -5 sin it ( 2 ) ; then the curve required is y = y^ + y t , i. e., it is the sum of two curves of different periods. The period of y = 4 sin / is 2ir, while the period of y = -5 sin 2t is 27T -V_J X = 4sint \^- --^T\ \ Fig. 214. Complete period of curve y = 4 sin t + '5 sin 2 t. Therefore the curves must be plotted between / = o and / = lit to give the full period of the resultant curve, so that there will be one period of curve (i) and two of curve (2). The curves are now dealt with separately, because, being of different periods, values suitable for the one would not be so for the other. For curve (i) period = 2n-, and amplitude = 4. The two curves must be plotted to the same scales. The simple sine curve " template " already mentioned would serve for curve (i), but curve (2) must be previously adjusted in scale to make it possible to apply the " template." It may be sometimes easier to set out the work as follows instead of using a template : Curve (i). Values of / need only be taken between o and ^ 372 MATHEMATICS FOR ENGINEERS Curve (2). Values of t need only be taken between o and ; there- fore take values one-half of those in the previous case, so that the calculation is simplified. Curve (i) Curve (2) t sin t y x = 4 sin t o o o 2 198 792 4 385 i-54 7854 707 2-828 960 819 3-276 i-i 891 3-564 i'4 985 3*94 I-57I I'D 4-0 t zt sin 2t y a = -5 sin 2/ o o o I 2 198 099 2 4 385 193 3927 7854 707 354 48 96 819 4i 55 i-i 891 446 7 i'4 985 493 7854 I-57I I-O 5 These curves are plotted as shown in Fig. 214, and the resultant curve is obtained by adding corresponding ordinates, paying careful attention to the signs. One further example of this compounding of curves will be given. Example 23. The current in an electric circuit is given by C = 50 sin 628*, whilst the voltage is given by V = 148 sin (628^ + -559). Plot curves to represent the variations in the current, voltage and power at any time. i Fig. 215. Variations in Current, Voltage and Power in Electric Circuit. Dealing with the three curves in turn : (See Fig. 215.) Curve (i). This is the curve of current. 27T C = 50 sin 628*, and the periodic time = ^x = -01 sec THE PLOTTING OF DIFFICULT CURVE EQUATIONS 373 Plot the curve = sin T from o to 2ir, and then multiply the vertical scale by 50 and divide the horizontal by 628. Curve (2), the curve of voltage = 148 sin 628Tj = 148 sin T provided that T = 628"]?! and Tj = t + -00089. This is the first curve with its axis moved to the right a distance of -00089 sec. and with all ordinates multiplied by -4- or 2-96. Thus AB = 2-96 x ab. Curve (3), the curve of power, is obtained by multiplying correspond- ing ordinates of curves (i) and (2). Confusion is avoided by plotting curve (2) along a different hori- zontal axis from that used for (i). The reader will find it convenient to draw out the simple sine curve on tracing paper to a scale convenient for his book or paper, and to use that as a template ; much time and labour being saved by this means. Curves for Equations of the Type y = e~* x sin(6x+c). In plotting such a curve it is not wise to select values of x and then calculate values of y directly : it is easier to split the function up into y 1 = e~ ax and y 2 = sin(&#-f-c), and plot the curves representing these equations separately, obtaining the final curve y = y"iXy 2 by multiplication of ordinates. The forms of the two component curves are already known. They must, however, be plotted to the same horizontal scale, which should always be a scale of radians (if an angle is measured along the horizontal) or one of seconds (if time is measured along the horizontal). Example 24. Plot the curve y = e~** sin (5*+ 2-4), showing two complete waves. Let y = yixy, where y t = e~* x and y, = sin (5*4-2-4). To avoid any trouble with the scales, this example is worked in full, i. e., templates are not used. It will be slightly more convenient to deal first with curve (2). Curve (2) y, = sin (5* + 2-4) = sin 5(#+'4 8 ) = sin 5X where X = x + -48 Hence the vertical axis through the zero of x in Fig. 216 will be 48 unit to the right of that for X ; hence, since the second scale has to be used again in the plotting, the construction vertical axis must be chosen -48 unit to the left of some convenient starting-line. y, = sin 5X, the period being -^ = 1-256 radians 374 MATHEMATICS FOR ENGINEERS Hence values of x need only be taken between o and or to -314. 4 X . . o 04 08 I57 1 192 22 28 314 5 X . . 2 4 7854 96 I -I i'4 I-57I sin 5 X . o 198 385 707 819 891 985 I This curve can now be plotted, reckoning the horizontal scale from the construction vertical axis; and then the zero is shifted to its correct position, -48 unit to the right, as shown in Fig. 216. Fig. 216. Curve of y = e - J* sin (5* -f 2-4). Curve (i) yi = "". For one complete wave of curve (2) v = 1-256, and therefore for two waves it will be more than sufficient if values of x are taken up to 3. The table of values is as follows : X . 4 8 1-2 1-6 2-O 2-4 2-8 3-0 \x or X o 2 4 6 8 1-0 1-2 1-4 i'5 ~ x = yi I 819 67 '549 449 3 68 301 247 223 A word of explanation regarding this table is necessary. Consider the value x = 1-2 ; then to find y, the value of e~* x1 ' 2 or e~' 6 must be found. Therefore X = -6 is read in the first column of Table XI at th.7 end of the book, and e~' 6 is read off in the third column. This curve can now be plotted, always to the same horizontal scale as that chosen for curve (2), but not necessarily to the same vertical scale. In this example, however, the same scale is convenient for both. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 375 Curve (3), or y = y x x y, can next be obtained by selecting corre- sponding ordinates of the two curves and multiplying them together. When x = 1-06, y^ = -58 and y t = i ; hence in this case the par- ticular product of y>i and y a has the same value as y lt and accordingly the vertical scale chosen for curve (3) is advisedly that for (i), so that the curve when plotted touches the curve (i) at its highest points. Glancing at the curve (3) we observe that the amplitude is now diminished in a constant ratio, although the period remains the same, i. e., there is some damping action represented. If a condenser discharges through a ballistic galvanometer and deflections left and right are taken, then by plotting the readings a curve is obtained (naturally of a very small period) of the char- acter of curve (3). The logarithm of the ratio of the ampli- tudes of successive swings is called the logarithmic decrement of the galvanometer. For the case considered, the ratio of consecutive amplitudes is -,-}*+l-256 f -\x v fl'256 '-~ = - ~ - = * 1 ' 256 = 3-5 (approx.) .'. logarithmic decrement = log 3-5 = 1-253. Again, imagine a horizontal metal disc within a fluid, hung by a vertical wire. If the wire is twisted and then released, the disc oscillates from the one side to the other. Measurements of the amplitudes of the respective swings demonstrate the facts that (a) the ratio of the amplitude of one swing to the amplitude of the preceding swing is constant for any fluid, and (b) this ratio is less for the more viscous fluids. Thus if the disc osciUated in air, the successive swings would be very nearly alike as regards amplitude ; or, in other words, the motion is practically simple harmonic, and its representation in the usual manner gives a sine curve. If the medium is water or thick oil, the motion is represented by a curve like No. (3) in Fig. 216, but the damping effect would be much more marked in the case of the oil. Exercises 39. On the Plotting of Graphs Representing Trigonometric Functions. 1 Write down the amplitudes and periods of the following func- tions : i "cos 4 * : * sin (3*- 4): 51* sin 314' (< 1S in seconds); "*XA?^*1-* the amplitude and also the period of this function- 376 MATHEMATICS FOR ENGINEERS 3. The range of a projectile fired with velocity V at elevation A is "y"2 gjn 2A given by ' . Plot a curve to show the range for angles of elevation up to 45, the velocity of projection being 1410 ft. per sec. 4. On the same diagram and to the same scales plot the curves y t = 2 sin x and y t = 5 sin \x, and also, by addition of ordinates, the curve y = 2 sin x + 5 sin \x. 5. A crank rotates in a right-hand direction with angular velocity 10, starting from the inner dead centre position. To a time base draw a curve whose ordinates give the displacement of a valve, the connecting-rod (or valve-rod) being many times as long as the crank. The travel of the valve is to be i|*. 6. Plot the curve s = 2-83 sin(4< -016) for one complete period, the angle being in radians. 7. Plot the curve y -81 cos 3$ for a complete period. 8. Plot the curve 5^ = 4-72 tan 4$ for a complete period. 9. The current from an alternator is given by C = 15 sin iirft, and the voltage by E = 100 sin (zirft n). If the frequency / is 40 and n (the lag) = -611, draw curves of current and E.M.F., and by multi- plication of corresponding ordinates plot the curve of power. 10. The acceleration A of the piston of a reciprocating engine is given by A 2 1 f /I , COS 20} A = 47r 2 n 2 ri cos -\ Plot a curve to give values of the acceleration for one complete revo- , ,. ,, connecting-rod length lution when r = crank radius = i ft., m = ^ = 10, crank length n = R.P.S. = 2. 11. The displacement y of a certain slide valve is given by y = 2 -6 sin (0+ 32) + -2 sin (20+ 105). Plot a curve to give the displacement for any angle between o and 360. 12. Plot the curve y = e-'^sin $x, showing two complete waves. 13. Plot a curve to give the displacement x of a valve from its centre position when x = 1-2 cos pt 1-8 sin pt and p = angular velocity of the crank, which revolves at 300 R.P.M. 14. Plot the curve y = 5 cosec 0, showing a complete period. 15. What is the period of the curve 7^ = 2-8 sec3#? Plot this curve. 16. An E.M.F. wave is given by the equation E = 150 sin 3i4/ + 50 sin 942^. Draw a curve to show the variation in the E.M.F. for a complete period. 17. The " range " of an object from a point of observation is found by multiplying the tangent of the observed angle by the length of the base. Draw a curve to give ranges for angles varying from 45 to 70, the measured base being two chains long. Graphic Solution of Equations. The application of purely algebraic rules will enable us to solve simple or quadratic equa- tions. Equations of higher degree, or those not entirely algebraic, THE PLOTTING OF DIFFICULT CURVE EQUATIONS 377 can best be solved by graphs ; and in some cases no other method is possible. The general plan is to first obtain some approximate idea of the expected result, either by rough plotting or by calculation, and to then narrow the range, finally plotting to a large scale the portion of the curve in the neighbourhood of the result. Occasionally the work is simplified by plotting two easy curves instead of the more complex one. Example 25. Solve the equation e 3 * $x z 17 = 0. The equation may be written e 3 * = $x z +17. Then if the two curves y x = e sx and y t = $x z + 17 are plotted, their point or points of intersection will give the value or values required. Tabulating : For Curve (i) y l = e 3x . For Curve (2) y t = $x z +17. X 3# or X e* = y t o o i 5 i'5 4-48 i-o 3 20-09 1-5 4'5 90 2 6 404 X 5* 2 + I 7 y o 0+17 17 5 1-25+ 17 18-25 i 5 +17 22 i'5 11-25 + J 7 28-25 2 20 + 17 37 3 45 + J 7 62 We conclude from an examination of these tables that y t and y t are alike when x has some value between i-o and 1-5. Values of x are next taken between i-o and 1-5 ; thus : X 3* = X Vi 5 *2+I7 y\ i-o I-I 3-0 3'3 2O-I 27-1 5 + J 7 6-05 + 17 22 23-05 Therefore, the solution is evi- dently between i-o and i-i; hence plot the two curves for these values and note the point of intersection (see Fig. 217). This is found to be at the point for which x = 1-032, and therefore x = 1-032 is one solution, and as the curves intersect at one point only, it is the only solution. Numerous examples of this, method of solution of equations 1-02 104. 106 108 Fig. 217. Solution of e* x 5** 17 = o. MATHEMATICS FOR ENGINEERS occur in connection with problems in hydraulics. As an example take the following : Example 26. Water flows at 7-45 cu. ft. per sec. through a pipe of diameter d ft., and the loss of head in 10 miles is 350 ft. The co- efficient of resistance is /= -005(1 -\ -A Find the diameter of the pipe, given that Head lost = i where 2gm d m - 4 Area of pipe = ~d z Then the velocity = Z&. = 7J45 x area ird z _ and Substituting for / d x 64-4 x d* _ 40 x 5280 x 9-48 x 9-48 , 350 x 64-4 = 8 3 8/. o. from which d 6 4-19^ -35 To solve this equation, we know that no negative values need be taken ; hence as a first approximation Let y = d 6 4-19^ -35 Then d d- - 4-igd- -35 y o o I I 2 64 - o - -35 - 4-19 - '35 - 8-38 - -35 - '35 -4-54 55-27 Since d = i makes y negative and d = 2 makes y positive the value of d that makes y = o must lie between i and 2 and nearer to i. For d = 1-5, y = (i'5) 6 -(4'i9Xi-5)--35 = 4'76- Thus the required value of d is between i and 1-5. If d = 1-3, y = -98, and we see that the required value is between i -3 and 1-5. Plot the values of y for the values of d 1-3 and 1-5, as THE PLOTTING OF DIFFICULT CURVE EQUATIONS 379 in Fig. 218, allowing a fairly open scale for d, and join the two points by a straight line. The inter- section of this line with the axis of d gives the value of d required, which is seen to be Example 27. The length of an arc is 2-67*, and the length of the chord on which it stands is 2 -5*. Find the angle sub- tended at the centre of the circle. [This question has refer- ence to the length of sheet metal in a corrugated sheet.] Arc = radius x 6 radians. Now Also where is in rd = 2-67 and r = ^ L . 6 1-25 . i'25 sin = *. t. e., r = ., 2 r ' sini 2 2-67 1-25 1-5 Fig. 218. Solution of Equation giving Diameter of Pipe. 2-67 or sin - = ^d = -480. 2 2-67 Making our first approxi- mation, taking 6 from o to 6 d_ 6 sin - 4680 2 2 O o 5 25 247 234 i-o 5 479 468 1-5 75 682 702 2-O i-o 842 936 2-5 1-25 945 1-17 3-0 I- 5 -998 1-403 3-14 i-57 i-o 1-47 0T* Fig. 219. Length of Metal in Corrugated Sheet. we see that the solution must lie between 6 = i-o and d = 1-5. When 6 = 1-2, sin - = -565, and -468$ 562. 380 MATHEMATICS FOR ENGINEERS Q Plotting the two curves in Fig. 219, y x = sin -, and y a = -468$ for values of 6 from 6 = i-o to 1-5; we note the point of intersection to be at 6 = 1-19. .*. 6 = 1-19 radians or 68-2. Exercises 40. On the Graphic Solution of Equations. 1. Find a value of x in terms of / to satisfy the equation 3# 3 3l z * + I 3 = o x being a distance from one end of a beam of length /. 2. Solve for z the equation I 3 $lz z z 3 = o when / = 10. 3. In order that a hollow shaft may have the same strength as a solid one the following equation must be satisfied *f D*-d* *f 16 * D 16 *' Writing x for -r this equation reduces to x* x i = o. Find the ratio of the diameters so that the given condition may be satisfied. 4. Find a value of d (a diameter) to satisfy the equation where r = 3-2, /= 6, P = 15. 5. Solve the equation e x = 4*. 6. Find values of x between 4 and +3 to satisfy the equation to los = 16 + 4* x z 7. Find a value of x between i and 5 to satisfy the equation x* log* x = 8 8. Solve for positive values of x the equation 50- ^ sin 4*= 1-8. (Note that the value of x must be in radians.) 9. Determine a value of x between o and it to satisfy the equation x 1 ' 5 3 sin x = 3 10. To find the height of the water in a cylindrical pipe so that the flow shall be a maximum it is necessary to solve the equation 6(2 3 cos 6) + sin 6 = o Find the value of 6 (radians) to satisfy this equation. 11. Solve for / in terms of L the equation 56/ 3 - uiL/ 2 + 72/L 2 - i 4 L 3 = o which occurs when finding the most economical arrangement of the three spans of a continuous beam ; I being the length of each of the end spans and L being the total span. 12. In finding the ratio of expansion r for a direct acting single cylinder steam engine of 14" diameter and 22" stroke, the equation i+log e r -389^ = o was obtained. Find the value of r to satisfy this equation. 13. The maximum velocity of flow through a circular pipe is reached when the angle 6 at the centre of the circular section sub- tended by the wetted perimeter has the value given by the equation Find this value of 0. sin 6 ; cos 6 = THE PLOTTING OF DIFFICULT CURVE EQUATIONS 381 14. Solve, for positive values of/ (the length of a link of a certain mechanism), the equation / 3 - I 9'5/ 2 + 42-5/ + 546 = o. 15. Forty cu. ft. per sec. are to pass through a pipe laid at a slope of i in 1500, the pipe to run half full. The velocity is given by^ . where m = -1530 and the quantity = -D 2 u I -4- 4 Simplifying and collecting these equations we arrive at the simpler form Find the value of D to satisfy this equation. 16. The bottom of a trapezoidal channel (the slope of the sides being 2 vertical to i horizontal) is 4 ft. wide. Find the depth of flow d, if the discharge is 12000 gallons per min., the slope is i in 500, and the coefficient of resistance is -006. /Equations reduce to 2 '3 2 (&*+*)* = V8 + 4-47^ 17. Find a value of r, the ratio of expansion, to satisfy the equation - 1083 log e r 225 = o 18. A hollow steel shaft has its inside diameter 3*. What must be the outside diameter so that the shaft may safely stand a torque of 200 tons ins., the allowable stress / being 5 tons per sq. in. ? Given that Torque 2/ (D* - 3 ) 32 v 19. Find a value of 6 (the angle of the crank from line of stroke) to satisfy the equation sin 6 6 n z sin 4 6 n* sin 2 + n* = o when n = 5. [Hint. Let X = sin 2 6 and then solve for X.] 20. An equation occurring in connection with the whirling of shafts is j cosh x H = o cos x Find a value of x between o and it to satisfy this equation. [Note that the values of cosh x should be taken from Table XI at the end of the book.] 21. Find the height above the bottom of a cylindrical tank of diameter 10 ft. at which a pipe must be placed so that the water will overflow when the tank is two-thirds full. Construction of PV (pressure-volume) and r<f> (temper- ature-entropy) Diagrams. It is impossible to proceed far in the study of thermodynamics without a sound working knowledge of the indicator and entropy diagrams of heat engines ; and to assist in the acquisition of this knowledge these paragraphs are addressed mainly to students of the theory of heat engines. Although we are not concerned in this volume with the full meaning of these 382 MATHEMATICS FOR ENGINEERS curves, we can deal with them as practical examples of graph- plotting. More can be learned about the advantages and useful- ness of an entropy diagram by actual construction and use than by absorbing the remarks of some one else, and taking for granted all that he says. Careful attention should, therefore, be directed to the following exercises, which should be worked out step by step by the reader. Example 28. Draw a PV diagram (Fig. 220) and also a T$ diagram (Fig. 221) for i Ib. of steam expanding from a pressure of 100 Ibs. per sq. in. absolute, to atmospheric pressure, the steam being dry and E B O E 4 6 12, - 16 "I/" 2O Fig. 220. Pressure- Volume or PV Diagram. saturated throughout. [Note. Since these diagrams are to be used for subsequent examples, they must be so constructed that the lowest pressure indicated is 5 Ibs. per sq. in. absolute.] To calculate for points on the expansion line BD in Fig. 220 steam tables must be used; the volumes (V) of i Ib. weight of steam for various pressures (P) between TOO Ibs. per sq. in. and 14-7 Ibs. per sq. in. absolute being read off from the tables and tabulated thus : P IOO 80 60 40 20 I 4 -7 V 4 '4 5'48 7-16 10-50 20 26-8 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 383 Horizontals through 100 and 14-7 on the pressure scale complete the diagram in Fig. 220. BD is the saturation or 100 % dryness curve. For the T$ diagram (Fig. 221) rather more calculation is necessary. The entropy of water at any absolute temperature T Fahrenheit = log* , if the entropy is considered zero at 32 F., i. e., at 461 + 32 or 493 F. absolute. For our example we require the " water " line from about 160 F. to 320 F., since these temperatures correspond approximately to pressures 5 and 100. Hence the range of T = 621 to 781 F. absolute, or, say, 620 to 780. The tabulation is next arranged as follows, it being noticed that log, = 6 493 -log, 493 - 2-303(log 10 r-log lo4 93) T Iog 10 r-log, 493 2-303 x column (2) = log, - 493 62O 660 7OO 750 780 2-7924 2-6928 2-8195 2-6928 2-8451 2-6928 2-8751 2-6928 2-8921 2-6928 0996 x 2-303 = -229 1267 x 2-303 = -292 1523 x 2-303 = -351 1823 x 2-303 = -42 1993 x 2-303 = -459 It is unwise to plot this line until the calculations for the " steam " line have been made. The width of the diagram, . e., from the water line to the steam line (a to b, r to /, etc., in Fig. 221), is always , where L is the latent heat at the temperature r considered. The -values of the latent heat are read from the steam tables and are set down thus : [Taking 460 instead of 461.] tF. r F. absol. L L T 160 620 IOO2 I-6I5 200 660 974 1-475 240 700 947 1-353 290 750 912 1-215 320 780 891 1-142 Hence the scale for entropy must be chosen so that the largest value may be shown, viz. 1-844, which is obtained by adding 1-615 to -229. Plotting the values of T, taken from the last two tables, to a hori- zontal base of Q, we obtain the water and steam lines, which are straight lines over short distances. The vertical scale may also be numbered to read pressures, which 384 MATHEMATICS FOR ENGINEERS may be obtained for the temperatures required from steam tables. Thus : T 788 753 710 672 622 P IOO 60 30 14-7 5 A horizontal through 100 on the scale of P gives the line ab (corre- sponding to AB on the PV diagram), and the intersection of the hori- zontal through 14-7 Ibs. per sq. in. with the steam or saturation line gives the point d. Example 29. On the T$ diagram (Fig. 221) draw the adiabatic line be, and also the constant volume line dcf, the latter on the assump- tion that qV is constant throughout the curve ; q being the dryness fraction and V the volume of i Ib. of dry steam. Draw also the corre- sponding line BC in Fig. 220, and the constant volume line DCF. The line DCF (Fig. 220) is a vertical through D, which meets the horizontal through 5 on the pressure scale in F; but certain calcula- tions are necessary before the line dcf (Fig. 221) can be drawn. As the pressure decreases, the volume increases. Thus at 14-7 Ibs. pressure the volume of i Ib. weight of steam = 26-8 cu. ft., while at 10 Ibs. pressure the volume of i Ib. weight is 38-4 cu. ft. Consequently if only 26-8 cu. ft. of steam are present at the lower pressure instead of the 38-4 cu. ft., the dryness of the steam must be - r , i. e., -698 ; 3'4 and accordingly the latent heat is only -698 of its true value. Hence if we make, on the horizontal through 10 Ibs. pressure, kx t = 'bq&kx (Fig. 221), the point x^ lies on the line of constant volume, viz. 26-8 cu. ft. At 5 Ibs. pressure, volume of i Ib. weight = 72-4 cu. ft.; hence , 26-8 wf = - ws = 'ZJiws 72-4 A number of points can be found in this manner, and the smooth curve through them, viz. dcf, is obtained. All adiabatics on the r<f> chart are vertical lines, so that be may be drawn. To draw the line BC in Fig. 220, proceed as follows : Select any convenient pressure, say 60, and calculate the value of the ft ratio -y in Fig. 221. Referring to Fig. 220, determine the position of T! on the horizontal through 60, RT T vti so that 1 = 1 and other points on the curve BC may be found in like manner. It should be noted that the adiabatic BC lies under the saturation curve BD, since the steam is not dry throughout the expansion ; and T?HT the dryness fraction at any pressure is the value of a ratio like -W- THE PLOTTING OF DIFFICULT CURVE EQUATIONS 385 Example 30. Draw the adiabatics through / and F, the final pressure being 5 Ibs. per sq. in. absolute. (Figs. 22l and 220!) Dryncss Fraction Enfropy Fig. 221. Temperature-entropy or r$ Diagram. The point /, on the constant volume line dcf, has already been fixed ; and a vertical through / gives the adiabatic ef. EF is obtained from BD in just the same way as BC was derived ; 4 r t t i. e.. -^r^ = -7 etc. lii rt C C 386 MATHEMATICS FOR ENGINEERS Example. 31. Draw the Rankine cycle for the case in which the steam is initially dry; and also for the case in which the steam at the commencement of the expansion has its dryness fraction = ae -j- ab. The initial and back pressures are 100 and 30 Ibs. per sq. in. absolute respectively. The Rankine cycle is made up of (i) expansion at constant pressure, (ii) adiabatic expansion, (iii) exhaust at constant pressure, and (iv) com- pression at constant volume. Thus the horizontals PL and pi (Figs. 220 and 221) must be drawn, and the Rankine cycle is given by the figures ABPL and abpl for the one dryness, and AEHL and aehl for the other. Example 32. Draw the common steam engine diagram with a toe drop from 30 Ibs. to 5 Ibs. per sq. in. absolute ; showing the case when the engine is jacketed and also that when there is no jacket. (See Figs. 220 and 221.) If the engine is jacketed, the steam expansion line lies along the Saturation curve, so that the diagram is ABMNX on the PV diagram (Fig. 220) and abmnw on the r^> chart (Fig. 221) ; the line tnn being a line of constant volume obtained in the same way as cf. If there is no jacket, the diagram is ABPQX in Fig. 220, and abpqw in Fig. 221; pq being a line of constant volume. Example 33 Calculate the dryness fraction from the entropy diagram for various temperatures, and thence plot on this diagram the " quality " curve for the adiabatic be (Fig. 221). At 100 Ibs. pressure the dryness fraction is i, whilst at 60 Ibs. ft pressure the dryness fraction = -y ; and at 30 Ibs. pressure the dryness Ip fraction = fr-. Selecting some vertical line as the base set off hori- zontals to represent these various dryness fractions, taking -9 as the base of the curve : thus the position of y represents the dryness at 60 Ibs. pressure. A curve through the points so obtained is the quality curve. - Example 34. Calculate the values of the exponent in pv n = C for the expansions represented by BC and EF, Fig. 220. For the line BC p = 100 when v = 4-44 p = 13 when v = 26-8 also log p + n log v = log C Thus log 100 + n log 4-44 = log C log 13 + n log 26-8 = log C or 2 + -6474% = log C and 1-1139 + 1-4281^ = log C whence by subtraction -8861 = -780771 8861 In like manner the exponent for the expansion EF is i -06. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 387 We may compare these values with those given by Zeuner's rule; viz. = 1-035 + 'I? where q is the initial dryness. For BC q = i and therefore n = 1-035 + ** = i'i35 For EF q = -332 and therefore n = 1-035 + -0332 = 1-068. Constant heat lines may be plotted on the r<f> diagram ; but before showing how this may be done, we must indicate what is meant by the term " constant heat line." If steam is throttled by being passed through an orifice its dryness is greater than it would be if the expansion were free. Thus in Fig. 223, at the temperature r t the dryness fraction = ~? and not =J~ as for DC* DC, adiabatic expansion ; and the line BCj is known as a line of constant heat. hh Line, of Constant" Heat Fig. 222. Constant Heat Lines. Four cases of the drying effect of expansion without doing external work, known as " throttling," are possible, these being represented by (a), (b), (c) and (d) in Fig. 222. Case (a) illustrates the expansion of a mixture of water and steam from temperature ij to temperature r 2 . At the commence- ment of the expansion the dryness fraction of the mixture is q lt its latent heat is L x and its sensible heat h lt while q z , L 2 and h z are the corresponding quantities at the. temperature T Z . Then, since the heat content is unchanged in which equation q lt L 1( A lt L 2 and h 2 would be known, and thus q 2 could be calculated. Case (b) is that of water being dried, thus becoming a mixture of steam and water. The equation here is A! = ? 8 L 2 + h r 388 MATHEMATICS FOR ENGINEERS Case (c) is that of dry saturated steam becoming superheated, and for this change ^i + L! = h z + L 2 + -5(1-, T 2 ) T, being the temperature to which the steam is raised by the throttling ; r s r 2 thus being the degrees of superheat (only obtained internally). In Case (d) steam of a certain wetness is completely dried by expansion under constant heat. (Any further throttling would naturally superheat.) For the change shown in the diagram ?iLi + hi = L 2 + &2 = 1115 7*2 + 2 2 60 = 1055 + -3/! 2 from which equation t z , the temperature at which the steam is just dry, can be found. From a consideration of the foregoing cases it will be seen that lines of constant heat appear in either the " saturated area," viz. the area between the water and steam lines, or the " superheated area," viz. the area beyond the steam line; and these two cases will be dealt with in the following examples : Example 35. Steam -3 dry at 400 F. expands to 150 F., being dried by throttling. Draw the constant heat line representing this expansion. If T! and r a are the absolute temperatures, and h t and h t are the sensible heats T 1 -T-* r -* g -fr-4 To draw the line of constant heat it is necessary to calculate the dryness fraction at various temperatures. From the equation = 2 In this equation q lt L x , and / 4 are known, whilst values of t t may be assumed and values of L 2 calculated therefrom, or taken from steam tables. Now /! = 400, L x = 835, and q { = -3 Then, taking convenient drops of temperature, say 50 or 100, a table may be arranged as follows : t, L 2 <i-<i + fcLi ?l 400 835 o + 251 3 300 905 100 + 251 388 20O 975 2OO + 251 462 15 IOIO 250 + 251 495 THE PLOTTING OF DIFFICULT CURVE EQUATIONS 389 [X = 835 and ?1 L t = -3 x 835 = 251 ; also ^ = -388, Jll = -462 ^*^O y I j and = - The line of constant heat (Fig. 223) may be drawn after points such LK as K have been determined ; K being so placed that ^^r- =* -388. Example 36. On the r<f> chart (Fig. 223) plot the line of constant heat for superheated steam, which is dry at 350 F. Lines of CorisfanTHeaF . I I I 1 1 1 I 1 1 1 1 O -2 A- -6 -8 1-0 1-2 I 4 1-6 1-8 Fig. 223. T<)> Diagram showing Constant Heat Lines. This example is a numerical illustration of Case (c), Fig. 222, and hence we must use the equation -i = h t + L a + '5( r * ~ r i) ^ By transposition _ ^i + LI h t L, r, = 2(A X h 2 + L! L 2 ) + r, (absolute temp.) or /, =2(< 1 -/,+ L 1 -L t ) + / I (F.temp.) We know that * x = 350, L x = 870 ; and it is convenient to take drops of temperature of 50 F. 390 MATHEMATICS FOR ENGINEERS Then the table for the calculation is arranged in the following manner : < L 2 Li fj-f.+Lj-Jt, 2 x column (4) t 350 870 870 + o 35 300 905 870 50- 35 30 330 250 940 870 100 70 60 310 2OO 975 870 150 - 105 90 290 Horizontals through these temperatures meet the constant pressure lines (drawn on all charts, the equation being $ = Kp log e , i. e., the T O curve is of the same character as the " water" line) through 350, 300, etc. (on the " steam " line), at points on the line required ; join these and the line db is obtained (Fig. 223). Example 37. Steam of -2 dryness at 266 F. is dried further by the addition of heat and then allowed to expand through an orifice down to 200 F., where it is 6-9 % wet. Find the number of heat units added at 266 F. This may be worked by calculation, or by use of the chart. (a) By calculation. L at 266 F. = 929. Let x heat units be added, and then the dryness at the end of the addition of heat x = -- h '2 929 Let this dryness = q t q t at 200 F., = 93-1% = -931 L a = 975 Then Also But 20 (^9 + ' 2 ) 929 + 266 = (* 931 i e., x = 907 66 186 = 655. (b) By use of chart. Draw the constant heat line MN (Fig. 223), ,, fRM 1 starting from M. j -jrg = '931 f Then QN = -9 rank, or the heat units added = -9 x (460 + 266), i. e., x = 655 heat units. Construction of PV and r$ Charts for Engines other than Steam ; e.g., The Stirling, Joule and Ericsson Engines. Example 38. Trace the PV and T< diagrams for the Stirling engine working between 62 F. and 1000 F., the ratio of expansion being 3 to i. (Work with i Ib. weight of gas.) The PV diagram consists of two constant volume lines together with two isothermals. See Fig. 224. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 391 Starting from the point A, the pressure = 14-7, r = 461 -f 62 = 523, and the volume (read off from the steam tables) = 13-14 cu. ft. To find the position of the point B : It is true for all values of p, v and T that = constant. At B the temperature is 1000 F., or 1461 F. absolute : also the volume is 13-14, hence *. e., so that the point B is fixed. 14-7x13-14x1461., 3-14 x 5 2 3 40 _ B Stirling. 15 K> 10 15 80 5 -y 30 55 Fig. 224. PV Diagram for Stirling Engine. For the isothermal BC, pv = constant, and since p B = 41-1 and Va _ 13-14, the value of the constant is 41-1 x 13-14 = 540. Using the equation pv = 540, points on BC may be found thus : If p = 30 v = 18; p = 20, v = 27, etc. ; and the isothermal must be continued' until C is reached, the volume at C being three times that at B, . e., f = 3 x 13-14 = 39'42- CD is vertical ; and also T D 540 x 523 _ "f 1461 x 39-42 so that the position of D is fixed. 39 2 MATHEMATICS FOR ENGINEERS The constant for the isothermal DA is 4-91 X 39-42 = 193 '. and accordingly the points on the line may be obtained. To draw the T<|> diagram (Fig. 225) Suppose the entropy is zero at the start. Then points on the line ab are calculated from the equation <t> = K c lo&. , where K = specific heat at constant 523 volume = -1691. 4> = -1691 log e -^- = -1691 x 2-303 (Iog 10 r -Iog 10 523) = -39(log 10 r-log 10 523) J500L /300. 1100 90O . 5oo < O -05 / -15 '2 Fig. 225. r<p Diagram for Stirling Engine, and the table of values reads as follows : T log M r-10g lo5 2 3 39 x column (2)=* 7OO 2-8451 - 2-7185 049 IOOO 3*0 - 2-7185 1096 I2OO 3-0792 - 2-7185 1405 1461 3-1647 - 2-7185 174 The position of c is fixed, since the work done = r ^ and thus the distance be = log,, r 774 53-2 x log* 3 _ JJ _toj _ . 774 The lines be and ad are parallel, and cd is the curve ab shifted to the right a horizontal distance be; and thus the diagram can be completed. THE PLOTTING OF DIFFICULT CURVE EQUATIONS 393 Example 39. Plot PV and r$ diagrams for the Joule engine, when the compression pressure is 60 Ibs. per sq. in. and the lower temper- ature is 62 F. Work with i Ib. weight of the gas, and take for the adiabatics pv 1 ' 11 = C. Dealing with the PV diagram (Fig. 226) : At C the pressure = 14-7, the volume = 13-14 cu. ft., and r = 523 : hence p v v = 14-7 x 13-14 = 193. The point A, at pressure 60, is on the isothermal through C; then />Af A = P<,v = 193 whence V A = -^ = 3-22 40-- 20 - 10 Fig. 226. PV Diagram for Joule Engine. For the adiabatic AD pv 1 '* 1 = K (say) so that K = 60 x 3-22 1 ' 41 log K = log 60 + 1-41 log 3-22 = 1-7782 + (1-41 x -5079 = 2-4943 K = 312-1. Hence points on the line AD may be found from pv l ' tl = 312-1. The pressure at D = 14-7, and the volume = V 14^7" 8 '73 2 - Also- t^ = P^ 3-22 x 60 = 347 . 7 F. absolute. 394 MATHEMATICS FOR ENGINEERS For the adiabatic CB, the constant=/> u 1 " 41 =i4-7 x i3-i4 l ' 41 = 555-1 ; and thus this line may be drawn. Substituting the values of p, U A , T A , p B , 1^(4-845) in the equa- tion P*v p t v A 7OO t- 5 - 2 = -, T B is found TB TA . to be 787-7 F. abso- / lute. 600 For the r$ diagram (Fig. 227) : Starting from the point c, draw the horizontal through it; this being the iso- thermal for 523 F. absolute. The distance 50O 400 i 2375 log. 347-7 787-7. or the 30 1 O Fig. 227. T<(> Diagram for Joule Engine. 540 ratios of the tempera- tures being the same. Points on the line ab are obtained from the equation <t> = -2375 log, ~, as also are those on cd ; the latter values of $ being measured back- wards, i. e., towards the left of the diagram. The tabulation for this calculation would be arranged as in the previous example, so that there is no need for a detailed list of values here : and the diagram is completed by the verticals cb and 1o Q d ^ 18 24 y 3o 36 Fig. 228. PV Diagram for Ericsson Engine. Example 40. Plot PV and r<p diagrams for the Ericsson engine, when working between 62 F. and 1000 F., the compression pressure being 60 Ibs. per sq. in. absolute. (Work with i Ib. weight of the gas.) ^o _ THE PLOTTING OF DIFFICULT CURVE EQUATIONS 395 The calculation is left as an exercise for the reader; but his results may be checked from Figs. 228 and 229. In Fig. 228 AB and CD are isothermals, the equations to which are P v = 193 and pv = 540 respectively /5oo soo -/ Fig. 229. O / T(J> Diagram for Ericsson Engine. Exercises 41. On the Construction and Use of the PV and r<j> Diagrams. 1. Construct a r<t> chart, the temperature range being 120 F. to 380 F. ; and by the use of this chart solve the problems in Exercises 2 to 6. 2. Steam -42 dry at 350 F. expands adiabatically to 140 F. What is now its dryness fraction ? 3. Three hundred heat units are added to a sample of steam dry at 310 F. Find the dryness after the addition of the heat. The steam is now allowed to expand by throttling to 185 F. ; find the number of heat units that must be added so that the steam becomes dry saturated at this lower temperature. 4. Draw the Carnot cycle, the upper pressure being 150 Ibs. per sq. in. absolute, and the lower being 14-7 Ibs. per sq. in. absolute. 5. Show on the chart constant volume lines for volumes 5, 10, 15 and 20 cu. ft. respectively. 6. Draw constant heat lines in the superheat area for steam dry saturated at 250 F. and 65 F. respectively. 7. Draw on a PV diagram the adiabatic line mentioned in Exercise 2, working with i Ib. of steam. The equation of this expansion line being PV* = C, find the value of n (a) Directly from the diagram. (b) Using Zeuner's rule, viz. n= 1-035 + ' J <7> <7 being the initial dryness. 8. Draw constant-dry ness lines for dryness fractions of -2 and -3 respectively. 9. Calculate the dryness fraction for which the constant-dryness line is straight; assuming that L = 1437 "jr and <f> r = log,: CHAPTER X THE DETERMINATION OF LAWS IT is often necessary to embody the results of experiments or observation in concise forms, with the object of simplifying the future use of these results. Thus the draughtsman concerned with the design of steam engines might collect the results of research concerning the connection between the weight of an engine and its horse-power, and then express the relation between these variable quantities in the form of a law. He might, however, prefer to plot a chart, from which values other than those already known might be read off. The object of this chapter is to show how to fit the best law to correlate sets of quantities : and before proceeding with this chapter the reader should refer back to Chapter IV, where a method of finding a law connecting two quantities was demonstrated. The results of the experiments there considered gave straight lines as the result of directly plotting the one quantity against the other, and from the straight line the law was readily determined. The values of the quantities obtained in experiments, except in special cases, do not give straight lines when plotted directly the one against the other, but, by slight changes in the form of one or both, straight lines may be obtained as the result of plotting. The general scheme then is to first reduce the results to a " linear " or " straight-line " equation, to plot the straight line and then to calculate the values of the constants. The general equation of the straight line may be stated as Y = aX + b or (Vertical) = a (Horizontal) + 6 where a is the slope of the line. It is the only " curve " for which the slope is constant ; hence the reason for our method of procedure. e. g., suppose we know that two quantities P and Q are con- nected by an equation of the form THE DETERMINATION OF LAWS 397 We can rewrite this as I = aQ + b where P = P 3 and Q = Q 2 and this equation is then of the straight-line form. Therefore by plotting P against Q a straight line must result. Conversely, if the plotting -of P 3 against Q 2 gives a straight line the equation must be of the form , P 3 = aQ 2 + b. In dealing with the results of any original work there will probably be no guide as to the form of equation, and much time will therefore be spent in experimenting with the different methods of plotting until a straight-line form is found. Sometimes the shape of the curve plotted from the actual values themselves will give some idea of the form of the equation, but a great deal of experi- ence is needed before the various curves can be distinguished with certainty. It will be found of great value to work according to the scheme of substitutions here suggested, for by the judicious use of the method much of the difficulty will be removed. Thus small or large letters stand for the original quantities, and large or " bar " letters respectively stand for the corresponding " plotting " quantities. e. g., we are told that given values of x and y are connected by an equation of the type y = bx 2 + c. If we write Y for y and X for x z the equation becomes Y = 6X + c which is of the straight-line form required. The change here made is extremely simple but very effective. Again, suppose the equation H = aD n is given as the type. Seeing that a power occurs we must take logs : thus log H = log a + n log D. As this equation stands, it is not apparent that it is of the straight-line form; but by rewriting Has H = A + nU where H (H bar) = log H, A = log a and D = log D, it is seen to be of the standard linear form. We shall deal in turn with the various types of equation that occur most frequently. MATHEMATICS FOR ENGINEERS Laws of the Type y = a + -; y = a + bx z , etc. Example I. The following quantities are connected by a law of the form y ax 3 + b X o 2 5 9 IO y -8 -5 3i 212 291 Test the truth of this statement and find the values of a and 6. If we write the equation y = ax 3 + 6 as Y = aX + b, which is permissible provided that Y = y and X = x 3 ; then if the law is true, a straight line should result when Y is plotted against X. 300 250 100 50 810 100 ZOO MO 4OQ 5OO 6OO ^OO Boo 9OO tOOO Fig. 230. Determination of Law for Equation of y = ax 3 -f b type. Hence the table for the plotting reads ; X = x 3 o 8 125 729 IOOO y = y -8 -5 3i 212 291 Plotting these values, as shown in Fig. 230, we find that a straight line passes well through the points ; and therefore the statement as to the form of equation is correct. Selecting two convenient points on the curve X = 80 when Y = 15! and X = 890 when Y = 26oJ Inserting values 260 = 8900+6 (i) 15= 8oa+fc (2) THE DETERMINATION OF LAWS Subtracting Substituting in (2) 399 245 = 8ioa a = -302 15 = 24-2 + b b = 9-2 Y=. 3 o2X+(- 9 . 2 ) i. e., y = '$O2X 3 9-2. Alternatively, a and b might be found from the graph; since a = slope = g^ = -302 ; and b = intercept on vertical axis through o of X = - 9-2. .*. Y = -302X4- ( 9-2) and y -yzx* 9-2. 65 60 CP 55 50 I 234567 Amperes - A Fig. 231. Law connecting Volts and Amperes of Electric Arc. Example 2. An electric arc was connected up in series with an adjustable resistance. The following readings of the volts V and the amperes A were taken, the length of arc being kept constant and the resistance in the circuit being varied : V 67 63 59-7 58 56 53-8 52-2 Si'4 A i-95 2-46 3 3'44 3-96 4 '99 5'95 7 Find the law connecting V and A. 400 MATHEMATICS FOR ENGINEERS By plotting V against A, as in Fig. 231, a curve is obtained which shows clearly that the connection between V and A must be of an inverse rather than a direct character, since A increases as V decreases. Hence a good suggestion is to plot ^ against V, or, in other words, to assume an equation of the form Rewriting this equation as V = b + equation for a straight line. we see that this is the Fig. 232. Law connecting Volts and Amperes of Electric Arc. The plotting table will then be as follows : V 67 63 59'7 58 56 53-8 52-2 5i-4 A 513 407 333 291 253 2 168 143 The values of A, i. e., reciprocals of values of A, are obtained from the slide rule. To do this, invert the slide so that the B scale is now adjacent to the D scale. Then the product of any number on the B scale with the number level with it on the A scale equals unity, i. e., if the numbers are read on the B scale, their reciprocals are read on the A scale. The plotting of V against A gives a straight line (see Fig. 232). THE DETERMINATION OF LAWS 401 Selecting two sets of values and Y = 64-5 when Inserting values 64-5 = b + -456 . V = 52 when X = -15"! 5 = -45/ (i) Subtracting 12-5 = -y c = 41-7 and by substitution in (i) 64-5 = 6+1875 whence b = 45-75 = V = 45-75 + Notice that this problem could have been attacked in a slightly different way. A Multiplying through by A AV = 6A + c but the product of amps and volts gives watts (W). /. W = 6A + c. Therefore a straight line results if the power (watts) is plotted against the current (amperes). The table for the plotting would then read : A i'95 2-46 3 3-44 3-96 4*99 5-95 7 W = AV 130-5 155 179-1 199-4 221-5 269 3ii 359-8 and thence the procedure is as before. Laws of the Type y = ax". If there is no guide to the form of equation, it is most usual to assume it to be y = ax*, or, in more special cases, y = ax n + b ; this latter form embracing those already discussed. To avoid the quite unnecessary expenditure of time in searching for the form, this will be indicated before each example or set of like examples. If y = ax n , then, by taking logs log y = log a + n log x or Y = A + nX the large letters being written for the logs of the corresponding small ones. This last form is the equation of a straight line, the co-ordinates of the points thereon being X and Y, i. e., log x and log y. Accord- ingly, if corresponding values of two quantities are given, and it is D D 4 oz MATHEMATICS FOR ENGINEERS thought that they are connected by an equation of the type with which we are now dealing, a new or " plotting " table must be made, in which the given values are replaced by their logarithms. These must next be plotted, and if a straight line passes through or near the points, the form of equation is the correct one. The values of the constants n and a may be found, as before, by either of two methods : (a) by simultaneous equations, or (b) by working directly from the graph. 4 60 4-78 4-76 4-74 E 4-72 4-70 4-ee 4-66 08 62- 4-4 4-6 4-8 5-O 5-2 5-4 Fig. 233. Endurance Tests on Mild Steel Rods. 5-6 To illustrate by an example : Example 3. In some endurance tests on mild steel rod the following results were obtained : Maximum skin ^ stress F in Ibs. \ per sq. in. . . J 45200 47500 48700 49000 52100 54000 56750 58700 60150 64800 1 Revolutions to \ tracture R . . / 420000 223300 207300 186200 128600 85400 69000 45000 40000 '23200 4 Find the connection between F and R in the form F = aR n . In the log form or where log F = log a + n log R F = A + nR F = log F, A = log a and R = log R THE DETERMINATION OF LAWS The table of values reads : 403 F = logF . . 4-655I 4-6767 4-6875 4-6902 4-7168 47324 4 - 7540 4-7686 4-7793 4-8116 R = log R . . 5-6232 53489 5-3166 5-2700 5-1093 4-93I5 4-8388 4'6532 4-6021 4-3655 Plotting from this table, we see from Fig. 233 that a straight line passes well through the points. To find the values of n and a : By method (a). Select two convenient points on the line, giving the values F = 4-68 when K = 5-36 F = 4-76 when R = 4-74 4'76 = A+4-7 4 n ' . . . . (i) 4-68 = A+536 (2) 08 = -62 and Inserting values Subtracting whence n = 08 62 = 129 Substituting 129 in place of n in equation (i) 4-76 = A + (- -129 x 4-74) A = 5-37 but A = log a and therefore a = antilog of 5-37 = 234400 /. F = 234400R-' 129 By method (6). F = A + nR Hence if R be plotted horizontally n is the slope of the resulting line. In measuring the slope, ordinary scales must be used, since the question of logs does not arise at all; and from the equation it is observed that n is a small letter, and therefore represents a number and not a log. A is the intercept on the vertical axis through the zero of the R scale, and since the zero of any log scale is the reading corresponding to i, A is the intercept on the vertical axis through i on the scale of R. Obviously in the example under notice, it would be impossible to show this axis on the diagram, at the same time choosing a reasonable scale for R ; and consequently method (a) is the better. The slope of the line = -^- = -129. .'. n = -129 In many practical examples it is only the value of the exponent that is of importance, so that only the slope of the line is required. The slide rule may be used to great advantage in this connection, since its scales are scales of logarithms : and therefore there is no need to consult the log tables, for the logs of the given quantities 404 MATHEMATICS FOR ENGINEERS are plotted directly from the rule. After plotting, the slope is calculated, both horizontal and vertical distances being measured in centimetres or in inches, the scales on the rule being used : this slope is the value n. Note. If the B scale of the rule is used for both horizontal and ,, , difference of vertical vertical measurements, then the slope = -77^ -^, the difference of horizontal same units being employed for both lengths. If, however, a more open scale is required, say, for the vertical, i. e., the B scale is used for the horizontal and the C scale for the vertical, then the vertical difference must be divided by 2 before comparing with the horizontal difference. 4-OG5 4-77 1-2 Fig. 234. Hardness Tests of Mild Steel. Example 4. As a result of some tests for hardness, on mild steel, the following figures were obtained ; Pressure (tons per\ inch width) . . / 1-2 2-09 2-50 2-925 3-i8 4-065 4-46 477 Indentation (ins.) . 0045 0065 0085 0105 on 0145 0155 0165 If i = indentation in inches, p = tons per inch width, and c is a constant for the material, ci = p*. Find the value of n. THE DETERMINATION OF LAWS 405 For the actual plotting, shown in Fig. 234, the C scale of the slide rule was used along both axes, and therefore n = slope = ^ = i. For d = p n In the log form log c + log i = n log p or log i = n log p log c 1 = P_C , vertical difference ., T . = horizontal difference' lf Z 1S P lotted vertically and P hori- zontally. Laws of the Type y = ae bx , where e = 2-718, the base of natural logs. We have already seen that many natural phenomena may be expressed mathematically by an equation of the type y = ae bx ; so also is it possible that an equation of this type may best fit a series of observations so as to correlate them. If y = ae 1 " then log y = log a + bx log e and, since log e is a constant and equal to -4343, logy = log a + -43436* or Y = A + Cx where Y = log y, A = log a, and C = -43436. Y = A + Cx is the equation of a straight line of slope C, and whose intercept on the vertical axis through the zero of the hori- zontal scale is A; provided that Y, *'. e., log y, is plotted against x. In the cases in which this law applies we have to employ both direct and log values in the same plotting, and hence there is little advantage in using the slide rule ; in fact, it seems better to take the logs required from the tables only. Also, in finding the con- stants, simultaneous equations must be formed and solved. Example 5. The following are the results of Beauchamp Tower's experiments on friction of bearings. The speed was kept constant, corresponding values of the coefficient of friction and the temperature being shown in the table : t 1 20 no IOO 90 80 70 60 f- 0051 0059 0071 0085 OIO2 0124 0148 Find values of a and 6 in the equation p = ae bt for the set of results given. M = ae bt In the log form log /* = log a + bt log e = log a + -43436* or M = A + Ct where M - log /*, A = log a, and C = -43436. 406 MATHEMATICS FOR ENGINEERS Hence the plotting table reads : t 1 20 no IOO 90 80 70 60 M = log fj. 3-7076 3-7709 3-8513 3.9294 2-0086 2-0934 2-1703 In plotting the values of M it should be remembered that 3-7076 is 3 + '7076, and that therefore the marking for 3-7076 on the vertical scale is above that for 3, to the extent of -7076 unit. 2-2 _ 5-8 . 3-7 6o 70 80 90 100 HO /2O Fig. 235. Experiments on Friction of Bearings. Plotting these values, as in Fig. 235, we find the straight line that best fits the points. Selecting two sets of values of M and I viz., M = 2-13 when t = 65 \ and M = 3-73 when t= 115) we substitute these values in the equation M = A + Ct. Thus 2-13 = A+ 656 (i) 3-73 = A+ii 5 C (2) Subtracting -40 = 506 but Substituting for C in (i) whence and 50 C = -43436 C b = 4343 .008 '4343 = 0184. 2-13 = A+(- 1-195) A = 1-325 a = antilog of A = -2113 M = -zuie-' 018 * 1 THE DETERMINATION OF LAWS 407 Laws of the Type y = a + bx -f ex 2 . Suppose that given values of x are plotted against those of y and instead of the straight line a fairly well-defined curve suits them best. The curve is most likely to be a portion of some parabola, if not of the types of the two previous paragraphs. Its equation may then be of the form y = a + bx + ex 2 + dx? , any terms of which may be absent. This case thus includes types already discussed (e. g., y = a + bx 2 , and y = a + dx 3 ). If nothing is stated to the contrary, and it is thought that the curve is some form of parabola, it is usually sufficiently accurate to assume as its equation y = a-\-bx-\- ex 2 . In this equation there are three constants a, b and c\ and to determine them in any case three equations must be stated. If, then, the equation is to be of this type, plot the given values, sketch in the best smooth curve to pass well amongst the points, and select three convenient points on this curve : the three equa- tions can now be formed and solved in the manner indicated in Chapter II. If possible, one point should be on the y axis, for then x = o and y = a -f o + o ; or the value of y is such that the value of the unknown a is found directly. Example 6. Readings were taken as follows in a calibration of a thermo-electric couple : Temperature C. (T) . . o 490 840 1003 E.M.F. (microvolts) (E) . o 3152 5036 5773 Find (a) a formula connecting E and T in the form E = a + 6T + cT 2 and hence (6) an expression, enabling values of T to be calculated from any value of E. The plotting of the values from the tq,ble is shown in Fig. 236. Selecting three sets of values E = 150 when T = o E = 2600 when T = 400 and E = 5800 when T = 1000. a = 150 {for 150 = a + o + o} 5800 = 150+ 10006+ io e c (i) 2600 = 150 + 4006 + 16 x io*c (2) Multiplying (i) by 4 and (2) by 10 and subtracting 23200 = 600 + 40006 + 4,ooo,oooc. 36000 ~ 1500 -f- 40006 + 1,600,000? 408 MATHEMATICS FOR ENGINEERS Subtracting 2800 = 900 + 2,400,0005 370 =c 2,400,000 c = -00154 Substituting in (i) 5800 + 150 = iooo& 1540 whence b 7-49- E = 150 + 7'49T OOI54T 2 . 6ooo_ 000 /ooo _ Zoo 4oo 600 800 /ooo Fig. 236. Calibration of a Thermo-Electric Couple. To find an expression for T, solve the quadratic ooi 54 T 2 - 7'49T+ (150- E) = o. _ _ 7-49 V$6 '00616(150 E) Thus 00308 = 2430 325^/55-08 + -oo6i6E. Equations of Types other than the Foregoing. Very occa- sionally one meets with laws in the form y = a + bx n , y = b(x -\- a) n , y = a -f- be, or y = ax"z m . These may be dealt with in the following manner : (a) Type y = a + bx*. This may be written : y a = bx or Y = bx and is of the type already discussed ; but for the change from the pne form to the other to be effective, the value of a, roust be known, THE DETERMINATION OF LAWS 409 a is the value of y when % = o, so that if possible the curve with y plotted against x should be prolonged to give this value ; and it is worth while to sacrifice the scale to a certain extent to allow of this being done. Otherwise select two points on the curve, draw the tangents there, and measure their slopes. Let the slopes be s x and s 2 when x has the values # x and x z respectively. Then n, b and a can be calculated from logs, logs, , > = , - + i log *! log x t a = y l (b) Type y = b(x + ). If X = x -f- a, then y = 6X n , a standard type already discussed. When y = o, x-\- a = o or # = a, so that the value of x where the curve crosses the x axis is a. Values of b and n can then be found in the ordinary way. An alternative, but rather tedious, method is as follows : Select three sets of values of x and y, viz. * lf x z , x 3 , and y lf y 2 and y s . A _ i ~ log (*!+*)- log (* 3 + ) Then Y = A, because log y l = log b + n log (*,. + a) log y 2 = log 6 + n log (* 2 + a) log y 3 = log b + n log (*, + a) Whence by subtraction log y x log y 2 = {log (*,. + a) log (* a + a)} and log yj - log y 3 = n{log (*! + a) - log (x 3 + a)} {By division w is eliminated.} For various values of a plot values of (Y A) until this equals o ; thus the required value of a is found : and values of n and b can now be obtained by logarithmic plotting. (c) Type y = a + Plot y against x; select two points on the curve and draw the tangents there ; call the slopes of these s l and s,. 410 MATHEMATICS FOR ENGINEERS Then = log 5j - log S 2 4343(*i - * 2 ) b - *i " ^^ M'T (d) Type y = ax n z m . The method of dealing with this form of equation will be demon- strated in the following examples : Example 7. Assuming that the loss of head A in a unit length of pipe in which water is flowing with a mean velocity v can be expressed in the form h cv 3 ~d- n find the numerical values of c and expressed in feet and second units for a pipe of 4* diameter and 28 ft. long, using the experimental data of the annexed table : Loss of head in feet 58 1-064 1-635 Discharge in Ibs./min. I55<> 2138 2690 The corresponding values of h, i. e., loss per foot, will be found by dividing the first line in the table by 28, and are -0207; -0381; -0584 respectively. To find the velocity 1550 Ibs. per min. = -^ cu. ft./sec. 60 x 62-4 = -415 cu. ft./sec. Area of 4* diam. pipe = -0873 sq. ft. Velocity = _ = 4 . 75 ft./sec. Similarly, when Q = 2138, v = 6-54, and when Q = 2690, v 8-22. Now h = cv 3 ~ n d- n log A = logc+ (3 -n) \ogv-nlog_d f d = _'3333\ = log c + (3 ri) log v n x i -5228 \log d Y-5228J = logc+ (3 n)\ogv+ -477W ......... (i) Selecting two convenient points on the curve shown in Fig. 237, which is obtained by plotting h against v h = -03 when v = 5-8 h = -047 when v = 7-3 and substituting in (i) we have the equations 2-6721 = log c+ (3- ) x -8633 + - 4 7 7 n ...... ( 2 ) 2-477i=logc+(3-) X 7634+-477W < , , , , , (3) THE DETERMINATION OF LAWS Subtracting -195 = (3 w) -0999 = -3 -in In = -3 - -195 = -105 n = 1-05. Substituting in (2) 2-6721 = log c + (1-95 x -8633) + (-477 x 1-05) = log c + 1-682 + -501 log c = 4-489 .'. c = -0003083 Hence h = -0003083 411 rfl-05 OE 4-5678 Fig. 237. Experiment on Loss of Head in Pipe. Alternatively, we might have proceeded from (i) in the following manner : Plot log h against log v ; find the slope of the resulting straight line, this being the value of 3 n; find also the intercept on the vertical axis through o of the horizontal scale which gives the value of log c n log d, in which everything is known except c, and then calculate the value of c. Example 8. During experiments on the loss of head in a 6* diam. pipe on a measured length of 10 ft. the following observations were made : Experiment. Quantity (Gals, per min.). Loss of head (ins.). I 294 I- 7 2 2 441 3-66 3 588 6-14 4 735 9-18 412 MATHEMATICS FOR ENGINEERS Assuming that the loss of head in feet per foot run = m + n = 3, find values of n, p. and m. m = 3 n and that 02- -01 A- 5 6 789 10 Fig. 238. Experiments on Loss of Head in 6"-diameter Pipe. d = 6", area = -196 sq. ft. 294 gals, per min. = ^ . cu. ft. per sec. 6-24 x 60 = -785 cu. ft. per sec. Hence Similarly Q 294 441 588 735 V 4 6 8 10 Each value of loss of head (in ins.) must be divided by 10 x 12 to bring it to feet per foot, so that our final table reads : v (ft. per sec.) 4 6 8 10 h (ft. per foot) 0143 0305 0512 0765 THE DETERMINATION OF LAWS 413 Plot h against v (Fig. 238) and select two convenient points on the grapn, viz. V = 5, h= -022-| v = 8-5, h = -057/ XT v n f d = '5 \ Now- A = te^r I log d = 1-699 [ I =-. 3 oiJ log h = log p + n log v + (n - 3) log d. Substituting the above values 2 7559 = log /i + -9294 + (3 - n) x -301 (i) 2 '34 2 4 = log/i + -699 +(3 w) X -301 (2) Subtracting 2304 Substituting in (i) 27559 = log/* + 1-672 + -361 log p, = 4723 p. = -0005284 j,l-8 Hence h = -000528-^ Exercises 42. On the Determination of Laws. [In the following exercises it should be understood that " finding the law " means finding the constants in the equation.] 1. Find the law to express the following results of a test on an arc lamp, in the form W = m + A where W = watts = volts x amps. V (volts) . 65 72 62 68 64 66 68-4 A (amps) 8-5 5 9-2 8 9-0 10-5 6-5 2. The law connecting p, and v, for the following figures, has the form Find this law, which connects /x the coefficient of friction between belt and pulley, with v the velocity of the belt in feet per minute. V 500 IOOO 2OOO 4OOO 6000 r 29 33 38 45 51 MATHEMATICS FOR ENGINEERS 3. The working loads for crane chains of various diameters are given in the table. Find a law connecting W and d of the form W = a + bd*. i 3 * f J I i 8 Load on chain W (tons) 20 45 Si 1-27 1-83 2-49 3-25 4. Bazin gives the following results on the discharge over a weir; H being the head and m being a coefficient : H 164 328 656 984 1-312 1-64 1-968 m 448 432 421 417 414 412 409 If m = a + u , find the law connecting m and H. rl 5. The table of allowance for the difference / between the hypo- tenusal and horizontal measurements per 66 ft. chain in land surveying is given for various angles of slope a : a . . 5 6 7 8 9 IO 15 20 25 30 35 40 / (links) . 4 6 7 i 1-2 i'5 3'4 6-0 9'4 13-4 18-1 23H The connection between / and a can be expressed by a law of the form / = 6o a . Find this law. 6. The following table gives the weight W of cast-iron pedestals for various diameter of shaft d : d(it.) . i i \ 1 i 2 W (Ibs.) 18-005 18-017 18-138 18-464 19-1 26 Find a law of the form W = ad s + b to connect W and d. 7. The results of experiments at Northampton Institute with model aeroplanes were as follows : Space (ft.) . i 2-4 4-4 6 7-6 1 1 -2 I 5 -6 20-4 Time (sees.) 2 4 6 7 8 i-o 1-2 i'4 Find the law connecting S and t in the form S = Kt n . 8. Find a law connecting horse power H with speed v in the form H = av n , the following values being given : V 2O'I 24-9 30-2 H I54 2135 3850 THE DETERMINATION OF LAWS 415 9. Given the following values of torque T, and angle of twist 6, find a law connecting these quantities in the form T = a6 n . T (Ibs.in.) 800 850 900 95 IOOO 1050 IIOO 1150 I2OO 6 (degrees) 10-4 12-53 15-41 19-2 23-67 29-28 35-58 42-49 5 I-2 10. If d = diam. of rivet, / = thickness of plate, and d = al n , find values of a and to agree with the figures : d ft I I ift 'ft 11. The following are results of a test on a Marcet Boiler : *F 320 315 3ii 307-5 303 300 297 293 287 281 277 Gauge pressure 88 80 75 70 6 4 60 55 50 45 40 35 271 265 258 251 244 240 30 25 20 15 12-5 10 Find a law connecting the absolute temperature T (t + 460) and the absolute pressure p (gauge + 15) in the form r = ap n . 12. h and v are connected by a law of the form h = av n . Find this law if corresponding values of h and v are as in the table : V 8-04 11-67 14-43 17-41 I9-90 h 3-03 . 6-n 9-07 12-21 15-62 13. As a result of Odell's experiments on the torque required to keep a paper disc of diam. 22* rotating at various speeds we have the following : TorqueT(lbs.ins.) 33 56 875 1-29 1-76 2-4 R.P.M.(n) . . . 400 500 600 700 800 900 Assuming that T = an m , find the values of a and tn. 14. The following figures were obtained in a calibration test of the discharge of water through an orifice : Head H . 2-2 1-8 i'4 i-i 8 6 Quantity Q 8-9 8-03 7-23 6-4 5'5 4-85 The law connecting H and Q has the form Q = aH. Find this law. 416 MATHEMATICS FOR ENGINEERS 15. Find a law, of the form v = aH n , connecting the values : H 25 40 60 IOO 150 250 350 V 1119 1414 1732 2238 2740 3535 4180 16. / and t are connected by a law of the form I = at z + b. Find this law when corresponding values of / and / are : t 1-87 1-76 1-67 1-61 1-49 1-27 i-ii 79 I 34'5 30 28 25 21 16 12 6 17. The resistance R of a carbon filament lamp was measured at various voltages V, with the following Jesuits : V (volts) 62 64, 66 68 70 72 74 76 78 R (ohms) 73 72'7 72-1 71-7 70-7 70-4 70-5 69-7 69-2 80 82 84 86 88 90 92 94 67-8 68-4 67-7 67-2 67-2 66-6 66-3 66-2 Find the values of a and b in the equation R = ^ + 6. 18. The following are results of a test on a loo-volt carbon filament lamp. Find values for a and b as for Ex. 17 above. V (volts) . 54 60 65 70 75 80 85 90 95 IOO A (amps) 7 79 86 94 1-04 i'ii 1-2 i'3 i'4 i'5 (Values of R must first be calculated from R = -r ) 19. The difference between the apparent and the true levels owing to the curvature of the earth are given by Distance in\ feet d j 300 600 900 1 200 1500 1800 2400 3000 3900 Difference of\ level h (ins.)J 026 -103 231 411 643 925 1-645 2'57 4'344 i Find a law for this having the form h = Kd n . 20. If pv n = C, find n and C from the given values :- V I 2 3 4 5 p 205 II 4 80 63 5 2 THE DETERMINATION OF LAWS 417 21. y and x are connected by a law of the form y = ax* + bx + c. Given that X o 4 10 y 15 16-8 18-75 find values of a, b and c. 22. Coker and Scoble give the following results of a test on a thermo- electric couple : Hot junction temp. T (C.) o 327 419 657 E.M.F. E (millivolts) . 015 3-84 4'5 6-32 Find the coefficients in the equation E = a+bT+cT z . 23. Find a law connecting E and T, in the form E = a+6T+cT 2 , for the case in which T o 490 840 1003 1283 E o 3-152 5-036 5*773 6-382 24. The results of some experiments by Edge with a Napier car were Area of wind -re- } sisting surface > 42 38 34 32 28 24 22 18 16 12 A (sq. ft.) J Speed V in m.p.h. 47-9 52-9 54 55'5 57-6 62-5 64-2 70-3 75 79 The law fitting these results has the form A = a+bV+cV 2 ; find this law. 25. Given the equation R = a+&V-fcV 2 , and a table of the corresponding values of R and V, find the values of a, b and c. R o 9-3 21 35 V 16 14 12 10 26. The velocity of the Mississippi river was measured at various depths with the results : Proportional depth"! D below surface / i 2 3 4 5 6 Velocity (ft. per sec.) 3-195 3-23 3-253 3-261 3-252 3-228 3-181 7 8 9 3-127 3-059 2-976 If v and D are connected by a law of the form v = a+bD+cD*. find this law. E E 4i8 MATHEMATICS FOR ENGINEERS 27. Find values of a and b in the equation y = ae bx for the following case : X i i*5 2 2-5 3 3-5 4 4'5 y 13-28 15-04 ^SS I9-80 23-11 26 30-5 34'4 In Exercises 28 to 30 the law is T = 20^. 28. 29. SO. T 22-2 24-66 28-86 35-56 524 1-047 1-833 2-880 T 23-4 27-38 34-66 47H4 524 1-047 1-833 2-880 T 24-66 30-42 41-64 63-26 6 524 I-0 4 7 1-833 2-880 Find Find p. Find 31. Find values of a and b in the equation y = ae bx when values of y and x are as in the table : X 2-30 3-10 4 4-92 5-91 7-2 y 33 39-i 50-3 67-2 85-6 125 32. The following particulars were obtained from an experiment on the flow through a V notch. Determine a formula connecting the quantity Q with the head H for the notch (Q = aH n ) Quantity (cu. ft. per sec.) I-I2 88 72 17 Head (ft.) 900 815 7 = 7 "422 CHAPTER XI THE CONSTRUCTION OF PRACTICAL CHARTS IT has been seen that the correlation of two variables consti- tutes a graph. If two or more interdependent variables are plotted on the same axes so as to solve by intercepts problems of all con- ditions of related variability, the result is a chart. Charts may be classified as (a) correlation charts or graphs, (b) ordinary intercept charts, or (c) alignment charts. X Y 30- 2-7 --,L 1-52 3A 4 5 6 5. . units H 8 IO 12 IS 20 3O AO 50 Number Fig. 239. Chart giving Fifth Roots. Correlation Charts may be regarded as forms of the graphs already treated, but specially adapted for particular circumstances. The modification in the construction of the graph frequently con- sists of the substitution of a straight line in place of a curve, the former being far the easier to draw, and when powers occur, this necessitates logarithmic plotting. Example I. Construct a chart to read the fifth roots of all numbers up to 100. Along OX and OY in Fig. 239 mark out log scales, using the B scale of the slide rule for both directions. The scale of numbers being along 420 MATHEMATICS FOR ENGINEERS OX, extend this axis to show 100 at its highest reading. Set off OA = 5 units, say 2$", and set off AB = I unit, i. e., y. Join OB and produce to C. Then to find the fifth root of 38, erect a perpendicular through 38 on the horizontal scale to meet OC and project horizontally to meet OY in D, i. e., at the reading 2-07 : then '^38 = 2-07. 2 3^56 8/0 14- 20 Fig. 240. Chart to show Values of t> )-41 . The value of the exponent is thus the slope of the line, and hence this method can be used to great advantage when the power is somewhat awkward to handle otherwise. Example 2. In calculating points on an expansion curve, it was required to find values of v 1 ' 41 , v ranging from i to 30. Construct a chart by means of which the value of i; 1 ' 41 for any value of v within the given range can be determined. THE CONSTRUCTION OF PRACTICAL CHARTS 421 In Fig. 240 draw the axes OX and OY at right angles, and starting from i at the point O set out log scales along both axes ; the same scale of the slide rule being used throughout. Make OM = i unit of length and MN = 1-41 units of length (i. e., actual distances) : join ON and produce to cover the given range. Then for v = 5, w 1 ' 41 = 9-7, the method of obtaining this value being indicated on the diagram. If it be desired to have a more open scale along one axis, allow- ance must be made in the following way : Referring to Example i, suppose that the B scale of the slide rule is used for the scale of numbers and the C scale of the rule for the scale of roots. Then the slope of the line O^ (Fig. 239) must be made = and not . The scale for roots for this case is shown to the left of the diagram, viz., along O^j. Ordinary Intercept Charts. A combination of two or more graphs is often of far greater usefulness than the separate graphs, since intercepts can then be read directly and from the one chart. Intercept charts may take various forms, and the following examples illustrate some of the types : Example 3. Construct a chart to give the horse-power transmitted by cast-iron wheels for various pitches and at various speeds. The speeds vary from 100 to 1500 ft. per min. and the pitch from in. to 4 ins. Working with the units as stated, and allowing for the whole pressure to be carried by any one tooth at a time, the formula reduces to 110 V i) 2 This formula might be written as H = p z x or H = ^xV, sc that if p is constant H oc V or if V is constant H oc p 2 . We may thus draw on one diagram (see Fig. 241) a number of graphs : for on the assumption that p = 2, say, H = 4_ = -0364 V, and this relation may be represented by a straight line. By varying * other lines may be obtained, and as they are all straight lines passing through the origin (for H = o when V = o) only one point on each need be calculated, though as a check it is safer to make the calculation for a second point. E. g., when p = * and V = 440, H = i, giving a point on the 1 Plot values of V vertically and H horizontally. Join the origin to the 422 MATHEMATICS FOR ENGINEERS point for which H = i and V = 440 and produce this line to cover the given range. Indicate that this is the line for pitch = \" . For p = 2" and V = 440, H = 16. Hence join the origin to the point (16,440) ; produce this line and mark it for p-= 2*. By two simple calculations in each case a number of such lines may be drawn, say for each \" difference of pitch. To use the chart. To find the H.P. transmitted when the pitch is 3j" and the velocity is 560 ft. per min. : Draw a horizontal through 560 on the V scale to meet the sloping line marked p = si", and project from the point so obtained to the scale of H, where the required value, viz., 54, is read off. 1200 1KX) >IOOO 0900 800 Jf 700 600 5oo 4OO 3OO aoo loo /v I < if /I Values of H to 50 50 7O 9O 12O I4Q 16Q /8O Fig. 241. Chart giving H.P. transmitted by Cast-iron Wheels. 200 Again, if the pitch is ij*, what speed is necessary if 3^ H.P. is to be transmitted ? Draw a vertical through 3-5 on the H scale to meet the line marked p = if* and project to the vertical scale, meeting it in V = 125. In an exactly similar fashion a most useful chart might be con- structed to give values of the rectangular moments of inertia for rectangular sections of various sizes. Since I (moment of inertia of a rectangular section) = -^bh 3 , then I oc b if h is constant. Then for each value of h a straight line can be drawn, and the chart can be used in the same way as before. Example 4. Construct a chart to give the diameters of crank shaft necessary, when subjected to both bending and twisting actions, the THE CONSTRUCTION OF PRACTICAL CHARTS 423 greatest stress allowable in the material being 6000 Ibs. per sq. in. Given that Equivalent twisting moment and also X. T, = M+ VM 2 +T 2 -^/D 3 i6 7 where/ = 6000 and D = diam. of shaft in inches. Although there are three variables, viz., M, T and D, one simple chart suffices ; it being constructed in the following manner : Referring to Fig. 242, select an axis OY near the centre of the page, and along this axis set out the scale of torque in Ibs. ins. Along the .300,000 z 7 7 A Diam X 100,660 sopoo 1 2 3 4 5 6 Fig. 242. Chart to give Diameters of Crank-shaft subjected to Stresses. horizontal axis OX indicate a scale for diameters, taking the maximum value as 6-^". Two of the three variables may be combined by the following device : Suppose T = 75000 Ibs. ins. and M = 125000 Ibs. ins. ; then set off along OP a distance to represent T, using the same scale as along OY; make OL to represent M. With centre L and radius LP strike an arc to cut OY in R. Then OR = T,, since OR = OL+LR = OL+LP = OL+ V(LO)+(OP) = M+ VM 2 +f 2 = T e . Now T and D are connected by an equation which can be repre- sented by a curve, and irX6oooD 8 16 = H76D 1 . 424 MATHEMATICS FOR ENGINEERS For this curve the plotting table is D i 2 3 4 5 6 D 3 i 8 27 64 I 2 5 216 T e = iiyGD 3 1176 9420 31800 75300 147000 254000 By plotting T e against D complete the chart. For use : let it be asked what diameter of shaft is required which is to be subjected to a bending moment of 125000 Ibs. ins. and a twisting moment of 75000 Ibs. ins. Set off OP = 75000 and OL = 125000 : with centre L and radius LP strike the arc PR. Draw the horizontal through R to meet the curve and thence project vertically to the scale of D, where the diameter is read as 6-12 ins. Again, if M = 20000, and T = 30000, then D = 3-57, the method of obtaining this value being as before. A chart representing an equation similar to that in Example i might be constructed in a slightly different and better manner; thus : Example 5. Construct a chart to show the quantity of water flowing through pipes of various diameters, the velocity of flow also varying. Let Q! = quantity in cu. ft. per sec. = area in sq. ft. x velocity of flow in ft. per sec. then Q t quantity in cu. ft. per min. = 6oQi and Q = quantity in Ibs. per min. = 60 x 62-4 x area in sq.ft. X velocity in. ft. per sec. so if the diameter is given in inches and the rate of flow in ft. per sec. area X velocity 60 x 62-4 x ird 2 v ^- 4XM4 = where d = diam. of pipe in inches, and v = velocity of flow in ft. per sec. We will assume a maximum diameter of 6 ins., and a maximum velocity of 10 ft. per sec. Draw two axes at right angles in Fig. 243, the vertical axis being in the middle of the horizontal. Along OX! indicate a scale of diameters, the range being o to 6, and along OX indicate a scale of quantities, the range being o to 7500, to include the maximum value of Q, viz. 7350, the value of the product 2O'4x6 2 xio. Along OY set out values of 20-4^2, the maximum value being 20-4x36= 735; and draw the curve having the equation y = 20 -4^, a table for which is : 60 x 62-4 d o i 2 3 4 5 6 20-4^2 o 20 -4 81-6 183-6 326 5io 735 thus obtaining the curve OA. THE CONSTRUCTION OF PRACTICAL CHARTS 425 In the right-hand division of the diagram lines must be drawn of various inclinations, the slopes depending on the values given to v. E. g., if v = 2, when the value of y (i. e., zo'^d*) is 500, the value of Q is 1000, therefore join the origin to the point for which Q = 1000, y = 500, and mark this as the line for v = 2. The diagram is com- pleted by the lines for v = i, 3, 4 . . . . . 10. Use of the chart. To find the discharge when the pipe is 2 \" diam. and the velocity of flow is 5 ft. per sec. : Erect a perpendicular from 2j on the d scale to meet the curve OA; then move across on the horizontal till the line for v = 5 is met ; and a vertical from this point on to the scale of Q gives the required value, viz. 637 Ibs. per min. pC' lily sec Quantity. Q i . i . i . i i . ! , i Inches OOO SOOO 400O SOOO 600O 7OOOV Lbsper* minute Fig. 243. Chart to give Quantity of Water flowing through Pipes. Again, if the quantity is 3000 Ibs. per min. and the velocity is 9 ft. per sec., to find the diameter : Erect a perpendicular through 3000 on the Q scale to meet the line marked v = 9 : draw a horizontal through this point to cut the curve, and finally drop a perpendicular on to the scale of diameters. The diameter required is seen to be 4*. If desired, the scale of Q may be modified to show values of Q (cu. ft. per sec.) or Q, (cu. ft. per min.). Example 6. The weight in Ibs. of a cylindrical pressure tank with flat heads (allowing for manhole, nozzles, and rivet-heads) may be expressed, approximately, by W = ioDT(L+ D), where L = length in feet, D = diam. in feet, and T = thickness of shell in sixteenths of an inch. Construct a chart to show weights for tanks of any diameter up to 5 ft. and lengths up to 30 ft. ; the maximum thickness of metal to be *. 426 MATHEMATICS FOR ENGINEERS Let W = ioDT(L+D) = W/T, . e., Wj = ioD(L+D). On the left of the diagram (see Fig. 244) no notice is taken of the thickness, i. e., W x is plotted against (L+D) for various values of D. A number of straight lines result, since WiW (L+D). Along OXi indicate the scale from o to 35 for (L+D), and along OY the scale for W x from o to 1750. The scale along OX will be that for W, the maximum value required being 8x1750, i. e., 14000 Ibs. For the left-hand portion. Suppose D = 2, then for L = 30 Wi = loxax (30+2) = 640. Join the origin to the point for which (L+D) = 32 and W\ = 640, and mark this as the line for D = 2. Proceed similarly for other values of D. , &? I , . iff 4<X>0 6060, 8OOO 10OOO Lbs. Fig. 244. Chart to give Weights of Pressure Tanks. For the right-hand portion. Suppose T = ", i. e., $". When W x = looo, W = W\T = 1000 X 8 = 8000. Join the point for which W = 8000, Wt = 1000 to the origin, and mark this as the line for T = $*. Draw lines for T = similar manner. *, etc., in a To use the chart. Let it be required to find the weight of a tank of length 18 ft. and of diameter 4 ft., with thickness of shell f. Here (L+D) = 18+4 = 22. Hence erect an ordinate through 22 on the scale of (L+D) to meet the line for = 4; draw a horizontal to meet the line for which T = f * ; then project to OX, and the value of W is read off as 5250 Ibs. THE CONSTRUCTION OF PRACTICAL CHARTS 427 Again, what will be the length of the tank, of diameter 4 J ft., the thickness of shell being J*. and the weight 7000 Ibs. ? Erect a perpendicular through 7000 on the scale of W to meet the sloping line for which T = J", and draw a horizontal to meet the line for which D = 4-5. A perpendicular through this point cuts OX t in the point for which L+D = 38-5, but as D = 4-5, then L must = 34 ft. Example 7. The next chart involves a considerable amount of calculation, which, however, once done serves for all cases. We wish to find the volume of water in a cylindrical tank for various depths and various lengths. Preliminary calculation. Let the depth of the water be h (Fig. 245). Then OC = r h, or, taking the radius as i ft., i - h. Let L DOC = -, then cos = L 2 21 245. E. g., for h = i-/* cos- = i -i = -9 = cos 25 50' 2 = 25 50', .., = 5 i4o' Now, the area of the cross-section of the water = area of segment 2 6 sin Q = 2 where & is expressed in radians A i. e., 6 (radians) = (degrees) Hence our table, giving areas of cross-section for different heights, may be arranged as follows ; h being expressed as a fraction of the radius /; COS - 2 6 2 0* e (radians) sin Q 0-sin0 Area I O o o o o 2 8 3652' 7344' 1-287 960 327 164 4 6 538' io6i6' I-855 960 895 448 6 4 662 5 ' I 3 2 5 0' 2-316 733 1-583 792 9 i 84 16' i68 3 2' 2-94 199 2-741 i-37i 1-2 2 IOI32' 203 4' 3-545 -'392 3-837 1-919 **5 ~'5 120 240 4-186 -866 5- 52 2-5.46 1-7 7 I3426' 2685 2 ' 4-70 -'999 5-699 2-85 2-0 i 1 80 360 6-284 o 6-284 3-142 428 MATHEMATICS FOR ENGINEERS Plot a curve with h horizontally and areas vertically, as in Fig. 246. Now volume = area x length and for a length of 10 ft. and area 3 sq. ft. the volume = 30 cu. ft. Hence join the origin to the point for which V = 30, A = 3, and mark this as the line for / = 10. Add other lines for different values of / as before. If the chart is to be made perfectly complete, a number of curves must be drawn in the left-hand portion, one for each separate value of the diameter. For diam. = 4 ft., ordinates of the curve would be (-] \ \ Valu 46 \ 2-7 Vo/i. 456 2 1-6 t-S -0 .* Q to SO SO 40 SO 60 Fig. 246. Chart giving Volume of Water in Cylindrical Tanks. i. e., four times those of the curve for d = 2 as already drawn. This tends to cramp the scale, so that it is preferable to work from the one curve and to multiply afterwards, remembering that the variation will be as the squares of the diameters. E. g., if diam. = 2 ft., h = -46 ft., and / = 5 ft., then vol. = 2-7 cu. ft., the lines for this being shown on the diagram. But if the diam. = 6 ins., h = -46 x radius, and / = 5 ft., then vol. = 2-7 x (|) 2 = -169 cu. ft. Again, if h = 1-72 x radius, diam. = 5 ft., and length = 16 ft., to find the volume proceed as indicated on the diagram. The volume for 2 ft. diam. is 45-6, so that the volume for 5 ft. diam. 45-6 x (1)*= 285 cu. ft. THE CONSTRUCTION OF PRACTICAL CHARTS 429 The following construction may reasonably be introduced as a chart : Example 8. Resistances of 54 and 87 ohms respectively are joined in parallel ; what is the combined resistance of these ? This question may be worked graphically in the following manner- Draw OA and OB, Fig. 247, lines making 120 with one another. Along OA set off a distance to represent 54 ohms, thus obtaining the point E, and along OB set off OF to represent 87 ohms to the same scale. Bisect the angle AOB by the line OC. Join EF to intersect OC at D. Then OD measures, to the same scale as that used along OA and OB, the combined resistance, and it is found to be 33-2 ohms. Alignment Charts. In these charts two or more variables are set out along vertical axes, which are so spaced, and for which the scales are so chosen, that complicated formulae may be evaluated by the simple expedient of drawing certain crossing lines. Then for the same connection between the variables, one chart will give all possible values of all of them within the range for which the chart is designed. Thus transposition and evaluation of formulae become unnecessary; and, in fact, the charts can be used in a perfectly mechanical manner by men whose knowledge of the rules of transposition is a minimum. Referring to our work on straight line graphs, we see that the general equation of a straight line is Y = aX+6. By suitably choosing the values of a and b we may write this equation in the form AX+BY = C; and it is with the equation in this form we wish to deal. Plotting generally is to most minds connected inseparably with two axes at right angles : that is certainly the easiest arrangement of the axes when two variables only are concerned. Suppose, 43 o MATHEMATICS FOR ENGINEERS now, that three, four, or even eight or nine variables occur; then our method fails us, and in such a case it is found that vertical axes only can be used with advantage. It is not our intention to fill the book with alignment charts, for examples of these intensely practical aids may be found in the technical periodicals; what is intended is that the theory of the Fig. 248. Principle of Alignment Charts. charts should be grasped, so that any one can construct a chart to suit his own particular needs and conditions. Let us consider firstly the simplest case, viz. x-{-y = c, or, as we shall write it, +u = c (u and v being adopted for the sake of clearness, since both the u and the v axes are to be vertical, whereas axes for x and y are horizontal and vertical respectively). Draw two verticals AE and BF (Fig. 248) any convenient dis- tance apart, and let AE be the axis of u and BF be the axis of v. Draw also the horizontal AB, which is to be the line on which the zeros of the scales along the u and v axes lie. Assume some value for c and calculate values of and v for two cases ; set off along AE these values of u to a scale of / t units per inch, and along BF these values of v to a scale, say, of l z units per inch. Let AH represent the value of u when v has the value THE CONSTRUCTION OF PRACTICAL CHARTS 431 represented by BK, and AM the value of corresponding to the value of v represented by BN. Join HK and MN to intersect at G, and through G draw a vertical GC, which will be referred to through- out as the mid-vertical. Then AH represents the first value of ; call it u 1 ', and BK represents the first value of v x ; call it v r Similarly AM and BN represent 2 and v z respectively, and since u-\-v = c for all values of u and v, !+*>! = c and 2 +v 2 = c - AH, AM, BK, and BN are actual distances on the paper, hence / x x AH = MJ, /jXAM = u 2 , J 2 xBK = v 1( and / 2 xBN = v 2 . Substituting in the equations u 1 -\-v 1 = c and 2 +v 2 == c > (/ 1 xAH) + (/ 2 xBK)=c ........ (i) and (/ x xAM) + (/ a xBN) =c ........ (2) From the figure AH = AM+MH .......... (3) BK = BN-NK .......... (4) By multiplication of (3) by l t and (4) by l z , we obtain the equations AHX/ 1 =(AMX/ 1 ) + (MHX/ 1 ) ....... (5) BKx* a =(BNx/ 2 )-(NKx/ 2 ) ....... (6) By similar figures MH AC XTV MHxCB , . NK = CB' Whence NK= -AC~ ..... (7) Add equations (5) and (6), then (AHx/,)+(BKx/i) =(AMx/ 1 )+(MHx/ 1 ) + (BNx/ 2 )-(NKx/ 2 ) and by substitution for NK its value found in equation (7) / ~\ MHxCB AC *. e., by substitution from (i) and (2) - x/ t ) Hence Muf^-xl) must equal zero, so that either \ AL/ / CB , MH = o or 'i- x/2 = - 432 MATHEMATICS FOR ENGINEERS Accordingly, since MH is not zero X'i ........... (8) Let the lengths AB, AC and CB be represented by m, m 1 and itoyt 1 w 2 respectively, then equation (8) may be written 1 1 = - .. m, L j u -i W 2 ^i whence - - = , , , and by similar reasoning - = , ,* , m /i+/2 m ^1+^2 Any pairs of values of w and v to suit the equation w+w = c might have been chosen, and the same argument might have been applied, so that as long as the scales for the u and v axes and the constant c remain the same, the ratio * will hold, i. e., there can m z only be the one mid-vertical. Also G will be a fixed spot, since it is vertically over C, and any one crossline satisfying the equation M+V = c will give the position of G. The length of GC is thus fixed. Let it represent the constant c to some scale, say the scale of 1 3 units per inch. A relation between / 3 , /j and / 2 can now be found. GC is an actual length, representing c to the scale of 1 3 units per inch Substituting in (i) and (2) (/iXAHJ-H^xBK) = / 3 xGC and (/! X AM) + (l t x BN) = / x GC. Calculate the value of v when u = o, and plot BL to represent this value ; join AL, then this line passes through G, by the argu- ment already given. When u = o, v = c, so that BL actually represents c, or BLx/2 = c- But GCx/s also = c BLx/ 2 = GCx/ 3 . AC By similar triangles = .-= x BL x /, m ' = THE CONSTRUCTION OF PRACTICAL CHARTS 433 Now - 1 = -A m or /. = /!+/, . <?., the scale along the mid- vertical is the sum of the scales along the outside axes. The student of mechanics may be helped by the analogy of the case of parallel forces. If weights of Wj and W 2 are hung at the ends of a bar of length /, their resultant W 3 is the sum of the separate weights, and acts at a point which divides the length into two parts in the inverse proportions of the weights. Thus, in Fig. 249, if C is the point of action of the resultant W 3 AC_W a A _ C _ B A C _ B L -- j- I -- H U-m,-4*-m.T^ { *fc ** *' T, *L \ Fig. 249. Fig. 24Qa. This is exactly the same kind of thing as we have in connection with the scales along the three axes, for we may replace W^ W 2 and W 8 by l lt 1 2 and l a respectively, and we get the bar loaded as in Fig. 2490. We can now proceed to the more general case, viz., that in which the equation is au-\-bv = c. Use may be made of the same diagram (Fig. 248) as that used for the simpler equation, viz., u+v = c. To do this, however, the scale of u must be opened out " a " times, and that of v opened " b " times ; the distance BL, which formerly represented c, now f* representing T, since it shows the value of v when u is zero. Accordingly, if l\ and 1' 2 are the new scales along AE and BF l\=- 1 and ^=4* a b Hence, the scale along GC = Ii+l 2 and * = *w 2 / x a/! FF 4 34 MATHEMATICS FOR ENGINEERS l\ and /' 2 would be the actual scales used. For a general statement, therefore, we can regard these as /j and l t and the scale along GC as 1 3 ; so we sum up our results in the forms m l bl 2 where J lf 1 2 and 1 3 are the actual scales used. * These results might also be summarised in the following way : If the general equation is au + bv = c, then the scale of c (along 10' o A D B Fig. 250. Alignment Chart for Equation 41* + 6 = 30. the mid-vertical) = " a " times the scale of + " b " times the scale of v, and the division of AB at C is such that CB _ a times the u scale AC ~ b times the v scale To illustrate by some numerical examples : Let us first deal with the equation 4M+6v = 30. To construct a chart for this equation, draw two vertical lines, as in Fig. 250, fairly well apart, say 6" (this distance being simply a matter decided by the size of the paper and the degree of accuracy desired). Number from the same horizontal line scales for and v, and let the two vertical scales be equal in value, viz. /j = J a = 2 (units per inch). THE CONSTRUCTION OF PRACTICAL CHARTS 435 Then the mid-vertical must be so placed that ^ l = ^ w a/ 4X2 or i.e., MI = xm = x6" = 3-6* ^ D or the mid- vertical is 3-6" distant from the axis of . Also the scale along the mid-vertical is fixed, since / 3 is given by /!+&/ 2 , i. e., 1 3 = (4x2) + (6x2) = 20, or i* represents 20 units. If AB is the horizontal on which the zero of the scale of u and also that of the scale of v lies, number from the point D the scale along the mid-vertical, and indicate the marking for the constant term in the equation, viz., 30. If u = 7-5, v = o, and it will be noticed that if a line is drawn from 7-5 on the scale of through the point C (30 on the mid- vertical), it intersects the axis of v at the point B, *. e., at the point for which v = o. Similarly, if v = 5, then u = o, and the line joining 5 on the axis of v to o on the axis of u passes through the point C. Hence if a value of u, say, is given, the value of v to satisfy the equation 4+6y = 30 can be readily obtained by drawing a straight line through that given value of u and the point C, and noting its intersection with the axis of v : e. g., to find the value of v when = 3 : join 3 on the axis of u to C and produce to cut the axis of v ; read off this value of v, viz., 3, and this is the solution required. As an illustration of the fact that the alteration in the value of c alone alters the position of the point C on the mid-vertical and not the position of the mid-vertical, let us deal with the equa- tion 4+6v = 18. Working with the same scales, join 4-5 on the axis of u to o on the axis of v, since if u = 4-5, v = o. This line passes through the point C x numbered 18 on the mid-vertical. To find the value of v when u = o, join o on the axis of u to 18 on the mid-vertical and produce to cut the axis of v in the point 3 ; then the required value of v is 3. Example 9. Construct a chart to read values of / in the formula I "jd+'OO^D, where t = thickness at edge of a pulley rim, d = thick- ness of belt, and D = diameter of pulley, all in inches, d is to range from -i* to -5* and D from 3* to 10*. Construction of the chart (see Fig. 251). Draw two verticals, say 5* apart (as in the original drawing for Fig. 251). Let values of d be 436 MATHEMATICS FOR ENGINEERS set out on the left-hand vertical. The range of d being -4*, let 4* repre- sent this value, so that i* = -i unit or / x = !. The range of D is 7", so let 3^* represent this, so that i* = 2 units or /, = 2. Also a = -7 and b = -005, so that /, = o/x+Wj = (-7X-i) + (-005x2) = -07+ -oi = -08 i.e., i" = -08 unit of *, along the mid -vertical. To fix the position of the mid-vertical m 2 _ al-i _ -7X-I _ _7 m^ ~~ bl t ~ -005x2 ~~ i so that the mid-vertical is J x 5", i. e., -625 from the axis of d. Chart Full Size. Fig. 251. Alignment Chart giving Thickness at Edge of Pulley Rim. The zero on the t scale will lie on the line joining the zero on the axis of d to that on the axis of D. We are not, however, bound to enlarge the diagram to allow this line to be shown ; in fact, in a great number of cases the line of zeros or virtual zeros is quite outside the range of the diagram. As a matter of convenience let -i on the d scale and 3 on the D scale be on the same horizontal ; then, since / = -085 when d = -i and D = 3, this horizontal will cut the mid-vertical at the point to be numbered -085. The scales along the three axes can now be set out, and the chart is complete. Use of the chart. To find the value of t when d = -5 and D = 3, join -5 on the d scale to 3 on the D scale to intersect the mid-vertical in the point -365; then the required value of t is -365. Again, if THE CONSTRUCTION OF PRACTICAL CHARTS 437 E > = 6 and t = -325, the value of d is found by joining 6 on the axis > to -325 on the axis of t and producing the line to cut the axis of d in -421 ; the required value of d thus being -421. To carry this work a step further : Most of the formula encoun- tered in practice contain products, many in addition containing powers and roots. By taking logs, the multiplications are con- verted to additions, and the methods of chart construction already detailed can be applied with slight modifications. To deal with a simple case, by way of introduction : Chart giving Horse-power supplied to Electric Motor. Example 10. Construct a chart to give the horse-power supplied to an electric motor, the amperage ranging from 2 to 12 and the voltage from no to 240. [Watts = Amps x Volts and H.P. = Watts \ \ 746 ; Taking initials to represent the quantities W = AV and H = ^X 746 or 746H = AV. Taking logs throughout log 746 + log H = log A + log V. Let log 746 + log H = C, then if for log A we write K and for log V we write V, the equation becomes A + V = C, which is of exactly the same form as an + bv = c, where a = b = i. Hence / 8 = al t +bl t = li+l t In order that the scale along the mid-vertical may be the sum of the scales along the outside axes, the mid-vertical must be so placed that it divides the distance between the outside axes in the inverse proportion of the scales thereon. By the scales, it must now be clearly understood that i* represents so many units of logarithms and not units of the actual quantities. Slide rule scales will often be found convenient for small diagrams. If the B scale is used, 9-86* (the length from index to index) would represent 2 units (i. e., log 100 log i), whilst if the C scale is used, 9-86* would represent i unit. If a log scale is not used, the best plan is to tabulate the numbers, their logarithms, and corresponding lengths, before indicating the scales on the diagram. One setting of the slide rule will then serve for the conversion of the logs to distances, according to the scales chosen. In this case A varies from 2 to 12, i. e., log A varies from -301 to 1-0792, a range of about -8 units ; and a fairly open scale will result if i* = J unit is chosen, t. e., l^ = -2. MATHEMATICS FOR ENGINEERS For V the range is no to 240, so that the range in the logs is 2-0414 to 2-3802, or about -35 unit ; and accordingly let l t = -i. Then l a = li+l z = -2+-I = -3 w a /! -2 2 and - -r = = ! j a ! i In the original drawing (Fig. 252) m, the distance between the outside axes, was taken as 6"; hence m 1 = of 6*, i. e., 2", or the mid-vertical must be placed 2* from the axis of A. 240 220 200 190 180 170 I6O 150 140 130 110 110 Fig. 252. Chart giving H.P. supplied to Electric Motors. Preliminary tabulation for the graduation of the outside axes reads : For the A axis. A 2 2 -l > a V5 4 C log A 3OI 3Q7Q 4771 5441 6O2I 6OOO Diff . of logs o OQ7 176 243 3OI ^08 Actual distance from\ base line (ins.) . . j" O 485 88 1-22 I'S 1 1-99 6 7 8 9 10 ii 12 7782 8451 9031 9542 1-0 1-0414 1-0792 477 '544 602 653 699 740 -778 2-39 2-72 3-01 3-27 3'5 37 3-89 THE CONSTRUCTION OF PRACTICAL CHARTS 439 The marking for 2 is first fixed, and then all distances are measured from that : thus to find the position of the point to be marked 4, log 4 log 2 = -301, and since i" = -2 units the actual distance from 301 2 to 4 must be ^ , viz., 1-51*. For the V axis V no 1 20 T-JQ 140 T en 160 *5 U loeV . 2*0414 2*O7Q2 2-TT3Q 2*1461 2- 1 76 1 Diff . of logs o O378 O72S IO47 T5J.7 1627 Actual distance from\ base line (ins.) . ./ 378 725 1*047 1-347 1-627 170 180 190 200 2IO 22O 230 240 2*2304 2-2553 2*2788 2-3010 2*3222 2*3424 2*3617 2*3802 1890 2139 2374 2596 2808 3010 3203 3388 1-890 2-139 2-374 2*596 2*8o8 3-oi 3-203 3-388 The fourth line in the latter table is obtained by division of the third line by *i, since /, = !. The scales can now be indicated along their respective axes, and the mid-vertical may be drawn. It is not convenient in this par- ticular example to join the zero of each of the outside scales, which would necessitate the axes being extended to show i on the A scale and i on the V scale, since log i = o. If such a line were drawn, however, it would be the line on which the virtual zero of the scale of H would lie. Then the virtual zero would be ^ since when A = V = i H = IXI . It is, therefore, the best plan to locate some 746 convenient point on the mid-vertical to serve as a zero. Thus, join 5 on the A scale to 149*2 on the V scale; and mark the point of intersection of this line with the mid-vertical as i, since H = 440 MATHEMATICS FOR ENGINEERS For other graduations, tabulate thus : H I 1-5 2 3 8 1 CT H o 1761 3OI 4771 1^9031 Diff of logs o 1761 301 4771 -0969 Actual distance from i -o (ins.) o 586 I'D i'59 -32 6 5 4 3 2 1-7782 1-699 I-602 1-477 T-30I 2218 -301 398 -523 -699 738 i 1-32 -1-74 -2'33 The fourth line is obtained from the third by division by -3, since /, = '3. Marking in these numbers along the H axis, the chart is complete. Use of the chart. To find the H.P. supplied if the current is 4 amps and the pressure is 170 volts. Join 4 on the A scale to 170 on the axis of V : this line passes through -91 on the H axis, and therefore the required value of H is -91. Again, if H = -6 and current = 2-5, what is the voltage? Join 2-5 on axis of A to -6 on the H axis, and produce the line to cut the V axis in 179; therefore V = 179. {It should be noted that the chart is not crowded with figures, because clearness is desired. Charts to be used frequently, and from which great accuracy is desired, should be drawn to a much larger scale.} At a first reading one may be tempted to comment on the length of calculation necessary to perform what is, after all, a very simple operation : it must be borne in mind, however, that (a) a most simple example has been chosen as an illustration, and (b) a chart once constructed by this method may be used very many times in a perfectly mechanical way. So many formulae contain powers, that we must now investigate the effect of the exponents on the scales, etc., of these charts, and the modification in the construction due to them. Flow of Water through Circular Pipes. Example n. If water is flowing through a pipe of diameter d inches, at the rate of v ft. per sec., then the quantity Q in Ibs. per sec. is obtained from THE CONSTRUCTION OF PRACTICAL CHARTS 441 Transposing In the log form- . e., Q ^34~ log Q - log -34 = log d 2 + log v log Q - log -34 = 2 log d + log v Pi g. 253 ._Chart giving the Flow of Water through Circular Pipes Let C = log Q - log -34. then log t> + 2 log d = C or V+aD = C i. e., in comparison with the standard form, a = i and b = 2. 442 MATHEMATICS FOR ENGINEERS Assume the range of pipe diameters to be i* to 9*, and the range of the velocity of flow to be i to 10 ft. per sec. Then the same scale will be convenient for both axes. Let / x = / a = 7^7, i. e., use the B scale of the slide rule. Now 4'93 2X- 4'93 IX 4'93 so that if m is taken as 6* (as in the original drawing for Fig. 253), m = -x6*, i. e., 4*, or the mid-vertical is 2" removed from the axis of v. 3 Also- I. - * 1+ W. + axjj - - -607, or i* = -607 unit along the axis of Q. Draw the axes of v, Q and d and graduate the outside ones, using the B scale of the slide rule. In Fig. 253 the I of each scale is on a horizontal, but it is quite immaterial where the graduations begin. To select a starting-point on the mid-vertical, join 10 on the axis of v to i on the axis of d, and call the point of intersection with the mid-vertical G. Now, Q = 34^ 2 w, and therefore for the particular values of d and v chosen, Q = -34Xi 2 x 10 = 3-4. G is therefore at the position to represent 3-4 Ibs. per sec. The table for the graduation of the mid-vertical will then be : Q. V4 3 2 c 8 IO loe O . C32 4.77 3OI 600 on 3 I Diff . of logs o CIS '231 l67 371 468 Distance above or below G o O9 38 27 61 77 15 20 30 40 5 80 IOO 1-176 I-30I 1-477 1-602 1-699 1-903 2 644 -769 945 1-07 1-167 i-37i 1-468 i -06 1-26 i'55 i'75 1-92 2-25 2- 4 I The fourth line is obtained by multiplying the third by 1-64 or by dividing it by -607, since l a = -607. Use of the chart. Find the discharge through a pipe of 9* diam. when the flow is at the rate of 2 ft. per sec. Join 2 on the axis of v to 9 on the axis of d, to intersect the axis of Q at 55 ; then the required quantity is 55 Ibs. per sec. Again, what diameter of pipe is required if the discharge is 30 Ibs ./sec. THE CONSTRUCTION OF PRACTICAL CHARTS 443 and the rate of flow is 3 ft./sec. ? Join 3 on the v scale to 30 on the Q scale and produce the line to cut the axis of d in 5-4; then the required diameter is 5-4*. To iUustrate the question of scales further, consider the follow- ing cases : Example 12. Show how to decide upon the scales for the chart giving the values of T, / and d in the equation T = ^fd*, referring to the torsion of shafts. T The equation may be written - - = fd 3 , i. e., 5-iT =fd 3 , and by taking logs throughout log 5-1 + log T = log/ + 3 log d. Write this log/+3logd = C, then F+ 3 D = C (the large letters being written to represent logs). Thus a = i and 6 = 3. Hence if li = 5. sa y. and I* = 2, /, = al^+bl^ = (iX5)+(3X2) = n w i bl t 3x2 6 6 and -? = * = a = _ or m x = of w. m t alj. 1x5 5 ii Similarly for pv n = C, where n may have values such as -9, 1-37, 1-41, etc. Here log p + n log v = log C t. e., P + wV = C so that a i, b = n. Hence l a = (ixlj+(nxl a ) = l t +nl t m, bl t nl t and _* = -J = _? W 2 0/| /! Questions involving more complicated formulae can be dealt with by a combination of charts. From the above work it will be seen that when three axes are employed, three variables may be correlated, or one axis is required for each variable. However many variables occur, they may be connected together in threes, so that the graph work is merely an extension of that with the three axes. Chart giving Number of Teeth in Cast-iron Gearing. Example 13. To construct a chart giving the number of teeth necessary for strength in ordinary cast-iron gearing. Given that T = ~j^r where T = No. of teeth in wheel, N = revs, per min. H = H.P. transmitted, p = pitch. 444 MATHEMATICS FOR ENGINEERS H so that Tp 3 = C (i) and also ^p = C ' (2) i . e., two charts can be constructed, and by suitably choosing the scales and the positions of the axes the charts may be made interdependent. For chart (i), let / t = -i- unit of T and let /, = unit of p, i. e., use the B scale of the slide rule for both the T and the p axes. N 5 - Fig. 254. Chart giving Number of Teeth necessary in Cast-iron Gearing. Then, since logT+ 3 log/> = log C, /, = 4 4 4'93 Also - = 4-93^3 BO that m 1 = 3 of m. _l_ i 4 4'93 Draw two axes for T and p respectively 4* apart, as drawn in the original. drawing for Fig. 254, and also put the mid-vertical 3* from the axis of T. The last is simply a connecting-link between charts (i) and (2), and therefore no graduations need be shown upon it. THE CONSTRUCTION OF PRACTICAL CHARTS 445 Along the axis of T mark off readings (using the B scale of the slide rule) for, say, 6 up to 80 and along the axis of p, readings from i to 8. Join 2 on the paxis to 30 on the axis of T, and note the point of intersection with the mid-vertical ; this must be marked 240, since 30x2' = 240. For chart (2), we already have the mid-vertical and its scale. We must now proceed to find scales for the axes of H and N, t. e., the usual process is reversed. Suppose the range of H is 5 to 100 and that of N is 20 to 150, and we decide to use the same scale for both, say l t . TJ /-> Then ~ N = 79! or log H - log N = log C - log 791 '. e., a = i, whilst 6 = i. If, however, the numbering for N is placed in the opposite direction to that for H, we may say that b = i. Hence /, = / 4 +/ 4 = 2/ 4 or / 4 = -?- = = -406 also * t. e., the mid-vertical, which has already been drawn, must be midway between the axes of H and N. For convenience let m l = m t = 4*. Then for N the tabulation is as follows : N 20 on dO CQ 60 log N T -7QI I '4.77 I '60 2 T-fiQQ I'778 Diff . of logs o 176 3O I 308 477 Distance from mark for 20 o 432 74 9 8 I-I7 70 80 90 100 150 2OO I-845 1-903 1-954 2 2-176 2-30I 544 602 653 699 875 I i'34 1-48 1-61 1-72 2-16 2-46 To obtain the fourth line from the third divide by -406, for / 4 = -406. Some little trouble may arise in the placing of the marking for 20 conveniently : thus in our case we have marked 20 fairly high up on the paper. Join any point on the N axis, say 150, to the 240 on the mid -vertical, and produce this line to cut the axis of H in the point A. We must now find the reading for A. C = 240 and C = ^ 9 * -, but N = 150 TT 150x240 hence H = * ^* = 4 5'5- Thus we can graduate the axis of H from 45-5 as zero. 446 MATHEMATICS FOR ENGINEERS The table for the graduation of the H scale is : H 4 V5 <; 10 15 20 2<; loeH 1-658 699 i 1-176 1-301 1-398 Diff. of logs. . . o -'959 -658 482 -357 260 Distance from A o 2-36 1-62 -1-19 -88 64 3 35 40 5 60 80 100 1-477 1*544 1-602 1-699 1-778 I-903 2 181 114 056 041 120 245 342 445 -28 14 i 295 602 84 / = -406, so that the fourth line is obtained by dividing the figures in the third line by -406. Use of the chart. Suppose N = 20, T = 20, p = 5, and the value of H is to be found. Join 20 on the T axis to 5 on the axis of p ; and let this line intersect the mid-vertical at G. Join 20 on the N axis to G, and produce the line to cut the axis of H in 63. Then the required value of H is 63. Again, if H = 15, N = 80, and p = 2, we are to find the value of T. Join H = 15 to N = 80 to cut the mid-vertical at D. Join p = 2 to D, and produce to cut the axis of T in 18-6 : then the required value of T is 1 8-6. The lines must join values of either T and p or H and N, because the chart was so constructed. Exercises 43. On Alignment Charts. 1. Construct a chart giving values of u and v to satisfy the equation 2M+7V = 52, the range of v being 2 to 12. 2. Construct a chart to give values of u and v to satisfy the equation I-2V -6414 = -85, u ranging from 5 to 20. (To allow for the minus sign, either the mid- vertical may be placed outside the axes of u and v, as for unlike parallel forces, or the number- ing on the u scale may be downward, whilst that on the v scale is upward.) 3. The thickness of boiler shell necessary if the working pressure is p Ibs. per sq. in., the diameter of the boiler is d inches, and the allow- able stress is /Ibs. per sq. in., is found from t = &j. Taking the value of /as 10000, construct a chart to give values of t, the range of diameter being i'-6* to 6 ft., and the pressure varying from 40 to 150 Ibs. per sq. in. What is the thickness when the diameter is 2 '-3* and the working pressure is 85 Ibs. per sq. in. ? If the thickness is J" and the diameter is 4'-6*. what is the working pressure ? THE CONSTRUCTION OF PRACTICAL CHARTS 447 4. According to the B.O.T. rule the permissible working pressure in a boiler having a Fox's corrugated steel furnace is P = -^, where / = thickness of plate in sixteenths of an inch and D is the internal diameter in inches. Construct a chart to give values of P for boilers of diameters ranging from 2 ft. to 5 ft., the thickness of the shell varying between -fa" and *. 5. The diameter in inches for a round shaft to transmit horse-power 3/rr H at N revs, per min. (for a steel shaft) is given by d = 2-g\/ =^. If N varies from 15 to 170 and H from J to 10, construct a chart to show all the diameters necessary within this range. 6. For tinned copper wire the fusing current C is found from C = 6537<Z 1 ' 403 , where d is the diameter in inches. Construct a chart to read the diameter of wire necessary if the fusing current is between 22 and 87 amperes. 7. Hodgkinson's rule for the breaking load for struts is _ 1 where d = diameter in inches and L = length in feet, A being a con- stant. Construct a chart to give the breaking load for cast-iron struts with rounded ends, the diameters ranging from 2* to 15* and the lengths from 6 ft. to 20 ft. The value of A for solid cast-iron pillars with rounded ends is 14-9. 8. Construct a chart to give the points on an adiabatic expansion line of which the equation is pv l ' n = 560, the range of pressure being 14-7 Ibs. per sq. in. to 160 Ibs. per sq. in. 9. The coefficient of friction between a certain belt and pulley was 32. If the angle of lap varies from 30 to 180, construct a chart to give the tensions at the ends of the belt, the smaller tension varying from 50 Ibs. wt. to 100 Ibs. wt. Given that T = te?*, ft being the coefficient of friction, and 6 being the angle of lap in radians. [Note that the scales for T and t will be log scales, but that for 6 will be one of numbers only.] 10. If P = safe load in tons carried by a chain, d = diameter of stock, and / = safe tensile stress, then for a chain with open links P = - 4 <* 2 /. If/ varies between 4 and 10 tons per sq. in., and the diameter of the stock ranges from J* to 2", construct a chart to give the safe load for any combination of / and d. CHAPTER XII VARIOUS ALGEBRAIC PROCESSES, MOSTLY INTRODUCTORY TO PART II Continued Fractions. Consider the fraction i 3+; or, as it would usually be written as a continued fraction 1 J _g._ 4 2+ 3+ 5+ 7 Its true value would be found by simplification, working from the bottom upwards; thus, 5+= ^ 39 7 39' 131 131 "39" and -L = *3i 301 301 39 39 131 131 o* o o/ 02 \N? of Converge nh X 2 3_L 4 i. e., the true value of the fraction, known as a continued fraction, is ^ Fi &' 2 55- 301 Conversely, if the resulting fraction ^1 is given, various approxi- mations can be made for it by taking any portion of the continued fraction, the correct order being maintained. VARIOUS ALGEBRAIC PROCESSES 449 Thus, for example, a first approximation would be -, which is T Q too large ; the second approximation is = -, which is too 2+| 7 small, but is nearer the correct value. The approximations, or convergents, are alternately too large and too small, but the error becomes less as more terms of the fraction are taken into account. To illustrate this fact a curve is plotted dealing with the above fraction, in which the ordinates are errors and the abscissae the numbers of the convergents. (Fig. 255.) JO J Occasionally in engineering practice a fraction such as -*- occurs which is not convenient to deal with practically, so that one seeks for some more convenient fraction which is a fair approxi- mation to that given. The following example introduces such a case : Example i. A dividing head on a milling machine is required to be set for the angle 19 25'!* with great accuracy. For the Brown & Sharpe dividing heads, 40 turns of the crank make one revolution of the spindle, and there are three index plates with number of holes as follows 15, 16, 17, 18, 19, 2o^| 21, 23, 27, 29, 31, 33V 37. 39, 41. 47, 49 Thus one turn of the index crank would give an angle of ~ -, i. e., 9. 4 Evidently two turns will be required for 18, and i25'i* has then to be dealt with. Expressing this as a fraction of 9, the proportion of one turn is found. , _ Now i25'i" = -- mms. . u 5101 5101 hence the fraction of one turn required = ^ x Q x ^ = ^35 We wish to reduce this fraction to its best approximation having a denominator between 15 and 49, according to the numbers of holes as above. Proceed as though finding the G.C.M. of 5101 and 32400. Thus 5101)32400(6 I794)5 IOI ( 2 281)1513(5 108)281(2 65)108(1 22, etc. G G ^o MATHEMATICS FOR ENGINEERS Then the continued fraction is _i_ j__ JL _i_ JL T - 6+ 2+ 1+ 5+ 2 + l + ist convergent = g, 2nd = ^-, 3 rd = ' 4th = 108' 5th = 235 ' these being found by simplification of the fraction, a method which is a trifle laborious. The 3rd convergent might have been found from the 2nd in the following way Numerator of 3rd convergent = {numerator of 2nd convergent X denominator of last fraction added} + {numerator of ist con- vergent x numerator of last fraction added}. Denominator of 3rd convergent == {denominator of 2nd convergent X denominator of last fraction added} + {denominator of ist convergent x numerator of last fraction added}. In this case 12 I ist convergent = ^, 2nd = and the next fraction = - x i) + (i x i) 3 , 3 rd convergent = x +6 x Now from the and and 3rd convergents the 4th convergent may be 2 ^ found ; for the 2nd convergent = , the 3rd = , and the next fraction = . (3 x 5) + (2 x i) 17 To obtain the 5th convergent : the 3rd convergent = , the 17 I 4th = ^, and the next fraction = - ,, ,, (17 x 2) + (3 x i) 37 .*. the 5th convergent = 7-^ : , '. = -^- L (108 x 2) + (19 x i) 235 For the purpose of the question we require the convergent with denominator between 15 and 49 : the only one is . Therefore, it would be best to take two complete turns together with 3 holes on the i9-hole circle. The error in so doing is very small. Thus 5101 , .. , 3 "* = ' I 5744 whilst = -15790 32400 J/ ^ 19 i.e., the error is 46 in 15744 or ^ x 100 % J 5744 = 292 % too large. VARIOUS ALGEBRAIC PROCESSES 451 Example 2. Find a suitable setting of the dividing head to give 2i'45*. No. of turns of index crank = 88 2I/ 45* = 2I2 ? = Q I7<>7 9 2160 ^2160 589 = g2_* 720 Hence 9 complete turns are necessary together with ^-? of a turn. To find a convenient convergent for ^5_? : 720 589)720(1 65)131(2 1)65(65 /. The fraction = -L -L -L i+ 4+ 2+ 65 The ist convergent = -, the 2nd convergent = 4. so that the 3rd convergent = J4 * 2) + (i x i) = _o_ ^ x 2) + (i x i) ii also the 4 th convergent = / 9 X 6 ? + ( f x ') = 5?? (n x 65) +(5x1) 720 Thus the best convergent for our purpose = , and 27 holes on the 33-hole circle would give this ratio. Therefore, 9 complete turns together with 27 holes on the 33-hole circle are required. An interesting example concerns the convergents of IT. Example 3. To 5 places of decimals the value of IT is 3-14159 : what fractions may be taken to represent this ? 14159 3-14159 = 3- ^ -^ J ~ */ ^ */ -\ 'IOOOOO 14159)100000(7 887)14159(15 5289 854)887(1 33)854(25 194 29)33(1 4 i i I.e., rr - 3 + - 7+ 15+ i+ 25+ 452 MATHEMATICS FOR ENGINEERS 22 The ist convergent = 3, the 2nd convergent = , (22 x 15) +(3x1) 333 and hence the 3 rd convergent = \ 7 x ^ + x J - g| a. X l) + (22 X l) 355 the 4 th convergent = x ^ + \ y x ,/ = ff| the 5 th convergent - J355 x 25) + (333 x i) = 9208 (113 x 25) + (106 x i) 2931 The values of these convergents in decimals are 3, 3-14286, 3-14151, 3-14159 -K and 3-14159 , respectively. A rule often given for a good setting of the slide rule for multiplica- tion or division by IT is : Set 355 on the one scale level with 113 on the other, etc. The reason for this is seen from the above investigation; 5^ as a value for TT being far more accurate than, say, 113 7 Partial Fractions. Consider the fractions - , - , and their sum. X 4 2# 7 To find their sum, i. e., to combine them to form one fraction, the L.C.D. is found, viz., (x 4) (2* 7) or 2x 2 15* + 28; the numerators are multiplied by the quotients of the respective de- nominators into the L.C.D. , and the results are added to form the final numerator. Thus- 2 I 4 ' x 4 ' 2x 7 2x 2 15*4-28 8x 30 The fraction last written may be spoken of as the complete 2 A. fraction, for which - and - are the partial fractions. x 4 2* 7 It is often necessary to break up a fraction into its partial fractions : they are easier to handle, and operations may be per- formed on them that could not be performed on the complete fraction. To resolve into partial fractions, proceed in the manner out- lined in the following examples : Example 4. Resolve , * ~ 3 5 into partial fractions. 2X Z 1$X + 28 8* 30 8*- 30 A (2*-7)(*- 4 ) (X - 4) where A and B have values to be found. VARIOUS ALGEBRAIC PROCESSES 453 Reduce to a common denominator, (zx 7) (# 4), and calling this D 8* -30 _ A( 2 * - 7) + B(* - 4) D D Equating the numerators 8* - 30 = A(2* - 7) + B(* - 4 ). This relation must be true for all values of x : accordingly let x = 4, this particular value being chosen so that the term containing B, vanishes, and one unknown only remains. Then 32-30 = A (8 - 7) + B( 4 - 4) or 2 = A. Now let the term containing A be made to vanish by writing 3$ in place of x Then 28 - 30 = A(7 - 7) + B( 3 i - 4) - 2 = - B B = 4 /. the fraction = *-4 2 *~7 Example 5. Express ~- as a sum of two or more fractions. The numerator and denominator are here both of the same degree ; in such cases divide out until the numerator is of one degree lower than the denominator. Now suppose ^ = C with D remainder then the fraction ~ = C+ ^ Applying to our example, by actual division the quotient = - and the remainder = ^ : hence the fraction = *- -f- ,/J. m 3 3 6\5 X ) Example 6. (a) Find the sum of 4 7* (2* + I) 5 (X + I) 8 (X + I) and (b) resolve ~" 24 * ~T, 12 * ? 2 into P artial fractions. * f*A* " (a) 2*+ i 5(^4- i) 2 x+ i 4 X 5^ + I) 2 - 7Ar(2^r + i) - 3 X 5(* + l)(2* + l) 5(2^+ i)(x+ i) 8 2o^- + 40^ + 20 - 14** - 7* - 30*' - 45* ~ I 5 D - 24** - 12* + 5 454 MATHEMATICS FOR ENGINEERS (6) To resolve ~ 4 ., .% into partial fractions, therefore, it is necessary to consider the possibility of the existence of (x + i) as a denominator, in addition to (x + i) 8 , for (x + i) is included in (x + i) 2 . Let the fraction = [Bx is written in place of B, so that the numerator shall be of degree one less than the denominator, i. e., all terms of the numerator, when over the same denominator, will then be of the same degree.] Thus the fraction = A(* + ,)* + 5 B*( 2 * + i) + 5 C(a* + x)(* + z) Equating numerators - 2 4 * a - 12* + 5 = A(* + i) 2 + 5B*(2* + i) + 5C(2X + i)(* + i) Let x = i {i.e., terms containing (#+i) are thus made to vanish} /. -24+12+5 = + 5 B(-i)(-i)+o Let x = {i. e., 2X+i = o} -6+6+5 = A A = 20. The numerators must be identically equal, i. e., term for term r therefore the coefficients of # 2 must be equated. Thus 24 = A + loB + loC = 20 14 + loC -I for A = 20 and B = 2 /. zoC = 30 C = -3 /. the fraction 2 . , . - . .. -- _ 5(2^+1) 5(*+i) 2 x+i 4 7* 3 Example 7. Resolve . _ ^,~^^ x , x into partial fractions. Let the fraction = . A . + B * + C (2X - 3) (x* + 5* + 9) _ A(* 2 + 5 x + 9) + (B* +C)(zx - 3) Equating the numerators 9*-i7 = A(# 2 + 5* + 9) + (B* + C)(2x - 3) VARIOUS ALGEBRAIC PROCESSES 455 Let- Then = | i. e., let 2* - 3 = o A 14 75 Equating the coefficients of x*, and as no terms on the L.H.S. contain # 2 , its coefficient = o, o = A + 2B=-H + 2B /. 2B = H 75 B = J7 75 Equating the coefficients of x on the two sides of the equation _ 7 66 75 ~~75~ 7* + 383 14 /. the fraction j- f - ; r -. ? 75 (** + 5* + 9) 75(2* - 3) Limiting Values, or Limits. Let it be required to find the value of the fraction ^T 1- when x = i. AX |~^~~ J ^>y 2 O When x =*= i, ^-r '-=- if x be replaced by i. 4# -\-x 5 o We can give no definite value at all to -; it might indicate anything, and therefore we must find some other method for dealing with cases such as this. Let us calculate the value of the fraction F when x is slightly less than i, say when x has the value -9 : Then F = -i' When x has a value nearer to i, say -95 456 MATHEMATICS FOR ENGINEERS Now let us take values of x slightly in excess of i. ~ 2-12 -i F = . ._ . - = = -2174. When x = 1-05, When x = 1-1, F = 4-41+1-05-5 2-2 2 46 4-84+I-I-5 Therefore for values of x in the neighbourhood of i the fraction has perfectly definite values, and consequently it is un- reasonable to suppose that there is no value of F for x = i. If we plot a curve, as in Fig. 256, of F against '225 x, we see from it, assuming that it is continuous (and there is nothing to negative this supposition) that the value of F when x = i is 2222. We say, then, that the limiting value of F when x approaches i is -2222, or 2X2 22 . L 4* 2 +*-5 = -2222. Fig. 256. To obtain this value without the aid of a graph we might take values of x closer and closer to i and see to what figure the value of F was tending e. g., when x = -99, F = -2232 when x = -995, F = -2227. This method, besides being somewhat laborious, is not definite enough. As an alternative method : if x does not actually equal i but differs ever so slightly from it, (x i) does not equal o, and there- fore we may divide numerator and denominator by it. (x i) 2 Thus F= (*-i)( 4 *+5) (, As x approaches more and more nearly to i, this last fraction becomes more nearly = - and in the limit when x = i, F = -. 9 9 Later on we shall see that this method of obtaining a value or limit by " approaching " it is of great utility and importance, VARIOUS ALGEBRAIC PROCESSES 457 Example 8. Corresponding values of y and x are given in the table : X 3'9 3'94 3-97 4-02 4-05 4'i y 30-42 31-04 3I-52 32-32 32-80 33-62 Required the probable value of y when x = 4. When x has values slightly under 4, those of y are increasing fairly uniformly; thus for an increase of x from 3-9 to 3-94 (i. e., -04) the increase of y is -62, or the rate of increase is , i,e., 15-5, and .04 whilst x increases a further -03 unit, y increases -48 unit, or the rate of change of y compared with x is , i. e., 16. Thtis y is increasing at a rather greater rate as the value of x increases. This is confirmed by dealing with values of x greater than 3-97 : we might tabulate the differences of x and of y thus : Change in x. Change in y. Rate of change of y. 80 3-97 to 4-02, i. e., 05 80 = 16 'O5 4-02 to 4-05, i. e., 03 48 48 _ -2- = 16 03 82 4-05 to 4-1, . e., 05 82 ^05" Therefore, as nearly as we can estimate, 54 when x has the value 4, 11 *< y has a value very slightly over '-^| of -80, 53 i. e., slightly more than 48 above its value \J JG o s & f the value of y when x = 4 is most probably 3i-52+'4 8 t. e-, 32. si This result is further illustrated by the graph (see Fig. 257). JQ : Example 9. Find the value of 2# 2 + 1 8* ^ / + 28 5-95 ] w 4 -4-O5 ^ - Fig. 257- hen x = -2, I3* 3 + ?6# a ?- 76* - t 458 MATHEMATICS FOR ENGINEERS TH * -R 2(* 2 *2 in F= +13^-38^- 80) ~(x+2)(6x* + x- 40) [(x + 2) is tried as a factor, use being made of the Remainder Theorem, to which reference is made on p. 55.] . F _ (*+7) -2+7 . = 5 6# 2 +# 40 24240 1 8 = ^ when x = 2. Io (# + a) 4 x* Example 10. Find the limiting value of - - - - when a = o. By direct substitution of o for a we again arrive at the indeter- minate form -. o Proceeding along other lines F = (x + a)* - x l _ x* + 4* 8 a + 6* 8 a 8 + 4*a 3 + a 4 - x* a a 4# 3 a + 6# 2 a 2 + 4x0? + a* ^ . . a If a is to equal o, and the value of F is then required, this value must differ extremely slightly from the value if calculated on the assumption that a is infinitely near to o but not exactly so. If a is not zero, we may divide by it then F = 4* 3 + 6x*a + 4x0* + a*. Hence, the limiting value to which F approaches as a is made nearer and nearer to zero is 4# 3 , for all the terms containing a may be made as small as we please by sufficiently decreasing a. L L(x + a) 4 x* , . ,, , , . , . 5 - ' - = 4# 3 is the abbreviation recognised for the statement : " The limiting value to which the fraction ^ a ' ~ x a approaches as x approaches more and more nearly to a, is 4**." Example n. Find the limiting value of '-'^j- when 6 = o, it being given that sin 6 = 6 - 6 - + -^1 - . ( 6 te^Z measured:! 6 120 I in radians J Adopting this expansion sing _ 6 ^ 120 2 6 ~e~ ~ 6 and L sm ^ _ T as terms containing 2 and higher powers of 0->o 6 6 must be very small compared with i. VARIOUS ALGEBRAIC PROCESSES 459 This result is of great importance : for small angles we may replace the sine of an angle by the angle itself (in radians). This rule is made use of in numerous instances. Thus when determining the period of the oscillation of a compound pendulum swinging through small arcs, an equation occurs in which sin is replaced by 6 ; the change being legitimate since 0, the angular displacement, is small. Exercises 44. On Continued Fractions, ete. 1. Find the first 4 convergents of 8-09163. By how much does the 3rd convergent differ from the true value ? 2. Find the 5 th convergent of - ^ |. 3. Convert 4--- into a continued fraction. What is the 3rd convergent ? 4. Express as a continued fraction the decimal fraction -08172. 5. Using the dividing head as in Example i, p. 449, an angle of 59i4 / 5* is required to be marked off accurately. How many turns and partial turns would be required for this ? 6. Similarly for an angle of 73 2'i9*. 7. Similarly for an angle of 5i9 / 3i'- 8. It is desired to cut a metric screw thread on a lathe on which the pitch of the leading screw is measured in inches. To do this two change wheels have to be introduced in the train of wheels to give the correct ratio. If i cm. = -3937*, find the number of teeth in each of the additional wheels, i. e., find a suitable convergent for the decimal fraction -3937. On Partial Fractions. 9. Express 3*+ 8 Q^ a sum or difference of simpler fractions. x 2 + 7* + 6 10-16. Resolve the following into partial fractions 2 -H 3* +5 12 *(* + ') 10 ' 6*+i9*+i5 ** ~ 3* - 88 " ** - 3* + 2 6x* - 9* + 3Q 14 - 22** - 179* ~ 240 l6 ' (x- 5 )(x*+2X- 8) ' 6* 3 + is* 2 - 57* - 126 4C 3* + 2 16 2 * ~ 3. 1J>> x 3 + 2x* - x - 2 * (x - 3)(* 2 + 3* + 3) On Limiting Values. 17. Find the limiting value of * 2 * + g when x = - I. L x s + 3*2 ijx + 14 *. ^ ^ ^2 + 2Ar _ 8 19. Show exactly what is meant by the statement T x* - 6ax + 5 g2 _ _ji ^^o x z + gax ioa a n 20. Determine the limiting value of the sum of the series 16, 8, 4, 2, etc. 460 MATHEMATICS FOR ENGINEERS 21. A body is moving according to the law, space = 4 x (time) 3 . By taking small intervals of time in the neighbourhood of 2 sees., and thus calculating average velocities, deduce the actual velocity at the end of 2 sees. x- x 3 22. If e? = i + x H 1 h . . . . ; find the limiting value of the fraction when x = o. 6 Z 6* 6 3 6 6 23 If cos 6 = i 1 ; and sin 6 = 6 1 1.2 T 1.2.3.4 1.2.3 I-2.3-4-5 Find i > sin 6, -L cos 6, and by combination of these results 0->o 0-> \ ^ . an . Hence show that no serious error is made when calling 9^K> 6 the taper of a cotter the angle of the cotter. 24. Find L *-*" Permutations and Combinations. Without going deeply into this branch of algebra, we can summarise the principal or most useful rules. By the permutations of a number of things is understood the different arrangements of the things taken so many at a time, regard being paid to the order in these different arrangements. By the combinations of a number of things is understood the different selections of them taken so many at a time. e.g., a firm retains 12 men for their mo tor- van service. There are 6 vans and 2 men are required for each, I to be the driver. By simply arranging the men in pairs, a number of groups or com- binations is obtained. But if the first pair might be sent to any one of the 6 vans, i. e., if regard is paid to the arrangement of the pairs, and if also either of any pair might drive, we get further arrangements. We are then dealing with permutations. To make this example a trifle clearer : let the men be repre- sented by A, B, C, D, etc. Then the different selections of the 12, taken 2 at a time, would be A and B, A and C, A and D . . ., B and C, B and D . . ., C and D . . ., and so on. But A and B might be in the ist van or in any of the others, so that a number of different arrangements of pairs amongst the vans would result. Also A might drive or B might, so that the arrangements in the vans themselves would be increased. As we might write it for one van, the different arrangements would be A (driver) and B, or B (driver) and A. To find a rule for the number of permutations of n things taken r at a time. If one operation can be performed in n ways and (when that has been performed in any one of these ways), a second operation VARIOUS ALGEBRAIC PROCESSES 461 can then be performed in p ways, the number of ways of perform- ing the two operations in conjunction will be nxp: e.g., suppose a cricket team possesses 5 bowlers; then the number of ways in which a bowler for one end can be chosen is 5. That end being settled, there are 4 ways of arranging the bowler for the other end. For each of the 5 arrangements at the one end there can be 4 at the other end, so that the total number of different arrangements will be 5x4, i. e., 20. Suppose a choice of r things is to be made out of a total of n to fill up r places. Then the 1st place can be filled in n ways. For the 2nd place (the ist being already filled) choice can only be made from (n i) things; hence the number of different ways in which the ist and 2nd can together be filled is n(n i). The ist, 2nd and 3rd together can be filled in n(n 1)( 2) ways, and so on, so that all the r places can be filled in n(n i) (n 2) ... to r factors. When there are 3 factors, the last = (n 2) = (n 3+1) When there are 4 factors, the last = (n 3) = (n 4+1) .*. When there are r factors, the last = (n r+i) .*. The number of permutations of n things taken r at a time For shortness this product is often written n r . If n places are to be filled from the n things the number of possible ways = n P n = n n = (i) (n 2) .... ( +2)(n n+i) = n(n I)(M 2) .... 2.1 i.e., is the product of all the integers to n : this is spoken of as factorial n and is written \n or n \ Thus- " factorial 4 " = [4 = 1.2.3.4 = 24. To find the number of combinations of n things taken r at a time, written B C f : Obviously n C r must be less than n P r , because groups of things may be altered amongst themselves to give different permutations. For groups of r things, the number of different arrangements in each group must be \r_ (r things taken r at a time) ; hence the number of permutations must = [r X the number of combinations or P r = = n(ni)(n2) .... (n r+i) I 2.3 .... r 462 MATHEMATICS FOR ENGINEERS If both numerator and denominator are multiplied by \n r i. e., by 1.2.3 ( ') then nr _n(n i) .... (n r+i)x(n r) .... 2.1 from which we conclude that C r = n Cn- r a result often useful. The number of permutations of n things taken n at a time when \n b of them are alike and all the rest are different = Hr \L The number of permutations of n things taken r at a time when each thing may be repeated once, twice, . . . . r times in any arrangement = n r . The total number of ways in which it is possible to make a selection by taking some or all of n things = 2 rt i. Example 12. Find the values of 8 P 2 , 9 C 3 and 15 C U . P, = 6(6 - i) = 30 9, = [3 1-2-3 c u = I5u or III 154 III HI 14 If 15.14.13.12. = J ^ = 1365. 1.2.3.4 ^ When w and r are nearly alike (as in this last case) and n Cr is required, we use the form n C r = n Cn_r ; and the arithmetical work is thus reduced. Example 13. There are six electric lamps on a tramcar direction board ; find the number of different signs that may be shown by these. If the lamps all show the same coloured light, the question resolves itself into finding the total possible arrangements of 6 lamps when any number of them are lighted. Thus if 6 lamps are on, there is only one arrangement possible. If 5 lamps are on, these can evidently be placed amongst the six places in six different ways ; or, in other words, the number of arrangements in this case is 6 C B or 'C^ ["Q. = n Cn_ f ]. If 4 lamps only are to be switched on, the possible arrangements will be *C t , i. e., 6 C, . e., 15. Similarly the numbers of arrangements for the cases of 3, 2 and i VARIOUS ALGEBRAIC PROCESSES 463 lamp on will be fl C,, 6 C, and 6 C X respectively : hence the total number of different arrangements giving the different signs will be I + 6 C S + 8 C 4 + C S + 6 C, + "Ci = 1 + 6+15 + 20+15 + 6 = (. This result could also have been obtained by making use of the rule given on p. 462 for the total number of ways in which it is possible to make a selection by taking some or all of n things. Thus total = 2 n i = 2' i = 641 = 63. If the lamps had been of different colours the number of different signs would be greatly increased, since the different sets of the above could be changed amongst themselves. Example 14. Twelve change wheels are supplied with a certain screw-cutting lathe ; find the number of different arrangements of these, 4 being taken at a time, viz. for the stud, pinion, lathe spindle, and spindle of leading screw. In this case the order .in which the wheels are placed is of conse- quence ; hence we are dealing with Permutations. As there are 12 to be taken, 4 at a time, the total number of arrange- ments = 12 P 4 = 12.11.10.9 = 11880. The Binomial Theorem. By simple multiplication it can be verified that (x+a)* = x z +2ax+ a* (x+a)* = (x+a)' = It is necessary to find a general formula for such expansions; (x+a) is a two-term or binomial expression, and the expansion of (x+a}* is performed by means of what is known as the Binomial Theorem. For simple cases, such as the above, there is no need for the theorem, but for generality it is desirable that some rule should be found. The expansion of (x+ a)' 3 could certainly be found by writing it as , * , and then performing the division, an endless series resulting, but it would be a painfully laborious process. Suppose the continued product of (x+a)(x+b)(x+c) .... to n factors is required, n being a positive integer. The ist term is obtained by taking x out of each factor, giving x. The 2nd term is obtained by taking x out of all brackets but one, and then taking one of the letters a, b, c .... out of the remaining bracket. The 2nd term thus = x n ~ l (a+b+c+d .... to n terms). The 3rd term is obtained by taking x out of all brackets but two, and combining with the products of the letters a, b, c . . . . taken two at a time. 464 MATHEMATICS FOR ENGINEERS The 3rd term thus = x*~ z (ab + ac + ad + . . . .+ be + . . . . to, say, P terms). p is then the number of combinations of n letters taken two at a time ., [2 1.2 so that the 3rd term is found. In the same way any particular term may be found. Example 15. Write down the value of the product ist term = x* (i. e., x is taken out of each bracket). 2nd term = x 3 { 2+4+6 7} = x 3 (x being taken out of all brackets but one). 3 rd term = ^(-2)x(+ 4 )+(-2)x(+6)+(-2)x(-7) = # 2 {-8 12+14+24 28-42} = 52**. 4 th term = *{(- 2 )(+ 4 )(+6)+(-2)(+6)(-7)+(+ 4 )(+6)(-7) + (-2) (+4) (-7)}- = #{-48+84168+56} = 76*. 5 th term = (-2)(+ 4 )(+6)(- 7 ) = 336. 7) = **+* 3 -5 2 * 2 -7 Now let b = c = d= . . . . = a, then (x+a)(x-\-b)(x-{-c) .... to n factors, becomes (#+)". Then ist term of the expansion = x n the 2nd term of the expansion = x*~*(a + a -f- a . . . . to n terms) = nx n ~ 1 a the 3rd term of the expansion = x n ~ z (a 2 -}-a 2 -\-a 2 ... to n C a terms) n(n i) _ = -J -- 'x n -*a? 1.2 Similarly, the 4th term of the expansion n(ni)(n2) . ^\ _ /_\ _ / v W ~ >/7 ** 1.2.3 = x . . . l.J Thus the indices of x and a together always add up to n, that of x decreasing by one each term. The numerical coefficients can be remembered in a somewhat similar fashion; the numerator VARIOUS ALGEBRAIC PROCESSES 465 having a factor introduced which is one less than the last factor in the preceding numerator, whilst the denominator has an addi- tional factor one more than the last factor in the preceding de- nominator, i. e., a kind of equality is preserved. The proof here given is of an elementary character, and only applies when n is a positive integer, but it can be proved that the theorem is true for all values of n, integral or fractional, positive or negative. To find an expression for any particular term in the expansion : The 3rd term = 2 x n ~ 2 a 2 , i.e., is distinguished by the 2's throughout, and is on that account called term (2+1) or T ( J + D The I4th term is thus written T (1J + 1) Putting the terms in this form we are enabled to write down at a glance, i. e., without full expansion, any particular term desired. e. g., the 6th term = T (6+1) = ^ x 5 a 5 . is The (r + i) th term is usually taken as the general term, and it is given by or Example 16. Find the 8th term of the expansion of (x 2y) 10 . Here n = 10 ^ x = x I in comparison with the standard form. and a = 2y ) Hence T 8 = T (7+1) = ^V- '(-*?)' -V(_ 2 j,)i [for "C 7 = "C lt _ 7 = "CJ Li = i^!*s Example 17. Expand (a 3&) 1 to 4 terms. [Whenever n is fractional or negative the expansion gives an infinite series, and therefore it is necessary to state how many terms are required.] Comparing with the standard form x = a a = (-36) -* H II 466 MATHEMATICS FOR ENGINEERS Hence the expansion 1.2.3 * - a- -fx -Jx Example 18. Expand f 3m - j to 3 terms. Here x = 3m, a = , n = 4 Hence the expression 1.2 4.3~ 8 m- 5 , 4 X 5 X - * - - 5 2 x 25 , , Sim* I2I5W 6 3645m 6 The method of setting out the work in these examples (Nos. 17 and 18) should be carefully noted ; the brackets inserted helping to avoid mistakes with signs, etc. Thus in the evaluation of n(n i) when n = 4 one is very apt to write down the result straight away as 4X 3, whereas its true value is ( 4)( 4 i), i. e., -f 20. Example 19. In the Anzani aero engine the cylinder is " offset," t. e., the cylinder axis does not pass through the axis of the crank shaft, but is " offset " by a small amount c. The length of the stroke is given by the expression V(l + r)* c* V(l r) 2 c 2 , where / = length of connecting-rod and v = length of crank. Show that stroke = zr\ i + J , 2 _ 2 [ Dealing with the expression V(l + r) z c 2 , we may rewrite it as {(/ + y) 2 c 2 }* and then expand by the binomial theorem. Thus {(I + r)* - c 2 }* = {(/ + r) 2 }* + {(l + r) z }l~ l x (-c 2 ) (/ + r) . -f terms containing as factors the fourth and higher powers of c; these terms being negligible, since c\ c 6 , etc., are very small. VARIOUS ALGEBRAIC PROCESSES 467 In like manner it can be shown that Hence stroke =(< + r) - - Comparing this result with the length of the stroke of the engine if not offset, we see that there is small gain in the length of the stroke ; the increase being the value of rc a -i- / 2 r 1 . Use of the Binomial Theorem for Approximations. Let us apply the Binomial Theorem to obtain the expansion for (!+*). Writing i in place of x, and % in place of a, in the standard form / \n i n(ni) , n(n 1)( 2) , . (i+#) n = i+nx-\ s -- '-x 2 -\ *- - '- - 'x 3 -\- . . . , 1.2 1.2.3 If x is very small compared with i, then x z , x 3 , and higher powers of x will be negligible in comparison. Hence (l+jr) B = l+flx when x is very small. Example 20. In an experiment on the flow of water through a pipe the head lost due to pipe friction was required. The true velocity was 10 f.p.s., but there was an error of -2 f.p.s. in its measurement. What was the consequent error in the calculated value of the head lost, given that loss of head oc (velocity) 2 ? Let He = calculated loss of head. H = Kv 2 = K(io + -2)* {v being the measured velocity} = K x io a (i + -02) 2 Making use of the above approximation H = iooK(i + -02 x 2) = iooK(i + -04) But true head lost = K x io 8 = looK error = iooK x -04 or 4 %. Example 21. Find the cube root of 998. 998 = 1000 2 = 1000(1 -002) /. cube root of 998 = 998* = iooo*(i -002)* = io (i J x -002) = io (i -0007) = 9-993- 468 MATHEMATICS FOR ENGINEERS Example 22. Find the value of 1005*. 1005 = 1000 (i+ -005) 1005* = iooo 4 (i+-oo5)* = iooo[i+(4X-oo5)] = IO la XI'O2. With a little practice one can mentally extract roots or find powers for cases for which these approximations apply e. g., Vcfi = 9-9 For 98 differs from 100 by 2, hence its square root differs by \ of -2, *. e,, -i from 10. Similarly, (i-O3) 3 = 1-09 very nearly. Further instances of approximation are seen in the following : (i+*)(i+y) = i+x+y+xy = i+x+y when x and y are small (i-\-x)(i-\-y)(i-\-z) = i+x+y+z when x, y and z are small Example 23. Find the value of g85 X 5 ' 8 1004 F _ 1000(1 '015) x 5(1 + 'Qi6) 1000(1 + -004) = 5(1 -015 + -016 -004) = 4-985. Example 24. If / = measured length of a base line in a survey L = correct or geodetic length, i. e., length at mean sea-level h = height above mean sea-level at which the base line is measured and r = mean radius of the earth Then ^ = -JL, / r + h und it is required to find a more convenient expression for L. T lr I L = -' Whence L - since h is very small compared with r. VARIOUS ALGEBRAIC PROCESSES 469 Exponential and Logarithmic Series. (i\"* i+- ) w/ -i) JL + m(m-i)(m-2) x 1 1.2 W a . 1.2-3 f -^ -f V ~\ m\ w/ w\ wA m) ~~ 1.2.3 = I + I+ ^^ + 1.2.3 + Suppose now that m is increased indefinitely, then - etc., nt tn become exceedingly small, and may be neglected. Hence when m is infinitely large / . i\ m . i , i (i+is) ..i + i + j5 + 5 +.... This is the case of compound interest with the interest very small but added to the principal at extremely short intervals of time. The letter e is written for this series [If it is any aid to the memory this statement may be written 1,1,1,1, -i = \o + |i + J2 + J3 +' '] (I\ mx i-\ ) would be e* if m were infinitely large. . But i\ m ' i . mx(mxi) i . mx(mxi)(mx2) - L - X X = T.+X++.-+ .... (when m is very large) 2 3 47 o MATHEMATICS FOR ENGINEERS To obtain a more general series, i.e., one for a*, where a. has any value whatever, let a = e k , so that log* a = k. Then a 1 = e** The series for e kx can be obtained from that for e* by writing kx in place of x. llv\* lbv\Z Then a* = and substituting for k its value we arrive at the important result This is known as the Exponential Series. A further series may be deduced from this, by the use of which natural logarithms can be calculated directly ; common logarithms being in turn obtained from the natural logs by multiplying by the constant -4343. For let a = I + y Then by employing the exponential series It is now required to obtain a series for log e ( i -f- y), which can be done by equating coefficients on the two sides. The left-hand side may be expanded by the Binomial Theorem, giving x(x i)(x 2)y 3 . - - -- ' Now x occurs in every term except the first, and the coefficient of x in the second term = y. v 2 The third term is %(x 2 y z xy z ) ; and the coefficient of x is 2 The fourth term is ^(^y 3 3# 2 y 3 -f 2#y 3 ) ; and the coefficient , . y 3 of x is 3 yZ y3 ,y4 Hence the coefficients oi x = y -f- -- -f and, equating the coefficients of x on the two sides 10ge(l+^)=^-^+^-^+ ...... (I) which is known as the logarithmic series. In the form shown, however, it is not convenient for purposes of calculation, because the right-hand side does not converge VARIOUS ALGEBRAIC PROCESSES 471 rapidly enough; and a huge number of terms would need to be taken to ensure accurate results. In the expansion for log,(i+y) let y be replaced by y; then log, (i-y) = - y - y ~- __ ....... ( 2 ) Subtracting the two series, . e., taking (2) from (i) log* (i+y) - log, (i-y) = but log, (i+y) - log, (i-y) = log, - hence log, 2-j = 2 (y + + ^ + ....) Now let L_ ! be denoted by , i. e., m my-= n+ny nin or y = - w+n +K i w m of"" i 1/ni n\ 3 I/in /i\ 5 then log,, = 2 { s ^ s ( ; + -( - - + . . . , n Im+n 3\fn+/i/ 5V/n+n/ which is a series well adapted for the calculation of logs. Example 25. To calculate log, 2. Let m = 2, n = i, and thus y = J then log, 2 = 2J- + (-x a) 4 (~ x ~5j + [ = -6930 (which is one wrong in the 4th decimal place; and this error would have been remedied by taking one more term of the series) . An equally convenient series would be obtained by writing , i+y . i for T -i. ' e - y = , . ., - i y' 2w+i Then Thus if = -6930 as before. and this latter form is slightly easier to remember. 472 MATHEMATICS FOR ENGINEERS To obtain log, 3 let n = 2. Then = -40546 but log* | = log, 3 log* 2 2 40546 = log, 3 6931 log, 3 = 1-0986. Again, log 4 = 2 X log, 2 and log, 5 can be obtained by using the series for log, - when n = 4, and the value of log, 4, so that a table of natural logs could be compiled : in fact, this is the way log tables are made. The corresponding common logs are found by multiplying the natural logs by -4343. Example 26. The " modified area " A, a term occurring in con- nection with the bending of curved beams, is given by for a rectangular section of breadth 6 and depth d. Show that this can be written as R _ . might be written as - -j- and is therefore of the form, Hence R in this formula is the radius of curvature of the beam, and hence if the beam is originally straight R = oo , so that ^ = o and the expres- sion for A reduces to bd, i. e., the area of the section. VARIOUS ALGEBRAIC PROCESSES 473 Exercises 45. On the Binomial Theorem, etc. 1. Write down the 5th term in the expansion of (a 6) 7 . 2. Expand (20 + $c) 11 to 4 terms. 3. Find the aoth term of the expansion of (3* y) 23 . * T? j ( m 2\ 8 , 4. Expand ( -- I to 4 terms. 5. Write down the first 5 terms of the expansion of (a 2)"*. 6. Find the 7th term of (i - j^" 7. Expand to 3 terms (2 # 2 )*. 8. Expand to 4 terms ($a + 4c)~*. 9. Write down the 3rd term of (a 26)"* 10. Expand ^/ i -^ sin 2 # to 4 terms, and hence state its / /length of connecting-rod \ approximate value when I- ,, \ is large. a \ length of crank ) On Permutations and Combinations. 11. In the Morse alphabet each of our ordinary letters is repre- sented by a character composed of dots and dashes. Show that 30 distinct characters are possible if the characters are to contain not more than 4 dots and dashes, a single dot or dash being an admissible character. 12. Find the number of ways in which a squad of 12 can be chosen from 20 men. (a) When the squad is numbered off (i. e. t each man is distinguished by his number). (6) When no regard is paid to position in the line. 13. Find the values of 15 C 1S , 12 P 4 , 5 P . On Approximations. 14. Use the method of p. 467 to obtain the value of (-996)*. 15. Evaluate ^^ * 2 'f 3 3 X '" 8 by the same method. \'997l 16. State the approximate values of (a) (ioo2) 8 ; (6) (-9935) 7 : ( c ) (i - -oo6) 5 ; (d) (10 + -17) X -995 X 4-044- /I tan 4*\ a , 17. The maximum efficiency of a screw = ( ) , where $ \l ~}~ i tin -j <p/ is the angle of friction, t. e., tan = p. Show that this may be written in the form T ~ ** if u is small. i +/* On Series. 18. Find series for the expression cosh x, i.e., (^ ^ J and for sinh x, i.e., (- - - j 474 MATHEMATICS FOR ENGINEERS 19. Find, by means of a series, the value of log e 4 correct to 3 places of decimals. T> 20. Express ^ - as a series. What is the approximate value of K. + y this fraction when y is small compared with R ? 21. A cable hanging freely under its own weight takes the form % of a catenary, the equation of which curve is y = c cosh -, c being the . horizontal tension value of the ratio weight per foot run Express y as a series, and thence show that if the curve is flat it TT iv2 may be considered as a parabola, having the equation y = -\ jj- 22. By substituting -5 for x in Newton's series calculate the value of re correct to 3 places of decimals. Determinants. When a long mathematical argument is being developed, as occurs for example when certain aspects of the stability of an aeroplane are being considered, it frequently happens that the coefficients of the variable quantities become very involved ; and in such cases it is often convenient to express the coefficients in " determinant " form. This mode of expression is also utilised for the statement of some types of equations, for by its use the form of equation and its solution are suggested concisely and the attention is not distracted from the main theme of the working. Thus when dealing with the lateral stability of an aeroplane in horizontal flight the equation occurred AX 4 -f BX 3 + CX 2 -f DA + E = o where A, B, C, etc., were all solutions of other equations and hi some cases rather long expressions. For example, A had the form a z b z c 4 and E was equal tog' sin Q(l i n 1 / 2 w 2 ) S cos ^i n 3~ n z^)- To avoid writing these expressions in their expanded form, they were expressed thus A = -c- -c 2 b* and E = g sin 6 g cos 6 and it will be shown that from these " determinant " forms the expansions may easily be obtained. Before proceeding to illustrate the use of determinants it is necessary to define them and to show how they may be evaluated. VARIOUS ALGEBRAIC PROCESSES 475 Let D = a b c d f g h k I then D is called the determinant of the quantities a b c . . . I, and a determinant of the third order since there are three columns and three rows; its value being found according to the following plan : The letter a occurs both in the first row and in the first column : take this letter and associate it with the remaining columns and rows, thus [It will be observed that / g k I is a determinant of the second order and it is termed the minor determinant of a.] Then the value of the minor of a is found by multiplying f by I and subtracting from it the product k by g. a Thus = l k and = a(fl-sk} =A. In like manner the minor containing the products of b is b and for c g h I d f h k = b(dl - gh) = B = c(dk -fh) = C. Then the value of the full determinant = D = A - B + C = a(fl - gk) - b(dl - gh) + c(dk -fh). To avoid the minus sign before the second term the letters might be written out as follows a b c f h k I h k and the one sequence could be maintained, thus D = a(fl - gk) + b(gh - dl) + c(dk - hf) Similarly, for a determinant of the fourth order D = m r 476 MATHEMATICS FOR ENGINEERS g h k m n p r s t - b f h k I n p q s t f g k I m p q r t -d f g h I m n q r s each of these determinants of the third order being evaluated in the manner previously explained. Example 27. Evaluate the determinant 235 6 4 2 319 D = 2[ 3 6 _ ( _ 2)] _ 3[ _ 54 _ ( _ 6)] + 5 [_6-l2] = 76 4 144 90=130. Example 28. Evaluate the determinant D = 2 41 2 3653 -122 3 4 82 4 653 -4 353 + I 3 63 42 3 65 -223 -123 -I -2 3 I 2 2 824 424 4 84 4 82 -6)-5(-8-2 4 ) + 3(-4-i6)}- 4 {3(8-6)-5(- 4 -i2) + 3(~2-8)} + i{ 3 (-8-2 4 )-6(- 4 -i2) + 3 (-8 4 8)} 4 2{ 3 (- 4 -i6)-6(-2-8) + 5 (-8 + 8)} = 224224 4 4 = - It will be observed that all the numbers in the second column are the same multiple of the corresponding numbers in the first column; and it can be proved that when this is the case the determinant is equal to zero. Example 29. A number of equations in a long investigation reduced to the determinant form * + -15 - '3 -30 6 #45 100* o -i x* 4 = o Express this in the form necessary for the solution of the equation. The determinant = (* + -I5){(# + 5)(* a + 9*) + iox} + -3{-6* 2 -|- 5-4*} 3O{ -06} = ** 4- I4'i5* 3 + 57-28*2 4. 9.87* + 1-8 and thus the equation is ** 4 I4-I5* 3 4- 57-28*2 4 9-87* 4 1-8 = o. VARIOUS ALGEBRAIC PROCESSES 477 Solution of Simultaneous Equations of the first degree by the determinant method. Equations containing two or more unknowns may be readily solved by setting them in a determinant form and proceeding according to the following scheme : To solve the equations 5* 4y = 23 Write the equations as $x 4y 23 = o and set out in the determinant form x y i 5 -4 -23 375 the last column containing the constants. x v i Then = -y = x minor " y minor i minor '' 20 + 161 whence and 35 + 12 * "25 + 69 ' 35 + 12 x = 3 and y = 2. Example 30. Solve the equations ^ax cy = 6 a $bx + 2ay = a* x and y being the unknown quantities. Set out thus y I c b 2 20, a 2 x 4 a Then ca whence and 8a 2 _ a z c + 2ab 2 * ~ 8a 2 + tfc y= 8a 2 + 3&c. Example 31. Solve the equations 2a 56 + 4^ = 28 o + lib $c = 41 3a 2& c = 3 473 MATHEMATICS FOR ENGINEERS Set out thus Then Thus a b c i 2-5 428 i ii 5 41 3 - 2 - i - 3 -b c I a minor b minor c minor i mnor - 5(15 + 41) - 4( - 33 + 82) - 28( - ii 10) i 15) 2-33) -6 and 2(15 + 41) - 4 ( - 3 - 123) - 28( - i + 15) c 2( - 33 + 82) + 5( - 3 - 123) - 2 8( - 2 - 33) 112 I 112 whence Exercises 46 On Determinants. Evaluate the determinants in Nos. 14. 1. 3. 5'4 - 6 -3 - 5 354 1-5 2-5 2 -3 75 2. R 2 when R = 3-6 R! = 7-2 R 2 = 710 R 3 = 220 4. 2351 3246 8-4 3-5 2 I 6 2 5. A coefficient C in an equation was expressed as C = U0 -}- Zn M u Evaluate this when Z w = 3, M w = 2-5, Mg = 200, Z 3 = 9, U = I, X u = !, Xg = -5, M M = O, KJB = 20, XK, = -2, Z tt = I. 6. Solve the equation a 53=0 2 a + 2 i 3 1-6 -4 Using the determinant method solve the equations in Nos. 7-10. 7. n# 4y = 31 8. 8a b = 20 2X + 37 =28 ioa + 76 = 71 9. 4* 5y + 7* = 14 10. za + 36 + 5c = 4-5 # 7 5.2=11 96 ioa = 39-3 ANSWERS TO EXERCISES Exercises 1 1. 1-2 2. -0009 3. 150 4. 60 5. 10-8 6. 1-2 7. -000225 8. 28-5 9. -009 10. -052 11. 93 12. -161 13. 2-7 14. -9 15. 22 16. -56 17. -031 18. 7-4 19. 2-5 20. 6-3 Exercises 2 *&' v> -^ 11 I 2 *7 ', 9V; ~t 2. 2; -i y 32' 8 ' 384; 958* 3- "8 7x /^ ' 4 7 5ii f%^T 8ix*y 2 * . i9& ! 6*343 9 7. 1-41^ 8. a a~a~ 9. 1-6 10 jj Pt V 2 Pl v l /Cf, 1 -" might be written \ 1 #if i n x fj 1 -", etc. J ~ i-n Exercises 3 1. 589-5 2. 246-5 3. -02138 4. 57-03 5. -0005423 6. 116700 7. 12-34 8. 19-63 9. -06664 10. 1-924 11. 244-4 12. 29-14 13. 1618 14. -00009506 15. 49-64 16. 3-114 17. -0001382 18. -6874 19. -02231 20. -2777 21. 3642 22. 1-669 23. -00001509 24. -3352 25. -001155 26. 3-841 x io 6 27. 20-17 28. -2421 29. 4-814 30. 7-21 ix io- 14 31. -07041 32. 971-8 33. io 34. 85-8 35. 5-418 36. 32-75 37. 220-4; 1369 38. 3-29 39. 1-98 40. 5400 41. 4-6 42. 500-77 43. 1-315 ; -591 44. 356 45. 36; 39-2 46. -0935 47. 22-21 48. 400 49. 80-072 50. 1-434 51. -506 52. -479 53. 5130 54. 15-72 55. 46,500 56. 2 -8 1 57. 3-5 x io 5 58. 1-392 59. 1-016 x io 6 60. -284 61. -128 62. t = &, * - ft, B - x&, T - ft, - 7 Exercises 4 1. Ibs. per sq. in. 2. 1-68 3. 2-79 4. 5-44 f 1 I2ttt^ 5. 2-785 6. Ibs. 7. Stress = - 75 g 10. Incorrect as wri-H-on T-T P = ' Sir 1 cu. ft. per sec. ' 396000 12. 746 48o MATHEMATICS FOR ENGINEERS Exercises 5 1. i 7 a 2. - 2-285 3. 3-62 4- 7'54 5. 6 93 6. 75 7. 7 d ioa 2 b 8 -+ t 9. 30-1 x io 10. () L = (T- 9+T- -*li (b) 933 11. 1205 12. i A Ti 1 13. 5-01 14. E = 2- 5C 15. I 97 16. R AJU 126-5 17. 1-67 18. q = l{ ^(A,-^) + A .-A} 19. n cells. 20. 22-85 21. i5 43 22. d= 3/I-274P/ V f 23. 6 '9 24. foj 4915 \ 25. () 2Wh -\ Z\ TT 5j o XV t 467 J (&) Xl/C 294 I ' : VV 1 26. 3 27. 2-4 28. 80 29. 1493 30. 3 3 31. Ibs. ft. must be first brought to Ibs. ins . ; 6-46 ins 32. 200,000 33. 2-57 34. 27-3 ft. J ~\ k 2 1 , \ ^ (*)|= 26.6 7 1 W'-^j^l 3? ^ C = 85> ,M ^_ 36 ' W -00675 f 37 ' (W E _ 9CK (6) ^- - 2-4 J - 3K + 38. r + ^ 39. 7oolbs./ a * 40. 1-37 2 It> 41. A's speed =10 m.p.h. ; B's speed = 15 m.p.h. 42. 6-31 43. M = 221-4; H = -17 M {Multiply equations together ; thus, = x MH = M 2 } 5 - 57 ' 7 " D = lx) c(-2.h 47. y-- ^ Exercises 6 2. a=7 3. w=i-8 4. * = &= 2 n = 2-6 y = 5. 9. * = 1-24 6. # = 12 y = 3-59 y = 20 p = 9 10. E = 4- 5 = 7 7. * = 5 y = 3 8. a = 1-2 6 = 5.7 c= 4-8 11. a = 1160, b = 69000 ~ 4 12. E = -I2W+4-6 13. V = 42+^ 14. 1 = -953B+ 1-284: 4-81 15. a = -916 16. R = 4-714 6 = -191 a = -00707 17 T. = TTTC '?/ ' nfi'7 1 R * 7 ' \/area ANSWERS TO EXERCISES 481 ._ ioi-6\/area . length + 97 20. E = -i6 4 -i + 7- 3 o 9 T+-ooo326T 21. /= i4-8--ooooi38(* - 6o) a 22. /= i6-i--oooo26(/ - 6o) a 23. W = 8-28 + n* . , i7oo/ total steam per hour \ 24. w = 10-3 + -y- ( w = - - ^ - and must be first \ 1 V ' calculated / 25. 8610 of iron; 7000 of copper 26. = 99-8; K=i 27. A = 120, B = 140, C = 160 28. m = 44-71, m t = 31-54 29. 311-1 tons of saltpetre ; 388-9 tons of ginger. Exercises 7 1. (X+22)(X- 4) 2. (*-!!)(* -8) 3. (#-5)(* 21) 4. (2a-5& 2 )( 4 a a + 25&* + ioa& 2 ) 5. (8* - 1 1) (3* + 4) 6. ( 5 a - 36) (56 - a) 7. (a + 96) (a - 56) 8. (3* - 7 y)(^x - i 5 y) 9. (8 + *)(n 3*) 10. 2(5W an) (2m - 5**) 11. - g - (9&r/- i6/ 3 + 5/Ar a -24^r 3 ) 12. ^ 13. (94* + 32 1) (x - 3) 14. 15. 6a z b(a + 2c)(ga - 25c) 16. (20 - 36 + 4c)(2a 36 17. 8(2C 2 + Ja 3 6) ( 4 c + ia 6 6 2 - a 3 6c 2 ) 18. {R 2 + Rr + r 2 } 19. (* + 7 )(* - i) (2* - 5) 20. (^ - i)( 3 / + 7 )(2p + 5) 21. 199 x 23(2 + 6 4) = 18308 22. 14,130 23. 12 (* - 3) (* - 4) (* + 2) 24. - a ."^ 66 J cTT 9 r 4(^+ 2) o R 3(3* a ~ 4*- 6) ^ 5 - 5 1^^) ^' 2(^-2^-8) 21 (a- 6) 2g 560 - 327* + 99* 2 - I20* 3 ' 4(9 - I4&) ' 20(3* - 5) (3*+ io)(2* - 7) ort ii __ 29. -or -407 30. 37'8 31. 32. 3*(* + 9)(* - 7) ; (8 - 9*) (3 + 8x) ; (5* 33. (x + 8)(x - i}(x 2 + 7X + 26). {Hint : Let X = x* + jx + 6.} Pas(i8s+2 5 ) n _ _ 34 ' 35 * U t; 3(6 + ) Exercises 8 1. 4 or i 2. 2-5 or 3 3. 4-13 or 1-13 4. 4 or -i 5. 2-83 or --83 6. --278 -381; 7. 3 8. 4-23 or -2-43 9. --125 1-219; 10. 2-421 x io 5 or 2-379 x io 5 11. 2-75 or 457 12. 28-98 or 1-03 13. 3-89 x io 4 or 2-97 x io* 14. 57-5 or 56-5 15. 23 (18 has no meaning here) II 482 MATHEMATICS FOR ENGINEERS nr 16. /, = - _ ab VaW - 24t 2 v 2 + Satgv ** u ~ 6t 19. 120 or 13-3. (Divide all through by 75 x io 6 first) 20. 155 or 32 21. -845 (2-845 h as no meaning here) 22. v g$Vmi 23. 5 or 7-5 24. 80 or 90 25. -2ii and -789 of span from one end 26. 13 or 10-42 27. 1-475 28. 6-55 or 3-05 29. 100 ft. Exercises 9 1. *= 7 or-3 1 2. o = ^or 3 I 3. p = T 3\ y = | or 2 6 9 J y = *- or V q = 2) 4. x = 2\ 5. a = -6 or i \ 6. 3-63 or 2-3 y = 5/ b = 2-8 or 2) 7. -68* 8. 5 9. 765 (the work is shortened by dividing by 25-6 straight away) 10. 197 11. 27-4 12. I 13. ^ 14. 9-22 15 7-37 16. m = -01277; n = -0026 17. 3-9 or 15-8 18. 6-763 Exercises 10 1. 91-1 nautical miles 2. 22-63* 3. 70-7 5. 3 1 '-9* 6. (a) i in 12-5; (b) i in 12-46 7. 15 ft. 8. AB= 12-37'; BC= 12-15"; AC = 21-63" 9. 72-4 10. 26 11. 405 grey; 340 red 12. 480 13. 51-2 14. 132 15. 2-36" 16. 6110 Ibs. 17. 8-66"; -307 18. 5-11 tons 19. 65-80* 20. ^3 175. od. 21. 7-24 sq. ins. 22. (b) 12-25; ( c ) 4'> (<0 5"86 Exercises 11 1. 3-44 sq. ins. 2. 10-2*; 52-1 sq. ins. 3. 6 sq. ins.; 2-4 ins. 4. -2374 5. 12-82 sq. ins. 6. -536 7. 58-5 sq.ft.; 18-07 ft. 8. 137-5 sq. ft. 9 - 3'6i ft. 10. 301-5 sq. ft. 11. 13-25 sq. ins. 12. 47-7 sq. ins. 13. 69-5 14. -52 amp. 15. 6-42 ins. 16. 9-13 sq. ins. Exercises 12 1. 22-4" 2. 29-1 ft. 3. 161 sq. ft 4. 1-46; 2-14; 3-33; -35; -06 5. 2-51* 6. 1-571" 7. 2-9 ft. 8. 18" 9. 1380 ft. per sec. 10. 31-23* 11. 8-92* 12. 66 13. 970 14. 5935 15. 27-13 sq. ins. 16. 3'-2 J* 17. -196 sq. in. 18. 12080 Ibs. 19. 4816 Ibs. 20. 3-76 miles 21. 12-8* 22. 16-65 sq. ins. 23. 1010 sq. ft. 24. -0294 ohm 25. 2-68 26. 24-378*; 66 and 36 27. 214 sq. ins. 28. 60 ohms ANSWERS TO EXERCISES 483 Exercises 13 1- c i = 475*; h = -36* 2. c, = 44-72*, h = 20* 3. 7 = 62-8"; A = 4'97* 4. 6-12*; 2-07 sq. ins. 5. 9-06 sq. ins.; 60 6. i-ii'; 3-67' 7. 2-n*; 3-33*; 1-29 sq. ins.; -56* 8. 13-1 9. 1686 10. 29-8 sq. ins. 11. 6076 12. 5-8* 13. B = A/fRT 14. 141 15. i '-26' 16. -375* Exercises 14 1. 4-5"; 36^"; 8-48" 2. 1-8"; i"; 7-2:3"; -55" from base 3. 2-48"; 12" 4. 473"; 8"; 1-8"; 25-14"; 25-91*; 25-57' 5. 66-30*; 3'88*; 29-6"; 29-75*; 29-3* 6. 3340* 7. 30-4 ft. tons; 3-38 ft. tons 8. 623 yds. 9. 3*; ii* 10. 7000 Exercises 15 1. 689 cu. ft. 2. -28; 7-75 3. -3125* 4. 144 5. -833 6. 5-13 cu. ft. 7. 6-91 ; 25900 8. 41-1 9. 52600 10. 47-94 cu. ft.; 6715153. 11. 23-85 sq.ft.; 40-17 sq.ft.; 19-3 cu. ft. 12. 70 13. -53 (watt = volts x amps) 14. 31-85 Ibs. 15. 245 Ibs. 16. 9 17. 852 sq. ft. 18. 14* 19. i 5 '-6" 20. 508000 21. -0006 22. 1-74* 23. 1-83 x io- 10 ohms 24. 137 25. 12-55* Exercises 16 1. 593 cu. ins.; 321 sq. ins.; 13-55 ins. 2. 20-4* 3. 200 sq. ft. 4. 13-4 cu. ins. 5. 1592 Ibs. 6. 26-1 ft.; 581 sq. ft. 7. 4'O3"; 10-69" 8. 367 cu. ins. 9. 14520 sq. ft. ; 70420 cu. ft. 10. 773'8; 967-4 Ibs. 11. 173-8 sq. ins. ; 234-3 12. 213-5 cu. ft. 29,890 Ibs. 13. 241 tons 14. -389" 15. 155 cu. ins. ; 40-2 Ibs. 16. 559 cu. ins. ; 243 sq. ins. 17. 4-63* 18. 2*; 5*; 6-12* 19. 105 sq. ins.; 138 sq. ins. 20. U59sq. ins.; 2530 cu. ins. Exercises 17 1. 160 sq. ins.; 191 cu. ins. 2. 8-3; 518 Ibs. 3. 8-8 cu. ins. 4. 8610 Ibs. 5. -1033 6. 5-44 cms. 7. 100-4 sq. ins. ; 151 cu. ins. 8. 4-2* 9. 636 10. 559 sq. ft. 11. -0941* 12. 1440 sq. yds. 13. 1-082* 14. 15,520 sq. ft. 15. 7-59* from vertex 16. 104-6 sq. ins. 17. 14-7; 2-45 18. 53-51 acres 19. 77* 20. 16-1, 47-3, 27-6, 27-6 sq. ins. 21. 406 Ibs. 22. 72 ft. 23. 1-3* Exercises 18 1. 175 sq.ft.; 100 cu. ft. 2. 326 sq. ins.; 244 cu. ins. 3. 847 sq. ins.; 1057 cu. ins. 4. 136 sq. ins.; 98-2 cu. ins. 5. Paraboloid = J cylinder 6. 90-2 cu. ins. 7. 2-02 Ibs. 484 MATHEMATICS FOR ENGINEERS Exercises 19 1. 6-44 Ibs. 2. 2630 Ibs. 3. 1278 Ibs. 4. 960 Ibs. 5. 272 Ibs. 6. 372 7. 1-16 tons 8. 171 Ibs. 9. i -84 Ibs. 10. 761 Ibs. 11. 45-5 Ibs. 12. 10-25 tons 13. 5-08 Ibs. 14. 19-55 lbs - 15. 28-2 Ibs. 16. 93*5 Ibs. 17. 3-59 Ibs. 18. 258 Ibs. 19. 6-47 Ibs. Exercises 20 1. -7'; 74-5 tons 2. 30800; 650; 25000 3. 420 Ibs. per min. 4. 3400; if 5. 1440 8. 339; 55 !! 11850 tons 12. -317 13. ssmins.; 45 14. 63mins.; 42 22. 54000 lbs./D* 24. -8% low 28. 4-10 o'c. Exercises 21 2. 250 4. Slope = -375; intercept = 2-375 8. 5-78" 11. Slope = 2-5 if V is plotted along horizontal 12. m= 1-8, n= 2-6 13. x = i, y o 14. x = 1-24, y 3-59 15. x = -43, y = 2-33 16. x = 3-18, y = - 4-75 17-55 Exercises 22 1. -392 2. -31 3. 30-2 x io 6 4. 17-9 x io 6 5. I = -8576 + 4-71 6. d l = -84^ -03 7. di = -95^ -07 8. T== 51-70 +7 9. T = 35300 10. R = 78V + 86 11. R = -772V+64-5 12. R = 2-5V + 75 13. R = i, a = -004 14. R = 1-125, a = -00452 15. I = -00232* 96 16. 32 17. 29-25 x io 6 Exercises 23 1. Vertex downward 2. Vertex upward 7. Total weight = 50^ + 5/ 2 14. 6 or i 15. 2-67 or 3-5 16. 1-44 or 7-65 17. Divide throughout by io 4 : 9-22 or -12 18. 17-1 19. (a) 4-9"; (6) 5-5" 20. # = 3-64'; h =1-83' 21. 7-7 air to i of gas 22. 5-5 23. e 55, effy. = -5 24. 15-23 knots; 948 25. 40; -69 26. 2-1 27. Assume some value for v : u = - ;i 28. 8-33; - 29. 2" 30. 2 rows of 8 31. 6 34. i, 2 or 1-5 35. -2, 2-25 or 3 36. 1-2, 4-6 or 1-6 37. -2, -5 or -8 38. max- at x = 3, min"- at x = + 2 39. x=-ziil 40. 1-475 41- 5'6 Exercises 24 1. 88-1 Ibs. 4. 10-89 tons 8. 2-3 pence 12. 22 knots 16. -01088' 2. 2-37 3. v = 66-3 Vr; 1195 5. 10970 Ibs. 6. 5" 7. -028 cm 9. 13-75 ohms 10. 246 11. 80 13. 841 14. 4-27 15. 533 M 17. Cost oc ~ ANSWERS TO EXERCISES 485 Exercises 25 1. 28; 72 2. 200> 3. 105 4. 3'7. 4'6 ... i 5. 12, 15, 18 or 9, 16-5, 24 6. 24 7. & [ 155. 6d. ; 377 los. od. 8. 10; 8160 9. 15-5 ^ 10. 592 ft. ; 1 6 sees. 11. 3-15 p.m. 12. 2-074 13. 53-33 14. 2, 3. 44 15. 835-2 . 16. 5-5 17. 10-081 ; -821 18. 20 Ibs. 19. 7-52, 1 8, etc. 20. a = 2 b = o, c = = i; 8040 21. 25 days 22. 983 in. 23. 4-284; 6-116; 8- 734:12-48 17-8; 25H3 36-31 ', 5*' 84; 74-03; 105-5 Exercises 26 1. 3-06; 1-1569; 2-192 2. 5-071; 4-28; 1-4 3. o; -081 ; -194; 285; -55 775 4. 1-301 5. 923 6. 09877 7. -00005445 8. 4-612 9. 264 10. i -086 11. 1546 12. 11-03 13. -07784 14. 3-3 x io- 26 15. 26, 560 16. 47-2 17. 75-4 18. 370 19. 38-2 20. 123; -109 21. -401 22. 1518 23. -0336 24. 475 25. -0391 26. 528 27. <t> w = = '39* $ = = I-796 28. 4000 29. -638 30. 24-3 % 31. -296 32. 325 33. -103 34. 357 35. 4-48 36. 000334 37. 38-4 38. 8-51 39. 4-44 40. 71-5 41. -65 42. j= 1-875; T = = 445, * = 237 43. Pv 1 ' 06 = 392 44. 1-47 45. o or 7-28 46. 2-16 47. o or 1-368 48. -0955 49. 481-5 50. -033 51. V = 222VH 52. -01895 53. 4-6 54. 5380 55. i-48xio 8 56. -605 57. 1-44 58. 7965 59. 61-6 60. -2 61. n 1-405; C = 502 62. 81300 63. 47610 64. 1-115 65. 66. Thl. Disch. Cd Thl. Disch. Cd 53 66 120-5 .731 118 672 154-7 708 183 727 171 658 22O-4 711 67. r = -00356^- 5 68. h = -I538*; 1 - 8 f Take logs of both equations and solve as a pair of simul-^ by. 3 -07 1 taneous equations. 70. 8-41 71. F = -00277V 1 ' 9 1. 2. 5. 8746; 9756; 17*13' 7. 3342' 12. 52-8 00305 455 17. 21. Exercises 27 "3443 > '5 2 3 I '2309 9641; 1-6139; -2826 3. 3-82 4. -397 6. 252 ; o (the case of wattless current) 8. 143-3 9. 20 10. 28 6' 11. 13. 1340 14. 6750 15. 3150 16. 18. 61,200 ft. 19. 23-4 20. 268' 22. -8923 23. 16850 24. 25!' 181 825 486 MATHEMATICS FOR ENGINEERS Exercises 28 1. a = 11-65"; 6 = 43-47" 2.6 = 8-72"; c = 14-83* 3. a = 48-3"; 6 = 43-5* 4. a = 66-73"; c = 74-8; 5. a = 22-14"; b = 16-08* 6. a = 57-66"; c = 92-63" 7. a = 20-8"; 6=10-72" 8. 6037 ft. ; 2927 ft. 9. 8 8' 10. 30-6 ft. 11. 78 12. 14 4i' 13. 2" 14. (a) 14 16'; (6) 20 53' 15. 2-38" 16. 2856' 17. AB = 5 oxAD 18. A = o, 50 R.B. of BC = 35-5 S.E. B = 33-9, 77-8 R.B. of CD = 82-5 S.W. C = 74-6, 20-5 R.B. of DA = 23 N.W. D = 15-2, 13 Area = 2700 n' 19. A = 10, 20 R.B. of BC = 67 S.W. B = 19-05, 14-8 R.B. of CA = 18 N.W. C = 12-58, 12-06 Area = 29 20. 3 chns. 49 links 21. 235 i' 22. 73-6 ft. 23. -2901" 24. -121* Exercises 29 1. -8988 -.6157 -6157 -4384 ; +.7880 ; 7880 -2-0503 -.7813 + 7813 2. --9903 -6157 + 8480 +1392 ; + 7880 ; -5299 7813 i -6003 3. --3289 -3242 -9953 -9444 ; + 9460 ; -0979 + 3482 -3427 + 10-17 4. --7570 -8m 7513 6534 ; +5850 ; 6600 + 1-1585 -1-3865 1-1383 5. --9I35 -6374 9218 4067 ; + 7705 ; 3877 +2-2460 + 8273 2-3772 7. 124 36' 8. 120 55' 10. i4946' or 2 >ioi4' 9171 3987 2-2998 6. 7265; oo ; -1625 9. 1 1 9 30' Exercises 30 1. c = 4-89", A = 34 25 / , C = 67 5 ' 2. A = 8o52', 6 = 59-46", c = 63-04" 3. B = 4446' or I35i4', A = io824' or i756', a = 11-93* or 3-87' 4. B = 4042', a = 8-84", c = 8-25" 5. A = 5343' or I26i7', B = 8oi7' or 743', 6 = 12-61 or 1-72 6. a = 9-54 ft., B = 37 4 7', C = 685 7 ' 7. A = 8o6', B = 48i8', C = 5i36' 8. c = 21-97", B = 2i29', A = 283i' 9. B = 4 i 4 2'or I38i8', C=io9i8'or I2 4 2 / , c = 8-31" or 1-94* 10. C = 42 or 138, A = 108 or 12, a = 9779 or 2138 11. C = 6940', B = 59 3o', a = 830 12. A = I0333' or 527', C = 4 o57' or i393', a = 64-62 or 6-311 ANSWERS TO EXERCISES 487 13. A = 863i', B = 4557', c = 16-11 14. C = 43V, A = 8i2i', a = 47-28; area = 637 15. 19-46 ft. 16. 8 30' 17. 55. 87, 38 18. 18-75 Ibs.; 58 19. OB = 1-224 chns., OC = -3236 chn., BE = -7673 chn., CF = i 667 chns. 20. 5636' 21. Jib = 26-2 ft.; tie = 17-4 ft. 22. 1191 ft. 23. 647^.; 374ft. 24. 53-2 ft. 25. 2083 ft. 26. AB = 2983 links ; 767 links 27. BG = 74-96 chns.; CH = 74-14 28. AB = 527-4"} 29. AB (horiz.) = 607-5 yds. \ DC = 475-3 I Hnks Diff. of level = 129-3 yds./ BC = 774-6 J 30. r = 473-3' 1 31. AP = 983' (AC = nSo'U BE = BF = 126-7' h BP = 9 6 7' CF = CG = 473-3' J CP = 919' J 32. I233f.p.s.; 29 36' 33. diam. = -506 34. 106-4; 93' 35. 9-06 to i 36. 60-3 sq.ft.; 422 Ibs. 37. 7-46"; 10-65" 38. 10-3"; 14-5" 39. 244 sq.ft. 40. 2286; 2912 41. 17-25 sq. ins. Exercises 31 1. cos A = -893, tan A = -504 3. cos (A+B) = -442, sin (A-B) = -298 4. tan (A+B) = 36-7, tan (A-B) = -536 ^ - tan a . 6 p = ' i + n tan a ' 3 2nrpfji 7. P = W (sin a + /i cos a) 8. 1-162 9. 4-99 sin (5* + i) 10. 238-5 sin (50* -576) " R = -os'' 12 ' ' I89: I0 42 ' 13. E = I2i-6sin (i2on-f + 1-022) Exercises 32 1. cos 2 A = -566; tan 2 A = 1-455 2. sin 2A = -7962; cos 2A = -605 3. |(i + cos 28) 4. sin 2B = 2 cos B\/i-cos a B; -731 A A 5. sin - = -161 ; cos = -987; sin 3A = -8236 2 2 A 6. cos 4A = -616; tan = -114 7. 2-5(1 - cos 4*) 8. 7-85 (sin 189 - sin 131) = 7-85 (- sin 9 - sin 49) 9. 2 sin gt {cos 6t + sin it} 10. sin A = -261 ; tan A = -270 11. 505 x sin 2a; 53 12. (a) 2cos323o' sin I53o'; (6) - 2 cos 423o' cos 383o'; (c) 24 sin 50 sin 45 13. 25-91; 63-73 14. -333 or- 1-25 15. 9'4 {'995 - cos U"/' A cos - = -991 -2701 J MATHEMATICS FOR ENGINEERS Exercises 33 1. 30 or 150 2. 45. ^S ' 225 or 315 3. 120 or 240 4. 88' or 188 8', I5326' or 33326' 5. 120 or 240 6. 538', 191 32', I26 5 2' or 34828' 7. 30 or 150, 210 or 330 8. 45, 7i34'. 225 or 2 5 i 3 4 ' 9. 30 or 150 10. 3545' or I44i5' 11. 45, 225, i6i34' or 34i34' * 2 - 45 or 225 13. 45, 225, i826' or I9826' 14. 30, 150, 210 or 330 15. o or 45 16. o or 120 17. 27^' or 2433o' 18. 4856' or I5639' 19. I46 ig or 326I9' 20. o; 180; 2o56' or 339 V 21. 57' Exercises 34 2. 5-0018 3. 151 ft. 5. E cosh x *J - + ^rV sinh x lj ^ 8. -93 9. i852' 11. -86 12. 19-4 15. 1-928 + 2-298; Exercises 35 1. 318 2. 51-7 Ibs. per sq. in. 3. 38-35 sq.ft.; 575 cu. ft. 4. 765 sq. ft. 5. 7231 sq. ft. 6. 269 ft. 7. 430 sq. ft.; 2190 cu. ft. per sec. 8. 6850 sq. ft. 9. 730 10. 8-72 sq. ins. 11. 60-5 Ibs. per sq. in. Exercises 36 1. 168750 cu. yds. 2. 8350 tons 3. 244000 tons 4. 44920 sq. yds. 5. 5-21 x io 6 galls. 6. 40 ft.; 51-43 ft. 7. 12020 cu. yds. 8. 96-6 ft.; 61-9 ft.; 11600 cu. yds. 9. 26-5 ft.; 17-66 ft.; 33-5 ft.; 22-35 ft -.' 27-5 ft.; 18-33 ft.; 184, 325, 202 sq. ft.; 1375 cu. yds. 10. 28-25 anc * 43'3 ft- from the centre line Exercises 37 14. The table of values would be arranged thus : 1. 1-2552; 2-1293 4. 40-54 ft. ; 156-6 ft. 6. io45' 7. -00383 14. 44-09 6 log sin 6 cos 6 i -84 cos 6 I = A A log sin 6 + log P log/> P 15. Treat ( i z ) as a constant multiplier iioo\ &*) 16. Values of R and V are as follows : V 10 20 30 40 55 R 2-5 3-21 4-74 6-9 9-61 14-6 17. 2-9 18. Values of v and 77 are as follows : Y 2 3 5 7 10 12 1 962 968 961 947 934 932 ANSWERS TO EXERCISES 489 19. latus rectum = 2-5; vertex is at (2-75, 8-42) 20. 4-27 tons per sq. in. ; 23 Exercises 38 1. -404 6. 62-2 Ibs. 8. Plot E = cosh#(# ranging from o to 500 Vgr) and then alter both scales Exercises 39 1. Amplitudes: 8; -2; 51-8; -116; -91 T> J If 27r /- Periods: ; ; -02; -0102; 36-9 23 2. Amplitude = -4 .ic -r. j .r 2jr o Period = if Period as for cosine curve : - or 120 Exercises 40 1. -{Assume some convenient value for /}. x = -403^ 2. * = 5*3 3. i -221 4. 4-58 5. -36 or 2-17 6. 1-9 or -2-45 7. 2-79 8. -143 or -333 9. 2-66 10. 5-37 (308) 11. / = -3 5 L 12. 7-9 13. 4-49 rad. (257) 14. 10-42, 13 15. 5-523' 16. 1-484' 17. 2-9 18. 6-005* 19. 796'34* 20. 1-87. (Plot the curves y l = cosh x and y a = = sec x and note the point of intersection.) 21. 6-34 ft. Exercises 41 2. -454 3. -3341 560 7. 1-043; i-o?? 9. -22. [Hint. Let <*> = a + br ; also <t> = log -J- + g ^ 437 ~ T ^ ; and 4OX T solve for q.] Exercises 42 1. W = 47 + 6o-5A 2. /i= -2 + -oo4\/t/" 3. W = 3-28^ 4. m = - 4 o 5 +^H 5. /=-oi 4 82 6. W=ri<P + i8 7. S = lo-gi/ 1 ' 51 8. H = -0955^ u 9. T = 435 ' 238 10. <Z=i"2v7 11. T = 5 4 i'079 12. h = -0724W 1 ' 8 13. T= 1-29 x io- 7 . 2 ' 46 14. Q = 6-nH- 48 15. y = 224\/H 16. / = io/ a 17. a = 1205, b = 53-5 18. a = 1320, 6 = 54-4 19 ' h = g.gx'io' 20< W = ' 87 ' C = 2 5 21. a = 14-9, b = -58, c = -02 22. E = -I + -OI32T -00000583T 2 23. E = -15 + -OO795T 'Ooooo2iT a 24. A = 144-6 2-7631; + -OI384U 2 25. R = 160 I6-4V+-4V 2 26. v = 3-195 +'452D - 27. a = 10, b = -277 28. -2 29. -3 30. -4 31. y = i85- 26 * 82. Q = i-sH 2 ' 5 490 MATHEMATICS FOR ENGINEERS Exercises 43 3. -115*; 92-6 Ibs./n* 5. Write equation log d- log 2-9= JlogH- JlogN or D = JH - |N Exercises 44 1. 8, 8^, 8f, 8^ ; -009% too low 2. i i i i. 31 10+ 2+ 15+ i+ i+ 7' 325 4 x _I_ JL JL_ 12+ 4+ 4+ 1 + 5. 6gf, say 6 complete turns with 19 holes on 33 hole circle 6. 8 complete and 2 holes on 18 hole circle (approx.) 7. 10 holes on 17 hole circle 8. 50 to 127 91.2 ^ A 4 j j * * x+i ' x+6 2* +3 3* +5 x ii 13 3 , 5 2 4 X I X 2 4 5 '*+4 #-5 #-2 2 15 5 . i 2X+ 7 x 3 3(. X + 2) * 6(AT i) 2(# + i) F 17 6 18 7 3(* + 2) io 6 a 8 c 3 32 21. 48 22. i 23. I 24. b* 5 Exercises 45 A m 8 vn'n t jm*n z i^m 5 n 3 7 I 7 2 55*> / 256 40 100 125 8 I 8c 8oc 2 ' 3 . a 2 sin 2 12. (a) 20.19.18 . . . 10.9; (6) 125970 13. 105; 11880; 120 14. -984 15. 2-074 16. (a) i -oi x io 15 ; (b) -9545; (c) -73; (d) 40-92 18. cosh x = i + r^ + ~ + ... ; sinh x = x + p + + . . f ~~ y y* y 3 jT ~~ Exercises 46 1. 45 2. 5904 3. o 4. 1728 5. 795 6. -4372 or 2-449 T. x = 5, y 6 8. a = 1-5, & = 8 9- * = 5, y = 4, z = 2 10. a = 1-2, & 5-7, c = 4-8 MATHEMATICAL TABLES TABLE I. TRIGONOMETRICAL RATIOS Angle. Chord. Sine. Tangent. Co-tangent. Cosine. De- grees. Radians. o o o i 1-414 1-5708 90 i 2 3 4 0175 0349 0524 0698 017 035 052 070 0175 0349 0523 0698 0175 0349 0524 0699 57-2900 28-6363 19-0811 14-3007 9998 9994 9986 9976 1-402 1-389 1-377 1-364 1-5533 1-5359 1-5184 1-5010 89 88 87 86 5 0873 087 0872 0875 11-4301 9962 I-35I I-4835 85 6 8 9 1047 1222 1396 1571 105 122 140 157 1045 1219 1392 1564 1051 1228 1405 1584 9-5I44 8-1443 7-U54 6-3138 9945 9925 9903 9877 1-338 1-325 1-312 1-299 1-4661 1-4486 1-4312 I-4I37 84 83 82 81 10 1745 174 1736 1763 5-67I3 9848 1-286 1-3963 80 ii 12 13 H 1920 2094 2269 2443 192 2O9 226 244 1908 2079 2250 2419 1944 2126 2309 2493 5-I446 4-7046 4-33I5 4-0108 9816 9781 9744 9703 1-272 1-259 J-245 1-231 1-3788 1-3614 1-3439 1-3265 79 78 77 76 IS 2618 261 2588 2679 3-7321 9659 1-218 1-3090 75 16 17 18 19 2793 2967 3142 3316 278 296 313 33 2756 2924 3090 3256 2867 3057 3249 3443 3-4874 3-2709 3-0777 2-9042 9613 9563 95" 9455 1-204 1-190 1-176 1-161 1-2915 1-2741 1-2566 1-2392 74 73 72 7i 20 3491 347 3420 3640 2-7475 9397 1-147 1-2217 70 21 22 23 24 3665 3840 4014 4189 364 382 399 416 3584 3746 3907 4067 3839 4040 4245 4452 2-6051 2-4751 2-3559 2-2460 9336 9272 9205 9135 I-I33 1-118 1-104 1-089 1-2043 1-1868 1-1694 1-1519 69 68 67 66 25 4363 433 4226 4663 2-1445 9063 1-075 I-I345 65 26 27 28 29 4538 4712 4887 5061 450 467 484 501 4384 454 4695 4848 4877 5095 5317 5543 2-0503 1-9626 1-8807 1-8040 8988 8910 8829 8746 i -060 1-045 1-030 1-015 1-1170 1-0821 1-0647 64 3 62 61 30 5236 518 5000 5774 1-7321 8660 I-OOO 1-0472 60 31 32 33 34 54" 5585 5760 5934 534 551 568 585 5150 5299 5446 5592 6009 6249 6494 6745 1-6643 1-6003 1-5399 1-4826 8572 8480 8387 8290 985 970 954 939 1-0297 1-0123 9948 9774 59 58 57 56 35 6109 601 5736 7002 1-4281 8192 .923 9599 55 36 37 38 39 6283 6458 6632 6807 618 635 651 668 5878 6018 6157 6293 7265 7536 7813 8098 I-3764 1-3270 1-2799 1-2349 8090 7986 7880 7771 908 892 877 861 9425 9250 9076 8901 54 53 52 51 40 6981 684 6428 8391 1-1918 7660 845 8727 5 41 42 43 44 7156 733 7505 7679 700 717 733 749 6561 6691 6820 6947 8693 9004 9325 9657 1-1504 1-1106 1-0724 1-0355 7547 7431 73M 7193 829 813 797 781 8552 8378 8203 8029 49 48 47 46 45 7854 765 7071 I'OOOO I-OOOO 7071 765 7854 45 Cosine Co-tangent Tangent Sine Chord Radians Degrees Angle 491 492 MATHEMATICAL TABLES TABLE II. LOGARITHMS 1 2 3 4 5 6 7 8 9 123 456 789 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 4 9 13 4 8 12 17 21 26 16 20 24 30 34 38 28 32 37 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 12 4 7 11 15 19 23 15 19 22 27 31 35 26 30 33 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 11 3 7 10 14 18 21 14 17 20 25 28 32 24 27 31 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 3 7 10 3 7 10 13 16 20 12 16 19 23 26 30 22 25 29 14 1461 1492 1523 1653 1584 1614 1644 1673 1703 1732 369 369 12 15 18 12 15 17 21 24 28 20 23 26 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 369 368 11 14 17 11 14 16 20 23 26 19 22 25 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 358 358 11 14 16 10 13 15 19 22 24 18 21 23 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 368 267 10 13 15 10 12 15 18 20 23 17 19 22 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 267 257 9 12 14 9 11 14 16 19 21 16 18 21 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 247 246 9 11 13 8 11 13 1C 18 20 15 17 19 20 3010 3032 3054 3076 3096 3118 3139 3160 3181 3201 246 8 11 13 16 17 19 21 22 23 84 3222 3424 3617 3802 3243 3444 3636 3820 3263 3464 3655 3838 3284 3483 3674 3856 3304 3502 3692 3874 3324 3522 3711 3892 3345 3541 3729 3909 3365 3560 3747 S927 3385 3579 3766 3945 3404 3598 3784 3962 246 246 246 345 8 10 12 8 10 12 7 9 11 7 9 U 14 16 18 14 15 17 13 15 17 12 14 16 25 8979 3997 4014 4031 4048 4065 4082 4099 4116 4133 235 7 9 10 12 14 15 26 27 28 29 4150 4314 4472 4624 4166 4330 4487 4639 4183 4346 4502 4654 4200 4362 4518 4669 4216 4378 4533 4683 4232 4393 4548 4698 4249 4409 4564 4713 4265 4425 4579 4728 4281 4440 4594 4742 4293 4456 4009 4757 236 236 235 134 7 8 10 689 689 679 11 13 15 11 13 14 U 12 14 10 12 13 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 134 679 10 11 13 31 82 83 34 4914 5051 5185 5315 4928 5065 5198 5328 4942 5079 5211 6340 4955 5092 5224 5353 4969 5105 5237 5366 4983 5119 5250 5378 4997 5132 5263 5391 5011 6146 6276 5403 5024 5159 5289 5416 6038 5172 5302 5428 134 134 134 134 678 678 668 568 10 11 12 9 11 12 9 10 12 9 10 11 S5 5441 5453 6465 6478 6490 5502 5514 6527 6539 5551 124 567 9 10 11 36 37 38 39 5563 5682 6798 6911 5575 5694 5809 5922 5587 6705 5821 5933 5599 5717 5832 5944 5611 5729 5843 5955 5623 5740 5855 5966 6635 5752 5866 6977 5647 6763 6877 5988 5658 5775 6888 5999 5670 5786 5899 6010 124 123 123 123 567 567 567 457 8 10 11 8 9 10 8 9 10 8 9 10 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 123 456 8 9 10 41 42 43 44 6128 6232 6335 6435 6138 6243 6345 6444 6149 6253 6355 6454 6160 6263 6365 6464 6170 6274 6375 6474 6180 6284 6385 6484 6191 6294 6395 6493 6201 6304 6405 6503 6212 6314 6415 6513 6222 6325 6425 6522 123 123 123 123 456 456 456 456 789 789 789 789 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 123 456 789 46 47 48 49 6628 6721 6812 6902 6637 6730 6821 6911 6646 6739 6830 6920 C656 6719 6839 6928 6665 6758 6848 6937 6675 6767 6857 6946 6684 6776 6866 6955 6693 6785 6875 6964 6702 6794 6884 6972 6712 6803 6893 6981 123 123 123 123 456 455 445 445 778 678 678 678 50 6990 G998 7007 7016 7024 7033 7042 7060 7059 7067 123 345 678 MATHEMATICAL TABLES TABLE II. (contd.) 493 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 789 51 7076 7084 7093 7101 7110 7118 7126 7156 7143 7162 52 53 54 7160 7243 7324 7168 7261 7332 7177 7259 7340 7185 7267 7348 7193 7275 7356 7202 7284 7364 7210 7292 7373 7218 7300 7380 7226 7308 7388 7235 7316 7396 l 2 a 123 123 345 346 346 677 667 667 55 7404 7412 7419 7427 7435 7443 7461 7459 7466 7474 132 346 667 56 7482 7490 7497 7505 7513 7530 7528 7636 7643 7551 67 58 59 7559 7634 7709 7566 7642 7716 7574 7649 7723 7582 7657 7731 7589 7664 7738 7597 7672 7745 7604 7679 7752 7612 7686 7760 7619 7694 7767 7627 7701 7774 122 112 113 346 344 344 667 667 567 60 7782 7789 7796 7803 7810 7818 7825 7833 7839 7846 61 62 63 64 7853 7924 7993 8062 7860 7931 8000 8069 7868 7938 8007 8076 7875 7945 8014 80S2 7882 7952 8021 8089 7889 7959 8028 8096 7896 7966 8035 8102 7903 7973 8041 8109 7910 7980 8048 8116 7917 7987 8055 8122 112 112 112 113 344 334 334 334 566 566 666 656 65 8129 8136 8143 8149 8156 8162 8169 8176 8182 8189 112 334 666 63 67 68 69 8195 8261 8325 8388 8202 8267 8331 8395 8209 8274 8338 8401 8215 8280 8344 8407 8222 8287 8351 8414 8228 8293 8357 8420 8235 8299 8363 8426 8241 8306 8370 8432 8248 8312 8376 8439 8254 8319 8332 8445 112 112 113 112 334 334 334 234 556 666 456 466 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 112 234 466 71 72 73 74 8513 8573 8633 8692 8519 8579 8639 8698 8525 8585 8645 8704 8531 8591 8651 8710 8537 8597 8657 8716 8543 8603 8663 8722 8549 8609 8669 8727 8555 8615 8675 8733 8561 8621 8681 8739 8567 8627 868(5 8745 113 113 113 112 234 234 234 234 455 466 456 456 75 8751 8766 8762 8768 8774 8779 8785 8791 8797 8802 112 233 466 76 77 78 73 8808 8865 8921 8976 8814 8871 8927 8982 8820 8876 8932 8987 8825 8882 8933 8993 8831 8887 8943 8998 8837 8893 8949 9004 8342 8899 8954 9009 8848 8904 8900 9015 8854 8910 8965 9020 8859 8915 8971 9025 112 112 112 113 233 233 233 233 455 445 446 446 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 112 233 446 81 82 83 84 9085 9138 9191 9243 9090 9143 9196 9248 9096 9149 9201 9253 9101 9154 9206 9258 9106 9159 9212 9263 9112 9165 9217 9269 9117 9170 9222 9274 9122 9175 9227 9279 9128 9180 9232 9284 9133 9186 923S 9289 112 112 112 112 233 233 233 2 33 445 446 445 445 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 112 233 446 86 87 88 89 9345 9395 9445 9494 9350 9400 9450 9499 9355 9405 9455 9504 9360 9410 94CO 9509 9365 9415 9465 9513 9370 9420 9469 9518 9376 9425 9474 9523 9380 9430 9479 9528 9385 9435 9484 9533 9390 9440 9489 9538 112 Oil Oil Oil 233 223 223 223 445 344 344 344 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 Oil 223 344 91 92 93 91 9590 9638 9635 9731 9595 9043 9689 9736 9600 9647 9094 9741 9605 9652 9699 9745 9609 9657 9703 9750 9614 9661 9708 9754 9619 9666 9713 9759 9624 9671 9717 9763 9628 9675 9722 9768 9633 9680 9727 9773 1 1 Oil Oil Oil 223 223 223 223 344 344 344 344 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9813 Oil 223 344 96 97 98 99 9823 9868 9912 9956 9827 9872 9917 9961 9832 9877 9921 9965 9836 9881 9926 9969 9841 9886 9930 9974 9845 9890 9934 9978 9850 9894 9939 9983 9854 9899 9943 9987 9859 9903 9948 9991 9863 9908 9952 9996 1 1 Oil Oil Oil 223 223 223 223 344 344 344 334 494 MATHEMATICAL TABLES TABLE III. ANTILOGARITHMS 1 2 3 4 5 6 7 8 9 123 456 789 00 1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 001 111 222 01 02 03 04 1023 1047 1072 1096 1026 1060 1074 1099 1028 1052 1076 1102 1030 1054 1079 1104 1033 1057 1081 1107 1035 1059 1084 1109 1038 1062 1086 1112 1040 1064 1089 1114 1042 1067 1091 1117 1045 1069 1094 1119 001 001 001 Oil 111 111 111 112 222 232 222 222 05 1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 Oil 112 222 06 07 08 09 1148 1176 1202 1230 1161 1178 1205 1233 1163 1180 1208 1236 1156 1183 1211 1239 1159 1186 1213 1242 1161 1189 1216 1246 1164 1191 1219 1247 1167 1194 1222 1260 1169 1197 1225 1253 1172 1199 1227 1256 Oil Oil Oil Oil 112 112 112 112 222 222 223 223 10 1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 Oil 112 223 11 12 13 14 1288 1818 1349 1380 1291 1321 1352 1384 1294 1324 1365 1387 1297 1327 1358 1390 1300 1330 1361 1393 1303 1334 1365 1396 1306 1337 1368 1400 1309 1340 1371 1403 1312 1343 1374 1406 1315 1346 1377 1409 Oil Oil Oil Oil 122 122 122 122 223 223 233 233 15 1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 Oil 122 233 16 17 18 19 1445 1479 1514 1549 1449 1483 1517 1562 1452 1486 1521 1556 1455 1489 1524 1560 1459 1493 1528 1663 1462 1496 1531 1567 1466 1500 1635 1570 1469 1603 1538 1674 1472 1507 1542 1578 1476 1510 1546 1581 Oil Oil Oil Oil 122 122 122 192 233 2 3 S 233 333 20 1585 1689 1592 1596 1600 1603 1607 1611 1614 1618 Oil 122 333 21 22 23 24 1623 1660 1698 1738 1626 1663 1702 1742 1629 1667 1706 1746 1633 1671 ino 1750 1637 1675 1714 1754 1641 1679 1718 1758 1644 1683 1722 1762 1648 1687 1726 1766 1652 1690 1730 1770 1656 1034 1734 1774 Oil 1 1 Oil 1 1 222 222 222 222 333 333 334 334 25 1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 Oil 222 334 26 27 28 29 1820 1862 1905 1950 1824 1866 1910 1954 1828 1871 1914 1959 1832 1875 1919 1963 1837 1879 1923 1968 1841 1884 1928 1972 1845 1888 1932 1977 1849 1892 1936 1982 1854 1897 1941 1986 185S 1901 1945 1991 Oil Oil Oil Oil 223 223 223 223 334 334 344 3 4 i 30 1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 Oil 223 344 31 32 33 34 2042 2089 2138 2188 2046 2094 2143 2193 2051 2099 2148 2198 2056 2104 2153 2203 2061 2109 2158 2208 2065 2113 2163 2213 2070 2118 2168 2218 2075 2123 2173 2223 2080 2128 2178 2228 2034 2133 2183 2234 Oil Oil Oil 119 223 233 223 233 344 344 344 445 35 2239 2244 2249 2254 2259 2265 2270 2276 2280 2286 112 233 445 35 37 38 89 2291 2344 2399 2455 2296 2350 2404 2460 2301 2355 2410 2466 2307 2360 2415 2472 2312 2366 2421 2477 2317 2371 2427 2483 2323 2377 2432 2489 2328 2382 2438 2495 2333 2388 2443 2500 2339 2393 2449 2506 112 112 112 112 233 233 233 233 445 445 445 455 40 2512 2518 2523 2529 2535 2541 2547 2563 2559 2564 112 234 465 41 42 43 44 2570 2630 2692 2754 2576 2636 2698 2761 2582 2642 2704 2767 2688 2649 2710 2773 2594 2655 2716 2780 2600 2661 2723 2786 2606 2667 2729 2793 2612 2673 2735 2799 2618 2679 2742 2805 2624 2685 2748 2812 112 112 112 112 234 234 334 334 455 456 456 456 45 2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 112 334 556 46 47 48 49 2884 2951 SO-20 3090 2891 2958 3027 3097 2897 2965 3034 3105 2904 2972 3041 3112 2911 2979 3048 3119 2917 2985 3055 3126 2924 2992 3062 3133 2931 2999 3069 3141 2938 3006 3076 3148 2944 3013 3083 3155 112 112 112 112 3 S 4 334 344 344 566 556 566 666 MATHEMATICAL TABLES TABLE III. (contd). 495 1 2 3 4 5 6 7 8 9 123 456 789 50 3162 3170 3177 3181 3192 3199 3206 3214 3321 3223 113 344 567 51 52 53 54 3236 8311 3388 3467 3243 3319 3396 3475 3251 3327 3404 3483 3258 3334 3412 3491 3266 3342 3420 3499 3273 3350 3428 3508 3281 3357 3436 3516 3289 3365 3443 3524 3296 3373 3451 3532 3304 3381 3459 3540 1 3 2 133 133 132 345 345 345 346 567 667 667 667 55 3548 3656 3565 3573 3581 3589 3597 3606 3614 3623 133 346 677 56 57 58 59 3631 3715 3802 3890 3639 3724 3811 3899 3648 3733 3819 3903 3656 3741 3828 3917 3664 3750 3S37 3926 3673 3758 3846 3936 3681 3767 3855 3945 3690 3776 3864 3954 3698 3784 3873 3963 3707 3793 3883 3972 133 133 123 123 345 346 446 466 678 678 678 678 60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 123 466 678 61 62 63 64 4074 4169 4266 4365 4083 4178 4276 4375 4093 4188 4285 4385 4102 4198 4295 4395 4111 4207 4305 4406 4131 4217 4315 4416 4130 4227 4325 4426 4140 4236 4335 4436 4150 4246 4345 4446 4159 4256 4355 4457 133 123 133 123 466 466 466 456 789 789 789 789 65 4467 4477 4487 4498 4608 4619 4529 4639 4550 4660 123 456 789 66 67 68 69 4571 4677 4786 4898 4581 4688 4797 4909 4592 4699 4808 4920 4603 4710 4819 4932 4613 4721 4831 4943 4624 4732 4842 4965 4634 4742 4853 4966 4645 4753 4864 4977 4656 4764 4875 4989 4667 4775 4887 5000 123 123 123 133 456 457 467 567 7 9 10 8 9 10 8 9 10 8 9 10 70 6012 6023 6035 6047 5058 C070 6082 6093 5105 6117 134 667 8 9 11 71 72 73 74 5129 5248 6370 6495 5140 6260 5383 5508 5152 5272 5395 6621 6164 5284 6408 5534 6176 5297 6420 5546 5188 5309 6433 5559 6200 6321 6445 6572 5212 6333 5458 5585 5224 6316 5470 6598 5236 6358 5483 6610 134 124 134 134 567 567 568 568 8 10 11 9 10 11 9 10 11 9 10 12 75 5623 6036 6649 5662 6676 6689 5702 5715 6728 6741 134 678 9 10 12 76 77 78 79 6754 6888 6026 6166 6768 5902 6039 6180 6781 6916 6053 6194 5794 5929 6067 6209 6808 5943 6081 6223 6821 5957 6095 6237 6834 6970 6109 6252 6848 6984 6124 6266 5861 5993 6138 6281 6875 6012 6152 6295 134 134 134 134 578 678 678 679 9 11 12 10 11 13 10 11 13 10 11 13 80 6310 6324 6339 6353 6368 6383 6397 6413 6427 6442 134 679 10 12 13 81 82 83 14 6457 6607 6761 6918 6471 6622 6776 6934 6486 6637 6792 6950 6501 6653 6808 6966 6516 6668 6823 6982 6531 6683 6839 6998 6546 6699 6855 7015 6561 6714 6871 7031 6577 6730 6887 7047 6592 6745 6902 7063 335 235 235 236 689 689 689 6 8 10 11 12 14 11 12 14 11 13 14 11 13 15 85 7079 7096 7112 7129 7146 7161 7178 7194 7211 7228 236 7 8 10 12 13 15 86 87 83 89 7244 7413 7586 7762 7261 7430 7603 7780 7278 7447 7621 7798 7295 7464 7638 7816 7311 7482 7656 7834 7328 7499 7674 7852 7345 7616 7691 7870 7362 7534 7709 7889 7379 7551 7727 7907 7396 7568 7745 7925 336 236 345 245 7 8 10 7 9 10 7 9 11 7 9 11 12 13 15 12 14 16 12 14 16 13 14 16 CO 7943 7962 7980 7998 8017 8036 8054 8072 8091 8110 346 7 9 11 13 15 17 91 92 93 4 8128 8318 8511 8710 8147 8337 8531 8730 8166 8356 8551 8750 8185 8375 8670 8770 8204 8395 8590 8790 8222 8414 8610 8810 8241 8433 8630 8831 8260 8453 8650 8851 8279 8472 8670 8872 8299 8492 8690 8892 246 246 246 246 8 9 11 8 10 13 8 10 12 8 10 12 13 15 17 14 15 17 14 16 18 14 16 18 95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 246 8 10 12 15 17 19 f6 97 98 99 9120 9333 9550 9772 9141 9354 9672 9795 9162 9376 9594 9817 9183 9397 9616 9840 9204 9419 9638 9863 9226 9441 9661 9886 9247 9462 9683 9908 9268 9484 9705 9931 9290 9506 9727 9954 9311 9528 9750 9977 246 247 247 267 8 11 13 9 11 13 9 11 13 9 11 14 16 17 19 15 17 20 16 18 20 16 18 20 496 MATHEMATICAL TABLES TABLE IV. NAPIERIAN, NATURAL, OR HYPERBOLIC LOGARITHMS Number. I 1 2 3 4 5 6 7 8 9 0-1 1*6974 7927 8797 9598 0339 1029 1674 2280 2852 3393 02 2-3906 4393 4859 5303 5729 6i37 6529 6907 7270 7621 0-3 0-4 7960 1-0837 8288 1084 8606 1325 8913 1560 9212 1790 9502 2015 9783 2235 0057 245 0324 2660 0584 2866 05 3068 3267 346i 3651 3838 4022 4202 4379 4553 4724 06 0-7 4892 6433 5057 6575 5220 6715 538o 6853 5537 6989 5692 7"3 5845 7256 5995 7386 6i43 7515 6289 S 43 Mean Differences. 0-8 7769 7893 8015 8i37 8256 8375 3492 8007 8722 8835 0-9 8946 9057 9166 9274 938i 9487 9592 9695 9798 9899 123 456 789 1-0 o-oooo 0100 0198 0296 0392 0488 0583 0677 0770 0862 1-1 0953 1044 "33 1222 1310 1398 1484 157 1655 1740 9 17 26 35 44 52 61 70 78 1-2 1823 1906 1989 2070 2151 2231 2311 2390 2469 2546 8 1624 32 40 48 56 64 72 1-3 2624 2700 2776 2852 2927 3001 3075 31483221 3293 7 1522 30 37 45 52 59 67 1-4 3365 3436 3507 3577 3646 371 3784 38533920 3988 71421 283541 48 55 62 1-5 455 4121 4187 4253 43i8 4383 4447 45" 4574 4637 6 13 19 26 32 39 45 52 S 8 1-6 4700 4762 4824 4886 4947 5008 5068 5128 5188 5247 6 12 18 24 30 36 42 48 55 1-7 53o6 5365 5423 548i 5539 5596 5653 571 5766 5822 6 ii 17 24 29 34 4 46 52 1-8 5878 5933 5988 6043 6098 6152 620662596313 6366 5 " 16 22 27 32 38 43 49 1-9 6419 6471 6523 6575 6627 6678 672967806831 6881 5 II5 2O 26 31 36 41 46 2-0 6931 6981 73i 7080 7129 7178 7227 7275 7324 7372 5 IOI 5 20 24 29 34 39 44 2-1 7419 7467 75H 756i 7608 7655 7701 7747 7793 7839 5 914 1923 28 33 37 42 22 7885 793 7975 8020 8065 8109 8154 81988242 8286 4 913 18 22 27 31 36 40 2-3 8329 8372 8416 8459 8502 8544 8587 8629 8671 8713 4913 17 21 26 30 34 38 2-4 8755 8796 8838 8879 8920 8961 9002 9042 9083 9123 4 8 12 16 20 24 29 33 37 2-5 9163 9203 9243 9282 9322 936i 9400 9439 9478 9517 4 8 12 16 20 24 2731 35 26 9555 9594 9632 9670 9708 9746 9783 9821 9858 9895 4811 15 19 23 26 30 34 2-7 9933 9969 0006 0043 0080 0116 0152 0188 0225 0260 47" 15 l8 22 26 29 33 2-8 1-0296 0332 0367 0403 0438 0473 0508 0543 0578 0613 4 7" 14 18 21 25 28 32 2-9 0647 0682 0716 0750 0784 0818 0852 0886 0919 0953 371 14 17 20 242731 3-0 0986 1019 1053 1086 1119 "5i 1184 1217 1249 1282 3 7 10 13 16 20 23 26 30 3-1 I3H 1346 1378 1410 1442 *474 1506 1537 1569 1600 3 6 10 13 16 19 22 25 29 32 1632 1663 1694 1725 1756 1787 1817 1848 1878 1909 369 12 15 18 21 25 28 3-3 1939 ig6c 2OOO 2030 2060 2090 2119 2149 2179 2208 369 12 15 18 21 24 27 34 2238 2267 2296 2326 2355 2384 2413 2442 2470 2499 369 12 1417 2O 23 26 3-5 2528 2556 2585 2613 2641 2669 2698 2726 2754 2782 368 II 1417 20 22 25 36 2809 2837 2865 2892 2920 2947 2975 3002 3029 3056 3 5 8 ii 14 16 19 22 25 3-7 3083 3110 3137 3164 3i9i 3218 3244 3271 3297 3324 3 5 8 ii 13 16 19 21 24 3-8 3350 3376 343 3429 3455 348i 3507 3533 3558 3584 3 5 8 10 13 16 18 21 23 3-9 3610 3635 3661 3686 37" 3737 3762 3788 3813 3838 3 5 8 1013 15 18 20 23 40 3863 3888 3913 3938 3962 3987 4012 4036 4061 4085 257 10 12 15 17 2O 22 4-1 4110 4134 4 J 59 4183 4207 4231 4255 4279 433 4327 257 IO 12 14 17 19 22 4-2 4351 4375 4398 4422 4446 4469 4493 45i6 454 4563 257 9 12 14 16 19 21 4-3 4586 4609 4633 4656 4679 4702 4725 4748 477 4793 257 9 " 14 16 18 21 4-4 4816 4839 4861 4884 4907 4929 495i 4974 4996 5i9 247 9 " 13 16 18 20 4-5 504 1 5063 5085 5107 5129 5i5i 5173 5195 5217 5239 247 9" 13 15 18 20 46 5261 5282 5304 5326 5347 5369 5390 5412 5433 5454 246 9 " 13 15 17 !9 4-7 5476 5497 55i8 5539 556o 558i 5602 5623 5644 5665 246 8 ii 13 15 1719 4-8 5686 577 5728 5748 57 6 9 5790 5810 5831 5851 5872 246 8 10 12 14 16 19 4-9 5892 5913 5933 5953 5974 5994 6014 6034 6054 6074 246 8 IO 12 14 16 18 50 6094 6114 6134 6i54 6174 6194 6214 6233 6253 627^ 246 8 IO 12 14 16 18 MATHEMATICAL TABLES TABLE IV (contd.) 497 51 52 53 54 55 56 57 53 59 60 61 62 63 64 65 66 67 68 69 70 71 7-2 7-3 7-4 75 76 7-7 7-8 79 80 8-1 8-2 83 84 8-5 8-6 8-7 8-8 8-9 90 91 92 93 94 95 96 97 98 99 10 7048 70 7228 7246 7263 7281 745 7422 7440,7457 7475 7492 1-6292 6312 6332 6351 6371 6390 6409 6429 6448 6467 648/|65o6'6525!6544 6563 6582 6677:6696,6715673467526771 f - ---'---- 648/|65o6'6525!6544 6563 6582 6601 662OJ6639 6658 69196938:69566975699370117029 66 7084 7102 7120 7138 7156 7174:7192,7210 7750 7767 7783 7800 7817 7834 7918 7934 7951 7968 7984 8001 62 8278 8294 8310 8326 8342 8358 8374^390 8437 8453 8469 8485 8500 8516 8532 8547 8083 8099 8116 8132 8148 8165 8181 819718213 8229 82 4 6|82 ' 84068421 8 503'8579 8594J86ioJ8625 8641 871887338749)876418779 8795 8810 8825|884o|8856 887188868901 902190369051 6 8 9 Mean Differences. 123456 789 246 246 679o68o8|6827J6846|2 4 6 246 245 75097527|7544;75i| 769977167733' 7851 7868 7884 7901 8017 8034 8050 8067 7299731773347352,7370,738712 4 5 245 235 235 235 235 235 235, 235 235: 235 234 234 1 3 4 i 3 4 i 3 4 i 3 4 i 3 4 i 3 4 134 8976 8991 9006 9066*9081 9095 9110 9125 9140^155 9169 9184 9199 92139228 9243 9257 9272 9286,9301 9315 9330,9344,935919373 9387 942 94i,943i 9445 9459 9473 9488,9502 9516 953 9545 9559 9573 9587 9601 9615^629 9643 9657 9671 9685 9699^7139727 9741 9755!97 6 9'9782!9796]98io 9824 9838,985119865 loooi 0412 0425^438 0451,0464 oi49'oi62 0176 0189 0202 0216 0229 0242 0255 0268 0281 029510308 0321(0334 0347 0360 0373 03860399 0477 0490 0503 0516 0528 U^X^ \J^4.^) \J^^V,\JI^ J. 0541 0554 l 0567Jo58o'o592 0605 061810631 0643^0656 066906811069410707,071907320744 0794 0807:0819 0832 0844 o -- ~ Q<; ~' 1033 1748 1861 1972 2721 28242834 2 2-3026 KK 0919 0931 0943 0956 0968 0980 0992 1041 1054 1163 1175 1282 1401 1412 1424 151815291541 1066 IO78|IO9O IIO2 1187 1199 I2II I759I77 18721883 1983 1994 1552 1645 1656 1668 1679 1691 1782 1894 . 075707690782 857 0869 O882 0894 O9O6 III4 1223 1235 1342 1354 1436144814591471 1564 1679 I793I804 1905 1917 1928 1939 I950I96I 2083 2094 2105 2116 2127 2138 2149 2159 2170 2181 2192 22O3 221 4 222512235 2246 2257 2268 2279 2289 2300 23IIJ2322|2332 2343 2354 2365 2375 2386 2397 2481 2492 2502 2513 2523 2534 2544 2555 2565 2576 2586 2597 2607 261826282638 2844 2854 2732 2742 2752 2762 2773 2783 2793 2803 2814 925 2935 2946 2956 2966 2976 2986 2996 3006 3016 15761587 I7O2 18161827 2OO6 2OI7 2O28 2O39 2050 2O6l 2072 2649 2659 2670 268O 2690 2701 1005 1126 113 1305 1483! 1599 1610 1622 171317251736 2865 2875 2885 2895 2905 2915 1029 2711 134 i 3 4 i 3 4 i 3 4 i 3 4 i 3 4 i 2 4 i 2 4 i 2 4 i 2 4 2 4 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 8 IO 12 8 IO 12 8 9 n 7 9 ii 7 9 ii 7 9 ii 7 9 10 7 9 10 7 8 10 7 8 10 7 8 10 6 8 10 6 8 10 689 689 689 679 679 679 679 678 678 5 7 8 5 7 8 5 7 8 5 7 J 578 568 568 567 567 5 6 7 567 567 567 567 567 56? 467 467 4 5 I 456 456 456 456 456 456 456 498 MATHEMATICAL TABLES TABLE V. NATURAL SINES. i I Q 0' 00 6' 0-1 12' 02 18' 0-3 24' 0-4 30' 05 38' 06 42' 0-7 48' 08 54' 09 Mean Differences. 1' 2' 3' 4' 5' oooo 0017 0035 0052 0070 0087 0105 OI22 0140 0157 3 6 9 12 15 1 0175 0192 0209 0227 0244 0262 0279 0297 0314 332 3 6 9 12 15 2 349 0366 0384 0401 0419 0436 454 0471 0488 0506 3 6 9 12 15 3 0523 054 1 0558 0576 0593 0610 0628 0645 0663 0680 3 6 9 12 15 4 0698 0715 0732 075 0767 0785 0802 0819 0837 0854 3 6 9 12 14 5 0872 0889 0906 0924 0941 0958 0976 0993 IOII 1028 3 6 9 12 14 6 1045 1063 1080 1097 "15 1132 1149 1167 1184 1201 3 6 9 12 14 7 1219 1236 1253 1271 1288 1305 1323 1340 1357 1374 3 6 9 12 14 8 1392 1409 1426 1444 1461 1478 1495 1513 1530 J 547 3 6 9 12 14 9 1564 1582 1599 1616 i633 1650 1668 1685 1702 1719 3 6 9 12 14 10 1736 1754 1771 1788 1805 1822 1840 1857 1874 1891 3 6 9 ii 14 11 1908 1925 1942 1959 1977 1994 2OII 2O28 2045 2062 3 6 9 ii 14 12 2079 2096 2113 2130 2147 2164 2181 2198 2215 2233 3 6 9 ii 14 13 2250 2267 2284 2300 2317 2334 2351 2368 2385 2402 3 6 8 ii 14 14 2419 2436 2453 2470 2487 2504 2521 2538 2 554 2571 3 6 8 ii 14 15 2588 2605 2622 2639 2656 2672 2689 2706 2723 2740 3 6 8 ii 14 16 2756 2773 2790 2807 2823 2840 2857 2874 2890 2907 3 6 8 u 14 17 2924 2940 2957 2974 2990 3007 3024 3040 3057 3074 3 6 8 ii 14 18 3090 3107 3123 3MO 3156 3i73 3190 32O6 3223 3239 3 6 8 ii 14 19 3256 3272 3289 3305 3322 3338 3355 3371 3387 3404 3 5 8 ii 14 20 3420 3437 3453 34 6 9 3486 352 35i8 3535 3551 3567 3 5 8 ii 14 21 3584 3600 3616 3633 3649 3665 3681 3697 37M 373 3 5 8 ii 14 22 3746 37 62 3778 3795 3811 3827 3843 3859 3875 3891 35 ? ii 14 23 3907 3923 3939 3955 3971 3987 4003 4019 435 45i 3 5 8 ii 14 24 4067 4083 4099 4H5 4131 4M7 4163 4179 4195 4210 3 5 8 ii 13 25 4226 4242 4258 4274 4289 4305 432i 4337 4352 4368 3 5 8 ii 13 28 4384 4399 4415 4431 4446 4462 4478 4493 4509 4524 3 5 8 10 13 27 454 4555 4571 4586 4602 4617 4633 4648 4664 4679 3 5 8 10 13 28 4695 4710 4726 4741 4756 4772 4787 4802 4818 4833 3 5 8 10 13 29 4848 4863 4879 4894 4909 4924 4939 4955 4970 4985 3 5 8 10 13 30 5000 5015 5030 5045 5060 5075 5090 5105 5120 5135 3 5 8 10 13 31 5150 5165 5180 5195 5210 5225 524 5255 5270 5284 2 5 7 10 12 32 5299 5314 5329 5344 5358 5373 5388 54 2 5417 5432 2 5 7 10 12 33 5446 54 6 i 5476 5490 5505 5519 5534 5548 5563 5577 2 5 7 10 12 34 5592 5606 5621 5635 5650 5664 5678 5693 577 5721 2 5 7 10 12 35 5736 5750 5764 5779 5793 5807 5821 5835 5850 5864 2 5 7 9 12 36 5878 5892 5906 5920 5934 5948 5962 5976 5990 6004 2 5 7 9 12 37 6018 6032 6046 6060 6074 6088 6101 6115 6129 6i43 257 9 12 38 6157 6170 6184 6198 6211 6225 6239 6252 6266 6280 257 9 ii 39 6293 6307 6320 6334 6347 6361 6 374 6388 6401 6414 247 9 ii 40 6428 6441 6455 6468 6481 6494 6508 6521 6534 6547 247 9 ii 41 6561 6574 6587 6600 6613 6626 6639 6652 6665 6678 247 9 ii 42 6691 6704 6717 6730 6743 6756 6769 6782 6794 6807 2 4 6 911 43 6820 6833 6845 6858 6871 6884 6896 6909 6921 6934 246 8 ii 44 6947 6959 6972 6984 6997 7009 7022 7034 7046 759 246 8 10 45 7071 7083 7096 7108 7120 7133 7145 7157 7169 7181 246 8 10 MATHEMATICAL TABLES 499 TABLE V. (contd.) 8 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. 1 0-0 0-1 02 0-3 0-4 0-5 0-6 0-7 08 09 1' 2' 3' 4' 5' 45 7071 7083 7096 7108 7120 7133 7145 7i57 7169 7181 246 8 10 46 7i93 7206 7218 7230 7242 7254 7266 7278 7290 7302 246 8 10 47 7314 7325 7337 7349 736i 7373 7385 7396 7408 7420 246 8 10 48 7431 7443 7455 7466 7478 7490 75oi 75i3 7524 7536 2 4 6 8 10 49 7547 7559 757 758i 7593 7604 7615 7627 7638 7649 24689 50 7660 7672 7683 7694 7705 7716 7727 7738 7749 7760 24679 51 7771 7782 7793 7804 78i5 7826 7837 7848 7859 7869 2 4579 52 7880 7891 7902 7912 7923 7934 7944 7955 7965 7976 24579 53 7986 7997 8007 8018 8028 8039 8049 8059 8070 8080 23579 54 8090 8100 8m 8121 8131 8141 8151 8161 8171 8181 2357 8 55 8192 8202 8211 8221 8231 8241 8251 8261 8271 8281 23578 56 8290 8300 8310 8320 8329 8339 8348 8358 8368 8377 23568 57 8387 8396 8406 8415 8425 8434 8443 8453 8462 8471 23568 58 8480 8490 8499 8508 8517 8526 8536 8545 8554 8563 23568 59 8572 8581 8590 8599 8607 8616 8625 8634 8643 8652 3467 60 8660 8669 8678 8686 8695 8704 8712 8721 8729 8738 3467 61 8746 8755 8763 8771 8780 8788 8796 8805 8813 8821 3467 62 8829 8838 8846 8854 8862 8870 8878 8886 8894 8902 3457 63 8910 8918 8926 8934 8942 8949 8957 8965 8973 8980 3456 64 8988 8996 9003 9011 9018 9026 9033 9041 9048 9056 3456 65 9063 9070 9078 9085 9092 9100 9107 9114 9121 9128 2456 66 9135 9143 9150 9157 9164 9171 9178 9184 9191 9198 2356 67 9205 9212 9219 9225 9232 9239 9245 9252 9259 9265 2346 68 9272 9278 9285 9291 9298 9304 93" 9317 9323 9330 2345 69 9336 9342 934 8 9354 936i 9367 9373 9379 9385 9391 2345 70 9397 943 9409 9415 9421 9426 9432 9438 9444 9449 2345 71 9455 9461 9466 9472 9478 9483 9489 9494 9500 9505 2345 72 95" 95i6 9521 9527 9532 9537 9542 9548 9553 9558 2334 73 9563 9568 9573 9578 9583 9588 9593 9598 9603 9608 2234 74 9613 9617 9622 9627 9632 9636 9641 9646 9650 9655 2234 75 9659 9664 9668 9673 9677 9681 9686 9690 9694 9699 1234 76 973 9707 9711 97*5 9720 9724 9728 9732 9736 9740 1233 77 9744 9748 975i 9755 9759 9763 9767 977 9774 9778 233 78 9781 9785 9789 9792 9796 9799 9803 9806 9810 9813 223 79 9816 9820 9823 9826 9829 9833 9836 9839 9842 9845 i 223 80 9848 9851 9854 9857 9860 9863 9866 9869 9871 9874 122 81 9877 9880 9882 9885 9888 9890 9893 9895 9898 9900 122 82 9903 9905 9907 9910 9912 9914 9917 9919 9921 9923 O 122 83 9925 9928 993 9932 9934 9936 9938 9940 9942 9943 O 112 84 9945 9947 9949 9951 9952 9954 9956 9957 9959 9960 112 85 9962 9963 9965 9966 9968 9969 997 1 9972 9973 9974 O O I I I 86 9976 9977 9978 9979 9980 9981 9982 9983 9984 9985 I I I 87 9986 9987 9988 9989 9990 9990 9991 9992 9993 9993 O O I I 88 9994 9995 9995 9996 9996 9997 9997 9997 9998 9998 O O O O O 89 9998 9999 9999 9999 9999 I-OOO I-OOO I-OOO I-OOO I-OOO O O O O O 90 I-OOO KK2 5oo MATHEMATICAL TABLES TABLE VI. NATURAL COSINES 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. 9 q 0-0 0-1 02 0-3 0-4 0-5 06 0-7 08 09 1' 2' 3' 4' 5' I-OOO ooo I'OOO I-OOO ooo I-OOO 9999 9999 9999 9999 o o o o o 1 9998 9998 9998 9997 9997 9997 9996 9996 9995 9995 o o o o o 2 9994 9993 9993 9992 9991 9990 9990 9989 9988 9987 O O I I 3 9986 9985 9984 9983 9982 9981 9980 9979 997 s 9977 I I I 4 9976 9974 9973 9972 9971 9969 9968 9966 9965 9963 O I I I 5 9962 9960 9959 9957 9956 9954 9952 9951 9949 9947 I I I 2 6 9945 9943 9942 994 9938 9936 9934 9932 9930 9928 O I I I 2 7 9925 9923 9921 9919 9917 9914 9912 9910 9907 9905 O I I 2 2 8 993 9900 9898 9895 9893 9890 9888 9885 9882 9880 O I I 2 2 9 9877 9874 9871 9869 9866 9863 9860 9857 9854 9851 O I I 2 2 10 9848 9845 9842 9839 9836 9833 9829 9826 9823 9820 II223 11 9816 9813 9810 9806 9803 9799 9796 9792 9789 9785 II223 12 9781 9778 9774 977 9767 97 6 3 9759 9755 9751 9748 II233 13 9744 974 9736 9732 9728 9724 9720 9715 9711 9707 II233 14 973 9699 9694 9690 9686 9681 9677 9673 9668 9664 II234 15 9659 9655 9650 9646 9641 9636 9632 9627 9622 9617 12234 16 9613 9608 9603 9598 9-593 9588 9583 9578 9573 9568 12234 17 9563 9558 9553 954 s 9542 9537 9532 9527 9521 95i6 12334 18 95" 9505 9500 9494 9489 9483 9478 9472 9466 9461 12345 19 9455 9449 9444 9438 9432 9426 9421 94*5 9409 943 12345 20 9397 9391 9385 9379 9373 9367 936i 9354 934 s 9342 12345 21 9336 9330 9323 9317 93" 934 9298 9291 9285 9278 12345 22 9272 9265 9259 9252 9245 9239 9232 9225 9219 9212 12346 23 9205 9198 9191 9184 9178 9171 9164 9157 915 9143 12356 24 9135 9128 9121 9114 9107 9100 9092 9085 9078 9070 2456 25 9063 9056 9048 9041 9033 9026 9018 9011 9003 8996 3456 26 8988 8980 8973 8965 8957 8949 8942 8934 8926 8918 3456 27 8910 8902 8894 8886 8878 8870 8862 8854 9846 8838 3457 28 8829 8821 8813 8805 8796 8788 8780 8771 8763 8755 34 6 7 29 8746 8738 8729 8721 8712 8704 8695 8686 8678 8669 13467 30 8660 8652 8643 8634 8625 8616 8607 8599 8590 8581 13467 31 8572 8563 8554 8545 8536 8526 8517 8508 8499 8490 23568 32 8480 8471 8462 8453 8443 8434 8425 8415 8406 8396 23568 33 8387 8377 8368 8358 8348 8339 8329 8320 8310 8300 23568 34 8290 8281 8271 8261 8251 8241 8231 8221 8211 8202 2357 35 8192 8181 8171 8161 8151 8141 8131 8121 8111 8100 23578 36 8090 8080 8070 8059 8049 8039 8028 8018 8007 7997 23579 37 7986 7976 7965 7955 7944 7934 7923 7912 7902 7891 24579 38 7880 7869 7859 7848 7837 7826 7815 7804 7793 7782 24579 39 7771 7760 7749 7738 7727 7716 775 7694 7683 7672 24679 40 7660 7649 7638 7627 7 6l 5 7604 7593 758i 757 7559 24689 41 7547 7536 7524 7513 75oi 7490 7478 7466 7455 7443 2 4 6 8 10 42 7431 7420 7408 7396 7385 7373 736i 7349 7337 7325 246 8 10 43 7314 7302 7290 7278 7266 7254 7242 7230 7218 7206 246 8 10 44 7193 7181 7169 7157 7H5 7133 7120 7108 7096 7083 246 8 10 45 7071 759 7046 734 7022 7009 6997 6984 6972 6959 246 8 10 MATHEMATICAL TABLES TABLE VI (contd.) 501 j w p Q 0' 00 6' 0-1 12' 0-2 18' 0-3 24' 0-4 30' 0-5 36' 0-6 42' 0-7 48' 0-8 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 45 7071 7059 7046 7034 7022 7009 6997 6984 6972 6959 246 8 10 46 6947 6934 6921 6909 6896 6884 6871 6858 6845 6833 246 8 ii 47 6820 6807 6794 6782 6769 6756 6743 6730 6717 6704 246 9 ii 48 6691 6678 6665 6652 6639 6626 6613 6600 6587 6 574 2 47 9 ii 49 6561 6547 6534 6521 6508 6494 6481 6468 6455 6441 2 4 7 9 ii 50 6428 6414 6401 6388 6374 6361 6347 6334 6320 6307 247 9 ii 51 6293 6280 6266 6252 6239 6225 6211 6198 6184 6170 257 9 ii 52 6157 6i43 6129 6115 6101 6088 6074 6060 6046 6032 2 57 9 12 53 6018 6004 5990 5976 5962 5948 5934 5920 5906 5892 2 5 7 9 12 54 53/8 5864 5850 5835 5821 5807 5793 5779 5764 5750 2 5 7 9 12 55 5736 572i 5707 5693 5678 5664 5650 5635 5621 5606 2 5 7 10 12 56 5592 5577 5563 5548 5534 5519 555 5490 5476 546i 2 5 7 10 12 57 5446 5432 5417 5402 5388 5373 5358 5344 5329 53M 2 5 7 10 12 58 5 2 99 5284 5270 5255 5240 5225 5210 5195 5180 5165 2 5 7 10 12 59 5150 5135 5120 5105 5090 5075 5060 5045 5030 5015 3 5 8 10 13 60 5000 4985 4970 4955 4939 4924 4909 4894 4879 4863 3 5 8 10 13 61 4848 4833 4818 4802 4787 4772 4756 4741 4726 4710 3 5 8 10 13 62 4695 4679 4664 4648 4633 4617 4602 4586 457i 4555 3 5 8 10 13 63 454 4524 4509 4493 4478 4462 4446 4431 44i5 4399 3 5 8 10 13 64 4384 4368 4352 4337 4321 4305 4289 4274 4258 4242 3 5 8 ii 13 65 4226 4210 4i95 4179 4163 4147 4i3i 4"5 4099 4083 3 5 8 ii 13 66 4067 4051 435 4019 4003 3987 3971 3955 3939 3923 3 5 8 ii 14 67 3907 3891 3875 3859 3843 3827 3811 3795 3778 3762 3 5 8 ii 14 68 3746 3730 3714 3697 3681 3665 3649 3633 3616 3600 3 5 3 ii 14 69 3584 3567 3551 3535 35i8 3502 3486 3469 3453 3437 3 5 8 ii 14 70 3420 3404 3387 3371 3355 3338 3322 3305 3289 3272 3 5 8 ii 14 71 3256 3239 3223 3206 3190 3173 3156 3MO 3123 3107 3 6 8 ii 14 72 3090 374 3057 3040 3024 3007 2990 2974 2957 294 3 6 8 ii 14 73 2924 2907 2890 2874 2857 2840 2823 2807 2790 2773 3 6 8 ii 14 74 2756 2740 2723 2706 2689 2672 2656 2639 2622 2605 3 6 8 ii 14 75 2588 257 1 2554 2538 2521 2504 2487 2470 2453 2436 3 6 8 ii 14 76 2419 2402 2385 2368 2351 2334 2317 2300 2284 2267 3 6 8 ii 14 77 2250 2233 2215 2198 2181 2164 2147 2130 2113 2096 3 6 9 ii 14 78 2079 2062 2045 2028 2OII 1994 1977 1959 1942 1925 3 6 9 ii 14 79 1908 1891 1874 1857 1840 1822 1805 1788 1771 1754 3 6 9 ii 14 80 1736 1719 1702 1685 1668 1650 1633 1616 1599 1582 3 6 9 12 14 81 1564 1547 1530 1513 1495 1478 1461 1444 1426 1409 3 6 9 12 14 82 1392 1374 1357 134 1323 I35 1288 1271 1253 1236 3 6 9 12 14 83 1219 I2OI 1184 1167 1149 1132 i"5 1097 1080 1063 3 6 9 12 14 84 1045 1028 IOII 0993 0976 0958 0941 0924 0906 0889 3 6 9 12 14 85 0872 0854 0837 0819 O8O2 0785 0767 0750 0732 0715 3 6 9 12 14 86 0698 0680 0663 0645 0628 0610 0593 0576 0558 054 i 3 6 9 12 15 87 0523 0506 0488 0471 0454 0436 0419 0401 0384 0366 3 6 9 12 15 88 0349 0332 0314 0297 0279 0262 0244 0227 0209 0192 3 6 9 12 15 89 0175 0157 0140 OI22 0105 0087 0070 0052 0035 0017 3 6 9 12 15 90 oooo 502 MATHEMATICAL TABLES TABLE VII. NATURAL TANGENTS. Q 0' 00 6' 0-1 12' 02 18' 0-3 24' 0-4 30' 0-5 36' 0-6 42' 0-7 48' 0-8 54' 09 Mean Differences. 1' 2' 3' 4' 5' oooo 0017 0035 0052 0070 0087 0105 OI22 0140 oi57 3 6 9 12 15 1 0175 0192 0209 0227 0244 0262 0279 0297 0314 0332 3 6 9 12 15 2 0349 0367 0384 0402 0419 437 454 0472 0489 0507 3 6 9 12 15 3 0524 54 2 559 577 0594 0612 0629 0647 0664 0682 3 6 9 12 15 4 0699 0717 734 0752 0769 0787 0805 0822 0840 0857 3 6 9 12 15 5 0875 0892 0910 0928 0945 0963 0981 0998 1016 1033 3 6 9 12 15 6 1051 1069 1086 1104 1122 "39 "57 "75 1192 1210 3 6 9 12 15 7 1228 1246 1263 1281 1299 1317 1334 1352 137 1388 3 6 9 12 15 8 1405 1423 1441 1459 T 477 1495 1512 1530 1548 1566 3 6 9 12 15 9 1584 1602 1620 1638 1655 1673 1691 1709 1727 1745 3 6 9 12 15 10 1763 1781 1799 18*17 1835 1853 1871 1890 1908 1926 3 6 9 12 15 11 1944 1962 1980 1998 2016 2035 2053 2071 2089 2IO7 3 6 9 12 15 12 2126 2144 2162 2180 2199 2217 2235 2254 2272 2290 3 6 9 12 15 13 2309 2327 2345 2364 2382 2401 2419 2438 2456 2475 3 6 9 12 15 14 2493 2512 2530 2549 2568 2586 2605 2623 2642 266l 3 6 9 12 16 15 2679 2698 2717 2736 2754 2 773 2792 2811 2830 2849 3 6 9 13 16 16 2867 2886 2905 2924 2943 2962 2981 3000 3019 3038 3 6 9 13 16 17 3057 3076 3096 3"5 3134 3153 3172 3I9T 3211 3230 3 6 10 13 16 18 3249 3269 3288 3307 3327 334 6 3365 3385 3404 3424 3 6 10 13 16 19 3443 3463 3482 3502 3522 3541 356i 358i 3600 3620 3 7 io 13 16 20 3640 3659 3679 3699 37*9 3739 3759 3779 3799 3819 3 7 io 13 17 21 3839 3859 3879 3899 3919 3939 3959 3979 4000 4O2O 3 7 io 13 17 22 4040 4061 4081 4101 4122 4142 4163 4183 4204 4224 3 7 io 14 17 23 4245 4265 4286 4307 4327 4348 4369 4390 44" 4431 3 7 i 14 17 24 4452 4473 4494 4515 4536 4557 4578 4599 4621 4642 4 7" 14 18 25 4663 4684 4706 4727 4748 4770 4791 4813 4834 4856 4 7 ii 14 18 26 H877 4899 4921 4942 4964 4986 5008 5029 5051 5073 4 7 " !5 18 27 5095 5"7 5139 5161 5184 5206 5228 5250 5272 5295 4 7 n, 15 18 28 5317 534 5362 5384 5407 5430 5452 5475 5498 5520 4 8 ii 15 19 29 5543 5566 5589 5612 5635 5658 5681 574 5727 5750 4 8 12 15 19 30 5774 5797 5820 5844 586 7 5890 59M 5938 596i 5985 4 8 12 16 20 31 6009 6032 6056 6080 6104 6128 6152 6176 6200 6224 4 8 12 16 20 32 6249 6273 6297 6322 6346 6371 6395 6420 6445 6469 4 8 12 16 20 33 6494 6519 6 544 6569 6594 6619 6644 6669 6694 6720 4 8 13 17 21 34 6 745 6771 6796 6822 6847 6873 6899 6924 6950 6976 4 9 13 17 21 35 7002 7028 7054 7080 7107 7133 7 r 59 7186 7212 7239 4 9 13 18 22 36 7265 7292 7319 7346 7373 7400 7427 7454 7481 75 08 5 9 14 18 23 37 7536 7563 7590 7618 7646 7673 7701 7729 7757 7785 5 9 14 18 23 38 7813 7841 7869 7898 7926 7954 7983 8012 8040 8069 5 9 14 19 24 39 8098 8127 8156 8185 8214 8243 8273 8302 8332 8361 5 io 15 20 24 40 8391 8421 8451 8481 8511 8541 8571 8601 8632 8662 5 io 15 20 25 41 8693 8724 8754 8785 8816 8847 8878 8910 8941 8972 5 io 16 21 26 42 9004 9036 9067 9099 9131 9163 9195 9228 9260 9293 5 ii 16 21 27 43 9325 9358 939i 9424 9457 9490 9523 9556 9590 9623 6 II 17 22 28 44 9657 9691 9725 9759 9793 9827 9861 9896 993 9965 6 ii 17 23 29 45 I -OOOO 0035 0070 0105 0141 0176 O2I2 0247 0283 0319 6 12 18 24 30 MATHEMATICAL TABLES 503 TABLE VII. (contd.) g 9 Q 0' 00 6' 0-1 12' 02 18' 03 24' 0-4 30' 05 36' 0-6 42' 0-7 48' 08 54' 09 Mean Differences. 1' 2' 3' 4' 5' 45 oooo 0035 0070 0105 0141 0176 0212 0247 0283 0319 6 12 18 24 30 46 0355 0392 0428 0464 0501 0538 0575 0612 0649 0686 6 12 18 25 31 47 0724 0761 0799 0837 0875 0913 0951 0990 1028 1067 6 13 19 25 32 48 1106 "45 1184 1224 1263 1303 1343 1383 1423 1463 7 13 20 27 33 49 1504 1544 1585 1626 1667 1708 1750 1792 1833 1875 7 14 21 28 34 50 1918 1960 2OO2 2045 2088 2131 2174 2218 2261 2305 7 14 22 29 36 51 2349 2393 2437 2482 2527 2572 2617 2662 2708 2753 8 15 23 30 38 52 2799 2846 2892 2938 2985 3032 3079 3127 3i75 3222 8 16 24 31 39 53 3270 33i9 3367 34i6 3465 35M 3564 3613 3663 3713 8 16 25 33 41 54 3764 3814 3865 39i6 3968 4019 4071 4124 4176 4229 9 17 26 34 43 55 4281 4335 4388 4442 4496 4550 4605 4659 47i5 4770 9 18 27 36 45 56 4826 4882 4938 4994 5051 5108 5166 5224 5282 5340 10 19 29 38 48 57 5399 5458 5517 5577 5637 5697 5757 5818 5880 594i 10 20 30 40 50 58 6003 6066 6128 6191 6255 6319 6383 6447 6512 6577 ii 21 32 43 53 59 1-6643 6709 6775 6842 6909 6977 745 7"3 7182 7251 ii 23 34 45 56 60 1-7321 7391 7461 7532 7603 7675 7747 7820 7893 7966 12 24 36 48 60 61 1-8040 8115 8190 8265 8341 8418 8495 8572 8650 8728 13 26 38 51 64 62 1-8807 8887 8967 9047 9128 9210 9292 9375 9458 9542 14 2 7 41 55 68 63 1-9626 9711 9797 9883 997 0057 oi45 0233 0323 0413 15 29 44 58 73 64 2-0503 0594 0686 0778 0872 0965 1060 "55 1251 1348 16 31 47 63 78 65 2-1445 1543 1642 1742 1842 1943 2045 2148 2251 2355 17 34 51 68 85 66 2-2460 2566 2673 2781 2889 2998 3109 3220 3332 3445 18 37 55 73 92 67 2-3559 3673 3789 3906 4023 4142 4262 4383 4504 4627 20 40 60 79 99 68 2-475I 4876 5002 5129 5257 5386 5517 5649 5782 59i6 22 43 65 87 108 69 2-6051 6187 6325 6464 6605 6746 6889 7034 7179 2326 24 47 7 1 95"9 70 2-7475 7625 7776 7929 8083 8239 8397 8556 8716 878 26 52 78 104 131 71 2-9042 9208 9375 9544 97H 9887 0061 0237 0415 0595 29 58 87116145 72 3-0777 0961 1146 1334 1524 1716 1910 2106 2305 2506 32 64 96 129 161 73 3-2709 2914 3122 3332 3544 3759 3977 4197 4420 4646 36 72 108 144 180 74 75 3-4 8 74 3-732I 5io5 7583 5339 7848 5576 8118 5816 8391 6059 8667 6305 8947 6554 9232 6806 9520 7062 9812 41 81 122 163 204 46 93 139 186 232 76 4-0108 0408 0713 1022 1335 1653 1976 2303 2635 2972 77 4'33i5 3662 4 OI 5 4374 4737 5107 5483 5864 6252 6646 78 4-7046 7453 7867 8288 8716 9152 9594 0045 6504 0970 79 5 -I 44 6 1929 2422 2924 3435 3955 4486 5026 5578 6140 80 5-67I3 7297 7894 8502 9124 9758 0405 To66 1742 2432 81 6-3138 3859 4596 535 6122 6912 7720 8548 9395 0264 Mean differences are 82 7-II54 2066 3002 3962 4947 5958 6996 8062 9158 6285 no longer suffici- ently accurate, 83 8-1443 2636 3863 5126 6427 7769 9152 0579 2052 3572 since the ilitler- 84 85 9-5M n-43 9-677 n-66 9-845 11-91 10-02 12-16 IO-2O 12-43 10-39 12-71 10-58 13-00 10-78 13-3 10-99 13-62 II-2O I3-95 ences vary cou- siderably along each line. 86 14-30 14-67 15-06 I5-46 15-89 16-35 16-83 I7-34 17-89 18-46 87 19-08 19-74 20-45 21-20 22-02 22-90 23-86 24-90 26-03 27-27 88 28-64 30-14 31-82 33-69 35-80 38-19 40-92 44-7 47-74 52-08 89 57- 2 9 63-66 71-62 81-85 95-49 114-6 143-2 191-0 286-5 573-0 90 oo 504 MATHEMATICAL TABLES TABLE VIII. LOGARITHMIC SINES. V 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. 8> 6 0-0 0-1 02 0-3 0-4 0-5 06 0-7 0-8 0-9 1' 2' 3' 4' 5' oo 3-2419 5429 7190 8439 9408 O2OO 0870 1450 1961 1 2-2419 2832 3210 3558 3880 4i79 4459 4723 497i 5206 2 5428 5640 5842 6035 6220 6397 6567 6731 6889 7041 3 7188 7330 7468 7602 773i 7857 7979 8098 8213 8326 4 8436 8543 8647 8749 8849 8946 9042 9i35 9226 93i5 16 32 48 64 So 5 9403 9489 9573 9655 9736 9816 9894 9970 0046 OI2O 13 26 39 52 65 6 1-0192 0264 0334 0403 0472 0539 0605 0670 0734 0797 ii 22 33 44 55 7 0859 0920 0981 1040 1099 "57 1214 1271 1326 I 3 8l 10 19 29 38 48 8 1436 1489 1542 1594 1646 1697 1747 1797 1847 1895 8 17 25 34 42 9 1943 1991 2038 2085 2131 2176 2221 2266 2310 2353 8 15 23 30 38 10 2397 2439 2482 2524 2565 2606 26 47 2687 2727 2767 7 14 20 27 34 11 2806 2845 2883 2921 2959 2997 3034 3070 3107 3M3 6 12 19 25 31 12 3179 3214 3250 3284 33i9 3353 3387 3421 3455 3488 6 ii 17 23 28 13 3521 3554 3586 3618 3650 3682 3713 3745 3775 3806 5 ii 16 21 26 14 3837 3867 3897 3927 3957 3986 4 OI 5 4044 4073 4IO2 5 10 15 20 24 15 4130 '4158 4186 4214 4242 4269 4296 4323 4350 4377 5 9 14 18 23 16 4403 4430 4456 4482 4508 4533 4559 4584 4609 4 6 34 4 9 13 17 21 17 4659 4684 4709 4733 4757 4781 4805 4829 4853 4876 4 8 12 16 20 18 4900 4923 4946 4969 4992 5015 5<>37 5060 5082 5104 4 8 ii 15 19 19 5126 5M8 5170 5192 5213 5235 5256 5278 5299 5320 4 7 ii 14 18 20 5341 536i 5382 542 5423 5443 5463 5484 5504 5523 3 7 10 14 i? 21 5543 5563 5583 5602 5621 5641 5660 5679 5698 5717 3 6 10 13 16 22 5736 5754 5773 5792 5810 5828 5847 5865 5883 59oi 3 6 9 12 15 23 5919 5937 5954 5972 5990 6007 6024 6042 6059 6076 3 6 9 12 15 24 6093 6110 6127 6144 6161 6177 6194 6210 6227 6243 3 6 8 ii 14 25 6259 6276 6292 6308 6324 6340 6356 6371 6387 6403 3 5 8 ii 13 26 6418 6 434 6449 6465 6480 6495 6510 6526 6541 6556 3 5 8 10 13 27 6570 6585 6600 6615 6629 6644 6659 6673 6687 6702 2 5 7 10 12 28 6716 6730 6744 6759 6 773 6787 6801 6814 6828 6842 2 5 7 9 12 29 6856 6869 6883 6896 6910 6923 6937 6950 6963 6977 2 4 7 9 ii 30 6990 7003 7016 7029 7042 755 7068 7080 793 7106 2 4 6 9 ii 31 7118 7131 7M4 7156 7168 7181 7 J 93 7205 7218 7230 2 4 6 8 10 32 7242 7254 7266 7278 7290 7302 73H 7326 7338 7349 2 4 6 8 10 33 7361 7373 7384 7396 7407 74i9 7430 7442 7453 7464 2 4 6 8 10 34 7476 7487 7498 7509 7520 7531 7542 7553 7564 7575 24679 35 7586 .7597 7607 7618 7629 7640 7650 7661 7671 7682 24579 36 7692 7703 7713 7723 7734 7744 7754 7764 7774 7785 23579 37 7795 7805 78i5 7825 7835 7844 7854 7864 7874 7884 23578 38 7893 7903 7913 7922 7932 794i 7951 7960 7970 7979 23568 29 798^ 7998 8007 8017 8026 8035 8044 8053 8063 8072 23568 40 8081 8090 8099 8108 8117 8125 8i34 8i43 8152 8161 13467 41 8169 8178 8187 8195 8204 8213 8221 8230 8238 8247 3467 42 8255 8264 8272 8280 8289 8297 8305 8313 8322 8330 34 6 7 43 8338 8346 8354 8362 8370 8378 8386 8394 8402 8410 3457 44 45 8418 8495 8426 8502 8433 8510 8441 8517 8449 8525 8457 8532 8464 8540 8472 8547 8480 8555 8487 8562 3456 2456 MATHEMATICAL TABLES 505 TABLE VIII. (contd.) j ff 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. Q 00 01 02 0-3 0-4 0-5 0-6 0-7 08 0-9 1' 2' 3' 4' 5' 45 46 47 48 49 1-8495 8569 8641 8711 8778 ' OO 00 00 00 OO IVJ V) ONOl Ol OO HI 4> VI O 4>. 00 OOVJ W 8510 8584 8655 8724 8791 8517 8591 8662 8731 8797 OO 00 OO CO 00 OOv} O>Ol Ol o oo o\\o to 4>. (X>\O OOOi 8532 8606 8676 8745 8810 OO 00 OOO 00 OOVI ON O>Ol M Ol OO HI 4* VJ HI OO OO O 8547 8620 8690 8758 8823 8555 8627 8697 8765 8830 8562 8634 8704 8771 8836 12456 12456 12356 12346 1 2 3 4 5 50 8843 8849 8855 8862 8868 8874 8880 8887 8893 8899 1 2 3 4 5 51 52 8905 8965 8971 8917 8977 8923 8983 8929 8989 8935 8995 8941 9000 8947 9006 8953 9012 8959 9018 1 2 3 4 5 1 2 3 4 5 53 9023 9O29 9035 9041 9046 9052 9057 9063 9069 9074 12345 54 9080 9085 9091 9096 9101 9107 9112 9118 9123 9128 1 2 3 4 5 55 9134 9139 9144 9149 9155 9160 9165 9170 9175 9181 1 2 3 3 4 56 9186 9191 9196 9201 9206 9211 9216 9221 9226 9231 1 2 3 3 4 57 9236 9241 9246 9251 9255 9260 9265 9270 9275 9279 12234 58 9284 9289 9294 9298 9303 9308 9312 9317 9322 9326 12234 59 "9331 9335 9340 9344 9349 9353 9358 9362 9367 937 1 11234 60 '9375 9380 9384 9388 9393 9397 9401 9406 9410 9414 11234 61 9418 9422 9427 9431 9435 9439 9443 9447 945i 9455 11233 62 '9459 94 6 3 9467 947 1 9475 9479 9483 9487 9491 9495 1 233 63 '9499 9503 9507 95X0 9514 9522 9525 9529 9533 1 233 64 '9537 9540 9544 9548 9551 9555 9558 9562 9566 9569 i 223 65 '9573 957 6 9580 9583 9587 9590 9594 9597 9601 9604 i 223 66 9607 9611 9614 9617 9621 9624 9627 9631 9634 9637 i 223 67 9640 9643 9647 9650 9653 9656 9659 9662 9666 9669 i 223 68 9672 9675 9678 9681 9684 9687 9690 9693 9696 9699 122 69 9702 9704 9707 9710 9713 9716 9719 9722 9724 9727 O 122 70 9730 9733 9735 9738 9741 9743 9746 9749 975i 9754 O 122 71 '9757 9759 9762 9764 9767 9770 9772 9775 9777 9780 O I I 2 2 72 9782 9785 9787 9789 9792 9794 9797 9799 9801 9804 I I 2 2 73 9806 9808 9811 9813 9815 9817 9820 9822 9824 9826 I I 2 2 74 9828 9831 9833 9835 9837 9839 9841 9843 9845 9847 I I I 2 75 9849 9851 9853 9855 9857 9859 9861 9863 9865 9867 I I I 2 76 9869 9871 9873 9875 9876 9878 9880 9882 9884 9885 I I I 2 77 9887 9889 9891 9892 9894 9896 9897 9899 9901 9902 O I I I I 78 9904 9906 9907 9909 9910 9912 9913 9915 9916 9918 01 II 79 9919 9921 9922 9924 9925 9927 9928 9929 9931 9932 O I 80 '9934 9935 9936 9937 9939 994 9941 9943 9944 9945 O O I 81 9946 9947 9949 995 9951 9952 9953 9954 9955 9956 O O I 82 9958 9959 9960 9961 9962 9963 9964 9965 9966 9967 O I 83 9968 9968 9969 997 9971 9972 9973 9974 9975 9975 OOO I 84 9976 9977 9978 9978 9979 9980 9981 9981 9982 9983 O I 85 9983 9984 9985 9985 9986 9987 9987 9988 9988 9989 O O 86 9989 9990 9990 9991 9991 9992 9992 9993 9993 9994 O O 87 '9994 9994 9995 9995 9996 9996 9996 9996 9997 9997 O O 88 '9997 9998 9998 9998 9998 9999 9999 9999 9999 9999 o o o o o 89 9999 9999 oooo oooo oooo oooo oooo oooo oooo OOOO o o o o o 90 O'OOOO 506 MATHEMATICAL TABLES TABLE IX. LOGARITHMIC COSINES V 0' 6' 12' 18' 24' 30' 36' 42' 48' 54' Mean Differences. & Q 0-0 0-1 0-2 0-3 0-4 0-5 0-6 0-7 0-8 0-9 1' 2' 3' 4' 5' 5*OOOO oooo oooo oooo oooo oooo oooo oooo oooo 9999 O O 1 1-9999 9999 9999 9999 9999 9999 9998 9998 9998 9998 O O O 2 9997 9997 9997 9996 9996 9996 9996 9995 9995 9994 O O O O O 3 4 9994 9989 9994 9989 9993 9988 9993 9988 9992 9987 9992 9987 9991 9986 9991 9985 9990 9985 9990 9984 O O O O O 00000 5 9983 9983 9982 9981 9981 9980 9979 997 s 9978 9977 O O O I 6 7 9976 9968 9975 9967 9975 9966 9974 9965 9973 9964 9972 9963 9971 9962 997 996i 9969 9960 9968 9959 O O I I 001 I 8 9953 9956 9955 9954 9953 9952 9951 9950 9949 9947 001 I 9 9946 9945 9944 9943 9941 994 9939 9937 9936 9935 001 I 10 9934 9932 993 1 9929 9928 9927 9925 9924 9922 9921 001 I 11 9919 9918 9916 9915 9913 9912 9910 9909 9907 9906 Oil I 12 9904 9902 9901 9899 9897 9896 9894 9892 9891 9889 Oil I 13 9887 9885 9884 9882 9880 9878 9876 9875 9873 9871 Oil 2 14 9869 9867 9865 9863 9861 9859 9857 9855 9853 9851 Oil 2 15 9849 9847 9845 9843 9841 9839 9837 9835 9833 9831 Oil 2 16 9828 9826 9824 9822 9820 9817 9815 9813 9811 9808 O I I 2 2 17 9806 9804 9801 9799 9797 9794 9792 9789 9787 9785 O I I 2 2 18 9782 9780 9777 9775 9772 9770 9767 9764 9762 9759 O I I 2 2 19 9757 9754 975i 9749 9746 9743 9741 9738 9735 9733 O I I 2 2 20 9730 9727 9724 9722 9719 9716 97 J 3 9710 9707 9704 O I I 2 2 21 9702 9699 9696 9693 9690 9687 9684 9681 9678 9675 O I I 2 2 22 9672 9669 9666 9662 9659 9656 9653 9650 9647 9643 II223 23 9640 9637 9634 9631 9627 9624 9621 9617 9614 9611 II223 24 9607 9604 9601 9597 9594 9590 9587 9583 958o 957 6 II223 25 9573 9569 9566 9562 9558 9555 9551 9548 9544 9540 II223 26 9537 9533 9529 9525 9522 95i8 9514 95io 9507 9503 II233 27 9499 9495 9491 9487 9483 9479 9475 9471 9467 9463 II233 28 9459 9455 945i 9447 9443 9439 9435 9431 9427 9422 II233 29 9418 9414 9410 9406 9401 9397 9393 9388 9384 9380 II234 30 9375 9371 9367 9362 9358 9353 9349 9344 934 9335 II234 31 9331 9326 9322 9317 9312 9308 9303 9298 9294 9289 12234 32 9284 9279 9275 9270 9265 9260 9255 9251 9246 9241 12234 33 9236 9231 9226 9221 9216 9211 9206 9201 9196 9191 12334 34 9186 9181 9175 9170 9165 9160 9155 9149 9144 9i39 12334 35 9134 9128 9123 9118 9112 9107 9101 9096 9091 9085 12345 36 9080 9074 9069 9063 9057 9052 9046 9041 9035 9029 12345 37 9023 9018 9012 9006 9000 8995 8989 8983 8977 8971 12345 38 8965 8959 8953 8947 8941 8935 8929 8923 8917 8911 12345 39 890" 8899 8893 8887 8880 8874 8868 8862 8855 8849 12345 40 8843 8836 8830 8823 8817 8810 8804 8797 8791 8784 12345 41 8778 8771 8765 8758 8751 8745 8738 8731 8724 8718 12356 42 8711 8704 8697 8690 8683 8676 8669 8662 8655 8648 12356 43 8641 8634 8627 8620 8613 8606 8598 8591 8584 8577 12456 44 8569 8562 8555 8547 8540 8532 8525 8517 8510 8502 12456 45 8495 8487 8480 8472 8464 8457 8449 8441 8433 8426 I345 6 MATHEMATICAL TABLES 507 TABLE IX (contd.) 1 ! 0' 00 6' 0-1 12' 0-2 18' 0-3 24' 0-4 30' 05 36' 06 42' 0-7 48' 08 54' 09 Mean Differences. 1' 2' 3' 4' 5' 45 ^8495 8487 8480 8472 8464 8457 8449 ' 8441 8433 8426 1 3 4 5 6 46 8418 8410 8402 8394 9386 8378 8370 8362 8354 8346 13457 47 8338 8330 8322 8313 8305 8297 8289 8280 8272 8264 1 3 4 6 7 48 8255 8247 8238 8230 8221 8213 8204 8195 8187 8178 ** i / 13467 49 8169 8161 8152 8M3 8134 8125 8117 8108 8099 8090 13467 50 8081 8072 8063 8053 8044 8035 8026 8017 8007 7998 23568 51 7989 7979 7970 7960 795i 794i 7932 7922 79i3 7903 23568 52 7893 7884 7874 7864 7854 7844 7835 7825 7815 7805 23578 53 7795 7785 7774 7764 7754 7744 7734 7723 77 J 3 773 23579 54 7692 7682 7671 7661 7650 7640 7629 7618 7607 7597 24579 55 7586 7575 7564 7553 7542 7531 7520 7509 7498 7487 24679 56 7476 7464 7453 7442 7430 7419 7407 7396 7384 7373 246 8 10 57 736i 7349 7338 7326 7314 7302 7290 7278 7266 7254 246 8 10 58 7242 7230 7218 7205 7193 7181 7168 7*56 7M4 7*31 246 8 10 59 7118 7106 7093 7080 7068 7055 7042 7029 7016 7003 246 9 II 60 6990 6977 6963 6950 6937 6923 6910 6896 6883 6869 2 4 7 9 ii 61 6856 6842 6828 6814 6801 6787 6773 6759 6744 6730 2 5 7 9 12 62 6716 6702 6687 6673 6659 6644 6629 6615 6600 6585 2 5 7 10 21 63 6570 6556 6541 6526 6510 6495 6480 6465 6449 6434 3 5 8 10 13 64 6418 6403 6387 6371 6356 6340 6324 6308 6292 6276 3 5 8 ii 13 65 6259 6243 6227 6210 6194 6177 6161 6144 6127 6110 3 6 8 ii 14 66 6093 6076 6059 6042 6024 6007 5990 5972 5954 5937 3 6 9 12 15 67 5919 5901 5883 5865 5847 5828 5810 5792 5773 5754 3 6 9 12 15 68 5736 57 J 7 5698 5679 5660 5641 5621 5602 5583 5563 3 6 10 13 16 69 5543 5523 5504 5484 5463 5443 5423 5402 5382 53^1 3 7 i H 17 70 534i 5320 5299 5278 5256 5235 5213 5192 5170 5M8 4 7 ii 14 18 71 5162 5104 5082 5060 5037 5 OI 5 4992 4969 4946 4923 4 8 ii 15 19 72 4900 4876 4853 4829 4805 4781 4757 4733 4709 4684 4 8 12 16 20 73 4 6 59 4 6 34 4609 4584 4559 4533 4508 4482 4456 443 4 9 13 17 21 74 '443 4377 4350 4323 4296 4269 4242 4214 4186 4158 5 9 14 18 23 75 4130 4102 4073 4044 4015 3986 3957 3927 3897 3867 5 10 15 20 24 76 3837 3806 3775 3745 37*3 3682 3650 3618 3586 3554 5 ii 16 21 26 77 3521 3488 3455 3421 3387 3353 3319 3284 3250 3214 6 ii 17 23 28 78 3179 3143 3107 3070 3034 2997 2959 2921 2883 2845 6 12 19 25 31 79 2806 2767 2727 2687 2647 2606 2565 2524 2482 2439 7 14 20 27 34 80 2397 2353 2310 2266 2221 2176 2131 2085 2038 1991 8 15 23 30 38 81 1943 1895 1847 1797 J 747 1697 1646 1594 1542 1489 8 17 25 34 42 82 1436 1381 1326 1271 1214 "57 1099 1040 0981 0920 10 19 29 38 48 83 0859 0797 0734 0670 0605 0539 0472 0403 0334 0264 ii 22 33 44 55 84 0192 OI2O 0046 9970 9894 9816 9736 9655 9573 ^489 13 26 39 52 65 85 2-9403 9315 9226 9135 9042 8946 8849 8749 8647 8543 16 32 48 64 80 86 8436 8326 8213 8098 7979 7857 773i 7602 7468 7330 87 7188 7041 6889 6731 6567 6397 6220 6035 5842 5640 88 5428 52O6 497 1 4723 4459 4179 3880 3558 3210 2832 89 2419 I96l M50 0870 O2OO 9408 8439 7190 5429 2419 90 -00 5 o8 MATHEMATICAL TABLES TABLE X. LOGARITHMIC TANGENTS. - E be Q 0' 00 6' 0-1 12' 0-2 18' 0-3 24' 0-4 30' 0-5 36' 0-6 42' 0-7 48' 0-8 54' 09 Mean Differences. 1' 2' 3' 4' 5' -00 3-2419 5429 7190 8439 9409 02 oo 0870 M50 1962 1 2-2419 2833 3211 3559 3881 4181 4461 4725 4973 5208 2 543 1 5643 5845 6038 6223 6401 6571 6736 6894 7046 3 7194 7337 7475 7609 7739 7865 7988 8107 8223 8336 4 8446 8554 8659 8762 8862 8960 9056 9150 9241 933i 16 32 48 64 81 5 9420 9506 9591 9674 9756 9836 9915 9992 0068 oi43 13 26 40 53 66 6 I-02I6 0289 0360 0430 0499 0567 0633 0699 0764 0828 II 22 34 45 56 7 0891 0954 1015 1076 "35 1194 1252 1310 1367 1423 10 20 29 39 49 8 1478 1533 1587 1640 1693 1745 1797 1848 1898 1948 9 17 26 35 43 9 1997 2046 2094 2142 2189 2236 2282 2328 2374 2419 8 16 23 31 39 10 2463 2507 2551 2594 2637 2680 2722 2764 2805 2846 7 14 21 28 35 11 2887 2927 2967 3006 3046 3085 3123 3162 3200 3237 6 13 19 26 32 12 3275 3312 3349 3385 3422 3458 3493 3529 3564 3599 6 12 18 24 30 13 3634 3668 3702 3736 3770 3804 3837 3870 3903 3935 6 II 17 22 28 14 3968 4000 4032 4064 4095 4127 4158 4189 4220 425 5 10 16 21 26 15 4281 43ii 4341 43?i 4400 443 4459 4488 4517 4546 5 10 15 20 25 16 4575 4603 4632 4660 4688 4716 4744 477i 4799 4826 5 9 14 19 23 17 4853 4880 4907 4934 4961 4987 5014 5040 5066 5092 4 9 13 18 22 18 5118 5143 5169 5195 5220 5245 5270 5295 5320 5345 4 8 13 17 21 19 537 5394 5419 5443 5467 549i 55i6 5539 5563 5587 4 8 12 16 20 20 5611 5634 5658 5681 574 5727 5750 5773 5796 5819 4 8 12 15 19 21 5842 5864 5887 5909 5932 5954 5976 5998 6020 6042 4 7 " 15 19 22 6064 6086 6108 6129 6151 6172 6194 6215 6236 6257 4 7 ii 14 18 23 6279 6300 6321 6341 6362 6383 6404 6424 6445 6465 3 7 10 14 17 24 6486 6506 6527 6547 6567 6587 6607 6627 6647 6667 3 7 10 13 17 25 6687 6706 6726 6746 6765 6785 6804 6824 6843 6863 3 7 10 13 16 26 6882 6901 6920 6939 6958 6977 6996 7015 7034 753 3 6 9 13 16 27 7072 7090 7109 7128 7146 7165 7183 7202 7220 7238 3 6 9 12 15 28 7257 7275 7293 73ii 7330 7348 7366 7384 7402 7420 3 6 9 12 15 29 7438 7455 7473 7491 7509 7526 7544 7562 7579 7597 3 6 9 12 15 30 7614 7632 7649 7667 7684 7701 7719 7736 7753 7771 3 6 9 12 14 31 7788 7805 7822 7839 7856 7873 7890 7907 7924 7941 3 6 9 ii 14 32 7958 7975 7992 8008 8025 8042 8059 8075 8092 8109 3 6 8 ii 14 33 8125 8142 8158 8i75 8191 8208 8224 8241 8257 8274 3 5 8 ii 14 34 8290 8306 8323 8339 8355 8371 8388 8404 8420 8436 3 5 8 ii 14 35 8452 8468 8484 8501 8517 8533 8549 8565 8581 8597 3 5 8 ii 13 36 8613 8629 8644 8660 8676 8692 8708 8724 8740 8755 3 5 8 ii 13 37 8771 8787 8803 8818 8834 8850 8865 8881 8897 8912 3 5 8 10 13 38 8928 8944 8959 8975 8990 9006 9022 9037 9053 9068 3 5 8 10 13 39 9084 9099 9H5 9130 9146 9161 9176 9192 9207 9223 3 5 8 10 13 40 9238 9254 9269 9284 9300 9315 9330 9346 936i 9376 3 5 8 10 13 41 9392 9407 9422 9438 9453 9468 9483 9499 95H 9529 3 5 8 10 13 42 9544 9560 9575 9590 9605 9621 9636 9651 9666 9681 3 5 8 10 13 43 9697 9712 9727 9742 9757 9773 9788 9803 9818 9833 3 5 8 10 13 44 9848 9864 9879 9894 9909 9924 9939 9955 9970 9985 3 5 8 10 13 45 O'OOOO 0015 0030 0045 0061 0076 0091 0106 0121 0136 3 5 8 10 13 MATHEMATICAL TABLES 509 TABLE X. (contd.) V b of Q 0' 00 6' 0-1 12' 0-2 18' 0-3 24' 04 30' 05 36' 0-6 42' 0-7 48' 0-8 54' 0-9 Mean Differences. 1' 2' 3' 4' 5' 45 oooo 0015 0030 0045 0061 0076 0091 0106 OI2I 0136 35 8 10 13 46 -0152 0167 0182 0197 O2I2 0228 0243 0258 0273 0288 35 8 10 13 47 -0303 0319 0334 349 0364 0379 0395 0410 0425 0440 35 8 10 13 48 -0456 0471 0486 0501 0517 0532 547 0562 0578 0593 35 8 10 13 49 -0608 0624 0639 0654 0670 0685 0700 0716 0731 0746 35 8 10 13 50 0762 0777 793 0808 0824 0839 0854 0870 0885 0901 35 8 10 13 51 0916 0932 0947 0963 0978 0994 IOIO 1025 1041 1056 35 8 10 13 52 -1072 1088 1103 1119 "35 1150 1166 1182 1197 1213 35 8 10 13 53 -1229 1245 1260 1276 1292 1308 1324 1340 1356 1371 35 8 ii 13 54 -1387 1403 1419 1435 i45i 1467 1483 1490 1516 1532 35 8 ii 13 55 -1548 1564 1580 1596 1612 1629 1645 1661 1677 1694 35 8 ii 14 56 -1710 1726 1743 1759 1776 1792 1809 1825 1842 1858 35 8 ii 14 57 -1875 1891 1908 1925 1941 1958 1975 1992 2OO8 2025 36 8 ii 14 58 -2042 2059 2076 2093 2IIO 2127 2144 2161 2178 2195 36 9 ii 14 59 -2212 2229 2247 2264 228l 2299 2316 2333 2351 2368 3 6 9 12 14 60 -2386 2403 2421 2438 2456 2474 2491 2509 2527 2545 36 9 12 15 61 -2562 2580 2598 2616 2634 2652 2670 2689 2707 2725 3 6 9 12 15 62 -2743 2762 2780 2798 2817 2835 2854 2872 2891 2910 36 9 12 15 63 -2928 2947 2966 2985 3004 3023 3042 3061 3080 3099 36 9 13 16 64 -3"8 3i37 3i57 3176 3196 3215 3235 3254 3274 3294 3 6 10 13 16 65 -33I3 3333 3353 3373 3393 3413 3433 3453 3473 3494 3 7 10 13 17 66 -3514 3535 3555 357 6 3596 3617 3638 3659 3679 3700 3 7 10 14 17 67 -3721 3743 3764 3785 3806 3828 3849 3871 3892 39M 4 7 ii 14 18 68 -3936 3958 3980 4002 4024 4046 4068 4091 4ii3 4136 4 7 ii J 5 19 69 -4158 4181 4204 4227 4250 4273 4296 4319 4342 4366 4 8 12 15 19 70 -4389 4413 4437 4461 4484 4509 4533 4557 458i 4606 4 8 12 16 20 71 4630 4 6 55 4680 4705 4730 4755 4780 4805 4831 4857 4 8 13 17 21 72 4882 4908 4934 4960 4986 5013 5039 5066 5093 5120 4 9 13 18 22 73 5147 5174 5201 5229 5256 5284 5312 5340 5368 5397 5 9 14 19 23 74 5425 5454 5483 5512 5541 5570 5600 5629 5659 5689 5 10 15 20 25 75 5719 5750 5780 5811 5842 5873 5905 5936 5968 6000 5 10 16 21 26 76 6032 6065 6097 6130 6163 6196 6230 6264 6298 6332 6 ii 17 22 28 77 6366 6401 6436 6471 6507 6542 6578 6615 6651 6688 6 12 18 24 30 78 6725 6763 6800 6838 6877 6915 6954 6994 733 7073 6 13 19 26 32 79 7113 7 X 54 7*95 7236 7278 7320 7363 7406 7449 7493 7 14 21 28 35 80 7537 758i 7626 7672 7718 7764 7811 7858 7906 7954 8 16 23 31 39 81 8003 8052 8102 8152 8203 8255 8307 8360 8413 8467 9 17 26 35 43 82 8522 8577 8633 8690 8748 8806 8865 8924 8985 9046 10 20 29 39 49 83 9109 9172 9236 9301 9367 9433 95i 9570 9640 9711 ii 22 34 45 56 84 9784 9857 9932 0008 0085 0164 0244 0326 6409 0494 13 26 40 53 66 85 1-0580 0669 0759 0850 0944 1040 1138 1238 I34i 1446 16 32 48 64 81 86 I-I554 1664 1777 1893 2012 2135 2261 2391 2525 2663 87 1-2806 2954 3106 3264 3429 3599 3777 3962 4155 4357 88 1-4569 4792 5027 5275 5539 5819 6119 6441 6789 7167 89 1-7581 8038 8550 9130 9800 0591 1561 2810 4571 758i 90 -00 5 io MATHEMATICAL TABLES TABLE XI. EXPONENTIAL AND HYPERBOLIC FUNCTIONS X e* .-* cosh x ' sinhx tanhx 2 2 e*+e * 1 1-1052 9048 1-0050 1002 0997 2 1-2214 8187 I-020I 2013 1974 3 1-3499 7408 1-0-153 3045 2913 4 1-4918 6703 1-0811 4108 3799 5 1-6487 6065 1-1276 5211 4621 6 1-8221 -5488 1-1855 6367 537 7 2-0138 4966 1-2552 7586 6044 8 2-2255 4493 1-3374 8881 6640 9 2-4596 4066 I-433I I-0265 7163 10 2-7183 3679 I-543I I-I752 7616 11 3-0042 3329 1-6685 1-3357 8005 1-2 3-3201 3012 1-8107 1-5095 8337 13 3-6693 2725 1-9709 1-6984 8617 1-4 4-0552 2466 2-1509 1-9043 8854 15 4-4817 2231 2-3524 2-1293 9051 16 4-9530 2019 2-5775 2-3756 9217 1-7 5-4739 1827 2-8283 2-6456 9354 1-8 6-0496 1653 3-1075 2-9422 9468 19 6-6859 1496 3-4 J 77 3-2682 9563 20 7-3891 1353 3-7622 3-6269 9640 2-1 8-1662 1225 4-M43 4-0219 9704 22 9-0251 1108 4-5679 4-457I 9758 23 9-9742 1003 5-0372 4-937 9801 24 11-0232 0907 5-557 5-4662 9837 25 12-1825 0821 6-1323 6-0502 9866 26 13-4638 0743 6-7690 6-6947 9890 2-7 14-8797 0672 7-4735 7-4063 9910 2-8 16-4446 0608 8-2527 8-1919 9926 29 18-1741 0550 9-1146 9-0596 9940 30 20-0855 0498 10-068 10-018 31 22-1980 0450 11*122 11-076 9959 32 24-5325 0408 12-287 12-246 9967 33 27-1126 0369 13-575 I3-538 9973 3-4 29-9641 334 14-999 14-965 9978 35 33-II55 0302 16-573 16-543 9982 36 36-5982 0273 I8-3I3 18-285 9985 3-7 40-4473 0247 20-236 20-211 9988 3-8 44-7012 0224 22-362 22-339 9990 39 .49-4024 O2O2 24-711 24-691 9992 4-0 54-5982 0183 27-308 27-290 9993 4-1 60-3403 Ol66 30-178 30-I62 9995 4-2 66-6863 0150 33-351 33-336 9996 43 73-6998 0136 36-843 9996 4-4 81-4509 0123 40-732 40-719 9997 4-5 90-0171 OIII 45-OI4 45-003 9997 4-6 99-4843 OIOI 49-747 49-737 9998 4-7 109-9472 0091 9998 4-8 121-5104 0082 60-759 60-751 9999 4-9 50 134-2898 148-4132 0074 0067 67-149 74-210 67-141 74-203 9999 9999 INDEX A 2 B 2 , factors of, 52 A 3 - B 3 , , 53 A 3 + B 3 , ,; , 53 Abbreviations, i Abscissa, 159 Addition formulae in Trigonometry 273 Adfected quadratic, 60 Algebraic fractions, Addition of, 57 , Multiplication of, 56 Alignment chart with four variables, 443 charts, Choice of scales for, 434 involving powers of the variables, 440 , Principle of, 429 Allowance for depreciation, 211 " Ambiguous " case in the solution of triangles, 260 Amplitude of sine functions, 361 Amsler planimeter, 300 Angle of elevation, 239 of regular polygon, 88 Angles of any magnitude, Ratios of, 251 Angular velocity, 363 Annulus, Area of, 93 Antilogarithms, 16 Approximation, by use of the binomial theorem, 467 for the area of a circle, 92 for products and quotients, 6 for square roots, 8 for the volume of a cylinder, 1 1 1 Arc, Height of elliptic, 105 , Height of circular, 97 , Length of circular, 98 Area of annulus, 93 of circle, 90 of ellipse, 104 of fillet, 132 of indicator diagram, 87 of irregular polygon, 87 of irregular quadrilateral, 87 of parabolic segment. 106 of parallelogram, 84 of rectangle, 79 Area of regular polygon, 88 of rhombus, 84 of sector of circle, 101 of segment of circle, 101 of trapezoid, 85 of triangle, 79, 80, 267 Areas of irregular curved figures, by averaging boundaries, 305 computation scale, 306 counting squares, 305 graphic integration, 312 mid-ordinate rule, 308 planimeter, 300 Simpson's rule, 310 trapezoidal rule, 307 Asymptotes of hyperbola, 108. 349 Bearing, Reduced, 244 , Whole-circle, 245 Binomial theorem, 463 Boussinesq's rule for the perimeter of an ellipse, 105 Calculation of co-ordinates in land- surveying, 244 Cardan's solution for cubic equation?, 67 Catenary, 217, 292, 357 Ccntroid, Definition of, 129 Centroids, Positions of, 130 Characteristic of a logarithm, 14 Charts, Alignment, 429 Correlation, 419 Intercept, 421 Circle, Arc of, 98 Area of, 90 Chord of, 97 Circumference of, 90 Sector of, 101 Segment of, 101 Coffin averager and planimeter, 303 Combinations, 460 Complement of an angle, 233 Complex quantities, 294 Compound interest, 208 periodic oscillations, 369 5*2 INDEX Computing scale, 306 Cone, Frustum of, 117 , Surface area of, 116 , Volume of, 116 Constant heat lines, 387 volume lines, 384 Constants, Useful, 4 Construction of regular polygons, 88 Continued fractions, 448 Convergents of ir, 451 Co-ordinates, Calculation of, 244 , Plotting of, 159 Correlation charts, 419 Cosine rule for the solution of tri- angles, 256 Cubic equations, Solution of, 67 Curves of type y = ax n , 336 y = ae bt , 352 y = e~ ax sin (bx + c), 373 Cutting, Section of, 321 , Volume of, 324 Cylinder, Surface area of, in , Volume of, in D Definitions, i Depreciation allowance, 211, 343 Determinants, 474 Determination of laws, 396 Difference of two squares, Factorisa- tion of, 52 Dividing head problem, 449 E Efficiency curves, Plotting of, 151 Ellipse, Area of, 104 , Equation of, 344 , Height of arc of, 105 of stress, 345 , Perimeter of, 105 Embankment, Section of, 321 , Volume of, 326 Equation of time, 370 of a straight line, 162 Equations, Cubic, 67, 181 , Graphic solution of, 376 , Quadratic, 61, 176 , Quadratic, with imaginary roots, 67 , Simple, 31 , Simultaneous, 43, 46, 164 , Simultaneous quadratic, 70 , Surd, 74 , Trigonometric, 287 - to conic sections, 344 equilateral triangle, Area of, 82 Equivalent acute angle, 252 Ericsson engine diagrams, 394 Euclid, Propositions of, 4 Expansion curves for gases, 338 Exponential series, 470 Factor theorem, 55 Factorisation. Method of, 51 Factors, i Fathom, 3 Fillet, Area of, 132 , Centroid of, 130 Formula for solution of cubic equa- tions, 67 of quadratic equations, 64 Fractions, Addition of algebraic, 57 , Continued, 448 , Multiplication of algebraic, 56 , Partial, 452 Frustum of cone, 117 Function, 2, 161 Graph of a sine function, 359 tangent function, 366 Graphic integration, 312 solution of equations, 3-76 of quadratic equations, 1 76 of simultaneous equations, 164 Graphs, Introduction to, 148 of quadratic expressions, 174 Guldinus, Rules of, 129 H Homogeneous equations, 73 Hyperbola, Definition of, 108 , Equation of, 348 Hyperbolic functions, 290 Hypotenuse of right-angled triangle, So Imaginary quantities, 67 Independent variable, 161 Indices, 10 Intercept charts, 421 Inverse trigonometric functions, 297 j, Meaning of, 67 Joule engine diagrams, 393 Knot, 3 K Latus rectum of parabola, 106 Laws of machines, 166 of type y = a + -, 398 y = ax n . 401 y = ae*, 405 INDEX 513 Laws of type y = a + bx + ex*, 407 y = a + bx", 408 y = b(x + a)", 409 y = a -f- fee"*, 409 y = ax"z m , 410 L.C.M., Finding the, 51 Length of chord of a circle, 97 Limiting values, 455 Logarithm, Definition of, 12 Logarithmic decrement, 375 - equations, 224 series, 470 Logarithms, Napierian, 216 of trigonometric ratios, 247 , reading from tables, 13 Log-log scale on the slide rule, 337 If Mantissa of logarithm, 14 Maximum and minimum values, 183 Mensuration, 79 et seq. Mid-ordinate rule, 308 N Napierian logs. 13 , Calculation of, 216, 471 , reading from tables, 216 Parabola, Area of segment of, 106 , Definition of, 106 , Equation of, 347 : . Length of arc of, 106 Parabolic segment, Centroid of, 130 Parallelogram, Area of, 84, 268 Partial fractions, 452 Period of sine functions, 361 Permutations, 460 Planimeter, Use of the Amsler, 300 , Use of the Coffin, 303 Polygon, Area of irregular, 87 , Area of regular, 88 . Construction of regular, 88 Prism, Surface area of, no , Volume of, no Prismoidal solid, Volume of, 319 Products of IT, 94 Progression, Arithmetic, 201 , Geometric, 205 PV diagrams, 381 et seq. Pyramid, Frustum of, 117 , Surface area of, 115 , Volume of, 115 Q Quadrant of circle, Centroid of, 130 Quadratic equations, Solution of, by completion of square, 61 Quadratic equations. Solution of. by factorisation, 61 , , by graphs, 176 . , by use of formula, 63 , , on the drawing-board, 176 Quadratic expressions, Plotting of, 1 74 Quadrilateral, Area of irregular, 87 , Centroid of, 130 R Radian, 99 Ratios of multiple and sub-multiple angles, 279 , Trigonometric, 232 Rectangle, Area of, 79 Reduced bearing, 244 Remainder theorem, 55 Reservoir, Volume of, 332 Rhombus, area of, 85 Right-angled triangle, Relation be- tween sides of, 80 , Solution of, 239 " Roots " of a quadratic equation, 61 " s " rule for area of triangle, 80 Sector of circle. Area of, 101 Segment of circle, Area of, 101 Semicircular arc, Centroid of, 130 area, Centroid of, 130 perimeter, Centroid of, 130 Series, 200 , Exponential, 470 for calculation of logs, 471 , Logarithmic, 470 Similar figures, 122 Simple harmonic motion, 365 Simpson's rule, 310 Sine curves, Plotting of, 359 et seq. rule for the solution of triangles, 256 Slide rule, Area of circle by, 92 , Log-log scale on, 337 . Reading of logs from, 17 , Reading of trigonometric ratios from, 242 . Special markings on, 17 , Uses of, for plotting log quantities, 403, 419 , , in solution of tri- angles, 261 , , Volume of cylinder by, in Solution of triangles, 255 et seq. Sphere, Surface area of, 120 , Surface area of zone of, 120 , Volume of, 120 , Volume of segment of, 121 , Volume of zone of, 120 Square measure, 3 514 INDEX Sterling engine, Diagrams for, 390 Sub-normal of parabola, 106 Sum curve, 312 Surd equations, 75 Surds, Rationalisation of denomin- ators of, 74 Surface area, for cuttings and em- bankments, 331 of cone, 116 of cylinder, 1 1 1 of frustum, 117 of prism, no of pyramid, 115 of sphere, 1 20 Surveyor's measure, 87 Table of areas and circumferences of circles, 127 of areas and circumferences of plane figures, 144, 145 of earthwork slopes, 319 of signs of trigonometric ratios, 253 of volumes and surface areas of solids, 146, 147 of weights of earths, 319 of weights of metals, 132 Tables of weights and measures, 3 Terms, i Transposition of a factor in an equa- tion, 33 of term in an equation, 32 T< diagrams, 381 et seq. Trapezoid, Area of, 85 , Centroid of, 130 Trapezoidal rule for area of irregular curved figure, 307 Triangle, Area of, 79, 267 , Lettering of, 80 , Right-angled, relation between sides of, 80 Triangles, Solution of, 255 et seq. Trigonometric equations, 287 ratios, 232 from slide rule, 242 from tables, 234 Turning-points of curves, 183 U Units, Investigation for, 26 Variation, 193 Vectors, 295 Velocity ratio of machine, 169 Volume of cone, 116 of cylinder, in of frustum of cone or pyramid, "7 of prism, no of prismoidal solid, 319 of pyramid, 115 of reservoir, 332 of segment of sphere, 121 of sphere, 120 of wedge-shaped excavation, 321 of zone of sphere, 120 W Wedge-shaped excavation, Volume of, 321 Weights and measures, Table of. 3 , Calculation of, 132 et seq. of earths, Table of. 319 of metals. Table of, 132 Whole circle bearing, 245 Zero circle of planimeter, 96, 302 Zone of sphere, Surface area of, 120 Volume of, 120 THE DIRECTLY-USEFUL (D.U.) 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