r REESE LIBRARY OF THE UNIVERSITY OF CALIFORNIA. ,/9O . No. succession No. ' 2963 _-___,____- I ! I *wg mis '$ 'M' : :- j - ; MACHINISTS' AND DRAFTSMEN'S HANDBOOK: CONTAINING TABLES, RULES AND FORMULAS, WITH NUMEROUS EXAMPLES EXPLAINING THE PRINCIPLES OF MATHEMATICS AND MECHANICS AS APPLIED TO THE MECHANICAL TRADES INTENDED AS A REFERENCE BOOK FOR ALL INTERESTED IN MECHANICAL WORK. BY PEDER LOBBEN, MECHANICAL ENGINEER. Member of the American Society of Mechanical Engineers and Worcester County Mechanics Association. Honorary Member of the N. A. Stationary Engineers. Non-Resident Member of the Franklin Institute. NEW YORK: NOSTRAND COMPANY, 23 MURRAY and 27 WARREN STREETS. 1900. COPYRIGHT, 1899, BY PEDER LOBBEN. PREFACE. It is the author's hope and desire that this book, which is the outcome of years of study, work and observation, may be a help to the class of people to which he himself has the honor to belong, the working mechanics of the world. This is not intended solely as a reference book, but it may also be studied advantageously by the ambitious young engineer and machinist; and, therefore, as far as believed practical within the scope of the work, the fundamental principles upon which the rules and formulas rest are given and explained. The use of abstruse theories and complicated formulas is avoided, as it is thought preferable to sacrifice scientific hair- splitting and be satisfied with rules and formulas which will give intelligent approximations within practical limits, rather than to go into intricate and complicated formulas which can hardly be handled except by mathematical and mechanical experts. In practical work everyone knows it is far more important to understand the correct principles and requirements of the job in hand than to be able to make elaborate scientific demonstra- tions of the subject; in short, it is only results which count in the commercial world, and every young mechanic must remem- ber that few employers will pay for science only. What they want is practical science. Should, therefore, scientific men, (for whom the author has the greatest respect, as it is to the scien- tific investigators that the working mechanics are indebted for their progress in utilizing the forces of nature), find nothing of interest in the book, they will kindly remember that the author does not pretend it to be of scientific interest, and they will therefore, in criticizing both the book and the author, remember that the work was not written with the desire to show the reader how vulgarly or how scientifically he could handle the subject, but with the sole desire to promote and assist the ambitious young working mechanic in the world's march of progress. P. LOBBEN. NEW YORK, October, 1899. 82963 NIVERSIT yCS IRotes on flfoatbematics. A Unit Js any quantity represented by a single thing, as a magnitude, or a number regarded as one undivided whole. Numbers are the measure of the relation between quanti- ties of things of the same kind and are expressed by figures. Numbers which are capable of being divided by two without a remainder are called even numbers. 2? 4, 6, 8, etc., are even numbers. Numbers which are not capable of division by two without giving a remainder are called odd numbers. 1, 3, 5, 7, 9, etc., are odd numbers. A number which can not be divided by any whole number but itself and the number 1 without giving a remainder is called a prime number. 1, 2, 3, 5, 7, 11, 13, 17, 19, etc , are prime numbers. All numbers that are not prime are said to be composite numbers, because they are composed of two or more factors ; 4, 6, 8, 9, 10, 12, etc., are composite numbers. Whole numbers are called integers. Whole numbers are also called integral numbers. A mixed number is the sum of a whole number and a fraction. The least common multiple of several given numbers is the smallest number that can be divided by each without a remainder. For instance, the least common multiple of 3, 4, 6, and 5 is 60, because 60 is the smallest number that can be divided by those numbers without a remainder. Signs. + (plus) is the sign of addition. (minus or less), is the sign of subtraction. The signs -f and are also used to indicate positive and negative quantities. X (times or multiply) is the sign of multiplication, but in- stead of this sign, sometimes a single point (.) is used, especially in formulas; in algebraic expressions very frequently factors are written without any signs at all between them. For in- stance, aKb or a.b or ab. All these three expressions indicate that the quantity a is to be multiplied by the quantity b. 2 NOTES ON MATHEMATICS. -^- (divided by) is the sign of division. = (equal). When this sign is placed between two quanti- ties, it indicates that they are of equal value. For instance : 4 + 5 + 2 = 11 8 X 12 = 96 100-4- 5 = 20 . (decimal point) signifies that the number written after it has some power of 10 for its denominator. ' " means degrees, minutes and seconds of an angle. " means feet and inches. a' a 1 ' a 1 " reads a prime, a second, a third. a\ az as reads a sub 1, a sub 2, a sub 3, and is always used to designate corresponding values of the same element. n _ \/ This is the radical sign and signifies that a root is to be extracted of the quantity coming under the sign ; this may be square root, cube root, or any other root, according to what there is signified by the number prefixed in place of the letter n . For instance : \/reads square root, \/reads cube root, \/reads fourth root, V reads fifth root, \/64~= 8, because 8 X 8 = 64 3 _ V 64 = 4, because 4 X 4 X 4 = 04 4 _ V 81 = 3, because 3X3X8X8 = 81 The sign that a quantity is to be raised to a certain power is a small number placed at the upper right hand corner of the quantity ; this number is called the exponent. For in- stance, 7 2 signifies that 7 is to be squared or multiplied by itself, that is : 72 = 7 x 7 = 49 73 =7x7X7 = 343, etc. > braces, [ ] brackets, ( ) parentheses, signify that the quantities which they include are to be considered as one quantity. For instance : 35 (8 + 6) is equal to 35 14 = 21. In this case the parenthesis indicates that not only 8, but the sum of 8 + 6 is to be subtracted from 35. - (vinculum or bar) is a straight line placed over two or more quantities, indicating that they are to be operated upon as one quantity. For instance, \/25 + 11. The vinculum attached to the radical sign indicates that the square root shall be extracted from the sum of 25+11, which is the same as the square root of 36. In an expression as ' the bar indicates that 3 X b NOTES ON MATHEMATICS. 3 the sum of 35+154-22 shall be divided by the product of 3X8 which is the same as 72 divided by 24. Whenever a number or a quantity is placed over a line and a number or a quantity is placed under the same line it always indicates that the number or quantity over the line shall be divided by the number or quantity under the line. Such a quantity is called a fraction. The quantity above the line is called the numerator, and the quantity below the line is called the denominator. A frac- tion may be either proper or improper. The fraction is proper when the numerator is smaller than the denominator; for instance, -?; but improper if the numerator is larger than the denominator, for instance, V = 1^. A fraction can always be considered simply as a problem in division. Formulas. A formula is an algebraic expression for some general rule, law or principle. Formulas are used in mechanical books, because they are much more convenient than rules. Generally speaking, the knowledge of algebra is not required for the use of formulas, because the numerical values corre- sponding to the conditions of the problem are inserted for every letter in the formula except the letter representing the unknown quantity, which then is obtained by simple arithmet- ical calculations. It is generally most convenient to begin the interpretation of formulas from the right-hand side ; for instance, the formula for the velocity of water in long pipes is : v = 8.02 J h. d. f. /. In this formula v represents the velocity of the water in feet per second. h represents the "head"* in feet. d represents the diameter of the pipe in feet. f represents the friction factor determined by experiments. / represents the length of the pipe in feet, and 8.02 is a constant equal to the square root ol twice acceleration due to gravity. Assume, for instance, that it is required to find the velocity of the flow of water in a pipe of 3 inches diameter ( % foot ) ; the length of the pipe is 1,440 feet, the "head" is 9 feet, and the friction factor is 0.025. Inserting in the formula these numerical values, and for convenience writing the diameter of the pipe in decimals, we have: * In hydraulics the word " head " means the vertical difference between the level of the water at the receiving end of the pipe and the point of discharge, or its equivalent in pressure. See Hydraulics, page 413. NOTES ON MATHEMATICS. v = 8.02 X J 9 X0.25 * 0.025 X 1440 Solving the problem step by step we have : v = 8.02 X J 2 - 25 >~3' v = 8.02 X Vo.0626" V 8.02 X 0.25. v = 2.005 feet per second. In mechanical formulas, if not otherwise specified, it is always safe to assume the letter g to mean acceleration due to gravity, usually taken as 32.2 feet or 9.82 meters. In formulas relating to heat the letter / usually signifies the mechanical equivalent of heat = 778 foot pounds of energy ; but in formulas relating to strength of materials the letter J usually signifies the polar moment of inertia, and the letter / the least rectangular moment of inertia. The letter x always expresses the unknown quantity. The following Greek letters are also used more or less. The letter TT, called pi, is used to signify the ratio of the circumference to the diameter of a circle, and is usually taken as 3.1416. 2, called sigma, usually signifies the sum of a number of quantities. The letter A, called delta, usually signifies small increments of matter. The letter 0, called theta, or the lefter $, called phi, usually signifies some particular angle, sometimes also the coefficient of friction. But all these letters may be employed to express anything, although it is usually safe, if not otherwise specified. to expect their meaning to be as stated. It is always customary to express known quantities by the first letters in the alphabet, such as a, <, c, etc., and unknown quantities by such letters as x,y,z, etc. arithmetic Addition. All quantities to be added must be of the same unit; we can not add 3 feet + 8 inches + 2 meters, without first reducing these three terms either to feet, inches or meters. The same also with numbers. Units must be added to units, tens to tens, hundreds to hundreds, etc. EXAMPLE. 318 + 5 + 38 + 10 + H5 = 486 Solution : 318 5 38 10 115 486 = Sum. Subtraction. Two quantities to be subtracted must be of the same unit. In subtraction, the same as in addition, the units are placed under each other, and units are subtracted from units, tens from tens, hundreds from hundreds, etc. EXAMPLE. 2543 1828 = 715 Solution: 2543 . . Minuend. 1828 . . Subtrahend. 715 .. Difference. Subtrahend -}- Difference = Minuend. (S) O ARITHMETIC. Hultiplication. A quantity is multiplied by a number by adding it to itself as many times' as the number indicates. EXAMPLE. 314 X 3 = 314 + 314 + 314 = 942 Solution : 314 .. Multiplicand 3 . . Multiplier 942 . , Product. Product Multiplicand = Multiplier. Product Multiplier = Multiplicand. Division. The quantity or number to be divided is called the dividend. The number by which we divide is called the divisor. The number that shows how many times the divisor is contained in the dividend is called the quotient. EXAMPLE. 6852 -4- 3 = 2284 Solution: 3)6852(2284 6852 . . Dividend. 6 3 . . Divisor. 2284 . . Quotient. 8 Divisor X Quotient = Dividend, 6 25 24 12 12 ARITHMETIC. 7 FRACTIONS. Addition. Fractions to be added must have a common denominator ; thus we cannot add % + % + X + 3 /i unless they be reduced to a common denominator instead of the denominators two, three and four ; in other words, we must find the least common multiple of the numbers 2, 3 and 4, which is 12. Thus we have: i = A t-A f| = 2A = 2i EXAMPLE 2. Add : A + 1+ J + A + I 4- T 4- I + t The common denominator is found in the following manner: Write in a line all the denominators, and divide with the prime number, 2, as many numbers as can be divided with- out a remainder. The numbers that cannot be divided without a remainder remain unchanged, and these together with the quotients of the divided numbers, are written in the next line below. Repeat this operation as long as more than one num- ber can be divided without remainder, then try to divide by the next prime number, and so on. These divisors and all those numbers remaining undivided in the last line are multi- plied together, and the product is the least common denominator. 2)_J0 * n 9 7 p 9 2) *"? "p 3 7 jt 9 2) g ? 1 337?9 3) 211 3?71? 211 11713 The common denominator is thus : 2X2X2X3X2X7X3= 1008 Thus 1008 is the least common multiple of 16, 8, 4, 12, 7 and 9. 8 ARITHMETIC. The principle of this solution can probably be better understood by resolving these numbers into prime numbers, and also resolving 1008 into prime numbers ; we then find that 1008 contains all the prime numbers necessary to make 16, 8, 4, 12, 6, 7 and 9. Prime numbers in 1008 are 2 2 2 2 7 3 3 " 16 " 2 2 2 2 < 8 " 2 2 2 " " " 4 " 2 2 " " " 12 " 2 2 3 " " " 6 " 2 3 " " " 7 " 7 " " " 9 " 3 3 Solution of Example 2 : 1008 T ^ 63 X 7 = 441 | 126 X 5 = 630 i 252 X 1 = 252 -^2 84 X 7 = 588 | 168 X 5 = 840 | 144 X 5 = 720 % 126 X 3 = 378 112 X 4 = 448 Subtraction. When fractions are to be subtracted, they must first be reduced to a common denominator, the same as in addition. EXAMPLE. | ; J must be reduced to | i = I EXAMPLES. No.l. f-^f-f^f No. 2. &-i=f&-if = H No. 3. - = - = FRACTIONS. 9 Multiplication. Fractions are multiplied by fractions, by multiplying numer- ator by numerator and denominator by denominator ; thus : t X j\ = f J = 3\ The correctness of this rule can easily be understood if we consider these two fractions as two problems in division. -* X i 7 ! will then be 3 divided by 8 and the quotient multiplied by 7 and the product divided by 12; thus, 3 is to be multiplied by 7 and the product is to be divided by 8 times 12. Therefore : _ 3 X7 1 X 7 __ r < '** ~ 8 X n - 8X~4 - A mixed number may first be reduced to an improper frac- tion and then multiplied as a common fraction, numerator by numerator and denominator by denominator. For instance: 3| X f == f X f = Y == 2 f A fraction may be multiplied by a whole number by multi- plying the numerator and letting the denominator remain un- changed. For instance : i 7 ! X 2 = If = l T 2 f = U This must be correct, because we may consider 7 as indicat- ing the quantity and 12 as indicating what kind of quantity in exactly the same sense as we may say 7 dollars or 7 cents ; if either of those were multiplied by 2 the product would, of course, be either dollars or cents respectively, and for the same reason 7 twelfths multiplied by 2 must be 14 twelfths. A fraction may also be multiplied by a whole number, by dividing the denominator by the number and letting the numer- ator remain unchanged. For instance : T r j X 2 = | '-, 11, because ^ is equal to , so must T ^ X 2 EXAMPLES. No. 1. 31 X f = V- X f = f No. 2. ljxU = fXf = f No. 3. AX1 = A No. 4. lXA = X& = IO FRACTIONS. Division. A fraction is divided by a fraction by writing the fractions after each other, then inverting the divisor (that is, changing its numerator to denominator and its denominator to numer- ator), proceed as in multiplication. For instance : t * i'= * X 'fs?-ff = t The reason for this rule can very easily be understood, when we consider the fractions as problems in division. That is to say, 5 shall be divided by 8 and the quotient is to be divided by one-fourth of 3. But if the quantity f is divided by 3 instead of one-fourth of 3, we must, of course, multiply the quotient by 4 to make the result correct. Therefore : A fraction may be divided by a whole number by dividing the numerator by the number and letting the denominator re- main unchanged. For instance : A fraction may be divided by a whole number by multiply- ing the denominator by the whole number and letting the numerator remain unchanged. For instance : Mixed numbers are reduced to improper fractions the same as in multiplication ; they are then figured the same as if they were proper fractions. EXAMPLES. No. 1. No. 2. . No. 3. No. 4. No. 5. In No. 4 it will be understood that f divided by 6 must be f F , because T ^ is exactly a sixth of $. In No. 5, also, it will be understood that if - 1 / is divided by 4, the quotient must be , because 4 is one-fourth of 16. DECIMALS. I I To Reduce a Fraction of One Denomination to a Fraction of Another Fixed Denomination, and Approx- imately of the Same Value. In mechanical calculations, on drawings, and on other oc- casions, it is very frequently necessary to reduce fractions of other denominations to eighths, sixteenths, thirty-seconds, or sixty-fourths. This may be done by multiplying the numerator and the denominator of the given fraction by the number which is to be the denominator in the new fraction, then dividing this new numerator and denominator by the denominator of the given fraction. EXAMPLE. Reduce % to eighths, sixteenths, thirty-seconds, sixty- fourths, or to hundredths. |f = -*- or ft approximately. f xi g = f| - = 3L or } approximately. |x| 21 '- | = AJ = i>7^~ or f 1 approximately. I x| 42 2 ^i }2* -^2. or j|-J approximately. f xi 66 2 U l = 100 - or xVV approximately. Thus f, instead of f , is considerably too small, namely, but f| is a great deal nearer, only T ^ too large, and 0.67 is too large. DECIMALS. In decimal fractions the denominator is always some power of ten, such as tenths, hundredths, thousandths, etc. The denominator is never written, as it is fixed by the rule that it is 1 with as many ciphers annexed as there are figures on the right-hand side of the decimal point. y z = 0.5 = five-tenths = fV % = 0.25 = twenty -five hundredths = T ^ y% = 0.125 = one hundred and twenty-five thousandths = 1 1^ = 1.5 = one and five-tenths = IfV 1# = 1.25 = one and twenty-five hundredths = 1 T 2 ^, etc. 1 2 DECIMALS. Figures on the left side of the decimal point are whole num- bers. When there are no whole numbers, sometimes a cipher is written on the left side of the decimal point, but this is not always done, as it is common with many writers not to write anything on the left side of the decimal point when there is no whole number. Thus: X niay be written .5 X " " " .25 % " " " .125 It is, however, preferable to fill in a cipher on the left- hand side of the decimal point when there is no whole number, as by so doing the mistake of reading a decimal for a whole numoer is prevented. To Reduce a Vulgar Fraction to a Decimal Fraction. Annex a sufficient number of ciphers to the numerator, divide the numerator by the denominator, and point off as many decimals in the quotient as there are ciphers annexed to the numerator. EXAMPLE. Reduce ^ to a decimal fraction. Solution : 8 ) 7.000 ( 0.875 64 60 56 40 40 00 Thus, % is equal to the decimal fraction 0.875. DECIMALS. Fractions Reduced to Exact Decimals. A A A A .015625 .03125 .046875 .0625 tt A if T 5 * .265625 .28125 .296875 .3125 it ii II T 9 * .515625 .53125 .546875 .5625 n it it .765625 .78125 .796875 .8125 A A A 1 .078125 .09375 .109375 .125 tt u II 1 .328125 .34375 .359375 .375 i* II 1 .578125 .59375 .609375 .625 u H II 1 .828125 .84375 .859375 .875 A A H A .140625 .15625 .171875 .1875 3-i' H II TV .390625 .40625 .421875 .4375 tt i If H .640625 .65625 .671875 .6875 .5 7 but the ratio between 12 and 3 is 4. The ratio be- tween the circumference of a circle and its diameter is TT or 8.1410, but the ratio between the diameter and the circumfer- ence is ^- or 0.3183, etc. This is the sense in which the word is used in this book, as this seems to agree with the common cus- tom with most mechanical writers. The term ratio is also sometimes applied to the difference of two quantities as well as to their quotient ; in which case the former is called arithmetical ratio, and the latter geometrical ratio. (See Progressions, page 68.) PROPORTION. In simple proportion there are three 'known quantities by which w r e are able to find the fourth unknown quantity ; there- fore proportion is also called "the rule of three", and it is either direct or inverse proportion. It is called direct proportion if the terms are in such ratio to one another that if one is doubled then the other will also have to be doubled, or if one is halved the other must also be halved. For instance, if 50 pounds of steel cost $25, how much will 250 pounds cost? 50 Ibs. cost $25; 250 must cost - 50 X ' = $125. 50 This is direct proportion, because the more steel we buy, the more money we have to pay. In inverse proportion the terms are in such ratio that if one is doubled the other is halved, or if one is halved the other is doubled. PROPORTION. I 7 EXAMPLE. Eight men can finish a certain work in 12 days. How many men are required to do the same work in 3 days ? Here we see that the fewer days in which the work is to be done, the more men are required. Therefore, this example is in inverse proportion. In 12 days the work was done by 8 men ; therefore, in order to do the work in 3 days it will require - L2 = 32 men. It requires 4 times as many men because the work is to be done in one quarter of the time. Compound Proportion. A proportion is called compound, if to the three terms there are combined other terms which must be taken into considera- tion in solving the problem. A very easy way to solve a compound proportion is to (same as is shown in the following examples) place the .con- ditional proposition under the interrogative sentence, term for term, and write x for the unknown quantity in the inter- rogative sentence ; draw a vertical line ; place x at the top at the left-hand side ; then try term for term and see if they are direct or inverse proportionally relative to .r, exactly the same way as if each term in the conditional proposition and the correspond- ing term in the interrogative sentence were terms in a simple rule-of-three problem. Arrange each term in the interrogative sentence either on the right or left of the vertical line, according to whether it is found to be either a multiplier or a divisor, when the problem, independent of the other terms, is considered as a simple rule-of-three problem. After all the terms in the interrogative sentence are thus arranged, place each corresponding term in the conditional proposition on the opposite side of the vertical line. Then clear away all fractions by reducing them to improper frac- tions, and let the numerator remain on the same side of the verti- cal line where it is, but transfer the denominator to the opposite side. Now cancel any term with another on the opposite side of the vertical line; then multiply all the quantities on the right side of the vertical line with each other. Also multiply all the quantities on the left side of the vertical line with each other. Divide the product on the right side by the product on the left, and the quotient is the answer to the problem, EXAMPLE 1. A certain work is executed by 15 men in G days, by work- ing 8 hours each day. How many days would it take to do the same amount of work if 12 men are working 7^ hours each day? PROPORTION. Solution : 15 Men 6 Days 8 Hours. 12 " x " i 2 " ty* 8 Days. EXAMPLE 2. A steam engine of 25 horse power is using 1500 pounds of coal in 1 day of 9^ working hours. How many pounds of coal in the same proportion will be required for 2 steam engines each having 30 horse power, working 6 days of 12% hours each day ? Solution : 1 Machine 25 Hp. 1,500 pounds 1 Day 2 " 30 x "6 " x \ im 300 1 2 hours. 28,800 pounds of coal. EXAMPLE 3. A piece of composition metal which is 12 inches long, % l / 2 inches thick and 4> inches wide, weighs 45 pounds. How many pounds will another piece of the same alloy weigh, if it meas- ures 8 inches long, 1% inches thick and 6% inches wide? 12" 8 2 1 long, 3X" .r 7 %y 2 $ /4 l * 1 thick, 4^" wide, " 6% " 45 ? ? I 3 X / ?' 4 1 ? 1 45 pounds. x " 1 2 45 22^ pounds. INTEREST. 19 INTEREST. The money paid for the use of borrowed capital is called interest. It is usually figured by the year per 100 of the principal. Simple Interest. Simple interest is computed by multiplying the principal by the percentage, by the time, and dividing by 100. What is the interest of $125, for 3 years, at 4% per year? Solution : 125 X 4 X 3 = 15 100 In Table No. 1, under the given rate per cent., find the interest for the number of years, months, and days-; add these together, and multiply by the principal invested, and the product is the interest. EXAMPLE. What is the interest of $600, invested at 6%, in 5 years, 3 months, and (> days? Solution : $1.00 in 5 years at 6% = 0.30 " " 3 months " " = 0.015 " " 6 days " " = 0.001 0.316 600 = Principal. $189.60 = Interest 20 INTEREST. >> I } 1 o I Ci C5 CO GO IT t O?OC^'^ I CCCC oooooooooooooooooooooo o o o o o o o o o o o o o o o o o o o o o o O *"^ C^ CC "^ *O ^O 1^* CO O T~H C^ CC "^ *O CO J^* QC O ^" (T^J o o o o o o o o o o o o o o o o o o o o o o " PO ^ IC^O tNOO O\ O INOO a o - t>.oo INTEREST. 21 o o^ ooooooooooo ._ I if? O .t O (t O O O O O f? * Islliiillsi - 20 O <-H O 000 05 * 5> co *- Q oo 5f -2 S TO S S co OT^Olr-O'NOt^ T _l T _ t ^-(^-((M / NiMfN OOOOOOOO ^ IftO l>00 O\ O - ^ oooooooooo i-i'NCO-^iOsOl-OOOlO O O O O O >O O O O r-l'FH% ? The interest is added to the principal at the end of each year. Solution : Principal and interest at the end of first year, 105 XJ 100 Principal and interest at the end of second year, 105 X 315 - #330.75. Principal and interest at the end of third year, 105 X 330.75 100 = #347.2875, = #347.29 = Amount. When compound interest for a great number of years is to be calculated, the above method of figuring will take too mucli time, and the following interest tables, No. 2 and No. 3, are computed in order to facilitate such calculations. In Table No. 2, under the given rate per cent., and opposite the given number of years, find the amount of one dollar in- vested at that rate for the time taken. Multiply this by the principal invested and the product is the amount. EXAMPLE. #400 is invested at 5% compound interest for 17 years, com- puted annually. What is the amount? Solution : In Table No. 2, under 5%, and opposite 17 years, we find 2.292011. Multiply this by the principal. Thus: 2.292011 400 916.8044 = #910.80 = Amount. COMPOUND INTEREST. COCO'Ni-l CO O >.1 O Ci 'X 'M Oi-Ht^TfOt~>O'Mi iCt'Xl'-l'-X 04 04 , tt ~J ^ E O 3 *- c g* I s p >> IS J= ^ ^ "^ ^ 1"^* Oi ^H ^ i^ CO ^^ i^ COCOCi( l--OClCiCOCOOC5OGOr- OOO500CNOO t O CO -M 7-1 X X .^ Cl O Ci CO CO Cl CO -f Tft J^ Tf O ^H O"*! (NOSOf-HCO l^C:-tOCOCiCO>CCOO'MXCOOCOT-i-tOfN <-t 'O ri C: I- O 1- CO i- i CO '-C CO X CO C: -f CO O O C: O 00 d 1- ioWgoOtrG6G8tiQK50&t00^iH 01 r? t^ t^ o co o co co 01 co O co t- co c: - aq GO cs cs q q o 01 c c: -t c: to -t co -t >.o x n i f T-I co co c: -M o c: 01 co O -f x 01 -^ o o o c - 01 01 co co co -t -t -o o to to \N o o to co a x x \ sQ G^ 1- ?C X -t -t C: Nco*M^TCpTto T-I i co cr. --c co o t ff co *-i T-iCO^t^5i~-CtO51 7-1 O(? : lXCO'*f X Ci 74 'N >C \N 1-1 ~t t^* CO T - " "^ *"^ - ^* (N O t O *J I O "4" X '7^1 O O >^ Ci -f Ci O O tC 7-1 X ^ O 1 (N 26 INTEREST. Table No. 4 gives time in which money will be doubled if it is invested either on simple or compound interest, compounded annually. TABLE No. 4. SIMPLE INTEREST. COMPOUND INTEREST. % Years. Days. % Years. Days. 2 50 2 35 1 *# 40 2# 28 30 8 33 120 3 23 162 8# 28 206 W 20 54 4 25 4 17 240 4/ 2 22 80 4X 15 168 5 20 5 14 75 6 16 240 6 11 321 7 14 103 7 10 89 8 12 180 8 9 2 9 11 40 9 8 16 10 10 10 7 98 11 9 33 11 6 231 12 8 120 12 6 42 Results of Saving Small Amounts of Honey. The following shows how easy it is to accumulate a fortune, provided proper steps are taken. The table gives the result of daily savings, put in a savings bank paying 4 per cent, per year, computed semi-annually : Savings per Day. Savings per Mo. Amount in 5 years. Amount in 10 years. Amount in 15 years. Amount in 20 years. Amount in 25 years. .05 .10 .25 .50 .75 $1.00 $ 1.20 2.40 6.00 12.00 18.00 24.00 $ 78.84 157.68 394.20 788.40 1,182.60 1,576.80 $ 174.96 349.92 874.80 1,749.60 2,624.40 3,499.20 $ 292.07 584.15 1,460.37 2,920.74 4,381.11 5,841.48 $ 434.88 869.76 2,174.40 4,348.80 6,523.20 8,697.60 $ 608.94 1,217.88 3,044.74 6,089.48 9,133.22 12,178.96 Nearly every person wastes an amount in twenty or thirty years, which, if saved and carefully invested, would make a family quite independent; but the principle of small savings has been lost sight of in the general desire to become wealthy. EQUATION OF PAYMENTS. 2 7 EQUATION OF PAYMENTS. When several debts are due at different dates the average time when all the debts are due is calculated by the following rule: Multiply each debt separately by the number of days be- tween its own date of maturity and the date of the debt earliest due. Divide the sum of these products by the sum of the debts ; the quotient will express the number of days subsequent to the leading day when the whole debt should be paid in one sum. EXAMPLE. A owed to B the following sums: $250 due May 12, $120 due July 19, $410 due August 1(5, and $60 due September 21, all in the same year. When should the whole sum be paid at once in order that neither shall lose any interest? Solution : May 12 $250 May 12 to July 11) is 08 days ; 120 X 68 = 8160 May 12 to Aug. 16 is 96 days; 410 X 96 = 39360 May 12 to Sept. 21 is 132 days; 60 X 132 = 7920 $840 ) 55440 = 65.9 66 days after May 12 will be July 17. When several debts are due after different lengths of time, the average time is calculated by this rule : Multiply the debt by the time ; divide the sum of the products by the sum of the debts, and the quotient is the time when all the debts may be considered due. EXAMPLE. A owed B $600, due in 7 months ; $200 due in one month, and $700 due in 3 months. When should the whole debt be paid in one sum in order that neither shall lose any interest? Solution : (500 X 7 = 4200 700 X 3 = 2100 200 X 1 = 200 1500 ) 6500 = 4J4 months. NOTE : If the debts contain both dollars and cents the cents may, if such refinement is required, be considered as deci- mal parts of a dollar, but practically in such problems the cents may be omitted in the calculation. PARTNERSHIP, or calculating of proportional parts, is the calculation of the parts of a certain quantity in such a way that the ratio between the separate parts is equal to the ratio of certain given numbers. 2 8 PARTNERSHIP. EXAMPLE 1. A composition for welding cast steel consists of 9 parts of borax and one part of sal-ammoniac. How much of each, borax and sal-ammoniac, must be taken for a mixture of 5 Ibs.? Solution : T ^ x 5 = l / 2 Ibs. borax. TO X 5 = YZ Ib. sal-ammoniac. EXAMPLE 2. An alloy shall consist of 1(50 parts of copper, 15 parts of tin and 5 parts of zinc. How much of each will be used for a cast- ing weighing 360 Ibs.? Solution : 160 iffy X 360 = 320 Ibs. of copper. 15 . ro X 360 = 30 Ibs. of tin. 5 T! o X 360 = 10 Ibs. of zinc. 180 EXAMPLE 3. Four persons A, B, C and D, are buying a certain amount, of goods together. A's part is $500, B's, $100, C's, $250, and D's, $150. On the undertaking they are clearing a net profit of $120. How much of this is each to have ? Solution : 500 A's Part = tVA X 120 = $60 100 B's " = T Vo^ years ; the net profit of the undertaking is $2,300. How much is each to have of the profit ? Solution: A, 2000 X 2 = 4000 B, 3000 X 2^ =_ 7500 11500 A's Part is T$& X 2300 $ 800 B's " " rtV o years, while A has only had his capital invested 2 years. The ratio is, therefore, not #2,000 to $3,000 but #4,000 to $7,500, because $2,000 in 2 years is equal to $4,000 in one year, and $3,000 in 2^ years is equal to $7,500 in one year. SQUARE ROOT. When the square root is to be extracted the number is di- vided into periods consisting of two figures, commencing from the extreme right if the number has no decimals, or from the decimal point towards the left for the whole numbers and towards the right for the decimals. (If the last period of deci- mals should have but one figure then annex a cipher, so that this period also has two figures, but if the period to the extreme left in the integer should happen to have only one figure it makes no difference; leave it as it is.) Ascertain the highest root of the first period and place it to the right of the number as in long division. Square this root and subtract the product of this from the first period. To the remainder annex the next period of numbers. Take for divisor 20 times the part of the root already found* and the quotient is the next figure in the root, if the product of this figure and the divisor added to the square of the figure does not exceed the dividend. To the difference between this sum and the dividend is annexed the next period of numbers. For divisor take again 20 times the part of the root already found, etc. Continue in this manner until the last period is used. If there is any remainder, and a more exact root is required, ciphers may be annexed in pairs and the operation continued until as many decimals in the root are obtained as are wanted. EXAMPLE 1. Extract the square root of 271,441. Solution : \/27jl4 41 5 2 = 25| 20 X 5 = 100 ) 214 100 X 2 + 2 2 = 204 20 X 52 = 1040 ) 1041 1040 X 1 + I 2 1041 = 521 0000 Thus: \/27M4i = 521, because 521 X 521 = 271,441. * If this divisor exceeds the dividend, write a cipher in the root; annex the next period of numbers, calculate a new divisor, corresponding to the increased rpot, and proceed as explained. 30 CUBE ROOT. EXAMPLE 2. Extract the square root of 26.6256. Solution : \/266256 =5.16 20 X 5 = 100 ) 162 100 X 1 + I 2 = 101 20 X 51 = 1020 ) 6156 1020 X 6 + 6 a = 6156 0000 CUBE ROOT. When the cube root is to be extracted, the number is divided into periods consisting of three figures. Commencing from the extreme right if the number has no decimals, or from the decimal point, toward the left, for the whole number, and toward the right for the decimals. (If the last period of deci- mals should not have three figures, then annex ciphers until this period also has three figures, but if the period to the extreme left in the integer should happen to consist of less than three figures it makes no difference ; leave it as it is.) Ascer- tain highest cube root in the first period and place it to the right of the number, the same as in long division. Cube this root and subtract the product from the first period. To the remainder annex next period of numbers. For the divisor in this number take 300 times the square of the part of the root already found,* and the quotient is the next figure in the root, if the product of this figure multiplied by the divisor and added to 30 times the part of the root already found, multiplied by the square of this quotient and added to the cube of the quotient, does not exceed this dividend. To the difference between this sum and the dividend is annexed the next period of numbers. For divisor take again 300 times the square of the part of the root already found, etc. Continue in this manner until the last period is used. If there is any remainder from last period, and a more exact root is required, ciphers may be annexed three at a time, and the operation continued until as many decimals are obtained in the root as are waated. * If this divisor exceeds the dividend, write a cipher in the root, annex the next period of numbers, calculating a new divisor corresponding to the increased root, and proceed as explained. CUBE ROOT. 31 EXAMPLE l. Extract the cube root of 275,894,451. Solution : X/2751894 6 8 = 216| 300 X 6 2 == 10800 ) 59894 10800 X 5 + 30 X 6 X 5 2 + 5 3 = 58625 451 300 X 65 2 = 1207500 ) 1269451 1267500 X 1 + 30 X 65 X I 2 + I 3 = 1269451 651 0000000 Thus : 3 \/275,894,451 = 651, because 651 X 651 X 651 = 275,894,451. EXAMPLE 2 : Extract the cube root of 551.368. Solution : 3 \/551 8 3 = 512 8.2 300 X 8 2 = 19200 ) 39368 19200 X 2 + 30 X 8 X 2 2 + 2 3 =_39368 00000 The square root of a number consisting of two figures will never consist of more than one figure, and the square root of a number consisting of four figures will never consist of more than two figures; hence, the rule to divide numbers into periods consisting of two figures. The cube root of a number consisting of three figures will never consist of more than one figure, and the cube root of a number consisting of six figures will never consist of more than two figures ; hence, the rule to divide the numbers into periods consisting of three figures. There will always be one decimal in the root for each period of decimals in the number of which the root is extracted. This relates to both cube and square root. The root of a fraction may be found by extracting the separate roots of numerator and denominator, or the fraction may be first reduced to a decimal fraction before the root is extracted. The root of a mixed number may be extracted by first re- ducing the number to an improper fraction and then extracting the separate roots of numerator and denominator, or the number may be first reduced to consist of integer and decimal fractions, and the root extracted as usual. 32 CUBE ROOT. Radical Quantities Expressed without the Radical Sign. The radical sign is not always used in signifying radical quantities. Sometimes a quantity expressing a root is written as a quantity to be raised into a fractional power. For instance : >/16~may be written 16^. This is the same value ; thus, Vl6~= 4 and 16* = 4. 3 V 27 may be written, 27* = 3. 3 3 8t = V8 2 = > V 64 =4. The denominator in the exponent always indicates which root is to be extracted. Thus, 8^ will be square 8 and extract the cube root from the product. EXAMPLE. *,_ 4 __ 16l = V 16 3 = V 4096 = 8. Thus, cube 16 and extract the fourth root of the product. RECIPROCALS. The reciprocal of any number is the quotient which is ob- tained when 1 is divided by the number. For instance, the reciprocal of 4 is X 0-25 5 the reciprocal of 16 is T a ff = 0.0625, etc. Frequently it is a saving of time when performing long division to use the reciprocal, as multiplying the dividend by the reciprocal of the divisor gives the quotient. For instance, divide 4 by 758. In Table No. 6 the reciprocal of 758 is given as 0.0013193. Multiplying 0.0013193 by 4 gives 0.0052772, which is correct to six decimals. When reducing vulgar frac- tions to decimals the reciprocal may be used with advantage. For instance, reduce j to decimals. In Table No. 6 the recip- rocal of 64 is given as 0.015625, and 15 X 0.015625 = 0.234375, which is the decimal of . IMPORTANT. Whenever the exact reciprocal is not ex- pressible by decimals the result obtained by its use is, as explained above, only approximate. SQUARES, CUBES, ROOTS, AND RECIPROCALS. 33 TABLE No. 5. Giving Squares, Cubes, Square Roots, Cube Roots, and Reciprocals of Fractions and flixed Numbers, from J to 10. n V n 3 VT ^T 1 n A 0.000244 0.0000038 0.125 0.25 64 A 0.000977 0.0000305 0.17678 0.31496 32 A 0.002196 0.000103 0.21651 0.36056 21.3333 A 0.003906 0.000244 0.25 0.39685 16 A 0.006104 0.000477 0.27951 0.42750 12.8 A 0.008789 0.000823 0.30619 0.45428 10.6667 A 0.011963 0.001308 0.33072 0.47823 9.1428 i 0.015625 0.001953 0.35355 0.5 8 A 0.01977 0.00278 0.375 0.52002 7.1111 A 0.02441 0.00381 0.39528 0.53861 6.4 a 0.02954 0.00508 0.41458 0.55599 5.8182 A 0.03516 0.00659 0.43301 0.57236 5.3333 H 0.04126 0.00838 0.45069 0.58783 4.9231 A 0.04785 0.01047 0.46771 0.60254 4.5714 if 0.05493 0.01287 0.48412 0.61655 4.2666 j 0.06250 0.01562 0.5 0.62996 4 ti 0.07056 0.01874 0.51539 0.64282 3.7647 A 0.07910 0.02225 0.53033 0.65519 3.5556 H 0.08813 0.02616 0.54482 0.66709 3.3684 A 0.09766 0.03052 0.55902 0.67860 3.2 H 0.10766 0.03533 0.57282 0.68973 3.0476 H 0.11816 0.04062 0.58630 0.70051 2.9091 H 0.12915 0.04641 0.59942 0.71097 2.7826 i 0.14062 0.05273 0.61237 0.72112 2.6667 H 0.15258 0.05960 0.625 0.73100 2.56 li 0.16504 0.06705 0.63738 0.74062 2.4615 II 0.17798 0.07508 0.64952 0.75 2.3703 jV 0.19141 0.08374 0.66144 0.75915 2.2857 H 0.20522 0.09303 0.67314 0.76808 2.2069 H 0.21973 0.10300 0.68465 0.77681 2.1333 H 0.23463 0.11364 0.69597 0.78534 2.0645 i 0.25 0.12500 0.70711 0.79370 2 34 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n w 2 * vr
s4si 23.68544 8.24747 0.0017825 562 315844 177504328 23.70654 8.25237 0.0017794 563 316969 178453547 23.72762 8.25726 0.0017762 564 318096 179406144 23.74868 8.26215 0.0017730 52 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n W 2 n 3 VT fr 1 n 565 319225 180362125 23.76973 8.26703 0.0017699 566 320356 181321496 23.79075 8.27190 0.0017668 567 321489 182284263 23.81176 8.27677 0.0017637 568 322624 183250432 23.83275 8.28163 0.0017606 569 323761 184220009 23.85372 8.28649 0.0017575 570 324900 185193000 23.87467 8.29134 0.0017544 571 326041 186169411 23.89561 8.29619 0.0017513 572 327184 187149248 23.91652 8.30103 0.0017483 573 328329 188132517 23.93742 8.30587 0.0017452 574 329476 189119224 23.95830 8.31069 0.0017422 575 330625 190109375 23.97916 8.31552 0.0017391 576 331776 191102976 24 8.32034 0.0017361 577 332929 192100033 24.02082 8.32515 0.0017331 578 334084 193100552 24.04163 8.32995 0.0017301 579 335241 194104539 24.06242 8.33476 0.0017271 580 336400 195112000 24.08319 8.33955 0.0017241 581 337561 196122941 24.10394 8.34434 0.0017212 582 338724 197137368 24.12468 8.34913 0.0017182 583 339889 198155287 24.14539 8.35390 0.0017153 584 341056 199176704 24.16609 8.35868 0.0017123 585 342225 200201625 24.18677 8.36345 0.0017094 586 343396 201230056 24.20744 8.36821 0.0017065 587 344569 202262003 24.22808 8.37297 0.0017036 588 345744 203297472 24.24871 8.37772 0.0017007 589 346921 204336469 24.26932 8.38247 0.0016978 590 348100 205379000 24.28992 8.38721 0.0016949 591 349281 206425071 24.31049 8.39194 0.0016920 592 350464 207474688 24.33105 8.39667 0.0016892 593 351649 208527857 24.35159 8.40140 0.0016863 594 352836 209584584 24.37212 8.40612 0.0016835 595 354025 210644875 24.39262 8.41083 0.0016807 596 355216 211708736 24.41311 8.41554 0.0016779 597 356409 212776173 24.43358 8.42025 0.0016750 598 357604 213847192 24.45404 8.42494 0.0016722 599 358801 214921799 24.47448 8.42964 0.0016694 600 360000 216000000 24.49490 8.43433 0.0016667 601 361201 217081801 24.51530 8.43901 0.0016639 602 362404 218167208 24.53569 8.44369 0.0016611 603 363609 219256227 24.55606 8.44836 0.0016584 604 364816 220348864 24.57641 8.45303 0.0016556 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 53 n n 2 n 3 */n 3 v n 1 n 605 366025 221445125 24.59675 8.45769 0.0016529 606 367236 222545016 24.61707 8.46235 0.0016502 607 368449 223648543 24.63737 8.46700 0.0016474 608 369664 224755712 24.65766 8.47165 0.0016447 609 370881 225866529 24.67793 8.47629 0.0016420 610 372100 226981000 24.69818 8.48093 0.0016393 611 373321 228099131 24.71841 8.48556 0.0016367 612 374544 229220928 24.73863 8.49018 0.0016340 613 375769 230346397 24.75884 8.49481 0.0016313 614 376996 231475544 24.77902 8.49942 0.0016287 615 378225 232608375 24.79919 8.50404 0.0016260 616 379456 233744896 24.81935 8.50864 0.0016234 617 380689 234885113 24.83948 8.51324 0.0016207 618 381924 236029032 24.85961 8.51784 0.0016181 619 383161 237176659 24.87971 8.52243 0.0016155 620 384400 238328000 24.89980 8.52702 0.0016129 621 385641 239483061 24.91987 8.53160 0.0016103 622 386884 240641848 24.93993 8.53618 0.0016077 623 388129 241804367 24.95997 8.54075 0.0016051 624 389376 242970624 24.97999 8.54532 0.0016026 625 390625 244140625 25 8.54988 0.0016000 626 391876 245314376 25.01999 8.55444 0.0015974 627 393129 246491883 25.03997 8.55899 0.0015949 628 394384 247673152 25.05993 8.56354 0.0015924 629 395641 248858189 25.07987 8.56808 0.0015898 630 396900 250047000 25.09980 8.57262 0.0015873 631 398161 251239591 25.11971 8.57715 0.0015848 632 399424 252435968 25.13961 8.58168 0.0015823 633 400689 253636137 25.15949 8.58622 0.0015798 634 401956 254840104 25.17936 8.59072 0.0015773 635 403225 256047875 25.19921 8.59524 0.0015748 636 404496 257259456 25.21904 8.59975 0.0015723 637 405769 258474853 25.23886 8.60425 0.0015699 638 407044 259694072 25.25866 8.60875 0.0015674 639 408321 260917119 25.27845 8.61325 0.0015649 640 409600 262144000 25.29822 8.61774 0.0015625 641 410881 263374721 25.31798 8.62222 0.0015601 642 412164 264609288 25.33772 8.62671 0.0015576 643 413449 265847707 25.35744 8.63118 0.0015552 644 414736 267089984 25.37716 8.63566 0.0015528 54 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n ft 2 n 3 VrT v n 1 n 645 416025 268336125 25.39685 8.64012 0.0015504 646 417316 269586136 25.41653 8.64459 0.0015480 647 418609 270840023 25.43619 8.64904 0.0015456 648 419904 272097792 25.45584 8.65350 0.0015432 649 421201 273359449 35.47548 8.65795 0.0015408 650 422500 274625000 25.49510 8.66239 0.0015385 651 423801 275894451 25.51470 8.66683 0.0015361 652 425104 277167808 25.53429 8.67127 0.0015337 653 426409 278445077 25.55386 8.67570 0.0015314 654 427716 279726264 25.57342 8.68012 0.0015291 655 429025 281011375 25.59297 8.68455 0.0015267 656 430336 282300416 25.61250 8.68896 0.0015244 657 431649 283593393 25.63201 8.69338 0.0015221 658 432964 284890312 25.65151 8.69778 0.0015198 659 434281 286191179 25.67100 8.70219 0.0015175 660 435600 287496000 25.69047 8.70659 0015152 661 436921 288804781 25.70992 8.71098 0.0015129 662 438244 290117528 25.72936 8.71537 0.0015106 663 439569 291434247 25.74879 8.71976 0.0015083 664 440896 292754944 25.76820 8.72414 0.0015060 665 442225 294079625 25. 78749 8.72852 0.0015038 666 443556 295408296 25.80698 8.73289 0.0015015 667 444889 296740963 25.82634 8.73726 0.0014993 668 446224 298077632 25.84570 8.74162 0.0014970 669 447561 299418309 25.86503 8.74598 0.0014948 670 448900 300763000 25.88436 8.75034 0.0014925 671 450241 302111711 25.90367 8.75469 0.0014903 672 451584 303464448 25.92296 8.75904 0.0014881 673 452929 304821217 25.94224 8.76338 0.0014859 674 454276 306182024 25.96151 8.76772 0.0014837 675 455625 307546875 25.98076 8.77205 0.0014815 676 456976 308915776 26 8.77638 0.0014793 677 458329 310288733 26.01922 8.78071 0.0014771 678 459684 311665752 26.03843 8.78503 0.0014749 679 461041 313046839 26.05763 8.78935 0.0014728 680 462400 314432000 26.07681 8.79366 0.0014706 681 463761 315821241 26.09598 8.79797 0.0014684 682 465124 317214568 26.11513 8.80227 0.0014663 683 466489 318611987 26.13427 8.80657 0.0014641 684 467856 320013504 26.15339 8.81087 0.0014620 SQUARES, CUBES, ROOTS, AND RECIPROCALS. ss n n 2 n 3 Vn vnr 1 n 685 469225 321419125 26.17250 8.81516 0.0014599 686 470596 32282885(5 26.19160 8.81945 0.0014577 687 471969 324242703 26.21068 8.82373 0.0014556 688 473344 325660672 26.22975 8.82801 0.0014535 689 474721 327082769 26.24881 8.83229 0.0014514 690 476100 328509000 26.26785 8.83656 0.0014493 691 i 477481 329939371 26.28688 8.84082 0.0014472 692 478864 331373888 26.30589 8.84509 0.0014451 693 480249 332812557 26.32489 8.84934 0.0014430 694 i 481636 334255384 26.34388 8.85360 0.0014409 695 483025 335702375 26.36285 8.85785 0.0014388 696 484416 337153536 26.38181 8.86210 0.0014368 697 ; 485809 338608873 26.40076 8.86634 0.0014347 698 487204 340068392 26.41969 8.87058 0.0014327 699 488601 341532099 26.43861 8.87481 0.0014306 700 490000 343000000 26.45751 8.87904 0.0014286 701 491401 344472101 26.47640 8.88327 0.0014265 702 492804 345948408 26.49528 8.88749 0.0014245 703 494209 347428927 26.51415 8.89171 0.0014225 704 495616 348913664 26.53300 8.89592 0.0014205 705 497025 350402625 26.55184 8.90013 0.0014184 706 498436 351895816 26.57066 8.90434 0.0014164 707 499849 353393243 26.58947 8.90854 0.0014144 708 501264 354894912 26.60817 8.91274 0.0014124 709 502681 i 356400829 26.62705 8.91693 0.0014104 710 504100 357911000 26.64583 8.92112 0.0014085 711 505521 359425431 26.66458 8.92531 0.0014065 712 506944 360944128 26.68333 8.92949 0.0014045 713 508369 362467097 26.70206 8.93367 0.0014025 714 509796 363994344 26.72078 8.93784 0.0014006 715 511225 365525875 26.73948 8.94201 0.0013986 716 512656 367061696 26.75818 8.94618 0.0013966 717 514089 368601813 26.77686 8.95034 0.0013947 718 515524 370146232 26.79552 8.95450 0.0013928 719 516961 371694959 26.81418 8.95866 0.0013908 720 518400 373248000 26.83282 8.96281 0.0013889 721 519841 374805361 26.85144 8.96696 0.0013870 722 521284 37(5367048 26.8700(5 8.97110 0.0013850 723 522729 377933067 26.88866 8.97524 0.0013831 724 524176 379503424 26.90725 8.97938 0.0013812 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n W 2 n 3 V n V-T 1 n 725 525625 381078125 26.92582 8.98351 0.0013793 726 527076 382657176 26.94439 8.98764 0.0013774 727 528529 384240583 26.96294 8.99176 0.0013755 728 529984 385828352 26.98148 8.99589 0.0013736 729 531441 387420489 27 9 0.0013717 730 532900 389017000 27.01851 9.00411 0.0013699 731 534361 390617891 27.03701 9.00822 0.0013680 732 535824 392223168 27.05550 9.01233 0.0013661 733 537289 393832837 27.07397 9.01643 0.0013643 734 538756 395446904 27.09243 9.02053 0.0013624 735 540225 397065375 27.11088 9.02462 0.0013605 736 541696 398688256 27.12932 9.02871 0.0013587 737 543169 400315553 27.14771 9.03280 0.0013569 738 544644 401947272 27.16616 9.03689 0.0013550 739 546121 403583419 27.18455 9.04097 0.0013532 740 547600 405224000 27.20291 9.04504 0.0013514 741 549081 406869021 27.22132 9.04911 0.0013495 742 550564 408518488 27.23968 9.05318 0.0013477 743 552049 410172407 27.25803 9.05725 0.0013459 744 553536 411830784 27.27636 9.06131 0.0013441 745 555025 413493625 27.29469 9.06537 0.0013423 746 556516 415160936 27.31300 9.06942 0.0013405 747 558009 416832723 27.33130 9.07347 0.0013387 748 559504 418508992 27.34959 9.07752 0.0013369 749 561001 420189749 27.36786 9.08156 0.0013351 750 562500 421875000 27.38613 9.08560 0.0013333 751 564001 423564751 27.40438 9.08964 0.0013316 752 565504 425259008 27.42262 9.09367 0.0013298 753 567009 426957777 27.44085 9.09770 0.0013280 754 568516 428661064 27.45906 9.10173 0.0013263 755 570025 430368875 27.47726 9.10575 0.0013245 756 571536 432081216 27.49545 9.10977 0.0013228 757 573049 433798093 27.51363 9.11378 0.0013210 758 574564 435519512 27.53180 9.11779 0.0013193 759 576081 437245479 27.54995 9.12180 0.0013175 760 577600 438976000 27.56810 9.12581 0.0013158 761 579121 440711081 27.58623 9.12981 0.0013141 762 580644 442450728 27.60435 9.13380 0.0013123 763 582169 444194947 27.62245 9.13780 0.0013106 764 583696 445943744 27.64055 9.14179 0.0013089 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 57 n n 2 n 3 VT 3 Vn 1 n 765 585225 447697125 27.65863 9.14577 0.0013072 766 586756 449455096 27.67671 9.14976 0.0013055 767 588289 451217663 27.69476 9.15374 0.0013038 768 589824 452984832 27.71281 9.15771 0.0013021 769 591361 454756609 27.73085 9.16169 0.0013004 770 592900 456533000 27.74887 9.16566 0.0012987 771 594441 458314011 27.76689 9.16962 0.0012970 772 595984 460099648 27.78489 9.17359 0.0012953 773 597529 461889917 27.80288 9.17754 0.0012937 774 599076 463684S24 27.82086 9.18150 0.0012920 775 600625 465484375 27.83882 9.18545 0.0012903 776 602176 467288576 27.85678 9.18940 0.0012887 777 603729 469097433 27.87472 9.19335 0.0012870 778 605284 470910952 27.89265 9.19729 0.0012853 779 606841 472729139 27.91057 9.20123 0.0012837 780 608400 474552000 27.92848 9.20516 0.0012821 781 609961 476379541 27.94638 9.20910 0.0012804 782 611524 478211768 27.96426 9.21303 0.0012788 783 613089 480048687 27.98214 9.21695 0.0012771 784 614656 481890304 28 9.22087 0.0012755 785 616225 483736625 28.01785 9.22479 0.0012739 786 617796 485587656 28.03569 9.22871 0.0012723 787 619369 487443403 28.05352 9.23262 0.0012706 788 620944 489303872 28.07134 9.22653 0.0012690 789 622521 491169069 28.08914 9.24043 0.0012674 790 624100 493039000 28.10694 9.24434 0.0012658 791 625681 494913671 28.12472 9.24823 0.0012642 792 627264 496793088 28.14249 9.25213 0.0012629 793 628849 498677257 28.16026 9.25602 0.0012610 794 630436 500566184 28.17801 9.25991 0.0012594 795 632025 502459875 28.19574 9.26380 0.0012579 796 633616 504358336 28.21347 9.26768 0.0012563 797 635209 506261573 28.23119 9.27156 0.0012547 798 ! 636804 508169592 28.24889 9.27544 0.0012531 799 638401 510082399 28.26659 9.27931 0.0012516 800 640000 512000000 28.28427 9.28318 0.0012500 801 641601 513922401 28.30194 9.28704 0.0012484 802 643204 515849608 28.31960 9.29091 0.0012469 803 644809 517781627 28.33725 9.29477 0.0012453 804 646416 519718464 28.35489 9.29862 0.0012438 SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 W 3 V n V^T 1 n 805 648025 521660125 28.37252 9.30248 0.0012422 806 649636 523606616 28.39014 9.30638 0.0012407 807 651249 525557943 28.40775 9.31018 0.0012392 808 652864 527514112 28.42534 9.31402 0.0012376 809 654481 529475129 28.44293 9.31786 0.0012361 810 656100 531441000 28.46050 9.32170 0.0012346 811 657721 533411731 28.47806 9.32553 0.0012330 812 659344 535387328 28.49561 9.32936 0.0012315 813 660969 537367797 28.51315 9.33319 0.0012300 814 662596 539353144 28.53069 9.33702 0.0012285 815 664225 541343375 28.54820 9.34084 0.0012270 816 665856 543338496 28.56571 9.34466 0.0012255 817 667489 545338513 28.58321 9.34847 0.0012240 818 669124 547343432 28.60070 9.35229 0.0012225 819 670761 549353259 28.61818 9.35610 0.0012210 820 672400 551368000 28.63564 9.35990 0.0012195 821 674041 553387661 28.65310 9.36270 0.0012180 822 675684 555412248 28.67054 9.36751 0.0012165 823 677329 557441767 28.68798 9.37130 0.0012151 824 678976 559476224 28.70540 9.37510 0.0012136 825 680625 56 J 51 5625 28.72281 9.37889 0.0012121 826 682276 563559976 28.74022 9.38268 0.0012107 827 683929 565609283 28.75761 9.38646 0.0012092 828 685584 567663552 28.77499 9.39024 0.0012077 829 687241 569722789 28.79236 9.39402 0.0012063 830 688900 571787000 28.80972 9.39780 0.0012048 831 690561 573856191 28.82707 9.40157 0.0012034 832 692224 575930368 28.84441 9.40534 0.0012019 833 693889 578009537 28.86174 9.40911 0.0012005 834 695556 580093704 28.87906 9.41287 0.0011990 835 697225 582182875 28.89637 9.41663 0.0011976 836 698896 584277056 28.91366 9.42039 0.0011962 837 700569 586376253 28.93095 9.42414 0.0011947 838 702244 588480472 28.94823 9.42789 0.0011933 839 703921 590589719 28.96550 9.43164 0.0011919 840 705600 592704000 28.98275 9.43538 0.0011905 841 707281 594823321 29 9.43913 0.0011891 842 708964 596947688 29.01724 9.44287 0.0011876 843 710649 599077107 29.03446 9.44661 0.0011862 844 712336 601211584 I 29.05168 9.45034 0.0011848 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 59 n n 2 n 3 Vn ^ 1 n 845 714025 603351125 29.06888 9.45407 0.0011834 846 715710 605495736 29.08608 9.45780 0.0011820 847 717409 (507645423 29.10326 9.46152 0.0011806 848 719104 609800192 29.12044 9.46525 0.0011792 849 720801 611900049 29.13760 9.40897 0.0011779 850 722500 014125000 29.1547(5 9.47268 0.0011765 851 724201 010295051 29.17190 9.47640 0.0011751 852 725904 618470208 29.18904 9.48011 0.0011737 853 727009 020(550477 29.20610 9.48381 0.0011723 854 729310 622835804 29.22328 9.48752 0.0011710 855 731025 025020375 29.24038 9.49122 0.0011696 850 73273(5 027222010 29.25748 9.49492 0.0011682 857 734449 029422793 29.27456 9.49861 0.0011669 858 730104 031028712 29.29164 9.50231 0.0011655 859 737881 033839779 29.30870 9.50600 0.0011641 800 739000 030050000 29.32576 9.50969 0.0011628 801 741321 03X277381 29.34280 9.51337 0.0011614 802 743044 64050392S 29.35984 9.51705 0.0011601 868 744709 042735047 29.37686 9.52073 0.001 1587 H54 740490 044972544 29.39388 9.52441 0.0011574 805 748225 647214025 29.41088 9.52808 0.0011501 800 749950 049401890 29.42788 9.53175 0.0011547 807 751089 051714303 29.44486 9.53542 0.0011534 8(58 753424 053972032 29.40184 9.53908 0.0011521 809 755101 050234909 29.47881 9.54274 0.0011507 870 750900 058503000 29.49570 9.54640 0.0011494 871 758041 000770311 29.51271 9.55006 0.0011481 872 700384 663054848 29.52905 9.55371 0.0011408 873 702129 665338617 29.54057 9.55736 0.0011455 874 703870 667627624 ^9.56349 9.56101 0.0011442 875 765025 669921875 29.58040 9.56466 0.0011429 870 767376 672221376 29.59730 9.56830 0.0011410 877 769129 674526133 29.61419 9.57194 0.0011403 878 770884 676830152 29.63106 9.57557 0.0011390 879 772641 679151439 29.64793 9.57921 0.0011377 880 774400 681472000 29.66479 9.58284 0.0011364 881 776161 OS3797841 29.68164 9.58647 0.0011351 882 777924 686128968 29.69848 9.59009 0.0011338 883 779688 088405387 29.71532 9.59372 0.0011325 884 781450 690807104 29.73214 9.59734 0.0011312 6o SQUARES, CUBES, ROOTS, AND RECIPROCALS. n n 2 n 3 V n v^T 1 n 885 783225 693154125 29.74895 9.60095 0.0011299 886 784996 695506456 29.76575 9.60457 0.0011287 887 786769 697864103 29.78255 9.60818 0.0011274 888 788544 700227072 29.79933 9.61179 0.0011261 889 790321 702595369 29.81610 9.61540 0.0011249 890 792100' 704969000 29.83287 9.619 0.0011236 891 793881 707347971 29.84962 9.62260 0.0011223 892 795664 709732288 29.86637 9.62620 0.0011211 893 797449 712121957 29.88311 9.62980 0.0011198 894 799236 714516984 29.89983 9.63339 0.0011186 895 801025 716917375 29.91655 9.63698 0.0011173 896 802816 719323136 29.93326 9.64057 0.0011161 897 804609 721734273 29.94996 9.64415 0.0011148 898 806404 724150792 29.96665 9.64774 0.0011136 899 808201 726572699 29.98333 9.65132 0.0011123 900 810000 729000000 30 9.65489 0.0011111 901 811801 731432701 30.01666 9.65847 0.0011099 902 813604 733870808 30.03331 9.66204 0.0011086 903 815409 736314327 30.04996 9.66561 0.0011074 904 817216 738763264 30.06659 9.66918 0.0011062 905 819025 741217625 30.08322 9.67274 0.0011050 906 820836 743677416 30.09983 9.67630 0.0011038 907 822649 746142643 30.11644 9.67986 0.0011025 908 824464 748613312 30.13304 9.68342 0.0011013 909 826281 751089429 30.14963 9.68697 0.0011001 910 828100 753571000 30.16621 9.69052 0.0010989 911 829921 756058031 30.18278 9.69407 0.0010977 912 831744 758550528 30.19934 9.69762 0.0010965 913 833569 761048497 30.21589 9.70116 0.0010953 914 835396 763551944 30.23243 9.70470 0.0010941 915 837225 766060875 30.24897 9.70824 0.0010929 916 839056 768575296 30.26549 9.71177 0.0010917 917 840889 771095213 30.28201 9.71531 0.0010905 918 842714 773620632 30.29851 9.71884 0.0010893 919 844561 776151559 30.31501 9.72236 0.0010881 920 846400 778688000 30.33150 9.72589 0.0010870 921 848241 781229961 30.34798 9.72941 0.0010858 922 850084 783777448 30.36445 9.73293 0.0010846 923 851929 786330467 30.38092 9.73645 0.0010834 924 853776 788889024 30.39737 9.73996 0.0010823 SQUARES, CUBES, ROOTS, AND RECIPROCALS. 6 1 n n 2 n 3 \fn~
" in the tables. The difference in this case is given in the table as 119. EXAMPLE 2. Find logarithm to 1892.5. Solution : The mantissa of the number 1892 is given in the table as 276921. The difference is given as 229. The index for a num- ber consisting of four integers is 3. Thus : Log. 1892 3.276921 0.5 X 0.000229 = 0.0001145 = 115 Log. 1892.5 = 3.277036 EXAMPLE 3. Find logarithm to 85673. Solution : The mantissa for the number 85670 is given in the table as 932829. The difference is given in the table as 51. The index for a number consisting of five integers is 4. When an increase of 10 in the number increases the log- arithm 0.000051 an increase of 3 must increase the correspond- ing logarithm 0.3 times 0.000051. Thus: Log. 85670 = 4.932829 0.3 X 0.000051 = 0.0000153 = 15 Log. 85673 = 4.932844 These calculations (or- interpolations as they are usually called) are based upon the principle that the difference between the numbers and the difference between their corresponding logarithms are directly proportional to each other. This, how- ever, is not strictly true ; but within limits, as it is used here, it is near enough for practical results. To Find the Number Corresponding to a Given Logarithm. EXAMPLE 1. Find the number corresponding to the logarithm 2.610979. Solution : Always remember when looking for the number not to consider the index, but find the mantissa 610979 in the table. In the same line as this mantissa, under the heading " N," is 408, and on the top of the table in the same column as this mantissa is 3; thus, the number corresponding to this mantissa is 4083 and the in4ex of the logarithm is 2 \ consequently the LOGARITHMS. 75 number is to have three figures on the left-hand side of the decimal point; thus, the number corresponding to the logarithm 2.610979 will be 408.3. EXAMPLE 2. Find the number corresponding to the logarithm 3.883991. Solution : This mantissa is not in the table. The nearest smaller mantissa is 883945, and to this mantissa corresponds the number 7055. The nearest larger mantissa is 884002, and to this corresponds the number 7656. Thus, an increment in the mantissa of 57 increases the number by 1, but the difference between the mantissa 883945 and 883991 is 46, therefore the number must increase if =0.807. Number of Log. 3.883945 = 7655 Difference 0.000046 = 0.807 Number of Log. 3.883991 = 7655.807 Addition of Logarithms. (MULTIPLICATION.) Where the logarithms of the factors have positive indexes, add as if they were decimal fractions, and the sum is the log- arithm corresponding to the product. EXAMPLE 1. Multiply 81 by 65 by means of logarithms. Solution : Log. 81 = 1.908485 Log. 65 = 1.812913 3.721398 and to this mantissa corresponds the number 5265. The index is 3; therefore the number has no decimals, as it consists of only four figures. To Add Two Logarithms when One Has a Positive and the Other a Negative Index. EXAMPLE 2. Multiply 0.58 by 32.6 by means of logarithms. Solution ; Log. 0.58 = 9,763428 10 , 38.8 as 1.513218 11,270646 ~~10 76 LOGARITHMS. This reduces to 1.276646 and to this logarithm corresponds the number 18.908. This mantissa, 276646, cannot be found in the table, but the nearest smaller mantissa is 276462, and the differ- ence between this and the next is found by subtraction to be 230, and the difference between this and the given mantissa is 184. Thus: Given logarithm 1.276646 To the tabulated log. 1.276462 corresponds 18.90 Difference 0.000184 gives _i _i Thus, logarithm 1.276646 gives number 18J908 To Add Two Logarithms, Both Having a Negative Index. Add both logarithms in the same manner as decimal frac- tions, and afterwards subtract 10 from the index on each side of the mantissa. EXAMPLE. Multiply 0.82 by 0.082 by means of logarithms. Solution : Log. 0.82 = 9.913814 10 Log. 0.082 = 8.913814 10 18.827628 20 By subtracting 10 on each side of the mantissa this logar- ithm reduces to 8.827628 10 and to the mantissa 827628 corre- sponds the number 6724, but the negative index 8 10 indicates that this first figure 6 is not a whole number, but that it is six-hundredths ; therefore a cipher must be placed between this 6 and the decimal point in order to give 6 the right value according to the index; thus, to the logarithm 8.827628 10 corresponds the number 0.06724. Subtraction of Logarithms. (DIVISION.) Logarithms are subtracted as common decimal fractions. To Subtract Two Logarithms, Both Having a Positive Index. EXAMPLE. Divide 490 by 70 by means of logarithms. Solution : Log. 490 = 2.690196 Log. 70 1.845098 0.845098 LOGARITHMS. 77 and to the mantissa of this logarithm corresponds the number 7 or 70 or 700 or 7000, etc., in the table of logarithms, but the index of this logarithm is a cipher ; therefore the answer must be a number consisting of one figure, thus it must be 7. To Subtract a Larger Logarithm From a Smaller One. This is the same as to divide a smaller number by a larger one. Before the subtraction is commenced add 10 to the index of the smaller logarithm (that is, to the minuend) and place 10 after the mantissa, then proceed with the subtraction as if they were decimal fractions. EXAMPLE. Divide 242 by 367 by means of logarithms. Solution : Log. 242 = 2.383815 = 12.383815 10 Log. 367 = 2.564666 9.819149 10 and to the mantissa of this logarithm corresponds, according to the table, the number 6594, but the negative index, 9 10, indicates it to be 0.6594. Thus, 242 divided by 367 = 0.6594. Multiplication of Logarithms. (INVOLUTION.) To multiply a logarithm is the same as to raise its corre- sponding number into the power of the multiplier. Logarithms having a positive index are multiplied the same as decimal fractions. Thus : Square 224 by means of logarithms. Solution: 2 X log. 224 = 2 X 2.350248 = 4.700496 = 50176 Logarithms having a negative index are multiplied the same as decimal fractions, but an equal number is subtracted from both the positive and the negative parts of the logarithm, in order to bring the negative part of the index to 10. EXAMPLE 1. Square 0.82 by means of logarithms. Solution : 2 X log. 0.82 = 2 X (9.913814 10) = 19.827628 10, and subtracting 10 from both the positive and the negative parts of the logarithm, the result is 9,827628 10 ; this gives the num- ber 0.6724, 7 8 LOGARITHMS. EXAMPLE 2. Raise 0.9 to the 1.41 power. Solution : 1.41 X log. 0.9 = 1.41 X (9.954243 10) = 14.035483 14.1 In this example 10 cannot be subtracted from both parts of the logarithm, but 4.1 must be subtracted in order to get 10, after the subtraction is performed. The logarithm will then read 9.935483 10, which corresponds to the number 86195, and the negative index, 9 10, makes this 0.86195. Division of Logarithms. (EVOLUTION.) To divide a logarithm is the same as to extract a root of the number corresponding to the logarithm. Logarithms having a positive index are divided the same as common decimal fractions. EXAMPLE. Extract the cube root of 512 by means of logarithms. Solution : ** 512 = 2.70927 = 30 O and the number corresponding to this logarithm is 8, 80, 800, 8,000, etc., but the index of this logarithm is a cipher ; there- fore the answer must be a number consisting of one integer, consequently it must be 8. To Divide a Logarithm Having a Negative Index. Select and add such a number to the index as will give 10 without a remainder for the quotient in the negative index on the right-hand side of the mantissa after division is per- formed. EXAMPLE 1. Extract the square root of 0.64 by means of logarithms. = 19^18-20 2 2 and to this logarithm corresponds the number 0.8". EXAMPLE 2. Extract the cube root of 0.125 by means of logarithms. LOGARITHMS. 7 9 Solution : log. 0.125 _ 9.0961)110 _ 29.0969130 _ 9 69g97 _ 1Q 33 3 and to this logarithm corresponds the number 0.5. EXAMPLE 3. Extract the 1.7 root of 0.78. Solution : log. 0.78 9.892095 10 1.7 1.7 We cannot here, as in previous examples, add a multiple of 10 to the index on each side of the mantissa, but 7 must be added in order that the negative quotient shall be 10 after the division is performed. Thus: 9.892095-10 10.892005 -17 1.7 1.7 and to this logarithm corresponds the number 0.864. Short Rules for Figuring by Logarithms. MULTIPLICATION. Add the logarithms of the factors and the sum is the logar- ithm of the product. DIVISION. Subtract divisor's logarithm from the logarithm of the divi- dend and the difference is the logarithm of the quotient. INVOLUTION. Multiply the logarithm of the root by the exponent of the power and the product is the logarithm of the power. EXAMPLE. Log. 86 2 = 2 X log. 86 = 2 X 1.934498 = 3.868996 and to this logarithm corresponds the number 7396. EVOLUTION. The logarithm of the number or quantity under the radical sign is divided by the index of the root, and the quotient is the logarithm of the root. 80 LOGARITHMS. EXAMPLE. 4 log. 2401 3.380392 Log. y 2401 = - "^j = | = 0.845098 and this logarithm corresponds to the number 7. EXPONENTS. The logarithm of a power divided by the logarithm of the root is equal to the exponent of the power. EXAMPLE. 8* = 64 x x = log.S 1.80618 0.90309 x = 2 The logarithm of a quantity under the radical sign divided by the logarithm of the root is equal to the index of the root. EXAMPLE. x _ 8 = \/512 x = l 8' 512 x = 2.70927 0.90309 x =3 The reason for these last rules may be understood by re- ferring to the rules for Involution and Evolution ; for instance : 86 2 = 7396, and this expressed by logarithms is: 2 X log. 86 = log. 7396. Therefore: lo ?' 7896 = 2. log. 86 FRACTIONS. The logarithm of a common fraction is found, either by first reducing the fraction to a decimal fraction, or by taking the logarithm of the numerator and the logarithm of the denominator and subtracting the logarithm of the denominator from the log- arithm of the numerator; the difference is the logarithm of the fraction, LOGARITHMS. 8 1 EXAMPLE. Log. % = log. 3 log. 4 Log. 3 = 0.477121 = 10.477121 10 Log. 4 0.602060 Thus, log. % = 9.875061 10 This is also the logarithm of the decimal fraction 0.75. RECIPROCALS. Subtract the logarithm of the number from log. 1, which is 10.000000 10, and the difference is the logarithm of the reci- procal. EXAMPLE. Find the reciprocal of 315. Solution : Log. 1 = 10.DOOOOO 10 Log. 315 2.498311 Log. reciprocal of 315 = 7.501689 10 To this logarithm corresponds the decimal fraction 0.0031746, which is, therefore, the reciprocal of 315. Simple Interest by Logarithms. Add logarithm of principal, logarithm of rate of interest, and logarithm of number of years ; from this sum subtract log- arithm of 100. The difference is the logarithm of the interest. EXAMPLE. Find the interest of #800 at 4% in 5 years. Solution : Log. 800 = 2.90309 Log. 4 = 0.60206 0.69897 40412 Log. 100 = 2.00000 Log. interest = 2.20412 = $160 = Interest. #2 LOGARITHMS. Compound Interest by Logarithms. When the interest, at the end of each period of time, is added to the principal the amount will increase at a constant rate; and this rate will be the amount of one dollar invested for one period of the time. For instance : If the periods of time be one year each, then $30 in 3 years at 5 % compound interest will be : $30 X 1.05 = $31.50 at the end of first year. $31.50 X 1.05 = $33.075 at the end of second year. $33.075 X 1.05 = $34.73 at the end of third year. This calculation may be written : $30 X 1.05 X 1.05 X 1.05 = $34.73 which also may be written $30 X (1.05) 3 = $34.73. Thus, compound interest is a form of geometrical progres- sion, and may be calculated by the following formulas : a =p X r" Log, a == n X log. r + log. p Log. p = log. a n X log. r log. r Log. r = !og. n p = Principal invested. n = The number of periods of time. a = The amount due after n periods of time. r = The amount of $1 invested one period of time. NOTE. The quantity r is always obtained by the rule : Divide the rate of interest per period of time by 100, and add 1 to the quotient. EXAMPLE. What is the amount of $816 invested 6 years at 4% com- pound interest? LOGARITHMS. 83 Solution by formula : Log. a = n X log. r -j- log. p Log. a = 6 X log. 1.04 + log. 816 Log. a = 6 X 0.017033+ 2.911690 Log. a = 0.102198 + 2.911690 Log. a = 3.013888 a = $1032.49 = Amount. EXAMPLE. If $750 is invested at 3% compound interest, how many years will it take before the amount will be $950. Solution by formula: n= log- a log. r lo &- 95 ~ l - 75 log. 1.03 2.977724 - 0.012837 EXAMPLE. A principal of $3750 is to be invested so that by compound interest it will amount to $5000 in six years. Find rate of interest. Solution by formula : Log. r = Lo r = n 3-698970 3.574031 6 Log. r = 0.020823 r = 1.0491 Rate of interest = lOOr 100 = 100 X 1.0491 100 = 4.91% ; or 5 % per year (very nearly). Discount or Rebate. When calculating discount or rebate, which is a deduction upon money paid before it is due, use formula: Log. p = log. a n X log. r 84 LOGARITHMS. EXAMPLE. A bill of $500 is due in 3 years. How much cash is it worth if 3% compound interest should be deducted. Log. p = log, a n X log. r Log.p = 2.698970 3 X 0.012837 Log.p = 2.698970 0.038511 Log.p = 2.660459 p = $457.57 = Cash payment. NOTE. Such examples may be checked to prevent miscal- culations, by multiplying the result (the cash payment), by the tabular number given for corresponding number of years and percentage of interest in table on page 23 ; if calculations are correct, the product will be equal to the original bill. For in- stance, 457.57 X 1.092727 = 499.99909339 = $500.00. Thus, the calculation in the example is correct. Sinking Funds and Savings. If a sum of money denoted by , set apart or saved dur- ing each period of time, is put at compound interest at the end of each period, the amount will be : a = b at the end of the first period. a = b + br at the end of the second period. a = b + br + bi* at the end of the third period. At the end of n periods the last term in this geometrical series is br n ~ l and the first term is , while the ratio is r. The sum of the series is the amount which according to the rules for geometrical progression (see page 69) will be : r 1 r 1 EXAMPLE. At the end of his first year's business a man sets apart $1200 for a sinking fund, which he invests at 4% per year. At the end of each succeeding year he sets apart $1200 which is invested at the same rate. What is the value of the sinking fund after 7 years of business ? Solution : = 1200 X (1.04 7 1) 1.04 1 a = 1200 X 0.31593 0.04 a = $9477.90 LOGARITHMS. 85 EXAMPLE. A man 20 years old commences to save 25 cents every working day, and places this in a savings bank at 4% interest, computed semi-annually. How much will he have in the bank when he is 36 years old? (NOTE. 25c. a day = #1.50 a week = 26 X $1.50 = $39 in six months. 4 % per year = 2 % per period of time ; 36 20 = 16 = 32 periods of time). Solution by formula: r 1 = 39 X (1.02 32 1) 1.02 1 = 39 X (1.8845 1) 0.02 a = 39 X 0.8845 X 50 a = 1724.775 = $1724.77 = Amount. Thus, in 16 years a saving of 25c. a day amounts to $1724.77. If the money is paid in advance of the first period of time the terms will be : a = br at the end of the first period. a = br + br* at the end of the second period. a = br + br* -\- br* at the end of the third period. At the end of n years the last term in this geometrical series is br 11 and the first term is br, while the ratio is r. The sum of the series is the amount, which, according to rules for geometrical progressions (see page 69), will be : r r ~~ br r 1 _ br (r n _ 1} r 1 EXAMPLE. Assume that the man mentioned in previous example, instead of commencing to save money when 20 years old, already had $39 to put in the bank at 4 % the first period of time, and that he always kept up paying $39 in advance semi- annually. How much money would he then save in 16 years ? 86 LOGARITHMS. Solution : = 39 X 1.02 X (102 32 1) 1.02 1 _ 39 X 1.02 X 0.8845 ~002- a = 1759.27 Thus, by paying the money in advance semi-annually, he will gain (1759.27 1724.77) = #34.50. If a principal denoted by P is invested at a given rate of compound interest, and successive smaller or larger equal pay- ments denoted by b are made at the end of each period of time so that they will commence to draw interest at the beginning of the following period at the same rate as the principal, the formula will be : but for logarithmic calculations it is more convenient to denote the rate of interest by y % and the formula will read : a = p X r n + ~J~ 100 a = j xr * + 100*(r-l) y b _ aypy X r n 100 r n 100 r n = . a.y + 100* py + 100 6 4- 100 b = log. r lo? * y + 10 * *' p y + 100 ^ - - - S - NOTE. Using these formulas it must be understood that represents the number of periods of time that the principal is invested, and that this first period is considered to be the period at the end of which the first payment, b, is made. EXAMPLE. A man has $50 in a savings bank and he also puts in $25 every month, which goes on interest every 6 months ; the bank pays 4% interest, computed semi-annually. How much money LOGARITHMS. &7 can he save in 5 years in this way? (NOTE. 4% per year = 2% per 6 months, or per period of time, and $25 a month = $150 every 6 months, or per period of time. The interest is computed semi-annually ; therefore 5 years = 10 periods of time). Solution by formula : y a = 50 X 1.02- + 100 X 150 X (102 "-I) a = 50 X 1.219 + 100 X 150 X 0.219 2 a = 60.95 + 1642.50 a = $1703.45 = Amount. The original sum of $50 has increased to $60.95, and the monthly payments amounted to $1500. The last six payments did not draw any interest, as they were deposited in the last six months of the fifth year and would commence to draw inter- est at the beginning of the sixth year if the amount had not been withdrawn. EXAMPLE. A man has $800 invested at 5%. How much must he save and invest at the same interest every year in order to increase it to $3000 in five years ? Interest is computed annually. Solution by formula : b _ ay py X r* 100 r* 100 b __ 3000 X 5 800 X 5 X 1.05 s 100 X 1.05 5 100 b _ 15000 1.2763 X 4000 100 X 1.2763 100 15000 5105.2 b 127.63 100 9894.8 27.63 b = 358.118 = $358.12 to be paid in each year. The total payments will be : 800 + 5 X 358.12 = $800 + $1790.60 = $2590.60. The rest of the amount is accumulated interest. The last pay- ment is made at the end of the fifth year ; therefore this money does not draw interest. 88 LOGARITHMS. EXAMPLE. A man calculates that if he had $1800 he would start in business. He has only $120, but is earning $15 a week and figures that he can save half of his weekly earnings. He puts his money in a savings bank, where it goes on interest every six months, at the rate of 4% a year. How many years will it take him to save the required amount? (NOTE. $7.50 a week == 26 X 7> = $195 in six months, and 4% per year = 2% per six months, or per period of time). Solution by formula : log. n = log. log, r 1800X2 + 100X195 _ 120 X 2 + 100 X 195 log. 1.02 , 3600 + 19500 l S' 240 + 19500 log. 1.02 One period = 6 months ; 8 periods = 4 years ; therefore, under these conditions it takes four years to save this amount of money. If a certain sum of money is withdrawn instead of added, at the end of each period of time, the formula on page 86 will change to : y Every letter denotes the same value as it had in the formula on page 86, except that b represents the sum withdrawn instead of the sum added. EXAMPLE. A man has $5000 invested at 5% interest compounded an- nually, but at the end of each year he withdraws $200. How much money has he left after six years ? Solution : a = 1.05" X 5000 - 100 X 200 X (1.05* - 1) 5 a = 1.34 X 5000 - 10 X X ' 34 a = 6700 1360 a = $5340 = Amount. LOGARITHMS. 89 If the deducted sum, , exceeds the interest due at the first period of time, the amount a will become smaller than the prin- cipal/, and in time the whole principal will be used up. This will be when : 100(r 1) P X r = - -y- This transposes to 100 = log. -py 1003 100 b _ py log.r EXAMPLE. A principal of $5000 is invested at 4% per year, but at the end of each year $600 is withdrawn. How long will it take to use the whole principal ? 100 X 600 _ l S- 100 x 600 5000 X 4 'log. 1.04 60000 60000 20000 n = log. 1.04 log. 1.50 log. 1.04 0.176091 0.017033 n = 10.3 years. Paying a Debt by Instalments. This same formula applies also in this case; for instance: A man uses $1500 every year toward paying a debt of $10,000, and 5% interest per year. How long will it take to pay it? 100 X 1500 ' 100 X 1500 10000 X 5 n = log. 1.05 150000 100000 n= log. 1.05 log* 1.5 log. 1.05 = . 0.176091 0.021189 n = 8.3 years, 9 o LOGARITHMS. 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OS 05 GO 00 GO ^1-COCDCO CO CO CO CO tO OS CD <3O tO iO tO tO tO tO iO iO O iO tO tO iO CO O CO O CO "f "^H OO CO-t^cMO^- tOCMOSCOCO >OCOTHO1CO TtO tO tO tO tO tO iO tO tO GO T-H CO CO ^M Oi ^O 3 tOOSTHCMCMOGOCO COi-HOGOi- tOCOTHGOCO ^tii-HXtO'N OSCD.T^XtO T-H t- *0 GO Tfi ^NOXtOCO THOSI t^(M OXtOCOrH XCO-fr-iOS J^TfifMOil^ TH '2^ 'M CO -t tO >O CO I GO OS Ci O ' CM O iO CO *- X X OS 10 >0 >0 iO >0 tO iO to (N OS >O O CO O CO CM TH OS TH OS L >O CM OX TH TH i^l^ GO OS OS O rH C^^CO^flO O CD t- GO GO O IT- GO CO i- rfl O iO OS rH O O OS -* GO QOGOCD"^O O GO rH O rHOGOOSOS GOlO(Nt-rH _ _ GO r ^* |^ O CO CO OS rH ^ t-^ rHOSCO^r-t GOCOCOOGO iOOCOCDt-QO co co cococot-r~ t-t-t-t^. t- t- i t- 1- S Od t-(MOCOb- CD-^rHI-rH GOC<|lOGOrH -^1 t O 0^1 >O (M OS 1- ^ rH TOrHQO OCOOt->O (M OS 1 Tf rH OSOO rH(MCOCO-^ O O CO t GO r- oc 05 o H r- r- r- 00 QO 00 tQ tQ IQ >C 10 >4 O O 1Q IQ no LOGARITHMS. o 10 OS OS OS OS CO GO GO CO ^ C iO GO CO GO 00 GO GO GO O CO OS CO Tj< O O CO o IM os co co CO t i CO OS 00 CO GO CO CO -# GO O TH CM t- I GO GO GO O Jt "HH -rH GO O O TH 2<) CM os os os os os o CM cs to co CO HH H< >o CD os os os os os t- Jt- 1 t- 1 GO OS OS CO CO O CM -H/i CO GO CO O 1 ^ ^-H OS O O TH CM t- CO 00 CO CO (N GO CM lO t- n co Tt 10 o GO GO CO CO GO t- i o co os co t- .x os os -* i-H CO iO (M CO t- t- GO GO GO i^ r- i- "* t- O ~ O T I co 4t ^t >o cb os os os os - lO Jt OS TH i-i CD i l Tf CO ^ CO t GO lO 'M OS fN CO rfH -t 00 GO GO 00 OS O TH CM CM CO -H CO lO CM CO t- Jt- CO ~ GO 00 00 GO . . -t CO CO CO 'N COCOOt- HH -H CO O (M O-r-triCM CO-*-tccO OS OS OS OS OS OS OS OS OS HH CM OS IO OS ^H t- co r os H -^ O 1~ CO GO r- ^- co co 10 t- - 1- t- 1- -H CO CO CO *H OS ^H CO lO t- QO O iM OS GO CO CO TH CO CO O TH CO "^ CO Tt< rH CO >O CM CO TJH tn O GO OO OO GO 00 t- 1- 1- Jt- 1- - t- 1- i t- TH O t- ^ OS CO CO CM CM ( CM OS CO CO O CO CO ^f lO CO OS OS OS OS OS ;M o GO ^ os CO CD iC C ^t 4 TH GO ift C O CO 1-1 OS CO <* 1-1 OS CO O t -rf ^H t ^ i I X " CC JO I "t I-H CO >O i-H CO O vTJ t- 1 CO OS OS I-H I-H 71 CO OSOSOiOSOS OOOOO ~ t CO CO CO GO CO O CO t- CO iO co i~i os i >o OS CO 71 OS CO CO -t O O CO X X X X X Tf X TH CO "tf "* CO i-H CO O X O CO O i~ ^t< I-H X CO O CO CO O i CO ~ t^~ X X OS O X X X X X S X X X X X IT* X OS X 1^* "^ I~H CO O *^ CO Jt^* i^* CO "^ -fTlOX CO'ti-HOSCO COOt-^^H iO7lOSO (MO5CO71CS COCOOSCOCO OCO t-t-XOSOS ^^^^ XX X X X X X XOi-ii-HOS t-COOSCOCO OSXCO-*i-l OSt~--t71OS _ , . ._ t- rf 1-t X ^O i-H X C 71 X lO 71 OS O 71 CO -t C i-O CO t- 1 X OS OS O T-I T- C t- -t 1 1-1 X CO 1 GO X OS OS OS OS OS T^COCOi-l CO --I OS 1 O i -^ O t -- f i i 71 CO CO -t 1 O O GOCOCO COCOCCGO 'TlOOCOt- OS 1 CO ^ OS CO CO O O i i 71 CO CO CO CO GO CO X CO CO OS iO OS (M CO71OOS1 CD ooo ooooo oo XXX XXXXX XX O i I -H C1 O CO I-H -^ t OS OS OS t- 1^* "^ 71 OS CO CO O t~* "^ CO CO O CO CO ^^ IT* CO C5 * ^ OS O I-H T-I 71 CO X X X X X X "* t- OS O OS ;e co 7i 71 o X X X X X 1 ^H O X 1- 10 CO 71 O 51 TjH O t -t l-H CO O 5O-^(M7-1 CO-t^tO ooo o o o o CO i X CO COOt^i-l ' >a 71 X .O iTl OS O 71 OS I- CO X OS O O r-n 71 O O>OiOOO KO >O 't 1 ^fl "rt O O X CO ococo t r- co os os XXXXX XXXXX XXXXX XXXXX XXXXX O CO rH 1C X -* X CO t- rH J C^l 1 ^ XXXXX XXXXX XXXXX XXXXX XXXXX 10 OS rH ,_, f^. TJI ^, t^ ^ -^ 10 co co X -to r- co o 2 co os o o rH ri CM co rH 1 C- COXCOXCN t-(T1O OS OS lO C^l X "^t CO t X X OS C<1 'N T<1 C. LOGARITHMS. CS - O rH rH 7 X X X X XXXXX XXXX 7-1 X "* rH t- CO OS 1C rH i i Ift COCO^^lC XXXXX t X t CO o co co os rH rH (M 7-1 CO CO CO CO XXXX co -t co co co co co X X X X X r- cs o o cs CO OS CO CO X o i 1 2J CO CO CO CO CO X X X 'X X X >C ^t i-l CO CO 1 Cl Ci O rH OO -tO rH rH CM CO m m m m xx xx ci -t x CM in CO CM Ci CM 00 "* -. 00 CO 00 00 ^.ninm m in ,n $ o ! $ S X X OO X X XXXXX XX XXX XXXXX rH O in r~ rH t- CO 00 00 00 00 r- co x CM in i- x ci x t- co -^t ~t in in o o o o in co ci in> i 1 1 co ci o r- o ;o i i^- x x ci o o o in >n o m in >n ~~c so XXXX XXXXX O X O CO Cl -t X rH -f iC O ci o -M -t m i- x o i-i CM co O t- CO Cl O T-I I- -t O D CM j ;O JO 1- 1^- OO Gl Ci O rH -^ CM C? XXXXX XXXOOX X X CO 1^ X X Ci CO CM X -t O O -t m in :c i- i - X Ci X 1- in in in m o XXXXX XXXXX X <> -t rH t- CMOOCM-t 01 *>** S^gS SI GO -t O O I i X X X 00 00 X _S c2J2 OCMt 1 -t CDXC1X1- r-t CM CM CO CC CO CO CO CO CO (M GO -t O O CMX-tOSC m o ;c i- i^- x x ci o o m in m m m in o in ^ ^ XXXXX XXXXX '-t "^t CO 1 ' Ci i-l CM CO 't -t ^N ^^ KJ i ( t^* co ci in OOi-( CMCMCOCOrfi m in in ci o i i in ci i i -H co m so i ci X -* O SO CM X t- t- OO Cl Ci " OOXX XXXOOX XOOXOOX O ** O X Cl X X ^ CO " X -* O CM -t iO t- in co t t X X X X 00 t O CM COCOCMi-HX ^HCO^ ia t- GO GO Tt< O SO CM X O CO O X X 00 X CM-HC1^>CM GO t x Ci x co co co co H M ^ MJ ?Or-XO HNCO^IWS OrCOSO H OOOCO OOOOH HHHHH ^r^i-lTHW W LOGARITHMS. O O O O Ci CD CD CD CD o Ci Ci Ci Ci O O5 Ci Ci Ci SSS 88gg 5i CD to o o CD CD X CO CO GO CO CO 'M O t- 00 Ci O ^t* CO O CO ~ GO GO 00 00 GO GO 00 OO i co os HH cs CO OS t- -* > CO TH t- CO CO t- t- 00 ^ t O CO CO c5i os t- ^ TH OS -t O CO CO 'X OS O O TH r- t GO oo GO 00 GO GO 00 CO Tf< TH CO TH OS 00 C CO OS lO i- co OS HH o TH (N TH t- co ~ "* O O CO GO 00 GO 00 GO OO 'GO GO 00 00 OS rH CO "* ^ HH ,H t- CO OS * O 10 TH -- xxxxx xxxx CO O CO iO CO CO ^ TH GO O 00 -O CO OS OS rH OS CO O t- t- X X OS xxxxx xxxxx CO TH X -^ OS TH OS CO "<* TH X CO CS uT5 TH O CO CO t X X X X X X "* X TH CO iO OS CO ^* TH X CO (N X HH OS X' OS OS O O x X X X X X CO CO iO O TH CO -H CM OS CO Ol O i O O O -* t- OS -T-I (N OrHlr-GO GO i-t T^ GO CO t- I-H O O rH t^ CM 00 CO CO OS "^ GO t^ 'N 00 CO CO'tOSiO (MCOCO-^ os os o os OO 00 GO GO GO i-H CO i-H -^ t- CO t- (N CD O I-H ?D (N I CO I-H T-I (M <^ CO O5 OS OS OS OS '^^ ut^ OS * 00 GO GO GO '-IOSO (N GO CO t O CO>?COI:-1 OS "^ O CO i^ 1^ <^l GO CO GO CO GO CO Ot ^~- Ct OS OS OS OS OS OS OS OS OS OS CO GO CO COOOGOOCGO GOOOOOGOOO (N t- (N C OS CO T-I O OS CO ^H N CO O COOCOt-t- COKt>OSi~it- COCOCOt-O OCOi-Ht- (Mi SOOO i-HCO^Hi-CM t- CM GO CO GC OI-HI-H CMCMOOCO't -tiOiOCOCC OSOSOS OSOSOSOSOS OSOSOSOSOS O i-l CN (?q CM co o co I-H co I-H t- i-l -^ CO liOGOrH o 1-1 co 1-1 os os os os os os os os osososcsos COi-HOSCOCN GOCOt-i-H^ t-GOOSOO Tjn CO 1-1 O OS CM'-OOSCOCO OSC^OOSCM ^osooo CO OS CO o rt -rt O >O CO OS OS OS OS OS OS OS OS OS OS S^IiOCOi I rtit-OCMO i-H CO(Mt^(MCO COCO-^OS"<* i i-lO OS CM iO t- X Ci iO 1 O 71 -t I- Ci CO O t- Ci T-I CM -t CCt-CiO-H CO-^iOCCI^ -t Ci iO O iO O C CD t CD TH 1 71 t- 71 t- CM X CO X CO X CO X 1- L X Ci Ci O O i TH CI 71 CO CC -t ~f iO O CO CO t- t- X X Oi Oi T-H T-I i 7-1 71 71 71 71 71 71 71 71 71 71 71 71 CM CI 71 71 71 71 71 CiCiCiCiCi CiCiCiCiCi CiCiCiCiCi CiCiCiOiCi OiCiCiOiCi i O iO O CC CD Ci TH 71 CM 7} O O -t< X TH CO iO I t- O CI iO t- Ci T- -t CD X O CI -T iO I Ci O 71 O CD X Oi O T-H CM ar -f Ci -t Ci -t O >C O C i CO T-H CO T-H CD CM 1 CM 1 CM t- CM X CO X i I r- X X Ci O O 71 71 CO CO -F -* lOOCDCOt- t- X X Ci Ci CI CI CI CI CI 71 71 CM CM CM 71 71 71 T-H "* CD X T- CC iC 1- Ci T- CC iC CO X T-I 1~ 71 1 71 X CC X CO X "^ Ci -t Ci "* t- t X X Ci Ci O O T-H -H 7-1 71 CO CO Tf iO iC CD 1- TH T-H T-I T T-H CI 71 71 71 71 71 71 71 CM 71 71 71 71 71 Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Oi Ci Oi Ci Ci Oi Ci Ci Ci Ci COX^XCM CDX-HT1CO O T CO ^ CD i X O T-H 71 0,00,00 .OOCOT.CO Ci Ci Ci Ci Ci T-I O t- "* T-I 1 CM I -H -ft Ci T-I 71 x ci T co icccxcii-i coco-rcoi X CO Ci "^ Ci ""^" Oi "^ O iO O iC O iC 71 71 71 71 71 71 71 71 71 7~1 71 71 71 Ci Ci Ci Ci Ci Oi Ci Oi Ci Ci' Ci Oi Ci 1 i CO O X O CM -*t t Ci T-H 71 f iO J '*. ~ t 1-4 t- 71 i"- 71 t CO X CO X 1 t X X Ci Ci O O i TH T-ITICOCO^ TH TH T-I TH TH71717171 7I7171CM7 Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci Oi Oi Ci Tt" Oi ^t 1 Ci "^ O 1C iO CD CD t- I- X Ci Ci 50 7^ 71 CO 71 CM <71 CM OiCiOi CiCiOiCiCi CO X CC t O CC CO t- X CO Xt-CO-*rH X-^Oi^X Ot-CiO O CO iC I"- Ci TH CO O t- Ci T-H CO Tjn CD i Ci O CM CO 'f O I COT-H CDT-^CDTli CM t 71 X CO XCOXCOCi ^Oi^tCi-t lOO^-^H CMCMCOCO-^ TfijCiCCDCD r- t CO 00 OS CM 71 CM CM 71 CM 71 71 7^1 71 Ci Oi Ci OS Oi Oi Oi Oi Ci Ci l Ci Ci Ci CT. Ci O CO O >C X T CC iC X O CC iC I- O CM -t Ci iC O C O CD -H to CD I X X Ci Ci O Q Ci Ci Ci Ci Ci O-*CMCi OTOXCMt OCOCDXCi Cl-tCOt" Cii iTl-^iO Ir XCiO-H C^ CM CM 71 _. _. _. .. _. _. _. _. _. O r- X O 00 X X 00 n <*" w n co w X X X O *- X O ct x x x I2O LOGARITHMS. cs x t- o co o r~ co x co X OS O ri o (7 co co ;c co co i - i co co co co cc o o -f CS -t CS -t< OS -t CS -f OS -tOS-tCS-f uO O CC CC I i X X OS OS O O i I-H ^ ZZ OS OS OS C5 C5 X 1 CO iO .w --. o o r-i T-I rM ,ncoc< co co co co co OCOCOX OSOOCSX ^OOl O O CO CO 1-- 1-- X X OS OS co co co co co co co co co cc O5OSOSOSOS O5C5OSCSO OS (0 rH 1^ l t CM oococcr i^-xx co co co co co co co co OS OS OS OS OS OSOSOS OS O O r-i rH CO T^ ^t 1 ^ TjH OS OS OS OS CS OS C<1 CO CO rt< O rH CO 1-1 CO 3 OS OS C5 C5 OS ** OO -M lO b- 10 10 co cc co rH CO rH CC rH iO O O O t- co co co co co CS OS OS OS OS coo - - co i-t co i-i - 00 X' OS co co co co co os os os os os O G CO X i- X X OS OS ^f OS -t co co co co co OS OS OS OS OS CO t- rH -^ CO O O rH T I T I CO CO CO CO CO OS OS OS OS OS XOSOOOS COCC^T-lX rHrHfNO O O O O O O O -O r-XXC535 OOrHrH O H N W ^J 10 90 00 00 00 00 00 LOGARITHMS. 121 Q" O . T-( O O5 (N -t t- CO -t* rt T^ ^ Tt< -f O >O O O O O O O OS OSOSCSOSOS CSCSCSCSO5 CSCSOS fMGO'MCOO CO O t OS O OCS 22SiSSg SiS33 S^cooo 000.60 71 CO CO -t ^ >O O CO CO 1 t- X X CS CS O O 1-1 i-t iN ^ Tf^ r^ -^ -^ ^ -^ -^ -^ -^ ^H --^ -^ --^ T^ iO i^ iO *O Q oscsoscso oscsososos cscscsoscs cscscsoscs . . ____ . . . _ i CO 1-1 CO r-iCO-^ CO CO Tt< "* O iO CO cs cs os cs CSCSGSOSCS O TH O O X CO CS X I"- CO -t CO o o o o o o t X X OS OS os C^7,^ 2:2 3*3x3 -t 1 '-i t CO 7-1 I-H OS X S8^-H o o o >c o cs c: cs os cs cs (N (M ' CS X O 71 Ci O O O OS CO CO OS 71 -H OS X t- CO "!f CO -N O X I- O CO O 10 OS -f CS -f CS -* Ci ^ GO CO X CO - i-t i-H CM 74 CO CO -t -*OS-HGO(N lOCOO^H o^t-tcoco 7^11^-0 O O iO O O O >O O O (N o^ CO "^f "^ cscsososcs cscsoos OCSCO CO OS i OCOrH'M Ci i h ^ ci o -^ i Ci -t X CO 'X CO l~ -M l~ OJ O O t"* JO CN O 1^~ ^f ^ Ci CO CO X CO X (M l~ C^l CO T-H 11 I I ilili t-COCi-t O CO O X O CO O 1~ i? O 1C O >O Ci Ci Ci Ci Ci i CO 1^ 1-1 -* ?C CO O i-< TH T-H O - O CM O i -* i-t Ci ^ CO O I- X CO X CN l-^ M 1- T-H CO -H Oi O O i-t T- t (M OJ CO CO 't ,0 CO CO CO CO CO -O CO CO CO Ci Ci Ci Ci Ci Ci Ci Ci CS Ci T-H GS i 10 oo TH ci r- O^Oi-*Ci Tt^X-CO Ci Ci Ci Ci Ci Ci Ci Ci Ci Ci CO CO OCOCOCi-H iO CO O X iO fM Ci t t- ^ t _ ^H ;o T-H iO O "ococococo cocoiococo OiOiCiCSCi OCi'CiCiCi O o 10 10 >o >ra io 10 o Ci Ci Ci Ci Ci Oi Ci Ci Ci Ci i >O Ci ^N Oi TH 5<1 CO CO CO lO CO rH Oi X CO X CO t JO >O *O iO iO Oi Ci Ci Ci Ci (70 -H Ci t >O T (XCOCiCC t-O(NOX COCOT-^XCO >O >O iO O O O CO CO CO CO Oi Ci Oi Oi Ci Oi Ci Ci Ci Oi 33o< 3Sii?i 000 q;^;5-Se K^ to en ro o ^ tc CO ^ rH I- Ci i-H CN CO CO rH X 1C 'M O t 1 < O Ci -fi CI -t X CO X co co -ti T* ra o 01 fNi ^a ^ iiiHi Tt< iO O CO CO 1~ t- X X OS iO iO >O iO O O O tO O O OS Oi Oi Ci Oi Oi Oi 01 OS 05 -t 1^ O 0 O 10 O CD --0 CO O OiCiCiOiCS OiCiCiCiCi t- X Ci 01 Ci O t h c X _- . .. CO 1"- "M t- T-I T-H (CV| (CV| CO CO "^ "H^ lO lO CO occcococo cococo-occ OiCiCiCiCi CiCiCiCiCi __ OS X t- TJH io iO co co r~* t"~ o o o o o o o o o o OS 05 OS 05 OS OS OS OS CS OS iO 01 CO CO X Oi CO Tfi T-H X CO CO O CO GO CO I O iO CO CO 1 I" CC I-H CO ^ r-t 1 TO O ^H -H -N CM TO t- t- 05 Ci Ci O O _ _ - - - 1-- t t t-- r 1 t I"- I"- J OS OS OS CS CS OS CS CS OS CS OS OS CS OS OS O ~f ?O CO -H CO -t O CO CO CD 1 OS CO TO O 1- CO O CO TO CS CO ON) 'X >!7 -H t~ TO CS O -H .. 1 OI I- OJ CO *- CO C ' - CS -t CS TO CO CO JC^ 7OiO CS CO 3s X^ SS^ rHrHO^OiCO TO-t-*>OlO COCbCOt-t^ OSCSCSCS OSC5CSOSOS OSOSCSOSOS OSOSOSOSOS OSOSCSCSOS O-l O CO CO CO 1^ C^l CO ^i CO p i-H JO CS !N Tf CO Tf< OS Tf^ CO CO CO O^l O O H i-l IN !M CO OS OS OS CS OS OS OS OS OS OS OS OS OS OS OS *M O GO O OS >O ^ ^.^ t.v ''> ^r "^> ii* 1 *i> K^ ^^ ^i? IT" IT" CSOSCSOS OSOSOSOSOS OSOSOSOSOS OSOSCSOSOS CO X CM 1 - , i-l i-t >>l CM CO CM O T^ (MCOCSO1'* Ot^COGOGO O I-H CO co r- 5O CO CO X CO 00 Oi Oi Oi Oi Oi -* co r i i Oi -t Oi -fi Oi Si TO co ~# ~* CO X X X X Oi Oi Oi Oi Oi GO CO '-O Oi Oi Oi Oi Oi O -H CO -H CO 'N I CO GO co r sq O O5 11 X CO Oi Ti O O ^H CO -H ^ 'M 1^- 'N I"- d ! ^\ CO ^H CO -H iO C- -f OiOiOiOiOi O5OiOiOiO5 OiOiOiOiOi 'OiOiOiOiOi OiOiOiOiOi 82i3 5^i^ ^ii?S?2 Si^^ v 02 co ^ o ^s?, 1 s gi S' s s O ^ N M <* >-5 i-. f. i>. r~ r^ i-. C5 CS CS CS C5 LOGARITHMS. I2 S 33333 CO CO CM TH Q OS CO t i l O OS OS CO X O "* X OS OS OS X O >O -f CO T-I OS t -- f O H^OSCOt-O CO "^ X OS O ' O OS 1 OCOo O OS CO X JN O I-H O OS CO i iO OS CO i f X -N ^ C~. " 'C "^* X CO t- S2 0.0198 1/27 0/2- ',90 1.62 0.4187 1.77 0.5710 .03 0.0 96 1.28 0/2469 153 0.4253 1.78 0.6766 .04 0:i92 1.29 0.2546 1.54 0.4318 1.79 0.5822 .05 00488 1.30 0.2624 1 55 0.4383 1.80 0.5878 .06 0.05S3 1.31 2700 1.56 0.4447 1.81 0.5933 .07 0.0677 1.32 0.2776 157 0.4511 1.82 06988 .08 00770 1.33 0.2S52 1.58 0.4574 1.83 0.6043 .09 00862 1.31 0.2927 1.59 0.4637 1.84 0.6098 .10 0.0953 1.35 03001 1.60 0.4700 1.85 0.6152 .11 0.1044 1.36 0.3075 1.61 0.4762 1.86 0.6206 .12 0.1133 1.37 O.:<148 1.62 0.4824 1.87 0.6259 .13 0.1222 1.38 0.3-J21 1.63 0.4886 1.88 0.6313 .14 01310 1.39 3-J'.i: .64 0.4947 1.89 0.6366 .15 0.1398 1.40 03365 .65 0.5008 1.90 0.6419 .16 0.1484 1.41 0.3436 .66 0.5068 1.91 0.6471 .17 0.1570 1.42 0.3507 .67 0.5128 1.92 0.6523 .18 0. 1 655 1.43 0.3577 .68 0.5188 1.93 0.6575 .19 0.1740 1.44 0.3646 .69 05247 l.v4 H627 1 .20 0.1823 1.45 0.3716 .70 0.5306 1.95 0.6678 .21 0.1906 1.46 0.3784 J.71 0.5365 1.96 0.6729 .22 0.1988 1.47 0.3853 1.72 0.5423 1.97 0.6780 .23 0.2070 1.48 0.3920 1.73 0.5481 1.98 O.H831 .24 0.2151 1.49 3988 1.74 0.5539 199 O.H881 1.25 0.2231 1.50 0.4055 1.75 0.5596 2.00 0.6931 HYPERBOLIC LOGARITHMS. 127 N Log. N - Log. N Log. N Log. >.oi 0.6981 ?.1l 0.870ft 2.S1 1.0332 3.21 1.1063 2.02 0.7031 2.42 0.883S 282 10367 3.22 1.16-4 2 03 0.70SO 2.43 0>x;n 2.83 10403 3.23 1.1725 2.04 0.7129 244 0.89^0 2.84 1.0438 3.24 1.1756 2.05 0.7178 2.45 0.8961 2.85 1.0473 3.25 1.1787 2.06 0.7227 2.46 0.9002 2.86 1.0508 3.28 .1817 2.07 0.7275 2.47 0.9042 2.87 .0543 3.27 .1848 2.08 0.7324 2.48 0.90S3 2.88 .Oft78 3/28 .1878 2.09 0.7372 2.49 0.9123 2.89 .0613 3.29 .1909 2.10 0.7419 2.50 0.9163 2.90 .0647 3.30 .1939 2.11 0.7467 2.51 0.9203 2.91 .0682 3.31 .1969 2.12 0.7514 H.62 0.9243 2.92 .0716 332 .2000 2.13 0.7561 2.53 0.9282 2.93 0750 3.33 .2030 2.14 0.7608 254 0.93-J2 2.94 .0784 3.34 .2060 2.15 0.7655 2.55 09361 2.95 .0818 3.35 .2090 2.16 0.7701 2.56 0.91PO 2.96 .0852 336 .2119 2.17 0.7747 2.57 0.9439 2.97 .0886 3.37 .2149 2.18 07793 2.58 0.9478 2.98 .0919 3.38 1.2179 2.19 0.7839 2.59 0.9517 2.99 .0953 3.39 1.2208 2.20 0.7885 260 0.9555 3. .0986 3.40 1.2238 2.21 0.7930 261 0.9594 3.01 .1019 3.41 1.2267 2.22 0.7975 2-62 0.9632 3.02 .1053 3.42 1.2296 2.23 08020 2-63 0.9670 3.03 .1086 3.43 1/23-26 2.24 0.8065 2'64 0.9708 3.04 1119 3.44 1.2355 2.25 0.8109 265 0.9746 3.05 .1151 3.45 1.2384 2.26 0.8154 2.66 0.9783 3.06 .1-184 3.46 1.2413 2.27 0.8198 2.67 098-21 3.07 .1217 3.47 1.2442 2.28 0.8242 2.68 0.98'.8 3.08 .l'_M9 3.48 1.2170 2.29 0.8286 2.69 0.9895 3.09 .U'S2 3.49 1.24'.9 2.30 0.8329 270 0.9933 3.10 .1314 3.50 1.2528 2.31 0.8372 2.71 0.9969 3.11 .1346 3.51 1.2556 2.32 0.8416 2.72 1.0006 3.12 .1378 3.52 1.2:,85 2.33 0.8459 2.73 1.0043 3.13 .1410 3.53 1.2613 2.34 0.8502 2.74 1.0080 3.14 .1442 3.54 1.2641 2.35 0.8544 2.75 1.0116 3.16 .1474 3.55 1.2669 2.36 0.8687 2.76 1.0152 3.16 .1506 3.56 1.2H98 2.37 0.8629 2.77 1.0188 3.17 .1537 3.57 1.2726 2.38 0.8671 2.78 1.0225 3.18 .1569 3.58 1.2754 2.39 0.8713 2.79 1.0260 3.19 .1600 3.59 1.2782 2.40 0.8755 2.80 1.0296 3.20 .1632 3.60 1.2809 128 HYPERBOLIC LOGARITHMS. N Log. N Log. N Log. N Log. 3.61 1.2837 4.01 1.3888 4.41 1.4839 481 1.5707 362 1.2865 4.02 1.3913 4.42 1.4801 4.82 1.5728 3.63 1 2892 4.03 1.3938 4.43 1.4884 483 1.5748 364 1.29'JO 4.04 1.3962 4.44 1.4907 4.84 1.5769 3.65 1.2947 4.05 1.3987 4.45 1.4929 4.85 1.5790 3.06 1.2976 4.06 1.4012 4.46 1.4951 4.86 1.5810 3.67 1.3002 4.07 1.4036 4.47 1.4974 4.87 1.5831 3.68 1.3029 4.08 1.4061 4.48 1.4996 488 1.5851 3.69 1.3056 4.09 1.4085 4.49 1.5019 4.89 1 572 3.70 1.3083 4.10 1.4110 450 1.5041 4.90 1.5892 3.71 1.3110 4.11 1.4134 4.51 1.5063 4.91 1.5913 3.72 1.3137 4.12 1.4159 4.52 1.5085 492 1.5933 3.73 1.3164 4.13 1.4183 4.53 1.5107 4 93 1.5953 374 1.3191 414 1.4207 4.54 1.5129 4 94 1.5974 3.75 1.3218 4.15 1.4231 4.55 1.5151 4.95 1.5994 3.76 1.32-14 4.16 1.4255 4.56 1.5173 4 96 1.6011 3.77 1.3271 4.17 1.4279 4.57 1.5195 4.97 1 GO;U 3.78 1.3297 4.18 1.4303 4.58 1.5217 4.98 1.6054 3.79 1.3324 4.19 1.4327 4.59 1.5239 4.99 1.6074 3.80 1.3350 4.20 1.4351 4.60 1.5261 5. 1.6094 3.81 1.3376 4.21 1.4375 4.61 1.5282 6.01 1.6114 382 1.3403 4.22 1.4398 4.62 1.5304 5.02 1.6134 3.83 1.3429 423 1.4422 4.63 1.5326 5.03 1.6154 3.84 1.3455 4.24 1.4446 4.64 1.6347 5.04 1.6174 3.85 1.3481 4.25 1.4469 4.65 1.5369 5.05 1.6194 3.86 1.3607 4.26 1.4493 4.66 1.5390 5.06 1.6214 3.87 1.3533 4.27 1.4516 4.67 1.5412 5.07 1.6233 388 1.3558 4.28 1.4540 4.68 1.5433 5.08 1.6253 3.89 1.3584 4.29 1.4563 4.69 1.5454 5.09 1.6273 3.90 1.3610 4.30 1.4586 4.70 1.5476 5.10 1.6292 3.91 1.3635 431 .4609 4.71 1.5497 6.11 1.6312 3.92 1.3661 4.32 .4633 4.72 1.5518 5.12 1.6332 3.93 1.3686 4.33 .4666 473 1 .5539 5.13 1.6351 3.94 1.3712 4.34 .4679 4.74 1.6.)60 5.14 16371 3.95 1.3737 4.35 .4702 4.75 1 5581 5.16 1.6390 3.96 1.3762 4.36 .4725 4.76 1.5602 5.16 1.6409 397 1.3788 4.37 .4748 4.77 1.5623 5.17 1.6429 3.98 1.3-13 4.38 .4770 4.78 1.56U 5.18 1.6448 3.99 1 3N38 4.39 .4793 4.79 1 r,65 8.87 2.1827 9.27 2 2268 9.67 2.2690 8.48 2.1377 8.88 2.1838 9.28 2.2279 9.68 2.2701 8.49 2.1389 8.89 2.1849 9.29 2.2289 9.69 2.2711 8.50 2.1401 8.90 2.1861 9.30 2.2300 9.70 2.2721 8.51 2.1412 8.91 2.1872 9.31 2.2311 9.71 2.2732 8.52 2.1424 892 2.1883 9.32 2.2322 9.72 2.2742 8.53 2.1436 8.93 2.1894 9.33 2.2332 9.73 2.2752 8.54 2.1448 8.94 2.1905 9.34 2.2343 9.74 2.2762 8.55 2.1459 8.95 2.1917 9.35 2.2354 9.75 2.2773 8.56 2.1471 8.96 2.1928 9.36 2.2364 9/76 2.2783 8.57 2.1483 8.97 2.1939 9.37 2.2375 9.77 2.2793 8.58 2.1494 8.98 2.1950 9.38 2.2386 9.78 2.2803 8.59 2.1606 8.99 2.1961 9.39 2.2396 9.79 2.2814 8.60 2.1618 9. 2.1972 9.40 2.2407 9.80 2.2824 8.61 2.1629 9.01 2.1983 9.41 2.2418 9.81 2.2834 862 2.1541 9.02 2.1994 9.42 2.2428 9.82 2.2844 8.63 2.1552 9.03 2.2006 9.43 2.2439 9.83 2.2854 8.64 2.1564 9.04 2.2017 9.44 2.2450 9.84 2.2865 8.65 2.1576 9.05 2.2028 9.45 2.2460 9.85 2.2875 8.66 2.1587 9.06 2.2039 9.46 2.2471 9.86 2.2885 8.67 2.1599 9.07 2.2050 9.47 22481 9.87 2.2895 8.68 21610 9.08 2.2061 9.48 2.2492 9.88 2.2905 8.69 2.1622 9.09 2.2072 9.49 2.2^.02 9.89 2.2915 8.70 2.1633 9.10 2.2083 9.50 2.2513 9.90 2.2925 8.71 2.1645* 9.11 2.2094 9.61 2.2523 9.91 2.2935 8.72 2.1fio6 9.12 2.2105 9.52 2.2534 9.92 2.2946 8.73 2.1668 9.13 2.2116 9.53 2.2544 9.93 2.2956 8.74 2.1679 9.14 2.2127 9.54 22555 9.94 2.2966 8.75 2.1691 9.16 2.2138 9.55 2.2565 9.95 2.2976 876 2J702 9.16 2.2148 9.66 2.2576 9.96 22986 8.77 2.1713 9.17 2.2159 9.67 2.2686 9.97 2.2996 8.78 2.1725 9.18 2.2170 9.58 2.2597 9.98 2.3006 8.79 2.1736 9.19 2.2181 9.59 2.2007 9.99 2.3016 8.80 2.1748 9.20 2.2192 9.60 2.2618 10. 2.3026 '32 HYPERBOLIC LOGARITHMS. N Log. N Log. N Log. N Log. 10.1 2.3126 14.1 2.6462 18.1 2.8959 22.1 3.0956 10.2 2.3225 14.2 2.6532 18.2 2.9014 222 3 1001 10.3 2.3322 14.3 2.6602 183 29069 22.3 31046 10.4 2.3419 14.4 2.6672 18.4 2.9123 22.4 3.1090 10.5 2.3515 14.5 2.6741 18.5 2.9178 22.5 3.1135 10.6 23609 14.6 2.6810 18.6 2 9231 226 3.1179 10.7 23703 14.7 2.6878 18.7 2 9285 22.7 3.1224 10.8 23796 14.8 2.6946 18.8 2.93;38 22.8 3.1267 10.9 23888 14.9 2.7013 18.9 2.9M1 22.9 3.1311 11 2.3979 15. 2.7080 19. 2.9444 23. 3.1355 11.1 2.4070 15.1 27147 19.1 2.9497 23.1 3.1398 11.2 2.4160 15.2 2.7213 19.2 2.9549 232 3.1441 11.3 2.4249 15.3 2.6279 19.3 2 9601 23.3 3 1484 11.4 2.4337 15.4 2.7344 19.4 2.9(553 23.4 31527 11.6 2.4424 15.5 2.7408 19.5 2.9704 23.5 3.1570 11.6 2.4510 15.6 2.7472 19.6 2.9755 23.6 3.1612 11.7 2461)6 15.7 2.7536 197 2.9806 23.7 3.1655 11.8 2 4681 158 2.7600 198 2.9856 23.8 3.1697 11.9 2.4765 15.9 2.7663 19.9 2.9907 23.9 3.1739 12 2.4849 16. 2.7726 20. 2.9957 24. 3.1780 12.1 2.4932 16.1 2.7788 20.1 30007 24.1 3.1822 12.2 2.5014 16.2 2.7850 20.2 3.0057 24.2 3.1863 12.3 2.5096 16.3 2.7912 20.3 3.0106 24.3 3.1905 12.4 2.5178 16.4 2.7973 20.4 3.0155 24.4 3 1946 12.5 2.5259 16.5 2.8033 20.6 3.0204 24.5 3.1987 12.6 2.6338 16.6 2.8094 20.6 3.0253 24.6 3.2027 12.7 25417 16.7 2.8154 20.7 3.0301 24.7 3.2068 12.8 2.5495 16.8 2.8214 20.8 3.0349 24.8 3.2108 12.9 2.5572 16.9 28273 20.9 3.0397 24.9 3.2149 13 2.5649 17. 2.8332 21. 3.0445 25. 3.2189 13.1 2.5726 17.1 2.8391 21.1 3.0493 25.5 3.2387 13.2 2.5802 17.2 2.84 19 21.2 3.0540 26. 3.2581 133 2.5877 17.3 2.8507 21.3 3.0587 26.5 32771 13.4 2.5962 17.4 2.8565 21.4 3.0634 27. 3.2968 13.6 2.6027 17.5 2.8622 21.5 3.0680 27.6 3.3142 13.6 2.6101 17.6 2.8679 21.6 3.0727 28. 3.3322 13.7 2.6174 17.7 2.8735 21.7 3.0773 28.6 3.3499 13.8 2.6247 17.8 2.8792 21.8 3.0819 29. 33673 13.9 2.6319 17.9 28848 21 9 3. OH 65 29.5 3.3844 14. 2.6391 18. 2.8904 22. 3.0910 30. 3.4012 Weights anb Measures. The yard is the standard unit for length in the United States and Great Britain. To determine the length of the yard, a pendulum vibrating seconds in a vacuum at the level of the sea in the latitude of London, with the Fahrenheit thermometer at 62, is supposed to be divided into 301,393 equal parts ; 360,- 000 of these parts is the length of the standard yard. Actually, the standard yard in both the United States and Great Britain is a metallic scale made with great care and kept by the respec- tive governments, and from this standard other measures of length have oeen produced. The standard unit of weight in the United States and Great Britain is the Troy pound, which is equal in weight to 22.2157 cubic inches of distilled water at 62 Fahrenheit, the barometer being 30 inches. The Troy pound contains 5,760 Troy grains; the Pound Avoirdupois, which is the unit of weight used in commercial transactions and mechanical cal- culations in the United States and Great Britain, is equal to 7,000 Troy grains. In the United States the standard unit of liquid measure is the wine gallon, containing 231 cubic inches or 8.3389 pounds avoirdupois of distilled water at a temperature of its greatest density (39-40 F). In the United States the standard unit for dry measure is the Winchester Bushel, containing 2150.42 cubic inches. In Great Britain the standard measure for both liquid and dry substances is the Imperial Gallon, which is defined as the volume of 10 pounds avoirdupois of distilled water, when weighed at 62 Fahrenheit with the barometer at 30 inches. The Im- perial Gallon contains 277.463 cubic inches. The Imperial Bushel of 8 gallons contains 2219.704 cubic inches. Long Measure. 12 inches = 1 foot = 0.30479 meters. 3 feet = 1 yard = 0.91437 meters. 5>^ yards = 1 rod or pole = 16^ feet = 198 inches. 40 rods = 1 furlong = 220 yards = 660 feet. 8 furlongs = 1 statute or land mile = 320 rods = 1760 yards. 3 miles = 1 league = 24 furlongs = 960 rods. 5280 feet = 1 statute or land mile = 1.609 kilometer. 1 geographical or nautical mile = 1 minute = ^ degree. (I33-) 134 WEIGHTS AND MEASURES. As adopted by the British admiralty,* a nautical mile is 6080 ft. I nautical mile = 1.1515 statute or land miles. 1 statute or land mile = 0.869 nautical miles. Square Measure. 1 square yard = 9 square feet = 0.836 square meters. 1 square foot = 144 square inches = 929 square centi- meters. 1 square inch = 6.4514 square centimeters. A section of land is 1 mile square = 640 acres. 1 acre = 43,560 square feet = 0.40467 hectare. 1 square acre is 208.71 feet on each side. Cubic fleasure. 1 cubic yard = 27 cubic feet = 0.7645 cubic meters. 1 cubic yard = 201.97 (wine) gallons = 7.645 hectoliter. 1 cubic foot = 1728 cubic inches = 28315.3 cubic centi- meters. 1 cubic foot == 7.4805 (wine) gallons = 28.315 liters. NOTE. 1 cubic foot contains 6.2355 imperial (English) gal- lons. A cord of wood = 128 cubic feet, being 4X4X8 feet. A perch of stone = 24^ cubic feet, being IQ% X 1 1 A X 1 foot, but it is generally taken as 25 cubic feet. Liquid Hea&ure. 1 pint = 28.88 cubic inches. 2 pints = 1 quart = 57.75 cubic inches = 0.9463 liter. 4 quarts = 1 gallon = 231 cubic inches = 3.7852 liters. NOTE. 1 imperial (English) gallon is 277.274 cubic inches. Dry Measure. 1 standard U. S. bushel = 2150.42 cubic inches. 1 standard U. S. bushel = 4 pecks. 1 peck = 2 gallons = 8 quarts. 1 gallon = 4 quarts = 268* cubic inches. 1 quart = 2 pints = 67| cubic inches. 100 bushels (approximately) = 124> cubic feet. 80 bushels (approximately) = 100 cubic feet. Avoirdupois Weight. (Used in business and mechanical calculations.) 1 pound = 16 ounces = 0.45359 kilograms. 1 ton = 2240 pounds. A short ton is 2000 pounds. * See Machinery, page 23, Sept., 1897. WEIGHTS AND MEASURES. Troy Weight. (Used when weighing gold, silver and jewelry.) 1 pound = 12 ounces = 0.37324 kilogram. 1 ounce = 20 pennyweights = 1.0971 ounces avdp. Apothecaries' Weight. 1 pound = 1 pound troy weight = 12 ounces. 1 ounce = 8 drachms. 1 drachm = 3 scruples. 1 scruple = 20 grains. 135 Weights of Produce. The following are the weights of certain articles of produce: Pounds per bushel. Wheat, 60 Corn in the ear, 70 Corn shelled, 56 Rye, 56 Buckwheat, 48 Barley, 48 Pounds per bushel. Oats, 32 Peas, 60 Ground peas, 24 Corn meal, 48 Malt, 38 White beans, 60 Pounds per bushel. White potatoes, 60 Sweet potatoes, 55 Onions, 57 Turnips, 57 Clover seed, 60 Timothy seed, 45 The fletric System of Weights and Heasures. The unit in the metric system is the meter. The length of (one ten-millionth part) 1 () ooo odd the meter was intended to be iv , uuuuuw , of the length of a quadrant of the earth through Paris, which is the same as TcrWooi 41.27 9 228.6 % 15.87 1% 44.45 10 254 ! ,\ 17.46 17/ S 47.62 11 279.4 % 19.05 |j 2 50.80 12 304.8 138 WEIGHTS AND MEASURES. Table of Reduction for Pressure per Unit of Surface. 1 kilogram per sq. centimeter = 14.223 pounds per sq. inch. 1 kilogram per sq. centimeter = 0.968 atmosphere. 1 pound per sq. inch = 0.0703 kilograms per sq. centimeter. 1 pound per sq. inch = 0.068 atmosphere. Table of Reduction for Length and Weight. 1 kilogram per kilometer 3.548 pounds per mile. 1 kilogram per meter = 0.672 pounds per foot. 1 pound per mile = 0.282 kilograms per kilometer. 1 pound per foot = 1.488 kilograms per meter. Weight of Water (4 C.) 1 cubic cm. weighs 1 gram. 1 cubic inch weighs 0.036125 pounds = 16.386 grams. 1 liter weighs 1 kilogram = 2.2046 pounds. 1 quart weighs 2.0862 pounds 0.9463 kilograms. 1 cubic meter weighs 1000 kilograms = 2204.6 pounds. 1 cubic foot weighs 62.425 pounds = 28.32 kilograms. Measure of Water. 1 kilogram measures 1 liter =1.057 quarts. 1 kilogram measures 0.353 cubic feet = 61.03 cubic inches. 1 pound measures 0.01602 cubic ft. = 0.454 liter. 1 pound measures 27.68 cubic ins. = 453.59 cubic centimeters. SPECIFIC GRAVITY. The specific gravity of a body is its weight as compared with the weight of an equal volume of another body which ^ is adopted as a standard. For all solid substances, water at its maximum density (4 C.) is the usual standard. For instance, the specific gravity of zinc is 7 ; this simply means that one cubic foot of zinc is 7 times as heavy as one cubic foot of water. One cubic foot of water weighs 62.425 pounds. There- fore, by multiplying the specific gravity of any solid body by 62.425 its weight per cubic foot is obtained. In the metric system of measure and weight, one cubic centimeter of water weighs one gram; therefore the table of specific gravity will also directly give the weight of the material in grams per cubic centimeter, in kilograms per cubic decimeter, or in 1000 kilo- grams (the so-called metric ton) per cubic meter. WEIGHTS AND MEASURES. 139 TABLE No. 9. Specific Gravity, Weights and Values. METALS. Metric. Kilog. per ;ubic dec. or specific gravity American. Pounds per cubic inch. Pounds per cubic foot. Approximate value per pound. Water . . . Gold (24 k) . Platinum . . Silver . . . Wrought iron Cast iron . . Tool Steel . Zinc .... Antimony . . Copper . . . Mercury . . . Tin . . . Aluminum . Lead . 1 19.361 21.531 10.474 7.78 7.21 7.85 7 6.72 8.607 13.596 7.291 2.67 11.36 0.036125 0.697 0.775 0.377 0.28 0.26 0.284 0.252 0.242 0.31 0.489 0.262 0.096 0.408 62.425 1208 1344 654 485 450 490 437 419 537 849 455 166 708 $299.70 122.00 12.14 0.015 0.008 0.10 0.10 0.12 0.15 0.25 0.05 TABLE No. 10. Specific Gravity and Weight of Medium Dry Wood. VARIETY. Metric. Kilog. per cubic dec. specific gravity. American. Pounds per cubic foot. Birch Ash Beech Oak Ebony Lignum-vitas . . Spanish mahogany Hickory Spruce Pine Pitch pine .... 0.60 to 0.80 0.50 to 0.80 0.60 to 0.80 0.60 to 0.90 1.19 1.33 0.85 0.50 0.50 0.40 to 0.80 0.80 37.5 to 50 31 to 50 37.5 to 50 37.5 to 56 74 83 53 32 32 25 to 50 50 140 WEIGHTS AND MEASURES. TABLE No. ii. Specific Gravity and Weight per Cubic Foot of Various Haterials. (The weight may vary according to the properties of the material). MATERIALS. Metric. Kilog. per cubic dec. specific gravity. American. Pounds per cubic foot. Asphalt .... Brick Gray granite . . Red granite . . Limestone . . . Sand Portland cement Brickwork . . Slate Glass Emery .... Grindstone . . Coal Porcelain . . . Lime . 1.4 1.6 to 2 2.4 2.5 to 3 2.7 1.5 1.26 1.75 2.8 2.52 4.0 2.4 1.5 2.4 0.96 87 100 to 125 150 157 to 187 168 94 78 110 175 157 250 150 64 150 60 TABLE No. 12. Specific Gravity and Weight of Liquids. LIQUIDS. > Metric. Kilog. per cubic dec. Kilog. per liter. American. Pounds per cub. inch. Pounds gallon. Water . . . Sea water . Sulphuric acid Muriatic acid Nitric acid . Alcohol . . . Linseed oil . Turpentine . Petroleum Machine oil . 1 1.03 1.841 1.2 1.217 0.833 0.94 0.87 0.878 0.9 1 1.03 1.841 1.2 1.217 0.833 0.94 0.87 0.878 0.9 .03 .841 .2 .217 0.833 0.94 0.87 0.878 0.9 0.036125 0.037 0.067 0.043 0.044 0.03 0.034 0.031 0.032 0.0324 8.33 8.55 15.48 9.93 10.16 6.93 7.85 7.16 7.39 7.5 WEIGHTS AND MEASURES. 141 To Calculate Weight of Casting from Weight of Pattern. When pattern is made from pine and no nails used, .the rule is : Multiply the weight of the pattern by 17 and the prod- uct is the weight of the castings. When nails are used in the pattern, multiply its weight by a little less, probably 15 or 16. When the pattern has core prints, their weight must be calculated and also the weight of what there is to be cored out in the casting, which must all be deducted. This mode of cal- culating the weight of castings is, of course, only approxima- tion, but it is frequently very useful. Weight of an Iron Bar of any Shape of Cross Section. A wrought iron bar of 1 square inch area of cross section and one yard long weighs 10 pounds. Therefore, the weight of wrought iron bars of any shape, as, for instance, railroad rails, I beams, etc., may very conveniently be obtained by first mak- ing a correct, full size drawing of the cross section and measur- ing its area by a planimeter, which gives the area in square inches. Multiply this area by 10 and the product is the weight in pounds per yard; or multiply the area by 3.33 and the product is the weight in pounds per foot. To Calculate Weight of Sheet Iron of any Thickness. One square foot of wrought iron, 1 inch thick, weighs very nearly 40 pounds (40.2 pounds) and one square foot y, which is joffo thick, weighs 1 pound. Therefore, a practical rule for quick calculation of the weight of sheet iron is: Divide the thickness of the iron as measured by a micrometer calliper in thousandths of inches by 25, and the quotient is the weight in pounds per square foot. To Calculate the Weight of Metals Not Given in the Tables. Find the weight of wrought iron, and multiply by the fol- lowing constants : Weight of wrought iron X 0.928 = cast iron. " X 1.014 = steel. " " " " X 0.918 = zinc. " " " " X 1.144 = copper. ' ' X 1.468 ~ lead, 142 WEIGHTS AND MEASURES. To Calculate the Weight of Zinc, Copper, Lead, etc., in Sheets. First find the weight by the rule given for sheet iron, and multiply by the constant as given in the above table, and the product is the weight of each metal in pounds per square foot. To Calculate the Weight of Cast Iron Balls. Multiply the cube of the diameter in inches by 0.1377, and the product is the weight of the ball in pounds. Thus : WD^y. 0.1377. D = 1. D = diameter of ball in inches. W = weight of ball in pounds. In metric measure, multiply the cube of the diameter in centimeters by 0.003775, and the product is the weight of the ball in kilograms. Thus : IV = M* X 0.003775. M 6.422 X W = weight in kilograms. M diameter of ball in centimeters. TABLE No. 13. Weight of Round Steel per Lineal Foot. Steel weighing 489 pounds per Cubic Foot. Diameter in inches. Weight Per Foot. Diameter in Inches. Weight Per Foot. Diameter in Inches. Weight Per Foot. A .0104 1 TV 3.011 2 */s 12.044 .042 y& 3.375 i/ 13.503 3_ .094 T 3 6 3.761 N 15.045 1 A .167 X 4.168 16.67 A .261 A 4.595 X 18.379 .375 && 5.043 20.171 A .511 Tff 5.512 H 22.047 \/ .667 H 6.001 3 24.005 T 9 * .844 9 6.512 # 26.048 1.042 ff 7.043 X 28.173 8 1.261 7.596 30.382 1.5 X 8.169 \/ 2 32.674 it 1.761 8.702 ft 35.05 2.042 7 A 9.377 % 37.508 if 2.344 it 10.013 40.05 l 2.667 2 10.669 4 42.675 WEIGHTS AND MEASURES. TABLE No. 1 4. Weights of Square and Round Bars of Wrought Iron in Pounds per Lineal Foot. (Iron weighing 480 pounds per cvibic foot). Thickness or Dinmeter in Inches. Weight of Square Bar One Foot Long. Weight of Round Bar One Foot Long, Thickness or Diameter in Inches. Weight of Square Bar One Foot Long. Weight of Round Bar One Foot Long. VlO .013 .010 2 %e 21.89 17.19 % .052 .041 % 22.97 18.04 %e .117 .092 H1 6 24.08 18.91 % .208 .164 % 25.21 19.80 r /io .320 .256 13 /io 26.37 20.71 % .469 .368 % 27.55 21.64 7 /10 .638 .501 "Ke 28.76 22.59 X .833 .654 3 30 23.56 /16 1.055 .828 Vie 31.26 24.55 % 1.302 1.023 % 32.55 25.57 U /10 1.576 1.237 3 /16 33.8-7 26. 60 % 1.875 1.473 % 35.21 27.65 I: HG 2.201 1.728 5 /16 36.58 28.73 % 2.551 2.004 % 37.97 29.82 lr /io 2.930 2.301 7 /16 39.39 30.94 1 3.333 2.618 I/ /2 40.83 32.07 VlO 3.763 2.955 %e 42.30 33.23 % 4.219 3.313 % 43.80 34.40 Yio 4.701 3.692 ll /16 45.33 35.60 % 5.208 4.091 % 46.88 36.82 %e 5.742 4.510 13 /16 48.45 38.05 % 6.302 4.1)50 % 50.05 39.31 7 /10 6.888 5.410 lr /16 51.68 40.59 % 7.5 5.890 4 53.33 41.89 %0 8.138 6.392 Mo 55.01 43.21 % 8.802 6.913 % 56.72 44.55 Hio 9.492 7.455 */ic 58.45 45.91 % 10.21 8.018 % 60.21 47.29 i-Tio 10.95 8.601 %e 61.99 48.69 % 11.72 9.204 % (53.80 50.11 lr /io 12.51 9.828 He 65.64 51.55 2 13.33 10.47 X 67.50 53.01 He 14.18 11.14 %G 09.39 54.50 % 15.05 11.82 % 71.30 56 3 /10 15.95 12.53 Vl6 73.24 57.52 % 16.88 13.25 % 75.21 59.07 5 /16 17.88 14 13 /10 77.20 60.63 % 18.80 14.77 % 79.22 62.22 !/io 19.80 15.55 ir /io 81.26 63.82 X 20.83 16.36 5 83.33 65.45 144 WEIGHTS AND MEASURES. TABLE No. i4. (Continued). Thickness or Diameter in Weight of Square Bar One Foot Weight of Round Bar One Foot Thickness or Diameter in Weight of Square Bar One Foot Weight of Round Bar One Foot Inches. Long. Long. Inches. Long. Long. 5 I/LQ 85.43 67.10 7 \ 169.2 132.9 X 87.55 68.76 175.2 137.6 3 /16 89.70 70.45 % 181.3 142.4 X 91.88 72.16 5 ^ 187.5 147.3 5 /16 94.08 73.89 193.8 152.2 X 96.30 75.64 3^ 200.2 157.2 %6 98.55 77.40 % 206.7 162.4 X 100.8 79.19 8 213.3 167.6 %6 103.1 81.00 X 226.9 178.2 X 105.5 82.83 X 240.8 189.2 ll^Q 107.8 84.69 255.2 200.4 % 110.2 86.56 9 4 270.0 212.1 13 /16 112.6 88.45 i/ 285.2 224.0 X 115.1 90.36 i/ 300.8 236.3 15 /16 117.5 92.29 % 316.9 248.9 6 120.0 94.25 10 333.3 261.8 X 125.1 98.22 3 350.2 275.1 X 130.2 192.3 X 367.5 288.6 % 135.5 106.4 % 385.2 302.5 X 140.8 110.6 11 403.3 316.8 X 146.3 114.9 i/ 421.9 331.3 % 151.9 119.3 i/ 440.8 346.2 X 157.6 123.7 3X 460.2 361.4 7 163.3 128.3 12 480. 377. TABLE No. 15. Weight of Flat Iron in Pounds per Foot. Inches Vl6 % 3 /16 X 5 /16 % 7 /10 X % % % 1 X 0.11 0.21 0.32 0.42 0.53 0.63 0.73 0.84 % 0.13 0.26 0.40 0.53 0.66 0.79 0.92 1.06 1.31 % 0.16 0.32 0.47 0.63 0.79 0.95 1.11 1.26 1.58 1.90 % 0.18 0.37 0.55 0.74 0.92 1.11 1.29! 1.48 1.85 2.22 2.58 0.21 0.42 0.03 0.84 .05 1.26 1.47 1.68 2.11 2.53 2.95 3.37 IX 0.24 0.47 0.71 0.95 .18 1.42 1.66 1.90 2.37 2.84 3.32; 3 79 IX 0.26 0.53 0.79 1.05 .32 1.58 1.84 2.11 2.63 3.16 3.68 4.21 1% 0.29 0.58 0.87 1.16 .45 1.74 ! 2.03 2.32 2.89 3.47 4.05 463 IX 0.32 0.63 0.95 1.26 .58 1.90 2.21 2.53 3.16 3.79 4.42 5.05 IX 0.34 0.68 1.03 1.37 .71 2.05 2.39j 2.74 3.42 4.11 4.79 5.47 1% 0.37 0.74 1.11 1.47 1.84 2.21 2.58 2 95 3.68 4.42 5.16J 5.89 1% 0.40 0.79 1.18 1.58 1.97 2.37 2.76 3.16 3.95 4.74 5.53 6.32 2 0.42 0.84 1.26 1.68 2.11 2.53 2.95 3.37 4.21 5.05 5.89 6.74 TABLE No. 16. Sizes of Numbers of the U. S. Standard Gage for Sheet and Plate Iron and Steel. (Brown & Sharpe Mfg. Co.) Number of Gage. Approximate Thickness in Fractions of an Inch. Approximate Thickness in Decimal Parts of an Inch. Weight Per Square Foot in Ounces Avoirdupois. Weight Per Square Foot in Pounds Avoirdupois. 0000000 X .5 320 20.00 000000 If .46875 300 18.75 00000 0000 1 .4375 .40625 280 260 17.50 16.25 000 H .375 240 15. 00 H .34375 220 13.75 * .3125 200 12.50 1 .28125 180 11.25 2 IT .265625 170 10.625 3 *% .25 160 10. 4 8 .234375 150 9.375 5 A .21875 140 8.75 6 If .203124 130 8.125 7 8 I .1875 .171875 120 110 7.5 6.875 9 A .15625 100 6.25 10 & .140625 90 5.625 11 % .125 80 5. 12 A .109375 70 4.375 13 A .09375 60 3.75 14 A .078125 50 3.125 15 16 f .0703125 .0625 45 40 2.8125 2.5 17 * .05625 36 2.25 18 A .05 32 2. 19 T T6<* .04375* 28 1.75 20 8 3 * .0375 24 1.50 21 aVcr .034375 22 1.375 22 * .03125 20 1.25 23 71T5" .028125 18 1.125 24 .025 16 1. 25 ~5%~5 .021875 14 .875 26 T6T7 .01875 12 .75 27 28 f .0171875 .015625 11 10 .6875 .625 29 9 tftf .0140625 9 .5625 30 1 "8% .0125 8 .5 31 <54]) .0109375 7 .4375 32 tflb .01015625 W .40625 33 db .009375 6 .375 34 5h .00859375 5^ .34375 35 db .0078125 5 .3125 36 9 T2TTF .00703125 4K .28125 37 rttjf .006640625 4# .265625 38 db .00625 4 .25 (145) 146 WEIGHTS AND MEASURES. TABLE No. 1 7. Different Standards for Wire Gage in Use in the United States. Dimensions of Sizes in Decimal Parts of an Inch. (Brown & Sharpe Mfg. Co.) Number of Wire Gage. American or Brown & Sharpe. Birmingham or Stubs' Wire. Washburn & Moen Mfg. Co. Worcester, Mass. Trenton Iron Co., Trenton, N.J. g |S wg -d jj tfl'jj E> Number of Wire Gage. 000000 46875 000000 00000 .45 4375 00000 0000 .46 .454 .3938 .4 40625 0000 000 .40964 .425 .3625 .36 .375 000 00 .3648 .38 .3310 .33 34375 00 .32486 .34 .3065 .305 .3125 1 .2893 .3 .2830 .285 ' .227 .28125 1 2 .25763 .284 .2625 .265 .219 .265625 2 3 .22942 .259 .2437 .245 .212 .25 3 4 .20431 .238 .2253 .225 .207 .234375 4 5 .18194 .22 .2070 .205 .204 .21875 5 G .16202 .203 .1920 .19 .201 .203125 6 7 .14428 .18 .1770 .175 .199 .1875 7 8 .12849 .165 .1620 .16 .197 .171875 8 9 .11443 .148 .1483 .145 .194 .15625 9 10 .10189 .134 .1350 .13 .191 .140625 10 11 .090742 .12 .1205 .1175 .188 .125 11 12 -080808 .109 .1055 .105 .185 .109375 12 18 .071961 .095 .0915 .0925 .182 .09375 13 14 .064084 .083 .0800 .08 .180 .078125 14 15 .057068 .072 .0720 .07 .178 .0703125 15 10 .05082 .065 .0625 .061 .175 .0625 16 17 .045257 .058 .0540 .0525 .172 .05625 17 18 .040303 .049 .0475 .045 .168 .05 18 10 .03589 .042 .0410 .04 .164 .04375 19 20 .031961 .035 .0348 .035 .101 .0375 20 21 .028462 .032 .03175 .031 .157 .034375 21 22 .025347 .028 .0286 .028 .155 .03125 22 23 .022571 .025 .0258 .025 .153 .028125 23 24 .0201 .022 .0230 .0225 .151 .025 24 25 .0179 .02 .0204 .02 .148 .021875 25 20 .01594 .018 .0181 .018 .146 .01875 26 27 .014195 .016 .0173 .017 .143 .0171875 27 28 .012641 .014 .0162 .016 .139 .015625 28 20 .011257 .013 .0150 .015 .134 .0140625 29 30 .010025 .012 .0140 .014 .127 .0125 30 WEIGHTS AND MEASURES. TABLE No. 17. (Continued). jiJ 111 1 -s 111 3 11 11 c jj fjj ^S 11 u gj= li s e f 3 !*"_ c/}"w |l ** a 1 jp 6 21 C/3 t> 31 .008928 .01 0132 .013 .120 .0109375 31 32 .00795 .009 .0128 .012 .115 .01015625 32 33 .00708 .008 .0118 .011 .112 .009375 33 34 .006304 .007 .0104 .01 .110 .00859375 34 35 .005614 .005 .0095 .0095 .108 .0078125 35 36 .005 .004 .0090 .009 .106 .00703125 36 37 .004453 .0085 .103 .006640625 37 38 .003965 .008 .101 .00625 38 39 .003531 .0075 .099 39 40 .003144 .007 .097 40 TABLE No. 1 8. Weight of Iron Wire in Pounds per 1 00 Feet. No. of Wire Gage. American or Brown & Sharpe. Birmingham or Stubs' Wire. No. of Wire Gage. American or Brown & Sharpe. Birmingham or Stubs' Wire. 0000 56.074 54.620 19 0.341 0.467 000 44.4683 47.865 20 0.270 0.324 00 35.265 38.200 21 0.214 0.271 27.9(5(5 30.634 22 0.170 0.207 1 22.178 23.850 23 0.135 0.165 2 17.588 21.373 24 0.107 0.128 3 13.948 17.776 25 0.0849 0.106 4 11.061 15.010 26 0.0673 0.0858 5 8.772 12.826 27 0.0534 0.0678 6 6.956 10.920 28 0.0423 0.0519 7 5.516 8.586 29 0.0335 0.0447 8 4.375 7.214 30 0.0266 0.0381 9 3.469 5.804 31 0.0211 0.0265 10 2.751 4.758 32 0.0167 0.0214 11 2.182 3.816 33 0.0132 0.0169 12 1.730 3.148 34 0.0105 0.0129 13 1.372 2.391 35 0.00836 0.00662 14 1.088 1.825 36 0.006(52 0.00424 15 0.868 1.372 37 0.00525 16 0.684 1.119 38 0.00416 17 0.542 0.891 39 0.00330 18 0.430 0.636 40 0.00262 TABLE No. 10. Decimal Equivalents of the Numbers of Twist Drill and Steel Wire Gage. (Brown & Sharpe Mfg. Co.) Size in Size in Size in Size Size Size in No. Deci- No. Deci- No. Deci- No. in No. in No. Deci- mals. mals. mals. Deci- mals. Deci- mals. mals. 1 .2280 15 .1800 29 .1360 42 .0935 55 .0520 68 .0310 2 .2210 16 .1770 30 .1285 43 .0890 56 .0465 69 .02925 3 .2130 17 .1730 31 .1200 44 .0860 57 .0430 70 .0280 4 .2090 18 .1695 32 .1160 45 .0820 58 .0420 71 .0260 5 .2055 19 .1660 33 .1130 46 .0810 59 .0410 72 .0250 6 .2040 20 .1610 34 .1110 47 .0785 60 .0400 73 .0240 7 .2010 21 .1590 35 .1100 48 .0760 61 .0390 74 .0225 8 .1990 22 .1570 36 .1065 49 .0730 62 .0380 75 .0210 9 .1960 23 .1540 37 .1040 50 .0700 63 .0370 76 .0200 10 .1935 24 .1520 38 .1015 51 .0670 64 .0360 77 .0180 11 .1910 25 .1495 39 .0995 52 .0635 65 .0350 78 .0160 12 .1890 26 .1470 40 .0980 53 .0595! 0(5 .0330 79 .0145 13 .1850 27 .1440 41 .0960 54 .0550 67 .0320 80 .0135 14 .1820 28 .1405 1 TABLE No. 20. Decimal Equivalents of Stubs' Steel Wire Gage. ( Brown & Sharpe Mfg. Co.) . Size o Size -I II 8> Size hjj Size t* Size -S O Size 4- in . ^ in J*^ in o in ^ in in 3 Deci- jS Deci- Z ^ Deci- & Deci- 55.S Deci- " Deci- mals. 1 mals. mals. mals. mals. mals. z .413 H .266 11 .188 29 .134 47 .077 65 .033 Y .404 G .261 12 .185 30 .127 48 .075 66 .032 X .397 F .257 13 .182 31 .120 49 .072 67 .031 w .386 E .250 14 .180 32 .115 50 .069 68 .030 V .377 D .246 15 .178 33 .112 51 .066 69 .029 u .368 C .242 16 .175 34 .110 52 .063 70 .027 T .358 B .238 17 .172 35 .108 53 .058 71 .026 S .348 A .234 18 .168 36 .106 54 .055 72 .024 R .339 1 .227 19 .164 37 .103 55 .050 73 .023 Q .332 2 .219 20 .161 38 .101 56 .045 74 .022 P .323 3 .212 21 .157 39 .099 57 .042 75 .020 O .316 4 .207 22 .155 40 .097 58 .041 76 .018 N .302 5 .204 23 .153 41 .095 59 .040 77 .016 M .295 6 .201 24 .151 42 .092 60 .039 78 .015 L .290 7 .199 25 .148 43 .088 61 .038 79 .014 K .281 8 .197 26 .146 44 .085 62 .037 80 .013 J .277 9 .194 27 .143 45 .081 63 .036 I .272 10 .191 28 .139 46 .079 64 .035 In using the gages known as Stubs' Gages, there should be constantly borne in mind the difference between the Stubs Iron Wire Gage and the Stubs Steel Wire Gage. The Stubs Iron Wire Gage is the one commonly known as the English Standard Wire, or Birmingham Gage, and designates the Stubs soft wire sizes. The Stubs Steel Wire Gage is the one that is used in measuring drawn steel wire or drill rods of Stubs' make, and is also used by many makers of American drill rods. . t s ft 8 t TRIGONOMETRY. 157 The Trigonometrical Table and Its Use. Table No. 21 gives sine, cosine, tangent, and cotang, to angles from to 90 degrees with intervals of 10 minutes. For sine or tangent find the degree in the left-hand column and find the minutes on the top of the table. For instance, sine to 18 40' = 0.32006. If cosine or cotangent is wanted, find the degree in the column at the extreme right and the minutes at the bottom of the table. For instance, cotang 48 10' = 0.89515. As the table only gives the angles and their trigonometrical functions with 10-minute intervals, any intermediate angle must be calculated by interpolations. For instance, find sine of 60 15' 10". Solution : Sine 60 20' 0" = 0.86892 Sine 60 10' 0" = 0.86748 Difference of 10' 0" = 0.00144 60 15' 10" 60 10' 0" = 5' 10' = 310 seconds and a difference of 10' = 600" increases this sine 0.00144. Therefore a difference of 310 seconds will increase the sine. 310 * t M = 0.00074 and sine 60 10' 0" 0.86748 Therefore sine 60 15' 5" = 0.86822 IMPORTANT. During all interpolations concerning the trigonometrical functions, remember the fact that if the angle is increasing both sine and tangent are also increasing, and corrections found by interpolations must be added to the num- ber already found ; but as the cosine and cotangent decrease when the angle is increased, for these functions the corrections must be subtracted. Interpolations of this kind are not strictly correct, as neither the trigonometrical functions nor their logarithms differ in pro- portion to the angle. The error within such small limits as 10 minutes is very slight. When very close calculations of great distances are required, tables are used which give the functions with less difference than 10 minutes; but for mechan- ical purposes in general these interpolations are correct for all ordinary requirements. It is very seldom in a draughting office or a machine shop that any angle is measured for a difference of less than 10'. To Find Secant and Cosecant of Any Angle. Divide 1 by cosine of the angle and the quotient is secant of the same angle. Divide 1 by sine of the angle and the quotient is cosecant of the same angle. I to TRIGONOMETRICAL TABLES. 00 00 00 CO 00 00 CO 00 CO OO L t t s . r-io?iOOOOO'Mr7Ot^O: g >Q G: -f X ?M CO c: -._i O I- Cl O > O 00 Q Or-ccO'^o7)Oi^co opoo < ^' i i i .I i iiMCNicN|cccoeococcco bbobbbo'bbbo'bbbbbbbbbbbb CM -^ O t- OS i C lCOvOlTC3iO(M"*iOj>- TRIGONOMETRICAL TABLES. 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Ct O C^ - OS - -T- -- GOai Oi' IOI-H?DC: a-^r^t- ^ 01 os oi as os os os o: o a ; os TRIGONOMETRICAL TABLES. 00 00 GO CO (M i lCOCiOas-"fCCCOGOCOl^COOiQCO->OCOCCCO>OOOCOOit-l^-OiCOO5GOO^ lr-^(MO5t-O-iCOiOCOGCO-COC-' 'Jt- 0Or-G005O'-HiMCO^001r-GOOSO i c^i c^' i c> O 49 00 t T i O i~"i "^ '""' ' ^ t->- GO 00 s OCVICDC^OO^ ii r- 1 i 4 C *O-'CO rH CM CO CO GO -<* t~ -^ot^t^coco ooccoos CCOCO HCO-HCOlOJ>" ^- J> t^* ^O CM C^l CO OS i~H O t^* CO *Q 00 CO CO O *"^ CO OS OS 00 !> *^Hi Irfi^tii-HOO^OSCCCX; >OOO' i^COOO-- iCO*Ot I CO O GOGOcdcOCOOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOS T-lr-I^HrHr-lT-(^-li-iT-(T-((MC^lM TRIGONOMETRICAL TABLES. CC GO C^ t~- O". I O O5 O OS OcDt--OOCr. o^^loT-T'ti - CD CD CD CD CD tD CD !> t"^ r~- t-* Ir t t^- t~- r 00 00 GO GO 00 00 os' os' os os o; oi OS os' OS os os' os' oi Os' OS os os cs os' os' 05 os' g O C^ O* OS O *-; GO GO CO 00 00 00 C^ Ci Cw C^ Gw G^ "*^ CD O 00 =0 05 I I JD ^a J ! 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OOOOOOOOOOOSOSOSCT-OSOSOSOSOSOSC: OS CO fi 2 COCOt^(MOSi-HOOCOCOQ^OOSCOOCO(MOOOOSCO' I T-^ T-^ iju i^j <^^j i^v u*; ^f ILJ CT^J w> "^r r^- <^^ ^^ ^^ GO >O ^ *O 00 CS W CO !> l>- t^ Is* 00 00 QO QO 00 OS OS OS OS OS OS OS ui OS OS OS OS OJ OS OS OS o: OS 05 OS OS OS OS os OS o; OS OS OS OS OS OS O: OS OS OS) OS OS OS OS OS OS O OS OS OS OS OS OS O^ OS OS OS OS OS OS OS OS OS OS OS OS OS OS O^ O^ OS O^ OS OS OS OS OS w - OS OS OS Oi OS OS* 05 05 OS OS OS OS' OS* OS OS OS OS* OS OS O: OS* OS Os' OS* OJ tr~ **s ^4 iw uu <^ u^ 'n 1 ^^ -*~- ^ "^ CO *^ C CO t^* GO 00 C^ OS OS ^^ CO !> t'- t*" !> 00 OO QO 00 00 O^ Oi OS OS OS OS OS OS OS OS OV OS OS OS OS OS OS OS O* OS OS OS OS OS OS CS OS OS OS OS OS OS OS OS OS OS OS OS OS OS OS* OS OS OS OS OS 05 OS OS OS OS OS OS OS* OS OS 172 TRIGONOMETRICAL TABLES. 1 3 (M CO O OS CO>O ft T-IOiCDO5OiOOiiOOieCi li-H OCCiOT i>O-*CT. O it-00-* i^- GO GO GO GO GO O^ O^ C^ OS OS OS OS C^ OS OS OS CS OS OS OS OS OS s ^ B ! t-OOQOOOCOGOOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOSOS 82S OSO.OSOSOS g^S TRIGONOMETRICAL TABLES. 173 COt-COCDI^OCCl^CXDr- CO lO CO OO Q ^^ i-t-t-r-~t-t-.ao6ooboooooooiosasciO5 OS OS OS 05 01 O !> CD CO CO i vv uu T-^ vj ^r % OS O5 OS Ol" OS* OS* OS O5 O5* OS O5' OS OS OS* Ol' O5 OS* OS* OS Ol* O5 O5 " -2 x^* *^ i~*~ yu >u) i> >i? <*-> i"^ W3 icj iju oa ou u^> ^^ (f <*-> crj) 113 1^- O^ C*&C*a*&ci&&&C*C*G*0*&G*&0*C*aio:>G*Gi O5 CO CM O O o o^ c> *o ^o w ^2 ^^ -_ - -- ,- . ..^-^^ T^ ^ CO O CQ ^J* * OO CO iQ t^ Oi *"H CO O lr^- 00 ^^ C^l CC O iO 00 ^^ ^ CO ^ CO tr-* Oi CO Kr\ CDcDcqcDt-. t-lr-t-t-00000000000005050105050505 a OOt^COCOCDOiiOOiCOOi^CO^COOOiOC 1-" 00 00 00 CO GO CO OO Oi Oi Oi Oi Oi Oi OS h3 9 1 ? *^ *^T P? T ~^ ^ "^ ^ ^O Oi ^ CO ^t 1 CD t~^ Oi T*^ ^-1 ^J* *O t^* OO O b OS os" o: os oi oi o: OS oi oi oi oi OS oi oi oi oi OS os' oi oi os 174 TRIGONOMETRICAL TABLES. t-DiO-*eOO^CO(M CO^COt-?Oi--^ >t---i Angled Triangles. FIG. 18. Right-angled triangles (see Fig. 18) may be solved by the following for- A mulas : Solving for Any Side. A = CXsmea = *X tang. *=. = -* B = C X cos. a = A X cot. a = - A sec. tang. # c = Solving for Any Function or for Any Angle. Sin. a A c Tang. a A B Cos. a B c Cot. a B ~ A Sin. a = cos. b Tang. a = cot. b Sin. b = cos. a Tang. b = cot. a Sin. b B ~ ry Sin. b = sin. - = sin. a C A C C Sin. c = sin. # = sin. b A B # = 180 (b + c) b = 180 (a + c) c =180 (a+b) Solving for Area. Area sin - c X A x B sin, a X C X ^? _ sin, b X A X C 222 EXAMPLE 1. Find the length of the side C (see Fig. 20) when angle a 20 38' 12", angle c 117 48' 5", and side A 12.75 feet long. NOTE. The angle c exceeds 90, therefore the supplement of the angle must be used, which is 180 117 48' 5" 62 11' 55". Thus the solution : c _ 12.75 X sin. 62 11' 55" sin. 20 38' 12" c _ 12.75 X 0.88456 0.35243 C 32 feet long. TRIGONOMETRY. 179 EXAMPLE 2. Find the length of the side B (see Fig. 20) when angle b is 41 33' 43", side C is 32 feet and side A is 12.75 feet. In this example two sides and their included angle are given and the third side is required ; therefore the formula B \/A 2 + C 2 2 A C cos. b must be used. Solution : B Vl2.75 2 + 32 2 2 X 12.75 X 32 X 0.748238 B VH86.562 610.562 B \/576 = 24 feet long. EXAMPLE 3. Find the length of the side B when side A is 12.75 feet long, angle b is 41 33' 43" and angle c is 117 48' 5' . ( See Fig. 20). In this problem one side and its two adjacent angles are given ; therefore it can not be solved directly by any of the preceding formulas, but the first thing to do is to find the angle opposite to side A. Thus: Angle a = 180 (41 33' 43" + 117 48' 5") = 20 38' 12". The side B may be found by the formula A sin. b ** sin. a Solution : A sin. 41 33' 43" B = sin. 20 38' 12" B = 12 ' 75 * 0.66343 0.35242 B = 24 feet long. EXAMPLE 4. Find length of the side C when B is 24 feet long, angle c is 117 48' 5" and the side A is 12.75 feet long. ( See Fig. 20). Solution : C = A* + J5* 2A cos. c C Vl2?75 2 24 2 2 X 12.75 X 24 X ( 0.4664) C Vl 62.56 + 576 + 285.44 C Vl02T = 32 feet long. NOTE. In this example the cos. of 117 48' 5" is used, which, in numerical value, is equal to cos. of 62 11' 55" = 0.4004, but cos. in the second quadrant is negative (see page 154): therefore cos. 117 48'' 5" = ( 0.40045) and the essential sign of the last product after it is multiplied by this negative cos. must change from to -f ( See Algebra, page 63). I So TRIGONOMETRY. EXAMPLE 5. Find the length of the side A when C is 32 feet long, angle a is 20 38' 12" and angle c is 117 48' 15". NOTE. Supplement to c is 62 11' 55". Solution : _ Csin. a A = sin. c 32 X 0.35242 0.88456 A 12.75 feet long. In this example, as in the preceding one, we use the sup- plement of the angle in obtaining its function, but here it has no influence on the signs because sin. is positive as well in the second as in the first quadrant. EXAMPLE 6. Find angle a in Fig. 20, when A is 12.75 feet, B is 24 feet and C is 32 feet. Solution : _ 576+ 1024 162.5625 C S - a ^ -1636- cos. a 0.93583 Angle a 20 38' 12" EXAMPLE 7. Find angle ft, Fig. 20, by the same formula. Solution : co, = - + --* , _ 12.75 2 + 32 2 24 2 2 X 12.75 X 32 cos, ft = 610 ' 5625 816 cos. ft = 0.748238 Angle ft = 41 33' 43" TRIGONOMETRY. l8l EXAMPLE 8. Find angle c, Fig. 20, by the same formula. Solution : cos e _ 12.75 2 + 24 2 32 2 COS. C = 2 X 12.75 X 4 738.5625 1024 612 cos. c = 0.46640 Supplement to angle c = 62 11' 55", and angle c = 117 48' 5" NOTE. The negative cosine indicates that it is in the sec- ond quadrant, therefore the angle is over 90. The angle corresponding to this cosine is the supplement of angle c. To obtain angle c, the angle of its supplement must be subtracted from 180. EXAMPLE 9. Find angles a, b and c in Fig. 20, when side A is 12.75 feet, B 32 feet, and C 24 feet. cos.^ 32 ^ 242 - 12 - 75 " cos. a = 2 X 32 X 24 1437.4375 1536 cos. a = 0.93583 Angle a = 20 38' 12" Angle b may be found by the formula : n sin. b = sin. a - A sin. b = sin. 20 38' 12" A sin. b = 0.35244 X - 32 12.75 sin. b = 0.35244 X 1.8824 sin. b = 0.66343 Angle b = 41 33' 43" 1 82 TRIGONOMETRY. Angle c may be found by the formula: c = 180 (a + b) c = 180 (20 38' 12" + 41 33' 43") c = 180 62 11' 55" c = 117 48' 5" EXAMPLE 10. Find the area of a triangle (see Fig. 20), when it is known that side A is 12.75 feet, side B is 24 feet, and the including angle C= 117 48' 5". Solution : Sin. to supplement of 117 48' 5" = sin. 62 11' 55" = 0.88456. Area = gjn. C X ^ X Z? Area = 0-88456X^12.75X24 = 135 34 square cet EXAMPLE 11. Find angle c and the sides X and y in the triangle, Fig. 23. Solution : c = 180 (40 -f- 60) = 80 F.Q. 23 The side ;r=:?5_Xsm.40 y sin. 60 25 X 0.64279 0.86603 x 16 - 06975 " 0.86603 X 18.556 meters long. By the use of logarithms the side X is solved thus : Log. X log. 25 -f log. sin. 40 log. sin. 60. Log.X= 1.39794 + (9.80806710) (9.93753110). Log.X= 1.268476 X 18.556 meters long. The side y = 25 X sin ' G 80 sin. 60 _ 25 X 0.98481 0.86603 y 28.429 meters long. By the use of logarithms the sidejj/ is solved thus: Log. y = log. 25 -f log. sin. 80 log. sin. 60. Log. y 1.39794 + (9.99335110) (9.93753110). Log. y = 1.45376 y 28.429 meters long. TRIGONOMETRY. 183 EXAMPLE 12. Find angles c and b and the length of the side X in Fig. 24. 35 Sin c 42 X 0-80902 35 Sin. c = 0.97082 ^ ^^ Angle c = 76 7' 26" Angle = 180 (54 0' 0" + 76 T 26") = 49 52' 34" c . , y 35 X sin. 49 52' 34" bide X = = ^^ 0.80901 By means of logarithms the side ^T is solved thus : Log. X log. 35 -f /^. sin. 49 52' 34" log. sin. 54. Log. X^ 1.544068 + (9.88346310) (9.90795810). Log.X 1.519573 X 33.08 meters long. NOTE. The angle c is obtained by interpolation thus : In the table of trigonometrical functions the sine 0.97100 corre- sponds to the angle 76 10' and the sine 0.97030 corresponds to the angle 76. Thus, a difference of 0.00070 in the sine gives a difference of 10' 600" in the angle. The sine to angle c is 0.97082 The nearest less sine in the table is 0.97030 corresponding to angle 76 0' 0". Difference, 0.00052 Therefore when an increase in sine of 0.00070 corresponds to an increase of 600" in the angle, an increase of 0.00052 will increase the angle 60 X - 00052 == 446" = Q 7' 26" 0.00070 thus, the angle corresponding to the sine 0.97082 must be 76 7' 26". 184 GEOMETRY. PROBLEMS IN GEOMETRICAL DRAWING. cf /, e, d, To divide a straight line into a given number of equal parts. (See Fig. 1). Given line a b, which is to be divided into a given number of equal parts. Draw the line b c, of indefinite length, and point off from b the re- quired number of equal parts, as h,g, join c 1 and 0, and draw the other lines parallel to c 1 a. To erect a perpendicular at a given point on a straight line. FIG. 1 FIG. 2 (See Fig. 2). Given line a b, and the point x. The required perpendicular is x y. 5 Solution : '1 '2 With x as center and any radius, as x 1, cut the line a b at 1 and 2. With 1 and 2 as centers and with a radius somewhat greater than 1 to x, describe arcs intersecting each other at y. Draw x y. This will be the required perpendicular. From a given point without a straight line to draw a per- pendicular to the line. (See Fig. 3). Given line a b and the point c. The required perpendicular is x. Solution : With the point c as center and any radius as c 1, strike the arc 1 to 2. With 1 and 2 as centers and any suitable radius, describe arcs intersecting each other at , lay the straight edge through points n and c and draw the perpendicular x. To erect a perpendicular at the extremity of a straight line. FIG. (See Fig. 4). li n The required perpendicular is x. Solution : From any point, as c, with radius as a c, draw the circle. From point of intersection, #, through center, c, draw the diameter n p. From the points, through the point of intersec- tion at p, draw the perpendicular x. __, The correctness of this con- struction is founded on the principle that inside a half circle no other PROBLEMS IN GEOMETRICAL DRAWING. '85 FIG angle but an angle of 90 can simultaneously touch three points in the circumference when two of these points are in the point of intersection with the diameter and the circumference and the third one anywhere on the circumference of the half circle. The pattern maker is mak- ing practical use of this geometrical principle, when he by a common car- penter's square is trying the correctness of a semi-circular core box, as shown in Fig. 5. FIG. 6 FIG. 7 Draw a line parallel to a given line. (See Fig. 6). Given line a b. The required line x y. Solution : Describe with the compass a from the line a b, the arcs 1 and 2 ; draw line x y, touching these arcs. To divide a given angle into two equal angles. The given angle, a b c, is divid- ed by the line b d. Solution : With b as center and any radius, as b 1, describe the arc 1 to 2. With 1 and 2 as centers and any suitable radius, describe arcs cutting each other at d. Draw line b d, which will divide the angle into two equal parts. To draw an angle equal to a given angle. Given angle a b c. Construct angle x y z. With b as center and any radius, as b 1, describe the arc 1 to 2. Using y as center and without alter- ing the compass, describe the arc /, intersecting y z. Measuring the distance from 2 to 1 on the given angle, transfer this measure to the 2 arc /, through the point of intersection. Draw the line y *, and this angle will be equal to the first angle. NOTE. Angles are usually measured by a tool called a pro- tractor, looking somewhat like Fig. 9 or 10, usually made from metal, and supplied by dealers in draughting instruments. A FIG. 8 i86 PROBLEMS IN GEOMETRICAL DRAWING. protractor may also be constructed on paper and used for measuring angles, but it should then always be made on as large a scale as convenient. FIG. 9 FIG. 10 To draw a protractor with a division of 5. (See Fig. 10). Construct an angle of exactly 90 degrees, divide the arc into nine equal parts, then each part is 10 C ; divide each part into two equal parts and each is 5. Prove that the sum of the three angles in a triangle consists of 180. (See Fig. 11). a F|C - 11 Solution: In the triangle a b c, extend the base line to i. Draw the line o p, parallel to the side a b, thereby the angle g will be equal to the angle d, and the angle h must be equal to angle c. The angle/is one angle in the triangle and /+ g -\- h = 180, therefore / + d + c must also be 180. To draw on a given base line a triangle having angles 90, 30 and 60. (See Fig. 12). Given line a b, required triangle is a, c, b. Solution : Extend the line a b to twice its length, to the point e. With e and b as centers strike arcs intersecting each other and erect the perpendicular a c. With b as center and any radius as /, draw the arc / m. With / as center and with the same radius, describe arc intersecting at m. From b through point of intersection at m, draw line b intersecting the perpendicular at c. This will complete the triangle. FIG. 12 PROBLEMS IN GEOMETRICAL DRAWING. l8 7 To draw a square inside a given circle. (See Fig. 13). Solution : Draw the line a b through the center of the circle. From points of intersection at a and b, describe with any suitable radius arcs inter- secting at;* and ///. Draw through the points the line c d. Connect the points of intersection on the circle and the required square is constructed. FIG. 13, To draw a square outside a given circle. (See Fig. 14). Solution : Draw lines a b and c d, and from points of intersection at b and r, describe half circles; their points of intersection determine the sides of the square. To draw a hexagon within a given circle. (See Fig. 15). Apply the radius as a chord succes- sively about the circle; the resulting figure will be a hexagon. FIG. 15- FIG. 16. To inscribe in a circle a regular polygon of any given number of sides. Solution: Divide .360 by the number of sides, and the quotient is the number of degrees, minutes, and seconds contained in the center angle of a triangle, of which one side will make one of the sides in the polygon. For instance, draw a hexagon by this method. (See Fig. 10). 360 = 6QO 6 i88 PROBLEMS IN GEOMETRICAL DRAWING. FIG. 17 To find the center in a given circle. (See Fig. 17). Solution : Draw anywhere on the circumfer- ence of the circle two chords at ap- proximately right angles to each other, bisect these by the perpendiculars x and ^, and their point of intersection is the center of the circle. FIG. 18 To draw any number of circles between two inclined lines touching themselves and the lines. (See Fig. 18). Solution : Draw center line ef. Draw first circle on line i g. From point of intersection between this circle and the center line draw the line //, perpendicular to a b. Describe with a radius equal to ^, the arc intersecting at g\ draw line^ 1 1 1 , parallel to g i, and its point of intersection with the center line gives the center for the next circle, etc. FIG. -Xr-4 1- To draw a circle through three given points. (See Fig. 19). The given points are #, , and c. Solution : From a and b as centers with suitable radius, describe arcs inter- secting at e e. Draw a line through these points. From b and c as cen- ters, describe arcs intersecting at d d; draw a line through these points. The point where these two lines intersect is the center of the circle. FIG. 20 To draw two tangents to a circle from a given point without same circle. (See Fig. 20). Given point , and the circle with the center . The required tangents are a d, and a b. Solution : Bisect line n a. With c as center and radius a c, describe X^ SE ; NIVERSF PROBLEMS IN GEOMETRICAL DRAWING. the arc b d through the center of the circle. The points of intersection at b and d are the points where the required tan- gents a b and a d will touch the circle. F/G FIG. 22. To draw a tangent to a given point in a given circle. (See Fig. 21). Given circle and the point h, x y is required. Solution : The radius is drawn to the point h and a line constructed perpendicular to it at the point h. This perpendicular, touching the circle at ^, is called a tangent. To draw a circle of a certain size that will touch the perphery of two given cir- cles. (See Fig. 22). Given the diameter of circles a, b, and c. Locate the center for circle , when centers for a and b are given. Solution : From center of a, describe an arc with a radius equal to the sum of radii of a and c. From b as center, describe another arc using a radius equal to the sum of the radii of b and c. The point of intersection of those two arcs is the center of the circle c. NOTE. This construction is useful when locating the center for an intermediate gear. For instance, if a and b are the pitch circles of two gears, c would be the pitch circle located in correct position to connect a and b. To draw an ellipse, the longest and shortest diameter being given. The diameters a b and c d are given. The required ellipse is constructed thus : (See Fig. 23). From c as center with a radius a n, describe an arc f l f. The points where this arc intersects a b are foci. The distance fn is divided into any number of parts, as 1, 2, 3, 4, 5. With radius 1 to b, and the focus/ as center, describe arcs 6 and 6 1 ; with the same radius and with f 1 as center describe arcs 6 2 and 6 3 . With radius 1 to a and/ 1 as center, describe arcs intersecting at 6 and 6 1 ; with the same radius and with / as center, describe arcs intersecting at 6 2 and 6 8 . Continue this operation for points 2, 3, etc., and when all the points for the circumference are in this FIG. 23 190 PROBLEMS IN GEOMETRICAL DRAWING. way marked out, draw the ellipse by using a scroll. It is a property with ellipses that the sum of any two lines drawn from the foci to any point in the circumference is equal to the largest diameter. For instance : e + fe, = ab, or/6 1 6 1 , = a b. Cycloids. Suppose that a round disc, c, rolls on a straight line, a b, and that a lead pencil is fastened at the point r; it will then describe a curved line, a, /, r, ;/, b. This line is called a cycloid. (See Fig. 24). This supposed disk is usual- ly called the generating circle. The line a b is the base line of the cycloid and is equal in length to TT times m r, or practically 3.1416 times the diameter of the generating circle. The length of the curved line a, /, r, , 3, is four times r m, (four times as long as the diameter of the generating circle). A circle rolling on a straight line generates a cycloid. (See Figs. 24 and 25). A circle rolling upon another circle is generating an epicycloid. (See Fig. 26). A circle rolling within another circle generates a hypo- cycloid. (See Fig. 27). To draw a cycloid, the generating circle being given. Solution : Divide the diameter of the rolling circle in 7 equal parts. Set off 11 of these parts on each side of a on the line d e. This will give a base line practically equal to the circumference. Divide the base line from the point a into any number of equal parts; erect the perpendicu- lars, with center-line as centers and a radius equal to the radius of the generating circle describe the arcs. On the first arc from d or e set off one part of the base line. On the second arc set off two parts of the baseline; on the third arc three parts, etc. This will give the points through which to draw the cycloid. PROBLEMS IN GEOMETRICAL DRAWING. 191 To draw an epicycloid (see Fig. 26), the generating circle a and the fundamental circle B being given. Solution FIG. 26 Concentric with the circle B, describe an arc through the center of the generating circle. Divide the circumference of the generating circle into any number of equal parts and set this off on the circumference of the circle B. Through those points draw radial lines extending until they intersect the arc passing through the center of the generating circle. These points of inter- section give the centers for the different positions of the gener- ating circle, and for the rest, the construction is essentially the same as the cycloids. In Fig. 26, the generating circle is shown in seven different positions, and the point , in the circumfer- ence of the generating circle, may be followed from the position at the extreme left for one full rotation, to the position where it again touches the circle B. To draw a hypocycloid. (See Fig. 27). The hypocycloid is the line generated by a point in a circle rolling within another larger circle, and is constructed thus: (See Fig. 27). FIG- 27. Divide the circumference of the gener- ating circle into any number of equal parts. Set off these on the circumference of the fundamental circle. From each point of division draw radial lines, 1, 2, 3, 4, 5, 6. From 11 as center describe an arc through the center of the generating circle, as the arc c d. The point of intersection between this arc and the radial lines are centers for the different positions of the gener- ating circle. The distance from 1 to a on the fundamental circle 1 is set off from 1 on the generating circle in its first new position ; the distance 2 to a on the fundamental circle is set off from 2 on the generating circle in its second position, etc. For the rest, the construction is substantially the same as Figs. 25 and 26. NOTE. If the diameter of the generating circle is equal to the radius of the fundamental circle, the hypocycloids will be a straight line, which is the diameter of the fundamental circle. 192 PROBLEMS IN GEOMETRICAL DRAWING. Involute. An involute is a curved line which may be assumed to be generated in the following manner : Suppose a string be placed around a cylinder from a to ^, in the direction of the arrow (see Fig. 28), and having a pencil attached at b ; FIG. 28 keep the string tight and move the pencil toward r, and the involute, b c, is generated. To draw an involute. Solution : From the point b, (see Fig. 28) set off any number of radial lines at equal distances, as 1, 2, 3, 4, 5. From points of intersection draw the tangents (perpendicular to the radial lines). Set off on the first tangent the length of the arc 1 to b\ on the second tangent the arc 2 to b, etc. This will give the points through which to draw the involute. To draw a spiral from a given F,G. 29 point,*. Solution: Draw the line a b through the point c. Set off the centers r and S, one-fourth as far from c as the distance is to be between two lines in the spiral. Using r as center, describe the arc from c to 1, and using ,5* as center, describe the arc from 1 to 2 ; using r as center, de- scribe the arc from 2 to 3, etc. FIG. Conical Sections. If a cone (see Fig. 30), is cut by a plane on the line a b, which is parallel to the center line, the section will be a hyperbola, If cut by a plane on the line c d, which is parallel to the side, the section will be a parabola. If cut by a plane on the line^, which is parallel to the base line, the section will be a circle. If cut by a line, e /, which is neither parallel to the side, the center- line nor the base, the section will be an ellipse* MENSURATION. 193 MENSURATION. If each side in a square (see Fig. 1) is two feet long, the area of the figure will be 4 square feet; that is, it contains four squares, each of which is one square foot. Thus the area of any square or rectangle is FIG. 1 calculated by multiplying the length by the width. EXAMPLE 1. What is the area of a piece of land having right angles and measuring 108 feet long and 20 feet wide ? Solution : 108 X 20 = 2160 square feet. -2 :feet EXAMPLE 2. What is the area in square meters of a square house-lot 30 meters long and 30 meters wide. Solution : 30 X 30 = 900 square meters. ( Square meter is frequently written m* and cubic meter is written m s ). A square inscribed in a circle is half in area of a square outside the same circle. Divide the side of a square by 0.8862, and the quotient is the diameter of a circle of the same area as the square. The Difference between One Square Foot and One Foot Square. One foot square means one foot long and one foot wide, but one square foot may be any shape, providing the area is one square foot. For instance, Fig. 1 is two feet square, but it con- tains four square feet. One inch square means one inch long and one inch wide, but one square inch may be any shape, provided the area is one square inch. One mile square means one mile long and one mile wide, but one square mile may have any shape, provided the area is one square mile. Area of Triangles. The area of any triangle may be found by multiplying the base by the perpendicular height and dividing the product by 2. 194 MENSURATION. EXAMPLE. Find the area of a triangle 16 inches long and 5 inches per- pendicular height. Solution : Area = 5 X 16 = 40 square inches. The perpendicular height in any triangle is equal to the area multiplied by 2 and the product divided by the base. The area of any triangle is equal to half the base multiplied by the perpendicular height. The perpendicular height of any equilateral triangle is equal to one of its sides multiplied by 0.866. The area of any equilateral triangle may be found by mul- tiplying the square of one of the sides by 0.433. EXAMPLE. f Find the area of an equilateral triangle when the sides are 12 inches long. Solution : Area = 12 X 12 X 0.433 = 62.352 The side of any equilateral triangle multiplied by 0.6582 gives the side of a square of the same area. The side of any equilateral triangle divided by 1.3468 gives the diameter of a circle of the same area. To Figure the Area of Any Triangle when Only the Length of the Three Sides is Given. RULE. From half the sum of the three sides subtract each side separately; multiply these three remainders with each other and the product by half the sum of the sides, and the square root of this result is the area of the triangle. EXAMPLE. Find the area of a triangle having sides 12 inches, 9 inches and 15 inches long. Solution : Half the sum of the sides = 18 Area = V(18 12) X (18 9) X (18 15) X 18 Area = \/6 X 9 X 3 X 18 Area = \/2916 Area = 54 square inches. MENSURATION. 195 To Find the Height in any Triangle when the Length of the Three Sides is Given. (See Fig. 2). The base line is to the sum of the other two sides as the difference of the sides is to the difference between the. two parts of the base line, on each side of the line measuring the perpendicular height. If half this dif- ference is either added to or subtracted c from half the base line, there will be obtained two right-angled triangles, in which the base and hypothenuse are known and the perpendicular may be calculated thus : Using Fig. 2 for an example, and adding half the difference to half the base line, this may be written in the formula: -*/~^2 f ( a 4- b} X (a b} \ c \ 2 x \ a* f - ^ : + I V V 2c 2 / RULE. Multiply the sum of the sides by their difference and divide this product by twice the base; to the quotient add half the base; square this sum (that is, multiply it by itself); subtract this from the square of the longest side, and the square root of the difference is the perpendicular height of the triangle. EXAMPLE. In the triangle, Fig. 2, the sides are: c = 12 inches. a = 9 inches. b = 6 inches ; find the perpendicular height x. _ / (9 + 6) X (9 - 6) 12_ 2 X 12 2 x ~ 81 7.875 2 x- = J 81 62.015 x= J 18.985 x = 4.357 inches. 196 MENSURATION. To Find the Area of a Parallelogram. Multiply the length by the width, and the product is the area. NOTE. The width must not be measured on the slant side, but perpendicular to its length. To Find the Area of a Trapezoid. Add the two parallel sides and divide by two ; multiply the quotient by the width, and the product is the area. (See Fig. 3). FIG. 3. EXAMPLE. h 7 feet. -^ F . nd the area of a trapezo j d (Fig. 3). Solution : Area = ' "*" v X 4 32 square feet. NOTE. The correctness of this may be best understood by assuming the triangle b cut off and placed in the position a, and the trapezoid will be changed into a rectangle 8 feet long and 4 feet wide. The area of any polygon may be found by dividing it into triangles and calculating the area of each separately, and the sum of the areas of all the triangles is the area of the polygon. The Area of a Circle. The area of a circle is equal to the square of the radius multiplied by 3.1416, which written in a formula is, Area = 3.1416 r*. The area of a circle is also equal to the FIG. *. square of the diameter multiplied by 0.7854, which may be written, Area = 0.7854 a a = 180/ 57.2956 / 3714167- " r A = Area of sector. r = radius of sector. a = number of degrees in arc. /= length of arc in same units as A and r. EXAMPLE. The arc of the sector (Fig. 6) is 60 no. and the radius is 6 feet. Find area. 360 TT r 2 60 Area ) ?* 2 7T 860 Area = 60 X 6 X 6 X 3 - 1416 36~0 Area = 18.849 square feet. If the length of the arc is known instead of the number of degrees, multiply the length of the arc by the length of the radius, divide product by 2, and the quotient is the area of the sector. The correctness of this rule will be understood by the rule for area of circles, explained under Fig. 4. MENSURATION. I 99 To Find the Length of Arc of a Segment of a Circle. The length of the arc may be calculated by the formula,* 7 8c C 1-= Length of arc, a fb \ c = Length of chord from a tof ( ( See Fig. 7). C = Length of chord from a to b ) RULE. Multiply the length of the chord of half the arc by 8 ; from the product subtract the length of the chord of the arc ; divide the remainder by 3, and the quotient is the length of the arc. When chord and height of segment are known, the chord of half the arc is calculated thus : Chord of half the arc = f^/ n i _|_ ^2 h = Height of segment (see d f, Fig. 7). n = Half the length of chord (see a d or b d, Fig. 7). When only the radius and the height of the segment are known, the length of the chord of the whole arc expressed in these terms will be : 2 X *J 2 r /t _ /fc 2 The chord of half the arc will be : r fr Therefore the length of the arc will be : \/2r h 2 ) I = 8 X r h /= length of arc (afb, Fig. 7). h = height of segment ( df, Fig. 7 ). r radius of circle (c f, Fig. 7). To Find the Area of a Segment of a Circle. (See Fig. 7). Ascertain the area of the whole sector and from this area subtract the area of the triangle, and the rest is the area of the segment. EXAMPLE. Find the the area of the segment when the radius is 9 inches and the arc 60. * This formula is called " Huyghens's approximate formula for circular arcs," but it is so close that it may for any practical purpose be considered absolutely cor- rect for arcs having small center angles; for center angles as large as 120 , the result is only one quarter of one per cent, too small, and even for half a circle the result is scarcely more than one per cent, small as compared to results calculated by taking TT as 3.1416. FIG. 7 2OO MENSURATION. Solution : Area .of segment = A *JL 0.433 r 2 360 A = 60X9X9X3.1416_ g Q 360 A 42.4116 35.073 A = 7.3386 square inches. In this example the arc was 60, consequently the triangle is equilateral ; therefore its area is found by the formula 0.433 r 2 . ( See area of equilateral triangles, page 194). NOTE. When the segment is greater than a semicircle, calculate by preceding rules and formulas the area of the lesser portion of the circle ; subtract it from the area of the whole circle. The remainder is the area of the segment. To Find the Radius Corresponding to the Arc, when the Chord and the Height of the Segment Are Given. RULE. Add the square of the height to the square of half the chord ; divide this sum by twice the height, and the quotient is the radius. In a formula this may be written : I radius c b or cf ( See Fig. 7). n = half the chord = db h height df The above rule and formula may be proved by rules for right-angled triangles; thus, c b or r equals hypothenuse, and ;/, or half the chord, equals perpendicular, and c d, which is equal tor h, is the base. From the rule that the square of the hypothenuse is equal to the sum of the square of the base and the square of the perpendicular, we have : 2 rh = The perpendicular height of the triangle is always equal to the radius minus the height of the segment. ( See triangle a b c, and height, df, Fig. 7). MENSURATION. 201 TABLE No. 23 Areas of Segments of a Circle. The diameter of a circle = 1, and it is divided into 100 equal parts. h D Area. k D Area. h D Area. 0.01 0.001329 0.18 0.096135 0.35 0.244980 0.02 0.003749 0.19 0.103900 0.36 0.254551 0.03 0.006866 0.20 0.111824 0.37 0.264179 0.04 0.010538 0.21 0.119898 0.38 0.273861 0.05 0.014681 0.22 0.128114 0.39 0.283593 0.06 0.019239 0.23 0.136465 0.40 0.293370 0.07 0.024168 0.24 0.144945 0.41 0.303187 0.08 0.029435 0.25 0.153546 0.42 0.313042 0.09 0.035012 0.26 0.162263 0.43 0.322928 0.10 0.040875 0.27 0.171090 0.44 0.332843 0.11 0.047006 0.28 0.180020 0.45 0.342783 0.12 0.053385 0.29 0.189048 0.46 0.352742 0.13 0.059999 0.30 0.198168 0.47 0.362717 0.14 0.066833 0.31 0.207376 0.48 0.372704 0.15 0.073875 0.32 0.216666 0.49 0.382700 0.16 0.081112 0.33 0.226034 0.50 0.392699 0.17 0.088536 0.34 0.235473 Table No. 23 gives the areas of segments from 0.01 to 0.5 in height when the diameter of the circle is 1. The area of any segment is computed by the following rule: Divide the height of the segment by the diameter of its corresponding circle. Find in the table in column marked -jj- the number which is nearest, and multiply the corresponding area by the square of the diameter of the circle, and the product is the area of the segment. EXAMPLE. Figure the area of a segment of a circle, the height of the segment being 12 inches and the diameter of the circle 40 inches. Solution : 12 divided by 40 = 0.3 In the column marked p find 0.3; the corresponding area is 0.198168. The area of the segment is 40 X 40 X 0.198168 = 317.0688 square inches, or 317 square inches, 202 MENSURATION. To Calculate the Number of Gallons of Oil in a Tank. EXAMPLE. A gasoline tank car is standing on a horizontal track, and by putting a stick through its bung-hole on top it is ascertained that the gasoline stands 15 inches high in the tank. The diameter of the tank is GO inches and the length is 25 feet. How many gallons of gasoline are there in the tank? Solution : 15 divided by 60 is 0.25 In Table No. 23, the area corresponding to 0.25 is 0.153546. Area of cross section of the gasoline is 60 X 60 X 0.153546 = 552.7656 square inches. Twenty-five feet is 300 inches ; the tank contains 300 X 552.7656 165829.68 cubic inches. One gallon is 231 cubic inches. The tank contains 165829.68 divided by 231 = 717.88, or 718 gallons. ' NOTE. If the tank is more than half full, figure first the cubical contents of the whole tank if full, then figure the cubical contents of the empty space and subtract the last quantity from the first, and the difference is the cubical contents of the fluid in the tank. Circular Lune. The circular lune is a crescent-shaped figure bounded by two arcs, as a b c and adc. (Fig. 8). Its area is obtained by first finding the area of the segment a dc (having c% for center of the circle), then the area of the segment a b c (having c\ for center of circle), then by subtracting the area of the last segment from the area of the first ; the difference is the area of the lune. A practical example of a circular lune is the area of the opening in a straight-way valve when it is partly shut. FIG. 9. Circular Zone. Circular Zone. The shaded part, a b c d, of the figure is called a circular zone. Its area is ob- tained by first finding the area of the circle and then subtracting the area of the two segments ; the difference is the area of the zone. When the zone is narrow in proportion to the diameter, its area is ob- tained very nearly by following the rule : Add line a b or c d to the diameter of the circle, divide the sum by 2 and multiply MENSURATION. 203 the quotient by the width of the zone, and the product is the area. To Compute the Volume of a Segment of a Sphere. RULE. FIG. 10. Square half the length of its base, and | 8- -I , multiply by 3. To this product add square ^^ ^ \-jf- of the height. Multiply the sum by the "\ height and by 0.5230. /\^ /\1 EXAMPLE. ( Find volume of the spherical segment \ shown in Fig. 10; base line is 8" and \ height is 2". Solution : Segment of a Sphere. Volume = v = (3 X 4 2 -f 2 2 ) X 2 X 0.5236 v = (3 X 16 + 4) X 2 X 0.5236 v = 52 X 2 X 0.5236 v = 54.4544 cubic inches. To Find the Volume of a Spherical Segment, when the Height of the Segment and the Diameter of the Sphere are Known. RULE. Multiply the diameter of sphere by 3, and from this product subtract twice the height of segment. Multiply the remainder by the square of the height and the product by 0.5236. EXAMPLE. The segment (Fig. 10) is cut from a sphere 10 inches in diameter and it is 2 inches high. Figure it by this last rule. Solution : Volume = v = (10 X 3 2 X 2) X 2 2 X 0.5236 v = (30 4) X 4 X 0.5236 v = 26 X 4 X 0.5236 i/ = 54.4544 square inches. To Find the Surface of a Cylinder. RULE. Multiply the circumference by the length, and to this pro- duct add the area of the two ends. A cylinder has the largest volume with the smallest surface when length and diameter are equal to each other. 2O4 MENSURATION. To Find the Volume of a Cylinder. RULE. Multiply area of end by length of cylinder, and the product is the volume of the cylinder. EXAMPLE. What is the volume of a cylinder 4 inches in diameter and 9 inches long? Solution : Area of end r 2 TT Volume = r 2 TT / 2 X 2 X 3.1416 X 9 = 113.0976 cubic inches. To Find the Solid Contents of a Hollow Cylinder. RULE. Find area of end according to outside diameter ; also find area according to inside diameter ; subtract the last area from the first and multiply the difference by the length of the cylinder. Formula : Area = (/? 2 r 2 ) TT / R = Outside radius. r Inside radius. / = Length of cylinder. EXAMPLE. Find the solid contents of a hollow cylinder of 6 feet outside diameter, 4 feet inside diameter and 5 feet long. Solution : Solid contents = x (3 2 2 2 ) X 3.1416 X 5 x (9 4) X 3.1416 X 5 x- 5 X 3.1416 X 5 x = 78.54 cubic feet. FIG. 10. ^--~ To Find the Area of the Curved Surface of a Cone. (See Fig. 10). RULE. Multiply the circumference of the base by the slant height and divide the product by 2 ; the quotient is the area of the curved surface. If the total surface is wanted, the area of the base is added to the curved area. Cone, MENSURATION. 20 5 If the perpendicular height is known, the length of the slant side or the slant height is found by adding the square of the per- pendicular height to the square of the radius and extracting the square root of the sum. Formula : Curved area = x = r = Radius of base. */= Diameter of base. h = Perpendicular height. To Find the Volume of a Cone. RULE. Multiply the area of the base by the perpendicular height, and divide the product by 3. By formula : Volume = r *" h To Find the Area of the Curved Surface of a Frustum of a Cone. (See Fig. 11). RULE. Add circumference of small end to circumference of large end, multi- ply this sum by the slant height and divide the product by 2. Formula : Curved area = (21?ir + 2r *) which reduces to Curved area = (^? -f- r ) * S If the perpendicular height instead of the slant height is known, we have : Curved area = (R -}- r) K R = Large radius. r = Small radius. Frustum of a Cone. _ r ^ _j_ j h = Perpendicular height, ^ = Slant height, 206 MENSURATION. To Find the Volume of a Frustum of a Cone. RULE. Square the largest radius ; square the smallest radius. Multiply largest radius by smallest radius ; add these three pro- ducts and multiply their sum by 3.1416 ; multiply this last product by one-third of the perpendicular height. Formula : Volume = (R* -f r 2 -f- Rr) K _A_ 3 EXAMPLE. Find the volume of a frustum of a cone. The largest diameter is 6 feet, the smallest diameter is 4 feet, and perpen- dicular height is 12 feet. Solution: Volume = x (3 2 + 2 2 + 3 X 2) X 3.1416 X 3 x (9 -f 4 -f- 6) X 3.1416 X 4 x = 19 X 3.1416 X 4 x 238.7616 cubic feet. NOTE. This rule will also apply for finding the solid con- tents of wood in a log. 12. Pyramid To Find the Area of the Slanted Surface of a Pyramid. (See Fig. 12). RULE. Multiply the length of the perim- eter of the base by the slant height of the side (not the slant height of the edge). Divide the product by 2, and the quotient is the area. To Find the Total Area of the Surface of a Pyramid. RULE. Find area of the slanted surface as explained above, and to this add the area of a polygon equal to the base of the pyramid. To Find the Volume of a Pyramid. RULE. Multiply the area^ of the base by one-third of the perpen- dicular height MENSURATION. 207 frlG. 13, Frustum Of a Pyramid To Find the Area of the Slanted Surface of a Frustum of a Pyramid. RULE. Add perimeter of the small end to the perimeter of the large end. Multiply this sum by the slant height of the side (not slant height of edge). Divide the product by 2. To Find the Total Area of the Surface of a Frustum of a Pyramid. RULE. Find the area of the slanted surface as explained above, and to this area add the area of the two ends. Their areas are obtained in the same way as areas of polygons. (See page 196). To Compute the Volume of a Frustum of a Pyramid. RULE. Multiply the area of the small end by the area of the large end, extract the square root of the product, and to this add the area of the small end and the area of the large end ; multiply the sum by one-third of the perpendicular height. Formula : Volume = A (a + A -f \f~A~a) 3 EXAMPLE. Find volume of a frustum of a pyramid when the area of the small end is 8 square feet, the area of the large end is 18 square feet and the perpendicular height is 30 feet. Of\ Volume = v= - 18 _j_ x/18 X 8 ) v = WX ( 8 + 18 + l44~ ) v = 10 X ( 8 -f- 18 -f 12 ) v = 10 X 38 v = 380 cubic feet. To Find the Surface of a Sphere. RULE. Multiply the circumference by the diameter. NOTE. The surface of a sphere is equal to the curved sur- face of a cylinder having diameter and length equal to the diameter of the sphere. 208 MENSURATION. To Find the Volume of a Sphere. RULE. Multiply the cube of the diameter by 3.1416, divide the product by 6 and the quotient is the volume of the sphere. Or, another rule is : Multiply the cube of the diameter by 0.5236 and the product is the volume of the sphere. EXAMPLE. Find the volume of a sphere 15'' diameter. Solution : 0.5236 X 15 X 15 X 15 = 1767.15 cubic inches. A sphere twice as large in diameter as another has twice the circumference, four times the surface, eight times the volume, and if of the same material will weigh eight times as much. To Compute the Diameter of a Sphere when the Volume is Known. RULE. Divide the volume by 0.5236 and the cube root of the quotient is the diameter of the sphere. To Compute the Circumference of an Ellipse. RULE. Add the square of the largest diameter to the square of the smallest diameter and divide the sum by 2; multiply the square root of the quotient by 3.1416. * EXAMPLE. Find the circumference of an ellipse. The largest diameter is 24 inches and the smallest diameter is 18 mches. Solution : Circumference = c= 3.1416 c = 3.1416^450 c 3.1416 X 21.2132 c 66.643 inches. To Compute the Area of an Ellipse. RULE. Multiply the smallest diameter by the largest diameter, and this product by 0.7854. * This Rule gives only approximate results. There is no known rule giving exact results. CIRCUMFERENCES AND AREAS OF CIRCLES. 209 TABLE No. 24. Giving Circumferences and Areas of Circles. Diameter. Circumfer- Area. Diameter. Circumfer- Area. ence. ence. , 0.0491 0.00019 H 2.1108 0.35454 l 0.0982 0.00077 iff 2.1598 0.37122 A 0.1473 0.00173 6 2.2089 0.38829 A 0.1904 0.00307 || 2.2580 0.40574 A 0.2454 0.00479 || 2.3071 0.42357 A 0.2945 0.00690 X 2.3562 0.44179 i ^ 0.3430 0.00940 ft 2.4053 0.40039 0.3927 0.01227 if 2.4544 0.47937 '^ 0.4418 0.01553 1? 2.5035 0.49874 A 0.4909 0.01918 it 2.5525 0.51849 H 0.5400 0.02320 If 2.6016 0.53862 A 0.5890 0.02761 H 2.6507 0.55914 if 0.0381 0.03241 If 2.6998 0.58004 A 0.0872 0.03758 2.7489 0.60132 I! 0.7303 0.04314 'ft 2.7980 0.62299 X 0.7854 0.04909 2.8471 0.64504 H 0.8345 0.05542 tf 2.8962 0.66747 0.8830 0.06213 it 2.9452 0.69029 if 0.9327 0.00922 H 2.9943 0.71349 fV ! 0.9818 0.07070 3.0434 0.73708 j? l r 1.0308 0.0845(5 I! 3.0925 0.76105 U 1.0799 0.09281 3.1410 0.78540 ] 1.1290 0.10144 IST 3.1907 0.81013 H 1-1781 0.11045 1 j 3.2398 0.83525 15- 1.2272 0.11984 l^ 3 f 3.2889 0.86075 J3 1.27(53 0.12902 1 r<> 3.3379 0.88664 1.8254 0.13979 1^ 3.3870 0.91291 rt 1.3744 0.15033 3.4361 0.93956 S 1.4235 0.10120 IA 3.4852 0.96660 11 1.4720 0.17258 1^6 3.5343 0.99402 ;!1 1.5217 0.18427 1 9 3.5834 1.02182 >^ 1.5708 0.19035 IA 3.0325 1.05001 -21- 1.0199 0.20881 18 3.6816 1.07858 11 1.0090 0.22100 iA 3.7306 1.10753 || 1.7181 0.23489 i*l 3.7797 1.13687 A i 1-7671 0.24850 3.8288 1.16659 H 1.8162 0.20250 i^f 3.8779 1.19670 if 1.8653 0.27088 IX 3.9270 1.22718 H 1.9144 0.29165 1^1 3.9761 1.25806 * 1.9635 0.30680 1A 4.0252 1.28931 ft 2.0126 0.32233 iff 4.0743 1.32095 H i 2.0017 0.33824 4.1233 1.35297 210 CIRCUMFERENCES AND AREAS OF CIRCLES. Diameter. Circumfer- ence. Area. Diameter. Circumfer- ence. Area. m 4.1724 1.38538 2^ 6.6759 3.5466 4.2215 1.41817 2fV 6.8722 3.7584 ill 4.2706 1.45134 2X 7.0686 3.9761 iH 4.3197 1.48489 2 T 5 6 7.2649 4.2 ill 4.3688 1.51883 2^ 7.4613 4.4301 4.4179 1.55316 2r 7 5 7.6576 4.6664 li 4.4670 1.58786 2^ 7.8540 4.9087 T 7 ff 4.5160 1.62295 2 T 9 6 8.0503 5.1573 19 4.5651 1.65843 2ft 8.2467 5.4119 - M 4.6142 1.69428 2H 8.4430 5.6727 it 4.6633 1.73052 2% 8.6394 5.9396 || 4.7124 1.76715 2|| 8.8357 6.2126 li 4.7615 1.80415 2# 9.0321 6.4918 1 7 4.8106 1.84154 2J-5 9.2284 6.7772 if 4.8597 1.87932 3 9.4248 7.0686 T 9 * 4.9087 1.91748 8& 9.6211 7.3662 if 4.9578 1.95602 3^ 9.8175 7.6699 it 3 2 5.0069 1.99494 B& 10.0138 7.9798 II 5.0560 2.03425 3X 10.2102 8.2958 # 5.1051 2.07394 3, 5 , ; 10.4066 8.6179 111 5.1542 2.11402 3^ 10.6029 8.9462 If* 5.2033 2.15448 3 T V 10.7992 9.2807 Mi 5.2524 2.19532 3^ 10.9956 9.6211 ill 5.3014 2.23654 3 r 9 . 11.1919 9.9678 ill 5.3505 2.27815 3^ 11.3883 10.3206 ill 5.3996 2.32015 8H 11.5846 10.6796 iff 5.4487 2.36252 3K 11.7810 11.0447 iK 5.4978 2.40528 3H 11.9773 11.4160 if! 5.5469 2.44843 3% 12.1737 11.7933 iff 5.5960 2.49195 3}| 12.3701 12.1768 iff 5.6450 2.53586 4 12.5664 12.5664 ifl 5.6941 2.58016 4 T V 12.7628 12.9622 iff 5.7432 2.62483 4^ 12.9591 13.3641 ill 5.7923 2.66989 4 T 3 3 13.1554 13.7721 ill 5.8414 2.71534 4^ 13.3518 14.1863 1$ 5.8905 2.76117 4 r 5 ,T 13.5481 14.6066 ill 5.9396 2.80738 4^ 13.7445 15.0330 i?! 5.9887 2.85397 4rV 13.9408 15.4656 iff 6.0377 2.90095 4^ 14.1372 15.9043 if! 6.0868 2.94831 4 r 9 ,5 14.3335 16.3492 Jff 6.1359 2.99606 4>/ 8 14.5299 16.8002 HI 6.1850 3.04418 4H 14.7262 17.2573 iff 6.2341 3.0927 4^: 14.9226 17.7206 2 6.2832 3.1416 4}| 15.1189 18.19 2A 6.4795 3.3410 4% 15.3153 18.6655 CIRCUMFERENCES AND AREAS OF CIRCLES. 211 Diameter Circumfer- ence. Area. Diameter Circumfer- ence. Area. HI 15.5116 19.1472 10# 32.9868 86.5903 5 15.7080 19.6350 10# 33.3795 88.6643 5M 16.1007 20.6290 IOK 33.7722 90.7625 5% 16.4934 21.6476 10^ 34.1649 92.8858 5# 16.8861 22.6907 11 34.5576 95.0334 5^ 17.2788 23.7583 HH 34.9503 97.2055 5# 17.6715 24.8505 nx 35.343 99.4019 5% 18.0642 25.9673 ii# 35.7357 101.6234 5% 18.4569 27.1084 HM 36.1284 103.8691 6 18.8496 28.2744 IIH 36.5211 106.1394 6^ 19.2423 29.4648 ii# 36.9138 108.4338 6% 19.635 30.6797 11% 37.3065 110.7537 6H 20.0277 31.9191 12 37.6992 113.098 6^ 20.4204 33.1831 12tf 38.4846 117.859 6^ 20.8131 34.4717 12/2 39.2700 122.719 6% 21.2058 35.7848 12* 40.0554 127.677 6% 21.5985 37.1224 13 40.8408 132.733 21.9912 38.4846 18# 41.6262 137.887 ^ 22.3839 39.8713 13>^ 42.4116 143.139 7% 22.7766 41.2826 - 18M 43.1970 148.490 7>i 23.1693 42.7184 14 43.9824 153.938 7^ 23.5620 44.1787 14X 44.7678 159.485 7>g 23.9547 45.6636 U/ 2 45.5532 165.130 7% 24.3474 47.1731 14% 46.3386 170.874 7/8 24.7401 48.7071 15 47.1240 176.715 8 25.132S 50.2656 15X 47 9094 182.655 8^ 25.52:.r> 51.8487 15^ 48.6948 188.692 8% 2.->. 1)1 82 53.4561 15* 49.4802 194.828 8^ 26.3109 ; 55.0884 16 50.2656 201.062 S/2 26.7086 56.7451 16^ 51.051 207.395 8^ 27.0963 58.4264 16^ 51.8364 213.825 8% 27.489 60.1319 16% 52.6218 220.354 8/8 27.8817 61.8625 17 53.4072 226.981 9 28.2744 63.6174 17X 54.1926 233.706 $% 28.6671 65.3968 17^ 54.9780 240.529 9% 29.0598 (57.2008 17% 55.7634 247.450 9/8 29. 4525 69.0293 18 56.5488 254.470 9K 29.8452 70.8823 18% 57.3342 261.587 9^ 30.2379 72.7599 18# 58.1196 268.803 9% 30.6306 74.6619 18% 58.905 276.117 9% 31.0233 76.5888 19 59.6904 283.529 10 31.4160 78.5400 19% 60.4758 291.040 10# 31.8087 80.5158 19^ 61.2612 298.648 iox 32.2014 82.5158 19% 62.0466 306.355 10^ 32.5941 84.5409 20 62.8320 314.16 212 CIRCUMFERENCES ANt> AREAS OF CIRCLES. Diameter Circumfer- ence. Area. Diameter Circumfer- ence. Area. 21 65.9736 346.361 66 207.34 3421.19 22 69.1152 380.134 67 210.49 3525.65 23 72.2568 415.477 68 213.63 3631.68 24 75.3984 452.39 69 216.77 3739.28 25 78.540 490.87 70 219.91 3848.45 26 81.681 530.93 71 223.05 3959.19 27 84.823 572.56 72 226.19 4071.50 28 87.965 615.75 73 229.34 4185.39 29 91.106 660.52 74 232.48 4300.84 30 94.248 706.86 75 235.62 4417.86 31 97.389 754.77 76 238.76 4536.46 32 100.53 804.25 77 241.90 4656.63 33 103.67 855.30 78 245.04 4778.36 34 106.81 907.92 79 248.19 4901.67 35 109.96 962.11 80 251.33 5026.55 36 113.10 1017.88 81 254.47 5153.00 37 116.24 1075.21 82 257.61 5281.02 38 119.38 1134.11 83 260.75 5410.61 39 122.52 1194.59 84 263.89 5541.77 40 125.66 1256.64 85 267.04 5674.50 41 128.81 1320.25 86 270.18 5808.80 42 131.95 1385.44 87 273.32 5944.08 43 135.09 1452.20 88 276.46 6082.12 44 138.23 1520.53 89 279.60 6221.14 45 141.37 1590.43 90 282.74 6361.73 46 144.51 1661.90 91 285.88 6503.88 47 147.65 1734.94 92 289.03 G647.61 48 150.80 1809.56 93 292.17 6792.91 49 153.94 1885.74 94 295.31 6939.78 50 157.08 1963.50 95 298.45 7088.22 51 160.22 2042.82 96 301.59 7238.23 52 163.36 2123.72 97 304.73 7389.81 53 166.50 2206.18 98 307.88 7542.96 54 169.65 2290.22 99 311.02 7697.69 55 172.79 2375.83 100 314.16 7853.98 56 175.93 2463.01 101 317.30 8011.85 57 179.07 2551.76 102 320.44 8171.28 58 182.21 2642.08 103 323.58 i 8332.29 59 185.35 2733.97 104 326.73 1 8494.87 60 188.50 2827.43 105 329.87 8659.01 61 191.64 2922.47 106 333.01 8824.73 62 194.78 3019.07 107 336.15 8992.02 63 197.92 3117.25 108 339.29 9160.88 64 201.06 3216.99 109 342.43 9331.32 65 204.20 3318.31 110 345.58 9503.32 Strength of flfoaterfals. The strength of materials may be divided into Tensile, Crushing, Transverse, Torsional, or Shearing, and besides this, the elasticity of the material or its resistance against deflection must also be taken into consideration in figuring for strength. Tensile Strength. From experiments it is known that it it will take from 40,000 to 70,000 pounds to tear off a bar of wrought iron one inch square. Therefore we usually say that the tensile strength of wrought iron is from 40,000 to 70,000 pounds, according to quality. The average is 50,000 to 55,000 pounds. The tensile strength of any body is in proportion to its cross sectional area ; thus, if a bar of iron of one square inch area will pull asunder under a load of 40,000 pounds, it will take 80,000 pounds to pull asunder another bar of the same kind of iron but of two square inches area. The tensile strength is independent of the length of the bar, if it is not so long that its own weight must be taken into consideration. Table No. 24 gives the load which will pull asunder one square inch of the most common materials. No part of any machine should be strained to that limit. A high factor of safety must be used, sometimes from 4 to 30 or even more, which will depend upon the kind of stress the mem- ber is exposed to, as dead load, variable load, shocks, etc. Dif- ferent factors of safety are also used for different kinds of material. (See page 274). flodulus of Elasticity. The modulus of elasticity for any kind of material is usually defined as the amount of force which would be required to stretch a straight bar of one square inch area to double its length or compress it to nothing, if this were possible. But a more comprehensive definition is to say that the modulus of elasticity is the reciprocal of the fractional part of the length which one unit of force will, within elastic limit, stretch or com- press one unit of area. For instance, if the modulus of elasticity for a certain kind of wrought iron is 25,000,000, it means that it would take 25,000,000 pounds of pulling force to stretch a bar of (213) 214 STRENGTH OF MATERIALS. one square inch area to double its length, if this could possibly be done ; but it means also which is exactly equivalent that one pound of pulling force will stretch a bar of one square inch area one 25-millionth part of its length, or one pound compressive force will shorten the same bar one 25-millionth part of its length, and that two pounds of force will stretch or compress twice as much, three pounds thrice as much, etc. Strength of Wrought Iron. From experiments it is known that wrought iron can not very well be stretched or compressed more than one-thousandth part of its length without destroying its elasticity ; therefore if a bar of wrought iron has 25,000,000 as its modulus of elas- ticity, one pound will stretch it ^5 o oV OOIF of its length and it would take 25,000 pounds to stretch it ysVo of itslength. Thus, 25,000 pounds would then be said to be its strength at the limit of elasticity for that kind of iron ; 80 to 100 per cent, more will usually be the ultimate breaking load. The pull or load which such a bar can sustain with safety will depend a great deal on circumstances, but it must never exceed 25,000 pounds per square inch of area. It must not even approach this limit if the structure is of any importance or if the load is to be sustained for any lengtty of time, or if it is, besides the load, also exposed to shocks or jar. Strength of Cast Iron. Cast iron of good quality has a modulus of elasticity of 15,000,000 pounds, but if strained so it will stretch y^o o of its length its elasticity is usually destroyed. For instance, a bar of cast iron of one square incn area is exposed to tensile strain, its modulus of elasticity being 15,000,000 pounds and its elas- ticity being destroyed if it stretches ysV o of its length, what then would be its strength at limit of elasticity? One pound will stretch it ^-.^^^ of its length, therefore it must take 10,000 pounds to stretch it ysW of its length; thus we would say that 10,000 pounds is its strength at limit of elasticity. It is not always that cast iron is of as good quality as that; very frequently its elasticity is destroyed if it is exposed to a tensile stress of 6,000 pounds per square inch of area; thus the strength of cast iron at its limit of elasticity is often found to be only 6,000 pounds instead of 10,000 pounds. Besides, it is very often found that a pulling force of 10,000 pounds will stretcn a bar of one square inch area one twelve-hundredth part of its length, and this, of course, gives the modulus of elasticity 12,000,000 pounds. Frequently cast iron is of such quality that it cannot be stretched over ^^^ of its length before its elas- ticity is destroyed. Cast iron is very variable in quality, and STRENGTH OF MATERIALS. 215 especially so with regard to its tensile strength. Generally speaking, we may say that for cast iron the Modulus of elasticity is 12,000.000 to 15,000,000 pounds. Tensile strength at limit of elasticity, 5,000 to 10,000 pounds. Ultimate tensile strength, 10,000 to 20,000 pounds. Elongation Under Tension. The total stretch or elongation of any specimen when exposed to tensile stress within the elastic limit is directly pro- portional to the length of the specimen, but it is inversely proportional to the modulus of elasticity and the cross sectional area of the specimen. The following formulas may, therefore, be used in such calculations : F _ P X L p _ E X s X A s'xTA L P X L A PX L ~y^4~ ~7x~E _ E X sX A E = Modulus of elasticity in pounds per square inch. P Load or force in pounds acting to elongate the specimen. s Total stretch of specimen in inches in the length /,. L Original length of specimen in inches before force is applied. A Cross-sectional area of specimen in square inches. EXAMPLE. From experiments it is known that the modulus of elasticity for a certain kind of wrought iron is 28,000,000; what will then be the total stretch or elongation in a round boiler stay, \% inches in diameter and 6 feet long, when exposed to a stress of 5000 pounds? Solution : 1 1 4 inches diameter = 1.227 inches area (see table, page 209) C feet long = 72 inches. P X L s = X A 5000 X 72 28000000 X 1227 s = 0.0105 inches = total stretch in the stay. NOTE. As already stated, wrought iron can not be stretched as much as one-thousandth part of its original length without danger of destroying its elasticity; thus, for this stay, which is 72 inches, the limit of elasticity will be at a stretch of 0.072 inches ; therefore the stretch produced by a load of 5,000 pounds, which is calculated to be 0.0105 inches, is well within the safe limit. 2l6 STRENGTH OF MATERIALS. TABLE No. 25. Modulus of Elasticity and Ultimate Tensile Strength of Various Materials. MATERIALS. Modulus of Elasticity in Pounds per Square Inch. Ultimate Tensile Strength in Pounds per Square Inch. Modulus of Elasticity in Kilograms per Square Centimeter. Ultimate Tensile Strength in Kilograms per Square Centimeter. Cast steel . . . Bessemer steel . . Wrought iron bars Wrought iron wire Cast iron *.'..< Copper bolts . . . Brass 30,000,000 28.000,000 25,000,000 28,000,000 12,000,000 to 15,000,000 18,000,000 9,000,000 100,000 70,000 55,000 75,000 10,000 to 20,000 35,000 17,700 2,200,000 1,970,000 1,700,000 1,970,000 80,0000 to 1,000,000 1,200,000 630,000 7,000 4,930 3,850 5,250 700 to 1,400 2,400 1,200 Oak ,500,000 17,000 105,000 1,200 Hickory .... Maple ,400,000 ,100,000 20,000 15,000 98,000 77,000 1,400 1,000 Pitch pine .... Pine ,600,000 ,100,000 15,000 10,000 112,000 77,000 1,000 700 Soruce ,100,000 10,000 77,000 700 The two last columns in above table are calculated by the rule : One pound per sq. inch 0.07031 kilograms pei sq. centi- meter and the result is reduced to the nearest round number. Formulas for Tensile Strength. The ultimate tensile strength of any specimen is in propor- tion to its cross-sectional area, and is expressed by the following formula : P = A X S Side of a square bar -v? ^ _ P Diameter of a round bar 'V ; ^ o P = Force in pounds which will pull the specimen asunder. S = Ultimate tensile strength in pounds per square inch. (See Table No. 25). A Cross-sectional area of the specimen in square inches. * Very strong cast iron may have an ultimate tensile strength as high as 30,000 pounds per square inch. STRENGTH OF MATERIALS. 21 7 EXAMPLE. A piece of iron l / 2 inch square is tested in a testing machine and breaks at a total stress of 14,210 pounds. What is the ultimate tensile strength per square inch ? Solution : A bar y 2 inch square has a cross-sectional area of %" X % n is X square inch. S= !i?L = 56,840 pounds per square inch. EXAMPLE. What will be the breaking load for a wrought iron bar }i" X y & " when exposed to tensile stress, the ultimate tensile strength of the iron being 55,000 pounds per square inch, as given in Table No. 25, page 210 ? Solution : A bar y%" X y%" is B 9 r square inches in area. P fa x 55,000 = 7734 pounds, which will break the bar. In order to obtain the safe working stress introduce a suit- able factor of safety, from 5 to 10, according to circumstances, and calculate by the following formulas : P Xf=A X S \~P~X~f . v c Side of a square bar \ _ _ *. P = & A _ P X f Diameter of a round bar = -y f s ^0.7854 6* P Load in pounds. /=. Factor of safety. EXAMPLE. A load of 24,000 pounds is suspended on a round wrought iron bar. The ultimate tensile strength of the iron is 55,000 pounds per square inch. What should be the diameter of the bar to sustain the load, with 10 as the factor of safety ? Solution: A = 24000X1 = 4.363 square inches. 55000 In Table No. 24, we find the nearest larger diameter to be 2}i inches. The diameter may also be calculated directly by the fol- lowing formula : 2l8 STRENGTH OF MATERIALS. -J/'X/ 5" X 0.7854 ^24000 X 10 55000 X 0.7854 D = 2.358, or nearly 2^ inches diameter. To Find the Diameter of a Bolt to Resist a Given Load. RULE. Multiply pull in pounds by the factor of safety. Multiply the ultimate tensile strength of the material by 0.7854 ; divide this first product by the last and extract the square root from the quotient which will then be diameter of bolt at the bottom of the thread. P = T X 0.7854 X S X 0.7854 D' 2 X S X 0.7854 D = Diameter of bolt or screw in the bottom of the thread. P = Load or pull in pounds. y= Factor of safety. S = Ultimate tensile strength per square inch. NOTE. Bolts are frequently exposed to a considerable amount of initial stress, due to the tightening of nuts, which must always be allowed for when deciding upon the load to be considered when calculating their diameter. EXAMPLE. Find diameter of a bolt to sustain a load of 4,450 pounds, taking 10 as factor of safety and ultimate tensile strength of the iron to be 50,000 pounds per square inch. Solution : "4450 X _ /"4450 " ^ 50000 X 0.7854 D 1.064" in the bottom of thread ; thus, a 1 %" screw, standard thread, which is l^ s inches in diameter at the bottom of thread, will be the bolt to use. STRENGTH OF MATERIALS. 2 19 EXAMPLE 2. What size of bolt is required to sustain the same load as is mentioned in the previous example, if only 5 is wanted as a factor of safety? Solution : D - J 4450 X 5 * 50000 X 0.7854 D ~ \A).5G7 D 0.75 inch diameter in bottom of thread. Thus a J^-inch standard screw is too small, as that is only fij" in bottom of thread, but a 1-inch standard screw is suffi- cient, being ||-" in bottom of thread. To Find the Thickness of a Cylinder to Resist a Given Pressure. When the walls of cylinders are thin in proportion to their diameters use the formula : __ S X / x x/ t _ P X R X / P X/ P = Pressure per square inch. R = Radius of cylinder in inches. / = Thickness of cylinder wall in inches. f Factor of safety. S = Ultimate tensile strength of material. When cylinder walls are thick in proportion to the diameter, such as hydraulic cylinders, their thickness is usually figured by the formula : _ P X R s -cf>-f / Thickness of cylinder wall in inches. P = Pressure in pounds per square inch. 7? = Radius of cylinder. S Ultimate tensile strength. f= Factor of safety. EXAMPLE. ^ Find necessary thickness of a hydraulic cylinder of 10-inch inside diameter, made from cast-iron, to stand a pressure of 1000 22O STRENGTH OF MATERIALS. pounds per square inch, with 4 as factor of safety. The ulti- mate tensile strength of the iron is, by experiments, found to be 20,000 pounds per square inch. ( See Table No. 25). Solution : 10-inch diameter = 5-inch radius. f _ 1000 X 5 " 1000 X 5 5000 1000 5000 4000 IX inch. Strength of Flat Cylinder Heads. The American Machinist, in Question No, 147, March 22, 1894, gives the following formula for flat circular heads firmly fixed to the flange of the cylinder : / _ /2 X r z X P ' 3 X Si t == Thickness of cylinder head in inches, r = Radius of cylinder head in inches. P == Pressure in pounds per square inch. Si = Allowable working stress in the material. The allowable working stress may be taken as to A of the ultimate tensile strength and may, for cast iron, be from 1500 to 2500 and for wrought iron from 4000 to 6000. The above formula was used to calculate the thickness of a cast iron cylinder head of 30 inches diameter, to resist a pressure of 100 pounds per square inch. This formula is in that case con- sidered to give sufficient thickness, so that no ribs or braces are needed. The above formula may also be used for wrought iron, by selecting the proper value for Si. Assuming the tensile strength of wrought iron to be 44,000 pounds, and allowing a factor of safety of 8, the value of Si for wrought iron will be 5500. Strength of Dished Cylinder Heads. The American Machinist, in Question 183, April 12, 1894, ves the following formula for dished circular heads, firmly " to the flanges of the cylinder : * to 2,000 600 to 140 42 and 3 parts sand ) Brick work laid ) 240 16 to 17 in lime and sand ] Granite . 10,000 700 224 STRENGTH OF MATERIALS. FIG. 1 When a post or column is long compared to its diameter, its strength will decrease as the length is increased. Anyone will, from every-day observation, know that a short post will support with perfect safety a load which will break a long one. Short columns break under crushing, but long ones break under comparatively light load by the combined effect due to both crushing and flexure. It is, therefore, evident that the strength of long columns follows laws very different from those which apply to short ones. The form of the ends has also great influence on the strength of a column when under crushing and deflective stress. ( See Fig. 1). When both ends are round the column has least strength; if one end is round and one end flat it is stronger, but if both ends are flat and square with the center-line, it is strongest. The proportions are ap- proximately as 1, 2 and 3. Eccentric loading on columns will also have a very destruct- ive effect upon their strength. Theoretical calculations regarding the strength of columns and posts are difficult, and such empirical formulas as the well-known Hodgkingson's or Gordon's formulas are usually re- sorted to. The Hodgkingson formulas for long columns having square ends well fitted are : P = 99,000 X for solid cast iron columns. P = 99,000 X D ~,f for hollow cast iron columns. J~* P = 285,000 X D \ 2 for solid wrought iron columns. P = Breaking load in pounds. D = External diameter in inches. d= Internal diameter in inches. L = Length in feet. When the breaking load as calculated by these formulas exceeds one quarter of the crushing load of a short column of the same metal area, the result must be corrected by the for- mula; STRENGTH OF MATERIALS. PX C 225 - P 4- % C X A P\ = Corrected breaking load of column. C= Crushing strength of material (see Table No. 26). A = Metal area of column in square inches. IMPORTANT. Applying the last formula, the result, / must always be smaller than /'. Table No. 27 was calculated by the following formulas : ( 36000 Column I. Safe Load = 0.1 X < ~ ~fz ~ x (1 -f- (=T 2 - X 0.00025) 36000 X 0.07031 +(~ X 0.00025) Column II. Safe Load = 0.1 X 36000 Column III. Safe Load = 0.1 X Column IV. Safe Load = 0.1 X 1 + / X 0.0004 v/3* 36000 X 0.07031 0.0004) 80000 Column V. Safe Load = 0.1 X j 1 -f t X 0.0025 ) 80000 X 0.07031 Column VI. Safe Load = 0.1 X ) 1 + ( X 0.0025 ^ ( 80000 Column VII. Safe Load = 0.1 X Column VIII. Safe Load 0.1 X Column IX. Safe Load = 0.1 X Column X. Safe Load = 0.1 X 1 + ( _ x 0.0035 ) 80000 X 0.07031 1 +(^X 0.0035) 5000 I 1 + (^X 0.004) 5000 X 0.07031 { -, , ] h Z 2 226 STRENGTH OF MATERIALS. TABLE No. 27. Safe Load on Pillars Having Square Ends Well Fitted. (10 is used as Factor of Safety). U WROUGHT IRON. CAST IRON. WOOD. w m 1. Hollow Pillar Solid Pillar. Hollow Pillar. Solid Pillar. [opruce or white Pine) 3 .2 v '3-S xs | | f 'S> T3 a 3 2 MS '3 "= 5 60 3 1 W. 1 11 1 ll 1 . 3 I sal St? 3 L ~D~ i. ii. III. IV. V. VI. VII. VIII. IX. X. ~D~ 4 3585 252 3567 252 7692 540 7624 536 476 34 4 Q 3567 251 3549 250 7339 512 7105 500 438 30 6 8 3545 250 3510 247 6896 484 6536 460 398 28 8 10 3512 247 3462 243 6400 450 5926 417 357 25 10 12 3475 244 3404 239 5882 413 5314 374 317 22 12 14 3432 241 3338 235 5369 371 4745 333 280 20 14 16 3383 238 3266 230 4878 343 4226 297 248 17 16 18 3332 234 3187 224 4420 311 3749 264 218 15 18 20 3273 230 3103 218 4000 281 3333 234 192 13 20 22 3211 226 3018 212 3620 255 2969 208 170 12 22 24 3147 221 2926 206 3279 230 2653 186 151 10.6 24 26 3080 216 2834 199 2974 211 2376 167 135 9.5 26 28 3010 211 2741 192 2703 190 2137 150 121 8.5 28 30 2938 206 2647 186 2462 173 1928 136 109 7.6 30 32 2866 201 2554 180 2247 158 1745 123 98 6.9 32 34 2793 196 2462 173 2056 145 1585 111 88 6.2 34 36 2719 191 2370 167 1887 133 1440 102 81 5.7 30 38 2645 186 2288 160 1735 122 1321 93 74 5.2 38 40 2571 181 2195 154 1600 112 1212 85 67 4.7 40 42 2498 176 2111 148 1479 104 1113 78 62 4.3 42 44 2426 171 2029 143 1370 96 1028 74 57 4 44 46 2354 166 1950 137 1272 89 952 67 53 3.8 46 48 2284 161 1874 132 1183 83 882 62 49 3.4 48 50 2215 155 1800 127 1103 78 820 58 45 3.2 50 This table is intended, in connection with Table No. 24, to facilitate calculations for pillars of either wood or iron, and may be used with equal advantage for English or metric measures, provided both diameter and length are taken by the same system. For a round pillar divide the length by the. diameter, but for a square or rectangular pillar divide the length by the STRENGTH OF MATERIALS. 22 ^ smallest side. Find the quotient in the column headed -p- and find the safe load per square unit of area in the cor- responding column of the table. Multiply this by the metal area of the pillar, and the product is the safe load, with 10 as factor of safety, on a pillar well fitted and having square ends. For any other kind of ends and any other unit of safety, allow- ance must be made as explained on previous pages. EXAMPLE 1. Find the safe load in pounds, according to Table No. 27, for a round, hollow, cast-iron pillar five feet long, five inches outside and four inches inside diameter, having square ends well fitted and being evenly loaded. Solution : Five feet equals 60 inches, and 00 divided by 5 gives 12. In the first column, under the heading " Length divided by diameter, or smallest side, " is 12, and in that line, in the column headed ** pounds per square inch " for hollow cast-iron pillars, is 5882. The metal area of this pillar is obtained by subtract- ing the area of a circle four inches in diameter from the area of a circle five inches in diameter (see area of circles, page 196; Table, page 209), which is 19.68 12.57 = 7.06, or practically seven square inches, and seven times 5882 equals 41,174 pounds. EXAMPLE 2. Find the safe load in kilograms, according to Table No. 27, for a round spruce post 2 meters long and 20 centimeters in diameter. Solution : Two meters = 200 centimeters and 2 ^ s 10. The corre- sponding constant in the table is 25 kilograms. The area of a circle 20 centimeters in diameter is 814.2 square centimeters, and 25 times 314.2 = 7855 kilograms, as safe load. EXAMPLE 3. What would be the safe load on the same post if it had been 20 centimeters square, instead of round ? Solution : The length is the same; therefore the length divided by the side gives 10, as before, and the corresponding constant is 25 kilograms, but as the cross-sectional area in square centimeters is 20 X 20 = 400, the corresponding load will be 400 X 25 = 10,000 kilograms as the safe load. NOTE. It will be noticed that in figuring the strength of pillars according to this table, the strength of a square pillar will always be to the strength of a round pillar as 1 to 0.7854, while theoretically the strength of a square pillar compared to that of a round pillar will vary with the length, the extremes being 1 to 0.589 for extremely long pillars and 1 to 0.7854 for 228 STRENGTH OF MATERIALS. very short ones. This discrepancy is frequently unimportant in practical work, because pillars are usually comparatively short, and also because a high factor of safety is always used, but it is well to remember and provide for this fact in cases of very long pillars. Hollow Cast=Iron Pillars. By referring to the formulas and considering the laws governing the strength of pillars, it is seen that the strength of pillars increases very fast by increasing their diameter or their sides. In cast-iron pillars this is taken advantage of by making them large in diameter and coring out the stock on the inside. The thickness of the metal may be about T V of the diameter of the pillar. In small pillars it must be thicker in order to obtain good results when casting. A flange is cast on each end to form enough bearing surface, and the pillar is squared off very carefully so that both ends are square with the center-line. This is an important point, as the strength is enormously destroyed by squaring the ends carelessly and thereby bringing the load to act corner-ways on the pillar. Table No. 28 was calculated by the formula : 80000 X metal area >> ( J Safe Load = 0.1 X j 1 + (^X 0.0025) and the result obtained reduced to long tons (2240 pounds). Ten is thus used as a factor of safety ; both ends of the pillar are supposed to be square and evenly loaded. For other shapes of ends, mode of loading, or other factors of safety, propor- tional allowance must be made. For instance, if 15 is required as factor of safety, allow only two-thirds of the load given in the table. If the pillar has only one square end and one round end, allow only two-thirds as much load. If it has both ends rounded, or, which is the same, if the ends have only a very imperfect bearing, allow only one-third as much load. Weight of Cast=Iron Pillars. The weight of a cast-iron pillar may be calculated by the formula : W (D* d' 2 } X L X 2.45 W = Weight of pillar in pounds. D = Outside diameter in inches. d Inside diameter in inches. L = Length of pillar in feet. The weight given in Table No. 28 was calculated by this formula, and the length taken as one foot. STRENGTH OF MATERIALS. 2 29 TABLE No. 28. Safe Load on Round Cast-Iron Pillars. 11 -S (J ih * e 5 '" O.C.C Length of Pillars in Feet. External c eter in in Thicknei Metal in it < Q \s -J X / f o A L X a When a beam is laid in a horizontal position, fixed at both ends and the load evenly distributed over its whole length (see Fig. 11), the formula will be, P 12 X ^ * ^ L X a P Breaking load in pounds. .5" = Modulus of rupture, which is 72 times the weight, in pounds, which will break a beam one inch square and one foot long when fixed in a horizontal position, as shown in Fig. (>, and loaded at the extreme end, and which may be taken as follows : Cast-Iron, 36,000. Wrought Iron, 50,000. Spruce and Pine, 9,000. Pitch Pine, 10,000. These are the nearest values, in round numbers, of 72 times the average value of the constant given in Table No. 30. For the safe load, ^ may be taken as follows : For timber, 1,000 to 1,200 pounds. For cast-iron, 3,000 to 5,000 pounds. For wrought iron, 10,000 to 12,000 pounds. For steel, 12,000 to 20,000 pounds. L = Length of beam in inches. a = The distance in inches from the neutral surface of the section to the most strained fiber. / = Rectangular moment of inertia. The tables on pages 237 and 238 give the moment of inertia about the neutral axis X Y, and the distance a, for a few of the most common sections : (For explanation of moment of inertia and center of gravity see page 293). These formulas have the great advantage of being theoretic- ally correct for beams of any shape of cross-section, made from any material, providing the load is within the elastic limit of the beam, and a correct constant is used for 6" and the correct value obtained for the moment of inertia. STRENGTH OF MATERIALS. 237 -H -Y I 2 / H a - 6 (* B * j 12 H -Vf r _H*x B ~12~ a=*L 2 a 6 T 9 -aU 36 ai a 12 24 d 64 a = = 0.0982 Di D* H f ^(D* d*)^ 64 Z> 32 32 D STRENGTH OF MATERIALS. "T! B H* b -Yr X- 12 -Y.. tfB a = T" f

inches wide, and 8 feet long, supported under both ends and loaded at the center. Solution : J ,_CXBXH* _4 X 125 X 4^ X 8 X 8 8 P 18,000 pounds. EXAMPLE 3. Find the load which will break the beam mentioned in Example 2, if beam is laid flatwise. _ 4 X 125 X 8 X y 2 X 4^ P 10,125 pounds. In the first example the beam is square, 6" X 6" = 36 square inches, and its calculated breaking load is 13,500 pounds. In the second example the beam is rectangular, 8" X 4>" ' = 36 square inches, and laid edgewise its figured breaking load is 18,000 pounds. In the third example the same beam is laid flatwise, and its breaking load is only 10,125 pounds. Thus, by making a beam deep it is possible to secure great strength with only a small quantity of material, but the limit is soon reached where it will not be practical to increase the depth at the ex- pense of the width, because the beam will deflect sidewise and twist and break if it is not prevented by suitable means. The STRENGTH OF MATERIALS. 247 strongest beam which can be cut from a round log is one hav- ing the thickness 1% times the width. The stiff est beam cut from a round log has its thickness l T 7 o times its width. The best beam for most practical purposes which can be cut from a round log has its thickness l l / 2 times its width; for instance, 4 X 6, or 6 X 9, or 8 X 12, etc. The largest side in a beam having its thickness \ l / 2 times its width which can be cut from a round log is found by multiplying the diameter by 0.832. The diameter required in a round log to be large enough for such a beam is found by multiplying the largest side of the beam by L.2; for instance, the diameter of a round log to cut 6" X 9" will be 9" X 1.2 = 10.8 inches, or the diameter of a round log re- quired to cut 8" X 12" will be 12" X 1.2 14.4 inches, etc. To Calculate the Size of Beam to Carry a Given Load. Most frequently the load and the length of span are known and the required size of beam is to be calculated. For a rect- angular beam there would then be two unknown quantities, the width and the thickness, but if it is decided to use a beam having its thickness 1 l / 2 times its width, the thickness may be expressed in terms of the width. H Thickness. B Width. Use formula for rectangular beams, page 239, and it will read, C X (l/^ -B) 2 X B L This will reduce to, C X 2X X B 9 1 This will transpose to, C X 2# EXAMPLE. Find width and thickness of a spruce beam 10 feet long, when fastened at one end and required to carry, with 8 as factor of safety, a load of 1800 pounds at the other end, the thickness to be 1 l / 2 times the width. When the beam is to carry 1800 pounds, with 8 as a factor of safety, its breaking load is 8 X 1800 = 14,400 pounds. 248 STRENGTH OF MATERIALS. Solution : B B S inches in width. H\y z X 8" = 12 inches in thickness. The weight of the beam itself is not considered in this problem. To Find the Size of a Beam to Carry a Given Load When Also the Weight of the Beam is to be Considered. RULE. Calculate first the size of beam required to carry the load? then figure what such a beam will weigh and add half of this weight to the load, if the beam is fastened at one end and loaded at the other, or supported under both ends and loaded at the center, but add the whole weight of the beam to the weight of the load if the load is distributed along the whole length of the beam. Then figure the size of the required beam for this new load. EXAMPLE. Find width and thickness of a pitch pine beam to carry 2000 pounds, with 8 as factor of safety, and a span of 27 feet. The beam is supported under both ends and loaded at the center ; its own weight is also to be taken into consideration. Solution : Find the constant for pitch pine in Table No. 30 to be 150, and find the weight of pitch pine in Table No. 10 to be 50 pounds per cubic foot. When the beam is supported under both ends and loaded at the center it is four times as strong as if fastened at one end and loaded at the other; therefore, constant 150 is multiplied by 4. The load, 2000 pounds, multiplied by 8 as a factor of safety, gives 10,000 pounds as breaking load of the beam. ^ , 16000 X 27 j> = 2X X 150 X 4 3 B = 6.84 " = width, and 1# X 6.84" = 10.26" = thickness. The area is 6.84 X 10.26 = 70 square inches ; the weight per foot is 70 times 50 divided by 144, which equals 24.3 pounds, say 25 pounds. The weight of the beam is 25 X 27 = 675 pounds. This STRENGTH OF MATERIALS. 249 weight is distributed along the whole beam and, therefore, it does not have any more effect than if half of it, or 337 J^ pounds, was placed at the center, but as the beam is to be calculated with 8 as factor of safety, the weight allowed for the beam must be 337^2 X 8 = 2700 pounds. Thus, adding this weight to 16,000 pounds gives 18,700 pounds ; this new weight is used for calcu- ing the size of the required beam. J5 = 374 B = 7.2 inches = width, and IY 2 X 7.2 ff = 10.8 inches, thickness. This, of course is also a little too small, as only the weight of a beam 6.84 inches by 10.25 inches is taken into account, but if more exactness should be required the weight of this new beam may be calculated and the whole figured over again, and the result will be closer. This operation may be repeated as many times as is wished, and the result will each time be closer and closer, but never exact; but for all practical purposes one calculation, as shown in this example, is sufficient. EXAMPLE 2. Find width and thickness of a spruce beam to carry 4200 pounds distributed along its whole length. The span is 24 feet ; use 10 as factor of safety, and also allow for the weight of beam. The thickness of the beam is to be 1% times its width. Solution : 3 / 4200 X 24 X 10 > 2X X 125 X 8 B = 7.65 inches, and H 11.48 inches. iir u,. r u 7 - 65 X 11.48 X 24 X 32 Weight of beam = = 468 pounds. 144 Adding ten times the weight of the beam to ten times the weight to be supported, gives 46,680 pounds. 3 / 46680 X 24 2X X 125 X 8 3 V 497.9 /? = 7.93 inches, and H \% B= 11.9 inches, or prac- tically, a beam 8 inches by 12 inches is required. 250 STRENGTH OF MATERIALS. Crushing and Shearing Load of Beams Crosswise on the Fiber. Too much crushing load must not be allowed at the ends of the beams where they rest on their supports, as all kinds of wood has comparatively low crushing strength when the load is acting crosswise on the fiber. Approximately, the average ultimate crushing strength of wood, crosswise of the fiber, is as follows : White oak, 2000 pounds per square inch. Pitch pine, 1400 pounds per square inch. Chestnut, 900 pounds per square inch. Spruce and pine, 500 to 1000 pounds per square inch. Hemlock, 500 to 800 pounds per square inch. The safe load may be from one-tenth to one-fifth of the ultimate crushing load. When the wood is green or water- soaked, its crushing strength is less than is given above. EXAMPLE. How much bearing surface must be allowed under each end of the beam mentioned in Example 2, providing it also has 10 as a factor of safety ? The crushing strength of spruce crosswise on the fiber is 500 pounds, and using 10 as factor of safety, the load allowed per square inch must be only 50 pounds. The beam is 8 inches wide, and half of 4685 pounds is sup- ported at each end ; thus the length of bearing required under 2:}42 each end will be _ x g = 5.85 inches. Thus, the least bearing allowable should be about 6 inches long. When beams are heavily loaded and resting on posts, or have supports of small area, either hardwood slabs or cast-iron plates should be placed under their ends, in order to obtain sufficient bearing surface for the soft wood. The same care must be exercised when a beam is loaded at one point; the bearing surface under the load should at least be as long as the bearing surface of both ends added together. Short beams are liable to break from shearing at the point of support, especially when loaded throughout their whole length to the limit of their transverse strength. The ultimate shearing strength for spruce, crosswise of the fiber, is 3000 pounds per square inch (see page 273). Safe load may be 300 pounds per square inch. In the above example the beam is 8" X 12" 96 square inches, and its center load is 4085 pounds, or 2342 ]/ 2 pounds at 2342 1/ each end. The shearing stress is -^p = 24.6 pounds per square inch. Hence, the factor of safety against shearing is about 100, and there is not the least danger that this beam will give way under shearing; but such is not always the result* STRENGTH OF MATERIALS. 25 1 Round Wooden Beams. A round beam has 0.589 times the strength of a square beam of same length and material, when the diameter is equal to the side of the square beam. The area of a square beam compared to the area of the round beam is as 0.7854 to 1 ; there- fore it might seem as if that also should be the proportion be- tween their strength, which is the case for tensile, crush- i ig and shearing strength, but not for transverse strength or for deflection, because the material is not applied to such advantage in the round beam as it is in the square one. All preceding formulas for transverse strength of square beams may also be used for round beams if only constant C is multiplied by 0.589, or, say, 0.6. Thus, the formula for a round beam fastened at one end and loaded at the other will be : 0.6 C X Z> 3 NOTE. In a round beam, of course, it will be Z> 3 instead of //" 2 b for a rectangular one. EXAMPLE. Find the load in pounds which will break a spruce beam 12 feet long and inches in diameter when supported under both ends and loaded at the center. (Find constant C in Table No. 30.) Solution : 4 X 0.6 C X Z> 3 L 4 X 0.6 X 125 X 6 X 6 X 6 12 P = 5400 pounds. To Calculate the Size of Round Beams to Carry a Given Load When Span is Known. Where the load and span are known, the diameter of the beam is calculated, when fastened at one end and loaded at the other, by the formula : _ iP XLX factor of safety = >~~ 0.6 C RULE. Multiply together the load in pounds, factor of safety and length of span in feet, divide this product by six-teji<;hs of the 2 5 STRENGTH OF MATERIALS. constant in Table No. 30, and the cube root of this quotient is the diameter of the beam. EXAMPLE. A round spruce beam is fastened into a wall, and is to carry 1200 pounds on the free end projecting 4 feet from the wall, with 8 as a factor of safety, the weight of the beam not to be considered. Find diameter of beam. Solution : 1200 X 4 X 8 0.6 X 125 38400 75 V 512 D = 8 inches diameter. Load Concentrated at Any Point, Not at the Center of a Beam. If a beam is supported at both ends and loaded anywhere between the supports but not at the center (see Fig. 25), it will carry more load than if it was loaded at the center. With regard to breaking, the carrying capacity is inversely as the square of half the beam to the product of the short and the long ends between the load and the support. For instance, a beam 10 feet long is of such size that when it is supported under both ends and loaded at the center it will carry 1400 pounds. How many pounds will the same beam carry if loaded 3 feet from one end and 7 feet from the other ? Solution : ^ 1400 X 5 2 y 7X3 1400 X 25 21 FIG. 25. X= 1666% pounds. If weight of beam is also in- cluded in its center-breaking-load, the formula will be : P Breaking load (including weight of beam) if applied at the center in pounds. F = Half the length of the span. STRENGTH OF MATERIALS. 253 JF= Weight of beam. P\ = Breaking load applied at n. Load at n X distance b Load on Pier A = Load on Pier B = span Load at n X distance a span Beams Loaded at Several Places, FIG. 26. "i * -s 1 i 10-ffa When a beam is loaded at several places the equivalent center load and the load on each support may be calculated as shown in the following example: (See Fig. 26). The equivalent center load for a = 4 X 2 -- X 1000 = 555.6 Ibs. 12 X 12 The equivalent center load for b = 10 X 12 X 12 X 800 = 777.8 Ibs. The equivalent center load for c = 1 f X 8 X 900 = 800 lbs. The equivalent center load for d 19 X 5 X 300 = 197.9 lbs. 12 X 12 The equivalent center load for loads a, b, c and d is 2331.3 pounds. The load on Pier A = (5 X 300) + (8 X 900) + (14 X 800) + (20 X 1000) _ WQ2y lbs 24 The load on Pier B = (4 X 1000) + (10 X 800) + (16 X 900) + (19 X 300) = 1337 y lbs 24 NOTE. The sum of the load on supports A and B is always equal to the sum of all the loads ; therefore, by subtracting the, 254 STRENGTH OF MATERIALS. calculated load on B from the total load the load on A is ob- tained. By subtracting the calculated load at A from the total load, the load on B is obtained. To each load as calculated above for each support also add half the weight of the beam. To Figure Sizes of FIG. 23. Beams When Placed in an Inclined Position. Figure all calculations concerning the transverse strength from the dis- tance S, and leave the length L out of consideration. If the distance S cannot be obtained by measurement it may be found by multiplying L by cosine of angle a. DEFLECTION IN BEAMS WHEN LOADED TRANSVERSELY. Experiments and theory both prove that if the span is increased and the width of the beam increased in the same pro- portion the transverse strength of the beam is unchanged ; but such is not the case with its stiffness. If a beam is to have the same stiffness its depth must be increased in the same ratio as the span, providing the width is unchanged. Within the elastic limit of the beam the deflection is directly proportional to the load; that is, half the load produces half the deflection, out doubling the load will double the deflection. Deflection is proportional to the cube of the span ; that is, with twice the length of span the same load will, when the other dimensions of the beam are unchanged, produce eight times as much deflection. Deflection is inversely as the cube of the depth (thickness) of the beam. For instance, if the depth of a beam is doubled but the length of span and the width of beam is unchanged, the same load will produce only one-eighth as much deflection. Deflection is inversely as the width of the beam ; for instance, when a beam is twice as wide as another beam of the same material but all the other dimensions are unchanged, the same load will produce only half as much deflection. The deflection in a beam caused by various modes of load- ing is calculated by the following formulas : For beams laid in a horizontal position and loaded trans- versely, fastened at one end and loaded at the other : (See Fig. 6). 3X^X7 STRENGTH OF MATERIALS. 255 For beams laid in a horizontal position, fastened at one end and loaded thoroughout the whole length : (See Fig. 7.) P X Z 3 C* " 8 X E X S For beams laid in a horizontal position, supported under both ends and loaded at the center: (See Fig. 8). S 48 X E X I This formula may be transposed and used to calculate modulus of elasticity from the results obtained when specimens are tested for transverse stiffness. Deflection should be care- fully measured but the specimen must not be bent beyond its elastic limit ; the modulus of elasticity is calculated by the trans- posed formula : 48 X S X / For a square specimen / is (side of beam) 4 divided by 12. (See moment of inertia, page 237). (Also see rule for calculating modulus of elasticity, page 265). For beams laid in a horizontal position, supported under both ends and loaded uniformly throughout their whole length : (See Fig. D). S= 5 X P X L * 384 XE X r For beams laid in a horizontal position, fixed at both ends, and loaded at the center : (See Fig. 10). S= 192 XEXf For beams laid in a horizontal position, fixed at both ends and loaded uniformly throughout their whole length : (See Fig. 11). 384 X E X / In these formulas the definitions of the letters are : S = Deflection in inches. P = Load in pounds. L = Length of span in inches. E = Modulus of elasticity in pounds per square inch. / = Rectangular moment of inertia. (See pages 237-238). These formulas are applicable to any shape of section or material, when the load is within the elastic limit. 256 STRENGTH OF MATERIALS. For beams of symmetrical section it is more convenient to use the following equally correct but more practical formulas, by which the deflection is calculated directly from the size of the beam by simply using a constant obtained by experiment and reduced by calculation to a unit beam one foot long and one inch square, thus avoiding both the use of the modulus of elas- ticity and the moment of inertia. When beams are supported under both ends and loaded at the center, and the weight of the beam itself is not considered, the following formulas may be used for solid rectangular beams laid in a horizontal position : ? _ L 8 X P X c 3 _ H*XB L =. J// 3 X B XS ^ ' X B X c S X H* X B S X B L*X P B _ Z, 3 X P X c p _ H* X. B X S SXH* L*Xc S = Deflection in inches. ff= Thickness of beam in inches. B =. Width of beam in inches. L = Length of beam in feet. P = Load in pounds. c Constant obtained by experiment, and is the deflec- tion, in fractions of an inch, which a beam one foot long and one inch square will have if supported under both ends and loaded at the center; the average value for this constant is given in Table No. 31. For any other mode of loading, see rules and explanations on page 261. In previous formulas and rules, the weight of the beam itself was not considered. The deflection in a beam caused by its own weight when it is of rectangular shape and uniform size, and laid in a horizontal position, is obtained by the formula, Z 3 X # W X c H*XB When both the weight and the load are to be considered, the deflection in a solid rectangular beam laid in a horizontal position, supported under both ends and loaded at the center, is calculated by the formula, H*X B STRENGTH OF MATERIALS. 257 6* = Deflection in inches. L = Length of span in feet. P = Load in pounds. W Weight of beam in pounds. c Constant obtained by experiments, and is the deflec- tion in fractions of an inch, which a beam one foot long and one inch square will have if supported under both ends and loaded at the center, and may be found in Table No. 31. //= Thickness of beam in inches. B = Width of beam in inches. RULE. To the load add five-eighths of the weight of beam, mul- tiply this by the cube of the length of the span in feet, and multiply by a constant from Table No. 31. Divide this product by the product of the cube of the thickness and the width of the beam ; the quotient is the deflection in inches. The deflection in a beam supported under both ends and loaded evenly throughout is five-eighths of that of a beam supported under both ends and loaded at the center. Therefore, in the following formulas, the weight of the beam itself is multi- plied by five-eighths to reduce the effect of the weight of the beam to the equivalent of a load placed at its center. FOR SOLID SQUARE BEAMS. FIG. 28 FOR SOLID RECTANGULAR BEAMS. B H* -\ x FIG. 29 I B FOR HOLLOW SQUARE BEAMS. FIG. 3O STRENGTH OF MATERIALS. FOR HOLLOW RECTANGULAR BEAMS. FOR I BEAMS. FIG. 31 2 b h* b- FIG. 32 c^.g FOR SOLID ROUND BEAMS. h X W\ Z 3 Z* FOR HOLLOW ROUND BEAMS. 1.7<:CP + & W) Z 3 FIG. 33 FIG. 34 FOR SOLID ELLIPTICAL OR OVAL BEAMS. S = Dl X FIG. 35 FOR HOLLOW ELLIPTICAL OR OVAL BEAMS. 1.7 STRENGTH OF MATERIALS. S = Deflection in inches. L Length of span in feet. P = Load in pounds. W=- Weight of beam in pounds. c = Constant obtained from experiments, or may be ob- tained from Table No. 31. For meaning of the other letters, see figure opposite each formula. A round beam equal in diameter to the side of a square beam will deflect 1.698 times as much, and for convenience, when the deflection of a square or a rectangular beam, whether solid or hollow, is known, it may be multiplied by 1.7, and the product is the deflection of a corresponding round, oval, or elliptical beam of the same material and diameter and laid in the same relative position and loaded in the same manner as the calculated beam. It is well to remember that a round or elliptical beam weighs a little less than a square or rectangular one, when the sides and diameters are equal, and the deflection due to its own weight is, therefore, a little less. TABLE No. 3 1. Constant c, Giving deflection in inches per pound of load when the beam is supported under both ends and loaded at the center. MATERIAL. Constant c. MATERIAL. Constant c. Cast steel, Wrought iron, Machinery steel Cast-iron, 0.0000143 0.0000156 O.C000156 0.0000288 Pitch pine, Spruce, Pine, 0.00024 0.00035 0.00033 EXAMPLE. A beam 6" X 9" of pitch pine, 10 feet long, supported under both ends, is to be loaded at the center with one-tenth of its breaking load. Find the load and deflection. Solution : _ 9 2 X 6 X 4 X 150 10 X 10 Deflection will be, s= 10* X 2916^0.00024 = 699.84 = 260 STRENGTH OF MATERIALS. Therefore, if this beam had been curved 0.16 inch upward, by increasing its thickness on the upper side, it would have been straight after the load was applied.* In this example the weight of the beam itself is not considered either in figuring the strength or the deflection, because the beam is comparatively short in proportion to its width and thickness. The weight of the beam itself will only be about 200 pounds, and this will be of no account in propor- tion to the load that the beam will carry, with 10 as a factor of safety. The weight of the beam will increase its deflection cnly 0.006 inch. In such a beam the danger is probably greater from crushing of the ends at the supports, if it has not enough bear- ing surface. In long beams the weight of the beam must not be neglected, either in calculating safe load or in calculating deflection. EXAMPLE 2. A round bar of wrought iron is 5 feet long and 3 inches in diameter, and loaded at the center with 800 pounds. How much will it deflect ? A round bar of iron 3 inches in diameter and 5 feet long weighs 119 pounds. (See table of weights of iron, page 143.) Solution : e _ 5 3 X (800 -f # X 119) X 1.7 X 0.0000156 ~3*~ S = 0.0359 inch. Thus, such a shaft loaded with 800 pounds will deflect T - 7 of an inch in the length of 5 feet, or 60 inches. If the deflec- tion must not exceed ir ^ of the span (see page 266), then the greatest allowable deflection for this span would be 0.04 inch, and the calculated deflection is within this limit. NOTE. T3 W of the span is equal to a deflection of 0.008 inch per foot of length. EXAMPLE 2. A shaft of machinery steel, 11 inches in diameter and 6 feet between bearings, carries in the center a 12-ton fly wheel. How much deflection will the weight of the fly wheel cause? NOTE. Such shafts are usually considered as a beam sup- ported under both ends. (See formula for deflection in solid round beams, page 258.) Solution : 12 tons = 24,000 pounds. (Weight of shaft is not taken into consideration.) * This is a thing frequently done in practice. STRENGTH OF MATERIALS. 261 X 1.7 c 6 3 X 24000 X 1.7 X 0.0000156 II 4 e_ 216 X 24000 X 0.00002652 14641 S = 0.00939 inches. Thus, the calculated deflection caused by the fly wheel is a little less than T fa of an inch. The deflection per foot of span will be Q,-Qo^9_-3 9 which equals 0.001565 inch. EXAMPLE 3. Calculate the deflection of shaft mentioned in the previous example, when both the weight of fly wheel and the weight of shaft are to be considered. Solution : s _ 6 8 X (24000 -f H X 1920) X 1.7 X 0.0000156 _ 216 X 25200 X 0.00002652 o 14641 .S" = 0.00986 inch. Practically, the deflection is likely to be a little less than what is figured in the two previous examples, because if the hub of the fly wheel fits well on the shaft, it will stiffen it some. (It is a g^ood practice to make such shafts a little larger in diameter in the place where the hub of the wheel is keyed on ; this enlargement will then compensate for what the shaft is weakened by cutting the key-way.) The weight of the shaft may be obtained by considering a cubic foot of machinery steel to weigh 485 pounds, and a shaft 11 inches in diameter will then weigh 320.1 pounds per foot in length, and 6 feet will weigh 1920 pounds. Multiplying this by # gives 1200 pounds, to be added to the weight of the fly wheel, which gives 25,200 pounds. The weight of the shaft may also be found in the table of weight of round iron, page 144. To Calculate Deflection in Beams Under Different Modes of Support and Load. Constant c in Table No. 31 is the deflection in fractions of an inch per pound of load when a beam one foot long and one inch square is supported under both ends and loaded at the center, and when this constant for any given material is known, the deflection for beams subjected to other modes of fastening and loads may be calculated thus : For beams supported under both ends with the load dis- tributed evenly throughout their whole length, multiply c by # . 262 STRENGTH OF MATERIALS. For beams fixed at both ends and loaded at the center, multiply c by #. For beams fixed at both ends with the load distributed evenly throughout their whole length, multiply c by y%. For beams fixed at one end and loaded at the other, mul- tiply c by 16. For beams fixed at one end with load distributed evenly throughout their whole length, multiply c by 6. EXAMPLE. A square, hollow beam of cast-iron, 8 inches outside and 6 inches inside diameter, and 9-foot span, supported under both ends, is loaded at the center with 8000 pounds. How much will it deflect ? Solution : Weight of beam = 9 X 12 X (8 2 6 2 ) X 0.26 786 pounds. .S 1 = g3 X ( 800 + ^ x 786 ) x 0.0000288 8* 6* y _729 X 8492 X 0.0000288 4096 1296 s _ 178.291 2800 ^=0.064 inches. EXAMPLE. How much would this same beam deflect if the load had been distributed evenly throughout its whole span? Solution : S = L * (P + W "> 5/8 c s _ 9 3 X 8786 X ^ X 0.0000288 115.289 2800 .9 = 0.041 inch. EXAMPLE. A round cast-iron beam of 7 inches outside and 5 inches inside diameter is 4 feet between supports, with a load of 2000 pounds distributed evenly throughout its span. How much will it deflect, the weight of beam itself not being considered in the calculation ? STRENGTH OF MATERIALS. 263 Solution : o 4 3 X 2000 X 0.0000288 X 1.7 X 74 _ 5 4 S = 256 X 200 X 0.0000288 X 1.7 X ft 1776^ ^ = 0.0085 inch. In this example, 1.7 is used as a multiplier because the beam is round, and ^ because the load is distributed evenly throughout the length of the span. EXAMPLE. A fly wheel weighing 800 pounds is carried on the free end of a 3-inch shaft, 1 foot from the bearing. How much will the shaft deflect? This is the same as a round beam loaded at one end and fastened at the other; therefore, constant c is multiplied by 16 X 1.7. Solution : ~ L 8 P 1.7 c X 16 C* 1 X 800 X 1.7 X 0.0000156 X 16 S = 0.0042 inch. Previous calculations for breaking load and also for deflection are based upon a dead load slowly applied and not exposed to jar and vibrations. If the load is applied suddenly it will have greater effect toward breaking the beam than ff applied slowly. For instance, imagine a load having its whole weight hanging on a rope, like Fig. 37, just touching the beam but not actually resting upon it. If that rope was cut off suddenly this load would produce twice as much effect toward breaking the beam and would cause twice as much deflection as if it was loaded on gradually. A railroad train running over a bridge will, for the same reason, strain the bridge more when running fast than it would if running slow. FIG. 37 To Find a Suitable Size of Beam for Deflection. a Given Limit ol For a square beam supported under both ends and loaded at the center, use the formula : 264 STRENGTH OF MATERIALS. 4 '' //"3 P / Side of the beam -i/l-_ o A round beam supported under both ends and loaded at the center may be calculated by the formula : Diameter of beam = J^- 8 p 1 - 7 c O A rectangular beam supported under both ends and loaded at the center, and having its depth 1^ times its width, may be calculated by the formula : Depth or thickness of beam = 2S L = Length of span in feet. P = Center load in pounds. S= Given deflection in inches. c = Constant given in Table No. 31. NOTE. These three formulas are only approximate, as the weight of the beam itself is not considered ; but if necessary, after the size of beam is obtained, its weight may be calculated and five-eighths of it added to the center load, P\ and using the same formula again, another beam may be calculated for this new center-load, and this new calculation will give a beam only a mere trifle too small. Constants in Table No. 31 are for beams supported under both ends and loaded at the center. For any other mode of loading or fasten- ing, constant c must be multiplied according to rules on page 261. To Find the Constant for Deflection. If experiments are made upon rectangular beams, use formula, SH*B L* (P + ft W) EXAMPLE. Calculate the constant r, or deflection in inches per pound of load, for a beam of 1 foot span and 1 inch square, supported under both ends and loaded at the center, when experiments are made upon a pitch pine beam 40 feet long, 12" by 8'', weighing 1200 pounds and deflecting 1^ inches for a center-load of 500 pounds. Solution : _ 1.5 X 12 3 X 8 40 3 X (500 + % X 1200) c = 0.000259 inch. STRENGTH OF MATERIALS. 265 Modulus of Elasticity Calculated from the Transverse Deflection in a Beam. When experiments are made upon rectangular beams sup- ported under both ends and loaded at the center, the modulus of elasticity may be calculated by the formula, E = ST*B E = Modulus of elasticity. L = Length of span in inches (not in feet). P = Load in pounds. W = Weight of beam in pounds. S = Deflection of beam in inches. T= Thickness of beam in inches. B = Width of beam in inches. EXAMPLE. Calculate the modulus of elasticity for a pitch pine rec- tangular beam weighing 1200 pounds, 40 feet span, and 12" by 8", deflecting 1)4 inches for a center-load of 500 pounds. (This beam and conditions are the same as mentioned in the previous example for calculating constants.) Solution : E _ 480 8 X (500 +#X 1200) 4 X l l / 2 X 12 3 X 8 E _ 138240000000 82944 E 1,666,666 pounds per square inch. This deflection was obtained by actual experiments on a pitch pine beam of the dimensions given, and the calculated modulus of elasticity agrees fairly well with what is usually given by different authorities in tables of modulus of elasticity. When experimenting it is necessary to take the average of several experiments with different loads and to try the beam by turning it upside down, as very frequently it will then deflect a different amount under the same load. Care should be taken that the load is not so great as to strain the beam beyond its elastic limit. As long as the deflection increases regularly in proportion to the load, it is a sign that the elastic limit is not reached. It is very difficult to ascertain exactly when deflection will commence to increase faster than the load, because material is never so homogeneous but that the deflection will be more or less irregular, although by care and patience fairly good re- sults may be obtained, 266 STRENGTH OF MATERIALS. Allowable Deflection. The greatest amount of deflection which may be allowed in different kinds of construction can only be determined by prac- tical experience and good judgment of the designer. As a rule, in iron work the deflection is seldom allowed to exceed y^ of the span, which is equal to T ^, or 0.008 inch per foot of span. Line shaftings are sometimes allowed to deflect j^ of the distance between hangers which is equal to 0.01 inch per foot of span, but head shafts carrying large pulleys are generally not allowed to deflect more than 0.005 per foot of span. In woodwork, considerable more deflection is allowed than in iron structures. Beams in houses are frequently allowed to deflect -gfa, or even T | of the span; this is equal to 0.024 to 0.025 inch*, per foot of span. Woodwork to which machinery is to be fastened must never be allowed to deflect so much. Such woodwork must always be so stiff that it supports the machinery, and not vice versa; for instance, in beams or posts by which hangers and shafting are supported, it is not all-sufficient that they are strong enough, but they must also always be stiff enough. In factories it is very important that floor beams as well as beams supporting heavy shafting have sufficient stiff- ness as well as strength. Floors in factories are frequently loaded up to 300 pounds per square foot of surface. For floors in public buildings, which are never loaded with more than the weight of the people who can get room, the load will hardly ex- ceed 150 pounds per square foot of surface. Floors in tene- ment houses are seldom loaded more than 60 pounds per square foot. Slate roofs weigh about 8.5 pounds per square foot of surface. Snow may be reckoned, when newly fallen, to weigh 5 to 15 pounds per cubic foot, and when saturated with water it may weigh 40 to 50 pounds per cubic foot. Usual practice is to allow 15 to 20 pounds per square foot for snow and wind on roofs. TORSIONAL STRENGTH. The fundamental formula for torsional strength is, Pm = Twisting moment, and is the product of the length of the arm, m, in inches and the force, /*, in pounds. S Constant computed from experiments, and is some- times called the modulus of torsion ; its value usually agrees closely to the ultimate shearing strength per square inch of the material. J Polar moment of inertia (see page 297). a The distance in inches from the axis about which the twisting occurs to the most remote part of the cross section. * 0.025 is one-fortieth inch per foot of span. STRENGTH OF MATERIALS. 267 EXAMPLE 1. A round cast-iron bar 3 inches in diameter, is exposed to torsional stress; the length of the lever, OT, is 18 inches. Find the breaking force, P, in pounds when the modulus of torsion for cast-iron is taken as 25,000 pounds. Polar moment of inertia for a circle of diameter, . Flc ' 38- increases as the cube of the diameter. Shaft a is twice as large in diameter as shaft ft, and is, therefore, eight times as strong as , because 2 3 = 8. If shaft a had been three times as large as b it would have been 27 times as strong, because 3 s = 27 ; if shafts had been four times as large as , it would have been 64 times as strong, because 4 3 = 64. 268 STRENGTH OF MATERIALS. Therefore, if the constant corresponding to a load, which, applied to an arm one foot long will twist off or destroy a bar one inch in diameter, is found, the breaking load for any round shaft of the same material when under torsional stress may be easily calculated. The torsional strength (but not the torsional deflection in degrees) is independent of the length of the shaft. The strength depends only upon the kind and the amount of material, and the form of cross-section. A square shaft having its sides equal to the diameter of a round shaft will have ap- proximately 20% more strength than the round one, but it will take nearly 28% more material. A square shaft of the same area as a round shaft has approximately 15% less torsional strength than the round one. Thus: Formulas for torsional strength relating to solid round shafts will be : Pm P = Breaking load in pounds. D = Diameter of shaft in inches. ;// = Length in feet of the arm on which load P is acting. c = Constant, and it is the load in pounds which, when applied to an arm one foot long, will twist off or destroy a round bar one inch in diameter. This constant is obtained from ex- periments, and is given in Table No. 32. RULE. Multiply the cube of the diameter in inches by the constants, in pounds, divide this product by the length of the lever /, in feet, and the quotient is the breaking load in pounds. TABLE No. 32. Constant c. The ultimate torsional strength in pounds of a round beam one inch in diameter, when load is acting at the end of a lever one foot long. MATERIAL. Very Good. Medium Good. Poor. Cast Steel 2 000 1,000 600 Machinery Steel* . . . Wrought Iron .... Cast-iron 1,200 800 525 1,100 580 450 700 500 350 * Machinery steel or wrought iron may not actually break at this load, but it will deflect and yield so it will become useless. STRENGTH OF MATERIALS. 269 EXAMPLE 1. A wrought iron shaft is eight inches in diameter, and the force acts upon a lever two feet long. How much force must be applied in order to twist off or to destroy the shaft? Solution : P _ 8 3 X 580 _ 512 X 580 2 ~2~ EXAMPLE 2. A force of 870 pounds is acting with a leverage of four feet in twisting a wrought iron shaft. What must be the diameter of the shaft in order to resist the twisting stress, with 10 as a factor of safety? Solution : 3 r^ _ . / P *n X 10 X 4 X 10 580 D V60" = 3.914, or, practically, a 4-inch shaft. NOTE. Ten is used as a multiplier of the twisting moment, P m, because 10 is the factor of safety. Constant 580 is taken from Table No. 32. EXAMPLE 3. A round bar of cast-iron four inches in diameter is to be twisted off by a force of 3200 pounds. How long a leverage is necessary ? (c for cast-iron, in Table No. 32, is 450). Solution : EXAMPLE 4. Experiments are made upon a cast-iron round bar 2 inches in diameter with a leverage of 5X feet ; the bar is twisted off at a force of 832 pounds. Calculate constant c, or the force in pounds if acting with a leverage of one foot, which will break a round bar of the same material one inch in diameter. Solution : '^TT ^882X5X^4368 = 540 pounc , 5 . 270 STRENGTH OF MATERIALS. Hollow Round Shafts. In proportion to the amount of material used, around hollow shaft has more torsional strength than a solid shaft of the same diameter. This is because the fibers in any shaft exposed to twisting stress only offer resistance to the load in proportion to their stretch. Therefore, the fibers near the center are always in position to offer less resistance than the fibers more remote from the center. The formula for torsional strength in round hollow shafts will be : d* P Ultimate breaking load in pounds applied at a leverage of m feet. D Outside diameter of shaft in inches. d= Inside diameter of shaft in inches. m Length of lever in feet. c = Constant (same as for a solid shaft). Square Beams Exposed to Torsional Stress. The theoretical formula for twisting strength (on page 26(5) will apply to square as well as round beams. The proportional strength between a round and a square beam may, therefore, be compared by using that formula. Let S represent the side of a square beam and the polar moment of inertia is J S*. The distance from the center of the beam to the most remote fiber in a square beam is S -v %, and, dividing the polar moment of inertia by this distance, we have, sV y 2 Let D represent the diameter of a round beam. The polar moment of inertia is _! = 0.098 LP The distance from the center to the most remote fiber in the round beam is l / 2 D. Dividing the polar moment of inertia by this distance, we have -Q 98 ^ 0.195 D* Suppose, now, that S and D are equal, for instance. one inch ; the proportion in torsional strength between the two beams must be 0.23 divided by 0.19G, which equals 1.18. Thus, for square beams, use the formulas given for round beams, but multiply constant <:, in Table No. 32, by 1.2, and STRENGTH OF MATERIALS. 271 take the side instead of the diameter. The formula for tor- sional strength in a square beam will be : P _ ( Side ) 3 X 1.2 X c Length of leverage. c = Constant (same as for a round beam). P= Load in pounds. Side is measured in inches. Length of leverage is measured in feet. Torsional Deflection. The torsional deflection in degrees will increase directly with the length of the shaft and the twisting load, and inversely as the fourth power of the diameter of the shaft ; therefore, the formula for torsional deflection is : ?_c X m X L X P D^~ S = Deflection in degrees for the length of the shaft. m = Length of lever in feet. L = Length of shaft in feet. P = Load in pounds. D = Diameter of shaft in inches. c = Constant obtained from experiments for different kinds of material, and is the deflection in degrees for a shaft one inch in diameter and one foot long, when loaded with one pound on the end of a lever one foot long. The author of this book has made experiments on tor- sional deflection in wrought iron shafts two inches in diameter. The average deflection was l l / 2 degrees in 10 feet of length, when a load of 50 pounds was applied on a lever 5# feet long. Constant <:, as calculated from these experiments, will be 0.00914. Using this constant, the formula for torsional deflection for wrought iron will be : 9 L x m x p x - 00914 D* Machinery steel and wrought iron will deflect about the same. Cast-iron will deflect twice as much as wrought iron. A square bar will deflect 0.589 times as much as a round bar when side and diameter are alike. Formula for Torsional Deflection in Hollow Round Shafts. ~_L X m X PXc Lfi-d* D = Outside diameter in inches. d= Inside diameter in inches. All the other letters have the same meaning as explained under formulas for solid shafts. 272 STRENGTH OF MATERIALS. TABLE No. 33. Constant c. The torsional deflection in degrees per foot of length for a shaft of one inch side or diameter when loaded with one pound at the end of a lever one foot long : MATERIAL. Round Section. Square Section. Machinery Steel . . 00914 00538 \Vrought Iron 00914 00538 Cast-iron . 018 0106 Oak 0.795 0.468 Ash . 0.784 0.460 Pine and Spruce 1 35 79 EXAMPLE. A round bar of wrought iron 16 feet long and 3 inches in diameter is fastened at one end and the other is exposed to a twisting load of 1000 pounds, acting with 5 feet leverage. How many degrees will this load deflect the bar ? Solution : 16 X 5 X 1000 X 0.00914 S= 3 4 S= 731.2 81 S= 9 degrees. NOTE. From Table No. 33, it is seen that only steel and wrought iron are suitable for shafts exposed to torsional stress. Wrought iron is about twice as good as cast-iron, over 80 times better than oak, and about 150 times as good as pine. SHEARING STRENGTH. Sometimes force may act in such a manner that the material is sheared off. For instance, the rivets in a steam boiler are exposed to shearing stress (see Fig. 39) when the boiler is under steam pressure. FIG. 39. When holes are punched or bars of j iron are cut off under punching presses, ' the action of the punch in cutting off the - - material is shearing, and the resistance which the material offers is its ultimate shearing strength. The average ultimate shear- ing strength of wrought iron is 40,000 pounds per square inch, STRENGTH OF MATERIALS. 273 In cast-iron the ultimate shearing strength is usually between 20,000 and 30,000 pounds per square inch. In steel the ultimate shearing strength will vary from 40,000 to 80,000 pounds per square inch. The resistance offered to shearing is in proportion to the sheared area. Thus, it will take twice as much force to punch a hole two inches in diameter through a three-eighths inch plate as it would to punch a hole only one inch in diameter through the same plate, and it will take four times as much force to shear off a one-inch bolt as it would to shear off a one- half inch bolt, because the area of a one-inch bolt is four times as large as the area of a one-half inch bolt. EXAMPLE 1. How much force is required to shear off a wrought iron rivet of one-inch diameter if the shearing strength of the wrought iron is 40,000 pounds. Solution : One-inch diameter 0.7854 square inches; therefore the force required will be 0.7854 X 40,000 31,416 pounds. EXAMPLE 2. A wrought iron plate is one-quarter of an inch thick and the ultimate shearing strength of the iron is 40,000 pounds per square inch. How much pressure is required to punch a hole three-quarters of an inch in diameter ? Solution : The circumference of a ^-inch circle is 2.356 inches. The plate is X-inch thick; therefore the area of shearing surface, 2.3562 X X = 0.58905 ; thus, the force required will be 40,000 X 0.58905 = 23,562 pounds. TABLE No. 34. Shearing Strength Per Square Inch. MATERIAL. Pounds Per Square Inch. Steel . 45 000 to 75 000 Wrought Iron Rivets 35 000 to 55 000 Cast-iron 20000 to 30000 Oak, crosswise 4500 to 5500 400 to 700 4,000 to 5 000 Pitch Pine lengthwise 400 to 600 Spruce crosswise 3000 to 4000 Soruce. lengthwise 300 to 500 274 STRENGTH OF MATERIALS. FACTOR OF SAFETY. The factor of safety can only be fixed upon by the experi- ence and good judgment of the designer. It may vary from 4 to 40. In a temporary structure, when the greatest possible load to which it will be exposed is known, a factor of safety of four may be safe enough, but frequently a greater factor is necessary. Different factors of safety are also necessary for different materials; a different factor of safety may also be necessary in different parts of the same machine. The following Table, No. 35, is only offered as a guide in selecting factor of safety : TABLE No. 35. Factor of Safety. Dead Load, Variable Load, Machinery MATERIAL. such as build- ings contain- ing little or no such as bridges and slow- running Machinery in General. Exposed to hard usage, as Rolling Mills, machinery. machinery. etc. Steel, 5 7 10 15 Wrought Iron, 4 6 10 15 Cast-iron, 6 10 15 25 Brickwork, 15 25 30 40 Wood, 8 10 15 20 If a structure is exposed to stress alternately in one direc- tion and then in another, it is necessary to use a higher factor of safety than if it is only exposed to a steady stress one way. A comparatively small load, when applied a sufficient number of times, may break a structure or a machine, although it does not break it the first time. For instance, commence to hammer on a bar of cast-iron and it will break after several blows, although the last blow need not be any more powerful than the first one. It is the same way with anything else; it may break in time, although it is strong enough to resist the stress at the beginning; therefore, within practical limits, the larger the factor of safety the longer time the structure may last. NOTES ON STRENGTH OF flATERIAL. In steel, the crushing strength usually exceeds the tensile strength, but wrought iron has usually a little more tensile than crushing strength, and its shearing strength is about 80 per cent, of its tensile strength. Both steel and wrought iron are suitable to resist any kind of stress, and compared to other materials they are especially adapted for anything exposed to twisting and shearing stress. STRENGTH OF MATERIALS. 275 Cast-iron is variable ; it has usually five to six and a-half times as much crushing as tensile strength, and when loaded transversely it will deflect under the same load nearly twice as much as wrought iron. It is especially useful -for short pillars or anything exposed to crushing stress, where there is little danger of breakage by flexure; it is very much less reliable when exposed to tensile or torsional stress. Wood is not adapted to resist torsion, but is useful to resist tensile, crushing and transverse stress, also to resist flexure. It has nearly twice as much tensile as crushing strength ,; there- fore, it would seem specially well adapted, in all kinds of con- struction, to be the member exposed to tensile stress, but where wood and iron enter into construction together, iron is always used as the member to take the tensile stress and wood as the compressive member, because wood has suck low shear- ing strength lengthwise with its fibers that, with any kind of fastening at the ends, it will tear and split at the holes under comparatively little stress; but this difficulty is easily overcome when wood is used as the compressive member. Wood has comparatively low tensile and crushing strength crosswise on the fiber. This is well to remember with beams loaded transversely and laid on posts. The beams may be sufficiently strong, but under heavy load, if suitable precautions are not taken (see page 250) the top of the post may press into the beam, especially if the lumber is green. Stone has high crushing strength but low tensile strength, and, in consequence, very low transverse strength. It is very well adapted for foundations when supported and laid in such a way that its crushing strength comes into play, but when laid as a beam to resist transverse stress it is very unreliable, as it will break for a comparatively small load and it may break from a blow or jar. Brickwork is only suitable for crushing stress, and there is great difference in the strength of different kinds of brick. In calculating strength and stiffness in any kind of design- ing, it should be remembered that it is only possible to deter- mine the strength of any material by actual test, and that the tabular and constant numbers here given are only an average approximate. Mechanics, The science which treats of the action offerees upon bodies and the effect they produce is called Mechanics. Newton's Laws of Motion. The three fundamental principles of the relation between force and motion were first stated by Sir Isaac Newton, and are therefore called Newton's laws of motion. NEWTON'S FIRST LAW. All bodies continue in a state of rest or of uniform motion in a straight line, unless acted upon by some external force that compels change. NEWTON'S SECOND LAW. Every motion or change of motion is proportional to the acting force, and the motion always takes place in the direction of a straight line in which the force acts. NEWTON'S THIRD LAW. To every action there is always an equal and contrary re- action. Gravity. The natural attraction of the earth on everything on its surface which will cause any body left free to move to fall in the direction of the center of the earth is called the force of gravity. Acceleration Due to Gravity. If a body is left free to fall from a height, its velocity will not be constant throughout the whole fall, but it will increase at a uniform rate. It is this uniform increment in velocity which is called acceleration of gravity. It is usually reckoned in feet per second. A body falling free will at the end of one second have acquired a velocity of 32M$ feet, or, practically, 32.2 feet per second; but it has fallen through a space of 16.1 feet, because it started from rest and the velocity was increasing at a uniform rate until, at the end of the second, it was 32.2 feet per second ; therefore, the average velocity during the first. second can only be 16.1 feet. At the end of two seconds the velocity has increased to 64.4 feet per second and the space fallen (276) MECHANICS. 277 through is 64.4 feet, because the average velocity per second must be half of the final velocity ; therefore, the average velocity is 32.2 feet per second, and, as the time is two seconds the space will be 04.4 feet. At the end of three seconds the final velocity has increased to 3 X 32.2 96.6 feet per second and the space fallen through is -9 | ii x 3 = 144.9 feet, etc. This is supposing the body was falling freely in vacuum, but while the air will offer a resistance and somewhat reduce the actual, motion, the principle is the same. Acceleration due to gravity varies but little at different latitudes of the earth. At the equator it is calculated to be :J2.088 and at the pole 32.253 feet. Acceleration due to gravity decreases at higher altitudes,* but all these variations on the earth's surface are so small that they hardly need to be considered in any calculation concerning practical problems in mechanics. Velocity. The velocity of falling bodies increases at a uniform rate of 32.2 feet per second ; therefore, when commencing from rest, the final velocity in feet per second must be, RULE. Multiply the time in seconds by 32.2 and the product is the final velocity in feet per second ; or, multiply the height of the fall in feet by 64.4 and the square root ot the product is the velocity in feet per second. EXAMPLE. What final velocity will a body acquire in a free fall during seven seconds ? Solution : if 7 X 32.2 = 225.4 feet per second. Height of Fall. The average velocity per second is always half of the final velocity per second. Therefore the space fallen through in a given time is found by multiplying half of the final velocity by the number of seconds which produced that velocity. Thus, the formulas : h = f -^- = tQ.$v= v 0.5 /= -?! = = 2 2*2 * Above the surface of the earth the weight of a body is inversely propor. tional to the square of its distance from the center ot the earth. Below the surface of the earth the weight of a body is directly proportional to its distance from the center ot the earth. 278 MECHANICS. EXAMPLE. A fly-wheel has a rim speed of 48 feet per second. From hat height must a body drop to acquire the same velocity? w Solution : = = 85.78 feet. 2 g 64.4 644 Time. RULE. Divide the space by 16.1, and the square root of the quotient is the time ; or, divide given velocity by 32.2, and the quotient is the time. r S * 0.5 g EXAMPLE. How long a time does it take before a body in a free fall acquires a velocity of 100 feet per second ? Solution : ,-JL= ^ = 3.1 seconds. g 32.2 Distance a Body Drops During the Last Second. The space through which a body will drop in the last second is equal to the final velocity minus half of acceleration due to gravity. Therefore, this space is found by the formula: x = v y 2 g = g(t y 2 ) x = Space in feet which the body drops the last second of the fall. / = Time in seconds. v = Final velocity. g = Acceleration of gravity = 32.2 feet. h Height of fall in feet. EXAMPLE. A body has in a free fall obtained a final velocity of 40 feet per second. What space did it drop the last second ? Solution: x v y 2g 40 ^L 2 = 40 16.1 = 23.9 feet. EXAMPLE. A body was falling four seconds. How many feet did it drop the last second ? Solution : x g(t - */ 2 ) = 32.2 X (4 #) = 32.2 X 3.5 = 1 12.7 feet. MECHANICS. 279 TABLE No. 35. Time, Velocity and Height. =32.161 Feet. Time in Seconds. Velocity in Feet at the End of the Time. Height of Fall in Feet. Distance in Feet that the Body Drops in the Last Second. 1 32.161 16.08 16.08 2 64.322 64.32 48.24 3 96.483 144.72 80.40 4 128.644 257.28 112.56 5 160.805 402.00 144.72 Upward Motion. A body thrown perpendicularly upward with a certain velocity will continue the upward movement until it reaches the same height from which it would have to fall in order to get a final velocity equal to the starting velocity. Therefore, a body projected upward with a given velocity will return again witn the same velocity. This is theoretical in a vacuum, but actually the body neither continues to the theoretical height nor returns with a final velocity equal to the starting velocity, because the air will always offer considerable resistance. The greater the weight of a body, in proportion to its volume, the nearer the velocity, when it returns, will be equal to its starting velocity. EXAMPLE. A body is projected upward with a velocity of 45 feet per second. How high will it go before it stops and commences to drop again, the resistance of the air not being considered ? The solution of this problem is simply to find the theoretical height from which a body must drop to attain a final velocity of 45 feet, which is solved by the formula, *-- 2025 64.4 = 31.286 feet. Body Projected at an Angle. If a body is projected in the direction of the line d e (see Fig. 1), with an initial velocity per second equal to the distance from d to 1. no force acting after the body is started, it will con- tinue to move at constant velocity in a straight line indefinitely ; at the end of the first second it would be at 1, at the end of two seconds it would be at 2, at the end of the third second at 3, at the end of the fourth second at 4, etc.; but, on account of the force of gravity, the motion will be entirely different. The force of gravity acts on this body exactly as if it was falling in 280 MECHANICS. a vertical line. At the end of the first second the force of gravity has caused this moving body to drop 16.1 feet out of its path; therefore, instead of being at 1 at the end of the first second, it is at a point 16.1 feet vertically under 1 ; instead of being at 2 at the end of two seconds, it is at a point 2 X 2 X 16.1 = 64.4 feet vertically below 2; instead of being at 3 at the end of the third second, it is at a point 3 X 3 X 16.1 = 144.9 feet vertically below 3 ; and instead of being at 4 at the end of the fourth second, it is at a point 4 X 4 X 16.1 = 257.6 feet vertically below 4, etc. FIG. 1. When a body is projected in a vertical upward direction with an initial velocity of v feet per second, it proceeds to a z/ 2 height ~2~zr; therefore, when projected at an angle, a (see Fig. 1), with a velocity of i> feet per second, it will proceed to the ?/ 2 sin. 2 a height 2^ When a body is projected in a vertical upward dirction with a velocity of v feet per second, the time for ascent is - and the time for descent is equal to the time for ascent; therefore, v the total time will be 2 ; but when the body is projected upward at an angle of a degrees, the total time for ascent and 2 v sin. a descent will be o The horizontal distance, or the range from d to n, will be equal to the velocity in feet per second multiplied by the total MECHANICS. 28l number of seconds consumed in the ascent and descent, and this multiplied by cos. of the angle a; therefore, /2 T/ sin. a\ 2 7/ 2 sin. a cos. a Horizontal range = v ^ ) cos. a = _ but 2 X sin. a X cos. a is always equal to sin. of an angle of twice as many degrees as the angle a. Therefore, the formula z/ 2 sin. 2 a Deduces to horizontal range Thus, the following formulas will apply to bodies projected at an angle. ( See Fig. 1). The greatest possible height will be, * = ** The greatest possible range! will be, g The time in seconds will be, t _ 2 v sin, a _ g v =. Velocity in feet per second. g = Acceleration of gravity = 32.2. TO FIND THE HEIGHT TO WHICH A BODY CAN ASCEND. RULE. Multiply the velocity in feet per second by the sine of the angle (to the horizontal line), square this product and divide by 64.4, and the quotient is the height in feet. TO FIND THE LONGEST POSSIBLE RANGE. RULE. Multiply the square of the velocity in feet per second by sine of an angle of twice as many degrees as the angle of the throw (to the horizontal line), and divide by 32.2. The quotient is the longest distance the body can be thrown. TO FIND THE TIME OF FLIGHT. RULE. Multiply the velocity in feet per second by sine of the angle (to the horizontal line), and divide by 16.1. The quotient is the time in seconds. EXAMPLE. of 55 to the horizontal line, with an initial velocity of 120 feet per second. How high A body is projected at an angle j, with an initial velocity of 120 fee 282 MECHANICS. will it go ? How far will it go in a horizontal direction ? How many seconds will it take to finish the flight? Solution for height : T 2/ 2 sin. 2 a 11 -- - ^ _ 120 2 X sin. 2 55 h = -644 120 2 X 0.81915 2 64.4 h _ 14400 X 0.673 d4.4 h 150.5 feet. Solving for horizontal range : ^ __ -z/ 2 sin. 2 a g Twice the angle of 55 is 110 and sine of 110 will be sine of 70, because 180 110 = 70 ; therefore, sine of 110 equals sine of 70 in the second quadrant, and the solution will be : 120 2 X sin. 70 b = 32.2 r 14400 X 0.93969 322 " = 128 ' 4 feet Solving for time of flight: 0.5 g . _ 120 X sin. 55 16.1 . _ 120 X 0.81915 *- jTTj - = 6.1 seconds. EXAMPLE. A nozzle on a hose is placed at an angle of 28 to the horizontal line and the spouting water when leaving the nozzle has a velocity of 36 feet per second. How far will it theoretic- ally reach in a horizontal direction? Solution : Range = b = * sin ' 2 * g 36 2 X sin. 56 b = g , _ 1296 X 0.82904 feet ' MECHANICS. 283 EXAMPLE 3. A nozzle on a hose is placed at an angle of 38 to the horizontal line and is spouting water a distance of 40 feet in a horizontal direction. What is, theoretically, the velocity of the water when leaving the nozzle ? Solution: sin. 76 _ (40 X o2.2 o/j A r~~i. ,^ sprond 0.9703 NOTE. In Example 2 we multiply by sine of 56 degrees, because water is leaving the nozzle at an angle of 28 degrees, and twice 28 equals 56. In Example 3 we multiply by sine of 76 degrees, because twice 38 equals 76. See previous explanations. The greatest possible height will be reached if the body is thrown perpendicularly upward. The greatest possible range is obtained if the body is thrown at an angle of 45 and will then be : " ~g~ At an angle of 45 the horizontal range will be twice the greatest possible height which could have been reached if the body had been thrown perpendicularly upward. At this angle the horizontal range is four times the height. For an equal number of degrees over or under 45 degrees the horizontal range will be equal ; for instance, if a body is thrown out at an angle of 30 or 60 degrees, the horizontal distance is the same, but the height of ascension will be much more at 60 degrees than at 30 degrees. It is frequently useful to notice this in practical work. For instance, water under pressure is thrown the farthest distance in a horizontal direction from a hose when the nozzle is held at an angle of 45 degrees to the horizontal line. It is possible by the same pressure to throw water twice as far in a horizontal distance as in vertical height. Motion Down an Inclined Plane. A ball rolling along an incline, as n f-*^ F|G- 2 . a c (Fig. 2), will have the same velocity when it gets to c as it would have had if dropping freely from a to , supposing all friction to be left out of consideration. The average velocity will also be half of the final velocity, and the time used in the fall will be the distance a c (the length of the incline), divided by the average velocity per second. 284 MECHANICS. Body Projected in a Horizontal Direction Prom an Elevated Place. When a body is projected in a horizontal direction from a place which is higher than the one where it strikes the ground, the range in feet in a horizontal direction will be equal to the product of velocity in feet per second and the time in seconds which it will take for a body in a free fall to drop a distance equal to the difference in vertical height between the two places. Thus : VO~T ~JL s v = Initial velocity in feet per second. h = Vertical height in feet. g= Acceleration of gravity = 32.2 feet. EXAMPLE. Water spouts from a nozzle in a horizontal direction at a velocity of 30 feet per second and the nozzle is placed 12 feet above the ground. What is the horizontal range of the water ? Solution : Horizontal range = v -\j = 30 -^ 12 X2 = 22.45 feet. To Calculate the Speed of a Bursted Fly-Wheel from the Distance the Fragments are Thrown. The angle of 45 degrees is the one most favorable to the range ; therefore, suppose the fragments to leave the wheel at that angle and use the formula, z/ 2 Horizontal distance = b = which transposes to v = ^/^ g RULE. Multiply the horizontal distance by 32.2, and the square root of the product is the slowest possible rim-speed the wheel could have had at the time of the accident. EXAMPLE. A 30-foot fly-wheel bursts from the stress due to centri- fugal force, and fragments were thrown a distance of 300 feet from the place of accident. What was the slowest possible speed the wheel could have had at the time the accident occurred ? and what was the corresponding number of revolu- tions per minute? MECHANICS. 285 Solution : -v V300 X 32.2 = V 9660 = 98.3 feet per second. The length of the circumference of a 30-foot wheel is 94.25 feet, therefore the fly-wheel was running at a speed not less than 98.3 60 X 94^5 = 62 -6 revolutions per minute. This calculation does not prove that the wheel did not run faster than 62.6 revolutions per minute when it burst; it may have revolved a great deal faster, as it is not at all sure that any fragments left the wheel at an angle of 45 degrees, but it is certain that the speed of the wheel was not slower. Sometimes it may be pos- sible to settle upon the angle at which a certain fragment left the wheel by noticing traces and marks where it went, and, figuring from the angle and the range, a pretty fair idea of the bursting speed may be obtained. (See formula on page 283). Force, Energy and Power. Force is a pressure expressed in a push or a pull. Energy is the ability to do work. It is divided into poten- tial energy and kinetic energy. Potential energy is the ability of a body to perform work at any time when it is set free to do so. Kinetic energy is the ability of a moving body to do work when its motion is arrested. Kinetic energy is very frequently called " stored-up energy." Work is overcoming resistance through space. In the English system of weights and measures the common unit of work is the foot-pound. Power is the rate of doing work. Work is an expression entirely independent of time, but power always takes time into consideration. For instance, to lift one pound one foot is one foot-pound of work, no matter in what time it is done, but it takes 60 times as much power to do it in one second as it would lake to do it in one minute. Inertia. Inertia is the inability of dead bodies to change either their state of rest or motion. In order to bring about any change, either of motion or rest, dead bodies must always be acted upon by some outside force. Resistance due to inertia is the resistance which a dead body free to move presents to any external force acting tq change either its state of motion or rest. 286 MECHANICS. Mass. The mass of a body is the quantity of matter which it con- tains. By common consent the unit of mass is, in mechanics, considered to be that quantity of matter to which one unit of force can give one unit of acceleration in one unit of time ; therefore, when the weight of a body is divided by acceleration of gravity, the Quotient is the mass of the body. Thus : W = m X g - W Momentum. The product of the mass of a moving body and its velocity is called its momentum or, also, its quantity of motion. The unit for momentum is the product when unit of mass is multi- plied by unit of velocity per second. In mechanical calcula- tions, using English weights and measures, the unit of mass is weight divided by 32.2; therefore, unit of momentum will be: Weight of the moving body in pounds multiplied by velocity in feet per second and the product divided by 32.2. Thus : q = m X v m = mass = ; therefore, g q =. Momentum, or quantity of motion. W Weight of moving body in pounds. V Velocity of moving body in feet per second. g= Acceleration of gravity. is the formula by which the time in a free fall is obtained, and, consequently, the momentum of a falling body can also be expressed by the product of the weight of the body in pounds and the time in seconds during the fall. This product is usually called " time effect." Impulse. The product of the force and the time in which it is acting as a blow against a body is called impulse, and it is always of the same numerical value as the momentum of the moving body. MECHANICS. 287 Kinetic Energy. The kinetic energy stored in any moving body is always expressed in foot-pounds, by the product of the force in pounds acting to overcome the inertia of the body, and the distance in feet through which the force was acting in starting the body, and is always equal to the weight of the body multiplied by the square of the velocity and this product divided by twice the acceleration of gravity. Thus : K = W X t/ 2 2* K Kinetic energy in foot-pounds. IV = Weight of the body in pounds. v = Velocity of the body in feet per second. 2 g 64.4. In a free fall the height, /;, corresponding to a given ,0 velocity, is found by the formula, -^- ; therefore, K '= W X h. Thus, multiplying the weight of a moving body by the height which in a free fall corresponds to its velocity, the product will be the kinetic energy stored in the body. frTX-z/ 2 The formula K = ^ transposes to K ^ m v' 1 . Hence the simple rule : Multiply half the mass of a moving body by the square of its velocity in feet per second, and the product is the kinetic energy in loot-pounds stored in the body. The kinetic energy stored in any moving body always represents a corresponding amount of mechanical work which is required in order to again bring the body to rest. EXAMPLE. A body weighing 1610 pounds is moving at a constant velocity of 18 feet per second. How many foot-pounds of kinetic energy is stored in the body ? Solution : W X 7/ 2 1610 X 18 X 18 K = 2 = r eO == 8 ' 100 foot-pounds. If this moving body was brought to rest and all its stored energy could be utilized to do work it could lift 8,100 pounds one foot, or it could lift 81 pounds 100 feet, or any other combi- nation of distance and resistance which, when multiplied by one another, will give 8,100 foot-pounds. It is very important always to keep in mind a clear dis- tinction between work ?CN\ power, as power is the rate of doing work, and time must, therefore, always be considered in the question of power. For instance, when 33,000 foot-pounds of 288 MECHANICS. work is performed in one minute it is said to be one horse-power ; therefore, if this 32,400 foot-pounds of energy was utilized to do work and used up in one minute, it would do work at a rate of fi^lro M horse-power, but if utilized during a time of two minutes it would only do work at a rate of |i horse-power, or if utilized in a second the rate of work would be i* X 60 = 58 |f horse-power, etc. To Calculate the Force of a Blow. The force of a blow may be calculated by the change it produces. For instance, a drop-hammer weighing 800 pounds drops three feet, and compresses the hot iron on the anvil % inch. How much is the average force? (% inch = Vtsfoot). The kinetic energy stored in the hammer at the moment it commences to compress the iron is 800 X 3 = 2400 foot-pounds. The average force = 240 = 115,200 Ibs. Vis In the above example, friction is neglected. The shorter the duration of the blow the more intense it will be. Therefore the force of the hammer mentioned above, if, instead of striking against hot iron, compressing it % inch, had been struck against cold iron, compressing it only a few thou- sandths, the blow would have been as many times more intense as the duration of the blow had been shorter. Therefore it is entirely meaningless to say that a drop-hammer or any other similar machine is giving a blow of any certain number of pounds by falling a certain number of feet, because the in- tensity of the blow will depend upon its duration. Formulas for Force, Acceleration and Motion. From the laws of gravitation, it is known that when one pound of force acts upon one pound of matter it produces an acceleration of 82.2 feet per second each successive second as long as the force continues to act. From Newton's laws of motion, it is known that the motion is always in proportion to the force by which it is produced ; therefore, when one pound of force acts for one second upon 32.2 pounds of matter, it will produce an acceleration of one foot per second. Hence the following formulas: ;// Mass of the moving body, which is considered to be weight divided by 32.2. f= Constant force in pounds acting on a body free to move. G = Constant acceleration in feet per second due to the acting force, F. MECHANICS. 289 T= Time in seconds in which the force F acts upon a body free to move. i> final velocity acquired by the moving body in the time of T seconds. G in G = -?L T= ^L=.JL. vm=-FT ' T G T m -u^L2L m = J L F = 2^L T- vm m v T F When a moving body is arrested the product of the resist- ance and time is equal to its momentum. Thus: z> A = #= Constant resistance in pounds acting against the moving body. The average velocity of the moving body is half of the final velocity, and the space 'passed over by the moving body when acquiring the given velocity is half of the final velocity in feet per second multiplied by the time in seconds. Thus : F-* Sm - o_ " " 6* = Space in feet. The work in foot-pounds required to overcome the inertia of a given body when brought from a state of rest to a given velocity is equal to the kinetic energy stored in the moving body. Thus : K SF " ! " l '~ F>v T G m\ FIG. 7. a - 2 b = 2 X (2^) 2 = 7.5625 X (2X) 2 = 5.0625 c 2 X y 2 X (IX) 2 = 3.0625 d= 2 X ft X (IX) 2 = 1.5625 e = 2 X ft X ( X) 2 = 0.5625 / 2 X y 2 X ( X) 2 = 0.0625 ^=2 X # X ( X) 2 = 0.0625 // : = 2 X # X ( X) 2 = 0.5625 2 2 X y 2 X (IX) 2 = 1.5625 y = 2 X # X (IX) 2 = 3.0625 k 2 X Yz X (2X) 2 = 5.0625 1= 2 X Yz X (2X) 2 = 7.5625 Moment of inertia = 35^.75 (approximately). The correct value for the least rectangular moment of inertia for such a surface is obtained by the formula, (Depth) 3 X width and for Fig. 7 will be X 2 = 36. Thus, the 12 12 approximate rule gives results a trifle too small, but if the sur- face had been divided into smaller strips, the result would have been more correct. Radius of gyration for this surface, when rotating about the axis x- y, is : moment of inertia area = 1.78 inches. MECHANICS. 295 EXAMPLE 2. Find by approximation the rectangular moment of inertia for a surface, as Fig. 8, (the sectional area of an I beam) about the axis x y. When the beam is symmetrical, the neutral axis is at an equal distance from the upper and lower side, and the moment of inertia for the upper and lower half of the beam is equal ; consequently, when calculating momentof inertia for a surface like Figs. 8 and 7, it is only nec- essary to calculate the moment of inertia for half the beam, and multiply by 2 in order to get the moment of the whole beam. Solution : a = 3 X >/ 2 X /; = 3 X % X c =1 X Y 2 X d 1 X e = 1 X /= 1 X X (IX)' 2 = X ( K) -2 = X ( X) 2 = = 11.34375 nr 7.59375 = 1.53125 0.78125 0.28125 0.03125 Moment of inertia = 21.5625 Moment of inertia = 21.5625 for upper half, for lower half, for beam (approximately). Moment of inertia = 43.125 Area of cross-section of beam is 10 square inches. Radius of gyration = ^ 43 ' 125 = 2.07 inches. EXAMPLE 3. Find approximately the moment of inertia of a surface, as Fig. 9 (usual section for cast-iron beams), about the axis, x jr 9 passing through the center of gravity of the surface. In shapes of this kind the axis through the center of gravity is not at an equal distance from the upper and lower side, but it can be obtained experimentally by cutting a templet to the exact shape and size of the surface and balancing it over a knife's edge, or it maybe calculated by the principle of moments, as shown in this example. Divide the surfaces into three rectangles, the upper flange, the web and the lower flange. Assume some line as the axis, for instance, the line n m, which is the center line through the lower flange; multiply the area of each rectangle by the distance of its center of gravity from the axis n in, and add the products. Divide this sum by the area of the entire section, and the quotient is the distance between the center of gravity of the section and the axis n m. 296 MECHANICS. -2 FIG. 9. SOLVING FOR CENTER OF GRAVITY I (AREA.) (DISTANCE.) Area of upper flange = 2X1 = 2 square inches X 5 =10 Area of web 4X1=4 square inches X 2)4 = 10 Area of lower flange =4X1 = 4 square inches X =0 10 20 and 20 divided by 10 = 2'' which is the distance from the center of gravity of the lower flange to center of gravity of the section of the beam, or the neutral axis x y. SOLVING FOR MOMENT OF INERTIA : = 10.56250 = 7.56250 = 2.53175 X l / 2 X 2 d e = f= j 1 k = 4 / 4 X &] X xxx XXX (IX) 2 = X X ( X) 2 = X X ( X) 2 = l /2 X ( X) 2 = X x (ix) 2 = J A X 1.53225 0.78125 0.28125 0.03125 0.03125 0.28125 0.78125 6.12500 X (2#) 2 = 10.12500 Moment of inertia of beam = 40.6266 (approximately). Area of cross-section of beam = 10 square inches Radius of gyration of beam = - '- = 2.015 inches. MECHANICS. 297 FIG. 1O. Polar Moment of Inertia. The polar moment of inertia is a mathematical expression, used especially when calculating the torsional strength of beams, shafting, etc. It is very frequently denoted by the letter /. The polar moment of inertia is the sum of the products of each elementary area of the sur- face multiplied by the square of its distance from the center of gravity of surface. Suppose (in Fig. 10) that the area is divided into circular rings, as a, b, c, d, e, f,g,h, /, j\ k, /, ;//, , o,p, and the area of each ring multiplied by the square of its distance from the center, c ; the sum of all these products is the polar moment of inertia. The moment, calculated this way, will always be a trifle too small, but the smaller each ring is taken the more correct the result will be. If each ring could be taken infinitely small the result would be correct. The polar moment of inertia is equal to the square of the radius of gyration about the geometrical center of the shaft, mul- tiplied by the area of cross-section of the shaft; therefore, for a round, solid shaft (as the section shown in Fig. 10), the polar moment of inertia is always expressed by the formula : (Radius) 4 X * (Diameter) 4 X v 2 ' 32 For a hollow, round shaft, the polar moment of inertia is expressed by the formula, D = Outside diameter. d= Inside diameter. The fundamental principle for the polar moment of inertia for any shape of section is that, if two rectangular moments of inertia are taken, one being the least rectangular moment of inertia, about an axis passing through the center of gravity, and the other, the least rectangular moment, about an axis perpendic- ular to the first one, also through the center of gravity, the sum of those two rectangular moments is equal to the polar moment. In Fig. 10, the rectangular moment of inertia about the axis x y will be (diameter)* X TT and the rectangular moment about the axis *' y' will also be ( c * iameter)4 X -T ; thus the polar moment will be ( diameter ) 4 x T 298 MECHANICS. EXAMPLE. Find the polar moment of inertia and radius of gyration of a round shaft of 4" diameter. Solution : y= 32 4X 3.1416 32 Radius of gyration = ^/ Polar moment of inertia area of section. Radius of gyration = 25 - 1328 X 3.1410 Radius of gyration = 1.414 inches. The term, moment of inertia, as used in calculating stored energy in revolving bodies, is frequently and certainly more concisely called moment of rotation, and is a mathematical expression by which the effect of the whole mass (theoretically) is transferred to the unit distance from center of rotation. This term (moment of inertia or moment of rotation) is obtained by multiplying the square of radius of gyration by mass of moving body.* In English measure, mass is taken as ^ \ v of the weigh t of the revolving body, and the radius of gyration is always taken in feet. EXAMPLE. A solid disc of cast-iron, rotating about its geometrical center, is six feet in diameter and of such thickness that it will weigh 4073.3 pounds. What is its moment of rotation or moment of inertia? Radius of gyration = 3 X \/~f and (radius of gyration) 2 = 3 2 X $ Mass- . 126.5 32.2 Moment of rotation = 126.5 X 3 2 X % = 569.25. NOTE. In all such problems relating to stored energy in rotating bodies, the radius of gyration is usually taken in feet and not in inches, as jn previous examples of moment of inertia, when relating to strength of material. * Instead of multiplying the mass of the body by the square of radius of gyration in feet and calling the product moment of inertia, some writers multiply the weight of the body by the square of the radius of gyration in feet and call this product moment of inertia. Thus last expression for moment of inertia, of course, will have a numerical value of 32.2 times the first one. It does not make any differ- ence in the result of the calculation whether weight or mass is used, but the same unit must be adhered to throughout the whole calculation. MECHANICS. 299 Angular Velocity. When a body revolves about any axis, the parts furthest from the axis of rotation move the fastest. The linear 'velocity at a radius of one foot from the center of rotation is called the angular velocity of the body. It is usually reckoned in feet per second. The angular velocity of any revolving body is ex- pressed by the formula, F a = 2 TT n V* = Angular velocity in feet per second. n Number of revolutions per second. RULE. Multiply the number of revolutions per second by 6.2832, and the product is the angular velocity in feet per second. EXAMPLE. What is the angular velocity of a fly-wheel making 300 revolutions per minute? Solution : 300 revolutions per minute = 5 revolutions per second, therefore, angular velocity = 6.2832 X 5 = 31.416 feet per second. Angular velocity expresses the velocity at unit dis- tance from center of rotation and in English measure this unit is feet. As already stated, the moment of rotation is an expres- sion for the mass of the rotating body (theoretically) transferred to unit distance from center of rotation ; the product of angular velocity and moment of rotation will, therefore, be the momentum of the rotating body. The constant resistance which has to be exerted at unit radius in order to bring the body to rest in jT seconds will be : The resistance which has to be exerted at any radius of r feet to bring the body to rest in T seconds will be : R Resistance in pounds. F a Angular velocity in feet per second. / = Moment of rotation (also called moment of inertia). The constant force which has to be exerted at unit radius in order to bring the body from a state of rest to an angular velocity F a in T seconds will be: 300 MECHANICS. The constant force which has to be exerted at any radius, r, in order to bring the body from a state of rest to an angular velocity F a in T seconds will be : R = Constant resistance in pounds. F = Constant force in pounds. F a = Angular velocity in feet per second. / = Moment of rotation (also called moment of inertia). r = Radius in feet at which the force is applied. T ~ Time in seconds that the force is acting. EXAMPLE. A fly-wheel making 120 revolutions per minute and weigh- ing 483 pounds, is brought to rest in two seconds by a resistance acting at a six-inch radius. The radius of gyration of the fly- wheel is 1.2 feet. What is the average force exerted against the resistance during these two seconds ? Solution : 120 revolutions per minute = 2 revolutions per second. Angular velocity = 6.2832 X 2 = 12.5664 feet per second. Moment of rotation = 1.2 X 1.2 X 483 - =21.6 32.2 Radius of resistance, 6 inches = 0.5 feet. * = .12.6664 X 21.6 = 271 . 43 pounds . 2 X 0.5 If a rotating body is not brought to rest, but only reduced in speed to an angular velocity of F a , in T seconds, then the average force or resistance acting at unit radius is : F = _ (Fa F a i)7 The average force which has to be exerted at any radius at r feet to reduce the angular velocity to F a , in /"seconds will be : F = Fai)/ T r EXAMPLE. A fly-wheel on a punching machine weighs 644 pounds, its radius of gyration is 1^ feet, and it makes at normal speed 300 revolutions per minute, but when the machine .is punching the MECHANICS. 301 speed is in of a second reduced to a rate of 280 revolutions per minute. What average force has the fly-wheel communi- cated to the pitch-line of a 6-inch gear on the fly-wheel shaft ? Solution : The mass of the fly-wheel = 644 = 20 32.2 The moment of rotation = (1J^) 2 X 20 = 45 300 revolutions per minute = 5 revolutions per second. Angular velocity = 5 X 6.2832 = 31.416 280 revolutions per minute = 4% revolutions per second. Corresponding angular velocity = 4^ X 6.2832 = 29.3216 6-inch diameter of gear = 3-inch radius = % foot. F _ (31.416 29.3216) X 45 ix X F = 2.0944 X 45 X 5 X 4 = 1884.96 pounds. The kinetic energy in foot-pounds stored in the revolving body may be obtained by the formula : F a 2 X = kinetic energy. 2 Decreasing the angular velocity to Fi, the stored-up energy will also decrease to W X -. and the work done by the revolving body will be (Fa 2 - F a i 2 )X JL EXAMPLE 1. The moment of rotation in a fly-wheel is 1040 ; its angular velocity is 5 feet per second. What is the stored-up energy in the wheel ? Solution : Kinetic energy = 5 2 X !-*- = 13,000 foot-pounds. EXAMPLE 2. At certain intervals, when machinery is started, the angular velocity of this fly-wheel is reduced to ^/ 2 feet per second. How many foot-pounds of energy has the fly-wheel given up in helping to drive the machinery ? 3O2 MECHANICS. Solution : x = (5 2 (4>^) 2 ) X 520 x (25 20#) X 520 :r = 4% X 520 = 2470 foot-pounds of energy given out by the fly-wheel during this change of speed. EXAMPLE 3. How much stored energy is left in the wheel after its angu- lar velocity is reduced to 4> feet per second ? Solution : K= (Fa) 2 - X 520 = 20 # X 520 = 10,530 foot-pounds. The same result may be obtained by subtracting, thus : 13,000 2470 = 10,530 foot-pounds. Centrifugal Force. The centrifugal force is the force with which a revolving body tends to depart from its center of motion and fly in a direction tangent to the path which it describes. The centrip- etal force is the force by which a revolving body is prevented from departing from the center of motion. When the centri- fugal force exceeds the centripetal force the body will move away from the center of motion, but if the centripetal force ex- ceeds the centrifugal force, the body will move toward the center of motion. The centrifugal force in any revolving body is equal to the mass of the body (see page 286) multiplied by the square of its velocity, and this product divided by the radius of the revolving body. W X v 2 m X v 2 ~ 32.2 X r ~ r cf Centrifugal force in pounds. r = Radius in feet. -v = Velocity in feet per second. W =: Weight of moving body in pounds. m = Mass of moving body. F|G - 1 1 EXAMPLE. The weight #, in Fig. 11, is four pounds, and the length of the string is two feet; the weight is made to swing around the center c, three revolutions per second. What is the stress on the string due to centrifugal force? MECHANICS. 303 Solution : The distance from c, to the center of the ball is two feet, and making three revolutions per second, the velocity will be 2 X 3 X 3.1410 X 2 37.7 feet per second. cf= 4 X 37 ' 7 X 37 ' 7 = 88.2 pounds. 32.2 X 2 In metric measure, ^ 9.81 X r cf= Centrifugal force in kilograms. r = Radius in meters. v = Velocity in meters per second. W= Weight of moving body in kilograms. EXAMPLE. Suppose that the weight a, in Fig. 11, is five kilograms, swinging around the center, c, one revolution per second ; the distance from a to c isl}4 meters. What is the stress on the string due to centrifugal force ? Solution : The velocity will be 1.5 X 3.1416 X 2 9.4248 meters per second. 5 X 9.4248 X 9.4248 = ^ kilograms . 9.81 X iy 2 Friction. The resistance which a body meets with from the surface on which it moves is called friction. It is called sliding friction when one body slides on another ; for instance, a sleigh is pulled along on ice the friction between the runners of the sleigh and the ice is sliding friction. It is said to be rolling friction when one body is rolling on another so that new surfaces continually are coming into contact; for instance, when a wagon is pulled along a road, the friction between the wheels and the road is roll- ing friction, but the friction between the wheels and their axles is sliding friction. Sliding friction varies greatly between different materials, as everybody knows from daily observation. For instance, a sleigh with iron runners can be pulled with less effort on ice than on sand, even if the road is ever so smooth. This is because the friction between iron and ice is a great deal less than the friction between iron and sand, 304 MECHANICS. Coefficient of Friction. The ratio between the force required to overcome the re- sistance due to friction and the weight of a body sliding along a horizontal plane is called coefficient of friction. For instance, in Fig. 12 a piece of iron weighing 300 Ibs. rests on a horizontal plate b. A string fastened to , goes over a pulley, c. At the end of the string is applied a weight, d. If this weight is increased until the body a just starts to move along on b, and the weight is found to be 50 pounds, the co- efficient of friction will be - 50 = 0.166 300 6 When the weight of a moving body is multiplied by the coefficient of friction, the product is the force required to keep the body in motion. Of course, any pressure applied to the moving body, perpendicular to its line of motion, may be sub- stituted for its weight. For instance, the frictional resistance of the slide in a slide-valve engine is not due to the weight of the valve, but to the unbalanced steam pressure on the valve. In all cases the rule is : Multiply the coefficient of friction by the pressure perpen- dicular to the line of motion, and the product is the force required to overcome the frictional resistance. EXAMPLE. The coefficient of friction is 0.1, and the weight of the sliding body is 800 pounds. What force is required to slide it along a horizontal surface ? Solution : Force = 800 X 0.1 = 80 pounds. ' Rolling Friction. If the body, a, (see Fig. 12) was lifted up from the plane, b, high enough so that two rollers could be placed between a and , it would be found that the body would move with much less force than 50 pounds because, instead of sliding friction, as in the first experiment, it would be rolling friction. Suppose it is found that a commenced to move when the load, d, was four pounds, then the coefficient of friction for this particular case would be 4 - 0.0133 300 75 In these experiments the whole force at d is not used to move the load a, as a small part of it is used to move the pulley at c, but in order to make the principle plain, this loss has not been considered. MFCMANICS. _ 305 Axle Friction. The friction between bearings and shafts is frequently called axle friction. This, of course, is sliding friction, but owing to the fact that the surfaces in question are usually very smooth and well lubricated, the coefficient of friction is smaller than for ordinary slides. EXAMPLE 2. A fly-wheel weighs 24,000 pounds, the diameter of the shaft is 10 inches, and the coefficient of friction in the bearings is 0.08. What force must be applied 20 inches from the center in order to keep the wheel turning ? Resistance due to friction = 24000 X 0.08 = 1920 pounds. This resistance is acting at a radius of 5 inches, but the force is acting at a radius of 20 inches ; therefore, the required force necessary to overcome friction will be 1 - 2() -*J? 480 pounds. How much power is absorbed by this frictional resistance if the wheel is moving 72 revolutions per minute ? Solution : 72 X 20 X 2 X 3.1416 12 = 753.984 feet, and 753.934 X 480 361,912.32 foot-pounds and The space moved through by the force is Horse=Power Absorbed by Friction in Bearings. The horse-power absorbed by the friction in the bearings for any shaft may be figured directly by the formula, HP W/x / x n X 3.1416 X d S30"00"X 12 This reduces to : H-P =WXfXnXdX 0.000008 H-P Horse-power absorbed by friction. W = Load on bearings in pounds. d= Diameter of shaft in inches. /=. Coefficient of friction. n = Number of revolutions per minute. Calculating the previous example by this formula, we have: H-P = 24000 X 0.08 X 72 X 10 X 0.000008 = 11. 06 horse- power, which is practically the same as figured before. 3 o6 MECHANICS. Angle of Friction. Suppose, instead of using the string and the weight d (see Fig. 12), that one end of the plane is lifted until a commences to slide ; the angle between b and the horizontal line, when a com- mences to move, is called the angle of friction. The coefficient of friction may also be calculated from the angle of friction, thus : If the body commences to slide under an angle of a degrees, the coefficient of friction will be sm - a tang. a. Thus, the coefficient cos. a of friction is always equal to tangent of the angle of friction. Rules for Friction. 1. Friction is in direct proportion to the pressure with which the bodies are bearing against each other. 2. Friction is dependent upon the qualities of the surfaces of contact. 3. The velocity has, within ordinary limits, no influence on the value of the coefficient of friction. 4. Sliding friction is greater than rolling friction. 5. Friction offers greater resistance against starting a body than it does after it is set in motion. 6. The area of surfaces of contact has, within ordinary limits, no influence upon the value of the coefficient of friction, but if they are unproportionally large or small the friction will increase. TABLE No. 36. Coefficient of Friction. MATERIALS. SLIDES. BEARINGS. Well Lubri- cated. Not well Lubri- cated. Well Lubri- cated. Not well Lubri- cated. Cast-iron on wrought iron .... Cast-iron on cast-iron . . ... 0.08 0.08 0.08 0.10 0.16 0.16 0.20 0.20 0.05 0.05 0.05 0.05 0.075 0.075 0.075 0.075 Wrought iron on brass Wrought iron on wrought iron . . Friction in Machinery. When the surfaces are good the frictional resistance for slides may be assumed as 10 per cent., more or less, according to the conditions of the surfaces. It is always well not to take the coefficient of friction too small; it is better to be on the safe side and allow power enough for friction. In bearings for machinery, the frictional resistance ought not to absorb over six per cent. If more is wasted in friction, there is a chance for improvement. MECHANICS. 37 FIG. 13. 150 Ibs. Pulley Blocks. When friction is not considered, the force and the load will be equal in a single fixed pulley (as A, Fig. 13). Thus,. a single fixed pulley does not accomplish anything further than to change the direction of motion. In a single movable pulley (as -at B, Fig. 13), the force is equal to only half the load; thus, 75 pounds of force will lift 150 pounds of load, but the force must act through twice the space that the load is moved. The tension in any part of the rope in B is half of the load W\ thus, when the load is 150 pounds the tension in the rope is 75 pounds, when arranged at B, but it is 150 pounds when arranged! as at A. Fig. 14 shows a pair of single sheave pulley blocks in position to pull a car; when the blocks are arranged as at A, and friction is not consid- ered, a force of 100 pounds on the hauling part of the rope exerts a force of 300 pounds on the post, but only 200 pounds on the car; but, turning the blocks end for end, as shown at B, a force of 100 pounds on the hauling part of the rope exerts a force of 300 pounds on the car and 200 pounds on the post. This is a point well worth remembering when using pulley blocks. Suppose, for instance, that a man exerted a force of 100 pounds on the hauling part, and that it required 250 pounds of force to move the car ; if he used the pulley blocks as shown at A, his work would be useless, as far as moving the car is concerned, as he could not do it, but turning his blocks end for end he could accomplish the desired result. Always re- member whenever it is possible to have the hauling part of the rope coming from the movable block and pull in the same direction as the load is moving. FIG. 14. Friction in Pulley Blocks. In practical work, friction will have some influence, and, to a certain extent, change these results, because some of the ten- sion in the rope is lost by friction in each sheave the rope passes over, therefore the tension in each following part of the rope is always less than it was in the preceding part. This loss must be obtained from experiments. In good pulley blocks, having roller bearings, this loss is probably not more than 0.1, and we MECHANICS. get a useful effect of 0.9 of the force from one part of the rope to the next ; therefore, when friction is considered, the useful effect in the following cases will be : In single sheave blocks having the hauling part from the movable block (pulling with the load as in B, Fig. 14). W=F(\ + 0.9 + 0.9 2 ) W= F X 2.71 In single sheave blocks having the hauling part from the fixed block (pulling against the load as in A, Fig. 14), W = F (0.9 + 0.9 2 ) tV=FXl.1\ In double sheave blocks having the hauling part from the movable block, 0.9 2 + 0.9 3 + 0.9 4 ) W=FX 4.1 In double sheave blocks having the hauling part from the fixed block, W = F (0.9 + 0.9 2 + 0.9 3 + 0.9 4 ) IV=FX 3.1 Differential Pulley Blocks. In a differential pulley block (see Fig. 15), the proportion between the force and the weight, when friction is neglected, is expressed by the formula : F _ W X (R r) 2X7? The actual force required to lift a weight by such a pulley block is about three times the theoretical force, as calculated above. Inclined Plane. When a weight is pulled upward FIG. 16. on an inclined plane, as shown in Fig. 16, and the force F is acting parallel to the plane, the required force for moving the body will be F W X sin. a plus friction, and the perpen- *' dicular pressure P. against the plane will be IV X cos, . r _- eos .-a- FIG. 15. MECHANICS. 309 EXAMPLE 1. The weight, W, (Fig. 16) is 100 pounds ; the angle a is 30. What force, F, is required to sustain this weight, friction not considered ? Solution : Sin. 30 = 0.5 Thus: F W X sin. 30 = 100 X 0.5 = 50 pounds. EXAMPLE 2. What is the perpendicular pressure under conditions stated in Example 1 ? Solution : P = IV X cos. a = 100 X 0.86603 = 86.6 pounds. Therefore, the frictional resistance between the sliding body and the inclined plane will be only what is due to 86.6 pounds pressure ; in other words, the force required to over- come friction will be W X f X cos. a. EXAMPLE 3. What force is required to move the body mentioned in Example 1 when friction is also considered, taking coefficient of friction, F, as 0.15? Solution : F = W (sin. a + cos. a X /) .F = 100X (0. 5 -f- 0.86603X0. 15)= 100 X 0.6290 =62. 99 pounds. NOTE. This is the force required for moving the load. In order to put it in motion more force must be applied, varying according to velocity, but after motion is commenced the speed would be, under these conditions, maintained forever by this force of 62.99 pounds. When a load is moving down an inclined plane the force due to W X sin. a will assist in moving the body, and if the product W X sin. a exceeds the product W X cos. a X f the body will slide by itself. For instance, in the body mentioned in the previous example, the force required to overcome gravity, regardless of friction, is 50 pounds, and the force required to overcome friction is 12.99 pounds; thus, if the body should be let down the plane instead of pulled up, it would have to be held back with a force of 50 12.99 = 37.01 pounds. NOTE. When the incline is less than 1 in 35, cosine is so nearly equal to 1 that it may be neglected, and the force required to overcome friction may be considered to be the same as on a level plane. For instance, a horse is pulling a load and ascend- ing a gradient of 1 in 35 ; if the tractive force required to pull the load on a level road was 30 pounds and the weight of the load was 1400 pounds, when ascending the hill, the horse will first 310 MECHANICS. have to exert a force of 30 pounds, which is all due to friction, but beside that he must also exert a force of ^ times 1400 40 pounds ; thus the total pull exerted by the horse will be 70 pounds. Inclined Plane With the Force Acting Parallel to the Base. When the pressure is continually FIG. 17. acting in a line parallel to the base of ./ F" the incline, as F, (see Fig. 17) which J /?^/'^ * is frequently the case in mechanical F ^* ^2^^ ** movements, as for instance, in screws, /* *f some kinds of cam motions, etc., it /^ ^ _ will require more force to move the body than it would if the force was i* Cos a >i acting parallel to the incline. When force acts parallel to the base, as in Fig. 17, the force required to move the body, if friction is not considered, will be : F WK sm '^ = W X tang, a cos. a EXAMPLE 1. What force is required to move 100 pounds upward an incline of 30, as in Example 1, excepting that the force is acting parallel to the base instead of parallel to the incline ? Solution : FWY. tang. 30 F W X 100 X 0.57735 = 57.74 pounds. When both the friction and the weight of the body are con- sidered, the force required to move the body will be : /r_ W x s'm.a + (f X cos. a) (fX sin.rt) EXAMPLE 2. What force is required to move 100 pounds upward an in- cline of 30 (as in Example 1) if the force is acting parallel to the base line instead of parallel to the incline ; coefficient of friction is supposed to be 0.15? Solution : F _ 100 sin. 30 + (0.15 X cos. 30) cos. 30 (0.15 X sin. 30)~ p _ 100 x 0.5 + (0.15 X 0.86603) 0.86603 (0.15 X 0.5) yr^pox ' 5 + 0.1277045 0.86603 0.075 .F 100 X 0.7936 79.36 pounds. MECHANICS. 311 NOTE. From these calculations it is seen that it is more advantageous to apply the force parallel to the incline than parallel to the base. When force is applied parallel to the in- cline : The force required to overcome gravity 50 pounds. The force required to overcome friction = 12.99 pounds. Total force 62.99 pounds. When the force is acting parallel to the base : The force required to overcome gravity = 57.74 pounds. The force required to overcome friction = 21.62 pounds. Total force = 79.36 pounds. Screws. When friction is not considered, the force which may be exerted by a screw (see Fig. 18) will be : ~~ P R X 2?r W Weight of the load lifted, or force exerted, if the screw acts as a press. F = Acting force. R = Radius in inches at which the force acts. P = Pitch of screw in inches. FIG. 18 i Regarding friction in screws, the thread of a screw may be considered as an in- clined plane, of which the cos. is the middle circum- ference of the screw, the sin. is the pitch, and the force is acting parallel to the base. Hence the fol- lowing formula : F Force, acting at a radius of R inches. W Weight. P = Pitch of screw in inches. / = Coefficient of friction, usually taken as 0.15. R = Radius in inches at which the force is acting. r = Middle radius of screw in inches. d = Middle diameter of screw in inches. MECHANICS. EXAMPLE. Find the force required to act on a lever 30 inches long (see Fig. 18) in order to lift the load W, which is 8000 pounds ? The screw is ^-inch pitch and IX'inch middle radius ; coefficient of friction, 0.15. Solution : F = 8000 X - 5 + - 15 x 3-1416 X 2.5 1.25 2.5 X 3.1416 0.15 X 0.5 > 30 .F=8000 X X 0.0416 == 89.6 pounds. < .7 lo When the screw has V thread, the frictional resistance will be increased as -*- of the angle a (see Fig. 18), or equal to secant of half the angle of the thread. For United States standard screws the angle of thread is 60, half the angle is 30, and secant of 30 is 1.1547, and the formula will, for United States standard thread, become : - dK 1.15//> R All the letters having the same meaning as in the formulas for the square-threaded screws. The following table is calculated for square-threaded screws, the pitch of the screw being double that of the United States standard screw of same diameter. The depth of the thread is equal to its width. We see no good reason why the depth of a square-threaded screw should be, as frequently given in tech- nical books, of the pitch of the screw ; f J, as given in pre- vious tables, is more convenient, and also gives a little more wearing surface to the thread. The use of this table is so plain that it needs very little explanation. In the fourth column is the area of the outside diameter of the screw. In the fifth column, the sectional area of the screw at the bottom of the thread, which may be used in calculating the tensile and crush- ing strength of the screw. Subtracting the fifth column from the fourth gives the sixth column, which is the projected area of one thread ; this may be used in calculating the allow- able pressure on the thread, etc. The fourteenth column gives the tangential force which is required to act with a leverage of one foot in order to lift one pound by the screw if there was no friction. The fifteenth column gives the total tangential force required per pound of load when both load and friction are included. The sixteenth column gives the difference between the fourteenth and the fifteenth columns, and is the tangential force absorbed by friction alone. The coefficient of friction in both columns is assumed as 0.16. The last four columns in the table give the load or axial pressure which may be allowed on the screw corresponding to 200, 400, 600 and 1000 pounds pressure per square inch of projected area of screw thread when the length of the nut is twice the diameter of the screw. JO B3JB pajoafoad jo qoui aiEnbs jad s'punod 0001' Suipuodsajjoo aanssajd jBixy c;Si TH TH TH 'N" long. Completing the parallelogram by drawing lines c and d y the diagonal, x, will indicate the magnitude and direction of the resulting force. Suppose these two forces act in such direc- tions that when the parallelogram is completed and the diagonal drawn, it is, by measurement, found to be 4%." long = |f; then the result of the two forces, a and b, is a force of 76 pounds. In many cases, the result of force and stress in ma- chinery and structures may very conveniently be obtained in this way with much less labor than by calculation, and with accuracy consistent with good, legitimate practice, HORSE- POWER. 317 HORSE-POWER. The term horse-power, as applied in mechanical calcula' tions, is 33,000 foot-pounds of work performed per minute, or 550 foot-pounds of work per second. To Calculate the Horse-Power of a Steam Engine. RULE. Multiply the area of piston in square inches by the mean effective steam pressure, and this by the piston speed in feet per minute, and divide this product by 33,000. The quotient is the horse-power of the engine. Formula : Horse-power = 0.7854 Z* X X 8 , X 33000 D = Diameter of piston in inches. p = Mean effective steam pressure in pounds per square inch. s = Length of stroke in feet. n = Number of revolutions per minute. EXAMPLE. What is the horse-power of a steam engine of the following dimensions ? Cylinder, 20 inches diameter ; length of stroke, 3 feet ; number of revolutions per minute, 75 ; mean effective steam pressure in cylinder during the stroke, 60 pounds per square inch. Horse-power = 20 * X - 7854 X 2 X 3 X 75 X 60 Horse-power = 33000 314.16 X 450 X 60 33000 Horse-power 257.04 To Calculate the Horse=Power of a Compound or Triple Expansion Engine. RULE. Calculate the mean effective pressure of the steam (accord- ing to its number of expansions and initial pressure), and cal- culate the horse-power exactly as if it was a single cylinder engine of the same size as the size of the last cylinder. Another way is to take indicator diagrams of each cylinder, and calculate the power of each cylinder separately. 318 HORSE-POWER. To Judge Approximately the Horse=Power which may be Developed by Any Common Single Cylinder Engine. RULE. Square the diameter of the piston in inches and divide by 2 ; the quotient is the horse-power which the engine may develop. NOTE. This rule gives the exact horse-power, if the prod- uct of the piston speed in feet and the average pressure per square inch in the cylinder is 21,000. Horse=Power of Waterfalls. RULE. Multiply the quantity of water in cubic feet falling in a minute by 62.5 ; and multiply this by the height of the fall in feet; divide this product by 33,000, and the quotient is the horse-power of the waterfall. Or, multiply the quantity of water in cubic meters falling in a minute, by 1000, and multiply this by the height of the fall in meters ; divide the product by 4500, and the quotient is the horse-power of the waterfall. NOTE. The above rules give the gross power of the water- fall, but the useful effect of the fall is a great deal less and will depend on the construction of the motor. It may be only from 40% to 80% of the natural power of the waterfall. Animal Power. Under favorable circumstances, a horse can perform 22,000 foot-pounds of work per minute. For instance, a horse walking in a circle turning the lever in a so-called horse-power may exert a pull of 100 pounds, walking at a speed of 220 feet per minute. For the horse to work to advantage, the diameter of the circle ought to be at least 25 feet. Hauling a Load. The average speed when horses are used in hauling a load one way and returning without load the other way, allowing for necessary stoppages, may not be more than 175 feet per minute, and, in estimating, time must also be allowed for loading and unloading. Loads may vary from 1000 to 2000 pounds, accord- ing to the road. Commonly speaking, the force required to pull a loaded wagon on a good, level road increases in propor- tion to the load and decreases in proportion to the diameter of the wheels, and on soft roads it is less with wide tires than with narrow ones. The idea that a wagon having small wheels would be easier to pull up-hill than one having larger wheels is a fallacy. HORSE-POWER. 319 Power of Man. A man may be able to do work at a rate of 4000 foot- pounds per minute ; for instance, in turning a crank on a crane or derrick, a force of 15 pounds may be exerted on a crank, 18 inches long and, with 30 turns per minute, the work would be 4228 foot-pounds per minute. NOTE. In derricks, pulley blocks, jack-screws, etc., a large part of the expended power is consumed in overcoming friction. Power Required to Drive Various Kinds of Machinery. In the nature of the thing it is impossible from experiments on one machine to tell exactly what power it takes to run an- other similar machine, as there are so many different factors entering into the problem ; for instance, the speed and feed on the machine, the hardness of the stock it works on, the quality of the tools used, the kind of lubrication, etc. Therefore, such assertions are only approximations at the best. 16-inch engine lathe, back geared, % horse-power. 26-inch engine lathe, back geared, 1 % horse-power. Planer, 22" x 22" x 6 feet, l / 2 horse-power. Planer, 32" x 32" x 10 feet, % horse-power. Shaping machine, 10-inch stroke, # horse-power. 20-inch drill press, ]/ 2 horse-power. 26-inch drill press, back gear, boring a 3-inch hole, using boring bar, 1 horse-power. Plain milling machines (Lincoln pattern, No. 2), 1%. horse-power. Small Universal milling machines, % horse-power. Circular saws (for wood), 24" di- ameter (light work), 3> horse-power. Circular saws (for wood), 36" di- ameter (light work), 6 horse-power. Fan blower for cupola, melting four tons of iron per hour, 10 horse-power. Fan blower for five blacksmith fires, 1 horse-power. Drop hammer, 800 pounds, 8 horse-power. In machine shops and similar places, from 40% to 70% of the total power required is consumed in running the line shaft- ing and counter-shafts. An average of from 55% to 60% is probably the most common ratio. In exceptionally well-arranged establishments, under favor- able conditions, in light manufacturing it may be possible that only 30% of the power is consumed in driving line and counter shafting, and that 70% is used for actual work. 320 SPEED OF MACHINERY. SPEED OF MACHINERY. The peripheral velocity of circular saws ought not to exceed 10,000 feet per minute. Table No. 37 gives the number of revo- lutions per minute for circular saws of different diameters. TABLE No. 37- Diameter of saw in inches. 8 10 12 14 16 20 24 28 32 Number of revolutions per minute. 4500 3600 3000 2585 2222 1800 1500 1285 1125 Band Saws. Small band saws, such as are usually used in carpenter shops, have a velocity of 3600 feet per minute. The reason why band saws are run so much slower than circular saws is that if the band saw is given too much speed the blade will be pulled to pieces in starting and stopping. Drilling flachines for Iron. For drilling steel, the surface speed of a drill should not exceed 15 feet per minute; cast-iron, 22 feet; brass, 27 feet; malleable iron, 25 to 30 feet per minute. The feed will vary according to the hardness of the stock. In cast-iron a %" drill will drill a hole 1" deep in 125 revolutions. A l / 2 " drill will drill a hole I" deep in 120 revolutions. A 1" drill will drill a hole 1" deep in 100 revolutions. Lathes. Cast-iron may be turned at a speed of 32 feet per minute when Muchet steel is used for tools. Thus, lathes are usually calculated to have a velocity of about 30 to 32 feet on the slovy- est speed, supposing that as large a diameter as the lathe will swing is turned. For wood-turning the surface speed may be from 3000 to 0000 feet per minute ; but when the article to be turned is out of balance the speed must be considerably slower. Planers. Cast-iron is planed at a speed of 25 to 27 feet per minute ; wrought iron, 21 feet; steel. 16 feet per minute. A planer ought to return at least three times as fast as it goes forward. Hilling Machines. Rotating cutters working on Bessemer steel or other mate- rials of about equal hardness usually have a surface speed of SPEED OF MACHINERY. 321 about 40 feet per minute. Oil is used for lubrication. Cast- iron is milled without oil. Grindstones. When grindstones are used to grind steel and iron in manu- facturing, they work at a surface speed of 2000 to 2500 feet per minute, but grindstones for common shop use, to grind tools, chisels, etc., run at much slower speed. Emery Wheels and Emery Straps. Emery wheels and straps do good work at a speed of 5000 to 6000 feet per minute, but all such high-speed machinery, especially grindstones and emery wheels, must be used very carefully and special attention paid to the strength, so that they will not break under the stress of centrifugal force. Calculating Size of Pulleys. TO FIND SIZE OF PULLEY ON MAIN SHAFT. Multiply the diameter of pulley on counter-shaft by its number of revolutions per minute, and divide this product by the number of revolutions of the main shaft, and the quotient is the diameter of the pulley on the main shaft. EXAMPLE. A main shaft makes 150 revolutions per minute ; the counter- shaft has a pulley 9 inches in diameter and is to make 400 revolu- tions per minute. What size of pulley is required on the main shaft? Solution : Diameter of pulley 40 * 9 = 24 inches. 150 TO FIND SIZE OF PULLEY ON COUNTER-SHAFT. RULE. Multiply the diameter of pulley on the main shaft by its number of revolutions per minute, and divide this product by the number of revolutions of the counter-shaft; the quotient is the diameter of the pulley on the counter-shaft. EXAMPLE. The pulley on a main shaft is 30 inches in diameter and it makes 150 revolutions per minute ; the counter-shaft is to make 450 revolutions per minute. What size of pulley is required ? Solution : Diameter of pulley = 36 X 15 = 12 inches. 322 SPEED OF MACHINERY. TO FIND THE NUMBER OF REVOLUTIONS OF THE COUNTER SHAFT. RULE. Multiply the diameter of pulley on the main shaft by its number of revolutions per minute and divide this product by the diameter of pulley on the counter-shaft, and the quotient is the number of revolutions of the counter-shaft per minute. EXAMPLE. The pulley on a main shaft is 24 inches in diameter and makes 150 revolutions per minute, and the pulley on the counter- shaft is 15 inches in diameter. How many revolutions per minute will the counter-shaft make? Number of revolutions = = 240 revolutions per minute. 15 To Calculate the Speed of Gearing. In calculating the speed of gearing, use the same rules as for belting, but take the number of teeth instead of the diameter. EXAMPLE. The back gearing on a lathe consists of a gear and pinion of 8 pitch, 90 teeth and 32 teeth, and the other gear and pinion are 10 pitch, 120 teeth and 40 teeth. How many revolutions will the cone pulley make while the spindle makes one revolution ? Solution : Cone pulley makes = 9 ^ X 12 = 9 revolutions. o^ /\ 40 Efficiency of Machinery. Divide the energy given out by a machine by the energy put into the same machine ; multiply the quotient by 100, and the result is the per cent, of efficiency of the machine. A dynamo requires 15 horse-power, but the electrical power given out is only 12 horse-power. What is the efficiency? Solution : Efficiency = -If- X 100 = 15 A steam engine is to develop 60 horse-power net. What will be the gross horse-power if the efficiency is 75% ? Solution : Gross power = 60 * 10 = 80 horse-power. <5 CRANE HOOKS. CRANE HOOKS. 3*3 FIG. 3. Crane hooks, as shown in Figs. 1, 2 and 3, may be designed by the following formulas : P = Load in tons. D = Diameter of iron in inches. d\y 2 D S Standard screw of diameter r * Vl6 D When a rectangular iron plate is substituted for a washer, the bearing surface of the plate against the wood should at least be equal to the area of the washer, calculated by the above formula. Chain Links. (See Figure 4.) D = Diameter of iron. L = y 2 to5Z>. (For strength of chains, see page 222). FIG. 4. 3*4 CRANES AND DERRICKS. CRANES. Cranes and derricks are machines used for raising and lowering heavy weights. In its simplest form, a crane con- sists of three principal mem- bers : The upright post, the horizontal jib and the diagonal brace. (See Fig. 5). The weight P will produce tensile stress in the jib, compressive stress in the brace, and both compressive and transverse stress in the post. Tension in jib = P X x FIG. 5. Compression in brace = - y . PKk Stress in the upper bearing = e When the post is held at both ends, as in Fig. 5, it may, with regard to transverse strength, be considered as a beam of length t, fastened at one end and loaded at the other with a load equal to the force h X P The compression on the post caused by the load is equal to />. The downward pressure on the lower bearing is equal to the sum of the weight of the crane and the load which it supports. Proportions for a Two-Ton Derrick (Of the construction shown in Fig. 6) . Pulley blocks should be double-sheave (only single are shown in the cut). Circumference of manila rope, 3# inches. Mast, 8X8 inches, 26 feet long. Boom, 7X7 inches, 20 feet long. CRANES AND DERRICKS. 325 FIG. 6. Large gear, 72 teeth, 1- inch circular pitch, 2-inch face. Small pinion, 12 teeth, 1-inch circular pitch, 2-inch face. Crank shaft. \ l / 2 inches in diameter. Bearings, 2^ inches long. Crank, 18 inches long, Drum, 7 inches in diame- ter, 24 inches long. Drum- shaft, 2X inches in di- ameter. The drum and large gear are fitted and keyed to the drum shaft and also bolted together, thereby relieving this shaft from twisting stress. The radius of the drum added to the radius of the rope makes four inches, and the force is multiplied five times by the double-sheave pulley block; therefore, when Gear the friction in thec ran k mechanism is not consid- ered, the force required on the crank in order to lift 4400 pounds will be : F = -_L X _ 12 X 440 33 pounds, very nearly. 18 X 72 X 5 Thus, when two men are working the derrick (one at each crank), each man has to exert a force of \Q l / 2 pounds, but, including friction, each man probably exerts a force of 20 to 25 pounds, when the derrick is loaded to its full capacity. For very rapid work it is necessary to have four men (two on each winch-handle) to work the derrick, if it is kept loaded to its maximum capacity, but for ordinary stone work such a derrick is usually worked by two men. Stones as heavy as two tons are seldom handled, except where larger derricks and steam power are used. When the derrick is to be worked constantly, the limit of the average stress on the crank handle to be allowed for each man is 15 pounds. When working an 18-inch crank, 48 turns per minute, this corresponds to a force of 15 pounds acting through a space of a little over 220 feet 3300 foot-pounds of work per minute = jV horse-power. When the crank swings in a shorter radius a few more turns per minute may be expected, but experience indicates that an 18" radius is the most practical proportion. 326 BELTS. BELTS. Oak-tanned leather is considered the best for belting. The so-called "short lap" is cut lengthwise from the middle of the back of the hide, where it has the most firmness and strength. Single belting more than three inches in width is about T y thick, and weighs 15 to 16 ounces per square foot j when less than three inches in width it is usually -f.," thick and weighs about 13 ounces to the square foot. Light double belts, as used for dynamos and other ma- chinery having pulleys of comparatively small diameter, are about y thick and weigh about 21 ounces per square foot. Double "belting, as used for main belts, is a little heavier and weighs from 25 to 28 ounces per square foot. Belts as heavy as 30 ounces per square foot are frequently used, and are usually termed "heavy double." Large engine belts are sometimes made with three thicknesses of leather. Belts should be soft, pliable and of even thickness. When a belt is of uneven thickness and has very long joints, so that it looks as if it was partly single and partly double, it is very doubtful if it will do good service, for this is a sure sign that the thin and flimsy parts of the hide have been taken into the stock in making the belt. The ultimate tensile strength of leather belting is from 2600 to 4800 pounds per square inch of section. Thus, a leather belt T y thick will break at a stress of 500 to 900 pounds per inch of width. The lacing of belts will reduce their strength from 50 to 60 per cent.; therefore, when practicable, belts ought to be made endless by cementing instead of lacing. A belt will transmit more power, wear better and last longer, if it is run with the grain side next to the pulley. Belts should never be tighter than is necessary in order to transmit the power without undue slipping ; too tight belts cause hot bearings, excessive wear and tear, and loss of power in over- coming friction ; but, on the other hand, it is necessary to have a belt tight enough to prevent it from slipping on the pulley, be- cause if a belt slips there is not only a direct loss in velocity, but the, belt will wear out in a short time ; it is, therefore, very im- portant to use belts of such proportions that the power shall be transmitted with ease. Belts always run toward the side of the pulley which is largest in diameter (therefore pulleys are crowned, in order to keep the belt running straight). A belt will always run toward the side where the centers of the shafts are nearest together. Open belts will cause two shafts to run in the same direction, BELTS. 327 A crossed belt will cause the shafts to run in opposite direc- tions. If the distance between the shafts is' short, crossed belts will not work well. A short belt will wear out faster than a long one. Very long and heavy belts should be supported by idlers as well under the slack as under the working side ; if not, the weight of a long belt will cause too much stress on itself and also cause too much pressure on the bearings, as well on the driver as on the driven shaft. Belts should never, when it can be avoided, be run vertically, as the weight of the belt always tends to keep it away from the lower pulley, thereby reducing its transmitting capacity; the longer the belt the worse this is. Belts are most effective when they are run in a horizontal direc- tion and, whenever possible, the lower part of the belt should be the working part, as the slackness in the upper part, by its weight, will cause the belt to lay around the pulley for a longer distance, and this will, in a measure, increase its transmitting capacity; but if the upper part is the working part, the slackness in the lower part tends to keep the belt away from the pulleys, and thereby reduces its transmitting capacity. Lacing Belts. Figure 1 shows a good way of lacing belts ; a is the side run- ning next to the pulley and b is the outside. Holes should be punched and not made by an awl, as punched holes are less lia- ble to tear. The lacing is commenced by putting each end of the lace through holes 1 and 2 from the side next to pulley, and then continuing toward the edges, both sides simultaneously, FIG. 1. a w\ JL making a double stitch at the edges and sewing back again un- til holes 1 and 2 are reached ; and, lastly, by drawing each end of the lace through .r and y. Each stitch w'ill be double, except- 328 BELTS. ing- the middle one. The holes x and y, where the ends of the lacing are finally drawn through for fastening, are made by the belt awl and should always be made small, and the lacing, if laid out rightly, always enters these holes from the inside of the belt ; after it is pulled through, a small cut is made in the lacing on the outside, which will prevent it from drawing back again, then the ends are cut off about l / 2 " long, as shown in the figure at.r and j>. It is a bad practice to leave the lace-ends on the inside of belts, because they will then soon wear off, allowing the joint to rip. A 1-inch belt ought to have three lace-holes in each end. Length of lacing, 12 inches. A 2-inch belt ought to have three lace-holes in each end. Length of lacing, 18 inches. A 3-inch belt ought to have five lace-holes in each end. Length of lacing, 24 inches. A 4-inch belt ought to have five lace-holes in each end. Length of lacing, 32 inches. A 5-inch belt ought to have seven lace-holes in each end. Length of lacing, 40 inches. A 6-inch belt ought to have seven lace-holes in each end. Length of lacing, 48 inches. An 8-inch belt ought to have nine lace-holes in each end. Length of lacing, 60 inches. A 10-inch belt ought to have eleven lace-holes in each end. Length of lacing, 72 inches. A 12-inch belt ought to have thirteen lace-holes in each end. Length of lacing, 84 inches. Always have the row having the most holes nearest the end of the belt. Cementing Belts. When belts are cemented together, a 3-inch belt is lapped four inches and a 4-inch belt 4^ inches. In larger belts the lap is usually made equal to the width of the belt, but it may be made even shorter when the width of the belt is over 12 inches. The two ends are jointed together, so that the thickness is even with the rest of the belt. The American Machinist, in answer to Question No. 430, Dec. 5, 1895, says : " For leather belts take of common glue and American isinglass equal parts; place them in a glue pot and add water sufficient to just cover the whole. Let it soak 10 hours, then, bring the whole to a boiling heat, and add pure tan- nin until the whole appears like the white of an egg. Apply warm. Buff the grain of the leather where it is to be cemented; rub the joint surfaces solidly together, let it dry for a few hours, and the belt will be ready for use. For rubber belts take 16 parts gutta percha, 4 parts India rubber, 2 parts common caulk- er's pitch, 1 part linseed oil ; melt together and use hot. This cement can also be used for leather," BELTS. 329 Length of Belts. Small belts, such as 4 inches wide or less, will work well when the distance between the shafts is from 12 to 15 feet, larger belts when from 20 to 25 feet, and for large main belts 25 to 30 feet distance is satisfactory. Horse=Power Transmitted by Belting. A single belt weighing about 15 ounces per square foot is capable of transmitting one horse-power per inch of width, when running at a speed .of 800 feet per minute over pulleys of proper size, both of equal diameter. As one horse-power is 33,000 foot-pounds of work per minute, this will make the tension 38000 due to the power the belt is transmitting = ^ '= 41 y^ Ibs. per inch of width, but the total tension in the belt is, of course, considerably more per inch of width, because the belt must be tight enough to prevent its slipping on the pulley. For belts lighter than 15 ounces per square foot it is better to allow 1000 running feet per horse-power per inch of width of belt. For light double belts weighing 21 ounces per square foot, 600 running feet per horse-power per inch of width may be allowed. For double belts weighing 25 ounces per square foot, 500 running feet per horse-power per inch of width may be allowed. Hence the following formulas : For light single belts weighing less than 15 ounces per square foot, TT _ v X b JJX 1000 ~~Iobo~ ~^~ For single belts weighing 15 to 16 ounces per square foot, H -v X b HX 800 800"" v For light double belts weighing about 21 ounces per square foot, 7,X* //X600 <><)<) v For double belts weighing about 25 ounces per square foot, ff - "" X b t= //X500 500 v H = Horse-power. b Width of belt in inches. v Velocity of belt in feet per minute, which will be di- ameter of pulley in inches multiplied by 3.1416 and by the num- ber of revolutions per minute, and the product divided by 12, 33 BELTS. EXAMPLE 1. A double belt 10 inches wide, weighing 25 ounces per square foot, runs over 50-inch pulleys, making 240 revolutions per min- ute. How many horse-power will it properly transmit ? Solution : '. . , . , 50 X 3.1416 X 240 Velocity of belt = - 12 - = 3141.6 ft. per minute. 3141.6 X 10 ~ 500 -- = ^ 2 '^ norse 'P wer - EXAMPLE 2. One hundred horse-power is to be transmitted by a double belt weighing 25 ounces per square foot. The pulleys are 66 inches in diameter and make 150 revolutions per minute. What is the necessary width of belt ? Solution : Pulleys of 66 inches diameter, running 150 revolutions per minute, will give a belt speed of 15 X 3 ' 1416 X 66 _ 3591.8; say, 2592 feet per minute. 100 X 500 2592 - 19 - 3 inches; thus, a double belt 20 inches wide will do the work. EXAMPLE 3. A light single belt 4 inches wide, weighing 13 ounces per square foot, runs over pulleys of 36 inches diameter, making 100 revolutions per minute. How many horse-power may be trans- mitted? Solution : 36 X 3.1416 X 100 Velocity of belt = - ^ - = 942.48 ft. per minute. The belt is a light single belt and its transmitting capacity 4 X 942.48 will be, H = -- -- = 3.76992, about 3^ horse-power. To Calculate Size of Belt for Given Horse-Power when Diameter of Pulley and Number of Revolutions of Shaft Are Known. The following formulas may be used for calculating belt transmission, and will give results approximately consistent with previously given rules, but they are more convenient for use, as the velocity of the belt does not need to be first calculat- ed, but the velocity of the belt must not exceed the practical limit. BELTS. 331 This formula will do for either single or double leather belts with cemented joints (no lacing), of any weight from 12 to 30 ounces per square foot and of any width from one to thirty inches, when the pulleys are of suitable size to correspond with the thickness of the belt, and the diameter of both pulleys is equal or nearly so : _ d X n X b X w _ HX 50000 ~liOOOO~~ d ~ nXbXw H X 50000 _ H X 50000 H X 50000 H = Horse-power transmitted by the belt. d Diameter of pulley in inches. n = Number of revolutions per minute. b = Width of belt in inches. iv = Weight of belt in ounces per square foot. 50,000 is constant. EXAMPLE. Calculate Example 2 by the above formula. Solution: 100 X 50000 ^ ~ 06 X 150~X~25 = 20 '^ mcnes ' wnicn f r a U practical purposes, is the same as the result when calculated by the other rule. Wide and thin belts are unsatisfactory. It is far better when transmitting power to use double and narrow rather than single and wide belts. It is a very bad practice to run at too slow belt speed, and also to use pulleys of too small diam- eter. The smallest pulley for a light double belt should never be less than 12" in diameter, for a heavy double belt never less than 20" in diameter, and for a triple belt the pulley should not be less than 30" in diameter. To Calculate Width of Belt when Pulleys are of Unequal Diameter. When the pulleys are of different diameters the belt will lay around the smallest pulley less than ISO degrees, and the trans- mitting capacity of the belt is correspondingly reduced. The pressure on the pulley due to the tension of the belt will vary as the sine of half the angle of contact, and the adhesion of the belt to the pulley will vary as the pressure; consequently, also, the transmitting capacity of the belt will vary as the sine of half of the angle of contact, but it is usually advisable in practice to allow a little more on the width of the belt than is called for by this rule. A practical rule is : First calculate the width of the belt by the above rules and formulas, as though both pulleys had the same diameter, 332 BELTS. then multiply the result by the following constants, according to the arc of contact between the belt and the small pulley. When the arc of contact between the belt and the small pulley is 90 multiply by 1.60. 100 " " 1.45 140 multiply by 1.15 110 " " 1.35 150 " " 1.10 120 " " 1.25 160 " " 1.06 130 " 1.20 170 " " 1.04 EXAMPLE. The pulley on a dynamo is 15" in diameter, and it makes 1200 revolutions per minute. The driving pulley is so large that the belt only lays around the dynamo pulley for a distance of 150 degrees. What is the necessary width of a light double belt, weighing 21 ounces per square foot, when it takes 40 horse-power to run the dynamo ? Solution : If the arc of contact had been 180 degrees the belt would 40 X 50000 be b = 1200 X 15~X~2T : : inches wide, but as the arc of contact is not 180 degrees, but only 150 degrees, this width is multiplied by the constant 1.10, as given in the preceding table. Thus, the width of the belt will be 5.3 X 1.1 = 5.83 inches or, practically, a belt six inches wide is required. When belts are running in a horizontal direction, and the driven pulley and the driver are of equal diameter and finish, the belt will always, when overloaded, commence to slip on the driver, and when pulleys are of unequal size it is always more favorable for the belt when the driving pulley is the larger than when vice versa. To Find the Arc of Contact of Belts. Make a scale drawing of the pulleys and the belt, and measure the arc of contact from the drawing by means of a protractor, or the arc of contact in degrees on the small pulley for an open belt may be calculated by the formula: Cosine of half the angle = R ~ r R = Radius of large pulley in inches. r = Radius of small pulley in inches. / = Distance in inches between centers of the shafts. EXAMPLE. The distance between centers of two shafts is 16 feet ; the large pulley is 60 inches and the small pulley is 20 inches in diameter. What is the arc of contact of the belt? BELTS. 333 Solution : 16 feet = 192 inches. 60 inches diameter = 30 inches radius. 20 inches diameter = 10 inches radius. Cos. of half the angle = ( 30 10 ) .104 192 In tables of natural cosine (page 158), the corresponding angle is found to be 84 degrees, very nearly ; thus, the angle for arc of contact will be 2 X 84 = 168 degrees on the small pulley. On the large pulley the arc of contact will be 360 168 = 192 degrees. For a crossed belt the arc of contact is always the same on both pulleys, and it may be calculated by the formula : Cos. of half the angle = j A* = Radius of large pulley. r Radius of small pulley. / = Distance between centers. EXAMPLE. What will be the arc of contact for the belt on the pulleys in the previous examples if belt is run crossed instead of open ? Solution : Cosine of half the angle = 30 + 10 = 0.208 ; the 192 corresponding angle will be 180 77 = 103 degrees, and the arc of contact will be 103 X 2 = 206 degrees. Pressure on the Bearings Caused by the Belt. Approximately, the pressure on the bearings caused by the belt may be considered to be three times the force which the belt is transmitting. Therefore, the pressure may be calculated by the formula : p = 3 X 33000 X H v P = Pressure on the bearings due to pull of belt. H = Number of horse-power transmitted by the belt. v = Velocity of belt in feet per minute. EXAMPLE 1. A belt is transmitting 60 horse-power and its velocity is 900 feet per minute. What is the pressure in the bearings due to the belt? 334 BELT! Solution : 900 EXAMPLE 2. Suppose the diameters of the pulleys are increased until a belt speed of 3000 feet per minute is obtained. What will then be the pressure in the bearings caused by the belt when trans- mitting 60 horse-power ? Solution : p = 3 X 33000 X 60 = 1980 ^ 3000 By the above examples it is conclusively shown what a great advantage there is in using pulleys so large in di- ameter that proper belt speed is obtained. (See velocity of belts, page 337). The approximate pressure may also be very conveniently obtained from the width of the belt, thus: For light single belts, allow 1000 feet of belt speed per horse-power transmitted per inch of width of belt. The effective pull in such a belt will be 33 pounds per inch of width, and the pressure on the bearings due to the belt will accordingly be 33 X 3 = 99 pounds per inch of width of belt. For convenience, say 100 pounds pressure in the bearings per inch of width of such belts. For belts where 800 running feet are allowed per horse-power per inch of width of belt, this reasoning will give a pressure on the bearing equal to i23# pounds per inch of belt. For convenience, say 125 pounds pressure in the bearings per inch of width of such belts. For belts where 600 running feet are allowed per horse- power per inch of width, the pressure in the bearing is equal to 165 pounds per inch of width of belt, and where the belt is so heavy that only 500 feet of belt speed per horse-power per inch of width is allowed, the pressure in the bearings will be 198 pounds per inch of width. A good, practical rule, which can very easily be remembered, is, (when belts are in good order and have the proper size and the proper tension) : Multiply weight of belt in ounces per square foot by eight times' the width of the belt in inches, and the product is approximately the pressure in pounds upon the bearings caused by the belt. EXAMPLE. A belt is calculated with regard to the horse-power it has to transmit under a given velocity, and found to be 8-inch double belting, weighing 25 ounces per square foot. What pres- sure will it cause on the bearings when working at proper tension ? BELTS. 335 Solution, by the last rule : 1600 pounds. Solution, by the first rule: At a speed of 3000 feet per minute such a belt will transmit X 8 X) : formula : ~~500 ' ^ horse-power, and calculating the pressure by the p _ 3 X 33000 X H p = 3 X 33000 X 48 = 15g4 ^ 3000 Both rules give nearly the same result, and one is just as correct as the other, as all such figuring is nothing more than approximation at the best. The pressure on the bearings may be a great deal more than calculated above. Sometimes the pulleys are roughly made, belts are poor, and consequently the coefficient of friction between belt and pulley is small, and as the belt has to be a great deal tighter in order to do the work, the pressure on the bearing will be greatly increased. Very frequently, from pure ignorance or carelessness, belts sare made very much tighter than necessary, and enormous sums of money may be wasted in this way in large factories, as the steam engines, at the expense of the coal pile, have to furnish power not only to do the useful work, but also to overcome all the friction produced by such over-strained belts, hot bearings, etc. A belt will transmit more power over a good, smooth pulley than over a rough one. When pulleys are covered with leather a belt will transmit about 25% more power than it will when running over bare iron pulleys, and in transmitting the same power a much slacker belt may be used, thereby reducing the friction in the bearings. Special Arrangement of Belts. By the use of suitable guide pulleys it is possible to connect with belts shafts at almost any angle to each other. But experience is required and care must be exercised to do it suc- cessfully. When guide pulleys are used in order to change the direction of a belt, always remember that when the belt is run- ning the most pressure is thrown on the pulley guiding the working part of the belt. This pulley is, therefore, very liable to heat in its bearings, if not designed to have bearing surface enough and also to have proper means for oiling. 336 BELTS. Fig. 2 shows an arrangement by which the direction of motion of two shafts may be reversed, when the distance between the shafts is too short for the use of a crossed belt or when a crossed belt, for any other reason, cannot be used. Suppose pulley A to be the driver and to run in the direction of the arrow. C and D are guide pulleys, and the motion of the driven shaft B is in the op- posite direction to the shaft A. In this case the guide pulley C is on the working part of the belt and is the one to which^special attention must be paid in regard to heating. If the direction of shaft A is reversed, guide pulley D will be on the working part of the belt. Crossed Belts. If the distance between A and B (Fig. 2) had been long enough, it would have been preferable to reverse the motion of B by means of a crossed belt, instead of by the arrangement shown in Fig. 2. Crossed belts do not work well when running on pulleys small in diameter as compared to the width of the belt. Too short distance between the shafts must be avoided. Wide crossed belts are very unsatisfactory; therefore, instead of running one wide crossed belt it is preferable to use two belts, each of half the width, and run them on two separate pairs of pulleys. Such belts should be of equal thickness, and the pulleys should be crowned, well finished and of correct size, so that each belt will do its share of the work. Quarter =Turn Belts. Fig. 3 shows a so-called quarter- turn belt, used to connect two shafts when running at an angle and laying in different planes. The principal point to look out for is to place the pulleys (as shown in Fig. 3) so that the belt runs straight from the de- livering to the receiving side of each pulley. The pulleys shown in Fig. 3 are set right for belts running in the direction of the arrows. If the mo- tion is reversed, the belt will run off the pulleys. FIG. 3. BELTS. 337 Angle Belts. The belt arrangement shown in Fig. 4 is usually called an angle belt, and is used to connect two shafts at an angle. Either one, A or B, may be the driver, and there are two guide pulleys (one for F|G - 4 - , each part of the belt at C), one of which, of course, is on the driving part of the belt. Crossed belts, quarter-turn belts, and angle belts must never be wide and thin ; much better results are obtained by narrow, double belts than by wide, single ones. Angle belts and quarter-turn belts are frequently bothersome contrivances. Their running is sometimes improved by making a twist in the belt when joining its ends ; that is, lacing the flesh side of one end and the hair side of the other end on the outside. This will prevent one side of the belt from stretching more than the other. Slipping of Belts. Owing to the elasticity of belts, there must always be more or less slip or " creep " of the belts on the pulleys. Under favorable conditions it may be as low as 2%, but frequently the slip is more. Therefore, if two shafts are connected by belts, and both should have very nearly the same speed, the diameter of the driver should be at least 2% larger than the diameter of the driven pulley. When the driver is comparatively large in diameter and the driven pulley is small, it is advisable to have the driver from 2 to 5% over size, in order to get the required speed. Tighteners on Belts. If tighteners are used they should always be placed on the slack part of the belt. Velocity of Belts. Belts are run at almost all velocities from less than 500 to 5000 feet per minute, but good practice indicates that whenever possible main belts having to transmit quantities of power are run most economically at a speed of 3000 to 4000 feet per min- ute. At a higher speed both practice and theory seem to agree that the loss due to the action of the centrifugal force in the belt when passing around the pulley, and that the wear and tear is so great when the speed is much over 4000 feet per minute that there is not much practical gain in increasing the speed. But, as a general rule, whenever possible the higher the belt speed the more economical is the transmission as long as the belt speed does not exceed the neighborhood of 4000 feet per minute. 338 BELTS. Oiling of Belts. Belts should be kept soft and pliable and are, therefore, usually oiled with either neat's-foot oil or castor oil. Too much piling is hurtful, but the right amount of oiling at proper times is very beneficial to the action of the belt and will prolong its utility to a great extent. REMARKS. All previous rules for calculating belting are founded upon good, legitimate practice, but are only offered as a guide, as no rule can be given which will fit all cases. For instance, a belt may be amply large to transmit a given horse-power when running in a horizontal direction, but it may fail to do the same work if running in a vertical di- rection. A belt may be large enough to do its work when run- ning in a vertical direction over pulleys of unequal size with the large pulley on the lower shaft, but it may fail to do the same work satisfactorily with the large pulley on the upper shaft and the small pulley on the lower one. Leather belts should not be used where it is damp or wet, but rubber belting will usually give good service in such places. For information regarding rubber belts, see manufacturers' catalogues. WIRE ROPE TRANSMISSION. Transmitting power by wire ropes running at a high speed over grooved pulleys, or " telodynamic transmission." as it is also called, is the invention of the brothers Him of Switzerland. For long distances this mode of transmission is far cheaper than leather baiting or lines of shafting. Fig. 1 shows a section of a pulley as u^ed for this kind of transmission ; a is an elastic fill- ing, usually made from leather cut out and packed in edgewise. The groove is made wide, so that the rope will rest entirely against the packing and not touch the iron. This is different from transmission with hemp rope, which is made to wedge into the groove of the pulley. The diameter of the pulley in the groove, where the wire rope runs, ought to be at least 150 times the diameter of the rope ; the larger the better, so long as the velocity of the rope does not exceed 5000 feet per minute. The pulleys must run true and be in balance and in exact line with each other, and the WIRE ROPE TRANSMISSION. 339 shafts must be parallel. The distance between shafts should never be less than 60 feet and should preferably be from 150 lo 400 feet.* For distances longer than 400 feet, either carrying pulleys or intermediate jack shafts are generally used, although spans as long as (500 feet or more have been used, but only when it is possible to give the rope the proper deflection without its touching the ground. Usually the speed is from 3000 to 6000 feet per minute. Higher speed would be dangerous from the stress in cast-iron wheels due to centrifugal force. flj g 1& d ins. b ins. i 5 fi "f 1 u u H /"ins. I H H H S *t I g ins. H M H t ins. A ! i FIG. i. Tightening pulleys should not be used, because if the distance between centers of shaft is top short to give the proper tightness to the rope without a tightening pulley, wire rope transmission is not the form best adapted to the circumstances. Guide pulleys or idlers should be avoided as much as possible, but when necessary they should be as carefully made and put up as the main pulleys, and they ought not to be less than half the diameter of the main pulley if on the slack part, but of the same size if they are on the tight part of the rope. Wire rope for transmission is usually made from the best quality of iron, has seven wires to a strand and consists of six strands laid around a hemp core in the center of the rope. The diameter of the wire rope is from nine to ten times the diameter of each single wire. Never use galvanized rope for power transmission, but pre- serve the rope by painting with heavy coats of linseed oil and lampblack. * When distance between shafts is less than 60 feet, leather belts are prefer- able to wire rope. 340 WIRE ROPE TRANSMISSION. Transmission Capacity of Wire Ropes. A one-inch rope running 5000 feet per minute is capable of transmitting 200 horse-power. The transmitting capacity of the rope is in proportion to the square of its diameter, and the power transmitted by the rope when the velocity is less than 5000 feet per minute is practically in proportion to its velocity.* Hence the formula: H= d * X FX 20 which reduces to H = 0.04 X d* X V 5000 H = Horse-power transmitted. d Diameter of rope in inches. V Velocity of rope in feet per minute. EXAMPLE. How many horse-power may be transmitted by a wire rope y 2 inch in diameter running over proper pulleys at a velocity of 2500 feet per minute ? Solution : H= 0.04 X l / 2 X y 2 X 2500 = 25 horse-power. The pressure on the bearings will not be less than three times the force transmitted, and may be calculated thus : Pressure on bearings = S X "Q^e-power X 33000 Velocity in feet per mm. EXAMPLE. What will be the least pressure in bearings for a wire rope transmitting 150 horse-power at a velocity of 5000 feet per minute ? Pressure on bearings = 2Jli 50 _ >< _??_ 2 = 2970 pounds. 5000 If there is one bearing on each side at an equal distance from the pulley, the pressure on each bearing will be -2-y-ft = 1485 pounds. This is the calculated pressure, and represents what the pressure should be, but it is not certain that this is the actual pressure. It may be greatly increased by having the rope too tight, t * When the velocity of the rope exceeds 6000 feet per minute the stress caused by centrifugal force when the rope is bending around the pulley considerably reduces its transmitting capacity. This loss increases very fast above this speed, because the centrifugal force increases as the square of the velocity. It is very doubtful if there is practically any gain to run wire ropes at a speed exceeding 6000 feet per minute when wear and tear, loss due to centrifugal force, etc., are considered. t Sometimes a pulley is put on the free end of a line of shafting projecting through the wall and drawn by a wire rope outside the shop; this will do only when 3 comparatively small amount of power is to be transmitted. WIR ROPK TRANSMISSION. 341 The tension of the rope may be calculated from its de- flection when at rest (see Fig. 2), and for a rope running hor- izontally the usual formula is : (very nearly) P = Force in pounds at f. W = Weight of rope in pounds from d to f, which is half the span. b = Half the span in feet. a = Twice the deflection in feet. NOTE. (See Fig. 2.) If the length of the line a represents the weight of the part of the rope from */to/~, the length of the line .r represents the tension in the rope at/"; therefore the ten- sion will be as many times the weight as the length of line a is contained in the line x. EXAMPLE. The horizontal distance between two pulleys is 200 feet ; when standing still the deflection in a wire rope of %" diameter is 5 feet. What is the tension in the rope ? Solution : In Table No. 38 the weight of %" wire rope is given as 1.12 pounds per foot; therefore, 100 feet of %" rope will weigh 112 pounds. 112 X V 100 2 + 10* _ 112 10 = 1125.6 pounds. This is the tension in each part of the rope ; therefore the force against the pulley, due to the weight of the rope, is 1125.6 X 2 2251.2 pounds. If this is supported by a bearing on each side of the pulley, the pressure on each bearing, if both are the same distance from the pulley, will be 1125.0 pounds. 342 WIRE ROPE TRANSMISSION. The tension is increased by reducing the deflection. For instance, if the deflection is reduced to 4 feet the tension on the rope will be, P = U2 X p= 112JOOO.S = 140 4.2 pounds. Thus, the tension might be increased to any amount within the ultimate breaking strength of the rope. Deflection in Wire Ropes. When the rope is in motion the deflection will increase on the slack side and decrease on the tight side ; therefore, if the span is long the rope may touch the ground when running if the pulleys are not placed on sufficiently high towers. There is really nothing else which, within practical limits, determines the length of the span, which may just as well be 1000 feet, or even more, providing the proper deflection can be given to the rope without touching the ground. When possible the lower part of the rope should be the working side, but in a long span this is impossible, because, when running, the lower part of the rope would be tight and the upper part slack, causing the two parts of the rope to strike together, which must never be allowed. When the length of the span exceeds 35 times the diameter of the pulleys it is safest to have the upper part of the rope the working side and the lower part the slack side. When the lower part of the rope is the slack side, the least space allowable for the slack of the rope at the center of the span will (when the rope is as tight as given in Table No. 38), be obtained by the formula : Distance = 0.00015 X (span) 2 but, to allow for contingencies, it is better to have more room. When the lower side of the rope is the tight side, the rope will be clear from the ground when running if the space is 0.0001 X (span)' 2 . The deflection in the rope when standing still which will produce a pressure on the bearings and give tension enough to transmit the horse-power given in Table No. 38, may be cal- culated approximately by the formula : d 0.00009 X / 2 d = Deflection in feet. / = Distance between pulleys in feet. (See Fig. 2). EXAMPLE. The distance between the pulleys being 400 feet, find the greatest allowable deflection in the rope, when standing still, in order to transmit the horse-power given in Table No. 38. WIRE ROPE TRANSMISSION. 343 Solution : d = 0.00009 X 400 X 400 14.4 feet. When the rope is new it is always put on with more tension than is necessary to transmit the power, because new rope will stretch. It is, therefore, very important when de- signing such transmission to calculate the maximum pressure which the rope will exert on the bearings when put on with the least deflection ever wanted, and calculate size of bearings and shafting for pulleys according to this stress, with due considera- tion not only for strength but also for heat and wear. (See page 360 and page 367.) The correct amount to allow for stretch will vary with different kinds of rope and also with the tempera- ture. If a rope is spliced on a warm summer day it must be made slacker than if it was spliced on a cold winter day, as the length of the rope will be changed considerably by the difference in temperature ; the only guide is practical experience and good judgment. As a general rule, it may be safe to allow about half of the deflection as previously calculated when splicing a new rope, provided that the shafts and bearings are constructed so as to allow such tension. The rope is always strong enough. The splicing of the rope should be done by a man ex- perienced in that kind of work. The splice itself is usually made at least 240 times the diameter of the rope. TABLE No. 38. -Giving Suitable-Sized Pulleys for Different Sizes of Wire Rope, Weight of Rope, Horse-Power which Different Sizes of Wire Rope Hay Transmit at Different Velocities, the least Stress at which it may be done and the Least Corresponding Pressure on the Bearings ; also, the Ultimate Average Strength of Wire Rope. 1 i i 1 i S Horse-Power Trans- mitted at Different 1 | ll 1 1 1 O) H "c 1 1 Velocities. Jj 3 II !| J ^o M 1*6 V 3 1 3 3 .s s w-S w *C r^ cP^ 42 fi c g C .S .3 .S jj I s 1 ll V || I'D o S s| I s s' b fcl *o S* H^-^ 1 "* w '5 8, | S, 8, i i 1 f f 1 1 g d .2 ll go 1 1 1 1 1 tj} ^ . c *S5 c tfl _c A .2 .2 V a 'C s s P-4 ~ ~ ** o o $ Q Q ^ D C, O H H OH ^ S 5 y 0.21 4500 1S7 187 374 285.5 501 14 17 22 28 6 iV 0.23 6000 253 253 506 379.5 759 19 23 31 38 7 0.31 8000 330 330 660 495 990 25 30 40 50 8 fs 0.57 12000 517 517 1034 775.5 1551 39 47 62 78 10 & 0.92 18000 636! 6361272 954 1908 56 67 90 112 12 1.12 24000 1012 1012 2024 1518 3036 77 92 122 153 14 1 1.50 320001320,13201264011985 3960 100 120 160 200 344 wlRE ROPE TRANSMISSION. EXAMPLE. From a shaft running 150 revolutions per minute 100 horse- power is to be taken off by a wire rope. The velocity of the rope is to be 5000 feet per minute. What size of pulley and rope will be required ? Solution : In Table No. 38 it will be found that a ^-inch wire rope, run- ning at 5000 feet per minute, is capable of transmitting 112 horse- power ; thus, select a %-inch rope. The diameter of the pulley will be i5Q x 3 1416 = 10 ' 6 feetp In the table ' lt wil1 be found that a 10-foot pulley is the smallest advisable to run with a %-inch wire rope, therefore the pulley 10.6 feet in diame- ter is within the requirements. The next step is to calculate the pressure on the bearings. In the table it is found that the least pressure due to the transmission of 112 horse-power is 1908 pounds. This cannot be used in calculating sizes of shafts and bearings, but use the maximum pressure, which is calculated ac- cording to the allowable deflection in the rope, as explained on page 341. Also consider weight of pulley and shaft, then calculate size of shaft and bearings, with due consideration to strength, stiffness, wear, heat, etc. ( See pages 360-367.) Transmission of Power by Manila Ropes. Manila ropes are used more or less for transmission of power. In this country one continuous rope, going back and forth in separate grooves over the pulleys several times, is fre- quently used, and a tightening arrangement is placed on one of the slack parts, which automatically keeps the rope at the proper tension, regardless of changes due to weather or stretch due to wear. This arrangement has its advantages in keeping the rope at more even tension than is possible with the Euro- pean system, but the disadvantage is that if a break occurs the transmission is entirely disabled until it is repaired. The Euro- pean practice is to use several single ropes running in separate grooves side by side on the same pulley. This has the advan- tage that if one of the ropes should break it is usually possible to run undisturbed until there is a chance to repair it, because it is always advisable to have margin enough in the transmission capacity of the ropes so that the shaft will run satisfactorily, even if one rope is taken off. The disadvantage of this system is the difficulty in keeping all the ropes at equal tightness and getting them to pull evenly. Fig. 3 shows the usual shape of pulley used for manila ropes, which may be made from either wood or iron. The European practice is to use iron, but whichever material is used it is very important to have the sides of the grooves care- fully polished, as the rope rubs on the sides in entering and MANILA ROPE TRANSMISSION. 345 leaving the pulley and will wear out in a short time if the pulley is left as it comes from the lathe tool. Sand and blow- holes must also be avoided. The angle of groove is usually 45, and the rope is made to wedge into it, as shown in Fig. 3. The usual shape of grooves for guide pulleys is shown in Fig. 4. rt (D r> cx i = (18 + 6)_XJL20 _ 12 inches< (120 + 120) Thus, the sizes of each step, with regard to speed, should be 6, 12 and 18 inches, but with regard to belt tension these sizes have to be slightly altered. To Correct the Diameter of Stepped Pulleys so that the Belt will have the Same Tension on all the Steps. At first thought, it may seem as if the belt would have equal tension on each step when the sum of the diameters of the largest and the smallest steps of the two pulleys are equal to the sum of the diameters of the two middle steps ; but this is only correct if a crossed belt is used on the pulleys. For a two- step pulley it is also correct for either open or crossed belt, if both pulleys are of the same size ; but if the pulleys are of dif- ferent sizes, the diameter of the steps must be calculated for two steps as well as if there were more. It is evident from Fig. 2, that an open belt will be tighter over the largest and the smallest pulleys than it would be over the two middle pulleys, as the part a of the belt runs parallel to the center line and will be as long as the distance between centers, but the inclined line, b, will be as much larger as the distance dto e. (See Fig. 2). A convenient way to solve this is : First calculate pulleys that will give the required speed, and of such sizes that the sum of the diameters of the two steps which are to work together will be equal, then calculate the length of the belt when laying on the largest and smallest steps, with a given distance between PULLEYS. 353 the centers of the shafts. Then, by calculating the same way, try the belt on the other steps, which will then have to be cor- rected until the belt will fit each of the different pairs of steps. The length of the belt can be most conveniently calculated by the geometrical rule that the square of the perpendicular added to the square of the base is equal to the square of the hypothe- nuse. (See page 150.) The space between the centers of the shafts is considered as the base, and the difference in radius of the two corresponding steps is considered as the perpendicular, which are both known, and from this the length of the line b is calculated (see Fig. 2), which is considered as the hypothenuse. Assuming that the belt covers half the circumference of both pulleys, the length of the belt can be found by adding half the circumference of each step to twice the length of b. NOTE. This mode of calculation is not exactly correct, but is very well within practical requirements. The length of half the circumference of the pulley is most conveniently obtained by the use of Table No. 24, page 209, by dividing the circumference of the corresponding circle by 2. A practical rule is simply to calculate the distance from d to *, and for each T yinch the belt is found to be too [long, add aVinch to the diameter of the corresponding step on each pulley. For instance, the stepped pulleys in Example No. 2 are cal- culated so that they will give the required speed to the machinery when the three steps are 18, 12 and 6 inches in diameter and both pulleys are equal. Assume the distance between centers to be 5 feet. What will be the diameter of the middle step, after it has been corrected so that it will give the right tension to the belt? Solution : Five feet = 60 inches, and the difference between the radius of the corresponding steps is 9 3 = 6 inches. The distance from e to d will be : x \/60 2 + 6 2 60 = V 3636 60 = 60.3 60 = 0.3 Thus, each part of the belt will be 0.3" too long, or the whole belt will be 0.6" too long when on the middle step ; therefore, in order to make up for this, the middle step on each pulley must be increased /.," = fY' in diameter. Thus, the middle step on each pulley will be 12 T 3 ff inches instead of 12 inches in diameter ; but, as both pulleys are increased, this does not change the relative speed of the shafts when the belt is on the middle step, and the similarity of the pulleys is also preserved, which will admit that both may be cast from the same pattern. The square root of 3636 may be obtained by use of log- arithms (see page 71), thus : . log. 3636 3.560624 Log. X/3636 = -^ = % = 1-780312 and the number corresponding to this logarithm is 60.3. 354 PULLEYS. Stepped Pulleys for Back = Geared Lathes. On machinery having changeable reducing gearing, such as lathes, milling machines, etc., it is frequently the aim of the designer to arrange the speed of the counter and the diameters of the different steps of the cone pulley in such proportions that the same ratio of speed will be maintained on each step and also from the slowest speed, with back gears out, to the fastest speed, with back gears in. When the ratio of the back gearing is given, the ratio of speed for each step will be obtained by the formula : m s = V^ S = Ratio of speed for each step. m = Number of changes of speed on the cone pulley. 7? = Reduction of speed by the back gearing. EXAMPLE. The back gearing of a lathe reduces its speed 8 times. The cone pulley has 5 changes of speed. The largest diameter of cone pulley on the spindle is 10> inches. The cone pulley on the counter-shaft is to be of the same size as the cone pulley on the spindle, and an even ratio of speed is to be maintained throughout the whole range of the ten changes of speed. The slowest speed, when back gears are in, is 6 revolutions per. minute. Calculate the speed of the counter-shaft, the speed of the spindle for each change, and the diameter of each step on the cone pulley of the spindle. Solution : The ratio of speed for each step will be :. . /7T Ql = ! =: 0.18062 V 8 5 5 The corresponding number is 1.516. With back gears in, the speed of spindle : On first cone is 6 revolutions per minute. On second cone is 6 X 1.516 = 9 revolutions per minute. On third cone is 9 X 1.516 = 14 revolutions per minute. On fourth cone is 14 X 1.516 = 21 revolutions per minute. On fifth cone is 21 X 1.516 = 32 revolutions per minute. With back gears out, speed of spindle : On first cone will be 6X8= 48 revolutions per minute. On second cone will be 9X8= 72 revolutions per minute. On third cone will be 14 X 8 = 112 revolutions per minute. On fourth cone will be 21 X 8 = 168 revolutions per minute. On fifth cone will be 32 X 8 = 256 revolutions per minute. PULLEYS. The speed of the counter-shaft will be: N= V 48 X 256 = 112 revolutions per minute. As the speed of main lines in factories usually runs at some multiple of 10, we may, for convenience in getting even-sized pulleys for connections between counter and main shaft, in practical work, decide to run the counter-shaft 110 revolutions per minute. (When a pair of cone pulleys has an uneven number of steps, and are cast from the same pattern, the speed of the counter should be equal to the speed of the machine when the belt is run on the middle step). The diameter of the largest step of the cone pulley on the spindle is 10)4 inches. The corresponding step on the counter 1054 X 48 will be ~iJ(j = 4.581"; practically, 4%" diameter. The largest and smallest step on the counter-shaft will also be 10^ and 4^ inches in diameter. Any of the intermediate steps on the spindle may be cal- culated by the formula : - (P + <*) x N = 9 . 065 . practically , = 5.932 ; practically, 6 in. Thus, assuming the counter-shaft to run 110 revolutions per minute, the speed of the spindle, with back gears out, on the five different steps will be : X 10 ^ 265 revolutions per minute. 4/ 110 X 9 = 165 revolutions per minute. 6 110 X _ 110 revo i u tions per minute. 110 X 6 = 73 revolutions per minute. * ^ = 47 revolutions per minute. 356 PULLEYS. When the back gears are in action the speed will be : Q/K fr = 33 > revolutions per minute. 8 = 20 >6 revolutions per minute. _J_ = 13% revolutions per minute. 8 rq _!_ = 9^6 revolutions per minute. 8 JZ = 5 #5 revolutions per minute. 8 These speeds are all within the practical requirements of the problem, and now the next operation is to modify the diam- eters slightly in order to get proper tension on the belt. (See page 352.) FLY-WHEELS. Fly-wheels are used to regulate the motion in machinery by storing up energy during increasing velocity, and giving out energy during decreasing velocity. Fly-wheels cannot perform either of these functions without a corresponding change in velocity. The rim of the wheel may be very heavy and moving at a high velocity, the change in speed may be small and hardly perceptible if the energy absorbed and given out is small, but there must always be a change in velocity to enable a fly-wheel to act. The common expression of gaining power by a heavy fly-wheel is very misleading, to say the least. There is no power gained by a fly-wheel but, on the contrary, considerable power is absorbed by friction in the bearings when a shaft is loaded with a heavy fly-wheel, (see example in calculating fric- tion, page 305). Nevertheless, a fly-wheel performs a very use- ful function in machinery by storing up energy when the supply exceeds the demand and giving it out at the time it is needed to do the work. ( For momentum of fly-wheels see example, page 300. For kinetic energy, see example, page 301 ). Weight of Rim of a FIy=Wheel. The weight of a rim of a cast-iron fly-wheel will be : Wd^y. 0.7854 X D X 3.1416 X 0.26 ; this reduces to, W=D X d 2 X 0.64 D = Middle diameter of rim in inches. d Diameter of section of rim in inches. W = Weight in pounds. FLY-WHEELS. 35 7 EXAMPLE. A round rim of a fly-wheel is 4 inches in diameter and the middle diameter of the -wheel is 36 inches. What is the weight of the rim ? Solution: ^=36X4X4X 0.64 = 369 pounds. For a rim of rectangular section the weight will be : W Width X thickness X D X 3.1416 X 0.26 IV = Width X thickness X D X 0.816 EXAMPLE. The width of the rim is six inches, the thickness is two inches, and the middle diameter of the rim is 48 inches. What is the weight of the rim ? Solution : W 2 X 6 X 48 X 0.816 = 470 pounds. Centrifugal Force in Fly-Wheels and Pulleys. Pulleys are not only liable to be broken by the stress due to the action of the driving belt, but in fast-running pulleys and fly-wheels the stress due to centrifugal force is far more dan- gerous. This stress increases as the square of the velocity and directly as the weight, therefore there is a limit to the velocity at which fly-wheels and pulleys can be run with safety. Generally speaking, increasing the thickness of the rim does not increase its strength, because the total tensile strength, the total weight of the rim, and, consequently, also the centrifugal force, increase in the same proportion; but it has great influence upon the strength of the wheel to have the ma- terial in the rim distributed to the best advantage. At the same time it is very important to construct the rim and arms of such proportions that the initial stress due to uneven cooling in the foundry, is avoided. The common formula is : Mass X (velocity) 2 Centrifugal force = - ---- -v. ---- n X r X 2 K Ma ^ Weight w ( n x Velocit r X 2 TT V 32.2 Therefore, ct - - V 60 / 60 32.2 X r W X 2 X r 2 X 0.01096628 ^ ' 32.2 X r <:/ = W X 2 X r X 0.00034 ^/ = Centrifugal force in pounds. 358 FLY-WHEELS. W = Weight of revolving body in pounds. n = Number of revolutions per minute. r Middle radius of pulley rim in feet. Thus, for any body 'whose center of gravity swings in a circle of one foot radius, at a speed of one revolution per min- ute, the centrifugal force will be 0.00034 times the weight of the body. EXAMPLE. The rim of a fly-wheel is five feet in middle radius and weighs 8000 pounds. It makes 75 revolutions per minute. What is the stress due to centrifugal force? Solution : Centrifugal force = 8000 X 75 2 X 5 X 0.00034 = 76500 pounds. PIG 3> This is the total centrifugal force ~ i *w tending to burst the rim, (see arrows in /\ A /\ Fig. 3); the force tending to > tear the Xs^ ^A rim asunder in any two opposite points a ._L* 1-6 as a, b, is 3 ^Q x 2 = 12175 pounds. \+ ^J The next question is : Has the sec- X * \7 tion of the rim tensile strength enough to ^ 3 Trr _ 144 Z>3 w D = Diameter of shaft in inches. L = Distance between hangers in feet. W = Transverse load in pounds, supposed to be at the middle, between the hangers. 144 = Constant for wrought iron, and 100 to 120 may be used as constant for cast-iron, with 10 as factor of safety for transverse strength^ SHAFTING. 361 Formula 1, expressed as a rule, will be: Multiply the distance between hangers, measured in feet, by the transverse load in pounds ; divide this product by 144, and the cube root of the quotient will be the diameter of the shaft in inches, calculated with 10 as factor of safety for transverse strength. Shaft not Loaded at the Middle Between the Hangers. When a shaft is not loaded at the middle of the span, but somewhere toward one of the hangers, it will carry a heavier load, with the same degree of safety, than it would if loaded in the middle, and the ratio is in inverse proportion as the square of half the distance between hangers to the product of the short and the long ends of the shaft. For instance, a shaft is six feet between hangers and loaded at the middle. What would be the difference in transverse strength if it was loaded two feet from one hanger and four feet from the other? 3X3 = 9 and 2X4 = 8. Thus, find the transverse load for a shaft when loaded in the middle, multiply by 9 and divide by 8, and the quotient is the load which the same shaft will carry with the same degree of safety against transverse stress, if loaded two feet from one end and four feet from the other. This rule only applies to the transverse strength, and not to the transverse stiffness of the shaft. For different shapes of shafts and different modes of loading, see beams, pages 243-244. When shafts are heavily loaded near one hanger, and the hanger on the other side of the pulley is further off, most of the load is thrown on the bearing nearest to the pulley, and this bearing is, therefore, liable to heat and to cause trouble, even if the shaft is both stiff and strong enough. ( See reaction on the support of beams, page 252). Transverse Deflection in Shafts. The transverse deflection in a shaft may be calculated by the formula: L = L*WC ~L*~C S Deflection in inches. D = Diameter of shaft in inches. L = Length of span in feet. W = Load on middle of shaft in pounds. C Constant = 1.7 X constant in Table No. 31, and for wrought iron or Bessemer steel may be taken as 0.00002652. 362 SHAFTING. Constant C may be calculated from experiments by the formula, C= T^W S = Deflection in inches noted in the specimen, when supported under both ends and loaded transversely at the middle between supports. D = Diameter of specimen in inches. L = Distance between supports of specimen in feet. W= Experimental load in pounds. EXAMPLE. A round specimen placed in a testing machine, supported under both ends and loaded at the middle with 2000 pounds, deflects 0.1 inch. The diameter of the specimen is two inches and the distance between supports is three feet. Calculate constant C for this kind of material. Solution : c _ 0.1 X 2* 3 3 X 2000 C = L6 = 0.0000296 inch. 54000 Thus, the deflection for this kind of material is 0.0000296 inch per pound of load, applied at the middle, between supports, for a round bar one inch in diameter and one foot between supports. Allowable Deflection in Shafts. The distance between the hangers must always be deter- mined with due consideration to the allowable transverse deflec- tion in the shafting, especially when the shaft is loaded with large pulleys and heavy belts, remembering that the deflection increases directly with the transverse load and with the cube of the length between the bearings, (see page 254). The allowable transverse deflection in shafting ought not to exceed 0.006 to 0.008 inch per foot of span ( see page 266). A beam of wrought iron one foot long and one inch square, when supported under both ends and loaded at the middle, will deflect 0.0000156 inch per pound of load, (see Table No. 31, page 259),and a round beam deflects 1.7 times as much as a square beam, when the diameter and side are equal. A round shaft, one inch in diameter and one foot long, when loaded at the middle with 144 pounds will, therefore, deflect 144 X 1.7 X 0.0000156 = 0.00382 inch. Thus, this load does not give more than an allowable de- flection. But, suppose the distance between bearings is doubled and the load decreased one-half; the ultimate strength of the shaft will be the same, but the deflection will be 72 X 1.7 X 2 3 X 0.0000156 = 0,01528 = 0.0764 inch per foot. SHAFTING. 363 This calculation shows plainly how very necessary it is to have bearings near the pulleys where shafts are loaded with heavy pulleys and large belts. There is nothing more liable to destroy a shaft than too much deflection, because the shaft is, when running, continually bent back and forth, and at last it must break. The fact must never be lost sight of that strength and stiffness are two entirely different things and follow entirely different laws; therefore, after calculations are made for strength, the stiffness must also be investigated, as stiffness is a very important property in shafting. The best way to over- come too much transverse deflection is to shorten the distance between the bearings. Of course, increasing the diameter of the shaft will also overcome deflection, but shafting should never be larger in diameter than necessary, because the first cost increases with the \yeight, which increases as the square of the diameter, and the frictional resistance will also increase with the increased diameter ; consequently, also, the running expenses. Torsional Strength of Shafting. Shafting may be considered as a beam fastened at one end and having a torsional load applied at the other end equal to the pull of the belt on an arm of the same length as the radius of the pulley. In Table No. 32, page 268, constant c is given as 580 pounds for wrought iron. The formula for twisting stress, as explained under beams ( see page 267) is, P = D * c D = JZ*I m \ ~ c ;;/ = the length of the lever or arm in feet, and will here be the radius of the pulley and be denoted by r. The length of the shaft has no influence on its torsional strength, but only on its angle of torsional deflection (see page 268). Using 10 as factor of safety, the formula will be : 58 D = Diameter of shaft in inches. r = Radius of pulley in feet. W= Pull of belt in pounds. 58 = Constant, with 10 as factor of safety = Vio X 580, taken from Table No. 32, page 268. Frequently it is more convenient to calculate the torsional strength of shafting according to the number of horse-power the shaft is to transmit ( see page 317 ). In the above formula, assume W to be 58 pounds, r to be one foot, and D will be one inch. That is, a shaft one inch in diameter is strong enough to resist, with 10 as factor of safety, 364 SHAFTING. a torsional load of 58 pounds acting on an arm one foot long. Assuming this 58 pounds to act on the rim of a pulley of one foot radius, two feet in diameter, and making one revolution per minute, it will transmit power at a rate of 58 X 6% = 364 f foot- pounds per minute ; but one horse-power is 33,000 foot-pounds per minute, and if the shaft should transmit one horse-power it 33000 must make 0^44 90.52 revolutions per minute. Hence the practical formulas for torsional strength of shafting : 90 H X 90 90 D Diameter of shaft in inches. H = Number of horse-power transmitted by the shaft. 11 =. Number of revolutions made by the shaft per minute. 90 Constant, using 10 as factor of safety, and assuming the torsional strength to be as given in Table No. 32. Torsional Deflection in Shafting. In constructing different kinds of machinery it is frequently necessary to consider the torsional deflection. The formula for torsional deflection for wrought iron ( see page 271) will be : c _ 0.00914 X m L P >4 This will transpose to s _ 48 H L '~n D S = Deflection in degrees. H ' Number of horse-power transmitted. 0.00914 X 33000 _ 48 Constant ; calculated thus, 2X3 1416 L = Length of shaft in feet between the force and the resistance. n = Number of revolutions made by the shaft per minute. D = Diameter of shaft in inches. EXAMPLE. How many degrees is the deflection of a shaft two inches in diameter, 50 feet long, making 300 revolutions per minute and transmitting 15 horse-power, applied at one end and taken off at the other? Solution : 48 HL _ 48 X 15 X 50 _ ^ - ~^~D~ ~ 300 X 2* ~ 7/2 SHAFTING. 365 Classification of Shafting. Shafting may be divided into three different kinds. First. Shafts where the main belts are transmitting the power, or so-called " Jack Shafts." Such shafts must have their boxes as near the pulleys as possible. For torsional strength their diameter may be calculated by the formula, D = V * (See Table No. 40.) Second. Common shafting in shops and factories, where the power is taken off at different places for driving machinery. Such shafts ought to be supported by hangers as given in Table No. 43, and their supports must also be reinforced by extra hangers, if necessary, where an extraordinary large pulley or heavy belt is carried. For torsional strength the diameter of such shafts may be calculated by the formula, = \--^- (See Table No. 41.) Third. Shafting having practically no transverse stress, but used simply to transmit power from one place to another. Such shafts ought to be supported by hangers according to Table No. 43, and the diameter may be calculated by the formula, - ^ \ H X 50 (See Table No. 42.) TABLE No. 40. Giving Horse-Power of Main Shafting at Various Speeds. III Revolutions per Minute. !-i 60 80 100 125 150 175 200 225 250 275 300 1V 2.6 3.4 4.3 5.4 6.4 7.5 8.6 9.7 10.7 11.8 12.9 2 3.8 5.1 6.4 8 9.6 11.2 12.8 14.4 16 17.6 19 5.4 7.3 9.1 11 13 16 18 21 23 25 27 2 y 7.5 10 12.5 15 18 22' 25 28 31 34 37 2% 10 13 16 21 25 29 33 37 42 46 50 3 13 17 21 27 32 38 43 49 54 59 65 16 22 27 34 41 48 55 62 69 76 82 35^ 20 21 34 43 51 60 68 77 86 94 103 337 25 :!4 42 53 63 74 84 95 105 116 126 4 30 41 51 64 77 IK) 102 115 128 141 154 5 2 43 60 58 80 73 100 91 125 109 150 128 175 146 164 200 1225 182 250 201 275 219 300 366 SHAFTING. TABLE No. 41. Giving Horse=Power of Line Shafting at Various Speeds. Ill Revolutions per Minute. 11 SO.H 100 125 150 175 200 225 250 275 300 325 350 400 1* 6 7.4 9 10.4 12 13.4 15 16.4 18 19.4 21 23.8 i# 7.3 9.1 10.9 12.7 14.5 16 18 20 22 23.8 25 29 2 8.9 11.1 13 15.5 17.7 20 22 24 27 19 31 35 2/1? 10.6 13.2 16 18.5 21 24 27 29 32 34 37 42 2X 12.6 15.8 19 22 25 28 32 35 38 41 44 50 2^ 15 18.6 22 26 30 33 37 41 44 48 52 59 2^ 17 22 26 30 35 39 43 48 52 56 61 69 2% 23 29 34 40 46 52 58 64 69 75 81 92 3 30 37 45 52 60 67 75 82 90 97 105 120 3X 38 47 57 67 76 86 95 105 114 124 133 152 3/^ 48 59 71 83 95 107 119 131 143 1 155 L67 190 3|^ 58 73 88 102 117 132 146 161 L76 190 205 234 4 71 89 107 125 142 160 178 196 213 231 249 284 TABLE No. 42. Giving Horse=Power of Shafting Used Only for Transmitting Power. isi Revolutions per Minute. fil 100 125 150 175 200 225 250 275 300 32E 35( 400 ll/2 6.7 8.4 10 11.8 13.5 15.1 16.8 18.E 20.3 22 2c 27 1.^$ 8.6 10.7 12.8 15 17.1 19.3 21.4 23.6 25.8 28 3C 34 1% 10.7 13.4 16 18.7 21.5 24 26.8 29.4 32.1 35 3-5 43 i^ 13.2 16.5 19.7 23 26.4 29.7 33 36.2 39.5 43 4fc 52 2 16 20 24 28 32 36 40 44 48 52 56 64 2/^ 19 24 29 33 38 42 j 48 53 57 62 67 76 2X 23 28 34 40 45 51 57 63 68 74 80 91 2^ 27 33 40 47 54 60 67 74 80 87 94 107 2^ 31 39 47 55 62 70 78 86 94 102 109 125 2|^ 41 52 62 73 83 93 104 114 125 132 146 166 3 54 67 81 94 108 121 135 148 162 175 189 216 3% 69 86 103 120 137 154 172 189 206 223 240 275 l# 86 107 128 150 171 193 114 236 257 279 300 343 SHAFTING. 367 Distance Between the Bearings. Jack shafts should always have bearings as near the pulleys as possible. Ordinary line shafts, as given in Table No. 41, and shafts for simply transmitting power, may have the distance between the hangers as given in the following table : TABLE No. 43- Diameter of Shaft in Inches. IX tol# 2 to2^ 2^ to 4 Distance between bearings in feet 6/ 2 8 10 to be transmitted from A to B The gears on shafts A and B FIG. 1. Shafts for Idlers. Shafts for idlers ( see C, Fig. 1 ) have very little torsional stress and the distance between the bearings may also be very short, so that even with a great transverse load such a shaft may be of comparatively small diameter as far as requirements for strength is concerned. In such a shaft there is great danger of trouble from hot bearings ; therefore, in designing, it is very important to make its diameter and the length of the bearing of such proportions that excessive pressure per square inch of bearing surface is avoided. EXAMPLE. Twenty-five horse-power is through idler C. (See Fig. 1). are 36 inches in diameter and make 40 revolutions per minute. What is the necessary diameter of shaft C, which is supported by two bearings one foot apart and carrying a gear 48 inches in diameter placed at the mid- dle between the bearings. Solution : The velocity on pitch line of gear A will be 40 X 36 X 3.1416 = ^ f minute 12 25 horse-power = 33,000 X 25 825,000 foot-pounds. The pressure at the pitch line of A transferred to Cwill be 825000 = 2188 pounds. 377 SHAFTING. The reaction at the pitch line between C and B, is also 2188 pounds; therefore, the total pressure (besides the weight of C, which is omitted in this calculation ) on both bearings will be 2 X 2188 = 4376 pounds and the pressure of each bearing of C will be 2188 pounds. Allowing a pressure on the bearings of 100 pounds per square inch, the necessary bearing surface will be 218S - = 21.88 square inches for each bearing. J.UU Assuming the length of the bearing to be twice its diameter, D X 2 D 21.88 ja-21.88 Calculating the size required with regard to transverse strength by the formula on page 360, 3 = J \ -J1X487U =8.12 inches. 144 Thus, a shaft 3.3 inches in diameter is of ample size for strength. The surface velocity of this shaft will be, 8.3 X 3.1416 X 40 = 34 . feet per minute, and at that velocity a pressure of 100 pounds per square inch of bearing surface is very safe from liability of heating if the bearing is well made and amply provided with oil. Proportion of Keys. The breadth of the key is usually made to be one-fourth of the diameter of the shaft, and the thickness to be one-sixth of the diameter of the shaft. Keys and key-ways are usually made straight and should always be a very good fit sidewise. Frequently set-screws are used on top of keys in mill gearing. Sometimes in heavy ma- chinery keys are made tapering in thickness, usually one-eighth inch per foot of length. A corresponding taper is made in the depth of the key-way in the hub. Key-ways in shafts are always made straight. For light and fine machinery taper keys are never used* SHAFTING. TABLE No. 44. Dimensions of Couplings for Shafts. (All dimensions in inches.) Diameter of Shaft. Dimensions of Couplings. Diameter of Number of d D L / ** 2 2 3 2 4 2 5 2 3 5 2 6 7* 9 * 7 8 9 10 13 2 14/2 15* 17 4 6 2 6^: 7# 8* n i H X 8 ^ i 8 4 4 4 4 5 5 6 6 1 1"- 1 ^ ^ i i m^ j ^ . BE^ - d H VRINQS. A satisfactory rule is to make the length of the bearing for line shafting six times the square root of the diameter of the shaft. EXAMPLE. What is a suitable length of bearings for a shaft of four inches diameter? Solution : Length of bearing = C X *ST~= 12 inches. Some designers make the length of the bearing four times its diameter. Area of Bearing Surface. The projected area of any bearing is always considered as its bearing surface. Thus, the length of the bearing multiplied by the diameter of the shaft gives the area of bearing surface. For instance, the length of the box is twelve inches and the diameter of the shaft is four inches ; the area of bearing surface is 12 X 4 = 48 square inches. Allowable Pressure in Bearings. The allowable pressure per square inch of bearing surface will depehd on the surface speed of the shaft and the condition 370 BEARINGS. of the bearing, arrangements for oiling, etc. For common line shafting from two to four inches in diameter, not making over 200 revolutions per minute, a pressure not exceeding forty pounds per square inch ought to work well. Greater pressure or greater speed may make it difficult to keep the bearings cool. EXAMPLE. What pressure may be allowed on a bearing twelve inches long and four inches in diameter ? Solution : Pressure = 4 X 12 X 40 = 1920 pounds. In well constructed machinery there should not be any trouble from heating, if the surface velocity and the pressure in the bearings does not exceed the values given in the following table : METRIC MEASURE. ENGLISH MEASURE. Kilograms per Square Centimeter. Surface Velocity in Meters per Minute. Pounds per Square Inch. Surface Velocity in Feet per Minute. 5 12 20 100 50 20 75 180 300 300 150 60 The bearings for machinery in general are constructed in various ways and of different proportions, according to the de- signer's judgment, but it is a well-known fact that high-speed machinery must have longer bearings than slow-speed ma- chinery. The length of the bearing will usually vary from one and one-half to six times the diameter. When the shafts are small (less than two inches in diameter), and the speed is from 100 to 1000 revolutions per minute, the following empirical formula may be used as a guide : L = length of bearing, d diameter of bearing, n = num- ber of revolutions per minute. FIG. 1. BEARINGS. 37* FIGURE 1 shows a cheap, solid cast-iron box used for com- paratively small and less important shafts. Dimensions, suita- l)le for bearings from one to two inches in diameter, are given in the following table : (All Dimensions in Inches.) 55 73 bJ3 bJO C C ^ (11 **"" >* 25/^ 3A 2A 334 6 4 6^4 5/8 m FIGURE 2 shows a babbitted split box suitable for shafts from one to four inches in diameter, and running at a com- paratively slow speed. FIG. 2. d= Diameter of shaft. a 2)^ X d. b\#Kd. c 3 X d -f- % inch. k 4 X in c h ; S = %<* + % inch; = !# = 135 -L iQ5 = 3" inches diameter. , 2 X 32 X 105 135 4- 105 ~ mcnes diameter. After the diameter of the gears is calculated, the pitch is decided upon according to the power the gears have to transmit. Frequently the pitch will have to be altered somewhat, and such gears sometimes have teeth of very odd pitch, in order to obtain the right number of teeth to give the required ratio of speed. The ratio between the number of teeth in the gears may always be seen from the ratio of speed between the two shafts. For instance, in the above example, the ratio of speed between the shafts is 13 %os, which, reduced to its lowest terms, is % ; therefore, the number of teeth in the two gears may be any multiple of 9 and 7, respectively. For instance, 8 X 9 = 72 teeth for the large gear, and 8 X 7 = 56 teeth for the small gear ; or, 10 X 9 = 90 teeth for the large gear, and 10 X 7 = 70 teeth for the small gear, etc. The dimensions of teeth may be calculated according to rules given on page 375. Diametral Pitch. The diametral pitch of a gear is the number of teeth to each inch of its pitch diameter. In cut gearing it is always customary to calculate the gears according to diametral pitch. When gears are calculated according to circular pitch the corre- sponding circumference of the pitch circle is usually an even number, but the diameter will generally be a number having cumbersome fractions, and therefore the distance between the centers of the gears will be a number having fractions which may be very inconvenient to measure with common scales. This is because the circumference of a circle divided by 3.1416 is equal to its diameter and the diameter multi- plied by 3.1416 is equal to the circumference. When gearing is calculated according to diametral pitch this trouble is entirely avoided, as this directly expresses the number of teeth on the circumference of the gear according to its pitch diameter. For instance, "six diametral pitch" means that there are six teeth on the circumference of the gear for each inch of pitch diameter. Thus, a gear of six diametral pitch and forty-eight teeth will be eight inches pitch diameter. A gear of " eight diametral pitch " means that the gear has eight teeth per 378 GEAR TEETH. inch of pitch diameter. A gear of " ten diametral pitch " means that the gear has ten teeth per inch of pitch diameter. A gear of " twelve diametral pitch " means that the gear has twelve teeth per inch of pitch diameter, etc. Thus, the pitch diameter and, consequently, the distance between the centers, will be a number which may be conven- iently measured, and the dimensions of tooth parts are also much more easily calculated by this system. Rules for Calculating Dimensions of Gears According to Diametral Pitch. The pitch diameter is obtained by dividing the number of teeth by the diametral pitch. EXAMPLE. What is the pitch diameter of a gear of 48 teeth, 16 pitch? Solution : 48 divided by 16 = 3, therefore the pitch diameter is 3 inches. The number of teeth is obtained by multiplying the pitch diameter by the diametral pitch. EXAMPLE. What is the number of teeth in a gear of 5 inches pitch diameter and 12 pitch ? Solution : 5 X 12 = 60, therefore the gear has 60 teeth. The whole diameter of a spur gear is obtained by adding 2 to the number of teeth and dividing the sum by the diametral pitch. EXAMPLE. What is the whole diameter of a gear blank for 68 teeth, 10 pitch ? Solution : 68 I 2 Whole diameter = : = 7 inches. The number of teeth is obtained by multiplying the whole diameter of the gear by the diametral pitch and subtracting 2 from the product. EXAMPLE. The whole diameter of a gear blank is 8 inches ; it is to be cut 10 diametral pitch. Find the number of teeth. Solution : Number of teeth = (8 X 10) 2 = 78. The diametral pitch is obtained by adding 2 to the number of teeth and dividing by the whole diameter. GEAR TEETH. 379 EXAMPLE. A gear has 64 teeth and the whole diameter is \Q l / 2 inches. What is the diametral pitch ? Solution : Diametral pitch = r- = 4. Thus, the gear is 4 diametral pitch. NOTE. The term diameter of a gear usually means ter of pitch circle. The distance between the centers of two spur gears is ob- tained by dividing half the sum of their teeth by the diametral pitch. EXAMPLE. What is the distance between centers of two gears of 48 and 64 teeth and 8 diametral pitch ? Solution : Distance = 4 f j"^ 4 = 7 inches. 2i X o The circular pitch is obtained by dividing the constant 3.1416 by the diametral pitch. EXAMPLE. What is the circular pitch of a gear of eight diametral pitch ? Solution : Circular pitch = . 0.393 inch. o The thickness of the tooth on the pitch line is obtained by dividing the constant 1.5708 by the diametral pitch. EXAMPLE. What is the thickness of the tooth on the pitch line of a gear of 6 diametral pitch ? Solution : 1 P Thickness of tooth = ^ , = 0.262 inches. 6 The working depth of the tooth is obtained by dividing 2 by the diametral pitch. The clearance at the bottom of the teeth is T V of the thickness of the tooth on the pitch line. The whole depth to cut the gear is obtained by dividing the constant 2.157 by the diametral pitch. EXAMPLE. Find the depth to cut a gear of 8 diametral pitch. Solution : Depth = 2 -* 37 - = 0.27 inch. o 380 GEAR TEETH. The whole depth is nearly equal to 0.0806 times the circular pitch. The use of the following tables will facilitate calcula- tions regarding dimensions of teeth in diametral pitch. TABLE No. 48. Comparing Circular and Diametral Pitch. Diametral Pitch. Circular Pitch. Circular Pitch. Diametral Pitch. 2 1.571 inch. l l / 2 inch. 2.094 2^ 1.257 " 1A 2.185 3 1.047 " 2.285 3^ 0.898 " 1A 2.394 4 0.785 " IX 2.513 5 ' 0.628 " 1 3 2.646 6 0.524 " i/^ 2.793 7 0.449 " 1A 2.957 8 0.393 " i 3.142 9 0.349 " if 3.351 10 0.314 " ?i 3.590 11 0.286 " il 3.867 12 0.262 " X 4.189 14 0.224 " 4.570 16 0.196 " $& 5.027 18 0.175 " _9 5.585 20 0.157 " J& 6.283 22 0.143 " A 7.181 24 0.131 " y% 8.378 26 0.121 " A 10.053 28 0.112 " V ' 12.566 30 0.105 " fV " 16.755 32 0.098 " r/ U 25.133 TABLE No. 49. Giving Dimensions of Teeth Calculated According to Diametral Pitch. Diametral Pitch. Depth to be Cut in Gear. Thickness of Tooth on Pitch Line. Diametral Pitch. Depth to be Cut in Gear. Thickness of Tooth on Pitch Line. 2 1.078 in. 0.785 in. 12 0.180 in. 0.131 in. 2/^ 0.863 0.628 14 0.154 0.112 3 0.719 0.523 16 0.135 0.098 3/2 0.616 0.448 18 0.120 0.087 4 0.539 0.393 20 0.108 0.079 5 0.431 0.314 22 0.098 0.071 6 0.359 0.262 24 0.090 0.065 7 0.307 0.224 26 0.083 0.060 8 0.270 0.196 28 0.077 0.056 9 0.240 0.175 30 0.072 0.052 10 0.216 0.157 32 0.067 0.049 11 0.196 0.143 GEAR TEE To Calculate the Number of Teeth when Distance Be- tween Centers and Ratio of Speed is Given. Select for a trial calculation, the diametral pitch which seems most suitable for the work. Calculate the sum of the number of teeth in both gears corresponding to this pitch by multiplying twice the distance between their centers by the diametral pitch selected. The number of teeth in each gear is obtained by the follow- ing formula : ._ n X A i N+n T= Number of teeth in large gear. / = Number of teeth in small gear. N=. Number of revolutions of small gear. n = Number of revolutions of large gear. A = Number of teeth in both gears. EXAMPLE. The center distance between two shafts is 15 inches. The small gear should make 126 and the large gear, 90 revolutions per minute. Calculate the number of teeth in each gear, if 8 diametral pitch is wanted. Solution : The number of teeth in both gears is 2 X 15 X 8 = 240. 126 + 90 t = 90 x 24 = 100 teeth. 126 + 90 Frequently it is impossible to get gears of the desired pitch to fit within the given center distance and to give the exact ratio of speed. Some modifications must then be made ; either the exact ratio of speed must be sacrificed, the pitch must be changed, or the distance between centers must be altered. NOTE. The ratio of the number of teeth in the gears can be seen from the ratio of the speed. For instance, in the above ex- ample the ratio of speed is %26, which, reduced to its lowest terms, is % ; therefore, the number of teeth in the two gears may, with regard to speed ratio, be any multiple of 5 and 7, respectively, but in order to fit the given center distance and also to be 8 pitch, they must be 100 and 140, which is 20 X $ = 100 and 20 X 7 = 140. 82 GEAR TEETH. FIG. 1 The shape of gear teeth is usually either Involute or Cycloid ( also frequently called Epicycloid ). The shape of a cycloid looth for a rack is four equal cycloid curves, which may be con- structed, so to speak, by letting the generating circle a, ( sec Fig. 1 ) roll along on the pitch line of the rack, both above and below. / Cycloid gears have the curve out- side the pitch circle formed by an Epi- cycloid (see Fig. 20, page 191) and the curve inside the pitch circle by a Hypocycloid. The curves al- ways meet on the pitch line in both gears and racks. The theoretical requirements for correct form of Epi- cycloid gear teeth are that the face of the teeth of one gear and the flank of the teeth of the other gear must be produced by generating circles of the same diameter. The diameter of the generating circle is limited by the size of the smallest gear or pinion in the series of gears which are con- structed to run together, because if the generating circle is as large in diameter as half the pitch diameter of the gear, the hypocycloid will be a straight line ; thus, the flank of the tooth will be a straight radial line. If the generating circle is larger than half the pitch diameter of the gear, the result will be a weak and poor tooth with under-cut flank. When the same size of generating circle is used for gears of different diameters but of the same pitch, all such gears will work correctly together, and for this reason it is possible to construct interchangeable gears having cycloid teeth. If the diameter of the generating circle is equal to half the diameter of the smallest gear in the set, this gear will have teeth with radial flanks but all the other gears and the rack will have double- curved teeth. Fig. 1 shows a rack drawn to >-inch circular pitch ; the generating circle is 0.98 inch diameter, which is equal to half of the pitch diameter of a gear of 12 teeth and y 2 inch circular pitch. All gears of the same pitch having 12 teeth or more, con- structed by the same generating circle in the same manner as the rack, will match and be interchangeable with the rack, and will also match and be interchangeable with each other. GEAR TEETH. 383 When internal teeth are constructed by the above method, the difference between the number of teeth in the internal gear and its external pinion must never be less than 12; practically, it is better to limit the difference to 15 or 20 teeth. As interchangeability is seldom required for internal gear- ing, such gears and their mates are generally constructed together and the designer chooses a generating circle of suitable size to give the shape of tooth he considers best, and he may also vary the size of the driving or the driven gear so as to reduce con- tacts when the teeth are approaching each other, etc., according to his own judgment and experience. The difference in pitch diameter of the internal gear and its pinion should never be less than the sum of the diameters of the generating circles, and the diameter of the generating circle of the flanks for the pinion should never be larger than half the pitch diameter, but it should, preferably, be smaller. As a rule, fillets at the bottom of the teeth are not used in internal gears, but if used they should be very small. In order that gears constructed with cycloid teeth should run smoothly, it is very important to have the distance between centers correct, so that the pitch lines will exactly meet each other. For this reason, there are many kinds of machinery where cycloid gears should not be used : for instance, for change gears on lathes, involute teeth as far more suitable. When making patterns, the shape of one tooth is usually carefully drawn on a thin piece of sheet metal, either brass or iron ; this is then filed out and used as a templet in tracing the other teeth on the pattern. Sometimes a fly-cutter is made according to this constructed tooth, and all the teeth in the pat- tern are cut on an index machine or a gear cutting machine ; but if such. a machine is not available, the next best way is to set out the pitch line of the gear on this templet and also the center line of the tooth, radially towards the center, then draw the pitch line on the pattern, space off each tooth carefully with a pair of dividers and draw the center line on each tooth prolonged across the rim radially in the direction of the center of the gear, then lay the templet carefully on each of these spacings, making the pitch line and the center line of tooth on the templet to exactly match the pitch line and center line of the tooth drawn on the pattern, then trace around the templet and get the shape of one tooth ; then move the templet to the next spacing and trace the next tooth, and so on for all the teeth on the gear. For small patterns it is convenient to fasten the templet to a strip of metal long enough to reach from the teeth to the cen- ter of the gear wheel, placing a point in the center of the gear, drilling a hole in the strip and letting it swing around this point, then after all the teeth are spaced off on the pattern the tem- piet is swung from one tooth to the other and all the teeth are traced by the templet. This method has the advantage that 384 GEAR TEETH. it will mark all the teeth exactly alike, because ,the templet being fastened to this strip, can not easily get out of position. The distance from the pitch line of the templet to the cen- ter hole in the strip must be laid off according to the shrinkage rule, and is, of course, in numerical value equal to the pitch radius of the gear, which should always be calculated and given on the drawing. When gear patterns are less than six inches in diame- ter it is preferable not to allow anything for shrinkage, as the moulder will usually rap the pattern about as much as the cast- ing will shrink in cooling. When a pair of cycloid gears are constructed without con- sidering interchangeability with other gears of the same pitch, it is customary to choose a generating circle having a diameter equal to three-fourths of the radius of the pitch circle of the small gear, providing this gear has 24 or more teeth. A large generating circle probably reduces the friction in a small measure but gives teeth of less strength. The largest gen- erating circle used ought never to exceed the radius of the pitch circle of the small gear. Decreasing the generating circle will probably increase friction somewhat in the gears, but it gives teeth of greater strength. The smallest generating circle used in practice is equal to half the diameter of the pitch circle of a gear having 12 teeth of the same pitch as the gear to be con- structed. Many eminent mechanics consider it preferable never to use a generating circle smaller than half of the pitch diameter of a gear of 15 teeth. Cycloid gears are mostly used in large cast gears of one- inch circular pitch or more. Sometimes the driving gear is made of slightly larger di- ameter, and the teeth spaced at a correspondingly greater pitch than the theoretically calculated size. This is done in order that the teeth shall not rub on each other on the approaching side, but only touch as they are passing the center line and commence to slide away from each other. This will make the gears less noisy, but probably gears made in this manner will wear faster, as there are fewer points of contact, although this may be offset by the fact that the friction between the teeth when they are meeting and pushing onto each other is more injurious than the friction produced when they are sliding away from each other. The same idea is sometimes employed when constructing bevel gears, in order to make them run quietly. This mode of sizing gears is not, as a rule, used in modern gear construction, but it is a point well worth remembering, because if either of two gears is over or under size, the gear of over-size should always be used as the driver, and the gear of under-size should always be the driven ; never vice versa. This will apply as well to involute as to epicycloid gears. GEAR TEETH. 3*5 Involute Teeth. Suppose a strap is fastened at a and b on the two round discs in Fig. 2. If the disc b is turned in the direction of the arrow, the strap will move in a straight line from c, toward d. This motion will cause the disc a to rotate with exactly the same surface speed as the disc b, but in the opposite direction. Suppose, further, that to the under side of the disc a (see Fig. 3) is fastened a piece of sheet brass p, or other suitable material of somewhat larger diameter than disc a, and that a scratch awl is fastened in the strap at the point /// ; then by turning the disc b in the direction of h to b, and the strap moving with it, being kept tight by the resistance of disc a, the scratch awl will trace on the brass plate the curve from in to h, but if the discs are moving in the opposite direction, the scratch awl will trace the curve from ;// to K, Take an- other brass plate and do the same thing with the other disc, and a similar curve will be produced. In these two brass plates the stock may be filed away carefully, following the curves as shown in Fig. 4. The discs are laid to match each other and free to FIG. 2. FIG. 4. FIG. 3. swing on their centers ; turning the disc a in the direction of the arrow, it will give motion to b, and both discs will move with the same speed in exactly the same manner as if they were connected by the strap as shown in Fig. 2. 386 GEAR TEETH. The curve on these two discs represents the form of a gear tooth in the involute system.* The line h g, Fig. 4, is called the line of pressure or the line of action. The circles, P and P, are the pitch circles. The line B R, shows the direction of motion of the teeth at the moment they are passing the center line, c c. Approximate Construction of Involute Teeth. It will be noticed that the line of pressure, h g, forms an angle with the line B R. This angle is usually taken as 14)4 degrees. This makes the diameter of the base circle, ^, (see Fig. 5) equal to 0.968 times the diameter of the pitch circle. The base circle g g, in Fig. 5, corresponds to the disc in Fig. 3, and the line of pressure in Figs. 5 and 6 corresponds to the strap in Fig. 2. The line of pressure, h g, Fig. 5, is 75 > degrees to the center line,/Y. A perpendicu- lar is erected from the line h g, through the center, c. Using the point of in- tersection at i as center, the tooth is drawn simply by a circular arc. This will, in prac- tical work for small gears hav- ing more than twenty teeth, cor- respond nearly enough to the true involute, which was illustrated by means of the strap, disc and scratch awl, as explained in Figs. 2, 3 and 4. When the gear has less than twenty teeth, and is constructed by circular arcs, as shown in Fig. 5, the top of the tooth will be too thin ; but the top of the tooth will be too thick to clear in the rack, if the true involute curve is used. When the teeth are of true involute curve, a smaller gear than twenty-five teeth will not run freely in a rack having straight teeth slanting 14^ degrees. (See Figs. 6 and 7). Therefore, *The way to actually draw this curve on paper by means of drawing instru- ments is explained on page 192. This way explained here, using the disc on the strap, is merely for illustrating and explaining principles, and serves well for that purpose, but would be inconvenient to use in actual construction of gear teeth. In actual work one tooth is carefully constructed, and templets and cutters are made and used, as was explained for Cycloid Gears, pa^e 383. GEAR TEETH. when a gear has less than twenty-five teeth it is necessary to round the teeth somewhat outside the pitch circle. By making either a drawing or a templet, it is very easy to see how much to round FIG. 6. Involute Teeth (Cast.) the teeth to make them clear in the rack. In interchangeable sets of cut involute gears it is customary to cut the rack with a cutter shaped for a gear of 185 teeth. This will make the teeth in the rack slightly curved instead of straight, as shown in Fig. 6, and this will also make it possible to construct the small gears in an interchangeable set nearer to a true involute, and still have them run freely in the rack. 3 88 GEAR TEETH. When gear teeth are constructed as shown in Figs. 6 and 7, the line g h, is 75^ degrees to the line c f, and the line c /, is degrees to the line c f. (See Fig. 5). FIG. 7. Involute Teeth (Cut). The line h g, will always be tangent to the base circle which is concentric to the pitch circle. The diameter of the base circle is always 0.9G8 times the diameter of the pitch circle. The circle forming the shape of the tooth must always have its center on the circumference of the base circle, and its diameter will be one-fourth of the pitch diameter of the gear. As shown in Fig. 2, the same circle gives the form of tooth for coarser or finer pitch. When gears are drawn by this method the pitch circle is divided into as many teeth and spaces as there are to be teeth in the gear ; then the form of the tooth is simply struck by the dividers, always using the periphery of the base circle as center, and always taking the distance in the dividers equal to one-fourth of the radius of the pitch circle. The diameter of the base circle is 0.968 times the diameter of the pitch circle, because cosine of 14^ degrees is 0.96815. The diameter of the circle forming the shape of the tooth is 0.25 times the diameter of the pitch circle of the gear, because sine of 14^ degrees is 0.25038. If the line of pressure is laid at any other angle GEAR TEETH. 389 than 14^ degrees, all these other proportions will also change. Fig. shows a pattern for gears and rack constructed with necessary clearance as used for cast gears. All tooth parts are of the same dimensions as used for cycloid gears as given on page 375. Fig. 7 shows a cut gear and rack constructed in the same manner. The advantages of the involute system of gears are in the strength of teeth, and also that the gears will trans- mit uniform motion and run satisfactorily, even if the distance between centers should be slightly incorrect. Width of Gear Wheels. Gears with cast teeth are usually made narrower than gears with cut teeth. In spur gears with cast teeth it is customary to make the width of the gear four to five times the thickness of the teeth, or twice the circular pitch. Width of Gears With Cut Teeth. The following rule is recommended by Brown & Sharpe Mfg. Co. in their " Practical Treatise on Gearing": Divide eight by the diametral pitch, and add one-fourth inch to the quotient; the sum will be the width of face for the pitch required. EXAMPLE. What width of face is required for a gear of four pitch ? Solution : Face = f + ]- = 2% inches. For change gears on lathes where it is desirable not to have faces very wide, the following rule may be used : Divide four by the diametral pitch and add one-half inch. By the latter rule a four-pitch change gear would have but a 1^-inch face. BEVEL GEARS. Fig. 8 is a diagram showing how to size bevel gear blanks. First, lay off the pitch diameters of the two gears, which may be calculated according to diametral pitch or to circular pitch ; second, draw the pitch line of teeth ; third, lay off on the back of the gear the line a b, square to the " pitch line of teeth ;" fourth, on the line a b, lay off the dimensions of the teeth exactly in the same manner as if it was for a spur gear. If the gear is calculated according to circular pitch, find dimensions of teeth by formulas on page 375, but if tne gear is calculated according to diametral pitch, find dimensions of teeth in Table No. 49. 39 feEVEL GEARS. Make the drawing carefully to scale ( full size preferable whenever possible ), and measure the outside diameter as shown in the diagram. FIG. 8. To Calculate Size of Bevel Gear for a Given Ratio of Speed. Ascertain the ratio of speed in its lowest terms. Multiply each term separately by the same number, and the products give the number of teeth in each gear. EXAMPLE. Two shafts are to be connected by bevel gears, one shaft to make 80 revolutions and the other 170 revolutions per minute. Find the number of teeth in the gears. BEVEL GEARS. 391 Solution : For Ratio = = 8 /i7. instance, multiplying by 6, the large gear on the shaft making 80 revolutions will have 17X6 = 102 teeth. The small gear on the shaft making 170 revolutions will have 8 X 6 = 48 teeth. Assuming that on account of room it is necessary to use smaller gears, a smaller multiplier may be used, but if it is desir- able to have larger gears, use a larger multiplier. Decide on the pitch of the gears according to the work they are required to do. Make a scale drawing and get the dimensions as explained on page 390. Dimensions of Tooth Parts in Bevel Gears. Fig. 9 shows a sectional drawing of a pair of bevel gears of sixteen diametral pitch, 18 teeth in the small gear and 30 teeth in the large gear. The pitch diameter of the small gear is - = iVs inches. The pitch diameter of the large gear is f f = 1% inches. FIG. 9. The addendum of the teeth on the back at a is T V inch, the same as for a spur gear of 16 diametral pitch. The thickness 392 BEVEL GEARS. and the total depth to cut the gear at a are 0.098 inch and 0.135 inch, respectively. These dimensions are found in Table No. 40, as if it was a spur gear of 10 diametral pitch. All the dimensions of the tooth decrease gradually toward b, as the whole tooth is sup- posed to vanish in a point in the center at c. The dimensions of the teeth at b may be calculated and are always in tke same proportion to the dimensions at a as the distance c b is to the distance c a ; thus, if the length of the tooth from a to b is made one-third of the length of the distance c X the distance b c is two-thirds of the distance a c, and, consequently, all the dimen- sions of the tooth at b are two-thirds of the dimensions at a. Instead of calculating the size of the teeth at b, the dimensions may be obtained by careful drawing. The depth of the tooth at the smallest end is then measured directly at b, but the thickness is measured at /; the distance / h is laid off equal to b d. The length of the tooth from a to b is to a certain extent arbitrary, but a good rule is seven inches divided by the diame- tral pitch, but never longer than one-third of the distance from a to c. EXAMPLE. What is the proper length for the teeth of a bevel gear of 8 diametral pitch ? Solution : Seven inches divided by 8 % inch, if the gears are of such diameters that this will not make the length of the teeth more than one-third of the distance from a to c. Form of Tooth in Bevel Gears. Extend the line a (see Fig. 9), until it intersects the axial center line of the gear, as at h ; use h as the center, and the shape of tooth at a for the large gear is constructed as if it was a spur gear having a pitch radius as large as a h. The shape of the tooth at b is constructed in the same way, by extending the line ^, (which always the same as line a, is square to the pitch line of the tooth) until it intersects: the axial center line of the gear, as at d. Using d as center, the shape of the tooth is constructed as if it was a spur gear having a pitch radius equal to db. The shape of the teeth of the small gear is obtained in the same way, which is shown by the drawing. The form of tooth is shown to be approximately involute, constructed as explained for spur gears, page 386. Measuring the back cone radius, a h, of the large gear, it is found to be f f inch, and the diameter will be f- 1 inch ; thus, the shape of the tooth at a for the large gear will be the same as the shape of the tooth in a spur gear of 58 teeth, sixteen diametral pitch. BEVEL GEARS. 393 Measuring the back cone radius of the small gear, it is found to be f ^ inch, and the diameter will be f | inch ; conse- quently the shape of tooth at a for the small gear is the same as the shape of tooth in a spur gear of 21 teeth, sixteen diametral pitch. Therefore, if this pair of gears is to be cut by a rotary cutter having a fixed curve, a different cutter is required for each gear. When, in a pair of bevel gears, both gears are of the same size and have the same number of teeth, and their axial center lines are at right angles, they are called miter gears, and one cutter, of course, will answer for both gears. One cutter will also answer in practice when the difference of the back cone radius of a pair of gears is so small that it comes within the limit of one cutter as used for spur gears of the same size. Bevel gears may also be made with cycloid form of teeth, but when- ever cut by rotary cutters, as usually employed in producing small bevel gears of diametral pitch, the involute form of tooth should always be used. Cutting Bevel Gears. When bevel gear teeth are correctly formed, the tooth curve will constantly change, from one end of the tooth to the other. Therefore, bevel gears of theoretically correct form cannot be produced by a cutter of fixed curve ; but, practically, very satisfactory results are obtained in cutting bevel gears of small and medium size in this way. When a regular gear-cutting machine is not at hand, the Universal milling machine is a very convenient tool for cutting bevel gears of moderate size, and is used in the following way : First, see that the gear blank is turned to correct size and angle, and adjust the machine to the angle corresponding to the bottom of the teeth in the gear. The correct index is set ac- cording to the number of teeth in the gear. Adjust the cutter to come right to the centre of the gear, cut the correct depth as marked on the gear at a (see Fig. 9), according to Table No. 49, and when the machine is adjusted to the correct angle, and the correct depth is cut at a, the correct depth at b will, as a matter of fact, be obtained. Second, when a few teeth are cut in the gear (two or three) bring, by means of the index, the first tooth back to the cutter. By means f the index, rotate the gear, moving the tooth toward the cutter ; but, by the slide, move the gear sidewise away from the cutter, until the cutter coincides with the space'at b ; then cut through from a to b. This operation will widen one side of the tooth space at a. Note the position of the machine, and, by the use of the index and slide, return the cutter to its central position and in- 394 BEVEL GEARS. dex into the next space, and rotate the other side of the tooth toward the cutter as much as the first side ; but, by the slide, the gear is moved sidewise away from the cutter until the cutter coincides with the space at b; then cut through on this side from a to b. Thus, by repeated cutting on each side alternate- ly, one tooth is backed off equally on both sides and measured by a gage, until the correct thickness on the pitch-line at a, according to Table No. 49, is obtained. Be very careful to have the machine set over the same amount on each side of the tooth, or else the tooth will be askew. Third, when one tooth, thus by trial, is correctly cut, note the position of the machine and cut all the teeth through on one side, then set over to the other side in exactly the same position as was found to be right for the first tooth ; cut through again and the gear is finished. Thus, when the correct position of the machine is obtained, any number of gears of the same size and same pitch may be cut, by simply letting the cutter go through twice. NOTE. As already stated, bevel gear cutting in this way is only a compromise at the best, but by careful manipulation and good judgment an experienced man is able to do a very creditable job. A cutter is usually selected of the same curve as is correct for a spur gear corresponding to the back cone radius of the gear. Thus, it may be thought that the shape of the tooth should be the shape of the cutter, but by investi- gation it will be found that, on account of the " backing off," the teeth will be of a little more rounding shape at the large end than corresponds to the cutter; therefore, when the gear has few teeth, less than 25, it is usually preferable to make the shape of the cutter to correspond to a gear a little larger than would be called for by the back cone radius of the bevel gear to be cut; but when the gear has more than 25 teeth, a cutter of shape corresponding to the back cone radius of the gear will give fDod results. For instance, in the pair of bevel gears shown in ig. 9 the back cone radius of the large gear calls for a cutter corresponding in shape to a cutter for a spur gear of 59 teeth, 16 diametral pitch ; and this shape of cutter will, after the teeth are backed off, make the teeth a trifle too round at the large end, and a trifle too straight on the small end, but if the teeth are not too long the job will be very satisfactory. The back cone radius of the small gear calls for a cutter corresponding in shape to a cutter for a spur gear of 21 teeth, 16 diametral pitch, but when the teeth are backed off they will be a little too rounding on the large end ; therefore a better result is obtained by selecting a cutter having a shape corre- sponding to a little larger spur gear ; for instance, a gear of 24 teeth. Such a cutter will give the teeth a better shape on BEVEL GEARS. 395 the large end, although it may be necessary to round the teeth a little, outside the pitch line on the small end, by filing. Of course, a spur gear cutter cannot be used for cut- ting bevel gears, because, although it may have the correct curve, it would be too thick. The thickness of a bevel gear cutter must be at least 0.005 inch thinner than the space be- tween the teeth at their small end. Large bevel gears are made on theoretically correct prin- ciples by planing on specially constructed machines. WORMS AND WORM GEARS. Fig. 10 shows a worm and worm gear. 7TF f= Pitch diameter of gear. f = Smallest outside diameter. = Largest outside diameter. a = Outside diameter of worm. b = Pitch diameter of worm. c = Diameter of worm at bottom of thread. The pitch and diameter of worm screws are usually of such proportions that for single-thread the angle of the teeth on the gear is from two to three degrees. This angle is most conven- iently obtained by drawing a diagram as shown in Fig. 10. WORMS AND WORM GEARS. Draw a line ////, equal to 31 times the length of line b', this line will be equal to the length of the circumference of the pitch diameter of the screw. Erect the perpendicular, m o, equal to the pitch of the screw. Connect the points /and o by the line / ^ degrees. CAUTION. When cutting a worm gear, be careful and not lay the angle of the teeth in the wrong'direction. The diameter of worm gears is usually calculated according to circular pitch, for convenience in cutting the worm with the same gears as used for ordinary screw cutting in a lathe. When a worm gear has comparatively few teeth, the flank of the tooth will be undercut by the hob; to prevent this in a measure, it is customary to have the blank somewhat over size, so that from five-eighths to three-fourths of the depth of the tooth may be outside the pitch line. The form of teeth is usually involute, and the thread on a worm screw is constructed of the same shape as the teeth in a rack. Fig. 11 shows the shape of tooth and the tabh gives the dimensions of finishing tool for the most common pitches. The surface speed of a worm screw ought not to exceed 300 feet per minute. Table No. 50 is calculated by the following formulas. (See Fig. 11.) P = Circular pitch. A'= " p a = P X 0.3183 d P X 0.3683 S = P X 0.5 b = P X 0.31 C=P X 0.335 k = P X 0.1 WORMS AND WORM GEARS. 397 TABLE No. 50. Giving Proportions of Parts for Worms and Worm Gears, Calculated According to Circular Pitch. (See Fig. 11.) UUOM. JO ip J3AO qoq jo J3)3UIBIp til _ >O O O iO CO CO GO -M O t~ CO O 1~ O CN l-SDO-*CO''N in C/3 C/3 ^ c/5 & CO CO c^ in 2 % 160 40 1 *// ^7 140 40 1 V^ 2/ 120 40 ji5 % 100 40 120 32 1 1 80 40 96 32 1/^3 60 40 72 32 120 40 % 1% 50 40 60 32 100 40 i/ 2 40 40 .48 32 80 40 100 40 % 2% 40 50 48 40 64 40 80 40 i/ 3 40 60 48 48 64 48 60 36 48 24 2 /7 g^/ 40 70 48 56 64 56 60 42 48 28 4 40 80 24 32 40 40 50 40 48 32 % 4i/ 40 90 24 36 32 36 40 36 48 36 M> 5 20 50 24 40 32 40 j 40 40 48 40 60 30 % 6 20 60 24 48 32 48 1 40 48 48 48 60 36 8 20 80 24 64 24 48 30 48 48 64 JCO 48 rV 10 ' 20 100 24 80 24 60 24 48 24 40||50 50 Reduction of Speed by Worm Gearing. In a single-threaded worm screw one revolution of the worm moves the gear one tooth ; in a double-threaded worm screw one revolution of the worm moves the gear two teeth, and in a triple- threaded worm screw one revolution of the worm moves the gear three teeth. A great deal ol work is lost by friction by using worm gearing, frequently from 50 to 75 per cent., some of which could be saved by using a ball bearing to take the end thrust of the worm. The efficiency is also increased by using a worm of double, triple or quadruple thread, because this in- creases the angle of the teeth in the wheel and the efficiency of the mechanism is increased by increasing the angle until it reaches 20 to 25, when it rapidly falls off again. Calculating the Size of Worm Gears. EXAMPLE. Find dimensions of a worm gear having 68 teeth, ^ inch pitch, cut teeth. Make the pitch diameter of the single-thread worm six times the pitch of the worm, Use Table No, 50, WORMS AND WORM GEARS. 399 Solution : In Table No. 47 the pitch diameter of a gear of 68 teeth of one-inch pitch is given as 21.65 inches; thus, the pitch diameter for a gear of 68 teeth of |<-inch circular pitch will be 21.65 X 0.75 = 16.788 inches. In column a, of Table No. 50, the addendum for %-inch circular pitch is given as 0.2387 inch ; fhis is multiplied by 2, because it is to be added on both sides of the gear. Thus, the smallest outside diameter of the gear is 16.738 + 0.2387 X 2 = 17.215 inches; or, practically, 17^ inches. If the gear is to be made hollow to correspond to the "curve at bottom of thread of the worm, make a scale drawing as shown in Fig. 10, and make line^-, 17^ inches ; from this drawing the largest outside diameter may be obtained by measurement. The diameter of the worm on the pitch line was to be six times the pitch = 6 X %" = 4^ inches. The addendum for the thread on the worm can be obtained from Table No. 50, column *z, and is 0.2387. The outside diameter of the worm will be 4.5 -f 2 X 0.2387 = 4.977 inches, or, practically, 4f inches. The cutter to be used in roughing out the gear should have a curve of involute form corresponding to a spur gear cutter for 68 teeth, and its thickness ought to be at least 0.005 inch less than the width of space as given in column S of Table No. 50. Therefore the thickness on the pitch line of the roughing cutter will be 0.37 inch. The angle of the teeth may be obtained from a drawing as shown and explained in Fig. 10, or it may be calculated thus : circular pitch Tangent of angle S = , pitch circumference In Table No. 21 the corresponding angle is given as 3 degrees, very nearly. The depth to which the gear should be cut is given in col- umn D as 0.515 inch. The gear is finished with a hob, as described below, which is allowed to cut until it touches the bottom of the spaces in the gear. The outside diameter of the hob should be larger than the outside diameter of the worm, in order that the teeth in the hob may reach the bottom of the spaces in the gear and leave clearance for the worm, and at the same time leave the gear tooth of the proper thickness on the pitch line. This increment is obtained in column , Table No. 50, and for %-inch pitch is 0.075 inch ; thus, the outside diameter of the hob is 0.075 inch larger than the outside diameter of the worm, or 4.977 + 0.075 = 5.052 inches. The angle of the finish- ing threading tool for both worm and hob is 14)4 degrees, making the angle of space 29, as shown in Fig. 11. The clear- ance angle of the threading tool must be a little more than the angle of the thread. 4oo WORMS AND WORM GEARS. The width of the threading tool at the point is given in Column b, Table No. 50, as 0.2325 inch. The depth of the space to be cut in the worm is given in Column /?, as 0.515 inch. The diameter of the worm at the bottom of the thread will be : 4.977 2 X 0.515 = 3.947 inches. The depth of the space to be cut in the hob is given in col- umn h in Table No. 50 as 0.5525 inch. The diameter of the hob at bottom of thread will be : 5.052 2 X 0.5525 = 3.947 inches. Thus the only difference in size between the hob and the worm is in the outside diameter and in the depth of the cut. Both may be finished by the same tool, as the diameter at the bottom of the thread and the thickness of the teeth at the pitch line should be the same for both hob and worm. Elliptical Gear Wheels. Elliptical gear wheels are sometimes used in order to change a uniform rotary motion of one shaft to an alternately fast and show motion of the other. See Fig. 12. The pitch line is constructed and calculated the same as the circumference of an ellipse. (See page 189.) The gear is constructed involute the same as for spur gears. If the differ- ence between the minor and the major diameters is large it may be necessary to construct the teeth of different shapes at differ- ent places on the circumference; in other words, the whole cir- cumference of the gear cannot be cut with the same cutter. A cutter of the same pitch, of course, but corresponding to a larger diameter of gear, must be used where the curve of the pitch line is less sharp. The centers of the shafts are in the foci of the ellipse. If two elliptical gear wheels, made from the same pattern, or cut together at the same time, on the same arbor, are to work together they must have an uneven number of teeth so that a space will be diametrically opposite a tooth, as will be seen from Fig. 12, SCREWS. 4OI SCREWS. "Pitch," "Inch Pitch" and "Lead" of Screws and Worms. The term " pitch of a screw," as commonly used, means its number of threads per inch, while the " inch pitch " is the dis- tance from the center of one thread to that of the next. For instance, a one-inch screw of standard thread is usually said to be an " eight pitch" screw, because it has eight threads per inch of length ; but it might more correctly be said to be a screw of >-inch pitch, because it is >-inch from the center of one thread to the center of the next. The " lead " of a worm or a screw means the advancement of the thread in one complete revolution ; therefore, in a single- threaded screw, the inch pitch and the lead is the same thing, but in a double or triple-threaded screw the inch pitch and the lead are two different things. The " lead " in a double-threaded screw will be a distance equal to twice the distance from the center of one thread to the center of the next, but in a triple-threaded screw the lead is three times the distance from the center of one thread to the center of the next. Screw Cutting by the Engine Lathe. When the stud and the spindle run at the same speed ( which they usually do) the ratio between the gears may always be obtained by simply ascertaining the ratio between the num- ber of threads per inch of the lead-screw and the screw to be cut. EXAMPLE. The lead-screw on a lathe has four threads per inch and the screw to be cut has \l l / 2 threads per inch ( one-inch pipe- thread). Find the gears to be used when the smallest change gear lias 24 teeth and the gears advance by four teeth up to 96. The ratio of the number of threads per inch of the two screws is as 4 to \\y z . As the smallest gear has 24 teeth and the gears all advance by four teeth, this ratio of the screws must be mul- tiplied by a number which is a multiple of 4 and which, at least, gives the smallest gear 24 teeth. For instance, multiply by 8 and the result is 8 X 11 ^ = 92 teeth for the gear on the lead- screw ; 8 X 4 = 32 teeth for the gear on the stud. Cutting flultipIe-Threaded Screws or Nuts by the Engine Lathe. Calculate the change gears as if it was a single-threaded screw of the same lead. Cut one thread and move the tool the proper distance and cut the next thread. 4O2 SCREWS. The most practical way to move the tool from one thread to another, when cutting double-threaded screws or nuts, is to select a gear for the stud or spindle of the lathe having a number of teeth which is divisible by two, and when one thread is cut make a chalk mark across a tooth in this gear onto the rim of the intermediate gear ; count half way around the gear on the stud and make a chalk mark across that tooth; drop the swing plate enough to separate the gears, pull the belt by hand until the oppo- site mark on the gear on the stud comes in position to match the chalk mark on the intermediate gear; clamp the swing plate again and the tool is in proper position to cut the second thread. When triple threads are to be cut, select a gear for the spindle or stud whose number of teeth is divisible by three, and in changing the tool from one thread to the next, only turn the lathe enough so that the gear on the stud moves one-third of one revolution. If, for any reason, it should be inconvenient to make this change by the gear on the stud, the change may be made by the lead-screw gear. The intermediate gear is first released from the gear on the lead-screw, which is then moved ahead the proper number of teeth, and again connected with the intermediate gear. The proper number of teeth to move the gear on the lead-screw is obtained by the following rule : Multiply the number of teeth in the gear on the lead-screw by the number of threads per inch of the lead-screw ; divide this product by the number of threads per inch of the screw to be cut, and the quotient is the number of teeth that the gear on the lead-screw must be moved ahead. EXAMPLE. A square-threaded screw is to have ^-inch lead and triple thread. The lead-screw in the lathe has two threads per inch, and the gear on the lead screw has ninety-six teeth. How many teeth must the gear on the lead-screw be moved, when changing from one thread to the next ? Solution : A screw of J^-inch lead with triple thread has six threads 2 X 96 per inch, therefore the gear must be moved ^ = 32 teeth, in order to change the tool from one thread to the next. U. S. Standard Screws. Fig. 1 shows the shape of thread on United States stand- ard screws. The sides are straight and form an angle of sixty degrees, and the thread is SCREWS. 403 flat at the top and bottom for a distance equal to one-eighth of the pitch, thus the depth of the thread is only three-fourths of a full, sharp thread. (See Fig. 1.) Fig. 2 shows the shape of the Whitworth (the English) sys- tem of thread. As compared with the American system, the principal difference is in the angle between the sides of the thread, which is fifty- five degrees, and one-sixth of the depth of the full, sharp thread is made rounding at the top and bottom. There is also a difference in the pitch of a few sizes. The common V-thread screws have the angle of thread of sixty degrees, the same as the United States standard screws, but the thread is sharp at both top and bottom. This style of thread is rapidly, as it should be, going out of use. The prin- cipal disadvantages of this thread are that the screw has less tensile strength, and it is also very difficult to keep a sharp- pointed threading tool in order. Diameter of Screw at Bottom of Thread. The diameter of screws at the bottom of thread is obtained by the following formulas : United States Standard Screws: n For V-threaded screws : n For Whitworth screws : n d= Diameter of screw at bottom of thread. D = Outside diameter of screw. n = Number of threads per inch. 1.299 is constant for United States standard thread. 1.733 is constant for sharp V-thread. 1.281 is constant for Whitworth thread. 404 SCREWS. TABLE No. 52. Dimensions of Whitworth Screws. Diameter of Screw in Inches. H Number of Threads per Inch. 40 24 20 18 16 14 12 11 10 9 8 7 Diameter of Number of Screw in Threads per Inches. Lich. Diameter of Screw in Inches. 3% 4 4% 5 Number of Threads per Inch. Diameter of Tap Drill. The diameter of the drill with which to drill for a tap is, if we want full thread in the nut, equal to the diameter of the screw at the bottom of the thread, and is, therefore, obtained by the same formulas. However, in practical work it is always ad- visable to use a drill a little larger than the diameter of the screw at the bottom of thread, because in threading wrought iron or steel the thread will swell out more or less, and a few thou- sandths must be allowed in the size when drilling the hole. In drilling holes for tapping cast-iron, a little larger drill is used, because it is unnecessary in a cast-iron nut to have exactly full thread. Table No. 53 gives sizes of drill for both wrought and cast-iron, which give good practical results for United States standard screws. Table No. 53 gives sizes of hexagon bolts and nuts. The size of the hexagon is equal to 1)4 times the diameter of bolt + ^-inch ; the thickness of head is equal to half the hexagon. The thickness of nut is equal to the diameter of the bolt. When heads and nuts are finished they are ^-inch smaller. The table is calculated by the following formulas : d = D 1.299 C = 1.155 A B = 1AUA F=D SCREWS. 405 TABLE No. 53. Dimensions of U. S. Standard Screws. X u s* 0.185 0.240 0.294 0.345 0.400 0.454 0.507 0.620 # 0.731 0.838 0.939 1.065 1.158 1.284 1.389, 1.490 1.615 1.711 1.961 4tf i^ 2 '-V 5# 2.175 2 A 2.425 2X 2.629 2 d 2.8792 j 3.1003i^ 3.3173 3.567 3.798 3j 4.028 4 4.255 4-, > 4.480 4^ 4.730'4^ 4.963 4f| 5.2035|| 5.4236*1 0.0269 0.0452 0.0679 0.0935 0.1257 0-1619 0.2019 0.3019 0.4197 0.5515 0.6925 0.8892 1.0532 1.2928 1.5153 1.7437 2.0485 2.2993 3.0203 3.7154 4.6186 5.4284 6.5099 7.5477 8.6414 9.9930 11.3292 12.7366 14.2197 15.7633 17.5717 19.2676 21.2(U7 23.0978 0.0062 0.0069 0.0078 0.0089 0.0096 0.0104 0.0114 0.0125 0.0139 0.0156 0.0178 0.0178 0.0208 0.0208 0.0227 0.0250 0.0250 0.0278 0.0278 0.0313 0.0313 0.0357 0.0357 0.0384 0.0417 0.0417 0.0435 0.0455 0.0476 0.0500 0.0500 0.0526 0.0526 0.0556 406 SCREWS. NOTE. In finished work, the thickness of the head of the bolt and the nut is equal, and is Vie of an inch less than the diameter of the bolt. Columns B and C in Table No. 53 are very useful for many purposes ; for instance, in selecting size of counter-bore when finishing castings, to give bearing for screw heads : in turning blanks which are afterwards to be cut into square or hexagon heads, etc. Table No. 54. Coupling Bolts and Nuts. (Hexagon). (All Dimensions in Inches) kl > en _: QJ Q) '"C <-< " rt s b 2| Across Across Across Across bflOJ .2 ^2 the the the the 8* Q P Flats. Corners. Flats. Corners. 1-1 X 20 A y, 3/8 Ji 5< A 18 IJ TV 5/^ T*V 3/8 16 A /^ tl 3/8 A 14 5/8 II A i TV 13 y\ II 5/8 li y 2 A 12 H H IJ A H 11 IJlj j| ly 1 ^ M* 10 i IA ^ ll^ ^ 4 9 1/8 HI 1/8 HI ^ i 8 1* 1/i i 1/8 7 13/8 HI 13/8 HI i x /6 *# 7 l 1 ^ Hi 1^ % l /8 i/< Eye 'Bolts. It is very customary to weld an eye to a lag screw (see Fig. 3) to use in handling heavy weights in shops. The following table (No. 57) gives the holding power of lag screws or eye bolts when screwed into spruce timber a little over the full length of thread. The suitable size of bit for the thread is also given in the table. TABLE No. 57- Diameter of Screw. Diameter of Bit. Load at Which the Screw Pulled Out. Safe Load. 1 inch y inch 16,000 Ibs. 2,000 Ibs 7/8 " " 9,000 " 1,125 " 34 5/8 " 7,000 " 875 " 5/8 " X " 6,000 750 " # " 3/8 " 3,500 " 437 " 3/8 " -& " 1,900 237 X " & " 700 87 " 408 SCREWS. TABLE No. 58. Giving the Average Weight in Pouncj per 100 Square Head ( iim let = Pointed Lag Screws. LENGTH in Inches. ir A" 3/8" 7 " #" ft" " y." X* 1" 15* 254 H 7 10 2 3^2 H 8i 12 17 24 27^ 2^ 4j^ S* 9| 14 19 26 31 3 43^ 11 16 21 28 34 51 zy* 5 k 8* 121 18 24 31 38 55 A. 53/ 9* 14 20 26 34 42 60 85 112 4*^ 6^ 10i 151 22 28 37 46 65 91 121 5 7 Hi 17 24 32 40 50 70 97 130 5% ~*% 12 t 18* 26 34 43 54 76 103 140 6 8 20 28 36 46 58 81 110 150 6^ 21^ 30 38 49 62 86 117 100 7 23 32 41 52 65 92 125 170 7 241 34 44 55 69 97 132 iso 8 26 36 47 58 73 103 140 190 & l /2 77 108 148 lOO 9 81 113 156 2ir 9/^j 85 118 164 22' 10 89 123 172 23d Size of * ^ 6 33 * <> N, SJH l/ts ^ RfcO -r- 1 Head in X X X X X X X X X Inches. X MM |w T 1 W H * GIMLET-POINTED LAG SCREW. EXAMPLE. What is the weight of 8 lag screws 6" long and in diameter ? Solution: Under the heading ]/ 2 -inch, in the line with 6 in the column of length, is the number 36. Thus, 100 lag screws of this size will average to weigh 36 pounds, and one such screw will weigh 0.36 pound ; 8 such screws will weigh 0.36 X 8 = 2.88 pounds. SCREWS. 409 French System of Standard Threads. In the French system of standard screws the thread has an angle of 60, with flat top and bottom. (The French system is in this respect identically the same as the United States stand- ard thread.) The pitch and the diameter are given in millimeters (see Table No. 59). The form of thread is an equilateral tri- angle. The diameter of tap drill (or diameter of bolt or screw at bottom of the thread) is obtained in the following way : The height of an equilateral triangle is obtained by multiplying its base by 0.86603 (this number is sin. of 60). Thus, assuming that the base = 1, and taking off one-eighth of the depth at top and bottom, that is reducing the depth of the thread one-fourth, 86603 the remaining depth will be 0.86603 -^ =0.64952 ; mul- tiplying this by 2, to allow for the depth of the thread on both sides of the screw, the constant will be 2 X 0.64952 = 1.29904, which for all practical purposes may be reduced to 1.3. Thus, when the pitch of the screw is 1 millimeter, the diameter of the screw at the bottom of the thread is 1.3 millimeters less than the outside diameter of the screw. Therefore, the diameter of the screw at the bottom of the thread may always be calculated by the simple rule : Multiply the pitch in millimeters by 1.3, and subtract the product from the outside diameter of the screw; the remainder is the diameter of the screw at the bottom of the thread, which is the same as the diameter of the tap drill, given in Table No. 59. TABLE No. 59. French Standard Screws. (All Dimensions in Millimeters). Diameter of Screw. Pitch of Screw. Diameter of Tap Drill. Diameter of Screw. Pitch of Screw. Diameter of Tap Drill. 6 1 4.70 36 4 30.80 8 1 6.70 42 4.5 36.15 10 1.5 8.05 48 5 41.50 12 1.5 10.05 56 5.5 48.85 14 2 11.40 64 6 56.20 1(5 2 13.40 72 6.5 63.55 18 2.5 14.75 80 7 70.90 20 25 16.75 88 7.5 78.25 22 2.5 18.75 96 8 85.60 24 8 20.10 106 8.5 94.95 26 3 22.10 116 9 104.30 28 3 24.10 12(5 9.5 1 13.65 30 3.5 25.45 ISO 10 123.00 32 3.5 27.45 148 10.5 134.35 4io SCREWS. FIG. German System of Standard Threads. In the German system of standard threads the angle is 53 7' 47 /; . The reason for adopting such an odd angle is that the form of thread is a triangle, having its base equal to its height, and the top angle of such a triangle is 53 7' 47". The thread in this system is also made flat at top and bottom equal to one-eighth of the pitch (see Fig. 4). The diameter of the screw at the bottom of the thread is, in this system, calculated by this rule : Multiply the pitch in millimeters by 1.5, subtract this pro- duct from the outside diameter of the screw and the remainder is the diameter of the screw at the bottom of the thread, which is the same as the diameter of the tap drill given in Table No. 60. TABLE No. 60. German Standard Screws. (All Dimensions in Millimeters). g gj ~ "0= ~ H4 S 'o - Sft Sis "8 Q >H . 2 a 2 ?, K p^ 3 Ml *- O/) 3 bJD M 3 bfl ui tJO 3 b i-t bjo So IH ftfj 3 be 5 M "T N C/3 $ & * un Cfl en j 24 120 20 80 24 80 20 ! 40 1.5 24 80 30 80 36 80 30 40 2 20 100 24 80 24 60 30 60 24 40 40 40 2.5 20 80 24 64 24 48 30 48 30 40 50 40 3 24 80 27 60 24 40 30 40 36 40 60 40 3.5 28 80 21 40 28 40 35 40 42 40 70 40 4 32 80 24 40 i 32 40 40 40 48 40 80 40 4.5 27 60 27 40 36 40 45 40 54 40 5 30 60 30 40 40 40 50 40 60 40 5.5 33 60 i 33 40 44 40 55 40 66 40 6 36 60 36 40 48 40 60 40 72 40 6.5 39 60 39 40 52 40 65 40 7 28 40 42 40 56 40 70 40 7.5 30 40 45 40 60 40 8 32 40 48 40 64 40 8.5 34 40 51 40 9 36 40 54 40 9.5 38 40 57 40 10 40 40 60 40 10.5 42 40 11 44 40 11.5 46 40 12 48 40 li NOTES ON HYDRAULICS. 413 NOTES ON HYDRAULICS. Hydraulics is the branch of engineering treating on fluid in motion, especially of water, its action in rivers, canals and pipes, the work of machinery for raising water, the work of water as a prime mover, etc. Pressure of Fluid in a Vessel. When fluid is kept in a vessel the pressure will vary directly as the perpendicular height, independent of the shape of the vessel. For water, the pressure is 0.434 pounds per square inch, when measured one foot under the surface. The pressure in pounds per square inch may, therefore, always be obtained by multiplying the head by 0.434. The head corresponding to a given pressure is obtained by either dividing by 0.434 or multi- plying by 2.304. EXAMPLE. What head corresponds to a pressure of 80 pounds per square inch ? Solution : 80 X 2.304 = 184 feet. Velocity of Efflux. The velocity of the efflux from a hole in a vessel will vary directly as the square root of the vertical distance between the hole and the surface of the water. For instance, if an opening is made in a vessel four feet, and another 25 feet, below the sur- face of the water, and the vessel is kept full, the theoretical velocity of the efflux will be nearly 16 feet and 40 feet per second respectively, friction not considered ; or, in other words, the velocity will be as 2 to 5, because \^4 = 2 and \^26 = 5. The velocity of efflux in feet per second may always be calculated theoretically by the formula : v = 8.02 X Constant 8.02 is \S2g= \/64.4, and v velocity of efflux. h = Head in feet. Table No. 63 gives the theoretical velocity of efflux and the static pressure corresponding to different heads, and is calcu- lated by the following formulas : ^ ~ 644 * =* VTX 64.4 v = V P X 2.3 X 64.4 v = V P X 148 P = h X 0.434 h P X 2.3 414 NOTES ON HYDRAULICS. TABLE No. 63. Head, Pressure, and Velocity of Efflux of Water. Head in Feet. Pressure in Pounds per Square Inch. Velocity in Feet per Second. Head in Feet. Pressure in Pounds per Square Inch. Velocity in Feet per Second. h P i> h P V 0.1 0.0434 2.54 19 8.246 35 0.2 0.0868 3.59 20 8.68 35.9 0.25 0.1082 4.01 25 10.82 40.1 0.3 0.1302 4.39 30 13.02 43 0.4 0.1736 5.07 35 15.19 47.4 0.5 0.217 5.67 40 17.36 50.7 0.6 0.2604 6.22 45 19.53 53.8 0.7 0.3038 6.71 50 21.7 56.7 0.75 0.3255 6.95 55 23.87 59.5 0.8 0.3472 7.18 60 26.04 62.1 0.9 0.3906 7.61 65 28.21 64.7 1 0.434 8.02 70 30.38 67.1 1.25 0.5425 8.95 75 32.55 69.5 1.5 0.651 9.83 80 34.72 71.8 1.75 0.7595 10.6 85 36.89 73.9 2 0.868 11.4 90 39.06 76.1 2.25 0.9735 12 95 41.23 78.2 2.5 1.082 12.6 100 43.4 80.2 2.75 1.1905 13.3 110 47.74 84.2 3 1.302 13.9 120 52.08 87.68 3.25 1.4102 14.4 130 56.78 91.5 3.5 1.519 15 140 61.06 94.7 3.75 1.6375 15.5 150 65.1 98.3 4 1.736 16 160 69.44 101.2 4.25 1.8445 16.5 170 73.78 104.5 4.5 1.953 17 180 78.12 107.2 4.75 2.0615 17.5 190 82.46 110.4 5 2.17 17.9 200 86.8 113.5 6 2.604 19.6 225 97.35 120 7 3.038 21.2 250 108.2 126 8 3.472 22.8 275 119.05 133 9 3.906 24.1 300 130.2 139 10 4.34 25.4 325 141.05 144 11 4.774 26.6 350 151.9 150 12 5.208 27.8 375 163.75 155 13 5.678 28.9 400 173.6 160 14 6.106 30 425 184.45 165 15 6.51 31.1 450 195.3 170 16 6.944 32.1 475 206.15 174 17 7.378 33.1 500 217 179 18 7.812 34 550 238.7 188 NOTES ON HYDRAULICS. 415 Velocity of Water in Pipes. The theoretical velocity of water discharged from a pipe is calculated by the same formula as is used in calculating veloci- ties of falling bodies. ( See page 277). v = \/2 g h i> = Theoretical velocity of efflux per second. h = Head. 2g= 64.4 if v and h are reckoned in feet. 2 g = 19.64 if v and h are reckoned in meters. If the water, besides the pressure due to the head, is also acted upon by some additional pressure, for instance, steam, the theoretical velocity of the discharge is obtained by the formula, P = Pressure in pounds per square inch. The constant 0.434 is used because a column of water one foot high will exert a pressure of 0.434 pounds per square inch; thus, by dividing by 0.434, we actually convert the pressure into its corresponding head in feet. All other quantities in this formula are, of course, taken in English units. NOTE. By head is always meant the vertical height in feet, or its equivalent in pressure expressed in feet. Table No. 63 gives the theoretical velocity of the discharge and the pressure corresponding to different heads. The theoretical velocity is never obtained in practice, be- cause part of the total head is used to overcome the resistance at the entrance of the pipe, and part is used to overcome the frictional resistance to the flow of the water in the pipe. Thus, only a part of the total head is left to give velocity to the water, therefore the velocity of the water at dis- charge will only be what is due to the velocity head, after deduc- tions are made for resistance at the entrance and for friction in the pipes. In short pipes, the resistance at the entrance to the pipe is comparatively the larger loss, but in long pipes the fric- tional resistance is the larger. When both the resistance at the entrance and the friction in the pipe are considered the formula will be : v = Velocity of discharge in feet per second. 2 -=64.4 L = Length of pipe in feet. d= Diameter of pipe in feet. f= Coefficient of friction, which is obtained from experi- ments, and will vary according to conditions, from 0.01 to 0.05. It is usually in approximate calculations taken as 0.025. 41 6 NOTES ON HYDRAULICS. EXAMPLE. Find the velocity of discharge from a pipe six inches in diam- eter. The head is 16 feet and the length of the pipe is 100 feet, and coefficient of friction 0.025. Solution : (NOTE. 6 inches = 0.5 foot.) 64.4 X 16 :.5 -f 0.025 X -7TT 6.5 v = 12.6 feet per second. In Table No. 64 the quantity of water discharged per min- ute by a pipe six inches in diameter, when the velocity is one foot per second, is 88.14 gallons. Thus, the quantity of water deliv- ered when the velocity is 12.6 feet per second, is 12.6 X 88.14 = 1110.6 gallons per minute. When the length of the pipe is more than 4,000 diameters tne velocity of the water may be calculated by the formula, and the quantity is obtained by multiplying the velocity by the constants given in Table No. 64. EXAMPLE. Find the velocity of efflux from a water pipe of three inches diameter and 1200 feet long, having a head of six feet, assuming coefficient of friction as 0.025. Solution : 0.25 Discharge in gallons per minute : q = 22.03 X 1.79 = 39.4 gallons per minute. NOTES ON HYDRAULICS. 417 TABLE No. 64. Quantity of Water Discharged Through Pipes in One flinute, when Velocity of Efflux is One Foot per Second. -o * & J a r^- s *0 | o *; v '*, 60 1- i v ^ S j ^1 8 ^ rt P C t"t ^ c3|2| i^l fl J! s |l| IS c.s 1 8 5 a Is -S ><% 2^1 1 0. 1 u "la? J w-2 o ih-aS E.S. 1 si, let J 5> o Jl"o Ss sfi a fc j> t2 o,i> [2 . 'J c 5fc rt (2 cnj ^ c a c"* 5 S & a JS"" Q 5 % 0.0104 0.00008 0.0048 0.036 20 1.6666 2.182 130.90 979 }| ; 0.0208 0.00033 j 0.0198! 0.150 21 1.7500 2.405 144.32 1079 % 0.03120.00076! 0.0456 0.342 22 1.8333 2.640 158.39 1185 X 0.04160.00136! 0.0816J 0.612 23 1.9166 2.885 173.11 1294 % 0.0624 0.00306J 0.1836 1.380: 24 2.000 3.142 188.50 1410 1 0.0833 0.00545 0.3272 2.448 25 2.0833! 3.409 204.53 1530 lj 0.1042 0.00852 0.5094 3.828 26 2.1667 3.687 221.22 1655 0.12500.01227 0.7362 5.508 27 2.2500 3.976 238.56 1784 2 s 0.1667 0.0218 1.309 9.792 28 2.3333 4.276 256.56 1919 2i 0.2083 0.0341 2.045 15.30 29 2.4166 4.578 275.22 2058 3 0.25000.0491 2.945 22.03 30 2.5000 4.909 294.52 2203 ^1 0.29110.0668 4.008 29.99 31 2.5822 5.241 314.49 2352 4 0.3333 0.0873 5.238 39.17 32 2.6667 5.585 335.10 2506 41 0.37500.1104 6.626 49.58 33 2.7500 5.939 356.37 2666 5 0.4166 0.1364 8.181 61.20 34 2.8333 6.305 378.302829 6 0.5000 0.1963 11.781 88.14 35 2.9166 6.681 400.88 '2999 7 0.5822 0.2673 16.035 119.9 36 3.0000 7.069 424.14 3173 8 0.6667 0.3491 20.914 156.7 37 3.0833 7.467 448.02 3351 9 (0.75000.4418 26.507 198.3 38 3.1667 7.876 472.56 3535 10 0.83330.5454 32.725 244.8 39 3.2500 8.296 497.75 3724 11 0.9166!0.6597 39.597 296.2 40 3.3333 8.727 523.60 3918 12 1.00000.7854 47.124 352.5 41 3.4166 9.168 550.11 4115 13 1.0833 0.9218 55.305 413.7 42 3.5000 9.621 577.27 4318 14 1.1667 1.069 64.141 479.8 43 3.5822 10.085 605.09 4526 15 1.2500 1.227 73.631 550.8 44 3.6667 10.559 633.56 4739 16 1.3333 1.396 83.776 626.4 45 3.7500! 11. 045 662.68 4961 17 1.4166 1.576 94.575 707.4 46 3.8333 11.541 692.46 5180 18 1.5000 1.768 106.03 793.2 47 3.9166 12.048 722.90 5408 19 1.5822 1.969 118.14 883.8 48 4 12.566 753.98 5640 41 8 NOTES ON STEAM. NOTES ON STEAM. When water is heated and converted into steam of atmos- pheric pressure, one cubic foot of water will make 1646 cubic feet of steam. (The common expression that ** a cubic inch of water makes a cubic foot of steam " is not strictly correct, as a cubic foot contains 1728 cubic inches.) The specific gravity of steam at atmospheric pressure, when compared with water is, therefore. - = 0.000608. The weight of one cubic foot of steam at atmospheric pressure will, therefore, be 0.000608 X 62.5 = 0.038 pounds. At any other pressure the weight per cubic foot of steam is given in Table No. 65. Saturated steam is steam at the temperature of the boiling point which corresponds to its pressure. Saturated steam does not need to be wet steam, as the word saturated does not mean that the steam is saturated with water, but it means that it is saturated with heat ; that is to say : the temperature under the given pressure cannot possibly be any higher as long as the steam is in contact with water, because if more heat is added more water will be evaporated, and if the volume is kept con- stant, as in a steam boiler, both the pressure and temperature will increase simultaneously. High pressure steam is steam the pressure of which greatly exceeds the pressure of the atmosphere. Low pressure steam is steam the pressure of which is less than the atmosphere, and also steam having a pressure equal to, or not greatly above, the atmospheric pressure. Wet steam is steam which contains water held in suspen- sion mechanically. Dry steam is steam which does not contain water held in suspension mechanically. Super-heated steam is steam which is heated to a tempera- ture higher than the boiling point corresponding to its pressure. It cannot exist in contact with water, nor contain water, and resembles a perfect gas. Vertical boilers with tubes through the steam space (such as the Manning boiler) give slightly super- heated steam ; but if steam is to be super-heated to any consid- erable extent it must be passed through a super-heater, which usually is in the form of a coil of pipes subjected to the hot gases in the uptake from the boiler. The sensible heat of steam is the temperature which can be measured by a thermometer. The latent heat of steam is that heat which is absorbed when water of any given temperature is changed into steam of the same temperature. When water is evaporated under pressure the sensible heat will increase and the latent heat will decrease. For instance, at atmospheric pressure the sensible heat is 212 degrees, and the latent heat of evaporation is 966 B. T. U., but at 100 pounds NOTES ON STEAM. 419 absolute pressure the sensible heat is 327.9 degrees, while the latent heat of evaporation is only 883.1 B. T. U. (See steam table, No. 65.) TABLE No. 65. Properties of Saturated Steam. ill. fM W ffrf !!& l^fi Total Heat in B. T. U. per Pound of Steam from Water at 32 Degrees F. &3 3HJ g* w ^ 3W.S .a!~o . i!*S III* Slli .Sfa o% S3 W *&* Cubic Feet of Steam from 1 Cubic Foot of Water at 62 Degrees F. i 102.1 1112.5 1042.9 330.36 0.0030 20800 2 126.3 1119.7 1025.8 172.80 0.0058 10760 3 141.6 1124.6 1014 117.52 0.0085 7344 4 153.1 1128.1 1006.8 89.36 0.0112 5573 5 162.3 1130.9 1000.3 72.80 0.0138 4524 6 170.2 1133.3 995 61.52 0.0163 3813 7 176.9 1135.3 990 52.62 0.0189 3298 8 182.9 1137.2 985.7 46.66 0.0214 2909 9 188.3 1138.8 982.4 41.79 0.0239 2604 10 193.3 1140.3 978.4 37.84 0.0264 2358 11 197.8 1141.7 975.3 34.63 0.0289 2157 12 202 1143 972.2 31.88 0.0314 1980 13 205.9 1144.2 970 29.57 0.0338 1842 14 209.6 1145.3 968 27.61 0.0362 1720 14.7 212 1146.1 966 26.36 0.0380 1646 15 213.1 1146.4 964.3 25.85 0.0387 1610 16 216.3 1147.7 962.6 24.32 0.0411 1515 17 219.6 1148.3 960.4 22.96 0.0435 1431 18 222.4 1149.2 957.7 21.78 0.0459 1357 19 225.:} 1150.1 956.3 20.70 0.0483 1290 20 228 1150.9 952.8 19.72 0.0507 1229 25 240 1154.6 945.3 15.99 0.0625 996 30 250 1157.8 937.9 13.46 0.0743 838 86 259.3 1160.5 931.6 11.65 0.0858 726 40 267.3 1162.9 926 10.27 0.0974 640 45 274.4 1165.1 920.9 9.18 0.1089 572 50 281 1167.1 916.3 8.11 0.1202 518 55 287.1 1169 912 7.61 0.1314 474 60 292.7 1170.7 908 7.01 0.1425 437 65 298 1172.3 904.2 6.49 1538 405 70 302.9 1173.8 900.8 6.07 0.1648 378 75 307.5 1175.2 897.5 5.68 0.1759 353 80 312 1170.5 894.3 5.35 0.1869 333 85 316.1 1177.9 891.4 5.05 0.1980 314 90 320.2 1179.1 888.5 4.79 0.2089 298 95 324 1180.3 885.8 4.55 0.2198 283 100 327.9 1181.4 883.1 4.33 0.2307 270 105 331.3 1182.4 881.7 4.14 0.2414 257 420 NOTES ON STEAM. TABLE No. 65. (Continued). Absolute Pres- sure in Pounds per Square Inch. 1 =- s ^* fP! Total Heat in B. T. U. per Pound of Steam from Water at 32 Degrees F. Latent Heat of Evaporation in B. T. U. per Pound of Steam. 4*. S WT3 6 IS|l >-* u Weight in Pounds per Cubic Foot of Steam. Cubic Feet of Steam from 1 Cubic Foot of Water at 62 Degrees F. 110 334.6 1183.5 878.3 3.97 0.2521 247 115 338 1184.5 875.9 3.80 0.2628 237 120 341.1 1185.4 873.7 3.65 0.2738 227 125 344.2 1186.4 871.5 3.51 0.2845 219 130 347.2 1187.3 869.4 3.38 0.2955 211 135 350.1 1188.2 867.4 3.27 0.3060 203 140 352.9 1189 865.4 3.16 0.3162 197 145 355.6 1189.9 863.5 3.06 0.3273 190 150 358.3 1190.7 861.5 2.96 0.3377 184 160 363.4 1192.2 857.9 2.79 0.3590 174 170 368.2 1193.7 854.5 2.63 0.3798 164 180 372.9 1195.1 851.3 2.49 0.4009 155 190 377.5 1196.5 848 2.37 0.4222 148 200 381.7 1197.8 845 2.26 0.4431 141 In the preceding table the first column gives the absolute pressure, which is gage pressure plus 14.7 pounds, or, for ordinary practice, reckon as 15 pounds. For instance, when the gage pressure is 80 pounds per square inch, the correspond- ing absolute pressure is, for all practical purposes, 95 pounds per square inch, and the corresponding temperature is given in the second column in the table to be 324 degrees Fahr. The total number of British thermal units (B. T. U.) re- quired to convert each pound of water from 32 degrees Fahr. into steam of any given pressure is given in the third column. For instance, each pound of water of 32 degrees converted into steam of 95 pounds per square inch absolute pressure has received 1180.3 B. T. U. The fourth column gives the number of British thermal units (B. T. U.) of heat required to change one pound of water of the temperature given in the second column into steam of the same temperature ; which also is the number of heat units given up by one pound of steam when it is condensed to water of the same temperature as the temperature of the steam with which it is in contact. For instance, the table gives the latent heat of evaporation of steam at 95 pounds absolute pres- sure to be 885.8 B. T. U.; therefore, if steam of 95 pounds pres- sure per square inch is condensing into water in a steam pipe where steam and water are in contact, so that the temperature cannot drop below that due to the pressure, and the pressure is NOTES ON STEAM. 421 maintained at 95 pounds per square inch, the temperature of the water from the condensed steam will be 324 degrees, the oame as the temperature of the steam, but each pound of steam as it is condensing will give out 885.8 British thermal units of heat. The fifth column gives the number of cubic feet of satu- rated steam which will weigh one pound at the given pressure and temperature. The sixth column gives the weight of one cubic foot of saturated steam of corresponding given tempera- ture. For instance, one cubic foot of steam at 95 pounds per square inch absolute pressure will weigh 0.2198 pounds, and 100 cubic feet of steam of 95 pounds per square inch absolute pres- sure will weigh 100 X 0.2198 = 21.98 pounds. In other words, it will take 0.2198 pounds of water to give one cubic foot of steam at 95 pounds absolute pressure, and it will require 21.98 pounds of water to make 100 cubic feet of steam of 95 pounds absolute pressure. The seventh or last column gives the relative volume of steam at the given pressure as compared with water at 32 degrees F. For instance, one cubic foot of water will give 1046 cubic feet of steam at atmospheric pressure, but one cubic foot of water gives only 219 cubic feet of steam at 125 pounds absolute pressure. Steam Heating. In the ordinary practice of heating buildings by direct radiation the quantity of heat given off by the radiators or steam pipes will vary from 1 ^ to 3 heat units per hour per square foot of radiating surface for each degree of difference in tempera- ture; an average of from 2 to 2^ is a fair estimate. One pound of steam at about atmospheric pressure contains 1146 heat units, and if the temperature in the room is to be maintained at 70, while the temperature of the pipes is 212, the difference in temperature will be 142 degrees. Multi- plying this by 2#, the emission of heat will be 229^ heat units per hour per square foot of radiating surface. Dividing 229)4 by 1146 gives 0.2 pounds of steam condensed per hour, per square foot of radiating surface. From this may be estimated the re- quired size of boiler, as the boiler must always be capable of generating as much steam as the radiators are condensing. A rule frequently given is to have one square foot of heating sur- face in the boiler for every 8 to 10 square feet of radiating sur- face and one square foot of grate surface for every 350 to 500 square feet of radiating surface. One pound of coal is required per hour per 30 to 40 square feet of radiating surface. When steam is used for heating dwelling-houses, one square foot of radiating surface is required per 40 to 80 cubic feet of space, according to location, number of windows, etc. As a 422 NOTES ON STEAM. general rule, one square foot of radiating surface is sufficient for heating 40 to 60 cubic feet of air in outer or front rooms, and 80 to 100 cubic feet in inner rooms. The following rule may be used as a guide for different conditions : One square foot of radiating surface is sufficient for' heating 60 to 80 cubic feet of space in dwellings, schools and offices ; 75 to 100 cubic feet of space in halls, store houses and factories ; 150 to 200 cubic feet of space in churches and large auditoriums. In heating mills 1^-inch steam pipes are generally used, and one foot of pipe is allowed per 90 cubic feet of space to be heated. Value of Low Pressure Steam for Heating Purposes. When steam at atmospheric pressure is condensed into water at a temperature of 212, each pound of steam gives up 966 B. T. U. of heat, but if steam of 100 pounds gage pressure (115 pounds absolute) is condensed into water at 212 degrees, each pound of steam must give up 1004 B. T. U., which is only 38 heat units more than steam of atmospheric pressure. Hence it is evident that for heating purposes there is no advantage in using steam of high pressure ; one pound of exhaust steam, only a pound or two over atmospheric pressure, is almost as valuable an agent for heating purposes as live steam at 100 pounds pres- sure direct from the boiler. Hot Water Heating in Dwelling Houses. One square foot of heating surface is required per 30 to 60 cubic feet of space heated. Quantity of Water Required to Make any Quantity of Steam at any Pressure. The weight of water required to make one cubic foot of steam at any pressure is the same as the weight of one cubic foot of steam as given in the sixth column in Table No. 65. Therefore, the weight of water is obtained by multiplying the number of cubic feet of steam required by the weight of one cubic foot, as given in the table. EXAMPLE. How much water will it take to make 300 cubic feet of steam at 100 pounds absolute pressure ? Solution : One cubic foot of steam at 100 pounds pressure is given in the table as weighing 0.2307 pounds, therefore 300 cubic feet will weigh 300 X 0.2307 = 69.21 pounds of water. One cubic foot of water may, for any practical purpose, be reckoned to weigh 62 ^ pounds and one gallon of water may be NOTES ON STEAM. 423 taken as &fo pounds, Therefore 69.21 pounds divided by 62.5 gives 1.1 cubic feet, or 69.21 pounds divided by 8.3 gives 8.34 gallons. At atmospheric pressure one cubic foot of steam has nearly the weight of one cubic inch of water, and the weight increases very nearly as the pressure ; therefore, for an approximate estima- tion, if no steam tables are at hand, it is well to remember the rule: Multiply the number of cubic feet of steam by the absolute pressure in atmospheres, and the product is the number of cubic inches of water required to give the steam. NOTE. In all such calculations for practical purposes, a liberal allowance must be made for loss and leakage. Weight of Water Required to Condense One Pound of Steam. The following formula gives the theoretical amount of water required to condense one pound of steam : /2 t\ W = Weight of water required per pound of steam con- densed. // = Number of heat units above 32 in one pound of steam at the pressure of exhaust. This temperature is obtained from Table No. 65. t\ = Temperature of water when entering the condenser. / 2 = Temperature of water when leaving the condenser. / 3 = Temperature of the condensed steam when leaving the condenser and entering the air-pump. EXAMPLE. Steam of four pounds absolute pressure is exhausted into a surface condenser. The temperature of the condensed steam when leaving the condenser and entering the air-pump is 120. The temperature of the cold water when entering the con- denser is 65. The temperature when leaving the condenser is 105. How many pounds of condensing water is needed per pound of steam condensed ? NOTE. In the steam table, page 419, the total number of heat units above 32 per pound of steam of four pounds absolute pressure is given as 1128. Solution : _ 1128 + 32 120 105 65 1040 IV -- = 26 pounds of water per pound of steam. 40 424 NOTES ON STEAM. In a jet condenser the steam and the water are mixed to- gether, and, therefore, the condensed steam and the water when leaving the condenser are of equal temperature, and the formula will change to /2 = Temperature of mixture. t\ = Temperature of water when entering condenser. The other letters have the same meaning as in the previous formula. EXAMPLE. Steam of three pounds absolute pressure in exhausted into a jet condenser. The temperature of the cold water entering is 60. The temperature of the mixture leaving the condenser is 110. How many pounds of water are needed per pound of steam condensed? NOTE. In the steam table, page 419, the total number of heat units above 32 per pound of steam of three pounds pres- sure is given as 1124.0 or, for convenience, say 1125. Solution : 1125 + 32 110 lA/ * 110 CO W = = 20.1 pounds. Weight of Steam Required to Boil Water. An approximate rule is to allow that one pound of steam is condensed for every five pounds of water to be heated to the boiling point. It does not make much difference about the pressure of the steam, as long as it is a few pounds above atmospheric pres- sure ; for instance, one pound of steam at 10 pounds gage pres- sure when condensed into water at 212 will give up 973 heat units, and steam of 100 pounds gage pressure will give up 1003 heat units a difference of only 30 heat units in steam of 10 pounds gage pressure and steam of 100 pounds gage pressure. More correctly, the weight of steam required to boil one pound of water at 212 may be calculated by the formula, _ 212 A ~ //ISO And the weight of steam required to heat one pound of water to any temperature is obtained by the formula, *~ H + 32 / 2 NOTES ON STEAM. 425 x = Weight of steam required. H = Number of heat units above 32 in one pound of steam, as given in Table No. 65. /i = Temperature of water before heating. t% = Temperature of water after heating. Expansion of Steam in Steam Engines. When steam is expanded without doing work and prac- tically without losing heat by radiation, it will become super- heated, but if it is doing work, as in a steam engine, it will lose heat during expansion. According to the best authorities, the pressure varies in- jo < versely as the ^power of the volume, if heat is neither added 'nor taken* awayT>y any outside source during the time the steam is being expanded in the steam engine cylinder. This is called adiabatic expansion of steam. The pressure varies inversely as the 1 Vio power of the vol- ume, if the steam is kept dry at the temperature of saturation, during expansion, by means of a steam jacket outside the cylinder. When the pressure is considered to vary inversely as the volume it is called isothermal expansion. The isothermal curve is not exactly the correct curve to represent the expansion of steam, but it is the theoretical curve usually drawn on the indicator diagram, because it is so easy to handle and is also very nearly correct. The following formula gives the mean effective pressure according to isothermal expansion. M. E. P. = + hyp.log.r p.log.r\ __ Absolute terminal pressure = PI X Pi = Absolute initial pressure. r = Ratio of expansion. PZ = Absolute back pressure. M. E. P. Mean effective pressure. Hyp. log. (hyperbolic logarithm), see page 126. Table No. 66 gives the terminal and the mean effective pressure of steam expanded under any of these three different conditions. 426 NOTES ON STEAM. TABLE No. 66. Constants for Calculating Mean Terminal Pressure of Expanding Steam. and Kept Dry at Tem- V ,3 At Constant Tem- perature. (Isothermal Expan- perature of Satura- tion. (Expansion in a Condensing by Work- ing in a Cylinder. (Adiabatic Expan- w C/2 sion). Steam Jacketed Cy- sion). D V linder) . '5 03 jv V g 9 I i 3 2 I 1*^ ti 9 0.1111 0.355 0.0968 0.340 0.0862 0.327 Ho 10 0.1000 0.330 0.0865 0.315 0.0774 0.303 Hi 11 0.0909 0.309 0.0782 0.293 0.0696 0.282 Hi 12 0.0833 0.290 0.0713 0.275 0.0631 0.264 Vis 13 0.0769 0.274 0.0655 0.259 0.0578 0.248 M.4 14 0.0714 0.260 0.0605 0.245 0.0533 0.234 15 15 0.0667 0.247 0.0563 0.232 0.0494 0.222 16 0.0625 0.236 0.0526 0.221 0.0459 0.211 ^7 17 0.0588 0.225 0.0493 0.211 0.0429 0.201 %8 18 0.0556 0.216 0.0463 0.202 0.0403 0.192 H9 19 0.0526 0.208 0.0438 0.193 0.0379 0.184 20 0.0500 0.200 0.0415 0.185 0.0359 0.177 NOTES ON STEAM. 4.27 To Find the Mean Effective Pressure by the Preceding Table. Find the constant in the column corresponding to the con- ditions of expansion, and to the given cut-off. Multiply this by the absolute initial pressure, and the product is the average pressure. Subtract the back pressure and the remainder is the mean effective pressure. EXAMPLE. Find the mean effective pressure for isothermal expansion when the engine is cutting off at one-quarter stroke. The initial pressure is 90 pounds absolute. The absolute back pressure is 18 pounds. Solution : M. E. P. = 90 X 0.596 18 = 53.64 18 35.64 pounds. NOTE. All such calculations must be made from absolute pressure (not gage pressure), and when determining the cut-off the clearance must be considered. Clearance. The clearance of an engine is usually expressed as a per- centage of the piston displacement. The space between the piston and the cylinder head at the end of the stroke, also the cavities due to the steam ports, must be included in considering clearance. In high-class Corliss engines the clearance does not exceed 2 l /z to 5 per cent., but in common slide-valve engines the clear- ance may go as high as 5 to 15 per cent. When clearance is taken into account the actual ratio of expansion is R = Actual ratio of expansion. r = Nominal ratio of expansion. c = Clearance, expressed as a fractional part of the length of the stroke. EXAMPLE. The nominal ratio of expansion is 4, and the clearance is 5 per cent. What is the actual ratio of expansion? NOTE. 5 per cent, is %oo = Vso = 0.05 of the stroke. Solution : R _ 1 + 0.05 -- + 0.05 R = -j-- = 3.5 = Actual ratio of expansion. O.o 428 SHOP NOTES. SHOP NOTES. Weight of a Grindstone. Multiply the constant 0.064 by the square of the diameter in inches and this product by the thickness in inches; the result is the weight of the grindstone in pounds. EXAMPLE. Find the weight of a grindstone 30 inches in diameter and six inches thick. Solution : Weight = 0.064 X 30 X 30 X 6 = 346 pounds. Lathe Centers. In this country lathe centers are universally made 60 de- grees, but in Europe the most common practice is to make lathe centers 90 degrees. Morse Taper. The Morse Taper, which is so universally used for the shanks of drills and other tools, is given in TABLE No. 67. Horse Taper. No. of Taper. Standard Plug Depth. Diameter of Plug at Large End. Diameter of Plug at Small End. Taper per Foot. 1 2ys inch. 0.475 inch. 0.369 inch. 0.600 inch. 2 9 9 <* 0.7 " 0.572 0.602 " 3 Q 3 it ^16 0.938 0.778 " 0.602 " 4 1.231 " 1.02 0.623 5 5 T \ " 1.748 " 1.475 " 0.630 " 6 iy 4 " 2.494 2.116 " 0.626 For very complete information regarding the Morse Taper, see American Machinist, May 14, 1896. Jarno Taper. Inthe"Jarno Taper" the number of the taper gives the length of the standard plug in half-inches, and it gives the diame- ter of the small end in tenths of inches and the diameter of the large end in eighths of inches. For instance, a No. 8 "Jarno Taper " is four inches long, one inch diameter at large end, and 0.8 inch diameter at small end. The taper, of course, is 0.6 inch per foot for all numbers. This is a very convenient sys- tem, and deserves adoption for its merits. The same taper is SHOP NOTES. 429 also very well adapted to the metric system, as 0.6 inch per foot is equal to 0.05 millimeter per millimeter. The following table is given to illustrate the system. The table could be extended to as large size tapers as are required for any work. TABLE No. 68. Jarno Taper. Number of Taper. Length of Taper. Diameter of Large End of Taper. Diameter of Small End of Taper. 1 Y* ^ = 0.125 A = 0.1 2 1 % = 0.250 J =0.2 3 l/^ 3^ 0.375 A = 0.3 4 2 y 2 = 0.500 A = 0.4 5 2^ 5/6 = 0.625 # =0.5 6 3 Y = 0.750 1 =0.6 7 3^ 7^ = 0.875 A = o- 7 8 4 1 = 1.000 * =0.8 9 4>^ 1# = 1.125 T 9 * = 0.9 10 5 li^ = 1.250 1 = 1.0 This system of taper is described by "Jarno " in the Amer- ican Machinist, October 31, 1889. Marking Solution. Dissolve one ounce of sulphate of copper (blue vitriol) in four ounces of water and half a teaspoonful of nitric acid. When this solution is applied on bright steel or iron, the surface immediately turns copper color, and marks made by a sharp scratch-awl will be seen very distinctly. A Cheap Lubricant for Milling and Drilling. Dissolve separately in water 10 pounds of whale-oil soap and 15 pounds of sal-soda. Mix this in 40 gallons of clean water. Add two gallons of best lard oil, stir thoroughly, and the solu- tion is ready for use. Soda Water for Drilling. Dissolve three-fourths to one pound of sal-soda in one pail full of water. Solder. Ordinary solder is an alloy consisting of two parts of tin and one part of lead, and melts at 360. Solder consisting of two parts of lead and one part of tin melts at 475. For tin work use resin for a flux. 43 SHOP NOTES. Soldering Fluids. Add pieces of zinc to muriatic acid until the bubbles cease to rise, and the acid may be used for soldering with soft solder. Mix one pint of grain alcohol with two tablespoonfuls of chloride of zinc. Shake well. This solution does not rust the joint as acids are liable to do. When soldering lead use tallow or resin for a flux, and use a solder consisting of one part of tin and \y z parts of lead. Spelter. Hard spelter consists of one part of copper and one part of zinc. A softer spelter is made from two parts of copper and three parts of zinc. A spelter which will flow very easily at low heat consists of 46% of Copper, 46% of Zinc, and 8% of Silver. When making any of these different kinds of spelter, melt the copper first in a black lead crucible and then put in the zinc after the copper has cooled enough to furnish just sufficient heat to melt the zinc, but not enough to burn it. Stir with an iron rod and after the metals have compounded and the compound is still molten, pour upon a basin of water. The metal in striking the water will form into small globules or shot and will so cool, leaving a coarse granular spelter ready for use. When pouring the metal let a helper keep stirring the water with an old broom. Alloy Which Expands in Cooling. Melt together nine pounds of lead, two pounds of antimony and one pound of bismuth. This alloy may be used in fastening foundation bolts for machinery into foundation stones. In such cases, collars or heads are left on the bolts and after the hole is drilled in the stone a couple of short, small holes are drilled at an angle to the big hole ; when the metal is poured in, it will flow around the bolts and also into these small holes, and it is almost impossible for the bolt to pull out. CAUTION. When drilling holes in stone, water is always used, but this must be carefully dried out by the use of red-hot iron rods before the melted metal is poured in. If this pre- caution is not taken the metal will blow out, making a poor job, and it may also cause accident by burning the hands and face of the man who is pouring it in. Shrinkage of Castings. General rule : % inch per foot for iron. S /IQ inch per foot for brass. In small castings the molder generally raps the pattern more than the casting will shrink, therefore no shrinkage is al- lowed. Frequently castings are of such shape that the pressure SHOP NOTES. 431 of the fluid iron on some part of the mould is liable to make the sand yield a little and thereby cause the casting to be as large as, or even larger than the pattern. All such things a practical pattern maker takes into consideration when allowing for shrinkage in patterns. Case Hardening Wrought Iron and Soft Steel. Bone dust specially prepared for the purpose, or burnt leather scrap, is placed in a cast-iron box, together with the article to be hardened. Cover the top of the box with plenty of the hardening material in order to keep the air out. Heat the whole mass slowly in a furnace to a red heat from two to five hours in order that it may be uniformly and thoroughly heated through. A few iron rods about %