LIBRARY OF THE UNIVERSITY OF CALIFORNIA. Class SHOP PROBLEMS IN MATHEMATICS BY WILLIAM E. BRECKENRIDGE, M.A. CHAIRMAN OF THE DEPARTMENTS OF MATHEMATICS IN THE STUYVESANT HIGH SCHOOL AND THE STUYVESANT EVENING TRADE SCHOOL, NEW YORK CITY ; DEPUTY EXAMINER IN SHOP MATHEMATICS FOR THE INTERNATIONAL COMMITTEE OF THE Y. M. C. A. SAMUEL F. MERSEREAU, B.A. CHAIRMAN OF THE DEPARTMENT OF WOODWORKING, STUYVESANT HIGH SCHOOL^ NEW YORK CITY, PRACTICAL CARPENTER AND BUILDER CHARLES F. MOORE, B.S. CHAIRMAN OF THE DEPARTMENT OF METAL WORK, STUYVESANT HIGH SCHOOL ; INSTRUCTOR IN MACHINE-SHOP PRAC- TICE IN THE STUYVESANT EVENING TRADE SCHOOL, NEW YORK CITY GINN AND COMPANY BOSTON NEW YORK CHICAGO LONDON COPYRIGHT, 1910, BY WILLIAM E. BRECKENRIDGE, SAMUEL F. MERSEREAU, AND CHARLES F. MOORE ENTERED AT STATIONERS' HALL ALL RIGHTS RESERVED 710.6 gtftengum GINN AND COMPANY PRO- PRIETORS BOSTON U.S.A. PKEFACE The aim of this book is twofold : first, to impart to the student information in regard to shops and shop materials, including the names of the parts of machines used in wood working and metal working.; second, to give a thorough training in the mathematical operations that are useful in shop practice and science. A book of this size cannot be a treatise, but can deal only with the most important points in shop work. The cuts of the lathes, planers, band saw, and milling machine will be found useful in teaching the names of the parts of these machines and in illustrating the mathematical problems. To the problems sufficient theory has been added to make the book useful as a textbook in courses ^in shop mathematics. For training in mathematical operations a certain amount of drill is necessary on abstract work problems without content -+ to concentrate attention on the operations and develop speed and accuracy in computation. For this pur- pose the latter part of the book is devoted to a Review of Calculation. It is desirable that some of this abstract work should be studied every day in connection with the concrete problems. The slide rule has been treated at length because the authors believe that it should be in more common use. Emphasis has been laid on short methods and on checks. In the chapter on Ratio and Proportion athletic fields have been used as interesting subjects of work. The scope of the work on the informational side may be seen from the Table of Contents. On the mathematical side 208529 iv SHOP PROBLEMS IN MATHEMATICS the problems vary from the addition of fractions, which must be reviewed by all men who enter trade schools, to the trigonometry that is used in the more difficult parts of machine work. The arrangement is by subjects in order that a student may select those problems in which he is most interested, and that a teacher may readily arrange a course to meet the needs of his particular school. It is suggested that the work of the shop be correlated with that of the mathematical classroom. For example, when a student in a manual train- ing high school is beginning to handle boards in the shop, he is commencing the study of algebra in his mathematical classroom. Let him be assigned some problems in the chap- ter on Board Measure, together with a review of work in fractions. From these problems he will acquire consider- able information that will be useful in the shop, and from the formulas in board measure he will understand the meaning of letters as symbols of general number. If I and t stand for the length and thickness of the board that the student has just sawed off in the shop, they will have a definite meaning to him. The authors hope that the book will be found useful in any school where there are shops. It is adapted for use in the shop or the mathematical classroom, or in both. A student of this book should develop an appreciation of the value of mathematics in practical work, since all prob- lems in the book are based on actual shop practice. He should learn to make a formula, solve it for any letter, and apply it intelligently. As he progresses the approximate nature of the results in real problems should become evi- dent. He should learn that real problems very rarely "come out even," that there is usually an allowance necessary for waste, and he should develop a certain judgment as to what that allowance is. He should find that different degrees of PREFACE v accuracy are desirable in different kinds of work, a result correct to "feet being sufficiently accurate in certain pieces of work, while in others, measurements are required to ten- thousandths of an inch. In some cases where it seems desirable for the student to become familiar with several systems of notation in common use, different forms have been used in different parts of the book ; for example, 3 ft. 6 in. and 3' 6". The book is designed to correlate the work of the mathe- matical classroom with that of the departments of mechanic arts and science. This correlation is impossible without the help of the teachers in these departments. Their cordial cooperation in this school is hereby acknowledged. The authors especially acknowledge their indebtedness to Dr. Ernest R. von Nardroff for valuable suggestions, and to Mr. F. H. Law for corrections in English. CONTENTS CHAPTER PAGE I. BOARD MEASURE 1 II. HOUSE BUILDING, GENERAL CONSTRUCTION, HEIGHTS OF TREES AND OTHER MEASUREMENTS 17 III. PULLEYS, BELTS, AND SPEEDS 62 IV. AREAS, VOLUMES, AND WEIGHTS OF SOLIDS ; TURNED WORK 74 V. PATTERN MAKING AND FOUNDRY WORK ; WEIGHTS OF CASTINGS FROM WEIGHTS OF PATTERNS 87 VI. LENGTH OF STOCK FOR FORCINGS, STRENGTH OF FORCINGS, ALLOWANCE FOR SHRINK FITS .... 102 VII. SPEEDS OF PULLEYS, SHAFTS, AND GEARS Ill VIII. CUTTING SPEED AND FEED 110 IX. MICROMETER, VERNIER, AND TAPERS . 133 X. THREAD PROPORTIONS, GEARING FOR SCREW CUT- TING, INDEXING 147 XI. GEAR PROPORTIONS AND SPIRALS 166 XII. THE UNIVERSAL GRINDER AND THE GAS ENGINE . . 183 REVIEW OF CALCULATION WITH SHORT METHODS XIII. FRACTIONS 186 XIV. PERCENTAGE,. AVERAGES. SQUARE ROOT 193 XV. RATIO AND PROPORTION 199 XVI. MENSURATION 213 XVII. FORMULAS . . 237 XVIII. SOLUTION OF RIGHT TRIANGLES BY NATURAL FUNC- TIONS 249 XIX. SHORT METHODS 253 XX. THE SLIDE RULE 260 TABLES 274 INDEX 279 vii "^R^JEI^ ^ Of THE UNIVERSITY OF SHOP PROBLEMS MATHEMATICS CHAPTER I BOARD MEASURE 1. A board foot. A foot in board measure, or a board foot, means a piece of lumber having an area of 1 sq. ft. on its flat surface and a thickness of 1 in. or less. U 12" -4- 12." -^ /2"J<- /2" *U 12" J* FIG. 1. BOARD FEET 2. Feet. The term " feet " is generally used for " board feet," except in places where it is likely to be misunder- stood, when " board feet," " square feet," " linear feet," etc., are designated. 3. Rough stock. The term " rough stock " ordinarily means lumber roughly gotten out for some purpose, but having its dimensions a little larger than is actually re- quired, to allow for planing, truing up, etc. 4. Dressed. The term " dressed " has very much the same meaning as " planed," and is sometimes used in connection with a piece that has any of its surfaces planed, as " dressed on one or both sides." But for convenience in stating 1 2 SHOP PROBLEMS IN MATHEMATICS problems, etc., we shall use the term to designate lumber planed on all four of its surfaces, unless otherwise stated. 5. Allowance for dressing, i.e. planing. If lumber is dressed, it loses in size the amount taken off in shav- ings. Usually for stock 1^ in. or more in thickness the loss is about | in. on each surface planed. Hence a piece PLANER (By permission of the American Woodworking Machinery Co.) 8 in. wide and 2 in. thick when rough becomes 7| in. wide and If in. thick when dressed, if planed on all four surfaces. For lumber less than 1^ in. thick the amount taken off in planing is usually about in. for both surfaces. This varies, however, according to the size and nature of the stock. In ordinary lumber the loss of size due to planing is not taken into consideration when the lumber is sold. The pur- chaser expects to lose the difference between the rough and dressed lumber, and orders enough to make up this loss. BOARD MEASURE 3 6. Width of rough lumber. In measuring the width of common rough lumber, a fraction of an inch equal to or greater than \ is counted as a whole inch, while a fraction less than 1 is neglected. For example, a board 6^ in. wide would be called 7 in. wide, a board 6f in. wide would be called 7 in. wide, and one G| in. wide would be called 6 in. wide. 7. Surfaced. The term " surfaced " is usually applied to boards or planks that are planed on one or both sides. 8. Jointed. The term "jointed" has reference to lumber planed on its edges. It is also used to designate pieces that are made straight on the edges. 9. Allowance in cutting logs. Logs are supposed to be cut a few inches longer than the length actually required, to allow for bruising and other damage done to the ends in the lumbering operations. The amount allowed is usually sufficient to permit of squaring the ends of the lumber and still have the pieces long enough. However, it is well to make sure of the lengths when purchasing, or. to make a small allowance, when planning certain kinds of work, by reducing the dimensions a trifle. 10. Lumber. The term " lumber " is generally applied to pieces not more than 4 in. the smaller way, usually thickness. 11. Timber. The term "timber" is applied to pieces more than 4 in. the smaller way. It is also applied to trees stand- ing in the forest, as "one hundred acres of timber." It is sometimes used in speaking of the quality of wood, as " a piece contains good timber." 12. Board and plank. Any piece of lumber under 1^ in. thick is usually called a " board " ; any piece from 1^ in. to 4 in. thick is called a " plank." The use of these terms differs, however, in various localities. 13. Rule for board feet. From section 1 it is evident that the rule for calculating board feet is as follows : SHOP PROBLEMS IN MATHEMATICS To find the number of board feet in a piece of lumber, multiply the number of square feet in its fiat surface by the number of inches in thickness, counting a thickness less than one inch as an inch. EXAMPLE 1. Find the number of board feet in a piece of lumber 1 in. thick, 9 in. wide, and 14 ft. long. ~^ x 1 - 10i ft. EXAMPLE 2. Find the number of board feet in a piece of lumber i in. thick, 8 in. wide, and 12 ft. long. 8x12 12 X 1 = 8 f t. \ in. in thickness is counted as 1 in. EXAMPLE 3. Find the number of board feet in a piece of timber 7 in. thick, 8 in. wide, and 18 ft. long. (See Fig. 2.) 14. Table for calculating board measure. The following simple table familiarized will be found very useful in cal- culating board measure : FOR LUMBER 1 IN. OR LESS IN THICKNESS Pieces 3 in. wide contain as many feet as they are feet long. Pieces 4 in. wide contain 1 as many feet as they are feet long. Pieces 6 in. wide contain | as many feet as they are feet long. Pieces 9 in. wide contain f as many feet as they are feet long. Pieces 12 in. wide contain as many feet as they are feet long. Pieces 15 in. wide contain \\ as many feet as they are feet long. Pieces 16 in. wide contain \\ as many feet as they are feet long. BOARD MEASURE 5 ORAL EXERCISE According to the above table determine orally the num- ber of feet in pieces having the following dimensions : 1. 1 in. by 3 in. by 12 ft. 13. 8 in. by 9 in. by 20 ft. 2. 1 in. by 3 in. by 16 ft. 14. 7 in. by 9 in. by 14 ft. 3. i- in. by 4 in. by 12 ft. 15. 12 in. by 12 in. by 12ft. 4. i in. by 4 in. by 18 ft. 16. 14 in. by 15 in. by 16ft. 5. in. by 4 in. by 14 ft. 17. 7 in. by 18 in. by 20 ft. 6. 2 in. by 4 in. by 12 ft. 18. Two pieces each 3 in. 7. li in. by 3 in. by 16 ft. b J 6 in. by 16 ft. 8. li in. by 3 in. by 16 ft. 19 - Two pieces each 4 in. 9. If in. by 4 in. by 12 ft. b ? 6 in ' b ^ 14 ft ' 10. 5 in. by 4 in. by 18 ft. 20 ' Two P ieces each 6 in ' by 6 in. by 13 ft. 11. 1 in. by 6m. by 14 ft. 12. 2 in. by 9 in. by 16 ft. 15. Reducing inches to decimals of a foot. EXAMPLE. Re- duce 2 ft. 5 in. to feet. 2 ft. 5 in. = 2^ ft. = 2.4 ft. For the reduction of a fraction to a decimal, see sect. 162. EXERCISE Reduce to feet and decimals of a foot : 1. 6 in. 5. 7 in. 9. 3 ft. 4 in. 2. 3 in. 6. 11 in. 10. 4 ft. 10 in. 3. 9 in. 7. 1 ft. 6 in. 11. 6 ft. 7 in. 4. 5 in. 8. 2 ft. 3 in. 16. Boards of the same length but of different widths (short method). When several boards of the same length but of 6 SHOP PROBLEMS IN MATHEMATICS different widths are to be measured, it is sometimes con- venient to add all the widths together and figure the lot as one board. A good way to add these is to run off, on a tapeline, the width of each board separately until all are measured ; the last figure reached on the line will be the sum of the widths of all the boards. 17. Standard lengths of lumber. In selecting lumber it should be borne in mind that in most sections the standard lengths are 10, 12, 14, 16, 18 ft., etc. If it is cut to a spe- cial length, it always costs more. In the Adirondacks one of the standard lengths of spruce is 13 ft. PROBLEMS 1. How many board feet in a piece of lumber 1 in. thick, 12 in. wide, and 5 ft. long ? (See Fig. 1.) 2. How many board feet in a piece of lumber 1^ in. thick, 11 in. wide, and 16 ft. long ? (See Fig. 3.) ~\nj v ' FIG. 3. BOARD MEASURE 3. A piece 10 ft. long, 2 in. thick, and 11 in. wide con- tains how many board feet ? 4. How many board feet of stock are required to con- struct a platform 8 ft. 6 in. square, if the stock is 1^ in. thick and we allow three board feet for waste in squaring up the ends of the boards ? 5. A piece of stock is ^ in. thick, 9| in. wide, and 16 ft. long. How many board feet does it contain ? (See sect. 6.) 6. How many board feet in a piece 18 ft. long, 1\ in. wide, and \\ in. thick ? (See sect. 6.) BOARD MEASURE 7 7. A piece of lumber is 10 in. wide at one end and 8 in. wide at the other. If it is 12 ft. long and 1 in. thick, how many board feet does it contain ? 8. How many board feet in a stick 6 in. thick, 8 in. wide, and 14 ft. long ? 9. A board | in. thick, 8 in. at one end, 12 in. at the other, and 13^ ft. long contains how many board feet ? 10. A piece of 2^-in. plank is 18 ft. long, 9 in. wide at one end, and 12 in. wide at the other. How many feet does it contain ? 11. Five pieces of lumber (see Fig. 4) are 3, 4, 7, 9, and 5 in. wide respectively. What is the total number of feet in the lot, if the boards are 1 in. thick and 10 ft. long ? t*M ? 9 FIG. 4. SHOKT METHOD FOR BOARDS OF DIFFERENT WIDTHS 12. A pile contains nine boards whose widths are 3, 4, 5, 7, 9, 11, 13i, 15, 16^ in. respectively. If the boards are 16 ft. long and we saw off 1 in. on each end of every board for squaring, how many square feet of floor will the boards cover ? 13. A stack of lumber 7 ft. 5 in. wide and 9 ft. long is composed of 90 layers of boards 1 in. thick, placed edge to edge. How many feet does the stack contain ? 8 SHOP PROBLEMS IN MATHEMATICS 14. How many feet in a wagon load of lumber 14 ft. long, 3 ft. 2 in. wide, and 24 in. high, if the boards are in. thick ? o 15. The distance between the bolster stakes of a lumber wagon is 3 ft. 2 in. If we put on a load of planks, dressed on two sides only, that were 2 in. thick when rough, and the load is 16 ft. long and 35 in. high, how many board feet does the load contain ? 16. The back stop for a baseball diamond is constructed of 9-in. boards I in. thick and 16 ft. long. If it is 12 ft. high and 16 ft. long, how many boards does it require ? How many board feet are required for its construction, if we have 3 uprights 18 ft. long and 3 braces 12 ft. long, the uprights and braces to be 2 in. by 4 in. ? 17. At $35 per M, how much will it cost for lumber for a platform 20 ft. wide at one end, 25 ft. wide at the other, and 40 ft. long, the lumber to be 2 in. thick, and 1 per cent to be added for waste in squaring off the ends of the planks ? 18. A platform 12 ft. wide and 36 ft. long is constructed of If -in. by 9|-in. hemlock planks (dressed). If we add l per cent for waste in squaring, how much lumber must we buy for the platform ? (See sect. 5.) 19. The foundation for the floor of a factory building is built up of If-in. by 5f-in. by 12-ft. dressed hemlock lum- ber spiked together sidewise into one solid piece (see Fig. 5). If the building is 108 ft. wide and 145 ft. 10 in. long, how many pieces of lumber does the foundation require, and how many board feet do they contain, making no allowance for waste ? 20. How many planks (dressed) are required to con- struct a solid girder 49 ft. long, 9| in. deep, and 8| in. wide, if the planks are 1| in. thick, 9| in. wide, and 14 ft. long, and are spiked together flatwise ? How many BOARD MEASURE 9 board feet does the girder contain, if we add one board foot to every hundred for squaring ? (See Fig. 5.) 21. At $40 per M, how much will enough hard maple lumber cost to floor a bridge 16 ft. wide that spans a river 1000 ft. wide, the planks to be 3 in. thick and laid at an angle of 45 degrees (diagonally) ? Add 3 board feet to every 100 for waste in cutting the planks on a slant. FIG. 5. FACTORY FOUNDATION 22. The sidewalk on the bridge in Problem 21 is 6 ft. wide and is laid of hard-pine planks If in. by 7| in. by 12 ft. long (dressed). How many planks does the walk require, and how many feet do they contain, making no allowance for waste in squaring ? 23. A man wishes to lay a dressed-plank walk 4 ft. wide and 150 ft. long ; the planks before being dressed were 1^ in. by 9 in. by 16 ft. long. How many board feet of lumber must he buy, if he puts 3 sleepers 2 in. by 4 in. under the planks, to which to nail, allowing for no waste ? 24. If there are 29 4-in. spikes in a pound, how many pounds are required to build the above walk, if 6 spikes are used in each piece of plank ? 10 SHOP PROBLEMS IX MATHEMATICS 25. Find the cost of laying a board walk on the side and end of a city corner lot 40 ft. wide and 120 ft. long, with the following specifications : Width of walk, 4 ft. Boards to be li in. by 7f in., dressed and laid cross- wise. Boards to be placed \ in. apart. 3 sleepers placed underneath, 3 in. by 4 in. lOd.'wire nails to be used (70 nails in a pound, at 5 cents per pound), and 6 nails to be driven into each board. The lumber to cost $35 per M and the labor to cost $10. FIG. 6. DRAWING BOARD FIG. 7. Box 26. A drawing board 23 in. wide, 31 in. long, and f in. thick is constructed of 5 pieces joined edge to edge (see Fig. 6). Allowing J in. on the width of each piece for jointing and ^ in. all around the outer edge for truing up, how many board feet are required to construct the board without the cleats ? 27. A covered box (Fig. 7) is constructed of 1-in. stock. Its outside dimensions are as follows : Height, 14 in. Width, 24 in. Length, 42 in. At $70 per M, what will be the cost of lumber for the box, allowing 1 sq. ft. for waste in squaring ? BOARD MEASURE 11 28. How many board feet of lumber can be loaded into a car 36 ft. long and 8 ft. wide, if the lumber is f in. thick and is piled 6 ft. high ? 29. How many feet of lumber will be required for every 12 ft. of fencing constructed of 1-in. stock, the fence to be 7 ft. high, allowing 2 posts 4 in. by 4 in. by 10 ft. long, and 2 scantlings 2 in. by 4 in. by 12 ft. long for each 12 ft. ? 30. What width and length of surfaced boards, | in. thick, would you select for a tight, plain, covered box 3 ft. 11 i in. long, 13i in. high, and 23^ in. wide, outside measure? The edges of the pieces must be jointed before the box is nailed together ; hence \ in. will be taken from their width. 31. How many feet are required for the box mentioned in Problem 30, and how much lumber is actually wasted in its construction ? 32. The inside dimensions of a box without a cover are 9| in. deep, 2 ft. 9 in. wide, and 2 ft. 10 in. long. State, first, what size of boards you would select ; second, how many feet they would contain; third, how many feet of lumber the box would actually contain when finished, the lumber to be surfaced and to be f in. thick. 33. A man wishes to build a picket fence on the side of his lot. The distance between the two ends of the fence, if measured on a level, is exactly 150 ft. Because of a hill in the middle of the lot over which the fence must pass he finds its length to be 204 ft. How much will the fence cost according to the following specifications ? Pickets 3 in. wide and placed 3 in. apart (cost 5 cents each). Two rails on which to nail pickets, 2 by 4 in. (cost $24 per M). Two posts every 12 ft. (at 25 cents each). 12 SHOP PROBLEMS IN MATHEMATICS Each picket to receive 4 nails (8d. wire ; 95 nails in a pound, ,at 5 cents per pound). The rails to have 2 spikes in each post (20d. wire ; 28 spikes in a pound, at 4 cents per pound). Labor to cost $12. As an aid to solving this problem, it is suggested that the student draw a diagram of a fence running over a hill, and also of one on the level between the same two points, and compare the number of pickets. Pickets are placed plumb. 34. A load of lumber consists of l|-in. boards 12 in. wide and 16 ft. long, placed edge to edge. How many feet does the load contain, if it is 36 in. wide and 30 in. high ? (Use a short method.) 35. A carload of lumber consists of 3 tiers of 2*-in. planks, the tiers being 12 ft. long, 8 ft. 2 in. wide, and 8 ft. 4 in. high. How many feet does the car contain, if we deduct 2 in. from the width of each layer for cracks between the edges of the planks ? 36. A stack of 1^-in. boards is 16 ft. long, 12 ft. wide, and 13 ft. 5 in. high. To allow for seasoning, each layer of boards is separated from the one below by 3 pieces, each 1 in. by 6 in. by 12 ft. long (one at either end, and one in the middle running crosswise of the stack). If we deduct 3 in. from the width of the stack for cracks between the edges of the boards, and count in the crosspieces, how many feet does the stack contain ? 37. How many feet would there be in the stack of lum- ber mentioned in Problem 36, if the boards were 1^ in. thick and the cross pieces 4 in. instead of 6 in. wide ? 38. A class in manual training makes 36 ink trays (see Fig. 8) out of ash, each tray being constructed of two pieces, the base and the top. When finished the base is j| in. thick, 5f in. wide, and 8| in. long ; and the top T % in. thick, 3i in. BOARD MEASURE 13 wide, and 5| in. long. Allowing 1 in. on the length and J- in. on the width of each piece for rough stock, how much will lumber for the trays cost at $80 per M for $-in. stock and $50 per M for f-in. stock ? 39. Find the cost for material for 72 tool racks, if each rack requires the following : 1 piece 2 in. by 2 in. by 15 in. white wood at $70 per M. 1 piece | in. by 4 in. by 15 in. white wood at $60 per M. 3 flat-head screws li in., # 7, at 10 cents per gross. FIG. 8. PEN AND INK TRAY 40. At $60 per M for white wood, $80 per M for ash, 15 cents per gross for 1-in. ^ 6 round-head screws, and 10 cents per gross for burrs (washers), what will the material for 100 drawing boards cost, the boards to be constructed as follows ? Size of board, 14 in. by 17 in. by 1 in. thick. To be constructed of 4 pieces, jointed edge to edge. To have 2ash cleats f in. thick, 1 ^ in. wide, and 13 in. long. Each cleat to be fastened to board with 5 round-head screws, with burr under head. Cost of glue for each board, 2 cents. Allow i in. on the width of each of the 4 pieces in the board for jointing, and 1 in. on the length for truing up ; also I in. on the width and 1 in. on the length of the cleats for rough stock. 18. A formula. A formula is a brief statement of a rule. For example, " The area of a rectangle is equal to the base multiplied by the altitude " may be stated R = b x a. 14 SHOP PROBLEMS IN MATHEMATICS EXERCISE If b is the number of board feet, t the number of inches thick, w the number of inches wide, and I the number of feet long, write the formula for b in terms of , w, and L State the rule which this formula represents. Using the formula, find b in pieces of lumber having the following dimensions : 1. 61 in. by 8 in. by 14 ft. 2. 4 in. by 10 in. by 12 ft. 3. 2 in. by 8 in. by 16 ft. 4. 3 in. by 9 in. by 14 ft. 5. 2 in. by 9 in. by 10 ft. 6. i in. by 9 in. by 16 ft. 7. | in. by 7 in. by 10 ft. 8. I in. by 10 in. by 14 ft. 9. in. by 9 in. wide at one end, 15 in. wide at the other, and 12 ft. long. (Take w equal to the average width.) 19. Method of com- puting surface measure approximately. A handy method to use in calcu- lating the square feet in a rectangular sur- face, where the dimen- sions are in feet and inches, is shown in the following : Take, for example, a rectangle 3 ft. 7 in. by 4 ft. 6 in. (Fig. 9). If we begin at one corner, lay off feet and draw lines as shown in FIG. 9. APPROXIMATE SURFACE MEASURE the figure? we get 12 sq. ft., represented by the letter a, with 7 in. over on one side and 6 in. over on the other. a a a b ft -^ ^-7^. p O rf - ^ /^ BOARD MEASURE 15 ACTUAL WORK EXPLANATION 3 ft. 7 in. 4 ft. x 3 ft. = 12 sq. ft. (Area indicated by a) 4 6 4 ft. x 7 in. = f 8 sq. ft. 12 = 2sq.ft.and T \sq.ft. (b,c,d,e) 2 4 6 in. x 3 ft. = || sq. ft. 1 6 = 1 sq. ft. and ^ sq.ft. (f,g,h) 3 6 in. x7in. = 42 sq. in. 16 2 =y\sq. ft. and T | 4 : sq. ft. (i) The student will observe that the first column contains square feet and the second twelfths of a square foot. A remainder equal to 6 sq. in. ( T | sq. ft.) or more is counted as ^ sq. ft. and added to the second column. Thus, in the example, T 4 - sq. ft. is counted as T ^ sq. ft. and added to the second column, giving a sum of {f sq. ft., which is equal to 1 sq. ft. and ^ S( l- ^. Adding 1 sq. ft. to the first column, we have 16 sq. ft. and ^ sq. ft. or 16^ sq. ft. This method is accepted by the United States government as a standard method on all such material as imported granite and marble. EXERCISE By the above method find the number of square feet and fractions of a square foot in surfaces having the following dimensions : 1. 1 ft. by 1 ft. 6 in. 10. 12 ft. 11 in. by 10 ft. 5 in. 2. 1 ft. by 1 ft. 9 in. 11. 16 ft. 4 in. by 12 ft. 7 in. 3. 2 ft. by 2 ft. 6 in. 12. 17 ft. 8 in. by 9 ft. 11 in. 4. 4 ft. by 3 ft. 4 in. 13. 10 ft. by 9 ft. 5. 5 ft. by 3 ft. 9 in. 14. lift ft. by 7}i ft. 6. 2 ft. by 11 in. 15. 12J, ft. by 16 ft. 9 in. 7. 3 ft. 6 in. by 1 ft. 6 in. 16. 16f ft. by 16| ft. 8. 5 ft. 6 in. by 3 ft. 7 in. 17. 18 ft. 10 in. by 13f ft. 9. 10 ft. 8 in. by 6 ft. 5 in. 18. 20 ft. 6 in. by 16J ft. 16 SHOP PROBLEMS IN MATHEMATICS 19. 17 ft. 7 in. by 17^ ft- 22. 40 ft. 7 in. by 26 ft. 20. 30 ft. 7 in. by 20 ft. 9 in. 23. 29 ft. 4 in. by 20 ft. 7 in. 21. 25 ft. 4 in. by 35 ft. 3 in. 24. 33 ft. by 15 ft. 4 in. PROBLEMS 1. How many feet of lumber will be required to floor a platform 16 ft. 4 in. long by 11 ft. 6 in. wide, the lumber to be 1 in. thick, if no allowance is made for waste ? 2. The foundation of a house is 22 ft. 4 in. by 40 ft. 6 in. How much |-in. surfaced lumber will it take for the lower subfloor, if we cut out a place for the cellar stairs 3 ft. 3 in. by 7 ft. 7 in., making no allowance for waste ? 3. The distance from the bottom of the sill to the top of the plate on a cottage house is 16 ft. 10 in. If the house is 25 ft. 6 in. long and we deduct for openings (windows and doors), what will be the cost of sheathing one side with -in. surfaced lumber at $35 per M ? (See Fig. 14.) 4. At 15/ per square foot, what will be the cost of laying a cement sidewalk 4 ft. 11 in. wide and 50 ft. 8 in. long ? 5. At $36 per M, how much will hemlock lumber cost for the following platform ? Size of platform 16 ft. 11 in. long and 14 ft. wide. Supports 3 in. by 6 in. by 14 ft., placed 2 ft. 5 in. from center to center, running crosswise (rough). Floor to be of IJ-in. by 9|-in. boards (dressed). 6. A drawing board is 1 ft. 11 in. wide and 2 ft. 7 in. long. What is the size of its surface in feet ? CHAPTER II HOUSE BUILDING, GENERAL CONSTRUCTION, HEIGHTS OF TREES AND OTHER MEASUREMENTS FLOORING 20. Flooring. The term " flooring " is sometimes used to designate the floor of a building after it is laid; but for brevity we shall use the term, unless otherwise stated, to mean common matched flooring lumber before it is laid. FIG. 10. FLOORING 21. Allowance for matching and waste. When estimating flooring, ceiling, or other lumber that is matched, i.e. has a tongue-and-groove joint (see Fig. 10), enough stock must be added to make up for the amount cut away from the width in matching. This amount varies from J in. to f in. on each board, according to the size and nature of the stock matched. Because of the uneven nature of lumber and different methods of working it, it is impossible to formulate any 17 Of THt UNIVERSITY 18 SHOP PROBLEMS IN MATHEMATICS exact mathematical rule showing the relation of the feet in a lot of lumber before and after it is matched. However, it can be estimated approximately, and near enough for all practical purposes. A little flooring is always wasted in squaring the ends, cutting up, etc. To offset this the mechanic usually adds a few feet to his bill, enough, in his estimation from ex- perience, to make up. For the sake of convenience, as well as for the sake of fixing the subject in the student's mind, an allowance for this purpose will be added in the follow- ing problems. The amount allowed varies from one to ten per cent, according to the size and nature of the flooring. For flooring, ceiling, etc., from about 21 in. up to 5i in. in width, the amount added for matching is generally one fourth. For the so-called " Adams " flooring or pieces, usually about 1^ in. wide, the amount added is one third. For example, if a common floor to be laid is 12 ft. square, the amount of flooring required is 144 ft. plus of 144, or 36 ft., making in all 180 ft. If the floor to be laid is of pieces 1^ in. wide, the amount required is 144 ft. plus 48 ft., or 192 ft. 22. Face. In most localities it is customary to give the lumber dealer the actual size of the surface to be covered, called " face," and let him add the proper amount for matching. PROBLEMS 1. How much |-in. flooring is required to lay a floor 16 ft. by 18 ft., if we add 3 per cent for squaring the pieces ? 2. How much hard-pine flooring | in. thick is required for a floor 13 ft. 6 in. by 14 ft. 10 in., 4 per cent being allowed for waste ? 3. The foundation of a house is 26 ft. by 30 ft. At $60 per M, what will be the cost of flooring for one floor, if HOUSE BUILDING 19 the flooring extends within 5 in. of every outside edge, on account of the studding, and there is an opening for cellar stairs 3 ft. 4 in. by 8 ft. 2 in.? Allow 3 per cent for squaring. (See Fig. 14.) 4. A kitchen floor is 10 ft. 6 in. at one end, 9 ft. 2 in. at the other, and 11 ft. 7 in. long. If it is laid with hard- maple flooring 1| in. wide, what will the lumber cost at $60 per M? Add 4 sq. ft. for waste. 5. The floor of a veranda is laid of stock li in. thick. If the size of the floor is 7 ft. 11 in. wide and 22 ft. 6 in. long, how many feet of flooring does it require, allowing 2 per cent for squaring ? (See sect. 5.) 6. The veranda on a corner of a house runs 18 ft. 6 in. across the front and 13 ft. 4 in. down the side. If it is 7 ft. 6 in. wide and we use l|-in. flooring, how many feet must we buy, 2 per cent being added for squaring ? 7. How many squares (100 sq. ft.) of floor are there in a four-story house 27 ft. wide by 41 ft. 4 in. long within walls, deducting from every floor an opening 10 ft. by 4 ft. 4 in. for stairs; and how many feet of common hard- pine flooring does it require, adding 2 per cent for squar- ing, if the flooring is in. thick ? 8. The floor of a dancing pavilion is octagonal in form. If it is 12 ft. on a side, how much hard-pine flooring | in. thick and 2^ in. wide is required, allowing 3 per cent for waste in squaring? 9. A dealer's bill for flooring was $200. If the price was $45 per M, state, first, how many feet of lumber (flooring) there were; second, how many square feet of floor was covered ; and third, what was the length of one side of the octagonal floor laid ? No waste. 10. At $45 per M, what will be the cost of flooring li in. wide for a room 13 ft. 9 in. wide and 16 ft. 7 in. long, the 20 SHOP PROBLEMS IN MATHEMATICS room to have a bay window with octagonal corners at one end 2 ft. deep and 6 ft. long across the front, allowing 5 per cent for waste in cutting ? 11. How many feet of flooring are required for one floor of a circular tower, the radius of which is 7 ft. 4 in. within walls ? Add 5 per cent for squaring. 12. A room 14 ft. 7 in. wide and 15 ft. 8 in. long has a circular window extending from one corner. The radius of the window is 4 ft. 10 in. within walls, and its center is at the point where the corner of the room would be if the window were not there. Allowing 5 per cent for waste in squaring up the pieces, how many feet of flooring 1^ in. wide are required for the room ? 13. With ^-in. ceiling, at $30 per M, how much will the lumber cost to ceil the four walls and overhead of a room 12 ft. 6 in. by 18 ft. 8 in. by 10 ft. 2 in. high, if there are three openings for doors 3 ft. 4 in. by 7 ft. 6 in. each, and 2 openings for windows 3 ft. 7 in. by 7 ft. 2 in. each, the ceiling to be 2 in. and over wide, and 25 ft. to be added for waste in squaring ? 14. The dining room of a house is to be floored with white-oak flooring f in. thick and 1^ in. wide. It is also to be wainscoted 3 ft. up from the floor with 3-in. white oak matched stock J in. thick. If we deduct for 2 doors 3 ft. 4 in. each, and for 2 windows running within 18 in. of the floor, each 3 ft. 6 in. wide ; if the room is 11 ft. 5 in. wide and 16 ft. 3 in. long, and we add 5 per cent for waste, how many feet of flooring and wainscoting are required ? 15. The sides of an automobile garage are covered with 3-in. white pine beaded ceiling -J in. thick. The size of the house on the ground is 12 ft. by 16 ft. 9 in., the roof is a common gable (see Fig. 15), the peak rises above the level of the plates 4 ft., and the corner posts are 9 ft. long. HOUSE BUILDING 21 If we take out 2 openings 2 ft. 10 in. by 4 ft. 5 in. each for windows, and one opening 8 ft. wide and 8 ft. high for a door, how many feet of ceiling are required ? Add 16 sq. ft. for waste. SIMPLE STAIR BUILDING 23. Rise. The " rise " of a flight of stairs is the perpen- dicular distance the stairs are to rise, e.g. the distance be- tween any two floor levels of a building (see Fig. 11). In QJ -9 or Upper Flo or L in e Width of Riser XT Run r Lower Floor Line 1- Run * FIG. 11. STAIRS other words, it is the sum total of the widths of all the risers (see Fig. 12). 24. Nosing. The treads of a flight of stairs almost always project over the front of the risers. This projection is called " nosing." Its amount is about the same as the thickness of the tread (see Fig. 12). 25. Run. The "run" of a flight of stairs is the distance, measured on a level, from the foot of the stairs to a point directly under their upper end (see Fig. 11). It is the hori- zontal space the stairs take up, and is the sum of the widths of the treads minus their nosing plus the width of one tread. 22 SHOP PROBLEMS IN MATHEMATICS 26. Number of risers and treads. There is always one more riser than treads, one floor counting as a tread. 27. Number of steps. To get the desired number of steps for any common flight of stairs, divide the rise of the stairs by the even inch nearest the desired height of the steps. This generally gives a whole number and -a fraction. Now, in turn, divide the rise again by the whole number obtained above, disregarding the fraction, and the result will be the required width of each riser. (The whole number obtained above will be the number of steps.) For example, the rise of a flight of stairs is 9 ft. 8 in., and it is desired to have the steps as near 7 in. high as possible. 9 ft. 8 in. reduced to inches and divided by 7 equals 16f ; now if we, in turn, divide the 9 ft. 8 in. by 16, we have 7| in., the width of each riser or the height of each step. The width of the treads is obtained by dividing the run of the stairs by the number of steps or risers and then adding to this the amount allowed for nosing. PROBLEMS 1. The risers (Fig. 11) of a flight of stairs are 7 in. wide. How many steps will there be, if the rise of the stairs is 3ft. 6 in.? 2. The height of the first floor of a house above the cellar bottom is 8 ft. 2 in., and the run of the cellar stairs is 8 ft. If there are 13 steps, what will be the width of both the treads and the risers, allowing 1 in. for nosing ? 3. If the height of a veranda floor is 31^ in. from the ground, and we make the risers as nearly as possible 7 in., how many risers are required, and what will be their width ? 4. What is the length of a stair string, if the rise is 8 ft. 10 in. and the run is 8 ft. 2 in., making no allowance for waste in cutting ? (See Fig. 12.) HOUSE BUILDINJ& 23 5. The rise of a flight of stairs is 10 ft. 6 in. and the run is 11 ft. How many feet of lumber are required for the two strings, if they are | in. thick, 10 in. wide, and we allow IL ft. on the length of each string for cutting on a slant ? 6. The distance between the first and second floors of a house is 12 ft. 4 in. and the run of the stairs is 13 ft. 8 in. FIG. 12. STAIRS If we make the risers as near 7 in. as possible, what will be the number of treads and their width, if they project over the risers 1J in., and what will be the number and width of the risers ? 7. How many feet of lumber are required for the steps (treads and risers) of a front stoop the height of which is 4 ft. 1 in., the run 4 ft. 4 in., and the length 7 ft. 10 in., the risers to be of J-in. stock, and the treads to be of IJ-in. stock and to project over the risers 1^ in.? Allow 1 in. on 24 SHOP PROBLEMS IN MATHEMATICS either end of every piece for squaring, and J in. on the width for jointing up. 8. With oak lumber at $80 per M, what will the lumber cost for the following flight of stairs ? Eise, 13 ft. Eun, 14 ft. 7 in. Steps to be as near 7 in. as possible. Steps to be 3 ft. 11 in. long. Strings to be -J in. thick by 11 in. wide. Treads to be 1 J in. thick and to project over risers 1 J in. Eisers to be -J in. thick. Allow 1 in. on the length of the treads and risers for squaring, J- in. on the width of all pieces for jointing, and 1^- ft. on the length of each string for cutting on a slant. 9. The distance between floors in a house is 11 ft. 8 in. The conditions are such that the stairs must go part way to a platform and then turn and go up at right angles the rest of the way. The height of the platform above the first floor is 4 ft. 4^- in. How many steps does each flight require (the riser to be as near 7 in. as possible), what is the width of the risers in each flight, and what is the run of each flight, if the treads are lOf in. wide and project over the risers 1J in. ? SIMPLE FRAMEWORK 28. Ordering rougher lumber. In ordering rougher lumber only standard lengths are generally considered. This some- times involves the purchase of a few more feet than are actually required. The expense, however, of having the lumber cut to exact length would be greater than buying a few extra feet on the length of some pieces that are too long. Take, for example, floor joists that are 13 ft. 6 in. when placed ; the nearest standard length is 14 ft. It would HOUSE BUILDING 25 be cheaper to buy the 14-ft. lengths and throw away the extra 6 in. than to order the pieces cut 13 ft. 6 in. In either case the same amount of lumber would probably be charged for, besides the extra work in cutting off the pieces. When practicable, long pieces can be obtained and cut in two for short ones. This, however, can be done with profit only up to a certain limit, as all lengths above 16 ft. FIG. 13. HOUSE FOUNDATION increase in price in proportion as the length increases. In some localities the price increases with the length above 12ft. PROBLEMS 1. How many feet are required for the sills (see Fig. 13) of a building 20 ft. wide by 40 ft. long, the timber to be 6 by 8 in. by 16 ft. long, if we deduct 1 ft. from the length of each 16-ft. piece for splices ? 2. The floor foundation (sills and floor joists) of a build- ing is 16 ft. 6 in. by 24 ft., the sills are 6 by 8 in., the floor joists (see Fig. 13) are 2 by 10 in., and are placed across the 26 SHOP PROBLEMS IN MATHEMATICS narrow way of the building, 16 in. from center to center. What standard lengths of lumber would you select, and how many feet are required for the foundation, if splices on the sills take up 1 ft. on every piece, no piece to be over 16 ft. long ? 3. With framing lumber at $30 per M, how much will the lumber cost for the sills and lower floor joists of a FIG. 14. FRAME OP HOUSE house, with the following dimensions ? State also the lengths of the lumber you would select, nothing to be over 16 ft. long. Size of house to outside of sills, 25 ft. by 44 ft. 6 in. Size of sills, 6 by 8 in. Two rows of floor joists placed 16 in. from center to center, running across the narrow way of the building, HOUSE BUILDING 27 with one end resting on a girder in the center of the cellar. Splices to be 1 ft. long on sills. The studding (see Fig. 14) and floor joists in common house work are placed 16 in. from center to center so that standard laths (4 ft. long) can be used without waste. 4. The partition in a house is 8 ft. high and 16 ft. long. If it is studded with 2- by 4-in. pieces placed 16 in. from center to center, how many pieces are required ? How many feet do these pieces contain, and what standard- length stock would you select ? 5. If the studding are 3 by 4 in. by 18 ft., and are placed 16 in. from center to center, how many are required for one side of a house 36 ft. long; and how many feet do they contain, allowing 10 extra studding for double corners and double studs at the windows ? 6. At $30 per M, what will be the cost of frame lumber for the walls and floors of the following building, and how many pieces of each kind should there be? Size of building on the ground, 16 ft. by 20 ft. Height of building from bottom of sills to top of plates (see- Fig. 14), 16 ft. 8 in. Sills to be 6 in. square and no longer than 16 ft. for each piece. Lower floor joists to be 2 by 8 in., placed 16 in. from centers. Joists for 2 upper floors, 2 by 6 in., placed 16 in. from centers. Studding for 4 sides, 2 by 4 in., with 16 extra for double corners and windows. Plates to be 2 by 4 in., doubled. Add twenty by 6 in. by 12 ft. boards for stays, braces, etc. 28 SHOP PROBLEMS IN MATHEMATICS ROOF WORK 29. Length of rafters. In common roof construction the lengths and various cuts of rafters are usually obtained on the steel square without the aid of much mathematical calculation. However, knowledge of the mathematical principles involved and of their application is of in- estimable value to the mechanic. FIG. 15. COMMON GABLE ROOF 30. Rise. The " rise " of a rafter is the height of the ridge of the roof above the level of the plates. If the rafters are raised and project over the plate, as shown in Fig. 15, the rise is the distance indicated in the figure. 31. Run. The " run " of a rafter is the distance from where the rafter intersects the outer edge of the plate to a point on a level with the plates, directly under the upper end of the rafter (see Fig. 15). It is usually one half the width of the building. 32. Pitch. The term "pitch" means the slant or slope of a roof (see Fig. 16).- The terms " 1 pitch," " J pitch," " \ pitch," etc., mean that the height of the ridge of a roof is equal to 1, ^, |, etc., the width of the building or span of the roof. For example, let the width of the HOUSE BUILDING 29 building be 18 ft.; if the pitch of the roof is ^, the rise will be 6 ft. In order to figure out rafter problems in roof con- struction it is absolutely necessary that the student understand the principles underlying the solution of right-angled triangles (for which see sect. 44-46). PROBLEMS 1. The rise of a lean-to roof (see Fig. 17) is 4 ft. and its run is 12 ft. What length of the rafters ? 2. What length of rafters is quired, and what is the pitch common gable roof (see Fig. 15), if the width of the building is 24 ft. and the rise of the roof 8 ft., allowing the rafters to pro- ject 18 in. over the side of the build- ing at the eaves ? 3. If the rafters are 2 by 6 in., how many feet of lumber will they contain for a lean-to roof with a 13-ft. run and a 6-ft. rise, the width of the lean-to being 14 ft., and the rafters being placed 2 ft. from center to cen- ter and project- ing over at the eaves 18 in. ? (See sect. 28.) 4. The width of FIG. 17. LEAN-TO ROOF a house is 22 ft. MIL I I I I I I I I 15 14 13 IZ II 10 9 8 7 Run 6 5" 4 3 FIG. 16. PITCH tt j (f) = 2 en oO 30 SHOP PROBLEMS IN MATHEMATICS and its length is 34 ft. If the rise of the roof is 10 ft., and the rafters are 2 by 5 in., placed 2 ft. from center to center, FIG. 18. HIP ROOF and project over at the eaves 8 in., how many feet are re- quired for the rafters of the roof ? 5. The rise of a hip rafter (see Fig. 18) is 10 ft. and its run is 15 ft. What is its length ? - Run H 6. A building is 28 ft. square and the rise of its roof is 11 ft. What is the length of its hip rafters ? HOUSE BUILDING 31 7. How many feet are there in 4 hip rafters, 2 by 8 in., if the building is 20 ft. square and the rise of the roof is 8 ft., allowing 20 in. for eaves projection ? 8. The width of a straight-gable house is 25 ft. What is the length of the rafters if the roof has ^ pitch and the eaves projection is 18 in. ? How many feet do the rafters contain if they are 2 by 5 in. ? 9. How many feet of lumber are required for the rafters of a roof that has pitch and a span of 31 ft., the length Valley Rafter FIG. 20. VALLEY ROOF of the building being 48 ft., the rafters 2 by 5 in., placed 2 ft. from center to center, and having an eaves projec- tion of 22 in. ? 10. A gambrel-roof house (see Fig. 19) is 24 ft. wide, the first or bottom set of rafters has pitch, and the top set ^ pitch. If the break or joint in the roof comes 7 ft. in from the side of the building, how long is each set of rafters ? 11. A house is built in the form of an L, the width of each part being 20 ft. What is the length of the required valley rafters if the roof has pitch ? (See Fig. 20.) 32 SHOP PROBLEMS IN MATHEMATICS 12. In the above problem what will be the length of the jack rafters (see Fig. 20) if we begin at the top and place them 2 ft. from center to center ? In getting the length of a jack rafter of a hip roof, the thickness of the hip rafter can generally be disregarded if the measurement is taken between the extreme points of the jack rafter. 13. An octagonal tower is 4 ft. on a side. How long must the corner or hip rafters be if the rise of the roof is 10 ft. ? 14. How many feet are necessary for the hip rafters of the roof of a hexagonal tower 5 ft. on a side and having a perpendicular height of 11 ft., the rafters to be 2 by 4 in. and to project over at the eaves 20 in.? 15. A tower is circular in shape. Its diameter is 14 ft. and its roof has pitch. If the rafters are 2 by 5 in. and we place them 30 in. apart on the plate, how much lumber do they contain ? 16. The width of a house is 28 ft. and the rise of the roof is 12 ft. What are the lengths of the jack rafters for the hip, if we place them 2 ft. from center to center, beginning at the bottom end of the hip rafter and letting them project 18 in. at the eaves ? 17. A house is built in the form of the letter T. The large or top part of the T is 22 ft. wide and has a \ pitch roof ; the smaller part is 18 ft. wide and has a roof that rises 8 ft. What is the length of the common, valley, and jack rafters, the jack rafters to begin at the bottom end of the valleys and to be placed 2 ft. from center to center ? (See note below.) In actual practice enough must be added to the length of jack rafters for valleys to allow for the slant, on the lower end of the jacks, that fits the valley rafter. The actual measurement for valley jacks is generally taken from the lower short corner on the lower end to the extreme point on the upper end, the thickness of the valley being disregarded. HOUSE BUILDING 33 SHINGLING 33. Standard shingles are figured on a basis of 16 in. long and 4 in. wide, and are estimated by the thousand. They are put up in bunches of 250 each.; hence 4 bunches to the thousand. The 16-in. shingle, when laid, is usually exposed to the weather about 5 in. (called " 5 in. to the weather "). 34. Number of shingles. There are various methods of calculating the number of shingles for a roof, the most common of which are the following : (1) Find the number of squares (100 sq. ft.) in the roof ; divide this number by 1^ and multiply by 1000. EXAMPLE. A roof is 25 ft. long and 20 ft. wide. 25 x 20 = 5 squares. x 1000 = 4000, the number of shingles required. (2) Multiply the number of square feet in the roof by 1\. EXAMPLE. The length of the ridge of a roof is 32 ft. and the length of the rafters on either side is 18 ft. 32 x 2 x 18 x 7 J = 8640, the number of shingles required. (3) As it takes approximately 3 bunches (250 shingles each) to cover a square, multiplying the number of squares in the roof by 3 will give the number of bunches of shingles required. EXAMPLE. A roof contains 12 squares. 12 x 3 = 36 bunches, or 9000 shingles. The above methods give a few more shingles than are actually needed. Usually some are broken or are poor, or the bunches run a little short ; hence it is well to buy a few extra, to be sure of having enough. 34 SHOP PROBLEMS IN MATHEMATICS (4) Where roofs are straight it is sometimes convenient to get the number of courses and multiply by the number of shingles in a course. EXAMPLE. The length of the ridge of a roof is 25 ft. and the rafters are 15 ft. on either side. Required the number of shingles. 25 x 3 = 75 shingles in a course. To find the number of courses (5 in. to the weather), since there are 12 courses for every 5 ft., and it is ,30 ft. over the roof, or 6 times 5, we therefore multiply 12 by 6, obtaining 72, or the number of courses. Hence 25 x 3 x (30 -5- 5) x 12 = 5400 shingles. It will be noticed that method (4) gives the exact number required. To this must be added the extra required for waste, etc. PROBLEMS 1. A lean-to roof is 18 ft. wide and 15 ft. long. How many shingles laid 5 in. to the weather are required ? (By method (4).) 2. Find the number of shingles required for a roof, if the length of the ridge is 32 ft., the length of the rafters 15 ft., and the shingles are laid 5 in. to the weather. (By method (3).) 3. The roof of a bungalow has a 10-ft. run, a 5-ft. rise, and an eaves projection of 2 ft. If the ridge of the house is 25 ft. long, find how many shingles are required, if they are laid 4 in. to the weather. (Use method (3).) 4. How many shingles are required to cover a hip roof, if the building is 25 ft. square and the rise of the roof is 10 ft., the shingles to be laid 5 in. to the weather and the eaves projection to be 18 in.? (Use method (2).) 5. A cottage has four gable ends, each 18 ft. at the base and 9 ft. high. If we lay the shingles 4^ in. to the HOUSE BUILDING 35 weather, how many are required to cover the 4 gable ends ? 6. A house is built in the form of the letter T; the length of the ridge of the top of the T is 35 ft., and the rafters are 17 ft. long, with an additional eaves projection of 20 in. ; the length of the ridge of the stem of the T is 20 ft., and its rafters also are 17 ft. long, with an additional 20-in. eaves projection. If the width of each part of the house is 24 ft., how many shingles, laid 5 in. to the weather, will cover the roof ? 7. A tower 10 ft. square has a hip roof with a 10-ft. rise. With shingles at $4.50 per M, what will be the cost of shingling the roof, the shingles to be laid 5 in. to the weather, the roof to have an eaves projection of 24 in., and the cost of nails and the labor of laying the shingles to be $2.50 per M ? (By method (3).) 8. An octagonal tower is 12 ft. across between any two parallel sides, and has an eaves projection of 20 in. If the rise is 9 ft. and we lay the shingles 5 in. to the weather, how many are required for the roof ? (By method (2).) 9. A square house 32 ft. on a side has a veranda with a circular corner all the way across the front and halfway down the side. If the length of the rafters is 11 ft. and we lay the shingles 4 in. to the weather, how many are required for the roof? (By method (3).) 10. How many shingles laid 5 in. to the weather will be required to cover the roof of a hexagonal church steeple which is 6 ft. on a side at the base and 30 ft. high ? (By method (3).) 11. A round tower is 11 ft. 6 in. in diameter and the rise of the roof is 10 ft. If we lay the shingles 5 in. to the weather, how much will shingles for the roof cost at $4.50 per M? 36 SHOP PROBLEMS IN MATHEMATICS 12. At $5 per M, what will be the cost of special 4-in. shingles for the gable end of a house that is 26 ft. wide and that has a half-pitch roof, the shingles to be laid 5 in. to the weather ? 13. A pavilion is hexagonal in shape; the rise of the roof is 10 ft. and the length of the hip or corner rafter is 18 ft. If the shingles cost $4.50 per M and are laid 5 in. to the weather, what will be the cost of shingling the pavilion if the nails and labor of laying costs $2.50 per M ? BRICKWORK 35. Size and shape of bricks. Bricks vary in size and shape according to the locality in which they are made and the use to which they are to be put. (1) Ordinary red bricks are approximately 2^ in. thick, 3| in. wide, and 8 in. long. In estimating the number of these bricks for a wall, 1\ bricks are generally figured per square foot of wall surface 1 brick thick. (2) The so-called " Norman " bricks are 2J in. thick by 4 in. wide by 12 in. long, and are figured 4J bricks per square foot of wall. (3) " Roman " bricks are 1 in. by 4 in. by 12 in., and are estimated at 7 bricks per square foot of wall. (4) The standard face bricks are 2 in. thick, 4 in. wide, and 8 in. long. They also are estimated at 7 bricks per square foot of wall. 36. Measuring brickwork. Brickwork is generally meas- ured by the square foot, square yard, etc., of wall (surface) so many bricks thick. For example, a 4-in. wall would be 1 brick thick, an 8-in. wall 2 bricks thick, and so on. Brick- work is also estimated by cubic measure, in which case a wall 12 in. thick is figured to contain 22 bricks for every square foot of wall face, and an 8-in. wall 15 bricks. This HOUSE BUILDING 37 can be done only when the bricks throughout the wall are of uniform size. When openings for doors, windows, etc. , are few and small com- pared with the size of the wall, they are not taken into account in estimating the number of bricks. If large, they should be figured out. In ordinary brickwork the amount of mortar used is from 12 to 18 bushels per 1000 bricks. PROBLEMS 1. How many common bricks are required for a wall 8 ft. high, 26 ft. 6 in. long, and 2 bricks thick ? 2. A 12-in. wall is faced with Norman bricks and filled in with common red brick. If the wall is 18 ft. 7 in. high and 41 ft. 6 in. long, how many of each kind of brick are required? 3. How many Roman bricks are required to veneer a a house 1 brick thick up to the second floor, the size of the house being 26 ft. 6 in. by 35 ft., if we deduct 200 sq. ft. for door and window openings and the veneer work is 10 ft. high ? 4. A 16-in. wall is faced on both sides with standard-face brick and filled with common brick. If the wall is 60 ft. long and 31 ft. high, and has 8 openings for windows 3| ft. by 7 ft., how many of each kind of brick are required ? 5. If common red bricks cost $8 per M, how much will it cost to lay a wall 18 ft. high by 49 ft. long by 12 in. thick, the labor and mortar to cost $12 per M? 37. Chimneys. Common brick chimneys are generally des- ignated as so many bricks long and so many bricks wide, outside measure. It usually takes from 4| to 5 bricks for 1 ft. in height. Hence, to find the number of bricks required for a chimney, multiply the number of bricks in one course by 4t or 5, as the case may be, and this number again by the Iiei'jJit of the chimneij in feet. 38 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. How many bricks are required for a chimney 2 bricks square and 10 ft. high ? (5 bricks to the foot.) 2. A chimney is 2 bricks long and 2 bricks wide. If it is 35 ft. high, how many bricks does it require ? (5 bricks to the foot.) 3. A chimney 36 ft. from the bottom of the cellar to the roof is laid of common red bricks. Above the roof it pro- jects 4 ft. and is laid of Roman bricks. If it is 5 com- mon bricks long and 2 common bricks wide (40 in. by 16 in.) and has 3 flues separated by 2 brick partitions with bricks placed flatwise, how many of each kind of brick are re- quired ? (The common bricks lay 5, and the Koman bricks 7, to the foot.) LATHING AND PLASTERING 38. Wood laths. Common wood laths are about f in. thick, from li in. to 1^ in. wide, and 4 ft. long. They are generally put up in bunches of 50 laths each, although in some localities 100 make a bunch. Every bunch of 50 is figured to cover about 25 sq. ft. Every lath requires at least 4 nails, and there are about 480 3d lath nails to the pound. 39. Plastering. Plastering is usually done at so much per square yard of wall surface. 40. Common mortar. Common mortar is generally com- posed of five parts sand, one part lime, and one half part hair. A bushel of common mortar will cover about 3 sq. yd. with two coats. 41. Patent mortar. In addition to common mortar there are also several kinds of patent mortar now in use. Their covering capacities vary with the nature of their materials and the kinds of work on which they are used. HOUSE BUILDING 39 PROBLEMS 1. How many laths will it take to cover both sides of a partition that is 9 ft. high and 14 J ft. long ? 2. A room is 12 ft. high, 20 ft. 6 in. wide, and 34 ft. 8 in. long. How many laths and how many pounds of lath nails does it require for the overhead and side walls, if we deduct 3 squares for openings ? 3. The first floor of a house is divided into 4 rooms as follows : 12 by 14 ft., 10 by 10 ft., 8 by 10 ft., and 12 by 12 ft. respectively. If the rooms are 9 ft. high and we take out 350 sq. ft. for door and window openings, how much will it cost to lath the rooms, the laths costing $5 per 1000, the nails 5/ per pound, and the work of putting on 25 cents per 100 laths ? 4. If it takes about 18 laths per square yard, how many laths and lath nails are required for the side walls of a room 32 ft. wide, 46 ft. 8 in. long, and 16 ft. high, there being 8 windows and 4 doors, averaging 25 sq. ft. each, and a wainscoting all around, 4 ft. up from the floor ? 5. How much common mortar is required to plaster two coats on the four side walls and ceiling of a room that is 12 ft. wide, 13 ft. long, and 9 ft. high, if we deduct 100 sq. ft. for window and door openings ? 6. With sand at $2 per cubic yard, lime at $1.50 per barrel containing 3 bushels, and hair at 60 cents per bushel, what will be the cost of mortar for two-coat work for the side walls and ceilings of the following rooms ? 1 room 12 ft. by 16 ft. by 10 ft. high. 1 room Hi ft. by 12 ft. 8 in. by 10 ft. high. 1 room 9 ft. by 11 ft. 6 in. by 10 ft. high. Each room has one door opening 3 ft. by 7^ ft. and two windows 3 ft. by 7 ft. each. 40 SHOP PROBLEMS IN MATHEMATICS 7. The price of a certain patent plaster is $11 per ton; it has a covering capacity of 120 sq. yd. per ton. If the mason charges 15 cents per square yard for labor and we deduct 400 sq. ft. for door and window openings, what will be the cost of plastering the side walls of a three-room summer cottage, the sizes of which are as follows ? 1 room 13 ft. by 23 ft. by 11 ft. high. 1 room 10 ft. by 12 ft. by 11 ft. high. .1 room 9 ft. 4 in. by 12 ft. by 11 ft. high. 2 closets 3 ft. by 5 ft. 6 in. by 11 ft. high. 8. A mason takes the contract to furnish all materials and do the work involved in lathing and plastering a six- room flat, for 50 cents per square yard. The sizes of the rooms of the flat are as follows : 1 room 81 ft. by 12 ft. by 9 ft. high. 1 room 15 ft. by 11 ft. 4 in. by 9 ft. high. 1 room 10 ft. by 11 ft. 4 in. by 9 ft. high. 1 room 13 ft. by 9^ ft. by 9 ft. high. 1 room 12 ft. 8 in. by 15 ft. by 9 ft. high. 1 room 7 ft. by 11 ft. by 9 ft. high. 1 hall 3 ft. by 16 ft. by 9 ft. high. 3 closets, including bathroom, averaging 4 ft. by 6 ft. by 9 ft. high. If 100 sq. yd. are deducted for door and window openings, how much does the mason receive for the job ? 9. If, in Problem 8, the laths cost $4 per thousand, the sand $2 per cubic yard, the hair 60 cents per bushel, the lime $1.60 per barrel, containing 3 bushels, the lath nails 5 cents per pound, with 450 nails to a pound and 4 nails to every lath, the labor of lathing 25 cents per 100, and the labor of plastering for the two coats 18 cents per square yard, does the mason make or lose money, and how much ? HOUSE BUILDING 41 STONEWORK, EXCAVATING, CAPACITIES OF BINS 42. Approximate cubical contents. The following method is useful in finding the cubical contents of rectangular solids where a few inches may be disregarded, as in the excavating of earth, in brickwork, cementwork, certain kinds of stonework, etc. For finding the area of one surface of the solid see sect. 19. Multiply this area by the other dimension of the solid in the same manner as before. EXAMPLE. A rectangular solid is 3 ft. 7 in. wide, 4 ft. 6 in. long, and 2 ft. 8 in. thick. What is its cubical con- tents ? ACTUAL WORK EXPLANATION 3 ft. 7 in. 4 ft. x 3 ft. = 12 sq. ft. 4 6_ 4 ft. x 7 in. = f | sq. ft. 12 =2 sq. ft. and T 4 ^ sq. ft. 2 4 6 in. x 4 ft. = j| sq.ft. 16 = 1 sq. ft. and , T \ sq. ft. 3 6 6in.x7in. = 42sq. in. 16 2 = T 3 5 sq.ft. and T | ? sq.ft. 28 6 in the third column is called 1 32 and added to the second column, 4 giving a sum of j| sq. ft. 1 sq. 10 8 ft. and T \ sq. ft. Adding the 1 sq. 1 ft. to the first column, we have 43 cu. ft. T V cu. ft. 16 sq. ft. 2 ft. x 16 sq. ft. = 32 cu. ft. 2 ft. x T 2 2 sq. ft. = T \ cu. ft. 8 in. x 16 sq. ft. = T \ ft. x 16 sq. ft. = iff. cu. ft. = 10 cu. ft. and T 8 ^ cu. ft. 8 in. x T \ sq. ft. = T \ ft. x T \ sq. ft. = T ^ cu. ft. = T ^ cu. ft. and T | 3 cu. ft. 42 SHOP PROBLEMS IN MATHEMATICS Adding the second column, we have jf cu. ft. = 1 cu. ft. and yL cu. ft. Adding 1 cu. ft. to the first column, the result is 43 T T ^ cu. ft. PROBLEMS By the above method find the approximate cubical contents of rectangular solids having the following dimensions : 1. 1 ft. 6 in. by 1 ft: 4 in. by 2 ft. 2. 1 ft. 4 in. by 2 ft. 3 in. by 2 ft. 2 in. 3. 2 ft. 5 in. by 2 ft. 3 in. by 3 ft. 3 in. 4. 4 ft. 7 in. by 5 ft. 9 in. by 3 ft. 3 in. 5. 6 ft. 7 in. by 7 ft. 9 in. by 4 ft. 10 in. 6. 3 ft. 3 in. by 6 ft. 5 in. by 12 ft. 11 in. 7. 20 ft. 6 in. by 30 ft. 8 in. by 9 ft. 9 in. 8. 30 ft. 4 in. by 19 ft. 11 in. by 10 ft. 2 in. 9. 36 ft. 7 in. by 27 ft. 8 in. by 12 ft. 9 in. 10. 23 ft. 5 in. by 18 ft. 7 in. by 8 ft. 7 in. 11. How many cubic feet of stone mason work in a wall 16 in. thick, 9 ft. 3 in. high, and 40 ft. 6 in. long ? 12. If a ton of coal contains 35 cu. ft., how many tons will a bin 4 ft. 3 in. wide, 20 ft. 4 in. long, and 5 ft. 8 in. high hold ? 13. At $6.75 per ton, how much will a car of anthracite coal cost if the car is 3 ft. 3 in. deep, 8 ft. 1 in. wide, and 33 ft. long ? 14. A man wishes to build a potato bin that will hold 40 bushels. It is to be 6 ft. 8 in. long and 3 ft. 6 in. high, and a bushel contains 1| cu. ft. What must be the width of the bin ? 15. Susquehanna granite weighs approximately 169 Ib. per cubic foot. What will be the weight of a block that is 4 ft. 6 in. long, 3 ft. 8 in. wide, and 2 ft. 5 in. thick ? NERAL CONSTRUCTION 43 16. How many cubic yards in a bank of earth 12 ft. 8 in. wide, 21 ft. 6 in. long, and 7 ft. 6 in. deep ? 17. A cellar wall is 23 ft. 6 in. wide by 35 ft. 8 in. long, outside measure, and 8 ft. 6 in. high. If the wall is 16 in. thick, how many cubic yards of mason work does it contain ? 18. If a bushel of potatoes weighs 60 Ib. and contains approximately li cu. ft., how many bushels can be loaded into a car that is 33 ft. long and 8 ft. 2 in. wide, inside measure, the potatoes to be piled 6 ft. 6 in. high ? What will be the weight of the carload ? 19. The depths of earth taken in various points of a cellar that is to be excavated are 4 ft. 2 in., 5 ft. 6 in., 7 ft., 6 ft. 6 in., 3 ft. 10 in., 7 ft. 4 in., 8 ft. 8 in., and 9 ft. re- spectively. If the cellar is to be 60 ft. 6 in. wide and 90 ft. 10 in. long, how many cubic yards of earth must be re- moved ? (Take the average depth of the cellar.) 20. At $1 per cubic yard, what will be the cost of blasting out the rock from a cellar that is 22 ft. 6 in. wide by 46 ft. 6 in. long, the depth of the rock at various points being 4 ft., 2 ft. 6 in., 3 ft. 4 in., 6 ft. 8 in., 7 ft., 5 ft., and 4 ft. 6 in. ? 43. Amount of lumber for bins, boxes, and furniture. Before working out the following problems the student should read Chapter XVI. PROBLEMS 1. A man wishes to build a coal bin 5 ft. deep and 7 ft. square on the outside, each side to be stayed by three 2- by 4-in. pieces, 6 ft. long, and the bottom to be laid on four pieces of 2- by 4-in. stock, 7 ft. long. If the boards used are 1 in. thick, how many feet of lumber are required for the bin? 2. If there are 35 cu. ft. of anthracite coal in a ton, how much 1-in. lumber will it take to line a coal bin to hold 6 tons, the width of the bin to be 7 ft. and its length 8 ft. ? 44 SHOP PROBLEMS IN MATHEMATICS 3. A man wishes to put up a billboard 8 ft. wide and 30 ft. long, using matched lumber in. thick. The frame- work for the board consists of 2 pieces of 2- by 4-in. stock, running lengthwise, and 6 pieces of 2- by 4-in. stock, run- ning crosswise or up and down. If he makes no allowance for waste on the scantling and adds 25 ft. for waste and squaring to the matched stuff, how much of each kind of lumber must he buy for the board ? 4. A grain bin with a hopper bottom is lined with 1-in. boards of double thickness ; the bin is 16 ft. deep down to the hopper part and is 13 ft. square ; the hopper has a per- pendicular height of 5 ft. If we allow 25 ft. for waste in FIG. 21. HOPPER cutting the boards, how much lumber is required for the bin ? (See Fig. 21.) 5. A manual-training bench top is 15 in. wide, 4 ft. long, and 2| in. thick. To prevent warping it is constructed of 1^-in. by 3-in. pieces (in the rough stock) glued together. Allowing in. both on the thickness and width and 1 in. on the length of each piece for truing up, how much lum- ber is required for the top ? (See Fig. 22.) 6. A cabinet box is 4 in. high, 6 in. wide, and 10 in. long, inside measure. If the side and end pieces are constructed of stock | in. thick, and the top and bottom of stock f in. thick, how much stock is required for its construction, GENERAL CONSTRUCTION 45 allowing \ in. on the width and f in. on the length of every piece for rough stock ? 7. How much lumber is necessary to construct a mission umbrella rack having 4 corner posts If in. square and 2 ft. long, 2 crosspieces on each of the 4 sides J by 2 by 12 in., and a -in. board in the bottom, allowing ^ in. on the width and thickness and 1 in. on the length of every piece for rough stock ? 8. A quartered-oak mission table top is 1^ in. thick, 30 in. wide, and 47 in. long when finished. If it is constructed of four boards jointed edge to edge, how many feet does it contain, allowing at least in. on the width of every piece FIG. 22. BENCH TOP for jointing and 1 in. on the length of every piece for tru- ing up ? What size of boards would you select ? 9. With oak lumber at $75 per M, what will be the cost of lumber for 36 footstools 12 by 16 by 8 in. high, the rough stock for every stool to be as follows ? 4 legs 2 by 2 by 9 in. long. 2 rails J by 3 by 15 in. long. 2 rails J by 3 by 11 in. long. This does not provide for the top, which is usually upholstered. 10. A threefold screen is 58 in. high and 54 in. wide. Every panel has 2 wooden stiles and 2 rails : the stiles are 1 in. by If in. ; the bottom rails are 1 in. by 4 in.; and the top rails are 1 in. by If in. when finished. If the rails go into 46 SHOP PROBLEMS IN MATHEMATICS the stiles with a tenon 1 in. long, state the number of pieces required, the size of every piece (rough stock), and the num- ber of feet of lumber required for the screen. 11. With ash lumber at $85 per M, how much will the wood cost for a porch swing 24 in. high, 24 in. wide, and 60 in. long, the rough stock to be as follows ? 4 posts 3 by 3 by 25 in. long. 2 rails 1 by 4J by 26 in. 2 rails 1^ by 4^ by 62 in. 2 rails Ij- by 3 by 26 in. 10 bottom pieces -J by 6 by 22 in. 5 slats for back and side -j- by 10 by 15 in. 12. A quartered-oak library table is 45 in. long, 30 in. wide, and 30 in. high ; the top when finished is 1J in. thick ; the legs are 2J in. square ; the 4 rails (2 side and 2 end) are J by 5f in. ; the legs set back from the end of the top 3 in. and from the side If in. ; the rails go into the legs with a IJ-in. tenon. Required the size of every piece of rough stock, the number of pieces, and the number of feet of lumber (the top to contain 4 pieces). 13. At $100 per M, what will quartered-oak lumber cost for 10 armchairs, to be constructed as follows : Size of chairs, 31 in. high, 25 in. deep, and 26 in. wide. To have 4 posts 2J in. square when finished full height. FIG. 23. ARMCHAIR GENERAL CONSTRUCTION 47 To have 4 rails 1 in. by 5 in. when finished. To have 3 rails 1 in. by 2 in. when finished, one at the top of the back and one at the top of either side. To have 4 slats for side and back in. by 6 in. by 15 in. long (rough stock). (See Fig. 23.) 14. What is the smallest rectangular piece of stock from which a hexagonal jardiniere-stand top 6 in. on a side can be cut, allowing ^ in. on the width and 1 in. on the length of the piece for rough stock ? 15. A corner shelf is 16 in. on each of its shorter sides. What is the smallest piece of rectangular stock from which it can be cut, allowing \ in. on the width and 1 in. on the length for rough stock ? 16. A boy wishes to construct an octagonal table top 12 in. on a side. What size of board must he buy, the top to contain 4 pieces and to be 1 in. thick when finished ? Allow \ in. on the width of every piece and 1 in. on the length for jointing and truing up. 17. How much lumber is required for a magazine cabinet 40 in. high, 14 in. wide, and 10 in. deep when finished, the rough stock to be as follows : 2 uprights 41 in. high by 10^ in. wide by 1 in. thick. 4 shelves 15 in. long by 9 in. wide by 1 in. thick. 2 cross rails 15 in. long by 4 in. wide by 1 in. thick. 44. Finding a square equivalent to two squares. To find the side of a square that has an area equal to the areas of two given squares, denote the sides of the given squares by s and ! and the side of the required square by S, as in Fig. 24. If a right triangle is constructed with the legs equal respectively to s and s lt the hypotenuse S will be the side of the required square, since S 2 = s 2 + sf. 48 SHOP PROBLEMS IN MATHEMATICS This calculation may be made by any of the following methods : (1) On the steel square, laying off s and s x on the sides of the square and measuring the diagonal S from the ends of s and s lf (See Fig. 16.) (2) On graph or cross- section paper, using any convenient unit. (3) Drawing to scale a right triangle, given the two legs s and s x and measuring S. (4) By the form- ula S 2 = s 2 + s?, substituting the given values of s and s x and solv- ing for S. EXAMPLE. Two grain bins of the same depth have bottoms 9 ft. and FIG. 24. SQUARE EQUIVALENT TO Two SQUARES spectively. What must be the size of the square bottom of a third bin of the same depth that will hold as much as the two given bins ? Sl = 11. Substituting in the formula S 2 s 2 -f- sf, S 2 = 81 + 121 = 202. S = 14.21 ft. = 14 ft. 3 in. 45. Finding a circle equivalent to two circles. Since the areas of circles are to each other as the squares of their GENERAL CONSTRUCTION 49 diameters, the methods of section 44 may be used to find the diameter of a circle that is equivalent to two given circles. Denoting the given diameters by d and d ly and the re- quired diameter by Z>, we have the formula D 2 = d 2 -f- d?, from which D may be found by any of the methods of section 44. EXAMPLE. Two branches of an iron pipe are respectively 2 in. and 3 in. in diameter. What must be the diameter of the pipe into which they empty, in order that the water may be carried off if the velocity of the water in all three pipes is the same ? FIG. 25. CIRCLE EQUIVALENT TO Two CIRCLES Substituting in the form- ula D 2 = d 2 + d}, D 2 = 4 + 9 = 13. D = 3.6 in. Since the nearest standard size of iron pipe above 3.6 in. is 4 in., the result should be called 4 in. 46. Finding a square equivalent to the difference of two given squares, or a circle equivalent to the difference of two given circles. The methods of sections 44 and 45 may be used when s, s 1? d, or d 1 is the unknown quantity. If the work is done by drawing to scale, as in Fig. 25, a right triangle must be constructed, given the hypotenuse and one side. If the formula is used, C 2 _ C-2 C 2 s o ij or d' 2 = D 2 d*. 50 SHOP PROBLEMS IN MATHEMATICS EXAMPLE. A lead water pipe 2 in. in diameter discharges into a tank. The water leaves the tank by two lead pipes, one of which is \ in. in diameter. What must be the diame- ter of the other in order that the water may leave the tank as fast as it runs in, if the velocity of the water in all three pipes is the same? d' 2 = D 2 - d} = 4 - 1 = 3. d = 1.7 in. Since the nearest standard size of lead coil pipe above 1.7 in. is If in., the result should be called If in. SOME STANDARD SIZES OF PIPE Lead coil pipe j, ^, >, |, |, 1, li, 1|, If, 2. Lead waste pipe U, 2, 3, 3J, 4, 4-|, 5, 6. Iron pipe |, 1, f , |, f , 1, li, 1J, 2, 2J, 3, 3J, 4, 4J, 5, 6, 7, 8, 9, 10, 11, 12. PROBLEMS 1. The two branches of an iron sewer pipe are respec- tively 2 and 3| in. in diameter. What must be the diame- ter of the iron pipe into which they empty, in order that the sewage may be carried off ? 2. How long must the side of a square plot of ground be, to contain as much land as two other square plots whose sides measure 100 ft. and 150 ft. respectively ? 3. Find the diameter of a circular flower bed that will be equal in area to two circular flower beds having diame- ters of 15 ft. and 20 ft. respectively. 4. A lead water pipe 2 in. in diameter discharges into a tank. The water leaves the tank by two lead waste pipes, the diameter of one of which is 1 in. What must be the GENERAL CONSTRUCTION 51 diameter of the other pipe in order that the water may leave the tank as fast as it runs in ? 5. A vessel 12 in. deep and 15 in. in diameter holds about one bushel. Each of two other vessels is 12 in. deep, the diameter of one being 7 in. What must be the diameter of the other in order that the two together may hold as much as the first ? 6. Two grain bins are 8 ft. and 7 ft. square on the bottom respectively. What must be the size of the square bottom of the third in order that it may hold as much as the other two, the depth of all three being the same ? 7. Water is conducted into a tank through two lead coil pipes having diameters of f in. and If in. respectively. Find the standard size of the lead waste pipe that will allow the water to run out as fast as it runs in. If all pipes are iron, supply the missing -diameters indi- cated in the following table : D d (fi 8. 9. 10. 11. i H 2* 41 ** 2 10 3| 12. Two branch iron sewer pipes, each 6 in. in diameter, empty into a third pipe ; farther down the line another branch 8 in. in diameter runs into the sewer. What must be the diameter of the iron pipe below this point in order to carry off the sewage? 13. The side of a square plot of ground measures 10 rd. Find the side of a square plot twice as large. Show that the ratio of the side of the second plot to the side of the first plot is V2 = 1.414. 52 SHOP PROBLEMS IN MATHEMATICS HEIGHTS OF TREES OR OTHER OBJECTS 47. First method. By the isosceles right triangle. The ob- server stands with his eye at A. ABC is an isosceles right triangle. Evidently h, the height of the tre,e, is equal to d, the distance from the observer to the tree, plus a, the height of the observer's eye above the ground (Fig. 26). WRITTEN EXERCISE 1. If a = 6, d = 50', find h. 2. If a = 5' 6", d = 45' 9", find h. 3. If a, = 5i, d = 90, find 48. Second method. If a triangle is used in which BC = 2 AC, show that h = 2 d + a. FIG. 26. HEIGHTS or TREES FIRST METHOD WRITTEN EXERCISE 1. If a = 5' 4", d = 30', find h. 2. If a = 5' 3", d = 25', find h. 3. If a = 5 f , d = 42' 6", find h. HEIGHTS OF TREES 53 49. Third method. To find h, the height of the tree, meas- ure the shadow d, set up a stake h' f measure its shadow d', and then, using the proportion - = , solve for h. ct ct EXAMPLE. Find hiid = 30', h' == 5', d' = 6'. , r h 5 / 30~6' / ; 5 X 30 / 6 / = 25. / 4 / / c / / / / AL. 1C 1 / 1 / f / /. * / , ! .__*._. \*d'4 1*- - ^ J FIG. 27. SECOND METHOD FIG. 28. THIRD METHOD WRITTEN EXERCISE Find h if d, h', and d' have the values indicated. d h' d' 1. 45' 3" 5' 3' 4" 2. 42' 6" 5' 6' 8" 3. 22' 4" 5' 10' 3" 4. 21' 5" 5' 15' 6" 54 SHOP PROBLEMS IN MATHEMATICS 5. The shadow of a stake driven into the ground perpen- dicularly is 2 ft. and the height of the stake is 3 ft. If the shadow of a near-by tree is 70 ft. long, how high is the tree? 6. The shadow of a tree is 40 ft. long. If the shadow of a stake 4 ft. high is 2 ft. 3 in., what is the height of the tree ? 50. Fourth method. Set two poles in a line with the tree and perpendicular to the earth, as in Fig. 29. With the eye at E, a point on one pole, sight across the other pole to the base and to the top of the tree. Have an assistant mark the points where the lines of vision cross /\ the second pole at P and P'. FIG. 29. FOURTH METHOD Measure EP, EA, and PP'. Then A, the height of the tree = EA x EP WRITTEN EXERCISE Find h if EA, PP', and EP have the values indicated. EA PP' EP 1. 44' 3' 2' 2. 33' 5' 6" 2' 3" 3. 52' 6" 5' 3" 5' 4. 24' 4" 5' 8" 2' 4" 5. 25' 8" 6' 1'6" 6. 40' 3" 5' 4" 3' 7. 38' 5" 5' 10" 4' 3" HEIGHTS OF TREES 55 51. Fifth method. The observer walks on level ground to a distance TE from the foot of the tree, about equal to its estimated height TT'. He then lies on his back, stretched at S* full length, with his eye at E and his feet at A. An assistant marks the point B on a perpendicular ,-'' staff erected at his feet, ,-'' the exact point where his line of vision to the top of the E -'-'--- tree crosses the FlG - 30 - FlFTU METHOD staff. The distances BA, EA, and ET are measured. BA X ET Then TT' = EA EXERCISE Find TT' in the following problems : BA ET EA I. 4' 2" 52' 4" 6' 2. 5' 3" 45' 2" 5' 10" 3. 8' 4" 30' 5" 5' 6" 4. 10' 6" 23' 7" 5' 4" 52. Sixth method. The followin; method gives a rough approximation to the height of a tree : The distance ylZ? is marked /' ,. on the tree as high as the woodman can reach. / Then, stepping back, E^l a pencil PP' is held vertically with the FIG. 31. SIXTH METHOD 56 SHOP PROBLEMS IN MATHEMATICS eye at E, as in the figure. The pencil is raised with .P' transferred to P, and the point C marked with the eye. This is repeated till the top of the tree T is reached. d, d', d", d"', d iv are approximately equal. The same method applies to finding the approximate diameter of the tree at any given point above the ground. From the height and diameter a rough approximation can be made of the amount of lumber in the tree. ORAL EXERCISE 1. If d = 6' 6" and the pencil is raised 8 times before the top of the tree is reached, find the height of the tree. If n is the number of times the pencil is raised, find the height of the trees in the following problems : 2. 6' 2" 6 3. 6' 6" 8 4. r-l i 7 5. 7' 3" 4 NOTE. For a description of instruments used in forestry for meas- uring heights of trees, the student is referred to "The Woodman's Handbook," Part I, by H. S. Graves, published as Bulletin No. 36 by the U. S. Dept. of Agriculture, Bureau of Forestry. 53. To measure the distance across a river gorge, etc. For the principle of right triangles the student is referred to sections 44-46. PROBLEMS 1. Two trees, T and C, stand directly opposite each other on the banks of the river. A man takes a right-angled tri- angle ACS, with equal sides AC and BC, and marks the line CD on the bank at right angles to CT. Then he walks DISTANCE ACROSS A RIVER 5T along the line CD until, placing the triangle to his eye in a horizontal position, he sees in range of the line of the hy- potenuse A'B 1 the tree Ton the opposite bank, and in range of the line of the side A'C' the tree C on the same side as himself. If the distance from where he stands to the tree on the same side is 700 ft., how far is it across the river ? FIG. 32. DISTANCE ACROSS A RIVER 2. A man wishes to know the distance across a gorge. He drives a stake on one bank and takes for his observa- tion point on the other bank a round stone. He then takes a right-angled triangle, the sides of which are as 2 is to 9, places it in a position similar to that of Problem 1, and walks up the bank until the hypotenuse ranges with the stone on the opposite bank and the short side ranges with the stake. If the distance from where he stands to the stake is 46 ft., what is the width of the gorge ? 3. With a right-angled triangle with sides as 3 is to 7J, placed as in Problem 1, we find the distance from our start- ing point on the bank of a river to a tree on the opposite side to be 460 ft. We take a stand at the starting point and with two straight, thin strips, get the range of the top of 58 SHOP PROBLEMS IN MATHEMATICS the tree with one of the strips and the range of the bottom with the other, thus forming an acute angle at the eye with the strips. We now measure a distance on the lower strip (say 3 ft. from the angle) and the perpendicular from this point to the upper strip, which we find to be 3 in. What is the height of the tree on the opposite bank ? HEIGHTS OF CLIFFS 54. By falling bodies. The number of feet passed over by a body falling from a state of rest is equal to 16.08 times the square of the number of seconds during which the body falls. (1) Denoting this space by S and the number of seconds by t, write the formula for S in terms of t. (2) A stone dropped from a cliff reaches the ground in 4 seconds. Find the height of the cliff. (3) From a point on the Palisades a stone is dropped, reaching the river in 3 seconds. How high is the point above the river ? (4) A stone dropped from the top of a building reaches the ground in 2f seconds. Find the height of the building. MISCELLANEOUS PROBLEMS 1. A man holds a ruler vertically between his thumb and finger at arm's length, and places it in such a position that he can see the top of a tree over the top of the ruler, and the bottom of the tree on a line with the top of his thumb. He finds the distance from his eye to the top of his thumb to be 2 ft. 4 in., and the distance the ruler pro- jects above his thumb to be 9 in. If from the base of the tree to where he stands the distance is 60 ft., how high is the tree ? 2. What would be the height of a church steeple, accord- ing to the above method, if the distance from the top of MISCELLANEOUS PROBLEMS 59 the ruler to the top of the man's thumb is 2 ft., the dis- tance from his eye to the top of the thumb is 1 ft. 9 in., and the distance from his eye to the ground directly below the steeple is 100 ft. ? 3. The height of a building on one bank of a river is 55 ft. By holding a ruler on the opposite bank according to Problem 1, we find when the top of the ruler ranges with the top of the building, and the top of the thumb with the bottom of the building, the ruler projects above the thumb 6|- in., and the distance from the thumb to the eye is 23 in. What is the width of the river ? 4. The legs of a right-angled triangle are to each other as 3 is to 5. With the triangle placed as in sect. 47, what will be the height of the tree if the eye is 4 ft. 6 in. from the ground and 45 ft. from the base of the tree ? 5. The trunk of a tree when trimmed is 48 ft. long. At the small end it is 8 in. in diameter and at the large end 20 in. If we cut the tree into four 12-ft. lengths, what is the largest square stick of timber in even inches that can be obtained from every length ? 6. If we saw the above tree into ^-in. boards, allowing ^-in. waste in sawdust for every cut, how many feet of lumber shall we obtain ? 7. The height of a tree to the first branch is 40 ft., its diameter at the base is 26 in., and at the branch is 16 in. Allowing 2 in. on the diameter for the bark, what is the largest square stick of timber in even inches that can be sawed from the tree, using it only to the branch? How many feet does it contain ? What proportion of the wood of that part of the tree used does the timber contain ? 8. What is the largest square stick of timber in even inches that can be sawed from a log 16 in. in diameter ? How many board feet does it contain if it is 14 ft. long ? 60 SHOP PROBLEMS IN MATHEMATICS 9. How many 1-in. boards can be sawed from the above timber if ^ in. is allowed for sawdust every time the saw goes through ? How many |-in. boards ? How many -in. boards ? 10. If a California sequoia tree measures 32 ft. in diame- ter at the base and 2 ft. in diameter 300 ft. from the ground, measurements taken inside the bark, how many feet of 1-in. boards will it make, deducting 3 in. from the diameter for slabs and for sawdust ? (Use the average diameter.) 11. How many rails averaging 3 in. square and 10 ft. long would this tree make, there being no waste in slabs and sawdust ? How long would it take a man to make the rails if he made 200 a day ? 12. If a stick of timber is 16 in. square, how far must we measure each way from the corner to find the points to which to chamfer in order to make the stick a perfect octagon ? 13. The distance between any two parallel sides of an octagonal table top is 3 ft. What is the length of one of its sides ? 14. On a piece of work 2 ft. square the distance to be measured each way from a corner in order to make the work octagonal is 7 in. How far from the corner must we measure to make octagonal a piece of work that is 17 in. square? 15. The top of a derrick for hoisting stone is 32 ft. high. How many feet of wire rope are required for its 4 guy ropes if they are fastened to its top and anchored to points on the ground 90, 100, 84, and 124 ft. respectively from the base ? (Add 12 ft. for fastenings.) 16. The height of a house from the ground to the eaves is 21 ft. If a man sets the foot of a ladder 12 ft. from the house, how long must the ladder be to project 1 ft. above the eaves ? a El CHAPTER III PULLEYS, BELTS, AND SPEEDS PULLEYS AND SPEEDS 55. Driving pulley. The driving pulley (Fig. 35) is the pulley that transmits power to the belt, or, in other words, causes the belt to move. 56. Driven pulley. The driven pulley is the pulley to which the power is transmitted by the belt or, in other words, the pulley that is moved by the belt. 57. Speed. The term " speed " as applied to pulleys means the number of revolutions they make per minute, and is usually designated by the letters K. P. M. (revolutions per minute). 58. Rule for speeds of pulleys. The diameter of the driving pulley multiplied by its speed is equal to the diameter of tit e driven pulley multiplied by its speed. 59. Cutting speed. The term " cutting " or " surface " speed in connection with a lathe means the number of linear feet measured on the surface of the work that passes the edge of a cutting tool in one minute. 60. Rule for cutting speed. Multiply the number of feet in the circumference of the work being turned by the number of revolutions per minute, and the result ivill be the cutting speed in feet per minute. For example, if the circumference of a piece of work that is being turned is 20 in., and the work makes 400 R. P. M., the cutting speed is 400 x f = 666| ft. per minute. PULLEYS, BELTS, AND SPEEDS 63 PROBLEMS 1. Let c equal the circumference of the work and C the cutting speed. Write the formula for C in terms of c and E. P. M. ; also solve for c and K.' P. M. 2. Let D represent the diameter of the driving pulley, d the diameter of the driven pulley, S the speed of the driving pulley, and s the speed of the driven pulley. Write the formula for D in terms of d, S, and s. Solve for d, S, and s. 3. The diameter of a turned cylinder is If in. What is its circumference? 4. What is the diameter of a turned cylinder the cir- cumference of which is 11 in.? 5. Find the circumference of a pulley whose radius is 10J in. ? 6. The circumference of a pulley is 66^ in. What is its radius ? 7. The diameter of a driving pulley is 12 in. and its speed is 400 R. P. M. What is the speed of the driven pulley whose diameter is 3 in. ? 8. The Pv. P. M. of the cone pulley (see Fig. 33) on a wood-turning lathe is 2500 and the diameter of one step is 5 in. What must be the diameter of the driving pulley if its R. P. M. is 450? 9. The driving pulley on a shaft is 42 in. in diameter and makes 20 revolutions per minute. How many revolu- tions will the driven pulley make if its diameter is 3 ft. ? 10. The diameter of the driving pulley is 9 in. and its speed is 1000 R. P. M. At what speed will the driven pulley run if its diameter is 4 in.? 64 SHOP PROBLEMS IN MATHEMATICS 11. If the speed of an 18-in. pulley on the driving shaft is 300 E. P. M., what is the speed of the driven shaft which has a 12-in. pulley belted to the driver ? 12. The diameter of a piece that is being turned is 2 in. If the piece makes 2500 revolutions per minute, what is its cutting speed ? 13. The surface speed of a turning piece of work is found to be 3000 ft. per minute. If its diameter is 4 in., what is its E. P. M. ? 14. What is the diameter of a cylinder, if its E. P. M. is 2500 and its surface speed 1500 ? 15. Sometimes we increase the speed of the lathe while turning the same piece of work. Explain why the wood cuts off faster when we do this. 16. Using the same speed, we turn a piece down from 7^ in. to 2| in. in diameter. What is the ratio of the cut- ting speeds of the two diameters ? 17. A mechanic is turning on the flat side of a disk 8 in. in diameter; his turning chisel is placed 2 in. from the center point. What is the difference in the rate of the cut- ting or surface speed, if he moves his chisel f in. towards the edge of the disk, and what are the two cutting speeds if the E. P. M. is 1600 ? 18. If a bicycle wheel is 30 in. in diameter, how many times will it go around in running a mile ? 19. A locomotive drive wheel makes 290 revolutions in running a mile. What is its diameter ? 20. The main driving wheel of an engine is 12 ft. 6 in. in diameter and revolves at 96 E. P. M. ; it is belted to a 48-in. pulley on the main line shaft. Find the speed of the shaft. 21. A 36-in. pulley making 143 E. P. M. is belted to an- other making 396 E. P. M. Find the diameter of the latter. PULLEYS, BELTS, AND SPEEDS 65 22. A certain grindstone will stand a surface or rim speed of 800 ft. per minute. At how many R. P. M. could it run if its diameter is 4 ft. 8 in.? 23. The cutting speed of a turning chisel when held at the rim of a revolving disk is a. What is the cutting speed of the tool if it is moved 1 in. towards the center, the diameter of the disk being d ? 24. The R. P. M. of a wooden disk that is being turned is 2500 and its diameter is 6 in. To what speed must the lathe be changed to make the surface speed on the edge of the disk the same when the diameter is turned down from 6 to 3J in. ? 25. The surface or rim speed of a grindstone is 500 ft. per minute and its diameter is 30 in. If the line shaft to which it is belted makes 300 R. P. M. and the driving pulley on this shaft is 3 in. in diameter, what must be the diameter of the driven pulley on the grindstone shaft? 26. An emery wheel is 16 in. in diameter and its surface speed is 2000 ft. per minute. What must be the diameter of the driving pulley on the line shaft if the R. P. M. of the shaft is 250 and the diameter of the pulley on the emery-wheel shaft is 4| in. ? 27. A motor pulley is 8 in. in diameter and the motor makes 1200 R. P. M. If this speed is increased to 1600 R. P. M., what must be the size of the pulley in order to give the belt the same speed ? 28. The splicing of a belt connecting two equal pulleys travels through the air at the rate of 2000 ft. per minute. At what speed must the pulleys run if they are 20 in. in diameter ? 29. The surface speed of an emery wheel is 3000 ft. per minute and its R.P. M. is 600. What is its diameter ? 66 SHOP PROBLEMS IN MATHEMATICS 30. A band saw (Fig. 34) runs over two wheels, each 32 in. in diameter. If the band saw is 16 ft. long and the speed of the wheels is 600 R. P. M., what is the cutting speed of the band saw ? 31. A circular saw (Fig. 35) makes 3000 R. P. M. and its diameter is 16 in. If the speed of the upper and lower wheels of a band saw is 450 R. P. M. and the length of Wheel Guard Star fi 09 FIG. 34. BAND SAW (By permission of The Crescent Machine Co.) the saw 18 ft., what must be the diameter of the band-saw wheels in order that its cutting speed may be one third that of the circular saw ? 32. A table top 30 in. in diameter is being turned. What must its R. P. M. be if its surface speed at the outer edge is equal to that of a cylinder whose diameter is 2^ in. and whose R. P. M. is 2500 ? PULLEYS, BELTS, AND SPEEDS 67 33. The front wheel of a carriage is 3 ft. 4^- in. in diam- eter. How many turns does it make in traveling a distance of 5 mi. ? 34. Two boys sitting side by side on wooden horses ride on a merry-go-round. The distance from the outer horse to the center is 16 ft. and the distance between the horses 2 ft. If the merry-go-round makes 30 turns, which boy rides farther, and how much ? Fence Driving Pulle FIG. 35. CIRCULAR SAW TABLE (By permission of The Crescent Machine Co.) 35. How many revolutions per minute are made by a locomotive driving wheel that is 6 ft. in diameter, if the train is running 40 mi. an hour ? 36. If a locomotive wheel 4 ft. in diameter goes around 9000 times in running 20 mi., how much distance is lost by slipping ? 68 SHOP PROBLEMS IN MATHEMATICS LENGTHS OF BELTS 61. Practical measurement of belts. Although in actual practice the length of a belt is obtained by measuring with a tapeline, or otherwise, after the pulleys are in place, a knowledge of how the length is obtained mathematically is of no small value in the study of machinery. 62. Rule for lengths of belts. To find the length of a belt divide the sum of the diameters of the two pulleys by 2 and multiply the quotient by 3=; to this add twice the distance a I FIG. 36. PULLEYS OF EQUAL DIAMETERS If I is the length of the belt, show that I =7rD + 2. between the center of the two pulleys and the result will be the length of the belt. This rule applies only to the cases of open belts in Fig. 36 and Fig. 37. Belts are open unless cross belt is stated. 63. Degree of accuracy. The above method is absolutely correct only when the pulleys are equal in diameter (Fig. 36). The error obtained, however, when the pulleys are of different diameters (Fig. 37) (unless there is too great a difference) is so slight that it can be neglected, for all practical purposes, when dealing with ordinary leather, rubber, or canvas belts. 64, Exact lengths of belts. If it is desired to find the exact length of belts when the pulleys are of unequal diam- eters, it can be done by trigonometry (see sect. 249). PULLEYS, BELTS, AND SPEEDS 69 PROBLEMS 1. The distance between the centers of two pulleys of equal diameters (Fig. 36) is 12 ft. and their diameter is 21 in. How long must the connecting belt be ? 2. Each of two pulleys has a diameter of 24 in. If the connecting belt is 28|- ft. long, what is the distance between the centers of the pulleys ? 3. If the distance between the centers of two pulleys of equal diameters is 11^ ft., and the length of the connecting belt is 30 ft., what is the diameter of the pulleys ? a FIG. 37. PULLEYS OF UNEQUAL DIAMETERS WHERE THE DIFFERENCE is SLIGHT Show that I = TT -- I- 2 a. 4. How much belting is required to connect two pulleys 24 and 18 in. in diameter respectively (Fig. 37), the dis- tance between centers being 9J ft.? 5. If the diameter of the driving pulley on an engine is 56 in., the distance between its center and the center of a driven pulley on a line shaft is 18 ft., and the length of the connecting belt is 48 ft., what is the diameter of the driven pulley ? 6. A motor is belted to a pulley on a circular-saw bench. If the length of the belt is 17 ft. and the diameters of the 70 SHOP PROBLEMS IN MATHEMATICS two pulleys are 10 and 4 in. respectively, what is the dis- tance between the centers of the pulleys ? 7. The cone pulley on a lathe has three steps with 3, 5, and 7-in. diameters respectively, and the corresponding cone pulley on the shaft over the lathe has its largest step 14 in. in diameter. What must be the diameter of the other two steps on the larger pulley in order to have the same belt work on all three steps of both pulleys ? J I FIG. 38. PULLEYS OF UNEQUAL DIAMETERS WHERE THE DIFFERENCE is GREAT Showthat 1 = 2 A/I' 1 ) + a 2 H TT - 8. A driving pulley on a motor is 5 in. in diameter and the driven pulley on the line shaft is 36 in. in diameter. If the distance between the centers is 8 ft., how long must the connecting belt be ? 9. Two pulleys are belted together to run in opposite directions (cross belt, Fig. 39) ; the diameter of each is 16 in., and their distance between centers is 10 ft. What length of belt is required ? 10. Find the diameter of the driving pulley on a motor, if the distance between its center and the center of the driven pulley on the lathe is 22 in., the length of the connecting belt is 60 in., and the diameter of the driven pulley is 3 in. PULLEYS, BELTS, AND SPEEDS 71 11. Two pulleys run with cross belt (Mg. 40). The diam- eter of one is 30 in. and the diameter of the other is 16 in. What length of belt do they require if the distance between centers is 13^ ft. ? a A i I Fio. S..). PULLEYS OF EQUAL DIAMETERS WITH CROSS BELT Show that I = 2 VZ> 2 -f a 2 -f irD. 12. Two pulleys of equal diameters are connected with a cross belt. If the length of this belt is 29 ft. and the FIG. 40. PULLEYS OF UNEQUAL DIAMETERS WITH CROSS BELT Show that I = 2\l[ ) + a 2 + TT distance between centers of the pulleys is 7 ft. 5 in., what is the diameter of the pulleys ? 13. The diameters of two pulleys are to each other as 3 is to 5, and the distance between their centers is 8 ft. If 72 SHOP PROBLEMS IN MATHEMATICS the diameter of the larger pulley is 25 in., what is the length of the connecting belt? 14. What is the length of the connecting belt if the dis- tance between the two pulleys is 15 ft. and the diameters of the pulleys are 24 in. and 8 in. respectively ? 15. The diameters of two pulleys are to each other as 9 is to 10. If the diameter of the smaller pulley is 18 in. and the distance between the centers is 19 ft., what is the length of the connecting belt ? 16. The diameters of two pulleys are 16 in. and 18 in. respectively, and the distance between their centers is 11^ ft. If the belt is 8 in. wide, how many square feet of belting are required ? How many square feet if the belt is 6 in. wide ? 17. The distance between centers of two pulleys is 12 ft. and the smaller of the pulleys is 16 in. in diameter. What is the length of the belt if the diameters of the pulleys are to each other as 2 is to 3 ? 18. How much leather (surface measure) will it take for a belt 6 in. wide to connect two pulleys whose diameters are 16 in. and 20 in. respectively, the distance between cen- ters being 9 ft. ? 19. At $1 per square foot, what will be the cost of leather belting for the following : Belt to be crossed and 5 in. wide. Distance between centers, 15 ft. 4 in. Small pulley, 18 in. diameter. Diameters of pulleys in the ratio 5 : 6. COILS OF BELTING, ETC. 65. Rule for belting. To find the number of feet of belting in a roll, add the diameter of the roll to the diameter of the hole left in the center, divide the sum by 2, multiply by y, and multiply this result by the number of coils in the roll. PULLEYS, BELTS, AND SPEEDS 73 EXAMPLE. A roll of belting contains 60 coils ; its outside diameter is 36 in. and the diameter of the hole in the center is 6 in. How many feet does the roll contain ? op I f OO OO 22LT.2 x -- x 60 = 21 x -^ X 60 = 3960 in. = 330 ft. PROBLEMS 1. At $1 per linear foot, what will be the cost of a roll of belting whose outside and inside diameters are 30 and 5 in. respectively, the number of coils being 40 ? 2. The length of a piece of belting is 80 ft., its width is 6 in., and its thickness T \ in. Find how many coils it will make and what will be the diameter of the roll if the hole left in the center is 4 in. in diameter ? 3. A cable ^ in. in diameter is wrapped around a roller 12 in. in diameter. How many coils will be required to reach the bottom of a shaft 120 ft. deep ? 4. A hoisting windlass drum or roller is 18 in. in diameter on one end and 9 in. in diameter on the other. A rope has one end winding on the 18-in. end of the drum. When it has passed around a tackle-block pulley below, it runs up and winds around the 9-in. end of the drum in the opposite direction. If the windlass makes 25 turns per minute, how fast does the load fastened to the tackle block rise ? CHAPTER IV AREAS, VOLUMES, AND WEIGHTS OF SOLIDS ; TURNED WORK SHEET-METAL WORK 66. Formulas for sheet-metal work. For formulas of sur- face and cubic measure that apply to problems in sheet- metal work the student is referred to Chapter XVI. PROBLEMS 1. A galvanized-iron measure (Fig. 41) 14 in. in diameter and 14 in. deep holds very nearly a bushel. Allowing 2 in. on the width of the side piece ^- -^^ for the lock at the bottom and the roll at the top, 1 in. on the length for the side seam (lock), and 1 in. on the diameter of the bottom for the lock, how much metal is required for its con- struction ? 2. If the above measure were 16 in. in diameter, how deep would it have to be to hold a bushel ? (A bushel contains 2150 cu. in.) 3. If the measure mentioned in Problem 1 were 9^ in. deep, what would its diameter have to be to hold a bushel and a half ? 74 FIG. 41. BUSHEL MEASURE AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 75 4. If we make the depth of the above measure 20 in., what must be its diameter to hold a bushel? Would it contain more or less metal, and how much ? 5. A copper clothes boiler has semicircular ends ; its extreme length is 21 in., its width 11^ in., and its depth 13 in. If we allow 1^ in. on the width of the side piece for lock and top roll, 1 in. on the length for the side lock, and ^ in. all around the bottom piece for the lock, how much copper is required to construct the boiler ? 6. If the boiler above were 11 in. wide and 21^ in. long and held the same amount, what would be its depth ? Would it contain more or less copper, and how much ? 7. A galvanized-iron water pail is 10^- in. in diameter at the top, 8^ in. at the bottom, and 9^ in. in slant height. If we allow 14^ in. extra for bottom FIG. 42. CIRCULAR HOPPER lock and top roll, 1 in. for side lock, and 1 in. on the diameter of the bottom piece for lock, how much metal is required for its construction ? 8. A tin basin is 16 in. in diameter at the top and 14 in. in diameter at the bottom. How much tin is required for its construction if its slant height is 4^ in. ? Allow 1 in. extra on the length and width of two pieces for. the side and 1 in. on the diameter of the bottom piece for locks. 9. How much tin is required for the above basin if its perpendicular height or depth is 3% in.? 10. A circular hopper (Fig. 42) is constructed of two pieces of galvanized iron ; its perpendicular height is 20 in., its large diameter is 40 in., and its small diameter is 3^ in. How much metal is required for its construction if 1 in. is added to each piece for lock ? 76 SHOP PROBLEMS IN MATHEMATICS 11. A funnel (Fig. 43) is 6 in. in diameter at the top and 1 in. at the bottom, its slant height is 4|- in., the spout is 1 in. in diameter at the larger end and \ in. at the smaller, and its slant height is 3J in. Find the amount of tin required if we allow ^ in. on the length and width of each piece for locks. (See Fig. 44 for laying out pattern for funnel.) 12. A cone-shaped hood for a chimney is a in. in diameter at the bottom and has a slant height of b in. Find the amount of metal FIG. 43. FUNNEL FIG. 44. PATTERN FOR FUNNEL in the hood if 1 in. is added to the length of the piece for a side lock. 13. If the perpendicular height- of the above hood is 11 in. and a equals 28 in., how much metal is required for the hood, allowing 1 in. on the side for lock ? 14. If a sheet of tin is 20 by 28 in., how many disks for can bottoms 3f in. in diameter will it make, and how much tin will be wasted ? AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 77 15. How much tin (surface measure) is required for 1 gross of tin cans 3f in. in diameter and 4| in. high, allow- ing f in. on the diameter of the end pieces, ^ in. on the length, and 1 in. on the height of the side pieces for locks ? 16. If iron ^ in. thick weighs 5 Ib. per square foot, what will be the weight of iron for the following tank : extreme length, 4 ft. ; width, 2 ft. ; depth, 2 ft. ; ends to be true semicircles, 2 in. extra allowed on the length and 1 in. extra on the width of the side piece, and 1 in. extra all around the edge of the bottom piece for locks (seams) ? 17. The size of sheets of tin is 20 by 28 in. If we allow 1 in. on both the length and width of each sheet for lock, how much tin (how many sheets) is required to cover a roof 36 ft. wide and 50 ft. long ? At 5 cents per square foot, what is the cost of the tin ? 18. A certain make of tin shingles 9 by 14 in. when laid is exposed to the weather 8 in. by 12 in. At $6 per square (100 sq. ft.), what will be the cost of covering a roof 26 ft. wide and 34 ft. long ? How many shingles willit take ? 19. If sheet lead T ^ in. thick weighs 3.69 Ib. to the square foot, how many pounds will be required to line the four sides and bottom of a tank 3 ft. 6 in. deep, 4 ft. wide, and 8 feet long ? 20. Copper ^ in. thick weighs 2.888 Ib. per square foot. A man cuts from a sheet of this copper 3 equal disks, each tangent to all the others and 12 in. in diameter. What is the weight of the piece left in the center after the disks are cut out ? 21. If number 20 copper weighs 1.5855 Ib. per square foot, what will be the weight of enough metal to line a range reservoir that is 30 in. long, 10 in. wide, and 11 in. deep, allowing J sq. ft. for locks and waste ? 78 SHOP PROBLEMS IN MATHEMATICS 22. A copper teapot is 9f in. in diameter at the bottom, 7 in. at the top, and 10 in. deep. Allowing 40 sq. in. for locks and waste, how much metal is required for its con- struction without a cover ? TANKS, RESERVOIRS, CISTERN^, ETC. 67. Exact measure. To find the exact cubic contents of any containing vessel, the student is referred to Chap- ter XVI. PROBLEMS 1. If a can is 5 in. square on the bottom, how high must it be to hold a gallon ? (A gallon contains 231 cu. in.) 2. A reservoir tank on a range is 20 in. long and 10 in. wide. How deep must it be to hold 10 gal. ? 3. How much sheet lead is required to line a tank 5 ft. square on the bottom, the tank to hold 300 gal. ? 4. A tank 1 ft. square on the bottom and 18 in. high holds how many gallons ? 5. A vessel 7 in. in diameter and 6 in. deep holds how much ? 6. A can that is 10 in. in diameter must be how deep to hold 5 gal. ? 7. If a milk can is 24^- in. high, what must be its diame- ter to hold 40 qt. ? 8. An ash can is 18 in. in diameter and 30 in. high. How many bushels does it hold ? (A bushel contains 2150 cu. in.) 9. If the depth of a galvanized measure is 12 in., what must be its diameter to hold a bushel ? 10. What must be the diameter of a vessel that is 8 in. deep, to hold a bushel ? to hold a half bushel ? to hold a peck ? AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 79 11. A boy lets a string with a weight on its end down into a cistern that is circular in shape ; when the weight strikes the bottom and the string is drawn up he finds it wet 3 ft. and 5 in. If the cistern is 5 ft. in diameter, how many gallons of water does it contain ? 12. A vertical water pipe is 2 in. in diameter, inside meas- ure. What is the pressure per square inch on the bottom end of the pipe, if water weighs 62.5 Ib. per cubic foot and the length of the pipe is 33 ft. ? 13. A standpipe is 40 ft. high and 16 ft. in diameter, inside measure. What is the pressure in pounds on the bottom of the pipe when it is f full ? 14. What must be the diameter of a cistern 7 ft. deep to hold 35 bbl. ? (A barrel contains 31^ gal.) 15. A copper tank is 11^ in. wide, 13 in. deep, and 21 in. long. If the ends are semicircular in form, how many gallons does it hold ? 16. A silo is 16 ft. in diameter and 30 ft. high. How much matched lumber J- in. thick is required to cover its outside, allowing 50 ft. for waste in squaring off the ends of the boards ? 17. If a water boiler is 12 in. in diameter, how high must it be to hold 30 gal. ? 18. If a water boiler is 5 ft. high, what must be its diameter to hold 52 gal.? 19. A water boiler is 22 in. in diameter and holds 100 gal. What is its height ? 20. If the boiler mentioned in Problem 19 is made of T Vin. iron, what will be its weight if we add 1 in. for the lap joint (lock) on the side and 2 in. to the diameter of the end pieces for the same purpose, jVin. iron weighing 12.673 Ib. per square foot and the rivets 10 Ib. extra ? 80 SHOP PROBLEMS IN MATHEMATICS 21. A flat tin roof is 20 by 30 ft. and a conductor pipe leads from the roof to a cistern 8 ft. in diameter. When it began to rain the cistern was empty, and there were 4 ft. of water in it when the rain stopped. How many inches of rain fell ? 22. If the above were a common gable roof 25 ft. long, with 19J-ft. rafters having a half pitch, what would be the number of inches of rainfall necessary to produce the same amount of water in the cistern ? 23. If a ^-pitch roof is 31 ft. long and has rafters 19 ft. 6 in. long, how many, gallons of water will run into the cistern if the rainfall is in. ? 24. A railroad water tank is 16 ft. in diameter and 16 ft. high. How many gallons does it hold ? how many barrels? (31* gal.) 68. Formulas for areas and volumes. Before working the problems in this chapter the student should read Chapter XVI. PROBLEMS 1. A cylinder 3 in. in diameter and 6 in. long is being turned from a block 3 in. square and 7 in. long. How much wood is cut away in shavings ? 2. A manual-training class is composed of 30 boys, and every boy turns up a cylinder If in. in diameter and 6 in. long from a piece of rough stock 2 in. square and 7 in. long. At $70 per M, how much will the lumber cost and how much of the wood is actually used in the cylinders ? 3. A disk 5 in. in diameter and 1 in. thick is turned from a piece of stock 5J in. square and 1 in. thick. How much of the wood is wasted in shavings ? 4. A croquet ball is 3f in. in diameter and the mallet heads are 2 in. in diameter and 5f in. long. If we allow AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 81 f in. on the length and J in. on the other two dimensions of every piece for rough stock, how many feet of hard maple lumber is required for a set of 8 balls and mallet heads ? How much of the lumber is wasted in shavings ? 5. What will be the weight of a bowling ball 6 in. in di- ameter that is to be turned from a rectangular block 6^ in. square and 7 in. long, the weight of the block being 13J Ib. ? 6. A lignum- vitse bowling ball is 8 in. in diameter and weighs 10 Ib. What was the weight of the cube from which it was turned, if we allow J in. on each of the three dimensions for rough stock ? 7. A dumb-bell is turned from a rectangular piece of wood 3^ by 3 by 12 in. The balls on the end are 3 in. in diameter, and are connected by a handle 1^ in. in diameter and 4 in. long. Find how much wood is cut away in shavings, and how much the dumb-bell weighs, if the piece be- fore being turned weighed 5-J- Ib. 8. What is the weight of a cast-iron sphere 5 in. in diam- eter, if cast iron weighs .26 Ib. per cubic inch ? 9. How many cubic inches in a hollow cast-iron sphere 6 in. outside and 4 in., inside diameter ? 10. What is the weight of the above sphere, if the specific gravity of cast iron is 7.202 ? 11. At n cents per pound, what will be the cost of a brass sphere d inches in diameter ? 12. A ring (Fig. 45) has a cross section 1 in. square and an outside diameter of 6 in. What is the volume of the ring ? FIG. 45. RING WITH SQUARE CROSS SECTION 82 SHOP PROBLEMS IN MATHEMATICS 13. The cross section of the rim of a flywheel is 2^ by 3 in. If its outside diameter is 30 in. and the inside diam- eter 23 in., what is the volume of the rim ? 14. In Problem 13 the inside diameter is 23 in., the width of the riin 2^ in., and its depth 3^ in. If the diameter were the same and the rim were built the other way, with the width 3^- in. and the depth 2^ in., would there be more or less metal ? How much ? 15. The weight of a cast-iron ring with a square cross section is 200 Ib. and its mean diameter is 35 in. Find the outer diameter of the ring and its volume. 16. The outer and inner diam- eters of the rim of a cast-iron handwheel are 13^ and 12 in. respectively. What is the weight of the rim if its cross section is a circle ? (Fig. 46.) 17. The area of the circular cross section of the rim of a FIG. 46. RING WITH CIRCULAR . _ . . . _ . - CROSS SECTION cast-iron flywheel is 20f in. and its weight is 300 Ib. What are its inner and outer diameters ? 18. The rim of a cast-iron handwheel has an octagonal cross section f in. on a side. If its volume is 65 cu. in., what are its inner and outer diameters ? 19. A brass ring having an inner diameter of 4 in. and a cross-section diameter of 1 in. is melted down and run into a ball. What is the diameter of the ball ? 20. A brass ring 1 sq. in. in cross section and 5 in. out- side diameter is melted and run into a ring with a 1-in. cross-section diameter and the same outside diameter. How much brass will be left ? AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 83 PROBLEMS TANKS 1. A wooden tank is 6 ft. square at the bottom and 5 ft. square at the top. If it is 4 ft. 6 in. deep, how many gallons does it hold ? 2. A wooden tank is 6 ft. 5 in. in diameter at the top, 6 ft. 10 in. in diameter at the bottom, and 4 ft. deep, inside measure. How many barrels does it contain ? (A barrel contains 31J gal.) 3. If a wooden churn holds 7 gal., is 20 in. deep, and 10 in. in diameter at the bottom, what must its diameter be at the top ? 4. If the above churn was 24 in. deep, 12 in. in diameter at the bottom, and 9 in. in diameter at the top, how many gallons would it hold? 5. What would be the depth of a churn that would hold 15 gal., if its top and bottom diameters were 10 in. and 14 in. respectively ? 6. The staves of a water tank are 3f in. wide at the bottom and 3^ in. at the top. If the tank is 8 ft. in diam- eter at the bottom, what is its diameter at the top and how many staves are required ? 7. A water tank 14 ft. in diameter is built of 2-in. by 5-in. staves 12 ft. long in the rough. If we allow in. on the width of every stave for jointing, how many staves are required and how much lumber do they contain ? 8. How much paint will it take to give the above tank two coats all over, if a gallon of paint will cover 500 sq. ft. ? 9. A cylindrical flume constructed of wooden staves is ^ mi. long and 3 ft. in outside diameter. Find how many staves 2 in. by 4 in. by 10 ft. are required for its 84 SHOP PROBLEMS IN MATHEMATICS construction, if we allow in. on the width of each stave for jointing. 10. At $35 per M, what will be the cost for lumber for the flume mentioned in Problem 9 ? HEAPS 11. A farmer has a conical-shaped heap of grain on his granary floor ; the height of the heap is 4 ft. and the diam- eter of the circle covered by its base is 10 ft. If he puts the grain in a bin 4 ft. square, how deep will it be and how many bushels will he have ? 12. An amount of pea coal is thrown up against the side of a bin. The highest point of the coal is 6 ft., where it touches the side of the bin, and the slant height of the heap is 9 ft. If the base of the heap is semicircular in shape, how many tons does it contain, allowing 36 cu. ft. per ton ? (This holds approximately for the ordinary sizes of anthracite coal.) 13. A sack of grain containing 2 bu. is emptied out on a level floor. The heap naturally takes the form of a cone. If its highest point is 14 in., what is the diameter of the circle on the floor covered by the grain ? 14. How many bushels of wheat are there in a heap thrown into the corner of a bin, if the highest point of the corner is 4 ft. 2 in. and the distance of the edge of the heap from the corner on the floor is 6 ft. 6 in.? CONTENTS OF BARRELS, CASKS, ETC. 69. To compute the contents of barrels, casks, etc. There are two methods for computing the contents of barrels, casks, etc. The first is approximate, the second more nearly exact. AREAS, VOLUMES, AND WEIGHTS OF SOLIDS 85 (1) Add the smaller and the greater diameters together (the head and bung diameters) and divide by 2. This gives the mean diameter. We can then proceed as though the barrel were a common c}dindrical vessel with a diameter equal to this mean diameter. (2) Multiply the square of the mean diameter in inches by .0034, and this product by the length of the barrel in inches. The result will be the number of gallons. 70. Degree of accuracy. It is evident that the more curva- ture there is in the staves of a barrel the greater error there will be in using the first method. It is also evident that because the curvature of the staves in some barrels and casks is greater than in others, no general rule can be formed that will be correct in all cases. The second method, however, will be found to be accurate enough for all practi- cal purposes. 71. Gallons in a barrel and in a hogshead. A common barrel is said to contain 31^ gal. and a hogshead 63 gal. As a matter of fact they are made in a number of sizes, to meet the various uses to which they are put. The dimensions given in the following problems are inside dimen- sions, unless otherwise stated. PROBLEMS 1. If the head diameter of a barrel is 17^ in., its bung diameter 20 in., and its length 2^- ft., how many gallons does it hold ? (By first method.) 2. A barrel has a bung diameter of 24 in., a head diameter of 20 in., and is 30 in. long. How many gallons will it hold? (By second method.) 3. A barrel holds 40 gal. If its length is 27.9 in. and its head diameter 19 in., what is its bung diameter ? (By first method.) 86 SHOP PROBLEMS IN MATHEMATICS 4. If a barrel holds 61 gal., has a length of 34 in., and a bung diameter of 24 in., what is its head diameter ? (By second method.) 5. The ordinary sugar barrel is about 19 in. in head diameter, 23 in. in bilge diameter, and 28 in. long. How many of these barrels will it take to hold a half ton of pea coal ? (By first method.) 6. How many gallons of vinegar in a cask 19 in. in head diameter, 23 in. in bung diameter, and 30 in. long ? (By second method.) 7. A man has 10 bbl. of potatoes for which he wishes to build a bin. The barrels are 18 in. in head diameter, 21 in. in bung diameter, and 30 in. long. If he builds his bin 4 ft. 6 in. by 5 ft. 4 in., how deep must it be to hold the potatoes, and how many bushels does he have ? (By first method.) 8. A dealer delivers 3 bbl. of coal ; the head and bung diameters of the barrels are 19 and 23 in. respectively and their length is 28 in. How much ought he to receive for the coal if he charges $7.50 per ton ? (By first method.) CHAPTER V PATTERN MAKING AND FOUNDRY WORK ; WEIGHTS OF CASTINGS FROM WEIGHTS OF PATTERNS 72. Weight of metal. Foundrymen usually have special rules by which they estimate the weights of castings from the weights of patterns (see table of weights). When the castings have small openings, their cores and core prints need not be considered in weighing the patterns. When, however, the castings involve large dry sand cores, the approximate amount of metal required can be obtained by multiplying the weight of the dry sand core by the proper factor (see table) and deducting this amount from the weight obtained for the casting as though it were to be solid. It will readily be seen that the above rules are applicable only in cases where the patterns are of solid wood throughout. As most of the larger patterns are constructed more or less hollow, it is impossible to formulate any general rule that will apply to all cases. 73. Allowance for shrinkage and waste. In the tables given in the back of the book the factors by which to multiply are a little larger than figure out from the actual weights of the various woods and metals. The extra amount is added to allow for any shrinkage of the metal in melting, waste in sprues, etc., or variation in the weight of wood in the patterns. PROBLEMS 1. A solid cast-iron cylinder is 3 in. in diameter. What length must we make the pattern to have the casting weigh 15 Ib. ? 87 88 SHOP PROBLEMS IN MATHEMATICS 2. If window weights are 1^ in. in diameter, how long must we make the pattern for 7 Ib. weights ? 9 Ib. ? 3. The casting for a solid brass roller weighs 50 Ib. What will be the weight of a cherry pattern for the roller if the specific gravity of cherry is .715 and that of brass 7.82 ? ^Balancing Point Core Print FIG. 47. RETURN BEND 4. A boy wishes to make a pattern for a cast-iron dumb- bell that will weigh 20 Ib. If the handle connecting the two balls is 4 in. long and 1^ in. in diameter, what must be the size of the balls ? 5. The mean circumference of a return-bend pattern (Fig. 47) is c, and the diameter of a cross section of the open- ing is 2 r. Find I, the length of the core prints required for the balanced core, the diameter of which is d. Make the volumes of the parts of the core outside and inside of the casting equal. 6. A saw-bench top is made of cast iron and weighs 136 Ib. What will be the weight of its pattern if it is con- structed of white pine ? PATTERN MAKING AND FOUNDRY WORK 89 TT 7. The pattern for a hand wheel 12^ in. in diameter is made of white pine and weighs \ Ib. What will the casting weigh if it is made of cast iron ? (See table of weights.) 8. How much cast iron must be melted for 10 pulleys with 8f in. face and 20| in. diameter, if the pattern is made of cherry and weighs 8 Ib.? 9. The side spindle pulley for a planer and matcher is cast iron ; the weight of its mahogany pattern is 2 Ib. How much metal must be melted for 36 pulleys, if If Ib. is deducted from each for the , , hole cored out for the shaft ? 10. A lever-arm pattern made of white pine weighs 1^ Ib. In the larger end there is an open- ing 3^ in. in diameter that re- quires a core 5 in. long. If we make 12 iron castings, how much iron must we have ? 11. A pattern for a rectan- gular iron casting is 3 by 3 by 10 in. What is the weight of the casting, cast iron weigh- ing 26 Ib. per cubic inch? 12. The pattern for a hollow brass cylinder is 10 in. long, 2^ in. in diameter, and has core prints 1^ in. in diameter on each end. If brass weighs 488.75 Ib. per cubic foot, what is the weight of the casting ? 13. What is the weight of an engine-crank disk (cast iron), if its pattern is 4 in. thick and 30 in. in diameter, 20 Ib. being deducted for openings ? 14. The pattern for a brass bushing is made of mahogany and weighs 1 Ib. If the specific gravity of mahogany is .560 and that of brass 7.82, what is the weight of the cast- ing, the core equaling ^ of the volume of the pattern ? FIG. 48. SURFACE PLATE 90 SHOP PROBLEMS IN MATHEMATICS 15. At 3 cents per lb., what will be the cost of 25 cast- iron surface plates 16 in. by 18 in. by f in. thick, ribbed f in. deep, as shown in Fig. 48 ? 16. The pattern for a brass bushing weighs 3 lb., and the core prints are 3J in. in diameter and 10 in. long. What is the weight of the bushing if the pattern is made of cherry, the specific gravity of which is .715 ? 17. The entire set of pat- terns for a gas engine weighs 3^ lb. ; the dry sand cores F,a. 49. HOLLOW CHOCK Wei h ^ lb ' H W mUCh WlU one set of castings weigh ? (See table for weight of cores.) 18. The pattern for a hollow chuck (Fig. 49) on the lathe is 2g- in. in diameter and 2J in. long. If the diameter for the core prints for the opening is If in., how many pounds of cast iron will be required for 12 chucks ? 19. The pine pattern of a small surface plate weighs J lb. How much would the I casting weigh ? (See table of j,|o- weights.) 20. If the pattern for the face plate shown in Fig. 50 is made | of cherry and weighs J lb., how [^ Y/Q ,| much cast iron will be required Jr. FIG. 50. FACE PLATE to- make 36 castings ? (Cherry weighs 38 lb., cast iron 450 lb. per cubic foot.) 21. If brass weighs .31 lb. per cubic inch and mahogany weighs 51 lb. per cubic foot, how many pounds of brass will be required for a solid brass roller if the mahogany pattern weighs 4^ lb. ? 2 ->'^ PATTERN MAKING AND FOUNDRY WORK 91 22. If white pine weighs 34 Ib. per cubic foot and cast iron weighs 450 Ib., how much iron must be melted to make a solid casting from a pattern that weighs 8 Ib.? from a pattern that weighs 12 Ib. ? from a pattern that weighs 15 Ib. ? (No allow- ance for waste.) 23. The pattern for a cast- iron flanged pulley weighs 1| Ib. If the pattern is made of cherry and there is an opening through the casting 1^ in. in diameter and 6^ in. long, how much metal is re- quired ? (Disregard the core print.) FIG. 51. CONE PULLEY 24. How much cast iron must be melted for 50 cone pulleys like those in Fig. 51? PRESSURE IN MOLDS 74. Bursting of molds. It is important that the student should be able to figure out the pressure exerted upon the mold by different metals, in order to take proper care in preventing the molds from bursting or leaking. 75. Laws of pressure. The laws of pressure are : (1) Fluid pressure at a point in a liquid at rest is the same in all directions. (2) The total pressure exerted on any surface is equal to the product of the area of that surface, the weight of a cubic inch of that liquid, and the height of the liquid above the center of the surface. If the given surface is irregular, the center of gravity must be taken instead of the center of surface. 92 SHOP PROBLEMS IN MATHEMATICS EXAMPLE. Denoting the pressure by P, area by A, weight per cubic inch by w, the height of the column of liquid by A, find the pressure exerted on the surface db (see Fig. 52) by the melted iron when h = 5", w = .26 lb., and A = 13" x 10" = 130 sq. in. P = 130 x .26 x 5 = 171 lb'. Therefore a force of 171 lb. tends to lift the cope from the drag. Find the pressure exerted sidewise on ac. The center of gravity of this surface is at G, 11" below the top. Strip ac is 10" wide and 13" deep ; then P = 130 x .26 x 11 J = 388.70 lb. ; i.e. 388.7 lb. is pressing toward the left and the right, tending to force the flask apart sidewise. Simi- larly, the pressure on cd may be found by taking FIG. 52. ANGLE IRON IN MOLD h = 18" and e = 1". 76. Sprue. The channel through which the metal is let into the mold is called a sprue. The top of the sprue is sometimes spoken of as a pouring basin. (See Fig. 52.) 77. Pressure head. The distance from the top of the metal in the mold to the level of the sprue is called the pressure head. To insure sound castings the pressure head should be sufficiently great to drive out all gases produced by the hot metal. These gases escape through the sand. (See Fig. 52.) 78. Cope and drag. The flask in which a mold is made is generally in two parts. The upper part is called the cope. The lower part is called the drag. PATTERN MAKING AND FOUNDRY WORK 93 PROBLEMS 1. Denoting the pressure in the mold by P, the area by A, the weight per cubic inch by w, and the height of the column of the liquid by h, write the formula for P in terms of A, w, and h. 2. The top of a surface plate measures 16" x 18". What is the upward pressure exerted on the cope if the cope is 8" deep ? (See Fig. 48.) 3. A large face plate 18" in diameter has a rim 1J" wide. The distance from the top of the mold to the top of the face plate is 9". Find the up- ward pressure. What is the total lateral (side) pressure exerted by the rim ? 4. A core is made as shown in Fig. 53. The distance of the surface a below the top of the liquid iron is 6", and of the surface b, 6f ". Is the balance of pressure upward on b or downward on a ? FIG. 53. SHOWING CORE IN MOLD J 5. A block of the size shown in Fig. 54 is to be cast. If the depth of the drag is 7", which face should be up to give the least lifting pressure ? 6. A water pipe 3 ft. in out- side diameter, 10 ft. long, and 1" thick is cast on end. The top of the pipe is 1 ft. below the top of the pouring basin. Find the total pressure exerted on the bottom of the mold and the total side pressure tend- ing to burst the mold. FIG. 54. BLOCK 94 SHOP PROBLEMS IN MATHEMATICS 7. Given a crucible measuring 10" in diameter at the top, 11" at the middle, 9" at the bottom, and 14" high on the inside. Approximately, how many pounds of brass will it hold, brass weighing .29 Ib. per cubic inch. How many pounds of iron, weight per cubic inch .260 ? How many pounds of lead, weight per cubic inch .410? 8. A graphite crucible is 4^" inside diameter at the top, 3" at the bottom, and 7^" high. What approximate weight of brass will it hold ? THE CUPOLA AND BLAST FURNACE 79. Total charge. The total weight of iron put into the cupola during the heat is called the " total charge." 80. Overcharge. The weight of iron in excess of what is required to pour the molds is called the " overcharge." 81. Shrinkage. The total loss of weight of iron by burn- ing and otherwise is the shrinkage. The shrinkage is found by deducting from the total charge the weight of the cast- ings delivered, the sprues, poor castings, and the excess of iron after all the molds have been poured. The ratio of fuel to metal is valuable in figuring out costs and in finding the quantities of fuel for subsequent charges. The ratio varies greatly with different cupolas and with different kinds of work.. PROBLEMS 1. A cupola measures 24" inside diameter and 8" from the sand bottom to the bottom of tuyeres. How many pounds of iron will it hold ? For parts of cupola, see Fig. 55. 2. How many times would a cupola like the preceding have to be tapped to draw off 4800 Ib. of iron ? Allow a 15 per cent margin in the bottom of the cupola for slag and safety. PATTERN MAKING AND FOUNDRY WORK 95 3. What diameter and height should a ladle be to hold the metal from one of these taps ? The ratio of the height to the diameter of the ladle may be 1^ to 1. FIG. 55. CUPOLA 4. A cupola melts 43,000 Ib. of iron with 5400 Ib. of fuel. What is the ratio of fuel to iron ? 5. A cupola measures 34" inside diameter and 14" from bottom plate to bottom of tuyeres. Allow 4" for the bed 96 SHOP PROBLEMS IN MATHEMATICS lining. How many pounds of melted iron will this cupola hold before it is necessary to tap ? 6. A cupola melts 3650 Ib. of iron with 550 Ib. of fuel. What is the ratio of fuel to iron ? 7. It is desired to melt a charge of 2400 Ib. of iron. How much fuel would be required at the' ratio found in Problem 6 ? 8. Allowing $20 a ton for pig iron and $4.75 a ton for coke, what value is represented in the materials mentioned in each of the two preceding problems ? 9. A cupola melts 8800 Ib. of iron with 850 Ib. of coke. What is the ratio of fuel to iron ? 10. The first charge of fuel in a cupola is 500 Ib. and the first charge of iron is 1500 Ib. What is the ratio ? 1 1. The second charge of fuel is 60 Ib. and of iron 1200 Ib. What is the ratio ? The ratio of total charges is shown in Problem 9. 12. A cupola melts 6800 Ib. of iron with 720 Ib. of coke. What is the ratio of fuel to iron ? 13. A cupola melts 12,000 Ib. of iron with 2200 Ib. of coke. What is the ratio of fuel to iron ? 14. A cupola melts 37,700 Ib. of iron with 3550 Ib. of coke. What is the ratio of fuel to iron ? Problems 15, 16, 17, and 18 are based on the same cupola. 15. A cupola measures 25" inside diameter and 8"(average) from sand bottom to bottom of tuyeres. How much melted iron can it hold before the iron runs out through the tuyeres ? 16. The thickness of the fire-brick lining is 8", inside diameter 25", and height of fire-brick lining 15 ft. Deduct 4 sq. ft. for charging door. Find the number of brick 9" x 4" x 2" necessary to line the cupola. PATTERN MAKING AND FOUNDRY WORK 97 17. The first charge of fuel was 400 Ib. and of iron 1600 Ib. ; the second charge of fuel 100 Ib. and of iron 900 Ib. What was the ratio of fuel to iron in the first and second charges ? 18. The total iron melted was 3400 Ib. and the fuel used was 610 Ib. What was the total ratio of fuel to iron ? 19. The first charge of fuel in a Colliau cupola is 250 Ib. and the first charge of iron 1000 Ib. What is the ratio of fuel to iron ? 20. The second charge of fuel is 70 Ib. and of iron 700 Ib. What is the ratio ? 21. The total charge of fuel is 390 Ib. and of iron 2400 Ib. What is the final ratio of fuel to iron ? 22. A charge is to be made up of different kinds of iron as follows : Emporium, 25 per cent; Saxon, 20 per cent ; scrap, 55 per cent. Find weights of each in a total charge of 3600 Ib. 23. A total charge of iron is composed of 8500 Ib. of Genesee pig iron, 8500 Ib. of Susquehanna, and 9400 Ib. of scrap. W T hat is the percentage of each ? If the 1st, 2d, 3d, and 4th charges of Genesee are respectively 2000, 1050, 1100, 1000 Ib., what will the weights of scrap be if added as in the above ? 24. We desire to make the following mixture for our foundry castings : Tonawanda No. 1, 30 per cent ; Erie No. 2, 25 per cent ; and scrap, 45 per cent. What weights will be needed of each for a heat of 38,700 Ib. ? 25. A cupola charge consists of 26,600 Ib. of iron costing $16 a ton and 4500 Ib. of coke costing $4.25 a ton. What is the cost per pound of iron ? 26. In Problem 25, if it requires a melter at $3 a day and two helpers, each at $2 a day, what will be the total cost per pound for material and labor ? 98 SHOP PROBLEMS IN MATHEMATICS 27. A molder with one helper 'makes one mold a day. If the casting weighs 500 Ib. and the wages of both molder and helper are $5.25 a day, what must be charged in order to give 1 cent per pound profit ? 28. A molder, without a helper, earning $4 per day, makes 20 molds in a day ; every mold yields 11 Ib. of good castings. If the cost of material is the same as that found from Problem 26, and the manufacturer wishes to make 1 cent a pound profit, what must he charge per pound for these castings ? 29. A blast furnace requires 2420 Ib. of hard coal to produce 1 T. of iron. The ore used contains 59.54 per cent iron. What proportion of coal to iron ore must be charged into the furnace ? (Use long ton of 2240 Ib.) 30. A blast furnace produces 1357 tons of iron every week. How much coal and iron ore would it consume ? 31. A blast-furnace charge may be 4000 Ib. of ore con- taining 60.7 per cent of iron, 900 Ib. of limestone, and 2300 Ib. of coke. What is the percentage of each per ton of pig iron produced? (Long ton.) 32. At $1.50 a ton for coke, what would be the cost of fuel sufficient to melt a day's run of 200 T. of iron, when made from the above ore ? (Long ton.) 33. How much iron ore, limestone, and coke would be needed to produce 200 T. of iron from the above mixture ? (Long ton.) COST OF MATERIAL AND LABOR 82. Estimating. In estimating the time required to do work or " jobs " in manufacturing plants, experience shows the estimator about how much time should be allowed for every*operation performed in his shop. To ascertain the PATTERN MAKING AND FOUNDRY WORK 99 cost of a piece of work, add to the cost of material the cost for labor in every operation, the fixed or office charge, and the profit. In machine work, for example, on any one job there will be lathe work (turning and thread cutting), milling, drill- ing, etc., the time for all of which must be estimated if a price is to be named before the job is completed. PROBLEMS 1. Soft solder is composed of 2 parts tin and 1 part lead. How many pounds of each are required to make 25 Ib. of solder ? 2. If a solder is composed of H parts tin and 1 part lead, how many pounds of each are required for 25 Ib. of solder ? 3. At 35 cents a pound for tin and 4 cents a pound for lead, which of the compositions in Problems 1 and 2 is cheaper ? 4. Silver solder is 1 part copper to 2 parts silver. How much of each is in 6 oz. of solder ? 5. If silver solder had the mixture of 1 part brass, 1 part copper, and 19 parts silver, how much of each would there be in 6 oz. of solder ? 6. At 15 cents a pound for brass, 17 cents for copper, and 55 cents an ounce for silver, which of the compositions given in Problems 4 and 5 is cheaper ? 7. From a solid bar 1^" in diameter a bolt is turned with the following dimensions: f" in diameter; 4" long under the head; head f" thick and l T y square. What proportion of stock is wasted ? 8. What will be the cost of a solid brass sphere 5 in. in diameter, if copper costs 20 cents per pound and zinc 6 cents ? The weight of brass is 488.75 Ib. per cubic foot and the proportion is 2 of copper to 1 of zinc. 100 SHOP PROBLEMS IN MATHEMATICS 9. Brass contains ,65 per cent copper and 35 per cent zinc. If copper is selling at 17 cents a pound and zinc at 5 cents, for what ought brass to sell when 2 cents a pound is allowed for casting the brass ? 10. If bronze contains 85 per cent copper, 9 per cent zinc, and 6 per cent tin, what is it worth per pound when copper sells for 18 cents, zinc for 6 cents, and tin for 35 cents ? 11. A machinist receiving $3.25 per day is given a 10 per cent increase. A few months later he is cut 10 per cent. What is his wage now ? Why is it not the same as before ? 12. A shop has been paying its men for ten hours' work, but it is now obliged to pay them the same day's wage for an eight-hour day. What per cent increase does each man receive ? 13. What would men receiving the following per hour in a ten-hour day get per hour in an eight-hour day, the total day's wage to be the same : 20 cents, 22 J, 25, 27 J, 30, 33, 35, 37 ? 14. Make a table of wages based on 37^ cents per hour for every successive 20 inin. of an eight-hour day, that is, 20 min., 40 min., 1 hr., 1 hr. and 20 min., etc. 15. If the time cards for a piece of work show 3 hr. and 15 min. lathe work, 12 hr. and 15 min. planing, and 45 min. vise work, what is the total cost of the job ? Allow $2.75 for lathe work, $2.45 for planing, and $2.30 for vise work per day of 8 hr. 16. If a man makes 300 bolts at 3 cents each, 175 at 2J cents, and 560 at 3^ cents, what are his total earnings ? 17. If a man works 5 da., 9 hr. a day and 5f hr. over, what will his week's wages be at 33^ cents an hour ? 18. What wages per day of 9 hr. would be equivalent to $27.25 per week ? PATTERN MAKING AND FOUNDRY WORK 101 19. If a man works 65^ hr. in one week, 54 of which is regular day work at 29| cents an hour, and the remainder is overtime, counted as time and one half, what are his week's wages ? 20. An apprentice receives $6.50 a week. In that time he makes 55 large cap screws. A skilled mechanic at 30 cents an hour (nine-hour day) can make 25 of the same pieces a day. Whose labor is cheaper ? 21. A screw made of brass has a head T 7 ff " in diameter by J" thick, and a body T 3 ff " in diameter by T \" long. What will be its approximate weight ? The weight of a cubic inch of brass is 0.31 Ib. 22. What per cent of the material for the screw described in Problem 21 is wasted if the same is made from a ^" round bar of brass, allowing 1" in length for cutting off? This does not allow for thread cut on screw. 23. If brass is worth 19 cents a pound and the scrap is worth 6 cents a pound, what will the material cost for 1000 screws ? 24. If 875 screws are made in a day on an automatic screw machine at a labor and machine cost of 87 cents, what does each screw cost ? Count in cost of material as found above. 25. The speed lathe has four speeds, 1200, 750, 400, and 250 R. P. M. Find the step which will give to a " brass rod a surface speed of 90 ft. a minute (approximately). Solve for a ", 1", 1|", and 2" rod. OF TH UNIVERSITY Of CHAPTER VI LENGTH OF STOCK FOR FORCINGS, STRENGTH OF FORCINGS, ALLOWANCE FOR SHRINK FITS 83. Length of stock. In making calculations for the amount of stock required for a given forging, the weight of the forging must be found, and then an equivalent weight of the size of bar stock to be used must be ascer- tained. In some cases a small additional amount of mate- rial must be allowed, as, for example, in making a weld, an allowance in length equal to the thickness of stock is generally necessary. 102 LENGTH OF STOCK FOR FORCINGS 103 FIG. 56. ANGLE IRON 84. Use of center lines. In calculating weights, all dimen- sions should be taken from the axes or center lines of the forging and not from the outside or inside meas- urements. For example, in finding the length of stock to make a right angled bend, as in Fig. 56, the length of the center line is 4^" + 3f " = 8". The student should try to divide the forging into sections of straight, curved, or circular parts for ease in _J . ^ calculation. -r "^ A convenient rule to remem- L 4 ber in finding lengths of circu- lar forgings is the following: To the inside diameter add one thickness (of stock) and multiply by 3 ; or, if greater accuracy is required, multiply by 2 T 2 (TT). PROBLEMS 1. Find the length of stock for a flat ring J" thick, f " wide, and 3^" in inside diameter. 2. A |" bolt is to be forged 6" long under the head. How much extra stock must be allowed for the head ? The size of a bolt head (in the rough) is equal to \\ x diameter of bolt + J". This applies to both square and hexagonal heads. A finished head measures the same. The thickness of the head is equal to the diameter of the bolt. 3. What allowance must be made for the hexagonal head of a f " bolt ? 4. A swivel (see Fig. 57) is to be forged from y ' square stock. How much stock is required ? 5. A piston ring having a diameter of 22" is -J-" thick and J" wide. What will it weigh at .26 Ib. per cubic inch ? FIG. 57. CHAIN SWIVEL ' 104 SHOP PROBLEMS IN. MATHEMATICS 6. A crank (see Fig. 58) must be forged T \" larger all over than the dimensions given. What will the crank weigh in the rough ? 7. If the crank is forged from a bar 2|" square, how long a piece of stock will be required ? 3 ^ ' I 1 1 Hi FIG. 58. DOUBLE-THROW CRANK 8. If these cranks, when made in lots of 15, are forged at cents per pound, what would one crank cost ? 9. Find the length of 1^" square stock necessary to forge the crank shown in Fig. 59, all the dimensions given being finished measurements. Allow ^" for finish. 10. The leg of an. andiron (see Fig. 60) is made from 1J" square iron. Find the length of stock required. r 2 : - sW - / / i f- MOO -1- * tid u- je: FIG. 59. SINGLE-THROW CRANK 11. A disk 3^" in diameter and 1" thick is required for the machine shop. The largest round bar in the shop is 2" in diameter. How long a piece of this bar must be cut off to make the disk ? 12. A ring 10" in outside diameter, -J" thick, and 3" wide is required. The nearest size of flat bar in stock is 2|-" x !" How long a piece of the bar must be used ? LENGTH OF STOCK FOR FORCINGS 105 13. An eyebolt, shown in Fig. 61, must be forged from 1" X 1J" stock. How long a piece will be needed ? 14. A flat ring has the following dimensions : inside diameter 6", outside diameter 8" thickness ". Select a suitable size of stock, not over 4" wide, and ascertain the amount required. Add about \ the width of stock to the length calculated, to allow for welding. 15. In every 2000 Ib. of coal there are 117.5 Ib. of ash. What is the per cent of ash ? 16. A welded iron stay may be counted 75 per cent as strong as the original bar. Allowing the tensile strength of a 1" round bar to be 35,000 Ib. per square inch, what will be its strength after it has been welded ? 17. What must be the size of a welded bar, to be as strong as a 1" bar before welding? A bar is 1" in KH ''H 7 diameter at the weld. How small may it be outside the weld to be of the same strength ? FIG. 61. EYEBOLT 18. An axle is 7 ft. long and averages 5" in diameter. Compute its weight, assuming steel to weigh 489 Ib. per cubic foot. FIG. 62. GROUP OF FORCINGS 106 LENGTH OF STOCK FOR FORCINGS 107 19. Find the length of stock necessary to make each of the forgings shown in Fig. 62. Select sizes of stock which most nearly conform with the largest part of the forging. 20. A T rail (see Fig. 63) 30 ft. long weighs 80 Ib. to the yard. How many rails in a ton of 2240 Ib. ? What would each rail cost at $27.60 per ton (2240 Ib.)? 21. What would the T rails cost for a section of railroad 2 mi. long ? 22. If a 70-lb. rail instead of an 80-lb. were used, how much would be saved in a 1000-mi. railroad ? 23. If a girder rail (see Fig. 63) weighs 104 Ib. to the yard and is 60 ft. long, what is its entire weight ? What does it cost at $34.70 a ton Flu ' c3 ' T AND GlRDER RAILS (2240 Ib.) ? 24. What would be the cost of 10 mi. of railroad made from the girder rail ? 25. How much more expensive is the 1000-mi. railroad made with the girder rail than if made with the T rail ? 26. The long arm of a shear is 26" and the short arm 9". If a force of 1000 Ib. is applied at the end of the long arm, what sized square bar would the shear cut when placed 6" from the pivot ? What width of a f " flat bar would the shear cut ? 27. A punch has a leverage of 2f " to 16". Allowing the shearing force of iron to be 50,000 Ib. per square inch, how much force applied at the end of the long arm is required to punch a -J-" hole ? How much work is done if the plate is y thick ? Work done is equal to the force multiplied by the number of feet through which the force acts. 108 SHOP PROBLEMS IN MATHEMATICS 28. A blacksmith's shear and punch has a leverage on the cutting arm of 10 to 1. What force at the power end would be necessary to shear a bar of iron T ff " x l"? Shearing stress for iron is 50,000 Ib. per square inch. How much force for stock J" x |", ft" X |", f" X Ifc", i" X 1"? What is the work done in each case ? 29. Soft steel contains approximately 0.37 per cent of silicon. In 1 T. of steel how much is silicon ? 30. One grade of tool steel contains 1.2 per cent carbon. How many pounds of carbon are contained in 200 Ib. of this steel ? 31. A Bessemer converter contains 10 T. of liquid iron to which it is desired to add 0.5 per cent manganese. How much ferromanganese containing 20 per cent of manganese must be added ? 32. A chrome-steel armor plate containing 1.75 per cent of chromium and weighing 8 T. is to be cast. How much ferrochrome containing 65 per cent of chromium must be used? 33. Nickel steel for automobile construction contains 3.40 per cent of nickel. If a ladle holds 3700 Ib. of steel, how many pounds of nickel must be added ? 34. Nickel costs 30 cents a pound. How much more would the materials cost in 37,000 Ib. of nickel steel than in the same amount of mild steel at $20 a ton (2240 Ib.). Nickel steel is 3.5 per cent nickel. 35. A nickel-chrome steel contains .45 per cent of chro- mium and 1.25 per cent of nickel. How much of each (chromium and nickel) are contained in 12 T. (2000 Ib.) of this composition? 36. A 15-lb. tool steel die is to be hardened. When ready to be put into the water its temperature is 1450 F. How LENGTH OF STOCK FOR FORCINGS 109 large must the tank be in order that the temperature of the water (when cooling the die) may not rise above 100 F. ? The temperature of the water at first is 50 and the specific heat of steel is .117. 37. Suppose the above die is to be hardened in oil in which the temperature may rise to 300 F. How many gallons of oil would be required ? The specific heat of the oil is .310. 38. If 500 steel gears weighing -^ Ib. each are hardened by cooling from a temperature of 1475 F. to 100 F. in one day, how much heat will be carried away by the cool- ing substance ? How many pounds of coal would be required to provide this amount of heat if there were 14,000 heat units in each pound ? How many pounds if but 60 per cent of the heat were available ? 39. How many pounds of steel would a 10-gallon pail of water harden if we allow the water to rise from 50 to 180 F. ? The steel is to be cooled from 1350 to the temperature of the water, and the specific heat of water is .117. 85. Shrink and force fits for shafts and pulleys. In a loose fit the diameter of the shaft may be from .01" to .02" smaller than the diameter of the hole. For a bearing the difference between shaft and pulley may be from .003" to .006" per inch of diameter. For a sliding fit the shaft may be .001" undersize per inch of diameter. For an easy driving fit the hole and shaft* may be as nearly the same size as possible. For driving fits .001" per inch of diameter is a fair allow- ance. The shaft in this case must be larger than the hole. For pressed fits the allowance is the same as for driving fits. For shrink fits the shaft may be from .001" to .002" oversize per inch of diameter. It is understood that the variation is generally made in the shaft and not in the hole, as the latter is often determined by the size of the reamer. 110 SHOP PROBLEMS IN MATHEMATICS The term " nominal size " is used to indicate the size given on the drawing or some standard dimension, i.e. ", i", II", etc. PROBLEMS 1. A driving-wheel tire is to be shrunk 'upon the wheel center. The tire is bored smaller than the outside of the wheel center by .001" for every inch of diameter. How much smaller should a 60" tire be turned if its rim is 2^" thick ? 2. How much shorter would the inner circumference of the tire be than the outer circumference of the wheel center ? 3. The hub of a car wheel is bored to 5.377". How much larger should the axle be turned for a pressed fit ? 4. A shaft is 1.4375" in diameter. To what size must a collar be bored to give a solid fit ? 5. A shaft nominally 3.777" slides freely in a pulley. What should be the exact size of each ? 6. Using the formula .001 d + .0005 = allowance for press fits, where d = diameter in inches, find the diameters of shafts of the following nominal sizes : f", 1", 1+", 1 J", li", If", li". 7. Iron expands 0.0000066 of its length for each degree of temperature (Fahrenheit) increased. If a steel shaft 5 ft. long at 70 F. is heated, in turning, to 300 F., how much longer will it be when hot ? 8. A line shaft is 100 ft. long at the temperature of 40 F. How long would it be at 80 or summer temperature ? 9. How much space should be left between T rails, 30 ft. long, laid in the winter at a temperature of 40 F. ? CHAPTER VII SPEEDS OF PULLEYS, SHAFTS, AND GEARS Before reading this chapter the student should be familiar with sections 55-60 on the speeds of pulleys. The problems assume a knowledge of the general processes of machine work, but a brief description of some terms is given below. SPEEDS OF PULLEYS AND SHAFTS 86. Back gears. All engine lathes are provided with back gears (see Figs. 64 and 73), to increase the number of speeds obtainable without an overin- crease in the size of the cone pulley. 87. Direct drive. When the power is transmitted directly from belt to cone pulley to lathe spindle, the term " di- rect drive " is used. 88. Back-gear drive. When the power is transmitted from belt to cone pulley to back gears, thence to spindle, the term "back-gear drive" is used. 89. Countershaft. The countershaft is a short shaft con- taining tight and loose pulleys and is placed between the main line shaft and the machine. Its purpose is to change the speed and to permit a convenient location of machines. By counterspeed is meant the speed of the countershaft. 90. Cone pulleys in geometrical progression. The sizes of cone pulleys are so arranged as to make the increase in 111 ^E Back Gear 307: L 88r. = ^__ n 90, 1 1H ^z Lathe L 32 T _ Spindle Cone Pulley* FIG. 64. BA Civ G BARS 112 SHOP PROBLEMS IN MATHEMATICS speeds agree as closely as possible with some geometrical progression. In the case of gear drives for lathes the speeds are arranged by some geometrical ratio. For example, sup- pose the speeds of a lathe ran as follows : 25, 35, 49, 70, 98, 140. By dividing any one speed by the one preceding we obtain the geometrical ratio 1.4. This ratio is not exact, but is approximately correct. PROBLEMS 1. A motor running at 1800 R. P. M. with a driving pulley of 4^" is to drive a line shaft, which hangs 18" be- low the ceiling, at 200 R P. M. Can it be driven directly ? Why ? 2. Ascertain all the pulleys necessary to drive this line shaft through an inter- mediate or jack shaft. 3. There are three driven pulleys 14" in diameter on a No. 1 milling-machine countershaft. The two forward-driven pulleys are to run 90 and 120 R P. M. and the reverse 105 R P. M. The line shaft runs at 200 R P. M. Find the diameters of the line-shaft pulleys. 4. The milling-machine countershaft pulleys driven from a line shaft are 14" in diameter and are to run 100, 200, and 150 E. P. M. respectively. Find the diam- eters of pulleys for the line shaft running at 200 R P.M. 5. The largest step on the cone pulley of the milling machine is 10" in diameter and its mate on the counter- shaft is 5J". Find the speed when the countershaft re- volves 120 times a minute (see Fig. 65). 6. The cone-pulley diameters of a No. 1 milling machine Lathe r: Sp.nd/et - - FIG. 65. CONE PULLEYS are 10", 5-J-", and the corresponding countershaft SPEEDS OF PULLEYS, SHAFTS, AND GEARS 113 diameters are 5J-", 7", 8", 10". Find the speed for each step, considering the two forward-driven pulleys of the countershaft as running at 100 and 200 R. P. M. respec- tively. There will be eight speeds. 7. To what geometrical progression does this series of speeds most nearly conform ? 8. The back-gear ratio for milling machine No. 1 is 1 to 8. Find the eight speeds with back gears in. Are they in the same geometrical progression as in Problem 7 ? 9. It is desired to drive a countershaft of a 12" engine lathe at 170 R. P. M. The size of the driven pulley on the counter is 8". The motor speed is 1800 R. P. M. and its pulley is 4-J-" in diameter. Find the two pulleys for the line shaft. The pulleys on the line shaft cannot be over 37" in diameter. 10. Work out the formula for 'Problem 9, denoting the drivers by D and D 1 and the driven pulleys by d and d. 11. A countershaft of a lathe makes 170 R, P. M. The diameters of the steps of the cones of both the lathe and the countershaft are as shown in Fig. 65. Ascertain the spindle speed for each step. 12. With the back gears out, a lathe spindle makes 112 R. P. M. ; with them in, it makes 12 R. P. M. What is the ratio of the direct drive to back-gear drive ? 13. Ascertain the approximate ratio in the geometrical progressions of the following pulley speeds : (a) 5, 8, 13, 20, 33; (b) 61, 99, 154, 242, 402. 14. Ascertain the approximate ratio in the geometrical progression of the following speeds, taken from a variable- speed motor drive : 54, 59, 64, 88, 108, 126, 144, 160, 172, 192, 230, 280, 354. 114 SHOP PROBLEMS IN MATHEMATICS FIG. 66 GEARING 15. Considering the two drives in Problems 13 and 14, which arrangement is preferable, and why ? 16. A countershaft, driving a planer, runs at 350 R. P. M. ; the forward driving pulley is 20" in diameter and the back driving pulley 8". What is the ratio of forward to back drive ? GEARS 91. Driving gear. A driving gear is one that transmits the power to another gear. 92. Driven gear. A driven gear is a gear to which the power is transmitted. If gear A (Fig. 67) is the driver, then gear C is the driven gear, and gear B is an intermediate. If the 75-tooth gear (Fig. 68) is the driver, then the 60-tooth gear is also a driver, and therefore the 30-tooth gear and the 25-tooth gear are the driven gears. 93. Intermediate gear. An intermediate gear is a gear used to transmit the power from one gear to another where these are too far apart to mesh. The speed is not affected. 94. Train of gears. Three or more gears meshing together are called a train of gears. One or more of these is always an inter- mediate and is not considered in calculating the speed. 95. Rule for speed. (1) The speed of any two meshing gears is inversely proportional to the number of their teeth. Or in the case of a train of gears, as in the case of pulleys, the following rule may be used : C FIG. 67. TRAIN OF THREE GEARS SPEEDS OF PULLEYS, SHAFTS, AND GEARS 115 (2) The continued product of the R. P. M. of the first driver and the number of teeth in every driving gear is equal to the continued product of the R. P. M. of th e last driven gear and the number of teeth in every driven gear. PROBLEMS FIG. 08. TRAIN OF FOUR GEARS 34 80 FIG. 69. GEARING FOR POWER FEED 1. Denoting the speed of the driver and driven gears by S and s respectively, and the number of teeth by N and n, write the formula for s in terms of 5, TV, and n. 2. Denoting the R.P.M. of the first driver by 36 S and the number of teeth in the drivers by TV, TV X , ? and TV 2 , the R. P. M. of the last driven gear by s, and the number of teeth in the driven gears by n, n^ and n 2 , write the formula for N in terms of S, TV 1? TV 2 , and s, n, n^ n 2 . 3. Solve formula in Problem 2 for n and s. 4. In Fig. 66 the gear of 60 teeth revolves 114 times per minute. Find the speed of the driven gear. 5. Gear A (Fig. 67) has 69 teeth and revolves 85 times per minute. Gear C must revolve 172J times per minute. Find the number of teeth in C. 6. The gear with 75 teeth (see Fig. 68) has a speed of 40 R.P.M. Find the speed of the gear of 25 teeth. FIG. 70. GEARING 116 SHOP PROBLEMS IN MATHEMATICS 7. Gear A (Fig. 69) revolves 80 times per minute. Find the speed of shaft F. 8. Select a gear in place of the 60-tooth gear that will cause F to turn twice around while A turns around 80 times. 9. Select a "gear in place of A (Fig. 69) so that F will re- volve Jj as fast as the shaft on which A is mounted. Drill Spindle FIG. 71. BEVEL AND WORM n 24 36 42 46 10. Shaft A (Fig. 70) makes 240 R. P. M. Find the speed of shaft B. 11. Compute the ratio of speeds between shafts A and B if gear A (Fig. 70) had 34 teeth. 12. The drill spindle D (Fig. 71) makes 110 R,. P. M. Find the speed of the worm-wheel shaft E. 13. One turn of the worm wheel E (Fig. 71) raises spindle D 2". Find the amount that the spin- dle D is raised by the worm wheel in turning once around. Fig. 72 illustrates the gear- ing for the feed of a 22" drill press. Gear A on the drill spindle drives the train of gears down to the worm wheel W. On the same shaft with this worm wheel is a spur-gear meshing with a rack secured to the drill spindle, by which the spindle is fed into the work. All the gears in cluster P are fast to the shaft Q, but those in cluster are arranged in such a way that any FIG. 72. GEARING FOR 22" DRILL PRESS SPEEDS OF PULLEYS, SHAFTS, AND GEARS 117 one may be made fast to shaft N. It then becomes the driver. Hence four different feeds may be given the spindle A for every different speed of the driving pulley. One turn of worm wheel W raises the drill spindle 5J". 14. From the data given in Fig. 72 find the four feeds of the spindle in decimals of an inch per revolution. 15. This same drill has eight spindle speeds, 10, 16, 26, 44, 74, 120, 200, 300. Find the four feeds per minute for each of the spindle speeds. E 1 1 Bac < G 5sT c 9 ars 1 dtir 1 . m s C, \ 89 T N i D 897 H \ n\ hp F 1 ?: \ 20 T Spindle M 58 T| R FIG. 73. DOUBLE-BACK GEARS Double-back gearing. A lathe having a double-back gear is con- structed as shown in Fig. 73. Sleeve M, carrying gears F, D, and jR, revolves freely on the lathe spindle L and is driven by a motor through gear E. Sleeve N is keyed to the back-gear shaft S, but slides freely from left to right and carries gears E and C. So either gear E or C can be put in mesh with F or D respectively, and the power is transmitted from the motor through E to F to E to G to H and L, or through R to D to C to G to H and L. With the back gears E and C both out of mesh, the spindle L is connected with R by a pin or bolt in the side of gear H. 118 SHOP PROBLEMS IN MATHEMATICS 16. If sleeve M revolves at 500 R. P. M., what will the speed of L be when C and D are in mesh ? when E and F are in mesh ? 17. What is the ratio of direct to indirect drive when C and D are in mesh ? when E and F are in mesh ? 18. If gear R has 50 teeth and the gear on the motor spindle has 30 teeth, what will be the speed of L for the following motor speeds using back gears E-F : 400, 500, 600 R.P. M.? CHAPTER VIII CUTTING SPEED AND FEED 96. Cutting speed. Cutting speed has been defined in section 59, in connection with the lathe where the work revolves. But in some cases the work does not revolve. For example, on the planer the work moves in a straight line, but the cutting speed is still the rate at which the work passes the tool. On the shaper the tool moves instead of the work and the cutting speed is the rate at which the tool passes the work, while on the milling machine the cutter revolves and its surface speed is also the cutting speed. 97. Regulation of the cutting speed. The cutting speed depends on four things : (1) the kind of material to be cut ; (2) the feed of the tool ; (3) the depth of cut ; and (4) the kind of tool used. In solving the problems that follow, the pupil must select certain speeds and feeds as given in Tables I and II. The roughing cut or cuts are the heavy cuts taken to reduce the size of work to within fa" or J/' of the correct size. The finishing cut is the cut that reduces the work to the exact size required ; it is seldom more than fa" deep. TABLE I. CUTTING SPEEDS MATERIAL REVOLUTIONS PER MINUTE Lathe Planer Miller Drill Steel or wrought iron Cast iron .... Tool steel .... Brass 25 to 45 40 to 60 20 to 40 100 to 120 60 to 80 20 to 30 25 to 30 25 to 30 40 to 60 20 to 40 100 to 120 30 to 40 40 to 60 20 to 30 100 to 120 Phosphor bronze 119 i<- N) ** $ ^ ^S' i h-H^ HiJ.^ 120 CUTTING SPEED AND FEED 121 In the above table the higher speeds may be used for the lighter cuts and smaller diameters. The cutting speeds given here are for carbon steel. For " high- speed " tools the cutting speed and feed may be nearly doubled. For drills the peripheral speed is assumed as the cutting speed. TABLE II. FEEDS MATERIAL FEED IN INCHES PER REVOLUTION Lathe Planer Miller Drill Steel or wrought iron Cast iron .... Tool steel .... Brass TV to & T 1 * tO ^ aV tO Jo 1-V to A sV to i iV to \ .008 to .028 .008 to .040 .008 to .020 .010 to .040 .005 to .015 .008 to .015 .005 to .010 010 98. Feed. There are several ways of expressing the feed. (1) The feed is the distance in inches that the tool ad- vances along the work at every revolution or stroke. (2) Feed is sometimes given as the number of turns which the work makes while the tool advances 1 inch, as " 50 or 60 turns to the inch." This is also equivalent to -^ G " or ^L" per revolution. (3) Feed may be given in inches per minute, as is often the case in connection with the milling machine. This last method is very convenient in estimating the time necessary to finish a given cut. The feed when thus given can also be expressed as in (1) if the R. P. M. are known. 99. Rule for feed. A simple rule for feed is : The feed is the distance moved by the tool divided by the number of revolutions made by the work while the tool is moving that distance. For convenience, the distance moved is generally taken as 1 ". For example, a lathe spindle revolves 75 times while the carriage moves 1". Then the feed is 7 y. Or, a milling cutter revolves 65 times while the table moves 1". Then 122 SHOP PROBLEMS IN MATHEMATICS the feed is ff y. Or, the shaper makes 35 strokes while the table (that holds the work) moves 1". Then the feed is ^". PROBLEMS 1. Denoting the feed by F, the number of revolutions by N, and the distance that the tool moves .by D, write the formula for F in terms of D and N. 2. Solve this formula for D and N. 3. A lathe tool moves 2" along the work in 1 min. and the speed of the lathe is 400 E. P. M. Find the feed. 4. A piece of work revolves 60 times while the tool moves f". Find the feed. 5. A planer makes 35 strokes while the tool moves 1-J-". Find the feed. 6. If a milling cutter revolves 100 times while the table moves 1", what is the feed ? 7. A milling cutter has a feed of .017" per revolution and turns at 160 E. P. M. What is the feed per minute ? 8. A high-speed drill f " in diameter drills through 237 pieces of cast iron 3" thick without being resharpened. A carbon drill of the same diameter drills only 55 pieces. How many feet in length does each drill cut ? 9. By what per cent is the high-speed drill better than the carbon drill ? 10. How many pounds of iron does each of the above drills remove? 11. A high-speed drill of $" diameter revolving at 250 revo- lutions per minute may be fed T J^" per revolution. How long would it take to drill through a piece of iron 1J" thick ? 12. If the proper feed per revolution for a if" drill is .008", how many revolutions will the drill make in passing through a piece of iron 3 T ^" thick? CUTTING SPEED AND FEED 123 13. At a speed of 268 revolutions per minute, how long will it take to drill through the piece of cast iron mentioned in Problem 12 ? 14. A high-speed drill of f " diameter is required to drill f " pieces of cast iron in 28 sec. What feed and how many revolutions per minute would you have to give the drill? Allow a cutting speed of 60 ft. a minute for drills. 15. A cylinder 3 ft. 10" long and 9" in diameter is turned to 81" in a lathe with a cutting speed of 25 ft. a minute and a feed of T L". How long will it take to make the roughing FIG. 74. SINGLE-THROW CRANK cut ? How long will it take to make the finishing cut, using a feed of "? 16. A boring bar is to be 35" long and If" in diameter. Allow a cutting speed of 50 ft. and feeds of y and ^\" respectively for roughing and finishing. Find the time re- quired to complete the work, using If" stock. 17. A lot of 16 spindles, 15 " long, are to be turned down from 2" to 1" in diameter for 15" of their length. How long will it take with a cutting speed of 40 ft., a feed of Jj" (roughing) and 3 y (finishing), and a depth of cut not to exceed ^"? 18. Estimate the time necessary to finish the crank shown in Fig. 74. Owing to the opening between the arms, either a low cutting speed or a light cut must be used. 19. A taper ring-mandrel " at the small end and 1^" at the large end is turned from a If" round bar. Estimate 124 SHOP PROBLEMS IN MATHEMATICS the time required to do the work if the surface speed em- ployed is 40 ft. a minute, the feed y, and the depth of cut does not exceed ^". 20. Estimate the time required to make the phosphor- bronze bearing shown in Fig. 75. Time must be allowed for setting the taper attachment and setting for" the thread. 21. If a milling cutter revolves 85 times per minute with a table feed of 6Jg " per minute, what is the feed per revolu- tion ? What is the feed per tooth with a 10-tooth cutter ? 22. A cutter revolves at 203 R. P. M. with a table feed of 9^" per minute. Find the cutting speed for a f" end mill r/MiMrtfi __ with 10 teeth. Find UF :::::: miflO Llj^ i \ !! , I ^ 1 S " -13 Tap 4" FIG. 77 .4. EXERCISES m MACHINE WORK 126 2;s- r/6*! FIG. 77 J5. EXERCISES IN MACHINE WORK 127 128 SHOP PROBLEMS IN MATHEMATICS 34. With a cutting speed of 35 ft. a minute for the angu- lar cutter and feeds of .050" and .032" for roughing and finishing cuts respectively, find the total time for the 4 cuts (2 on each side) for the angle. 35. Find the total time for completing the cross slide, allowing 2 min. for the setting of the work' on the milling- machine table. 36. Estimate the time required to finish the pieces shown in Fig. 77. These are to be made in lots of 50. The time required to set up the machines ready for the various operations should be divided by 50 and added to the time necessary to do the actual machining. Suitable cutting speeds and feeds may be selected from the preceding tables. Those exercises requiring threads to be cut on the lathe should be omitted until the chapter on Thread Cutting has been taken up. With the exception of the pulley, the stuffing box, and the right and left coupling, the pieces are all made of machine steel. 100. Cutting speeds on the planer and shaper. The calcula- tions of cutting speeds on the planer and shaper are some- what more difficult than on other machines on account of their forward and reverse movements, arid because the re- verse is generally faster than the forward stroke. Henceforth the term "stroke " will mean the complete forward and backward movement, except when the reference is to the length of the stroke. EXAMPLE. If a shaper makes 22 strokes in 1 min., the reverse is 2 to 1 of the forward stroke, and the length of the stroke is 1 ft., what is the cutting speed ? If the length of stroke is 1 ft., then the total distance traveled in one direction in 1 min. is 22 x 1 ft., or 22 ft. But only f of the time is used in the forward movement, so that the cutting speed per minute is 22 ft. x f , or 33 ft. (In other words, if the shaper travels 22 ft. in f min., in 1 min. it would travel 22 ft. x }, or 33 ft.) CUTTING SPEED AND FEED 129 Now suppose the length of the stroke is \ ft. instead of 1 ft. The total distance traveled is 22 x i ft., or 11 ft., in | of 1 min., or at the rate of 16^ ft. per minute. It is evident that to obtain the same cutting speed for all lengths of stroke the E. P. M. should be increased for shorter and decreased for longer strokes. The cutting speed of the shaper varies with the length of the stroke ; on the other hand, the length of stroke does not depend upon the number of strokes per minute. In the case of the planer the longer its stroke, the fewer strokes it will make per minute, but the cutting speed is constant, no matter what the length of the stroke may be. EXAMPLE. A planer makes 12 strokes per minute, 2 ft. in length, and the reverse is 2 to 1. Find the cutting speed. The total distance traveled in one direction is 12 x 2 ft., or 24 ft. The actual time taken in moving this distance is f min. Then the cutting speed is 24 ft. x f , or 36 ft. per minute. PROBLEMS 1. A shaper makes 36 strokes per minute and has a 2 to 1 reverse. What is its cutting speed with a 16" stroke ? 10" stroke? 2" stroke? 2. Ascertain the number of strokes per minute that would be necessary to give a cutting speed of 36 ft. per minute for each of the following lengths of stroke of the above shaper: 2", 6", 10", 14". 3. Estimate the time necessary to finish upon the shaper the slide shown in Fig. 76, considering that it can be done in two cuts, roughing and finishing, allowing 1^ min. for setting the head for the dovetail and 1 min. for turning over. Allow a cutting speed of 30 ft. per minute. 4. A planer makes 10 strokes per minute ; the length of the stroke is 4 ft. and the reverse 2 to 1. Find the cutting speed. 130 ~ 00 d Q rH bc.2 .3 rd js n ^ >_ a 26 * a C4 *-l.2 0) fl ^30 A 2 43 SH ~ B 43 a ~ il M S *> ij_j 2 ?fe a^ OJO Cri ,0^^^^ WOHfe^ gf-ailllPPK^pi li&llJIilsl-gju-fiS S2S : n : g : G-;n b 5? *a 3 S"* g gg 6U S o3 .^VB . ^ i ^3 >, a I ia ? i ii T3 S rl S 5 * TJ S rt n T- . -H .3 o> as a o^3 > Q< "-" o ll|-g|il ifJsP 1 P ^^ ^3 V|ag -^g^-^-ag -33 .5 & ifS^llrilwjJfi! 1 illlltHliltiltllltl!' gi . [Sg 1 COO O JL b ^ il| > SH l 11" 25" 27" TF > t > 16 !>> 1* ' 3* > 64 ' 5. With the micrometer of Problem 2 a piece of work measures 0.3895". What is its true diameter? 6. Apiece of work is to be finished exactly .6375". What reading should be made, if the micrometer when closed reads .0004" less than ? 7. A piece of steel which is to measure .995" is heated to 215 F. while grinding and then calipers .996". What will it measure at 65 F.? (Coefficient of expansion = .0000066" for each degree.) 8. A space between two finished surfaces of iron was found to be filled exactly by 5 sizing pieces measuring respectively .375", .050", .020", .010", .002". How large was the space ? A B FIG. 80. SECTION OK MICROMETER CALIPERS 9. One surface of a jig must be just .387" higher than another. This surface is to be measured by piling up siz- ing blocks to the distance required and then making the surface level with these blocks. Select the required pieces from the following: .375", .150", .100", .050", .020", .014", .012", .011", .010", .005". 10. Select from the above sizing blocks the required num- ber to give the distances as follows : .437", .529", .233", .137". On account of the size of the divisions on the thimble D it becomes easy to estimate to one quarter of a thousandth (.00025") with only a slight error. For example, in Fig. 80, A , the reading is nearly .4065", and in Fig. 80, B, it is nearly .4348". 136 SHOP PROBLEMS IN MATHEMATICS Some micrometers have an additional set of lines on the barrel, called a "vernier" (see following section), for accurate reading to ten thousandths. Since holding the micrometers in the hand, as the work- man usually does, may cause them to vary the readings by at least .0002", the use of the vernier is of value only with the most delicate handling. THE VERNIER 103. Vernier. The vernier is an instrument for measuring very small distances. It was named for its inventor, Pierre Vernier, who first used it in 1631. 104. The vernier of the micrometer. A micrometer reading to thousandths of an inch may be made to read to ten thousandths of an inch by the addition of a vernier on the i ~ c A barrel. Ten divisions on the vernier correspond to nine di- visions on the thimble. There- fore the difference between the .xx c /\ width of one division on the ver- nier and one division on the FIG. 81. VERNIER ,, . , , ,-, ,, -,. thimble is one tenth of a di- vision on the thimble, or one ten-thousandth (.0001) of an inch. For example, in Fig. 81 the third line from on the thimble coincides with the zero line of the vernier. The next line on the thimble is distant from the next line of the vernier by one tenth of a division on the thimble. The next two lines on the thimble and vernier respectively are separated by two tenths of a thimble space, the next two by three tenths, etc. In opening the micrometer every space on the thimble represents an opening of one thousandth of an inch. If the thimble is turned to the left, so that the first division line on the vernier coincides with the fourth division line of the thimble, the micrometer has been opened one tenth of one thousandth, or one ten-thousandth, of an inch. If the second MICROMETER, VERNIER, AND TAPERS 137 division line of the vernier coincides with the line marked 5 on the thimble, the tool has been opened two ten-thou- sandths, etc. Hence, to read the vernier of a ten-thousandths microm- eter, observe the line on the barrel that coincides with a line on the thimble. If the line is marked 1, the reading is one ten- thousandth ; if the line is marked 2, the reading is two ten-thou- sandths, etc. T7< T T7<' 00 F 4-T, FlG . 82. VERNIER EXAMPLE. In Fig. 82, 6 on the vernier coincides with a division line on the thimble. The vernier reading, therefore, is .0006". If the reading of the micrometer to thousandths is .325", the complete reading is .3256". TAPERS 105. Taper. A piece of machine work is said to be tapered when it gradually increases in diameter or thickness. It may be round as in a lathe center, or flat as in a taper gib. The amount of taper can be given only when the increase is uniform (see Fig. 83). 106. Method of expressing tapers. Taper is most com- monly given in inches per foot. It is the difference be- tween the diameters (in inches) of the ends of a piece of work 1 ft. long. To ascertain the amount per foot when the taper is actually more or less than a foot in length the difference in diameter for 1 ft. must be calculated. Flat tapered work may be given in the same way. Generally when the taper is excessive its value is given in degrees, as in the case of bevel gears, lathe centers, etc. (see Fig. 88). The taper is also sometimes expressed in fractions of an inch per inch of length, as T ^" to 1". 138 SHOP PROBLEMS IN MATHEMATICS This method is often more convenient for the workman, saving him the trouble of changing the amount of the taper per foot to the per-inch basis. 107. Length of tapers. In turning tapers, if the tail- stock is to be offset, then the actual length of the work or the mandrel, and not merely the length of the tapered part T r of the work, must be - ' A considered in the cal- culations. If the taper attachment is to be used, it is only neces- sary to find the correct B "jy J_ * < / l" >- _ < 3" - * ^ " FIG. 83. TAPERS taper in inches per foot and set the taper attachment to that. No attention need be paid to length of work or mandrel. 108. Method of calculating tapers. (a) Where the tailstock is to be offset. CASE 1. In piece A, shown in Pig. 83, the difference in diameter between the large and the small ends is If" 1" = f ". Take one half of this difference for the offset of the tailstock, or i of f " = f". CASE 2. In piece B the taper does not extend from end to end, but it must be treated as though it did extend the MICROMETER, VERNIER, AND TAPERS 139 whole length and the difference in diameter at the ends calculated. The difference in diameter in 3" of length is 1|" _ 1" = f .". In 1" of length the difference in diameter would be ^ of f " = y\". If the piece is 10" long, then 10 x fV = 2i", which is the total difference in diameter at the ends. If the piece were 7" long, then 1\ x T V = If V- For the offset one half of 2^-" must be taken. (b) When the taper attachment is to be used. Eem ember that in setting the taper attachment the taper per foot must be obtained. In A (see Fig. 83) the total difference in diameters is f "; then the taper per inch is -J of f ", or " per inch. In 12" it would be 12 x i", or 3" per foot. In B the taper per inch was found to be ^\"; then the taper per foot is 12 x f\ " 3", no matter what is the length of the whole piece. Having found the taper per foot, it is only necessary to set the taper attachment to that amount. PROBLEMS 1. Make a simple formula for the offset of the tailstock in taper turning, letting D equal the large diameter, d the small diameter, and s the offset. 2. A' lathe center is T V in diameter at the small end. If the taper were |" per foot, what would be the diameter 4" from the small end ? 3. Change the following from taper per inch to taper per foot : .0833", .0625", ,052", .0416", .0312", .0208", .0104". 4. A lathe center is 1.02" in diameter at the small end. If the taper per foot is .623", what is the diameter 4" from the end ? 5. What difference in diameter would there be at the large end of two pieces 4" long, if each was T y at the small end and one had a taper of " to the foot, the other .602" to the foot ? 140 SHOP PROBLEMS IN MATHEMATICS o| "r\J \ : -T L 2. ._*__ 2" * -1 6. How much larger would one of two pieces be if both were f " at the small end, 5" long, and one were cut to a |" taper, the other to a Morse taper No. 2 ? Morse taper No. 2 is .602" per foot. 7. A crank for a gas engine is 25" long and is to have a taper of f " to the foot turned at one end. How much must the tailstock be offset ? 8. A piston rod has a taper 5" long, cut at one end. At the small end of the taper the rod measures 1.792" and the large diameter of the piston rod is 2". What is the taper per foot ? 9. The tailstock of a lathe can be offset a distance of 2|". What is the largest taper that can be cut on a bar 10" long? 2 ft. long? 3 ft. long ? 10. What is the greatest length that the piston rod in Problem 8 could be and still have its taper cut by a lathe with a tailstock which can be offset not more than 2" ? 11. A plug gauge T y in diameter enters a hole of un- known taper a distance of 2|". A plug of T 9 ^" in diameter enters 2.4". Find the taper per foot. 12. A lathe center is \%" in diameter at the small end and 3" from the end it is |". Find the taper per foot. 13. Stock for a milling-machine arbor is 14" long. It is to be tapered for a length of 6" on one end to " per foot. How much must the tailstock be offset ? 14. A lathe spindle bearing as shown in Fig. 75 is to be turned on a mandrel 8" long. Find the offset of the tail- stock. If the mandrel were 6" long, what would be the offset ? FIG. 84. TAPER BUSHING MICROMETER, VERNIER, AND TAPERS 141 15. What is the taper per foot of the bearing shown in Fig. 75 ? 16. The bushing shown in Fig. 84 is to be turned on a mandrel 7" long. Find the offset of the tailstock. 17. What is the taper per foot of the piece shown in Fig. 84? 18. Stock for lathe centers has been cut off 5J" in length. How much must the tailstock be offset to cut a taper of T Y'per foot? 1_ FIG. 85. LATHE SPINDLE 19. By use of the table of natural functions find the approximate included angle for the following tapers per fnnf 7 " 9 " 11" 13" 15" )0t . T % , T % , T ^ , Tff , Tff . 20. To what included angles are the sleeves of the fol- lowing diameters equal ? Large diameter Inches Small diameter Inches Length Inches 2 1 2 3 1 3 3 2 3 3 2 1 4 1 2 4 1 u 4 1 i 21. Find the taper per foot of the long taper of the lathe spindle shown in Fig. 85. 142 SHOP PROBLEMS IN MATHEMATICS 22. Find the included angle of the abrupt taper of the lathe spindle shown in Fig. 85. 23. A pulley with a 6" face is " larger in diameter at the middle than at the edges. Find the taper per foot. 24. Find the offset of tailstock when the pulley of Problem 23 is turned on a mandrel 11" long. The increase in diameter toward the middle of the face of a pulley is called crowning. It causes the belt to keep in the center of the pulley. 25. A pulley of 1" face is crowning -fa". What is the taper per foot ? Find the offset of the tailstock when the pulley is turned on a 6" mandrel. I Line of L^the Centers I _J I FIG. 86. METHODS OF GRADUATING THE BASES OF COMPOUND SLIDE RESTS 109. Slide rests. The correct setting of slide rests of lathes or the crossheads of planers and shapers for cutting tapers or angles depends largely upon a clear understanding of the relations of angles one to another, such as the com- plement and supplement of an angle ; also that the three angles of a triangle are equal to 180, and that the two acute angles of a right triangle are equal to 90. The complement is equal to 90 minus the given angle. The supplement is 180 minus the given angle. EXAMPLE. The complement of 33 is 57 ; of 45, 45 ; of 69, 21. The supplement of 120 is 60; of 175, 5; of 25, 155. MICROMETER, VERNIER, AND TAPERS 143 110. Setting of angle. The angle to which a slide rest must be set in order to turn a given angle depends on two things : (1) how the angle is given in relation to the center line of the lathe ; (2) at what place on the slide rest the zero point is taken. A slide rest is generally set at zero when it is in line with the cross-feed screw, or, in other words, at right angles to the center line of the lathe. Therefore : If the required angle is measured from a line parallel to the cross-feed, the slide must be set to the same angle as given on the draiving. If the included angle is given, the slide rest must be set to the complement of one half the included angle. (See Figs. 86, 87, and 88.) 111. Graduations on slide rests. The diagrams shown in Fig. 86 will give some idea of the different ways in which M ' N PR S T FIG. 87. ANGLES FOR COMPOUND SLIDE REST the graduations are put upon the circular bases of com- pound slide rests. Notice that except in E the zero marks on the upper and lower slides coincide when the slide is set at right angles to the center line of the lathe. All angles which the student may meet in his work may be resolved into one of the cases shown in Fig. 87. For example, slides A, B, C, D may be set directly to the angle M. Slide E must be set to the complement of M, or 55. Slide D may be set direct to angle N, but for slides A, B a new zero must be made on the lower slide at a, and the slides brought around until 65 45, or 20, as shown by the arrow, is opposite a. The arrow on C shows the setting 144 SHOP PROBLEMS IN MATHEMATICS i H FIG. 88. BEVEL GEARS AND LATHE CENTER of that slide for angle N. Slide E is set at the comple- ment of 65, or 25, as shown by the arrow on E. Angles O and P may be set di- rectly on slide E, bpt on slides A, B, C, D their com- plements must be found, 30 and 75 respectively. The setting is then made as in M and N. In the case of included angles the student may find the cor- rect reading to which to set the slide rest by constructing an auxiliary triangle, as at F' (see Fig. 88). In the case of bevel gears the angles given in G are the required angles, while those given in H are not. By constructing the auxiliary triangles, as at F', the correct angles may be easily obtained. On the shaper and planer the slide is always vertical when the scale is at 0. Then the scale in Fig. 89 A would indicate that the slide was set to an angle of 15 to the vertical. It follows that Whatever angle is given on the drawing, the correspond- ing angle with the vertical should be found and the slide set to that angle. FlG " S9A ' PLANER CROSS SLIDE In using the universal head on the universal cutter grinders the determination of the correct angles is often confusing. Then a sketch showing clearly the different angles will help much in obtaining a correct solution. MICROMETER, VERNIER, AND TAPERS 145 PROBLEMS 1. A triangle has two of its angles 45 and 70 respec- tively. What is the third angle ? 2. A right triangle has one angle equal to 57. What is the value of the third angle ? 3. A triangle has two equal angles and an obtuse angle of 127. Find the value of the other angles. 4. A perpendicular line is dropped from the vertex of the obtuse angle (Problem 3) to the opposite side. Find the value of the angles of the two triangles thus made. 5. Find the angles to which the top slide of a compound rest should be set in order to turn the tapers found in Problem 20, sec- tion 108. 6. A lathe tail stock can be set over either way 3" from the center line. What is the FIG. 89 B. LATHE CROSS SLIDE greatest taper that can be turned on a piece of work 6" long ? 8" ? 10" ? 12" ? 7. To what included angles (approximately) are these tapers equal ? 8. At what angle should the compound rest be set in order to turn tapers similar to those found in Problem 2 ? 9. To what angle must the compound rest be set to turn an included angle of 30 ? 40 ? 45 ? 55 ? 65 ? 90 ? 10. A dovetail which is to be cut on a planer makes 60 with the horizontal. To what angle should the slide rest be set ? 146 SHOP PROBLEMS IN MATHEMATICS 11. Give the correct setting for angles R, S, T, and U (see Fig. 87) upon the graduated scales A, B, C, D, E (Fig. 86). 12. Give the correct angles for setting a planer cross- head for the angles shown in Fig. 90. 6Q A~ ~~Z?" T *W\ 87 ^7 VTT 757 FIG. 90. DOVETAILS 13. Give the correct setting for the angles shown in Fig. 88. 14. To what angle should the compound rest be set in order to turn the abrupt taper on the lathe spindle shown in Fig. 85 ? CHAPTER X THREAD PROPORTIONS, GEARING FOR SCREW CUTTING, INDEXING THREAD PROPORTIONS 112. Number of threads per inch. A screw is generally determined by the diameter of its body and the number of threads per inch ; also by the shape of the thread. 113. Pitch. The distance from the top of one thread to the top of the next is called the pitch. It is the reciprocal of the number of threads per inch. 114. Lead. The distance which a screw would advance when given one complete turn is called the lead. In single threads the lead is equal to the pitch. 115. Depth and double depth. The perpendicular distance from the top to the bottom of the thread of a screw is called the depth. Twice this distance is the double depth. 116. Root diameter. The diameter at the bottom of the thread, or the outside diameter minus the double depth, is called the root diameter. 117. Calculations for depth. It is very important to be able to find the depth of any thread, for upon that value depends the cutting of all threads and the size of all tap drills. The depth of a V -thread is equal to the altitude of an equilateral triangle whose sides are equal to the pitch of the thread (see Fig. 92). This altitude is very easily found by geometry or trigonometry. Considering 1" as a standard or basis of 147 FIG. 91. ENGINE LATHE (By permission of The Hendey Machine Co.) 1. Lathe bed. 2. Shears or ways. 3. Lead screw and feed rod. 4. Reverse rod.* 5. Gear box and rocker handle. 6. Gear box and rocker handle. 7. Stop rod and dogs.* 8. Apron. 9. Carriage. 10. Handwheel for longitudinal feed. Power-feed stud in center. 11. Power cross-feed clutch. 12. Lead screw half nut. 13. Cross-feed screw and ball crank. 14. Bridge of carriage. 15. Cross slide of carriage. 16. Clamping screw to hold car- riage. 17. Tool post. 18. Ring and wedge. 19. Top slide of compound rest. 20. Screw stop for thread cut- ting.* 21. Swivel seat of compound rest. 22. Cone pulley. 23. Bearings. 24. Back gears. 25. Headstock. 26. Small face plate. 27. Live spindle and center. 28. Fore gear and lock pin. 29. Tailstock. 30. Tailstock saddle. 31. Handwheel for tail spindle. 32. Tailstock clamp. 33. Tail-spindle binder. 34. Tail spindle and center. 35. Rack for moving carriage. * Parts marked with star are peculiar to this lathe. 148 THREAD PROPORTIONS 149 measurement, we may find a value, C (really the depth of a thread of 1" pitch), such that the following rule may be derived : For a screw of any pitch the depth of thread is the same fractional part of C as the given pitch is of the base pitch 1". FIG. 92. V-THREAD EXAMPLE. Find the depth of a thread of -J" pitch whose angle is 55. By trigonometry the altitude a of the triangle A (Fig. 92), whose side P is equal to 1", is found thus : and a = .5 .5206 Tan ^ = - = .5206, 2i a Tan 27| = .5206, = .96" = C" for a 55 sharp thread. Then the depth of a thread of " is of .96", or .12". For a U.S. standard-shaped thread the same principle holds. The angle is the same as for a V-shaped thread, namely 60. Since -J P (see Fig. 93) is taken off the top and ^ filled in at the bottom, its depth is only f of that of a V-thread, that is, C' = f C. FlG - 93 U ' S - STANDARD THREAD Hence, by remembering C for a screw of 1" pitch, the depth, double depth, and root diameter of any V-shaped or U.S. standard-shaped thread may be easily found. 150 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. A screw has 12 threads per inch. What is its pitch ? What is its lead ? 2. A lathe lead screw has 2^ threads per inch. What is its lead ? 3. A certain pipe thread has 11^ threads per inch. What is its pitch ? 4. Find the pitch for the following number of threads per inch : 3, 3^, 4, 4^-, 5 7 5^-, 6, 6. 5. Find the number of threads per inch for the following leads : r, t", r,i", F, r, i". 6. Denoting the number of threads per inch by TV, the body diameter by D, and the pitch by P, write a formula for P in terms of N. 7. Solve the formula in Problem 6 for N. 8. Find the value of C for a screw of 1" pitch. (Use the geometrical method.) 9. Make a table for both V- and U.S. standard threads as follows : THREAD PROPORTIONS, V-SHAPE Number of threads per inch Pitch Double depth THREAD PROPORTIONS, U.S. STANDARD Number of threads per inch Pitch Double depth THREAD PROPORTIONS 151 10. The depth of a certain thread is .025". What is the root diameter of a screw whose outside diameter is 1^" ? 11. If the double depth of a thread is .108", to what size must a nut be bored to receive a screw whose outside diameter is J" ? 12. Find the depth of thread for the common V-thread rifr>ifQ 1 " 1 " 1 " 1 " 1 " 1 " 1 1" 1" 1" 1" pltCiieS ^0 > I*g > iV > T4 > T3 ) T2 9 10) 8 > ff > 5 ? 13. Find the depth of thread for the pitches given in Problem 12, but for a U.S. standard-shaped thread. 14. Ascertain the root diameter of a screw 1^" in outside diameter and with 7 threads per inch ; with 10 threads ; 12 threads ; 40 threads. The common method of expressing the size of a screw is ^"-20. It means that the screw body is J" in diameter and has 20 threads to the inch. 15. Find the root diameter of the following V-shaped threads : J-20, &-!*, f "-16, T V'-14, J-12, $"-11, f "-10, |"-9, l"-8. 16. Find the root diameter of the following U.S. standard threads : J-20, T V'-18, J"-16, T V '~14, J"-13, T ^"-12, |"-11, "-10, |"-9, l"-8, lJ"-7, lJ"-6. Tabulate. 17. Allowing 30,000 Ib. to the square inch for the ten- sile strength, find the load which a V-shaped screw, f "-10, would hold. 18. Find the tensile strength of a U.S. standard bolt f "-11, allowing 30,000 Ib. per square inch. 19. What per cent stronger is a screw which is threaded with a U.S. standard-shaped thread than one with a V- shaped thread ? The point of a tool for cutting U.S. standard-shaped threads is flattened an amount equal to one eighth of the pitch of the screw to be cut. 152 SHOP PROBLEMS IN MATHEMATICS A. Worm Thread 20. Find the width of the point of a tool for cutting a U.S. standard thread of i" pitch; J" pitch; ^" pitch; ^" pitch. 21. Find the width of the point of a tool for cutting a U.S. standard-shaped thread (double) of 10 threads per inch ; 8 threads per inch ; 6 threads per inch. 22. Find the tap drill for a 1" U.S. standard screw. 23. Find the tap-drill size for a 14-20 tap. The size of No. 14 wire is .242" and the thread is V-shape. 24. Find the tap drill for a 12-24 tap. The size of No. 12 wire is .215" and the thread is U.S. standard shape. 25. It is desired to cut 20 threads to the inch on a shaft 1.375" in diameter. What will be the root diameter ? 26. If an inside thread of 18 to the inch is to be cut in a nut to be fitted to a 1.75" shaft, what is the smallest hole that can be bored in the nut? 27. Denoting the outside diameter of a screw by D, the number of threads by JV, and the root diameter by d r , make a formula for d r in terms of D, N, and a constant C. 28. A U.S. standard bolt 2" in diameter has 4|- threads per inch% How much larger should a bolt with V-shaped threads be in order to be as strong as the U.S. standard ? Square threads are shaped so that the depth of tooth, width of tooth, and width of space between the teeth are equal. Hence the depth of thread is equal to i P (see Fig. 94). B. Square Thread FIG. 94 THREAD PROPORTIONS 153 29. A square-threaded screw has 4J threads to the inch. How wide must the point of a tool be made to cut this thread ? 30. A shaft 3-J-" in diameter has a square thread cut upon it, 1^ threads to the inch. What is the root diameter ? 31. A nut is to be made to fit a square-threaded screw of JQ" pitch and of " diameter. What width of tool should be used ? What diameter should the hole be bored ? 32. Compare the shearing strength of the threads of a U.S. standard screw and of a square-threaded screw, diam- eter of the shaft being 1 j-", with 6 threads per inch. For the proportions of a worm thread see Fig. 94. 118. Proportions of an Acme thread. An Acme thread is very similar to a worm thread, but has the following pro- portions : Let D a = outside diameter. N = number of threads per inch. Root diameter of screw = D n I -- h .02 ). \N ) Outside diameter of tap = D a + .02. Depth of thread = - ----- h .01. Width of point of tool for screw = '~^ - .0052. Width of screw thread at top = : PROBLEMS 1. Denoting the outside diameter by D a , the root diam- eter by d r , and the pitch by P, write a formula for the root diameter, d rj of an Acme screw in terms of D a and P. 2. An Acme thread is to be cut on a 2-J-" shaft. If there are to be 3 threads per inch, what will be the root diameter ? 154 SHOP PROBLEMS IN MATHEMATICS 3. A nut is to be threaded to fit a screw If" in diameter, with 4| threads per inch. How large a hole must be bored in the nut, allowing .050" for clearance ? 4. Find the width of the point of an Acme-thread tool for 5 threads per inch ; 4 threads ; 3 threads ; 1^ threads. 119. Double threads, triple threads, etc,' Oftentimes the depth of a thread upon a shaft is great enough to materially reduce its strength. When an extreme lead is required for a screw and the shaft must not be materially weakened, a double, or triple, or even quadruple thread may be cut. In a double thread the 3.2JA2J. . _ , , . pitch and depth are only one half of that of a single thread of the same lead. Similarly, for a triple thread the pitch and depth are only one third of that FIG. 96. SINGLE AND TRIPLE THREADS of a single thread of the or THE SAME LEAD same i ead Fig. 95 shows a single- and a triple-threaded screw of the same lead. Note the excessive depth in the single thread. Also note the three endings of the triple thread. PROBLEMS 1. A double-threaded screw has 12 threads per inch. What is its pitch ? What is its lead ? 2. Find the double depth of the screw in Problem 1. 3. A triple-threaded screw has 15 threads per inch. What is its pitch ? What is its lead ? 4. A double-threaded screw has 9 threads per inch. What is the lead and the equivalent single thread ? 5. A double-threaded screw of 6 threads per inch and 'a triple-threaded screw of 9 threads per inch have the same GEARING FOR SCREW CUTTING 155 lead. What per cent stronger would a 1^" shaft be with a triple than with a double thread cut upon it ? Unless otherwise stated the U.S. standard-shaped threads should be taken. 6. Compare the strength of a screw when cut with a single, double, or triple thread of " lead on a 1" shaft, allowing a tensile strength of 60,000 Ib. per square inch. 7. A double- and a triple-threaded screw have " lead. What number of threads do they have per inch? What are their depths of threads ? 8. A nut is to be bored out preparatory to cutting a triple thread of J" lead for a 3" shaft. What size should the hole be made ? 9. Find the root diameter of a quadruple-threaded screw of f " lead and 1" outside diameter. 10. Find the lead for a double-threaded screw which will have the same depth as : single threads of yL" pitch, " pitch, I" pitch, T -y pitch respectively; triple threads of y 1 ^" pitch, |" pitch, J" pitch respectively. For the method of setting tools for multiple threads see the fol- lowing sections. GEARING FOR SCREW CUTTING 120. Simple gearing. A simple geared lathe is illus- trated in Fig. 96. The change gears are the gears on S and L. Gear 7 is used only to connect S and L, and does not appear in the calculations. The reverse gears, as shown in both Figs. 96 and 97 in their two different positions, are simply for the purpose of changing the direction of rotation of gear L, and hence of the lead screw. Thus by reversing these gears a right- or left-hand thread may be cut. The first thing to do in calculating gearing is to see that spindle S' makes the same number of revolutions as S. If they do not revolve 156 SHOP PROBLEMS IN MATHEMATICS the same, then equal gears should be placed on S and L and the lathe tested to see what screw would be cut. This screw should then be used as the calculating lead screw instead of the lead screw on the machine. (By "screw" we mean the number of threads per inch.) 121. The rules for calculating the gears for thread cutting. RULE I. The number of threads per inch on the screw to be cut is to the number of threads per inch on the lead screw Reverse Gears Intermediate Gear on Screw FIG. 96. END VIEW OF LATHE HEADSTOCK as the number of teeth in the gear on the lead screw is to the number of teeth in the gear on spindle or stud; Screw to be cut _ Gear on lead screw Lead screw Gear on spindle For example, find the gears for cutting a thread of -J" pitch on a lathe whose lead screw has 5 threads per inch. GEARING FOR SCREW CUTTING 157 -J" pitch is equivalent to 8 threads per inch. Our ratio is 8 Gear on lead screw 5 Gear on spindle But since there are no gears of 5 or 8 teeth we are obliged to multiply the terms of the ratio by some num- ber, as 4, 5, G, or 7. Compound Gearing Then FIG. 1>7. END VIEW OF LATHE HEADSTOCK 8 6 48 Gear on lead screw 5 X 6 30 Gear on spindle If gears of 48 or 30 teeth are not available, then take some other number than 6 as a multiplier. RULE II. If the screw to be cut is a larger number than the lead screw, then the smaller gear goes on the spindle, or vice versa. 158 SHOP PROBLEMS IN MATHEMATICS 122. Compound gearing. When by simple gearing the number of teeth in the gears run up to over 120, com- pounding is resorted to. That is, an additional gear, C (see Fig. 97), is put upon a sleeve along with D, or a small bracket, C' (see Fig. 96), holding two gears, may be swung into mesh between S and /. The change gears in Fig. 97 are A, C, D, and /,. The rules for calculating the compound gears for thread cutting are as follows : T-i- 7 7 /. Screw to be cut . RULE III. Divide tlie ratio of - - into two Lead screw ratios, the terms of which may be increased Inj a multiplier to such numbers as will represent available gears. RULE IV. In compounding, if the denominator of one ratio is taken as a driving gear, then the denominator of the other ratio will be a driving gear and the numerators will be driven gears, or vice versa. For example, find, the gearing necessary to cut 22 threads per inch on a lathe with a lead screw of 5 threads per inch. Our ratio is ^ 2 = y X ?. Bracket C" (see Fig. 96) has two gears fixed upon it, generally of 60 and 30 teeth ; therefore one of our ratios should be multiplied by such a number as to give 60 and 30. The required gears then are and . It is not so easy to determine their relative positions on the lead screw or spindle as in simple gearing, but it may be arrived at according to Rules II and IV. Since 22 is larger than 5, the gears 30 and 30 will be the drivers and gears 66 and 60 the driven gears. In Fig. 97 gears A and D are drivers and C and L are driven gears. If the ratios obtained at first do not represent available gears, new ratios must be taken. GEARING FOR SCREW CUTTING 159 123. Metric threads. A similar procedure may be fol- lowed in rinding the gears necessary to cut a metric thread without a metric lead screw. For example, find the gears necessary to cut a thread of 1.5 mm. pitch on a lathe with a lead screw of 5 threads per inch. This pitch must first be reduced to threads per inch. 1" = 25.4 mm. ; 25.4 hence -T-TT- = 16} | threads per inch. 16}*- 254 Our ratio is ' -=^-' o 75 Since in these calculations the figure 254 will always appear no matter what the pitch, and since its factors are 127 and 2, it follows that 127 should always be selected as the term of one ratio. Hence our gears are -^V and f g. Then 25 and 60 are the drivers and 127 and 40 are the driven gears. /PROBLEMS Find the gears necessary to cut a screw of 10 threads per inch on a lathe with a lead screw of 5 threads per inch. Select from the following gears : 25, 30, 35, 40, 45, 50, 55, 60, 65, 69, 70, 80, 90, 100, 120. 2. The lead screw of a lathe has 6 threads per inch. What gears will be necessary to cut 8 threads ? Do not use a gear of less than 24 teeth. 3. On the lathe of Problem 1 what is the largest num-/ ber of threads that can be cut by simple gearing ? The gear of 120 teeth must be used as an intermediate. 4. The standard thread for f " pipe is ll. per inch. What gears are necessary to cut this thread with a lead screw of 160 SHOP PROBLEMS IN MATHEMATICS 6 threads per inch ? 5 threads per inch ? 4 threads per inch ? Use 69 as the gear on. the lead screw. 5. A lathe with a lead screw of 5 threads per inch is supplied with gears as given in Problem 1. Can a screw of 2 threads per inch be cut on this lathe ? 6. With the gears given in Problem 1,. find the correct gears for cutting 24 threads by compounding. 7. The calculating lead screw of a lathe has 16 threads per inch, and the lathe is equipped Avith the following gears : 32, 32, 36, 40, 44, 46, 48, 52, 56, 64, 72, 80, 96, 120. Select the sets of gears necessary to cut 4 threads ; 5 threads ; 6 threads. 8. Using the lathe as described in Problem 7, select proper gears for cutting 8 threads ; 20 threads ; 36 threads. 9. The calculating lead screw of a lathe is 8. Select gears from the following suitable to cut 11^ threads per inch : 24, 24, 28, 32, 32, 36, 40, 44, 48, 52, 56, 64, 69, 72, 80, 96. 10. From the gears in Problem 9 select the sets of gears necessary to cut the following threads per inch: 1, 14-, 2, 3, 4, 5, using as the compound the gears ||. 11. What is the smallest thread that can be cut by simple gearing, using the lathe and gears of Problem 9 ? 12. The ratio of the K. P. M. of the lathe spindle to those of the stud (see Fig. 96) is 4 to 3. Find the calculating lead screw if the real lead screw has 6 threads per inch. Place equal gears on stud and lead screw. 13. Find the gears necessary to cut a metric screw of 4 mm. pitch on a lathe with a lead screw of " pitch. I" = 25.4 mm. 14. Find the number of threads per inch which is equal to 1 mm. pitch ; 1.5 mm. pitch ; 2 mm. pitch ; 2.5 mm. pitch ; 3 mm. pitch. INDEXING 161 c 15. Find the gears necessary to cut a metric screw of 2 nun. pitch on a lathe with a lead screw of 5 threads per inch. Do not use a gear of less than 25 teeth. 16. How many metric threads could be cut by the fol- lowing gears : 127, 25, 40, 60 ? What are their pitches ? 17. Find the gears necessary to cut a metric screw of 3.5 mm. pitch on a lathe with a lead screw of 6 threads per inch. 124. Indexing. It is frequently necessary to space equal distances upon a surface, as holes in a circle or teeth upon a gear, or to mill an hexagonal or square nut. Locating the drill or cutter for these purposes is properly called indexing. This work is most frequently done upon a milling machine, although the principle may be applied to any machine. 125. Indexing by use of a gear. The teeth of a gear may furnish the means of indexing. This gear may be put on the cross- feed screw of a shaper or planer, and thus enable one to cut a rack, etc., upon these machines. A disk with a circle of equally spaced notches may be placed upon the back end of a lathe spindle and indexing done upon the lathe. 126. Index head. This head consists essentially of a base B (see Fig. 98), a spindle (7, revolved by a worm wheel W and worm. Arm A and the index plate I are con- nected to the worm shaft. The index plate generally has a number of circles of holes called index circles, each circle having a different number of holes, as 31, 33, 37, 39, etc. It is by means of these index circles that we are enabled to get an exact fraction of a revolution, as J, this being FIG. 98. INDEX HEAD 162 SHOP PROBLEMS IN MATHEMATICS 13 holes in a 33 circle ; or f f , being 25 holes in a 27 circle. Sometimes an index plate is attached directly to spindle C ; then each circle will index a few numbers directly. For instance, a circle of 30 holes would index 2, 3, 5, 6, 10, 15, 30. 127. Method. It takes forty revolutions of arm A to turn spindle C around once. Hence it can be seen that a rela- tively large movement of arm A produces only a small rotation ( ? L) of spindle C. For instance, if we rotate A T L of one revolution, C will revolve ^ of one revolution; or if we rotate A 13^ times, then C will revolve ^ of 13, or ^ of one revolution, so that by moving arm A this amount 3 times in succession we are able to obtain 3 equal spaces in one revolution. RULE I. Divide 40 by the number to be indexed and the result will be the turns required for arm A. RULE II. The denominator of the fraction determines the index circle to be used\ RULE III. If the fraction is a small one, its terms must be multiplied by some number which will give an available circle. For example, find the number of turns necessary to cut 22 teeth in a gear. *tf lit = ly 9 ! number of turns. T 9 T X I = If. T \ X g - J. There are two index circles which may be used in this ca^e, namely 22 or 33. PROBLEMS V 1. An index plate on spindle C (see Fig. 98) has 24 holes. What numbers can be indexed by it ? 2. A certain milling machine is supplied with three in- dex plates containing the following circles : 15, 16, 17, 18, 19, 20, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49. Find INDEXING 163 the index circle and the number of turns necessary to index these numbers : 5, 7, 9, 10, 11, 13, 28, 29, 88, 98, 100, 260. 3. A reamer is to have 8 teeth, unequally spaced one to the other. Present a scheme by which this can be done on the index head. 4. Find the different numbers up to 100 which can be indexed by using the 33 circle. 5. What numbers from 1 to 300 can be indexed in com- mon by the 27 and 33 circles ? 6. Find the three circles which will index in common the following numbers : 3, 15, 30, 59, 61. A straight piece of iron or steel with gear teeth cut in it is called a rack. 7. Several teeth in a rack of .3142" circular pitch must be recut upon a shaper whose cross-feed screw has 4 threads per inch. A gear of 86 teeth is mounted on the end of the screw for indexing. How many turns will be necessary to move the tool from one tooth to the next ? A slight approximation must be made. 8. What error in thousandths of an inch is made in the spacing of each tooth by the approximation made in Prob- lem 7 ? 9. Would a gear of 100 teeth give a more accurate spacing for the rack in Problem 7 than a gear of 86 teeth ? Why? 10. What gear among the following would give more accurate spacing for the rack in Problem 7 than the gear of 86 teeth : 25, 30, 35, 40 ? What error in thousandths would this gear make in the spacing of each tooth ? 11. The following gears are included with the equipment of a lathe : 25, 30, 35, 40, 45, 50, 55, 60, 65, 69, 70, 80, 90. Find a gear which may be used for spacing the teeth of a 164 SHOP PROBLEMS IN MATHEMATICS rack of .7854" circular pitch upon the shaper of Problem 7 ; .5236" circular pitch ; f " circular pitch ; 3 diametral pitch ; 16 diametral pitch. The circular pitch is equal to TT (V) divided by the diametral pitch. 12. A punch shaped as shown in Fig. 99 is to have the three surfaces, A, B } C, milled with a plain" mill. Find the index circla which may be used and the number of turns necessary to bring B into a horizontal position after milling A. Also find the turns necessary to bring C into the hori- zontal position after milling B. 128. Setting the tool in cutting double threads, etc. In double, triple, or quadruple threads, after cutting the first thread the tool must be put into position for the next thread. This can be done , 60' in several ways. (1) By changing the tail of the dog in FlG 99 END V , EW the face plate. The face plate must have OP PUNCH slots accurately milled for this method. (2) By careful measurement and resetting of the tool. (3) By withdrawing gear 7 (see Fig. 96) and revolving gear S , ^, or ^ of a complete revolution and again placing / in mesh with S. This moves the spindle but not the carriage, and can only be used when the gear S is divisible by 2 for double threads or 3 for triple threads, etc. (4) By calculating how many turns of gear L would be necessary to move the carriage the required distance and by withdrawing / while L is being turned this amount. PROBLEMS 1. Find the gears necessary to cut a double-threaded screw of " lead on a lathe with a lead screw of 5 threads per inch. Use 5 as a multiplier. INDEXING 165 2. Can the tool be set for the second thread by the third method described in section 128 ? 3. A triple-threaded screw of 6 threads per inch is to be cut 011 a lathe with a lead screw of 4 threads per inch. Find the gears required when the smallest gear has 30 teeth. 4. A lathe is set up with the 70-tooth gear on the screw and the 28-tooth gear on the stud. What thread would be cut if the lead screw were of " lead ? 5. A double-threaded screw of i" pitch is to be cut on a lathe having a lead screw of " pitch. Select the proper gears from the following : 25, 30, 35, 40, 45, 50, 60, 75, 90. 6. Can the tool be set for the second thread by the third method in section 128 ? Why ? 7. Find the number of turns of gear L (see Fig. 96) neces- sary to set the tool for the second thread in Problem 1. 8. A triple thread of " pitch is to be cut on a lathe having a lead screw of " pitch. Select the proper gears from those given in Problem 5. 9. Find the number of turns of the gear on the lead screw in order to set the tool to the correct position for the second thread (see sect. 128). 10. How much must the tool be advanced in setting it for the second thread on a double-threaded screw of 3 threads per inch ? 4 threads ? 6 ? 7 ? 8 ? 11. How much must the tool be advanced in setting it for the second thread on a triple-threaded screw of 3 threads per inch ? 4 threads ? 5 ? 6 ? 7 ? CHAPTER XI GEAR PROPORTIONS AND SPIRALS CALCULATIONS OF GEAR TEETH 129. Speed ratio. The estimating of all gearing is based upon such theoretical cylinders (for spur gears) or cones (for bevel gears) as will produce the desired speed ratio when rolled upon each other, no slip being considered. 130. Pitch circles and pitch diameters. The circumfer- ences of these cylinders are called pitch circles and their diameters pitch diameters. In gearing, the pitch circle is located practically halfway between the top and bottom of the tooth. 131. Spur gears. Spur gears are gears cut on a cylinder with teeth parallel to the axis. 132. Spiral gears. Spiral gears have their teeth cut at an angle to the axis of the cylinder. 133. Bevel gears. Bevel gears are gears cut upon a con- ical surface. 134. Diametral pitch. The number of teeth in a gear per inch of its pitch diameter is called the diametral pitch. 135. Circular pitol The distance from the center of one tooth to the center of the next measured on the pitch circle is called the circular pitch, and is equal to - J J d 136. Addendum. The distance from the pitch circle to the outside of the tooth is called the addendum. It is equal to 1 divided by the diametral pitch, or (see sect. 145). l X > *-- > J-ff J ^2? > ^ ) - 1 ? > -^ > - 1 J * > I J B 5 2 T% > 2 157 45. Find the whole depth + ' for each of the above "d *d diametral and circular pitches. 46. Find the thickness of the tooth on the pitch line for the above pitches. Tabulate the results obtained in Problems 43-46, as follows : SAMPLE TABLES TOOTH PROPORTIONS Circular pitch Whole depth Thickness of tooth Equivalent diametral pitch GEAR PROPORTIONS AND SPIRALS 173 TOOTH PROPORTIONS Diametral pitch Whole depth Thickness of tooth Equivalent circular pitch A. Square and Hexagon TRIGONOMETRY IN THE SHOP 147. Some of the problems in this section may be solved by arithmetic, but more easily by trigonometry. Others can be solved only by trigo- nometry, for which see section 249. The method of finding by arithmetic the largest square or hexagon which can be milled on a cy- lindrical piece of work, or what size to turn up a piece in order to be able to mill a certain- sized square upon it is as follows : 148. For a square. The diagonal of a square is equal to the diameter of the smallest cylinder upon which that square can be made. The diagonal of a 1" square is equal to the square root of 2, or 1.414". Using this number 1.414" as a constant C, to find the diagonal of any square in inches or to find the side of B. Milling of a Square FIG. 100 FIG. 101. MILLING MACHINE (By permission of Brown and Sharpe Mfg. Co.) PARTS OF THE MILLING MACHINE 1. Spindle clutch, for driving arbor. (For driving arbor or collet hav- ing clutch collar.) 2. Spindle adjusting knob. (For ob- taining a tine adjustment of spindle by hand.) 3. Spindle-change feed lever. (Moves sliding quill gears for obtaining fast or slow series of speeds.) 4. Spindle-speed tumbler gear, sliding knob. (For moving tumbler gear into position to engage any one of the cone of gears.) Spindle-speed tumbler gear, self- locking lever. (Throws tumbler gear in and out of mesh with any of the cone of gears.) Feed lever. (Gives feeds per revo- lution of spindle, or in inches per minute.) Feed tumbler gear, sliding knob. (For moving tumbler gear into position to engage any one of the cone of gears.) Feed tumbler gear, self-locking lever. (Throws tumbler gear in 174 and out of mesh with any of the 29. cone of gears.) 9. Feed - changing lever. (Upper.) (Moves sliding quill gears to ob- 30. tain a fast or slow series of feeds.) 10. Feed -changing lever. (Lower.) (Moves sliding quill gears to ob- 31. tain an additional series of fast and slow feeds.) 32. 11. Feed reverse lever. (Starts, stops, and reverses all automatic table 33. feeds.) 12. Transverse-feed trip lever. (Con- trols automatic transverse move- 34. ment of table.) 13. Vertical-feed trip lever. (Controls automatic vertical movement of 35. table.) 14. Longitudinal-feed trip lever. (Con- 36. trols automatic longitudinal movement of table; also re- verses same.) 37. 15. Transverse-feed handwheel. (For use when adjusting or feeding 38. table by hand.) 16. Vertical-feed handwheel. (For use when adjusting or feeding table 39. by hand.) 17. Table quick-return handle. (For 40. use when adjusting or feeding table by hand ; also furnishes 41. means for quick return of table.) 18. Transverse-feed clutch knob. (For throwing handwheel in and out 42. of action.) 19. Vertical-feed clutch knob. (For throwing handwheel in and out of action.) 43. 20. Longitudinal - feed safety stops. (To prevent damage to longitu- dinal-feeding mechanism in case the trip dog is on the wrong side 44. of the trip plunger.) 21. Longitudinal-feed trip dog. (Ad- justable. Throws feed out of ac- tion at any desired point.) 22. Transverse-feed trip dogs. (Adjust- 45. able. Throws feed out of action at any desired point.) 46. 23. Vertical-feed safety stop. (To pre- vent damage to vertical-feeding 47. mechanism when the trip dog is on the wrong side of the trip plunger.) 24. Vertical -feed trip dog. ( Adjust- 48. able. Throws feed out of action at any desired point.) 25. Trip-stop pin knob. (Allows table 49. to be fed only in the desired direction.) 50. 26. Adjusting screw. (For adjusting length of feed chain for wear.) 51. 27. Arm-clamping lever. (Clamps over- hanging arm at two places by one movement.) 52. 28. Friction - clutch lever. (Short.) (Used for throwing clutch in driving pulley in and out when 53. changing spindle speeds.) 175 Friction clutch lever. (Long.) (Used for stopping or starting machine.) Oil pocket for quick oiling. (De- livers oil to all bearings in spindle speed and feed cases.) Pad for motor bracket. (Facilitates application for motor drive.) Spiral head. (For use in indexing and the cutting of spirals, etc.) Footstock. (For supporting outer end of work when using spiral head.) Rapid indexing plate. (For rapid indexing of numbers, 2, 3, 4, 6, 8, 12, and 24.) Work driver. (For driving work when using centers.) Footstock center adjusting knob. (For adjusting center in relation to work arbor.) Arbor yoke. (Intermediate support for arbor.) Graduations on saddle. (Provides means for setting table to any required angle.) Spiral-head pan. (To hold spiral head when not in use.) Raising block. (To set spiral head in any angle or position on table.) Spiral-head center. (For use in dif- ferential indexing for all num- bers from 1 to 382.) Collet. (To provide means for hold- ing shanks of cutters and arbor with smaller tapers than spindle of machine.) Spiral-head change gears. (Used ^to obtain different leads with the spiral head, and differential indexing.) Table stops. (To govern travel of table when setting by hand ; al- so to form a positive lock when transverse and vertical feeds are not in use.) Knock-out rod. (Removes arbors and collets from spindle.) Center rest. (To support work car- ried by spiral heads.) Bushing with center. (For sup- porting outer end of arbor. In- terchangeable with bushing in Arbor-holding nut. (For locking arbor or collet and clutch to Index plates. (Used for differen- tial and ordinary indexing.) Thread guard. (To protect thread on end of spindle.) Chuck. (For holding round stock either in machine or spiral-head spindle.) Machine-spindle chuck plate. (To hold chuck when used on ma- chine spindle.) Vise. (For holding work.) 176 SHOP PROBLEMS IN MATHEMATICS the square when the diagonal is given we have the follow- ing rule : The diagonal of any square is equal to the side of that square multiplied by C, or the side of any square is equal to its diagonal divided by C. ^ 149. For a hexagon. The distance across the corners of a hexagon is equal to the diameter of the smallest cylinder upon which that hexagon can be made. The distance across the corners of any hexagon is equal to twice its side. By a simple proposition in geometry it can be proved that the distance across the corners of a hexagon is equal 2 to the distance across the flats multiplied by i=, or 1.156. Letting C' stand for this value 1.156, we have the rule: The diameter of the smallest cylinder upon which a re- quired hexagon can be made, is equal to the product of the distance across the flats and the constant C'. In order to find the length of the side of a square or the distance across the flats of a hexagon which can be milled upon any given cylinder, divide its diameter by C or C". PROBLEMS 1. Denoting the side of the square by s, the diagonal by d, and the constant by C, write a formula for d in terms of s and C. 2. Denoting the side of a hexagon by h, the distance across the flats by I, and the constant by C', write a formula for h in terms of I and C'. 3. A shaft is .785" in diameter. What is the largest square that can be milled upon it ? GEAR PROPORTIONS AND SPIRALS 177 4. How wide would the flats (parallel sides) of a hexag- onal head be when milled on a shaft .895" in diameter? What would be its distance across the flats ? 5. A square .687" on a side must be milled on a shaft. What is the smallest diameter to which the shaft may be turned ? 6. A f " square is to be milled on a |" shaft How many thousandths must the shaft be reduced on each side? (See Fig. 100.) 7. One cut on all sides of B (Fig. 100) has been taken and the work now measures .733". How many thousandths should the table of the milling ma- chine be raised in order to finish the square to f " ? When it is necessary to drill FlG - 102. JIG. PLATE holes with great accuracy in a piece of work which cannot be swung in a lathe, a milling machine may be used. Various methods of measuring the distances for the location of these holes may be employed. The longitudinal and transverse feeds of the milling-machine table have dials graduated to thousandths of an inch, and a reasonable degree of accuracy in measuring movements of the table can be obtained by use of these dials. Also by knowing the number of threads per inch of the feed screws, dials may be devised for special work done in large quantities. (See later examples.) The longitudinal feed of the table is in the direction of the length of the table ; the transverse feed is at right angles to the length of the table. 8. Holes A and B (Fig. 102) are to be drilled on the milling machine. In order to drill A, how much movement of the table will there be in each direction after drilling B ? 178 SHOP PROBLEMS IN MATHEMATICS 3-37 FIG. 103. JIG PLATE 9. Find table movements for drilling A, B, and C when angle CBA is 100 (Fig. 103). 10. Five holes, one at every corner of a pentagon of the size shown in Fig. 104, are to be drilled on a milling machine. Starting from A, give the lateral and vertical movements in thousandths of an inch of the milling- machine table for all of these holes. 11. Two plugs are located 3.773" apart, measuring to the outside of the plugs. If plug A (Fig. 105) is .375" in diameter and B is .625", what is the exact distance be- tween their centers ? 12. Given two plugs by which to locate the centers of two holes 1.987" apart. What should they measure over all, if the plugs are f" in diameter? 13. It is desired to locate points A, B, C accurately, as shown in Fig. 106. Select diameters for the plugs at A, B, and C such that when they are tangent A to C and C to B, the correct distances 1.975" and 1.365" will be given. Distance AB is first obtained as in Problem 12. 14. A disk is 10|" in diameter. Flo< 105 . SPACING HOLES Find the distance necessary to set a pair of dividers in order to space off 7 sides ; 8 sides ; 10 sides ; 13 sides. 15. Six holes, one in every corner of a hexagon, are to be drilled on the milling machine. The distance across the FIG. 104. PENTAGON GEAR PROPORTIONS AND SPIRALS 179 flats of the hexagon is 1.25". Find the lateral and vertical movements of the milling-machine table for every hole. 16. The vertices of an equilateral triangle are to be lo- cated by 3 plugs, each measuring .750" in diameter. The sides of the triangle are to be 4.75". What distance over all must be measured on the plugs ? ^'"y\ Vvt B -2.365 FIG. 106. JIG PLATE -475'- FIG. 107. ACCURATE SPACING 17. It is desired to turn up a plug which will just touch the plugs in Problem 16 when they are correctly located. What should be its diameter ? (See Fig. 107.) 18. An eccentric is to give a rocker arm A (Fig. 108) a motion of 15. If the throw of the eccentric is 6", how long must the arm be ? 19. A shaft is to have an oscillating motion of 12. If the arm on the shaft is 5" long, what is the throw of the eccentric? FIG. 108. ECCENTRIC AND ROCKER SHAFT 20. A rocker arm is 10" long, and a Second arm is 7" long, to which the eccentric rod is attached (see Fig. 108). What throw must an eccentric have to move the end of the first arm 3"? To what angle would that be equal ? 180 SHOP PROBLEMS IX MATHEMATICS 21. Allowing at least 1" of metal around the eccentric shaft which is 2" in diameter, how large must the eccentric be to meet the conditions in Problem 20 ? 22. A slide valve has a motion of 3". What throw must an eccentric have to produce this motion through a bell crank one arm of which is 7|" and the other 2|-", if the valve is connected to the long arm ? SPIRALS 150. Spirals. The spiral as used in the machine shop is, correctly speaking, a helix, but in the following problems we shall follow the common usage. 151. Angle of spiral to axis of work. The angle which the spiral makes with the axis of the work is easily found. Take, for example, JJ the angle which a thread makes with the axis of the cylinder upon Jj which it is cut. |*-Lead-| FIG. 109. ANGLE OF SPIRAL Lay off upon one side of a right angle to any convenient scale a distance equal to the lead of the spiral ; upon the other side of the right anglej^y off a distance equal to the circumference of the cylinder. Connect the ends of these lines. The angle a (Fig. 109) is the angle which the spiral makes with the axis of the work. This is useful in showing GEAR PROPORTIONS AND SPIRALS 181 at what angle the side of the tool should be ground for clearance in cutting the thread. Angle /3 gives this clear- ance. It applies to the angle which the flute of a drill makes with the axis of the drill, and is useful in setting the milling- machine table for cutting the spiral of drills and cutters. For the definition of lead, see section 114. PROBLEMS 1. A " twist drill has a spiral of 1 turn in 7.58". What angle does the edge of the flute make with the axis of the drill ? 2. A 1-J-" twist drill has a spiral of 9.759" to one turn. What angle does the edge of the flute make with the axis of the drill ? 3. Two spiral gears mesh at an angle of 75. Their velocity ratio is 7 to 8 and the angles of the spiral gears are proportional to the ratios. What are the angles ? I i FIG. 110. SHOWING MILLING-MACHINE TABLE 4. A shaft 3" outside diameter has a square thread " deep. What angle does the top edge of the thread make with the axis of the shaft ? What angle does the bottom of the thread make ? 5. Work out a formula for the angle of a spiral. 6. The shafts of two spiral gears which mesh together make an angle of 90. If both gears are equal, what angle does the teeth make with the shaft ? 182 SHOP PROBLEMS IN MATHEMATICS 7. What spiral (number of turns to one inch) would this angle give on a gear with a pitch diameter equal to 3"? 2"? 1"? 8. A f " drill is to be milled with a spiral of 6.5" to one turn. At what angle must the milling table be set ? 9. A 1" drill has a spiral of 9" to one turn. What angle does its flute make with the axis of the drill ? 10. A |" drill is cut with the table set at 20. What is the lead of the drill ? What are also the leads in inches of drills of the following diameters : ", f ", f " ? 11. It is desired to have the angle of the teeth of a 3" spiral mill make about 15 with its axis. What should be its lead ? 12. The milling machine is geared to cut an 8" spiral. If the teeth are cut on a 1" cylinder, what angle will they make with the center line ? what angle for a 1" cylinder? 2"? 3"? 4"? 13. In a worm and worm wheel the worm has a lead of i" and its pitch diameter is 1J". At what angle across the face of the worm wheel will the teeth be cut ? Make a sketch of a worm and worm wheel. Notice the relations between the angle of the worm and the angle of the teeth on the face of the worm wheel. 14. A worm has a pitch diameter of 1" and a lead of J". What is the angle between the teeth of the worm wheel and its axis ? 15. A worm has a lead of 0.333" and a pitch diameter of 1-J-". At what angle must the milling table be set in order to gash out the teeth in the worm wheel ? 16. What is the worm-wheel angle for a worm of 6 turns per inch, the pitch diameter being 1J" ? What for a worm of 5 turns ? 4 turns ? 3J turns ? CHAPTER XII THE UNIVERSAL GRINDER AND THE GAS ENGINE THE GRINDER 152. The angle of clearance. In backing off the edges of a cutter or reamer the edge makes a certain angle a with the tangent to the cutter at the cutting point (see Fig. 111). Tanyent fo -J \ ^ Cutter FIG. 111. BACKING-OFF OF CUTTER This angle may be more or less than 7, according to the work on which the reamer is to be used. The angle A may be ground by the face of the grinding wheel, and for approxi- mate results we may consider that the tangent line T and the face of the wheel coincide for the short distance b. 153. Application. If it is desired to back off a cutter to a certain number of degrees, say 7, with a wheel 5" in diam- eter, a formula can be derived which will give the distance D, or the horizontal distance of the center of the cutter from the center of the grinding wheel. 183 184 SHOP PROBLEMS IN MATHEMATICS In Fig. Ill if we may consider tangent T and circle G to coincide for the width of the land to be ground, then the angle a = a'. Hence, by trigonometry, D = r sin a. Since r and a are given, sin a can be found in Table I and D can be calculated. EXAMPLE. Find the distance D for a grinding wheel 5"in diameter. D = 2.5 x sin a = 2.5 x .122 = .305". PROBLEMS 1. Find the value of D for grinding a clearance of 7 with a wheel 6" in diameter; with a wheel 2" in diameter; 3^"; 4"; 8". 2. Find the values of D for a 5 angle of clearance, using the same wheels as in Problem 1. 3. Find the values of D for a 10 angle of clearance, using the same wheels as in FlG 112 p ISTON Problem 1. R ING GAS ENGINE 154. The following problems are of general interest ; they occur in the design of a gas engine and are simple enough to permit of a place in this book. PROBLEMS 1. The ends of a piston ring are intended to overlap each other " (Fig. 112) to prevent leakage. To what diam- eter should a ring be turned so that when sawed apart and lapped the outside diameter will be 3^" ? 2. If the clearance space of a steam engine holds 1^ gal. of water, what is the clearance in cubic inches ? What is the per cent of clearance if the cylinder is 8" x 10"? UNIVERSAL GRINDER AND GAS ENGINE 185 3. If the clearance space of a gas engine holds 11 cu. in. of water, the diameter of the cylinder is 3\", and the stroke 3^", what is the per cent of clearance ? 4. A water jacket of a gas engine is f " wide, 4" long, and has an inside diameter of 4f ". How many quarts will it hold? . 5. Find the total area of the valve seat in Fig. 113. 6. If the total pressure exerted by a spring upon stem A is 93 lb., what pressure per square inch is exerted on. the valve seat ? 7. The maximum pressure ex- erted by the gas in a gas engine is about 270 lb. per square inch. What is the total pressure on top of the valve whose diameter is If"? FIG. 113. VALVE AND SEAT 8. What area of opening would be obtained by lifting the valves * " ? (See Fig. 113.) 9. At what velocity must gas enter a cylinder 4J" in diameter by 4i" long, and having a valve as shown in Fig. 113 in order to fill it in .037 of one second ? 10. The mean effective pressure in a gas engine is 75 lb., the stroke 4", the cylinder diameter 4J", and the number of power strokes per minute 400. Find the horse power from Pi A N the formula HP = where P is the mean effective ooUUU pressure, I the length of stroke, A the area of the cylinder, and N the number of power strokes. REVIEW OF CALCULATION WITH SHORT METHODS CHAPTER XIII FRACTIONS 155. Addition of fractions. EXAMPLE 1. Add ^ and T \. T V + fV = tt = EXAMPLE 2. Add f and |. t + I = A + if ^ if = 1 T V EXAMPLE 3. Add the mixed numbers 3f and 4f . 3f + 4* = 3if + 4H - 7|f *= 8/ T . ORAL EXERCISE !- 1 + i- 6. A + A. 2. + J. 7. i + A- 3. | + A- 8. 1\ + 4. 4. A + J. 9. 2^ + 31. 5. A + |. 10. 1J + 4f + 2J. WRITTEN EXERCISE ! T 5 ff + | + J. 3. 2| + 3} + 2/ . 2 - ^ + A + I- 4 - 5f + 3| + 7-^. 5- A + 2i- + f 186 FRACTIONS 187 6. How wide a floor space can be covered with 5 boards whose widths are respectively 101", 12 ", 9|", 6|", and 4"? 7. In laying pipe a plumber uses 4 pieces measuring respectively 5' 3J", 4' 2|", 10' 3", and 3' 7f". Find the total length of the pipe. 156. Subtraction of fractions. EXAMPLE 1. From subtract J. 5 1 4 2 I ~~ ff f 3- EXAMPLE 2. Subtract j\ from . I - T 3 5 _, i. 9 . A ^ r. 4. i - i. 7. 2^ i 5. 10. 31 4- 8. WRITTEN EXERCISE ! I - iV- 6. ^ - |. 2. 150 -- 5 V 7 - 4 i^ 6 ?- 3- A - if- s- II - sV 4. 9^ * 7 - 9 - 20 I * 18J. 5 - it Hh A- 10. A H- fi- ll. The dimensions of a box, inside measure, are 3' 4", 2' 6", and 1' 8". The dimensions of another box, inside measure, are 1' 6", 2', and 2' 3". The volume of the larger box is how many times as great as that of the smaller ? 161. Decimal fractions. EXAMPLE 1. Add 3.5, 4.03, 7.275, and .2. 3.5 4.03 7.275 .2 15.005 FRACTIONS 191 EXAMPLE 2. Subtract .42 from 5.3. 5.3 .42 4.88 EXAMPLE 3. Multiply 3.246 by 2.45. 3.246 2.45 16230 12984 6492 7.95270 EXAMPLE 4. Divide 7.285 by 3.5. Before dividing multiply dividend and divisor by such a number as will make the divisor an integer. 7.285 -f- 3.5 = 72.85 -s- 35. 72.85 70 285 280 35. 2.08+ 5 Check. 3.5 x 2.08 + .005 = 7.285. Check by casting out nines. EXAMPLE 5. Divide 5.3 by 4.235. 5.3 -r- 4.235 = 5300 -=- 4235. 5300. 4235 4235. 1.25+ 10650 8470 21800 21175 625 Check. 4.235 x 1.25 + .00625 = 5.3. 192 SHOP PROBLEMS IN MATHEMATICS WRITTEN EXERCISE 1. Add 4.25, 7.3, 6.002, and 5.372. 2. From 92.73 subtract 8.673. 3. Multiply 4.3126 by 52.7. 4. Divide 673.25 by 12.3 correct to .three places of decimals. 5. Divide 802.3 by 7.621 correct to three places of decimals. 6. The circumference of a circle is 3.1416 times the diameter. If the diameter is 24.5", find the circumference. 7. How many cubic inches in a box whose inside dimen- sions are 3.4", 8", 1.8"? 162. Reduction of a fraction to a decimal. EXAMPLE 1. Reduce fa to a decimal correct to three places. 3.000 28 28 .107 + 200 196 4 ORAL EXERCISE Express as decimals : 1. f. 3. i. 5. &. 7. |. 9. 2. i. 4. ^. 6. i. 8. &. 10. WRITTEN EXERCISE Reduce to decimals correct to three places : 1. |. 3. T \. 5. 2|. 7. 3ff. 9. 2. T V 4. Jf. 6. 8.34J. 8. CHAPTER XIV PERCENTAGE. AVERAGES. SQUARE ROOT 163. Important per cents. J = 50% J : i - 25% | = 75% 1 = 40% J = 60% * = 80% 164. Base. The base is the number of which the per cent is to be found. 165. Rate. The rate is the number of hundredths to be taken. 166. Percentage. The percentage is the result found by taking a certain per cent of the base. 167. Finding the percentage. The percentage is found by multiplying the base by the rate. EXAMPLE. 4 per cent of $200 == $200 x .04 (since 4 per cent = .04) = $8. 168. Finding the rate. The rate is found by dividing the percentage by the base. EXAMPLE. If you take an examination of 8 questions and answer 5 correctly, you have answered f, or 62^ per cent, correctly. In this case 8 is the base, 62^ per cent is the rate, and 5 is the percentage. 193 194 SHOP PROBLEMS IN MATHEMATICS 169. Finding the base. The base is found Inj dividing the percentage by the rate. EXAMPLE. A man receives 10 per cent increase in salary, amounting to $5 a month; i.e. $5 is 10 per cent of his former salary. 1 per cent is T V of $5, and 100 per cent is 100 x A of $5 = $50. This result. could have been secured more simply as follows : ' ORAL EXERCISE 1. Denoting the base by B, the rate by R, and the per- centage by P, state a formula for P in terms of B and R. 2. Solve the formula for B. 3. Solve the formula for R. 4. If you are receiving $100 a month and get 5 per cent increase, what will your salary be next month ? 5. If there are 200 school days in the year and you attend 150 days, what is your rate of attendance ? 6. An engineer receiving $100 per month has a 10 per cent increase in salary. What is his monthly salary ? Six months later he is cut 10 per cent. What is his final salary ? Why is this not the same as at first, he has had an increase of 10 per cent and a reduction of 10 per cent ? 7. A surveyor buying a transit instrument listed at $200 gets 40 per cent discount for cash. How much does he pay ? 8. 52 is 13 per cent of what number ? 9. 27 is 9 per cent of what number ? PROBLEMS 1. Find 17 per cent of $345. 2. Find 13J per cent of $523. AVERAGES 195 3. If your salary is $1200 a year and you receive an increase of $225, what is the per cent of increase? 4. $500 is 14 per cent of what number ? 5. On January 1, 1908, 1 deposit $400 in a savings bank paying 4 per cent. The bank adds the interest to the prin- cipal July 1 and January 1 of each year. How much can I draw on July 1, 1910? 6. If a man's savings are $150 on a salary of $1100, what per cent does he save ? 7. The premium on a life-insurance policy for $2000 is $52 per year. What is the rate paid for insurance ? 8. If you borrow $100 on a note of 3 months date.d to-day, and have it discounted on the same day at 6 per cent, what are the proceeds of the note? 9. $14.50 is 18 per cent of what number ? 10. What sum of money must be deposited in a savings bank paying 4 per cent, to yield a semiannual income of $1500? AVERAGES 170. Average. The average of two or more quantities is the result of adding the quantities and dividing the sum by the number of quantities ; e.g. to average 3, 8, 7, 6, and 9, divide their sum, 33, by 5, and the average is 6. The average may often be found without adding the numbers ; e.g. to find the average of 68, 70, 75, and 60, take any convenient number, as 60, for a base. 68-60- 8 70 - 60 = 10 75 - 60 = 15 33 33 + 4 = 8}. 60 -f 8J = 68|, the average. 196 SHOP PROBLEMS IN MATHEMATICS Again, find the average of 75, 90, 85, and 65. Taking 70 for a base, 75 - 70 = 5 90-70= 20 85-70= 15 65 - 70 = -_6 35 35-4 = 8f 70 + 8| = 78J ORAL EXERCISE Find the averages of the following : 1. 8, 7, 6, 4, 3, 9. 3. 10, 15, 20. 2. 7, 9, 10, 4. 4. 40, 45, 35, 50. 5. A student has marks on the four quarter years as follows : 75, 85, 68, 70. Find the average. 6. A man works 8 hr. Monday, 10 hr. Tuesday, 9 hr. Wednesday, 7 hr. Thursday, 11 hr. Friday, and 10 hr. Sat- urday. Find the average length of his day for the week. PROBLEMS 1. At different times during the day an indicator shows a pressure of steam of 175, 180, 140, 100, 65, 50, 40, and 45. Find the average or mean pressure. 2. What is the average speed of a train that travels 200 yd. in 10 sec., 400 yd. in 15 sec., and 75 yd. in 5 sec. ? 3. The maximum temperature at a certain mountain resort on seven successive days was 60, 65, 68, 70, 52, 66, and 72. Find the mean of maximum temperature for the week. 4. The minimum temperature at the same place for the seven days was 40, 42, 48, 51, 47, 50, and 52. Find the mean of minimum temperature. SQUARE ROOT 197 5. Five planks measure respectively 10' 3", 12' 4", 11' 6", 9' 8", and 10' 9". Find their average length. 6. A board is 18^" wide at one end and 11 " at the other. What is the average width ? 171. Square root. EXAMPLE 1. Find the square root of 2025. 2025(45 Check 16 45 2 = 2025 85J425 425 EXAMPLE 2. Find the square root of 138,384. 138384(372 Check 9 372 2 = 138,384 67)483 (Casting out nines) 469 0=0 742)1484 1484 EXAMPLE 3. What is the square root of 2.3 correct to two decimal places ? 2.30(1.51+ Check J_ 151 2 + 199 = 23,000 25 ) 130 (Casting out nines) 125 4 + 1 = 5 301 ) 500 301 199 ORAL EXERCISE Find the square root of : 1. 121. 3. 64. 5. 225. 7. .0016. 9. T V 2. 36. 4. 144. 6. .49. 8. .25. 10. . 198 SHOP PROBLEMS IN MATHEMATICS WRITTEN EXERCISE Find the square root of : 1. 4225. 2. 2809. 3. 3721. 4. 53361. 5. 21224449. Find the roots correct to two decimal places : 6. 3.1416. 7. 3. 8. 24.6. 9. 3840.12. 10. |$. (Eeduce to decimal before finding square root.) 172. Factoring method. EXAMPLE. Find the square root of the product of 27 x 9 x 32 x 6. V27 x 9 x 32 x- 6 = V3 3 x 3 2 x 2 5 x 2 x 3 = V3 6 X 2 6 = 3 3 x 2 3 = 216. ORAL EXERCISE 6. V10 x 5 x 14 x 7. 7. V18 x 10 x 5. 3. V6 x 3 x 2. 8. V36 x 49. 4. V125 x 5. 9. V35 x 7 x 2. 5. V7 x 14 x 2. 10. V21 x 14 x 6. CHAPTER XV RATIO AND PROPORTION 173. Ratio. The relation of one quantity to another of the same kind with respect to magnitude is called their ratio. It is expressed by dividing the first by the second, e.g. : The ratio of $2 to $3 is , or 2 : 3. The ratio of 4" to 8" is f , or , or 1 : 2. The ratio of 18 bd. ft. to 72 bd. ft. is }|, or ', written also 1 : 4. 174. Terms of a ratio. The numerator and the denomi- nator of the ratio are respectively the first and second terms of the ratio. 175. Antecedent. The first term is the antecedent. 176. Consequent. The second term is the consequent. ORAL EXERCISE Find the value of x in the following : ! *- 5 3 E = 3 5 *_ 2 . 7 *_ 3 9 8 _! *' 3 3 ' 5 5^5 8~4 ' x~ 8 .f-i. 4 . >i. .4 iXj. io.- A 4 62 63 x 2 x 11. A door measures 8' x 4 f . What is the ratio of the length to the width ? 12. There were 25 fair days in November, the rest being stormy. What was the ratio of fair days to stormy ? 13. The ratio of the height of a certain door to its width is 3 : 2. It is 4| f wide. How high is it ? 109 200 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. Measure (Fig. 114) AD and AB AD and find the ratio A -Z> 2. Measure (Fig. 115) h, b } h', and * |_ b'. Find the areas of the rectangles , , , . FIG. 114. RECTANGLE and their ratio. 3. Does a ratio exist between R and h ? Give the reason. 4. In a tennis court (Fig. 116) find the following ratios : Length of the double court _ AB ' Width of the double court ~ 17)' Length of the single court EF ' Width of the single court R h FIG. 115. RECTANGLES Area of the single court (GI ' v ' Area of the double court Length of a service line, MN Length of a base line, AD Perimeter of the single court ( f>\ * ' Perimeter of the double court 5. In a baseball diamond (Fig. 117) find the following ratios : Distance from first base to second ^ ' Distance from first base to third RATIO AND PROPORTION 201 Distance of the pitcher's box from the home plate Distance from the home plate to second base A / c^ B E r 1 F R F I / s. ^ \J ! . H C rr r J FIG. 116. TENNIS COURT C Area of the double tennis court i\ . . Area of the baseball diamond Perimeter of the baseball diamond ' Perimeter of the single tennis court FIG. 117. BASEBALL DIAMOND 6. In the basket-ball court (Fig. 118) find the ratios Length of the boundary line on one end Length of the side line 202 SHOP PROBLEMS IN MATHEMATICS to O 15 2.' FIG. 118. BASKET-BALL COURT Length of a side line Length of a diagonal , v Radius of the circle in the center (4 Radius of the circle from which free throws are made Width of the players' lane Distance of thrower from the basket 7. In a football field (Fig. 119) find the following ratios : * w idth of the field \ u 330' Length of the fiel t IT) ! "0 vO 4 3> 1 1 12 J i i c 5 >l "^ 1 -j q: c. O >o (1 \ i D FIG. 119. FOOTBALL FIELD RATIO AND PROPORTION 203 (c) Make up three other problems in ratios of lines on the football field. 177. Inverse ratio. The ratio 4 : 5 is the inverse of the ratio 5 : 4. In Fig. 120 the lever F is the fulcrum, P the power, W the weight, FB the weight arm, and AF the power arm. It P weight arm P w is a law of physics that = - , or = -In W power arm W p other words, the mechanical advantage of the lever is equal to the inverse ratio of its arms. FIG. 120. LEVER B EXAMPLE. What power will be neces- sary to balance a weight of 25 Ib. on the .If. above lever, if p = 5 and w = 4 ? P _4 25 ~ 5' 178. Separating in a given ratio. EXAMPLE. Divide $17 between A and B in the ratio 2 : 3. Let 2 x = the number of dollars that A receives, and 3 x = the number of dollars that B receives. Then 5 x = their sum, and $17 = their sum. 5 a = $17. x = $3f , or $3.40. 3 x = $10.20. Check. $6.80 + $10.20 = $17. 2 = (dividing by 3.40). 204 SHOP PROBLEMS IN MATHEMATICS ORAL EXERCISE 1. Divide 20 in the ratio 2:3; 3:1. 2. Divide 18 in the ratio 4 : 2. 3. Divide 100 in the ratio 4:1; 9:1. 4. A board 18 in. long is to be divided in the ratio 5 : 1. How far from each end is the point of division ? 5. In water there are 2 parts of hydrogen to 1 of oxygen. How many parts of oxygen are contained in 120 parts of water ? WRITTEN EXERCISE 1. Divide 200 in the ratio of 7 to 1; 5 to 2 ; 2 to 7. 2. In September the ratio of clear to cloudy days was 5:1. How many days were clear ? 3. If a line 4 f 6" long is divided in the ratio of 5 to 6, what is the length of each part ? 4. Divide a legacy of $25,000 between A and B, so that their shares shall be in the ratio of 2 to 5. 179. Reducing ratio to a percentage. A ratio reduced to a decimal becomes a percentage ; e.g. the ratio 2:3, or f , re- duced to a decimal becomes .66f, or 66f per cent. PROBLEMS 1. In a basket-ball league of ten schools each school plays a game with every other school. In games already played, if your school has won 5 games and lost 2, what is its per- centage, that is, the ratio reduced to a decimal, of games won to games played? 2. If another school has won 2 games and lost 7, find its percentage. 3. A third school finishes the series, having won 5 games and lost 4, while a fourth school has won 4 out of 7 games played. Which has the higher percentage ? RATIO AND PROPORTION 205 4. In a class of 27 students 22 passed an examination. Find the percentage of successful students. 5. In a class of students 23 passed and 5 failed. What percentage of students passed ? 6. In Problems 4 and 5 note which class makes the better record, and from this determine whether a ratio is increased or diminished by adding 1 to both numerator and denominator. 7. Which is greater, or > if a < b ? Reduce to a common denominator and compare the numerators. 8. Which is greater, ^ or 7 ~ ' if a, , or < the second, according as (6 a c) is positive, zero, or negative. Note that in the second case the ratio is always equal to 1. 10. In a civil service examination, from 31 candidates in English 24 men pass ; from 31 candidates in algebra 29 pass; from 26 candidates in geometry 24 pass. Compare the records of algebra and geometry and apply the formula of Problem 9. Here a = 24, & = 31, c = 5, and (& a c) < 5. 11. Check Problem 10 by finding the percentage of al- gebra and geometry. From this notice which record would have been better, to have had 5 more men pass or to have had 5 fewer candidates ? 12. In cases like Problem. 10 would it be possible to have (b - a - c) < ? 206 SHOP PROBLEMS IN MATHEMATICS 13. In the ratio f^f, which operation gives the greater ratio, adding 9 to the numerator or subtracting 9 from the denominator ? (Here (b a c) < 0.) 180. Proportion, proportionally, pro rata. These terms are sometimes used meaning "in the same ratio as"; e.g. (1) Between the center of the earth aiid its surface the weight of a body is proportional to its distance from the center of the earth. (2) A bonus of 10 per cent of their salaries was given to three engineers as a reward for completing their work ahead of time. Their salaries were respectively $2500, $1500, and FIG. 121. GRADE $1000 per year. Their total salaries being $5000, the first received T V of $2500, or $250; the second, T \ T of $1500, or $150; and the third, T V of $1000, or $100. This may be expressed in any of the following ways : (a) The amount each received was proportional to his salary. (b) The bonus was divided among the engineers propor- tionally, according to salary. (c) The bonus of 10 per cent was divided among the engineers. The first received 10 per cent of $2500, or $250, and the others pro rata. GRADE 181. Grade of roadbed. If a trolley roadbed rises h ft. in a horizontal distance of 100 ft., the ratio -rr is the grade. RATIO AND PROPORTION 207 If k = 2', the grade is T-p-r* or 2 per cent. When the rise 1UU is uniform and it is not convenient to measure 100 hori- zontal feet, the following method may be used : Fi multiplying by Id we have ad = be. Tf 3 5 If = > X 1 and 7x3, (Product of means = product of extremes.) 7x3 5 Hence the rule : (#) In any proportion the product of the means divided by the given extreme is equal to the required extreme. WRITTEN EXERCISE Find the value of x in Exercises 1-8. 1. x : 25 = 13 : 14. 5. 11 : 12 = 18 : x. 2. x : 7 = 12 : 17. 6. 125 : x == 206 : 305. 3. 8 : x = 5 : 11. 7. 144 : 105 = x : 24. 4. 7:9 = ^:14. 8. 1003 : 1800 = 27 : z. PROBLEMS 1. The Chicago, Milwaukee, & St. Paul R,ailroad uses for the prevention of scale in locomotive boilers an alkaline compound consisting of 3750 gal. of water, 2500 Ib. of caus- tic soda, and 1500 Ib. of soda ash. How many pounds of each chemical are necessary when a mixture composed of 1000 gal. of water is used ? How many pounds of each should be used to make 5000 gal. of the mixture? How many pounds of each are necessary if 1000 Ib. of caustic soda are used ? if 2000 Ib. of soda ash are used ? 2. A formula for a blue-print solution is as follows : 2 oz. citrate of iron and ammonia, 1.33 oz. red prussiate of potash, 210 SHOP PROBLEMS IN MATHEMATICS 16.5 oz. of water. Find the amount of other chemicals, if 3 oz. of citrate of iron and ammonia are used ; if 3.5 oz. of the potash are used j if 64 oz. of the solution are required. 3. How many ounces of pure silver must be melted with 200 oz. of silver 800 fine to make a bar 900 fine ? 4. How many ounces of gold must be melted with 10 oz. of gold 14 carats fine (if pure) to make an ingot 18 carats fine ? 5. How many pounds of copper should be melted in 94.5 Ib. of an alloy consisting of 3 Ib. of silver to 4 Ib. of copper, so that the new alloy shall consist of 7 Ib. of copper to 2 Ib. of silver? 6. If pig iron contains 93 per cent of pure iron, 3 per cent of carbon, 2 per cent of sulphur, and the rest consists of silicon, phosphorus, etc., how many pounds of pure iron are there in 3 tons of pig iron ? 7. An ingot of gold and silver weighs 13.20 oz. What is the weight of each of the two metals, supposing the gold and the silver in the ingot to have the same value, and gold to be worth 15J times as much as silver ? 8. Alcohol is received in the laboratory 0.95 pure. How much water must be added to a gallon of this alcohol so that the mixture shall be 0.5 pure ? (Check the result.) 9. How much water must be added to a 5 per cent solution of a certain medicine to reduce it to a 1 per cent solution ? 10. How many ounces of silver 700 fine and how many ounces 900 fine must be melted together to make 78 oz. 750 fine? 11. If bell metal is made of 25 parts of copper to 11 parts of tin, find the weight of each metal in a bell weigh- ing 1044 Ib. 12. Gunpowder being composed of $ sulphur, 75 per cent niter, and the balance charcoal, how many pounds of each are contained in 400 Ib. of powder? a-x *- RATIO AND PROPORTION 211 13. An architectural drawing of a house is made to the scale of | " = 1 ft. Find the number of square inches occupied on the drawing by a lot of land 120 ft. long and 80 ft. wide. 14. If the heating surface of a locomotive boiler is 1862 sq. ft., find the ratio of heating surface to grate surface when the grate measures 9 ft. 6" by 8 ft. 15. In drawing a picture of a tower which is 180 ft. high and 32 ft. in diameter, the diameter is to be represented by a line 4" in length. By how many inches should the height be represented ? 16. Find the ratio of the volume of a sphere 6" A I ^~i 1 B in diameter to the volume of a cylinder 6" in diam- , , . , FIG. 124. GOLDEN SECTION eter and 6" high. 17. A wooden pattern from which an iron casting is made weighs GA per cent as much as the iron. If the pattern weighs 35 Ib. 12 oz., how much does the casting weigh ? 18. If " on a map corresponds to 7 mi. of a country, what distance on the map represents 20 mi. ? 186. The golden section. If AR is divided at C so that , AB is divided in extreme and mean ratio and AC is the golden section between AB and CB. Denoting AB by a, A C by x, and CB by a x, show that x = a (approximately). Hence A C and CB are approximately in the ratio 3 : 2. The relation of the golden section is one of the most beautiful relations of size and was so used by the Greeks. It is a principle of art that the nearer the proportions of an article approximate 3 : 2, the ratio of the golden mean, the more beautiful the article is. Find examples of this ratio in familiar objects, such as the ratio of the height of the back of a chair to the length 212 SHOP PROBLEMS IN MATHEMATICS of the legs; the ratio of the length of an oblong picture frame to its width. " All good design depends on order and order is depend- ent on geometry." 187. Gear of bicycles. The gear of a bicycle means the diameter of an old-fashioned large wheel, which in one revolution would cover as much ground as a modern bicycle when the pedal makes one revolution. If G is the gear, C the number of cogs on the crank 2S c* sprocket, c the number on the rear sprocket, G = C where 28 is the standard diameter of the ordinary wheel. The standard number of teeth in the front and rear sprockets is as follows : Front sprocket, 20, 22, 24, 26, 28, 30. Rear sprocket, 7, 8, 9, 10, 11. PROBLEMS 28 C 1. Solve G = for C and c. c Fill out the blanks in the following table, and find C or c, correct to the nearest standard size. C c G 2. 20 7 3. 20 11 4. 30 7 5. 28 9 6. 8 84 7. 7 100 8. 24 70 9. 22 75 R CHAPTER XVI MENSURATION 188. Area of a rectangle. The area of a rectangle is equal to the. product of the base and altitude. PROBLEMS 1. If A stands for the area of a rectangle, a the altitude, and b the base, write the formula for A in terms of a and b. 2. Draw a rectangle, meas- ure a and b, and find A. 3. If a = 5", 6 = 81", find ,4. a 4. In a tennis court a = 36' and b = 78'. Find A. (See Fig. 116, sect. 176.) 5 . 5. Allowing 10' on the side FIG. 125. RECTANGLE lines and 15' on the back lines, find the area of the plot of ground necessary for such a court. 6. The diagonals of a rectangle are equal. How could this principle be used in determining whether the court as laid out was a true rectangle ? 7. In a right triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides (see sects. 44-46). If the hypotenuse is c, one side a, and the other b, write a formula for c in terms of a and b. 8. Using this formula, find correct to inches the length of a diagonal of (a) the double tennis court ; (b) the single court. 213 21-4 SHOP PROBLEMS IN MATHEMATICS 9. Is a triangle with c = 5, & = 3, ft = 4a right triangle ? with c = 10, a = 6, ft = 8 ? 10. How may this triangle be used in making the corner of the tennis court a right angle ? 11. Draw a plan of a plot of ground to contain four tennis courts, making what allowance you think necessary for space about the courts, specifying (a) the total area of ground required ; (ft) the length of each line ; (c) a method of laying out the lines in their order ; (d) a method of obtaining a right angle ; (e) a test for the accuracy of the work. 12. Copy this figure and in each rectangle write the area in terms of a and b. 13. From the above figure complete the fol- lowing formula : a FIG. 126. (a + 6) 2 Find the numerical area of the above figure by substituting in the formula : 14. If a = f, ft = 8". 15. If a = 6' 9", b = 3' 6". 16. If a = 0", 1 = 10". 19. If a = 2' 3", ft = l'4". 20. If a = 4|", b = 4|". 17. If a = 12', ft = O f . 18. If a = |", ft = 4J". 21. A baseball diamond is a square 90 ft. on a side. Find its area. What part of an acre is the diamond ? (160 sq. rd. = 1 acre.) MENSURATION 215 22. Draw a diagram of a baseball diamond. 23. Find the distance from home plate to second base correct to feet and inches. 24. Applying the test for the tennis court to the form of the diamond, what distances must be equal in order that the diamond shall be a rectangle ? 25. What further test is necessary to prove that the figure is a square ? 189. Perimeter. The perimeter of a figure is the sum of its sides. The perimeter of a rectangle is twice the sum of its base and altitude. PROBLEMS 1. In Fig. 125 write the formula for^>, the perimeter, in terms of a and b. FIG. 127. TRIANGLE FIG. 128. EQUILAT- ERAL TRIANGLE 2. Find the perimeter of the single tennis court (Sect. 176). 3. Find the perimeter of the double tennis court. 4. Find the perimeter of the baseball dia- mond. 6. Write the formula for the perimeter, p, of a square in terms of s, its side. 190. Area of a triangle. The area of a tri- angle is equal to one half the product of the ISOSCELES base and altitude. TRIANGLE 216 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. Denoting the area by A, the base by b, and the alti- tude by a, write the formula for A in terms of I and a. FIG. 130. RIGHT TRIANGLE FIG. 131. SCALENE TRIANGLE 2. Draw a triangle, measure b and a, and find A. 3. Fill out the blanks in the following table : b a A 1 25" 15" 2 6' 4" 6" 3 16' 3" 8' 4" 4 45' 7" 24' 11" 4. Find the area of a triangular plot of ground whose base is 50' 10" and altitude 15' 6". 191. Area of an equilateral triangle. The area of an equi- lateral triangle is equal to the square of a side multiplied by the square root of 3 divided by 4- PROBLEMS 1. Denoting the area by A and each side by a, express A in terms of a. 2. Draw an equilateral triangle, measure a side, and find A. 3. Find A correct to two decimal places : (a) Ifa = 4". (c) If a = 10' 3". (b) If a = 6' 6". (d) If a = 20' 7". (e) If a = 35' 10". MENSURATION 217 4. Write the formula for the perimeter, p, of an equi- lateral triangle in terms of a side, a. 5. With the values given in Problem 3, find p. 6. Find the area of a plot of ground in the form of an equilateral triangle, if one side measures 75' 9". 192. Area of any triangle in terms of its sides. The area of any triangle is equal to the square root of the product of 5, s a, s b, s c, where a, b, and c are the sides and s is half the perimeter. PROBLEMS 1. Denoting the area by A, write the formula for A in terms of a, b, c, and s. 2. Draw a triangle, measure a, b, and c, and find A cor- rect to two decimal places. 3. Fill out the blanks in the following table : a b c A I 5 6 9 2 13 14 15 3 9 10 17 Median FIG. 132. TRAPEZOID 4. Find the area of a triangular field whose sides are 40 rd., 58 rd., and 84 rd. 5. Find a formula for the pe- rimeter, PJ in terms of a, b, and c. 6. With the values given to a, b, and c in Problem 3, find j9. 7. Find the length of the side of a plot of ground in the shape of an equilateral triangle, to have the same area as the plot in Problem 4. 193. Area of a trapezoid. The area of a trapezoid is equal to the product of the altitude by one half the sum of the bases. 218 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. Denoting the area by A, the bases by b and b', and the altitude by a, write the formula for A in terms of b, b', and a. \-\ Q One half the sum of the bases is equal to the median. 2. Denoting the median by m, write the formula for A in terms of a and m. 3. Draw a trapezoid, meas- ure the bases and altitude, and find the area. 4. Check the preceding problem by measuring the median and using it to find FIG. 133. IRREGULAR FIELD , , the area. 5. Fill out the blanks in the following table : b b' a A c 1 V 4' 3' 2 6" 8" 5" S 30 rd. 40 rd. 15 rd. * 1'3" 2 / 8 // 18" 6. A field in the form of a trapezoid has for dimensions a = 25 rd., b = 75 rd., b' = 80 rd. Find the area. 7. Find the length of the side of a square field, to have the same area as that in Problem 6. 194. Irregular figures. Irregular figures are usually meas- ured by dividing them into triangles and trapezoids. The irregular field in Fig. 133 may be measured by draw- ing a diagonal DD\ then the perpendiculars AB, CE, FG, and HK, thus dividing the figure into four right triangles and two trapezoids whose areas are easily computed. MENSURATION 219 PROBLEMS 1. Compute the area of the field in Fig. 133, if the dimen- sions in rods are HK = 25, KD = 5, ED 15, BA 35, CE = 30, BE = 40, ED' = 10, FG = 38, EG = 6. How many acres ? (160 sq. rd. = 1 acre.) 2. Draw to scale an irregular figure of five sides, divide it into triangles and trapezoids, take the necessary measure- ments, and compute the area. 3. Do the same for a figure of six sides. 4. Do the same for a figure of seven sides. 5. Do the same for a figure of ten sides. 195. A polygon is a plane figure bounded by straight lines. The triangle and quadrilateral are polygons. FIG. 134. REGULAR POLYGONS 196. A regular polygon is a polygon that has equal sides and equal angles. The apothem of a regular polygon is the line drawn from the center of the polygon perpendicular to one of the sides, as h in the figure. The short diameter or the distance across the flats of a regular hexagon is the perpendicular distance between two opposite sides. It is equal to the diameter of the inscribed circle or to twice the apothem. The long diameter of a regular hexagon is the distance between two opposite vertices. It is equal to the diameter of the circumscribed circle. 197. The area of a regular polygon is equal to one half the product of the apothem and perimeter. 220 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. Denoting the area by A, the apothem by h, and the perimeter by p, write the formula for A in terms of h and p. 2. Draw a regular hexagon by using the radius of a circle as a chord six times, measure the apothem and side, and find the area. 3. How much surface is there in the hexagonal top of a jardiniere measuring 6" on a side ? In the case of the regular hexagon the apothem is always - V3 where a is the length of a side. 4. From the last problem show that the area of a regular hexagon in terms of its side a is A = - - V3. -i THE CIRCLE 198. In the circle ABC, ABC is the circumference, OC a radius, AB a chord, DE a diameter. PROBLEMS The circumference of a circle is equal to its diameter mul- tiplied by 3.1416 ( approximately}. 1. Denoting the circum- ference by c, the diameter by d, and 3.1416 by TT, write the formula for c in terms of TT and d. 2. Since d = 2 r } where r is the radius, write a formula for c in terms of TT and r. Find c in the following FIG. 135. CIRCLE 3. d = W". 4. d = 7'. problems (use TT 2 7 2 ) : 6. r = 3". 7. r=11.7'. MENSURATION 221 8. Find the circumference of a circular flower bed whose diameter is 20' 6". 199. The area of a circle is equal to ^ multiplied by the square of the radius. PROBLEMS 1. Denoting the area by A, 2 7 2 - by TT, and the radius by r, write (1) the formula for A in terms of TT and r; (2) the formula in terms of TT and d, the diameter. 2. Draw a circle, measure r, and find A. Fill out the blanks in the following tables : 3. 4. 1 10" 2 2' 6" i 25' 4" 4 30' 7" 6 50' 11" 1 7" 2 14" 3 60' 6" 4 2' 4" 5 r 7f" 5. A circular flower bed has a diameter of 25' 8". Find the area. 200. Length of an arc. An arc has the same ratio to the circumference that the angle at the center has to 360. Q/J Hence the arc AB = ^ - of the circumference 6bO 36 Trd PROBLEMS 1. Denoting the length of the arc by I and the number of degrees in the angle by N, write the formula for I in terms of N and d. 222 SHOP PROBLEMS IN MATHEMATICS 2. The minute hand of a clock is 4' long. How many inches does its extremity move in 20 minutes ? 3. Find the length of that part of a circular railway curve with a radius of one half a mile that subtends an angle of 36. 4. Solve the formula in Problem 1 for N. 5. Solve the formula in Problem 1 for d. 6. A pendulum swings through an angle of 36. The end describes an arc of 12 in. Find the length of the FlG - 136 - ANGLE -, T OF 36 pendulum. 201. Area of a sector. A sector is that part of a circle which is bounded by two radii and an arc. In Fig. 137 A OB is a sector. The area of a sector is equal to one half the product of its radius by its arc. It has the same ratio to the area of the circle that the central angle has to 360. PROBLEMS 1. Denoting the area of the sector by A and the number of degrees in the central angle by N } write the formula for A in terms of N and r. 2. A sector of a circle has a central angle of 27, and the radius of the circle is 30'. Find the area of the sector. 3. Find the area of the sector of the end A 1f"B of the circular head of a tank, if the length of FIG. 137. the arc is 28" and the diameter of the head of SECTOR AND ., , . . SEGMENT the tank is o r . 202. A segment of a circle is the part of the circle bounded by an arc and its chord. MENSURATION 223 203. The area of a segment ACE (Fig. 137) is equal to the difference between the area of the corresponding sector OACB and the area of the triangle OAB. These areas may be found by sections 201 and 190-192. In the following problems it will be useful to recall the geometrical principle that in a right triangle having one acute angle equal to 30, the side opposite the 30 angle is one half of the hypotenuse. B PROBLEMS 1. If angle A = 30 and EC = 1', then AB = 2'. AC can be found by sections 44-46. / 30' 2. If the right tri- , FIG. 138. 30 RIGHT TRIANGLE angle has one angle equal to 45, the triangle is isosceles; p e.g. if EC = 8, and Z A = 45, A C = 8. AB may then be found. Find the area of the segment in each of the following circles : 3. r = 10", Z = 60. /^$ /\ a 4. r = 8", Z O = 90. 8 5 r _ 2' 6" /- = 120 ^ 1 ' ' 204. An annulus or ring is the area, comprised between two circles that have the same center. Its area is the difference between the areas of the two circles and "may be denoted by A = TT (R 2 - r 2 ) = 7r(R + r) (R - r), where R and r are the respective radii of the two circles. 224 SHOP PROBLEMS IN MATHEMATICS PROBLEMS 1. The inner and outer diameters of an annulus are 8" and 10" respectively. Find the area of the annulus. 2. What is the area of the annulus be- tween two circles 9" and 7" in diameter ? 3. A circular grass plot 20' in diameter is surrounded by a walk 4' wide. Find the area of the walk. FIG. 140. PRISM 4. A circular pond 75' in diameter is sur- rounded by a path 10' wide. Find the area of the path. THE PRISM 205. The volume of a prism is equal to the product of its base and altitude. In the case of the rectangular block, the volume is the product of the three edges that represent the length, breadth, and height. PROBLEMS 1. Denoting the volume by V and the three dimensions by a, b, and c, write the formula for V in terms of a, b, and c. 2. The total area of a rectangular block is the sum of the areas of the six faces. FlG - 141. Denoting the total area by T, find T in RECTANGULAR , , BLOCK terms ot a, b, and c. 3. Show that the length of a diagonal of a rectangular block is Va 2 + b 2 + c 2 . 4. The volume of a cube is equal to the cube of an edge. Derive the formula for the volume of a cube from that of Problem 1 by making a = b = c. MENSURATION 225 5. Derive the formula for the total area of a cube. 6. Derive the formula for the diagonal of a cube from Problem 3 by making a = b = c. 1. Find the volume of a rectangular block 4" x 2" x 8". 8. Find the total area of the surfaces of this block. 9. What is the length of its diagonal ? 10. Find the volume of a cube 6' on an edge. 11. Find the total surface area of this cube. 12. Find the length of its diagonal. FlG - 142 - CUBE 13. How many gallons of water are contained in a tank in the form of a rectangular parallelepiped whose dimensions are 15' x 22' x 8' 6" ? (231 cu. in. = 1 gal.) 14. How many cubic yards of earth must be removed in digging a canal 1 mi. 1600 ft. long, 100 ft. wide, and 18 ft. deep ? 15. Find the volume of a brick measuring 8"x 4" x 2". 16. A room measures 15' x 20' x 10'. How many cubic feet of air does it contain? How many square yards of surface on the walls and ceiling? 17. In the room of Problem 16 find the length of a piece of string that will just stretch from one corner of the ceil- ing to the opposite corner of the floor. 18. How long an umbrella will go into a trunk measuring 32" x 17" x 21", inside measure, (a) if the umbrella is laid on the bottom ? (&) if it is placed diagonally between oppo- site corners of the top and bottom of the trunk ? 19. A railroad embankment one quarter of a mile long has a section in the form of an isosceles trapezoid (i.e. a trapezoid having the nonparallel sides equal) 300' at the base, 200' at the top, and 25' high. How many cubic yards of earth does the embankment contain ? 226 SHOP PROBLEMS IN MATHEMATICS 20. How deep must a cistern be to contain 800 gal., if its length and width are respectively 8' 4" and 4' 6" ? 21. A cubical tank 14" deep, open at the top, is to be lined with copper. How many square inches of copper must be used, not allowing for overlapping ? FIG. 143. STEEL PLATE 22. How many gallons of water will the tank contain ? 23. A cubical box has an edge measuring 4' 4", outside measure. Find the cost of painting the outside, including the top, at 10/ per square yard. 24. Find the weight of a box and lid made of wood -J-" thick, measuring on the outside 3 f x 2' x 2', if a cubic foot of the wood weighs 35 Ib. 25. A steel plate 2" thick is of the D form and dimensions of Fig. 143. If a cubic inch of steel weighs .28 Ib., find the weight of the plate. FIG. 144. PYRAMID THE PYRAMID 206. A pyramid is a solid whose base is a polygon and whose sides are triangles having a common vertex. The altitude of a pyramid is the distance from the vertex to the base. The volume of a pyramid is one third of the product of the area of the base by the altitude. MENSURATION 227 207. A regular pyramid is a pyramid whose base is a regu- lar polygon and the foot of whose altitude coincides with the center of the base. The slant height of a regular pyramid is the altitude of a lat- eral face or the distance from the vertex to a side of the base. 208. The lateral area of a regular pyramid is one half the product of the perimeter of the base and the slant height. 209. The frustum of a pyramid is that part of the pyramid included between the base and a plane parallel to the base. 210. Denoting the volume of the frustum of a pyramid by V, the altitude of the frustum by h, the upper base by b', and the loiver base by b, FIG. 145. REGULAR PYRAMIDS 211. The lateral area of the frustum of a regular pyramid is equal to one half the sum of the perimeters of the upper and lower bases multiplied by the slant height of the frustum. PROBLEMS 1. Denoting the volume of the pyramid by V, the base by b, and the altitude by //, write the formula for V in terms of b and h. 2. Denoting the lateral area of the pyramid by Z, the perimeter of the base by p, and the slant height by I, write the formula for L in terms of p and I. Find the lateral surface and volumes of regu- p YRAMID lar pyramids with the following dimensions : 3. Base a square containing 64 sq. ft., altitude 8 ft. FIG. 146. 228 SHOP PROBLEMS IN MATHEMATICS 4. Base a square 13" on a side, altitude 10". 5. Base an equilateral triangle 6" on a side, altitude 8". 6. Each face, including the base, an equilateral triangle 10" on an edge. 7. The base a square 15" on a side, lateral edge 16". 8. Base regular hexagon 6" on a side, altitude 8". /Tl 9. Find the volume of a pyramid / a. whose base is a rhombus 6" on a side, /(>o' \ and whose height is 6", if one angle _ FIG. 147. 60 EHOMBUS oi the rhombus measures 60 . In a right triangle where one acute angle is 60, the side opposite Vs this angle is equal to the hypotenuse multiplied by Thus in the /7> 2 figure a = 6 x = 3V3. 10. The Great Pyramid in Egypt is 480f ft. in height, and its base is a square measuring 764 ft. on a side. Find its volume in cubic yards. Find the volume of the frustums of pyramids having the following dimensions : 11. Lower base a square 4' on a side, upper base 2' on a side, altitude 3'. 12. Lower base an equilateral triangle 8" on a side, upper base 5" on a side, altitude 4". 13. Find the lateral area of the frustum of a square pyramid the edge of whose lower base measures 10", the edge of the upper base 7", and the slant height 6". THE CYLINDER 212. A cylinder of revolution or a right circular cylinder is the solid generated by the revolution of a rectangle about one of its sides as an axis. MENSURATION 229 FIG. 148. CYLINDER The rectangle ABCD revolving about AB as an axis generates the cylinder, as in the figure. The bases of the cylinder are evidently circles. 213. The volume of a cylinder of revolution is equal to the product of the base and altitude. 214. The lateral surface of a cylinder of revolu- tion is equal to the product of the circumference of the base by the altitude. 215. The total surface of a cylinder of revolu- tion is equal to the area of the bases added to the lateral area. 216. A frustum of a cylinder is the part in- cluded between a base and a plane oblique to the base and cutting the lateral surface. 217. The volume of a frustum of a cylinder of revolution is equal to the product of the base and the altitude to the base from the center of the oblique section. A R -4- C* T) In the figure the altitude is equal to - A 218. The volume of a hollow cylinder is the difference between the cylinders of its out- side and inside diameters. PROBLEMS 1. If V is the volume of the cylinder, r the radius of the base, and a the altitude, write the formula for Fin terms of r and a. FIG. 149. FRUSTUM *'..,,. OF A CYLINDER 2. If L is the lateral surface of the cylin- der, write the formula for L in terms of r and a. 3. Derive a formula for T, the total surface in terms of a and r. 4. If V is the frustum of a cylinder, write a formula for V in terms of r, h', and h" (see Fig. 149). 230 SHOP PROBLEMS IN MATHEMATICS Find the volumes, lateral areas, and total areas of cylin- ders of revolution having dimensions as follows : 5. Diameter of base 2', height 18". 6. Diameter of base 18' 6", height 50". 7. Find the volume of the frustum of a cylinder the radius of whose base is 25" and whose altitude from the center of the oblique section to the base is 30". 8. What is the weight of a hollow cylinder of cast iron 20' long and 1" thick whose outside diameter is 3' ? (1 cu. in. of cast iron weighs .26 Ib.) 9. How many gallons (231 cu. in.) of water will a cylindrical tank hold that is 15' in diameter and 25' high ? 10. How much will it cost to paint the tank in Problem 9 at 15/ per square yard ? 11. How many square inches are there in the curved surface of a wire -J-" in circumference and 100' long ? 12. Find the number of square yards in the surface of a smokestack 40' high and 3' in exterior diameter. 13. How many cubic feet of water are contained in 100 yd. of pipe whose internal diameter is 2" ? 14. If a cubic foot of marble weighs 173 Ib., find the weight of a cylindrical marble column 20' high and 25" in diameter. 15. Water is poured into a cylindrical reservoir 25' in diameter at the rate of 300 gal. a minute. Find the rate (number of inches per minute) at which the water rises in the reservoir. 16. Find the cost of digging a well 100' deep and 5' 6" in diameter at an average cost of $3 per cubic yard. 17. Find the weight of a copper tube in. in outside diameter, .05 in. thick, and 6 ft. long, if 1 cu. in. of copper weighs .318 Ib. MENSURATION 231 18. The larger diameter of a hollow cast-iron roller is 1' 10", the thickness of the metal is If", and the length 6'. Find the weight of the roller. THE CONE 219. A cone of revolution (usually called a cone) is the solid generated by the revolution of a right triangle about one of its legs as an axis. In the figure the re volution of the triangle A CB about A C as an axis generates the cone. The base is a circle, the altitude is h, and the slant height is I. 220. The volume of a cone is one third the product of the base and the altitude. 221. The lateral surface of a cone is equal to one half the product of the circumference of the base by the slant height. 222. A frustum of a cone is the part of a cone included between the base and a plane parallel to the base. 223. The volume of a frustum of a cone of revolution is equal to FIG. 150. FRUSTUM OF A CONE FIG. 151. CONE where h is the altitude and b, b' are the bases of the frustum. 224. The lateral surface of a frustum of a cone of revolution is equal to one half the product of the slant height by the sum of the circumferences of the bases. PROBLEMS 1. If V is the volume of a cone of revolution, r the radius of the base, and h the altitude, write the formula for V in terms of r and h. 232 SHOP PROBLEMS IN MATHEMATICS 2. Denoting the lateral surface of the cone by L and the slant height by I, write L in terms of r and L 3. Write the formula for L, the lateral area of a frustum of a cone of revolution, in terms of r, the radius of the upper base ; /', the radius of the lower base ; and I, the slant height. Find the volumes and lateral areas of cones of revolution having the following dimensions : Diameter of base Altitude Slant height 4. 6" 10" 5. 5' 4' 6. 14' 6" 12' 4" 7. 28" 25" 8. 8J W 6f" 9. Find the volume of the frustum of a cone of revolu- tion having 10" for the diameter of the lower base, 7" for the diameter of the upper base, and a height of 4". 10. Find the lateral area of the frustum of a cone of revolution the diameters of whose upper and lower bases are 4' and 8' respectively, and whose slant height is 3' 6". 11. A conical spire has a slant height of 60' and the perimeter of the base is 50'. Find the lateral surface. 12. Find the weight of a solid circular cone of cast iron whose height is 8" and the diameter of whose base is 5". 13. What length of canvas f yd. wide will be required to make a conical tent measuring 12' high and 16' in diam- eter at the base ? 14. Find the volume of a cone with an elliptical base, major axis 5", minor axis 3", and height 8". The area of an ellipse is equal to Trafc, where a and b are the semiaxes. MENSURATION 233 15. A tin funnel is to be constructed with the dimensions as follows : 10" in diameter at the top, 1" in diameter at the bottom, slant height 14", spout 1" in diameter at the larger end and -J-" at the smaller end, slant height 4". Allowing y on the length and width of each piece for locks, find the amount of tin required for the funnel. 16. A cone-shaped hood for a chimney is 30" in diameter and 13" high. Allowing 1" on the side for locking, what surface of metal is required for the hood ? 17. A copper teapot is 9|" in diameter FlG 152 S PHERE at the bottom, 8" at the top, and 11" deep. Allowing 42 sq. in. for locks and waste, how much metal is required for its construction without the cover ? THE SPHERE 225. A sphere is a solid generated by the revolution of a circle about a diameter. 226. The volume of a sphere is equal to V = - Trr 3 , where r is ike radius. or V = ) where d is the diameter. 6 227. The surface of a sphere is equal to FIG. 153. SEGMENT OF A SPHERE or S = 228. A segment of a sphere is the part included between two parallel planes. 229 . The volume of a spherical segment is - (Trr 2 + Trr' 2 ) + > where h, r, and r 1 are the dimensions represented in the figure. The volume of a spherical segment is thus equal to that of two cylinders (-Trr 2 and -Trr'' 2 ) and a sphere ( ) The cylinders have \2 2 / \ 6 / 234 SHOP PROBLEMS IN MATHEMATICS a common altitude, -, and the radii of their bases are r and r' re- spectively. The sphere has a diameter of h. 230. The volume of a spherical segment of one base is equal to Trk 2 I r - 231. A zone is the portion of the surface of a sphere in- cluded between two parallel planes. 232. Its area is 2irrh where r is the radius of the sphere and h the altitude, as in the figure above. 233. Ring. If the circle whose center is O (Fig. 154) be re- volved about AB as an axis, the figure formed is a ring. ED is the internal radius, CD the external A radius, and OD the mean radius. The ring may be considered as a cylinder bent in the form of a circular arc till c the ends meet. The mean length of this cylinder is evidently the circumference of the circle whose radius is OD. oo/i mi, * ' -i i FIG. 154. CIRCLE GEN- 234. The area of a ring is equal to ERATING A RlNG the product of the circumference of a cross section and the mean length. Denoting this area by A } the radius OE of the cross section by r, and the mean radius OD by R, 235. The volume of a ring is equal to the product of the area of a cross section and the mean length. Denoting this volume by V, 236. When the cross section of the ring is a rectangle the volume may be obtained by multiplying the area of a cross section by the mean length which is the circumference of a circle whose diameter is one half the sum of the inside and outside diameters of the ring. MENSURATION 235 PROBLEMS Find the volumes and surfaces of spheres having the following diameters : I. 7". 2. 4|". 3. 8' 6". 4. 10.2'. 5. 8000 miles. 6. Find the weight of a sphere of cast iron 5" in diameter. (1 cu. in. of cast iron weighs .26 Ib.) 7. How many cubic inches of water will a hemispherical bowl contain whose diameter is 15" ? 8. What is the weight of a sphere of steel 9" in diameter, if 1 cu. ft. of water weighs 1000 oz. and the specific gravity of steel is 7.8. 9. Show that the area of the curved surface of a hemi- sphere is twice that of its plane surface. 10. Find the weight of a spherical shell of cast iron whose external diameter is 8" and whose thickness is ^". II. What must be the diameter of a hemispherical basin to hold 1 gal. of water ? 12. How many yards of material 3.25 ft. 'wide are neces- sary to make a spherical balloon containing 1500 cu. ft. of gas ? 13. Show that a sphere is equal to two thirds of a cylin- der of the same diameter and height. 14. A copper clothes boiler has semicircular ends. Its extreme length is 20", its width 13", and its depth 10". If we allow 1^-" on the width of the side piece for lock and top roll, 1" on the length for the side lock, and ^" all around the bottom piece for the lock, how much copper is required to construct the boiler ? 15. Find the volume of a spherical segment where r = 10", r' = 4", and h = 6". 236 SHOP PROBLEMS IN MATHEMATICS 16. A bowl is in the form of a spherical segment of one base, with r = 14" and k = 5". How many gallons will it contain ? 17. How many square feet of surface in a zone whose altitude, h, is 12', on a sphere whose radius is 21'? 18. Find the area and volume of a ring, with a circular cross section, whose mean radius is 11 in. and whose in- ternal radius is 10 in. 19. The cross section of a solid ring of wrought iron is a circle of 6" radius and the inner radius of the ring is 4'. Find (1) the area of the surface; (2) the volume of the ring; (3) its weight if 1 cu. in. of wrought iron weighs .27 Ib. 20. The cross section of the rim of a cast iron fly wheel is a rectangle measuring 4 x 8 in. ; the inner diameter of the rim is 4'. Find (1) the area ; (2) the volume ; (3) the weight of the rim. CHAPTER XVII FORMULAS In this chapter sufficient work in algebra is given to enable a student to evaluate the common formulas used in the shop. 237. Law of precedence of signs. When the signs -f, , X, and H- occur in the same formula, the operations of multiplication and division must be performed before those of addition and subtraction. EXAMPLE. Find the value of 8 x 2 4 -;- 2. 238. Letters as numbers. The use of letters for numbers has been illustrated in sections 18, 47, 60, in Chapter XVI, and in other parts of the book. By referring to section 18 the student will observe how the same letter may be used to represent a certain number in one problem and another number in another problem. Thus in section 18, problem 1, I 14, and in problem 2, I = 12. 239. Factors. The factors of a number are the numbers that when multiplied together produce the number. Thus 2, 3, and 5 are the factors of 30. 240. Three methods of indicating multiplication. In using letters for numbers the product of a and b may be indicated by any one of the following methods : (1) a x b. (2) a b. (3) ab. 237 288 SHOP PROBLEMS IN MATHEMATICS 241. Terms. A term is a number whose parts are not separated by a plus or minus sign ; e.g. 6 ab is a term. a -f b is a number of two terms. a -f b c is a number of three terms. 242. Exponent. The exponent of a number is a number placed at the right and a little above another number called the base, and indicates how many times the base is to be taken as a factor. Thus 5 2 = 5 x 5 = 25. a 8 = a X a X a. 243. Coefficient. Any factor of a term is a coefficient of the rest of the term ; e.g. in 4 a, 4 is the coefficient of a ; in ab, a is the coefficient of b, and b is the coefficient of a; d + d' f in TT ? TT is the coefficient of Unless otherwise specified, the numerical coefficient is called the coefficient, as in 4 a. 244. Rule for evaluating an algebraic expression. (1) Substitute for every letter its numerical value. (2) Find the value of every term separately, performing the multiplication and division that is indicated. This follows the law of precedence of signs in section 237. (3) Combine the terms by performing the addition and subtraction that is indicated. (4) If a parenthesis occurs in the expression, its nu- merical value must be found before combining with num- bers outside this parenthesis. The meaning of the formulas used in the following examples may be found by referring to the sections indicated at the right. FORMULAS 239 Iwt EXAMPLE 1. Evaluate b = (Sect. 18) when I = 15', w = 1", and t = 2". 15-7-2 ~I2~ = 17*. EXAMPLE 2. Evaluate h = 2 d + a (Sect. 48) if d = 55' and a = 6'. h = 2 x 55 + 6 = 110 + 6 EXAMPLE 3. Evaluate F=^(7rr 2 +7rr' 2 )+^ (Sect. 229) if h = 4", r = 8", r' = 6", and TT = 2 7 2 . 32 = 4_4_Q 0- = J. = 662^- cu. in. (b) This problem may be worked as follows 32 200.+%^ 3 632 TT 3 s. a . cu. n, 240 SHOP PROBLEMS IN MATHEMATICS ORAL EXERCISE Evaluate the following formulas with the values indi- cated : 1. A = ab, if a = 10", b = 3". (Sect. 188) 2. h = ~, if d = 40', h' = 5', d 1 = 4', (Sect. 49) b-a,tfl = 10", a = 5". (Sect. 190) 4. A = a (Mjr^V if = 2", ft - 5", ft' - 7". (Sect. 193) 5. C = Trd, if TT = - 2 7 2 -, d = 21". (Sect. 198) 6. .4 = ?rr 2 , if TT = - 2 7 2 , r = 7". (Sect. 199) 7. F = aftc, if a = 3", ft - 5", c = 4". (Sect. 205) 8. d= Va 2 + ft 2 + c 2 , if a = 3", ft = 4", c = VTi". (Sect. 205, Problem 6) 9. V = ^B.h,\iB= 12", h = 4". (Sect. 220) 10. 5 = 4 Trr 8 , if TT = - 2 7 2 , r = 7". (Sect. 227) 11. F = ^-^.7rAr, if d=3 f , *! = !', ^=6. (Sect. 65) 12. ^ = ?rr 2 A, if r = 4", 7^ = 14". (Sect. 213) WRITTEN EXERCISE Evaluate the following formulas with the values indi- cated : 1. 5 = | gt*, ifg = 32.15, t = 4.5. (Sect. 54) 2. c 2 = a 2 + ft 2 , if a = 7", ft = 9". (Sect. 44) 3. a 2 = c 2 - ft 2 , if c = 10", ft = 8". (Sect. 46) 4. 7) 2 = ^ + d ^ ifd== 6 rfi = 7 ^ ^ gect 45 ) 5. d* = D 2 - d 2 , if D = 12", ^ = 9". (Sect. 46) 6. P = Awh, \i A = 125, w = .25, A = 10.5". (Sect. 75) FORMULAS 241 rj 8 = L, if 5 = 40, n = 25, TV = 75. (Sect. 95) ?i 8. F = - if D = 3", JV = 300. (Sect. 99) 9. l = ^-. 7r d,ifN= 22, d = 12". (Sect. 200) 10. A= 1 j-- Trr 2 , if N = 32, r - 15". (Sect. 201) ooU 11. C = c X E.P.M., if c = 18", R.P.M. = 300. (Sect. 60) 12. I = D ~t D ' TT + 2rf, if D = 3.5', D f = 2.5', d = 15'. . (Sect. 62) 13. F = - TTN, if cZ-24", d 1= =26", JV=6. (Sect. 65) 14. H.P. = ?r ? if P=8 lk ' l = ^"> A - 300. (Sect. 154, problem 10) 15. A = -Vs(s a)(s b)(s c), if a = 24 rd., ft = 30 rd., c = 36 rd. (Sect. 192) 16. ,4 =TT(R*- r 2 ), if ,R = 16", r = 12". (Sect. 204) 17. A=$(p+p l )8,tip = 20" > p l =15",8 = &'. (Sect.211) 18. V = , if d = 7". (Sect. 226) 19. G= l ' 0034 I, if ^ = 20", d 1 = 22", 1=30". (Sect. 69) 20. V=ih(b + b' + V^ 7 ), if A = 5", 6 = 7", 6' = 9". (Sect. 210) 245. Transposition of terms. If a line 12" long were divided into two parts 8" and 4" long respectively, the length of the line could be expressed by 8" + 4". Evidently (1) 12" = 8" -f 4". (2) 12" -8" = 4". 242 SHOP PROBLEMS IN MATHEMATICS If the length of the line were a and the two parts re- spectively b and c, instead of (1) we should have (3) a = b + c, and instead of (2) we should have (4) a -b = z. Observe that b has been changed from the right-hand side of the equation to the left-hand side by changing its sign from + to . From an inspection of (3) and (4), evidently b could be changed back to the right-hand side of the equation by changing the sign from to -f- . This changing from one side of the equation to the other is called transposing. Hence a term may be transposed from one side of the equation to the other by changing its sign. EXAMPLE 1. Solve the formula a -\- b c for a, i.e. find the value of a in terms b and c. (1) a + b = c. Transposing b to the right-hand side of the equation by changing its sign, we have (2) a = c b. EXAMPLE 2. Solve the formula 2 s = a -f- b + c for b. When no sign is given to a term, as in the case of 2 s in this example, the sign is understood to be + . (1) 2 5 = a + b + c. Transposing a and c, (2) 2 s - a - c = b, which is evidently the same as (3) I = 2 s - a - c. FORMULAS WRITTEN EXERCISE 243 Given the formula Solve for For meaning of formula see section 1. h= 2d+a 2d 48 2. k=2d+a a 48 3. c 2 = a 2 + V a 2 44 4. S 2 = S 2 + S 2 s 2 44 5. p = 2a + 2b 2 a 189 6. 2m*=b+b' b 193, note 7. p = a -f b + c a 192, prob. 5 8. d 2 = D 2 - d 2 D 2 46 9. d 2 = D 2 - d 2 4 46 10. , D + D' 22 2d 62 2 ' 7 246. How to find the formula for 6, if ab = c. If the cost of 3 apples == 6^, then the cost of an apple = / = 2/. Kepresenting the number of cents in the cost of the apple by a, In the same way, a = = 2. ab = c, c if In this equation a is the coefficient of b, and the solution consists in dividing both sides of the equation by the co- efficient of the unknown number. In a similar way, dividing ab = c by b, we have a = - 244 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 1. If A = ab (sect. 188), find the formula for a. ^ Dividing by b, = a, A or a = EXAMPLE 2. If DS = ds (sect. 58), solve for S. Dividing by D, S == EXAMPLE 3. The formula for the area of an ellipse is A = Trab where a and b are the semiaxes. Solve the formula for b. A Trab. Dividing by Tra, = b, Tra or b = WRITTEN EXERCISE Given the formula Solve for For meaning of formula see section 1. A = ab a 188 2. P = br b 167 3. P = br r 167 4. A =2 Trrh h 214 5. V = Trr^a a 213 6. L = jrda d 214 7. V=abc c 205 8. A = ma a 193, Prob. 2 9. N=P d D Pa 139 10. SNN-iNz = sun-in* S 95, Prob. 2 FORMULAS 245 247. Clearing of fractions. If we wished to make f a whole number, we could do it by multiplying by 3. The operation would be as follows : 2 v o _ o ? In the same way we may make the following fractions into whole numbers by multiplying by the number indicated: If we wish to change the formula A = y (Sect. 190) so that there shall be no fraction in it, we may multiply both sides of the equation by 2 and obtain 2^=yx?, or 2 A = bk. The advantage of this operation is now apparent, since we may solve for b or li by section 246. EXAMPLE 1. In the formula V = (sect. 206) clear of fractions and solve for h. V J*L ~ 3' Multiplying by 3, 3 V = bk. 3 V Dividing by b } = h, 3V or h = b 246 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 2. Solve the formula P c (sect. 146) for N. P = ^. c N Multiplying by N, NP C = TrD. Dividing by P c , N = - EXAMPLE 3. Solve I = TT D "t D - + 2d (sect. 62) for d. . Multiplying by 2, 21 = TT(D + D 1 ) + 4 d. Transposing, 2 I - TT(D + D 1 ) = 4 d. Di viding by 4, 2* EXAMPLE 4. Solve = for TV (sect. 95). * N In this case we clear of fractions by multiplying by the product of the two denominators. S n x sN = X sN. s N SN = ns. 7?*? Dividing by 5, N = S 248. When the unknown number has an exponent 2 or 3. EXAMPLE 1. Solve A = a 2 for a (sect. 188). .1 = a 2 . Taking the square root of both members of the equation, or FORMULAS 247 EXAMPLE 2. Solve V = a s for a (sect. 205, Problem 4). V=a s . Taking the cube root of both members, or a = The student should interpret the results of all transformed formulas. EXAMPLE 3. Solve V = 7rr 2 a for r (sect. 213). F = 7rr*a. Dividing by Tra, - = r 2 , or r 2 = Tra Taking the square root of both members of the equation, r = EXAMPLE 4. Solve V= 7J ^~ for d (sect. 226). _ : 6 Multiplying by 6, 6 V = Dividing by TT, - = d 3 , 6V or d* = 7T (or Taking cube root, d = Jj --- ^ 7T In the following exercise the answers may be checked by reversing the operations and solving the resulting formula for the letter that was first given. The student should interpret the meaning of the given formula and the result- ing formula. 248 SHOP PROBLEMS IN MATHEMATICS WRITTEN EXERCISE Given the formula Solve for For meaning of formula see section 1. t _P c PC 143 2 2. A - kp A 197 2 3. h dh/ K 49 ~~ ri 1 * j. . K 2 n AT" K o. 2 a.V oo 6. P _w 10 177 W~ p P _w 7. W~p P 177 8. P _w W 177 9. s= 16. 08 * 2 t 54 10. c 2 = a 2 + & 2 a 188, Prob. 7 11. -4 d 199 4 12. I _ N .nd N 200 360 ' 13. S = 47ZT 2 r 227 14. 5 - 7T#> d 227 15. OD-D+ 2 Pd 138 Pd 16. JV" r 201 ~ 360' 7 17. F=27T 2 r 2 # R 235 18. T7- _ 2 ^' + h" T 217 2 19. ^1 = ?r(jR 2 - r 2 ) R 204 4 20. F = - TIT 3 r 226 3 CHAPTER XVIII SOLUTION OF RIGHT TRIANGLES BY NATURAL FUNCTIONS 249. The trigonometric functions. If B is a point on the side of the Z. EAD and BC is drawn perpendicular to AD at C, forming the right triangle ABC, - is called the sine of the angle A. c - is called the cosine of the angle A. G - is called the tangent of the angle A. - is called the cotangent of the angle A. Ct - is called the secant of the angle A. C - is called the cosecant of the angle*^4. * The sine, cosine, tangent, etc., are called trigonometric functions of the angle A . &e. The value of each of these ratios may be found approximately by measuring a, b, and c, and rinding the numerical value of the ratios. A ' b C c' ~ If B is taken at any other point FlG 155 on A E, such as B', it can be shown by measurement or by similar triangles that the ratios -> -? etc., will have the same values as before, though a, b, and c will vary. 249 250 SHOP PROBLEMS IN MATHEMATICS i EC 1 In Fig. 156, sine A = -.- = = .5. AJ3 A Since the sine does not vary for a given angle, if the angle is given, the sine is determined. The sine, cosine, tangent, co- ,B' tangent, seoant, and cosecant are called natural functions of the angle, since for every J change in the angle there is a change in the value of the sine, cosine, etc. The table on page 279 gives the values of these functions for angles from to 90. In machine work, engineering, and surveying it is neces- sary to know how to use these functions in the solution of triangles. EXAMPLE 1. Given A = 35 30', c = 32 ; to find B, a, 1>. B = 90 - 35 30' = 54 30'. a c. a c sine A. From the tables, sin A = .5807 c = 32 11614 17421 - = cos A. c cos A = .8141 c= 32 a = 18.5824 = 18.6 16282 24423 I = 26.0512 Check. 32 2 = 1024 = 346 + 678 1024 = 1024 SOLUTION OF RIGHT TRIANGLES 251 EXAMPLE 2. Given A and a ; to find B, c, and b. = 90 - A. a a = sin A ; c = c sin. A - = cos A : b = c cos A. c Check as in Example 1. EXAMPLE 3. Given A and b ; to find B, c, and a. B = 90 - A. b b - = cos A : c c cos A a = sin A ; a = c sin A. Check as before. EXAMPLE 4. Given a and c; to find A, B, and &. a sin A = - c & = c sin ^4 . Check as before. EXAMPLE 5. Given a and b ; to find A, B, and c. a tan 4 = B = 90 - ,4. a c = Check as before. WRITTEN EXERCISE To solve a triangle means to find the unknown sides and angles. Solve and check the right triangles given : 1. ,4=42, c = 92.4. 2. A = 25, a = 56.1. 252 SHOP PROBLEMS IN MATHEMATICS 3. A =50, 5 = 17.3. 4. a 75, c = 85.5. (Find angles correct to degrees.) 5. a = 82, 5 = 64. (Find angles correct to degrees.) 6. = 74 30', a = 142.3. To find the sine of 74 30' ; to the sine of 74 add half the difference between the sine of 74 and 75. Thus the sine of 74 is .9613, and the sine of 75 is .9659. One half the difference is .0023, which, added to .9613, gives .9636 for the sine of 74 30". Similarly, for an angle of 15' use one fourth of the difference, and for 45' use three fourths. 7. B = 15 15', c=24.6. 8. yl = 6445', 5 = 18.4. 9. a = 16.2, 5 = 22.5. (Find angles correct to 15'.) 10. = 83 30', 5 = 125.7. 11. A disk is 10.5" in diameter. Find the distance neces- sary to set a pair of dividers in order to space off 5 sides (i.e. to inscribe a regular polygon of 5 sides in the disk). 12. Find the number of gallons of oil in a cylindrical tank 20' long, whose head has a diameter of 6', if a pole thrust through a hole in the top of the tank to the bottom is found, on being pulled out, to be covered with oil to the length of 2' 4". 13. From the preceding problem derive a formula for the volume of oil in a cylindrical tank in terms of d, the diam- eter of the head of the tank ; Z, its length ; and s, the part of the radius between the chord and the arc, i.e. the length of the pole that was wet with oil. 14. Derive a formula for the exact length of a belt around two pulleys in each of the following cases : (a) Open belt on pulleys of diameters D and d, the dis- tance between centers being a (see Fig. 38, p. 70). (5) Cross belt on the same pulleys (see Fig. 40, p. 71). For other problems involving trigonometry, see sections 147-149. CHAPTER XIX SHORT METHODS 250. Accuracy and speed. Most of the mathematical oper- ations that are used in the trades and in business are very simple, but in their use accuracy is absolutely essential and speed is desirable. Accuracy may be secured by practice and by checking all work. Speed may be improved by (1) a knowledge of short methods, (2) by noting the time required to perform the operation, and repeating the work, trying to reduce the time. 251. Addition. (1) When adding a column, try to see combinations that make 10, such as 6 and 4, 7 and 3, or 5, 4, and 1. (2) When numbers do not combine readily into 10, try for 9 or 11. (3) Practice adding columns that add to more than 199. (4) Check by adding in reverse order. (5) Learn to detect errors by dividing the column into several parts and adding each separately. (6) Time yourself when doing a problem. Repeat the addition with the same or a similar problem and try to reduce the time. (7) An error in copying, due to transposition of figures, is divisible by D. Thus 75 copied 57 makes an error of 18, which is divisible by 9. The proof of this is as follows for a number of two digits : (10 a + 6)-(o+ 106) = 9a-96 = 9(a -6). 253 254 SHOP PROBLEMS IN MATHEMATICS WRITTEN EXERCISE h k I m a 3478.92 9136.07 721.3 4278.91 b 6782.75 349.25 842.65 542.31 c 702.43 7865.50 384.99 465.78 d 685.21 2842.65 371.46 1942.85 e 4889.75 5600. 768.24 1392.99 f 4672.44 8929.48 3421.18 9276.49 9 3498.25 6854.35 8421.55 3498.46 Write in a column each combination of rows or columns indicated below, add, and check by adding in reverse order. I. a + b + c. 6. k + I. 2. d + e+f+g. 3. a -f- c -f- & + g. 4. i 4. d + /. 5. 7. k + m. 8. h + k -f I. 9. k + I + m. 10. A + k + Z 252. Multiplication. (1) To multiply by 10 move the deci- mal point one place to the right ; e.g. 25.67 x 10 = 256.7. (2) To multiply by 100 move the decimal point two places to the right ; e.g. 2.5673 x 100 = 256.73. (3) To multiply by 5 (V) divide by 2 and multiply by 10 ; e.g. 3842 x 5 = S^SL x 10 = 1921 x 10 = 19,210. (4) To multiply by 3 (y) divide by 3 and multiply by 10 ; e.g. 5670 x 3J = Mp X 10 = 1890 x 10 = 18,900. (5) To multiply by 2^ (-L-) divide by 4 and multiply by 10 ; e.g. 16 x 2 = ^ x 10 = 4 x 10 = 40. (6) To multiply by 50 (J-) divide by 2 and multiply by 100 ; e.g. 42 x 50 = -\ 2 - x 100 = 21 x 100 = 2100. SHORT METHODS 255 (7) To multiply by 25 ( L $&). divide by 4 and multiply by 100 ; e.g. 48 x 25 = V X 100 = 12 x 100 = 1200. (8) To multiply by 12| (i&) divide by 8 and multiply by 100 ; e.g. 24 x 12J = - 2 /- X 100 = 3 x 100 = 300. (9) To multiply by 33J (ijp) divide by 3 and multiply by 100 ; e.g. 27 x 33J = V x 100 - 9 x 100 = 900. (10) To multiply by 16 ( L $ a ) divide by 6 and multiply by 100 ; e.g. 42 x 16$ - V X 100 - 7 x 100 = 700. (11) To multiply by 14f (1^) divide by 7 and multiply by 100 ; e.g. 35 x 14f = *f x 100 = 5 x 100 = 500. (12) To multiply by 66| (f of 100) multiply by 100 and subtract of the product from it ; e.g. 39 x 66f = 39 x 100 - a ^QO = 2600. (13) To multiply by 333^ (1%-*) divide by 3 and multiply by 1000; e.g. 21 x 333 J - V X 1000 = 7000.. (14) To multiply by 166| (1%--) divide by 6 and multi- ply by 1000 ; e.g. 54 x 166 = -%* x 1000 = 9000. (15) To multiply by 125 (^%-^-) divide by 8 and multiply by 1000 ; e.g. 96 x 125 = * x 1000 = 12,000. (16) To multiply by 9 (10 1) multiply by 10 and sub- tract the multiplicand from the product ; e.g. 23 x 9 = 23 x 10 - 23 = 207. (17) To multiply by 11 (10 + 1) multiply by 10 and add the multiplicand to the product ; e.g. 34 x 11 = 34 x 10 + 34 = 374. (18) When one number is even and is between 10 and 20 multiply half of this number by double the other. e.g. 37 x 18 = 54 x 9 = 486. 256 SHOP PROBLEMS IN MATHEMATICS (19) To square a number of two figures with 5 in units place multiply the tens' figure by one more than itself and annex 25 to the product ; e.g. 65 2 = 6 x 7 with 25 annexed = 4225. The reason for this may be seen from the algebraic form (10a + 5) 2 . (10 a + 5) 2 = 100 a 2 + 100 a + 25 - 100 a (a + 1) + 25. (20) To square a mixed number with 1 for the fractional part, multiply the whole number by one more than itself and add J ; e.g. 7i 2 = 7 x 8 + J = 66J. This may be proved as follows : (a + i) 2 = a 2 + a + \ - a ( a + 1) + \. (21) When the numbers differ by one, to the square of the smaller add the smaller ; e.g. 15 x 16 = 15 2 + 15 = 240. (22) To multiply by a number that is one more or less than a multiple of 10, proceed as in the following examples : EXAMPLE 1. 28 x 19 = 28 x 20 - 28 = 560 - 28 = 532. EXAMPLE 2. 28 x 21 = 28 x 20 + 28 = 560 + 28 = 588. (23) To multiply by 15, multiply by 10 and add to the product of itself ; e.g. 48 x 15 = 480 -f 240 = 720. ORAL EXERCISE 1. 3.457x10. 10. 126 x 16}. 19. (35)*. 2. 23.6 x 100. 11. 357 x 14f . 20. ( 9 i) 2 - 3. 7234 x 5. 12. 27 x 66|. 21. 84 x 25. 4. 234 x 3f 13. 48 x 333^. 22. 34 x 19. 5. 92 x 2|. 14. 42 x 166}. 23. (75) 2 . 6. 7893 x 50. 15. 56 x 125. 24. ( 5 i) 2 - 7. 76 x 25. 16. 35x9. 25. (15i) 2 . 8. 272 x 12|. 17. 42 x 11. 26. 16x17. 9. 126 x 33i. 18. 13 x 14. 27. 34 x 15. SHORT METHODS 257 28. 53x 9. 35. 78 X 9. 42. (!' H ) 2 . 29. 674 x 5. 36. 54 X 66J. 43. 27 x 15. 30. 639 x 3i. 37. 84 x 33J. 44. 32 X 11. 31. 76 x 2f 38. 96 X 333J. 45. 16 X 19. 32. 92 x 25. 39. 64 X 125. 46. 27 x 33. 33. 66 x 16f. 40. 45 2 . 47. 72 X 25. 34. 504 x 14f . 41. m ff ! 48. 66 X 66|. 253. Division. (1) To divide by 10 move the decimal point one place to the left; e.g. 34.5 -f- 10 = 3.45. (2) To divide by 100 move the decimal point two places to the left ; ^ ^3 + i0 = 2.456. (3) To divide by 5 (-V -) multiply by 2 and divide by 10 ; . ., ... r 4olO X ^ oro e.g. 4315 -j- 5 = - -- = 863. (4) To divide by 3^ ( L-) multiply by 3 and divide by 10 ; e.g. 231 ^3^ = ^j^ = 69.3. (5) To divide by 2i (i) multiply by 4 and divide by 10 ; e.g. 52 + 2i = ^ = 20. 8 : (6) To divide by 50 multiply by 2 and divide by 100 ; r ' e .g. 378 + 50 = (7) To divide by 33J (iga) multiply by 3 and divide by 100 5 132 xS e.g. 132 + 33J = = 3.96. 258 SHOP PROBLEMS IN MATHEMATICS (8) To divide by 25 multiply by 4 and divide by 100; V- ^300.25 = (9) To divide by 16 multiply by 6 and divide by 100 ; ' e.g. 210. 16 = ^_2p = . 12 .60. (10) To divide by 14f multiply by 7 and divide by 100 ; e.g. 320 -,-l4=??_>p = 22 .4o. (11) To divide by 12| multiply by 8 and divide by 100; = 32.80. (12) To divide by 333i multiply by 3 and divide by 2321 x 3 e.g. 2321 H- 333J = ** * = 6.963. (13) To divide by 166 multiply by 6 and divide by e.g. 3200 + 166 = ~ = 19.20. (14) To divide a number by 66 add to the number of itself and divide by 100 ; 200 + 100 e.g. 100 t. >. ORAL EXERCISE 1. 4.532-10. 6. 427-5-50. 11. 30 - 12J. 2. 3.245-100. 7. 62 - 33^. 12. 2120 - 333J. 3. 3240 -f- 5. 8. 2300 -=- 25. 13. 700-3-166}. 4. 322-3i. 9. 306 -i- 16}. 14. 400 -5- 66}. 5. 33 -2|. 10. 41-f-14f. 15. 72 - 33J. SHORT METHODS 259 16. 350 -5- 25. 19. (55) 2 . 22. 36 x 25. 17. 200 -f- 16|. 20. (9) 2 . 23. 31 x 11. 18. 24-i-12. 21. 19x16. 24. 42x15. 25. Multiply 256 by 5. 33. 62 + 5. 26. Multiply 480 by 25. 34. 721 -5- 25. 27. Multiply 320 by 125. 35. 201 -f- 125. 28. Multiply 660 by 33. 36. 64 x 25. 29. Multiply 422 by 5. 37. 201| -*- 25. 30. Multiply 524 by 25. 38. 72 x 125. 31. Multiply 136 by 33. 39. (125) 2 . 32. Multiply 720 by 125. 40. (16i) 2 . CHAPTER XX THE SLIDE RULE ., 254. Use. The slide rule, also called the calculating rule and the guessing stick, is an instrument for rapid cal- culations in operations involving multiplication, division, squares and square root, cubes and cube root, natural trig- onometric functions, and logarithms of numbers. Addition and subtraction cannot be performed on this rule. 1 2. 3 A 5 6, 7 8 9. 10 20 30 40 50 60 70 GOX loo A 11 1 \ 1 11 1 R 'I 'I 1 1 1 I 1 1 2 3456 ' * 7 Q 9 10 20 30 40 JO 60 708030 MW 3 -4 5 t> 7 & 9 to u f 1 1 1 1 1 D l 1 3 4567 83/0 FIG. 158. SLIDE RULE The results are usually correct to three significant figures, which is sufficiently accurate for ordinary work. 255. Principles of its use. 10 2 = 100. Another method of stating this equation is The logarithm of 100 is 2. Similarly 10 8 = 1000. Hence the logarithm of 1000 is 3. Note that the logarithm is the exponent which is given to 10. 256. Law of multiplication. 100 x 1000 = 100,000, and 10 5 = 100,000, or log 100,000 = 5 -2 + 3 = log 100 + log 1000. 260 THE SLIDE RULE 261 It is a principle of loga- rithms, as may be seen in this example, that the loga- rithm of a product is the sum of the logarithms of the multiplicand and multiplier. Hence to multiply one num- ber by another add their log- arithms. 257 . Description of the slide rule. The slide rule consists of a grooved stick about 10" long, carrying a movable slide. There are four scales, A, B, C, and D, as in the fig- ure, divided proportionally to the logarithms of the num- bers that are marked upon them. Scale A is marked exactly like B, and scale C like D. EXAMPLE 1. Suppose we wish to multiply 2 by 3, us- ing the slide rule with the upper scales A and B. Move the slide to the right until 1 on scale B is under 2 on scale A. Then over 3 on B find the product 6 on A. Here we have added log 2 and log 3, obtaining log 6. If we use the lower scales C and D, move the slide to the right till 1 on scale C 262 SHOP PROBLEMS IN MATHEMATICS is above 2 on scale D. Then under 3 on C find Jbhe product 6 on D. 258. Finding numbers on the slide rule. ORAL EXERCISE 1. On scales A and B find the following numbers, check- ing by referring to Fig. 159 : 1.24, 1.85,' 1.97, 2.62, 9.45, 17.3, 8.68, 9.25. 2. On scales C and D find 1.24, 1.85, 2.62, and check by referring to Fig. 159. 259. When to use scales A and B. If only approximate results are desired and the numbers involved have a wide range of values use scales A and B. 48 r 1 35.5 ^1 \ 1 ! 1 7.4 FIG. 160. 4.8 x 7.4 = 35.5. SCALES A AND B. 260. When to use scales C and D. If greater accuracy is desired, use scales C and Z>. For shop work scales C and D are more often used than A and B. 261. Multiplication. EXAMPLE 1. Multiply 4.8 by 7.4 (see Fig. 160). Set 1 on B to 4.8 on A, and above 7.4 on B read 35.5 on A. 3.62 . I S= =" 2-35 S.5 FIG. 161. 2.35 x 3.62 = 8.50. SCALES C AND D EXAMPLE 2. Multiply 2.35 by 3.62 (see Fig. 161). Set 1 on C to 2.35 on D, and under 3.62 on C read 8.50 on D. EXAMPLE 3. Multiply 23.5 by 36.2. 23.5 x 36.2 = 2.35 x 3.62 x 100 = 8.50 x 100 (by Example 2) = 850. THE SLIDE RULE 263 EXAMPLE 4. Multiply 5.347 by 172.3. Taking each number correct to three significant figures, we multiply 5.35 by 172. Set 1 on C to 5.35 on D, and under 1.72 on C find 9.20 on D. 5.35 x 172 = 5.35 x 1.72 x 100 = 9.20 x 100 = 920. If the fourth significant figure is 5 or greater than 5, we increase the third figure by 1. If this occurs in both numbers, a more accurate result will be secured by increasing the third figure in only one num- ber. Thus 5.435 x 7.018 should be worked as 5.43 x 7.02. FIG. 162. 3.41 x 6.85 = 23.4. EXAMPLE 5. Multiply 3.41 by 6.85 (see Fig. 162). Set 1 on C to 3.41 on D, and 6.85 on C projects beyond the end of D. Set 10 on C to 3.41 on D, and under 6.85 on C find 2.34 on D. Setting 10 instead of 1 opposite the first number divides the result by 10. Hence we must multiply 2.34 by 10. The result is finally 23.4. In such a case the position of the decimal point may be found by inspection. If scales A and B are used and the slide projects to the left, use 100 as the index. EXAMPLE 6. Multiply .00246 x .0565. Using scales C and Z>, .00246 x .0565 = 2.46 x 10~ 3 x 5.65 x 10~ 2 = 2.46 x 5.65 x 10- 5 = 13.9xlO- 5 = .000139. 10-3 = : J_ 10 3 1000 264 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 7. Another method of obtaining the decimal point in Example 6. .00246 x .0565 is roughly .002 x .06 = .00012. From this we see that the significant figures 139 should be read .000139. EXAMPLE 8. Find the product of 4,2 x 65 x .34 x 2.6. Using scales C and D, set 10 on C to 4.2 on D and bring the runner to 6.5 on C. Bring 1 on C to the runner line, then the runner to 3.4 on C, 10 on C to the runner. Under 2.6 on C read the result, 241, on D. The position of the decimal point is determined by a rough calculation. 4.2 x 65 x .34 x 2.6 is roughly 4 x 60 x .5 x 2 = 240. Hence the result is 241. ORAL EXERCISE Using scales C and D, find the value of : 1. 2.32 x 3.14. 8. 4.51 x 7.94. 2. 3.78 x 5.26. 9. .00382 x .00475. 3. 1.79 x 3.25. 10. 9.84 x .000262. 4. 8.62 x 3.44. 11. 4673 x 842.5. 5. 2.3 x 3.16. 12. .641 x .032. 6. 32.5 x 25.2. 13. 8.2 x 32 x .45 x 6.4. 7. 6.375 x 281.3. 14. .75 x 22 x 6.5 x 8.65. 15. Find the circumferences of circles having diameters of 3', 4.5', 15". C = TT D. Set 1 on B to TT on A and above 3, 4.5, and 15 on B, read the circumferences on A. 16. Find the lateral area of a cylinder whose length is 16 f and the diameter of whose base is 7.5'. Lateral area = length x TT x diameter of base. THE SLIDE RULE 265 Using scales A and B, find values of : 17. 7.82 x 0.48. 19. 13.4 x 8.42. 18. 11.5 x 8.35. 20. 6.76 x 14.5. 262. Division. Since division is the reverse process of multiplication, it may be performed on the slide rule by reversing the operations for multiplication. EXAMPLE 1. Divide 6 by 3. Set 3 on B under 6 on A and over 1 on B, read the quotient 2 on A . In finding 6 on A it may be convenient to use the runner. EXAMPLE 2. Divide 35.5 by 4.8. Set 4.8 011 B to 35.5 on A and over 1 on B, read the quotient 7.4 on A. EXAMPLE 3. Divide 8.5 by 2.35, using scales C and D. Set 2.35 on C to 8.5 on D and under 1 on C, read the quotient 3.62 on D. EXAMPLE 4. Divide 850 by 23.5. This is the same as Example 3 except that the' decimal point is evidently after the 6, making the quotient 36.2. EXAMPLE 5. Divide 9.203 by 5.347. Using three significant figures, this becomes 9.20 -j- 5.35 = 172. EXAMPLE 6. Divide 2.34 by 3.45. Set 3.41 on C to 2.34 on D and opposite 10 on C, read the quotient 6.85 on D. If scales A and B are used, over the index 100 on J3, read the significant figures and place the decimal point by in- spection. EXAMPLE 7. .000139 -f-.0565=(l. 39 x 10~ 4 ) -=-(565 xlO~ 2 ) = (1.39-5- 5.65) x 10- 2 = . 246xlO- 2 = .00246. 266 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 8. Another method of obtaining the decimal point in Example 7. .000139 -f- .0565 is roughly .0001 -r- .05 = .002. From this we see that the significant figures 246 should be read .00246. ORAL EXERCISE Using scales A and B, find the value of : 1. 65.5 -f- 3.45. 3. 80.5-^1.75. 2. 100 -h 39.5. 4. 14.4 -- 16.5. 5. 1.63 -r- 3.1416 (3.1416 = TT). Using scales C and D, find the value of : 6. 5.48 H- 2.44. 13. 1.38 -=- 5.65. 7. 7.35-3.13. 14. 1670 -.375. 8. 6.13 - 4.62. 15. .00665 - .000144. 9. 9.65 - 8.45. 16. .285 -=- .075. 10. 10 - 3.1416. 17. 2580 -=- .187. 11. 3.64-6.45. 18. .00156-32.8. 12. 4.35 - 7.52. 19. .375 - .065. 20. .0000038537 - .0011887. Reduce to decimals, correct to three significant figures. Divide the numerator by the denominator. 21. T V 23. . 25. T V 27. |f. 29. }f. 22. if. 24. |if. 26. Jf. 28. if^ 30. 6jff. 263. Reciprocals. Two numbers are reciprocals if their product is equal to 1. Thus the reciprocal of 3 is -; the reciprocal of is | ; the reciprocal of ^ is 3. 264. Reciprocals by the inverted slide. If the slide be in- verted so that the C scale is in contact with the A scale, and the right-hand index (10) of the slide be placed in co- incidence with the left-hand index (1) of the A scale, every THE SLIDE RULE 267 number on the D scale multiplied by one tenth of the cor- responding number on the inverted C scale is equal to 1 ; e.g. 2 x .5 = 1. Hence, to find the reciprocal of a number, invert the slide, set 10 on the slide to 1 on scale A, and one tenth of every number on the inverted scale C will be the reciprocal of the corresponding number on scale D ; 2.5 e.g. the reciprocal of 4 is -3*** = .25. WRITTEN EXERCISE 1. Make a table of the reciprocals of the integers from 1 to 10, using the inverted slide. 2. Make a table of the reciprocals of numbers 1.5, 1.6, 1.7, etc., to 2.5 inclusive. 265. Combined multiplication and division. "24.5x46.5 EXAMPLE 1. Find the value of ^-77 ou^U To 2.45 on 7) set 3.62 on C. At 4.65 on C find the signifi- cant figures of the result 315 on D. To obtain the decimal point, 24.5 x 46.5 = 2.45 x 10 X 4.65 x 10 3620 3.62 x 1000 2.45 x 4.65 _!_ 3.62 X 10 = 3.15 x T V = .315. EXAMPLE 2. Find the value of - - O.Zi To 1.46 on D set 5.2 on C. 3.25 lies beyond the extrem- ity of D. Bring the runner to the index 10 on C, and move the slide till the other index coincides with the runner. This operation multiplies the result by 10 and does not 268 SHOP PROBLEMS IN MATHEMATICS affect the significant figures. Under 3.25 on C read the re- sult 913 on D. The position of the decimal point may be determined by the method of Example 1 or by making a rough calculation as follows : 1.46 x 3.25 _ 1.5 X 3 5.2 5 From this we see that the result should be 913. EXAMPLE 3. Find the value of 24.4 x 580 x 36.8 x 9.65 x 1.4 3.55 x 8450 x 5.15 x 1.24 At 2.44 on D set 3.55 on C; runner to 5.8 on C ; 8.45 on C to the runner; runner to 3.68 on C; 5.15 to the runner ; runner to 9.65 on C ; 1.24 on C to the runner. At 1.4 on C read the significant figures of the result 367. For the position of the decimal point, making a rough calculation, 8 20 x X X 10 X x l From this we see that the result is 36.7. _,. 32.6 x 51.5 x 945 EXAMPLE 4. Find the value of At 3.26 on D set 2.74 on C ; runner to 5.15 on C ; 4.8 on C to the runner ; 9.45 projects beyond T) ; runner to 1 on C ; 10 on C to runner ; runner to 9.45 on C ; 5.25 on C to the runner. Under the index 10 of C read the significant figures of the result 22.9. , , 30 x 50 x 900 Calculating roughly, 3 x 500 x 5Q - 18. Hence the result is 22.9. THE SLIDE RULE 269 It is evident that any series of multiplications and divi- sions may be reduced to the form of Example 3 or 4 by in- troducing the factor 1 in numerator or denominator. Thus 546 x 3.14 x 7.18 = 546 x 3.14 x 7.18 9.12 9.12 x 1 ORAL EXERCISE Find the value of : 2.18 x .0136 1 5.44 ' 3.66 x .33 x 6.25 7.85 x 1.25 2 2. ,-.,.. * . 23.3 ' 3.42 x 8.65 574 3. 28.6 x 5.8 x 7.14. 8. 264 x 436 2.34 x 4.82 3.14 x 5.6 x .49 9.38 x 1.36 ' 25.3 x .06 x .685 8.45 6.8 x 4.5 x 7.7 x 122 5- 4.65 x .0176 ' 8.3 x 5.2 x 26 x 12 266. Proportion. Problems in proportion are special cases of multiplication and division. Thus a\b = c:x. x = -- (See sect. 185.) Set a on C to b on D, and opposite c on C find x on D. Note that the first and third terms of the proportion are on one scale, while the second and fourth terms are on the other scale. EXAMPLE 1. Solve 16 : 25 = 18.5 : x. 25 x 18.5 T6~ Solve by the method of section 265. 270 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 2. Solve x : 25 = 16.5 : 18. 25 x 1.65 x = Solve as before. ORAL EXERCISE Find x in the following proportions : . 1. 25 : 35 : : 18.5 : x. 6. 67.5 : 59.4 : : 94.9 : x, 2. 1.5 : 2.6 : : 13 : x. 7. 81.7 : x : : 48.4 : 23. 3. 3.74 : 2.3 : : 52.1 : x. 8. 6.7 : 5.3 : : x : 4.2. 4. 2.65 : 4.82 : : 7.58 : x. 9. 3 : 17 : : x : 10. 5. x : 6.42 : : 11.2 : 14.4. 10. 5 : 19. : : x : 10. 267. Centimeters into inches. On the slide rule, by simple division, centimeters may be changed into inches. ORAL EXERCISE Convert into inches : (1 in. = 2.54 centimeters.) 1.2.18cm. 2.5.25cm. 3.7.99cm. 4.8.78cm. 5.4.65cm. 268. Squares. Over any number on D is found its square on A. Thus, setting the runner to 2 on D, 4 will be indi- cated on A. Conversely, 2 on D is the square root of 4 on A. CASE 1. When the number lies between 1 and 10, the runner set to the number on D will indicate the square on A. EXAMPLE 1. Square 1.35. The runner set to 1.35 on D indicates 1.82 011 .1. CASE 2. When the number does not lie between 1 and 10 : (1) change the number to one that is between 1 and 10 by moving the decimal point ; (2) square by Case 1 ; and (3) move the decimal point twice as many places as it was moved in (1) and in the opposite direction. THE SLIDE RULE 271 EXAMPLE 2. Square 75. 75 2 = (7.5) 2 x 10 2 = 56.25 x 100 = 5625. EXAMPLE 3. Square .00355. .00355 = (3.55) 2 x (10" 3 ) 2 = 12.6 x 10- 6 = .0000126. ORAL EXERCISE Find the squares of : 1. 4.75. 3. 9.65. 5. 24.2. 7. .0635. 9. .0346. 2. 1.14. 4. 376. 6. .56. 8. .0087. 10. 2250. 11. Find the area of a square field measuring 126 rd. on a side. 12. How many square yards are there in a circular plot of ground measuring 12.5 yd. in diameter ? The area of a circle may be found by multiplying TT by the square of the radius, or by multiplying .7854, the area of a circle 1 yd. in diameter, by the square of the given diameter. 269. Square root. Square root is the inverse operation of squaring ; hence the operations with the slide rule must be reversed. CASE 1. When the number lies between 1 and 100, the runner set to the number on A will indicate the square root on D. EXAMPLE 1. Find V38.5. The runner set to 38.5 on A indicates 6.2, the square root, on D. CASE 2. When the number does not lie between 1 and 100 : (1) change it to one lying between 1 and 100 by moving the decimal point an even number of places ; (2) find the square root by Case 1 ; and (3) move the decimal point one half as many places as it was moved in (1) and in the opposite direction. 272 SHOP PROBLEMS IN MATHEMATICS EXAMPLE 2. Find Vl22. Vl22 = VL22 x VlOO = 1.104 x 10 = 11.04 EXAMPLE 3. Find Vl220. V1220 = Vl^2 x VlOO = 3.49 x.10 = 34.9. EXAMPLE 4. Find V.00122. V.00122 = VK2 x VTF 1 = 3.49 x 10- 2 = .0349. ORAL EXERCISE Find the square roots of : 1. 48.5. 3. 485. 5. 2.58. 7. .375. 9. .0066. 2. 4850. 4. .0485. 6. 97.5. 8. .31416. 10. 1540. 11. How long must one side of a square field be to con- tain 5 A. ? 160 sq. rd. = 1 A. 12. Find the diameter of a circular cylinder 30" in height that will contain 5 gal. 231 cu. in. = 1 gal. REVIEW EXERCISE Find the value of : 5.35 x 425 x (64.5) 2 .45 x .072 Use scale C for the factor that is to be squared. 175 x V31 x VT8 x 250 29 x 13 x 841 Use scale B for the square roots. 3. 14.6 xV2. 4. 8 V3. THE SLIDE RULE 273 15.5 (.0247) 2 V3580' " (-0068) 2 ' 6 2 ' 1 9. Vl4 x 13 x 11 X 3. ' (8.4)* (18.4) 2 x 56.2. V640 x 1.54 (3.7) 2 124 x. 163 ' 11. The area of a triangle is V* (s a) (s b) (s c\ where s is % the perimeter, and a, b, and c are the sides. If a = 4:7 rd., b = 31 rd., and c = 22 rd., find the area of the field. 12. Using the slide rule, make a list of squares of inte- gral numbers between 20 and 30. 13. Make a list of square roots of integral numbers be- tween 70 and 80, correct to one place of decimals. TABLES TABLE I TABLE OF NATURAL SINES, TANGENTS, COSINES, AND COTANGENTS Degrees Sine Tangent Cotangent Cosine 8 1 90 1 2 3 4 .0175 .0349 .0523 .0698 .0175 .0349 .0524 .0699 57.2900 28.6363 19.0811 14.3007 . . .9998 .9994 .9986 .9976 89 88 87 86 5 .0872 .0875 11.4301 .9962 85 6 7 8 9 .1045 .1219 .1392 .1564 .1051 .1228 .1405 .1584 9.5144 8.1443 7.1154 6.3138 .9945 .9925 .9903 .9877 84 83 82 81 10 .1736 .1763 5.6713 .9848 80 11 12 13 14 .1908 .2079 .2250 .2419 .1944 .2126 .2309 .2493 5.1446 4.7046 4.3315 4.0108 .9816 .9781 .9744 .9703 79 78 77 76 15 .2588 .2679 3.7321 .9659 75 16 17 18 19 .2756 .2924 .3090 .3256 .2867 .3057 .3249 .3443 3.4874 3.2709 3.0777 2.9042 .9613 .9563 .9511 .9455 74 73 72 71 20 .3420 .3640 2.7475 .9397 70 21 22 23 24 .3584 .3746 .3907 .4067 .3839 .4040 .4245 .4452 2.6051 2.4751 2.3559 2.2460 .9336 .9272 .9205 .9135 69 68 67 66 25 .4226 .4663 2.1445 .9063 65 26 27 28 29 .4384 .4540 .4695 .4848 .4877 .5095 .5317 .5543 2.0503 1.9626 1.8807 1.8040 .8988 .8910 .8829 .8746 64 63 62 61 30 .5000 .5774 1.7321 .8660 60 31 32 33 34 .5150 .5299 .5446 .5592 .6009 .6249 .6494 .6745 1.6643 1.6003 1.5399 1.4826 .8572 .8480 .8387 .8290 59 58 57 56 35 .5736 .7002 1.4281 .8192 55 36 37 38 39 .5878 .6018 .6157 ,6293 .7265 .7536 .7813 .8098 1.3764 1.3270 1.2799 1.2349 .8090 .7986 .7880 .7771 54 53 52 51 40 .6428 .8391 1.1918 .7660 50 41 42 43 44 .6561 .6691 .6820 .6947 .8693 .9004 .9325 .9657 1.1504 1.1106 1.0724 1.0355 .7547 .7431 .7314 .7193 49 48 47 46 45 .7071 1.0000 1.0000 .7071 45 Cosine Cotangent Tangent Sine Degrees 274 TABLES 275 TABLE II WEIGHT OF CASTINGS FROM WEIGHT OF PATTERNS Factors by which to multiply weight of pattern to get weight of casting Material of pattern For cast iron For brass For copper For aluminum White pine . . . 16 12 l"l 1? 4| 4 * Mahogany .... Brass U 9 * 13* 1 U) Dry sand .... Aluminum . . . ? 5 3.4 5 3J U 1 TABLE III WEIGHTS OF MATERIALS Woods Lbs. per cu. ft. Woods Lbs. per cu. ft. Ash Birch . . 52 38 Lignum vitae . . Linden 83.3 377 Cherry 44 Mahogany 35 to 53 Chestnut 35 28 Maple Oak (white) 45 50 Elm Gum 41 40 Oak (red) .... Pine (white) 52 30 Hemlock . ... 24 Pine (yellow) 43 Hickory 45 to 53 Metals Metals A 1 160 Steel . ." . 499 Babbit Metal 457 Tin 4555 488 Zinc 437 Copper 545 Other material Cast iron Wrought iron . . . Lead 450 486 708.5 Water Granite .... Marble Brick and mortar . 62J 169 166 120. TABLE IV COMMON NAILS Size Length Approximate No. to Ib. Size Length Approximate No. to Ib. 2d 3d 4d 5d 6d 7d 8d 9d 1 inch 1 inch 1 inch If inch 2 inch 2 inch 2 inch 2| inch 876 568 316 271 181 161 106 96 lOd 12d 16d 20d 30d 40d 50d 60d 3 inch 3i inch 3| inch 4 inch 4J inch 5 inch 5 inch 6 inch 69 63 49 31 24 18 14 11 TABLE V MISCELLANY A gallon contains A bushel contains 231 cu. in. 2150 cu. in. A gallon of water weighs A barrel contains . . . 8.35 Ib. 31i gal. 276 SHOP PROBLEMS IN MATHEMATICS TABLE VI FOR FINDING THE UNKNOWN PARTS OF GEARS FROM THE KNOWN l To get Having Formula , The diametral pitch The circular pitch p 3.1416 The diametral pitch The diametral pitch The pitch diameter and the number of teeth . The outside diameter and the number Po *-S p, N+2 Pitch diameter . . of teeth The number of teeth and the diame- tral pitch d ~o D =K Pitch diameter . . The number of teeth and the outside ODxN Pitch diameter . . Pitch diameter . . Outside diameter . The outside diameter and the diame- tral pitch Addendum and the number of teeth . The number of teeth and the diame- D=OI>- Pd D=sN "N" + > n~n i> T