Class Book Copyright W. S6 COPYRIGHT DEPOSIT. In Press BY THE SAME AUTHOR A MANUAL OF THE METRIC SYSTEM Showing by comparisons the sim- plicity of the construction of the system and the ad- vantages to be obtained by its use A HAND BOOK OF General Instruction FOR MECHANICS CONTAINING USEFUL RULES AND MEMORANDA FOR PRACTICAL MEN BY FRANKLIN E. SMITH NEW YORK D. VAN NOSTRAND COMPANY 23 MURRAY AND 27 WARREN STS. I907 Copyright, 1906 By D. Van Nostrand Company COPYKI«aT 1907, ■T B. Van Nost&and Company , ?U3RARYofCONGRiIs| Two Cooles Received AUG 2o I90f Cooyrifht Entry CLASS 4 XXc./No, COPY B. T4 •Sco T/ie Plimpton Press Norwood Mass. U.S.A. PREFACE The primary object of the author in writing this book is to give the mechanic, who has not had educational advantages, a text-book explaining established rules for calculating in a clear, simple, and concise way, making him familiar with the various technical terms and their meaning, and to be in general such a course of instruction as to impart, in a simple manner, the required knowledge to enable him to read understandingly more advanced scientific works. In writing the book, no attempt has been made to limit the language to the rigid brevity of the average text-book on mathematics and physics, a very free style being used which will be more acceptable to the general reader. Franklin E. Smith. CONTENTS PART I Arithmetic i PAGE Numeration » 4 Notation 9 Addition 10 Subtraction 16 Multiplication 22 Division '. 40 Tables of Weights and Measures Used by Mechanics .- . . 63 Reduction 67 Fractions 72 Addition of Fractions . 85 Arithmetical Signs 85 Subtraction of Vulgar Fractions 94 Multiplication of Vulgar Fractions 97 Division of Vulgar Fractions 99 Decimal Fractions 103 Reduction of Decimal Fractions 108 Addition of Decimals 109 Subtraction of Decimals 110 Multiplication of Decimals Ill Division of Decimals 113 Proportion 119 Compound Proportion 126 Interest 131 Involution 137 Evolution .138 Cube Root 141 vi CONTENTS PART II Arithmetical Signs and Characters and Explanation of Solving Formula page Arithmetical Signs 149 Examples on the Use of the Addition Signs 151 Examples Performed where Addition and Minus Signs are Used 152 Examples Showing how Brackets are Used 154 Examples on Proper Use of Multiplication Sign .... 156 Examples on Proper Use of Division Sign 158 Examples on Signs Representing the Powers of Numbers . 161 Examples on Signs that Represent the Roots of Numbers . 162 Law of Signs in Multiplication 164 Law of Signs in Division 166 Formula 167 PART III Mensuration Table of Decimal Ecfuivalents 177 Explanation of Terms 177 The Circle 182 The Ellipse 184 The Triangle 187 The Rhomboid 192 The Trapezium 196 The Sphere 199 Volume Measure and Contents of Solids 200 Rectangular Solids 200 The Cylinder 201 The Pyramid 202 The Cone 204 The Frustum 205 CONTENTS vii PART IV Weight > Specific Gravity How the Dimensions, Measurements, and Weight of Different Shaped Vessels is Found and How the Weight of Different Parts is Found page Specific Gravity 211 Table of Specific Gravities of Liquids and Solids . . . .214 Circular Measure 219 Square Measurements 224 Circular Areas 229 Measurements and Weights of Tanks 234 Measurements of Circular Tanks ......... 238 How to Calculate the Weight of Different Materials and the Weight of the Different Parts of a Machine or Structure of any Kind 242 Table of Weights of Various Metals .251 PART V The Primary or Simple Machines The Lever 256 The Compound Lever 268 The Pulley 269 The Strength of a Rope 275 The Wheel and Axle 276 The Inclined Plane . 284 The Wedge 288 The Screw 290 PART VI Strength of Materials and Questions Relating to Stress Table Showing Ultimate and Elastic Strength of Materials . 308 Table Showing Safe Working Stress 309 Table of Factors of Safety 310 Tensial Stress 311 Compressive Stress 313 Shearing Stress 315 PART I ARITHMETIC ARITHMETIC Arithmetic is the science of numbers and has been known to man in various forms for all time. It is some- times called the ''science of calculation/' and when so termed the early period of the science is referred to when (calculi) pebbles were used by the savages to make the process of counting, computing, and estimating easy. In Arithmetic there are ten characters called figures: 1234567 89 Named: one two three four five six seven eight nine nought And each figure, with the exception of 0, represents a specific collection of single things called units. By unit is meant a single thing, as one cent, one dollar, one ounce, one pound, etc. Figure 1 represents one unit. Figure 2 represents a collection of two units. Figure 3 represents a collection of three units. The figure stands for nothing, and when alone ex- presses no units. A unit or collection of units is called a number, and a figure is a character used to represent a number; that is, figure 1 may be termed number 1, and figure 9 may be termed number 9. A number, then, is a unit or a collection of units Number 1 is called a unit. Number 9 is a collection of nine units. Number 2 is a collection of two units. Number 3 is a collection of three units, etc. 3 4 A HAND BOOK FOR MECHANICS By means of the figures 1, 2, 3, 4, etc., any known number may be represented. There are also six rules according to which arithmetical operations are performed, called Numeration. Notation. Addition. Subtraction. Multiplication. Division. The last four of these rules are called the fundamental rules of Arithmetic, and upon them depend the solution of all arithmetical problems. NUMERATION Numeration is the art of reading numbers which are written in figures. The art of naming the value of figures. Example. Naming the value of 9376267, for instance. Figures have two values called Simple value and Place value. The simple value of a figure is the value it has when standing alone, or on the right hand of a number of figures. In both cases the figure denotes the number of units it represents. For example, 2 standing alone, or when standing on the right hand of a number of figures, thus, 342, denotes in both two units or ones. The place value of a figure is the value it denotes when standing with other figures. And the place value of a figure depends upon its relative position to another figure. NUMERATION 5 For example, each figure of the following number, 1364, has a place value. And that value depends upon the number of places it is toward the left hand from the right-hand figure 4. Each removal of one place to the left increases the value of a figure ten times. For example, 4 standing at the left hand of 4, thus, 44, expresses ten times the value it does when standing alone, or in the right-hand place. Standing alone or on the right-hand place it denotes four units only, but being removed one place to the left, as above, increases its value ten times, and it then de- notes ten times four, which is forty, and the four on the right of it denotes four units. And the two figures then together express forty-four (forty-four units or ones). Again, by removing the figure 4 one place more to the left, making it then three places from the right thus, 444, its value is again increased tenfold. The first removal to the left increases it from 4 to 40, and the next or third removal increases its value ten times more, making four hundred, and the three figures together express four hundred forty-four. Each place has a name, as follows, reading from right to left: 03 73 PQ d \2 ° O S <** 03 7 8 an example of a complex fraction having simple fractions for numerator and denominator; 6 or — * 3| example of complex fraction having whole number for numerator and mixed number for denominator; or 51 example of complex fraction having mixed numbers for numerator and denominator. REDUCTION OF FRACTIONS Reduction of fractions is the process of changing frac- tional terms without altering their values. The "terms" of a fraction means its numerator and denominator. REDUCTION OF FRACTIONS 75 To reduce a fraction to its lowest terms, divide the numerator and denominator by any number that will divide them both without a remainder. Example. Reduce t 8 2 to its lowest terms. Thus: 4)_§_ 2 12 3 Here 4 will divide both numerator 8 and denominator 12 without a remainder. 4 goes into 8 two times and into 12 three times. Hence f is equal to t 8 2 In other words, ^ of an inch is the same part of an inch (or length) as § of an inch is, or § of an apple would represent a piece of an apple the same size as t 8 2 of the same apple. Therefore, by bringing the fraction T 8 2 to its lowest terms, f, the value of the fraction, has not been altered. The terms of the fraction were changed only. Reduction of fractions, then, is performed by dividing the numerator and denominator by any number that will divide them both without a remainder, and the successive results of the division are divided by any number that will divide them both without a remainder, until the operation of dividing can be carried no further. Example. Reduce ft to its lowest terms. 8)24 3 , 72 9 3 Here the number 8 will divide the numerator and denominator without a remainder, and by the division is obtained f . We next divide this result by 3, because 3 will divide both 3 and 9 without any remainder, and obtain J, which cannot be further divided; therefore, J is the fraction f f reduced to its lowest terms, and both fractions represent the same value. 76 A HAND BOOK FOR MECHANICS Example. Reduce iWA to its lowest terms. First divide the numerator and denominator by 12, and by the division the terms become 1W2. Then divide again by 12 and the terms tW* become M. 12 will again divide them and it becomes J, which is the lowest term. Consequently, J of an inch is the same length as tVsV? of an inch. Examples for Exercise Reduce t% to its lowest terms. Reduce to to its lowest terms. Reduce it to its lowest terms. Reduce if to its lowest terms. Reduce if to its lowest terms. Reduce tWs to its lowest terms. Reduce Iff to its lowest terms. To reduce a fraction of any denominator to a fraction having a greater denominator, that is, to bring a fraction to its largest terms, which is the reverse of the last rule, see how many times the denominator of the fraction in sight will divide into the larger denominator to which you are to bring it. Example. Bring J to a fraction whose denominator is 8. Proceed by dividing the denominator 2 of the fraction \ into the greater denominator 8, and multiply the numerator and denominator of the fraction \ in sight by the quotient obtained by the division. Here 2 goes into 8 four times. Then multiply the numerator and denominator of the fraction \ by the quotient 4, and by the multiplication is obtained |. Arts. i Arts. i Arts. § Arts. j Arts. i 3- Arts. i Ans. 236 4T7 REDUCTION OF FRACTIONS 77 Multiplying both terms of a fraction by the same number does not change its value, hence, 1 _ 4 2 — 8- That is, J of an inch would be equal to | of an inch. Example. Bring § to a fraction whose denominator is 15. Here 3 goes into 15 five times, then § becomes if- Examples for Exercise Bring f to a fraction whose denominator is 12. Arts. tV Bring J to a fraction whose denominator is 16. Ans. if • Bring f to a fraction whose denominator is 20. Ans. |§. Bring J to a fraction whose denominator is 8. Ans. f . When bringing a fraction to its largest terms, as in the above examples, the denominator of the fraction whose terms are to be changed must divide into the denominator to which it is to be changed without a remainder. To reduce a mixed number to an improper fraction, multiply the whole number by the denominator of the fraction, and to the product add the numerator, and place their sum over the denominator of the fraction. Example. Reduce 6 J to an improper fraction. Multiply the whole number 6 by the denominator 8 of the fraction, and add to the product obtained by the multiplication the numerator 7, saying 8 times 6 are 48 and 7 are 55. Place the sum 55 thus obtained over the denominator 8 of the fraction, as follows: 55 Then 5 gf represents the mixed number 6| when reduced Arts. 1 9 4 Ans. 47 s • Ans. 20 3 • Ans. 53 5 • Ans. 406 78 A HAND BOOK FOR MECHANICS to an improper fraction, and it shows that there are contained in 6J, 55 eighths. That is, there would be contained in 6J inches 55 eighths of an inch. For instance, in one inch there are 8 eights, therefore in 6 inches there are just six times as many eighths, which would be 48 eighths, to which add seven more eighths and you have 55 eighths. Examples for Exercise Reduce 4§ to an improper fraction. Reduce 5 J to an improper fraction. Reduce 6§ to an improper fraction. Reduce 101 to an improper fraction. Reduce 67f to an improper fraction. To reduce an improper fraction to a whole or mixed number, divide the numerator by the denominator, and if there is a remainder, place it over the denominator at the right of the whole number obtained by the division. Example. Change ff to a mixed number. Divide the numerator 24 by the denominator 16 as follows, and place the remainder ^ on the right of the whole number 1, obtained by the division. 16)24(11 ^6 1=1 n 2 Examples for Exercise Reduce - 1 ? 2 - to a whole number. Ans. 3. Change V - to a mixed number. Ans. 14$. Reduce *£■ to a whole number. Ans. 15. Change Vi 3 " to a mixed number. Ans. 48t4- REDUCTION OF FRACTIONS 79 To reduce a compound fraction to a single fraction: Example. Reduce J of J to a single fraction. Multiply all the numerators together for the new numerator (that is, for the numerator of the single frac- tion), and all the denominators together for the new denominator; then, if necessary, reduce the fraction obtained to its lowest terms. Bear in mind that the sign X is read " multiplied by" and means that the num- ber before it is multiplied by the number after it. Proceed now to reduce the above compound fraction to a simple, or single, fraction, as follows: I X i = U- Ans. As shown, the numerator 3 and 7 are multiplied to- gether for the new numerator, which is 21, and the denominators 4 and 8 are multiplied together for the new denominator, which is 32. We then have the fraction %\. Example. Reduce § of J of T V to a single fraction. As before, multiply all the numerators together for the new numerator, and all the denominators for the new denominator, and, if necessary, reduce the fraction obtained by the multiplication to its lowest terms, as follows: 2 17 U J_ . 3 X 5 X 16~pFl20 AnS - By multiplying all numerators together we get a new numerator, 14, and by multiplying the denominators together we get a new denominator, 240, both of which may be divided by 2 without a remainder, as shown. By a process of canceling we can also reduce a compound fraction to a single fraction. 80 A HAND BOOK FOR MECHANICS Rules for Canceling: Any numerator can be divided into any denominator, provided no remainder is left, and any denominator can be divided into any numerator, provided no remainder is left. Example. Reduce | of § of | to a single fraction. Thus: 8 ,2 ,4 Here the 3 in the first numerator and the 3 of the second denominator will cancel one into the other; these are therefore left out. Also the 4 of the first denomi- nator cancels with the 4 of the last numerator, and then we have 2 left as a numerator and 9 as a denominator. Thus : 2 Many problems may be considerably shortened by the process of cancellation, as the following example will show. Example. Reduce § of f of \% of fft to a single fraction. Thus: ? of ^ of M of ii = Z _JL P £ # m 3 X 2 X 34 204 3 £ 34 2 The process is as follows: The first numerator, 2, will go into 8, the denominator of the second fraction, 4 times; the denominator 18, of the third fraction, will go into the 90, the numerator of the last quantity, 5 times. The numerator 3 of the second fraction will go into the de- nominator 9 of the first fraction, 3 times. 5 will go into the denominator 170, of the last quantity, 34 times; 2 will go into 4 twice, and into the numerator 14 of the REDUCTION OF FRACTIONS 81 third fraction, 7 times. And as we cannot find any more figures that can be divided without leaving a remainder, we are at an end and the quantity left must be put into an expression. On examination, we have 7 left in the top row; this is put down at the end as the final numer- ator. On the bottom we have 3, 2, and 34; these multi- plied together give us 204, which is the final denominator. Any numerator and denominator may be divided by the same number, provided there be no remainder left, and the values obtained by the division are used in the place of the values that were canceled. Examples. Reduce f of §r to a single fraction. Here 8, the denominator of the first fraction, can be divided by 4, and 20, the numerator of the last quantity, can be divided by the same number without leaving any remainder, as shown. Thus: 5 3 ,?|_ 3X5 15 JB 31 ~ 2 X 31 "" 62 Am ' 2 Example. Reduce -ft- of A of tt to a single fraction Thus: 1 f A f M 7 _7_ J0 ffi 17 " 3 X 2 X 17 " 102 3 i 2 Examples for Exercise Reduce to their simplest forms the fractions: i of § of | of J This can be done by canceling as follows. Thus: 82 ^1 HAND BOOK FOR MECHANICS 12-341 7 of -z of 7 of - = - Arts. '1 p 4 5 o Reduce f of H of | to a single fraction. 3 3 ,16 4_ _3_ _3^ 5 16 ° 9 ~ 4 X 3 ~ 12 4 3 Reduce f of J of }* to a single fraction. Thus: 2 p , 1 , 1 "4 1 1 of 7 of — = /7 £ ^ 2x3 6 2 3 Reducing Whole Numbers to Improper Fractions Placing 1 for a denominator under a whole number reduces the whole number to an improper fraction. Example. Reduce 7 to an improper fraction. Place 1 under the whole number 7 and we have an improper fraction, whose numerator is 7 and whose de- nominator is 1. Thus: All whole numbers in a compound fraction must be converted into an improper fraction before multiplying the numerators together and the denominators together. Example. What is the value of f of f of 7. Notice that the last quantity,. 7, is a whole number. It is therefore converted into an improper fraction by placing 1 underneath it for a denominator, as follows: § 4 7 = _ 4X7 = 28 5 9 1 5 X 3 X 1 == lo)28(l T f 3 15 13 REDUCTION OF FRACTIONS 83 By first canceling and then multiplying the remaining numerators together for a new numerator, and then multiplying the remaining denominators together for the new denominator, we obtain the improper fraction ft, which is reduced to a mixed number. Then, as shown, the value of § of f of 7 = lit • All mixed numbers in a compound fraction must be reduced to an improper fraction before multiplying the numerators together and the denominators together. Example. Find the value of | of T 4 T of H of £§ of 5|. Notice the last value, 5f, is a mixed number, which, when reduced to an improper fraction is t 3 -. Then we have f of t 4 t of {$ of J§ of -£-. After canceling and multiplying together the remaining numerators for the new numerator, and the remaining denominators for the new denominator, we have, as shown: 5 n 1% w i 5 Examples for Exercise Reduce f of to of if to single fraction. Ans. #5. Reduce J of tt of 7 to mixed numbers. Ans. 4J|. What is § of tt of f of f. Ans. tIt- Change to a single fraction H of J of f of 2V of 8. Ans. 3T 3 o. Find the value of § of J of f of tt of li of 5|. Ans. f |. What is I of 2|. Ans. 2 T \. Reduce tr of ft of if of 9f to a whole number. Ans. 3. Reduction of Complex Fractions Process. Reduce first all whole numbers, if any, and all mixed numbers, if any, to improper fractions, and 84 A HAND BOOK FOR MECHANICS then reduce all compound fractions, if any, to single fractions. Then divide the numerator of the complex fraction by the denominator, according to the rule for the division of fractions. 3. Q Example. Reduce 3 toa single fraction. 4 This is an example of a complex fraction having for both terms single fractions. Knowing that the numera- tor of a fraction is the dividend and the denominator is the divisor, all that will be required to solve the above problem is to divide the numerator f by the denominator 3 4' a Thus 33 34121 8 '4~83~24~2 Am ' 3 Example. Reduce — to a single fraction. Thus: » 3_ jf = 3_453jj^24 5| V 1 " 8 1 45 45 The above example shows that the whole numbers and mixed numbers were first reduced to improper fractions. We then proceed with the operation as in former example. 3 Example. Reduce L \ 3 to a single fraction. Thus: f f 3 . 3 2 3 * 8 24 = n X - = — = 1-fj = 1| Arts. ioff | 7 8 7 3 21 As shown, the denominator \ of }, which is a compound fraction, is first reduced to a single fraction and then we proceed as before. Examples for Exercise i Reduce to single fraction f Ans. §. REDUCTION OF FRACTIONS 85 84 Reduce to single fraction — Ans. 11 J 4 33. 3 Reduce to single fraction ^f of \ Ans. §f § ^i, ^ Following now are the rules, with examples, which govern operations in Addition, Subtraction, Multiplica- tion, and Division of Vulgar Fractions. And note, when performing such operations, that all fractions contained in a problem are first reduced to their simplest form. Addition op Fractions Addition of vulgar fractions is the process of adding together two or more fractions, so that their value can be expressed in one sum. When adding together a number of fractions having the same denominator, add together the several numer- ators for the new numerator and reduce the fraction obtained in this way to its simplest form. ARITHMETICAL SIGNS As the following signs are used in performing operations in fractions, it is necessary that the pupil should under- stand them. The sign + is read " plus," and means that the number after it is to be added to the number before it; thus, 5+ 4 are 9. The sign — is read "minus," and means that the number after it is to be subtracted from the number before it; thus, 9 — 4 is 5. The sign X is read " multiply by," and means that the number after it is multiplied by the number before it; thus, 7 X 3 are 21. The sign -7- is read " divide by," and means that 86 A HAND BOOK FOR MECHANICS the number before it is to be divided by the number after it; thus, 12 -=- 6 are 2. The sign = is read " equal to," and means that the quantity after it is of the same value as the quantity before it; thus, 4 + 5 = 9. Addition of Fractions Example. Add together, or find the sum of, J, f , and f . Here we have three fractions to add together, all having the same denominator, 8. As before stated, the denominator tells us how many equal parts the whole thing has been divided into, and the numerator shows us how many of those parts we have. Therefore, in the example, we have in the first fraction 1 part of the whole thing, and in the second fraction we have 3 parts of the whole thing, and in the third fraction we have five parts of the whole thing. Therefore, it is evident, to find the number of parts contained in the three fractions, we add together the numerators 1, 3, and 5, the sum of which equals 9. The 9 thus obtained is the new nume- rator, and we have the fraction |, which, reduced to its simplest terms, equals 1J. Then, as shown, the sum of I, f, and f equals I or 1J. Example. Add together, or find the sum of, if, if, if, ie, ana i^ — 16 or iif J±ns. Here you have again a number of fractions to add together, all having the same denominator, and as each of the numerators represents a certain number of parts of the whole thing that is divided into 16 equal parts, it is evident, to get the sum of all the parts represented by the several numerators, these numerators are simply ARITHMETICAL SIGNS 87 added together and the sum is written over the common denominator, and the fraction, fi, thus formed, is re- duced to its simplest terms and 1M is the required sum. Examples for Exercise Add together fr, tt, t 5 t, tV, tt, and H- Arts. l}f . Add together t 4 t, tt, tt, tt, tt, and tt- Ans. 3 t 8 t. Add together f , f , J-, and J. Ans. 2. Add together 37, s 4 t, 3 6 t, st, and st- Ans. %\. To bring fractions having different denominators to fractions having one common denominator (that is, a denominator common to all the fractions) : First, put all the denominators down in a row. Second, cancel all the denominators that are alike, except one. That is, if there are three denominators alike, two of them would be canceled and one would be left. For instance, if in a row of denominators there hap- pened to be three fives, as follows, 5, 5, 5, two of them would be canceled. Thus: and one would be left. If there were four fives, three of them would be canceled. Third, then cancel all the denominators that will divide another without a remainder. By that is meant, the smaller denominators that will divide the larger are canceled, the larger one which can be divided is not altered, the smaller ones being left out (canceled), because they will divide the larger. For instance, if a row of denominators consisted of the numbers, 2, 3, 7, 5, 10, 11, the 2 and 5, which will divide the 10, would be canceled, as shown: % 3, 7, f>, 10, 11 88 A HAND BOOK FOR MECHANICS Having canceled all the numbers that are alike, except one, and struck off the numbers that will divide others without a remainder, the fourth thing to do is — Take all the remaining numbers and place them in a row. Fifth, then see if any number will divide two or more of them (not less than two); if so, divide by it and place in the quotient those numbers which cannot be divided. For instance, if the numbers left were 12, 7, 16, 5, they would be placed in a row as follows: 12, 7, 16, 5 It is seen that the number 4 will divide two of these numbers, 12 and 16, without any remainder, so divide by it and also place in the quotient the numbers 7 and 5, which cannot be divided by 4. Thus: 4 )12, 7, 16, 5 3, 7, 4, 5 And as no two of these quotient numbers, 3, 7, 4, 5, can be divided by any number, we divide no further. Multiply all the numbers in the quotient together and multiply their product by the divisor. In this case the numbers 3, 7, 4, 5 would be multiplied together, and the product 420 is multiplied by the divisor 4. The product 1680 obtained is the common denominator. That is, 1680 is the common denominator of a number of fractions whose denominators are 12, 7, 16, 5. It will often happen that two or more of the quotient numbers may, in turn, be divided by some number. When the case is such, proceed to divide as before. For instance, if the remaining numbers were 12, 16, 10, 35, we would first divide the 12 and 16 by 4 and place in the quotient the numbers 10 and 35, which ARITHMETICAL SIGNS 89 could not be divided by 4 without any remainder, as follows: 4)12, 16, 10, 35 5 ) 3, 4, 10, 35 3, 4, 2, 7 Now in the quotient 3, 4, 10, 35, there are two num- bers, 10 and 35, which can be divided by 5, so we place the 5 in the divisor and proceed to divide by it, placing in the quotient the figures 3 and 4, which cannot be divided, as shown. Then, as before, multiply all the quotient figures together and multiply their product by the divisors, and the result of the multiplication is the common denominator. In this case the numbers 3, 4, 2, 7 would be multiplied together, and the product 168 is multiplied by the divisors 5 and 4, and the result of the multiplication, 3360, is the common denominator of a number of fractions whose denominators are 12, 16, 10, 35. Lastly, divide the common denominator by the de- nominator of the first fraction and multiply the quotient by its numerator, and the product obtained is the new numerator required. Then, in short, to bring fractions having different de- nominators to fractions having one common denomina- tor, place all the denominators in a row; cancel all that are alike, except one. Also cancel any that will divide into another one without remainder. If there is any number that will divide two or more of those left, then divide by it, putting down those numbers that cannot be divided. Repeat this till all the numbers are prime numbers. A prime number is a number which cannot be divided without a remainder by any number except itself and unity. 90 A HAND BOOK FOR MECHANICS Then multiply all the prime numbers together and their product by the divisor; the result will be the com- mon denominator for all the fractions. Then divide the common denominator by the denominator of the first fraction and multiply the quotient by its numerator; its product is the new numerator required. Repeat this for each fraction. Example. J, f, §, f, i, f, T 7 2, A, i, and f. Bring these fractions to other fractions having a common denominator. All the denominators placed in a row: 2, 6, 3, 8, 3, 8, 12, 16, 8, 4 There are two figures 3, cancel one of them. Next the 2 and the 8 (two of them) and the 4 will go into the 16, therefore they must be canceled; the 6 and 3 also, because they will divide the 12. Then there only remain two, 12 and 16; place them as below and divide them by 4, because 4 will divide them both without a remainder. % % ^ & 0, & 12, 16, ft i 4 )12, 16 3, 4 Then multiply the 3 by the 4 equals 12 and multiply the 12 by the divisor 4 equals 48, and the 48 thus ob- tained is the common denominator. Lastly, bring each fraction to one having the denominator 48. 15 2 3 1 5 _7_ 5 1 3 2) 6) 3) 8) 3) 8? 12) 1^) 8) 4 Answer: 24 40 32 18 16 30 28 15 _6_ 3fi 48) 4?; 48> 48) 48) 48) 4¥) 48) 48) 48 This is done, as before explained, by dividing the com- mon denominator by the denominators of each of the frac- tions, and multiplying the quotient by the numerator. ARITHMETICAL SIGNS 91 The denominator 2 of the first fraction J goes into the common denominator, 48, 24 times. Multiply the numer- ator 1 by the quotient 24, and the product 24 is the new numerator. Do likewise with the remaining frac- tions, dividing the denominator of each fraction into the common denominator and multiply the quotient by its numerator, the product being the new numerator. The denominators of the several fractions which are brought to other fractions having a common denomi- tor must divide the "common denominator" an even number of times. To add together fractions having different denomi- nators : First bring them to fractions having one common denominator, then add together all the numerators found by dividing the several denominators with the common denominator and multiplying the quotient by the nu- merator, the number obtained by the addition being the new numerator, which place on the right of the several fractions and underneath it place the common denomi- nator. Example. Add together, |, §, f , tu, and J. 2 )0, 3, 4, 10, % 3, 2, 5 _2 10 J5 30 _2 60 common denominator. 48 + 40 + 45 + 42+30 205 o25 60 60 Add together §, §, f , and \. 3f* = 3AAns. 92 A HAND BOOK FOR MECHANICS 2 )2, k ft 3 & £> 6, 8 2 )6, 8 3, 4 J3 12 24 common denominator 16 + 18 + 20 + 21 _ 75 _ q3 _ Ql 24 - 24 -^24-^s As stated before, compound fractions must be reduced to single ones, whole numbers and mixed numbers to improper fractions, and then all are reduced to their lowest terms before finding the common denominator. Example. Add together 4f, 5f , and 7. Showing the mixed and whole numbers, $ 9 , V, t, reduced to improper fractions, 4 7 1 _7 7 _4 28 common denominator 133 + 164 + 196 493 _., __ — — __ 1 7JL1. 28 28 28 As shown, the mixed numbers 4 J and 5f, and the whole number 7, are reduced first to the improper fractions, V-, \ L , t- The common denominator, 28, is then found, and the different denominators are divided into it and the quo- tients obtained are multiplied by the different numerators. By dividing the common denominator by the first de- nominator, 4, and multiplying the quotient 7 by the numerator 19, we get the product 133. By dividing the ARITHMETICAL SIGNS 93 common denominator, 28, by the next denominator, 7, and multiplying the quotient 4 obtained by the nu- merator 41, we get the product 164. Then divide the common denominator by the last denominator, 1, and as 1 goes into 28 just 28 times, the quotient 28 is multi- plied by the numerator 7, the product of which is 196. As shown, the three products, 133, 164, and 196, are added together and their sum, which equals 493, is the new numerator, and it is placed on the right of the several fractions, with the common denominator, 28, underneath it, thus, 4 2 9 s 3 , which fraction is then reduced to its simplest form. Example. Add J of f to 6|. - of - to 6| When reduced to J^ 51 When reduced to im- single fraction 16 8 proper fraction Common denominator, 16 \ 3 + 102 105 = 6 T ^ 16 16 As shown, the compound fraction | of f is reduced to the single fraction ft, and the mixed number 6| to an improper fraction % 1 -. As shown, we then have ft and V to add together. Examples for Exercise Add together J, J, \, and f . Arts. 2J. Add together f, f , f , and f . Arts. 2f . Add together f, ft, ft, and f . Ans. 3 A nSt Example. Divide 14 by f. Thus: ARITHMETICAL SIGNS 101 14 + f = 14 X i = 98 3)98(32|. Arts. 9^ 8 A 2 3 Example. Divide 32 by t 9 t- Thus: ~9)352(39£. Arts. 27 82 81 1_ 9 All mixed numbers and whole numbers must be re- duced to improper fractions, and all compound fractions reduced to single fractions, before dividing. Example. Divide If by 1J. Reduce first the mixed numbers to improper fractions, then proceed to divide. Thus: 13 _^ 11 Above reduced to improper fractions. 1 -J- £L 4 • 8 Then 7^? = 7 x 8 = 56 4 ' 8 4 9 36 9 Example. Divide f by 1|, that is: 3 . 9 — 3 V 8 24--2 Avtd Example. Divide f of 2 by f of 7. Reduce first all whole numbers to improper fractions. 102 A HAND BOOK FOR MECHANICS Thus: | of f + | of 1 Reduce, then, above compound fractions to single fractions. Thus: | of f - | of 1 Above reduced to single fractions: TViAn 6 _i_ 2 1 4 * 4 _L IlfcJIl 6 -21 6 V 4 — 24 -_ 2 4 ' 4 4 A 21 84 7- Ans. Example. Divide 4f of \% by 3f of 3^. That is: That is: 30 n fl4 _t_ 1 5 n f 1 6 "7 U1 15 • 4 U1 5 WlJa i 3 4 _M of W_4 x ±_±_ t 1 12 " 12 ~ -. Ans. o Canceling, as shown above, reduces the dividend to \ and the divisor to V. Examples for Exercise \ + | Ans. f . | -T- J Ans. 2. | -f- i Ans. 1J. 3J -s- f Ans. 6J. of 12i - i of 84. Ans. 1031. 3 TT '5 £ of 81. « 2 5# ~^" ^5- Ans. ^6ff« 4f of if -s- 3| of 3J. Ans. f£. § of | -^ | of |. Ans. 18|. | of f of f ^ f of f of T V Ans. 3ff. DECIMAL FRACTIONS 103 DECIMAL FRACTIONS The term "decimal" is derived from the Latin word deci, or decimus, which means ten or tenth. The decimal system of figures is very old and in all probability originated from the use of the ten fingers as helps to count. It is supposed that in India there was in use, as early as a.d. 525, an imperfect form of our present decimal system. The system was not used in Europe till about 1200, when it was introduced through" the Arabians. According to the decimal system, ten or some multiple of ten units of a lower denomination make one unit of the next higher denomination. Our present money system, for instance, is a decimal system, because: 10 mills make one cent. 10 cents make one dime. 10 dimes make one dollar. 10 dollars make one eagle. According to the decimal system, a dot, thus (.), called the decimal point, separates the unit from the fractional parts of an expression. Thus, in the following decimal expression: 354.5 the 354 represents units and the .5 represents the frac- tional parts of a unit. A decimal fraction is one whose denominator is always 10 or 100 or 1000, or some other power (as it is called) of 10, but its numerator may be any number. For example, T V, T £o, Woo, to, too, are all decimal fractions. Strictly speaking, there should be as many figures in 104 A HAND BOOK FOR MECHANICS the numerator as there are cyphers in the denominator; and if there are not as many, they may be made so by attaching cyphers to the left of the given numerator. For example, the five fractions above would be: JL JUL 001 3 J)_6_ 10) 100) 1000) 10) 100* This being the case, the denominator can be done away with, a dot, thus (.), being placed before the numerator. A decimal fraction, then, is always represented by one number having a dot before it ; that is : "to is written .1 and is in value equal to one tenth of a whole number. to is written .7, and is in value equal to seven tenths of a whole number. tVo is written .01, and is in value equal to one hundredth of a whole number. toVo is written .001, and is in value equal to one thousandth of a whole number. So it will be seen that, in decimals, by placing a figure one place to the right makes it a tenth of what it was before, just as in whole numbers. Thus: 1. is one whole thing. .1 is one tenth whole thing. .01 is one hundredth whole thing. .001 is one thousandth whole thing, etc. If the fractions have a numerator other than 1, thus: T % is written .5, T Vo is written .27, and f\ 7 o 2 o is written .372. The first place on the right of the decimal point is always tenths, the second place on the right of the decimal point is always hundredths, and the third place on the right of the decimal point is always thousandths. Knowing, then, that a decimal fraction is expressed without its denominator, by means of the decimal point, DECIMAL FRACTIONS 105 and that it may be expressed in this way because the denominator of a decimal fraction is always 1, with as many places annexed as there are places in the deci- mal, proceed to express the following in decimal form: (1) T Vo Ans. .85 ; (2) T U Arts. .06 ; (3) T c°o o Ans. .009 ; (4) T Vo Ans. .75 ; (5) ftVo Ans. .201 ; (6) 349 WoW Ans. 349.000019 ; (7) 12 tVo 2 o Ans. 12.342 ; (8) T o oVWo Ans. .000012. To Bring a Decimal Fraction to a Vulgar Fraction Process. Put down the given decimal as a numerator, leave out the decimal point, then place underneath it a denominator 1 with as many cyphers after it as there are places in the given decimal. For instance, .25 to a vulgar fraction. Thus: 100 4* -tli I*' The operation was performed by omitting the decimal point and placing the denominator underneath the given numerator, then changing the fraction to its lowest terms. Example. Bring .875 to a vulgar fraction. Thus: _8_7_5_ — 17 5 3 5 — 7_ A ™ Q 1000 200 40 8 ^-'^. Example. Bring .87500 to a vulgar fraction. Thus: 87^00 — J7 500 — lioOO 3 000 __ 7.0 — 7 A ~, Q •°« indicates that the number which is before it is greater than the number which follows it. The sign < indicates that the number before it is less than the number which follows it. The sign ^ reads, or stands, for the words "is not equal to." The sign < stands for the words "is not less than." The sign > stands for the words "is not greater than." Examples on the Use of the Addition Sign The addition sign requires little notice, because when placed before a number it indicates that it is to be "added" to what has gone before. Thus, 3 + 9 means that 9 is to be added to 3, and 3 + 9 + 2 means that 9 is to be added to 3 and then 2 added to the result. Thus, in addition, the order of performing operations is from left to right. 152 A HAND BOOK FOR MECHANICS Examples for Exercise What is the sum of 2 + 6 + 9? Arts. 17. What is the sum of 15 + 3 + 1? Arts. 19. What is the sum of 7 + 9 + 6 + 8? Arts. 30. What is the sum of 8 + 2 + 3 + 1? Arts. 14. What is the sum of 6 + + 3 + 5 + 7? Ans. 21. Examples on the Use of the Minus Sign This sign requires little notice also, because when placed before a number it indicates that it is to be sub- tracted from what has gone before. Thus, 4 — 2 means that 2 is to be subtracted from 4, and 16 — 4 — 3 means that 4 is to be subtracted from 16 and then 3 is to be subtracted from the result. Thus, 16 — 4 = 12, and 12 - 3 = 9. Ans. Then in subtraction, the order of performing opera- tions is from left to right, as it was in addition. Examples for Exercise What is the result of 25 - 3 - 4? Ans. 18. What is the result of 19 - 2 - 10? Ans. 7. What is the result of 16 - 8 - 8? Ans. 0. What is the result of 100 - 50 - 25? Ans. 25. What is the result of 300 - 50 - 100 - 100? Ans. 50. Examples Performed where the Addition and Minus Signs are Involved Note when the first term of any quantity has no sign before it the + (plus) sign is always understood to be there. Example. What is the sum of 12 + 4 - 2 + 7 - 4. Here we add together all those terms which have the + sign actually before them as +4 + 7, and the term ARITHMETICAL SIGNS 153 which is understood to have the + sign before it, al- though the sign is not put down; in this case it is the 12. Then 12, 4, 7 added together make 23, and — 2 and — 4 added together make —6. The sum then becomes 23 - 6 = 17. Ans. Example. Find the sum of 24 - 16 + 4 - - 7 + Thus + 24 + 4 + 4 = + 32 And - 16 and - 7 = - 23 9 Ans. Example. What does the following come to? -6 + 3 + 2-5 Proceed as before by adding together all those terms which actually have the + sign before them, and then add together all those signs having the — sign; sub- tract, then, the smaller from the larger, but put before the remainder the same sign as that of the larger quantity. The above example, then, will be solved thus: + 3 + 2 = 5 sum of terms having the + sign — 6 — 5 = — 11 sum of terms having the — sign Then - 11 + A — 6 Ans. Example. -8 + 12 + 2-9 + 3-8 + 2-17 comes to what? Thus - 8 +12 - 9 +2 - 8 +3 - 17 + _2 - 42 +19 Then - 42 + 19 - 23 Ans, 154 A HAND BOOK FOR MECHANICS Note that the answers of both above examples have the — sign before them, because the sum of the quan- tities after the — signs are the larger. Examples for Exercise What do the following come to? 1. 8 — 12-2 + 9-3 + 8-2 + 17. Ans. + 23. 2. -4 + 3-7-8 + 2. Ans. - 14. 3. 5-17 + 3 + 2. Ans. - 7. 4. 25-6-2 + 7. Ans. + 24. 5. 10 - 25 + 3 + 15 - 17. Ans. - 14. Examples Showing how Brackets ( ) are Used Example. -3 + 4 + 6- (5 + 6) +8- (9-3) + (14 — 4 — 4) is equal to what? Process. Get rid of all the brackets first by putting the quantities within each set of brackets together so that they become a single number. Then to solve the above problem we proceed to deal with the terms inside each bracket first. Thus (5 + 6) = 11 (9 - 3) = 6 (14 - 4 - 4) = 6 Hence - 3 + 4 + 6 - (5 + 6) + 8 - (9 - 3) + (14 - 4 - 4) Becomes -3 + 4 + 6-11+8-6 + 6 Add now all the + signs together and all the — signs together, subtract the larger from the smaller and the result is the required answer. ARITHMETICAL SIGNS 155 Thus +4 - 3 + 6 - 11 + 8 - _6 + _6 - 20 + 24 - 20 + 4 Ans. Example. 7-8+6- (5-6 + 3) +9+ (9 -3 + 2 — 6 + 7) is equal to what? Thus 7-8 + 6- (5-6 + 3) +9+ (9-3 + 2-6 + 7) Equals 7-8+6-2+9+9 Adding all the + signs together we get 31 Adding all the — signs together we get 10 Thus 31 - 10 = 21 Ans. Note. When no sign is between the quantity outside the bracket and the bracket, it means that the quantity within the bracket is to be multiplied by the quantity outside. Example. 3(4 - 2 + 3) - 2 + 6 is equal to what? Here outside the brackets is 3, having no sign between it and the bracket, which means that after reducing the numbers within the brackets to a single number we multiply it by 3. Thus 3 (4 - 2 + 3) = 5 multiplied by 3 = 15 Hence 3 (4-2 + 3) -2 + 6 Equals 15-2 + 6 Adding all the plus quantities together, and all the minus ones separately, we get, 21 — 2 = 19 Ans. 156 A HAND BOOK FOR MECHANICS Example. 4 (3 + 2 - 6 + 4) + 2 (3 + 6 - 2) is equal to what? Thus '4 (3 + 2-6 + 4) +2 (3 + 6- 2) Becomes 3x4+7x2 Which is 12 + 14 Which added together is: 12 14 26 Arts. Examples for Exercise The following equals what? 1. - 2 + 12 + 3 - (9 + 4) + 3 - (7 - 5) + (6 - 2 + 7). Arts. 12. 2. 4 (3 + 2 - 6 + 4) + 2 (3 + 6 - 2). Ans. 26. 3. 4 (3 + 2 - 6 + 4) - 2 (3 + 6 - 2). Ans. - 2 4. 3 (4 - 2 + 3) + 2 - 6. Ans. 11. 5. - 2 + 6 + 2 - (4 + 3) + 4 - (7 - 5) + (7 - 1 + 2). Ans. 9. Examples on the Proper Use of the Multiplication Sign The multiplication sign, X, is placed between two numbers to indicate that the first number is to be mul- tiplied by the second. Thus, 12 X 6 means that 12 is to be multiplied by 6; also 12 X 6 X 3 means that 12 is to be multiplied by 6 and the result multiplied by 3. Sometimes the sign X is replaced by a point or dot which is placed in a line with the bottom of the quan- tities to be multiplied. Thus, 12 X 6 X 3, and 12.6.3, mean the same thing, namely, that 12 is to be multiplied by 6 and the result by 3. Because of the improper use of the multiplication sign ARITHMETICAL SIGNS 157 it causes more trouble than any of the other signs. For instance, in a quantity such as follows 7 + 2x4 we would be apt to say 7 and 2 are 9, which, multiplied by 4 = 36. Now this would be wrong. The first step would be to multiply the 2 by 4 and then to the result add the 7. Thug 7 + 2 x 4 Becomes 7 + 8 Equals 15 Arts. We learn, then, from the above, that the first step to take when the multiplication sign is involved, is to get rid of it in the manner shown. That is, always get rid of the multiplication sign first, unless brackets are used, in which case, we would get rid of all brackets first, and then dispose of the multiplication signs. Example. 4-3X2 + 6X4-2X3 + 8X2 is equal to what? Thus 4-3X2 + 6X4-2X3 + 8X2 Becomes 4-6 + 24-6 + 16 Now put the terms together as in addition and sub- traction, and we have 44 — 12 = 32 Arts. Examples. What does the following equal? -3 + 4 + 2X6X2- 3 + 4X6 Thus -3 + 4 + 2X6X2- 3 + 4X6 Becomes -3 + 4 + 24-3 + 24 + 4 + 24 + 24 = + 52 And -3+-3 --6 46 Arts. Example. What is the following equal to? 24 - (32 - 15) + 3 X (9 - 5) Process. Get rid of all the brackets and then get rid of the multiplication signs. 158 A HAND BOOK FOR MECHANICS The example, 24 - (32 - 15) + 3 X (9 - 5), becomes 24 — 17 % + 3 X 4. The brackets are here disposed of. Now dispose of the multiplication sign and we have 24 - 17 + 12. Proceed now as in addition and subtraction. 24 12 36 - 17 = 19 Arts. Examples for Exercise Find the value of the following: 1. 6 + 4 X (9 - 3) + 7 - 2 X (45 - 41). Arts. — 41. ' 2. 54 - (64 - 29) + 6 X (18 - 5). Arts. 97. 3. 24 - (32 - 15) + 3 X (9 - 5). Ans. 19. 4. -3 + 4 + 2X6X2-3 + 4X6. Ans. 46. 5. 4-3X2 + 6X4-2X3 + 8X2. Ans. 32. Examples on the Use of the Division Sign Example. 5 + 2X(7-2)-4-8-h(3 + 5) equals what number? Process. First get rid of all the brackets; secondly, get rid of the multiplication and division signs, then proceed as in addition and subtraction. Example. 5 + 2 X (7 - 2) - 4 - 8 -r- (3 + 5). Brackets left out, 5 + 2X5-4-8^8. Multiplication and division signs done away with. 5 + 10-4-1. + 5 + 10 = +15 -4 + - 1 = - _5 10 Ans. Example. 32 + 4 + 2. ARITHMETICAL SIGNS 159 Example. Thus 4)32 2)8 4 32-4 + 2. Thus 4)32 Ans. 8 _2 10 Ans. Example. What is 24 - 6 + 4 — 2 equal to? Thus 6)24 4 4 8-2 = 6 Ans. Example. What does the following amount to? (3 + 5) X 6 - 2 X (2 + 2) - 4 - 5 + 10 Thus 8X6-2X4-4-5 + 10 + 48 - 2 + 10 - 5 + 58 - 7 = 51 Ans. Example. 2X4—2 amounts to what? 8^-2 = 4 Ans. Sometimes division is represented thus: 4 + 2 6-4 where one quantity is placed above another with a line between them. When thus represented the value of the top quantity is to be divided by the value of the bottom quantity. That is, in the above example the 4 + 2 is to be divided by the 6 — 4. 160 A HAND BOOK FOR MECHANICS Thus And 6 - 2= 6 2 Example. - 4 Then 6-2 = 3 Ans. 5 + 4 X 6 is equal to how many? 2 + 3X6 Thus 5+4X6 = 29 And Example. 3 2 + 3 X 6 = 20 Thus 29 ~- 20 = lft 12 - 4 + (6 - 4) Ans. (6 - 4) + (16 - 8) Such examples as the above sometimes cause trouble. Note that the 3 before the fraction is placed on the level of the division line, and in this position it indicates that the whole fraction is to be multiplied by it. It will be remembered, however, when multiplying a frac- tion by a whole number, that the numerator of the fraction is multiplied only. Then in the above example the top line only is multiplied by the 3. 12 - 4 + (6 -- 4) - 8 + 2 = 10 10 Thus: 3 Then 3 + 8 = = 3 Ans. (6 - 4) + (16 - 8) = 2 10 30 10 = 10 The pupil would better now refer to vulgar fractions and become very familiar with the fact as there explained, that when multiplying a fraction by a whole number the numerator of the fraction is multiplied only. Example. 3 (12 - 4 + 6 - 4) - (6 - 4 + 16 - 8) is equal to how many ? Thus 3 (12 - 4 + 6 - 4) - (6 - 4 + 16 - 8) Becomes 10 -r- 10 3^ 30 -=- 10 = 3 Ans. ARITHMETICAL SIGNS 161 It will be remembered, when no sign is placed between a quantity and a bracket, that the quantity within the bracket is to be multiplied by the quantity without. Therefore, after reducing the quantity in the first set of brackets to 10, we multiplied by 3 and then divide by the sum of the quantities within the next brackets. The following example will impress this more fully upon the mind: Example. Find the value of 4 + 2 (2 + 6) - 3 + 2 (9 - 4), which is the same as4 + 2X(2 + 6)-3 + 2 X (9 - 4). Hence 4 + 2X8-3 + 2X5 4+16-3+10 4 16 10 30 - 3 = 27 Ans. Examples for Exercise Find the value of: L 7 + 7 - (6 - 2 ) (5 - 2) + 3 - 1 AnS ' *' 2. 14 + 6 - (2 + 4) -r (6 + 8 - 12). Ans. 17. 3. 9 + 4 X (7 - 3) - 8 - 16 -T- (9 - 5). Ans. 13. 4. 12 - 4 + (6 - 4) (6 - 4) + (16 - 8) Ans. 3. Exercises on the Signs Representing the Power of Numbers, as 4 2 , 4 3 , 4 4 , etc. The powers of a number are its square, cube, fourth power, fifth power, etc. 162 A HAND BOOK FOR MECHANICS Thus 4 2 is equal to 4 X 4 = 16 4 3 is equal to 4 X 4 X 4 = 64 4 4 is equal to 4 X 4 X 4 X 4 = 256 Especial names are given to 4 2 and 4 3 ; they are called respectively, the square and cube of 4. The small figure, or letter, placed above a quantity, to indicate the number of times that quantity is to be taken as a factor is called the " Index," or " Exponent." Thus, 4 2 means that the factor 4 is to be taken 2 times and 2 is called the index. Therefore, always multiply the given number that number of times by itself, as indicated by the " Index," and the continued product is the required power. That is, 4 to the eighth power (4 8 ) is 4X4X4X4X4X4X4X4 = 65538 Example. 4 2 + 6 3 is equal to what? 4 2 = 16 6 3 = 216 232 Ans. Example. 4 3 — 3 3 is equal to what? 4 3 = 64 3 3 = 27 Then 64 - 27 = 37 Ans. Exercises on the Sign that Represents the Roots of Numbers a/ and ^/ and \/~ A number which, when squared, is equal to some specific number is called a " square root," and is repre- sented by the symbol y/ and sometimes \V. Thus 6 is \/ 36, since 6 2 = 36, And a number which, when cubed, is equal to any specific number is called a "cube root," and is rep- resented by the symbol ^/ . Thus 4 is ^/ 64, since 4 3 = 64. ARITHMETICAL SIGNS 163 The sign (>/), which represents the root of numbers, was originally the first letter of the word radix, which word really means the source from which anything springs. The sign is now called the " radical sign/' and the radical sign is common to all roots. Thus, when it is required to express the square root of a number or quantity, we simply place this sign before it, as V 144, and when it is required to express the cube root of a number or quantity, the same sign is placed before the number with a 3 in the elbow, thus %/. Sometimes a number composed of two or more terms is to have its root expressed. In all such cases, place the radical sign in front and draw a line over the num- bers whose root is required as far as they extend. For example, \/3(4 + 2 + 3) means that the square root of the sum of those numbers under the line is to be expressed, also \/2 + 14 means that the square of the sum of the numbers underneath the line is required. Also ^/2 + 6 expresses that the cube root of the sum of the numbers under the line is required. Following is another way, which is not often used, for expressing that the root is required. i Thus, 16 2 means that the square root of 16 is required. i And (2 + 14) 2 means that the square root of the sum of the numbers within the brackets is required. Some- 3. times the power and root are combined, as 4\ This is read as the square root of 4 cubed. In all such cases the numerator figure represents the power, and the denominator figure represents the root. In the above example 4 cubed = 64 and the square root of 64 = 8. 164 A HAND BOOK FOR MECHANICS Law of Signs in Multiplication For those who are not acquainted with algebra it is difficult and very confusing to know how to determine the sign of the product. The rule by which the sign of the product is determined is called the "Law of Signs/' Rule 1. Plus and minus multiplied together always give minus. Rule 2. Minus and minus multiplied together always give plus. In other words, when multiplied together like signs give + and unlike signs give — . For example: -3 X + 2 = -6 Here minus 3 is multiplied by plus 2 and the sign of the product 6 is a minus sign, thus — 6, product. Because unlike signs multiplied together give — • (Rule 1.) Example. 8 + 4-3X(3-2 + 4)is equal to how much? First gather together the numbers inside the brackets and we have 8 + 4 _ 3 x (3 _ 2 + 4) Becomes 8 + 4 - 3 X +5 When a number has no sign before it, remember that the + sign is always understood to be before it, hence the 3 in the brackets is + 3, which, when added to the -f 4 = +7, from which — 2 is subtracted, and we have + 5 left. Therefore, when the numbers inside the brackets are gathered up we have + 5, which is to be multiplied by — 3. The sum then is: 8+4-3X +5 8 + 4-15= -3 Ans. ARITHMETICAL SIGNS 165 Example. What is the value of 8 + 14 - 6 X (10 - 14 + 6) Becomes 8 + 14-6X+2 Becomes 8+14-12 Becomes 8 + 14 = 22 - 12 = 10 Arts. Example. What is the product of — 6 X — 6? Thus -6X-6=+36 Am. Here minus 6 is multiplied by minus 6 and the sign of the product is plus. Because like signs multiplied together give + (Rule 2). Example. What is the product of + 6 X +6? + 6 X + 6 = + 36. Am. Example. 13 - 9 X 4 - 11 X (5 - 42 + 29) equals What? 13 -36- 11 X -8 This is an example of Rule 2, being — 11 X — 8 = + 88. Hence 13 - 36 + 88 = 65 Am. Example. What is. the value of 32 - 8 - 11 - 9 X (32 - 71 + 29) 4 - 11 - 9 X - 10 This again is an example of Rule 2, being minus 9 X minus 10. Thus - 9 X - 10 =+ 90. (Like signs give + .) Hence 4 - 11 + 90 = 83 Am. The + sign is not placed before the answer, 83, of the above example, because when no sign is before a quantity + is always understood to be there. 166 A HAND BOOK FOR MECHANICS Law of Signs in Division In division we have the same "Law of Signs" as in multiplication. Unlike signs give — . (Rule 1.) And like signs give + . (Rule 2.) That is, if the signs of the dividend and divisor are unlike, the sign of the quotient is minus ( — ), and if the signs of the dividend and divisor are like signs, the sign of the quotient is plus ( + ). For example: — 12 -= 4 = +3 Ans. The answer here is plus 3 because the sign of the dividend (-12) and the sign of the divisor (- 4) are like signs, and like signs give + . (Rule 2.) Example. + 12-+4=+3 Ans. (Rule 2.) Example. + 15 -s- — 3 = - 5 Ans. (Rule 1.) In the above example the quotient sign is minus, because the sign of the dividend and the sign of the divisor are unlike. Example. -16-h+4=-4 Ans. (Rule 1.) Example. (9 X 24 + 6) -(4 + 2x8- 131). 216 + 6 -4 + 16 - 131 + 222 -?■ - 111 This is now an example of Rule 1. + 222 h 111 = - 2 Ans. Miscellaneous Examples for Exercise 1. 42-7 + 6-2. Ans. 10. 2. 56 4- 4 - 3 X 6 + 4. Ans. 0. 3. 250 -f- 10 + 4 - 2 X (26 - 13). Ans. 3. 4. (52 - 8.14) X (8 + 4). Ans. 526.32. 5. (52 - 8.14) X 8 + 4. Ans. 350.92. ARITHMETICAL SIGNS 167 6. ( 42.4 - 12) 2 - 8 X 4. Arts. 21.76. 64 - 23 7. What is the value of (12 + 3) 3 - (21 - 4) 2 . Ans. . 3014. 8. (4.05) 2 . Ans. 8398.08. (i) 3 9. (6.012 + 0.050) X (.070 + |). Ans. 3.395. 10. (2.562 + 6.0002) - (0.0642 + 2.9808). Ans. 5.5172. 11. 3 + 4 - 2 X 4 - 3 X (8 - 28 + 6). Ans. 41. 12. 28 -^ 4 - 6 - 12 X (- 8 + 30 - 10). Ans. - 143. 13. 18 - 7 X 2 + 9 -T- 3 - (16 - 51 + 13). Ans. 15. 14. What is the value of 29.4 2 - \/25? Ans. 107420. .008 Formula A formula is a rule expressed in a brief and concise way, by the means of certain letters and arithmetical signs. The pupil, therefore, in order to read or solve a formula clearly and easily must be thoroughly acquainted with the arithmetical signs, and the operations which they indicate are to be performed. The reading and solving of formula may appear to those who have not had the early school advantages very difficult; however, for such, if the following rules which govern the reading and solving of formula gener- ally are given careful study and thought, many of the problems which the mechanic comes across in his daily walks of life, and which have no earthly meaning to him, will be clearly read and easily solved. Formulae are used for expressing general rules in mathematics and physics. For instance, the rule of estimating the nominal horse- power for ordinary condensing marine engines is the 168 A HAND BOOK FOR MECHANICS square of diameter of cylinder in inches multiplied by the number of cylinders and the product divided by 30. This rule may be expressed by the following formula: Example 1. p2 x N _ Am 30 D = Diameter of cylinder in inches. N = Number of cylinders. So it is seen by means of two or three letters and signs we can express a long rule. This fully explains the ob- ject of all formulae. Again, the rule for finding the nomi- nal horse-power of paddle steamers is to square the diameter of the piston and multiply by its velocity, and the product divided by 6000 gives the nominal horse- power. Now this rule may be expressed by the formula: D 2 X V = nominal horse-power Example 2. — Suppose, in the above example, the diameter of the piston is 24 inches, stroke 2 feet, revolution 44 per minute, what would the nominal horse-power be? Thus 24 2 X 176 = 15.981 nominal horse-power. " 6000 The above examples are given only to demonstrate that a formula which is composed of a few letters and signs will express a long rule, and the student is not supposed to try to solve same until he at least has become familiar with the following examples, which explain the solving of formulae. Examples Showing how Formulae are solved By referring to the above examples, 1 and 2, it will be seen that in a formula letters represent a value or some ARITHMETICAL SIGNS 169 dimension. Thus D 2 in example 1 represents a dimen- sion and N represents a number or quantity. And in example 2 it is seen that D 2 represents 24 2 and V repre- sents a number equal to 176. The first step to take, then, in solving a formula is to substitute figures for letters, then proceed as directed by the arithmetical signs of the formula. The following examples will clearly demonstrate the process of solving all formula?. Example. IfA=B + C-D + E-F, what must the value of A be when B = 12, C = 8, D = 5, E = 9, and F = 10? We first proceed by substituting the figures for the letters, thus: A = 12 + 8-5 + 9- 10 Then proceed as directed by the signs. A = 29 - 15 = 14 Ans. Example. IfX = A + B-C + D-F, what is the value of X when A = 20, B = 14, C = 9, D = 8, and F = 12? First substitute figures for the letters, thus: X = 20 + 14 - 9 + 8 - 12 Then proceed as in the arithmetical part. X =42-21 = 21 Ans. Example. If K = h X - i D + J C - f F, what is the value of K when A = 12, D = 24, C = 30, and F = 12? 170. A HAND BOOK FOR MECHANICS As A = 12, then § A = 6 And as D = 24, then \ D = 6 And as C = 30, then \ C = 6 And as F = 12, then f F = 9 Then K = 6-6 + 6-9 =12-15 = — 3 Arts. Example. IfK = 3A + 4B + 9C-7D-E + 3F, find the value of K when A = 12, B = 5, C = 2, D = 4, E = 12, F = 1. Here 3 A = 3 times 12 = 36; 4 B = 4 times 5 = 20; 9 C = 9 times 2 = 18; 7 D = 7 times 4 = 28; E = 12, and 3 F = 3 times 1 = 3. Hence K =36 + 20 + 18-28-12 + 3 = 77-40 = 37 Arts. Note when two or more letters are together, without any sign between them, it is always understood that they are to be multiplied together. Thus D N is the same as D X N A~B A X B Example. What is the value of N when N = AB + CD - EF; when A = 4, B = 5,C = 3,D = 4, E = 2, F = 10? N=4X5 + 3X4-2X10 = 20 + 12-20 = 32-20 = 12 Ans. Example. If M = AB - CD + EF - G, what is the. value of M when A = 2, B = 3, C = 4, D = 5, E = 6, F = 7, and G = 20? ARITHMETICAL SIGNS 171 M=2x3-4x5 + 6x7-20 = 6-20 + 42-20 = 48-40 = 8 Arts. Example. If A = C — f C j, find the value of A when C = 8, D = 20, C = 10. A = 8 - (20 - 10) 4 = 8-5-10 = 8-15 = — 7 Arts. Example. If N = C - (- - pY find the value of N when C = 8, S = 3J, P = 1|. N = 8 - (3i - 1J) 2 = 8 - (If - 1§) = 8-i = 7J Ans. Example. If M = 4 ABC - 3 ED - 5 EFG, what is the value of M when A = 2, B = 3, C = 4, D = 5, E = 6, F = 4, G = 2? M=4X2X3X4-3X4X 5- 5X6X4X2 = 96 - 60 - 240 = 96 - 300 = - 204 Ans. Note. In the above example there is no sign between the figures and the letters. The sign X is generally omitted between letters or be- 172 A HAND BOOK FOR MECHANICS tween a figure and a letter. Thus AB means the same as A X B and 4 AB the same as 4 X A X B. Example. If N = A B , what is the value of N, C - D when A = 6, B = 7, C = 16, D = 10? N = 6 X 7 = 42 16-10 6 = 7 Ans. B C Example. If M = A^ — rr what is the value of M D — E when A = 2, B = 3, C = 4, D = 16, and E = 8? „ n 3X4 2X3X4 24 M=2 T6~8 = —8 = ¥ = 3 Am ' Example. What is the value of 2 H — — — when A = 82 and a = 38? 82 + 38 -64 _jg_ = 2 + .056 = 2.028 Ans. 2000 2000 A "D _i_ p Example. What is the value of 2 -± when A = 20, B = 6, C = 16? 20-6+16 . 30 _ ^ 30 60 2 = 2 — = 2X- = -t = 4 4ns. 15 15 15 15 D 3 — d 3 Example. 6 : — W T hat is the value of this when A D = 14, d = 12, and A = 15? 14 3 - 12 3 2744 - 1728 n 1016 6096 6 — IT" =6 h5 6X -l5- = -l5~ = 406.4 Am. Example. A - (B ~ § }\~ K) X .003 N. What is the value of the above when B = 120, C = 32, T = 67, K = 32, and N = 3375, M = 120, R = 67? ARITHMETICAL SIGNS 173 (120 - 32) (67 - 32) Thus A = ^ 19r / V „ 7 ; X .003 X 3375. 120 — 67 _ (88X35) x 0Q3 x 3375 _ DO 3080 31185.000 = ~- X .003 X 3375 = ^ = 588.396 Ans. 53 od Note in the above example there is no sign between the brackets. Whenever this is the case (that is, when- ever there is no sign between two quantities) the two quantities must be multiplied together. S 2 Example. What is the value of jr~ when S = 32 and C = 32,X2? S*_ 32 X 32 1024 ihus 8C " 8 X 32 X 2 ~ 512 ~ 2 ' An8 ' PART III MENSURATION TABLE OF DECIMAL EQUIVALENTS 1 _ 8 .125 A = .01563 11- = .34375 ii =.65625 A - .03125 || = .39538 || = .67188 A = .04688 | =.375 i| = .6875 A = .0625 || = .39063 t| = .70313 1 _ 4 — .25 A = .07813 || = .40625 If = .71875 A = .09375 || = .42188 f| = .73438 e 7 ¥ = .10938 A - -4375 | =.75 3 _ 8 — .375 § =-125 || = .45313 || = .76563 6 9 4 - = .14063 i| = .46875 ff = .78125 A = -15625 || = .48438 || = .79688 1 _ 2 .5 || = .17188 i =.5 H = .8125 A - .1875 f| = .51563 || = .82813 || = .20313 II = .53125 || = .84375 5 _ 8 ~ .625 A = .21875 fl = .54688 || = .85938 If = .23438 T % = .5625 i =-875 3 _ 4 — .75 | =.25 f | = .57813 || = .89063 || = .26563 3-2 == .oyo/o ff = .90625 j2 r= .^Ol^O || = .60938 || = .92188 |f = .29688 f =.625 || = .9375 7 _ 8 .875 A = -3125 || = .3283 fi = .64063 f| = .95313 fl = .96875 f| = .98438 1 =1.00000 EXPLANATION OF TERMS The Point A point is that which has position but no dimension. An object having a position but no extension. A place having a specific position but no size. That is, a point has neither length, breadth, nor thickness, hence it has no dimension. 177 178 A HAXD BOOK FOR MECHANICS The Line The Straight or Right Line Straight means erect, tight, without bend, like a string, for instance, which is tightly stretched. A line is length without breadth. That is, a line is space of one dimen- sion (length). And a line which lies evenly between two extreme points is called a straight or right line. Xo matter what the position of the points so long as they are connected by a line without bend, that line is straight. Thus, the lines A B, CD, EF are all straight lines. Parrallel Lines Parallel means alike, similar to, having the same direc- tion or course. Hence parallel lines are lines which are alike, lines having the same direction or course and lying in the same plane as the lines A, B. Now if we draw another line, C, having the same EXPLANATION OF TERMS 179 direction and course as the lines A, B, it will be parallel to them, thus the three lines A, B, C, are all parallel lines. , . A B C The Vertical and Perpendicular Line A vertical and perpendicular line are one and the fame thing. The word vertical relates to the word vertea, which means the highest point or summit. A vertical line may be thought of as such a line as would be formed by a string or line which is suspended from some point overhead; such a line would be upright and would be perpendicular to the horizon. Thus the line A is vertical and is perpendicular to the line C D. The Diagonal and Oblique Line , Diagonal and oblique have practically the same mean- ing. Diagonal means across from angle to angle, oblique means aslant, or slanting. By diagonal line, then, is meant a slanting line. Thus 180 A HAND BOOK FOR MECHANICS The Horizontal Line By horizontal line is meant a line that is parallel to the horizon; such a line is neither vertical or inclined and is represented thus: ' Horizontal. The Curved Line A curved line is a line whose direction changes con- tinually along its path thus or The Convex Line A convex line is a curved, rounded, or arched line, which curves away from the line of view. Thus the curved part, C, of the figure A, C, B, is said to be convex, because it curves away from the ends, A and B, of the figure A, C, B. c A sphere, or circle, looked at from any point without, presents a convex surface or line. The Concave Line The concave line is the reverse of the convex line. Concave means incurved. Thus the curved part, E, of the figure C, E, K, is said to be concave because it curves in. EXPLANATION OF TERMS 181 A sphere, or circle, when looked at from any point within, presents a concave surface or line. Examples for Exercise What is the name of this line? Arts. Horizontal line. What is the name of that line which lies evenly between two points? Ans. Straight line. What are these lines called? IZZZZZIZ^IIZ Ans. Parallel lines. Ans. Vertical or perpen- dicular. Ans. Diagonal or oblique. Ans. Curved line. What is the name of these lines? What is the name of these lines? What is the name of this line? What is the difference between a convex and concave line? Ans. The convex is outcurved and the concave is incurved. MENSURATION Mensuration is the art of measuring length and volume, content, etc. For instance, the art of determining the length of a circle or the contents of a cylinder or other shaped vessels, and it is necessary that the pupil should become thoroughly familiar with, the names of the differ- ent shaped surfaces given in this chapter, and at any time be able to form a mental conception of same, be- cause mensuration treats of the measurement of such curved lines, surfaces, and solids. 182 A HAND BOOK FOR MECHANICS The Circle The circle is a plane figure formed by a curved line called the "circumference," (Fig. 1), and is such that all right lines drawn from a certain point within, called the " center," to the circumference, are equal to each other. A radius of a circle is any right line drawn from the center to the circumference, such as FlG - l C D or C A or C B. A diameter of a circle is a straight line drawn through the center and ending both ways at the circumference, as A B. To Find the Circumference of a Circle Rule. Multiply 3.1416 by the diameter. Example. What is the circumference of a circle whose diameter is 4 inches. Process: 3.1416 4 12.5664 inches. Arts. Example. What is the circumference of a circle whose diameter is 3J inches? L ROCESS '. rt -i a -t r* 3.1416 3.5 3+ = 3.5 157080 94248 10.99560 inches. Arts. (Almost 11 inches MENSURATION 183 Examples for Exercise 1. What is the circumference of a circle whose diameter is 4 inches? Arts. 12.5664. 2. What is the circumference of a circle whose diameter is 4.75? Ans. 14.922600. 3. What is the circumference of a circle whose diameter is 4f ? Ans. 14.922600. 4. What is the circumference of a circle whose diameter is 5.5? Ans. 17.27880. 5. What is the circumference of a circle whose diameter is 6.6? Ans. 20.73456. 6. What is the circumference of a circle whose diameter is 5.6? Ans. 17.59296. To Find the Area of a Circle By area is meant that which is confined within some specific surface space. For in- stance, the base, or site, on ^MIIIIIMII^k" ~~f" which a building stands. M\ . IK Thus, in the following exam- /[ pie, which is to find the area of a circle whose diameter is 3 v inches, the answer tells the \|i| iiijH \\W number of square inches con- xH i!i, : jB^ tained within a circle whose diameter is 3 inches, or the number of square inches that would be enclosed on a plane surface by a circle whose diameter is 3 inches. Example. Find the area of a circle whose diameter is 3 inches. (Fig. 2.) Rule. Multiply .7854 by the square of the diameter. Thus: 184 A HAND BOOK FOR MECHANICS 3 .7854 3 9 9 square of diameter 7.0686 square inches. Arts. Example. The diameter of a circle is 3.5, what is its area? 3.5 .7854 3.5 12.25 175 39270 105 15708 12.25 15708 7854 9.621150 square inches. Ans. Examples for Exercise 1. What is the area of a circle whose diameter is 5.5 inches? Ans. 23.758350. 2. What is the area of a circle whose diameter is 4 inches? Ans. 12.5664. 3. What is the area of a circle whose diameter is 3.75 inches? Ans. 11.04468750. 4. What is the area of a circle whose diameter is 2 inches? , Ans. 3.1416. 5. What is the area of a circle whose diameter is 6 J inches? Ans. 37.0910058750. 6. What is the area of a circle whose diameter is 4.6 inches? Ans. 16.619064. The Ellipse An ellipse may be called a flattened circle, the longest diameter of which is called the " transverse axis" (latus transversum) and the shortest is called the "conguatt axis." (Fig. 3.) MENSURATION 185 Fig. 3 To Find the Circumference of an Ellipse Rule. Multiply 3.1416 by half the sum of the two diameters. Example. What is the circumference of an ellipse whose diameters are 6 and 4 inches? 6 ^4 2)10 5 X 3.1416 = 15.7080 inches. Arts. Example. What is the circumference of an ellipse whose diameters are 3| and 7 J? 3i = 7i = 3.75 7.5 2 )11.25 5.625 3.1416 5.625 157080 62832 188496 157080 17.6715000 inches. Ans. To Find the Area of an Ellipse Rule. Multiply .7854 by the product of the two diameters. 186 A HAND BOOK FOR MECHANICS Example. What is the area of an ellipse whose diameters are 3| and 4j inches? (Fig. 4.) 3J = 3.5 .7854 4± = 4.5 15.75 175 39270 140 54978 Product 15.75 39270 7854 12.370050 square inches. Ans. What is the area of an ellipse whose diameters are 5 and 10 inches? * 5 .7854 10 50 50 39.2700 square inches. Ans. Examples for Exercise 1. What is the area of an ellipse whose diameters are 5f inches and 4J inches? Ans. 19.19321250. 2. What is the area of an ellipse whose diameters are 7 feet and 9 feet? Ans. 49.4802 square feet. 3. What is the area of an ellipse whose diameters are 3.5 feet and 7.5 feet? Ans. 20.616750. MENSURATION 187 4. What is the area of an ellipse whose diameters are 3.5 feet and 4.5 feet? Ans. 12.370050. 5. What is the area of an ellipse whose diameters are 5 inches and 10 inches? Ans. 39.2700 square inches. The Triangle A triangle is a figure formed by three right lines joined together end to end. The three lines are called its " sides." Thus a, b, c, are the sides of the triangle 1. (Fig. 5.) Fig. 5 Any three-cornered or three-sided figure body bounding a three-sided space is called a triangle. A triangle whose three sides are unequal is called a Scalene. (Triangle 1.) A triangle having two of its sides equal is called an " Isosceles Triangle." (Triangle 2.) And a triangle all of whose sides are equal is called an " Equilateral Triangle.'- (Triangle 3.) To Find the Area of a Triangle Rule. Multiply the base by half the perpendicular height. Example. What is the area of a triangle whose base is 4 feet and whose height is 4 feet? 188 A HAND BOOK FOR MECHANICS Half the height = 2 feet; thus 2x4 = 8 square feet. Example. Find the area of a triangle whose base is 12 inches and whose height is 8 inches. (Fig. 6.) Fig. 6 Half height = 4 inches; then 4 X 12 = 48 square inches Example. Find the area of a triangle whose base is 12 inches and whose height is 4 inches. (Fig. 7.) Fig. 7 Half the height = 2 inches; then 2 X 12 = 24 square inches. Example. Find the area of a triangle whose base is 15 feet and whose height is 9 feet. (Fig. 8.) Half the height = 4.5; then 4.5 X 15 feet. MENSURATION 189 15 ^5 75 60 67.5 square feet. Fig. 8 Examples for Exercise Find the area of triangles having the dimensions below. 1. Base 15 feet, height 4.5. Arts. 33.75 square feet. 2. Base 14.75 feet, height 61 feet. Arts. 46.09375 square feet. 3. Base 8 inches, height 4 inches. Ans. 16 inches. 4. Base 27.5 feet, height 5 feet. Ans. 68.75 square feet. The Square A square is an area bounded by four equal sides. (Fig. 9.) A figure all of whose sides are equal and all of whose angles are right angles. Fig. 9 190 A HAND BOOK FOR MECHANICS To Find the Area of a Square Rule. Multiply the base by the height, or the length by the breadth. Example. What is the area of a square whose base is 3 inches? (Fig. 9.) 3 3 9 square inches. Ans. Example. What is the area of a square whose base is 3.5 feet? 3.5 3.5 175 105 12.25 square feet. Examples for Exercise Find the area of the following squares: 1. Of a square whose base is 9 feet. Ans. 81 square feet. 2. Of a square whose base is 36 inches. Ans. 9 square feet. 3. Of a square whose base is 3 feet. Ans. 9 square feet. 4. What is the area of a square whose side is 2.5 feet? Ans. 6.25 square feet. The Oblong An oblong is a figure whose sides are perpendicular to its base and whose length is greater than its breadth. (Fig. 10.) MENSURATION 191 Fig. 10 To Find the Area of an Oblong Rule. Multiply the length by the breadth. Example. What is the area of an oblong whose length is 7 feet and whose height is 4 feet? (Fig. 10.) 7 = base 4 = height 28 square feet. Ans. Example. What is the area of an oblong whose base is 12 feet and height 4 feet? 12 _4 48 square feet. Ans. Examples for Exercise Find the area of the following oblongs: 1. Base 12 feet, height 6 feet. Ans. 72 square feet. 2. Base 5.5 feet, height 3.5 feet. Ans. 19.25 square feet. 3. Base 9f inches, height 5 inches. Ans. 48.75 square inches. 4. Base 15 feet, height 2.5 feet. Ans. 37.5 square feet. 192 A HAND BOOK FOR MECHANICS The Rhomboid A rhomboid is a four-sided figure whose opposite sides are parallel, whose dimensions are greater one way than another, but whose ends are not perpendicular to its base. To Find the Area of a Rhomboid Rule. Multiply the base by the perpendicular height. Example. Find the area of a rhomboid whose base is 9 feet and perpendicular height 3 feet. (Fig. 11.) Fig. 11 9 _3 27 square feet. Arts. Example. Find the area of a rhomboid whose base is 6 feet and whose height is 3^ feet. 6. 3J5 30 18 21.0 square feet. What is the area of a rhomboid whose base is 3 feet and whose perpendicular height is 4 feet? (Fig. 12.) MENSURATION 193 Fig. 12 3 _4 12 square feet. Arts. Example. Find the area of a rhomboid whose base is 2 feet and whose height is 5^ feet. (Fig. 13.) *—i Fig. 13 2. 10 10 11.0 square feet. Arts. 194 A HAND BOOK FOR MECHANICS The Trapezoid A trapezoid is a plain four-sided figure, having two of its opposite sides parallel and the other two not so. (Figs. 14 and 15.) Fig. 14 Fig. 15 To Find the Area of a Trapezoid Rule. Multiply half the sum of the two parallel sides by the distance between them. Example. What is the area of the following trapezoid? (Fig. 16.) U = I" J Fig. 16 Here, the lines 8" and 6" being the parallel sides, we multiply half their sum by the perpendicular distance between them, which is 3 inches. Thus: 8 _6 2)14 7 X 3 = 21 square inches. MENSURATION 195 Example. What is the area of a trapezoid whose parallel sides are 5.5 feet and 9.75 feet, and the distance between them is 2 feet? (Fig. 17.) Fig. 17 Thus; 9.75 5.5 2)15.25 7.625 X 2 = 15.250 square feet Examples for Exercise 1. What is the area of a trapezoid, the longer of the two parallel sides being 130 feet, the shorter 100 feet, and the height 80 feet? Ans. 9200 square feet. 2. What is the area of a trapezoid, the longer of the two parallel sides being 25 feet, and the shorter 23 feet, and the width being \\ feet? Ans. 30 square feet- The Trapezium A trapezium is any plane figure contained by four straight lines, no two of which are similar. (Figs. 18, 19, 20.) 196 A HAXD BOOK FOR MECHAMCS Figs. 18. 19. 20 To Fixd the Area of a Trapezium Rule. Divide the trapezium into two triangles, by connecting its opposite angles, making this line of division the base line of both triangles. Measure the perpendicu- lar height of each triangle from this base line. Then find the area of each triangle, add both together, and the sum is the area of the whole figure. Example. Find the area of the following trapezium. (Fig. 21.) Fig. 21 MENSURATION 197 Thus 2)4 2 X 12 =24 sq. ft., area of triangle C 2)3 1.5 X 12 = 18.0 sq. ft., area of triangle B 42.0 sq. ft. Ans. Here we first of all divide the figure in two by the dotted line (A); we then have two triangles, B and C, and this line (A) proves the base for both. We then find the area of each triangle separately, add them both together, and the sum 42.0 equals the area of the whole figure. Example. What is the area of the following trape- zium? (Fig. 22.) Thus 2)8 4X20= 80 2)4 2 x 20 = jtO 120 sq. ft. Ans. To Find the Surface of a Cylinder Rule. Multiply the diameter by 3.1416 and multiply the product by the height. 198 A HAND BOOK FOR MECHANICS Example. Find the number of square inches in sur- face of a cylinder of the following dimensions. (Fig. 23.) Fig. 23 Thus 4" = diameter And 10" = height Then 4" X 3.1416 = 12.5664 circumference 10 125.6640 sq. in. Arts. Example. What is the surface of a cylinder whose diameter is 9 inches and height 15 inches? 9 X 3.1416 = 28.2744 =• circumference Thus, 28.2744 X 15 = 424.1160, area of surface in square inches. MENSURATION 199 The Sphere A sphere is a body bounded by a curved surface, all parts of which are an equal distance from a certain point within, called the " center." The diameter of a sphere is a straight line drawn through its center, terminating both ways in the surface. The circumference of a sphere is the greatest distance around the sphere. (Fig. 24.) F v & ' Fig. 24 To Find the Surface of a Sphere Rule. Multiply 3.1416 by the square of the diameter. Example. What is the surface of a sphere whose diameter is 6 feet? Thus 6 _6 36 = sq. of diameter X 3.1416 = Answer. 3.1416 36 188496 94248 113.0976 area of surface in square feet. Arts. Example. What number of square inches of gold leaf will cover a sphere whose diameter is 18 inches? 18 18 144 18_ 324 square of diameter 3.1416 .324 125664 62832 94248 1017.8784 sq. in. Arts. 200 A HAND BOOK FOR MECHANICS The contents of a sphere are equal to the product of | of 3.1416 multiplied by cube of diameter. Example. How many cubic feet of gas will fill a spherical balloon whose diameter is 6 feet ? J of 3.1416 X6 3 = 113.097 cubic feet. 4ns. VOLUME MEASURE AND CONTENTS OF SOLIDS Rectangular Solids To Find the Contents of a Rectangular Solid A rectangular solid is a figure bounded by six rect- angles. (Fig. 25.) The dimensions of a rectangular figure are its length, breadth, and thickness, and the contents or volume of a rectan- gular figure is the space contained within its bounding surfaces. The volume or content of any rectangular is determined by the following rule: Rule. Multiply the length, breadth, and height together. Fig. 25 Example. What is the con- tent of a rectangular solid whose length is 3 feet, breadth 2 feet, height 2 feet? 3 2 6 _2 12 cubic feet. Ans. Example. How many cubic feet in a rectangular figure, each of whose sides measures 3 feet? VOLUME MEASURE AND CONTENTS OF SOLIDS 201 3 3 9 _3 27 cubic feet. Ans. Example. How many cubic feet of water will a vessel hold whose inside dimensions are, length 5 feet, breadth 4 feet, and depth 3 feet? 5 X 4 X 3 = 60 cubic feet of water The Cylinder A cylinder is a round body of a uniform diameter, whose bases are equal and parallel circles. (Fig. 26.) To Find the Cubic Contents of a Solid Cylinder Rule. Find the area of the base and multiply this by the height or length. Example. What are the cubic con- tents of a cylinder whose diameter is 2 feet and whose height or length is 7 feet? Thus 2 2 4 X .7854 is .7854 4 Fig. 26 3.1416 = area of base in sq. ft. 7 = height of cylinder 21.9912 cubic feet. Ans. 202 A HAND BOOK FOR MECHANICS Example. How many cubic feet of water will a cylindrical shaped vessel hold, whose inside dimensions are, diameter 4 feet, and height 1\ feet? 4 .7854 _4 16 16 47124 7854 12.5664 = number of sq. ft. in area of base 7.5 = height 628320 879648 94.24800 cubic feet. Arts. The Pyramid A pyramid is a body whose base is a polygon but whose sides are all triangles, meeting at one point (A), called the vertex of the pyramid. (Fig. 27.) Polygon means manycurves or many angles. A polygon, then, is a plane figure bounded by three, four, five, six, or any number of sides. Accordingly, the base of a pyramid may be triangular, square, pentagonal, etc., and pyramids are named for the figure or shape of their base. A pyramid having a three- curved or triangular base is called a trianglor pyramid, and one having a square base is called a square pyramid, Fig. 27 VOLUME MEASURE AND CONTENTS OF SOLIDS 203 and one having a five-sided or pentagonal base is called a pentagonal pyramid. To Find the Cubic Contents of a Pyramid Rule. Multiply the area of the base by one third of its perpendicular height (altitude). Example. How many cubic inches are there in a pyra- mid of the following dimensions: base 3 inches square, and height (altitude), 12 inches? 3 3 9 = square inches area of base 3)12 4 = one third of altitude 9 _4 36 cubic inches. Arts. Example. How many cubic inches are there in a pyramid whose base is a triangle, each of whose sides measures 3 inches and whose height is 15 inches? Thus: 3 15 = one half of perpendicular height 4.5 = area of triangle base 5 = one third of height of pyramid 22.5 = number of cubic inches. Ans. Process. We first find the area of the triangular base, according to the rule given for finding the area of a triangle, which we multiply by one third of the per- pendicular height of the pyramid, the product 22.5 thus obtained being the answer. 204 A HAND BOOK FOR MECHANICS The Cone A cone is a body whose base is a circle and whose convex surface tapers uniformly to a point, "A," called the vertex of the cone. (Fig. 28.) To Find the Cubic Contents of a Cone Rule. Multiply the area of the base by one third the perpendicular height. Example. Find the cubic inches in a cone whose base is 3 inches Fig. 28 diameter and whose height is 15 inches. .7854 9 7.0686 _5 35.3430 area in square inches of base. one third of perpendicular height of cone. cubic inches contained in cone. Example. How many cubic inches of water will a cone-shaped vessel hold, the diameter of whose base is 3^ inches and whose height is 12 inches? 3.5 .7854 3.5 12.25 175 " 39270 105 15708 12.25 15708 7854 9.621150 » 4 area of base. one third of perpendicular height. 38.484600 cubic inches of water. Arts. VOLUME MEASURE AND CONTENTS OF SOLIDS 205 The Frustum A frustum is that part of any solid figure which is between two planes, either parallel or inclined to each other. It is the part of a solid which remains after cutting off the top part by a plane parallel to the base. The "frustum" actually means, a piece; particularly, a remaining piece of something of which a part is lacking. A frustum of a cone, or pyramid, then, is the part which remains after cutting off the top by a plane parallel to the base. (Figs. 29 and 30.) Fig. 29 Fig. 30 To Find the Cubic Contents of a Frustum of a Cone Rule. Find the sum of the squares of the two diam- eters, add on to this the product of the two diameters, then multiply by .7854 and then multiply by one third the height. Example. Find the cubic contents of a frustum of a cone whose large diameter is 12 inches, small diameter 6 inches, and height 4 inches. 206 A HAND BOOK FOR MECHANICS 12 12 144 = square of large diameter 36 = sq. of small 36 [diameter. 180 = sum of squares of both diameters 72 = product of both diameters 252 .7854 12 6 72 = product of both diameters. 3)4 1.33 = one-third of thickness. 1008 1260 2016 1764 197.9208 1.33 5937624 5937624 1979208 263.234664 cubic inches. Ans. Example. How many cubic inches in the body shown in Figure 31? Fig. 31 VOLUME MEASURE AND CONTENTS OF SOLIDS 207 20 JO 400 = square of large diameter. 100 500 = sum of square of both 200 = product of both diamel 700 .7854 2800 3500 5600 4900 • 549.7800 2 = = one-third of thickness. 1099.5600 cubic inches. Ans. 10 10 100 = = sq. of small diameter. 20 10 200 = product of both diameters. 3)6 2 = one-third of thickness. To Find the Cubic Contents of a Frustum of a Pyramid Rule. The sum of the areas of the two bases added to the square root of their product, multiplied by one third of the altitude. Example. What are the contents of the frustum of a square pyramid whose height is 30 feet, and whose side at the base is 20 feet and at the top ten feet? (Fig. 32.) 208 A HAND BOOK FOR MECHANICS Fig. 32 Process: 20 X 20 = 400; 10 X 10 = 100; 400 X 100 V40000 = 200; 200 + 400 + 100 = 700; 30 -T- 3 = 10; 700 X 10 = 7000 cubic feet. = 40000; PART IV WEIGHT, SPECIFIC GRAVITY, HOW THE DIMEN- SIONS, MEASUREMENTS AND WEIGHT OF DIFFERENT SHAPED VESSELS ARE FOUND, AND HOW 7 THE WEIGHT OF DIFFERENT PARTS IS FOUND WEIGHT By weight, is meant the heaviness of a body, the downward force a body hasj which force is created by the action of a force called the " force of gravity," working upon all its particles. The " force of gravity" is that attraction which the earth exerts upon all bodies near it, tending to draw them toward its center. When this force is resisted it gives rise to pressure, which is called weight. Weight, there- fore, is due to and is the effect of resisted gravity. SPECIFIC GRAVITY In nature all substances have, under the same condi- tions, a weight " specific or peculiar" to themselves. This is due to the fact that like volumes of different materials contain variable amounts of matter. There is, for instance, more matter contained in a cubic inch of lead than in a like volume of wood, and the greater the density of the body or, in other words, the greater the amount of matter contained within any specific space, the greater the weight, for the reason that the action of gravity on all bodies is proportional to their volume or density. Therefore, under the same conditions, a cubic inch of lead will weigh more than a cubic inch of wood, and the comparative weights of equal volumes of different sub- stances are called their "specific" gravities; and the standard of reference is pure water at a temperature of 60° Fahrenheit. In other words, the " specific gravity" of a body is its 211 212 A HAND BOOK FOR MECHANICS relative weight, that is, the number of times it is " heavier" or "lighter" than a body of the same size of a different substance, and in order to determine the relative weights of equal sized bodies of different composition a standard is taken, which is pure water at the temperature of 60° Fahrenheit. There are different methods used for taking the specific gravities of bodies, one of which, called the "Hydrostatic Balance/' is here described. By this method a regular balance is used, having two scale pans. The body whose specific gravity is to be taken is placed on one of the scale pans, and on the other pan sufficient known weight is placed to exactly counterbalance it. This will give the weight of the body in the air. The body is then taken and suspended by a thin wire from the bottom of the scale pan and completely sub- merged in pure water. When thus submerged the body is counterbalanced by placing weights in the other scale pan and, because of the buoyancy of the water working on the body thus submerged, it will be seen to weigh less in water than in air. Now, in taking the specific gravity of solids, advantage is taken of the important fact that when a solid is wholly submerged in water it displaces a volume of that liquid exactly equal to its own volume, and the solid appears to lose its weight, that is, it is supported by the sur- rounding water with a force exactly equal to the weight of the water displaced; hence a body will weigh less in water than in air, and the difference of its weight in water from that of its weight in air must be the weight of an equal volume of water. For instance, if a piece of glass is found to weigh, in air, 577 grains, and when suspended by a fine wire from the bottom of the scale pan and immersed in a vessel of SPECIFIC GRAVITY 213 pure water it is found to weigh 399.4, the difference between the weight in air (577) and the weight in water (399.4) is the weight of the volume of water displaced by the glass. Therefore: 577.0 399.4 177.6 is the difference, which difference is the weight of the water displaced. Hence the rule for finding the specific gravities of solids: Weigh the solid in air and then in pure water (distilled water) and divide the weight of the body in air by the difference between the weights in air and water. Hence, if a piece of glass is found to weigh 1154 grains in air and 798.8 grains in water, the specific gravity of the glass will be determined by dividing 1154.0 by the difference between 1154.0 and 798.8. Thus: Ans. 1154.0 7S8.8 355.2 355.2)1154.0(3.248 10656 8840 7104 17360 14208 31520 28416 This shows us, then, that glass is 3.248, etc., times heavier than water. That is, a (cubic inch), for instance, of glass would weigh over three times as much as a cubic inch of water. Example. If a piece of marble weighs 48.0 grains in air and 31.0 grains in water, what is its specific gravity? 214 A HAND BOOK FOR MECHANICS 48.0 31.0 17.0 = difference between weight in air and water 17.0)48.0(2.823, etc. Arts. 340 1400 1360 400 340 600 510 90 The specific gravities of liquids are ascertained by an instrument called the " Hydrometer." A Table of Specific Gravities of Liquids and Solids NAME SPECIFIC GRAVITY Air 0.001228 Pure Water 1.0000 Sea Water , 1.029 Alcohol ." .79 Linseed Oil 0.9347 Olive Oil 0.9176 Solids Gold 19.3 Lead 11.4 Copper 8.767 Brass 8.384 Cast Iron (Average) 7 . 1 10 Wrought Iron (Average) 7 . 690 Steel 7.780 Tin 7.293 Zinc • 7.215 SPECIFIC GRAVITY 215 A cubic foot of pure water weighs 62.5 pounds. There- fore, to find the weight of anything contained in the above table, multiply 62.5 (the weight of a cubic foot of pure water) by the specific gravity of the given body. Example. What is the weight of a cubic foot of alcohol? .79 62.5 395 158 474 49.375 lbs. Arts. Example. What is the weight of 3 cubic feet of alcohol? .79 62.5 395 158 474 49.375 = weight of 1 cubic foot 3 148.125 = weight of 3 cubic feet Example. What is the weight of a cubic foot of cast iron? 7.110 62.5 35550 14220 42660 444.3750 lb. Ans. Example. What will 2 cubic feet of steel weigh? 216 A HAND BOOK FOR MECHANICS 7.780 62.5 38900 15560 46680 486.2500 lb. = weight of 1 cubic foot. 2 972.5000 lbs. Ans. Example. If one cubic foot of steel weighs 486 pounds, how many cubic feet are there in a ton (2000 pounds)? 486)2000 (4.115, etc., cubic feet of steel 1944 in a ton 560 486 740 •< 486 2540 2430 • 110 Example. How much does a cubic foot of wrought iron weigh? 62.5 7.690 56250 3750 4375 480.6250 lb. Ans. In practice 480 pounds is called the weight of a cubic foot of wrought iron. Example. How many cubic feet of wrought iron are there in a ton? SPECIFIC GRAVITY 480)2000(4.16 1920 800 480 217 3200 2880 3200 4.16 cubic feet = 1 ton. Ans. Example. What is the weight of a cubic inch of wrought iron? 1 cubic foot = 1728 cubic inches 480 pounds = weight of 1 cubic foot 1728)480.0(.27 of a pound. Ans. 3456 13440 12096 1344 Example. One cubic inch of brass weighs what ? 1728)521.75(.313. Ans. 5184 3350 1728 6220 Example. How many cubic inches of wrought iron weigh 1 pound? 480)1728(3.6 cubic inches = 1 lb. 1440 2880 2880 Example. How many cubic inches of fresh water weigh 1 pound? One cubic foot of fresh water weighs 62.5 pounds. 218 A HAND BOOK FOR MECHANICS 62.5)1728.0(27.656 1250 4780 4370 4100 3750 3500 3125 3750 3750 0000 27.656 cubic inches of fresh water = 1 pound. Ans. Example. How much will 2 cubic feet of lead weigh? 62.5 11.4 2500 625 625 712.50 = weight of 1 cubic foot '2 1425.00 = weight of 2 cubic feet From the above examples we learn how the weight of anything may be found when its volume and specific gravity are known. Further on examples will be given showing how to find the weight of things whose specific gravity is known, but whose dimensions are given only, by which its volume may be determined, but whose volume is not given. These examples are given so as to instruct the pupil in figuring the sizes of different parts of machinery, etc., and also to show how to figure the weights of the different parts. CIRCULAR MEASURE 219 CIRCULAR MEASURE The diameter of a circle is equal to the circumference divided by 3.1416. Example. What is the diameter of a circle whose circumference measures 12f inches? 12} = 12.75 Then 12.75 -f- 3.1416 is 3.1416)12.7500(4.058 125664 183600 157080 265200 251328 The diameter, then, of a circle whose circumference is 3J = 4.058 inches. Example. A piece of shafting measures 14^ inches around its circumference, — what is its diameter? Ans. 4.535, say 4^ inches. Example. What is the diameter of a shaft whose circumference measures 14J inches? Ans. 4.535. Example. What is the diameter of a shaft whose circum- ference measures 12.5664 inches around? Ans. 4 inches. Example. A tube is made of two plates hav- ing flanges A, B, as shown in Fig. 33. The outside diameter of tube is 3 feet and each flange measures 4 inches; what is the full length of each plate be- fore being bent? Fig. 33 220 A HAND BOOK FOR MECHANICS Reduce first the diameter (3 feet) to inches. Thus 3 X 12 = 36 inches Then 36 3.1416 216 36 144 36 108 113.0976 equals circumference in inches, to which add the 4 flanges, each of which is 4 inches long, equals 16 inches, then divide by 2, because there are two plates, and the quotient is the length of each plate. Thus: 113.0976 16 2 )129.0976 64.5488 inches = length of each plate About 5.379 feet. Ans. Example. The outside diameter of a tube which is made of three plates is 4 feet, each plate has a lap of 1 J inches, what is the full width of each plate? (Fig. 34.) If = 1.75 4 X 12 = 48 = diameter in inches 48. 3.1416 288 48 192 48 144 150.7968 = circumference in inches CIRCULAR MEASURE 221 Fig. 34 Now, as this tube is made of three equal sections, the whole circumference divided by 3 will equal the length of each section from A to B, and to this length must be added the length of the laps, If inches, and the sum will equal the whole width of each plate. Thus: 150.7968 + 1.75 = 52.0156 inches. Arts. Example. What is the full length in inches of a brass band, \ inch thick, which goes around a funnel whose diameter is 5i feet? The end of the band laps 3 inches. (Fig. 35.) Fig. 35 222 A HAND BOOK FOR MECHANICS 5J = 5.25 5.25 X 12 X 3.1416 = 197.920 circumference in inches of strap. Then 197.920 + 3 = 200.920 inches, or 16.743 feet, Example. If a circular plate, 5 feet in diameter, has to have some holes bored in it, 2\ inches from the edge of the plate to the center of the holes, what will be the diameter of the circle on which holes are bored? (Fig. 36.) Fig. 36 The circle on which the holes are bored is 2\ inches inside the outer diameter, of plate, and therefore five inches less in diameter. Diameter of plates = 5 feet = 60 inches Then 60 — 5 = 55 inches. Arts. l'IG. 61 Example. What is the full length around and across the base of an arched piece of metal like Figure 37. The arch equals \ circle. Divide circumference of CIRCULAR MEASURE 223 circle, whose radius is 3 inches (diameter 6 inches) by 2, to which add 6 inches (distance across base). Thus 3.1416 X 6 + 6 = 15.4248 inches. Ans. Example. What is the outside circumference of a ring whose outside diameter is 4 feet? And what is the inside circumference when the diam- eter is 2 J feet, and what width of ring? (Fig. 38.) Fig. 38 3.1416 X 4 = 12.5664 feet, outside circumference 3.1416 X 2.5 = 7.85400 feet, inside circumference 4 — 2.5 = 1.5 -T- 2 =.75 feet, width of ring. Ans. Example. The circumference of a flange is 60 inches and in it there are a number of holes 4 inches from the outside of flange to center of holes. What is the diameter of cir- cle on which holes are drilled? (Fig. 39.) The first thing we do is to find the p IG . 39 diameter of the flange, from which we subtract twice the distance from the outside of flange to center of holes. 224 A HAND BOOK FOR MECHANICS Thus 60 3.1416 4X2= Ans. 3.1416)60.0000(19.095 = diameter of flange 31.416 285840 282744 309600 282744 16856 19.095 - 8 = 11.095. Ans. Then 11.095 is the diameter of circle on which holes are drilled. Square Measurements Note: one dash after a dimension represents feet Thus 23' = 23 feet, and 6' = 6 feet And two dashes after a dimension represents inches. Thus 23" = 23 inches, and 6" = 6 inches Fig. 40 Example. Find the surface in square feet, of a sheet of metal of the following dimensions. (Fig. 40.) This may be done by adding the ends A and B to- gether and dividing by 2, which will give the mean width, which multiply by the length, 3', 6". Thus CIRCULAR MEASURE 2' 4" + 4" 32" 225 16 mean width Thus 16 X 3' 6" = 16 X 42" = 672 square inches, or 4.6 square feet, or it may be done by adding together the area of the triangle, A, E, C, and the area of the parallelogram, A, E, B, K. rpi *!.'*■ 1 2 ' X ^ 6 " 24 " X 42 " KiXA Ihus the triangle = , or = 504 square inches. And the parallelogram = 4" X 42" = 168 square inches 144 square inches == 1 square foot • Then 504 And 168 672 sq. in., or 4.6 sq. ft. Ans. 144)672(46. 5/o "960 864 96 Example. Find the measurement in square feet, of a sheet of metal of the following dimensions. (Fig. 41.) Answer: 10^ square feet. Fig. 41 226 A HAND BOOK FOR MECHANICS ^-2 = 10i Am. Example. What is the area of a sheet of metal of the following shape and dimensions. (Fig. 42.) Answer: 88.3575 square inches. Fig. 42 15" X 15" X .7854 = 88.3575 square inches. Ans. The area of the whole circle would equal 176.7150 square inches, therefore the plate being \ circle equals 88.3575 square inches. Example. What is the area cf a sheet of metal of the above shape and dimensions. (Fig. 43.) The circular end of plate equals \ circle. CIRCULAR MEASURE 227 Thus 5 25 X .7854 = 9.8125 sq. ft. Find now the area of the remainder of the sheet, which add to (9.8125) the area of the circular part. Thus 9.8125 + 70 = 79.8125 sq. ft. Ans. Example. Find the surface in square feet of a piece of plate of the following dimensions. (Fig. 44.) -36 Fig. 44 31.8087 *= Area of curved part. 288 = oblong part. 20.25 = triangular part. Thus 31.8087 + 288 + 20.25 = 340.058 square inches., or 2.361 square feet. Ans. Example. What is the number of square inches in the outside surface of a cylinder whose outside diameter is 5 inches and height 15 inches. (Fig. 45.) 228 A HAND BOOK FOR MECHANICS -5— Fig. 4o 3.1416 5 15.7080 = circumference 15 = height 785400 157080 235.6200 sq. in. Ans. Example. What is the number of square inches in the outside surface of an elliptical tube whose diameters are 2 and 4 inches and whose length is 12 inches? (Fig. 46.) Fig. 46 CIRCULAR MEASURE 229 3.1416 3 half sum of diameters 9.4248 = circumference 12 113.0976 sq. in. Ans. Circular Areas Example. Find the area of metal in a ring whose outer diameter is 13 inches and inner diameter 9 inches. (Fig. 47.) First find the area of the outer diameter. Next find the area of the inner diameter. Then their difference is the required area of metal. ± Fig. -.7 Thus 13 X 13 X .7854 = 132.7326, area of large diameter, and 9 X 9 X .7854 = 63.6174 , area of small diameter. Difference, 69.1152, area of metal. Example. What area of metal is there in a section 3 inches in diameter? (Fig. 48.) j— of a round bar of iron f 3 3 9 .7854 X 9 = 70.686 area _t_. Fig. 48 Example. What area of metal is there in a section of a tube whose inside diameter is 5 inches and the thickness of tube is \ inch? (Fig. 49.) 230 A HAND BOOK FOR MECHANICS In this case, as the inner diameter is 5 inches and the tube is J inch thick, the outside diameter must be 5" + i + 1 = 5.5. Jg THICK Fig. 49 Now find the difference between the two areas. 5 25 X .7854 = 19.6350 area of inside diameter. 5.5 5.5 275 275 30.25 X .7854 = 23.758350 area of outside diameter. 19.6350 4.123350 area. Ans. Example. The inner diameters of an elliptical tube are 3 feet and 4 feet, and the thickness of metal is J inch, what area of metal is there in a section of the tube? (Fig. 50.) As the inner diameters are 3 feet and 4 feet, that is, 36 inches and 48 inches, and the metal is J inch thick, the outside diameters must be CIRCULAR MEASURE 231 Conguatt axis (outside) \ + 36 + \ = 36^ Transverse axis (outside) } + 48 + \ = 48^ Inside diameters 3 feet And 4 feet 36.5 inches 48.5 inches 36 inches 48 inches Find the difference between the two areas and the answer is the area of metal in a section of the tube. Fig. 50 36.5 X 48.5 X .7854 36 X 48 X .7854 Arts. 1390.354 1356.971 Fig. 51 Example. If the outer diameter of a metal ring is 21 inches and the inner diameter is 19 inches, and the ring has 4 quarter-inch holes in it, what is the area ot the metal? (Fig. 51.) 232 A HAND BOOK FOR MECHANICS First find the area enclosed by the outer diameter, which is 346.3614 square inches. Next find the area enclosed by the inner diameter which is 283.5294 square inches. Then find their differ- ence. Thus: 346.3614 283.5294 62.8320 From which subtract the sum of the areas of the four ^-inch holes and the difference is the required area, \ = .25. .25 62.8320 .25 .1965000 125 62.6355 = are 50 .0625 .7854 2500 3125 5000 4375 .04908750 = area of \" hole 4 . 19635000 = area of 4 holes To Find the Diameter of a Circle when the Area is Given Rule. Divide area by .7854, then find the square of the quotient obtained, which will equal the diameter. Example. What is the diameter of a circle whose area is 190.0668 square inches? CIRCULAR MEASURE 233 .7854)190.0668(242 15708 32986 31416 15708 15708 2,42(15.5 1 25)142 125 305)1700 1525 17500 15.5 etc. inches. Ans Example. If the area of a shaft is 3.1416 square inches, what is its diameter? .7854)31416(4 4(2 inches. Ans. 31416 4 Example. If the area of a shaft is 6.2832 square inches, what is its diameter? .7854)62832(8 8(2.828 etc. 62832 4 48)400 384 562)1600 1124 5648)47600 45184 Example. If the area of a circular plate is 706.86 square inches, what is its diameter? Ans. 30 inches. Note. The diameter of a circle equals the square root of the area multiplied by 1.12838. However, the diam- 234 A HAND BOOK FOR MECHANICS eter is determined usually when the area is given by the process worked out in the example. Measurements and Weights of Tanks. Examples Showtng how to find the Cubic Capacity and Weight of Rectilinear and Circular Vessels and Tanks Note. The U. S. Standard gallon measures 231 cubic inches and contains 8| pounds of distilled water (pure water). A cubic foot of water (fresh) weighs 62 \ pounds and contains 1728 cubic inches, or nearly 1\ gallons U. S. standard. Fig. 52 Rule. To find the capacity of four-sided vessels in gallons, find the cubical contents by multiplying the length, breadth, and height in inches, and divide the product by 231. Example. If the inside dimensions of a tank are 4 feet wide, 3 feet deep, and 12 feet long, how many gallons of water will it hold? (Fig. 52.) 3 feet = 36 inches. 4 feet = 48 inches. 12 feet = 144 inches. CIRCULAR MEASURE 235 Then 36 X 48 X 144 = 248832 cubic inches. 231)248832(1077, etc., gallons. 231 1783 1617 1662 1617 Example. How many gallons will a tank of the following dimensions hold : 5 feet long, 4 feet deep, 9 feet wide? 5 feet = 60 inches. 4 feet = 48 inches. 9 feet = 108 inches. Then 60 X 48 X 108 = 311040 cubic inches. 231)311040(1346.49 gallons. 231 800 693 1074 924 1500 1386 1140 924 2160 2079 1346, etc., gallons. Ans. Example. How many gallons of oil will a tank 9 feet 3 inches long, 4 feet 5 inches wide, and 6 feet 2 inches deep, hold? 236 A HAND BOOK FOR MECHANICS 9 feet 3 inches = 111 inches. 4 feet 5 inches = 53 inches. 6 feet 2 inches = 74 inches. Then 111 X 53 X 74 = 435342 cubic inches. 231)435342(1884.597 gallons. 231 2043 1848 1954 1848 1062 924 1380 1155 2250 2079 1710 1617 93 1884, etc., gallons. Ans. Example. What weight of water is there in a tank of the dimensions of the preceding example? As 1 gallon of water weighs 8J pounds, 1884 gallons will weigh 1884 X 8J. 1884 8.3 5652 15072 15637.2 15637, etc., lbs. = weight. Ans. The best process, perhaps, for determining the weight of the contents of tanks is to multiply the cubic contents in feet by the weight of a cubic foot of the contents. CIRCULAR MEASURE 237 Example: What will be the, weight of fresh water in a tank, when full, of the following dimensions: 5 feet long, 4 feet wide, 9 feet deep? 5 X 4 X 9 = 180 cubic feet. Then as 1 cubic foot of fresh water weighs 62.5 pounds, 180 X 62.5 = weight of 180 cubic feet of fresh water. Thus 180 62.5 900 360 1080 11250.0 = weight. 11,250.0 1b. Ans. Example. A tank 5 feet long, 4 feet wide, and 9 feet deep, will hold how many pounds of sea water? One cubic foot of sea water = 64.3 lb. 5 X 4 X 9 = 180 cubic feet. Then 180 X 64.3 = 11.574.0 lb. weight. Example. What will a tank of linseed oil of the following dimensions weigh: 5 feet deep, 4 feet wide, and 9 feet long? w , . r 5X4X9 = 180 cubic feet. Specific gravity of linseed oil = . 9347 62.5 46735 18694 56082 Weight of cubic foot of oil = 58.41875 180 X 58.418 = 10,515.240 lb. Ans. Example. How many pounds of linseed oil will a tank 5 feet 2 inches by 4 feet 5 inches, and 2 feet 5 inches deep, hold? 238 A HAND BOOK FOR MECHANICS Reduce all dimensions to inches. We then have 62 X 53 X 29 = 95 . 294 cubic inches and 95.294 cubic inches = 55.14, etc., cubic feet. Then 55 . 14 etc. 58.418 weight of cubic foot of linseed oil 44112 5514 22056 44112 27570 3221 . 16852 lb. 3221 lb. Ans. Measurements of Circular Tanks Example. How many cubic feet of space in a cylin- drical tank whose diameter is 25 inches and height 12 inches? (Fig. 53.) This is the same as finding the volume of a cylinder. 25 diam. 25 125 .7854 50 625 625 diam. squared. 39270 15708 47124 490 . 8750 area of base. 12 inches high. 5890 . 5000 cubic inches in volume. Then 5890.5000 ^ 1728 = 3.408 cubic feet. Ans. Example. If the above cylinder is filled with linseed oil, what will be its weight? CIRCULAR MEASURE 239 1 cubic foot of linseed oil weighs 58.418 pounds. 58.418 3.408 467344 233672 175254 199.088544 199 lb. Ans. Fig. 53 How many gallons will the above cylinder hold? (Fig. 53.) To find the Capacity of Cylindrical Vessels in Gallons. Rule. Multiply the area in inches by the height in inches and divide product by 231. 25 X 25 X .7584 X 12 = 5890.5000 area in inches. 231)5890.5000(25.5 gallons. Ans. 462 1270 1155 1155 1155 240 A HAND BOOK FOR MECHANICS Example. How many cubic feet, and what will be the weight of, and how many gallons of water will a tank of the following dimensions hold: diameter 43 inches, and height 24 inches? 43 X 43 X .7854 X 24 = 34852.9104 cubic inches. Then 34852.9104 - 1728 - 20.111 cubic feet. Weight equals 20.111 X 62.5 = 1256.937 lb. Now we can find the number of gallons contained in the tank by either dividing 34852.9104 cubic inches, which is its contents in cubic inches, by 231, because there are 231 cubic inches in a gallon. Thus 34852.9104 -- 231 = 150.878 gallons. (Almost 151 gallons.) Or we can find the number of gallons contained in the tank by multiplying its cubic contents in feet (20.111) by 1\, because there are 1\ gallons in one cubic foot. Thus 20.111 7.5 100555 140777 150.8325 gallons. Arts. Example. How many gallons of water will a tank having the following dimensions hold: base diameter 5 feet, top diameter 3 feet, and height, 4 feet? (Fig. 54.) This is the first example of a frustum-shaped tank we have had, and the process for determining the cubic contents of a frustum was explained in Mensuration. First find cubic contents in inches, then reduce cubic contents in inches to cubic feet, and multiply by 7.5; product will be required answer. CIRCULAR MEASURE 241 36 X 36 X .7854 = 1017.8784 area in inches of top diameter. And 60 X 60 X .7854 = 2827.4400 area in inches of bottom diameter. 3845.3184 = sum of areas of both bases. 16 = i of perpendicular height in inches. 230719104 38453184 61525.0944 = area in cubic inches. 61525.0944 ^ 1728 = 35.604 cubic feet. Ans. Fig. 54 Example. How many gallons will the above tank hold? 35.604 X 7.5 = 267.03 gallons. 242 A HAND BOOK FOR MECHANICS Examples Explaining how to Calculate the Weight of Different Materials and the Weight of the Different Parts of a Machine Examples coming under this head will now offer very little difficulty in solving. To find the weight of any body, we proceed first by finding its area, and having obtained the area the weight is found by multiplying the weight of a cubic foot of the material by the area in feet. Or, if the area is given in inches, the weight will be found by multiplying the area in inches by the weight of a cubic inch of the material. Example. What is the weight of a cast-iron plate of the following dimensions? 4' wide 6" thick and 8' long. (Fig. 55.) Fig. 55 4 feet = 48 inches. 6 inches = 6 inches. 8 feet = 96 inches. Then 48 X 6 X 96 = 27648 cubic inches area The specific gravity of cast iron being 7.110 the weight, of a cubic foot of cast iron will equal 7.110 X 62.5 = 444.3750 1b. CIRCULAR MEASURE 243 And since a cubic foot weighs 444.3750 pounds, a cubic inch will equal 444.3750 + 1728 = .257 lb. Then, since one cubic inch equals .257, the weight of 27648 cubic inches is 27648 X .257 = 7105.536 lb. Answer: About 7105^ pounds will be the weight of the plate. Example. What is the weight of a cast-iron plate of the following dimensions? (Fig. 56.) Fig. 56 3 feet 6 inches = 42 inches. 5 feet = 60 inches. 4 feet = 48 inches. First find the area in cubic inches of whole piece. Then find the area of the whole, and subtract the latter from the former, and the answer will be the area in cubic inches of iron in the plate. Then multiply the area thus obtained by .257 (the weight of a cubic inch of cast iron) and the product will equal the weight. Thus: 244 A HAND BOOK FOR MECHANICS 60" X 48" X 42" = 120. 960 cu. inches area of plate. 24" X 24" X .7854 X 42" = 19.000 cu. inches area of hole. Difference 101.960 number of cu. inches. in plate. .257 weight of cu. inches of 713720 CaSt ir ° n ' 509800 203920 26103.720 weight of plate. 26.103 pounds equals weight of plate. Arts. Example. What will be the weight of a cast-iron column 20 feet long, outside diameter 14 inches, and inside diameter 8 inches? (Fig. 57.) 63971 lb. Ans. Ti I *- 1/ Fig. 57 14" X 14" X .7854 X 240" = 36945 etc., area in in. of out- side diameter. 8" X 8" X .7854 X 240" = 12063 etc., area in inches of inside diameter. Difference 24882 = number of cu. inches of iron. Then 24882 X .257 = 6394.674 lb. About 6397^ pounds weight of column. Ans. Example. What is the weight of a wrought-iron CIRCULAR MEASURE 245 shaft having two flanges, one of which is 3f inches thick and 1 foot 5 inches diameter, the other 2\ inches thick and 1 foot 4^ inches diameter, the shaft is 5^ inches diameter and 14 feet 2 inches over all? (Fig. 58.) 1475.327544 etc. lb. Ans. 17 X 17 X .7854 X 3.75 = 851.1772 area of flange 1. 16.5 X 16.5 X .7854 X 2.5 = 534.5628 area of flange 2. 170" - 3. 75" - 2.5" = 163.75" (163|") length of shaft. 5.5 X 5.5 X .7854 X 163.75 = area of shaft part. 5.5 5.5 8511772 275 5345628 275 38904298 3025 5276 . 1698 = whole area in inches. .7854 ^28 = weight of cubic inch of 12100 420093584 wrought iron. 15125 105523396 24200 1475.327544 lb. = weight. 21175 23.758350 163.75 118791750 166308450 71275050 142550100 23758350 3890.42981250 = area in inches of shaft part. In the above example we first find the area of flange 1 then the area of flange 2. We then find the area of the whole shaft minus the two flanges. The three areas are then added together and the sum thus found is the whole area of shaft, which multiplied by the weight of a cubic 246 A HAND BOOK FOR MECHANICS inch of wrought iron will give the whole weight of the shaft. i r> 1 ' bal X > f-m- m.i. Fig. 58 The weight of a cubic inch of wrought iron is obtained by multiplying 7.690, the specific gravity of wrought iron, by 62.5, the weight of a cubic foot of water (fresh). The product is the weight of a cubic foot of wrought icon, which, divided by 1728, gives the weight of a cubic inch, .285. Example. A circular brass plate is 6 feet 5 inches in diameter, 4 inches thick, and has five 4-inch holes in it. (Fig. 59.) What will it weigh? 5512.565 lb. Ans. 77 X 77 X .7854 X 4 = 18626.5464 area of whole plate in cubic inches. 4 X 4 X .7854 X 4 = ' 50.2656 area of one hole in cubic inches. 251.3280 = area of five holes in cubic inches. CIRCULAR MEASURE 247 Then 18626.5464 - 251.3280 = 18375.2184 etc., whole area minus area of holes. Fig. 59 Specific gravity of brass equals 8.384. Then brass per cubic foot equals 8.384 X 62.5 = 524 lb. Hence the weight per cubic inch is 524 -v- 1728 = .3 lb. Then the weight of 18375.2184 cubic inches is 18375.2184 X .3 = 5512.565 lb. Arts. Example. What is the weight of a brass cylinder whose inside diameter is 10 inches, the brass being \ inch thick and the cylinder 3 feet 5 inches long? (Fig. 60.) 202.8 etc. lb. - Arts. 10 X 10 X .7854 X 41 = 3220.1400 area of inside diam. 1 +;. 10 + J = 11 outside diameter. Then 11 X 11 X .7854 X 41 = 3896.3694 area of out- side diameter. Then area of outside diameter minus area of inside diameter equals area of metal. Thus 3896.3694 - 3220.1400 = 676, etc. 248 A HAXD BOOK FOR MECHANICS Then as one cubic inch of brass weighs .3 of a pound, 676 cubic inches weighs 676 .3 202.8 lb. Arts. Fig. GO Example. What is the weight of a brass cylinder whose length is 45 inches, outer diameter 12 inches, and thickness f inch? (Fig. 61.) 357.84 lb. Ans. mzzzzzzzzzzzzzzzm '////////////////7m 45 ^ Fig. 61 CIRCULAR MEASURE 249 Inner diameter equals 12 — f — f = 10.5 12 X 12 X .7854 X 45 = 5089.39 cu. in. of metal in out- side diameter. 10.5 X 10.5 X .7854 X 45 = 3896.56 eu. in. of metal in inside diameter. Then 5089.39 - 3896.56 = 1192.83 cubic inches of metal. Hence 1192.83 X .3 - 357.84 lb. weight. Example. Find the weight of the rim of a cast-iron fly-wheel whose outer diameter is 9 feet 6 inches, and inner diameter 9 feet 3 inches, and 12 inches wide. (A cubic inch of cast iron equals .257 of a pound.) (Fig. 62.) 1634.9 lb. Ans. Fig. G2 114 X 114 X .7854 X 12 = 122484.700 cu. in. of metal in outside diam. Ill X 111 X .7854 X 12 - 116122.960 cu. in. of metal inside diam. Then 122484.700 minus 116122.960 equals 6361.740 cubic inches of iron in rim. Hence 6361.740 X .257 = 1634.9 etc. lb., weight. Ans. 250 A HAND BOOK FOR MECHANICS Example. Find the weight of the rim of a fly-wheel whose outer diameter is 10 feet 9 inches, and inner diameter 8 feet 2 inches,- and width S inches. 11363.2 etc. lb. Ans. Example. What is the weight of a cast-iron cap of the following dimensions, if a cubic inch of cast iron weighs .257 of a pound? (Fig. 63.) First find the weight of the flanges A and B. then find the weight of cap, and the sum of both weights equals whole weight. Flanges. Outside Dimension. 6 X 2 X 36 X 2 2 12 36 20 X 9 X 36 9 180 36 72 36 10S0 540 432 2 = flanges. 6480 4032 864 area in eu. in of flanges. 2448 area in eu. in. cap, Inside Dimensions. 16 X 7 X 36 7 112 36 672 336 4032 Then 2448 + 864 X .257 = weight. S64 3312 .257 231S4 16560 6624 851 . 184 lb. Ans. (About 852 lbs.) Note. — Width of flange from A to C = 6" TABLE OF WEIGHTS 251 k ft" fr . /fe" j^g Fig. 63 TABLE OF WEIGHTS (The approximate weight of a cubic inch of different metals) 1 cubic inch of zinc weighs .252 lb. 1 cubic inch cast iron weighs .26 lb. 1 cubic inch wrought iron weighs .28 lb. 1 cubic inch of steel weighs .288 lb. 1 cubic inch brass weighs .3 lb. 1 cubic inch copper weighs .32 lb. 1 cubic inch lead weighs .41 lb. 1 cubic inch tin weighs .262 lb. (The approximate weight of a cubic foot of different metals) 1 cubic foot zinc weighs 428 lb. 13 oz. 1 cubic foot cast iron weighs 450 lb. 7 oz. 1 cubic foot wrought iron weighs 483 lb. 6 oz. 1 cubic foot steel weighs 487 lb. 12 oz. 1 cubic foot brass weighs 543 lb. 12 oz. 1 cubic foot copper weighs 547 lb. 4 oz. 1 cubic foot lead weighs 709 lb. 8 oz. 1 cubic foot tin weighs 455 lb. 11 oz. 252 A HAND BOOK FOR MECHANICS Weight in pounds of a square foot of different metals, in thickness varying by 1-16 of an inch. Name of Metals 5C go 43 co o C C .2 a 43 — +3 43 . c £ c c 1— 1 GO o3 CO +3 m CO a a o Q CO a GO GO 03 pq "3 CO a o T3 CO *1 i 1 6 2.3 2.3 2.5 2.9 2.3 2.6 2.7 2.4 3.7 1 5.0 4.7 5.1 5.8 4.7 5.3 5.5 4.8 7.4 3 1 6 7.5 7.0 7.6 8.7 7.0 8.2 8.2 7.2 11.2 1 10.0 9.4 10.2 11.6 9.4 11.0 10.9 9.6 14.9 5 1 6 12.5 11.7 12.8 14.5 11.7 13.7 13.7 12.0 18.6 3 8 15.0 14.1 15.3 17.2 14.0 16.4 16.4 14.4 22.3 T7T 17.5 16.4 17.9 20.0 16.4 19.2 19.1 16.8 26.0 1 2 20.0 18.7 20.4 22.9 18.7 21.9 21.9 19.3 29.7 9 1 6 22.5 21.1 25.0 25.7 21.1 24.6 24.6 21.7 33.4 5 g 25.0 23.5 25.5 28.6 23.4 27.4 27.3 24.1 37.1 11 1 6 27.5 25.8 28.1 31.4 25.7 30.1 30.0 26.5 40.9 3 4 30.0 28.1 30.6 34.3 28.1 32.9 32.8 28.9 44.6 13 1 6 32.5 30.5 33.2 37.2 30.4 35.6 35.0 31.3 48.3 "8 35.0 32.8 35.7 40.0 32.8 38.3 38.2 33.7 52.0 1 5 ] 6 37.5 35.2 38.3 42.9 35.1 41.2 41.0 36.1 55.7 1 40.0 37.5 40.8 45.8 37.5 43.9 43.7 38.5 59.4 Note. — The weight per square foot to any gage can easily be obtained from, the above table by multiplying the weight of a square foot of the metal ONE inch thick by the thickness of the gage in inches or parts of an inch. PART V THE PRIMARY OR SIMPLE MACHINES THE MECHANICAL POWERS Although the combinations of motions are many, they are all related to, and spring from, only two primary principles, namely, the principle of the " Lever" and the principle of the " Inclined Plane." The lever is the fundamental base of all circular or angular action, that is to say, the lever is the primary element of all action or motion which may be about an axis or center, and the inclined plane is the fundamental base of all rectilinear action, that is to say, the inclined plane is the primary element of all straight-line action or motion. The primary, or simple machines, sometimes called the mechanical powers, are seven in number, called: 1. The lever. 2. The inclined plane. 3. The cord. 4. The pulley. 5. The wheel and axle. 6. The wedge. 7. The screw. The first three, that is, the lever, the inclined plane, and the cord, are simple elements, that is, each is an individual element within itself, capable of performing some function. And the other ones are made up of these three; that is, the pulley, the wheel and axle, the wedge, and the screw are combinations of the first three powers, and as the law of each one of the powers is explained in detail, the relation of the "pulley" and the " wheel" 255 256 A HAND BOOK FOR MECHANICS and "axle" to the lever, and the relation of the wedge and screw to the inclined plane, will be clearly seen. THE SIMPLE MACHINES The Lever By lever is meant a stiff bar of any shape, either straight or bent, which is free to turn about a fixed point called the fulcrum (which means a stationary prop or support). When acted upon by two forces in opposite directions, one force acts as a power and the other as a load. Therefore, in the law of the lever three points must be considered: First. The fulcrum or point, about which the bar turns. Second. The point where the power is applied. Third. The point where the load, resistance, or weight is applied. Levers are divided into three classes. In the first class the fulcrum (F) is between the power (P) and the weight (W). (Fig. 64.) Fig. 64 P = Power. F = Fulcrum. W = Weight, in all figures. THE SIMPLE MACHINES 257 In the second class the weight is between the power and the fulcrum. (Fig. 65.) u) — 1 Fig. 65 In the third class the power is between the fulcrum and the weight. (Fig. 66.) D) ii 'I il u) Fig. 66 In all three classes, the law of the lever is the same. To Find the Power Rule. Multiply the weight by its distance from the fulcrum and divide by the distance of the power from the fulcrum. Example. How much power, acting at a point on a lever arm, 16 inches from the fulcrum, will balance a weight of 5 pounds located at a point 8 inches from the fulcrum? (Fig. 67.) 258 A HAND BOOK FOR MECHANICS 8X5 40 2J lb. pressure. 16 16 That is, the pressure exerted by a weight of 2\ pounds 16 inches from the fulcrum, would exactly counterbalance a weight of 5 pounds 8 inches from the fulcrum. tp -— tf*-ifc 1 -/i Fig. 67 Example. How much power will have to be exerted on a crowbar to raise a stone which weighs 200 pounds, if the crowbar is 5 feet long and the fulcrum is placed at a point 18 inches from the end nearest the weight? (Fig. 68.) Fig. 68 Here the leverages are 18 and 42 200 X 18 3600 42 42 = 85f lb. THE SIMPLE MACHINES 259 That is, a pressure power, or force, of 85f pounds would with the above lever, counterbalance - a weight of 200 pounds. Example. How much power, or force, will be exerted to raise a block of stone which weighs 50 pounds, by a lever of the second order, if the stone is placed 9 inches from the fulcrum and the lever measures 5 feet? (Fig. 69.) 50 w -5' Fig. 69 -H 51 X 9 ' 459 60 60 7-i^ lb ' 20 1U - Ans. This answer will appear confusing to some. For such, it will be clearly understood, if they will think of the 50-pound weight as being placed at the point B of the lever, then to raise the weight from that point it would take a force or power equal to 50 pounds. Now remove the weight from the point B to a point C, which is located in the middle of the lever, and a force, or power, of 25 pounds exerted at the point B would be sufficient to raise it. From this the following may be deduced. The nearer the weight is to the point A, or fulcrum, the less the power required at the point B to raise it. Example. What power, exerted at a point located 3 inches from the fulcrum, will balance a weight of 120 pounds placed 3 feet from the fulcrum? 260 A HAND BOOK FOR MECHANICS This is a lever of the third class. (Fig. 70.) 120 X 36 Fig. 70 4320 ~3~ 1440 lb. pressure. To Find the Weight Rule. Multiply the power by its distance from the fulcrum and divide by the distance of the weight from the fulcrum. Example. What weight located 36 inches from the fulcrum will be counterbalanced by a force of 50 pounds acting 6 inches from the fulcrum? (Fig. 71.) 6 X 50 300 36 36 = 84- Ans. Then a power pressure, or weight of 8 J pounds placed at a distance of 36 inches from the fulcrum will balance a weight of 50 pounds placed 6 inches from the ful- crum. THE SIMPLE MACHINES 261 61 zv 36"- 50 ~V \0 Fig. 71 Example. Let A, B, C, be a bent lever with a force equal to 40 pounds applied at A on the long arm, the length A, B, is 16 inches, and the length of the short arm, B, C, is 4 inches. What weight at C can be lifted? (Fig. 72.) Fro. 72 262 A HAND BOOK FOR MECHANICS 40 X 16 640 = 160 lb. Am. Example. Let A, B, C, be a bent lever with a force equal to 80 pounds applied at A on the short arm, the length A, B, is 12 inches, and the length B, C, is 96 inches. What weight at C can be lifted? (Fig. 73.) 80 X 12 960 96 = 9b" 10 lb. Arts. Fig. 73 Example. If a force of 112 pounds is applied 12 inches from the fulcrum, what weight at A, which is 36 inches from the fulcrum, can be lifted? (Fig. 74.) Fig. 74 THE SIMPLE MACHINES 112 X 12 1344 263 36 36 = 37+ lb. Am. To Find the Distance of the Power from the Fulcrum Rule. Multiply the weight by its distance from the fulcrum and divide by the power. Example. How far from the fulcrum will a power of 50 pounds have to be placed to raise a weight of 300 pounds, which is 9 inches, from the fulcrum? (Fig. 75.) Fig. 75 300 X 9 2700 50 50 = 54 inches. Arts. Example. Let A, B, C, be a bent lever with a power or force equal to 40 pounds applied at A. How long will A, B, have to be to lift a weight of 150 pounds if A, C, equals 7 inches? (Fig. 76.) Fig. 76 264 A HAND BOOK FOR MECHANICS 150 X 7 1050 40 40 26| inches. Ans. To Find the Distance of the Weight from the Fulcrum Rule. Multiply the power by its distance from the fulcrum and divide by the weight. Example. How far from the fulcrum will a weight of 500 pounds balance a power of 300 pounds placed 72 inches from the fulcrum? (Fig. 77.) Fig. 77 72 X 300 21600 500 500 = 43.2 inches. Ans. Weights Between Two Supports If a weight is attached to a beam which rests upon two supports, the beam acts as a lever of the second class, and the part of the whole weight carried by either sup- port may be found by considering one support as the power and the other as the fulcrum. If the weight rests in the middle of the beam it is obvious that each support will carry half the burden. But if, as shown in Fig. 78, the load or weight is placed a distance equal to one THE SIMPLE MACHINES 265 quarter the length of the beam from A, the support A will bear three quarters the weight and the support B one quarter. 3 Fig. 78 Example. A beam 16 feet long is resting on two supports, A and B, there is a weight of 4 tons hanging 5 feet from A. What share of this weight will be sup- ported by A and B respectively? (Fig. 79.) F-G. 79 Support A holds 2f tons. Support B holds 1^ tons. Arts. By the principle of the lever, the shorter arm must bear the greater weight and the longer arm the smaller weight. 266 A HAND BOOK FOR MECHANICS This fact was clearly demonstrated by Fig. 69. The following figure will clearly explain the solving of all such problems as the above. Consider A, B, Fig. 80, a lever of the second class, 16 feet long, with a weight of 4 tons placed 5 -feet from the fulcrum. How much power, or what pressure ex- erted at B, will exactly counterbalance it? jr*-s Fig. 80 5X4 20 iii a = — = 11 tons. Arts. 16 16 4 That is to say, a weight of 1^ tons arranged over a pulley C, as shown in figure, would exactly counterbalance a weight of 4 tons placed as in the above. Now place a support D under the lever at B, take away weight E letting lever rest on support D, it is obvious then that support D will be resisting a pressure of 1 J tons, or that D will be supporting a weight of 1\ tons. And as according to the law of the lever the shorter arm (in this case the distance from the fulcrum to weight) bears the greater weight, it is very clearly seen that the share of the whole weight supported at A must be 2f tons, because 4 — 1J ■= 2f. Example. A beam 19 feet long and weighing 700 pounds is resting on two supports A and B, there is a THE SIMPLE MACHINES 267 weight of 4 tons hanging 6 feet from A and another weight of 8 tons hanging 6 feet from B. What share of these weights and beam will be supported at A and B ? (Fig. 81.) Fig. 81 First find the share each support bears of weight 1. weight of weight 1, supported 6X8 48 _ = — = 2.63 tons 19 19 at A. Then as A supports 2.63 tons of the weight 1, B must support 5.37 tons of the weight 1. In the same manner find the share each support bears of weight 2. 6X4 24 = — =. 1.27 tons = weight of weight 2 at B. 19 19 s s Then as B supports 1.27 tons of weight 2, A must sup- port 2.73 tons of weight 2. And each support bears also half the weight of the beam equals 350 pounds. Therefore, A's share = 2.63 + 2.73 + 350 = and B's share = 5.37 + 1.27 + 350 = Therefore A's share equals 2.63 + 2.73 + 350; A's share in pounds equals 5260 + 5460 + 350 = 11,070 lb., about 5 tons, etc. And B's share equals 5.37 -f 1.27 + 350; B's 268 A HAXD BOOK FOR MECHANICS share in pounds equals 10,740 + 2540 + 350 = 13.630 = 6 tons 10 cwt. 20 lb. = about 6£ tons. The Compound Lever By compound lever is meant a combination of levers. Any system of two or more levers acting upon each other is called a compound lever. By the use of compound levers a very small force applied will sustain a great weight. And compound levers are used when and where it is inconvenient to use a single lever having a long arm. Figure 82 shows a system of levers called a compound lever. A \&i — // * 5@7 •J3 Fig. 32 With such a system of levers as the above, where the short arm of one works on the long arm of the other lever, a very small power applied at B would lift or bal- ance a great weight at A. To Fixd the Power To find the power exerted at B in a combination of levers arranged as in Fig. 82. to balance a weight of 440 pounds suspended from A. Rule. Multiply the weight by continued product of short arms and divide by continued product of long arms. Thus in the above Fig. 82, the short arms are 3 inches, THE SIMPLE MACHINES 269 3 inches, and 2 inches, and the long arms are 11 inches, 9 inches, and 8 inches. Therefore the power required at B to balance a load of 440 pounds at A is 440 X 3 X 3 X 2 7920 in A = = 10. Arts. 11X9X8 792 That is, a power of 10 pounds exerted at B will balance a weight of 440 pounds suspended from A, or 10 pounds suspended from B will balance a weight of 440 pounds sus- pended from A. To Find the Weight Rule. Multiply the continued product of the long arms by the power, and divide by the continued product of the short arms. Example. If there be such a combination of levers as represented in Fig. 82, with long arms of 11 inches, 9 inches, and 13 inches, and short arms of 2 inches, 3 inches, and 2 inches, what weight will be balanced by a power of 10 pounds? w • U4r 10 X 11 X 9 X 13 12870 in7Q - „ Weight = == = 1072.5 lb. 2X3X2 12 Then a pressure or power of 10 pounds exerted at A would balance a weight of 1072| pounds suspended from B. The Pulley The pulley is a wheel over which a cord, chain, or band is passed in order to transmit the force applied to the cord, chain, or band in another direction. The pulley is really a combination of the cord and a wheel, or of a cord and a number of wheels, or of a number of cords and a number of wheels. The pulley or wheel is introduced only to reduce fric- 270 A HAND BOOK FOR MECHANICS tion, and the usefulness of the pulley is dependent wholly upon the cord. Pulleys are said to be fixed or movable according as their blocks are fixed or movable. Figure S3 shows a fixed pulley. In this pulley the block A is fixed or stationary, and the wheel C, D. turns within it. O O Fig. S3 In Fig. 84 is shown a single movable pulley. In this pulley the block A is movable and the wheel C turns within it. Fig. 84 THE SIMPLE MACHINES 271 There is no power or mechanical advantages gained by a single rope acting over one or more single fixed pulleys, but there is often great advantage gained by their use, because, as the force applied to the cord is transmitted in another direction, a fixed pulley or com- bination of fixed pulleys enables us to change the direc- tion of the force. Thus it is easier for a man by the use of a fixed pulley to hoist a weight to a loft than it would be for him to carry the weight upwards over a flight of stairs. Work Done by the Pulley The pulley may be considered as a continuous series of levers with equal arms on one fulcrum or axis. Figure 85 represents a fixed pulley which corresponds t u * A \ --C 6 6 Fig. 85 to a lever of the first class, the center line A, B, C, repre- senting the elements of the lever in which B, the center, may be looked upon as being the fulcrum, and A, B, and A, C, the lever arms. It is here obvious that the single fixed pulley changes the direction of the force only, without modifying the 272 A HAND BOOK FOR MECHANICS intensity of the power. That is, a weight of 100 pounds suspended from A would exactly counterbalance a weight of 100 pounds suspended from C, because A, B, and B, C, are equal, but a force applied to the cord at A, causing it to move in a downward direction, would cause the cord at C to move in an upward direction. Combination of Pulleys Movable pulleys are generally used in combination with fixed pulleys. Fig. 86 shows a combination of one fixed pulley with one movable pulley. Fig. 86 The figure shows a cord attached to a beam at A, then led around a single movable pulley, and then up over a single fixed pulley; the free end B is where the power is applied, and attached to the movable pulley is a weight = W. In such an arrangement it is evident that the weight W is supported equally by the two parts C, D, of the cord which passes around the movable pulley, that is to say, the whole weight is supported by the cord and each of the two singles C and D of the cord bears one half the weight ; and since no power is gained by the fixed pulley K in this combination it is seen the force at B required to balance the weight W will be half the weight. That is, THE SIMPLE MACHINES 273 50 pounds suspended from B will balance a weight of 1 00 pounds suspended from the movable pulley, because each part of the cord C and D (called each single of the rope) supports one half the weight. In the combinations of pulleys in common use, several fixed pulleys are con- tained in one block, and in the other block are an equal number of movable ones. Figure 87 shows a combination of this kind having two fixed pulleys in the upper block and two movable pulleys in the lower one. In all such combinations of pulleys, one continuous cord passes through the system. The weight here when the lift begins is supported by 4 singles of the rope, and therefore the power required to balance the weight in this combination will be one fourth of the weight. That is, in this system the tension of the weight is equally distributed among the four parts of the cord which sustains the weight, hence a power applied at P will coun- terbalance four times its weight at W. From this, then, the following may be deduced: To Find the Power Rule. Divide the weight by the number of singles of rope. Fig. 87 Example. A piece of iron weighing 9 tons is lifted by a pair of blocks of two sheaves each, the rope is fast- ened to the upper block. What power is required to lift it? (Fig. 87.) 274 A HAND BOOK FOR MECHANICS 9 7 = 21 tons. Ans. 4 Example. A cylinder cover weighing 1100 pounds is lifted by a pair of blocks of two sheaves each, the rope is fastened to the upper block. What power is required to perform the work? (Fig. 87.) -^ = 275 lb. Ans. 4 The answers to above examples only tell the amount of power required at B, Fig. 86, and at P, Fig. 87, to evenly balance weights suspended as shown in figures. If it be required to raise the weights, additional power must be applied to overcome friction. Example. A pair of double blocks are used to lift a ton weight. What power must be applied if 20 per cent be lost by friction? Power required without friction = = 500 lb. With friction: If the power were 100 pounds (a cwt.), then the weight lifted would be only 80 pounds, as 20 per cent is lost. Hence, as 80 : 100 :: 500: power required. 100 8.0)50000(625. Ans. 48 20 16 40 40 Example. A pair of double blocks are used to raise a weight of 2 tons. What must be the power applied at THE SIMPLE MACHINES 275 the free end of the rope, if 20 per cent be lost by fric- tion? Weight in pounds = 2000 X 2 = 4000 lb. Power required without friction, = 1000. With friction, 80: 100 :: 1000: power required. 100 8.0)100000(1250. Ans. 8_ 20 16 40 40 The Strength of a Rope The strength of a rope depends upon its area or cir- cumference; the circumference called the girth being gen- erally used. The rule is that the girth or circumference squared and divided by 24 is equal to the weight that may be lifted expressed as tons. The divisor 24 varies with the quality of the rope. Example. It will be safe to lift how many tons with a rope whose girth is 14 inches? 14 2 -r- 24 = Si tons. Ans. Example. What weight may be lifted by a rope 2J inches in girth? 2J = 2.5. 2.5 2 -i- 24 = 260 lb. Ans. Example K. What weight may be safely lifted by a rope whose girth is 3 inches? 3 2 -T- 24 - | of a ton = 750 lb. 276 A HAXD BOOK FOR MECHANICS To Find the Breaking Stress of a Rope Rule. — Multiply the girth squared by .28, the product is the breaking stress in tons. Example L. — What is the breaking stress of a rope 3 inches in girth? 3 2 X .28 = 2.52 tons = 4500 lb. Note that Example K shows that it is only safe for a rope of 3 inches girth to lift 750 pounds, and that Example L shows that a rope whose girth is 3 inches will not break till a weight of 2.52 tons or 4500 pounds be applied. In practice the working load for round rope should not exceed a seventh of the ultimate strength and for flat rope one ninth. Therefore, if the ultimate strength of a rope whose girth is 3 inches equals 2.52 tons or 4500 pounds it would not be safe to put a working load on that rope exceeding one seventh of 4500 pounds, 642 pounds, although in practice often such a rope carries 700 to 800 pounds load. Example. — What is the breaking stress of a rope 4J inches in girth? 4.5 2 X .28 = 7.23 tons. Ans. The Wheel and .axle The wheel and axle may be likened to a couple of pulleys of different diameters united on one axis of which the larger is the wheel and the smaller the axle. As shown in Fig. 88 the combination consists of a wheel, or drum A mounted upon an axle B. The power is applied at the cord wrapped around the wheel A. and the weight or resistance suspends from a cord wrapped around the axle B. THE SIMPLE MACHINES 277 The wheel and axle is really a lever of the first class and may be treated as such. In which, as shown in Fig. 89, the center line A, C, represents the elements of the lever, D being the fulcrum about which the lever turns. Fig. 89 The radius D, F, of the wheel will be the leverage of the force applied, or power, and the radius E, D, of the axle the leverage of the weight. 278 A HAND BOOK FOR MECHANICS From this, then, we may deduce the following rules: To find the power: multiply the weight by the radius of the axle and divide by the radius of the wheel. Example. What power will be required to lift a weight of 300 pounds if the diameter of the drum or wheel is 12 inches and the diameter of the axle 6 inches. (Fig. 90.) Fig. 90 Radius of axle == 3 inches; radius of wheel = 6 inches. Ans. 3 X 300 900 , Kn = = 150. 6 6 That is, a pressure of 150 pounds at R will balance a weight of 300 at K. To find the weight: Multiply the power by the radius of the wheel and divide by the radius of the axle. THE SIMPLE MACHINES 279 Example. What weight will be balanced by a power of 75 pounds if the diameter of the wheel is 10 inches and the diameter of the axle 5 inches? (Fig. 91.) Radius of axle = 2.5 inches; radius of wheel = 5 inches. 75 X 5 375 = 150 lb. 2.5 2.5 Then a weight of 150 pounds at A will be balanced by a pres- sure of 75 at B. Example. The diameter of a steering wheel is 4 feet and the barrel is 14 inches diameter.. What resistance will be over- come if a man applies a force equal to 200 pounds. (Fig. 92.) 4 feet = 48 inches diameter, or 24 inches radius of wheel; 14 inches diameter = 7 inches radius of axle. Fig. 91 Fig. 92 280 A HAND BOOK FOR MECHANICS Then 20 ° X 2i , 687 lb. The Windlass The windlass is a combination of the wheel and axle in which a crank answers the purpose of the wheel. The windlass consists of an axle A, B, and a crank C, D, E, Fig. 93, by means of which the axle A, B, is turned. Fig. 93 As shown in figure, the crank is an arm standing per- pendicular to the axle. The part 2 of the crank is called the crank arm, and the part 1 is called the crank handle. The windlass is used for raising heavy weights and is operated by applying the power to the crank handle 1, the weight to be raised being made fast to the cord, which is fast to and wraps around the axle as same is made to revolve. THE SIMPLE MACHINES 281 Example. ■ What weight will be raised by a windlass, if a pressure of 150 pounds be applied to the crank handle? From center of axle to center of crank handle 15" and axle 10" diameter. l*i \ ^wfca 1 ' | Ik ^^^ A' iiiiii. limn Fig. 94 With this arrangement of windlass, when the handle is turned so as to wind up rope on the cylinder A it is at the same time unwound from the cylinder B, and at each revolution of the crank handle the rope is shortened only by the difference in the circumference of the cylinders. The greater the difference in the circumferences, the quicker the weight will move, and the less the difference THE SIMPLE MACHINES 283 the slower it will move, and by having them nearly equal the weight moves very slowly and great power is gained. Example. In a differential windlass the diameters of the drums are 1 foot and f of a foot, the length of the crank arm is 2 feet 3 inches and the power applied to the crank handle is equal to 90 pounds. What weight will be lifted? Rule. Multiply the length of crank arm in feet by the power, and divide by half the difference in feet of the the radii of the drums. Thus 2 feet 3 inches = 2\ feet = 2.25 feet, length of arm in feet. Then 2.25 X 90 = 202.50 -f- J difference in feet of radii = Arts. As the diameters of the drums are 1 foot and f feet (=9 inches) the radius of the larger drum will equal 6 inches and the radius of the smaller drum equals 4| inches. Then the difference between the radii is 6 — 4.5 = 1.5 inches difference between radii of drums. Then J of 1.5 = f = .75 of an inch. Now reduce .75 of an inch, which is \ the difference of the radii in inches, to the decimal of a foot. Thus 12). 75 (.0625 of a foot. 72 30 24 60 60 202 ^ Then, = 3240 lb. Ans. .0625 284 A HAND BOOK FOR MECHANICS The Inclined Plane The inclined plane is a flat surface inclined to the horizon. Any plane which forms with a horizontal plane any angle whatever excepting a right angle may be con- sidered an inclined plane. In Fig. 95 the line A. then, is an inclined plane because it is a surface inclining to the horizontal line B. The inclined plane is used for raising weights, and by referring to Fig. 95 it will be seen that less power would be re- quired to raise a given weight by the use of the sloping path formed by line A than would be required if surface A approached nearer to the perpendicular as shown in Fig. 96. The less the height of the plane in proportion to its length, or the less the angle of inclination, the greater the mechanical effect ; and the greater the height of the plane in proportion to its length the less will be the mechanical effect. The following explanation will more clearly explain this ; When a body rests on a horizon- tal plane, say for example on a table, the action of gravity tend- ing to draw it down is completely counteracted by the resistance of the plane on which it rests and the body remains at rest. It is not so, however, when a body is placed on an in- clined plane. In this case the force of gravity acts upon Fig. 96 THE SIMPLE MACHINES 285 the body in two ways: one perpendicular to the plane, which tends to force the weight through it, and the other parallel to the plane, which tends to draw the weight along parallel to it. The action of the force of gravity in the first place is counteracted by the resistance of the plane, whilst in the second place the body meeting no resistance in a line parallel to the plane it will be forced along in that direc- tion by the force of gravity acting parallel to the plane. And it is evident that the nearer the plane approaches to Fig. 97 a horizontal surface, the greater will be the portion of weight supported by the surface. Let the plane be elevated toward the perpendicular and it will support less and less of the weight, till when it reaches an exact perpendicular it will be supporting no part of the weight at all. Fig. 97 will demonstrate this still more clearly. In the figure is shown two weights of unequal sizes, A and B, connected by a cord C which passes over a pulley 286 A HAND BOOK FOR MECHANICS D. The larger weight A is resting on the inclined surface and is balanced by the smaller weight B. It is evident that the nearer the inclined surface on which the larger weight A is resting approaches the horizontal line E the greater will be the portion of the weight A supported by it, and consequently the less the weight at B will be re- quired to counterbalance A. Also, the nearer the inclined surface on which A is resting approaches a perpendicular the greater the weight at B will be required to counterbalance A. From this we learn that the small weight B will coun- terbalance a weight larger than itself which Tests on an inclined plane; and demonstrates the fact that by the use of the inclined plane a given weight can be raised by a power which is less than the weight itself. Rules for Calculating Work Done by the Inclined Plane To find the power: Multiply the weight by the height of the plane and divide by the length. Perpendicular height of plane p _ W v/ £ 2 i length of plane Example. What force is necessary to keep a weight of 100 pounds stationary on an inclined plane whose per- pendicular height is 4 feet and length of incline 14 feet, supposing there is no friction? Thus P = i^- 4 = 4™ 28f lb. 14 14 p = Power or force. Example. The length of an inclined plane is 19 feet, the perpendicular height 6 feet. What power will be re- THE SIMPLE MACHINES 287 quired to sustain a weight of 250 pounds, supposing there is no friction? (Fig. 98.) Fig. 98 ThusP = 250 X 6 1500 19 19 78.9 lb. Ans. To find the weight : Multiply the power by the length of the plane and divide by the height. W = PX L H W = Weight P = Power L = Length of plane H = Height Example. The length of an inclined plane is 15 feet, the perpendicular height 7 feet. What weight will a power of 78 pounds sustain, supposing there is no friction? 78 X 15 = 167.1 lb. Ans. Example. The length of an inclined plane is 15 feet, the perpendicular height 3 feet. What weight will be sus- tained by a power of 78 pounds? 288 A HAND BOOK FOR MECHANICS I^li 5 = I 170 = 390 lb. Ans. The Wedge The form of wedge in general use is the double wedge represented in Fig. 99, which may be likened to a pair of Fig. 99 inclined planes joined together back to -back along the dotted line A, B. There is, however, another form of wedge used called the single wedge, represented in Fig. 100, which may be likened to the inclined plane. Wedges are used where it is Fig. 100 required to exert a great force in a small space. No accurate estimate can be made of the work done by wedges as they are ordinarily used, but calculations worked out according to the following rules will give an approx- imate idea of the amount of work which may be performed by them. THE SIMPLE MACHINES 289 To Find the Power Rule. Multiply weight by thickness of wedge and divide by length of wedge. Example. A wedge 20 inches long and 3 inches thick is employed to lift a weight of 200 pounds. What power must be exerted? (Fig. 101.) to « 1 Fig. 101 200 X 3 ~~ 20 600 20 30 lb. pressure Example. If a wedge 12 inches long and 3 inches thick is employed to raise a weight of 1000 pounds, what pressure must be exerted? P = 1000 X 3 3000 12 12 = 250 lb. Arts. To Find the Weight Rule. Multiply the power by the length of the wedge and divide by the thickness. P = power P X L L = length ~~ T T = thickness W = weight Weight = 290 A HAND BOOK FOR MECHANICS Example. If a wedge is 15 inches long and 2 inches thick, what weight will it raise if the power applied is 100 pounds? W = 100 X 15 1500 = 750 lbs. 2 -2 Example. If a double wedge 12 inches long 4 inches thick splits a block when it is driven in with a force of 200 pounds, the pressure exerted upon the block by the wedge is how many pounds? (Fig. 102.) W 200 X 12 2400 = 600 lb. The Screw Fig - 102 The screw is essentially an in- clined plane wrapped around a cylinder, as may be seen by taking a triangular piece of paper and winding it around a cylindrical barrel. Take, for example, the inclined plane A, B, C (Fig. 103), and bend it in circular form resting on its base line Fig. 103 A, B; by so doing a helical plane 1, 2 will be formed as shown in Fig. 104, which figure shows the inclined plane A, B, C (Fig. 103) curled around the cylinder D. Here, then, it is seen that as the incline winds around THE SIMPLE MACHINES 291 the cylinder D it assumes a spiral shape more accurately called helical. The screw, then, is really a combination of inclined planes, because it consists of a solid barrel A Fig. 104 (Fig. 105) enveloped by a spiral projection called the thread B, which is nothing more than an inclined plane wound around a solid barrel. Fig. 105 It is not clearly determined how far the mechanical powers were known to the ancients. Without doubt they 292 A HAND BOOK FOR MECHANICS had a knowledge of the lever, the wheel and the axle, and the pulley, and it would seem because of the great en- gineering feats of construction and building performed by the early Egyptians, that they must have had a knowl- edge of the principles of the inclined plane; this is only surmised and not an established fact, but it seems very probable, in order to have moved the huge blocks of stone of which the pyramids are built, that they must have had an acquaintance with the inclined plane. Just when the inclined plane was first used in the form of a screw is not known. The earliest authentic mention we have of the screw being used in mechanics is in the writings of Archimedes. This philosopher and mechanician, about 236 B.C., in- vented a pumping screw or spiral cylinder for raising water. The device is now known as " Archimedes' screw," and is one of the most ancient contrivances for raising water. This screw consists of a metallic tube, wound in a spiral form around a solid cylinder or shaft, which is made to revolve by turning a handle. When placed at the proper angle, the water, as the handle is turned, will continue to flow into those parts of the tube that are brought successively below the shaft, till gradually it will be discharged at the top. (Fig. 106.) Apparently Archimedes did not, however, understand the inclined plane as now used, for he makes no direct mention of it in any of his writings, and there is no posi- tive evidence to show that it was included in the knowledge of mechanics possessed by the Romans. The inclined plane seems to have been invented by Galileo in the sixteenth century, because several writers of that time allude to the wedge and the screw, showing that a knowledge of these powers formed a part of the THE SIMPLE MACHINES 293 revival of physical science in which Galileo took a most prominent part at that time, even if he did not fully inspire it. However, Stevinnes, a mechanician of Hol- land, was the first to fully explain in a treatise the theory of the power of the inclined plane and screw. The first use of the screw was' in the screw jack, an appliance used for raising heavy weights. Fig. 106 The various modifications and applications of this power belong to the era of mechanical discovery of the present century. Work Done by the Screw Screws may be either single-threaded or double- threaded. If we assume that a screw consists of a cylin- der with a coil which forms the thread wound around it, we may easily define a double screw as a cylinder with two coils parallel to one another wound around it. The 294 A HAND BOOK FOR MECHANICS screw works into a solid, fitted with a thread to receive it, called the nut. The nut may be fixed or movable: when fixed the screw turns within it, and when the nut is movable the screw is fixed, the nut being made to turn upon it. Motion is imparted to the screw or the nut (whichever the case may be) by means of a lever at the extremity of which the power is applied. The longer the lever and the less the distance between the threads, the greater will be the force exerted at the point of resistance. Screws are said to be right-handed or left-handed. A right-handed screw is one which, passing through a fixed nut and turned toward the right hand, will advance into the nut. A left-handed screw is one which will, by turn- ing it in a direction toward the left hand, be made ad- vance into a fixed nut. A screw in one revolution will advance into a fixed nut a distance equal to its pitch, or a distance equal to the distance between the centers of two successive threads. That is to say, if the distance between the centers of two successive threads be I inch, the screw during one com- plete turn will advance into a fixed nut that distance. The distance between the centers of two successive threads is always called the pitch of the screw. The following rules give an approximate knowledge of work performed by the screw: To Find the Power Rule. Multiply the weight in pounds by the pitch and divide by the circumference of the circle described in inches by the handle employed to turn either the screw or nut one revolution. Example. If the pitch be f inches, and the lever 3 THE SIMPLE MACHINES 295 feet long, what power at A will be required to move a weight of 5 tons? (Fig. 107.) [ -«■ •a* A^ T I — ,~ ! n Zbunu* I I i^/^C* Fig. 107 ^ W* IK J P = Power Twice length of lever arm = 6 feet — 72 inches. Then 72 X 3.1416 = 226.1952 inches circumference of circle described by revolution of lever. 5 tons = 2000 X 5 = 10000 lb. Thus P = 10000 X f 226.1972 33.1 lb. 296 A HAND BOOK FOR MECHANICS Example. If the pitch of a screw be \ inch, and the lever be 3 feet long, what force would move 5 tons? Twice length of lever = 6 feet = 72 inches; 72 X 3.1416 = 226.1952 inches circumference of circle described by lever. Weight in pounds = 2000 X 5 = 10000 lb. 10000 X .25 2500 , , „ P = == = 11 lbs. 226.1952 226.1952 It will be noticed from the above examples, that the- oretically there is no limit to the power of the screw, for the distance between the threads may be made as small as you please, and the lever may be made as long as you please; the limits will be mechanical, viz.: the strength necessary to be given to the threads and the space neces- sary to be given to the lever. To Find the Weight Rule. Multiply power by the circumference in inches of the circle described, by the handle or lever employed to either turn the screw or the nut, and divide by the pitch. Example. If the distance between the threads be \ inch (= .25), and a force of 100 pounds be applied to the end of a lever 2 feet in length, what weight will be moved by the screw? Thus: Twice length of lever = 4 feet — 48 inches. 100 lb. X 48 X 3.1416 15079.6800 6031glb .25 ~ .25 The rule mostly used in practice for determining the weight that can be lifted by a screw, or the tension that can be given a bolt or stud by using a wrench to the nut, THE SIMPLE MACHINES 297 is the distance moved by the power or the hand in making one complete revolution of the wrench, divided by the pitch of the screw and the quotient multiplied by the power applied. From this deduct 70 to 75 per cent for friction. All machines, no matter how complicated or what the work is they perform, are combinations of these simple mechanical devices, or primary machines, arranged to- gether and combined so as to give such direction and velocity to the motion that the required work is per- formed. PART VI STRENGTH OF MATERIALS AND QUESTIONS RELATING TO STRESS STRENGTH OF MATERIALS AND QUESTIONS RELATING TO STRESS The mechanic, to work intelligently, should be ac- quainted with the strength of the different materials used in construction. He should also be familiar with the strains, due to the various forces, any of the great variety of materials used will stand before final rupture. And it is therefore necessary that he have a knowl- edge of those laws which govern the rules for calculating the extent of the strain due to the various forces to be dealt with, and know what to allow as a safe margin of resistance to carry all such forces, and thereby make the required provisions for the strength of parts. The external forces applied to materials tending to cause their rupture or alteration of form are called stresses. These are five in number, called: 1. Tensial stress. 2. Compressive stress. 3. Transverse stress. 4. Torsional stress. ' 5. Shearing stress. A force applied in the line of fiber having a tendency to lengthen the body is called the tensial or breaking stress. As shown in Fig. 108, the beam A has a tendency to lengthen because of the force exerted upon it by the weight B, and the resistance of A due to the load B to alter its form is the stress, — in this case the tensial stress. 301 302 A HAND BOOK FOR MECHANICS A force which pushes the parts together, tending to shorten or crush the body, is called the compressive stress. As shown in Fig. 109 the weight A exerts a stress, called compressive stress, upon the column B, tending to crush it. A force applied to a body in such manner as to cause it to twist is called the torsional stress. As shown in Fig. 110, by applying a power to the lever A in such manner STRENGTH OF MATERIALS 303 that it would cause the beam B, which is anchored at C, to twist. Fig. 109 Fig. 110 If a beam is resting on two supports and a weight is placed on the beam between the supports, it is called a transverse stress. (Fig. 111.) 304 A HAND BOOK FOR MECHANICS If the beam be fixed at one end and loaded at the other, the stress is a bending stress. (Fig. 112.) Fig. Ill Lastly, if the stress is such that it acts in a way that it is trying to cut through a body it is called a shearing stress. As shown in Fig. 113, where the cutters A, B, are cutting through the bar C. 6 Fig. 112 The immediate result of stress is strain. Every load which acts on a structure produces a change of form, which is called the strain due to the load. The strain may be either a vanishing or elastic change of form; that is, one which disappears when the load is re- STRENGTH OF MATERIALS 305 moved, or it may be a permanent change or set, which remains after the load is removed. Strain, then, is the alteration in shape as the result of stress. And in designing structures, care must be taken that all parts will have the required strength, so that under the greatest straining action there will be no sensible permanent change in form. The stress required to strain, change the form of or rupture the various materials in use is usually deter- mined by an apparatus called the Testing Machine, of L 3£l Fig. 113 which there are many different forms, and it may be here pointed out that the differently constructed machines of this class give widely different results, there being a great variation in the data indicated by them represent- ing the standard resistances of the various materials tested. Therefore, the careful mechanic, rather than depend upon the data given by the different authorities as being correct in representing the strength of the different materials, is obliged to himself test and determine the strength of the materials he proposes to use, and will 306 A HAND BOOK FOR MECHANICS regard given data rather as approximate and subject to change than as precedents to be adopted and blindly followed. It will be the object of this chapter to explain the rules for determining the strain of the approximate stresses due to the action of the different loads, to show by tables and otherwise the approximate strength of the various materials used, and how to make comparative safe pro- vision for sustaining all those forces which enter as fac- tors in construction of every nature. As cast iron, wrought iron, steel and wood, are the ma- terials mostly used in machine construction, special note had better be made of their various strengths, so that intelligent judgment may be used when these materials are employed to construct the parts of such sizes that they will safely stand the stresses due to the action of the dif- ferent loads, and thereby prevent straining and rupture. The stresses to which constructions and parts of con- structions are mainly subjected to are the Tensial, Com- pressive, and Shearing stresses. Stress is usually measured in pounds per square inch of sectional area. That is, the ultimate strength of - materials is usually deter- mined by the number of pounds a square inch of sectional area will undergo before final rupture. For instance, if a wrought-iron bar is 2 inches by 2 inches in sec- tion. (Fig. 113.) Thus the whole area of section is 4 square inches, and it is obvious that if the ultimate tensial strength of wrought T i Fig. 113 STRENGTH OF MATERIALS 307 iron, for instance, is 52,000 pounds per square inch of section, a bar of wrought iron whose sectional area is 4 square inches would be caused to break by a power of 52,000 X 4 = 208,000 pounds. The following tables are approximate of the ultimate strength and elastic strength of different materials com- monly used. Ultimate Strength By ultimate strength is meant the greatest possible load any piece of material will carry before fracture is produced. In other words, the smallest load which will cause the rupture of a piece of material is called the ultimate strength of that piece; that is, the stress in pounds per square inch which the piece can sustain just before rupture takes place. Elastic Strength All materials are more or less elastic and will undergo, when put under great stress, change of form which may be vanishing or permanent as before stated. When the load is such that it has a permanent influence upon the piece and sets it so that it will not return to its original form when the load is removed, the limit of elasticity of the material has been reached (which is the elastic strength), and beyond this limit the strain increases faster than the stress until rupture is produced. 308 A HAXD BOOK FOR MECHANICS TABLES TABLE NO. 1 Ultimate and Elastic Strength of Materials in Pounds per Square Inch NAME OF MATERIALS Cast Iron. ULTIMATE OR BREAKING STRENGTH Tension. „^??? „ Shearing pression & Wrought Iron . Soft Steel En- hardened . . . . Soft Steel hard- ened Cast Steel Un- tempered. . . . Cast Steel Tem- pered. . Copper Brass Gun Metal 10.800 90.000 20.000) to 67.000 52.000 52.000 to 67.000 60.000 150.000 to 100.000 "0 . 000 Phosphor Bronze. 120.000 120.000 120.000 84.000 84.000 84.000 to to to 150.000 150.000 150.000 100.000 150.000 70.000 33.000 58.000 58.000 17.500 10.500 10.500 23.000 23.000 23.000 to to to 52.000 52.000 52.0001 58.000 58.000 58.000 ELASTIC STRENGTH Tension ^^„?^" Shearing pression & 10.500 21.000 7.900 24.000 24.000 20.000 35.000 35.000 26.000 70.500 70.500 53.000 80.000 80.000 74.000 S4.000 90.000 145.000 4 . 300 3 .900 2 . 900 6.950 6.950 5.200 6.200 6.200 -4.150 19.700 19.700 14.500 STRENGTH OF MATERIALS 309 TABLE NO. 2 Table Showing Safe Working Stress in Pounds per Square Inch ordinary working stress NAME OF MATERIALS Cast Iron Wrought Iron . Soft Steel Un tempered . . . Cast Steel Un tempered. . . Copper Brass Gun Metal .... Phosphor Bronze. SAFE LIMITS IN POUNDS PER SQUARE INCH Tension 3.600 10.400 17.700 36.000 3.600 3.600 3.120 9.870 Com- pression 10.400 10.400 17.700 36 . 000 3.120 3.600 3.120 9.870 Shearing THEORETICAL LIMIT OF STRESS Tension 2.700 7.800 13.000 20.000 2.300 2.700 2.400 7.380 5.250 17.280 24.000 52.000 9.900 5.250 10.800 17.400 Com- pression 28.500 15.000 24.000 52.000 17.400 3.150 10.800 17.400 Tension Com- pression 3.000 9.000 13.000 36.000 5.500 3.000 6.000 9.700 As shown by foregoing tables, 1 and 2, there is a great margin of difference between the ultimate strength and the safe working limit allowed for the load or weight that the various metals will safely carry before straining. This allowance is made because it is impossible to de- termine all the forces which produce straining action; therefore, to insure the safety of a structure, we have to make all parts sufficiently strong to support, not only the aggregate amount of straining action due to the forces which are taken into consideration, but also to take care of all unforeseen contingencies due to neglected causes of straining action. Hence it is necessary that the mechanic be sufficiently informed so that he be able to determine the approximate 310 A HAND BOOK FOR MECHANICS stress on a structure, and also be able to determine the necessary strength of the different parts to prevent strain and thereby insure absolute safety. And although by practical experience and familiarity he perhaps gains the best knowledge of the intensity of straining actions due to the different forces, it is customary in practice, when determining the straining actions on a structure, to be more or less guided by various factors of safety. As just stated, it is impossible to take into considera- tion all the forces which produce a straining effect upon a structure. However, by multiplying those forces which are known to exist by a factor of safety, a rough and comparatively safe allowance is made for the straining action which is exerted by the unseen forces. And in this way is obtained the approximate stress on the struc- ture due to all causes. And the sizes and strength of the various materials to be used to carry those various stresses may be deter- mined accordingly. Following is a table of factors of safety, which are generally adopted in practice for various materials, under dead and varying or live load, and for machines subjected to sudden and frequent strains of short duration, known as shocks. FACTORS OF SAFETY Name of Materials Dead Load Live Load Materials Subject to Shocks Cast Iron 6 4 4 5 8 15 10-15 6 7 8-10 10 25 15-20 Wrought Iron 12 Steel 15 Copper 10-15 Timber 15 Masonry and Brickwork . . . 30 STRENGTH OF MATERIALS 311 It will be noticed that the factor of safety is less for dead loads than for live or varying loads, and greatest when the structure is subjected to shock. From which we learn that the stress due to the different loads varies; therefore the nature of the load to be car- ried must be carefully considered and provisions accord- ingly made to carry them. It being impossible to correctly determine the intensity of a load all the parts must be made to resist a much greater load than will be brought to bear on them at any time; therefore the expected load is multiplied by one of the various factors of safety. And this factor of safety varies according to the nature of the load, and con- sequently for the same materials under the influence of different loads, a greater or less factor of safety must be used. Tensial Stress If a bar of iron 1 inch square is torn asunder by a stress of 22 tons, what will be required to break a bar 3f" square? That is, if a square bar whose sides measure 1" is torn asunder by a stress of 22 tons, what will be required to break a square bar whose sides measure 3| " ? 3.75 2000 = 1 ton. 3.75 22 1875 4000 2625 4000 1125 44000 stress in 14.0625 sq. inches in section of bar. p ounds 44000 required 562500 to break 562500 bar 1 inch 618750.0000 lbs. to break the bar. Arts. square. Example. If a bar of cast iron 2 inches square is torn 312 A HAND BOOK FOR MECHANICS asunder by a stress of 100,000 pounds, what will be required to break a bar of 3J inches square? 3.5 3.5 175 105 12.25 sq. inches in section of bar to be broken. 2 2 4 sq. inches in section of bar which is broken. Then if a force of 100,000 pounds is required to break a bar of cast iron of 4 square inches sectional area, how much force will be required to break a cast-iron bar of 12.25 square inches sectional area more? Thus, 4 : 12.25 :: 100,000: Ans. 12.25 500000 200000 200000 100000 4 )1225000.00 306250 lb. to break the bar. Ans. Example. If a bar of wrought iron 2 inches in diam- eter is torn asunder by a weight of 70 tons, what is its breaking stress? It will be remembered that stress is measured in pounds per square inch of section; and that breaking stress means the stress in pounds per square inch of section required to break any material. Hence, the question of the above example is to deter- mine the stress in pounds required to rupture a square inch of (in this instance) wrought iron. STRENGTH OF MATERIALS 313 Thus 2 2 4 = square of diameter of bar. Then, .7854 4 3.1416 sq. inches in section of bar. Then if 70 tons is the breaking stress of 3.1416 square inches, what is the breaking stress of 1 square inch? Thus, 3.1416 : 1 ::70: Arts. 1 3.1416)70.0000(22 tons. Arts. 62832 71600 62832 That is, each square inch of section will stand before breaking a force of 22 tons, or 70 -r- (.7854 X 2 2 ) = 22.28 tons per square inch of section. Ans. Example. A piece of plate 9 inches long, 4 inches wide, and f inches thick is tested in line of its length and breaks at 72 tons. What is its breaking stress? Here the section is 4 X f = 3 square inches. And the breaking stress = 72 tons -~ 3 = 24 tons per square inch of its section. Ans. Compressive Stress Example. If a solid casting 4 inches diameter is crushed by a weight of 450 tons, what is the crushing strength of this specimen of cast iron? Thus: 314 A HAND BOOK FOR MECHANICS Area of section = .7854 X 4 2 = 12.5664 square inches. Stress = 450 + 12.5664 = 35 tons. 35 tons per square inch of its section. Am. Example. If a cast-iron column 5 inches by 5 inches holds up a weight of 25 tons, what is the stress in pounds per square inch on the column? 5 _5 25 square inches in section of column. 2000 = 1 ton 25 10000 4000 50000 = weight in pounds supported by the column. Then stress in pounds per square inch of sectional area - 50000 + 25 = 2000 lb. Ans. Example. If four solid cast-iron columns, each 4 inches square, hold up a cistern weighing 46 tons, whose inside dimensions are 12 feet long, 8 feet wide, and 6 feet deep, what is the stress per square inch on these columns when the cistern is one half full of water? (Fig. 114.) First find weight of water in cistern when full, which divide by 2 to obtain weight in pounds when half full. Add to this the weight of cistern when empty. Divide the sum obtained by the total sectional area of the col- ums (4 of them), and the quotient 1562 is the required answer. Cubic contents of cistern when full of water = 12 X 8 X 6 = 576 cubic feet. 1 cubic foot of water weighs 62.5 lb. Weight of water in cistern = 576 X 62.5 ^ 2 = 18,000 lb. STRENGTH OF MATERIALS 315 Total weight on columns in pounds = 92,000 (46 tons) and 18,000 = 100,000 lb. Total sectional area of columns -4X4X4 = G4 square inches. Stress per square inch = 100,000 - 64 = 1562 lb. Ans. XJ^ Fig. 114 Shearing Stress In cutting through a bar we cut through its section: then in shearing stress always find the sectional area. The shearing stress of wrought iron is usually con- sidered to be 52,000 pounds (23 tons) per square inch. That is, a stress of 52,000 pounds is required to cut through every square inch of sectional area. Example. What shearing stress will cut through a three-quarter inch rivet? 316 A HAND BOOK FOR MECHANICS .75 X .75 X .7854 = .44178750 sectional area of rivet, (Fig. 115.) Then, .44148650 X 52,000 lb. = 22.972 lb. Ans. Fig. 115 The foregoing examples will serve to demonstrate the process for determining the magnitude of the strain car- ried per square inch of section which may be caused by the action of the stresses. various 116 we Examplks Demonstrating the Process for Constructing Parts of Suffi- cient Strength to Carry Loads of Known Magnitudes. Example. A square wrought-iron bar is to carry a weight of 208,000 pounds, what must be its sectional area? (Fig. 116.) Let us employ a factor of safety of 4. That is, we will suppose the load which is to be sustained will be 4 times greater than 208,000, which is 816,000 pounds. As the piece of material in question is subjected to a tensial find by referring to table No. 1 that the ulti- STRENGTH OF MATERIALS 317 mate tensial strength of wrought iron is 52,000 pounds per square inch of section. Therefore, to find the required area of section, divide the load 816,000 pounds by the ultimate strength 52,000 pounds. 832,000 52,000 16 square inches. Arts. Example. What must be the sectional area of a square cast-iron block which is to sustain a weight of 160,000 pounds. (Fig. 117.) Fig. 117 Let us employ a factor of safety, say 5 for cast iron. That is, we will suppose the load which is to be sustained will be 5 times greater than 160,000, which is 800,000, pounds. As the piece of material in question is subjected to a compressive stress, we find by referring to table No. 1 that the ultimate compressive strength of cast iron is 90,000 pounds per square inch of section. To find, then, the required area of a section, divide the load, 800,000 pounds by 90,000. Thus = 8.888 square inches. Ans. 90,000 318 A HAND BOOK FOR MECHANICS Example. What must be the sectional area of a wrought-iron bolt to withstand a shearing stress of 80,000 pounds? (Fig. 118.) Fig. 118 Let us employ a factor of safety of 4 for wrought iron. That is, we will consider the load which is to be carried to be 4 times greater than 80,000 pounds that is, 320,000 pounds. As the piece of material in question is subjected to a shearing stress, we find, by referring to table No. 1, the ultimate shearing stress of wrought iron to be 52,000 pounds per square inch of section. To find, then, the area of a section capable of sustaining a shearing stress of 80,000 pounds, divide the load 320,000 by 52,000. 320,000 Thus, 52,000 6.153 square inches. Ans. Examples Showing the Process for Determining the Amount of Stress on Beams Example. If a beam 10 feet long fixed at one end has suspended from it a weight of 300 pounds, what is the stress at the point where the beam is fixed? (Fig. 119.) Rule. The intensity of the force is equal to the force multiplied by the length of the perpendicular to the direction of the force from a point in which the beam is supposed to be fixed. STRENGTH OF MATERIALS 319 Therefore, the stress on the beam at A in pounds equals 300 X 10 = 3000 lb. — "S: J'-V. .lo'. Q->. Fig. 119 Because the length of the perpendicular from A, the point at which the beam is fixed to the center of the force 300 equals 10 feet, and 300 X 10 = 3000 lb. Example. What is the stress at B, Fig. 120, when a weight of 300 pounds is suspended from the free end of beam? L — oh^=-. Fig. 120 320 A HAND BOOK FOR MECHANICS In this case a perpendicular drawn from the point at which beam is fixed to the center of line of the force is 9 feet, therefore the stress at B = 300 X 9 = 2700 lb. The stress throughout the length of a beam fixed at one end and loaded at the other is not uniform. That is, the stress varies at different parts of the beam. The intensity of the stress is greatest at the point where the beam is supposed to be fixed and gradually diminishes in intensity toward the load. For instance, there is more stress on the beam at A (Fig. 121), than at B, and the stress is greater at B than at C. Fig. 121 To Find the Stress in Foot Pounds of any Sec- tion of a Beam Rule. Multiply the load by its distance from the section. For instance, it is required to find the stress at a section 5 feet from the free end of a beam from where a load of 100 pounds is suspended. (Fig. 122.) Thus 100 X 5 = 500 lb. That is, the intensity of the stress on the beam at a point (section) five feet from the load, the load being 100 pounds, is 500 pounds. STRENGTH OF MATERIALS 321 £"- r-i Fig. 122 Example. What is the stress on a section of a beam 2 feet from the load, the beam being fixed at one end and carrying a load of 300 pounds? Thus, 300 X 2 = 600 foot pounds. The above examples fully demonstrate that the greatest stress is at the point where the beam is supposed to be fixed, and that it gradually diminishes toward the load. Therefore it is customary, when constructing beams of this nature, to make them strongest where the greatest stress comes. (Fig. 123.) 6 Fig. 123 322 A HAND BOOK FOR MECHANICS Example. If the greatest stress allowed on wrought iron be taken at 52,000 lbs. per square inch, what should be the sectional area at the wall A of a wrought-iron beam 66 inches long, supporting a weight of 2 tons suspended from its free end? (Fig. 124.) Fig. 124 2 tons = 4000 lb. 4000 X 66 = 264,000 lb. stress at A. Let us now employ a factor of safety of 4 for wrought iron. That is, consider the stress 264,000 at A to be 4 times greater = 1,056,000 lb. Then, 1,056,000 h- 52,000 = 20.31 square inches. Arts. Example. A beam 56 inches long fixed at one end supports a weight of 5000 pounds. What will be the stress on the beam at fixed end A, and what will be the stress at end B? (Fig. 125.) 5000 lbs. = stress at B. 5000 X 56 = 280,000 lb. stress at A. Example. A beam is 3 inches thick, 5 inches deep, and 56 inches long. What stress will be put on it per STRENGTH OF MATERIALS 323 Fig. 125 square inch of its section at B, by hanging a weight of 5000 pounds from its extremity? (Fig. 126.) S6- (A % B M^ Fig. 126 Thus 3 X 5 = 15 square inches of section. 5000 X 56 = 280,000 lb. stress at B. 280,000 -f- 15 = 12,000 lb. per square inch. Arts. Example. If a wrought-iron beam 66 inches long has to carry weight of 5000 pounds, what ought the sectional area of the beam be at its fixed end, and what ought its sectional area be at its extremity? (Fig. 127.) 324 A HAND BOOK FOR MECHANICS CC? «3 Fig. 127 $040 5000 X 66 X 4 = 1,320,000 lb. stress at B. 5000 X 4 = 20,000 lb. stress at A. 1,320,000 ~ 52,000 = 25.5 square inches required at B. 20,000 ■*- 52,000 = .36 square inches required at A. ANSWERS ANSWERS ANSWERS TO PAGE 8 1. Five. 2. Twenty-five. 3. Three hundred and forty-two. 4. One thousand six hundred and seventy-four. 5. Fifty-four thousand three hundred and forty. 6. Nine hundred and sixty thousand seven hundred and eighty. 7. Three million seven hundred and twenty-six thousand nine hundred and two. 8. Fifteen million nine hundred and eight thousand six hundred and sixty. 9. Three hundred and two million six hundred and seven thou- sand six hundred and six. 10. Five billion three million three thousand and three. 11. Thirty-two billion six hundred seventy-three million three hundred thousand three hundred. 12. Nine hundred and ninety-nine billion nine hundred and ninety-nine thousand nine hundred and ninety-nine. ANSWERS TO PAGE 9 1. Four thousand three hundred and sixty-four. 2. One thousand nine hundred and twenty-seven. 3. Nine thousand and nine. 4. Four hundred and thirty-four thousand six hundred and seventy-two. 5. Six million four hundred and ninety-seven thousand nine hundred and twenty-three. 6. Fifty-three million two hundred and ninety thousand six hun- dred and seventy-eight. 7. Five hundred million four hundred and ninety thousand and sixty-nine. 8. Five billion eight hundred and sixty-seven million three hun- dred and forty thousand and sixty-eight. 327 328 ANSWERS ANSWERS TO PAGE 10 1. 3,725. 2. 12,600 3. 322,006. 4. 6,050,020. 5. 38,420,350. 6. 222,835,130. ANSWERS TO PAGE 15 1. 100,677. 1. .26219. 2. 17,903,922. 3. 2,394,627. 4. 127,228,172. 5. 784,666,605. 2. 3. .8552072. .40750202. 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VAN NOSTRAND COMPANY Sole Agents for the United States PAGE MECHANICAL ENGINEERING. 1 CIVIL ENGINEERING 11 MARINE ENGINEERING, &c. . 19 MINING & METALLURGY 22 COLLIERY WORKING, Sec. ... 26 ELECTRICITY 28 ARCHITECTURE & BUILDING. 31 SANITATION & WATER SUP= PLY 35 CARPENTRY & TIMBER 36 PAGE DECORATIVE ARTS 38 NATURAL SCIENCE 40 CHEMICAL MANUFACTURES. 41 INDUSTRIAL ARTS 43 COMMERCE, TABLES, &c 49 AGRICULTURE & GARDEN= ING SO MATHEMATICS & ARITH- METIC 54 LAW & MISCELLANEOUS. ... 56 MECHANICAL ENGINEERING, ETC. THE MECHANICAL HANDLING OF MATERIAL. A Treatise on the Handling of Material, such as Coal, Ore, Timber, etc., by Automatic or Semi-automatic Machinery, together with the Various Accessories used in the Manipulation of such Plant, and Dealing fully with the Handling, Storing, and Warehousing of Grain. By G. F. Zimmer, A.M. Inst. C.E. 528 pages. Royal 8vo, cloth, with 550 Illus- trations (including Folding Plates) specially prepared for the Work Net $10.00 Contents: — Chapter I. Introductory. — II. Elevators. — III. Worm Conveyors. — IV. Push-Plate or Scraper Conveyors. — V. Trough Cable Conveyors. — VI. Band Conveyors. — VII. Metal Band Con- veyors. — VIII. Picking Belts or Tables with or without Lowering Ends or Shoots. IX. The Continuous Trough or Travelling Trough Conveyor. — X. Vibrating Trough Conveyors. — XI. Tightening Gear for Elevators and Conveyors, and Driving Power required for Dif- ferent Types of Conveyors. — XII. The Travelling or Tilting Bucket Conveyors. — XIII. Pneumatic Elevators and Conveyors. — XIV. Con- veyors DESIGNED FOR SPECIAL PURPOSES, INCLUDING THE BoLINDER TIMBER Conveyor, Coke Conveyors, and Casting Machines. — XV. Endless Chain and Rope Haulage. — XVI. Ropeways and Aerial Cableways, including Ropeways, Cableways, and Appliances for Coaling at Sea. — XVII. Unloading Appliances, including Methods of Discharging by means of Skips and Grabs. — XVIII. Discharging Vessels and Barges by means of Elevators. — XIX. Unloading by means of Specially Con- structed Self-emptying Boats and Barges. — XX. Unloading by means of Specially Constructed Self-emptying Railway Trucks. — XXI. Un- loading BY MEANS OF COAL TlPS. XXII. COLLIERY TlPPLERS. XXIII. Miscellaneous Loading and Unloading Devices. — XXIV. Automatic Loading Devices. — XXV. The Automatic Weighing of Material. — XXVI. Coaling of Railway Engines. — XXVI I. Coal-handling Plant 2 CROSBY LOCKWOOD & SOX'S CATALOGUE. for Gas-works, Potter Stations, Boiler-houses, etc. — XXVIII. Floor and silo Warehouses for Grains and Seeds. — XXIX. Coal Stores and Coal Silos. — XXX. High-level Cranes. — Index. HOISTING MACHINERY. An Elementary Treatise on. Including the Elements of Crane Con- struction and Descriptions of the Various Types of Cranes in Use. By Joseph Horner, A.M.T.M.E., Author of "Pattern-Making" and other Works. Crown Svo, with 215 Illustrations, including Folding Plates, doth S3.00 AERIAL OR WIRE=ROPE TRAMWAYS. Their Construction and Management. By A. J. Wallls-Tayleb, A.M.Inst. C.E. With SI Illust rat ions. 12mo, cloth S3. 00 "An excellent volume, and a very good exposition of the various systems of rope transmission in use and gives as well not a little valuable informa- tion about their working, repair, and management. We can safely recom- mend it as a useful general treatise on the subject." — Engineer. MODERN MILLING MACHINES. Their Design. Construction, and Working. A Handbook for Practical Men and Engineering Students. By Joseph Horner, A.M.I.Mech.E., Author of ' ' Pattern -Making," etc. With 269 Illustrations. Demy 8vo, cloth. [Just Ready.] S4.00 TOOLS FOR ENGINEERS AND WOODWORKERS. A Practical Treatise including Modern Instruments of Measurement. By Joseph Horner. A.M.Inst .M.E., Author of "Pattern-Making," etc. Demy, Svo, with 456 Illustrations S3. 50 Summary of Contents: — Introduction. — General Survey of Tools. — Tool Angles. — Sec. I. Chisel Group. — Chisels and Applied Forms fob Woodworkers. — Planes. — Hand Chisels and Applied Forms foe Metal Working.— Chisel- like Tools for Metal Turning. Planing, etc. — Shearing Action and Shearing Tools. — Sec. II. Examples of Scraping Tools. — Sec. III. Tools. — Relating to Chisels and Scrapers. — Saws. — Files. — Milling Cutters. — Boring Tools for Wood and Metal. — Taps and Dies. — Sec. IV. Percussive and Moulding Tools. — Punches. Ham- mers and Caulking Tools. — Moulding and Modelling Tools. — Miscel- laneous Tools. — Sec. V. Hardening. Tempering, Grinding, and Sharp- ening. — Sec. VI. Tools for Measurement and Test. — Standards of Measurement. — Squares. Surface Plates, Levels, Bevels. Protrac- tors, c. 3 THE MECHANICAL ENGINEERS' REFERENCE BOOK. For Machine and Boiler Construction. In Two Parts. Part I. Gen- eral, Engineering Data. Part II. Boiler Construction. With 51 Plates and numerous Illustrations. By Nelson Foley, M.I.N.A. Third Edition, Revised throughout, and much Enlarged. Folio, half- bound In Press Part I: Measures. — Circumferences and Areas, &c. — Squares, Cubes, Fourth Powers. — Square and Cube Roots. — Surface of Tubes. — Reciprocals. — Logarithms. — Mensuration. — Specific Gravities and Weights. — Work and Power. — Heat. — Combustion. — Expansion and Contraction. — Expansion of Gases. — Steam. — Static Forces. — Gravi- tation and Attraction. — Motion and Computation of Resulting Forces. — Accumulated Work. — Centre and Radius of Gyration. — Moment of Inertia. — Centre of Oscillation. — Electricity. — Strength of Materials. — Elasticity. — Test Sheets of Metals. — Friction. — Transmission of Power. — Flow of Liquids. — Flow of Gases. — Air Pumps, Surface Condensers, &c. — Speed of Steamships. — Propellers. — Cutting Tools. — Flanges. — Copper Sheets and Tubes. — Screws, Nuts, Bolt Heads, &c. — Various Recipes and Miscellaneous Matter. — With DIAGRAMS for Valve-gear, Belting and Ropes, Discharge and Suc- tion Pipes, Screw Propellers, and Copper Pipes. Part II : Treating of Power of Boilers. — Useful Ratios — Notes on Construction. — Cylindrical Boiler Shells. — Circular Furnaces. Flat Plates. — Stays. — Girders. — Screws. — Hydraulic Tests. — Rtvet- ing. — Boiler Setting, Chimneys, and Mountings. — Fuels, &e. — Exam- ples of Boilers and Speeds of Steamships. — Nominal and Normal Horse Power. — With DIAGRAMS for all Boiler Calculations and Drawings of many Varieties of Boilers. THE WORKS' MANAGER'S HANDBOOK. Comprising Modern Rules, Tables, and Data. For Engineers, Mill- wrights, and Boiler Makers; Tool Makers, Machinists, and Metal Workers; Iron and Brass Founders, etc. By W. S. Hutton, Civil and Mechanical Engineer, Author of "The Practical Engineer's Hand- book." Sixth Edition, carefully Revised and Enlarged. 8vo, strongly bound $6.00 £3^~ The Author having compiled Rules and Data for his own use in a great variety of modern engineering work, and having found his notes extremely use- ful, decided to publish them — revised to date — believing that a practical work, suited to the daily requirements of modern engineers, would be favorably received. "The Author treats every subject from the point of view of one who has collected workshop notes for application in workshop practice, rather than from the theoretical or literary aspect. The volume contains a great deal of that kind of information which is gained only by practical experience and is seldom written in books." — The Engineer. ' STEAM BOILER CONSTRUCTION. A Practical Handbook for Engineers, Boiler-makers, and Steam Users. Containing a large Collection of Rules and Data relating to Recent Prac- tice in the Design, Construction, and Working of all Kinds of Stationary, Locomotive, and Marine Steam-boilers. By Walter S. Hutton, Civil and Mechanical Engineer, Author of "The Works' Manager's Handbook," "The Practical Engineer's Handbook," &c. With up- wards of 500 Illustrations. Fourth Edition, carefully Revised and Enlarged. 8vo, over 680 pages, cloth, strongly bound $6.00 ^P~ This Work is issued in continuation of the series of handbooks written by the Author, viz: "The Works' Manager's Handbook" and "The Practical Engineer's Handbook," which are so highly appreciated by engineers for the practical nature of their information, and is consequently written in the same style as those ivorks. Contents: — Heat, Radiation, and Conduction, Non-conductin g Materials, and Coverings for Steam Boilers. — Composition, Calorific Power, and Evaporative Power of Fuels. — Combustion, Firing Steam Boilers, Products of Combustion, &c. — Chimneys for Steam Boilers,— 4 CROSBY LOCKWOOD &■ SOX'S CATALOGUE. Steam Blast. — FORCE Draught.— FeEd Water. — Effect of Heat on Water. — Expansion of Water by Heat. — Weight of Water at Differ- ent Temperatures. — Convection. — Circulation. — Evaporation. — Properties of Saturated Steam. — Evaporative Power of Boilers. — Priming, etc. — Water-Heating Surfaces of Steam Boilers. — Trans- mission of Heat. — Smoke Tubes. — Evaporative Powers and Effi- ciency of Boilers. — Water Capacity a\d Steam Capacity of Boilers. — Eire-Grates, Eire-Bredges. and Eire-Bars. — Power of Boilers. — Cylindrical Shells and Furnace-Tubes of Boilers. &c. Tests of Materials. — Strength and Weight of Boiler-Plates. — Effect of Temperature on Metals. — Rivet Holes. — Rivets. — Rivet Joints of Steam Bollers.— Caulking. — Ends of Cylindrical Shells. — Stays for Boilers, ire. — Steam Generators. — Description and Pro- portions of Cornish. Lancashire, and Other Types of Stationary Boilers. — Boiler Setting. — Multitubular Boilers. — Locomotive Boilers. — Portable Boilers. — Marine Boilers. — Vertical Boilers. — Water-tube Boilers. — Superheaters. — Cost of Steam Production. — Furnaces for Refuse Fuels. — Destructors, In Drawing, Templating, and Calculating Boiler Work, etc. By J. \ Courtney." Practical Boilermaker. Edited by D. K. Clark, C.E. Seventh Edition. 12mo, cloth .SO BOILERMAKER'S READY RECKONER. With Examples of Practical Geometry and Templating for the Ese of Platers, Smiths, and Riveters. Bv John Courtney. Edited bv D. K. Clark, M.Inst. C.E. Crown 8vo, cloth SI. 60 BOILERMAKER'S READY RECKONER & ASSISTANT. With Examples of Practical Geometry and Templating for the L se of Platers, Smiths, and Riveters. Bv John Courtney. Edited by D. K. Clark. M.Inst. C.E. Fifth Edition. 480 pp., with 140 Illustrations. Fcap. 8vo, half-bound S3.00 *** This Work consists of the t"-o previous-mentioned volumes. "Boiler- maker's Assistant - ' a^ " Boilermaker's Reapt Reckoner," found iOG*th& ftl Q™ Volume. MECHANICAL ENGINEERING, &c. 5 STEAM BOILERS. Their Construction and Management. By R. Armstrong, C.E. Illus- trated. Crown 8vo, cloth t Q() THE PRACTICAL ENGINEER'S HANDBOOK. Comprising a Treatise on Modern Engines and Boilers; Marine, Loco- motive, and Stationary. And containing a large collection of Rules and Practical Data relating to Recent Practice in Designing and Construct- ing all kinds of Engines, Boilers, and other Engineering Work. The whole constituting a comprehensive Key to the Board of Trade and other Examinations for Certificates of Competency in Modern Mechan- ical Engineering. By Walter S. Hutton, Civil and Mechanical En- gineer, Author of "The Works' Manager's Handbook for Engineers, " &c. With upwards of 420 Illustrations. Sixth Edition, Revised and Enlarged. Medium 8vo, nearly 560 pp., strongly bound $7.00 t=P™ This Work is designed as a companion to the Author's "Works' Manager's Handbook." It possesses many new and original features, and contains, like its predecessor, a quantity of matter not originally intended for publication, but collected by the Author for his own use in the construction of a great variety of Modern Engineering Work. The information is given in a condensed and concise form, and is illus- trated by upwards cf 420 Engravings; and comprises a quantity of tabulated matter of great value to all engaged in designing, constructing, or estimating for Engines, Boilers, and other Engineering Work. TEXT-BOOK ON THE STEAM ENGINE. With a Supplement on Gas Engines and Part II. on Heat Engines By T. M. Goodeve, M.A., Barrister-at-Law, Professor of Mechanics at the Royal College of Science, London; Author of "The Principles of Mechanics," "The Elements of Mechanism," &c. Fourteenth Edition. Crown 8vo, cloth $2.00 1 ' Professor Goodeve has given us a treatise on the steam engine which will bear comparison with anything written by Huxley or Maxwell, and we can award it no higher praise." — Engineer. A HANDBOOK ON THE STEAM ENGINE. With especial Reference to Small and Medium-sized Engines. For the Use of Engine Makers, Mechanical Draughtsmen, Engineering Students, and users of Steam Power. By Herman Haeder, C.E. Translated from the German, with additions and alterations, by H. H. P. Powles, A.M.LC.E., M.I.M.E. Third Edition, Revised. With nearly 1,100 Illustrations. 12mo, cloth $3.00 Summary of Contents: — Introduction. — Types of Steam Engines. — Details of Steam Engines. — Governors. — Valve Gears. — Condensers, Air-Pumps, and Feed-Pumps. — Examples of Engines of Continental Make, from Actual Practice. — Particulars of Engines by English Makers. — Compound Engines. — Indicator and Indicator Diagrams. — Calculations for Power and Steam Consumption. — Effect of Inertia on Reciprocating Parts of Engines. — Friction Brake Dynamometer — Sundry Details. — Boilers. — Index. " There can be no question as to its value. We cordially commend it to all concerned in the design and construction of the steam engine." — Mechanical World. THE PORTABLE ENGINE. A Practical Manual on its Construction and Management, for the use of Owners and Users of Steam Engines generally.- By William Dyson Wansbrough. 12mo, cloth $1.50 "This is a work of value to those who use steam machinery. . . . Should be read by every one who has a steam engine, on a form or elsewhere ,"— ■» Mark <£*,*. 6 0R05BY LOCKWOOD 6" SOX'S CATALOGUE. THE STEAM ENGINE. A Treatise on the Mathematical Theory of, with Rules and Examples for Practical Men. By T. Baker, C.E. 12mo, cloth (jO "Teems with scientific information with reference to the steam-engine." — Design and Work. THE STEAM ENGINE. For the use of Beginners. By Dr. Lardxer. 12mo, cloth. . . .(JO LOCOMOTIVE ENGINE DRIVING. A Practical Manual for Engineers in Charge of Locomotive Engines. By Michael Reynolds, M.S.E, TwellLh Edition. 12mo, cloth b °ards S2.00 ' ' We can confidently recommend the book, not only to the practical driver, but to even.- one who takes an interest in the performance of locomotive engines.'* — The Engineer. THE LOCOMOTIVE ENGINE. The Autobiography of an Old Locomotive Engine. By Robert Weathehburn, M.I.M.E. With Illustrations and Portraits of George and Robert Stephexsox. 12mo, cloth SI. 00 THE LOCOMOTIVE ENGINE AND ITS DEVELOPMENT. A Popular Treatise on the Gradual Improvements made in Railway Engines between 1S03 and 1903. By Clement E. Strettox, C.E. Sixth Edition, Revised and Enlarged. 12mo, cloth S2.00 "Students of railway history and all who are interested in the evolution of the modern locomotive will find much to attract and entertain in this volume." — The Times. THE MODEL LOCOMOTIVE ENGINEER, Fireman, and Engine-Boy. Comprising a Historical Notice of the Pioneer Locomotive Engines and their Inventors. By Michael Retx- olds. Second Edition, with Revised Appendix. 12mo, cloth. S2.00 "We should be glad to see this book in the possession of every one in the kingdom who has ever laid, or is to lay, hands on a locomotive engine." — Iron. LOCOMOTIVE ENGINES. A Rudimentary Treatise on. By G. D. Dempset, C.E. With large •Additions treating of the Modern Locomotive, bv D. K. Clark, M.Inst. C.E. With Illustrations. 12mo, cloth S.120 "A model of what an elementary technical book should be." — Academy. CONTINUOUS RAILWAY BRAKES. A Practical Treatise on the several Systems in Use in the United King- dom; their Construction and Performance. By M. Retxolds. 8vo, cloth 83.50 ENGINE-DRIVING LIFE. Stirring Adventures and Incidents in the Lives of Locomotive Engine- Drivers. By Michael Retxolds. Third Edition. 12mo, cloth. ,60 STATIONARY ENGINE DRIVING. A Practical Manual for Engineers in Charge of Stationary Engines. By Michael Retxolds, M.S.E. Seventh Edition. 12mo, cloth boards. S2.00 THE CARE AND MANAGEMENT OF STATIONARY ENGINES. A Practical Handbook for Men-in-charge. By C. Hurst. 12mo. ,50 MECHANICAL ENGINEERING, &c. 7 THE ENGINEMAN'S POCKET COMPANION and Practical Educator for Enginemen, Boiler Attendants, and Me- chanics. By Michael Reynolds. With 45 Illustrations and numer- ous Diagrams. Fifth Edition. Royal 18mo, strongly bound for Pocket wear $1.50 "A most meritorious work, giving in a succinct and practical form all the information an engine-minder, desirous of mastering the scientific principles of his daily calling, would require." — The Miller., THE SAFE USE OF STEAM. Containing Rules for Unprofessional Steam Users. By an Engineer. Eighth Edition. Sewed .25 "If steam-users would but learn this little book by heart, boiler explo- sions would become sensations by their rarity." — English Mechanic. STEAM AND MACHINERY MANAGEMENT. A Guide to the Arrangement and Economical Management of Machin- ery, with Hints on Construction and Selection. By M. Powis Bale, M.Inst.M.E. 12mo, cloth $1.00 GAS AND OIL ENGINE MANAGEMENT. A Practical Guide for Users and Attendants, being Notes on Selection, Construction, and Management. By M. Powis Bale, M.Inst.C.E., M.I.Mech.E. Author of "Woodworking Machinery," &c. 12mo, cloth $1.50 ON GAS ENGINES. With Appendix describing a Recent Engine with Tube Igniter. By T. M. Goodeve, M.A. 12mo, cloth $1.00 THE ENGINEER'S YEAR=BOOK FOR 1906. Comprising Formulae, Rules, Tables, Data, and Memoranda in Civil, Mechanical, Electrical, Marine, and Mine Engineering. By H. R. Kempe, M.Inst.C.E., Principal Staff Engineer, Engineer-in-Chief's Office, General Post Office, London; Author of "A Handbook of Elec- trical Testing," "The Electrical Engineer's Pocket-Book," &c. With 1,000 Illustrations, specially Engraved for the Work. 12mo, 950 pp., leather $3.00 THE MECHANICAL ENGINEER'S POCKET=BOOK. Comprising Tables, Formulae, Rules, and Data: a Handy Book of Ref- erence for Daily Use in Engineering Practice. By D. Kinnear Clark, M.Inst.C.E. , Fifth Edition, thoroughly Revised and Enlarged. By H. H. P. Powles, A.M.Inst.C.E., M.I.M.E. Small 8vo, 700 pp., leather. $3, QQ Summary of Contents: — Mathematical Tables. — Measurement of Surfaces and Solids. — English Weights and Measures. — French Metric Weights and Measures. — Foreign Weights and Measures. — Moneys. — Specific Gravity, Weight, and Volume. — Manufactured Metals. — Steel Pipes. — Bolts and Nuts. — Sundry Articles in Wrought and Cast Iron, Copper, Brass, Lead, Tin, Zinc. — Strength of Mater- ials. — Strength of Timber. — Strength of Cast Iron. — Strength of Wrought Iron. — Strength of Steel. — Tensile Strength of Copper, Lead, &c. — Resistance of Stones and other Building Materials. — Riveted Joints in Boiler Plates. — Boiler Shells. — Wire Ropes and Hemp Ropes — Chains and Chain Cables. — Framing. — Hardness of Metals, Alloys, and Stones. — Labour of Animals. — Mechanical Prin- ciples. — Gravity and Fall of Bodies. — Accelerating and Retarding Forces. — Mill Gearing, Shafting, &c. — Transmission of Motive Power. — Heat. — Combustion. — Fuels. — Warming, Ventilation, Cooking Stoves. — Steam. — Steam Engines and Boilers. — Railways. — Tram- ways. — Steam Ships. — Pumping Steam Engines and Pumps. — Coal Gas, Gas Engines, &c. — Air in Motion. — Compressed Air. — Hot- Air Engines. — Water Power. — Speed of Cutting Tools. — Colours. — Electrical Engineering. 8 CROSBY LOCKWOOD 6 s SON'S CATALOGUE. PRACTICAL MECHANICS' WORKSHOP COMPANION. Comprising a great Variety of the most useful Rules and Formula? in Mechanical Science, with numerous Tables of Practical Data and Cal- culated Results for Facilitating Mechanical Operations. By William Templeton, Author of "The Engineer's Practical Assistant," &c, &c. Eighteenth Edition, Revised, Modernised, and considerably Enlarged, by W. S. Hutton, C.E., Author of "The Works' Manager's Hand- book," &c. Fcap. 8vo, nearly 500 pp., with 8 Plates and upwards of 250 Diagrams, leather $2.50 ENGINEER'S AND MILLWRIGHT'S ASSISTANT. A Collection of Useful Tables, Rules, and Data. By William Temple- ton. Eighth Edition, with Additions. 18mo, cloth SI. 00 TABLES AND MEMORANDA FOR ENGINEERS, MECHANICS, ARCHITECTS, BUILDERS, &c. Selected and Arranged by Francis Smith. Seventh Edition, Revised, including Electrical Tables, Formula, and Memoranda. Waist- coat-pocket size, limp leather .60 THE MECHANICAL ENGINEER'S COMPANION. Of Areas, Circumferences, Decimal Equivalents, in inches and feet, mil- limetres, squares, cubes, roots, &c; Strength of Bolts, Weight of Iron, &c; Weights, Measures, and other Data. Also Practical Rules for Eneane Proportions. By R. Edwards, M.Inst.C.E. Fcap. 8vo, cloth. SI. 00 MECHANICAL ENGINEERING TERMS. (Lockwood's Dictionary of). Embracing those current in the Drawing Office, Pattern Shop, Foundry, Fitting, Turning, Smiths', and Boiler Shops, &c. Comprising upwards of 6,000 Definitions. Edited by J. G Horner, A.M.I.M.E. Third Edition, Revised, with Additions. 12mo, cloth S3.00 "Just the sort of handy dictionary required by the various trades engaged in mechanical engineering. The practical engineering pupil will find the book of great value in his studies, and every foreman engineer and mechanic should have a copy." POCKET GLOSSARY OF TECHNICAL TERMS. English-French, French-English ; with Tables suitable for the Archi- tectural, Engineering, Manufacturing, and Nautical Professions. By John James Fletcher. Fourth Edition, 200 pp. Waistcoat -pocket size, limp leather .60 IRON AND STEEL. A Work for the Forge Foundry, Factory, and Office. Containing ready, useful, and trustworthy Information for Ironmasters and their Stock- takers'; Managers of Bar, Rail, Plate, and Sheet Rolling Mills; Iron and Metal Founders; Iron, Ship, and Bridge Builders; Mechanical, Mining, and Consulting Engineers; Architects, Contractors, Builders, &c. By Charles Hoare, Author of "The Slide Rule," &c. Ninth Edition. 32mo, leather S2.50 WORKMAN'S MANUAL OF ENGINEERING DRAWING. By John Maxton, Instructor in Engineering Drawing, Royal Naval College, Greenwich. Eighth Edition. 300 Plates and Diagrams. 12mo, cloth S1.40 "A copy of it should be kept for reference in every drawing office." — En- gineering. PATTERN MAKING. Embracing the Main Types of Engineering Construction, and including Gearing, Engine Work, Sheaves and Pulleys, Pipes and Columns, Screws, Machine Parts, Pumps and Cocks, the Moulding of Patterns in Loam and Greensand, Weight of Castings, &c. By J. G. Horner, A.M.I.M.E. Third Edition, Enlarged. With 486 Illustrations. 12mo, cloth. S3.00 MECHANICAL ENGINEERING, &c. g SMITHY AND FORGE. Including the Farrier's Art and Coach Smithing. By W. J. E. Crane. 12mo, cloth $1.00 "The first modern English book on the subject. Great pains have been bestowed by the author upon the book; shoeing-smiths will find it both useful and interesting." TOOTHED GEARING. A Practical Handbook for Offices and Workshops. By J. Horner, A.M.I.M.E. Second Edition, with a new Chapter on Recent Practice. With 184 Illustrations. 12mo, cloth $2.25 MODERN WORKSHOP PRACTICE, As applied to Marine, Land, and Locomotive Engines, Floating Docks, Dredging Machines, Bridges, Shipbuilding, &c. By J. G. Winton. Fourth Edition, Illustrated. 12mo, cloth $1.40 DETAILS OF MACHINERY. Comprising Instructions for the Execution of various Works in Iron in the Fitting Shop, Foundry, and Boiler Yard. By Francis Campin, C.E. 12mo, cloth $1.20 ENGINEERING ESTIMATES, COSTS, AND ACCOUNTS. A Guide to Commercial Engineering. With numerous examples of Es- timates and Costs of Millwright Work, Miscellaneous Productions, Steam Engines and Steam Boilers; and a Section on the Preparation of Costs Accounts. By A General Manager. Second Edition. 8vo, cloth $4.50 MECHANICAL ENGINEERING. Comprising Metallurgy, Moulding, Casting, Forging, Tools, Workshop Machinery, Mechanical Manipulation, Manufacture of the Steam En- gine, &c. By Francis Campin, C.E. Third Edition. 12mo, cloth $1.00 LATHE=WORK. A Practical Treatise on the Tools. Appliances, and Processes employed in the Art of Turning. By Paul N. Hasltjck. Eighth Edition. 12mo, cloth $2.00 "Written by a man who knows not only how work ought to be done, but who also knows how to do it, and how to convey his knowledge to others." — Engineering. SCREW=TH READS, And Methods of Producing Them. With numerous Tables and com- plete Directions for using Screw-cutting Lathes. By Paul N. Hasluck. Author of "Lathe-work," &c. Sixth Edition. Waistcoat-pocket size. .60 "Full of useful information, hints and practical criticism. Taps, dies, and screwing tools generally are illustrated and their action described." CONDENSED MECHANICS. A Selection of Formula?, Rules, Tables, and Data for the Use of Engi- neering Students, &c. By W. G. C. Hughes, A.M.I.C.E. 12mo, cloth. $1.00 MECHANICS OF AIR MACHINERY. By Dr. J. Weisbach and Prof. G. Herrmann. Authorized Translation with an Appendix on American Practice by A. Trowbridge, Ph.B., Adjunct Professor of Mechanical Engineering, Columbia University. Royal 8vo, cloth. Net $3.75 io CROSBY L0CKW00D 6- SON'S CATALOGUE. PRACTICAL MECHANISM. And Machine Tools. By T. Baker, C.E. With Remarks on Tools and Machinery by J. Nasmyth, C.E. 12mo, cloth $1.00 MECHANICS. Being a concise Exposition of the General Principles of Mechanical Science and their Applications. By C. Tomlinson, F.R.S. 12mo, cloth .60 FUELS: SOLID, LIQUID, AND GASEOUS. Their Analysis and Valuation. For the use of Chemists and Engineers. By H. J. Phillips, F.C.S., formerly Analytical and Consulting Chemist to the Great Eastern Railway. Fourth Edition. 12mo, cloth. . ,gQ "Ought to have its place in the laboratory of every metallurgical estab- lishment and wherever fuel is used on a large scale." — Chemical News. FUEL, ITS COMBUSTION AND ECONOMY. Consisting of an Abridgment of "A Treatise on the Combustion of Coal and the Prevention of Smoke." By C. W. Williams, A. Inst. C.E. With extensive Additions by D. Kinnear Clark, M.Inst. C.E. Fourth Edition. 12mo, cloth $1.50 "Students should buy the book and read it, as one of the most complete and satisfactory treatises on the combustion and economy of fuel to be had . ' ' — Engineer. STEAM AND THE STEAM ENGINE, Stationary and Portable. Being an Extension of the Treatise on the Steam Engine of Mr. J. Sewell. By D. K. Clark, C.E. Fourth Edi- tion. 12mo, cloth $1.40 "Every essential part of the subject is treated of competently, and in a popular style." PUMPS AND PUMPING. A Handbook for Pump Users. Being Notes on Selection, Construction, and Management. By M. Powis Bale, M.Inst. C.E. , M.I.Mech.E. Fourth Edition. 12mo», cloth $1.50 " Thoroughly practical and clearly written." REFRIGERATION, COLD STORAGE, & ICE-MAKING. A Practical Treatise on the Art and Science of Refrigeration. By. A. J. Wallis-Tayler, A.M. Inst. C.E. , Author of "Refrigerating and Ice- Making Machinerv." 600 pp., with 360 Illustrations. Medium 8vo, cloth $4.50 Contents: — Chapter I. Introduction. — II. The Theory and Pracs tice op Refrigeration. — III. The Liquefaction Process. — IV. The Vacuum Process. — V. The Compression Process or System. — VI. The Compression Process (Continued). — VII. The Compression Process (Con- tinued). — VIII. Condensers and Water-Cooling and Saving Apparatus. — IX. The Absorption and Binary Absorption Process or System. — X. The Cold-Air System. — XI. Cocks, Valves and Pipe-Joints and Unions. — XII. Refrigeration and Cold Storage. — XIII. Refrigera- tion and Cold Storage (Continued). — XIV. Refrigeration and Cold Storage (Continued). — XV. Refrigeration and Cold Storage (Con- tinued). — XVI. Marine Refrigeration. — XVII. Manufacturing, In- dustrial and Constructional Applications. — XVIII. Ice-Making. — XIX. The Management and Testing of Refrigerating Machinery. — XX. Cost of Working. — XXI. The Production of Very Low Temper- atures. — XXII. Useful Tables and Memoranda. — Appendix. — Bibli' OGRAPHY OF REFRIGERATION. CIVIL ENGINEERING, SURVEYING, &c. n THE POCKET BOOK OF REFRIGERATION AND ICE- MAKING. By A. J. Wallis-Tayler, A.M. Inst. C.E. Author of "Refrigerating and Ice-making Machinery," &c. Third Edition, Enlarged. 12mo, cloth $1.50 REFRIGERATING & ICE=MAKING MACHINERY. A Descriptive Treatise for the Use of Persons Employing Refrigerating and Ice-making Installations, and others. By A. J. Wallis-Tayler, A.M.Inst.C.E. Third Edition, Enlarged. 12mo, cloth $3.00 "May be recommended as a useful description of the machinery, the proc- esses, and of the acts, figures, and tabulated physics of refrigerating." — En- gineer. MOTOR VEHICLES FOR BUSINESS PURPOSES. A Practical Handbook for those interested in the Transport of Passen- gers and Goods. By A. J. Wallis-Tayler, A.M.Inst.C.E. With 134 Illustrations. Demy 8vo, cloth [Just -published. ,] $3.50 MOTOR CARS OR POWER=CARRIAGES FOR COMMON ROADS. By A. J. Wallis-Tayler, A.M.Inst.C.E. 212 pp., with 76 Illustrations. 12mo, cloth $2.00 AERIAL NAVIGATION. A Practical Handbook on the Construction of Dirigible Balloons, Aero- stats, Aeroplanes, and Aeromotors. By Frederick Walker, C.E., Associate Member of the Aeronautic Institute. With 104 Illustrations. Large 12mo, cloth $3.00 STONE=WORKING MACHINERY. A Manual dealing with the Rapid and Economical Conversion of Stone. With Hints on the Arrangement and Management of Stone Works. By M. Powis Bale, M.Inst.C.E., M.I.Mech.E. Second Edition, enlarged. 12mo, cloth $3.50 "The book should be in the hands of every mason or student of stone- work." "A handbook for all who manipulate stone for building or ornamental purposes." FIRES, FIRE=ENGINES, AND FIRE BRIGADES. With a History of Fire-Engines, their Construction, Use, and Manage- ment; Foreign Fire Systems; Hints on Fire-Brigades, &c. By C. F. T. Young, C.E. 8vo, cloth $8.00 CRANES. The Construction of, and other Machinery for Raising Heavy Bodies "for the Erection of Buildings, &c. By J. Glynn, F.R.S. 12mo, cloth. .60 CIVIL ENGINEERING, SURVEYING, ETC. PIONEER IRRIGATION. A Manual of Information for Farmers in the Colonies. By E. 0. Maw- son, M.Inst.C.E., Executive Engineer, Public Works Department, Bombay. With Additional Chapters on Light Railways by E. R. Calthrop, M.Inst.C.E., M.I.M.E. Illustrated by numerous Plates and Diagrams. Demy 8vo, cloth $4.00 Summary of Contents: — Value of Irrigation, and Sources op Water Supply. — Dams and Weirs. — Canals. — Underground Water. — Meth- ods of Irrigation. — Sewage Irrigation. — Imperial Automatic Sluice Gates. — The Cultivation of Irrigated Crops, Vegetables, and Fruit Trees. — Light Railways for Heavy Traffic. — Useful Memoranda and Data. 12 CROSBY LOCKWOOD & SOX'S CATALOGUE. THE RECLAMATION OF LAND FROM TIDAL WATERS. A Handbook for Engineers, Landed Proprietors, and others interested in Works of Reclamation. By A. Beazely, M.Inst.C.E. 8vo, cloth. S4.00 "The book shows in a concise way what has to be done in reclaiming land from the sea, and the best way of doing it. Contains a great deal of prac- tical and useful information which cannot fail to be of service to engineers entrusted with the enclosure of salt marshes, and to landowners intending to reclaim land from the sea." — The Engineer. THE WATER SUPPLY OF TOWNS AND THE CON- STRUCTION OF WATER-WORKS. A Practical Treatise for the Use of Engineers and Students of Engineer- ing. By W. K. Botox, A. M.Inst.C.E., Consulting Engineer to the Tokyo Water-works. Second Edition, Revised and Extended. With numerous Plates and Illustrations. Super-royal Svo, buckram. S9.00 I. INTRODUCTORY. II. DIFFERENT QUALITIES OF WATER. III. QUAN- TITY of Water to be Provided. — IV. On Ascertaining whether a Pro- posed Source of Supply is Sufficient. — V. On Estimating the Storage Capacity Required to be Provided. — VI. Classification of Water- works. — VII. Impounding Reservoirs. — VIII. Earthwork Dams. — IX. Masonry Dams. — X. The Purification of Water. — XI. Settling Res- ervoirs. — XII. Sand Filtration. — XIII. Purification of Water by Action of Iron, Softening of Water by Action of Lime, Xatural Filtration. — XIV. Service or Clean Water Reservoirs — Water Towers — Stand Pipes. — XV. The Connection of Settling Reservoirs, Filter Beds and Service Reservoirs. — XVI. Pumping Machinery. — XVII. Flow of Water in Conduits — Pipes and Open Channels. — XVIII. Distribution Systems. — XIX. Special Provisions for the Ex- tinction of Fire. — XX. Pipes for Water-works. — XXI. Prevention of Waste of Water. — XXII. Various Appliances used in Connection with Water-works. Appendix I. By Prof. JOHN MILXE. F.R.S.— Considerations Con- cerning the Probable Effects of Earthquakes on Water-works, and the Special Precautions to be Taken in Earthquake Countries. Appendix II. By JOHN DE RIJKE, C.E.— On Sand Dunes and Dune Sand as a Source of Water Supply. THE WATER SUPPLY OF CITIES AND TOWNS. By William Humber. A. M.Inst.C.E., and M.Inst.M.E., Author of "Cast and Wrought Iron Bridge Construction," &c, &c. Illustrated with 50 Double Plates. 1 Single Plate. Coloured Frontispiece, and up- wards of 250 Woodcuts, and containing 400 pp. of Text. Imp. 4to, elegantly and substantially half-bound in morocco S45.00 List of Contents: — I. Historical Sketch of some of the means that HAVE BEEN ADOPTED FOR THE SUPPLY OF WATER TO CrTIES AND TOWNS. IL Water and the Foreign Matter usually associated with it. — III. Rainfall and Evaporation. — IV. Springs and the Water-bearing Formations of Various Districts. — V. Measurement and Estimation of the Flow of Water. — VI. On the Selection of the Source of Sup- ply. — VII. Wells. — VIII. Reservoirs. — IX. The Purification of Water. — X Pumps. — XI. Pumping Machinery. — XII. Conduits. — XIII. Distribution of Water. — XIV. Meters. Service Pipes, and House Fittings. — XV. The Law and Economy of Water-works. — XVI. Constant and Intermittent Supply. — XVII. Description of Plates. — Appendices, giving Tables of Rates of Supply. Velocities, &c, &c, together with specifications of several works illustrated. among which will be found: aberdeen. bldeford, canterbury, dundee, Halifax, Lambeth, Rotherham, Dublin, and others. RURAL WATER SUPPLY. A Practical Handbook on the Simply of Water and Construction of Water-works for small Count rv Disfricts. Bv Allan Greenwell, A .M.Inst.C.E., and W. T. Curry. A.M Inst.C.E., F.G.S. With Illus- te-»tiene, Seeesd Edition. Revised. 12mo. clotfe. ........... $2.QO CIVIL ENGINEERING, SURVEYING, &c. 13 WATER ENGINEERING. A Practical Treatise on the Measurement, Storage, Conveyance, and Utilization of Water for the Supply of Towns, for Mill Power, and for other Purposes. By Charles Slagg, A.M.Inst.C.E. Second Edition. 12mo, cloth $3.00 WATER WORKS, FOR THE SUPPLY OF CITIES AND TOWNS. With a Description of the Principal Geological Formations of England as influencing Supplies of Water. By Samuel Hughes. 12mo, cloth $1.60 POWER OF WATER. As applied to drive Flour Mills, and to give motion to Turbines, and other Hydrostatic Engines. By Joseph Glynn, F.R.S., &c. New Edition. Illustrated. 12mo, cloth gq WELLS AND WELL=SINKING. By J. G. Swindell, A.R.I.B.A., and G. R. Burnell, C.E. Revised Edition. 12mo, cloth ^gQ "Solid practical information, written in a concise and lucid style. The work can be recommended." HYDRAULIC POWER ENGINEERING. A Practical Manual on the Concentration and Transmission of Power by Hydraulic Machinery. By G. Croydon Marks, A.M.Inst.C.E. Second Edition, Enlarged, with about 240 Illustrations. 8vo, cloth. [Just Published. $3.50 Summary of Contents: — Principles of Hydraulics. — The Flow of Water. — Hydraulic Pressures. — Material. — Test Load. — Packings for Sliding Surfaces. — Pipe Joints. — Controlling Valves. — Platform Lifts. — Workshop and Foundry Cranes. — Warehouse and Dock Cranes. — Hydraulic Accumulators. — Presses for Baling and other Purposes. — Sheet Metal Working and Forging Machinery. — Hy- draulic Riveters. — Hand and Power Pumps. — Steam Pumps. — Tur- bines. — Impulse Turbines. — Reaction Turbines. — Design of Tur- bines in Detail. — Water Wheels. — Hydraulic Engines. — Recent Achievements. — Pressure of Water. — Action of Pumps, &c. HYDRAULIC MANUAL. Consisting of Working Tables and Explanatory Text. Intended as a Guide in Hydraulic Calculations and Field Operations. By Lowis D'A. Jackson, Author of "Aid to Survey Practice," "Modern Metrol- ogy," &c. Fourth Edition, Enlarged. 8vo, cloth $6.00 "The author has constructed a manual which may be accepted as a trust- worthy guide to this branch of the engineer's profession." — Engineering. HYDRAULIC TABLES, CO=EFFICIENTS, & FORMUL/E. For Finding the Discharge of Water from Orifices, Notches, Weirs, Pipes, and Rivers. With New Formula?, Tables, and General Informa- tion on Rain-fall, Catchment-Basins, Drainage, Sewerage, Water Sup- ply for Towns and Mill Power. By John Neville, C.E. , M.R.I.A- Third Edition, revised, with additions. Numerous Illustrations. 12mo, cloth. $5.00 "It is, of all English books on the subject, the one nearest to complete- ness." MASONRY DAMS FROM INCEPTION TO COMPLETION. Including numerous Formula?, Forms of Specifications and Tender, Pocket Diagram of Forces, &c. For the use of Civil and Mining En- gineers. By C. F. Courtney, M.Inst.C.E. 8vo, cloth $3.50 "Contains a good deal of valuable data. Many useful suggestions will be found in the remarks on site and position, location of darn, foundations and constuct'on." — Building News. 14 CROSBY LOCKWOOD & SON'S CATALOGUE. RIVER BARS. The Causes of their Formation, and their Treatment by "Induced Tidal Scour"; with a Description of the Successful Reduction by this Method of the Bar at Dublin. By I. J. Mann, Assist. Eng. to the Dublin Port and Docks Board. Royal 8vo, cloth $3.00 "We recommend all interested in harbour works — and, indeed, (hose con- cerned in the improvements of rivers generally — to read Mr. Mann's inter- esting work." — Engineer. DRAINAGE OF LANDS, TOWNS, AND BUILDINGS. By G. D. Dempsey, C.E. Revised, with large Additions on Recent Practice in Drainage Engineering by D. Kinnear Clark, M.fnst.C.E. Fourth Edition. 12mo, cloth $1.80 i SURVEYING AS PRACTISED BY CIVIL ENGINEERS AND SURVEYORS. Including the Setting-out of Works for Construction and Surveys Abroad, with many Examples taken from Actual Practice. A Hand- book for use in the Field and the Office, intended also as a Text -book for Students. By John Whitelaw, Jun., A. M.Inst. C.E. , Author of "Points and Crossings." With about 260 Illustrations. Demy 8vo, cloth $4.00 PRACTICAL SURVEYING. A Text-book for Students preparing for Examination or for Sur-vey- work in the Colonies. By C.eorge W. Usiel, A.M.Inst.C.E. Eighth Edition, thoroughly Revised and Enlarged, ry Aiex Beazeley, M.Inst.C.E. With 4 Lithographic Plates and 360 Illustrations. 12mo, cloth $3.00 SURVEYING WITH THE TACHEOMETER. A practical Manual for the use of Civil and Military Engineers and Sur- veyors, including two series of Tables specially computed for the Re- duction of Readings in Sexagesimal and in Centesimal Degrees. By Neie Kennedy, M.Inst.C.E. With Diagrams and Plates. Second Edition. 8vo, cloth -$4.00 "The work is very clearly written, and should remove all difficulties in the way of any surveyor desirous of making use of this useful and rapid instru- ment . ' ' — Nature. LAND AND ENGINEERING SURVEYING. For Students and Practical Use. Py T. Baker, C.E. Twentieth Edi- tion, by F. E. Dixon, A.M.Inst.C.E. With Plates and Diagrams. 12mo, cloth 80 AID TO SURVEY PRACTICE. For Reference in Surveying, Levelling, and Setting-out; and in Route Surveys of Travellers by Land and Sea. With Tables, Illustrations, and Records. By L. D'A. Jackson, A.MJnst.C.E. Second Edition. 8vo, cloth $5.00 LAND AND MARINE SURVEYING. In Reference to the Preparation of Plans for Roads and Railways; Canals, Rivers, Towns' Water Supplies; Docks and Harbours. With Description and Use of Surveying Instruments. By W. Davis Haskoli., C.E. Second Edition , Revised with Additions. Crown 8vo, cloth . $3.50 CIVIL ENGINEERING, SURVEYING, &c. 15 ENGINEER'S & MINING SURVEYOR'S FIELD B( £>K. Consisting of a Series of Tables, with Rules, Explanations of Systems and use of Theodolite for Traverse Surveying- and plotting the work with minute accuracy by means of Straight Edge and Set Square only; Levelling with the Theodolite, Setting-out Curves with and without the Theodolite, Earthwork Tables, &c. By W. Davis Haskoll, C.E. With numerous Woodcuts. Fifth Edition, Enlarged. J2mo, cloth. $4.50 "The book is very handy; the separate tables of sines and tangents to every minute will make it useful for many other purposes, the genuine traverse tables existing all the same." AN OUTLINE OF THE METHOD OF CONDUCTING A TRIGONOMETRICAL SURVEY. For the Formation of Geographical and Topographical Maps and Plans, Military Reconnaissance, LEVELLING, &c, with Useful Problems, Formulae, and Tables. By Lieut. -General Frome, R.E. ]c. 19 PNEUMATICS, Including Acoustics and the Phenomena of Wind Currents, for the use of Beginners. By Charles Tomlinson, F.R.S. 12mo, cloth. t QQ FOUNDATIONS AND CONCRETE WORKS. With Practical Remarks on Footings, Planking, Sand, Concrete, B^ton, Pile-driving, Caissons, and Cofferdams. By E. Dobson. 12mo. # gQ BLASTING AND QUARRYING OF STONE, For Building and other Purposes. With Remarks on the Blowing up of Bridges. By Gen. Sir J. Burgoyne, K.C.B. 12mo, cloth. .. . t QQ SAFE RAILWAY WORKING. A Treatise on Railway Accidents, their Cause and Prevention; with a Description of Modern Appliances and Systems. By Clement E. Stretton, C.E. Third Edition, Enlarged. 12mo, cloth $1.50 "A book for the engineer, the directors, the managers; and, in short, all who wish for information on railway matters will find a perfect encyclopaedia in 'Safe Railway Working.'" — Railway Review. MARINE ENGINEERING, SHIPBUILDING, NAVIGATION, ETC. MARINE ENGINES AND BOILERS. Their Design and Construction. A Handbook for the Use of Students, Engineers, and Naval Constructors. Based on the Work "Berechnung und Konstruktion der Schiffsmaschinen und Kessel," by Dr. G. Bauer, Engineer-in-Chief of the Vulcan Shipbuilding Yard, Stettin. Translated from the Second German Edition by E. M. Donkin, and S. Brian Donkin, A.M. I. C.E. Edited by Leslie S. Robertson, Secretary to the Engineering Standards Committee, M.I.C.E., M.I.M.E., M.I.N.A., &c. With numerous Illustrations and Tables. Thiek 8vo, cloth, [Just Published. $9,00 Summary of Contents:— PART I.— MAIN ENGINES.— Determina- tion op Cylinder Dimensions. — The Utilisation of Steam in the En- gine. — Stroke of Piston. — Number of Revolutions. — Turning Moment. — Balancing of the Moving Parts. — Arrangement of Main Engines. — Details of Main Engines. — The Cylinder. — Valves. — Various Kinds of Valve Gear. — Piston Rods. — Pistons. — Connecting Rod and Cross- head. — Valve Gear Rods. — Bed Plates. — Engine Columns. — Revers- ing and Turning Gear. PART II. — PUMPS. — Air, Circulating Feed and Auxiliary Pumps. PART III.— SHAFTING, RESISTANCE OF SHIPS, PROPELLERS.— Thrust Shaft and Thrust Block.— Tunnel- Shafts and Plummer Blocks. — Shaft Couplings. — Stern Tube. — The Screw Propeller. — Construction of the Screw. PART IV. — PIPES AND CONNECTIONS.— General Remarks, Flanges, Valves, &c— Under Water Fittings. — Main Steam, Auxiliary Steam, and Exhaust Piping. — Feed Water, Bilge, Ballast and Circulating Pipes. PART V. — STEAM BOILERS. — Firing and the Generation of Steam. — Cylindrical Boilers. — Locomotive Boilers. — Water-tube Boilers. — Small Tube Water-Tube Boilers. — Smoke Box. — Funnel and Boiler Lagging. — Forced Draught. — Boiler Fittings and Mountings. PART VI.— MEASURING INSTRUMENTS. PART VII.— VARIOUS DETAILS. — Bolts, Nuts, Screw Threads, &c. — Platforms, Gratings, Ladders. — Foundations. — Seatings. — Lubrication. — Ventilation of Engine Rooms. — Rules for Spare Gear. PART VIII. — ADDITIONAL TABLES. 20 CROSBY LOCKWOOD & SON'S CATALOGUE. THE NAVAL ARCHITECT'S AND SHIPBUILDER'S POCKET-BOOK Of Formulae, Rules, and Tables, and Marine Engineer's and Surveyor's Handy Book of Reference. By Clement Mackrow, M.I.N. A. Eighth Edition, carefully Revised and Enlarged. Ecap, leather. . Ael So.00 Summary of Contents:— Signs and Symbols, Decimal Fractions. — Trigonometry. — Practical Geometry. — Mensuration. — Centres and Moments of Figures. — Moments of Inertia and Radii Gyration — Al- gebraical Expressions for Simpson's Rules. — Mechanical Principles. — Centre of Gravity. — Laws of Motion. — Displacement, Centre of Buoyancy. — Centre of Gravity of Ships' Hull. — Stability Curves and Metacentres. — Sea and Shallow- water Waves. — Rolling of Ships. — Propulsion and Resistance of Vessels. — Speed Trials. — Sailing, Cen- tre of Effort. — Distances down Rivers, Coast Lines. — Steering and Rudders of Vessels. — Launching Calculations and Velocities. — Weight of Material and Gear. — Gun Particulars and Weight. — Standard Gauges. — Riveted Joints and Riveting. — Strength and Tests of Materials. — Binding and Shearing Stresses. — Strength of Shafting, Pillars, Wheels, &c. — Hydraulic Data, &c. — Conic Sec- tions, Catenarian Curves. — Mechanical Powers, Work. — Board of Trade Regulations for Boilers and Engines. — Board of Trade Reg- ulations for Ships. — Lloyd's Rules for Boilers. — Lloyd's Weight of Chains. — Lloyd's Scantlings for Ships. — Data of Engines and Ves- sels. — Ships' Fittings and Tests. — Seasoning Preserving Timber. — ■ Measurement of Timber. — Alloys, Paints, Varnishes. — Data for Stow- age. — Admiralty Transport Regulations. — Rules for Horse-power, Screw Propellers, &c. — Percentages for Butt Straps. — Particulars of Yachts. — Masting and Rigging. — Distances of Foreign Ports. — Tonnage Tables. — Vocabulary of French and English Terms. — English Weights and Measures. — Foregn Weights and Measures. — Decimal Equivalents. — Useful Numbers. — Circular Measures. — Areas of and Circumferences of Circles. — Areas of Segments of Circles. — Tables of Squares and Cubes and Roots of Numbers. — Tables of Logarithms of Numbers. — Tables of Hyberpolic Logarithms. — Tables of Natural Sines, Tangents. — Tables of Logarithmic Sines, Tangents, &c. WANNAN'S MARINE ENGINEER'S GUIDE To Board of Trade Examinations for Certificates of Competency. Con- taining all Latest Questions to Date, with Simple, Clear, and Correct Solutions; 302 Elementary Questions with Illustrated Answers, and Verbal Questions and Answers; complete Set of Drawings with State- ments completed. By A. C. Wannan, C.E., Consulting Engineer, and E. W. I. Wannan, M.I.M.E., Certificated First Class Marine Engineer. With numerous Engravings. Fourth Edition, Enlarged. 500 pages. 8vo, cloth S4.00 WANNAN'S MARINE ENGINEER'S POCKET=BOOK. Containing Latest Board of Trade Rules and Data for Marine Engineers. By A. C. Wannan. Third Edition, Revised, Enlarged, and Brought up to Date. Square 18mo, with thumb Index, leather S2.00 MARINE ENGINES AND STEAM VESSELS. By R. Murray, C.E. Eighth Edition, thoroughly Revised, with Addi- tions by the Author and by George Carlisle, C.E. 12mo, cloth . SI ,80 ELEMENTARY MARINE ENGINEERING. A Manual for Young Marine Engineers and Apprentices. By J. S. Brewer. 12mo, cloth .60 CHAIN CABLES AND CHAINS. Comprising Sizes and Curves of Links, Studs, &c, Iron for Cables and Chains, Chain Cable and Chain Making, Forming and Welding Links, Strength of Cables and Chains, Certificates for Cables, Marking Cables, Prices of Chain Cables and Chains, Historical Notes, Acts of Parlia- ment, Statutory Tests, Charges for Testing, List of Manufacturers of MARINE ENGINEERING, NAVIGATION, &c 21 Cables, &c, &c. By Thomas W, Traill, F.E.R.N., M.Inst.C.E., En- gineer-Surveyor-in-Chief , Board of Trade, Inspector of Chain Cable and Anchor Proving Establishments, and General Superintendent, Lloyd's Committee on Proving Establishments. With numerous Tables, Illus- trations, and Lithographic Drawings. Folio, cloth $15.00 THE SHIPBUILDING INDUSTRY OF GERMANY. Compiled and Edited by G. Lehmann-Felskowski. With Coloured Prints, Art Supplements, and numerous Illustrations throughout the text. Super-royal 4to, cloth $4.20 SHIPS AND BOATS. By W. Bland. With numerous Illustrations and Models. Tenth Edi- tion. 12mo, cloth .60 SHIPS FOR OCEAN AND RIVER SERVICE, Principles of the Construction of. By H. A. Sommerfeldt. 12mo. .60 AN ATLAS OF ENGRAVINGS To illustrate the above. Twelve large folding Plates. Royal 4to, cloth $3.00 NAVAL ARCHITECTURE. An Exposition of the Elementary Principles. By J. Peake. 12mo. doth $1.40 THE ART AND SCIENCE OF SAILMAKING. By Samuel B. Sadler, Practical Sailmaker, late in the employment of Messrs. Ratsey and Lapthorne, of Cowes and Gosport. Plates. 4to, doth $5.00 "This extremely practical work gives a complete education in all the branches of the manufacture, cutting out, roping, seaming, and goring. It s copiously illustrated, and forms a first-rate text-book and guide." SAILS AND SAIL=MAKING. With Draughting, and the Centre of Effort of the Sails. Weights and Sizes of Ropes; Masting, Rigging, and Sails of Steam Vessels, &c. "By R. Kipping, N.A. 12mo, cloth $1.00 MASTING, MAST=MAKING, AND RIGGING OF SHIPS. Also Tables of Spars, Rigging, Blocks; Chain, Wire, and Hemp Ropes, &c, relative to every class of vessels. By R. Kipping. 12mo, cloth, .80 SEA TERMS, PHRASES, AND WORDS (Technical Dictionary of) used in the English and French Languages (English- French, French-English). For the Use of Seamen, Engineers, Pilots, Shipbuilders, Shipowners, and Ship-brokers. Compiled by W. Pirrie, late of the African Steamship Company. Fcap, 8vo, cloth li m P $2.00 This volume will be highly appreciated by seamen, engineers, pilots, ship- builders and shipowners. It will be found wonderfully accurate and com- plete. SAILOR'S SEA BOOK: A Rudimentary Treatise on Navigation. By James Greenwood, B.A. With numerous Woodcuts and Coloured Plates. New and Enlarged Edition. By W. H. Rosser. 12mo, cloth $1.00 Is perhaps the best and simplest epitome of navigation ever compiled. PRACTICAL NAVIGATION. Consisting of the Sailor's Sea Book, by J. Greenwood and W. H. Rosser; together with Mathematical and Nautical Tables for the Working of the Problems, by H. Law, C.E., and Prof. J. R. Young $2 8Q 22 CROSBY LOCKWOOD & SOX'S CATALOGUE. NAVIGATION AND NAUTICAL ASTRONOMY, In Theory and Practice. By Prof. J. R. Young. 12mo, cloth.Sl.OO "A very complete, thorough, and useful manual for the young navigator." MATHEMATICAL TABLES, For Trigonometrical, Astronomical, and Nautical Calculations; to which is prefixed a Treatise on Logarithms, by H. Law, C.E. With Tables for Navigation and Nautical Astronomy. By Prof. J. R. Yocng. 12mo, cloth SI. 60 MINING, METALLURGY, AND COLLIERY WORKING. THE OIL FIELDS OF RUSSIA AND THE RUSSIAN PETROLEUM INDUSTRY. . . A Practical Handbook on the Exploration, Exploitation, and Manage- ment of Russian Oil Properties, including Notes on the Origin of Petro- leum in Russia, a Description of the Tneory and Practice of Liquid Fuel, and a Translation of the Rules and Regulations concerning Rus- sian Oil Properties. By A. Beebt Thompson. A.M.I.M.E., late Chief Engineer and Manager of the European Petroleum Company's Russian Oil Properties. About 500 pp. With numerous Illustrations and Photographic Plates, and a Map of the Balakhany-Saboontchy-Romany Oil Field. Royal 8vo, cloth Net § 7 ,50 MECHANICS OF AIR MACHINERY. By Dr.' J. Weisbach and Prof. G. Heebmaxx. Authorized Translation with an Appendix on American Practice by A. Trowbridge, Ph.B., Adjunct Professor of Mechanical Engineering, Columbia University. Royal 8vo, cloth [Just Published. Xet §3.75 Summary of Contents: — The Movement of Air. — Natcral and Arti- ficial Ventilation. — Blowing-Engines; — Vacuum Pumps; Tcteres; Hot-air Blast; Work Performed bt Blowers; Blast-reservoirs; Piston-blowers. — Compressors. — Rotary Blowers. — Fans. — Recent American Practice, d:c. MACHINERY FOR METALLIFEROUS MINES. \ Practical Treatise for Mining Engineers. Metallurgists, and Managers of Mines. Bv E. Henry Davies, M.E.. F.G.S. 600 pp. With Fold- ing Plates and other Illustrations. Medium 8vo, cloth 88.00 'Deals exhaustively with the many and complex details which go to make up the sum total of machinery and other requirements for the success- ful working of metalliferous mines, and as a book of ready reference is of the highest value to mine managers and directors." — Mining Journal. THE DEEP LEVEL MINES OF THE RAND, And their Future Development, considered from the Commercial Point of View. By G. A. Denny (of Johannesburg), M.N.E.I.M.E.. Con- sulting Engineer to the General Mining and Finance Corporation, Ltd., of London. Berlin. Paris, and Johannesburg. Fully Illustrated with Diagrams and Folding Plates. Royal 8vo, buckram §10.00 ''Mr. Denny by confining himself to the consideration of the future of the deep-level mines" of the Rand breaks new ground, and by dealing with the subject rather from a commercial standpoint than from a scientific one, appeals to a wide circle of readers. The book cannot fail to prove of very great value to investors in South African mines." — Mining Journal. MINING, METALLURGY, & COLLIERY WORKING. 23 PROSPECTING FOR GOLD. A Handbook of Practical Information and Hints for Prospectors based on Personal Experience. By Daniel J. Rankin, F.R.S.G.S., M.R.A.S., formerly Manager of the Central African Company, and Leader of African Gold Prospecting Expeditions. With Illustrations specially Drawn and Engraved for the Work. Fcap. 8vo, leather $3.00 " This well-compiled book contains a collection of the richest gems of use- ful knowledge for the prospector's benefit. A special table is given to accelerate the spotting at a glance of minerals associated with gold." — Min- ing Journal. THE METALLURGY OF GOLD. A Practical Treatise on the Metallurgical Treatment of Gold-bearing Ores. Including the Assaying, Melting, and Refining of Gold. By M. Eissler, M.Inst.M.M. Fifth Edition, Enlarged. With over 300 Illus- trations and numerous Folding Plates. Medium 8vo, cloth. . .$7.50 " This book thoroughly deserves its title of a 'Practical Treatise.' The whole process of gold mining, from the breaking of the quartz to the assay of the bullion, is described in clear and orderly narrative and with much fulness of detail." THE CYANIDE PROCESS OF GOLD EXTRACTION, And its Practical Application on the Witwatersrand Gold Fields and elsewhere. By M. Eissler, M.Inst.M.M. With Diagrams and Work- ing Drawings, Third Edition, Revised and Enlarged. 8vo, cloth, $3.00 "This book is just what was needed to acquaint mining men with the actual working of a process which is not only the most popular, but is, as a general rule, the most successful for the extraction of gold from tailings." — Mining Journal. DIAMOND DRILLING FOR GOLD & OTHER MINERALS. A Practical Handbook on the Use of Modern Diamond Core Drills in Prospecting and Exploiting Mineral-Bearing Properties, including Par- ticulars of the Costs of Apparatus and Working. By G. A. Denny, M.N.E.Inst.M.E., M.Inst.M.M. Medium 8vo, 168 pp., with Illustra- tive Diagrams $5.00 "There is certainly scope for a work on diamond drilling, and Mr. Denny deserves grateful recognition for supplying a decided want." — Mining Journal. GOLD ASSAYING. 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With numerous Illustrations and Folding Plates Fourth Edition. 12mo, cloth : $1.20 A FIRST BOOK OF MINING AND QUARRYING. By J. H. Collins, F.G.S. Crown 8vo, cloth .60 ASBESTOS AND ASBESTIC. Their Properties, Occurrence, and Use. By Robert H. Jones, F.S.A. Mineralogist, Hon. Mem. Asbestos Club, Black Lake, Canada. With Ten Collotype Plates and other Illustrations. Demy 8vo, cloth. $6.40 GRANITES AND OUR GRANITE INDUSTRIES. By George F. Harris, F.G.S. With Illustrations. 12mo, cloth. $1,00 MINERAL SURVEYOR AND VALUER'S GUIDE. Comprising a Treatise on Improved Mining Surveying and the Valuation of Mining Properties, with New Traverse Tables. By W. Lintern, C.E. Fourth Edition, enlarged. 12mo, cloth $1.40 "Contains much valuable information, and is thoroughly trustworthy."— Iron and Coal Trades Review. TRAVERSE TABLES. For use in Mine Surveying. By William Lintern, C.E. With two plates. Small crown 8vo, cloth Net $1.50 SUBTERRANEOUS SURVEYING. By T. Fenwick. 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A First Year's Course for Students. _ By Tyson Sewell, A.I.E.E., Assistant Lecturer and Demonstrator in Electrical Engineering at the Polytechnic, Regent Street, London. Third Edition, Revised and En- larged, including an Appendix of Questions and Answers. 460 pages, with 274 Illustrations. Demy 8vo, cloth. . . . [Just Published.] §3.00 Ohm's Law. — Units Employed in Electrical, Engineering. — Series, and Parallel Circuits; Current Density and Potential Drop in the. 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A Manual for the Design of Electrical Circuits. By Arthur Vaughan Abbott, C.E., Member American Institute of Electrical Engineers, Member American Institute of Mining Engineers, Member American Society of Civil Engineers, Member American Society of Mechanical Engineers, &c. With Ten Folding Diagrams and Sixteen Full-page Engravings. Fourth Edition, entirely Re-Written and Enlarged. Royal 8vo, cloth Net §5.00 ELECTRICITY AS APPLIED TO MINING. By Arnold Lupton, M.Inst.C.E., M.I.M.E., M.I.E.E., late Professor of Coal Mining at the Yorkshire College, Victoria University, Mining En- gineer and Colliery Manager; G. D. Aspinall Parr, M.I.E.E., A.M.I. M.E., Associate of the Central Technical College, City and Guilds of London, Head of the Electrical Engineering Department, Yorkshire College, Victoria University; and Herbert Perkin, M.I.M.E., Cer- tificated Colliery Manager, Assistant Lecturer in the Mining Depart- ment of the Yorkshire College, Victoria University. With about 170 Illustrations. Second Edition, Revised and Enlarged. Medium 8vo, cloth [Just Published. §4.50 Introductory. — Dynamic Electricity. — Driving of the Dynamo. — The Steam Turbine. — Distribution of Electrical Energy. — Starting and Stopping Electrical Generators and Motors. — Electric Cables. — Central Electrical Plants. — Electricity applied to Pumping and Hauling. — Electricity applied to Coal-cutting. — Typical Electric Plants Recently Erected. — Electric Lighting by Arc and Glow Lamps. — Miscellaneous Applications of Electricity. — Electricity as Compared with other modes of Transmitting Power. — Dangers of Electricity. ELECTRICITY, ELECTRICAL ENGINEERING, &c. 29 CONDUCTORS FOR ELECTRICAL DISTRIBUTION. Their Materials and Manufacture, The Calculation of Circuits, Pole-Line Construction, Underground Working, and other Uses. By F. A. C. Perrine, A.M., D.Sc. ; formerly Professor of Electrical Engineering, Leland Stanford, Jr., University; M.Amer.I.E.E. 8vo, cloth. Ne * $3.50 Conductor Materials. — Alloyed Conductors. — Manufacture of Wire. — Wire-Finishing. — Wire Insulation. — Cables. — Calculation of Circuits. — Kelvin's Law of Economy in Conductors. — Multiple Arc Distribution. — Alternating Current Calculation. — Overhead Lines. — Pole Line. — Line Insulators. — Underground Conductors. DYNAMO ELECTRIC MACHINERY: its CONSTRUCT TION, DESIGN, and OPERATION. By Samuel Sheldon, A.M., Ph.D., Professor of Physics and Electrical Engineering at the Polytechnic Institute of Brooklyn, assisted by H. Mason. B.S. In two volumes, sold separately, as follows: — Vol. I.— DIRECT CURRENT MACHINES. 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Met $2.00 30 CROSBY LOCKWOOD &> SON'S CATALOGUE. SUBMARINE TELEGRAPHS; Their History, Construction, and Working. Founded in part on Wun- schendorff's "Traits de Telegraphie Sous-Marine," and Compiled from Authoritative and Exclusive Sources. By Charles Bright, F.R.S.E. A.M.Inst.C.E., M.I.E.E. 780 pp., fully Illustrated, including Maps and Folding Plates. Royal 8vo, cloth $25.00 ELECTRICAL AND MAGNETIC CALCULATIONS. For the Use of Electrical Engineers and Artisans, Teachers, Students, and all others interested in the Theory and Application of Electricity and Magnetism. By Prof. A. A. Atkinson, Ohio University. 12mo, cloth. . . : $1.50 "To teachers and those who already possess a fair knowledge of their sub- ject we can recommend this book as being useful to consult when requiring data or formulae which it is neither convenient nor necessary to retain by memory." — The Electrician. THE ELECTRICAL ENGINEER'S POCKET=BOOK. Consisting of Rules, Formulae, Tables, and Data. By H. R. 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Royal 8vo, cloth $8.00 Summary op Contents: — General, Notes (including Points in Speci- fication Writing, The Order of a Specification, and Notes on Items often Omitted from a Specification). — Form of Outside Cover to a Specification. — Specification of Works and List of General, Condi- tions. — Preliminary Items (including Shoring and House Breaker). — Drainage (including Rain-water Wells and Reports). — Excavator (including Concrete Floors, Roofs, Stairs, and Walls). — Pavior. — Bricklayer (including Flintwork, River, and other Walling, Spring- water Wells, Storage Tanks, Fountains, Filters, Terra Cotta and Faience). — Mason. — Carpenter, Joiner, and Ironmonger (including Fencing and Piling). — Smith and Founder (including Heating, Fire Hydrants, Stable and Cow-house Fittings). — Slater (including Slatb Mason). — Tiler. — Stone Tiler. — Shingler. — Thatcher. — Plumber (in- cluding Hot-water Work). — Zincworker. — Coppersmith. — Plasterer. — Gasfitter. — Bellhanger. — Glazier. — Painter. — Paperhanger. — General Repairs and Alterations. — Ventilation. — Road-making. — Electric Light. — Index. PRACTICAL BUILDING CONSTRUCTION. A Handbook for Students Preparing for Examinations, and a Book of Reference for Persons Engaged in Building. By John Parnell Allen, Surveyor, Lecturer on Building Construction at the Durham College of Science, Newcastle-on-Tyne. Fourth Edition, Revised and Enlarged. Medium 8vo, 570 pp., with over 1,000 Illustrations, cloth, $3.00 SPECIFICATIONS FOR PRACTICAL ARCHITECTURE. A Guide to the Architect, Engineer, Surveyor, and Builder. Upon the Basis of the Work by A. Bartholomew, Revised, by F. Rogers. 8vo, cloth $6.00 SCIENCE OF BUILDING: An Elementary Treatise on the Principles of Construction. By E. Wyndham Tarn, M.A.Lond. Fourth Edition. 12mo, cloth. $1.40 ART OF BUILDING, Rudiments of. General Principles of Construction, Character, Strength, and Use of Materials, Preparation of Specifications and Estimates, &c. By Edward Dobson, M.Inst.C.E. Fifteenth Edition, revised by J. P. 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Second Edition, Revised. 470 pp., 12mo. fully Illustrated, leather S3. 50 LIGHTING BY ACETYLENE Generators, Burners, and Electric Furnaces. By William E. Gibbs, M.E. With 66 Illustrations. 12mo, cloth SI. 50 ENGINEERING CHEMISTRY. A Practical Treatise for the Use of Analytical Chemists, Engineers, Iron Masters, Iron Founders, Students and others. Comprising Methods of Analysis and Valuation of the Principal Materials used in Engineering Work, with numerous Analvses, Examples and Suggestions By H. Joshua Phillips, F.I.C.. F.C.S. Third Edition, Revised and Enlarged. 12mo, 420 pp., withJPlates and Illustrations, cloth, S-4.50 NITRO=EXPLOSIYES. A Practical Treatise concerning the Properties, Manufacture, and Analysis of Xitrated Substances, including the Fulminates, Smokeless Powders, and Celluloid. By P. Gerald Saxtord, F.I.C., Consulting Chemist to the Cotton Powder Company, Limited, &c. With Illustra- tions. 12mo, cloth S3. 00 A HANDBOOK OF MODERN EXPLOSIYES. 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Containing all known Methods of Anhydrous Analysis, many Working Examples, and Instructions for Making Apparatus. Bv Lieut. -Colonel W. A. Robs, R.A., F.G.S. Second Ed., Enlarged. feJSmo, cloth, S3. 00 THE MANUAL OF COLOURS AND DYE=\YARES. Their Properties. Applications, Valuations, Impurities and Sophistica- tions. For the Use of Dyers, Printers, Drysalters, Brokers, &c. By J. W. Slater. Second Edition, Revised and greatlv Enlarged. 12mo, cloth S3.00 " There is no other work which covers precisely the same ground. To students preparing for examinations in dyeing and printing it will prove exceedingly useful." — Chemical Xeus. INDUSTRIAL AND USEFUL ARTS. 43 THE ARTISTS* MANUAL OF PIGMENTS. Showing their Composition, Conditions of Permanency, Non-Per- manency, and Adulterations, &c, with Tests of Purity. By H. C. Standage. Third Edition. 12mo, cloth SI. 00 " This work is indeed multum-in-parvo , and we can, with good conscience, recommend it tc all who come in contact with pigments, whether as makers, dealers, or users. : — Chemical Review. INDUSTRIAL ARTS, TRADES, AND MANUFACTURES. THE CULTIVATION AND PREPARATION OF PARA RUBBER. By W. H. Johnson, F.L.S., F.R.H.S., Director of Agriculture, Gold Coast Colony, West Africa, Commissioned by Government in 1902 to visit Ceylon to Study the Methods employed there in the Cultivation and Preparation of Para Rubber and other Agricultural Staples for Market, with a view to Introduce them into West Africa. Demy 8vo, doth $3.00 Summary op Contents: — Introductory. — The Para Rubber Tree (Hevea brasiliensis) at Home and Abroad. — Cultivation of the Tree: — Propagation. — Site for Plantation. — Distance Apart to Plant the Trees. — Transplanting. — Cultivation. — Insect Pests and Fungoid Diseases. — Collecting the Rubber: — Various Methods Employed in Tapping Rubber Trees. — Flow of Latex Increased by Wounding the Tree. — How to Tap. — The Preparation of Rubber from the Latex: — Latex. — Various Methods Employed in the Prep- aration of Rubber. — Suggested Method for Preparing Rubber. — Scrap Rubber. — Yield of Para Rubber from Cultivated Trees: — Ceylon. — Malay Peninsula. — Gold Coast, West Africa. — Establish- ment and Maintenance of a Para Rubber Plantation: — Ceylon. — Malay Peninsula. — Commercial Value of the Oil in Hevea Seeds. RUBBER HAND STAMPS And the Manipulation of Rubber. A Practical Treatise on the Manu- facture of India-rubber Hand Stamps, Small Articles of India-rubber, The Hektograph, Special Inks, Cements, and Allied Subjects. By T. O'Conor Sloane, A.M., Ph.D. With numerous llllustrations. Square 8vo, cloth $1.00 PRACTICAL PAPER=/UAKING. A Manual for Paper-Makers and Owners and Managers of Paper-Mills. With Tables, Calculations, &c. By G. Clapperton, Paper-Maker. With Illustrations of Fibres from Micro-Photographs. 12mo, cloth, $2.50 THE ART OF PAPER=MAKING. A Practical Handbook of the Manufacture of Paper from Rags, Esparto, Straw, and other Fibrous Materials. Including the Manufacture of Pulp from Wood Fibre, with a Description of the Machinery and Appliances used. To which are added Details of Processes for Recover- ing Soda from Waste Liquors. By Alexander Watt. With Illus- trations. 12mo, cloth $3.00 A TREATISE ON PAPER. For Printers and Stationers. With an Outline of Paper Manufacture; Complete Tables of Sizes, and Specimens of Different Kinds of Paper. By Richard Parkinson, late of the Manchester Technical School. Demy 8vo, cloth $1,40 44 CROSBY LOCKWOOD &» SON'S CATALOGUE. THE ART OF SOAP=MAKING. A Practical Handbook of the Manufacture of Hard and Soft Soaps, Toilet Soaps, &c. Including many new Processes, and a Chapter on the Recovery of Glycerine from Waste Leys. By Alexander Watt. Sixth Edition, including an Appendix on Modern Candlemaking. 12mo, cloth $3.00 "A thoroughly practical treatise. We congratulate the author on the success of his endeavour to fill a void in English technical literature." — Nature, j "The work will prove very useful, not merely to the technological student, but to the practical soap boiler who wishes to understand the theory of his art." — Chemical News. THE ART OF LEATHER MANUFACTURE. A Practical Handbook, in which the Operations of Tanning, Currying, and Leather Dressing are fully Described, and the Principles of Tanning Explained. Together with a Description of the Arts of Glue Boiling, Gut Dressing, &c. By Alexander Watt. Fifth Edition, thoroughly Revised and Enlarged. 8vo, cloth Nearly ready, $4.00 ART OF BOOT AND SHOEMAKING, Including Measurement, Last-fitting, Cutting-out, Closing, and Making; with a Description of the most Approved Machinery employed. By J. B. Leno. 12mo, cloth # gQ "By far the best work ever written on the subject." — Scottish Leather Trader. COTTON MANUFACTURE. A Manual of Practical Instruction of the Processes of Opening, Carding, Combing, Drawing, Doubling and Spinning of Cotton, the Methods of Dyeing, &c. For the use of Operatives, Overlookers, and Manu- facturers. By John Lister, Technical Instructor, Pendleton. 8vo, cloth, $3.00 "A distinct advance in the literature of cotton manufacture." — Machinery "It is thoroughly reliable, fulfilling nearly all the requirements desired." Glasgow Herald. WATCH REPAIRING, CLEANING, AND ADJUSTING. A Practical Handbook dealing with the Materials and Tools Used and the Methods of Repairing, Cleaning, Altering, and Adjusting all kinds of English and Foreign Watches, Repeaters, Chronographs, and Marine Chronometers. By F. J. Garrard, Springer and Adjuster of Marine Chronometers and Deck Watches for the Admiralty. With over 200 Illustrations. 12mo, cloth $2 t QQ MODERN HOROLOGY, IN THEORY AND PRACTICE. Translated from the French of Claudius Saunier, ex-Director of the School of Horology at Macon, by Julien Tripplin, F.R.A.S., Besancon Watch Manufacturer, and Edward Rigg, M.A., Assayer in the Royal Mint. With Seventy-eight Woodcuts and Twenty-two Coloured Copper Plates. Second Edition. Super-royal, 8vo, cloth . . . . $15.00 Half -calf. $18.00 THE WATCHMAKER'S HANDBOOK. Intended as a Workshop Companion for those engaged in Watchmaking and the Allied Mechanical Arts. Translated from the French of Claudius Saunier, and enlarged by Julien Tripplin, F.R.A.S., and Edward Rigg, M.A., Assayer in the Royal Mint. Fourth Edition 12mo, clotb, .,,,,,, , . .$3.00 INDUSTRIAL AND USEFUL ARTS. 45 CLOCKS, WATCHES, & BELLS for PUBLIC PURPOSES. A Rudimentary Treatise. By Edmund Beckett, Lord Grimthorpe, LL.D., K.C, F.Jt.A.S. Eighth Edition, with new List of Great Bells and an Appendix on Weathercocks. 12mo, cloth $1.80 HISTORY OF WATCHES & OTHER TIMEKEEPERS. By James F. Kendal, M.B.H.Inst. ,Q{) boards; or cloth, gilt, $1.00 ELECTROPLATING & ELECTRO=REFINING of METALS, Being a new edition of Alexander Watt's "Electro-Deposition." Revised and Largely Rewritten by Arnold Philip, B.Sc, A.I.E.E., Principal Assistant to the Admiralty Chemist. 8vo, cloth. . . .$4.50 ELECTROPLATING. A Practical Handbook on the Deposition of Copper, Silver, Nickel, Gold, Aluminium, Brass, Platinum, &c, &c. By J. W. Urquhart, C.E. Fifth Edition, Revised. 12mo, cloth $2.00 ELECTRO=METALLURGY, Practically Treated. By Alexander Watt. Tenth Edition, enlarged and revised. With Additional Illustrations, and including the most Recent Processes. 12mo, cloth $1.40 GOLDSMITH'S HANDBOOK, Containing full Instructions in the Art of Alloying, Melting, Reducing, Colouring, Collecting, and Refining. The Processes of Manipulation, Recovery of Waste, Chemical and Physical Properties of Gold; Solders, Enamels, and other useful Rules and Recipes, &c. By George E. Gee. Sixth Edition. 12mo, cloth $1.20 SILVERSMITH'S HANDBOOK, On the same plan as the above. By George E. Gee. Third Edition. 12mo, cloth $1.20 *** The two preceding Works, in One handsome Volume, half-bound, en- titled "The Goldsmith's and Silversmith's Complete Handbook," $2.80 JEWELLER'S ASSISTANT IN WORKING IN GOLD. 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Comprising a Selection of Geometrical Problems and Practical Rules for Describing the Various Patterns Required by Zinc. Sheet-Iron, Copper, and Tin-Plate Workers. By Reuben Henry Warn. Piactical Tin-Plate Worker. New Edition, Revised and greatly Enlarged by Joseph G. Horner, A.M.I.M.E. 12mo, 254 pp.. with 430 Illustra- tions, cloth §3.00 SHEET METAL=WORKER'S GUIDE. A Practical Handbook for Tinsmiths, Coppersmiths, Zincworkers, &c , with 46 Diagrams and Working Patterns. By W. J. E. Crane. Fourth Edition. 12mo, cloth # gQ GAS FITTING: A Practical Handbook. By John Black. Revised Edition With 130 Illustrations. 12mo, cloth $1.00 " It is written in a simple, practical style, and we heartily recommend it." — Plumber and Decorator. TEA MACHINERY AND TEA FACTORIES. A Descriptive Treatise on the Mechanical Appliances required in the Cultivation of the Tea Plant and the Preparation of Tea for the Market. By A. J. Wallis-Tayler, A.M.Inst C.E. Medium 8vo, 468 pp. With 218 Illustrations .$10.00 Summary of Contents. Mechanical Cultivation or Tillage of the Soil. — Plucking or Gathering the Leaf. — Tea Factories. — The Dressing, Manufacture, or Preparation of Tea by Mechanical Means. — Artificial Wither- ing of the Leaf. — Machines for Rolling or Curling the Leaf. — Fer- menting Process. — Machines for the Automatic Drying or Firing of the leaf. — Machines for Non-Automatic Drying or Firing of the Leaf. — Drying or Firing Machines. — Breaking or Cutting, and Sort- ing Machines. — Packing the Tea. — Means of Transport on Tea Plan- tations. — Miscellaneous Machinery and Apparatus. — Final Treat- ment of the Tea. — Tables and Memoranda. FLOUR MANUFACTURE. A Treatise on Milling Science and Practice. By Friedrich Kick, Imperial Regierungsrath, Professor of Mechanical Technology in the Imperial German Polytechnic Institute, Prague. Translated from the Second Enlarged and Revised Edition. By H. H. P. Powles, A.M. Inst. C.E. 400 pp., with 28 Folding Plates, and 167 Woodcuts. Royal 8vo, cloth $10.00 ORNAMENTAL CONFECTIONERY. A Guide for Bakers, Confectioners and Pastrycooks; including a variety of Modern Recipes, and Remarks on Decorative and Coloured Work. With 129 Original Designs. By Robert Wells. 12mo, cloth, S3. 00 BREAD & BISCUIT BAKER'S & SUGAR=BOILER'S ASSISTANT. Including a large variety of Modern Recipes. With Remarks on the Art of Bread-making. By Robert Wells. Fourth Edition. 12mo, cloth .50 PASTRYCOOK & CONFECTIONER'S GUIDE. For Hotels, Restaurants, and the Trade in general, adapted also for Family Use. By R. Wells, Author of "The Bread and Biscuit Baker." .40 INDUSTRIAL AND USEFUL ARTS. 47 MODERN FLOUR CONFECTIONER. Containing a large Collection of Recipes for Cheap Cakes, Biscuits, &c. With remarks on the Ingredients Used in their Manufacture. By R. Wells 40 SAVOURIES AND SWEETS Suitable for Luncheons and Dinners. By Miss M. L. Allen (Mrs. A. Macaire), Author of "Breakfast Dishes," &c. 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With upwards of 100 Illustrations .50 CLOCK JOBBER'S HANDYBOOK. A Practical Manual on Cleaning, Repairing, and Adjusting. With upwards of 100 Illustrations .50 CABINET WORKER'S HANDYBOOK. A Practical Manual on the Tools, Materials, Appliances, and Processes employed in Cabinet Work. With upwards of 100 Illustrations. .50 *'Mr. Hasluck's thorough-going little Handybook is amongst the most practical guides we have seen for beginners in cabinet-work." — Saturday Review. WOODWORKER'S HANDYBOOK. Embracing Information on the Tools, Materials, Appliances, and Processes Employed in Woodworking. With 104 Illustrations. .50 COMMERCE, COUNTING-HOUSE WORK, &c. 49 COMMERCE, COUNTING-HOUSE WORK, TABLES, ETC. LESSONS IN COMMERCE. By Professor R. Gambaro, of the Royal High Commercial School at Genoa. Edited and Revised by James Gault, Professor of Commerce and Commercial Law in King's College, London. Fifth Edition. 12mo, cloth $1.40 THE FOREIGN COMMERCIAL CORRESPONDENT. Being Aids to Commercial Correspondence in Five Languages — English, French, German, Italian, and Spanish. By Conrad E. Baker. Third Edition, Carefully Revised Throughout. 12mo, cloth $1.80 FACTORY ACCOUNTS: their PRINCIPLES & PRACTICE. A Handbook for Accountants and Manufacturers, with Appendices on the Nomenclature of Machine Details; the Income Tax Acts; the Rating of Factories ; Fire and Boiler Insurance ; the Factory and Work- shop Acts, &c, including also a Glossary of Terms and a large numbei of Specimen Rulings. By Emile Garcke and J. M. Fells. Fifth Edition, Revised and Enlarged. Demy 8vo, cloth $3.00 MODERN METROLOGY. A Manual of the Metrical Units and Systems of the present Century. With an Appendix containing a proposed English System. By Lowis d'A. Jackson, A.M. Inst. C.E., Author of "Aid to Survey Prac- tice," &c. 8vo, cloth $5.00 SERIES OF METRIC TABLES. In which the British Standard Measures and Weights are compared with those of the Metric System at present in Use on the Continent. By C. H. Dowling, C.E. 8vo, cloth $4.00 IRON-PLATE WEIGHT TABLES. For Iron Shipbuilders, Engineers, and Iron Merchants Containing the Calculated Weights of upwards of 150,000 different sizes of Iron Plates from 1 foot by 6 in. by ± in. to 10 feet by 5 feet by 1 in. Worked out on the Basis of 40 lbs. to the square foot of Iron of 1 inch in thickness. By H. Burlinson and W. H. Simpson. 4to, half -bound. . . .$10.00 50 CROSBY LOCKWOOD 6- SON'S CATALOGUE. AGRICULTURE, FARMING, GARDENING, ETC. THE COMPLETE GRAZIER AND FARMER'S AND CATTLE BREEDER'S ASSISTANT. A Compendium of Husbandry. Originally Written by William Youatt. Fourteenth Edition, entirely Re-written, considerably En- larged, and brought up to Present Requirements, by William Fream, LL.D., Assistant Commissioner, Royal Commission on Agriculture, Author of "The Elements of Agriculture," &c. 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With Illus- trations. Second Edition, Revised. 12mo, cloth $1.00 DAIRY, PIGS, AND POULTRY. Vol. IV. OUTLINES OF MODERN FARMING. By R. Scott Burn. Woodcuts. 12mo, cloth .80 THE ELEMENTS OF AGRICULTURAL GEOLOGY. A Scientific Aid to Practical Farming. By Primrose McConnell. Author of "Note-book of Agricultural Facts and Figures." 8vo, cloth, $7.50 SOILS, MANURES, AND CROPS. Vol. I.— OUTLINES OF MODERN FARMING. By R. Scott Burn. Woodcuts. 12mo, cloth .80 FERTILISERS AND FEEDING STUFFS. Their Properties and Uses. A Handbook for the Practical Farmer. By Bernard Dyer, D.Sc. (Lond.) With the Text of the Fertilisers and Feeding Stuffs Act of 1893, The Regulations and Forms of the Board of Agriculture, and Notes on the Act by A. J. David, B.A., LL.M. Fourth Edition, Revised. 12mo, cloth .40 THE ROTHAMSTED EXPERIMENTS AND THEIR PRACTICAL LESSONS FOR FARMERS. Part I. Stock. Part II. Crops. By C. J. R. Tipper. 12mo, cloth $1.40 AGRICULTURE, FARMING, GARDENING, &c. 51 SYSTEMATIC SMALL FARMING. Or, The Lessons of My Farm. Being an Introduction to Modern Farm Practice for Small Farmers. By R. Scott Burn, Author of "Outlines of Modern Farming," &c. 12mo, cloth $2.40 THE FIELDS OF GREAT BRITAIN. A Text-Book of Agriculture. Adapted to the Syllabus of the Science and Art Department. For Elementary and Advanced Students. By Hugh Clements (Board of Trade). Second Edition, Revised, "with Additions. 18mo, cloth $1.00 OUTLINES OF MODERN FARMING. By R. Scott Burn. Soils, Manures, and Crops — Farming and Farming Economy — Cattle, Sheep, and Horses — Management of Dairy, Pigs, and Poultry — Utilisation of Town-Sewage, Irrigation, &c. Sixth Edition. In One Vol., 1,250 pp., half -bound, profusely Illustrated. $4.80 FARM ENGINEERING, The COMPLETE TEXT=B00K of. Comprising Draining and Embanking; Irrigation and Water Supply; Farm Roads, Fences and Gates; Farm Buildings; Barn Implements and Machines ; Field Implements and Machines ; Agricultural Survey- ing, &c. By Professor John Scott. 1,150 pp., half-bound, with over 600 Illustrations $4.80 DRAINING AND EMBANKING. A Practical Treatise. By John Scott, late Professor of Agriculture and Rural Economy at the Royal Agricultural College, Cirencester. With 68 Illustrations. 12mo, cloth .60 "A valuable handbook to the engineer as well as to the surveyor." — Land. IRRIGATION AND WATER SUPPLY: A Practical Treatise on Water Meadows, Sewage Irrigation, Warping, &c; on the Construction of Wells, Ponds, and Reservoirs, &c. By Professor J. Scott. 12mo, cloth 60 FARM ROADS, FENCES, AND GATES: A Practical Treatise fon the Roads, Tramways, and Waterways of the Farm ; the Principles of Enclosures ; and on Fences, Gates, and Stiles. By Professor John Scott. 12mo, cloth .60 BARN IMPLEMENTS AND MACHINES: Treating of the Application of Power to the Operations of Agriculture and of the various Machines used in the Threshing-barn, in the Stock- yard, Dairy, &c. By Professor John Scott. 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