yC-NRLF ^B M?3 SSI ^': f; k I'.: ijlliipji !il! ill !i:l! !i[|;l illj!! li '|i!ilii;,;;''i IN MEMORIAM FLORIAN CAJORI Digitized by tine Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/dooleymathOOdoolrich VOCATIONAL MATHEMATICS BY WILLIAM H. DOOLEY AUTHOR OP " VOCATIONAL MATHEMATICS FOB GIRLS," "textiles," ETC. REVISED D. C. HEATH & CO., PUBLISHERS BOSTON NEW YORK CHICAGO Copyright, 191 5, By D. C. Heath & Co. 2 B O CAJORI PREFACE The author has had, during the last ten years, considerable experience in organizing and conducting intermediate and sec- ondary technical schools. During this time he has noticed the inability of the regular teachers in mathematics to give the pupils the training in commercial and rule of thumb methods of solving mathematical problems that are so necessary in everyday life. A pupil graduates from the course in mathe- matics without being able to " commercialize " or apply his mathematical knowledge in such a way as to meet the needs of trade and industry. It is to overcome this difficulty that the author has prepared this book on vocational mathematics. He does not believe in doing away with the regular course of mathematics but in supplementing it with a practical course. This course may take the place of the first year algebra and the first year geometry in vocational classes in which it is not desirable to give the traditional course in algebra and geometry. This book may be used by the regular teacher in mathe- matics and by the shop teacher. It can be used in the shop in teaching mathematics and in providing drill problems upon the shop work. A course based upon the contents of the book should be provided before pupils finish their training, so that they may become skillful in applying the principles of mathe- matics to the daily needs of manufacturing life. In revising the manuscript the author has had the assistance of his teachers in the Lawrence Industrial School, the Lowell Industrial School, the Fall River Technical High School, and 9i6287 IV PREFACE of many other teachers, practical men, and manufacturing firms. Valuable material has also been obtained from standard hand- books, such as Kent's. To mention the names of all persons to whom the author is indebted is impossible. Acknowledgment should be made to the following persons and firms who have kindly consented to furnish cuts and information, and have offered valuable suggestions and problems : Mr. George W. Evans, Principal Charlestown (Mass.) High School; Dr. David Snedden; Mr. Charles E. Allen ; Mr. Fred. W. Turner ; Mr. Peter Gart- land. Principal South Boston High School ; Mr. John Casey, Worcester (Mass.) Trade School ; Mr. John L. Sullivan, Princi- pal Chicopee (Mass.) Industrial School; Mr. William Hunter, Fitchburg Industrial Course; Mr. J. Gould Spofford, Princi- pal Quincy Industrial School ; Mr. Edward K. Markham, Cam- bridge Technical High School; Principal Joseph J. Eaton, Saunders Trades School, Yonkers, N, Y. ; Miss Bessie King- man, Brockton (Mass.) High School ; Mr. E. W. Boshart, Director of Industrial Arts, Mt. Vernon, N. Y. ; Mr. G. A. Boate, Technical High School, Newton, Mass. ; Mr. G. R. Smith, Bradford, England ; Mr, H. R. Carter, Belfast, Ireland ; New York Central Lines, Apprenticeship Department ; Mr. H. E. Thomas, Tuskegee Institute, Ala. ; Brown & Sharp Co. ; Simond's Guide for Carpenters ; R. M. Starbuck & Sons ; Garvin Machine Co. ; The Crane Co. ; The William Powell Co. ; Crosby Steam Gage and Valve Co. ; Mr. Peter Lobben ; Mr. H. P. Faxon ; Bardons and Oliver ; Stanley Rule and Level Co. ; Pittsburgh Valve Foundry and Construction Co. ; Becker Milling Machine; Braeburn Steel Co.; Fore River Ship Build- ing Co. ; Engineering Workshop Machines and Processes ; American Injector Co. ; B. F. Sturtevaut Co. ; E. W. Bliss Co. ; Dillon Boiler Works; Whitcomb-Blaisdell Co.; Hoefer Drill Co. ; Bradford Lathe Co. ; and Detroit Screw Works. The author will be very thankful for any suggestions relating to the work, CONTENTS PART I — REVIEW OF ARITHMETIC OUXPTXR PAUS I. Essentials of Arithmetic 1 Fundamental Processes. Fractions. Decimals. Com- pound Numbers. Percentage. Ratio and Proportion. Involution. Evolution. II. Mensuration 62 Circle. Triangles. Quadrilaterals. Polygons. El- lipse. Pyramid. Cone. Sphere. Similar Figures. III. Interpretation of Results 78 Reading of Blue Print. Drawing to Scale. Methods of Solving Examples. PART II — CARPENTERING AND BUILDING IV. Woodworking 83 Board Measure. Short Methods. Tables. V. Construction 89 Frame and Roof. Lathing. VI. Building Materials 94 Mortar. Plaster. Bricks. Stone Work. Cement. Shingles. Slate. Clapboards. Flooring. Stairs. PART III — SHEET METAL WORK Vn. Die Cutting 107 Blanking Dies. Combination Dies. Standard Gauge. Tables. V VI CONTENTS PART IV — BOLTS, SCREWS, AND RIVETS CHAPTER PAGE VIII. Bolts, Screws, and Rivets 126 Kinds of Bolts. Rivets. Nails. Tacks. Kinds of Screws. Screw Threads. V-Shaped. U. S. Stand- ard. Acme Standard. Square. Micrometer. PART V — SHAFTS, PULLEYS, AND GEARS IX. Shafts and Pulleys . 146 Belting. Arc of Contact. Speed. Countershafts. X. Gearing 159 Pitch. Trains of Gears. PART VI — PLUMBING AND HYDRAULICS XL Plumbing and Hydraulics . . . . . 172 Tanks. Drainage Pipes. Weight of Lead Pipe. Capacity of Pipe. Atmospheric Pressure. Water Pressure. Water Traps. Velocity of Water. Water Power Density of Water. Specific Gravity. PART VII — STEAM ENGINEERING XII. Heat 195 Heat Units. Temperature. Value of Coal. Intrinsic Heat. Latent Heat. Value of Water and Steam. Steam Heating. XIIL Boilers 203 Kinds of Boilers. Tensile Strength. Safe Working Pressure. Boiler Tubes. Safety Valves. Super- heated Steam. Boiler Pumps. XIV. Engines 222 Kinds of Engines. Fly Wheel. Horse Power. Di ameter of Cylinder. Diameter of Supply Pipe. Steam Indicator. CONTENTS Vll PART VITI. ELECTRICAL WORK caAPTBB PA«B XV. CoMMKKriAi. Elkctricity 231 Amperes. Volts. Ohms. Ammeter. Voltmeter. Series and Parallel Circuits. Power Measurement. Resistance. Size of Wire. PART IX — MATHEMATICS FOR MACHINISTS XVI. Matkkials 251 Cast Iron. Wrought Iron. Steel. Copper. Brass. Weight of Steel. XVn. Lathes 255 Engine Lathe, (iear and Pitch. Adjusting Gears. Compound Lathe. To cut Double or Multiple Threads. Machine Speeds. Net Power for Cutting Iron or Steel. Surface Speeds.' XVIII. Planers, Shapers, and Drilling Machines . 268 Planer. Planer and Shaper. Drilling Machine. PART X — TEXTILE CALCULATIONS XIX. Yarns 276 Materials. Kinds. Weight. APPENDIX Metric System 291 Graphs 296 Formulas 298 Logarithms 315 Trigonometry 320 Tables of Formulas 330 Index 337 NOTE TO TEACHERS The author has found that it adds to the interest of the subject if the teacher will provide models, instruments, etc., in the class and explain the subject matter and problems in terms of the actual subject. If the pupils have not had a course in Algebra, it is better to take up the subject of Formulas in the Appendix before beginning the subject of Mensuration. viU VOCATIOIN^AL MATHEMATICS PART I — REVIEW OF ARITHMETIC CHAPTER I Notation and Numeration A unit is one thing ; as, one book, one pencil, one inch. A number is made up of units and teUs how many units are taken. An integer is a whole number. A single figure expresses a certain number of units and is said to be in the units column. For example, 5 or 8 is a single figure in the units column ; 63 is a number of two figures and has the figure 3 in the units column and the figure 5 in the tens column, for the second figure represents a certain number of tens. Each column has its own name, as shown below. 1 I S 1 3 8, 6 9 6, 4 O 7, 1 2 5 Reading Numbers. — For convenience in reading and writing numbers they are separated into groups of three figures each by commas, beginning at the right : 138,695,407,125. The first group is 125 units. The second group is 407 thousands. The third group is 695 millions. The fourth group is 138 billions. 1 VOfSATjLONAL MATHEMATICS ' ^^e; ^rec^ding fiui»be,r is read one hundred thirty-eight billion, six hundred ninety-five million, four hundred seven thousand, one hundred twenty-five ; or 138 billion, 695 million, 407 thousand, 125. Standard Mathematics Sheet. — To avoid errors in solving problems the work should be done in such a way as to show each step, and should make it easy to check the answer when found. A sheet of paper of standard size, 8| in. by 11 in., should be used. Rule this sheet as in the following diagram, set down each example with its proper number in the margin, and clearly show the different steps required for the solution. To show that the answer obtained is correct, the proof should follow the example itself. Standard Mathematics Sheet 8h in. 1. t It John Smith — 100 Vocational Arithmetic 10-2-m No. 10 1,203 20 2,672 2^ 31,118 23 480 10 39 19,883 55,395 Ans. 2. Proof: 3. The pupil should write or print his name and class, the date when the problem is finished, and the number of the problem on the Standard REVIEW OP ARITHMETIC 3 Mathematics Sheet. If the question contains several divisions or prob- lems, they should be tabulated — (a), (6), etc. — at the left of the prob- lems inside the margin line. Tiiere should be a line drawn between problems to separate them. Addition Addition is the process of finding the sum of two or more numbers. The result obtained by this process is called, the sum or amount. The sign of addition is an upright cross, +, called plus. The sign is placed between the two numbers to be added. Thus, 9 inches + 7 inches (read nine inches plus seven inches). The sign of equality is two short horizontal parallel lines, =, and means equals or equal to. Thus, the statement that 8 feet + 6 feet = 14 feet, means that six feet added to eight feet (or 8 feet plus 6 feet) equals fourteen feet. To find the sum or amount of two or more numbers. Example. — What is the total weight of a machine made up of parts weighing 1203, 2672, 31,118, 480, 39, and 19,883 lb., respectively ? [The abbreviation for pounds is lb.] The sum of the units column is 3 + 9 -f -1-8 + 2 + 3 = 26 units or 20 and 5 more ; 20 is tens, so leave the 6 under the units column and add the 2 tens in the tens column. The sura of the tens column is 2 + 8 + 3 + 8 + 1 + 7 + = 29 tens. 29 tens equal 2 hun- dreds and 9 tens. Place the 9 tens under the tens column and add the 2 hundreds to the hundreds column, 2 + 8 + 4 + 1+6 + 2 = 23 hundreds ; 23 hundreds are equal to 2 thousands and 3 hundreds. Place the 3 hundreds under the hundreds column and add the 2 thousands to the next column. 2 + 9 + 1+2 + 1 = 15 thousands or 1 ten-thousand and 6 thousands. Add the 1 ten-thousand to the ten-thousands column 1,203 2p 2,672 29 31,118 2;^ 480 Ip 39 19,883 55,395 lb. 4 VOCATIONAL MATHEMATICS and the sum isl+l-)-3 = 5. Write the 5 in the ten-thousands column. Hence, the sum or weight is 55,395 lb. Proof. — Repeat the process, beginning at the top of the right-hand column. Exactness is very important in arithmetic. There is only- one correct answer. Therefore it is necessary to be accurate in performing the numerical calculations. A check of some kind should be made on the work. The simplest check is to estimate the answer before solving the problem. A comparison can be made between them. If there is a great discrepancy then the work is probably wrong. It is also necessary to be exact in reading the problem. EXAMPLES 1. Write the following numbers as figures and add them : Seventy-five thousand three hundred eight ; seven million two hundred five thousand eight hundred forty -nine. 2. In a certain year the total output of copper from the mines was worth $58,638,277.86. Express this amount in words. 3. Solve the following: 386 -h 5289 -f 53666 + 3001 -f 291 + 38 = ? 4. On a shelf there were three kegs of bolts. The first keg weighed 203 lb., the second 171 lb., and the third 93 lb. How many pounds of bolts were there on the shelf ? 5. On the platform in an electrical shop there were a motor- generator and two motors. The motor-generator weighed 275 lb., one of the motors 385 lb., and the other motor 492 lb. What weight did the platform support ? 6. Solve the following : 6027 + 836 -f 4901 + 3,800,031 -f 28,639 + 389,661 = ? 7. Four coal sheds in a shop contained respectively 1498 lb., 4628 lb., 6125 lb., and 12,133 lb. What was the whole amount of coal in the shop ? REVIEW OF ARITHMETIC 5 8. An engineer ordered coal and found that the first load weighed 5685 lb., the second 5916 lb., the third 6495 lb., and the fourth 5280 lb. What was the total weight ? 9. Wire for electric lights was run around four sides of three rooms. If the first room was 13 ft. long and 9 ft. wide ; the second 18 ft. long and 18 ft. wide ; and the third 12 ft. long and 7 ft. wide, what was the total length of wire required? Remember that electric lights require two wires. 10. Find the sum : 46 lb + 135 lb. + 72 lb. + 39 lb. + 427 lb. + 64 lb. -h 139 lb. Subtraction Subtraction is the process of finding the difference between two numbers, or of finding what number must be added to a given number to equal a given sum. The minuend is the num- ber from which we subtract ; the subtrahend is the number sub- tracted; and the difference or remainder is the result of the subtraction. The sign of subtraction is a short horizontal line, — , called minuSf and is placed before the number to be subtracted. Thus, 12 — 8 = 4 is read twelve minus (or less) eight equals four. To find the difference of two numbers. Example. — A reel of wire contained 8074'. If 4869' were used in wiring a house, how many feet remained on the reel ? [Feet and mches are represented by ' and " respectively.] Minuend 8074 ft. Write the smaller number under the Subtrahend 4869 greater, with units of the same order in Pfmn 'ndpr S^H)5 ft ^^^ same vertical line. 9 cannot be taken from 4, so change 1 ten to units. The 1 ten that was changed from the 7 tens makes 10 units, which added to the 4 units makes 14 units. Take 9 from the 14 units and 6 units remain. Write the 5 under the unit column. Since 1 ten was changed from 7 tens, there are 6 tens left, and 6 from leaves 0. Write under the tens col- unm. Next, 8 hundred cannot be taken from hundred, so 1 thousand 6 VOCATIONAL MATHEMATICS (ten hundred) is changed from the thousands column. 8 hundred from 10 hundred leave 2 hundred. Write the 2 under the hundreds column. Since 1 thousand has been taken from the 8 thousand, there are left 7 thousand to subtract the 4 thousand from, which leaves 3 thousand. Write 3 under the thousand column. The whole remainder is 3205 ft. Proof. — If the sum of the subtrahend and the remainder equals the minuend, the answer is correct, EXAMPLES 1. Subtract 1001 from 79,999. 2. A box contained one gross (144) of wood screws. If 48 screws were used on a job, how many screws were left in the box ? 3. What number must be added to 3001 to produce a sum of 98,322? 4. A machinist had castings of different kinds on hand in the morning, weighing 6018 lb. He used during the day 911 lb. of castings. How many pounds were left ? 5. The value of electrical equipment produced by one firm in a certain year was $ 102,000,000, and in the previous year it was $93,000,000. What was the difference between the two years ? 6. In the coal shed of a factory there were 52,621 lb. of coal at the beginning of the week. Monday 6122 lb. were used ; Tuesday 5928 lb ; Wednesday 2448 lb. were received and 5831 lb. used, (a) How much coal was used during those days ? (6) How many pounds of coal were there in the shed on Thursday morning ? 7. Monday morning an engineer had 72 gallons of cylinder oil. Monday he used 8 gallons, Tuesday 12 gallons, and Wednesday 9 gallons, (a) How many gallons did he use ? (b) How many gallons of oil did he have on hand Thursday morning ? a 69,221-3008 = ? REVIEW OF ARITHMETIC 7 9. A reel contained 13,211' of wire; 1112' were used in wiring a house and 341' in wiring another house, (a) How many feet of wire were used for the houses ? (6) How many feet of wire were left on the reel ? 10. Several castings were placed on a platform. Their total weight was 1625 lb. One casting weighing 215 lb. and an- other weighing 75 lb. were taken away, (a) What was tlie weight of the castings taken away ? (b) What was the weight of the remaining castings ? Multiplication Multiplication is the process of finding the product of two numbers. The numbers multiplied together are called factors. The vudtipUcand is the number multiplied ; the multiplier is tlie number multiplied by; and the result is called the product. The sign of multiplication is an oblique cross, x , which means midtiplied by or times. Thus, 8x3 may be read 8 multiplied by 3, or 8 times 3. To find the product of two numbers. Example. — An iron beam weighed 24 lb. per foot of length. What was the weight of a beam 17 feet long ? Midtiplicand 24 lb. Write the multiplier under the multipli- jif jf J- -I- cand, units under units, tens under tens, ^ — — etc. 7 times 4 units equal 28 units, which ^^^ are 2 tens and 8 units. Place the 8 under ^4 the units column. The 2 tens are to be Product 408 lb. added to the tens product. 7 times 2 tens are 14 tens -j- the 2 tens are 16 tens, or 1 hundred and 6 tens. Place the 6 tens in the tens column and the 1 hun- dred in the hundreds column. 168 is a partial product. To multiply by the 1, proceed as before, but as 1 is a ten, write the first number, which is 4 of this partial product, under the tens column, and the next number under the hundreds column, and so on. Add the partial products and their sum is the whole product, or 408 lb. 8 VOCATIONAL MATHEMATICS EXAMPLES 1. An electrical job required the following labor : 5 men, 42 hours each ; 6 men, 4 hours each ; 12 men, 14 hours each ; 7 men, 12 hours each ; and 3 men, 6 hours each. Find the total number of hours time on the job. 2. Multiply 839 by 291. 3. A machinist sent in the following order for bolts : 12 bolts, 6 lb. each; 9 bolts, 7 lb. each; 11 bolts, 3 lb. each; 6 bolts, 2 lb. each; and 20 bolts, 3 lb. each. What was the total weight of the order ? 4. Find the product of 1683 and 809. ' To multiply by 10, 100, 1000, etc.y annex as many ciphers to the multiplicand as there are ciphers iii the multiplier. Example. — 864 x 100 = 86,400. EXAMPLES Multiply and read the answers to the following : 1. 869 X 10 a 100 X 500 2. 1011 X 100 9. 1000 X 900 3. 10,389 X 1000 10. 10,000 x 500 4. 11,298 X 30,000 11. 10,000 x 6000 5. 58,999 X 400 12. 1,000,000 x 6000 6. 681,719 X 10 13. 1,891,717 x 400 7. 801,369 x 100 14. 10,000,059 x 78,911 Division Division is the process of finding how many times one number is contained in another. The dividend is the number to be divided ; the divisor is the number by which the dividend is divided; the quotient is the result of the division. When a number is not contained an equal number of times in another number what is left over is called a remainder. REVIEW OF ARITHMETIC 9 The sign of division is -j-, and wlieii placed between two numbers signifies tliat the first is to be divided by the second. Thus, 66 -!- 8 is read 56 divided by 8. Division is also indicated by writing the dividend above the divisor with a line between. Thus, *^ ; this is read 56 divided by 8. In division we are given a product and one of the factors to find the other factor. To find how many times one number is contained in another. Example. — A machinist had 8035 rivets which he wished to arrange in groups of three. How many such groups did he have ? How many rivets did he have left over ? Quotient Place the numbers in the manner in- 2678 dicated at the left. 8 thousand is in the Divisor 3)8035 Dividend thousands column. The nearest 8 thou- g sand can be divided into groups of 3 is 2 OA (thousand) times, which gives (> thousand. ^^ Write 2 as the first figure in tlie quotient over 8 in the dividend. Place the 6 (thou- •^"^ sand) under the 8 thousand and subtract ; 21 the remainder is 2 thousand or 20 hundred. 25 8 is contained in 20 himdred 6 hundred 24 times or 18 hundred and 2 hundred re- R m ind ~T mainder. Write 6 as the next figure in the quotient. Add the 3 tens in the divi- dend to the 2 hundred or 20 tens, and 23 tens is the next dividend to be divided. 3 is contained in 23 tens 7 times or 21 tens with a remainder of 2 tens. Write 7 as the next figure in the quotient. 2 tens or 20 units plus the 6 units from the quotient make 25 units. 3 is contained in 26, 8 times. Write 8 as the next figure in the quotient. 24 units subtracted from 25 units leave a remainder of 1 unit. Then the answer is 2678 groups of rivets and 1 rivet left over. Proof. — Find the product of the divisor and quotient, add the re- mainder, if any, and if the sum equal the dividend, the answer is correct. 10 VOCATIONAL MATHEMATICS EXAMPLES 1. A strip of plate measures 85" in length. How many pieces 6" long can be cut from it? AVould there be a re- mainder ? 2. How many pieces 2" long can be cut from a brass plate 62' long, if no allowance is made for waste in cutting ? 3. If the cost of constructing 362 miles of railway was $4,561,200, what was the cost per mile ? 4. If a job which took 379 hours was divided equally among 25 men, how many even hours would each man w^ork, and how much overtime would one of the number haye to put in to complete the job ? 5. The " over-all " dimension on a drawing was 18' 9". The distance was to be spaced off into 14-inch lengths, beginning at one end. How many such lengths could be spaced ? How many inches would be left at the other end ? 6. If a locomotive consumed 18 gallons of fuel oil per mile of freight service, how far could it run with 2036 gallons of oil ? 7. If 48 screws w^eigh one pound, how many cases each containing 36 screws could be filled from a stock of 29 lb. of screws ? 8. A plate measures 68" in length. How many pieces 15" long can be cut from it ? How much will be left over? 9. A machinist desired to know the number of lots of 42 lb. each, contained in 3276 lb. of screws. Solve. 10. If a freight train made a distance of 112 miles in 8 hours, what was the average speed per hour ? 11. If the circumference of the driving wheel of a loco- motive was 22 feet, how many turns did it make in going 88 miles ? 12. Divide 38,910 by 3896. REVIEW OF ARITHMETIC 11 REVIEW EXAMPLES 1. A load of castings came to a machine shop. It was not desirable to weigh the castings on the wagon, so they were weighed in 6 lots as follows: 196 lb., 389 lb., 876 lb., 899 lb., 212 lb., and 847 lb. What was the total weight? 2. Five steel bars are placed end to end. If each bar is 29 ft. long, what is the total length ? 3. A steam fitter found that the weight per foot of 3{" wrought iron pipe is 9 lb. ; what is the weight of a piece of 3k" wrought iron pipe 12 feet long ? 4. An accident happened in a mill. A number of men were sent out to make the repairs. The following number of hours was reported : 8 men 10 hours each 3 men 5 hours each 4 men 65 hours each 4 men 21 hours each 7 men 14 hours each 6 men 11 hours each What was the total number of hours worked? 5. The consumption of water for a city during the month of December was 116,891,213 gallons and for January 115,819,729 gallons. How much was the decrease in consumption ? 6. An order to a machine shop called for 598 machines each weighing 1219 pounds. What was the total weight ? 7. If an I beam weighs 24 lb. per foot of length, find the weight of one measuring 16' 9" long. 8. Multiply 641 and 225. 9. Divide 24,566 by 319. 10. An order was sent for ties for a railroad 847 miles long. If each mile required 3017 ties, how many ties would be needed ? 11. How many gallons j)er minute are discharged by two pipes if one discharges 25 gallons per minute and the other 6 gallons less ? 12 VOCATIONAL MATHEMATICS Factors ' The factors of a number are the integers which when multi- plied together produce that number. Thus, 21 is the product of 3 and 7 ; hence, 3 and 7 are the factors of 21. Separating a number into its factors is (tailed factoring. A number that has no factors but itself and 1 is a prime number. The prime numbers up to 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23. A prime number used as a factor is 2i prime factor. Thus, 3 and 5 are prime factors of 15. Every prime number except 2 and 5 ends with 1, 3, 7, or 9. To find the prime factors of a number. Example. — Find the prime factors of 84. 2 )84 The prime number 2 divides 84 evenly, leaving the quotient 2)42 ^'^' which 2 divides evenly. The„next quotient is 21 which 3 divides, giving a quotient 7. 7 divided by 7 gives the last quotient 1 which is indivisible. The several divisors are the prime factors. So 2, 2, 3, and 7 are the prime factors of 84. 3)21 7)7 1 Phoof. — The product of the prime factors gives the number. EXAMPLES Find the prime factors : 1. 63 4. 636 7. 1155 2. 60 5. 1572 8. 7007 3. 250 6. 2800 9. 13104 Cancellation To reject a factor from a number divides the number by that factor ; to reject the same factors from both dividend and divisor does not affect the quotient. This process is called cancellation. This method can be used to advantage in many everyday cal- culations. Example, — Divide 12 x 18 x 30 by 6 x 9 x 4. REVIEW OF ARITHMETIC 13 1 2 ;2 16 By ^^^ method it is not Dividend ;S2 X ;? X 3P OA ^ .. . neccBsary to multiply be- Tk' • a w Q w 7 = ^^ Quotient. fore dividing. Indicate \^ ^ ' the division by writing r the divisor under the divi- 1 dend with a line between. Since 6 is a factor of 6 and 12, and 9 of 9 and 18, reapectively, they may be cancelled from both divisor and dividend. Since 2 in the dividend is a factor of 4 in the divisor it may be cancelled from both, leaving 2 in the divisor. Then the 2 being a factor of 30 in the dividend, is cancelled from both, leaving 15. The product of the uncancelled factors is 30. Therefore, the quotient is 30. Proof. — If the product of the divisor and the quotient equal the dividend, the answer is correct. EXAMPLES Indicate and find quotients by cancellation : 1. Divide 36 X 27 x 49 x 38 x 50 by 70 x 18 x 15. 2. What is the quotient of 36 x 48 X 16 divided by 27 x 24 X8? 3. How many pounds of washers at 50 cents a pound must be given in exchange for 15 pounds of bolts at 40 cents a pound ? 4. There are 16 ounces in a pound ; 30 pounds of steel will produce how many horseshoes, if each weighs 6 ounces ? 5. How many pieces of steel rod, each weighing 10 pounds, at 20 cents a pound, must be given in exchange for 10 bars of jY' iron rod, each weighing 5 pounds, at 4 cents per pound ? 6. Divide the product of 10, 75, 9, and 96 by the product of 5, 12, 15, and 9. 7. If 24 men, working 9 hours a day, can do a piece of work in 12 days, how many days will it take 18 men, working 8 hours a day, to do the work ? 14 VOCATIONAL MATHEMATICS Greatest Common Divisor The greatest common divisor of two or more numbers is the greatest number that will exactly divide each of the numbers. To find the greatest common divisor of two or more numbers. Example. — Find the greatest common divisor of 90 and 150. 90 = 2x3x5x3 2 )90 150 First Method 150 = 2 X3x5x5 5 )45 75 The prime factors com- Ans. 30 = 2 X 3 X 5 3)9 15 ^^^^ ^o both 90 and 150 "o K are 2, 3, and 5. Since 2 X 3 X 5 = 30 Ans. ^^^ greatest common di- visor of two or more num- 90)150(1 hers is the product of gQ their common factors, 30 TTT^NQp..-. is the greatest common ' ^^^ divisor of 90 and 150. bO Greatest Common Divisor 30)60(2 '^^^^^^ ^^^^^^ (jQ To find the greatest — common divisor when the numbers cannot jbe readily factored, divide the larger by the smaller, then the last divisor by the last remainder until there is no remainder. The last divisor will be the greatest common divisor. If the greatest com- mon divisor is to be found of more than two numbers, find the greatest common divisor of two of them, then of this divisor and the third num- ber, and. so on. The last divisor will be the greatest common divisor of all of them. EXAMPLES Find the greatest common divisor : 1. 270,810. 3. 504,560. 5. 72,153,315,2187. 2. 264, 312. 4. 288, 432, 1152. Least Common Multiple The product of two or more numbers is called a multiple of each of them; 4, 6, 8, 12 are multiples of 2. The multiple RKVIKW OF AHITHMKTIC 15 of two or more numbers is called the common multiple of the numbers ; 00 is a common multiple of 4, 5, 6. The least common multiple of two or more numbers is the smallest common multiple of the number; 30 is the least common multiple of 8, 5, 6. To find the least common multiple of two or more numbers. ExAMPLK. — Find the least common multiple of 21, 28, and 30. _.. , ,. ,. ^ , First Method -1=3x7 Take all the factors of the firet number, all of 28 = 2 X 2 X 7 ^h^ second not already represented in the first, etc. 30 = 2 X i} X 5 'f has, 3 x7x2x2x5 = 420i.C.3f. Second Method 2 )21 28 30 3)21 14 15 7 )7 14 5 2 X 3 x 7 X 1 x 2 X 5 = 420 X. (7. M. Divide any two or more numbers by a prime factor contained in them, like 2 in 28 and 30. Write 21 which is not divided by the 2 for the next quotient together with tl>e 14 and 16. 3 is a prime factor of 21 and 15 which gives a quotient of 7 and 5 with 14 written in the quotient undi- vided. 7 is a prime factor of 7 and 14 which gives a remainder of 1, 2 ; and 6 undivided is written down as before. The product 420 of all these divisors and the last quotients i^ the least common multiple of 21, 28, and 30. EXAMPLES Find the least common multiple : 1. 18, 27, 30. 2. 15, 60, 140, 210. 3. 24, 42, 54, 360. 4. 25, 20, 35, 40. 5. 24, 48, 96, 192. 6. What is the shortest length of rope that can be cut into pieces 32', 36', and 44' long? 16 VOCATIONAL MATHEMATICS Fractions A fraction is one or more equal parts of a unit. If an apple be divided into two equal parts, each part is one-half of the apple, and is expressed by placing the number 1 above the number 2 with a short line between : i A fraction always indicates division. In i, 1 is the dividend and 2 the divisor ; 1 is called the nuinerator and 2 is called the denominator. A common fraction is one which is expressed by a numerator written above a line and a denominator below. The nu- merator and denominator are called the terms of the fraction. A proper fraction is a fraction whose value is less than 1 ; its numerator is less than its denominator, as -|, -f, f, ^. An improper fraction is a fraction whose value is 1 or more than 1; its numerator is equal to or greater than its denominator, as f, ■}-f. A number made up of an integer and a fraction is a mixed number. Read with the word and between the whole number and the fraction : 4:^\, S^, etc. The value of a fraction is the quotient of the numerator divided by the denominator. EXERCISE Read the following : 1. I 3. 121- 5. ^ 7. 9^ 9. I 2. \i 4. ^ 6. ^ 8. 12^ Reduction of Fractions Reduction of fractions is the process of changing their form without changing their value. To reduce a fraction to higher terms. Multiplying the denominator and the numerator of the given fraction by the same number does not change the value of the fraction. REVIEW OF ARITHMETIC 17 Example. — Reduce | to thirty -seconds. The denominator must be multiplied by 4 to 5 ^1 _ ^ jins, obtain 32 ; so the numerator must be multiplied 8 4 32 by the same number so that the value of the fraction may not be changed. EXAMPLES Change the following : 1. f to 27ths. 6. ^^ to 75th8. 2. \^ to 60ths. 7. Jl to 144ths. 3. f to 40ths. 8. ^ to 168ths. - 4. Jto56ths. 9. |Jto522ds. 5. 3% to 50ths. 10. ^ to 9375ths. A fraction is said to be in its lowest terms when the numersr tor and the denominator are prime to each other. To reduce a fraction to its lowest terms. Dividing the numerator and the denominator of a fraction by the same number does not change the value of the fraction. The process of dividing the numerator and denominator of a fraction by a number common to both may be continued until the terms are prime to each other.' Example. — Reduce -f| to fourths. The denominator must be divided by 4 to give 12 _ 3 J, the new denominator 4 ; then the numerator must be 16 4 divided by the same number so as not to change the value of the fraction. If the terms of a fraction are large numbers, find their greatest common divisor and divide both terms by that. Example. — Reduce |Jff to fourths. (1) 2166)2888(1 (2) 2166^3 . 2166 2888 4 ^ ' O. a D. 722)2166(3 2166 18 VOCATIONAL MATHEMATICS EXAMPLES Reduce to lowest terms : 1- A 3- Hil s- H '• IS* 9- Hi 2- m *■ U 6. -rVj 8. iff 10. T-VA To reduce an integer to an improper fraction. Example. — Reduce 25 to fifths. 25times4 = l*i Ans. , '" ' "'f""'[!,^- ^ 25 there must be 5 5 25 tunes 5, or i|^. To reduce a mixed number to an improper fraction. Example. — Reduce 16^ to an improper fraction. i_ sevenths Since in 1 there are }, in 16 there must 112 be 16 times ^, or ^2^ 4 sevenths lp + f=^K 116 sevenths, = J-^. EXAMPLES Reduce to improper fractions : 1. 3J 3. 171 5. 13J 7. S59j% 2. 16^ 4. 12^ 6. 27t«^ 8. 482i| 9. 25-^ - 10. Reduce 250 to 16ths. 11. Change 156 to a fraction whose denominator shall be 12. 12. In $ 730 how many fourths of a dollar ? 13. Change 12 1 to 16ths. 14. Change 24f to ISths. To reduce an improj^er fraction to an integer or mixed number divide the numerator by the denominator. Example. — Reduce -\%^- to an integer or mixed number. 24 16)385 ^^ Since || equal 1, ^^ will equal as many times 1 as 16 is contained in 385, or 24^15 65 24-jL Ans. ^^^^^^^ 64 1 REVIEW OF ARITHMETIC 19 EXAMPLES Reduce to intecjers oi* mixed numbers : 1. H *• 'iV 7. VV 10. i^i . Wlieii fractions have the same denominator their denomi- nator is called a common denominator. Thus, |§, ^, ^j have a common denominator. The smallest common denominator of two or more fractions is their least common denominator. Thus, \^, ^«2, ^ become ^, |, J when changed to their least common denominator. The common denominator of two or more fractions is a common multiple of their denominators. The least common denominator of two or more fractions is the least common multiple of their denominators. Example. — Reduce } and f to fractipi^s having a common denominator. a w 6 _- 18 The common denominator must be a , 4 _ 20 common multiple of the denominators 4 ^ T ~ tT and 0, and since 24 i.s the product of the ? ~ Ti ^"^ t — TX denominators, it is a common multiple of them. Therefore. 24 is a common denominator of J and f . To reduce fractions to fractions having the least common denominator. Example. — Reduce |, |, and j^ to fractions having the least common denominator. 2^^ S 6 12 "^'^^ least common de- ^— — — nominator must be the 1 — z. least common multiple of 1 1 -^ the denominators 3, 6, 12, 2x3x2 = 12 L. r'. M. which is 12. f = -(8^ ; ^ = 1^ ; Vj = f^. Ans. l^'vide the least common multiple 12 by the denom- inator of each fraction, and multiply both terms by the quotient. If the 20 VOCATIONAL MATHEMATICS denominators should be prime to each other, their product would be theii least common denominator. EXAMPLES Reduce to fractions having a common denominator : 1- h i 5- f> I, I 4- f , tV, i 8- i> A, I, i Reduce to fractions having least common denominator : 1- I, J, A 5- 1,1, A, 4 2- I, h A 6. I, f , I, I ^- A> TO i 7. Which fraction is larger, 4- iA.A.A fori? Addition of Fractions Only fractions with a common denominator can be added. If the fractions have not the same denominator, reduce them to a common denominator, add their numerators, and place their sum over the common denominator. The result should be reduced to its lowest terms. If the result is an improper fraction, it should be reduced to an integer or mixed number. The least common multi- ExAMPLE. — Add I, |, and y\. 1. 2 )4 6 16 2}2__3__8 pie of the de- 13 4 48 Z/. C. M. nominators is o 3i5i9 36i40i27_i03 J*n, 48. Dividing this by the de- nominator of each fraction and multiplying both terms by the quotient give If, f§, f|. The fractions are now like fractions, and are added by adding their numerators and placing the sum over the common denomi- nator. Hence, the sum is ^5^/-, or 2j7j. REVIEW OF ARITHMETIC 21 Example. — Add 5J, 7y^, and 6^j^- ^i = ^a First find the sum of the fractions, 7^= TfJ- which is fj, or 1|^. Add this to the QJL = 6lr4t sum of the integers, 18. 18 -|- 1 J^ = i8|f=19H. Ana. ^m- EXAMPLES 1. Find the "over-all" dimension of a drawing if the separate parts measure -j^", f", ^", and -^"j respectively. 2. Find the sum of ^, J, J, j-f , and fj. a Find the sura of 3|, 4}, and 2^. 4. Four castings weigh respectively 8J lb., 5^ lb., llf lb., and 7| lb. What is their total weight ? 5. The diameter of two holes is 3|" and the distance be- tween the sides of the holes is J". What is the distance from the outside of one hole to the outside of the other ? 6. Two brass rods measure 8-j^" and 5^" What is their combined length ? 7. A board was cut into two pieces, one 8f" and the other 6^" long. If ^" was allowed for waste in cutting, what was the length of the board ? 8. Three pieces of rod contain 38^, 12^, and 53| feet re- spectively. What is their total length in feet? 9. Add: lOi, 7|, 11, i|. Subtraction of Fractions Only fractions with a common denominator can be sub- tracted. If the fractions have not the same denominator, reduce them to a common denominator, and write the differ- ence of their numerators over the common denominator. The result should be reduced to its lowest terms. 22 VOCATIONAL MATHEMATICS Example. — Subtract J from |. 5_2_i5._i2_ 3 The least common denominator of ^ _/ = i' ^]^. ^ I and I is 6. t = f, and f = ^. ^^ ^' ' Their difference is \. Example. — From 11^ subtract 5|. 11 1 ^ 10 8 When the fractions are changed to ^ 5^ __ ^5 their least common denominator, they ^ — ^ _ fii A ' ^^^ ^^^ ~ ^t- I cannot be subtracted *> ~ 2* • • from I, hence 1 is taken from 11 units, changed to sixths, and added to the f which makes f . 10| — 4| = 6|=6|. EXAMPLES 1. The distance between two holes is 5f " measured from the centers. If the holes are -^^" in diameter, what is the length of metal plate between them ? 2. From a steel bar 26|" long were cut the following pieces : one 7\", one 6^", one 3|" long. If after cutting these pieces, the length of the bar was 8f", what was the amount of waste in cutting? 3. A piece of steel on a lathe is 1" in diameter. In the first cut -^" were taken off, in the second cut g-\", in the third cut -jig-", in the fourth -^-q". What was the diameter of the finished piece ? 4. A bolt is 15^" long. How much must be cut from it to make it llf " long ? 5. In wiring a house five men work 14 hours ; one man works 1 hour and 20 minutes ; a second man works 2 hours and 15 minutes ; a third man works 5 hours ; and a fourth man, 4J hours. How many hours did the fifth man work ? 6. It is S\" between the centers of two holes of the same size. The distance between the sides of the holes is 1^" What k the diameter gf each hole ? REVIEW OF ARITHMETIC 23 7. There were 48^ gallons in the tank. First 4^ gallons were used, then 5^ gallons, and last 2'j gallons. How many gallons were left in the tank ? 8. \Vhat is the difference between ^ and JJ ? 9. What is the difference between 32J and .SJJ ? 10. An electrician had a reel of .SOO feet of copper wire. He used at various times 50^', 32 J', 1091', and 2737^'. How much wire was left? Multiplication of Fractions To multiply fnictiona, mnltiply the numerators together for the neiv numerator and multiply the denominators together /or the new denominator. Cancel when possible. The word of between two fractions is equivalent to the sign of multiplication. To multiply a mixed number by an integer, nmltiply the whole number and the fraction separately by the integer then add the prrxlucts. To multiply two mixed numbers, change each to an improper fraction and multiply. Example. — Multiply | by J. I multiplied by J is the same as 5 0/ 1. 3 and 5 are prime to each other so that answer is |. This method of solution is the same as multiplying the numerators together for a new numerator and the denominators for a new denominator. Cancellation shortens the process. Example. — Find the product of 124} and 5. 124} - If the fi-action and integer are mul- — ^ «; 4 — L5 — i ^^^ divisor becomes 11 halves and the dividend ^ ' 7364 halves. Multiplying both dividend and ""^TT Ans. divisor by the same number does not change the quotient. Dividing, the quotient is 6603^. A fraction having a fraction for one or both of its terras is called a complex fraction. To reduce a complex fraction to a simple fraction. 42 Example. — Reduce ^ to a simple fraction. || = ^ = ¥^V- = ¥x^=if Ans. Change 4f and 7^ to improper fractions, ^* and V, respectively. Per- form the division indicated with the aid of cancellation and the result will bejf EXAMPLES 1. Divide ^ by |. 7. 296-10i = ? 2. Divide ^^ by f . 8. 28,769 -^7f=? 3. Divide |f by \. •■?r' 4. Divide J by \. 5. Divide } by J. 6. 384J^5 = ? - -m-' 26 VOCATIONAL MATHEMATICS Drill in the Use of Fractions Addition 1. i + 4 =? 19. 1 8 + i =? . 37. A + i =? 2. i + \ =^ 20. i + 1 =? 38. A + i =? 3. i + i ='' 21. i + i =? 39. iV + i =? 4. i + TV = ? 22. i + ,v = ? 40. A + tV = ? 5. i+^=? 23. I + .v=? 41. 3V + A=? 6. i+^v=? 24. 1 + 6^4=? 42. A + 6V = ? 7. i+i =? 25. tV + i =? 43. A +4 =? 8. i + i =^ 26 tV + i =? 44. eV +i =? 9. i + i =? 27. tV + i =? 4S. A +i =? 10. i + iV = ? 28. tV + tV = ''' 46. A + tV = ? 11. i + A = ^ 29. -A + A=? 47. A +A=? 12. i+6V = '^ 30. t's + «V=? 48. A +A=? 13. l + i =? 31. i + 4 =? 49. * +1 =? 14. l + i =? 32. 3 + i =? SO. i + i =? 15. l + i =? 33. i + i =? 51. i + i =? 16. I + tV = ? 34. i + tV = ? 52. 1 + tV = ? 17. I + ^V = ? 35. 8 4 + A = ? 53. i +A=? 18. l + -6^=? 36. f + eV = ? 54. i +A=? Subtraction 1. i-i =? 8. i- -i =? 15. 5 8" -i =? 2. i-i =? 9. i- -i =? 16. i -tV = ? 3. i-i =? 10. \ -tV = ? 17. 1- -A=? 4. i-TV=? 11. \- -sV = ? 18. f- -Vt = ? 5. i-^V = ? 12. i- -^v=? 19. i- -f =? 6. i-^V = ? 13. -4 =-? 20. i- -A = ? 7. i-^=? 14. f -i =? 21. 1 ' -i =? REVIEW OF ARITHMETIC 27 22- i -.', = ? 23. i -3'l = '' 24. i -ff'4 = -" 25-4 -T%=-' 26. ^ -tV=- 27. A\--tV = '^ 28. -jV — tV = ^' 29- tV - /2 = ■•• 30. T*! 31. f 32. f 1 ITT 1 — .» -i = 1. 4 2. ^ 3- 4 4. * 6. i a i 9. i 10. J 11. J 12. i 13. I 14. f 15. I 16. J 17. » 18. I X J — . xi =? X^ =? XA = ? V 1 *> xi =? xi =? xi =? Xt', = '^ x,V = ^^ X^ = ? Xi =? X J =? xj .= ? xA = ? V 1. — 9 33. J -J = 34. 3 -l'« = -' 35. J -5>j = V 36. I -j^f = 37. U-jV = 38. i -A = 39. H-A = 40. H-A = 42. i?-s^ = 43. 4 -« = Multiplication 19. I X .V = ? __ 9 20. J xi =? 21. i xi 22. J X j5 23. i X^j = ? 1 s^ \ •> 24. i Xj't 44. 1 -A = 45. \ -i,= 46. t*-."!" 47. s't-j^ = 48. A-,'i = 49. i -i = 50. J -V = 51- i -i = 52. J -t'j = 53. J -A = 54. i -A = ? 37. ,V 38. ,>j 39. ^j 40. ,V 41. ^j 42. ^2 25. ^Vxi =? 43. ,V 26. T»jXi =? 44. ,>j 27. iVxi =? 28. AXt>j = ? 29. T'ffX^j = ? 30. T'irX5'i = ? 31. I X 4 = ? 32. J X } = ? 33. I Xj =? 34. J X ,V = ? 35. f X jV = '.' 36. J Xji = ? 45. sV 46. ,V 47. i^T 48. sV 49. I 50. i 51. J 52. J 53. i 54. J X4 =? X i =? Xi =? Xt'5 = ? Xi^l = V X ^I = V xi =? X} =? xi =? xA = "'' x^=? X «V = ■'' xi =? xi =? xi =? Xt',= V X j'l = '! XVj=? 28 VOCATIONAL MATHEMATICS Division 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. Decimal Fractions A power is the product of equal factors, as 10 x 10 = 100. 10 X 10 X 10 = 1000. 100 is the second power of 10. 1000 is the third power of 10. A decimal fraction or decimal is a fraction wliose denominator is 10 or a power of 10. A common fraction may have any number for its denominator, but a decimal fraction must always have for its denominator 10, or a power of 10. A decimal is written at the right of a period (.), called the decimal point. A figure at the right of a decimal point is called a decimal figure. 1. i^i =? 19. \ -^i =? 2. i^i =? 20. i -i =? 3. i^i =? 21. i ^i =? 4. i-^TV = ? 22. i ^t'^ = ? 5. i-^A = ? 23. i -A = ? 6. i^^V=? 24. i ^7V = ? 7. i^i =? 25. iV ^i =? 8. i^i =? 26. tV ^i =? 9. i^i =? 27. tV ^i =? 10. i^A = ? 28. tV -tV=? 11. i^A = ? 29. tV ^A = ? 12. i^A = ? 30. 1^6 ^F't = ? 13. I^i =? 31. 8 ^i =? 14. I^i =? 32. 1 -i =? 15. I^i =? 33. f ^i =? 16. I^tV = ? 34. * ^tV = ? 17. I^A = ? 35. f ^A=? 18. I^^ = ? 36. 1 ^TrV = ? A- -i =? ^v- -i =? A- -i =? Ti- -tV = ? W- -A = ? 7V- -A = ? ^v- -i =? .v^ -i =? A- -i =? A- -tV = ? A- -^\ = -f e'l- -A = ? J - -i =? i - -i =? i - -i =? i - -tV = ? i - -A = ? i - -A = ? REVIEW OF ARITHMETIC 29 A mixed decimal is an integer and a decimal ; as, 16.04. To read a decimal, read the decimal as an integer, and give it the denomination of the right-hand figure. To tvrite a deci- mal, write the numerator, prefixing ciphers when necessary to express the denominator, and place the point at the left. There must be as many decimal places in the decimal as there are ciphers in the denominator. EXAMPLES Read the following numbers: 1. .7 7. .4375 13. .0000054 19. 9.999999 2. .07 8. .03125 14. 35.18006 20. .10016 3. .007 9. .21875 15. .0005 21. .000155 4. .700 10. .90625 16. 100.000104 22. .26 5. .125 11. .203125 17. y.1032002 23. .1 6. .0625 12. .234375. la 30.3303303 24. .80062 Express decimally : 1. Four tenths. 2. Three hundred twenty-five thousandths. 3. Seventeen thousand two hundred eleven hundred-thou- sandths. 4. Seventeen hundredths. 6. Five hundredths. 5. Fifteen thousandths. 7. Six ten-thousandths. 8. Eighteen and two hundred sixteen hundred-thousandths. 9. One hundred twelve hundred-thousandths. 10. 10 milliouths. 11. 824 ten-thousandths. 12. Twenty-nine hundredths. 13. 324 and one hundred twenty-six millionths. 14. 7846 hundred-millionths. 30 VOCATIONAL MATHEMATICS 1 c 5_6 3 1 2 123 3 2 8 6 5 4 17. One and one tenth. 18. One and one hundred-thousandth. 19. One thousand four and twenty-nine hundredths. Reduction of Decimals Ciphers annexed to a decimal do not change the value of the decimal; these ciphers are called decimal ciphers. For each cipher prefixed to a decimal, the value is diminished ten- fold. The denominator of a decimal — when expressed — is always 1 with as many ciphers as there are decimal places in the decimal. To reduce a decimal to a common fraction. Write the numerator of the decimal omitting the point for the numerator of the fraction. For the denominator write 1 with as many ciphers annexed as there are decimal places in the decimal. Then reduce to lowest terms. Example. — Reduce .25 and .125 to common fractions. 1 Write 25 for the numerator and nK 25 'l^ 1 A 1 for the denominator with two O's, "" 100 ~~ ^00 ~~ 4 * which makes j^^-^ ; -^-^^ reduced to 4 lowest terms is \. 1 ^ oe- 125 J2f> 1 A '^25 is reduced to a common frac- .125 = = ^ - = - Ans. ^. . ^^ 1000 ^p0j^ ^ ^^^^ ^" ^^^ ^^"^^ ^^y- 8 Example. — Reduce .37^ to a common fraction. 37i has for its denominator 1 3 371 37i^i^l5^_jL = ? Ans "^^^^^^'"^"^^^^^^^«I^- 100 100 2 ;0p 8 ' This is a complex fraction 4 which reduced to lowest terms isf. REVIEW OF ARITHMETIC 31 EXAMPLES Reduce to cominon trartions : 1. .01)375 6. 2.25 11. .16J 16. m 2. .15625 7. 16.144 12. ..S3J 17. .66J 3. .015625 a 25.0000100 13. .06J 18. .36i 4. .609.375 9. 1084.0025 14. .140625 19. .83i 5. .578125 10. .12.V 15. .984375 20. .62^ To reduce a cominon fraction to a decimal. Annex decimal ciphers to the numerator and divide by the de- nominator. Point off from the right of the quotient as many places as there are ciphers annexed. If there are not figures enough in the quotient, prefix cijthers. The division will not always be exact, i.e. | = .142f or .142+. Example. — Reduce | to a decimal. .75 4)3.00 28 20 I = .76 EXAMPLES Reduce to decimals : 1. ^V 6. i u. A 16. H 21. -^iji 2. TiTT 7. w 12 ^icT 17. 16i 22. 25.12i 3. TSU 8. M 13. T2oJi 18. 66| 23. 33' 4. i 9. A 14. m 19. U 24. ^i 5. i 10 jh 15. i\ 20. a IT 25. Ti^ Addition of Decimals To add decimals^ write them ,so that their decimal points are in a column. Add as in integers, and place the point in the sum directly under the points above it. 32 VOCATIONAL MATHEMATICS Example.— Find the sum of 3.87,2.0983, 5.00831, .029, .831. 3.87 2 0983 Place these numbers, one under the other, with n fW'iS'?! decimal points in a column, and add as in addition of integers. The sum of these numbers should '^^^ have the decimal point in the same column as the •831 numbers that were added. 11.83661 Ans. EXAMPLES Find the sum : 1. 5.83, 7.016, 15.0081, and 18.3184. 2. 12.031, 0.0894, 12.0084, and 13.984. 3. .0765, .002478, .004967, .0007862, .17896. 4. 24.36, 1.358, .004, and 1632.1. 5. .175, 1.75, 17.5, 175., 1750. 6. 1., .1, .01, .001, 100, 10., 10.1, 100.001. 7. Add 5 tenths; 8063 millionths; 25 hundred-thousandths ; 48 thousandths; 17 millionths; 95 ten-millionths ; 5, and 5 hundred-thousandths ; 17 ten-thousandths. 8. Add 24|, 17}, .0058, 7i, 9^^. 9. 32.58, 28963.1, 287.531, 76398.9341. 10. 145., 14.5, 1.45, .145, .0145. Subtraction of Decimals To subtract decimals, ivrite the smaller number under the larger icith the decimal point of the subtrahend directly under the decimal point of the minuend. Subtract as in integers, and place the point directly under the points above. Example. — Subtract 2.17857 from 4.3257. Write the lesser number under the greater, 4.32570 Minuend y^^^y^ ^^^ decimal points under each other 2.17857 Subtrahend Add a to the minuend, 4.3257, to give it the 2.14713 Remainder same denominator as the subtrahend. Then subtract as in subtraction of integers. Write the remainder with decimal point under the other two points, REVIEW OF ARITHMETIC 33 EXAMPLES Subtract : 1. 69.0364-30.8691 = ? 3. .0625 - .03125 = ? 2. 48.7209 - 12.0039 = ? 4. .0001 1 - .000011 = ? 5. 10 - .1 -H .0001 = ? 6. From one thousand take five thousandths. 7. Take 17 hundred-thousandths from 1.2. 8. From 17.37^ take 14.16f 9. Prove that ^ and .500 are equal. 10. Find the difference between yYA ^°^ rlfH* Multiplication of Decimals To multiply decimals jnoceed as in integers, and give to the product as many decimal figures as there are in both multiplier and multiplicand. When there are not figures enough in the product, prefix ciphers. Example. — Find the product of 6.8 and .63. 6.8 Multiplicand .63 Multiplier ^-^ ^^ ^^® multiplicand and .63 the multiplier. "^777 Their product is 4,284 with three decimal figures, ^ the number of decimal figures in the multiplier and multiplicand. 4.284 Product Example. — Find the product of .05 and .3. .05 Multiplicand The product of .05 and .3 is .016 with a cipher .3 Multiplier prefixed to make the three decimal figures re- .015 Product quired in the product. EXAMPLES Find the products : 1. 46.25 X. 125 3. .015 x .05 2. 8.0625 X. 1875 4. 25.863 x 4i 34 VOCATIONAL MATHEMATICS 5. 11.11 X 100 8. .325 X 12i 6. .5625 X 6.28125 9. .001542 x .0052 7. .326 x 2.78 10. 1.001 x 1.01 To multiply btj 10, 100, 1000, etc., remove the point one place to the right for each cipher m the multiplier. This can be performed without writing the multiplier. Example.— Multiply 1.625 by 100. 1.625 x 100 = 162.5 To multiply by 200, remove the point to the right and multiply by 2. Example. — Multiply 86.44 by 200. 86.44. 2 17,288 EXAMPLES Find the product of : 1. 1 thousand by one thousandth. 2. 1 million by one millionth. 3. 700 thousands by 7 hundred-thousandths. 4. 3.894 x 3000 5. 1.892 x 2000. Division of Decimals To divide decimals proceed as in integers, and give to the quo- tient as many decimal figures as the number in the dividend ex- ceeds those in the divisor. Example. — Divide 12.685 by .5. The number of decimal figures in Divisor . 5)12.685 Dividend the quotient, 12.685, exceeds the num- 25.37 Quotient ber of decimal figures in tlie divisor, .5, by two. So there tnust be two deQi- pial figures in the quotient. REVIEW OF ARITHMETIC 35 Example. — Divide 899.552 by 192. When the divisor is an integer* the point in tlie dividend, and tli<» diviMfoii |H»r- formeoint in both divisor and dividend as many place.s U) the right as tlicre are decimal placcB in tlie divisor, which is equivalent to multiplyin<{ l)oth divisor and dividend by the same number and does not change the (}Uo. tient. Then place the point in the quotient as if the divisor were an integer. In this ex- ample, the multiplier of both dividend and divisor is 100. EXAMPLES Eind the quotients : 1. .0625 -.125 5. 1000 -I- .001 2. 315.432 -.132 6. 2.490 + . 136 3. .75 -.0125 7. 28000 + 16.8 4. 125 -5- 12^ a 1.225 + 4.9 9. 3.1416 + 27 10. 8.33 + 6 To divide by 10, 100, 1000, etc.^ remove tin- inimt nue place to the left for each cipher in the didaor. 7V> divide by 200, remove the point ttco places to the le/t^ and divide by 2. 36 VOCATIONAL MATHEMATICS EXAMPLES Find the quotients : 1. 38.64-10 6. 865.45-5000 2. 398.42-1000 7. 38.28-400 3. 1684.32-1000 8. 2.5-500 4. 1.155-100 9. .5-10 5. 386.54-2000 10. .001-1000 Parts of 100 or 1000 1. What part of 100 is 12^ ? 25 ? 33^ ? 2. What part of 1000 is 125 ? 250 ? 333^ ? 3. How much is i of 100 ? Of 1000? 4. How much is J of 100 ? Of 1000 ? 5. What is 1 of 100 ? Of 1000? Example. — How much is 25 times 24 ? 100 times 24 = 2400. 25 times 24 = |^ as much as 100 times 24 = 600. Ans. Short Method in Multiplication To multiply by 25, multiply by 100 and divide by 4 ; 33|, multiply by 100 and divide by 3 ; 16|, multiply by 100 and divide by 6 ; 121 multiply by 100 and divide by 8 ; 9, multiply by 10 and subtract the multiplicand ; 11, if more than two figures, multiply by 10 and add the multiplicand; 11, if two figures, place the figure that is their sum between them. 63 X 11 = 693 74 X 11 = 814 Note that when the sum of the two figures exceeds nine, the one in the tens place is carried to the figure at the left. REVIEW OF ARITHMETIC 37 EXAMPLES Multiply by the short process : 1. 81 by 11 =? 10. 68 by 16J = ? 2. 76 by 33i = ? 11. 112 by 11 = ? 3. 128 by 12J = ? 12. 37 by 11 = ? 4. 87 by 11 = ? 13. 4183 by 11 = ? 5. 19 by 9 = ? 14. 364 by 33^ = ? 6. 846 by 11 = ? 15. 8712 by 12J = ? 7. 88 by 11 = ? 16. 984 by 16J = ? 8. 19 by 11 = ? 17. 36 by 25 = ? 9. 846byl6J = ? la 30by333J=? Aliquot Parts of $1.00 The aliquot parts of a nuinber are the numbers that are exactly contained in it. The aliquot parts of 100 are 5, 20, 12J, 16f, 334, etc. The monetary unit of the United States is the dollar, con- taining one hundred cents which are written decimally. 25 cents = $ ^ = quarter dollar 33^ cents = $ ^ 50 cents = $ ^ = half dollar n cents = $^^ «i cents = $ 3^5 12^ cents = $ i 16J cents = $ i 10 mills 5 cents 10 cents 10 dimes = 1 cent, ct. = $ .01 or $ 0.01 = 1 " nickel " = $ .05 = 1 dime, d. = $ .10 = 1 dollar, $ = $1.00 10 dollars = 1 eagle, E. = $ 10.00 Example. — What will 69 drills cost at 16J cents each ? 69 drills will cost 69 x 16i cts., or 69 x $i=V = f Hf = -^11.60. 38 VOCATIONAL MATHEMATICS Example. — At 25^ a box, ho^v many boxes of nails can be bought lor $ 8.00 ? 8^i = 8xf = 32 boxes. Ans. Review of Decimals 1. For work on a job one man receives $ 13.75, a second man $ 12.45, a third man $ 14.21, and a fourth man $ 21.85. What is the total amount paid for the work ? 2. A pipe has an inside diameter of 3.067 inches and an outside diameter of 3.428 inches. What is the thickness of the metal of the pipe ? 3. At 21 cts. a pound, what will be the cost of 108 castings each weighing 29 lb. ? 4. A man receives $ 121.50 for doing a piece of work. He gives $12.25 to one of his helpers, and S 10.50 to another. He also pays $75.75 for material. How much does he make on the job ? 5. An automobile runs at the rate of 9^ miles an hour. How long will it take it to go from Lowell to Boston, a dis- tance of 26.51 miles ? 6. A f square steel bar weighs 1.914 lb. per foot. What will be the cost of 5000 feet of |" steel bars, if it costs $ 1.75 per 100 lb. ? 7. Which is cheaper, and how much, to have a 13i cents an hour man take 13J hours on a piece of work, or hire a 17^ cents an hour man who can do it in 9^ hours ? 8. On Monday 1725.25 lb. of coal are used, on Tuesday 2134.43 lb., on Wednesday 1651.21 lb., on Thursday 1821.42 lb., on Friday 1958.82 lb., and on Saturday 658.32 lb. How many pounds of coal were used during the week ? 9. If in the example above there were 10,433.91 lb. of coal on hand at the beginning of the week, how much was left at the end of the week ? RKVIEW OF ARITHMETIC 30 10. A foot length of }" round steel bar weighs 1.50^5 lb., 10-foot length of J" square steel bar weiglis 19.140 lb., 1-foot length of y^" round steel bar weighs 3.017 lb. What is the total weight of the three pieces ? U. An alloy is made of copper and zinc. If .(iO is copper and .34 is zinc, how many pounds of zinc and how many pounds of co})per will there be in a casting of the alloy weighing 98 lb. ? 12.. A train leaves New York at 2.10 p.m., and arrives in Philadelphia at 4.15 p.m. The distance is 90 miles. What is the average rate per hour of the train ? 13. The weight of a foot of ^^" steel bar is 1.08 lb. Find the weight of a 21-foot bar. 14. A steam pump pumps 3.38 gallons of water to each stroke and the pump makes 51.1 strokes per minute. How many gallons of water will it pump in an hour? 15. At 12^ cents per hour, what will be the pay for 23^ days if the days are 10 hours each ? Compound Numbers A number which expresses only one kind of concrete units is a simple number; as, 5 ])k., 4 knives, 6. A number composed of different kinds of concrete units that aie related to each other is a compound number ; as, 3 bu. 2 pk. 1 qt. A denomination is a name given to a unit of measure or of weight. A number having one or more denominations is also called a denominate number. Reduction is the process of changing a number from one denomination to another without changing its value. Changing to a lower denomination is called reduction descend- ing; as, 2 bu. 3 pk. = 88 qt. 40 VOCATIONAL MATHEMATICS Changing to a higher denomination is called reduction ascending ; as, 88 qt. = 2 bu. 3 pk. Linear Measure is used in measuring lines or distance. Table 12 in. (in.) = 1 foot, ft. 3 feet = 1 yard, yd. 5^ yards or 16^ feet = 1 rod, rd. 40 rods = 1 furlong, fur. 8 furlongs = 1 mile, mi. 320 rods, or 5280 feet = 1 mile. 1 mi. = 320 rd. = 1760 yd. = 5280 ft. = 63,360 in. Surveyors' Measure is used in measuring land. Table 7.92 inches = 1 link, li. 100 links = 1 chain, ch. 80 chains = 1 mile, mi. A chain, or steel measuring tape, 100 feet long, is generally used by engineers. The feet are usually divided into tenths instead of into inches. Square Measure is used in measuring surfaces. Table 144 square inches = 1 square foot, sq. ft. 9 square feet = 1 square yard, sq. yd. 30i square yards 1 , , , 160 square rods = 1 acre, A. 640 acres = 1 square mile, sq. mi. 1 sq. mi. = 640 A. = 102,400 sq. rd. = 3,097,600 sq. yd. Cubic Measure is used in measuring volumes or solids. Table 1728 cubic inches = 1 cubic'foot, cu. ft. 27 cubic feet = 1 cubic yard, cu. yd. 16 cubic feet = 1 cord foot, cd. ft. 8 cord feet, or 128 cu. ft. = 1 cord, cd. 1 cu. yd. = 27 cu. ft. = 46,666 cu. in. REVIEW OF ARITHMETIC 41 Liquid Measure is used in measuring liquids. Table 4 gills (gi.)= 1 P'"^ pt" 2 pints = 1 quart, qt. 4 quart* = 1 gallon, gal. 1 gal. = 4 qt. = 8 pt. = 32 gi. A gallon contains 231 cubic inches. The standard barrel is 31 J gal., and the hogshead 63 gal. Dry Measure is used in measuring roots, grain, vegetables, etc. Table 2 pints = 1 quart, qt. 8 quarts = 1 peck, pk. 4 pecks = 1 bushel, bu. 1 bu. = 4 pk. = 32 qt. = 64 pints. The bushel contains 2150.42 cubic inches; 1 dry quart contains 67.2 cu. in. A cubic foot is |J of a bushel. Avoirdupois Weight is used in weighing all common articles ; as, coal, groceries, hay, etc. Table 16 ounces (oz.) = 1 pound, lb. 100 pounds = 1 hundredweight, cwt. ; or cental, ctl. 20 c\n., or 2000 lb. = 1 ton, T. 1 T. = 20 cwt. = 2000 lb. = 32,000 oz. The long ton of 2240 pounds is used at the United States Custom House and in weighing coal at the mines. Measure of Time. Table 60 seconds (sec.) = 1 minute, min. 60 minutes = 1 hour, hr. 24 hours = 1 day, da. 7 days = 1 week, wk. 365 days = 1 year, yr. 366 days = 1 leap year. 100 years = 1 century. 42 VOCATIONAL MATHEMATICS Counting. Table 12 things = 1 dozen, doz. 12 dozen = 1 gross, gr. 12 gross = 1 great gross, G. gr. Paper Measure. Table 24 sheets = 1 quire 2 reams = 1 bundle 20 quires = 1 ream 5 bundles = 1 bale Reduction Descending Example. — Reduce 17 yd. 2 ft. 9 in. to inches. 1 yd. = 3 ft. 17 yd. = 17 X 3 = 51 ft. 51 + 2 = 53 ft. 1 ft. = 12 in. 53 ft. = 53 X 12 = 6.36 in. 636 + 9 = 645 in. Ans. EXAMPLES Reduce to lower denominations : 1. 46 rd. 4 yd. 2 ft. to feet. 2. 4 A. 15 sq. rd. 4 sq. ft. to square inches. 3. 16 cu. yd. 25 cu. ft. 900 cu. in. to cubic inches. 4. 15 gal. 3 qt. 1 pt. to pints. 5. 27 da. 18 hr. 49 min. to seconds. Reduction Ascending Example. — Reduce 1306 gills to higher denominations. 4 )1.306 gi. Since in 1 pt. there are 4 gi., in 1306 gi. 2 )326 pt. + 2 gi. there are as many pints as 4 gi. are contained 4 )163 qt.. times in 1306 gi., or .326 pt. and 2 gi. remainder. 40 gal. + 3 qt. In the same way the quarts and gallons are 40 gal. 3 qt. 2 gi. Ans. found. So there are in 1306 gi., 40 gal. 3 qt. 2gi. REVIEW OF ARITHMETIC 43 EXAMPLES Reduce to higher denominations : 1. Reduce 225,932 in. to miles, etc. 2. Change 1384 dry pints to higher denominations. 3. In 139,843 sq. in. how many square miles, rods, etc. ? 4. How many cords of wood in 3692 cu. ft. ? 5. How many bales in 24,000 sheets of paper ? A denominate fraction is a fraction of a unit of weight or measure. To reduce denominate fractions to integers of lower denominations. Change the fraction to the next lower denomination. Treat the fractional part of the product in the same ivai/, and so pro- ceed to the required denomination. Example. — Reduce 4 of a mile to rods, yards, feet, etc. ^ of 320 rd. ^ J-V^ rd. = 228| rd. ^of Vyd. = nyd=3Jyd. ^ f of 3 ft. = Of ft. f of 12 in. = 5^ in. =5f in. f of a mile = 228 rd. 3 yd. ft. 6f in. The same process applies to denominate decimals. To reduce denominate decimals to denominate numbers. Example. — Reduce .87 bu. to pecks, quarts, etc. .87 bu. .84 qt. Change the decimal fraction to the next lower denomination. Treat the decimal part of the product in the same way, and so proceed to the re- quired denomination. 3.48 pk. 1.68 .48 pk. 8 3.84 qt. pt. 3pk. , 3 qt. 1.68 pt. Ans. 44 VOCATIONAL MATHEMATICS EXAMPLES Reduce to integers of lower denominations : 1. J of an acre. 3. ^ of a ton. 2. .3125 of a gallon. 4. .51625 of a mile. 5. Change f of a year to months and days. 6. .2364 of a ton. 7. What is the value of ^ of 1| of a mile ? 8. Reduce |^ bu. to integers of lower denominations. 9. .375 of a month. 10. y^y acre are equal to how many square rods, etc. ? Addition of Compound Numbers Example. — Find the sum of 7 hr. 30 min. 45 sec, 12 hr. 25 min. 30 sec, 20 hr. 15 min. 33 sec, 10 hr. 27 min. 46 sec. The sum of the seconds = 154 sec. = 2 min. 34 sec. Write the 34 sec. under the sec. column and add the 2 rain, to the min. column. Add the other columns 50 39 34 in the same way. 60 hr. 39 min. 34 sec. Ans. Subtraction of Compound Numbers Example. — From 39 gal. 2 qt. 2 pt. 1 gi. take 16 gal. 2 qt. 3 pt. 3 gi. As 3 gi. cannot be taken from 1 gi., 4 gi. or 1 pt. are borrowed from the pt. column and added to the 1 gi. Subtract 3 gi. from the 5 gi. and the remainder is 2 gi. Continue in the same way until all are subtracted. Then the remainder is 22 gal. 3 qt. pt. 2 gi. hr. min. sec. 7 30 45 12 25 30 20 15 33 10 27 46 gal. qt. pt. gi. 39 2 2 1 16 2 3 3 22 3 2 22 gal. 3qt .2 gi. Ans. yd. ft. In. 4 2 8 8 39 4 39 yd .4i in. Ans. REVIEW OF ARITHMETIC 45 Multiplication of Compound Numbers Example. — Multiply 4 yd. 2 ft. 8 in. by 8. 8 times 8 In. = 64 in. = 5 ft. 4 in. Place the 4 in. under the in. column, and add the 5 ft. to the product of 2 ft. by 8, which equals 21 ft. = 7 yd. Add 7 yd. to the product of 4 yd. by 8 = 39 yd. Division of Compound Numbers Example. — Fiud ^ of 42 rd. 4 yd. 2 ft. 8 in. -^ of 42 rd. = 1 rd. ; re- mainder, 7 rd. = 38^ yd. ; add 4 yd. = 42} yd. ^5 of 42^ yd. = 1 yd. ; remainder, 7i yd., = 22i ft. = 24J ft. 5^ of 24 J ft. = ft. 24 i ft. =294 in. ; add 8 in. =302 in. ^ of 302 in. = 8Jf in. Ans. Difference between Dates Example. — Find the time from Jan. 25, 1842, to July 4, 1896. 1896 7 4 It is customary to consider 30 days 1842 1 25 to a month. July 4, 1896, i.s the 1896th 54 yr. 5 mo. 9 da. Ans. yr., 7th mo., 4th da., and Jan. 25, 1842, is the 1842d yr., Ist. mo., 25th da. Subtract, taking 30 da. for a month. Id. yd. ft. in. 36)42 4 2 8(1 rd. 36 i ^ H 35)24^(0 ft. 35 12 38i 294 + 4 + 8 35)42J yd. (1 yd. 36)iM(8|^ in. 35 280 7i 22 3 22i ft. 1 rd. 1 yd. Sji in. 12 46 VOCATIONAL MATHEMATICS Example. — What is the exact number of days between Dec. 16, 1895, and March 12, 1896 ? Dec. 15 Do not count the first day mentioned. Tliere Jan. 31 are 15 days in December, after the 16th. Jan- Feb. 29 uary has 31 days, February 29 (leap year), Mar. 12 and 12 days in March ; making 87 days. 87 days. Ans. EXAMPLES 1. How much time elapsed from the landing of the Pil- grims, Dec. 11, 1620, to the Declaration of Independence, July 4, 1776? 2. Washington was born Feb. 22, 1732, and died Dec. 14, 1799. How long did he live? 3. Mr. Smith gave a note dated Feb. 25, 1896, and paid it July 12, 1896. Find the exact number of days between its date and the time of payment. 4. A carpenter earning $2.50 per day commenced Wednes- day morning, April 1, 1896, and continued working every week day until June 6. How much did he earn ? 5. Find the exact number of days between Jan. 10, 1896, and May 5, 1896. 6. John goes to bed at 9.15 p.m. and gets up at 7.10 a.m. How many minutes does he spend in bed ? To multiply or divide a compound number by a fraction. To multiply by a fraction, multiply by the numerator, and divide the product by the d^nomiriator. To divide by a fraction, multiply by the denominator , and divide the product by the numerator. When the multiplier or divisor is a mixed number, reduce to an improper fraction, and proceed as above. REVIEW OF ARITHMETIC 47 EXAMPLES 1. How much is 4 of 16 hr. 17 min. 14 sec. ? 2. A field contains 10 acres 12 sq. rd. of lan«l, which is J of the whole farm. Find the size of the farm. 3. If a train runs 60 mi. 35 rd. 16 ft. in one hour, how far will it run in 12f hr. at the same rate of speed ? 4. Divide 14 bu. 3 pk. 6 qt. 1 pt. by J. 5. Divide 5 yr. 1 mo. 1 wk. 1 da. 1 hr. 1 min. 1 sec. by 3J. EXAMPLES 1. A time card on a piece of work states that 2 hours and 15 minutes were spent in lathe work, 1 hour and 12 minutes in milling, 2 hours and 45 minutes in planing, and 1 hour and 30 minutes on bench work. What was the number of hours spent on the job ? 2. How many castings, each weighing 14 oz., can be obtained from 860 lb. of metal if nothing is allowed for waste? 3. How many feet hmg must a machine shop be to hold a lathe 8' 6", a planer 14' 4", a milling machine 4' 2", and a lathe 7' 5", placed side by side? 3' 3" were allowed between the machines and between the walls and the machines. 4. How many gross in a lot of 968 screws ? 5. Find the sum of 7 hr. 30 min. 45 sec, 12 hr. 26 min. 30 sec, 20 hr. 15 min. 33 sec, 10 hr. 27 min. 46 sec. 6. If a train is run for 8 hours at the average rate of 50 mi. 30 rd. 10 ft. per hour, how great is the distance covered? 7. A telephone pole is 31 feet long. If 4 ft. 7 in. are under ground, how high (in inches) is the top of the pole above the street? 8. If 100 bars of iron, each 2}' long, weigh 70 lb., what is the total weight of 2300 ? 48 VOCATIONAL MATHEMATICS 9. If 43 in. are cut from a wire 3 yd. 2 ft. 6 in. long, what is the length of the remaining piece ? 10. If a rod of iron 18' 8" long is cut into pieces 6J" long and Jj" is allowed for waste in each cut, how many pieces can be cut ? How much remains ? 11. I have 84 lb. 14 oz. of salt which I wish to put into packages of 2 lb. 6 oz. each. How many packages will there be ? 12. If one bottle holds 1 pt. 3 gi., how many dozen bottles will be required to hold 65 gal. 2 qt. 1 pt. ? 13. How many pieces 5J" long can be cut from a rod 16' 8" long, if 5" is allowed for waste ? 14. What is the entire length of a railway consisting of five different lines measuring respectively 160 mi. 185 rd. 2 yd., 97 mi. 63 rd. 4 yd., 126 mi. 272 rd. 3 yd., 67 mi. 199 rd. 5 yd., and 48 mi. 266 rd. 5 yd. ? Percentage Percentage is a process of solving questions of relation by means of hundredths or per cent (%). Every question in percentage involves three elements : the rate per cent, the base, and the percentage. The rate per cent is the number of hundredths taken. The base is the number of which the hundredths are taken. The percentage is the result obtained by taking a certain per cent of a number. Since the percentage is the result obtained by taking a cer- tain per cent of a number it follows that, the percentage is the product of the base and the rate. The rate and base are always factors, the percentage is the product. Example. — How much is 8 % of $200? 8% of $ 200 = 200 X .08 = $ 16. (1) REVIEW OF ARITHMETIC 49 In (1) we have the three elements: 8% is the rate, 8200 is the base, and $ 16 is the percentage. Since $200 x .08 = $ 16, the percentage ; $ 16 -i- .08 = $ 200, the base ; and $ 16 -i- $ 200 = .08, the rate. If any two of these elements are given, the other may be found : Base X Rate = Percentage Percentage -5- Rate = Base Percentage -5- Base = Rate Per cent is commonly used in the decimal form, but many operations may be much shortened by using the common frac- tion form. 1 % = .01 = ^^ i%= .00^ or .005 10%= .10 = ^V 33J% = .33J = i 100 % = 1.00 = 1 8^ % = .08^ = .0825 12^ % = .12^ or .125 = ^ | % = .00^ = .00125 There are certain per cents that are used so frequently that we should memorize their equivalent fractions. 10%=^ 37-^%=! 75% = I 12i%=i 40% = J 80%= J 16i%=i 50% =i 83i%=J 20% =i 60%= I 87i%=J EXAMPLES 1. Find 75 % of $ 368. 2. Find 15% of S412. 3. 840 is 33 J % of what number ? 4. 615 is 15% of what number? 5. What per cent of 12 is 8 ? 50 VOCATIONAL MATHEMATICS 6. What per cent of 245 is 5 ? 7. What per cent of 195 is 39 ? 8. What per cent of 640 is 80 ? 9. What per cent of 750 is 25 ? 10. What per cent of 819 is 45 ? Trade Discount Merchants and jobbers have a price list. From this list they give special discounts according to the credit of the customer and the amount of supplies purchased, etc. If they give more than one discount it is understood that the first means the discount from the list price, the second denotes the discount from the remainder. EXAMPLES 1. What is the price of 200 No. 1 cleats at $ 36.68 per M. at 40% off? 2. Supplies from a hardware store amounted to $ 58,75. If 121 % were allowed for discount, what was the amount paid ? 3. A dealer received a bill amounting to $ 212.75. Succes- sive discounts of 75%, 15%, 10%, and 5% were allowed. What was amount to be paid ? 4. 2 % is usually discounted on bills paid within 30 days. If the following are paid within 30 days, what will be the amounts due? a. $2816.49 d. $1369.99 g. $4916.01 b. 399.16 e. 2717.02 h. 30.19 c. 489.01 /. 918.69 5. What is the price of 20 fuse plugs at $ .07 each, 30 % off? 6. Hardware supplies amounted to $ 127.79 with a discount of 40 and 15 %. What was the net price? REVIEW OF ARITHMETIC 51 7. Which is better, for a merchant to receive a straight dis- count of 95 % or a successive discount of 75, 15, 5 % ? 8. Twenty per cent is added to the number of workmen in a machine shop of 575 men. What is the number employed after the increase ? 9. A steam pressure of 180 lb. per square inch is raised to 225 lb. per square inch. What is the per cent of in- crease ? 10. If 31 out of 595 wheels are rejected because of defects, what per cent is rejected ? 11. A clerk's salary was increased OJ %. If he now receives $ 850, what was his original salary ? 12. A company lost 12|^ % of its men and had 560 left. How many men were there before ? Simple Interest Money that is paid for the use of money is called interest. The money for the use of which interest is paid is called the principal^ and the sum of the principal and interest is called the amojint. Interest at 6 % means 6% of the principal for 1 year; 12 months of 30 days each are usually regarded as a year in com- puting interest. Example. — What is the interest on $ 100 for 3 years at 6 % ? 8100 .06 $ 6.00 interest for one year. Or, yg^ x Jf *i x f = $ 18. Ans. 3 % 18.00 interest for 3 years. Ana. 8 100 -I- § 18 = $ 118, amount. Pnncfpal X Kate x Time = Literest. 52 VOCATIONAL MATHEMATICS Example. — What is the interest on $ 297.62 for 5 yr. 3 mo at6%? 1297.62 ni:sM o,, 1_ ^ $ 297.62 21 ^ 118750.06 ^ ,3^, ., 100 1 ^ 200 44643 892860 Note. — Final results should not include mills. $93.7503 Mills are disregarded if less than 5, and called another $ 93.75. Ans. cent if 5 or more. EXAMPLES 1. What is the interest on $ 586.24 for 3 months at 6 % ? 2. What is the interest on $ 816.01 for 9 months at 5 % ? 3. What is the interest on $ 314.72 for 1 year at 4 % ? 4. What is the interest on $ 876.79 for 2 yr. 3 mo. at 41 fo ? The Six Per Cent Method By the 6 % method it is convenient to find first the interest of $ 1, then multiply it by the principal. Example. — What is the interest of $ 50.24 at 6 % for 2 yr. 8 mo. 18 da. ? Interest on $; 1 for 2 yr. =2 x $ .06 = .$ .12 Interest on § 1 for 8 mo. =8 x $ .00^ = .04 Interest on $ 1 for 18 da. = 18 x $ .000^ = .003 Interest on $ 1 for 2 yr. 8 mo. 18 da. .$ .163 Interest on ^50.24 is 50.24 times $ .163 = .$8.19. Ans. Find the interest on $ 1 for the given time, and multiply it by the prin- cipal, considered as an abstract number. EXAMPLES Find the interest and amount of the following : 1. $ 2350 for 1 yr. 3 mo. 6 da. at 5 %. 2. $ 125.75 for 2 mo. 18 da. at 7 %. 3. $ 950.63 for 3 yr. 17 da. at 41 %. 4. $ 625.57 for 1 yr. 2 mo. 15 da. at 6 %. REVIEW OF ARITHMETIC 53 Exact Interest When the time includes clays, interest computed by the 6% method is not strictly exact, by reason of using only .'^0 days for a month, which makes the year only 360 days. The day is therefore reckoned as -^ of a year, whereas it is ^^-y of a year. To compute exact interest^ find the exact time in daya^ and coiir aider 1 day^s interest as ^^ of 1 yea^s interest. Example. — Find the exact interest of $ 358 for 74 days at 7 %. $:368 X .07 = $26.06, 1 year's interest. 74 days* interest is j\^ of 1 year's interest. 3Vir of $26.06 = $6.08. Ans. ' 1 100 366 EXAMPLES Find the exact interest of : 1. $324 for 15 da. at 5 %. 2. $253 for 98 da. at 4%. 3. $624 for 117 da. at 7%. 4. $ 620 from Aug. 15 to Nov. 12 at 6 %. 5. $ 153.26 for 256 da. at 5^ %. 6. S 540.25 from June 12 to Sept. 14 at 8 %. Rules for Computing Interest The following will be found to be excellent rules for finding the inter* est on any principal for any number of days. Divide the principal by 100 and proceed as follows: 2 % — Multiply by number of days to run, and divide by 180. 2^ % — Multiply by number of days, and divide by 144. 3 % — Multiply by number of days, and divide by 120. 3^ % — Multiply by number of days, and divide by 102.86. 54 VOCATIONAL MATHEMATICS 4 % — Multiply by number of days, and divide by 90. 5 % — Multiply by number of days, and divide by 72. 6 % — Multiply by number of days, and divide by 60. 7 % — Multiply by number of days, and divide by 51.43. 8 % — Multiply by number of days, and divide by 45. Savings Bank Compound Interest Table Showing the amount of $ 1, from 1 year to 15 years, with compound interest added semiannually, at different rates. Pkb Cent 3 4 5 6 7 8 9 h year 101 102 102 103 103 1 04 104 1 year 103 104 105 106 107 108 109 1^ yeare 104 106 107 109 110 1 12 1 14 2 yeai*s 106 108 1 10 1 12 1 14 116 119 2^ years 107 1 10 113 1 16 1 18 1 21 124 3 years 109 1 12 1 15 1 19 1 22 1 26 130 3^ years 110 1 14 118 122 127 131 136 4 years 1 12 1 17 121 126 131 1 36 142 4| years 1 14 1 19 124 130 136 142 148 5 years 1 16 121 128 134 141 148 165 5^ years 1 17 124 131 138 145 163 162 6 years 1 19 126 134 142 1 61 160 169 6^ years 121 129 137 146 156 166 1 77 7 years 123 131 141 1 51 161 1 73 185 7^ years 124 1 34 144 166 167 180 193 8 years 126 137 148 1 60 173 187 2 02 8| years 128 139 152 166 179 194 2 11 9 years 1 30 142 1 66 170 185 2 02 2 20 9^ years 132 146 1 59 176 192 2 10 2 30 10 years 1 34 148 163 180 1 98 2 19 2 41 11 years 138 1 54 1 72 191 2 13 2 36 2 63 12 years 142 160 180 2 03 2 28 2 56 2 87 13 years 147 167 1 90 2 16 2 44 2 77 3 14 14 years 1 61 1 73 1 99 2 28 2 62 2 99 3 42 16 years 1 66 1 80 2 09 2 42 2 80 3 24 3 74 REVIEW OF ARITHMETIC EXAMPLES Solve the following problems according to the tables given above : 1. What is the compound interest of $ 1 at the end of 8.1 years? 2. What is the compound interest of $1 at the end of 11 years ? 3. How long will it take $400 to double itself at 4%, compound interest ? 4. How long will it take $580 to double itself at 4^ %, compound interest? 5. How long will it take $615 to double itself at 6%, simple interest ? 6. How long will it take $784 to double itself at 5| %, simple interest ? 7. Find the interest of S684 for 94 days at 3 %. a Find the interest of $ 1217 for 37 days at 4 %. 9. Find the interest of $681.14 for 74 days at 4^ %. 10. Find the interest of $414.50 for 65 days at 5 %. 11. Find the interest of $384.79 for 115 days at 6 %. Ratio and Proportion Ratio is the relation between two numbers. It is found by dividing one by the other. The ratio of 4 to 8 is 4 -§- 8 = i. The terms of the ratio are the two numbers compared. The first term of a ratio is the antecedent^ and the second the con- sequent. The sign of the ratio is (:). (It is the division sign with the line omitted.) Ratio may also be expressed fraction- ally, as J/ or 16:4; or yV or 3 : 17. A ratio formed by dividing the consequent by the antece- dent is an inverse raJtio : 12 : 6 is the inverse ratio of 6 : 12. 56 VOCATIONAL MATHEMATICS The two terms of the ratio taken together form a couplet. Two or more couplets taken together form a compound ratio. Thus, * 3 : 6 = 23 : 46 A compound ratio may be changed to a simple ratio by tak- ing the product of the antecedents for a new antecedent, and the product of the consequents for a new consequent. Antecedent -i- Consequetit = Ratio Antecedent -;- Ratio = Consequent Ratio X Consequent = Antecedent To multiply or divide both terms of a ratio by the same number does not change the ratio. Thus 12 : 6 = 2 3x12:3x6 = 2 EXAMPLES Find the ratio of 1. 20 : 300 Fractions with a common de- 2. 3 bu. : 3 pk. nominator have the same 3 21-16 ratio as their numerators. 4. 12 : 4- 7 8_.16 28.7 15.30 • -^*^ • 4 '• IT • IT? TT- 7 5J 11 • TT 5- i = l a |:|,B:|,.2.5 6. 16:(?) = | Proportion An equality of ratios is a proportion. A proportion is usually expressed thus : 4 : 2 : : 12 : 6, and is read 4 is to 2 as 12 is to 6. A proportion has four terms, of which the first and third are antecedents and the second and fourth are consequents. The first and fourth terms are called extremes, and the second and third terms are called means. The product of the extremes equals the product of the means. REVIEW OP ARITHMETIC 57 Tojind an extreme, divide the product of the means Ini the given extreme. Tojind a mean, divide the proilact of the extremes by the given mean. EXAMPLES Supply the missing term : 1. 1:836::25:() 4. 10 yd. : 50 yd. : : $ 20 : ($ ) 2. 6:24::( ) : 40 5. $}:S3|::( ):5 3. ( ):15::60:6 Simple Proportion An equality of two simple ratios is a simple proportion. Example. — If 12 bushels of charcoal cost $4, what will 60 bushels cost ? 12 • 60 • • S4 • f.«5 ^ There is the same relation between the cost „ ■ , " '^'' '' of 12 bu. and the cost of 60 bu. as there is be- — -|— =$20. Am. tween the 12 bu. and the 60 bu. $4 is the third term. The answer is the fourth term. It must form a ratio of 12 and 60 that shall equal the ratio of § 4 to the answer. Since the third term is lass than the required answer, the first must be less than the second, and 12 : 60 is the first ratio. The product of the means divided by the given extreme gives the other extreme, or $ 20. EXAMPLES Solve by proportion : 1. If 150 fuses cost $ 6, how much will 1200 cost ? 2. If 250 pounds of lead pipe cost S 15, how much will 1200 pounds cost ? 3. If 5 men can dig a ditch in 3 days, how long will it take 2 men ? 4. If .4' men can shingle a shed in 2 days, how long will it take 3 men ? 5. The ratio of Simon's, pay to Matthew^s is |. Simon earns $ 18 per week. What does Matthew earn ? 58 VOCATIONAL MATHEMATICS 6. What will llf yards of cambric cost if 50 yards cost $ 6.75 ? 7. A spur gear making 210 revolutions per minute is en- meshed with a pinion. The gear has 126 teeth and the pinion has 42 teeth. How many revolutions does the pinion make ? 8. In a velocity diagram a line 3|" long represents 45 ft. What would be the length of a line representing 30 ft. velocity ? 9. How many pounds of lead and tin would it take to make 4100 pounds of solder if there are 27 pounds of tin in each 100 pounds of solder ? 10. It is necessary to obtain a speed reduction of 7 to 3 by use of gears. If the pinion has 21 teeth, how many teeth must the gear have ? 11. A bar of iron 3^ ft. long and |" diameter weighs 6.64 pounds. W^hat would a bar 4|^ ft. long of the same diameter weigh ? 12. In a certain time 15 workmen made 525 pulleys. How many pulleys will 32 men make in the same length of time ? 13. When a post 11.5 ft. high casts a shadow on level ground 20.6 ft. long, a telephone pole near by casts a shadow 59.2 ft. long. How high is the pole ? 14. The diameter of a driving pulley is 18". This pulley makes 320 revolutions per minute. What must be the diam- eter of a driving pulley in order to make 420 revolutions per minute ? 15. A ditch is dug in 14 days of 8 hours each. How many days of 10 hours each would it have taken ? 16. If in a drawing a tree 38 ft. high is represented by IJ", what on the same scale will represent the height of a house 47 ft. high? 17. What will be the cost of 21 motors if 15 motors cost $ 887.509 ? REVIEW OF ARITHMETIC 59 18. The main drive pulley of a machine is 6 inches in diam- eter and makes 75G revolutions per minute. A pulley on the line shaft is belted to a machine. What is the diameter of the line shaft pulley if the line shaft makes 252 revolutions per minute ? 19. If a pole 8 ft. high casts a shadow 4 J ft. long, how high is a tree which casts a shadow 48 ft. long ? Involution The product of equal factors is a power. The process of finding powers is involution. The product of two equal factors is the second power, or square, of the equal factor. The product of three equal factors is the third power, or cube, of the factor. 4^ = 4 X 4 is 4 to the second power, or the square of 4. 2* = 2 X 2 X 2 is 2 to the third power, or the cube of 2. 3^ = 3x3x3x3 is 3 to the fourth power, or the fourth power of 4. EXAMPLES Find the powers : 1. 5» 3. 1^ 5. (2^)2 7. 9^ 2. 1.1' 4. 2o' 6. 2* a .15« Evolution One of the eqiuil factors of a power is a root. One of two equal factors of a number is the square root. One of three equal factors of a number is the cube root of it. The square root of 16 = 4. The cube root of 27 = 3, The radical sign (^) placed before a number indicates that its root is to be found. The radical sign alone before a number indicates the square root. Thus, Vd = 3 is read, the square root of 9 = 3. 60 VOCATIONAL MATHEMATICS A small figure placed in the opening of the radical sign is called the index of the root, and shows what root is to be taken. Thus, -v^ = 2 is read, the cube root of 8 is 2. Square Root The square of a number composed of tens and units is equal to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. ten^ + 2 X te7is X units + U7iits^ Example. — What is the square root of 1225? 12^25( 30 + 5^35 Separating Tens% 302 = 900 into periods of 325 two figures 325 each, by a 2 X tens = 2 x 30 =60 2 X tens + units = 2 x 30 + 6 = 65 checkmark ('), beginning at units, we have 12 '25. Since there are two periods in tlie power, there must be two figures in the root, tens and units. The greatest square of even tens contained in 1225 is 900, and its square root is 30 (3 tens). Subtracting the square of the tens, 900, the remainder consists of 2 x (tens x units) + units. 325, therefore, is composed of two factors, units being one of them, and 2 x tens — units being the other. But the greater part of this factor is 2 X tens (2 x 30 = 60). By trial we divide 325 by 60 to find the other factor (units), which is 5, if correct. Completing the factor, we have 2 X tens + units =; 66, which, multiplied by the other factor, 5, gives 325. Therefore the square root is 30 + 5 = 35. The area of every square surface is the product of two equal factors, length, and width. Finding the square root of a number, therefore, is equivalent to finding the length of one side of a square surface, its area being given. 1. Length X Width —Area 2. Area -r- Length = Width 3. Area -^• Width = Jjength , REVIEW OP ARITHMETIC 61 Short Mkthod Example. — Find the square root of 1300.0990. 13'0«.(W9<5 (8rtJ4 BeginninK at the decimal point, separate the 9 number into i)eri X i C In this formula D equals the diameter and C the circum- ference, 4 4 Example. — What is the area of a circle whose radius is 3ft.? 4 ^=irx9 ^=^^ = ir9 = 28.278q. ft. Ans. i Example. — What is the area of a circle whose circumfer- ence is 10 ft. ? 2) = -^ A=Idx-C 3.1416 2 2 - X -^~ X 1 X 10 = -^^ = 7.1 sq. ft. Ans. 2 3.141« 2 3.1410 Area of a Ring. — On examining a flat iron ring it is clear that the area of one side of the ring may be found by subtracting the area of the inside circle from the area of the outside circle. Let D = outside diameter d = inside diameter A — area of outside circle a = area of inside circle (1) ^ = :^=.7854Z>* 4 64 VOCATIONAL MATHEMATICS (2) a = ^ = .7854(^2 (3) A-a = ^~^ Let B = area of circular ring = A — a ^ = — - ^ = - (Z>2 - d2w .7854 (Z>2 _ (^2) 4 4 4 ^ Example. — If the outside diameter of a flat ring is 9" and the inside diameter 7", what is the area of one side of the ring? ^=.7854 (D2 _(f2) B = .7854 (81 - 49) = .7854 x 32 = 25.1328 sq. in. Ans. Angles Mechanics make two uses of angles : (1) to measure a cir- cular movement, and (2) to measure a difference in direction. A circle contains 360°, and the angles at the center of the circle contain as many degrees as their corresponding arcs on the circumference. Angle FOE has as many degrees as arc PE. A right angle is measured by a quarter of the circumference of the circle, which ^ is 90°. The angle AOG is a right angle. The angle AC, made with half the cir- cumference of the circle, is a straight angle, and the two right angles, AOG and GOC, which it contains, are supplementary to each other. When the sum of two angles is equal to 90°, they are said to be complementary angles, and one is the com- plement of the other. When the sum of two angles equals 180°, they are supplementary angles, and one is said to be the supple- ment of the other. MENSURATION 65 The number of degrees in an angle may be measured by a protractor. The distance around a semicircular protractor is Protractor — Semicircular, having 180°. divided into 180 parts, each division measuring a degree. It is used by placing the center of the protractor on the vertex and the base of the protractor on the side of the angle to be 3G0° Protractor. A a is a circle divided into degrees. measured. Where the other side of the angle cuts the circular piece, the size of the angle may be read. EXAMPLES 1. What is the area of a circular sheet of iron 8" in diameter ? 2. What is the distance around the edge of a pulley 6" in diameter ? 66 VOCATIONAL MATHEMATICS 3. What is the area of one side of a flat iron ring 14" inside diameter and 18" outside diameter ? 4. A driving wheel of a locomotive has a wheel center of 56" in diameter ; if the tires are 3" thick, what is the circum- ference of the wheel when finished ? 5. Find the area of a section of an iron pipe which has an inside diameter of 17" and an outside diameter of 17|". 6. Name the complements of angles of 30°, 45°, 65°, 70°, 85°. 7. Name the supplements of angles of 55°, 140°, 69°, 98° 44', 81° 19'. 8. What is the diameter of a wheel that is 12' 6" in circum- ference ? Triangles A triangle is a plane figure bounded by three straight lines. Triangles are classified according to the relative lengths of their sides and the size of their angles. A triangle having equal sides is called equilateral. One having two sides equal is isosceles. A triangle having no sides equal is called scalene. If the angles of a triangle are equal, the triangle is equi- angular. If one of the angles of a triangle is a right angle, the tri- angle is a right triangle. In a right triangle the side opposite the right angle is called the hypotenuse and is the longest side. The other two sides of the right triangle are the legs, and are at right angles to each other. Equilateral Isosceles Scalene Right MENSURATION 67 Kinds of Triangles Right Triangles In a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides or legs. If the length of the hy- potenuse and one leg of a right triangle is known, the other side may be found by squaring the hypotenuse and squaring the leg, and extracting the square root of their dif- ference. Example. — If the hypotenuse of a right angle triangle is 30" and the base is 18", what is the altitude? J 302 = 30 X 30 = 900 182 = 18 X 18 = 324 900 _ 324 = 576 \/670=24". Ana. Areas of Triangles The area of a triangle may be found when the length of the thi'ee sides is given by adding the three sides together, divid- ing by 2, and subtracting from this sum each side separately. Multiply the four results together and find the square root of their product. 68 VOCATIONAL MATHEMATICS Example. — What is the area of a triangle whose sides measure 15, 16, and 17 inches, respectively ? 16 16 17 V24 X 9 X 8x7 = \/l2096 2^ V12096 = 109.98 sq. in. Ans. 24 - 15 = 9 24 - 16 = 8 24-17 = 7 Area of a Triangle = ^ Base x Altitude Example. — What is the area of a triangle whose base is 17" and altitude 10"? 1 ^ ^ = - X 17 X ;^ = 85 sq. in. Ans. EXAMPLES 1. A ladder 17 ft. long standing on level ground reached to a window 12 ft. from the ground. If it is assumed that the wall is perpendicular, how far is the foot of the ladder from the base of the wall ? 2. Find the area of a triangular sheet of metal having the base 81" and the height measured from the opposite angle 56". 3. Find the length of the hypotenuse of a right triangle with equal legs and having an area of 280 sq. in. 4. Find the length of a side of a right triangle with equal legs and an area of 72 sq. in. 5. Find the hypotenuse of a right triangle with a base of 8" and the altitude of 7". 6. What is the area of a triangle whose sides measure 12, 19, and 21 inches ? 7. What is the altitude of an isosceles triangle having sides 8 ft. long and a base 6 ft. long ? MENSURATION 69 Quadrilaterals Four-sided plane figures are called quadrilaterals. Among them are the trapezoid, trajyezium, rectangle, rhotubus, atid rhom- boid. SquA&B Rkctamolb Rhomboid Rhombus Trafbzium Trapezoid Parallelugbam Kinds of Quadrilaterals A rectangle is a quadrilateral which has its opposite sides parallel and its angles right angles. Its area equals the prod- uct of its base and altitude. A= ha A trapezoid is a quadrilateral having only two sides parallel. Its area is equal to the product of the altitude by one half the sum of the bases. A=(b-\-c)xl^a ^ In this formula c = length of longest side J ^ V b = length of shortest side / i \ a = altitude * A trapezium is a four-sided figure with no two sides parallel. The area of a trapezium is found by dividing the trapezium into triangles by means of a diagonal. Then the area may* be found if the diagonal and perpendicular heights of the triangles are known. 70 VOCATIONAL MATHEMATICS Example. — In the trapezium ABCD if the diagonal is 43' and the perpendiculars 11' and 17', respectively, what is the area of the trapezium ? 43 X V- 43 X H^ ¥- = ^f : 236i- sq. ft., area of ABC : 36oj sq. ft.,areaof ^Z>(7 602 sq. ft., total area Ans. To find the areas of irregular figures, draw the longest diagonal and upon this diagonal drop perpendiculars from the ver- tices of the figure. These perpendiculars will form trapezoids and right triangles whose areas may be determined by the pre- ceding rules. The sum of the areas of the separate figures will give the area of the whole irregular figure. Polygons A plane figure bounded by straight lines is a polygon. A polygon which has equal sides and equal angles is a regular polygon. The apothem of a regular polygon is the line drawn from the center of the polygon perpen- dicular to one of the sides. A five-sided polygon is a pentagon. A six-sided polygon is a hexagon. An eight-sided polygon is an octagon. The shortest distance between the flats of a regular hexagon is the perpendicular distance between two opposite sides, and is equal to the diameter of the inscribed circle. The diameter of the -circumscribed circle is the long diameter of a regular hexa- gon. The perimeter of a polygon is the sum of its sides. Pentagon Hexagon MENSURATION 71 The area of a regular polygon equals one half the product of the apothem and the perimeter. Formula .1 = ^(1/' In this formula P = perimeter a = apothem Ellipse Only the approximate circumference of an ellipse can be ob- tained. The circumference of an ellipse equals one half the product of the sum of two diameters and tt. If c/i = major diameter c/j = minor diameter C = circumference then c = 4±^7r The area of an ellipse is equal to one fourth the product of the major and minor diameters by tt. If A = area dj = major diameter dj = minor diameter then A = TT^ 4 EXAMPLES 1. Find the area of a trapezium if the diagonal is 93' and the perpendiculars are 19' and 33'. 2. What is the area of a trapezoid whose parallel sides are 18 ft. and 12 ft., and the altitude 8 ft. ? 3. What is the distance around an ellipse whose major diameter is 14" and minor diameter 8" ? 72 VOCATIONAL MATHEMATICS 4. In the map of a country a district is found to have two of its boundaries approximately parallel and equal to 276 and 216 miles. If the breadth is 100 miles, what is its area ? 5. If the greater and lesser diameters of an elliptical man- hole door are 2' 9" and 2' 6", what is its area ? 6. Find the area of a trapezium if the diagonal is 78" and the perpendiculars 18" and 27". 7. The greater diameter of an elliptical funnel is 4 ft. 6 in., and the lesser diameter is 4 ft. (a) What is its area? (b) How many square feet of iron will it contain if its height is 16 ft., allowing 4" for the seams ? 8. What is the area of a pentagon, whose apothem is 4i" and whose side is 5" ? Volumes The volume of a rectangular-shaped bar is found by multi- plying the area of the base by the length. If the area is in square inches, the length must be in inches. The volume of a cube is equal to the cube of an edge. The contents or volume of a cyliJidrical solid is equal to the product of the area of the base by the height. If S = contents or capacity of cylinder R = radius of base H= height of cylinder TT = 3.1416+ or ^/ (approx.) S = ttB^H Example. — Find the contents of a cylindrical tank whose inside diameter is 14" and height 6'. H=6' = 72" /S' = V X 7 X 7 X 72 = 11,088 cu. in. MENSURATION 73 The Pyramid The volume of a puramid eciuals one third of the product of the area of the base and the altitude. V=\ba The volume of a frustum of a pyramid equals the product of one third the alti- tude and the sum of the two bases and the square root of the product of the bases. The surface of a regular pyramid is equal to the product of the perimeter of the bases and one half the slant height. S=Px^sh The Cone A cone is a solid generated by a right triangle revolving on one of its legs as an axis. The altitude of the cone is the perpendicular distance from the base to the apex. The volume of a cone equals the product of the area of the base and one third of the altitude. or F= .2618 D'H Example. — What is the volume of a cone 1^" ^ in diameter and 4" high ? Area of base = .7864 x | 7.0686 = 1.7671 sq. in. F= ^6181)2^ = .2618 X I X 4 = 2.3662 cu. In. Ans, The lateral surface of a cone equals one half the product of the perimeter of the base by the slant height. 74 VOCATIONAL MATHEMATICS Example. — What is the surface of a cone having a slant height of 36 in., and a diameter of 14 in. ? 3^ 14 X V = 44" ^i-^^ = 702 sq. in. Ans. Frustum of a Cone The frustum of a cone is the part of a cone included between the base and a plane or upper base which is parallel to the lower base. The volume of a frustum of a cone equals the product of one third of the altitude and the sum of the two bases and the square root of their product. When H = altitude B^ = upper base B = lower base V=iH{B-hB'-\- VBB') The lateral surface of a frustum of a cone equals one half the product of the slant height and the sum of the perimeters of the bases. The Sphere The volume of a sphere is equal to 3 where B is the radius. The surface of a sphere is equal to The Barrel To find the cubical contents of a barrel, (1) multiply the square of the largest diameter by 2, (2) add to this product MENSURATION 75 the square of the head diameter, and (3) multiply this sum by the length of the barrel and that product by .2618. Example. — Find the cubical contents of a barrel whose largest diameter is 21" and head diameter 18", and whose length is 33". V= 1{D'^ X 2) + d'^] X L X .2618 31)798 .2618 21* = = 441 X 2 = 882 182 = = 324 .324 1206 33 3618 3618 10419.11 cu. in 10419.11 39798 281 = 46.10 gal. Ans. Similar Figures Similar figures are figures that have exactly the same shape. The areas of similar figures have the same ratio as the squares of their corresponding dimensions. Example. — If two boilers are 15' and 20' in length, what is the ratio of their surfaces ? J^ = I, ratio of lengths — = — , ratio of surfaces 42 16 One boiler is ^5 as large as the other. Ans. The volumes of similar figures are to each other as the cubes of their corresponding dimensions. Example. — If two iron balls have 8" and 12" diameters, respectively, what is the ratio of their volumes ? ^ = f , ratio of diameters = /y, ratio of their volumes. Am. One ball weighs ^ as much as the other. 76 VOCATIONAL MATHEMATICS EXAMPLES 1. Find the volume of a rectangular iron bar 8'' by 10" and 4' long. 2. Find the weight of a rectangular steel bar 31" x 49" x 3" thick, if the metal weighs .28 lb. per cubic inch. 3. The radius of the small end of a bucket is 4 in. Water stands in the bucket to a depth of 9 in., and the radius of the surface of the water is 6 in. (a) Find the volume of the water in cubic inches. (6) Find the volume of the water in gallons if a cubic foot contains 7.48 gal. 4. What is the volume of a steel cone 2i" in diameter and 6" high ? 5. Find the contents of a barrel whose largest diameter is 22", head diameter 18", and height 35". 6. What is the volume of a sphere 8" in diameter ? 7. What is the volume of a pyramid with a square base, 4" on a side and 11" high ? 8. What is the surface of a steel cone with a 6" diameter and 14" slant height ? 9. Find the surface of a pyramid with a perimeter of 18" and a slant height of 11". 10. Find the volume of a cask whose height is 3^' and the greatest radius 16" and the least radius 12", respectively. 11. What is the weight of a cast-iron cylinder 2.75" in diameter and 12j" long, if cast iron weighs 450 lb. per cu. ft. ? 12. How many gallons of water will a round tank hold which is 4 ft. in diameter at the top, 5 ft. in diameter at the bottom, and 8 ft. deep ? (231 cu. in. = 1 gal.) 13. What is the volume of a cylindrical ring having an outside diameter of 6 J", an inside diameter of 5^", and a height of 3|" ? What is its outside area ? MENSURATION 77 14. A sphere has a circumference of 8.2467". (a) What is its area? (6) What is its volume ? 15. If it is desired to make a conical oil can with a base 3.5" in diameter to contain \ pint, what must the approximate height be ? 16. What is the area of one side of a flat ring if the inside diameter is 2^" and the outside diameter 4J" ? 17. There are two balls of the same material with diameters 4" and 1", respectively. If the smaller one weighs 3 lb., how much does the larger one weigh ? 18. If the inside diameter of a ring must be 5 in., what must the outside diameter be if the area of the ring is 6.9 sq. in. ? 19. How much less paint will it take to paint a wooden ball 4" in diameter than one 10" in diameter ? 20. What is the weight of a brass ball 3J" in diameter if brass weighs .303 lb. per cubic inch ? 21. A cube is 19" on its edge, (a) Find its total area. (6) Its volume. 22. If the area of a J" pipe is .049 sq. in., what will be the diameter of a pipe having 8 times the area? 23. What is the weight of a cast-iron cylinder 2.75' in diameter and 12}' long, if the cast iron weighs 450 lb. per cubic foot ? 24. A conical funnel has an inside diameter of 19.25" at the base and is 43" high inside, (a) Find its total area. (6) Find its cubical contents. 25. If a bar 3" in diameter weighs 24.03 lb. per foot of length, what must be the weight per foot of a bar 3" square of the same material? CHAPTER III Reading a Blue Print Every skilled worker in wood or metal must know how to read a " blue print," which is the name given to working plans Screw Table Frame Simple Blue Prints or Working Drawings and drawings with white lines upon a blue background. The blue print is the language which the architect uses to the 78 WORKING DRAWINGS 79 builder, the machinist to the pattern maker, the engineer to the foreman of construction, and the designer to the workman, Through following the directions of the blue print the carpenter, metal worker, and mechanic are able to produce the object wanted by the employer and his designer or draftsman. Two views are usually necessary in every working drawing, one the plan or top view obtained by looking down upon the object, and the other the elevation or front view. When an object is very complicated, a third view, called an end or profile view, is shown. All the information, such as dimensions, etc., necessary to construct whatever is represented by the blue print, must be supplied on the draw- ing. If the blue print represents a machine it is necessary to show all the parts of the machine put together in their proper places. This is called an assembly drawing. Then there must be a drawing for each part of the machine, giving information as to the size, shape, and number of the pieces. Then if there are interior sections, these must be represented in section drawings. Drawing to Scale As it is impossible to draw most objects full size on paper, it is necessary to make the drawings proportionately smaller. This is done by making all the dimensions of the drawing a certain fraction of the true dimensions of the object. A draw- ing made in this way is called drawn to scale. Triangular Scale The dimensions on the drawing are designated the actual size of the object — not of the drawing. If a drawing were made of an iron bolt 25 inches long, it would be inconvenient to represent the actual size of the bolt, and the drawing might be made half or ipiarter the size of the bolt, but the length would read on the drawing 25 inches. 80 VOCATIONAL MATHEMATICS In making a drawing "to scale," it becomes very tedious to be obliged to calculate all the small dimensions. In order to obviate this work a triangular scale is used. It is a ruler with the different scales marked on it. By practice the student will be able to use the scale with as much ease as the ordinary ruler. QUESTIONS AND EXAMPLES 1. Tell what is the scale and the length of the drawing of each of the following : a. An object 14" long drawn half size. h. An object 26" long drawn quarter size. c. An object 34" long drawn one third size. d. An object 41" long drawn one twelfth size. 2. If a drawing made to the scale of |" = 1 ft. is reduced \ in size, what will the new scale be ? 3. The scale of a drawing is made \ size. If it is doubled, how many inches to the foot will the new scale be ? 4. On the yy scale, how many feet are there in ^d> inches ? 5. On the y scale, how many feet are there in 26 inches ? 6. On the ^" scale, how many feet are there in 21 inches ? 7. If the drawing of a bolt is made -J- size and the length of the drawing is 8^', what will it measure if made to scale 3" = lft.? 8. What will be the dimensions of the drawing of a machine shop 582' by 195' if it is made to a scale of yV = 1 ft. ? Arithmetic and Blue Prints. — Mechanics are obliged to read from blue prints. In order to verify the necessary dimensions in the detail of the work, it becomes necessary to do more or less addition and subtraction of the given dimensions. This involves ability to add, and subtract mixed numbers and fractions. METHODS OF SOLVING EXAMPLES 81 Methods of Solving Examples Every mechanical problem or operation has two distinct sides : the collecting of data and the solving of the problem. The first part, the collecting of data, demands a knowledge of the materials and conditions under which the problem is given, and calls for considerable judgment as to the necessary accurateness of the work. There are three ways by which a problem may be solved : 1. Exact method. 2. Rule of thumb method, by the use of a two-foot rule or a slide scale. 3. By means of tables. The exact method of solving a problem in arithmetic is the one usually taught in school and is the method obtained by analysis. Every one should be able to solve a problem by the exact method. The rule of thumb method. — Many of the problems that arise in industrial life have been met before and very careful judg- ment has been exercised in solving them. As the result of this experience and the tendency to abbreviate and devise shorter methods that give sufficiently accurate results, we find many rule of thumb methods used by the mechanics in daily life. The exact method would involve considerable time and the use of pencil and paper, whereas in cases that are not too complicateid the two-foot rule or the slide scale will give a quick and accurate result. In solving problems involving the addition and subtraction of fractions, by the rule of thumb, use the two-foot rule or steel scale to carry on the computation. To illustrate : if we desire to add \ and ^, place the thumb on \ division, then slide (move) the thumb along a division cor- responding to ^, and then read the number of divisions passed over by the thumb. In this case the result is ^. For fractions involving ^, ^^, ^, and j^ use the steel scale. The majority of machinists, carpenters, etc., use this method of sliding the thumb over the rule, in adding and subtracting inches and fractions of inches. 82 VOCATIONAL MATHEMATICS The use of tables. — In the commercial and industrial world the tendency is to do a thing in the quickest and the most economical way. To illustrate : hand labor is more costly than machine work, so wherever possible, machine work is substituted for hand labor. The same condition applies to the calculations that are used in the shop. The methods of performing calculations are the most economical — that is, the quickest and most accurate — that the ordinary mechanic is able to perform. Since a great many of the problems in calculation that arise in the daily experience of the mechanic are about standardized pieces of metal and repeat themselves often, it is not necessary to wot-k them each time if results are kept on file when they are once solved. This tiling is done by means of tables that are made from these problems. See pages 105, 117, and 121, for tables used in this book. PART II — MATHEMATICS FOR CARPENTERING AND BUILDING CHAPTER IV MEASURING LUMBER The carpenter or builder is often required to give an estimate of the cost of the work to be done for his prospe(;tive custom- ers. People who contemplate building have several estimates submitted to them by different builders and generally give the work to the lowest bidder. An architect usually draws plans of the building and from these plans the contractor or builder makes his estii^ate of the cost. In doing repairing or cabinet- making the carpenter makes his own plans and estimates. In order to make a proper estimate of the cost, one must know the market price of materials, the cost of labor, the amount of material needed, and the length of time required to do the work. Preparation of Wood for Building Purposes In winter the forest trees are cut and in the spring the logs floated down the rivers to sawmills, where they are sawed into boards of different thickness. To square the log, four slabs are first sawed off. After these slabs are off, the remainder is sawed into boards. As soon as the boards or planks are sawed from the logs, they are piled on prepared foundations in the open air to season. Each layer is separated from the one above by a crosspiece, called a strap, in order to allow free cir- culation of air about each board to dry it quickly and evenly. If lumber were to be piled up without the strips, one board upon another, the ends of the pile would dry and the center would rot. This seasoning or drying out of the sap usually lasts several months. *' Air dried " lumber is used for most building purposes except in buildings or places where there is a warm, dry atmosphere. 88 84 VOCATIONAL MATHEMATICS Wood that is to be subject to a warm atmosphere has to be artificially dried. This artificially dried or kiln-dried lumber has to be dried to a point in excess of that of the atmosphere in which it is to be placed after being removed from the kiln. This process of drying must be done gradually and evenly or the boards may warp and then be unmarketable. Definitions Board Measure. — A board one inch or less in thickness is said to have as many board feet as there are square feet in its surface. If it is more than one inch thick, the number of board feet is found by multiplying the number of square feet in its surface by its thickness measured in inches and fractions of an inch. The number of board feet = length {in feet) x width (in feet) x thick- ness (in inches). Board measure is used for plank measure. A plank 2" thick, 10" wide, and 15' long, contains twice as many square feet (board measure) as a board 1" thick of the same width and length. To measure a board that tapers, the width is taken at the middle, where it is one half the sum of the widths of the ends. Boards are sold at a certain price per hundred (C) or per thousand (M) board feet. The term lumber is applied to pieces not more than four inches thick ; timber to pieces more than four inches thick ; but a large amount taken together often goes by the general name of lumber. A piece of lumber less than an inch and a half thick is called a board and a piece from one inch and a half to four inches thick a plank. Rough Stock is lumber the surface of which has not been dressed or planed. The standard lengths of pieces of lumber are 10, 12, 14, 16, 18 feet, etc. In measuring and marking large lots of lumber in which there are a number of pieces containing a fraction of a foot, in case of one half foot or more, 1 is added, and in case of less than one half foot, it is disregarded. This is especially true with boards. CARPENTERING AND BUILDING 85 A board 1" x 4" x 16' contains 6 J feet (board measure). Two boards would be marked 6 ft., and every third board would be marked (i ft. So two boards may be exactly the same length and one marked 6 ft. and the other ft. In the purchasing of a single board there might be a small undercharge or overcliarge, but in large lots the average would be struck. EXAMPLES 1. How many board feet in a board 1 in. thick, 15 in. wide, and 15 ft. long ? 2. How many board feet of 2-inch planking will it take to make a walk 3 feet wide and 4 feet long ? 3. A plank 19' long, 3" thick, 10" wide at one end and 12" wide at the other, contains how many board feet ? 4. Find the cost of 7 2-inch planks 12 ft. long, 16 in. wide at one end, and 12 in. at the other, at S 0.08 a board foot. 5. At $ 12 per M, what will be the cost of 2-inch plank for a 3 ft. 6 in. sidewalk on the street sides of a rectangular corner lot 56 ft. by 106 ft. 6 in. ? Quick Method for Measuring Boards To measure boards 1" thick, multiply length in feet by width in inches and divide by 12, and the result will be the board measure in feet. For boards IJ" thick, add one quarter of the quotient to the result as above. For boards 1|" thick, add one half of the quotient to the r(!sult as found above. For plank 2" thick, divide by 6 instead of 12. For plank 3" thick, divide by 4 instead of 12. For plank 4" thick, divide by 3 instead of 12. For timber 6" thick, divide by 2 instead of 12. EXAMPLES 1. Find the number of board feet in 10 planks, 3" thick, 12" wide, 14' long. 86 VOCATIONAL MATHEMATICS 2. Find the number of board feet in 4 timbers, 8" thick, 10" wide, 17' long. 3. Find the number of board feet in 18 joists, 2" thick, 4" wide, 14' long. 4. Find the number of board feet in 16 beams, 10" thick, 12" wide, 11' long. 5. Find the number of board feet in 112 boards, J" thick, .8" wide, 14' long. Note. — Ordinarily fractions of a foot less than one half are omitted, but when the fraction is one half or larger it is reckoned as a foot. This is sufficiently accurate for all practical purposes. Board measure of one lineal foot of timber may be found from the following table : Contents (Board Measure) of One Lineal Foot of Timber 5r. T HICKNE88 IN InOIIES 2 3 4 5 6 7 8 9 10 11 12 13 14 18 3. 4.5 6. 7.5 9. 10.5 12. 13.5 15. 16.5 18 19.5 21. 17 2.83 4.25 5.66 7.08 8.5 9.92 11.33 12.75 14.17 15.58 17 18.42 19.83 16 2.67 4. 5.33 6.67 8. 9.33 10.67 12. 13.33 14.67 16 17.3 18.66 15 2.5 3.75 5. 6.25 7.5 8.75 10. 11.25 12.5 13.75 15 16.25 17.6 14 2.33 3.5 4.67 5.83 1 . 8.17 9.33 10.5 11.67 12.83 14 15.17 16. ()6 13 2.17 3.25 4.33 5.42 6.5 7.58 8.67 9.75 10.83 11.92 13 14.08 12 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12 11 1.83 2.75 3.67 4.58 5.5 6.42 7.33 8.25 9.17 10.08 10 1.67 2.5 3.33 4.17 5. 5.88 6.67 7.5 8.33 9 1.5 2.25 3. 3.75 4.5 5.25 6. 6.75 8 1.33 2. 2.67 3.33 4. 4.67 5.33 7 1.17 1.75 2.33 2.92 3.5 4.08 6 1. 1.5 2. 2.5 3. 5 .83 1.25 1.67 2.08 4 .67 1. 1.33 3 .5 .75 2 .33 CARPENTERING AND BUILDING 87 To ascertain the contents of a piece of timber, find in the table the contents of one foot and multiply by the length in feet of the piece. EXAMPLES By means of the above table find the board measure of the following : 1. One timber 8" x 9", 14' long. 2. One timber 8" X 11", 13' long. 3. One timber 9" x 10", 11' long. 4. One timber 8" x 10", 16' long. 5. One timber 7" x 9", 11' long. 6. Two planks 2" x 3", 10' long. 7. Two planks 4" x 4", 13' long. 8. Two timbers 10" x 11", 15' long. 9. Two timbers 10" x 18", 14' long. 10. Two timbers 8" X 16", 13' long. The weight per cubic foot of different woods can be readily seen from the following table : Weight of One Cubic Foot of Timber Wood Weight I'ER CiT. Ft. Wood Wbiuut PER Cv. Ft. White Pine . . Georgia Pine Hemlock . . . Cypress . . . Spruce . . . White Oak . . Red Oak . . . Maple. . . . 28 1b. 38 1b. 24 1b. 331b. 28 1b. 481b. 46 1b. 42 lb. Whitewood .... Ash Hickory . • . . . Chestnut Cedar Birch ....... Ebony Boxwood 30 lb. 45 1b. 48 1b. 36 1b. 39 1b. 411b. 76 1b. 70 lb. Lumber is bought and sold in the log by cubic measure. Lumber used for framing buildings and for building bridges, docks, and ships is also sold by the cubic foot. 88 VOCATIONAL MATHEMATICS The rule that is most extensively used for computing the contents in board feet of a log is as follows : Rule. — SvMract 4 inches from the diameter of the log at the small end, square one quarter of the remainder, and multiply the result by the length of the log in feet. EXAMPLES Find the board feet of lumber in the following logs : Diameter in inches ^- 2. 3. 4. 5. 6. 7. 8. 9. at small end: 12 13 14 15 16 17 18 20 24 Length in feet : 10 12 14 16 18 20 16 20 24 CHAPTER Y CONSTRUCTION Excavations. — After the })lans for a building are drawn by the architect and the work given to the contractor on a bid, usually the lowest, the work of excavating begins. In esti- mating excavations the cubic yard, or 27 cubic feet, is used. EXAMPLES 1. What will it cost to excavate a cellar that is 32' x 28' and 5' deep at 34 cents per cubic yai-d ? 2. What will the cost be of excavating a lot 112' by 58' and averaging 12' deep at $1.65 per yard ? 3. In excavating a tunnel 374,166 cubic feet of earth were removed. If the length of the tunnel was 492 ft. and th« width 39 ft., what was the height of the tunnel ? 4. How many cubic yards of earth must be removed to build a cellar for a house when the measurements inside the wall are 28' long and 16' wide, the wall being 1'.8" thick and 8' deep, with 2' of the wall above the ground level? 5. In making a bid on some excavating a contractor notes that the excavation is in the shape of a rectangle 8' deep, 11' wide at the top and bottom, and 483' long. What will it cost him to excavate it at 29 cents per cubic yard ? What must his bid be to make 10 % profit ? Frame and Roof After the excavation is finished and the foundation laid, the construction of the building itself is begun. On the top of the foundation a large timber called a sill is placed. The timbers 90 VOCATIONAL MATHEMATICS running at right angles to the front sill are called side sills ; The sills are joined at the corners by a half-lap joint and held together by spikes. a. Outside studding b. Rafters c. Plates il. Ceiling joists de. Second floor joists def. First floor joists g. Girder or cross sill h. Sills i. Sheathing j. Partition studs k. Partition heads l. Piers 1 Then the building has its walls framed by placing corner posts of 4" by 6" on the four corners. Between these corner posts there are placed smaller timbers called studding, 2" by 4", 16" apart. Later the laths, 4' long, are nailed to this stud- ding. The upright timbers are mortised into the sills at the bottom. When these uprights are all in position a timber called a plate is placed on the top of them and they are spiked together. On the top of the plate is placed the roof. The principal timbers of the roof are the rafters. Different roofs have a dif- 1 If made of brick or stone, '' shones " or "supports " if made of wood. CARPENTERING AND BUILDING 91 fereiit pitcli or slope — that is, form different angles with the plate. To get the desired pitch the carpenter uses the steel square. When we speak of the pitch of a roof we mean the slope or slant of the roof. A roof with oi>€ half pitch means that the height of the ridkn|wt i iHi ' iiviPjer but a few more should be added for waste, etc. Verify answer by use of table on page 98. Another method is : Since it takes approximately 3 bunches (250 shingles each) laid 5" to the weather to cover a square, multiplying the number of squares in the roof by 3 will give the number of bunches required. Example. — If a roof contains 50 squares, how many shingles will the roof need to cover it ? 50 X 3 = 150 bunches = 37,500 shingles, Ans. EXAMPLES 1. How much will it cost for shingles to shingle a roof 50 ft. by 40 ft., if 1000 shingles are allowed for 125 square feet and the shingles cost $ 1. 18 per bundle ? 2. Find the cost of shingling a roof 38 ft. by 74 ft, 4" to the weather, if the shingles cost $ 1.47 a bundle, and a pound and a half of cuti nails at $.0G a i)ound are used with each bundle. 100 VOCATIONAL MATHEMATICS 3. How many shingles would be needed for a roof having four sides, two in the shape of a trapezoid with bases 30 ft. by 10 ft., and altitude 15 ft., and two (front and back) in the shape of a triangle with base 20 ft. and altitude 15 ft.? (1000 shingles will cover 120 sq. ft.) Slate Roofing Slates make a good-looking and durable roof. They are put on with nails similarly to shingles. Estimates for slate roof- ing are made on 100 sq. ft. of the roof. The following are typical data for building a slate roof : A square of No. 10 x 20 Monson slate costs about 38.35, Two pounds of galvanized nails cost $.16 per pound. Labor, $S per square. Tar paper, at 2| cents per pound, 1^ lb. per square yard. EXAMPLES Using the above data, give the cost of making slate roofs for the following : 1. What is the cost of laying a square of slate ? 2. What is the cost of laying slate on a roof 112' by 44' ? 3. What is the cost of laying slate on a roof 156' by 64' ? 4. What is the cost of laying slate on a roof 118' by 52' ? 5. What is the cost of laying slate on a roof 284' by 78' ? Weight of Roof Coverings Every builder should know the weight of different roof coverings in order to make a roof strong enough to support its covering. Besides, allowance must be made for ice and snow. The weight in pounds of roof covering is usually expressed in pounds per 100 sq. ft., or square of roof. CARPENTERING AND BUILDING 101 Approximate WEUiiiT ov Roof Covkrinos Na.mk Wkioiit pkr 1()0 bq. pt. Sheathing, Pine 1 inch thick, yellow northern . . . 300 Sheathing, Pine 1 inch thick, yellow southern . 400 Spruce, 1 inch thick 200 Sheathing, Chestnut or Maple, I inch thick .... 400 Sheathing, Ash, Hickory, or Oak, 1 inch thick . . . 500 Sheet iron, ^\ inch thick 800 Sheet iron, ^5 inch thick, and laths 500 Shingles, Pine 200 Slates, \ inch thick 1)00 Skylights (Glas-s, f\ to I inch thick) 250-700 Sheet Lead 500-800 Cast Iron Plates, | inch thick 1500 Copper 80-126 Felt and Asphalt 100 Felt and Gravel 800-1000 Iron, Corrugated 100-2375 Iron, Galvanized Flat . . • 100-350 Lath and Plaster 900-1000 Thatch 650 Tin 70-125 Tiles, Flat 1500-2000 Tiles (Grooves and Fillets) 700-1000 Tiles, Pan 1000 Tiles, with Mortar 2000-3000 Zinc 100-200 EXAMPLES What is the approximate weight of: 1. 800 sq. ft. of \ inch slate ? 2. 1645 sq. ft. of flat tiles ? 3. 23.32 sq. ft. of tiles with mortar? 4. 3184 sq. ft. of lath and plaster ? 5. 2789 sq. ft. of sheathing pine 1" thick ? 6. 1841 sq. ft. of pine shingles ? 102 VOCATIONAL MATHEMATICS 7. 1794 sq. ft. of thatch? 8. 3279 sq. ft. of felt and gravel ? 9. 1973 sq. ft. of asphalt ? 10. 1589 sq. "ft. of skylight glass i in. thick ? Clapboards Clapboards are used to cover the outside walls of frame buildings. Most clapboards are 4 ft. long and 6 in. wide. They are sold in bundles of twenty-five. Three bundles will cover 100 square feet if they are laid 4" to the weather. To find the number of clapboards required to cover a given area, find the area in square feet and divide by IJ. One quarter the area should be deducted to allow for openings. EXAMPLES 1. How many clapboards will be required to cover an area of 40 ft. by 30 ft. ? 2. How many clapboards will be necessary to cover an area of 38' by 42' if 56 sq. ft. are allowed for doors and windows ? 3. How many clapboards will a barn 60 ft. by 50 ft. require if 10% is allowed for openings and the distance from founda- tion to the plate is 17 ft. and the gable 10 ft. high ? •; * ; \ ; : y r \ ' Flooring Mcrst floors in hou'seft, are made of oak, maple, birch, or pine, i^his flooring is grooved so that the boards fit closely together without cracks between them. The accompanying figure shows the ends of ^ ^ j— 1 pieces of matched flooring. Matched boards are ^ ^— '— ' also used for ceilings and walls. In estimating for matched flooring enough stock must be added to make up for what is cut away from the width in matching. This amount varies from |" to |" on each board ac- cording to its size. Some is also wasted in squaring ends, cutting up, and CARPENTERING AND BUILDING 103 fittinjj to exact lengths. A common floor is made of unmatched boards and is usually used ;us an under floor. Not more than \ is allowed for wastr. ExAMi'LK. — A room is 12 ft. square and is to have a floor laid of matched boards IV' wide; one third is to be added for waste. What is the number of scjuare feet in the floor ? What is the number of board feet required for laying the floor? 12 X 12 = 144 sq. ft. = area. 144 x J = 48 144 Ans. 144 192 board measure for matched floor. 192 Ans. EXAMPLES 1. How much J in. matched flooring 3" wide will be re- quired to lay a floor 16 ft. by 18 ft. ? One fourth more is al- lowed for matching and 3 % for squaring ends. 2. How much hard pine matched flooring J" thick and 1^" wide will be required for a floor 13' 0" X 14' 10" ? Allow J for matching and add 4 % for waste. 3. An office floor is 10' 6" wide at one end and 9' 6" wide at the other (trapezoid) and 11' 7" long. What will the material cost for a maple floor |" thick and 1^" wide at $60 per M, if 4 sq. ft. are allowed for waste ? 4. How many square feet of sheathing are required for the outside, including the top, of a freight car 34' long, 8' wide, and 7^' high, if 12^% is allowed for waste and overhang? 5. In a room 50' long and 20' wide flooring is to be laid; how many feet (board measure) will be required if the stock is J" X 3" and \ allowance for waste is made? Stairs The perpendicular distance between two floors of a building is called the rise of a flight of stairs. The width of all the steps is called the run. The perpendicular distance between steps is called the width of riser. Nosing is the slight projec- 104 VOCATIONAL MATHEMATICS tion on the front of each step. The board on each step is the tread. To find the number of stairs necessary to reach from one floor to an- other: Measure the rise first. Divide this by 8 inches/ which is the most comfortable riser for stairs. The run should be 8i inches or more to allow for a tread of 9| inches with a nosing of li inches. Example. — How many steps will be required, and what will be the riser, if the distance between floors is 118 inches ? 118 -- 8 = 14f or 15 steps. 118 -=- 15 = 7^1 inches each riser. Ans. Stairs EXAMPLES 1. How many steps will be required, and what will be the riser, (a) if the distance between floors is 8' ? (b) If the dis- tance is 9 feet ? 104 2. How many steps will be required, and what will be the riser, (a) if the distance between floors is 12' ? (b) If the dis- tance is 8' 8" ? Carpenters' Table of Wages To find the amount due at any rate from 30 cents to 55 cents per hour, look at the column containing the rate per hour and the column opposite that containing the number of hours, and the amount will be shown. Time and a half is counted for overtime on regular working days, and double time for Sundays and holidays. 1 Other distances may also be used if required. CARPENTERING AND BUILDING 105 X X i s 04 H 91 M B 5 J a S s M n i M s o H a ti X St !! i H 2 M > © s 1 %... $0 30 ♦0 15 ♦0 224 $0 80 ♦0 824 ♦Olfii ♦0 24g ♦0 324 ♦0 4:) ♦0 224 ♦0 83} ♦0 45 1... SO 80 45 60 824 824 48} 66 45 46 674 90 s... 30 60 90 1 20 824 65 974 1 80 46 90 1 85 1 1-0 s... 80 90 1 85 1 80 824 974 1464 1 95 45 1 36 2 024 2 70 4... 80 1 20 1 80 2 40 824 1 80 1 95 2 60 46 1 80 2 70 3 ro 6... 80 1 50 2 25 8 00 824 1 624 2 48} 3 26 46 2 26 8 874 4 60 6... 80 1 SO 2 70 8 60 824 1 95 2 924 3 90 46 2 70 4 06 6 40 7... 80 2 10 8 15 4 20 824 2 274 8 41J 466 46 8 16 4 724 6 30 8... 80 2 40 8 60 4 80 324 2 60 3 90 5 20 45 3 60 5 40 7 20 9... 80 2 70 4 05 540 324 2 924 4 38} 6 86 45 4 05 6 074 8 10 10 .. 80 8 00 460 6 00 824 8 26 4 874 650 46 460- 6 75 9 (K) %... $0 47i $0 28} $0 85| ♦0 474 10 50 ♦0 26 ♦0 874 ♦0 50 ♦0 56 ♦0 274 ♦0 41} ♦0 56 1... 474 474 714 95 50 60 76 1 00 55 56 824 1 10 2... 47* 95 1 424 1 90 50 1 00 1 50 2 00 55 1 10 1 65 2 20 8... 474 1424 2 13} 2 85 50 1 50 2 26 8 00 56 1 65 2 474 8 80 4... 474 1 90 2 85 8 80 50 2 00 8 00 4 00 55 2 20 8 80 4 40 6... 474 2 874 8 564 4 75 50 2 50 8 75 500 Ki 2 76 4 124 5 50 6... 474 2 85 4 274 5 70 50 8 00 4 60 6 00 56 8 80 4 96 6 60 7... 474 8 824 4 98} 6 65 50 8 50 5 25 7 W 55 3 85 6 774 7 70 8... 474 880 6 70 7 60 50 4 00 6 00 8 00 56 440 6 60 8 80 9... 474 4 274 6 41} 8 55 50 4 50 6 75 9 00 65 4 96 7 424 9 90 10... 474 4 75 7 124 9 50 50 5 00 7 50 10 00 Ki 5 50 8 26 11 00 EXAMPLEvS 1. Find the amount due a carpenter who has worked 8 hours regular time and 2 hours overtime at 55 cents per hour. 2. A carpenter worked on Sunday from 8 to 11 o'clock. If his regular wages are 45 cents per hour, how much will he receive ? 3. A carpenter received 55 cents an hour. How much money is due him for working July 4th from 8-12 a.m. and 1-4.30 p.m. ? 4. A carpenter works six days in the week ; every morning from 7.30 to 12 m. ; three afternoons from 1 to 4.30 p.m. ; two afternoons from 1 to 5.30 ; and one from 1 until 6 p.m. What will he receive for his week's wage at 50 cents per hour ? 106 VOCATIONAL MATHEMATICS Painting Paint, which is composed of dry coloring matter or pigment mixed with oil, drier, etc., is applied to the surface of wood by means of a brush to preserve the wood. The paint must be composed of materials which will render it impervious to water, or rain would wash it from the exterior of houses. It should thoroughly conceal the surface of whatever it is applied to. The unit of painting is one square yard. In painting wooden houses two coats are usually applied. It is often estimated that one pound of paint will cover 4 sq. yd. for the first coat and 6 sq. yd. for the second coat. Some allowance is made for openings ; usually about one half of the area of openings is deducted, for considerable paint is used in painting around them. Table 1 gallon of paint will cover on concrete . . 300 to 375 superficial feet 1 gallon of paint will cover on stone or brick work 190 to 225 superficial feet 1 gallon of paint will cover on wood . . . . 375 to 525 superficial feet 1 gallon of paint will cover on well-painted sur- face or iron 600 superficial feet 1 gallon of tar will cover on first coat ... 90 superficial feet 1 gallon of tar will cover on second coat . . 160 superficial feet EXAMPLES 1. How many gallons of paint will it take to paint a fence 6' high and 50' long, if one gallon of paint is required for every 350 sq. ft. ? 2. What will the cost be of varnishing a floor 22' long and 16' wide, if it takes a pint of varnish for every four square yards of flooring and the varnish costs $ 2.65 per gal. ? 3. What will it cost to paint a ceiling 36' by 29' at 21 cents per square yard ? 4. What will be the cost of painting a house which is 52' long, 31' wide, 21' high, if it takes one gallon of paint to cover 300 sq. ft. and the paint costs $1.65 per gallon ? PART III — SHEET AND ROD METAL WORK CHAPTER VII Blanking or Cutting Dies Many kinds of receptacles are pressed or cut from difTerent kinds of sheet metal, such as copper, tin, and aluminum. Cans, pots, parts of metal boxes, and all sorts of metal novelties are punched out of sheet metal most economically by the punch and die operated by the hydraulic press. So skillfully can die makers produce dies and punches to cut out articles that thou- sands of everyday necessities in the household are made by this method. Parts of watches, parts of automobiles, and parts of machinery are punched out. Some presses that operate the Blanking" or "Cutting" Diks Cutting dies consist of an upper " male" die or "punch," and the lower or "female" die. Circumstances determine whether any or how much "shear" shall be given to the cutting eecial nnu-hinery. In some rases it is prefer- able to fasten the steel dies in cast-iron chucks or die-beds by means of keys 107 108 VOCATIONAL MATHEMATICS or screws. This applies more particularly to small dies. Cutting dies may be made to tit any size and style of press. For cutting thick iron, steel, brass, and other heavy metals, both the die and punch should be hard and provided with strip- pers. Punch and Die without Stripper Punch and Die with Strippkr punch and die are run by foot power, but those most generally used are run by electricity. A blanking or cutting die is a metal plate or disk having an opening in the center used in a punching machine or press which is supplied with ample power and which also supports the metal from which pieces are punched. Dies are made in almost any size and shape for cutting flat blanks in tin, iron, steel, aluminum, brass, copper, zinc, silver, paper, leather, etc. Holes are punched in thick sheets of metal or in heavy plates by means of great pressure exerted by a hydraulic press. This pressure, in pounds, is usually about 60,000 times the area (expressed in square inches) of the surface cut out, or in other words about 60,000 lb. per square inch. EXAMPLES 1. How much pressure will be necessary for a hydraulic press to exert on a sheet of boiler plate ^y thick, if it is de- sired to punch holes ^" in diameter ? 2. How much pressure will it be necessary to use in order to punch holes f" in diameter out of \" boiler plate ? 3. How much pressure will it be necessary to use in order to punch jV holes out of j^" boiler plate? SHEET AND HOD METAL WORK 109 Combination Dies Double dies for blanking; and perforating are extensively used in the manufacture of washers, key blanks, electrical instruments, hardware, etc. The blanking and perforating punches act simultaneously. At a single stroke of the ])ress Double" Dies for Blankino and Perforating Tlie perforating and blanking punches act simultaneously in such a manner that the lioles are punched first, whereupon the strip or slieet being fed for- ward, the blank is cut out around them and the holes for the next blank perforate*! at the same /" stroke. In this manner a blank is < ' " ' forated at every stroke of tlie press. The same principle may be extended so as to punch a number of perforated blanks at a time. ate— ^ \ ^^^ is completely per- f } O ) O ( O ) Perfobatino Dibs with Stripper Plates 110 VOCATIONAL MATHEMATICS one punch perforates the holes and the other punch cuts out the metal between the holes punched at the previous stroke. These dies are usually so arranged that the finished article is automatically pushed out from the dies by the action of the springs. An expert operator can punch many thousands of pieces in a day. Example. — How much metal will be required for 2000 blacking box covers, 6" in diameter, and f deep ? 6" + f " + f" = 7i", diameter of one cover. .7854 X 7.5 X 7.5 = 44.1787 sq. in., area. 44.1787 X 2000 = 88,357.4 sq. in. = 613.6 sq. ft. Ans. EXAMPLES 1. How large must the blank be cut for a pail cover that is to be (a) 5" in diameter and |" deep ? (6) 7" in diameter and ly deep ? (c) 6" in diameter and V' deep ? (d) 8" in diameter and li" deep ? 2. How large must the blank be cut for the pail bottom in each of the above examples (a, b, c, and d) ? The blank must be i" larger in diameter than the diameter of the part in order to allow |" all around for forming into the sides. 3. How large must the piece be to form the sides of the pail in each of the above examples ? Pail (a) to be 6" high, (6) 9" high, (c) 8" high, and (d) 10'' high. Allow ^" in height, and for the lock seam allow f" in circumference. 4. How much metal will be necessary for 850 complete pails in example (a) above ? 5. How much metal will be necessary for 1500 complete pails in example (6) above ? 6. How much metal will be necessary for 840 complete pails in example (c) above ? 7. How much metal will be necessary for 1000 complete pails in example (d) above ? SHEET AND ROD METAL WORK 111 Mah I >i.\«; Dies 8. How many square feet of sheet copper will be required to make a rectangular tank 7' long, 3' wide, IV deep, allowing 10 S for WiKK (iAHOK Dimensions ofSiKos in Decinml Parts of an Inch BlRMINO- Nl-MBKR AmkRH AN IIAM, OR Waphhi'rn Impkriai. Stubs U.S. Nl'MBKR or WiRK OR Br«>wn Stibi* * MOEN WlKK Stkbi. Stani>ari» «>K VV'iKK (Jakje iS^ SlIARPK Iron WiBB Mr.j. Co. Galgk .464 WiRB KoR Plate itAvr.y. 000000 .46875 000000 00000 .... .... .... .432 .... .4375 00000 0000 .46 .454 .3938 .400 .40625 0000 000 .40964 .425 .3625 .372 .375 000 00 .3648 .38 .3310 .348 .34375 00 .32486 .34 .3065 .324 .... .3125 1 .2893 .3 .2830 .300 .227 .28125 1 2 .25763 .284 .2625 .276 .219 .265625 2 3 .22942 .259 .2437 .252 .212 .25 3 4 .20431 .238 .2253 .232 .207 .234375 4 5 .18194 .22 .2070 .212 .204 .21875 5 6 .16202 .203 .1920 .192 .201 .203125 6 7 .14428 .18 .1770 .176 .199 .1875 7 8 .12849 .165 .1620 .160 .197 .171875 8 9 .11443 .148 .1483 .144 .194 .15625 9 10 .10189 .134 .1350 .128 .191 .140625 10 11 .090742 .12 .1205 .116 . .188 .125 — 11 12 .080808 .109 .1055 .104 .185 .109375 12 13 .071961 .095 .0915 .092 .182 .09375 13 14 .064084 .083 .0800 .080 .180 .078125 14 15 .057068 .072 .0720 .072 .178 .0703125 15 16 .05082 .065 .0625 .064 .175 .0625 16 17 .045257 .058 .0540 .056 .172 .05625 17 18 .040303 .049 .0475 .048 .168 .05 18 19 .03589 .042 .0410 .040 .164 .04375 19 20 .031961 .035 .0348 .036 .161 .0375 20 21 .028462 .032 .03175 .032 .157 .034375 21 22 .025347 .028 .0286 .028 .155 .03125 22 23 .022571 .025 .0258 .024 .153 .028125 23 24 .0201 .022 .0230 .022 .151 .025 24 > 26 .0179 .02 .0204 .020 .148 .021875 25 26 .01594 .018 .0181 .018 146 .01875 26 27 .014195 .016 .0173 .0164 .143 .0171875 27 28 .012641 .014 .0162 .0149 .139 .015625 28 29 .011257 .013 .0150 .01.36 .134 .0140625 29 30 .010025 .012 .0140 .0124 .127 .0125 30 31 .008928 .01 .0132 .0116 .120 .0109375 31 32 .00795 .009 .0128 .0108 .115 .01015625 32 33 .00708 .008 .0118 .0100 .112 .009375 33 34 .006304 .007 .0104 .0092 .110 .00859375 34 35 .005614 .005 .0095 .0084 .108 .0078125 34 36 .005 .004 .0090 .0076 .106 .00703125 36 37 .004453 .... .... .0068 .103 .006640625 37 38 .003965 .0060 .101 .00625 38 39 .003531 .... .0052 .099 39 40 .003144 .0048 .097 40 The American or B. «k 8. frauge l« the standard for sheet brass, copper, or German silver, and for wire of the same material. The Binnin^'ham (range is use85 6.875 . 9 5-32 .15(525 6.350 6.25 ^ 8% per cent. 10 •M)4 .14062 5.715 5.(525 1 10 per cent. 11 1-8 .125 5.080 5.00 12 7-(54 .10937 4.445 4.375 13 3-32 .09374 3.810 3.75 14 5-64 .07812 3.175 3.125 15 9-128 .07031 2.857 2.812 1(5 1-16 .0625 2.540 2..'>0 17 9-1(50 .05625 2.28(5 2.25 18 1-20 .a5 2.032 2. \\) 7-160 .04375 1.778 1.75 20 ;i-«o .0.375 1.524 l.-W 21 11-.'V20 .03437 1.397 1.375 22 1-32 .03125 1.270 1.2.5 23 i>-320 .02812 1.143 1.125 24 1-40 .025 1016 25 7-320 .02187 1.389 !875 26 3-160 .01875 .762 .75 27 \um .01718 .698 .(587 28 1-64 .01562 .635 .623 29 9-(i40 .01406 .571 .5(52 30 1-80 .0125 .508 .5 31 l-iy^O .010t)3 .694 .437 .S2 13-1280 .01015 .413 .40(5 33 3-320 .00937 .381 .375 . 34 11-1280 .00859 ..349 ..343 35 5-640 .00781 .317 .312 .36 9-1280 .00703 .28.'> .281 37 17-2560 .00f5(ht .271 .2(55 38 1-160 .00625 .254 .25 122 VOCATIONAL MATHEMATICS Weights and Arbas of Round, Square, and Hexagon Steel Weight of one cubic inch = .2836 lb. Weight of one cubic foot = 490 lb. c« Abea = DlAM.« X .7854 1 Akea = Side2 X 1 Area = DiAM.2x. 866 c - Hound Sqtiare Ilea'agon Weight Per Inch Area Square Inches Circum- ference Inches Weight Per Inch Area Square Inches Weight Per Inch Area Square Inches 1-32 1-16 S-S2 1-8 .0002 .0009 .0020 .0035 .0008 .0031 .0069 .0123 .0981 .1963 .2995 .3927 .0003 .0011 .0025 .0044 .0010 .0039 .0088 .0156 .0002 .0010 .0022 .0038 .0008 .0034 .0076 .0135 5-32 8-16 7-32 1-4 .0054 .0078 .0107 .0139 .0192 .0276 .0376 .0491 .4908 .5890 .6872 .7854 .0069 .0101 .0136 .0177 .0244 .0352 .0479 .0625 .0060 .0086 .0118 .0154 .0211 .0304 .0414 .0540 9-32 5-16 11-32 3-8 .0176 .0218 .0263 .0313 .0621 .0767 .0928 .1104 .8835 .9817 1.0799 1.1781 .0224 .0277 .0335 .0405 .0791 .0977 .1182 .1406 .0194 .0240 .0290 .0345 .0686 .0846 .1023 .1218 13-82 7-16 15-82 1-2 .0368 .0426 .0489 .0557 .1296 .1503 .1726 .1963 1.2762 1.3744 1.4726 1.5708 .0466 .0.543 .0623 .0709 .1651 .1914 .2197 .2500 .0405 .0470 .0540 .0614 .1428 .1658 .1903 .2161 17-82 9-16 19-32 5-8 .0629 .0705 .0785 .0870 .2217 .2485 .2769 .3068 1.6689 1.7671 1.8653 1.9635 .0800 .0897 .1036 .1108 .2822 .3164 .3526 .3906 .0693 .0777 .0866 .0959 .2444 .2743 .3053 .3383 21-32 11-16 23-32 3-4 .0959 .1053 .1151 .1253 .3382 .3712 .4057 .4418 2.0616 2.1598 2.2580 2.3562 .1221 .1340 .1465 .1622 .4307 .4727 .5166 .5625 .1058 .1161 .1270 .1382 .3730 .4093 .4474 .4871 25-32 13-16 27-32 7-8 .1359 .1470 .1586 .1705 .4794 .5185 .5591 .6013 2.4543 2.5525 2.6507 2.7489 .1732 .1872 .2019 .2171 .6103 .6602 .7119 .7656 .1499 ! .1620 .1749 .1880 .5286 .5712 .6165 6631 29-32 15-16 31-32 1 .1829 .1958 .2090 .2227 .6450 .6903 .7371 .7854 2.8470 2.9452 3.0434 3.1416 .2329 ,2492 .2661 .2836 .8213 .8789 .9384 1.0000 .2015 .2159 .2305 .2456 .7112 .7612 .8127 .8643 1 1-16 1 1-8 1 3-16 1 1-4 .2515 .2819 .3141 .3480 .8866 .9940 1.1075 1.2272 3.3379 3.5343 3.7306 3.9270 .3201 .3589 .4142 .4431 1.1289 1.2656 1.4102 1.5625 .2773 .3109 .3464 .3838 .9776 1.0973 1.2212 1.3531 1 5-16 1 3-8 1 7-16 1 1-2 .3837 .4211 .4603 .5012 1.3530 1.4849 1.6230 1.7671 4.1233 4.3197 4.5160 4.7124 .4885 .5362 .5860 .6487 1.7227 1.8906 2.0664 2.2500 .4231 .4643 .5076 .5526 1.4919 1.6373 1.7898 1.9485 1 9-16 1 5-8 1 11-16 1 3-4 .5438 .5882 .6343 .6821 1.9175 2.0739 2.2365 2.4053 4.9087 5.1051 5.3014 5.4978 .6930 .7489 .8076 .8685 2.4414 2.6406 2.8477 3.0625 .5996 .6480 .6994 .7521 2.1143 2.2847 2.4662 2.6522 1 13-16 1 7-8 1 15-16 2 .7317 .7831 .8361 .8910 2.5802 2.7612 2.9483 3.1416 5.6941 5.8905 6.0868 6.2832 .9316 .9970 1.0646 1.1342 3.2852 3.5156 3.7539 4.0000 .8069 .8635 .9220 .9825 2.8450 3.0446 3.2509 3.4573 SHEET AND ROD METAL WORK 123 Wkiohts and Areas of Round, Square, and Hexagon Stebl. — Con- tinued Weight of one cubic inch » .2S8C lb. Weight of one cubic foot - 490 lb. 8« Area = DiAM.* X .7854 Area = Side* x 1 Arka»Diaii.*x .866 IE Hound 1 Square , Ilevagon 55 Weight Per Inch Are* Square Indies CHrcum- ference Inches Weight Per Inch Area Square I iiuhes Weight Per Inch Area Square Inches S 1-1« S 1-8 S S-16 11-4 .9475 1.0058 1.0658 1.1276 3.3410 3.5466 3.7583 3.9761 6.4795 6.6759 6.8722 7.0686 1.2064 1.2806 1.3570 1.4357 4.2539 4.5156 4.7852 5.0625 1.0448 1.1091 1.1753 1.2434 3.6840 3.9106 4.1440 4.3892 S 6-16 t S-8 S 7-16 % 1-S 1.1911 1.2564 1.3234 1.3921 4.2000 4.4301 4.6664 4.9087 7.2649 7.4613 7.6575 7.8540 1.5165 1.6569 1.6849 1.7724 5.3477 5.6406 5.9414 6.2500 1.3135 1.3854 1.4593 1.5351 4.6312 4.8849 5.1454 5.4126 S6-8 S S-4 S 7-8 S 1.5348 1.6845 1.8411 2.0046 5.4119 5.9396 6.4918 7.0636 8.2467 8.6394 9.0321 9.4248 1.9541 2.1446 2.3441 2.5548 6.8906 7.5625 8.2656 9.0000 1.6924 1.8574 2.0304 2.2105 5.9674 6.5493 7.1590 7.7941 S 1-8 S 1-4 S S-8 Sl-1 2.1752 2.3527 2.5371 2.7286 7.6699 8.2958 8.9462 9.6211 9.8175 10.2102 10.6029 10.9956 2.7719 2.9954 3.2303 3.4740 9.7656 10.5625 11.3906 12.2500 2.3986 2.5918 2.7977 3.0083 8.4573 9.1387 9.8646 10.6089 S 8-8 SS-4 S 7-8 4 2.9269 3.1323 3.3446 3.5638 10.3206 11.0447 11.7932 12.5664 11.3883 11.7810 12.1737 12.5664 3.7265 3.9880 4.2582 4.5374 13.1407 14.0625 15.0156 16.0000 3.2275 3.4539 3.6880 3.9298 11.3798 12.1785 13.0035 13.8292 4 1-8 4 1-4 4 S-8 4 1-S 3.7900 4.0232 4.2634 4.5105 13.3640 14.1863 15.0332 15.9043 12.9591 13.3518 13.7445 14.1372 4.8254 5.1223 5.4280 5.7426 17.0156 18.0625 19.1406 20.2500 4.1792 4.4364 4.7011 4.9736 14.7359 15.6424 16.5761 17.5569 4 6-8 4 S-4 4 7-8 8 4.7645 5.0255 5.2935 5.5685 16.8002 17.7205 18.6655 19.6350 14.5299 14.9226 15.3153 15.7080 6.0662 6.6276 6.7397 7.0897 21.3906 22.5625 23.7656 25.0000 5.2538 5.5416 5.8371 6.1403 18.5249 19.5397 20.5816 21.6503 8 1-8 6 1-4 6 S-8 6 1-S 5.8504 6.1392 6.4351 6.7379 20.6290 21.6475 22.6905 23.7583 16.1007 16.4934 16.8861 17.2788 7.4496 7.8164 8.1930 8.5786 26.2656 27.5624 28.8906 30.2500 6.4511 6.7697 7.0959 7.4298 22.7456 23.8696 25.0198 26.1971 6 8-8 6 S-4 6 7-8 6 7.0476 7.3643 7.6880 8.0186 24.8505 25.9672 27.1085 28.2743 1 17.6715 18.0642 18.4569 18.8496 8.9729 9.3762 9.7883 10.2192 31.6406 .33.0625 34.5156 36.0000 7.7713 8.1214 8.4774 8.8420 27.4013 28.6361 29.8913 31.1765 6 1-4 6 1-S 6S-4 7 8.7007 9.4107 10.1485 10.9142 30.6796 33.1831 35.7847 38.4845 19.6350 20.4204 21.2058 21.9912 11.0877 11.9817 12.9211 13.8960 39.0625 42.2500 45.5625 49.0000 9.5943 10.3673 11.1908 12.0351 33.8291 36.5547 39.4584 42.4354 Tl-1 8 12.5291 14.2553 44.1786 50.2655 23.5620 25.1328 15.9520 18.1497 56.2500 64.0000 13.8158 16.7192 48.7142 66.3169 Pupils should practice the use of tables in order to obtain accurate results quickly. Additional tables like these may be obulned from such standanl handbooks as Kents. Multiply above weights by .998 for wrought iron, .918 for ca.st iron. ^^Y.V^\ for cast brass, .1209 for copper, and 1.1748 for phos. bronze. 124 VOCATIONAL MATHEMATICS EXAMPLES By means of the table of weights and areas of round, square, and hexagonal steel, solve the following problems : 1. Find the circumference of a steel bar (a) -^^" in thickness ; (6) ^" in thickness ; (c) \l" in thickness ; {d) 111" in thick- ness; (e) 3|" in thickness. 2. Find the area of a steel bar (a) |^" in diameter ; (6) 1^^" in diameter; (c) 3|" in diameter; (d) 4^' in diameter; (e) 5 J" in diameter. 3. Find the weight per inch of a steel bar (a) ||'' in diameter; (h) ly^' in diameter ; (c) 2^^" in diameter. 4. Find the area of a square bar (a) f" per side ; (h) If" per side ; (c) 5^ per side ; (d) 3|'' per side. 5. Find the weight per inch of a square bar (a) j^' in thickness ; (6) f|" in thickness ; (c) Iff" in thickness. 6. Find the area of a hexagonal bar (a) i|" in thickness ; (6) 1\^" in thickness ; (c) 3J" in thickness. 7. Find the weight per in. of a hexagonal bar (a) |f" in thickness ; (6) |^'' in thickness ; (c) 2^^ in thickness. 8. Find the weight of a round bar of steel (a) 7' long -f^"_ in diameter ; (&) 11' long lyV in diameter ; (c) 16' long 2^%" in diameter ; {d) IV long 2J" in diameter ; (e) 13' long 4|" in diameter. 9. Find the weight of a round bar of steel (a) 8' long 2|" in diameter; (h) 14' long 1|" in diameter; (c) 11' long lif" in diameter. 10. Find the weight of a hexagonal bar of steel (a) 8' long 3f" in diameter; {h) T long 2J" in diameter; (c) 9' long Ifi" in diameter. 11. What is the weight of a square bar of wrought iron 16' long 4^" in thickness ? SHEET AND ROD METAL WORK 125 12. What is the weight of a hexagonal bar of wrought iron 18' long 2yV' i" diameter ? 13. What is the weight of a round bar of cast iron 14' long Si" in diameter? 14. What is the weight of a square bar of cast iron 13' long 1|" in thickness ? 15. What is the weight of a square bar of cast brass 15' long l|f " in thickness ? 16. What is the weight of a hexagonal bar of cast brass 8' long If" in diameter ? 17. What is the weight of a hexagonal bar of copper 7' long 2yV' in diameter ? la What is the weight of a round bar of copper 6' long l^y in diameter? PART IV— BOLTS, SCREWS, AND RIVETS CHAPTER VIII BOLTS The most common forms of fastenings are bolts, screws, rivets, pins, and nails. These are turned out in large numbers, usually by feeding long lengths of iron or steel rod into auto- matic machines. In order to make these fasteners of the right size one has to be familiar with the problems that are con- nected with them. The common bolt is made in many different sizes and is usually held in place by a nut screwed on the end. There are many different kinds of bolts for the different uses to which they are put. Rough Bolts The small diameter of a rough bolt head, that is, the dis-. tance across the flats, is 1^ times the diameter of the bolt, plus 1 inch ; or, it may be stated as follows : The diameter of a rough bolt head = li^Z) plus ^ inch, D being diameter of the bolt. Sometimes bolts or nuts are made from round stock and cut either square or hexagon. In such cases it is necessary to hud the proper diameter to which the stock must be turned in order that it may be milled to size. Let A represent the distance across the flats on the head of a flat bolt, and B the diameter of the round stock required to make a square-head bolt. Square Head 126 BOLTS, SCREWS, AND RIVETS 127 Wlieii the dimension A is given and it is necessary to turn a piece that will mill down to this size and leave full corners, multiply A by 1.414. The product is the desired size B, B = y/WTA^ Why ? B = y/¥A^ = AVi = 1.414 A When ^1, the distance across the flats, is given, and it is necessary to turn a piece that will mill down to this size and leave full corners, multiply A by 1.155. The product is the desired size {B) of the hexagon head. Let X 2 (ir =(fr-(fy -^^ 4 ■■ 16 4 4 52 = //-' -f 4 X2 3 J?^ = 4 X2 4 B^ 3 X2 B=2X^2_XV3=1.165X V3 3 Hexagonal Head Example, — To what diameter should a piece of stock be turned in order that it may have full corners when milled down six-sided to \\" across the flats? li" X 1.155 = 1.7325", diameter of the blank. Ans. EXAMPLES 1. What is the size of stock required to make the following bolts having hexagonal heads ? (a) Body .}" diam. (6) Body J" diam. (c) Body I" diam. (d) Body V diam. 2. What is the size of stock required to make the following bolts with square heads? (a) Body \" diam. (6) Body \" diam. (c) Body |" diam. {d) Body \" diam. 128 VOCATIONAL MATHEMATICS EXAMPLES 1. To what diameter should a piece of stock be turned so that it ma}^ have full corners when milled down square to 1|" across the fiats ? 2. To what diameter should a piece of stock be turned in order that it may have full corners when milled down six-sided to 1|" across the flats ? Sizes of Standard Hexagon Head Bolts Size of DiAM. OK Bolt Thickness OF Heads Hexagon or UlSTANCE Across Corners Threads Tap Drill In. In. Across the Flats In. In. Per Inch In. i i 1% 20 t\ t\ H u n 18 C M H ¥i 16 N tV M 11 If 14 S tV i 1 13 if T? n M V. 12 If H ii*. 1/^ 11 II f n i-A 10 f II IrV m 9 H if n ii 8 II M m 2/2 7 fi 1 2 h% 7 1/t h\ 2A n Ui ifV 2| 2| 6 . m 1/3. 2/. 2-it 5^ 111 If 2| •3A 5 H HI m 3M 5 If 2 1^ H 3t H 111 ^ H n 4t^. 41 1.962 = If ^ ^ iH H 4i 4 2.176 = 2/^ 2f 2i H 4|f 4 2.426 = 2j.V 3 2A ^ H 3^ 2.629 = 2f ^ Notice that size of hexagon is equal to diameter of bolt + \ diameter of bolt + I of an inch, and also that thickness of head is \ of hexagon in every case. The thickness of nut is equal to the diameter of bolt. BOLTS. SCREWS, AKD UIVKTS 129 3. To what diameter should a piece of stock be turned so that it may have full corners when milled down sc^uare to 2J" across the flats ? 4. To what diameter should a piece of stock be turned so that it may have full corners when milled down six-sided to 2 J" across the flats ? Size across Corners of Squares SiXK OF SqUARK, In. DiAUONAL Size of Sqitabr, In. Diagonal } .177 1.4141 ^. .205 1.590 . \ .354 1.768 ^ . .442 1.945 i .530 2.121 A .619 2.298 i .707 2.476 A .706 2.652 f .884 2 2.828 n .972 ^ 3.005 i 1.061 2\ 3.182 H 1.149 ^ 3.535 1 1.237 2f 3.889 H 1.326 3 4.243 EXAMPLES Use the table to obtain the size of bolts. 1. Find the distance across the corners of a hexagon head bolt (a) with a diameter of 1 J" ; (6) with a diameter of 2J" ; (c) with a diameter of 1\". 2. Find the diagonal distance across a square bolt with size ofsquare(a)-fV'; (/>) A"- 3. (a) If a bolt^heading machine has the following daily output, 23:^0, 2060, 1950, 2420, 2310, 2030, what is the average 130 VOCATIONAL MATHEMATICS daily output ? (6) What would be the daily wage at 13 cents per 100 bolts ? (c) The weekly wage ? 4. A blacksmith requires six pieces of steel of the following lengths: If", 2|", 2j\'\ ^\\", 1\^", 2". How long a piece of steel will be necessary to make them, if ^' is allowed for each in finishing ? 5. A machinist has to make five bolts from the same size bar. One bolt is to be 1-^' over all, another 2^", another 2-^%", another 3//', and the last 2f".. How much stock will he need if he allows ^" for each cut-off ? Rivets One of the simplest and most efficient metal fastenings which has been extensively used is the rivet. It resembles the bolt, but it can be removed only by chipping off the head, while the bolt can be taken off by removing the nut. Rivets, like bolts and nails, are quickly turned out by the thousand with the aid of automatic machines. EXAMPLES 1. What must be the length of a bolt under the head, to go through 9^y thickness of plank and allow IJ" outside for tak- ing a nut ? 2. Whgit must be the length of a bolt under the head, to go through 7^\" thickness of plank and allow If" outside for tak- ing a nut ? 3. (a) How many rivets can be made in a bolt machine, from a round iron rod 6' long, if each rivet requires 2 J" of bar ? (6) If the bar weighs | lb. per ft., how many rivets weighing \ lb. apiece can be made from it ? (c) How much waste will there be in each case ? 4. What is the total thickness of three plates riveted to- gether, each plate i^" thick ? What would be the total thick- ness of five such plates ? BOLTS, SCREWS, AND RIVETS 131 5. What is the thickness of a steel plate that is only | the thickness of a -j^" plate ? 6. A blacksmith and his helper made 192 bolt dogs in 18J^ hours. They received 6.} cents apiece for them, (a) How much did they both receive per hour? (b) If the blacksmith re- ceived G65 % of the money, how much did each receive per hour? 7. How many feet of round iron weighing 2.67 lb. per foot will be required to make 87 rivets weighing 2^ lb. apiece, not counting waste ? Give answer in feet and decimals of foot. 8. How many rivets weighing 7.J ounces each can be made from 15' of round iron weighing 1.5 lb. per foot ? How much waste will there be ? , 9. If the drawing of an armor bolt is made J size and the length of the drawing measures 7J", what will it measure if made to the scale of 3'' = 1' ? What is the actual length of the bolt? 10. If the drawing of an armor bolt is made \ size and the length of the drawing measures 9|", what will it measure if made to the scale of 4" = 1'? What is the actual length of the bolt? 11. The over-all length of a threaded bolt is 7f", the thick- ness of the head is I", and the other end of the bolt is threaded for a distance of 2^" ; what is the length of the shank between the under side of the head and the threaded part of the bolt ? Nails W^ooden objects are often held together by means of nails. There are two kinds of nails: cut and wire. The wire nails are more commonly used, as they penetrate the wood without splitting it as the cut nails do. They have different kinds of heads, according to the use for which they are intended. 132 VOCATIONAL MATHEMATICS The origin of the common terms " sixpenny," " tenpenny," etc., as applied to nails, though not generally known, is involved in no mystery. Nails have been made a certain number of pounds to the thousand for many years and are still reckoned in that way in England, a tenpenny behig a thousand nails to ten pounds, a sixpenny a thousand nails to six pounds, a twentypenny weighing twenty pounds to the thousand ; and in ordering buyers call for the three-pound, six-pound, or ten-pound variety, etc., until, by the Englishmen's abbreviation of "pun" for "pound," the abbreviation has been made to stand for penny, instead of pound, as originally intended. Length and Number of Cut Nails to the Pound SiZK in d 1 K ■< S ■< s ^ M < H f |in. 800 f lin. 500 2d 1 in. 800 1100 1000 376 3d li in. 480 720 760 224 4d 11 in. 288 623 368 180 398 5d If in. 200 410 130 6d 2 in. 1H8 95 84 268 224 126 96 7d 2^ in. 124 74 64 188 98 82 8d 21 in. 88 62 48 146 128 75 68 9d 2| in. 70 53 36 130 110 65 lOd 3 in. 58 46 30 102 91 55 28 12d 3Jin. 44 42 24 76 71 40 16d 31 in. 35 38 20 62 54 27 22 20d 4 in. 23 33 16 54 40 14^ 30d 4^ in. 18 20 33 m 40d 5 in. 14 27 ^ 60d 6^ in. 10 8 60d 6 in. 6^ in. 7 in. 8 in. 8 6 BOLTS, SCREWS, AND RIVETS 133 EXAMPLES 1. How many Jd nails are there in 3 pounds ? 2. How many Jd nails are there in 4 pounds ? 3. How many 2d nails are there in 2 pounds ? 4. How many 7d nails are there in 8 pounds ? 5. How many 16d nails are there in 1(> pounds ? 6. How long is (a) an 8d nail? (b) a 40d? (o) a 16d? (d) a 9d ? 7. How long is (a) a 7d nail? (6) a 5d ? (c) a 12d ? Tacks Tacks are used to fasten thin pieces of material to wood. They vary in form and size. The size is represented by a number, as 16 oz. tacks, 24 oz. tacks ; a No. 1 tack is called a one-ounce tack. Ndmbbr of Tacks in a Pound Title Length No. PER Lb. Title Lexotii N(». PER Lll. 1 ounce A inch 10,000 10 ounce H inch 1,600 IJ ounce y'j inch 10,606 12 ounce 1 inch 1,332 2 ounce i inch 8,000 14 ounce \i inch 1,143 2\ ounce A i"^^ 6,400 10 ounce 1 inch 1,000 3 ounce f inch 5,3.32 18 ounce U inch 888 4 ounce ^ inch 4,000 20 ounce 1 inch 800 6 ounce A inch 2,060 22 ounce ly^g inch 727 8 ounce f inch 2,000 24 ounce 1\ inch 666 EXAMPLES 1. How many tacks are there in 8 lb. of 1 oz. or No. 1 tacks ? 2. How many tacks are there in 13 lb. of 2^ oz. or No. 2^ tacks? 3. How many tacks are there in 7 lb. of 8 oz. or No. 8 tacks ? 134 VOCATIONAL MATHEMATICS 4. How many tacks are there in 6 lb. of 12 oz. or No. 12 tacks ? 5. How many tacks are there in 17 lb. of 20 oz. or No. 20 tacks ? Screws Pieces of wood and of metal are often fastened together by means of screws instead of by nails, especially if it is desirable to separate the parts at any time. Screws are made of iron, steel, or brass and have either a flat, fillister, or round head. When it is desired to have the head of the screw flush with the surface, the flat head type is used. Screws are made on auto- matic screw machines into which wire of various sizes may be fed. The machines are so constructed that they turn out large numbers of screv/s all complete in a very short time. i Mmmm i Q^ UiUMlliillig !^ iuiiimiiiiiirf i ^^^^^^MHH| ^Wfffff'^^B^^ ^^^^^^^^^^jg flat head round head fillister head Iron Machine Screws Screw threads are divided into two classes : first, those used for fastening ; and second, those used in large machines for communicating motion. The screw threads used for communi- cating motion with which the mechanic has to deal are pro- duced by a cutting process in which the thread is formed from the solid piece of stock by means of a single pointed cutting tool in a lathe. Screws used for fastening are made by means of taps and dies. The tap is a tool used to produce internal threads, and the die is a tool used to cut the external threads. The screw thread is applied in many ways, but the most com- mon use is that of fastening together the various parts of machines, etc. We find the mechanic using many different forms of bolts and screws to meet the needs of industries. In order to BOLTS. SCREWS. AND RIVETS 135 specify a particular f?rade of bolt or screw it is necessary to mention (a) shape or form of head, (b) pitch or number of threads to the inch, (c) shape of thread, (rf) outline of body, barrel or stem, (e) size of diameter, (/) direction of thread, as right or left hand, (g) length, (/i) material, as brass, iron, etc. There are four different-shaped threads in common use in the United States : 1. The V thread ; 2. The U. S. standard ; 3. The Acme standard or worm ; 4. The square thread. Sharp V Thread *PITCH- U. S. Standard Thread AcMR Standard or Worm Thread TTTTZy P/TCH Squabb Thread 1 Sometimes called modified square thread. 136 VOCATIONAL MATHEMATICS The different screws are formed by cutting a spiral groove around a cylinder. The projecting stock between the grooves is called the land or thread. There may be any number of threads ; to every groove there is an accompanying thread. RiGHT-ILvxD, Single V-Thread, 8 Threads to the Inch Nut A screw that has one thread is called single-threaded ; one having two threads, double-threaded; three threads, triple-threaded; etc. The cutting of the spiral groove or grooves is called cutting or threading a screw. A nut is a piece of iron or steel with a threaded hole which goes over a screw, and will turn off and on the screw. Lead of a Screw. — The distance that a thread advances in one turn is called the lead of the screw. In a single-threaded screw the lead is equal to the distance occupied by one thread ; and when the nut has made one complete turn, it has advanced one thread upon the screw. In a double-threaded screw the nut advances two threads with each complete turn ; when the lead is three threads, the nut advances three threads in one turn. In general, the lead can be divided by any number of threads, the advance of any one of these threads in one turn being always equal to the lead. One complete revolution of a single-threaded screw or the lead of a screw, if the screw had twelve threads to the inch, would be one twelfth of an inch. BOLTS, SCREWS, AND RIVETS 137 Threads to an Inch. — By placing a scale upon a screw as in the figures on page 130, the number of thread-windings or coils in an inch can be counted. These windings or coils are called threads, and the number of coils to an inch is called the number of threads to an indi. The thread commences at the root or bottom of the screw. To measure screws for the number of threads per inch, the measurement must begin at the point of the thread. Place the end of the scale in line with this portion and then count the number of threads within the one- inch line. Pitch. — T?ie distance from the center of one thread to the cen- ter of the next thread, measured in a line parallel to the axis, is tJie pitch of the thread, or the thread-pitch. Divide 1" by the number of threads to 1" and the quotient is the thread-pitch. The threads to 1" and the thread-pitch are reciprocals of each other. In a single-threaded screw the pitch is equal to the lead. In a double-threaded screw the pitch is half the lead; in a triple- threaded screw the pitch is \ the lead; and so on. When the thread inclines so as to be nearer the right hand at the under side, or clockwise, it is a right-hand thread. When the under side is toward the left, or counter clockwise, the thread is left- handed. Again, when a right-hand screw turns in a direction to move its upper side away from the eye, the thread appears to move toward the right; while a left-hand thread moves toward the left. The term turns to an inch means the number of times a screw must be turned around to advance one inch. If a screw makes four turns in advancing one ioch, its lead is \, and it has 4 turns to the inch. Divide one inch by the lead, and the quotient is the number of turns that the screw makes in ad- vancing one inch. If a screw does not advance exactly an inch in a whole number of turns, or if it does not advance some whole number of inches in one turn, it is said to have a frac- tional thread. In any screw divide any number of turns by 138 VOCATIONAL MATHEMATICS the number of inches occupied by these turns and the quotient will be the turns to an inch. Thus, a screw that turns 96 times in 12.005 inches, turns 96 -^ 12.005, or 7.9967 turns in one inch. EXAMPLES 1. How many turns to the inch has (a) a screw that ad- vances 25 inches in 200 turns ? (b) a single-threaded screw of 9 threads to the inch ? (c) a double-threaded screw with 8 threads to the inch ? (d) a 'double-threaded screw with 6 threads to the inch ? (e) a single-threaded screw with 24 threads to the inch ? 2. AVhat is the lead of a single-threaded screw if it has (a) five threads to the inch ? (6) eight threads to the inch ? (c) fifteen threads to the inch ? (d) twenty-eight threads to the inch ? 3. What is the lead of a double-threaded screw if it has eight threads to the inch ? 4. What is the lead of a single-threaded screw if it has twenty-two threads to the inch ? 5. What is the lead of a single-threaded screw if it has twenty-four threads to the inch ? 6. A jack screw has three threads to the inch ; how far does it move in ^ of a revolution ? 7. A jackscrew has three threads to the inch ; how far does it move in i of a revolution ? 8. What is the thread pitch of a single-threaded screw that has 18 threads to the inch ? 9. What is the thread pitch of a double-threaded screw that has 8 threads to the inch ? 10. Let each pupil have five different kinds of screws, and tell the number of threads to the inch of each screw. BOLTS, SCREWS, AND RIVETS 139 The Micrometer Accurate mathematical work in measuring diameter is done with the micrometer caliper. With this instrument thousandths of an inch may easily be found. The micrometer is easily adjusted, finely graduated, and has stamped on its yoke the fractions and decimal equivalents which may be needed in close measuring. The parts of the micrometer best known are the screw J the linhj the thimble. MiCKOMKTER CALIPER The screw of the micrometer is covered by the thimble to protect it from dust and wear. By turning the thimble we move the screw back and forward, increasing or decreasing the distance between the measuring points of the micrometer and so opening or closing the instrument for larger or smaller diameters. One complete revolution of the thimble changes the opening of the caliper .025, and as the pitch of the screw in the caliper is 40 per inch and the circumference of the thimble graduated into 25ths, the turn from one of these to the next makes the caliper opening .001. The heel is graduated in a straight line parallel with the screw length and conforms to the pitch of the screw, each division being .025 inch, and the fourth division, which is .100, is made on the frame with the figure 1, the eighth, with 2, etc. When the thimble is turned one complete revolution, 140 VOCATIONAL MATHEMATICS the screw advances one fortieth of an inch and one twenty-fifth of one fortieth is .001. In using the micrometer care must be taken to get the proper touch with the instrument or it may be crowded over with an error of one half thousandth more or less than the actual size of the work required. The microm- eter is a very delicate instrument and must be kept away from excessive heat or cold as expansion or contraction of the metal will cause it to become inaccurate. In close work the heat of the hand, when it is held too long, will change a micrometer reading. EXAMPLES 1. If the screw of a micrometer has 40 threads to the inch, how far will it move in (a) one complete revolution ; (b) ^ rev- olution ; (c) \ revolution ; (cZ) f revolution ; (e) J^ of a revo- lution ? 2. What is. the lead of the screw in the above micrometer ? 3. If the screw of a micrometer has 60 threads to the inch, what is the lead ? 4. In example 3, how far will it advance in (a) ^ revolu- tion ; (b) \ revolution ; (c) f revolution ? V-Shaped Thread The common V-shaped thread is a thread having its sides at an angle of 60 degrees to each other, and perfectly sharp, top and bottom. The objections to using this thread are that the^ top is so sharp that it is injured by the slightest accident and in using the taps and dies in making it the fine, sharp edge is quickly lost, causing constant variation in fitting. Formula: P= Pitch = No. of threads per inch JD = Depth = P X .8660 BOLTS, SCREWS, AND RIVETS 141 The diameter of the root (effective diameter) of the thread is found by multiplying the product above by 2, and then sub- tracting this double depth from the diameter of the screw. A formula is used to tind the size of a tap drill to use in connection with a tap of a given size which has a given number of threads per inch, as : Let 7= diameter of the tap, or size of the thread the nut is to tit; N= number of threads per inch ; S = size at root of the thread, or size of the tap drill. N Example. — What must the size be of a tap drill for a 1-inch V-thread tap or 1-inch bolt having 8 threads per inch ? According to the formula : ^ = 1 _ hl^ OT S = l- .216 or S = .784 inch, Ans. 8 By referring to the Table of Decimal Equivalents ^ the drill nearest in size to .784 is ^ inch, which will cut a trifle larger and therefore will be right. United States Standard Thread The United States Standard Thread has its sides also at an angle of 60 degrees to each other, but the top is cut off to the extent of one eighth of its pitch and the same quantity filled in at the bottom. The advantages claimed for this thread are that it is not so easily injured, that the taps and dies retain their size longer, and that the bolts and screws made with this thread are stronger and have a better appearance. This system has been recommended by the Franklin Institute of Philadelphia and it is often called the Franklin Institute Standard. Although the V-shaped thread is the strongest ^ See Appendix for Table of Decimal Equivalents. 142 VOCATIONAL MATHEMATICS form of screw thread, yet as the thrust between the screw and the nut is parallel to the axis of the screw, there is a tendency to burst the nut. So this form of thread is unsuitable for transmitting power. As \ of the height of the U. S. standard thread is taken from the top and ^ from the bottom, the thread is only J as deep as the V-form, so in the formula for finding the diameter at the root of the thread we use a numerator which is but f of the numerator used for the V-thread : f of 1.733 is 1.3. So the formula is Example. — What should the size of a tap drill be for a one- inch U. S. S. tap? As the U. S. standard is not only a thread of a certain form but also of a given pitch for each diameter of screw, by re- ferring to the table of United States Standard Screw Threads we find that one-inch screws have eight threads to the inch. S = l-— ov S = l- .1625 ovS= .8375 8 In the Table of Decimal Equivalents .8375 has no common fraction equivalent among the sizes given, but as f f is only .006 larger we would select a drill of that size. Formula : P = Pitch = ^ — - No. of threads per inch Z>=Depth = Px .6495 i^=Flat = - EXAMPLES Solve the following examples by the use of the table. 1. How many threads are there per inch of the U. S. S. screw with (a) I" diameter ? (b) f " diameter ? BOLTS, SCREWS, AND RIVETS 143 2. What is the diameter of a screw with the U. S. S. thread with (a) 5 threads per inch ? (6) 4^ threads per inch ? (c) 13 threads per inch ? Table of United States Standard Screw Threads DiAM. or SCRKW TiiRKAne I'KB Incu DiAM. OK SCRBW Threads i'kb Ikcii i in. 20 l}in. 6 A in. 18 n in. 6 i in. 16 If in. 6i Ai°- 14 If in. 5 J in. 13 l|in. 6 A in. 12 2 in. ^ f in. 11 2\ in-. n i in. 10 2Jin. 4 i in. 9 2f in. 4 l.in. 8 3 m. 3i IJin. 7 l^in. 7 Acme Standard Thread The Acme standard thread is an adaptation of the most commonly used worm thread and is intended to take the place of the square thread. It is more shallow than the worm thread but was the same depth as the square thread and is much stronger than the latter. The worm thread has sides with an angle of 29° ; the top is flat, .335 P, and the bottom is .31 P. The depth is .6866 P, the double depth being 1.3732 P; d=D- ^"^''^'^ , that is, the N diameter at the bottom of a worm thread is equal to the diameter of the worm minus 1.3732" divided by the number of threads to one inch. 144 VOCATIONAL MATHEMATICS Square Thread A square thread has parallel sides; the thickness of the thread and its depth are each one half the pitch : d = D — —, that is, the diameter at the bottom of a square thread is equal to the diameter of the screw minus one inch divided by the number of threads to one inch. The thrust on the square- threaded screw is parallel to the axis of the screw; couse- quently the frictional losses are not so great in this form as in the V form, but it is not so strong in the base as the V thread. EXAMPLES 1. What is the diameter of the root of a |" V-threaded bolt with 20 threads to the inch? 2. What is the diameter of the root of a |" U. S. S. threaded bolt with 16 threads to the inch? 3. What is the thread pitch of a V-threaded screw with 20 threads to an inch ? 4. What is the thread pitch of an Acme worm screw with 20 threads to an inch ? 5. What is the thread pitch of a square-threaded screw with 14 threads to an inch ? 6. What is the diameter of the root of a y\" V-threaded bolt with 18 threads to the inch ? 7. What is the diameter of the root of a |" V-threaded bolt with 14 threads to the inch ? 8. A U. S. S. screw has 20 threads. What is its thread pitch ? What is its depth ? 9. What is the diameter of the root of a f " U. S. S. screw with 20 threads to the inch ? 10. What is the thread pitch of a square-threaded screw with 8 threads to the inch? BOLTS, SCREWS, AND RIVETS 145 11. A 2 J" diameter bolt has a diameter of 1.962" at the root of the thread. What is the depth of the thread ? If the bolt has 4.V threads per inch, what is the pitch of the threads ? 12. What is the depth of a U. S. S. threaded screw of 7 threads ? 13. What is the diameter of the root of a V-threaded J" screw with 11 threads to the inch ? 14. What is the depth of a V-threaded screw with 11 threads to the inch? 15. What is the diameter of the root of a U. S. S. threaded bolt of 11 threads ? 16. What is the diameter of the root of a U. S. S. J" threaded bolt of 8 threads? 17. What is the thread pitch of a U. S. S. threaded screw of 9 threads ? 18. What is the depth of a worm-threaded screw with 16 threads to the inch ? 19. What is the depth of a square-threaded screw with 8 threads to the inch ? 20. What is the depth of a U. S. S. threaded screw with 8 threads to the inch ? 21. What is the depth of a U. S. S. 1 J" threaded bolt of 9 threads ? 22. What is the diameter of the root of a U. S. S. threaded screw of 9 threads ? 23. What is the depth of a V-threaded screw with 14 threads to the inch? 24. What is the thread pitch of a V-threaded screw with 16 threads to the inch ? 25. A U. S. S. threaded bolt has 9 threads to the inch. What is its thread pitch ? PART V — SHAFTS, PULLEYS, AND GEARING CHAPTER IX SHAFTS AND PULLEYS In a machine shop one notices at once the revolution of the shafting. There is one long cylindrical bar called the main line attached to the ceiling. The power that drives the ma- chinery is taken from this main line by means of pulleys and belts to smaller shafts called countershafts. The machinery is driven directly from the countershafts while the main line is driven from a motor or engine and flywheel. Shafts of different sizes are used according to the horse power required. ,To determine the horse power (H. P.) of a shaft, multiply the speed (revolutions per minute) by the cube of the diameter of the shaft, and divide the product by 84 for a steel shaft or by 160 for an iron shaft, and the quotient is the H. P. (For further discussion of horse power see pages 188 and 225.) EXAMPLES 1. What is the H.P. of a 2Jg" steel shaft having 280 revo- lutions per minute ? 2. What is the H.P. of a 2J" iron shaft having 245 revolu- tions per minute ? 3. What is the H.P. of a 2|" steel shaft having 290 revolu- tions per minute ? 4. What is the H.P. of a 2f^'' iron shaft having 350 revolu- tions per minute ? 146 SHAFTS AND PULLEYS 147 Belting Belts for transmitting ix)wer are divided into two general classes : leather belts and canvas belts. Both are sold by the foot. The material of the belt, the thickness, and width determine its value. Coils of belting need not be stretched out to measure their length (A). To do this first count the number of coils (N), measure the diameter of the hole in the center of the coil (d), and the outside diameter of the roll (/>). Then L = 0.1309 N(D + d) In this formula Z=: length in feet, and D and d diameter in inches. This rule is used in estimating. The formula is obtained as follows : — ^^— = average diameter of coil The length L = circumference of average diameter C = 2 ^-12 — = 0.1309 24 Substituting this in the formula ^^(-P + <^) = 0.1309 iV^(Z? -h d) EXAMPLES 1. How many feet of belting are there in a coil that has a diameter of 18", if the hole is 3" in diameter and there are 36 coils? 2. How many feet of belting are in a coil that has a 16" di- ameter, if the hole is 2^" in diameter and there are 38 coils? 148 VOCATIONAL MATHEMATICS Length of Belting on Pulleys To find the length of belting on pulleys, add together the diameters of the pulleys in inches and divide the sum by 2. Multiply this quotient by 3.25. Add this product to twice the distance in inches between the centers of the pulleys, and divide by 12. The final quotient is the length in feet of the belting on the pulleys. This is an approximation and applies to open belts only. Example. — The diameters of two pulleys are 24" and 12" respectively, and the distance between their centers is 108 inches. Find the length of the belting. 24" + 12" = 36" 18" X 3.25 = 58.50" ^J^!! ^ 22.8 feet 36" ^ 2 = 18" 58.5" +216"- = 274.5" 22.8' = 22' 11". Ans. EXAMPLES 1. What is the length of the belting connecting two pulleys having diameters of 18'' and 12" respectively, if the distance between their centers is 92" ? 2. What is the length of the belting on two pulleys having diameters of 16" and 22" respectively, if the distance between their centers is 86" ? 3. Find how many feet of belting are needed to make a belt to run over two pulleys each 30" in diameter if the distance between their centers is 13'. Arc of Contact In setting up machinery where there are pulleys, it is some- times desirable to find the arc of contact on the smaller pulley. The Boston Belting Company gives this rule, which holds when the pulleys are nearly of the same diameter : Divide the dif- ference between the diameters of the two pulleys by the dis- tance between the centers of the shafts, both being in the same SHAFTS AND PULLEYS 149 denomination ; multiply the quotient by 57, and subtract this product from 180 ; the result will be the number of degrees in the arc of contact. Multiply the entire circumference of the smaller pulley in feet by the degrees of the arc of contact as above, divide by 360, and the result "will be the number in feet of the arc of contact of the belt on the smaller pulley. EXAMPLES 1. Two pulleys, one 18" and the other 24" in diameter, are connected by a belt. If the distance between the centers of the shafts is 8' 6", what is the number of feet in the arc of contact of the belt on the smaller pulley ? 2. Find the number of lineal inches that a belt touches a pulley when the arc of contact is 240°, if the diameter of the pulley is 4 feet. 3. If the distance between the centers of two shafts is 9' 8" and the pulleys are 22" and 16", what is the number of feet of arc of contact of the belt on the smaller pulley ? D A common way for one shaft to drive another is by means of a belt running upon two pulleys, one on the driving shaft and the other on the driven, as in the above figure. In solving problems the pulley on the driving shaft is called the driver and the one on the driven shaft the driven. In order to install machinery and have it run at the proper speed, the relations between the driving and the driven pulleys 150 VOCATIONAL MATHEMATICS and the different methods of transmitting power from one shaft to another must be thoroughly understood. To make the shafts and pulleys run at the proper speed the correct diameter and circumference of the driving and the driven pulleys must be known. At every revolution of the driver the belt is pulled through a distance equal to the circumference of the driver; in moving a distance equal to the circumference of the driven pulley, the belt turns the driven pulley one revolution. When two pulleys are connected by a belt, their rim speeds are equal. If we divide the distance the belt has moved by the circumference of the pulley, the quotient gives the number of revolutions of the pulley. The smaller pulley revolves at the higher speed, a fact that is usually stated mathematically by saying that the revolutions of the pulleys are inversely proportionate to their circumferences. Example. — A pulley 12 inches in diameter makes 300 revo- lutions per minute. How fast is the rim traveling in feet per minute ? The circumference equals the diameter multiplied by 3.1416, or approx- imately 3|. ■*■ 12 X 3.1416 = 37.6992 inches circumference l^JiAlil^^ 3.1416' feet 12 Since it is running 300 revolutions per minute, 300 X 3.1416 = 942.48 feet per minute It is possible to express the operations of the above in a for- mula by letters : Let D = diameter of the pulley in inches C = circumference of the pulley in inches R = revolutions per minute (abbreviated R. P. M.) ^ = 3.1416 F= feet per minute that the rim travels (circumference speed) SHAFTS AND PULLEYS 151 Feet traveled per minute is equal to the circumference in inches, multiplied by revolutions per minute, and divided by 12. CR That is, F— — — , or substituting for C its equal ttZ), 1 w If we know the value of any two of the quantities F, D, or Bf we can find the others: (I) P = ^ (2) /> = ^ (3) 7J = 1-2,^ EXAMPLES 1. If a ten-inch pulley is making 300 revolutions per minute, how fast is a point on the rim traveling in feet per minute ? 2. The rim of a 14" pulley is running 1048 feet per minute. How many revolutions are made per minute ? 3. A pulley 18" in diameter makes 375 revolutions per minute. How fast is the rim traveling in feet per minute ? 4. What is the diameter of a pulley making 286 revolutions per minute if a point on the rim is traveling 984 feet per minute ? 5. A 9" pulley is making 198 revolutions por minute. How fast is a point on the rim traveling? 6. A pulley 32" in diameter is making 198 revolutions per minute. How fast is the rim traveling ? Speed The speed of any machine from the driving shaft or motor may be traced as follows : The circumference of a pulley equals its diameter multiplied by 3.1416. 152 VOCATIONAL MATHEMATICS This may be abbreviated when C stands for the circumfer- ence and D for the diameter : C = 3.1416 D or 7rZ> Let C = circumference of driver pulley C" = circumference of driven pulley D = diameter of driver pulley Z>' = diameter of driven pulley N = revolutions of driver pulley per minute jV' = revolutions of driven pulley per minute (C is read C prime.) C X N =z distance belt moves per minute on driver wheel C X A^ = distance belt moves per minute on driven wheel The distance represented by C x N h called rim speed of driver wheel. The distance represented by C" X N^ is called rim speed of driven wheel. Since the surface speeds are equal, (1) CxN^C^xN', (2) g = ^, Since C ^-jtB ovD^N[ ^ ^ D^N' C'=^7rD' UN N If we know any three of the above four quantities, the fourth can be found : If i) = ^^' N==^^ N D The proportion may be easily remembered by noting that the primes come together as middle terms: D:D'::N':N The above formula may be expressed in the form of rules: Whenever one pulley drives another we have four quantities to consider — the diameters of the two pulleys and the revo- SHAFTS AND PULLEYS 153 lutions per minute of the two pulleys. The above equation shows us that there is a definite relation between them. If we know any three of the quantities, we may find the fourth by transferring the factors of the equation. NoTK. — II is often difficult to remember these niles. In that case draw a sketch and place the given infonnation about each pulley. Let x represent the unknown quantity. Then multiply the two numbers known about one pulley and divide by the number given in the other pulley. The quotient will represent x. Example. — Find the size of a pulley on a countershaft that runs 120 revohitions per minute, if the diameter of the pulley on line shaft is 18" and runs 180 revolutions per minute. 18 : Z>' : : 120 : 180 120 D' = 18 X 180 2 Z>' = 54 D' = 27 EXAMPLES 1. The diameter of a pulley on the line shaft is 30" and it runs 158 revolutions per minute; the countershaft runs 400 revolutions per minute. What is the size of the pulley on the countershaft ? 2. What is the size of a pulley on a countershaft of an engine lathe if the diameter of the pulley on the line shaft is 15" and it runs 150 revolutions per minute while the counter- shaft runs 140 revolutions per minute ? 3. What is the size of a pulley on a countershaft of a planer if the diameter of the pulley on line shaft is 26" and it runs 305 revolutions per minute while the countershaft runs 793 revolutions per minute ? 4. What size pulley should be placed on the countershaft of a band saw if the diameter of the pulley on the main line shaft is 14", if it runs 250 revolutions per minute, and the counter- shaft runs 225 revolutions per minute ? 154 VOCATIONAL MATHEMATICS 5. A pulley 30" in diameter on a main shaft running 180 revolutions per minute is required to drive a countershaft 450 revolutions per minute.. What will be the diameter of the pulley on the countershaft ? Example. — Find the number of revolutions tha£ a counter- shaft is running if the line shaft runs 140 revolutions per minute, and the diameter of the pulley on it is 30" and the diameter of the pulley on the countershaft 12". Z> : D' : : .V : .V 30 : 12 : : N' : 140 2N' = 700 JV'=360. Ans. EXAMPLES 1. Find the number of revolutions per minute of a counter- shaft of an engine lathe if driven by an 18" pulley on the line shaft which runs 168 revolutions per minute, if the diameter of the countershaft pulley of lathe is 12''. 2. What is the speed of a countershaft of a speed lathe, if the pulley on the main shaft is 10" with 305 revolutions per minute and the diameter of the pulley on the countershaft is 4"? 3. Find the number of revolutions of a countershaft of a band saw if the pulley on the main shaft is 19" with 215 revo- lutions per minute and the diameter of the pulley on the countershaft is 16". 4. What is the speed of the countershaft of a cutting-off saw if the pulley on the main shaft is 15" with 230 revolutions per minute and the diameter of the pulley on the countershaft is 12"? 5. A pulley 30" in diameter making 180 revolutions per minute drives a countershaft with a 12" pulley. What is the speed of the countershaft ? SHAFTS AND PULLEYS 155 6. The main driving pulley of an engine is 12' G" in diam- eter and makes 96 revolutions per minute; it is belted to a 48" pulley on the main shaft. Find the speed of the latter. Example. — The line shaft of a machine shop runs 120 rev- olutions per minute, the diameter of the pulley on the counter- shaft is 15", and the countershaft runs 240 revolutions per minute. Find the size of the pulley on the main line. D:D'::N':.y |=^ D : 1 V : : 240 : 120 120 Z> = lo'' x240 D = 30". Ans. N'D' = ND EXAMPLES 1. The main line runs 160 revolutions per minute; the countershaft pulley is 9" in diameter and runs 320 revolutions per minute. What is the diameter of the pulley on the main line ? 2. A pulley 24" in diameter running 144 revolutions per minute is to drive a shaft 192 revolutions per minute. What must the diameter of the pulley be on the driven shaft ? 3. A driving shaft runs 140 revolutions per minute; the driven pulley is 10" in diameter and is to run 350 revolutions per minute. What must the diameter of the driving pulley be ? 4. A countershaft with a 12" pulley runs 450 revolutions per minute ; the revolutions of the main shaft are 180. What size pulley must be used on the main shaft ? 5. A main line runs 189 revolutions per minute and the countershaft pulley is 10" in diameter and runs 385 revolutions per minute. What is the size of the pulley on the main line ? 6. A pulley 36" in diameter running 168 revolutions per minute is to drive a shaft 212 revolutions per minute. What must be the diameter of the pulley on the driven shaft ? 156 VOCATIONAL MATHEMATICS 7. A driving shaft runs 184 revolutions per minute; the driven pulley is 12" in diameter and is to run 350 revolutions per minute. What must be the diameter of the driving pulley ? 8. What is the speed of the main shaft if the pulley is 12" and the revolutions per minute on the other shaft are 228 with a pulley 8" ? 9. What is the speed of a countershaft if the pulley is 11" and the main shaft runs 196 revolutions per minute with a 14" pulley ? • 10. A 14' flywheel running 98 revolutions per minute drives a 9" pulley. What is the speed of the pulley ? Countershafts or Jackshafts The first shaft belted off from a flywheel is often called a jackshaft. In the figure below the jackshaft carries the pulley d ; on the main line is the large pulley D and the small pulley F on the jackshaft. On another main line is the pulley /. Pulley B is the first driver and by means of pulley F on the jackshaft drives pulley d. Pulley d is the second driver. Pulley F is the first driven and pulley /is the second driven pulley. D Main Line Jackshaft Example. — The main shaft runs 160 R. P. M. (revolations per minute) ; D is 60" in diameter and drives F 140 E,. P. M. What is the diameter of 2^? The second main / is to run 186 R. P. M. What should be the diameter of/ if the 72" driver d is running 140 R. P, M, ? SHAFTS AND PULLEYS 157 An examination of the data in connection with D and / in the figure above will show that the relative speed,of /to D is XxDxd=^N'xFxf where N= number of revolutions of driver and N' = number of revolutions of driven That is, the continued product of the speed of the first driver and the diameters of all the drivers is equal to the continued product of the speed of the last driven by the diameters of all the driven pulleys. In this combination of driving f hy D there are six quantities, any of which can be found when we know the other five, by figuring from one shaft to the next step by step. EXAMPLES 1. The R.P. M. of a 26" driving pulley is 270. What are the revolutions per minute of an 18" driven pulley ? 2. If there is a shaft with a speed of 270 R. P. M. upon which there is a 26" pulley driving a 16" pulley, and on the shaft with the 16" pulley is a 30" pulley driving an 18" pulley, what is the speed of the shaft which carries the 18" pulley ? 3. A flywheel which is 30 feet in diameter drives a coun- tershaft by means of a pulley 6 feet in diameter ; the flywheel makes 50 R. P. M. What size of pulley must be used on the countershaft to give 300 R. P. M. to a pulley 2 feet in diameter? 4. If 3 flywheels, respectively 13', lOf , and 9' di- ameter have the same circumferential speed of 2750' per minute, how many revolutions per minute does each make ? 270 158 VOCATIONAL MATHEMATICS 5. A flywheel which is 40 feet in diameter drives a counter- shaft by means of a pulley 6^ feet in diameter ; the flywheel makes 54 R. P. M. What size of pulley must be used on the countershaft to give 341 R. P. M. to a pulley 3 feet in diameter? 6. The size of a main driving pulley is 20 feet in diameter on a shaft with a speed of 70 K. P. M., driving a pulley 4 feet in diameter and two other pulleys 5.6 and G.36 feet in diameter, respectively. What is the speed of each shaft ? CHAPTER X GEARING In a machine shop power is transmitted from one part of a machine to another part by means of tooth-shaped interlocking wheels. The train of toothed wheels for transmitting motion is called gearing. Gears are nothing but pulleys with teeth, and are made to drive one another by the teeth coming in contact with each other. If a small gear drives a larger gear, the larger gear will go more slowly than the smaller ; that is, the larger gear will make fewer turns in a minute. Just the reverse is true if a larger gear drives a smaller one. The num- ber of revolu- tions which a gear makes is always pro- portional to the number of its teeth. As in pul- leys, there are the driver and the driven gears. The driver may be dis- tinguished from the driven by examining the gears and notic- ing that it is the gear that is bright or worn on the front of the tooth — that is, the side of the wheel moving. The driven wheel is worn on the side away from the direction of the motion. Bkvel Gear Spur 6bab 160 VOCATIONAL MATHEMATICS In order to express the relation between the driver and driven gears it is necessary to use abbreviations. Let Z) = number of teeth in driver D' = number of teeth in driven N= number of revohitions of the driver N" = number of revolutions of the driven D:D' :: N': ^Vor DN= D'N' That is, the product of the teeth in the driver by its revolutions equals the tooth transits of the driver, which in turn equal the tooth transits of the driven or follower. If any three of these quantities are known, the fourth can be found. Example. — How many revolutions does a 12-tooth follower make to five revolutions of a 24-tooth driver ? 12iV^=24x6 iV=10 Ans. EXAMPLES 1. A driver has 98 teeth and its follower 42. How many revolutions will the follower make to one revolution of the driver ? 2. In Example 1 how many revolutions of the driver will drive the follower one revolution ? 3. How many teeth must a follower have in order to make three revolutions while a 96-tooth driver makes one ? 4. How many teeth must a gear have to revolve 16 times while a 60-tooth mate revolves 12 times ? 5. A 96-tooth gear drives a 48-tooth gear. What is the ratio of their speeds ? 6. A 48-tooth gear drives a 120-tooth gear. What is the ratio of their speeds ? 7. Two shafts are connected by gears, one of which turns 55 times a minute and the other 11 times a minute. If the small gear has 32 teeth, how many teeth has the larger gear ? GEARING 161 Pitch In order to solve problems connected with the use of gears it is necessary to know the different terms used in connection with gearing. The pitch circle is shown in the illustration and is the circle which runs around the teeth. It is the same size as the fric- Gear Teethe L Pitch r^ lAat Diameter tion rollers or cylinders would be if no teeth were there. When two spur gears roll together their pitch circles are con- sidered to be always in contact. The jy^tch diameter is the diameter of the pitch circle. The word diameter when applied to gears always means the pitch diameter. The circular pitch is the distance measured on the pitch circle from the center line of one tooth to the center of the next. This is illustrated in the diagram. In solving gearing problems the circular pitch is not nearly so important as the diametral pitch. If the distance from the center of a tooth to the center of tlie next tooth is i", the gear is I" circular pitch. The diametral pitch is the number of teeth for each inch of pitch diameter. For example, if a gear has thirty teeth and the pitch diameter is three inches, then the diametral pitch is .SO -=- .3 = 10, or a 10 diametral pitch gear. Using P for the diametral pitch, D for the pitch diameter, and ^V for the number of teeth, we liave a formula P = N ^ D. 162 VOCATIONAL MATHEMATICS To find the thickness of tooth at the pitch line when the diametral pitch is given, divide 1.57 by the diametral pitch. For example, if the diametral pitch is 3, divide 1.57 by 3, and the quo- tient, which is .523 inches, is the thickness of the tooth. To find the circular pitch when the diametral pitch is given divide 3,1416 by the diametral pitch. For example, if the diametral pitch is 4, divide 3.1416 by 4 and the quotient, which is .7854, is the circular pitch. Diametral pitch is found when the circular pitch is given by dividing 3.1416 by the circular pitch. Since the circumference of any circle is equal to 3.1416 times its diameter, every inch of diameter of any circle is equal to 3.1416 inches of circum- ference, or in this instance, for every inch of pitch diameter we have 3.1416 inches of circumference measured on the pitch circle. But the diametral pitch is by definition the num- ber of teeth for each inch of pitch diameter, and by. the above reasoning we can also say that the diametral pitch is the num- ber of teeth for each 3.1416 inches of circumference of pitch circle. Since the circular pitch is the distance from the center of one tooth to the center of the next, it will also be equal to 3.1416 of circumference of pitch circle divided by the number of teeth in that 3.1416. But tlie number of teeth in 3.1416 is equal to the diametral pitch, and therefore the distance from the center of one tooth to the center of the next, or the circular pitch, is equal to 3.1416 divided by the diametral pitch. Therefore, using P for circular pitch and P for diametral pitch, P = 3.1416 - F or P' = 5:111^. EXAMPLES 1. What is the diametral pitch of a gear liaviug (a) 56 teeth and a pitch diameter of 8" ? [b) 60 teeth and a pitch diameter of 12" ? GEARING 163 2. What is the thickness of the tooth of a gear having diametral pitch of (a) 8? (6) 4? (c) 14? 3. What is the eirciiUir pitch if the diametral pitch of geAris(a) 8? (6) 10? (c) 5? 4. What is the diametral pitch if the circular pitch of gear is (a) 1.5708? (b) .5230? (c) .2618*^ ((J) .7854? To find the pitch di- ameter when the number of teeth and the diame- tral pitch aregiven, divide the number of teeth by the diametral pitch. For example, if the number of teeth is 40 and the diametral pitch is 4, divide 40 by 4, and the quotient, which is 10", is the pitch di- ameter. To make a gear the outside diameter must first be known. If the diametral pitch and the number of teeth are given, the outside di- ameter may be found by the following rule: D = N-\-2 where D = outside di- ameter, N= number of teeth, and P= diametral pitch. 164 VOCATIONAL MATHEMATICS For example, if a gear has 24 teeth and its diametral pitch is 2, the outside diameter will be 13"; O = ^£+2 ^ 26 ^ ^g,, ^^^ ^ :^±2. 2 2' To find the diame^raZj[)iYc7i when the number of teeth and the diameter of the blank are given, add 2 to the number of teeth and divide by the diameter of the blank. For example, if the number of teeth is 40 and the diameter of the blank is 10^", add 2 to the number of teeth, making 42, and divide by 10^, and the quotient, which is 4, is the diametral pitch. To find the thickness of tooth at the pitch line when the cir- cular pitch is given, divide the circular pitch by 2. For example, if the circular pitch is 1.047", divide this by 2, and the quotient, which is .523, is the thickness of tooth. To find the number. of teeth when the pitch diameter and the diametral pitch ai-e given, multiply the pitch diameter by the diametral pitch. For example, if the pitch diameter is 10" and the diametral pitch is 4, multiply 10 by 4, and the product 40 will be the number of teeth in the gear. To find the whole depth of tooth divide 2.157 by the diametral pitch. For example, if the diametral pitch is 6, divide 2.157 by 6, and the quotient, .3595", is the whole depth of the tooth. Formulas for Determining the Dimensions of Gears bt Diametral Pitch P = the diametral pitchy or the number of teeth to one inch of diameter of pitch circle. D' = the diameter of pitch circle. D = the whole (outside) diameter or the diameter of the blank. JV = the number of teeth. V= the velocity. t = the thickness of tooth or cutter on pitch circle. D" = the working depth of tooth. GEAKING 165 / = the amount added to depth of tooth for rounding the corners and for clearance. D" 4-/ = the whole depth of tooth. r denotes the constant 3.1416. P' denotes the circular pitch or the distance from the center of one tooth to the center of the next on the pitch circle. FormuUm KjrnmpleH ' ^±1 = 71+^, or T?+_2_ 12. 6.166 ' 6fV = 1:^^ = 12. 6 6.166 X 72 ^g 72 + 2 = ^ = 6. 12 = 12 X 6 = 72. = 12 X 6.166 - 2, or 12 X 6j-\ - 2 = 72. 72 + 2 D P-- N = — D' DxX N-^2 D' N P X. = PD^ X. = PD-2 n. _.V+2 P 12 rr = 6 166, or (jj\. D=D'-h- = 6 + :p^,or 6 + .166=6.166. ^ = 1:51 = M[ = .130. • P 12 9 O O f = — = :^ = .013. 10 10 />"+/ = .166 + .013 = .179. rw TT 3.1416 nf,n ^ = P = -12- = •2^2- 3.1416 P = ^ = ^•"" :^ = 12. F .262 1 The examples placed opposite the formulas above are for a single gear of 12 pitch, 6.166" or Gft" diameter. 166 VOCATIONAL MATHEMATICS EXAMPLES Solve by using the above fonnulas and the rules on pages 161-165 : 1. If the number of teeth of a gear is 46 and the diametral pitch is 8, what will the pitch diameter be ? 2. If a gear is to have 54 teeth and the diametral pitch is 6, what will be the pitch diameter ? 3. If a gear has 38 teeth and the diametral pitch is 6, what is the pitch diameter ? 4. If the pitch diameter of a gear is 6" and the diametral pitch is 8, what will be the number of teeth? 5. How many teeth has a gear if the pitch diameter is 7f " and the diametral pitch is 6 ? 6. How many teeth has a gear if the pitch diameter is 8J" and the diametral pitch is 4? 7. If the pitch diameter of a gear is 12" and the diametral pitch 7, how many teeth has the gear ? 8. If the diametral pitch is 16 and the pitch diameter of a gear is 6J", how many teeth has the gear ? 9. What is the whole depth of the tooth if the diametral pitch of a gear is (a) 8? . (6) 4 ? (c) 7 ? (d) 10? 10. If a gear has 48 teeth and the diameter of the blank is 6y, find the diametral pitch. 11. If a gear has 61 teeth and the diameter of the blank is 101", find the diametral pitch. 12. If a gear has 35 teeth and the diameter of the blank is Syy, find the diametral pitch. 13. If a gear has 78 teeth and the diameter of the blank is 5", find the diametral pitch. 14. If a gear has 54 teeth and the diameter of the blank is 6|", find the diametral pitch. 15. What is the thickness of the tooth if the circular pitch of a gear is (a) .1848" ? (b) .1428" ? (c) 6.2832" ? (d) 2" ? GEARING 167 If one examines closely the movements of two gears on a machine, one will notice that all the teeth pass a given point at every revolution of the wheel. That is, the teeth marked Tty on the gear f^J lyO shown in the illustration pass an out- ^ ^) side point like P in making one com- 2'— CT 7^^ plete revolution. So when a tooth and (\ /O— * the space adjacent to it have passed a (j^ r\j given point like P, one transit of a tooth has occurred. There are as many tooth transits at every complete revolution of a gear as there are teeth in the gear. If we multiply the number of teeth by the number of revolutions, the product will be the number of transits. If this product (number of transits) be divided by the number of teeth, the quotient will be the revolutions of the gear. For example, if a 12-tooth gear makes CO transits, then it has made (^J = 5) five revolutions. Two shafts, D and Fy are to be connected by gears so that shaft D will make one revolution while shaft F makes two. To do this a gear must be put on shaft D having twice the number of teeth that the gear on shaft F has. If a gear hav- ing 24 teeth is put on shaft D, the gear on shaft F will have half as many. Each time the gear on D turns around once the 168 VOCATIONAL MATHEMATICS gear on F will turn twice ; that is, the 24 teeth on gear D will have to turn gear F twice in order to mesh with the 24 teeth on F. The relation or ratio of the speed of F to the speed of D is 2 to 1. This is called the ratio of the gearing. The ratios between the speeds and number of teeth can be written in the form of a proportion : 24:12:: 2:1 A^nd the number of teeth on gear D is to the number of teeth on gear F as the speed of F is to the speed of Z>. Train of Gears When two gears mesh, as in the illustration on page 167, one revolves in the opposite direction from the other. Three or more gears running together, as in the illustration above, are often called a train of gears. In a train of spur gears like these, one gear, J, which is called an intermediate gear, meshes with the other two gears, D and F, and causes the revolutions of both Z) and F to be made in one direction, while the inter- mediate revolves in the opposite direction. The intermediate does not change the relative speeds of D and F, so that they can be figured as explained on page 167. An intermediate gear is also called an idler. GEARING 169 We may calculate the nuinbsr of revolutions of any follower for any number of revolutions of a driver, in a train of gears, step by step. To find the number of revolutions of the last follower when the number of revolutions of the first driver and the teeth in all the gears are known, take the continued product of the revolutions of the first driver and all the driv- ing gears, and divide it by the continued product of all the followers ; the quotient is the number of revolutions of the last follower. In other words, the product of the revolutions of the first driver and the teeth of all the driving gears is equal to the continued product of the revolutions of the last follower and the teeth of all the driven gears. EXAMPLES 1. If a 60-tooth wheel is to mesh with one having 46 teeth and the 60-tooth gear makes 25 revolutions per minute, how many will the 46-tooth gear make ? 2. A 168-tooth gear drives a 28-tooth gear. What is the ratio of the gearing ? 3. What would be the ratio of the gearing in the example above if the 28-tooth gear were the driver ? 4. If the 28-tooth gear is making 48 revolutions per min- ute, how many revolutions per minute is the 168-tooth gear making ? 5. How could the gearing be changed to make the 28-tooth gear turn in the same direction as the 168-tooth gear ? 6. Two gears running together have a speed ratio of 7 to 1. If the smaller one turns 14 times, how many times will the larger one turn ? 7. If a 144-tooth gear makes one complete turn, how many turns will a 32-tooth gear make working with it ? 8. In the above example if the 32-tooth gear turned once, how many turns will the 144-tooth gear make ? 170 VOCATIONAL MATHEMATICS 9. A train of 3 gears has 69, 30, and 74 teeth. If the 69- tooth gear makes 100 revolutions per minute, how many revolutions per minute will the 74-tooth gear make ? Make a sketch showing the direction in which each gear turns. 10. A train of gears is made up of 6 gears having teeth as follows: 46, 60, 32, 72, 56, and 48; while the first gear in the train makes 10 turns, how many turns will the last gear make ? 11. What two gears will give a ratio of speeds so that the driver will make if as many turns as the follower ; that is, while the driver makes 13 turns the follower will make 14 ? 12. If 24 gears work in a train, in what direction will the last one turn if the first turns right-handed ? REVIEW EXAMPLES 1. Two gears working together have pitch circles 14^' and 26f " in diameter. What is the distance between their centers ? The distance between centers may be obtained from the following formulas : D' -\-d' h «= — ^r — '•<''' = 7r^ 2 2P where a — distance between centers d' = diam. of pitch circle of smaller gear h = number of teeth on both wheels 2. Two gears in mesh have pitch circles 17.082" and 31.3124" in diameter. What is their center distance? 3. Two gears working together are four pitch, the larger has 36 teeth and the smaller has 14 teeth. What is their center distance ? 4. The circular pitch of a planer table gear is 1". What should be the total depth to cut the tooth space ? 5. The circular pitch of a milling machine gear is |^" and it has 96 teeth. What is its outside diameter ? 6. A flue rattler gear is to be renewed. Its outside diam- eter is 34" and it has 100 teeth. What size shall we draw its pitch circle ? GEARING 171 7. Two gears mesh together, one has 28 teeth and an outside diameter of 2|" and the other 68 teeth and an outside diameter of 5|". Find the center distance. 8. A gear has a circular pitcli of |f ". What is the thick- ness of the tooth at the pitch line ? 9. A pitch circle is 18" in circumference. If the teeth are I" thick, what is the circular pitch ? 10. A gear is 10 pitch. What is the total depth of its teeth ? 11. The diametral pitch of a gear is 4. What is the cir- cular pitch ? 12. The circular pitch of a gear is .157. What is the di- ametral pitch ? 13. A gear is 12 pitch. What is its circular pitch ? 14. A gear is 20 pitch. If it has 105 teeth, what is its out- side diameter ? 15. A gear of 10 pitch has 44 teeth. Find the pitch diameter. 16. A gear has 28 teeth and a pitch diameter of 8. What is the pitch ? 17. A gear is set in a milling machine ready to cut the teeth, and if the pitch is to be one, what is the depth of the cut? la A gear is 14 pitch and has an outside diameter of 4|". Find the depth of cut for the teeth. 19. A gear has 75 teeth and is 18 pitch. What is the work- ing depth of the tooth ? 20. A pinion has 44 teeth and is 10 pitch. What is the out- side diameter ? 21. A lathe back gear has 108 teeth and is 5 pitch. What is its outside diameter ? PART VI — PLUMBING AND HYDRAULICS CHAPTER XI RECTANGULAR AND CYLINDRICAL TANKS Water Supply. — The question of the water supply of a city or a town is very important. Water is usually obtained from lakes and rivers which drain the surrounding country. If a lake is located in a section of the surrounding country higher than the city (which is often located in a valley), the water may be obtained from the lake, and the pressure of the water in the lake may be sufficient to force it through the pipes 'm&WW' Water Supply into the houses. But in most cases a reservoir is built at an elevation as high as the highest portion of the town or city, and the water is pumped into it. Since the reservoir is as high as the highest point of the town, the water will flow from it to any part of the town. If houses are built on the same hill with the reservoir, a stand-pipe, which is a steel tank, is erected on this hill and the water is pumped into it. Water is conveyed from the reservoir to the house by means of iron pipes of various sizes. It is distributed to the different parts of the house by small lead, iron, or brass pipes. Since water exerts considerable pres- sure, it is necessary to know how to calculate the exact pressure in order to have pipes of proper size and strength. 172 PLUMBING AND HYDRAULICS 173 EXAMPLES 1. Water is measured by means of a meter. If a water meter measures for tive hours 7()0 cubic feet, how many gal- lons does it indicate ? Note. — 231 cubic inches = 1 gallon. 2. If a water meter registered 1845 cubic feet for 3 days, how many gallon? were used ? 3. A tank holds exactly 12,852 gallons ; what is the capacity of the tank in cubic feet ? 4. A tank holds 3841 gallons and measures 4 feet square on the bottom ; how high is the tank ? Rectangular Tanks. — To find the contents in gallons of a square or rectangular tank, multiply together the length, breadth, and height in feet ; multiply the result by 7.48. I = length of tank in feet b = breadth of tank in feet h = height of tank in feet Contents = Ibh cubic feet x 7.48 = 7.48 Ibh gallons (Note. — 1 cu. ft. = 7.48 gallons.) If the dimensions of the tank are in inches, multiply the length, breadth, and height together, and the re.sult by .004829. 5. Find the contents in gallons of a rectangular tank having in- side dimensions (a) 12' x 8' x 8'; (b) 15" x 11" X 6" ; (c) 3' 4" X2'8"x8"; (d) 5'8"x4'3"x3'5"; (e) 3' 8" x 3' 9" x 2' 5". Cylindrical Tank. — To find the contents of a cylin- drical tank, square the diameter in inches, multiply this by the height in inches, and the result by .0034. d = diameter of cylinder h = height of cylinder Contents = d^h cubic inches x .0034 = d^h .0034 gallons 6. Find the capacity in gallons of a cylindrical tank (a) 14" in diameter and 8' high; (b) (>" in diameter and 5' high; 174 VOCATIONAL MATHEMATICS (c) 15" in diameter and 4' high; (d) 1' 8" in diameter and 5' 4" high; (e) 2' 2" in diameter and 6' V high. Inside Area of Tanks. — To find the area, for lining purposes, of a square or rectangular tank, add together the widths of the four sides of the tank, and multiply the result by the height. Then add to the above the area of the bottom. Since the top is usually open, we do not line it. In the following problems find the area of the sides and bottom. 7. Find the amount of zinc necessary to 'line a tank whose inside dimensions are 2' 4"x 10" x 10". 8. Find the amount of copper necessary to line a tank whose inside dimensions are 1'9" x 11" x 10", no allowance made for overlapping. 9. Find the amount of copper necessary to line a tank whose inside dimensions are 3' 4" x 1' 2" x 11", no allowance for overlapping. 10. Find the amount of zinc necessary to line a tank 2' 11" X 1' 4" X 10". Drainage Pipes In order to have pipes of a proper size to carry away the water from the roof of a building, it is necessary to know the number of gallons of water that will vbe drained. To find the number of gallons that will drain from a roof in a month, multiply the number of square feet of the roof by the average number of inches of rainfall per month, and multiply this prod- uct by .623. When the roof is not flat, or very nearly so, its area should be considered as the area which it actually covers. ^ EXAMPLES 11. A roof is 48' by 62'. How many gallons of water will drain from it in a month if the rainfall is 6" ? 12. A roof is 29' by 74'. How many gallons of water will drain from it if the rainfall is 4^^" ? PLUMBING AND HYDRAULICS 175 Note. — The rainfall per month in Massachusettjs varies from J" to 12'^ In calculating the following problems consider 10" rainfall as the average amount. 13. Find the number of gallons that will drain from a flat roof (a) 112' by 64'; (6) 88' by 49'; (c) 120' by 80'. 14. Find the number of gallons that will drain from a slant- ing roof each half of which is (a) 52' by 34' ; {b) 49' by 28' ; (c) 112' by 54'; (d) 57' by 33'; (e) 54' by 31'. Weight of Lead Pipe Pipes, particularly those of lead, are sold by weight, aiid this depends upon the diameter and thickness of the metal in the pipe. To find the weight of a length of pipe, subtract the square of the inner diameter in inches from the square of the outer diameter in inches, and multiply the remainder by the weight of 12 cylindrical inches. This product multiplied by the length in feet gives the required weight. Example. — What is the weight of 1450 feet of lead pipe, the outer diameter being J" and the inner diameter being ^"? D = outer diameter d = inner diameter I = leiijith of pipe 3.8698 lb. = weight of 12 cylindrical inches of lead pipe 1' long and 1" in diameter. Weight in lb. = (Z>- — d-) x 3.8698 x I Then ^^ = LJLL - J^ (sciuare of outer diameter) (8)2 8 X 8 64 ^ * - ^ VJ « = ^ ^ ^ = — ^ (square of inner diameter) (16)2 16 X 16 256 ^ ^ ^ 15 _ i5. = iL = .5742 (difference) 64 266 256 ^ ^ .6742 X 3.8698 x 1450 = 3221.94 lb. EXAMPLES 1. What is the weight of 364' of lead pipe, outer diameter }", inner diameter ^y ? 176 VOCATIONAL MATHEMATICS 2. What is the weight of 1189' of lead pipe, outer diameter If", inner diameter ly\" ? 3. What is the weight of 2189' of lead pipe, outer diameter 2f' ', inner diameter 2^'' ? 4. What is the weight of 112' of lead pipe, outer diameter 3", inner diameter 2^%" ? 5. What is the weight of 212' of lead pipe, outer diameter 3i", inner diameter 3^^" ? Capacity of Pipe Rules for Finding the Capacity in Gallons of a Foot of Pipe of any Diameter : 1. Find the cubical contents in inches, and divide by 231, C=D2x.7854x 12-- 231 2. Multiply the square of the inside diameter by .0408. C=D^X .0408 Rules for Findiyig the Capacity of a Pipe of any Length and any Diameter : 1. Multiply the number of gallons in a foot of the pipe by the number of feet in the length of the pipe. 0=Z)2 X .0408 X L 2. Find the cubical contents in inches, and divide by 231. ,(7=7>2 X .7854 X i.^231 Note. — In the above formulas, let C = the capacity of the pipe in gallons, D = the diameter in inches, and L = the length in feet. EXAMPLES 1. What is the capacity of a pipe having an inside diameter of (a) J inch, 16 feet in length ; (b) 1} inches, 20 feet in length ; (c) 1.^ inches, 1 foot in length; (d) 3 J inches, 1 foot in length; (e) 14 inches, 10 feet in length ; (/) 5 inches^ 22 feet in length ? PLUMBING AND HYDRAULICS 177 Number of U. S. Gallons in Round Tanks for onb foot in defth DiA. or Tanks Ft. 1«. No. U. S. Gals. CiTBic Ft. AND Area in 8q. Ft DlA. OF Tanks Ft. In. No. U.S. Gals. Cub in Ft. ANo Area IN 9«i. Ft. DlA. Of Tanks Fr. In. No. U.S. Gals. Cdhio Ft. ANi» Arka iH 8m. Ft. 1 5.87 .7S5 3 52.88 7.0 , O r 0° — "" — 17.8° c 1 II Thermometers 90 32° Example. — Convert 212 degrees F. to C. reading. 5f212°-32°) ^ 5(180°) ^ 900^ ^ j^^o q 9 9 9' Example. — Convert 100 degrees C. to F. reading. ^ ^ ^^^° + 32° = ^^ + 32° = 180° + 32° = 212° F. 6 6 HEAT 197 If the teinperatiire is below the freeziuj^ point, it is usually written with a minus sign before it : thus, 15 degrees below the freezing point is written — 15**. In changing — 15° ( -. into F. we must bear in mind the minus sign. Thus, F =— + 32^. F = ~^'^° ^ ^ + 32" = - 27-^ + 32° = 6*^' o 5 Example. — Change - 22° F. to C. C. = § (F. - 32°) C. = I (- 22° - 82°) = |( - 64°) = - 30° EXAMPLES 1. Change 36° F. to C. 6. Change 225° C. to F. 2. Change 89° F. to C. 7. Change 380° C. to F. 3. Change 289° F. to C. 8. Change 415° C. to F. 4. Change 350° F. to C. 9. Change 580° C. to F. 5. Change 119° C. to F. Value of Coal in Producing Heat ^ There "are different kinds of coal on the market. Some grades of the same coal give off more heat than others in burning. The heating value of a coal may be found in three ways : (1) By chemical analysis to determine the amount of carbon, (2) by burning a definite amount in a calorimeter and noting the rise of temperature of the water, (3) by actual trial in a steam boiler. The first two methods give a theoreti- cal value, the third gives the real result under the actual con- ditions of draft, heating surface, combustion, etc. EXAMPLES 1. If 125 lb. of ashes are produced from one ton of coal, what is the percentage of ashes in that coal ? 198 VOCATIONAL MATHEMATICS 2. Twelve tons of coal are burned per day, and twenty -two baskets of ashes, each weighing 65 lb., are removed; what per- centage of ashes does the coal contain ? 3. If twelve tons of coal are burned per day, and 1450 lb. of ashes are produced, what percentage of ashes does the coal contain ? 4. A quantity of coal is built into a rectangular stack 50 ft. long, 25 ft. broad, and 6 ft. high ; what is the weight of the coal, allowing 45 cubic feet per ton ? 5. If 92,400 lb. of coal are consumed in 60 hours and the engines regularly develop 480 1. H. P. (Indicated Horse Power), how much coal is consumed per H. P. per hour ? 6. With the price of coal at $3.25 per ton of 2000 lb., and the power produced earning a profit of 25 % on the cost of production, what would be the amount of profit in Ex. 5 when running full power ? Latent Heat By latent heat of water is meant that heat which water ab- sorbs or discharges in passing from the liquid to the gaseous, or liquid to solid state, without affecting its own temperature. Thus, the temperature of boiling water at atmospheric pressure never rises above 212 degrees F., because the steam absorbs the excess of heat which is necessary for its gaseous state. Latent heat of steam is the quantity of heat necessary to convert a pound of water into steam of the same temperature as the steam in question. To find the latent heat of steam, subtract ten times the square root of the gauge pressure from 977. Example. — Find the latent heat of steam at 169 lb. gauge pressure. Vim = 13 13 X 10 = 130 977 - 180 = 847 B. T. U. HEAT 199 EXAMPLES Find the latent heat of steam at the following gauge pressures : 1. 132 lb. gauge pressure. 6. 39 lb. gauge pressure. 2. 116 lb. gauge pressure. 7. 41 lb. gauge pressure. 3. 208 lb. gauge pressure. 8. 160 lb. gauge pressure. 4. 196 lb. gauge pressure. 9. 159 lb. gauge pressure. 5. 84 lb. gauge pressure. 10. 180J^ lb. gauge pressure. Heat Units Required to Produce a Given Pressure To lind the number of units of heat required to raise the temperature corresponding to one gauge pressure to that of another, find the square roots of the gauge pressures, subtract these values, and multiply by 14J. Example. — Find the number of heat units required to raise the temperature of 64 pounds gauge pressure to 169 pounds gauge pressure. \/(l4 = 8 13-8 = 6 14^x6 = 711 Vl«9 = 13 Approximately 72 B. T. U. Am. EXAMPLES Find the number of heat units required to raise the tempera- ture between the following gauges: 1. From 64 to 128 lb. gauge. 5. From 42 to 121 lb. gauge. 2. From 26 to 131 lb. gauge. 6. From 28 to 132 lb. gauge. 3. From 39 to 149 lb. gauge. 7. From 33 to 144 lb. gauge. 4. From 49 to 165 lb. gauge. 8. From 55 to 164 lb. gauge. Volume of Water and Steam » According to steam tables one cubic foot of steam at 100 pounds' pressure weighs 0.2307 lb., one cubic foot of water weighs 62^ lb., and one gallon of water may be taken as 8^ lb. 200 VOCATIONAL MATHEMATICS At atmospheric pressure one cubic foot of steam has nearly the weight of one cubic inch of water, and the weight increases very nearly as the pressure. Hence, to find the number of cubic inches of water required to make a certain amount of steam, multiply the number of cubic feet of steam by the absolute pressure in the atmosphere ; the product is the number of cubic inches of water required. In all such calculations for practical purposes, a liberal allowance must be made for loss and leakage. Absolute pressure is the total pressure, or the gauge pressure plus the atmospheric pressure (which at sea level is 14.7 lb. per sq. in.). EXAMPLES 1. How much water will it take to make 800 cu. ft. of steam at 10 lb. pressure ? 2. How much water will it take to make 3020 cu. ft. of steam at 65 lb. pressure ? 3. How much water will it take to make 4812 cu. ft. of steam at 8 lb. pressure ? 4. How much water will it take to make 512 cu. ft. of steam at 75 lb. pressure ? 5. How much water will it take to make 1213 cu. ft. of steam at 80 lb. pressure ? Solve the following problems according to the table on the next page : 6. What is the total steam pressure if the steam gauge reads 55 lb. ? 7. How many cubic feet of steam from 2 lb. of water at a steam gauge pressure of 65 lb. ? 8. ,What is the latent heat of 1 lb. of water at a total pressure of 75 lb. at 307.5° F. ? 9. What is the total heat required to generate 1 lb. of steam from water at 32° F. under total pressure of 90 lb. ? HEAT 201 Properties of Saturated Steam Pbbmdeb VOLUMB Total Heat Tempera- ture In Fahrenbeit Dejfrees Latent Heat in B. T. U. required to XT ToUl Compared with Water (^Hblc Ft. of Steam fi*»ttn 1 Lb. of peneralc 1 Lb. of Steum from Water at «tio under Conhtant Water I're.smire Heat unltH 16 212.0 nm 26.36 966.2 1146.1 5 20 228.0 1220 10.72 952.8 1150.9 10 25 240.1 9JK5 15.99 945.3 1154.6 16 30 260.4 838 18.40 937.9 1157.8 20 36 269.3 726 11.65 931.6 1160.5 26 40 267.3 640 10.27 926.0 1162.9 30 46 274.4 672 9.18 920.9 1165.1 86 60 281.0 618 8.31 910.3 1167.1 40 66 287.1 474 7.61 912.0 1169.0 46 60 292.7 437 7.01 908.0 1170.7 60 65 298.0 405 6.49 904.2 1172.3 66 70 302.9 378 6.07 900.8 1173.8 60 76 307.5 353 5.68 897.5 1175.2 66 80 312.0 333 5.35 894.3 1176.6 70 86 316.1 314 5.05 891.4 1177.9 76 90 320.2 298 4.79 888.5 1179.1 80 95 324.1 283 4.55 885.8 1180.3 86 100 327.9 270 4.33 888.1 1181.4 00 106 331.3 267 4.14 880.7 1182.4 96 110 334.6 247 3.97 878.3 1183.6 100 116 338.0 237 3.80 875.9 1184.5 110 125 344.2 219 3.50 871.5 1186.4 120 135 350.1 203 3.27 867.4 1188.2 130 146 355.6 liK) 3.06 863.5 1189.9 140 165 361.0 179 2.87 859.7 1191.6 160 166 366.0 169 2.71 866.2 1192.9 Steam Heating A steam heating system with steam having a pressure of less than 15 lb. by the gauge is called a low pressure system. If the steam pressure is higher than 15 lb. it is called a high pressure system. When the water of condensation flows back to the boiler by gravity alone, the apparatus is known as a gravity circulating system. When the boiler is run at a high pressure and the heating system at a low pressure, the con- densed steam must be returned to the boiler by a pump,. steam return trap, or injector. 202 VOCATIONAL MATHEMATICS The quantity of heat given off by the radiators of steam pipes, in the ordinary methods of heating buildings by direct radiation, varies from If to 3 heat units per hour per square foot of radiating surface for each degree of difference in temperature ; an average of from 2 to 2i is a fair estimate. One pound of steam at about atmosplieric pressure contains 1146 heat units, and if the temperature in the room is to be kept at 70"^ F., while the temperature of the pipes is 212 degrees, the difference in temperature is 142 degrees. Multiplying this by 2J, the emission of heat will be 319^ heat units per hour per square foot of radiating surface. A rule often given is to allow one square foot of heating surface in the boiler for every eight to ten square feet of radiating surface. In steam heating the following rule is used: To find the amount of direct radiating surface required to heat a room, basing the calculation upon its cubic contents, allow one square foot of direct radiating surface for the cubic feet shown in the following table. Proportion of Radiating Surface to Volume of Rooms Bathrooms or living rooms, with 2 or 3 exposures . . . 40 cu. ft. Living rooms, with 1 or 2 exposures 50 cu. ft. Sleeping rooms 65-70 cu. ft. Halls 50-70 cu. ft. Schoolrooms 60-80 cu. ft. Large churches and auditoriums 65-100 cu. ft. Lofts, workshops, and factories 75-150 cu. ft. The above ratios will give reasonably good results on ordi- nary work if proper judgment is exercised. EXAMPLES 1. How much radiating surface is necessary to heat a bath- room containing 485 cu. ft. ? 2. How much radiating surface is necessary to heat a bath- room 10^' X 5i' X 9' ? 3. How much radiating surface is necessary to heat a living room with three windows, and containing 2798 cu. ft. ? 4. How much radiating surface is necessary to heat a living room 18' X 16|-' X 10', with three windows? CHAPTKK XIII BOILERS There are two classes of boilers — water tube and fire tube. The difference between the two is that water flows through water tube boilers and the fire to heat the water is outside, while in the fire tube boiler the conditions are reversed. Return Tubular Boilers. — The boiler most widely used in America is the return tubular, which is a type of fire tube boiler. It is a closed tube, simple in construction, inexpensive to make, and easy to clean and repair. The first horizontal tubular boiler was an ordinary storage tank made of iron. Now horizontal tubular boilers, 16 to 20 feet long, and 4 to 8 feet in diameter, and even larger, are used and can withstand a pressure of 150 lb. per sq. in. Boilers up to fourteen feet in length are constructed of two plates, each forming the entire circumference of the boiler. Above fourteen feet long the shell is constructed of three plates which make the required length of the boiler shell. These plates are J, f, ^, or ^ in. thick, having from 45,000 to 86,000 lb. tensile strength. Tensile Strength. — The tensile strength is the pull applied in the direction of its length required to break a bar of boiler plate one square inch in area. Different pieces are taken from the various parts of the boiler plate, reduced to \ inch square, and subjected to pressure on a testing machina The average strength of the samples is thus obtained and multi- plied by 16 to determine the strength of one square inch. The tensile strength is usually stamped on the boiler steel. If it is not stami)ed on it, the tensile strength is 48,000 lb. 203 § as 45 03 g ^ a mi en ts as pd c3 O N o W BOILERS 205 If four samples give tests of 3998 lb., 4001 lb., 4001 lb., and 4000 lb., then the avera«;e is 4000. Therefore, for one square inch the tensile strength is 4000 x 10 (the numlu'r of quartt-r sq. in. in one aq. in.) = 64,000 lb. O ^######JJ i SlNOLK RiVETKD LAP JoiVT ; ^ ErFirTEXCY ABOUT 56% o #--### #H Double Riveted Lap Joint; Efficiency about 70% Triple Riveted Butt-strap Joint; Efficiency about 85% Safe Working Pressure. — In order to know the safe working pressure of a single riveted boiler, it is necessary to multiply one sixth of the tensile strength by the thickness of the shell, and divide this product by the inside radius of the shell. If 1 The efficiency of a riveted joint id' the ratio of the strength of the joint to that of the solid plate. 206 VOCATIONAL MATHEMATICS a boiler is double riveted, add 20 per cent to the safe pressure of a single riveted boiler of the same dimensions. Example. — What is the safe working pressure of a single riveted boiler having 72" diameter, i" shell, if the boiler plate has a tensile strength of 66,000 lb. ? I X 66,000= 11,000 11,000 X .5" = 5500 5500 H- 36" (radius) = 152| lb. approx. Ans. EXAMPLES 1. (a) What is the safe working pressure of a single riveted boiler of 60" diameter, j%'' shell, the T. S. being 60,000 lb. per sq. in. ? (b) What is the safe working pressure of a double riveted boiler of the same dimensions? 2. (a) What is the safe working pressure of a single riveted boiler of 54'' diameter, f " shell, the T. S. being 60,000 lb. ? (6) What is the safe working pressure of a double riveted boiler of the same dimensions ? 3. (a) What is the safe working pressure of a single riveted boiler of 48" diameter, f " shell, the T. S. being 60,000 lb. ? (b) What is the safe working pressure of a double riveted boiler of the same dimensions? 4. (a) What is the safe working pressure of a single riveted boiler of 42" diameter, j\" shell, the T. S. being 60,000 lb. ? (6) Of a double riveted boiler of the above dimensions? The Plate. — The diameter of the rivet used for boiler plate is generally double the thickness of the plate. The distance between the rivet holes of a boiler is found by dividing the area of the rivet hole by the thickness of the plate. The pitch or distance between the rivet hole centers in a boiler plate is found by dividing the area of the rivet hole by the thick- ness of the plate and adding the diameter of one hole. The pitch in a rivet hole is found, when the shearing strength is 208 VOCATIONAL MATHEMATICS known, by multiplying the area of the rivet hole by the shear- ing strength, then multiplying the thickness of the plate by the tensile strength, dividing the first product by this product, and adding one rivet hole diameter to the quotient. Example. — What thickness of plate should be used on a 40-inch boiler to carry 125 lb. pressure, if the tensile strength of the plate is 60,000 lb. ? 125 = steam pressure 6 = factor of safety 40 = diameter of boiler 20 = i diameter of boiler 60,000 = tensile strength of plate 126 X 6 = 750 750 X 20 = 15,000 — 5 — = .25 or i" thickness of plate. Ans. 60,000 * ^ The Boiler Inspection Department of Massachusetts recom- mends the following fornuila for determining the thickness of boiler plate : rp^Px Rx F.S. T.S, X % T = thickness of plate P = pressure E = radius (| diameter of boiler) F. S. - safety factor T. S. = tensile strength % = strength of joint Example. — What thickness of plate should be used on a 40-inch boiler to carry 125 lb. pressure, if the strength of the plate is 60,000 lb., using 50 % as the strength of the joint? r = ^f><^Qxe = |// sheet. Ans. 60,000 X .50 ^ Find the safe working of the same boiler with the above figures : Safe working pressure = 60,000 x .5 x .50 ^ ^35 lb. Ans. 20 X 6 BOILERS 209 Small Vertical Boilkr Size. — The engineer often has to calculate the size of a boiler to carry a definite steam pressure. The size of a single riveted l)oiler may be found by multi- 210 VOCATIONAL MATHEMATICS plying J of the tensile strength by the thickness of the shell, and dividing this product by the steam pressure. The quo- tient is the radius of the boiler. Multiply this radius by two to obtain the diameter (twice this radius equals the theoretical diameter) ; add one fifth of the diameter just found as a safety factor, and this sum gives the working diameter of a boiler that will safely carry the required pressure. Example. — What is the diameter of a ^Ijoiler that will with- stand 150 lb. pressure, if made of f" steel, 60,0U0 lb. T. S.? I X 60,000 = 10,000 10,000- X f = 3750 -^■^-^Jt = 2.j" tlieoretical radius 50' = theoretical diameter 50" 4- 10" = 60" working diameter. A7is. The Boiler Inspection Department of Massachusetts recom- mends the following formula for finding the diameter of a boiler when the pressure, thickness, tensile strength, and per cent are known. rf^ Tx T.S.x % 2 P.F. D = diameter of boiler T = thickness of plate T. S. = tensile strength 0/, = strength of joint P = pressure F = safety factor Example. — What is the diameter of a boiler having |" shell, allowing 50 per cent for the strength of the joint, with a tensile strength of 60,000 lb., when the factor of safety is 6 and the pressure of steam is 125 lb. ? ^ .l>.4- 7854 226 VOCATIONAL MATHEMATICS Diameter of Supply Pipe. — The diameter of the steam supply- pipe for a given engine may be calculated from the H. P. of the engine by dividing the H. P. by 6, and extracting the square root of the quotient. hTp. 0=4 6 Example. — What is the diameter of a steam supply pipe of a 216 H. P. engine.? 216 - 6 = 36 \/36 = 6", diameter of supply pipe. Ans. EXAMPLES 1. Find the diameter of the steam supply pipe of a 180 H. P. engine. 2. What is the H. P. of an engine whose area of piston is 114 sq. in., mean effective pressure 80, and velocity of piston 112 ft. per minute ? 3. What is the approximate diameter of a cylinder of an engine of 50 H. P. ? 4. What is the approximate H. P. of an engine the cylinder diameter of which is 28" ? 5. What is the effective area, for power calculation, of the piston of a steam engine, the bore of the cylinder being 28", and the diameter of the piston rod which passes through both ends of the cylinder 2" ? 6. What is the H. P. of an engine that can raise 3 tons of coal (1 ton = 2240 lb.) from a mine 289 ft. deep in 9 minutes? 7. What is the approximate H. P. of an engine the cylinder diameter of which is 16" ? 8. Find the diameter of the steam supply pipe of a 24 H. P. ENGINES 227 9. What is the approximate diameter of a cylinder of an engine of 80 H. P. ? 10. What is the approximate H. P. of an engine the cylinder diameter of which is 20" ? 11. What is the H. P. of an engine whose diameter of piston is 15", mean effective pressure 110, velocity of piston per minute 189' ? 12. How many pounds of water per half minute can an 8 H. P. fire pump raise to a height of 86 ft. ? 13. What is the effective area in square inches of the piston of a steam engine if the diameter of the cylinder is 20" and the diameter of the piston rod is 3" ? 14. What is the horse power of an engine that is required to pump out a basement 51' x 22' x 10' deep, full of water, in 20 minutes ? 15. (a) Find the diameter of the steam supply pipe of a 98 H. P. engine. (6) Find .the approximate diameter of a cylinder of an eugine of 48 H. P. (c) What is the approximate H.P. of an engine the cylinder diameter of which is 28" ? Steam Indicator In order to know the condition of the steam within the cylinder of an engine an indicator is used. This consists of a small cylinder containing a piston, the rod of which is en- closed in a spiral spring which opposes the motion of the piston. The piston rod, after i)assing through the top of the cylinder cover, is connected with a long light lever, on the end of which is a pencil. This pencil moves in a vertical straight line when- ever the piston moves. Another cylinder with an axis parallel to the first carries a paper drum, and this drum is connected to the crosshead of the engine by means of a cord and a reducing motion, so that 228 VOCATIONAL MATHEMATICS the movement of the drum is proportional to that of the cross- head. When the pipe between the indicator and the engine is closed by means of a cock, the pencil, when held against the drum, makes a horizontal line called an atmospheric line. If Cross Section of Steam Indicator the cock is opened, admitting to the small cylinder of the in- dicator the pressure that exists in the engine cylinder, the pen- cil will trace a figure, every point of which is at a height from the atmospheric line proportional to the number of pounds' pressure in the engine cylinder at every point in the stroke. ENGINES 229 Operating Power. — To find the power required to drive a certain machine when driven direct from the shaft or engine : Indicate the engine with the machinery running and calcu- late from the card. Then indicate the engine (from formula on p. 225) without the machinery running and from this obtain the H. P. The difference will give the power for operating. To indicate an engine means to place a steam engine indicator on the steam engine and record on the diagrams the pressure in the steam cyl- inder at every part of the stroke. The diagrams are measured by means of a planimeter or by means of ordinutes. The latter way may be done by dividing the diagram into 10 or 20 equal spaces by vertical lines. On these verticals measure the length between the back pressure line and mark each length on a long strip of paper. Dividing the sum of all these lengths by 10 or 20 will give the average length of ordinate which multiplied by the scale of the spring will give the mean effective pressure (M.E.P.). Diagram from Hartford Engine Cylinder, IH X 24 inches. B<»iler pressure, 87 pounds. Vacuum by gauge, 23i inches. 130 revolutions per minute. Scale, 50. The vertii-al lines from A are called ordinates. Example. — If the total length is 9" and the spring used is 40 lb., how is the mean effective pressure (M. E. P.) found ? 40 X .9 = 36 lb. M. E. P. Ans. 230 VOCATIONAL MATHEMATICS EXAMPLES 1 1. If the total length of the ordinates is 11" and the spring used is 56 lb., what is the M. E. P. ? 2. A 9 X 10 engine has a 48 lb. average pressure on the piston, the length of rod is 35", and the crank is 5". Find the maximum thrust at right angles. 3. Find the weight of a flywheel of an engine 12'' x 24", diameter of wheel 7', with 168 K. P. M. 4. What is the necessary steam lap of an engine with a 45" stroke, valve travel 5", and cut-off at half stroke, and valve i" lead ? 5. If the total length of ordinates is 10" and the spring used is 49 lb., what is the M. E. P. ? 6. A 12'' X 24" engine has a 50 lb. average pressure on the piston, the length of the rod is 59", and the crank is 12" ; what is the maximum thrust at right angles ? 7. What is the necessary steam lap of an engine with a 64" stroke, valve travel 7, cut-off at half stroke, and valve ^" lead ? 8. If the total length of ordinates is 8" and the spring used is 38 lb., what is the M. E. P. ? 9. Find the weight of a flywheel of an engine 24" x 60", the diameter of wheel being 24' and having 75 R. P. M. 10. If the total length of ordinates is 12" and the spring used is 61 lb., what is the M. E. P. ? 11. Find the weight of a flywheel of an engine 30" x 60", the diameter of the wheel 30', with 63 K P. M. 12. An engine 10" x 12" has a 35 lb. average pressure on the piston, the length of the rod is 42", and the crank is 6". Find the maximum thrust at right angles. 1 Use 10 ordinates in solving problems. PART VIII — MATHEMATICS FOR ELECTRICAL WORK CHAPTER XV COMMERCIAL ELECTRICITY Amperes. — What electricity is no one knows. Its action, however, is so like that of flowing water that the comparison is helpful. A current of water in a pipe is measured by the amount which flows through the pipe in a second of time, as one gallon per second. So a current of electricity is measured Water Analogy of Fall of Potential by the amount which flows along a wire in a second, as one coulomb per second, — a coulomb being a unit of measurement of electricity, just as a gallon is a unit of measurement of water. The rate of flow of one coulomb per second is called one ampere. The rate of flow of five coulombs per second is five amperes. Volts. — The quantity of water which flows through a pipe depends to a large extent upon the pressure under which it flows. The number of amperes of electricity which flow along 231 232 VOCATIONAL MATHEMATICS a wire depends in the same way upon the pressure behind it. The electrical unit of pressure is the volt. In a stream of water there is a difference in pressure between a point on the surface of the stream and a point near the bottom. This is called the difference or drop in level between the two points. It is also spoken of as the pressure head, " head " meaning the difference in intensity of pressure between two points in a body of water, as well as the intensity of pressure at any point. Similarly the pressure (or voltage) between two points in an electric circuit is called the difference or drop in pressure or the poteyitial. The amperes represent the amount of electricity flowing through a circuit, and the volts the pressure causing the flow. Ohms. — Besides the pressure the resistance of the wire helps to determine the amount of the current: — the greater the resistance, the less the current flowing under the same pressure. To the electrical unit of resistance the name ohm is given. A wire has a resistance of one ohm when a pressure of one volt can force no more than a current of one ampere through it. Ohm's Law. — The relation between current (amperes), pressure (volts), and resistance (ohms) is expressed by a law known as Ohm^s Law. This is the fundamental law of the study of electricity and may be stated as follows : An electric current flowing along a conductor is equal to the pressure divided by the resistance. Current (amperes) = - — — ^— — ^ Resistance (ohms) Letting /= amperes, -EJ = volts, i? = ohms, I=E-r- Rov I = ~ R E=IR R = ^ I ELECTRICAL WORK 233 Example. — If a pressure of 110 volts is applied to a re- sistance of 220 ohms, what current will flow ? / = ^ = li2 = 1 = .6 ami^re. Ans. R 220 2 Example. — A current of 2 amperes flows in a circuit the resistance of which is 300 ohms. ^Vhat is the voltage of the circuit ? IIi = E 2 X 300 = 600 volte. Ans. Example. — If a current of 12 amperes flows in a circuit and the voltage applied to the circuit is 240 volts, find the ijesistance of the circuit. ^=Ji ?i2 = 20 ohms. Ans. Ammeter and Voltmeter. — Ohm's Law may be applied to a circuit as a whole or to any part of it. It is often desirable to / Ammbtkb Voltmetke know how much current is flowing in a circuit without calcu- lating it by Ohm's Law. An instrument called an ammeter is used to measure the current. This instrument has a low resistance so that it will not cause a drop in pressure. A voltmeter is used to measure the voltage. This instrument has high resistance so that a very small current will flow through 234 VOCATIONAL MATHEMATICS it, and is always placed in shunt, or parallel (see p. 235) with that part of the circuit the voltage of which is to be found. Example. — What is the resistance of wires that are carry- ing 100 amperes from a generator to a motor, if the drop or loss of potential equals 12 volts ? Drop in voltage = IB I = 100 amperes Drop in volts = 12 B = ? ohms -f B- = — = 0.12 ohm. J 100 Ans. Example. — A circuit made up of incandescent lamps and conducting wires is supplied under a pressure of 115 volts. The lamps require a pressure of 110 volts at their termiiials + lead Wiring of Incandescent Lamp Circuit and take a current of 10 amperes. What should be the resist- ance of the conducting wires in order that the necessary cur- rent may flow ? Drop in conducting wireS = 115 — 110 = 5 volts Current through wires = 10 amperes E 5 B = — zz — = 0.5 ohm resistance. Ans. I 10 EXAMPLES 1. How much current will flow through an electromagnet of 140 ohms' resistance when placed across a 100-volt circuit ? 2. How many amperes will flow through a 110-volt lamp which has a resistance of 120 ohms ? 3. What will be the resistance of an arc lamp burning upon a 110-volt circuit, if the current is 5 amperes ? ELECTRICAL WORK 235 4. If the lamp in Example 3 were to be put upon a 150- volt circuit, how much additional resistance would have to be put into it in order that it might not take more than 5 amperes ? Motor Electric Road System 5. In a series motor used to drive a street car the resistance of the field equals 1.06 ohms; the current going througli equals 30 amperes. What would a voltmeter indicate if placed across the field terminals ? 6. If the load upon the motor in Example 5 were increased so that 45 amperes were flowing through the field coils, what would the voltmeter then indicate ? Series and Parallel Circuits Pieces of electrical apparatus may be connected in two ways. When the pieces are connected so that the current passes through them in a single path, they are said to be in series. Wmf\ - B Cells Connected in Parallel ^}\m\ Cells Connected in Series WTien the pieces are connected so that the current is divided between them, they are said to be in jxtrallel with one another.. The total resistance of a series circuit is equal to the sum of the resistances of the separate parts of the circuit. The total 236 VOCATIONAL MATHEMATICS voltage of a series circuit is equal to the sum of the voltages across the separate resistances. A /?i /?2 B Total resistance A to B = Ei -\- R.^- Example. — If there is a circuitof 240 volts and the lamps are of 240 ohms' resistance but made to carry only ^ an ampere, two lamps would have to be put in series in order to use them on the 240-volt circuit. The resistance of the two lamps in series would then be 480 ohms, the voltage of the circuit 240 volts, and the current by Ohm's Law I= — = — = - ampere. H 480 2 A i?i R-2 . B In any closed circuit, the algebraic sum of the products found by multiplying the resistance of each part by the current passing through it, is equal to the voltage of the circuit. Example. — If three lamps of 110 ohms' resistance are con- nected in series and take i ampere, the voltage of the circuit IiBi 4- IoIi2 + Is^i = (^x 110) + (i X 110) + (^x 110) = 55 + 55 -f 55 =: 165 volts. Ans. Example. — A current of 50 amperes flowed through a circuit when the voltage was 550. What resistance should be added in series with the circuit to reduce the current to 11 amperes ? _ . Resistance = ^■^^- = 11 ohms Resistance = ^f^- = 50 ohms Additional resistance = 50 — 11 = 39 ohms. Ans. Example. — The voltage required by 15' arc lamps connected in series is 900 and the current is 6 amperes. If the resistance of the connecting wires is 5 ohms, how much additional volt- age will be necessary so that the lamp voltage may not drop below 900 ? Drop in voltage in connecting wires = E = IE 6 X 5 = 30 volts = additional voltage necessary. Ans. ELECTRICAL WORK 237 Example. — The field coil of a motor having 4 poles is measured for voltage across the terminals and the following readings are taken : Voltage across line = 220 Voltage across coil No. 1 = 78.33 Voltage across coU No. 2 = 00.00 Voltage across coil No. 3 = 73.33 Voltage across coil No. 4 = 73.33 Current flowing =1.5 amperes. What is the total resistance and what is the trouble at coil No. 2 ? Total resistance = 7? = ^ = ^=^ = 140.6 ohms. Am. I 1.5 As thei-e is no drop across the terminal of coil No. 2, there is prac- tically no resistance and the current is not going around the coil but throusjli a path of extremely low resistance. EXAMPLES 1. If three electromagnets are connected in series and the resistances are 3, 5, and 17 ohms, respectively, what is the total resistance of this set ? 2. The field coils of a series motor have a resistance of 10 ohms and the armature has a resistance of 7 ohms ; what is the total resistance of the motor ? 3. («) What would be the total resistance of two 110-volt incandescent lamps placed in series across a 110-volt line if each lamp has a resistance of 220 ohms? {h) Would these lamps light on this voltage in this position ? (c) Why ? 4. Three coils are connected in series and have a resistance of 3, 5, and 8 ohms, respectively. What current will flow if the voltage of the circuit is 64 ? 5. Five arc lamps, each having a resistance of 4 ohms, are connected in series. The resistance of the connecting wires and the other apparatus is o ohms. What must be the voltage of the circuit so that a current of 10 amperes may flow? 238 VOCATIONAL MATHEMATICS 6. A current of 10 amperes was passing througli a circuit under a pressure of 550 volts. The circuit was made up of three sections connected in series, and the resistance of two sections was 8 and 12 ohms, respectively. What was the resistance of the third section ? Example. — If 5 electromagnets are arranged in series and marked A, B, C, D, and E, the resistance of the circuit is 45 ohms and the resistance of each coil is : ^, 5 ohms ; B, 10 ohms ; C, 7 ohms; D, 8 ohms; E, 15 ohms. How much E. M. F. would be required to cause 10 amperes to flow through the coils, and what would be the E. M. F. across the terminals of each coil ? Voltage across ^ = 50 Voltage across jB = 100 Voltage across C = 70 Voltage across i> = 80 Voltage across E = 150 450 10 X 45 = 450, total voltage. Ans. Example. — What E. M. F. is necessary to send a current through 10 field coils connected in series, if each has a resist- ance of 10 ohms and 3 amperes are required to produce the necessary magnetization ? 10 ohms = resistance of each coil 10 coils are in series 10 X 10 = 100 ohms, total resistance B = 100 ohms 7=3 amperes E = IB =3x100= 300 volts. Ans. In any closed circuit the algebraic sum of the products found by multiplying the resistance of each part by the current passing through it, is equal to the voltage of the circuit. This is practically an inverse statement of the law of series circuits. Example. — If we have five lamps of 110 ohms' resistance, connected in series and taking ^ ampere, the voltage of the circuit is : ELECTRICAL WORK 239 E = IxRi + hRi + 7,^3 + URi + /»/?» = i X 110 + i X 110 -f i X 110 + \ X 110 + i X 110 = 66 + 66 + 66 -h 65 -f- 66 = 276 volts. Ans. In a parallel circuit the voltage across each branch is the same as the voltage across the combination. The current is equal to the sum of the currents in the separate parts. The resistance is equal to the reciprocal of the sum of the conductances of the separate parts. Conductance is the reciprocal of resistance and is equal to — . The unit of conductance is mho (ohm spelled backwards). mho= — — or ohm = ohm mho Series and parallel circuits may be combined and may exist in the same circuit. In parallel circuits the reciprocal of the total resistance is equal to the sum of the reciprocals of the paralleled resistance. If Rq = resistance of circuit and rj, r2, and r^ = parallel resistances, ri _ A r^ B rg resistance between A and B = Rq. R n r^ 'i\ Example. — Suppose that ?•,, Vz, and r^ are lamps of 300 ohms' resistance each, then Bo 300 300 300 iWO 100 Bo = 100 ohms. Ans. Or, ri = 100 ohms J__J_ ,_L,_L_ 11L__L rj= 60 ohms Bo 100 60 300~300~30 rs = 300 ohms Bo = 30 ohms. Ans. When it is necessary to get tlie resistance of parallel circuits, it is often more convenient to use the sum of the conductances as the total conductance of the circuit. 240 VOCATIONAL MATHEMATICS Example. — Three resistances in parallel : 7*1 = 2 ohms ^2 = 5 ohms rg = 1 olim A I'n Conductance » H- .5 mho <^>H' : .2 mho <" rr : 1 mho 1.7 mho'conductanc Resistance of circuit AB = — - = .58^ Ans. EXAMPLES 1. Ten arc lamps of 200 ohms' resistance are connected in series and the voltage of the circuit is 300. How much current does each lamp take ? 2. What will be the E. M. F. necessary to supply 60 Thomp- son-Houston arc lamps arranged in series, the resistance of each lamp being 5 ohms when burning, making a total resistance of 300 ohms in the circuit, if the current required is 10 amperes ? 3. Four parallel circuits of 2, 4, 5, and 10 ohms' resistance, respectively, have 40 volts impressed upon their terminals, (a) What is the total current flowing ? (b) How much current flows in each branch ? 4. Three incandescent lamps of different sizes are placed in parallel on a circuit. These have respective resistances of 100, 150, and 300 ohms. What is the total current flowing through these lamps when the pressure applied is 100 volts? ELECTRICAL WORK 241 5. What is the total resistance of four incandescent lamps placed in parallel, if each lamp has a resistance of 220 ohms ? 6. What is the resistance of a shunt-wound generator, if the field and the armature are respectively 25 and 10 ohms ? 7. Two coils on a large electromagnet for lifting iron ore are connected in parallel and the resistance of each coil is 40 ohms. What is the whole resistance ? 8. The resistance of a line frou) a power house to a mill is 6 ohms; there are 50 lamps in the mill, each lamp having a re- sistance of 220 ohms. What is the total resistance from the power house ? 9. In an electric street car 4 heaters are all connected in series and each has a resistance of 20 ohms with a voltage of 600 across the circuit, (a) What is the total resistance of these ? (b) How many amperes will go through them ? Power Measurement The flow of an electric current has been compared to the flow of water through a pipe. The water current is measured by the number of gallons or pounds flowing per minute. A cur- rent of electricity is measured by the number of amperes or coulombs per second. When a gallon of water is raised a foot by means of a pump a certain amount of work is done. So when a coulomb of electricity is passed through a wire under the pressure of one volt, a certain amount of work is done. In the case of water the work done is measured in foot pounds. A foot pound is the work done in raising a weight of one pound through a distance of one foot. Work = Force x Distance When one coulomb of electricity is passed through a wire under a pressure of one volt, the amount of work done is called one joule. The ix)wer required to keep a current of water flowing is the product of the current in pounds per minute by distance in feet. 242 VOCATIONAL MATHEMATICS This gives the power in foot pounds per minute. Mechanical power is usually expressed in horse power (H. P.). The power required to keep a current of electricity flowing is the product of the current in amperes by the pressure in volts and is ex- pressed in watts. 1 H. P. = 746 watts 1000 watts = 1 kilowatt 1 kilowatt (K. W.) = 1.34 or li- H. P. Volts X Amperes ^j^.j^^^^^^ lOUO Let P = power in watts / = current in amperes E = pressure in volts then (1) F= IE (equation for power) but E=TB (Ohm's Law) therefore P= I (IB) or PB (by substitution) (2) P=I^B but 7 = ^ (Ohm's Law) B therefore P = ^ (E) or ^ (by substitution) B B (3) ■ -f thus P= IE= DB = E^ B To measure power a wattmeter is used, which is a combina- tion of a voltmeter and an ammeter. In order to find the amount of work done by a certain engine, it is necessary to know the time it has been running and the power it has been supplying, i.e. its rate of doing work. If the power is measured in horse power and the time in hours, the work is done in horse power hours. Similarly, if the power is measured in kilowatts and the time in hours, the work done is measured in kilowatt hours (K. W. H.). 1 H. P. H. = 0.746 K.W. H. 1 K. W. H. = 1.84 H. P. H. ELECTRICAL WORK 243 These units are too large to be used conveniently in all problems, so a smaller electrical unit called the watt secondf or joulCf is used. EXAMPLES 1. If the resistance of a circuit is 1 ohm and the current 30 amperes, what energy is expended in one half hour ? 2. With a potential difference of 95 volts and a current of 15 amperes, what energy is expended in 20 minutes? 3. If a current of 100 amperes flows for 2 minutes under a pressure of 500 volts, what is the work done in joules ? 4. If 12 incandescent lamps burn for 10 hours under a pressure of 110 volts, each lamp consuming J an ampere, how many kilowatt hours are used ? 5. Fifty horse power expended continuously for one hour will produce how many kilowatt hours ? 6. If 4000 watts are expended in a circuit, how much horse power is being developed ? 7. If 20 horse power of mechanical energy were converted into electrical energy, how many watts would be developed ? 8. If a current of 50 amperes flows through a circuit under a pressure of 220 volts, what is the power ? 9. If 200 watts are expended in a circuit by a current of 4 amperes, what is the voltage required to drive the current through the wire ? 10. If an incandescent lamp requires \ an ampere of current and the resistance of its filament is 220 ohms, how many watts are required for it ? Measurement of Resistance The amount of current in a circuit depends upon the voltage and upon the resistance. To control the current it is necessary to change one of these two factors. The resistance to the flow of water through a pipe depends upon the shape of the pipe and 244 VOCATIONAL MATHEMATICS its length. The electrical resistance of a conductor depends upon the nature of the metal from which the conductor is made, its size, its length, and the temperature. The greater the size of the conductor, the greater is its power for conducting elec- tricity, and therefore the less its resistance. The longer the wire, the less its conducting power, and therefore the greater the resistance. As the resistance of a large pipe is less than the resistance of a small one, so the resistance of a large wire is less than the resistance of a small one. Copper is the material generally used for wires, and its con- ductivity, or capacity for conducting current, is taken as the standard. The conductivity of pure copper is expressed as 100 % . Commercial copper usually has from 98 to 99 % con- ductivity. Other materials used in electric wires are iron, aluminum, brass, etc. Iron has a conductivity of about 16 % and brass of about 25 %. It would require an iron wire with over 6 times the cross section of a copper wire to give the same conductivity, and brass wire would have to be about 4 times as large in its cross section for the same conductivity. Generally it is assumed that all electric wires are copper. In measuring the length of wires the unit used is feet, while the cross section area is measured in circular mils; y^Vir ^^ an inch is called a m^7, and a round wire one mil in diameter is said to have a cross-section area of one circular mil. A wire 1 foot long, with a cross-section area of 1 circular mil, is called a mil-foot wire. The area of any wire in circular mils may be found by squaring the number of thousandths of an inch in the diameter. If B = resistance of wire in ohms L = resistance of wire D = diameter of wire in mils cP = area of wire in circular mils K = resistance of 1 mil foot in ohms called resistivity or specific resistance of material then B = ^ ELECTRICAL WORK 245 Note. — /T varies with the niAterial and the temperature. The resist- ance of 1 mil foot of soft copper wire at 60° F. is 10.4 ohms. Example. — What is the area of a wire 0.1 inch in diameter ? 0.1 inch = 100 thouBandths of an inch 100 X 100 = 10,000, number of circular ihils. Ans. Example. — What is the resistance at ordinary temi)erature of a copper wire 2500 ft. long with a cross-section area of 10,000 circular mils ? (K= 10.) ^^^^10^^^500^2.6 ohms. Aus. d^ 10000 EXAMPLES 1. What must be the diameter of a wire in mils in order that it may have a cross-section area of 200 circular mils ? 2. How many circular mils are there in a wire 50 mils in diameter ? 3. How many circular mils are there in a wire 150 mils in diameter ? 4. What is the diameter in inches of a copper wire which has a cross-section area of 20,000 circular mils ? 5. If it is desired to have a copper wire of J ohm resistance and 2000 ft. long, what must its cross section be ? 6. What pressure is required to force a current of 50 am- peres over a copper wire IGOO ft. long which has a cross- section area of 20,000 circular mils ? 7. A current is forced through a copper wire 2000 ft. long under a pressure of 50 volts. If the wire has an area of 5000 circular mils, what is the value of the current flowing? 8. If a wire 5000 ft. long carries a current of 5 amperes under a pressure of 100 volts, what is the cross-section area of the wire ? 246 VOCATIONAL MATHEMATICS 9. A wire 100 ft. long is in series with another wire 500 ft, long and the first wire is ^" in diameter. If the second wire has a cross-section area of 20,000 circular mils, what is the resistance of the circuit ? 10. How many amperes will a wire ^" in diameter carry if a wire 1000 mils in diameter will carry 650 amperes ? Size of Wire If an electrician wishes to know the size of a wire to carry a certain current a certain distance with a certain drop of voltage, he may ascertain it by substituting values in the foL lowing formula, which is called a two-wire formula, e CM = size of wire in circular mils D = distance from distribution 7= amperage e = drop or volts lost Example. — What size of wire will be required for a motor situated 85 ft. from the center of distribution, if the motor is 5 H. P., operating at difference of potential of 110 volts, allow- ing 3 % drop ? 85 ft. Ans. P=ET 1 H. P. = 746 watts 1=^ j^3730 g:= no X. 03=3.30 D= P = 746 X 6 = 3730 E=no 21.6x85x^^3^ 110 _ 21.6 X 85 X 3730 3.30 3.30 X 110 = 18865.7 wire. REVIEW EXAMPLES 1. What size of wire will be required for a 15 H. P. motor operating at 550 volts and situated 35 ft. from the center of distribution, allowing a 2 % drop ? ELECTRICAL WORK 247 2. J low many coulombs are delivered in a minute when the current is 17^ amperes ? 3. What is the current when 480 coulombs are delivered per minute ? 4. In what time will 72,000 coulombs be delivered when the current is 80 amperes? 5. What size of wire will be required for a 7^ H. P. motor operating at 220 volts and situated 65 ft. from the center of distribution and allowing a 5 % drop ? 6. If a current of 20 amperes flows through a circuit for 21.2 hours, what quantity of electricity is delivered? 7. How many ampere hours pass in a circuit in 2^ hours when the current is 1 6 amperes ? 8. What size of wire will be required for a 10 H. P. motor operating at 110 volts and situated 105 ft. from the center of distribution, allowing a drop of 3 volts? 9. If 200 coulombs of electricity are passed through an electrolytic vat each second under a pressure of 10 volts, how many joules of work are expended in an hour ? 10. What quantity of electricity must flow under a pressure of 5 volts to do 125 joules of work ? 11. If 10 coulombs do 10 joules of work flowing through a wire, what is the pressure ? 12. What must be the diameter of a wire in mils in order that it may have a cross section of 200 circular mils? 13. A wii-e 5000 ft. long carries a current of 6 amperes under a pressure of 100 volts. What is the cross-section area of the wire ? 14. A copper wire ^ inch in diameter and 100 ft. long is in series with another copper wire 500 ft. long with a cross section of 20,000 circular mils. What is the resistance of the circuit ? 248 VOCATIONAL MATHEMATICS 15. What is the resistance at ordinary temperature of a copper wire 2500 ft. long having a cross-section area of 10,000 circular mils ? 16. If it is desired to have a copper wire of i ohm resistance and 2000 ft. long, what must be its cross-section area? 17. A voltmeter which measures the pressure on a circuit registers 500 volts and the ammeter on the same circuit shows 25 amperes. What is the resistance of the circuit ? 18. An electromagnet has a resistance of 25 ohms, and there must be 41 amperes passing through it in order that the magnetism may be strong enough. What must be the voltage ? 19. What current is needed to light a 16 C. P. lamp, if the hot resistance of the lamp is 220 ohms and the voltage is 110? 20. A storage battery gives 2.3 volts and it is connected with a coil having a resistance of 25 ohms. What current will flow through the circuit if the internal resistance of the cell is zero? 21. What is the power necessary to drive a current of 500 amperes through a resistance of 5 ohms ? 22. How many watts of power are going to an electric motor if the voltage of the line is 500 and there are 7 amperes enter- ing the motor ? 23. How many H. P. are needed to run a dynamo that is lighting 258 lamps in parallel, if each lamp takes | an ampere at 110 volts ? 24. How many 40-watt electric glow lamps can be run on a 110- volt circuit with an expenditure of 48 amperes of current ? Brown and Sharpe Wire Table The unit chosen for this table is a copper wire .1 inch in diameter. This is called No. 10 wire and has the following characteristics: No. 10, B & S wire, diameter, .1 inch; area in circular mils, 10,000 cm.; resistance, per 1000 ft., 1 ohm; ELECTRICAL WORK 249 weight per 1000 ft., 31. o lb. Since this table was made, the standard has changed slightly, so that at present No. 10 wire is .1019 inch in diameter and the other vahies are changed l)roportionately, but for all ooiumercial work the values as originally given are sufficiently accurate. The table is so arranged that the area of the wire is doubled every three gauges down and halved every three numbers up. Example. — Find the area of No. 7 wire. No. 7 is three numbers below No. 10, whose area is 10,000 cm., so that its area is 2 x 10,000 = 20,000 cm. Area of No. 4 wire = 2 x 20,000 = 40,000 cm. Area of No. 13 wire = I x 10,000 = 5000 cm. The resistance is reduced to one half every three numbers down and doubled every three numbers up. The weight doubles every three numbers down and halves every three numbers up. Example. — R per 1000 ft. of No. 10 H & S wire equals 1 ohm R per 1000 ft. of No. 7 H & S wire equals .5 ohm R per 1000 ft. of No. 4 H & S wire equals .25 ohm R per 1000 ft. of No. 13 B & S wire equals 2 ohms R per 1000 ft. of No. 16 B & S wire equals 4 ohms Example. — TTper 1000 ft. of No. 10 B & S wire equals 31.5 lb. W per 1000 ft. of No. 7 B & S wire equals (k^ lb. IF per 1000 ft. of No. 4 B & S wire equals 126 lb. TKper 1000 ft. of No. 13 B & S wire equals 15.8 lb. IK per 1000 ft. of No. 16 B & S wire equals 7.9 lb. If the gauge number is not three or a multiple of three be- low or above No. 10, get the area, resistance, or weight desired which is less than the value for the wire required, and if it is one below the required number, multiply by 1.26, and if two below, by 1.59. 250 VOCATIONAL MATHEMATICS In computing the resistance and weight of cables the follow- ing formula is used : Resistance = R = in ohms per 1000 ft. cm. Weight = .00305 x cm. in lb. per 1000 ft. EXAMPLES Find the resistance, area, and weight of the following wires : 1. No. 16. 6. No. 14. 2. No. 13. 7. No. 15. 3. No. 7. 8. No. 10. 4. No. 3. > 9. No. 2. 5. No. 1. 10. No. 11. PART IX — MATHEMATICS FOR MACHINISTS CHAPTER XVI MATERIALS Every machinist should be familiar with the strength and other properties of the materials that he uses — such metals as cast iron, wrought iron, steel, copper, bronze, and brass. Cast Iron. — Since much of the machinist's work is on cast iron, he should know something of its nature and manufacture. Iron ore as found in the earth generally contains many impurities, such as silicon, sulphur, phosphorus, manganese, combined carbon, and graphitic carbon. To free the iron from the grosser impurities, the ore is crushed and mixed with coke and limestone and intense heat applied in a blast furnace. The melted iron, being heavier than the other materials, falls to the bottom of the furnace. When a sufficient quantity has accumulated, it is allowed to flow out of a tap hole into molds of sand. After it has cooled it is broken into lengths suitable to be remelted in foundries and made into iron castings. Such iron is called pig iron. During the process of smelting in the blast furnace, the liquid iron combines with a considerable quantity of carbon, sulphur, silicon, phos- phorus, and manganese from the impurities in the ore and coke. Some of the carbon combines with the iron chemically and forms iron carbide, while the remainder exists in the iron as a mixture of carbon and is known as graphite. The amount of carbon may weaken the iron by making it soft, and it may also make the iron too brittle to work. So the man in charge of the foundry must use his judgment in mixing different grades and quantities of pig iron to obtain a casting of the de- sired strength, hardness, toughness, and clearness of grain. Castings. — Machines are made of iron castings, forgings, steel parts, etc. Castings and forgings can be distinguished from each other by the appearance of the fractures in them. After the machines are designed and the wooden patterns made in the pattern shop, the patterns are sent to 251 252 VOCATIONAL MATHEMATICS the foundry, where aa impression of the machine is made in sand. Dur- ing this operation of molding, the sand is confined in an iron or wooden device of two or more parts called a flask. The lower or bottom part of the flask is the drag or nowel, while the top or upper part is the cope; and other parts are the checks. Sometimes pig iron and old scrap iron are melted together in a furnace called a cupola. The liquid iron is taken from the furnace in ladles and poured into different molds. As the hot iron flows into the mold and cools, it becomes solid and takes the shape of the mold. When the castings are removed from the mold they present a rough surface and have to be cleaned and smoothed or "machined" before they can be put to the use intended for them. They are cleaned in various ways — by means of emery wheels and revolving wire brushes, by being rotated in " tumblers," by chipping with pneumatic chisels, or by means of a sand blast. The scales on the castings are removed by wetting them with diluted sulphuric acid. This process is called pickling. After this the casting is attached to the plate or table of the machine tool that is to perform the necessary work upon it. Special devices are made to hold castings when they are being machined. Wrought Iron. — One of the valuable qualities of wrought iron is the comparative ease with which it can be united with another piece by weld- ing. When two pieces of wrought iron are heated to a white heat, they assume a viscous condition, and when hammered together become united. Wrought iron differs from cast iron in that it is capable of assuming any shape under the hammer. It is readily made from cast iron by heating in & puddling furnace. In this furnace the cast iron is subjected to great heat and constant stirring, which allows the carbon to pass off as a gas and the other impurities to rise to the surface, where they can be removed. When the impurities 'are removed the iron is hammered to remove particles of slag and then rolled in order to make it more compact. After this it is heated again and rolled into bars for different purposes. Wrought iron is sometimes case-hardened when it is used in machine parts that need to be harder than the common iron. After the piece has been finished and properly sized it is heated a bright red and the surface rubbed with prussiate of potash. When it has cooled to a dull red, it is immersed in water. Three parts of prussiate of potash and one of sal ammoniac is a good case-hardening mixture. The temperature (Fahren- heit) for cherry red is 1832°. If there are holes in such iron work, the hardening by this process reduces them slightly. Steel. — Steel is a form of iron which contains, as a rule, more car- bon and other elements than wrought iron and less than cast iron. MACHINISTS' WORK 253 There are many grades of steel, and each one is made by a 8i)ecial process, steel may be recognized by the appearance of a dark spot when nitric acid is placed on its surface. The darker the spot, the harder the steel. Iron, on the contrary, shows no sign when touched with nitric acid, (iootl steel will not stand a high heat, but will crumble under the hammer blow at a bright red heat, while at a moderate heat — a full dull red or cherry red — it may be drawn out to a fine edge tool. Steel that has once been overheated or burned cannot be restored. Steel for cutting-tools on lathes and i)laners should be drawn to a straw color, or 430^, while for wood tools, taps, and dies, dark straw color, or 41(f ; for chisels for chipping, brown yellow or SW ; for springs, dark purple, or 650°. Steel may be softened or annealed if heated to a low red and placed in a box of slaked lime and well covered, or in a box of fine bonedust, care being taken in either case to cover the piece all around and on top to a depth of not less than one and one half inches. Copper. — Copper is used to a great extent because it may be easily forged when cold. It may also be pressed into different shapes by means of molds. Its strength is greatly increased by hammering and rolling. It is used principally in wires and plates. Brass. — Brass is an alloy or mixture of copper and zinc. Its tensile strength is nearly equal to that of copper. Weight of Bars of Steel. — The weight of bars of steel, as they are usually made, is found by multiplying the area of the ci'oss section or end in inches by the length in inches, and multiplying the resulting nural^er of cubic inches in the bar by 0.3. This will give practically an accurate result, since all bars, unless otherwise ordered, will be rolled or hammered slightly "full" to the dimensions given, or a bar that is " full " to size may be trimmed to exact dimensions if neces- sary. Whereas, if the bar is slightly under size, it cannot easily be made larger. To find the weight of a triangular bar of steel, multiply the area of the base in square inches by the height in inches and then by 0.3. The base of the triangular bar is found by multi- plying the length of one of the bars or sections by one half the l)erpendicular height; that is, by the distance to the opposite vertex of the cross section. 254 VOCATIONAL MATHEMATICS EXAMPLES 1. What is the weight of a triangular bar of steel when the base contains 16 sq. in. and the height is 6 ft. ? 2. What is the weight of a triangular bar of steel when the base contains 13 sq. in. and the height is 4 ft. ? 3. What is the weight of a triangular bar of steel when the base contains 21 sq. in. and the height is lo feet ? 4. What is the weight of a triangular bar of steel when the base contains 23 sq. inches and the height is 17 feet ? 5. What is the area of the base or section of a triangular bar one side of which is 8 inches and the altitude 6 inches ? 6. What is the area of the base or section of a triangular bar one side of which is 11 in. and the altitude 9 in. ? 7. What is the weight of a triangular bar of steel one side of which and altitude of whose base or section are respectively 13 in. and 11 in., and the length of the bar 23 ft. ? 8. What is the weight of a triangular bar whose section is 24 sq. in. and whose length is 22 ft. ? 9. The weight of a cast-iron wheel is approximately sixteen times as heavy as the white pine pattern from which it is cast. What is the probable weight of a casting if the pattern for it weighs 2J pounds ? 10. A white pine pattern weighs 12.5 pounds. What will be the weight of an iron casting from it ? (Use data in Ex. 9.) 11.' If the weight of a brass casting is approximately fifteen and a half times that of its white pine pattern, what will be the weight of a casting if the pattern weighs 15 oz. ? 12. A white pine pattern weighs 1.75 pounds. What will be the weight of 50 brass castings made from it ? (Use data in Ex. 11.) 13. Since the shrinkage of brass castings is about ^ inch in 10 inches, what length would you make the pattern for a brass collar which is required to be 6 inches long ? CHAPTER XVII LATHES The Engine Lathe. — The lathe is one of the most important machines in the machine shop. There are different kinds of lathes, each adapted to certain kinds of work, the engine lathe being the most important. The motion of the tool is controlled by power speed; that is, the tool is moved automati- cally parallel with and at right angles to the center line of the lathe spindle. Most lathes are furnished with a series of clutch gears and lead screws by means of which threads of different pitch may be cut. All lathes have a series of stepped cones in order to obtain a variety of speeds which are neces- sary in order to work on hard and soft metals and to obtain constant surface speeds for different diameters. The slower speed makes the deeper cut. Size. — The size of the lathe is expressed by stating the length of the bed and the largest diameter it will swing on centers. The swing is found by measuring from the point of the headstock center to the ways on the bed and then multi- plying by 2. The English measurement is from the center to the way. The next important measurement is the length of the bed, which is the entire amount of distance the tailstock will move backward. If, however, accuracy is desired in this measurement, the figure given should be the distance between the two centers when the tailstock is in its extreme backward position, as the lathe will turn no longer piece than will go between centers. Gear and Pitch. — Lathes that will cut a thread the same pitch as the lead screw, with gears having the same number of teeth on the stud and screw, are called geared even. If a 255 256 VOCATIONAL MATHEMATICS lathe will not do this, find what thread will be cut with even gears on both stud and lead screw and consider that as the pitch of the lead screw. Adjusting Gears. — A simple or single-geared lathe is one having a straight train of gearing from its spindle to its feed Engine Lathe screw, excepting intermediate gears, which only serve as idlers to take up the distance between the driver and driven gears or spindle and screw gears. Index plates are usually found on lathes giving the change gear used for different threads, LATHES 257 but when threads are called for that are not indexed, or when those ending in fractions are to be cut, the machinist must make his own figures. Refer to the screw cutting table and see what number of threads to an inch are cut with equal gears. This number is the number of turns to an inch that we a^ume the lead screw has, no matter what its real number of turns to an inch is. Write above the line the number of turns to an inch of the lead screw and below the line the number of turns to an inch of the screw to be threaded, thus expressing the ratio in the form of a fraction, the lead screw being the numerator and the screw to be threaded the denominator. Now find an ecjual fraction in terms that represent numbers of teeth in available gears. The numerator of this new fraction will be the spindle or stud gear and the denominator the lead screw gear. The new fraction is usually found by multiplying the numerator and denominator of the first fraction by the same number. PlxAMPLE. — It is required to cut a screw having 11^ threads l)er inch. The index gives 48 to 48 cuts 4 threads per inch. _L X -* = — Hi 6 69 Put the 24-tooth gear on the stud and the tt9-tooth gear on the lead screw to cut 11^ threads per inch. EXAMPLES 1. What gears should be used to cut a screw having 16 threads per inch, if a 40-gear on the stud and an 80-gear on the screw will cut 8 threads to the inch ? 2. What geai's should be used to cut a screw having 3 threads per inch, if a 48-gear on the stud and a 56-gear on the screw will cut 14 threads to the inch ? * If multiplying by 6 will not give the gears available, use any other number. 258 VOCATIONAL MATHEMATICS 3. What gears should be used to cut a screw 32 threads per inch, if the pitch of the lead screw is 12 ? 4. What gears should be used to cut the following threads per inch, if the pitch of the lead screw is 12 ? a. 36 thread* d. 64 threads b. 42 threads e. 3\ threads c. 56 threads / 3^ threads g. 12 threads Compound Lathes. — The term compound applied to a lathe means that in its train of gearing from its spindle to lead screws there is a stud or spindle having two different sized gears, both connected in such a way as to change the link of revolution between the spindle and the lead screw to a different number of revolutions from that which would take place if the straight line of gears were used. First Method. — Write the number of turns to an inch of the lead screw as the numerator of a fraction and the turns of the screw to be threaded as the denominator. Factor this fraction into an equal compound fraction. Change the terms of this compound fraction either by multiplying or dividing into another equal compound fraction whose terms represent num- bers of teeth in available gears. Then the two terms in the numerator represent the number of teeth in the gears to be used as drivers and those in the denominator the gears to be used as driven gears. Example. • — It is required to cut a screw having 3^ inches lead or -^^ turns to an inch. The lead screw is 1^ inches lead or I turns to an inch. 3 ' 13 3x4 Multiply numerator and denominator by 5, 2 X (13 X 5) _ 2 X 65 1 X (12 X 6) 1 X 60 >r ,.. 1 * a' A • . V. o. (24 X 2) X 65 48 X 65 Multiply numerator and denominator by 24, (24 x 1) x 60 " 24x60 LATHES 259 The 48-tootli and the 05-tooth gears will be tlie drivers and the 24- tooth and GO-tooth gears the driven. NoTK. — Any multiplier may be used to obtain the gear that is avail- able. Second Method. — Another satisfactory method of working out the change gears is by proportion. If it is desirable to cut a screw having 10 threads per inch and the lead screw has 6 threads per inch, the first two terms of the proportion would be 10: 6. As a rule, the smallest gear in the gear box is used on the spindle, if it will serve the purpose, and as the number of teeth on this gear is generally a multiple of the number of threads per inch on the lead screw, in the present case it would probably be 24. As the number of teeth on the lead screw is to be the third term of the proportion, and as this is unknown, x is used to represent it, and then the pro- portion is 10: 6: : x: 24. By multiplying the first and fourth terms together and the second and third terms together, the result is 6x = 240. Then ic = 40, the number of teeth on the screw gear. If the lathe is compound geared, it is necessary to find the proportional speed of spindle and stud. If the stud makes three quarters of a revolution while the spindle makes a com- plete revolution, it is necessary to use a gear on the screw with but three quarters the number of teeth represented by x in the proportion. Example. — What gear should be used on the screw of a compound geared lathe with the stud turning only three quarters as fast as the spindle, in order to cut a screw having 13 threads per inch, if the lead screw has 6 threads per inch and the stud gear 48 teeth ? 13 : 6 : : a: : 48 8 Cancelling, 13 : ^ : : x : ^ = 104 Three quarters of 104 = 78 With a 78-tooth gear on the screw, a 48-tooth gear on the stud of the compound geared lathe will cut a screw having 13 threads per inch. 260 VOCATIONAL MATHEMATICS Note. — If the stud turned but one half as fast as the spindle, then a gear should be used on the screw with one half as many teeth as shown under the method for simple geared lathes. , QUESTIONS AND EXAMPLES 1. Is the lathe in the classroom simple or compound geared ? 2. What gears should be used to cut a screw having 18 threads per inch, if a 40-gear on the stud and an 80-gear on the screw will cut 8 threads to the inch ? 3. How many threads per inch has the lead screw ? Is it a square, V, or acme thread ? Is it right or left hand ? 4. What gears should be used to cut a screw having 6 threads per inch, if a 48-gear on the stud and a 56-gear on the screw will cut 14 threads to the inch ? 5. How many gears are there between the spindle gear and the lead screw gear ? 6. If the pitch of the lead screw is 16, what gears should be used to cut a screw with 38 threads per inch ? 7. Has this lathe reversing gears? If not, state briefly how the reversing is accomplished. 8. If the pitch of the lead screw is 18, what gears should be used to cut a screw with 44 threads per inch ? 9. Put even gears on the first change gear stud and lead screw and put a smooth round piece of wood on the lathe centers. Clamp a pencil on the tool post so that it will mark on the wood, then turn the lathe spindle until the carriage has moved 1 inch. How many threads did the pencil draw on the wood ? 10. What gears should be used to cut the following threads per inch, if the pitch of the lead screw is 18 ? a. 42 threads c. 16 threads h. 8 threads d. 5^ threads LATHES 261 11. What is the true pitch of the lead screw ? Is it the same as the actual pitch ? 12. A 1" bolt is to have 8 threads per inch cut in it. If a 56-tooth gear is on the lead screw, what gear must be put on the stud ? 13. A lathe has a feed rod turning at the same rate as the lead screw, while the carriage travels one quarter as fast as it would when screw cutting. If geared for 12 threads per inch and the feed shaft is used, what will be the feed in fractions of an inch per revolution ? 14. At 64 revolutions per minute (R. P. M.) how long will it take to make a roughing cut with y\" feed and a finishing cut with y\" feed, if both cuts are 21" long and 1 minute is allowed for changing tools ? To Cut Double or Multiple Threads In modern machine construction there are many studs, screws, and feed rods having threads for rapid travel, and instead of a single spiral thread, there are two and sometimes three spirals. If a machinist is called upon to replace or duplicate such a thread, the method would not be to cut the multiple threads at one time but to cut one thread at a time. If the pitch of the double thread is measured, the pitch of every second thread will be measured and the lathe set for 4 threads per inch. The cut of 4 threads is chased out to size and the lathe left geared and the tool unchanged, and by turn- ing the gears on the lathe spindle one half revolution, the tool position for beginning the second thread is gained. Before fixing the gear wheel position, the carriage should be reversed in the position of the cut, thus taking up all lost motion in screws, lathe nuts, and gears. Then make the exact position of the mesh on the teeth of the spindle and on the lead screw and count the number of teeth on the spindle gear, which equals one half the entire number, and make this 262 VOCATIONAL MATHEMATICS tooth. Now take off the spindle and turn the lathe one half a revolution, bringing the second marked tooth to the position of the first, and the lathe is then ready for cutting the second thread. It is necessary in cutting multiple threads to select a driving gear wheel having a number of teeth exactly divisible by the number of threads cut. Machine Speeds Eveiy casting is cut into a definite shape by removing a certain portion of it, called a cut or turning, by one of the machine shop tools. The casting must be cut in the most economical way and in the shortest pos- sible time. The tool that does the cutting must attain the highest pos- sible speed, but there is a limit to the speed of a tool on account of the heat generated as it moves against the casting. If too great speed is used, the heat generated takes the temper out of the tool, renders it use- less, and causes the casting to expand. The effect of the expansion on the casting is to make it no longer true. The machine that is doing the cutting should be accurate to .005 of an inch. The cutting capacity of a machine depends on (1) the speed of the cut, (2) the distance traversed by the tool in passing from one cutting portion to the next, (3) the depth of cut, i.e., the thickness of the strip removed from the casting. The volume of metal removed from a casting may be calcu- lated as follows: The volume of metal removed from good steel in one minute equals : Cutting speed (length) times the feed (width) times the depth of cut (thickness). Example. — If the speed of a cut is 18 ft. per minute, the feed .06 in., and the depth of cut is \ in., what is the volume of metal removed in one hour ? 18' X 12" X .06 X .25 = 3.24 cu. in. weight of 1 cu in. of good steel = .277 lb. 3.24 X .277 = 0.897 lb. .897 X 60 = 53.8 lb. in 1 hr. Ans. LATHES 263 Speeds for Different Metals. — Various materials, such as iron, copper, ami wrought iron, possess different standards of hardness. A hanler metal will wear away the tool and strain the machine more than a softer metal. Therefore, there are different sj^eeds for each metal. With car- bon steel cutting tools, the surface speed varies from 30 to 40 feet per minute for cast iron, wrouglit iron, and soft steel ; 15 to 26 feet for well annealed tool steel, and from 60 to 80 feet per minute for brass, while the speed for dies and taps varies from 12 to 18 feet per minute on steel, and from 30 to 50 feet per minute on brass, depending upon the quality of the metal and the shape of the piece. With cutting tools of high-speed steel these speeds, except for dies and taps, can be nearly doubled. Cast iron should be worked at a speed which is j^j to j'j of that for copper, or J to 1*5 of that for wrought iron. Net Power for Cutting Iron or Steel To find the net power for cutting cast iron and steel, mul- tiply the section of cut or chip in square inches by 230,000 pounds for steel, or 168,000 pounds for cast iron, to get the pressure on the tool ; and multiply this product by the cutting speed in feet per minute, and divide the result by 33,000 to obtain the horse power required. Example. — Steel is being cut with \" cut, 1-64" feed at a speed of 20' per minute. What is the H. P. ? \" X 1-64" = .0039 sq. in. .0039 X 230,000 lb. = 897 lb., pressure on tool 897 X 20 =-- 17,940 17,940 -T- 33,000 = .64 horse power. Ans. Example. — If the cutting speed at the rim of iron stock should be 40 feet per minute, at what speed should the lathe (spindle) be driven for a piece of stock 3" in diameter ? 3" diameter 8 X 8.1416 = 9.4248", nearly 9.6", or f ' 40' H- 1' 40 X J = ifft = 53 R. P. M. Ans. 264 VOCATIONAL MATHEMATICS EXAMPLES 1. What is the amount of metal removed in one hour from a casting with a cutting speed of 69' per minute and a feed of j\" and a depth of -f ? 2. As a tool will stand a cutting speed of 35 feet per minute when turning cast iron, how many revolutions per minute should the lathe spindle make when a piece of cast iron 8'' in diameter is being turned ? 3. Write the formula for the feed of a lathe tool, making JV"= number of revolutions, D = distance the tool moves, and i^=the feed. 4. A lathe tool moves 2.1-" along the work in one minute, and the speed of the lathe is 400 R. P. M. What is the feed ? (Use the formula.) 5. In turning up in the lathe a gun metal valve, 4 inches in diameter, it is desirable that the surface speed shall not exceed 45 ft. per minute. How many revolutions per minute may the w^heel make ? 6. Find the time required and the speed of the lathe in turning one 20-foot length of 3-inch wrought iron shafting, one cut, traverse 28 per inch, cutting speed 20 ft. per minute, no allowance being made for grinding or breaking of tools or for setting stays. 7. How many revolutions per minute may be made in turning np a steel shaft 6 inches in diameter, if the surface speed must not exceed 12 ft. per minute ? 8. A piece of tool steel 1|'' in diameter is turned in a lathe at 74 K. P. M. What is the cutting speed ? 9. What is the amount of metal removed in one hour from a casting with a cutting speed of 64 ft. per minute and a feed of jY and a depth of i" ? 10. If a tool will stand a cutting speed of 237 F. P. M. when LATHES 265 turning cast iron, how many R. 1*. M. should the hithe (spindle) make when a piece of cast iron iV in diameter is being turned ? 11. A valve yoke, stem 2" in diameter is being turned in a lathe. If the lathe spindle makes 50 H.P. M., what is tlie cutting speed in F. P. M. ? 12. What number of revolutions must a lathe spindle make to cut 15 ft. per minute, in turning up an iron shaft 7J inches in diameter ? 13. (a) With a tool steel that can stand a cutting speed of 30' per minute, how many revolutions per minute may a lathe be run in turning one cut off a piece of shafting 15" in diameter ? (b) To hold the same cutting speed, how many revolutions per minute would be required if the shaft were but 7^" in diameter ? 14. In turning a cast iron piston head 15 inches in diameter in the lathe, it is desired that the surface speed shall not ex- ceed 15 ft. per minute. How many revolutions per minute may the work make ? 15. How many revolutions per minute should the lathe spindle make in turning up a cast iron pulley 33J inches in diameter, at a cutting speed of 15 feet per minute ? Table of Surface Speeds The table on page 266 has been computed to facilitate the figuring of speeds for machines and is used as follows : Having first determined the proper surface speed, refer in the table to the column of revolutions per minute corresponding to this surface speed, and in this column opposite the diameter corresponding to that of the work under consideration will be found the required revolutions per minute of spindle or work. Other surface speeds than those for which the table is com- puted can be readily obtained from the table by multiplying or dividing, as the case may require. 266 VOCATIONAL MATHEMATICS Table of Surface Speeds 32^ 3TJ 47^ DiAM. Bevolntions per Mimitc ■i^ 3667.8 3973.5 4279.1 4584.8 4890.4 5196.1 6501.8 6807.4 6113.0 l\ 1838.9 1986.7 2139.5 2292.4 2445.2 2598.0 2750.9 2903.7 3056.5 ^ 1222.6 1324.5 1426.3 1528.2 1630.1 1732.0 1833.9 1936.8 2037.7 i 916.9 993.3 1069.8 1146.2 1222.6 1299.0 1375.4 1451.8 1628.3 A 733.6 794.7 855.8 916.9 978.0 1039.2 1100.3 1161.4 1226.6 t\ 611.3 662.2 713.2 764.1 815.7 866.0 916.9 967.9 1018.8 j\ 523.9 567.6 611.3 654.9 698.6 742.3 785.9 829.0 873.3 k 458.4 496.6 534.9 673.1 611.3 649.5 687.7 725.9 764.1 ^ 407.5 441.5 475.4 509.4 543.3 677.3 611.3 645.2 679.2 t\ 366.7 397.3 427.9 458.4 489.0 519.6 560.1 680.7 611.3 M 333.4 361.2 389.0 416.9 444.5 472.3 500.0 627.9 555.7 f 305.6 331.1 356.6 382.0 407.6 433.0 458.6 483.9 509.4 il 282.1 305.6 329.1 352.3 376.1 399.7 423.2 446.7 470.2 tV 261.9 283.8 305.6 327.4 349.3 371.1 392.9 414.8 436.6 M 244.5 264.9 285.1 305.6 326.0 346.4 366.7 387.8 407.5 i 229.2 248.3 267.4 286.5 305.6 324.7 343.8 362.9 382.0 ^ 203.7 220.7 237.7 254.7 271.6 288.6 306.6 322.6 340.0 f 183.4 198.6 213.9 229.2 244.5 259.8 276.0 290.3 306.2 ii 166.7 180.6 194.5 208.4 222.3 236.1 250.0 263.9 277.8 1 152.8 165.5 178.3 191.0 203.7 216.6 229.2 241.9 254.2 H 141.0 152.8 164.5 176.2 188.0 199.8 211.6 223.3 234.8 1 130.9 141.9 152.8 163.7 174.6 186.6 196.4 207.4 218.2 if 122.2 132.4 142.6 152.8 163.0 173.2 183.4 193.9 203.8 1 114.6 124.2 133.7 143.2 152.8 162.3 171.9 181.4 191.0 IrV 108.2 117.2 126.2 135.2 143.2 163.2 162.2 170.8 180.4 H 101.9 110.3 118.8 127.3 136.8 144.3 152.8 161.3 170.0 ii\ 96.8 104.8 113.0 121.0 129.2 137.0 145.0 152.8 161.2 H 91.7 99.3 106.9 114.6 122.2 129.9 137.6 145.1 153.1 lA 87.5 94.7 102.1 109.4 116.8 124.0 131.2 138.2 146.0 LATHES 267 EXAMPLES 1. If the cutting speed of soft steel is 25, find the speed (spindle) for ly stock. 2. If the cutting speed of soft steel is 25, find the speed (spindle) for f J^" stock. 3. If the cutting speed of tool steel is 75, find the speed (spindle) for ^4'' stock. 4. According to the table find the speed of (a) soft steel 1^" in diameter; (b) soft steel ^" in diameter; (c) tool steel 2 J" stock ; (d) tool steel 2V' stock ; (e) soft steel ^" in diameter ; (/) soft steel if" in diameter. CHAPTER XVIII PLANERS, SHAPERS, AND DRILLING MACHINES Planers. — The planer is usually one of the heaviest machines in the machine shop. Planing is rough and heavy v^ork and stiffness is needed in order to make the heavy cuts the planer usually has on castings and heavy forgings. In many shops much of this rough surface cutting is saved, however, by the use of a pickling solution of one part of sulphuric acid to eight parts of water. This is painted over the casting and, after four or five hours, washed off with clear water, removing the sand and hard grit from the surface. Machine Shop Planer Care must be taken, during the operation of a planer, that chips and dirt from the platen are not swept or allowed to blow into the V-ways, as this will cause injury to the machine. Great care must be taken, also, that oil is freely supplied to the machinery under the planer platen, as these parts are hidden and likely to be neglected. A planer 24 inches by 24 inches by 6 feet will consume an average of 0.085 horse power for every pound of cast iron removed per hour, and a consumption of 0.066 horse power for every pound of n)achinery steel removed per hour under good operation with sharp tools. The operator 268 PLANERS AND DRILLING MACHINES 269 of a planer may, by poor grinding and not setting his tools, waste much power in driving the machine. The cutting speed for planers is about the same as for lathe tools, the advantage of the planer being ihat heavier cuts can be produced on account of the rigid support of the platen. The principal objection to planing machines is that they perform useful work in only one half of the motion. When the work is drawn back, no cutting takes place. Then again, siiice the forces are com- pletely changed when the motion is changed, the slides upon which the table moves, suffer. In this way accurate work is difficult on account of the backlash. The return on capital invested in such machines is smaller than in the case of machines with continuous motion. Planer and Shaper. — The planer and shaper, and such modified ma- chines as the slotting machine and key seater, are in a distinct cla.ss. Their use is to machine and plane irregular surfaces that can be machined by a straight line cut. The cutting tools of the planer and shaper are practi- cally the same as those on the lathe. In the planer the work moves, and vertical and lateral feeds are given to the tools. In the sliaper the tools move over the work, lateral to the work and vertical to the tool. The shaping machine is designed for small pieces and short travels. It is nicely adapted for cutting grooves, slots, and dovetails. Shaper tools should be kept in the best condition as to shear, clearance, and cutting edge, as they are called upon to do accurate shaping of metal parts of machinery not possible on other machines. Shear is a certain amount of angle given the face of a tool, which throws its cutting edge forward into the metal to be cut. Tools without clearance drag and pull heavily through the metal. For work on the shaper, the student should have a good knowledge of the small try-square and the surface block. These tools are constantly used, the square showing when finished pieces are square with the shap- ing machine vise or when one cut is square with another. A universal bevel or a bevel protractor should also be used, enabling angles to be laid out for planing; A scriber and a four-inch outside caliper will enable the beginner to grasp the first operations of shaping and planing. The ram should be adjusted to proper length of stroke to save time. If a piece of work which measures two and three fourths inches is to be placed on the machine, one fourth inch is sufficient for the tool to enter and one eighth inch is even more than enough to allow the head to overreach. These additions then give us a total stroke of three and one eighth inches, and any amount over this loses time and causes unnece8.sary wear on the machine. 270 VOCATIONAL MATHEMATICS EXAMPLES 1. If a planer has a cutting speed of 30 ft. per minute and a return speed of 147 ft. per minute, what is the ratio of the cutting to the return speed ? 2. On a 36" planer the ratio of the cutting speed to the re- turn speed of the table is 1 to 2.94, with a cutting speed of 68 ft. per minute. What is the return speed ? 3. It is necessary to plane a bench block on the top and bottom surfaces. An equal amount of stock is to be removed from each side, and the thickness of the casting should be re- duced from ly% in. to If in. What thickness of stock should be cut from the top surface ? 4. A piece of work on the planer is 10 in. thick ; it is re- duced to a certain size in 5 cuts ; at the first cut the tool takes off ^ in. from the thickness ; then i in. ; g\ in., ^i^- in., and the fifth cut is ^ly in. What is the thickness of the finished piece ? 5. On a cast iron block 6 in. square^ a groove y\" wide and I" deep must be planed. The planer makes 12 strokes per minute and the down feed is set at -^" for every stroke. How long will it take to cut the groove ? Drilling Machines. — Machines for drilling holes in the different pieces made in a machine shop are divided into two general classes — vertical and horizontal. The vertical drilling machines include those with a number of drill spindles called multiple spindles. Besides, there are special machines of both classes, as portable drills, hand drills, etc. The most common form of drill is the vertical drilling machine or drill-press. The machine consists of a frame supporting the drill spindle and the drilling table, and an arrangement for feeding the tool into the work by hand or power. On this machine the work to be drilled is 1 The information 6" square is not necessary for the solution of the problem, but is given to add interest to it. The same thing applies to some of the problems in drilling on the following pages, where the size of hole is given but not required. PLANERS AND DRILLING MACHINES 271 Slwimq Head Drillino Machine 272 VOCATIONAL MATHEMATICS placed on the drilling table, and is held stationary by means of a clamp or vise, while the revolving drill is fed through the work by hand or by power. The feeding mechanism is similar to that of the lathe. It is more convenient, usually, to drill small work in a speed lathe and to use the drill-press on heavy work. There are two classes of drills — straight and twist ; the twist drill being the latest and most approved in the leading shops. Whenever a manu- facturer is making standardized pai-ts, that is, uniform parts of the same machine, day after day, he designs and builds special drilling machines, also special milling and planing machines, gauges called jigs^ and gauges to produce the standard or duplicate parts. Power is transmitted from the pulley on the shaft to the countershaft at the bottom of the machine. Here are tight and loose pulleys so that the machine may be stopped by shifting the belt from the tight to the loose pulley. Different materials have qualities which make it necessary to use different cutting speeds in order to remove the metal quickly and efficiently. In order to provide for this a cone pulley giving a number of different speeds is used. The cone pulley on the countershaft is the same as the cone pulley on the headstock. The power is transmitted from the headstock to the drill spindle by bevel gears. The power is transmitted from the headstock shaft to the feeding spindle — which in turn lets the drill spindle down . Reamers. — Drills cannot be relied upon to make holes as round, straight, smooth, or uniform in diameter as are required in the construc- tion of accurate machinery. To make these holes accurately a tool called a reamer is used. It has two or more teeth, either parallel or at an angle with each other. The periphery of a drill or reamer is the distance around the outside, and the peripheral speed is the distance a point in the circumference travels in a minute. It is equal to the number of turns that the tool makes in a minute multiplied by the circumference. The proper speed at which to run a drill depends upon the kind of drill, its size, and the material to be drilled. A large drill must run more slowly than a smaller one, the turns per minute becoming less the larger the size of the drill. For instance, a drill \" in diameter should make twice as many revolutions per minute as a I" drill and four times as many as one 2" in diameter, used on the same material. The peripheral speeds usually recommended for carbon steel drills are as follows : Wrought iron or steel, 30ft. per min. High speed drills, 60 to 70 ft. per min. Cast iron, 35 ft. per min. High speed drills, 60 to 80 ft. per min. Brass, 60 ft. per min. High speed drills, 100 to 140 ft. per min. PLANERS AND DRILLING MACHINES 273 The feed of a drill is the amount the drill enters the hole for each turn or revolution, and in good practice the feed may be set to give 1" depth of the hole for every 96 to 125 revolutions, according to the size of the drill, the material of which the drill is made and the kind of metal drilled. Example. — How long will it take a one-inch drill, making 134 R. P. M., with a feed of .012" per revolution, to drill a hole 1^" deep in cast iron ? — ^— = 126 rev. 1 rev. = — of a minute .012 134 125 rev. = 126 X -^ = — = .933 of a minute. Ans. 134 134 Example. — A piece of wrought iron 2.69" thick is to have two If" holes drilled through it. If the drill makes 112 R. P. M., what must the feed be to drill each hole in two minutes ? 112 X 2 = 224 revs, in 2 min. ^i, of 2.69" = .012" feed. Ans. Example. — A 2" drill is used in drilling an electrical gen- erator bed plate. If the drill is making 67 R. P. M., and is being fed to the work at the rate of .015" per revolution, how deep will the hole be when the drill has worked 4| minutes ? 67 X .016" = 1.006" depth per min. 1.005x4.5=4.5225". Ans. Example. — What will be the R. P. M. of a drill If" in diameter used for drilling out a lathe spindle 30.24" long, the feed being .015" per revolution, and the time given the mar chinist to do the job being 42 minutes ? 80.24 OA1A 2016 Aan r> njr a :5T5- = 2016 .^ = 48R.P.M. Ans. Example. — Conditions being the same as in the above ex- ample, what length of a spindle could be drilled in 50 minutes ? 48 X .016 = .720" per min. .720" X 50 = 36" in 60 min. Ans. 274 VOCATIONAL MATHEMATICS EXAMPLES 1. How long will it take a carbon steel drill to drill a J" hole through a piece of wrought iron l|f" thick, if the drill makes 105 E. P. M., with the feed at .011'' per revolution ? 2. With the feed at .015'' per revolution and the speed at 115 E. P. M., how long will it take to drill a hole with a 2" carbon steel drill through a brass bushing 4|" long? 3. A machinist using a high speed drill If" in diameter, is to drill 4 bolt holes in the base of an electrical generator, the holes to be 2\" long, the drill to make 164 R. P. M., with the feed at .018" per revolution. How long will it take him to do the job if 3 minutes are allowed for the setting for each hole ? 4. A I" drill working on brass is running at 306 E. P. M. and advancing at the rate of .015" per revolution. How long will it take to drill through a piece lyV' thick ? 5. The E. P. M. of a carbon steel drill ly%" in diameter is 113, and the feed is .013" per revolution. How much time will a machinist require to drill 18 holes in a cylinder head 1^" thick, allowing one minute for the setting of each hole ? 6. A high speed drill If" in diameter is advancing at the rate of .018" per revolution, making 229 E. P. M. while drill- ing an angle iron If" thick. How long will it take the drill to go through it ? 7. In Example 6, what is the periphery speed in feet per minute of the drill ? 8. A cast iron casing for a steam turbine is to have a series of holes li" in diameter drilled in it 2^" deep with a high speed drill making 250 E. P. M. What must the feed be per revolution to drill each hole in f of a minute ? 9. The holes in a face plate 4" thick are to be drilled with a 1-in. carbon steel drill which makes 134 E. P. M. (a) What feed will be required to drill through the plate in PLANERS AND DRILLING MACHINES 275 2J minutes? (6) What will be the periphery speed in feet per minute of the drill ? 10. What must be the R. T. M. of a l.J" drill, feeding' at the rate of .013" per revolution, to drill a hole 3.'/' deep in an engine bed in 3 minutes ? 11. A 2" drill is used in drilling an electrical generator bed platen. If the drill is making 87 R. P. M., and is being fed to the work at the rate of .015" per revolution, how deep will the hole be when the drill has worked 4 J minutes ? 12. What will be the R. P. M. of a 1^" carbon drill used in drilling out an engine lathe spindle 24.05" in 24.GG minutes ? What will be the total number of revolutions? (Same rate of feed as in Ex. 11.) PART X — TEXTILE CALCULATIONS CHAPTER XIX YARNS Worsted Yarns. — All kinds of yarns used in the manufacture of cloth are divided into sizes based on the relation between weight and length. To illustrate : Worsted yarns are made from combed wools, and the size, technically called the counts, is Roving or Yarn Scales These scales will weigh one pound by tenths of grains or one seventy- thousandth part of one pound avoirdupois, rendering them well adapted for use in connection with yarn reels, for the numbering of yarn from the weight of hank, giving the weight in tenths of grains to compare with tables. based upon the number of lengths (called hanks) of 560 yards required to weigh one pound. Thus, if one hank weighs one pound, the yarn will be number one counts, while if 20 hanks 276 TEXTILE CALCULATIONS 277 are required for one pound, the yarn is the 20's, etc. The greater the number of hanks necessary to weigh one pound, the higher the counts and the finer the yarn. The hank, or 500 yai'ds, is the unit of measurement for all worsted yarns. Lknotu for Worsted Yarns No. Yards PRR Lb. N.». Yabdh PKR Lb. No. Yariw PER Lb. No. Yardk PKR Lh. 1 2 3 4 560 1120 1680 2240 6 6 7 8 2800 3360 3920 4480 10 11 12 6040 6600 6160 6720 13 14 16 16 7280 7840 8400 8960 Woolen Yams. — In worsted yams the fibers lie parallel to each other, while in woolen yarns the fibers are entangled. This difference is due entirely to the different methods used Yarn Reel For reeling and measuring lengths of cotton, woolen, and worsted yams. in working up the raw stock. In woolen yarns there is a great diversity of systems of grading, varying according to the dis- tricts in which the grading is done. Among the many systems 278 VOCATIONAL MATHEMATICS are the English skein, which differs in various parts of Eng- land ; the Scotch spyndle ; the American run ; the Philadelphia cut; and others. In these lessons the run system will be used unless otherwise stated. This is the system used in New England. The run is based upon 100 yards per ounce, or 1600 yards to the pound. Thus, if 100 yards of woolen yarn weigh one ounce, or if 1600 yards weigh one pound, it is technically termed a No. 1 run ; and if 300 yards weigh one ounce, or 4800 yards weigh one pound, the size will be No. 3 run. The finer the yarn, or the greater the number of yards necessary to weigh one pound, the higher the run. Length for Woolen Yarns (Run System) No. Taros PER Lb. No. Yards PER Lb. No. Yards PER Lb. No. Yards PER Lb. f 200 400 800 1200 1 If 1600 2000 2400 2800 2 ^ 2f 3200 3600 4000 4400 3 H 4800 5200 . 5600 Raw Silk Yarns. — For raw silk yarns the table of weights is: 16 drams = 1 ounce 16 ounces = 1 pound 256 drams = 1 pound The unit of measure for raw silk is 256,000 yards per pound. Thus, if 1000 yards — one skein — of raw silk weigh one dram, or if 256,000 yards weigh one pound, it is known as 1-dram silk, and if 1000 yards weigh two drams the yarn is called 2-dram silk, hence the following table is made : 1-dram silk = 1000 yards per dram, or 256,000 yards per lb. 2-dram silk = 1000 yards per 2 drams, or 128,000 yards per lb. 4-dram silk = 1000 yards per 4 drams, or 64,000 yards per lb. TEXTILE CALCULATIONS 279 ■ Dramr i'br 1000 Yards Yards prr Pound Yardh prr Ounck 1 2r)12 yards of 60's cotton. 39. Find the weight in grains of 118 yards of 44's linen. 40. Find the weight in pounds of 315 yards of 32's linen. 41. Find the weight in grains of 84 yards of 25's worsted. 42. Find the weight in grains of 112 yards of 20's woolen. 43. Find the weight in grains of 197 yards of 16's woolen. 44. Find the weight in grains of 183 yards of 18's cotton. 45. Find the weight in grains of 134 yards of 28's worsted 46. Find the weight in grains of 225 yards of 34's linen. 47. Find the weight in pounds of 369 yards of 16's spun silk. 48. Find the weight in pounds of 484 yards of 18's spun silk. To find the Size or the Counts of Cotton Yarn of Known Weight and Length Example. — Find the size or counts of 84 yards of cotton yarn weighing 40 grains. Since the counts is the number of hanks to the pound, ^ X 84 = 14,700 yd. in 1 lb. 14,700 H- 840 = 17.5 counts. Ans. Rule. — Divide 840 by the given number of yards ; divide 7000 by the quotient obtained ; then divide this result by the weight in grains of the given number of yards, and the quotient will be the counts. 840 - 84 = 10 7000 -f- 10 = 700 700 -^ 40 = 17.5 counts. Ans. To find the Ran of a Woolen TJiread of Knoivn Length and Weight Example. — If 50 yards of woolen yarn weigh 77.77 grains, what is the run '/ 1«00 ^ 50 = 32 7000 -f- 32 = 218.75 218.75 - 77.77 = 2.812-run yarn. Ans. TEXTILE CALCULATIONS 285 Rule. — Divide 1600 (the number of yards per pound of 1- run woolen yarn) by the given number of yards ; then divide 7000 (the grains per pound) by the quotient; divide this (juotient by the given weight in grains and the result will be the run. To find the Weight in Ounces for a Oiven Number of Yards of Worsted Yarn of a Known Count Example. — What is the weight in ounces of 12,650 yards of 30's worsted yarn ? 12,650 X 16 = 202,400 202,400 ^ 16,800 = 12.047 oz. Ans, Rule. — Multiply the given number of yards by 16, and divide the result by the yards per pound of the given count, and the quotient will be the weight in ounces. To find the Weight in Pounds for a Oiven Number of Yards of Worsted Yai'n of a Known Count Example. — Find the weight in pounds of 1 ,500,800 yards of 40's worsted yam. 1,500,800 -4- 22,400 = 67 lb. Ans. Rule. — Divide the given number of yards by the number of yards per pound of the known count, and the quotient will be the desired weight. EXAMPLES 1. If 108 inches of cotton yarn weigh 1.5 grains, find the counts. 2. Find the size of a woolen thread 72 inches long which weighs 2.5 grains. 3. Find the weight in ounces of 12,650 yards of 2/30's worsted yarn. 4. Find the weight in ounces of 12,650 yards of 40's worste = 1.1283 xV]4 11. The number of miles in a given length, expressed, in feet, may be obtained from the formula 3f=.00019xF 12. The number of cubic feet in a given volume expressed in gallons may be obtained from the formula 0=. 13367 X Q 13. Contractors express excavations in cubic yards; the number of bushels in a given excavation expressed in yards may be obtained from the formula C=.0495x Y 14. The circumference of a circle may be obtained from the area by the formula C= 3.5446 X V2 15. The area of the surface of a cylinder may be expressed by the formula A = (^C X L) -\-2a When C = circumference L = length a = area of one end 16. The surface of a sphere may be expressed by the formula S=D'x 3.1416 17. The solidity of a sphere may be obtained from the formula S = D'x .5236 18. The side of an inscribed cube of a sphere may be ob- tained from the formula S = R X 1.1547, where S = length of side, R = radius of sphere. 304 VOCATIONAL MATHEMATICS 19. The solidity or contents of a pyramid may be expressed by the formula F S = Ax^, where A = area of base, F = height of pyramid. 20. The length of an arc of a circle may be obtained from the formula L = Nx .017453 E, where L = length of arc, N= number of degrees, a = radius of circle. 21. The horse power of a single leather belt may be deter- mined by the formula DRW HP = ^ , where D = diameter of pulle}^ in inches, W— width of belt in inches, M = revolutions per minute, HP = horse power transmitted. 22. The formula for finding the weight of an iron ball may be calculated by the following : Tr= 2)3x0.1377 23. The formula for finding the diameter of an iron ball when the weight is given is D = 1.936 Vl^ where D = diameter of the ball in inches, W= weight of ball in pounds. 24. The volume of a sphere when the circumference of a great circle is known may be determined by the formula 25. The diameter of a circle the circumference of which is known may be found by the formula FORMULAS 305 26. The area of a circle the circumference of which is known may be found Jby the formula 4 tr Coefficients and Similar Terms When a quantity may be separated into two factors, one of these is called the coefficient of the other ; but by the coefficient of a term is generally meant its numerical factor. Thus, 4 6 is a quantity composed of two factors 4 and 6 ; 4 is a coef- ficient of h. Similar terms are those that have as factors the same letters with the same exponents. Thus, in the expression, 6 a, 4 6, 2 a, 6 a6, 5 a, 2 6. 6 a, 2 a, 5 a are similar terms ; 4 6, 2 & are similar terms ; 5 ab and 6 a are not similar terms because they do not have the same letters as factors. 8 a&, 6 ah, 1 aft, 8 ah are similar terms. They may be united or added by simply adding the letters to the numerical sum, 17 aft. In the following, 8 ft, 6 ft, 3 aft, 4 a, aft, and 2 a, 8ft and 5ft are similar terms ; 3 aft and aft are similar terms ; 4 a and 2 a are similar terms ; 8 ft, 3 aft, and 4 a are dissimilar terms. In addition the numerical coefficients are algebraically added ; in subtraction the numerical coefficients are algebraically sub- tracted ; in multiplication the numerical coefficients are alge- braically multiplied ; in division the numerial coefficients are algebraically divided. EXAMPLES State the similar terms in the following expressions : 1. 5 a;, 8 OLc, 3 a;, 2 ax. 6. 15 ahc, 2 aJbc, 4 dbc^ 2 aft, 2. 8aftc^ 7c, 2aft, 3c, ^ah, 3aZ>. 9«*c. 7. Saj, 6a;, 13xy, 6x, 7y. 3. 2pq, 5p,Sq, 2p, 3g, ^pq. a 7y, 2y, 2xy, 3y, 2xy, 4. 4.V, 5y2, 2y,ir)2,52, 2.V2. ir . „ « 1ft r K A o ^ 2ir, 5xr*, -, irr», 2xr. 5. lo mn, m, 5 7i, 4 mny 2 m. J 306 VOCATIONAL MATHEMATICS Equations A statement that two quantities are equal may be expressed mathematically by placing one quantity on the left and the other on the right of the equality sign (=). The statement in this form is called an equation. The quantity on the left hand of the equation is called the left-hand member and the quantity on the right hand of the equation is called the right-hand member. An equation may be considered as a balance. If a balance is in equilibrium, we may add or subtract or multiply or divide the weight on each side of the balance by the same weight and the equilibrium will still exist. So in an equation we may perform the following operations on each member without changing the value of the equation : We may add an equal quantity or equal quantities to ea/ih memr her of the equation. We may subtract an equal quantity or equal quantities from each member of the equation. We may multiply each member of the equation by the same or equal quantities. We may divide each member of the equation by the same or equal quantities. We may extract the square root of each member of the equation. We may raise each member of the equation to the same poiver. The expression, A = ttR^ is an equation. Why ? If we desire to obtain the value of R instead of A we may do so by the process of transformation according to the above rules. To obtain the value of R means that a series of opera- tions must be performed on the equation so that R will be left on one side of the equation. (1) A = 7rJ?2 A (2) — = R^ (Dividing equation (1) by the coefficient of B^.) IT (^) -\/~ = -^ (Extracting the square root of each side of the equation.) FORMULAS 307 Methods of Representing Operations Multiplication The multiplication sign ( X ) is used in most eases. It should not be used in operations where the letter (x) is also to be em- ployed. Another method is as follows : 2.3 a. 6 2 a. 36 4a;. 5a This method is very convenient, especially where a number of small terms are employed. Keep the dot above the line, otherwise it is a decimal point. Where parentheses, etc., are used, multiplication signs may be omitted. For instance, (a + b)x{a — h) and (a -\- b'){a — b) are identical ; also, 2'(x — y) and 2{x —y). The multiplication sign is very often omitted in order to simplify work. To illustrate, 2 a means 2 times a ; 5 xyz means 5 • X • y • z ; x{a — b) means x times (a — b), etc. A number written to the right of, and above, another (a:*) is a sign indicating the special kind of multiplication known as involution. In multiplication we add exponents of similar terms. Thus, «* . a^ = a:»+' = ar* (ibc ' ab • aVj = a*bh The multiplication of dissimilar terms may be indicated. Thus, a-b ' C' x-y 'Z = abcxyz. Division The division sign (-*-) is used in most cases. In many cases, however, it is best to employ a horizontal line to indicate division. To illustrate, means the same as (a -\' b) -i- x-y (x — y) in simpler form. The division sign is never omitted. 308 VOCATIONAL MATHEMATICS A root or radical sign (V^> y/^) is a sign indicating the special form of division known as evolution. In division, we subtract exponents of similar terms. Thus, ar»H-aj' = - = a^-2 = a; a^lfif^ ^ a'^bc^ = ^^^^^ = a^bc. a^bc^ The division of dissimilar terms may be indicated. Thus, (abc) -i- xyz = xyz Substituting and Transposing A formula is usually written in the form of an equation. The left-hand member contains only one quantity, which is the quantity that we desire to find. The right-hand member contains the letters representing the quantity and numbers whose values we are given either directly or indirectly. To find the value of the formula we must (1) substitute for every letter in the right-hand member its exact numerical value, (2) carry out the various operations indicated, remem- bering to perform all the operations of multiplication and division before those of addition and subtraction, (3) if there are any parentheses, these should be removed, one pair at a time, inner parentheses first. A minus sign before a parenthesis means that when the parenthesis is removed, all the signs of the terms included in the parenthesis must be changed. Find the value of the expression 3d-\-b(2a-b-\- 18), where a = 5,b = 3. Substitute the value of each letter. Then perform all addition or subtraction in the parentheses. 3x6-1- 3(10 - 3 + 18) 16-^.3(28-3) 15 + 3(26) 15 4-75 = 90 FORMULAS 309 EXAMPLES Find the value of the following expressions : 1. 2 ^ X (2 + 3 ^) X 8, when J = 10. 2. 8 o X (6 — 2 a) X 7, when a = 7. 3. 8 6 + 3 c + 2 rt (a -f /> + c) - 8, when a = 9; 6 = 11 ; c = 13. 4. 8(a; + y), when a? = 9; y = 11. 5. 13 {x- y), when a; = 27 ; y = 9. a 24y-|-8z(2 + y)-3y, when jy = 8; z = ll. 7. Q{(^M-\-^N)-^2 0, when Jlf=4, xV=5, Q=6, = 8. a Find the value of X in the formula X = ^^^^^"^^^ when J»f = 11, iV^= 9, P = 28. 9. .c = ^.^L±^, when n = 5, m = 6, P = 8, Q = 7. 10. Find the value of T in the equation (x + y)(x-y) 11. 3a+4(6-2a + 3c)-c, when a = 4, 6 = 6, c = 2. 12. op — Sq(p + r — S) — qy when /) = 5, g = 7, r = 9, i5 = 11. 13. S^^t*-p'i-3(S-\-t + p)j whenp = 5, S=Sy t = 9. 14. a* - &>+ c», when a = 9, 6 = 6, c = 4. 15. (a + 6) (a + 6 - c), when a = 2, 6 = 3, c = 4. 16. (a* - b*) (a* + 6*), when a = 8, 6 = 4. 17. (c-» + cP) (c» - rf»), c = 9, (/ = 5. 18. Va* + 2 a6 + 6«, when a = 7, 6 = 8. 19. ^c*-61, when c = 5. 310 VOCATIONAL MATHEMATICS PROBLEMS Solve the following problems by first writing the formula from the rule on page 300, and then substituting for the answer. 1. How many electrical horse power in 4389 watts ? 2. How many kilowatts in 2389 watts ? 3. (a) Give the number of watts in a circuit of 110 volts and 25 amperes. (6) How many electrical horse power ? 4. What is the voltage of a circuit if the horse power is 2740 watts and the quantity of electricity delivered is 25 amperes ? 5. What is the resistance of a circuit if the voltage is 110 and the quantity of electricity is 25 amperes ? 6. What is the pressure per square inch of water 87 feet high? 7. What is the capacity of a cylinder with a base of 16 square inches and 6 inches high ? (Capacity in gallons is equal to cubical contents obtained by multiplying base by the height and dividing by 231 cubic inches.) 8. What is the length of a 30° arc of a circle with 16" diameter? 9. What is the area of a sector with an arc of 40° and a diameter of 18"? 10. What is the weight of the rim of a flywheel of a 25 H. P. engine ? 11. What is the area of a cylinder of a 50 H. P. engine with the piston making 120 ft. per minute ? 12. What is the H. P. of a 2ff'' shaft making 180 R. P. M. ? 13. What is the capacity of a pail 14'' (diameter of top), 11" (diameter of bottom), and 16" in height ? 14. What is the area of an ellipse with the greatest length 16" and the greatest breadth 10" ? FORMULAS 311 Interpretation of Negative Quantities The quantity or number — 12 lias no meaning to us according to our knowledge of simple arithmetic, but in a great many problems in practical work the minus sign before a number assists us in understanding the different solutions. To illustrate : FaBRSNHKIT TllF.BlIOllBTKR Cbntiorape Tiibrmombtkr Boilinf? point of Freezinjf point of ■ water 212* Roilinsr point uf water 82* 2. a. Freezing point uf ' water 100» On the Centigrade scale the freezing point of water is marked 0°. Below the freezing point of water on the Centigrade scale all readings are expressed as minus readings. — 30° C means thirty degrees below the freezing point. In other words, all readings, in the direction below zero are expressed as — , and all readings above zero are called -|-. Terms are quantities connected by a plus or minus sign. Those preceded by a plus sign (when no sign precedes a quan- tity plus is understood) are called positive quantities, while those connected by a minus sign are called negative quantities. 312 VOCATIONAL MATHEMATICS Let us try some problems involving negative quantities. Find the corresponding reading on the Fahrenheit scale cor- responding to — 18° C. F = I C + 32° F= |(- 18°)+32° Notice that a minus quantity is placed in parenthesis when it is to be multiplied by another quantity. F =- ip° + 32" = - 32f° + 32° ; F = - |°. The value — §° is explained by saying it is § of a degree below zero point on Fahrenheit scale. Let us consider another problem. Find the reading on the Centi- grade scale corresponding to — 40° F. Substituting in the formula, we have C = s (_ 40° _ 32C) ^ |(_ 72) = _ 40^ Since subtracting a negative number is equivalent to adding a positive number of the same value, and subtracting a posi- tive number is equivalent to adding a negative number of the same value, the rule for subtracting may be expressed as fol- lows: Change the sign of the subtrahend and proceed as in addition. For example, 40 minus — 28 equals 40 plus 28, or 68. 40 minus + 28 equals 40 plus — 28, or 12. — 40 minus 4- 32 equals — 40 plus — 32 = - 72. (Notice that a positive quantity multiplied by a negative quantity or a negative quantity multiplied by a positive quantity always gives a negative product. Two positive quantities multiplied together will give a positive product, and two negative quantities multiplied together will give a positive product.) To illustrate : 5 times 5 = 5 x 5 = 25 5 times — 5=r5x(— 5) = — 25 (-5) times (-5) = + 25 In adding positive and negative quantities, first add all the positive quantities and then add all the negative quantities FORMULAS 313 together. Subtract the smaller from the larger and prefix the same sign before the remainder as is before the larger number. For example, add : 2a, 5a, - Oa, 8a, -2a 2a + 6a -I- 8a = 16a; -6a-2a=-8a T5a-8a = 7a EXAMPLES Add the following terms : 1. 3Xf —X, 7 Xf AXj —2x. 2. 6y, 2y, 9y, -7y. 3. 9 aby 2 a6, 6 aft, — 4 aft, 7 a6, — 5 ab. Multiplication of Algebraic Expressions Each term of an algebraic expression is composed of one or more factors, as, for example, 2 ab contains the factors 2, a, and b. The factors of a term have, either expressed or understood, a small letter or number in the upper right-hand corner, which states how many times the quantity is to be used as a factor. For instance, ab\ The factor a has the exponent 1 understood and the factor b has the exponent 2 expressed, meaning that a is to be used once and b twice as a factor, ab'^ means, then, a X 6 X 6. The rule of algebraic multiplication by terms is as follows: Add the exponents of all like letters in the terms multiplied and use the result as exponent of that letter in the product. Multiplication of unlike letters may be expressed by placing the letters side by side in the product. For example : 2 a6 x .3 6^ = a6« 4ax86 = 12a6 Algebraic or literal expressions of more than one term are multiplied in the following way : begin with the first term to the left in the multiplier and multiply every term in the multi- plicand, placing the partial products underneath the line. Then 314 VOCATIONAL MATHEMATICS repeat the same operation, using the second term in the multi- plier. Place similar products of the same factors and degree (same exponents) in same column. Add the partial products. Thus, a -\- b multiphed by a — 5. a + h a — h a^+ ab- b-^ -ah a^ - b-^ Notice the product of the sum and difference of the quantities is equal to the difference of their squares. EXAMPLES 1. Multiply a-\-hhy a-\-h. State what the square of the sum of the quantities equals. 2. Multiply X — y hy X — y. State what the square of the difference of the quantities equals. 3. Multiply (j) + q)(p — q). 7. Multiply (x — y)(x — y). 4. Multiply (p -f q)(p + q). 8. (x -\-yy=? 5. Multiply (r + s){r - s). 9. {x - yf = ? 6. Multiply (a ± 6)(a ± 6). 10. (x + y){x-y) = ? LOGARITHMS Thk logarithm of a iuiiuImt to the base 10 is defined as the povrer to Nvhicli 10 must be raiseil in order to equal the number. Thus the logarithm (or log, as it is more generally written) of 10 Ls 1, for the first power of 10 is 10. The log of 100 is 2, for the second power of 10 is 100. Hence the log of a number between 10 and 100 is a number between 1 and 2, and is, therefore, 1 plus a decimal. The whole number 1 is called the characteristic, and the decimal part is called the mers 7.21, 72.1, and 721. 721. = 10 X 72.1 = 10* X 7.21. Therefore, log 721 = log 10 + log 72.1 = log 10* + log 7.21. (In order to multiply the numbers, we may add the logarithms of these numbers.) Therefore, log 721 = 1 + log 72.1 = 2 + log 7.21, (log 10 = 1 ; log 10* = 2). Therefore, we see that the logarithms of numbers made up of the same sequence of 315 316 VOCATIONAL MATHEMATICS digits, but differing in the position of the decimal point, differ only in the characteristic, the mantissce remaining the same. For this reason, in making tables of logaritlims, only the mantissse are given, the characteristics being added according to the rules above. The tables we are to use are made for three significant digits in the number and four digits in the mantissa of the log. To Find the Logarithm of a Number A. When the number has three significant figures. The first two figures are found in the column headed No. (pages 318-319), and the third figure in the top row. The mantissa is stated in the column and row so determined. To illustrate, find the log of 62.8. In the 62 row and in the col- umn under 8 is the mantissa, 7980. By Rule I we find the charac- teristic to be 1 ; therefore, the log of 62.8 is 1.7980. Find the log of .00709. The significant figures are 709; in the 70 row and in the column under 9, we find the mantissa 8506. By Rule II the characteristic is — 3. Therefore the log of .00709 is — 3.8506. To prevent confusion in the operations of arithmetic — 3 is usually written 3, so the log is stated to be 3.8506, or 7.8506 — 10. EXAMPLES Find log of 117 ; of 1280 ; of 16.5 ; of 2.09 ; of .721 ; of .0121. B. When the number has four or more significant figures. Ex. 1. — Find the log of 7987. The tables give only the mantissa of numbers of three figures, so the mantissa of 7987 cannot be found in the table ; 7987, however, lies between 7980 and 7990, hence its log is between the log of 7980 and the log of 7990. We find the mantissa 7980 to be 9020, and the mantissa of 7990 to be 9025. Now the difference between these mantissas is 5, while the difference between the two numbers, 7980 and 7990, is 10. But the difference between 7980 and 7987 is ^jj of the difference between 7980 and 7990, and the mantissa of 7987 will be /^ of the difference between the mantissa of 7980 and the mantissa of 7990. Therefore it will equal 9020 (the mantissa of 7980) plus yV of 5 (the difference between the mantissse of 7980 and 7990) : 5 x .7 = 3.5, hence the mantissa of 7987 is 9024. In such reckoning all decimal parts less than .5 are counted as 0, and all decimal parts greater than .5 are counted as 1 ; .5 is counted as either 1 or 0. Log 7987 is .3^024. LOGARITHMS 317 Ex. 2. — Find the log of 12.564. First find the mantissa. In looking for the mantissa the decimal {H)int need not be considered. Solution. — 12554 lies between 12500 and 12600. Mantissa of 12600 is 1004. Mantissa of 12.500 is 0969. 35 is the difference between the mantissa;. 12600 - 12500 = 100 = difference between numbers. 12554 — 12500 = 54 = difference between original numbers. The multiplier is f*^ = .54. 35 X .54 = 18.90 or 19. 0969 -f 19 = 0988. log 12.564 is 1.0988. Ans. EXAMPLES Find log of 17.89 ; of 2172 ; of 652.12 ; of 4213 ; of 3342000. C To find the number corresponding to a given logarithm. Ex. 1. — What number has the log 1.6085? The mantissa 6085 is found in the 40 row and in the column under 6. Tiie number corresponding to mantissa 6085 is therefore 406. The characteristic 1 states that there are two digits to the left of the decimal point. The number is therefore 40.6. Ex. 2.— What number has the log 5.8716? The mantissa 8716 is in the 74 row and in the column under 4. The characteristic 5 states that there are six digits to left of decimal point. The number is therefore 744000. Ex. 3. — What number has the log ,2.6538? The mantissa 6538 is not found in the tables, but lies between 6532 and 6542, hence the number (not considering the decimal point) lies between 4.50 and 451. The difference between the mantissa; of 4.50 and 451 is 10; the difference between the mantissa of 450 and of the number to be found is 6. The difference between 450 and 451 is 1. 1 X .6 = .6. The number corresponding to mantissa 65.38 is 4506, hence the numl)er corre.sponding to log 2.6538 is 4.50.6. EXAMPLES Find number corresponding to log 1.5481; to log 0.6681; to log 1.9559; to log 2.9324. 31S LOGARITHMS OF NUMBERS No. 1 2 3 4 7 8 9 10 OOCX) 0043 0086 0128 0170 0212 0253 0294 0334 0374 IX 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1 106 13 1 139 1173 1206 1239 1271 1303 1335 ■^3^1 1399 1430 14 1 461 1492 1523 1553 1584 1614 1644 1673 1703 1732 15 1 761 1790 1818 1847 1875 1903 1931 1959 1987 2014 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 43H 4330 4346 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 48CX) 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 501 1 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 55^3 5575 5587 5599 561 1 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5786 38 5798 5809 5821 5832 5843 5855 ^Hl 5888 5899 39 59" 5922 5933 5944 5955 5966 5977 5988 5999 6010 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 691 1 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7IOI 7110 7118 7126 7135 7143 7152 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 3 7356 7364 7372 7380 7388 7396 No. 1 2 4 5 6 7 8 9 LOGARITHMS OF NUMBERS 310 Ko. 1 2 3 4 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 745J 7459 7466 7474 56 74S2 7490 7497 7505 75U 7520 7528 7536 7543 755» 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 6x 7853 7S60 7808 7875 78S2 7889 7896 7903 7910 7917 6a 63 7924 7993 ?^ 7938 8007 7945 8014 7952 8021 ^51 7966 8035 7973 8041 c 79S7 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8II6 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 821S 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8::S7 8293 8299 8306 8312 8319 C8 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 7a 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76^ 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 8x 9085 9138 9090 9096 9101 9106 9112 9117 9122 9128 9»33 8a 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 ga 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 968s 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 S7 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 9 Ho. 1 1 2 ! 3 4 5 6 7 8 TRIGONOMETRY There are many problems in the shop that involve a knowledge of angles and the relation between the parts of triangles — angles and sides. Such problems include laying out angles, depth of screw threads, diagonal distance across bolts, etc. Trigonometry is the subject which deals with the properties and measurement of angles and sides of triangles, and is spoken of in the machine shop as simply " trig." Trigonometric Functions (Ratios). — Since the sum of all the angles of a triangle equals 180^, and since a right triangle is composed of a right angle and two acute angles, it follows that, if we know one acute angle, we may obtain the other by subtracting it from 90°. The angle found by subtracting a definite angle from 90° is called the complement of the given angle. The complement of 45° is 45°, for 90° - 45° = 45°. The complement of 30° is 60°, for 90° - 30° = 60°. The supplement of an angle is the difference between it and 180°. The supplement of 60° is 120°, for 180° - 60° = 120°. Examine a right angled triangle very carefully. Notice the position of the two acute angles and of the right angle. The longest side of the triangle, called the hypotenuse, is op- posite the right angle. The sides of the triangle called the legs, are the two smallest sides. These sides are perpen- dicular to each other, and the longer side is opposite the greater angle. Draw a right angle with sides re- .^ spectively 3 and 4 inches long. _ . _ „, , .1 , , , Right Angled Triangle Then draw the hypotenuse and meas- ure the length of it, which will be 5 inches. Divide the length of the side opposite Z A hy the hypotenuse, 3 ^ 5 = .6. Divide the length of the side adjacent to Z A by the hypotenuse, 4 -f- 5 = .8. Divide the length of the side opposite Z A hy the adjacent side, 3 -i- 4 = .75. 320 TRIGONOMETRY 321 These values are called ratios. The ratio of the length of the side opposite Z A to the length of the hypotenuse is called the sine oi ^ A. The ratio of the adjacent Z ^ to the length of hypotenuse is called the cosine of Z A. The ratio of the length of side opposite Z A to the length of the adjacent leg is called the tangent oi Z A. The ratio of the length of the adjacent side of Z ^4 to the opposite side is called the cotangent. The ratio of the length of the hypotenuse to the adjacent side of Z A IB called the secant. The ratio of the length of the hypotenuse to the opposite side of Z A is called the cosecant. These six ratios or constants represent a definite relation between the sides and angles of right triangles. Since all triangles may be divided into right angled triangles, by dropping a perpendicular from one of the vertices of the triangle, we might say these relations apply to all triangles. The same relations apply to all figures, since any figure may be divided into triangles, and then into right angled triangles. These definite relations between the sides and angles are called the functions of the angles, and are i*eally constants representing the fixed proportions between the sides and angles of a triangle. The exact values of these functions, or the proportion of the various parts of the triangles, have been figured out for every degree that will be used in daily practice. They are given in the tables for sine, cosine, and tangent at the end of this chapter. By means of a protractor construct an angle of 45°, Z CAB. At a point on ^ C 1" from A^ erect a perpendicular, CB. Connecting parts A and B by the line AB, we have a rt. A with BC = AC, because Z A — 45° and Z B = 45°. Measure accurately the length of all the sides of the triangle. Divide the length of side opposite Z A by the hypotenuse, — '- — .7. ^^ AB AC Divide the length of side adjacent to ^A by the hypotenuse, — = .7. AB 322 VOCATIONAL MATHEMATICS Divide the length of side opposite Z A by the side adjacent to ZA, AC Divide the length of the adjacent side by the side opposite ZA, BC Divide the length of the hypotenuse by the side adjacent to ZA, i^ = 1.4. AC Divide the length of the hypotenuse by the side opposite ZA, ^^1.4. BC If the work is carried out accurately, the above values should be obtained. Find the value of sine, cosine, and tangent when angles of 25^, 30°, 60°, 75°, are constructed. The reciprocal of the tangent of an angle {Z A) is called the co- tangent. The reciprocal of the sine of an angle (^Z A) is called the cosecant. The reciprocal of the cosine of an angle {ZA) is called the secant. If the angles of the triangles are always marked with capital letters, and the length of the sides opposite the angles are marked with small letters as shown below, it will be possible to abbreviate functions of the angle in terms of the sides. Trigonometrical Formulas, etc. Geometrical Solution of Right Angled Triangles. c = Va^ + 62 h = Vc^ - a2 a = Vc2 - 62 . a opposite side „ j sin A=- = -i-s- cos A c hypotenuse c hypotenuse A _a_ opposite side + j _ ^ _ adjacent side b adjacent side a opposite side . c hypotenuse ^^c«« a c hypotenuse sec ^ = - = --4-t^ ^^ cosec A = -= — •^^^-^~ — ^^ b adjacent side a opposite side THIGONOMETHY 323 Practical Value of a Trigonometric Ratio Since each function of an angle may be expressed as the ratio of two sides, it follows that if we know the size of an angle and the length of one side of a right triangle, we may determine the other sides by sub- stituting in the equation. Ex. If one angle of a right angled triangle is 30^, and the adjacent side is 25 inches, what is the length of the other leg ? D Since we desire the other leg we should ust (Formula involving two legs) Tangent ^A=^ 25 Ix>oking up tables for tangent .1, we find it to be .5773.") ; substitut- ing, we have .577 = — a = 14.4 25 EXAMPLES 1. The simplest application of trigonometry in the shop is the problem for finding the depth of a V screw thread. Since the angle at the depth of the screw thread is 60° and the sides are equal, an equi- lateral triangle is formed. Forming right angled triangles by drop- ping a perpendicular from the angle of the thread, we divide the angle of 00° into two angles of 30'^ each. If we consider the pitch 1 inch, then the sides are each an inch and the perpendicular divides the base into two parts of \ inch each. The perpendicular, or depth of the thread, is the cosine of the angle of 30°. Looking in the table for 30°, then across until we come to the column headed cosine, we find .8. This gives us the depth directly. 2. Find the side of the thread if the depth is 1 inch. 3. What is the depth of a V thread with 8 threads to the inch ? 4. What is the distance across the corners of a square bar 3^" on each side ? 324 VOCATIONAL MATHEMATICS 6. What size circular stock will be required to mill a square nut lj\" on a side? 6. What is the distance across the corners of a bar 2^" by |"? What angle does the diagonal make with the base? Natural Trigonometric Functions The tables on pages 326-329 give the numerical values of the trigonometric functions of angles between 0° and 90° with inter- vals of 10 minutes. For angles from 0° to and including 44°, read from the top of the table downwards; for angles from 45° to and in- cluding 89°, read from bottom of table upwards. The degrees and minutes are found in the column marked A or Angle, and the value of the function in the column marked by the name of the function. For instance, the value of the sine of 15° 40' is found, in the column marked sine and in the row 40 under 15, to be .2700. The value of the cotangent of 63° 10' is found (reading from the bottom of the table upwards, since the angle is between 45° and 89°), in the column marked cotangent (at bottom of table) and in the row marked 10 above 63, to be .5059. The values of the sine, tangent, and secant in- crease as the angle increases, while the values of the cosine, cotangent, and cosecant decrease as the angle increases, so that for the former functions the correction must be subtracted from the value given in the table. I. To find the value of a function of an angle which is not given in the table. Ex. 1. Find the value of sine 35° 21'. Angle 35 is given in table, but 21' is not. 21', however, lies between 20' and 30', hence the value of sine of 35° 21' will lie between the value of sine of 35^ 20' and the value of sine of 35° 30'. The value of sine of 35° 20' is given in table as .5783, and the value of sine of 35° 30' is .5807 ; the difference between these two values is .0024. The tabular difference in the angles (that is, the angular difference between 35° 20' and 35° 30') is 10', \\4iile the angular difference between the smaller angle given in the table (35° 20') and the angle of which we are trying to find the value (35° 21') is 1'. The correction which we will have to add to .5783, the value of sine of 35° 20', is j\ x .0024 = .00024. Hence the value of sine of 35° 21' is .5783 -1- .00024 = .57854. TRIGONOMETRY 325 Ex. 2. Find the value of cotangent 82° 11'. 82*' 41' lies between 82° 40' and 82° 50 cotan82°40' = .12869 cotan 82° oO' = .12574 Difference = .00295 The tabular difference l>etween angles 82° 40' and 82° 50' is 10'. The difference between the smaller angle and given angle (82° 41' — 82° 40') is 1'. The correction is therefore y^^ x .00295 or .000295, but since the last figure is 5 we called it .00300. This correction must be subtracted from the value of cotan 82° 40' since, as above, the value of the cotan decreases as the angle increases. cotan 82° 41' = .12869 - .00.3 = .12569 EXAMPLES Find value of tan 45° 19' ; cosine 32° 8' ; cotan 78° 51' ; sine 62° 37'. II. To find the value of an angle, when the value of a function is given. Ex. 1. Find the angle whose cotan is .6873. In the column marked cotan look for the numbers .6873. Since it is found in the column marked cotan at the bottom of the table, the degrees and minutes will be found at the right-hand side of the table, reading from the bottom upwards. It is in the row marked 30 above 5."). Hence, angle 55° 30' has the cotan .6873. Ex. 2. Find the angle whose cosine is .9387. This number is found in the column marked cosine at the top of the table, therefore the degrees and minutes will be found on the left of the table reading from top downwards. It is in the row marked 10 under 20. Hence, the angle 20"^ 10' has the cosine .9387. EXAMPLES Find the angle that has the sine .7642; cotan 1..5607; cosine .4746; tangent 1.5108. 326 NATURAL SINES AND COSINES A. Sin. Cos. A. Sin. Cos. |A. Sin. Cos. 0° lO' 20' 30' 40' 50' 1° 10' 20'- 50' 2° 10' 20' 30' 40; 50' 3° 10' 20^ 30; 40 50' 40 10' 20' 30; 40' 50' 5° icy 20' 30; 40' 50' 6° 10' 20' 30; 40' 50' 70 10' 20' 30' .000000 1. 0000 90° 50; 40' 30' 20' 10' 89° 50; 40' 30; 20' 10' 88° 50; 40' 30; 20' 10' 87° 50; 40' 30; 20' 10' 86° 50; 40' 30' 20' 10' 85° 50; 40' 30; 20' 10' 84° 50; 40' 30; 20' 10' 83° 50; 40' 30' 30; 40 50' 8° 10' 20' 30; 40' 50' 9° 10' 20' 30' 40' 50' 10° 10' 20' 30' 40' 50^ 11° 10' 20' 30; 40 50' 12° 10' 20^ 30; 50^ 13° 10' 20' ^^' 40^ 50' 14° 10' 20' 30^, 40' 50' 15° •1305 .1334 .1363 .9914 •99 1 1 .9907 30; 20' 10' 82° 50' 40' 30; 20' 10' 81° 50; 40' 30' 20' 10' 80° 50; 40' 30; 20' 10' 79° 50; 30^ 20' 10' 78° 50; 40' 30' 20' 10' 77° 50; 40 30 20' 10' 76° 50; 40' 30; 20' 10' 75° 15° 10' 20^ 30; 40' SO' 16° 10' 20' 30; 40 50' 17° 10' 20' 40^ 50' 18° 10' 20' 30; 40' 50' 19° 10 20' 30' 40' 50' 20° 10' 20' 30' 40; 50' 21° 10' 20' 30/ 40 50' 22° lo' 20' 30' .2588 •9659 75° 50; 40' 30; 20' 10' 74° 50; 40' 30; 20' 10' 73° 50; 40' 30' 20' 10' 72° 50' 40' 30' 20' 10' 710 40^ 30' 20' 10' 70° 40^ 30' 20' 10' 69° 5°,' 40' 30; 20' 10' 68° 50; 40' 30^ .002909 .005818 .008727 .011635 .014544 1. 0000 I.UUOU 1. 0000 .9999 •9999 .2616 .2644 .2672 .2700 .2728 .9652 .9644 .9636 .9628 .9621 .1392 ■9903 .1421 .1449 .1478 •1507 •1536 .9899 •9894 .9890 .9886 .9881 .017452 .9998 .2756 .9613 .02036 .02327 .02618 .02908 .03199 .9998 •9997 •9997 •9996 •9995 .2784 .2812 .2840 .2868 .2896 .2924 •9605 •9596 .9588 •9580 •9572 •9563 .1564 ■1593 .1622 .1650 .1679 .1708 •9877 .9872 .9868 .9863 •9858 •9853 .03490 •9994 .03781 .04071 .04362 .04653 •04943 •9993 .9992 .9990 .9989 .9988 .2952 .2979 .3007 •3035 .3062 •9555 •9546 •9537 •9528 •9520 •1736 .9848 •1765 .1794 .1822 •9843 .9838 •9833 .9827 .9822 .05234 .9986 .3090 •95" .05524 .05814 .06105 .06685 •9985 •9983 .9981 .9980 •9978 .3118 •3145 •3173 .3201 .3228 •9502 .9492 •9483 •9474 •9465 .1908 .9816 •1937 .1965 .1994 .2022 .2051 .9811 •9805 •9799 .9793 .9787 .06976 •9976 •3256 •9455 .07266 •07556 .07846 .08136 .08426 •9974 .9971 •9969 .9967 •9964 •3283 •331 1 •3338 •3365 •3393 •9446 •9436 .9426 •9417 •9407 .2079 .9781 .2108 .2136 .2164 •2193 .2221 •9775 •9769 •9763 •9757 .9750 .08716 .9962 .3420 •9397 .09005 .09295 .09585 .09874 .10164 •9959 •9957 •9954 •9951 .9948 •3448 •3475 .3502 •3529 •3557 •9387 •9377 •9367 •9356 •9346 .2250 •9744 .2278 .2306 •2334 •2363 .2391 •9737 •9730 .9724 .9717 .9710 •10453 •9945 •3584 •9336 .10742 .11031 .11320 .11609 .11898 .9942 •9939 •9936 •9932 .9929 .3611 •3638 •3665 .3692 •3719 •9325 •9315 •9304 •9293 9283 .2419 •9703 .2447 .2476 .2504 •2532 .2560 .9696 .9689 .9681 .9674 .9667 .12187 .9925 •3746 .9272 .12476 .12764 .13053 .9922 .9918 .9914 .3800 •3827 .9261 •9250 •9239 .2588 •9659 Cos. Sin. A. Cos. Sin. A. Cos. Sin. A. NATURAL SINES AND COSINES 327 A. 8in. Con. A. Sin. Cos. A. Sin. Coi.. i 50' lOf 20* 40' 50' 240 10' 2& ^i 40' 50' 25° 10' 20' 30' 40' 50' 26^ 10' 20' 30' 40; 50' 27° 10' 2cy 30' 40' so' 28° i2 _ -cF) See page 64. Area of a triangle See page 68. ^ = i Base X Altitude Area of a rectangle See page 69. A = ha Area of a trapezoid See page 69. ^=(& + c)xi a Area of a polygon See page 71. A = \aP Area of an ellipse 4 Circumference of an ellipse See page 71. c- 2 . Contents of a cylinder See page 72. S = ttB^H Volume of a pyramid F=i6a. Volume of a frustum of a pyramid 330 TABLE OF FORMULAS 331 Surface of a regular pyramid See page 73. Volume of a cone See page 73 Lateral surface of a coue See page 73. Volume of a frustum of a cone See page 74. Volume of a sphere 3 Surface of a sphere See page 74. Lateral surface of a frustum of a cone See page 74. S = \shx(P-\- P^) Volume of a barrel See page 75. V=(D^x 2)-{-d^x Xx.2618 Diameter of blank for square bolt See page 127. 5 = 1.414^ Diameter of blank for hexagonal bolt See page 127. B = 1.155 X Pitch of a screw with V-shaped thread See page 140. No. of threads per inch Depth for V-shaped thread See page 140. D=Px .8660 Size of tap drill for V-shaped thread See page 141. c» rp l.i3*t 332 VOCATIONAL MATHEMATICS Size of tap drill for U. S. Standard Thread See page 142. Depth of thread of U. S. Standard See page 142. D = Px .6495 Flat -r Acme Standard Thread See page 143. , n 1-3732 ^ = ^— ^ Square Thread See page 144. Belting See page 147. L = 0.1309 N{D-{-d) Pulleys See page 161. ttDR F = 12 irli irD Surface speed of pulleys See page 152. N " D Thickness of tooth of gearing See page 162. y^ 1.57 Diametral Pitch Circular pitch See page 162. ^p^ 3.1416 Diametral Pitch Diametral pitch See page 162. 3.1416 DP Circular Pitch TABLE OP FORMULAS 333 Dimensions of gears by diametrical pitch See page 105. p_ iV-h2 p_N .1.57 D ly p N+2 P P N=Piy N=PD-2 f=Tr. jy ^f= whole depth of tooth P= — P' =z — p, p Distance between centers of two gears See page 168. Volume of rectangular tank See page 173. LxBxHx 7.48 = V Volume of cylindrical tank See page 174. C = d^hx .0034 or .0034 dVi Weight in lb. of lead pipe See page 175. W= (D'-iP) X 3.8697 x I Cubical contents of a foot of pipe See page 176. C=I)^ X.7854 X 12 ^231 C^W X .0408 Capacity of a pipe of any length and any diameter See page 176. C^LC X.0408 xL C=Z>»x.7854x L-!-231 Pressure of water per sq. in. See page 188. P=h X 0.434 lb. per sq. in. 334 VOCATIONAL MATHEMATICS Head of water in feet See page 183. 0.434 0.434 ^ ^ Thickness of pipe h X .s T = 750 Velocity of water See page 186. F=V^ X 2500 ; 13.9 I X d Head to produce a given velocity See page 187. V xLx^ h=- ^ 2500 Twaddell scale into specific gravity See page 194. (5 X N) + 1000 ^^ 1000 ^^^ To change specific gravity into Twaddell scale See page 195. (sa X 1000) - 1000 ^ ij^g^^^^ .,.^^^^,1 5 To convert Centigrade to Fahrenheit See page 196. jP=— +32° 5 To convert Fahrenheit to Centigrade See page 208. (7=^(i^-32°) Thickness of boiler plate See page 208. T. S. X % Diameter of boiler See page 210. /; = 7^ X T. .S. X % ^ TABLE OF FORMULAS 335 Size of safety valve See page 21C. n=^K^^n^:^^ p D = J ^-^ ^ \(P+ 8.62) X. 7854 Horse power of an engine See page 225. jj J, _ AxPx V 33,000 //.P.(api)rox.)=|^ Diameter of cylinder See page 225. ^^ / 55()0 X H. P. V .7854 X V Diameter of supply pipe See page 226. z,-vv Electric current See page 232. I=E^Ii or / = ' R Power in watts See page 242. "-f - Resistance of wire in ohms See page 244. 7? ^^ Size of wire See page 246. e DI Resistance of cables See page 250. R = ^N}^ in ohms per 1000 ft. cm. Weight of cables See page 250. ir= .00305 X c. m. in lb. per 1000 ft. 336 VOCATIONAL MATHEMATICS Table of Decimal Equivalents of the Fraction of an Inch By 8ths, 16ths, 32ds, and 64ths 8ths 82d8 64th s 64ths Continued i = .126 xfV = 03125 ^\ = .015626 If = .515625 \ = .250 xf\ = .09376 /j = .046875 II = .546876 1 = .375 ,\ = .16625 ^^ = .078125 II = .578125 | = .500 xjV = .21875 ^\ = .109375 If = .609376 1 = .625 ^^ = .28126 g\ = . 140625 II = .640625 f = .750 11 = .34376 li = .171875 If = .671875 i = .876 If = .40625 II = .46875 II = .63126 If = .203125 ^1 = .234375 II = .265625 II = .703126 leths II = .734375 tV = .0625 If = .765625 ^5 = .1876 II = ..59375 If = .296875 II = .796875 j', = .3125 fl = .65625 II = .328125 If = .828125 ^ = .4375 If = .71875 If = .359375 11 = .859375 ^■, = .6625 If = .78125 II = .390625 II = .890625 jl = .6875 II = .84375 II = .421875 If = .921875 H = .8125 II = .90626 If = .453125 II = .953125 if = .9375 II = .96875 II = .484375 If = .984375 By 64ths ; from ^j to 1 inch ^ = .Q15626 II = .265625 If = .516625 If = .765626 -g\ = .031250 j% = .281250 II = .631250 If = .781260 ^^ = .046875 11 = .296875 II = .546875 II = .796875 ^5 = .062600 t\ = .312500 j% = .562500 II = .812500 ^\ = .078125 II = .328125 II = .678126 If = .828125 ^\ = .093750 ^1 = .343750 II = .693750 II = .843750 ^\ = .109375 If = .369375 II = .609375 II = .869375 1 = .125000 1 = .375000 1 = .625000 1 = .875000 ^% = .140625 II = .390625 II = .640625 II = .890625 ^\ = .156250 II = .406260 II = .656250 If = .906250 II =.171876 II = .421875 II = .671875 II = .921875 j\ = .187600 j\ = .437500 \l = .687500 II = .937500 If = .203125 If = .453125 II = .703126 II = .953125 j\ = .218750 It = .468750 If = .718750 II = .968760 if = .234376 II = .484375 II = .734376 If =.984375 1 = .250000 I = .500000 f = .760000 1 = 1.000000 INDEX Addition, 3 Compound Nunabere, 44 Decimals, 31 Fractions, 20 Adjusting Gears, 256 Aliquot Parts. 37 Alligation, 287 Alternate, 288 Ammeter, 233 Amperes, 231 Angles, 64 Complementary, 64 Right. 64 Straight, 64 Supplementary, 64 Antecedent, 55 Apothem, 70 Arc, 62 Arc of Contact, 148 Area. 70 Circles, 62 Irregular Figures, 70 Rectangles, 69 Rings. 63 Triangles. 70 Atmospheric Pressure, 181 Avoirdupois Weight, 41 Barometer, 181 Base, 48 Belting, 147 Bevel Wheels, 163 Mitre Wheels. 163 Blanking Dies. 107 Blue Prints, 78 Board Measure, 84. Boiler Plates. 200 Pumps, 218 Tubes, 212 Boilers, 203 Bolts, 126 Bracket, 298 Brass, 253 Brickwork, 95 Brown and Sharpe Wire Table, 248 Building. 83 Building Materials. 94 Buying Cotton, 286 Rags, 286 Wool, 286 Yarn, 286 Cancellation, 12 Capacity of Pipe, 176, 179 Capacity of Pump, 219 Carpentering, 83 Carpenter's Table of Wages, 104 Castings, 251 Cast Iron, 251 Cement, 97, 190 Centigrade Thermometer, 196 Circles, 62, 112 Circuits, 235 Parallel, 235 Series, 235 Circular Pitch, 161 Circumference, 62 Clapboards, 102 CoeflBcient. 304 Common Fraction. 16 Common Multiple, 15 Complex Fraction, 25 Compound Interest Table, 54 Compound Lathes, 258 Compound Numbers, 39 Cone, 73 Consequent, 56 387 338 INDEX Construction, 89 Copper, 253 Cosecant, 321 Cosine, 321 Cotangent, 321 Cotton Yarns, 279 Countershaft, 156 Cube Root, 59 Cubic Measure, 40 Cutting Dies, 107 Decimal Fraction, 28 Addition, 31 Division, 34 Multiplication, 33 Reduction, 30 Subtraction, 32 Denominate Numbers, 39 Denomination, 39 Denominator, 16 Density of Water, 191 Diameter, 62 Diameter of Cylinder, 225 Diameter of Supply Pipe, 225 Diametral Pitch, 161, 162 Die, 134 Division, 8 Division of Compound Numbers, 45 Decimals, 34 Fractions, 24 Drainage Pipes, 174 Drilling Machines, 270 Driven Pulley, 149 Driver, 149 Dry Measure, 41 Efficiency of Water Power, 189 Ellipse, 71 Engine Lathes, 255 Engines, 222 Equations, 305 Equilateral Triangle, 66 Evolution, 59 Exact Method of Solving Example, 81 Exactness, 4 Excavations, 89 Factoring, 11 Factors, 11 Fahrenheit Thermometer, 196 Flooring, 102 Formulas, 297 Fractions, 16 Addition, 20 Common, 16 Complex, 25 Decimal, 28 Division, 24 Improper, 16 Multiplication, 23 Proper, 16 Reduction, 16 Simple, 25 Subtraction, 24 Framework, 89 Friction in Water Power, 189 Frustum of a Cone, 74 Fusible Plug, 213 Gear and Pitch, 255 Gearing, 159 Girders, 90 Graphs, 295 Greatest Common Divisor, 14 Hand Hole and Blow-Off, 213 Heat Units, 195 Hexagon, 70 Horse Power, 224 Hydraulics, 172 Hypotenuse, 66 Idler, 168 Inside Area of Tanks, 174 Interest, 51 Interpretation of Negative Quanti- ties, 311 Involution, 59 Isosceles Triangles, 66 Jack Shaft, 156 Joule, 241 Kilowatt, 242 Latent Heat, 198 Lateral Pressure, 183 Lathes, 255 Lathing, 93 Laths, 90 Law of Squares, 179, 180 INDEX 339 Least Common Multiple. 14 Lever Safety Valve, 214 Linear Measure, 40 Linen Yarns, 279 Liquid Measure, 40 Logarithms, 315 Lumber, 84 Machine Speeds, 262 Manhole. 213 Mantissa, 315 Maximum Pressure, 223 Measurement of Resistance, 243 Measure of Time, 41 Mensuration. 62 Methods of Representing Operations, 306 By Means of Tables. 81 Exact Method, 81 Of SoUnng Examples. 81 Rule of Thumb Method, 82 Metric Equivalent Measures, 292 Measure of Capacity, 292 Length, 292 Surface, 292 Volume, 292 Weight, 293 Metric System, 291 Micrometer, 139 Mixed Decimal, 29 Mixed Number, 16 Mortar, 95 Multiplication, 7 Multiplication of Algebraic Expres- sions, 313 Multiplication of Compound Num- bers, 45 Decimals. 33 Fractions, 23 Multiple, 14 Multiplier, 7 NaUs. 131 Net Power for Cutting Iron or Steel, 263 Numerator, 16 Octagon, 70 Ohms, 232 Ohm's Law. 232 Operating Power, 228 Painting, 106 Paper Measure, 42 Parallelogram, 69 Parenthesis, 298 Pentagon, 70 Percentage, 48 Perimeter, 70 Pitch, 161 Pitch Circle, 161 Pitch Diameter, 161 Planers, 268 Planks, 84 Plaster, 95 Plates, 90 Plumbing. 172 Polygons, 70 Pop Safety Valve, 215 Power, 28, 59, 188 Power Measurement, 241 Prime Factors, 12 Proportion, 57 Protractor, 65 Pulleys, 146 Driven, 149 Driver, 149. Pyramid, 73 Quadrilaterals, 69 Quotient, 8 Radii, 62 Radius. 62 Rafters. 90 Raising Water, 188 Rate Per Cent. 48 Ratio, 55, 62 Raw Silk Yarns, 278 Reamers, 272 Rectangle, 69 Rectangular Tanks, 173 Reduction Ascending, 40 Reduction Descending, 39 Reduction of Decimals, 30 Reduction of Fractions, 16 Regular Polygon, 70 Remainder. 5. 8 Return Tubular Boilers, 203 Rhomboid. 69 Rhombus. 69 Right Angle. 64 Right Triangle. 66, 67 340 INDEX Rivets, 130 Roofing, 89 Root, 59 Rough Stock, 84 Rule of Thumb Method, 81 Safety Valves, 214 Safe Working Pressure, 205 Scale Drawings, 79 Scalene Triangle, 66 Screws, 134 Lead, 136 Pitch, 137 Threads, 137 Turns, 137 Secant, 321 Sector, 62 Shafts, 146 Shaper, 269 Sheet and Rod Metal Work, 107 Shingles, 98 Similar Figures, 75 Similar Terms, 304 Simple Fractions, 25 Simple Interest, 51 Simple Number, 39 Simple Proportion, 57 Sine, 321 Size of a BoUer, 209 Size of Lathes, 255 Size of Pump, 219 Slate Roofing,. 100 Soldering, 190 Specific Gravity, 192 Speed, 151 Speeds for Different Metals, 263 Sphere, 74 Spun Silk, 279 Spur Wheels, 163 Square Measure, 40 Square Root, 59, 60 Stairs, 103 Standard Gauge for Sheet Metal and Wire, 116 Steam Engineering, 195 Steam Heating, 200 Steam Indicator, 227 Steam Lap, 224 Steel, 252 Stonework, 96 Studding, 90 Substituting, 308 Subtraction, 5 Compound Numbers, 44 Decimals, 32 Fractions, 21 Subtrahend, 5 Superheated Steam, 217 Supplement, 64 Surveyors' Measure, 40 Tables containing Number of U. S. Gallons in Round Tanks for One Foot in Depth, 177 Tables for Sheet Metal Workers, 117, 118 Tables of Metric Conversion, 293 Tacks, 133 Tangent, 321 Taps and Dies, 134 Temperature, 196 Tensile Strength, 203 Textile Calculations, 276 Thickness of Pipe, 184 Threads, 140 Acme Standard, 143 Square, 144 U. S. Standard, 141 V-shaped, 140 Trade Discount, 50 Train of Gears, 168 Translating Formulas into Rules, 301 Rules into Formulas, 298 Transposing, 308 Trapezium, 69 Trapezoid, 69 Triangle, 66 Equilateral, 66 Isosceles, 66 Right, 66 Scalene, 66 Triangular Scale, 79 Trigonometry, 320 Uses of Tables, 81 Value of Coal to produce Heat, 197 Velocity through Pipes, 186 Velocity of Water, 186 Voltmeter, 233 Volts, 231 INDEX 341 Volume, 72 Cone. 73 Cube. 72 Cylindrical Solid. 72 Rectangular Bar. 72 Water and Steam. 200 Water Gauge. 213 Power, 189 Pressure. 182 Supply, 172 Traps. 185 Wattmeter. 242 Watts. 242 Weight of Bars of Steel. 253 Flywheel. 223 Lead Pipe, 175 Roof Coverings, 100 Weights and Areas, 120, 123 Woodworking. 8.3 Woolen Yarns. 277 Worsted Yarns. 276 Wrought Iron. 252 Yarns. 276 Cotton. 279 Linen, 279 Raw Silk. 278 Two or More Ply, 279 W^oolen, 277 Worsted. 276 121, 122. 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