mm ^ ■ BHHi ^^MM i ; r'.-ni) i ■ ^H H HflfwtttNcj ■ ■ H8HR sffflilHfc 3 ; l\i;.! ■ I liSSPLtiiil! ■ JM; , '\if. ■ i •''^-••vii 1 !,^'.- iS$ mm iVm m liSlH H IB IS H ■ Hi nuw tflSSil ■niH Class _L^Jij5j Book . M 2. COPYRIGHT DEPOSIT. ■• — —~^~^~ TEXTILE Applied Arithmetic r AND Calculations W c. BY W. McSWAIN Copyright 19 19 By Victor-Monaghan Company Published October, 19 19 ffEC:l7i3l9 ©CI.A5610 20 Foreword This work was undertaken after a thorough inves- tigation of the best means of providing education for the man who had either been deprived of this advantage, or who had been so employed as to have become dulled by not having to apply an early education to his daily task. Realizing that the best way to carry out this cam- paign was to consult those who would afterward benefit from this research,, the author arranged con- ferences with superintendents, overseers, second hands, section men and other employes. At these meetings it was found that the form of education most needed was that bearing on the necessary calculations appertaining to the textile industry. It was, however, found that, in most cases, the men realized that such calculations could not be suc- cessfully studied until a better knowledge of mathe- matics was obtained. The result was the idea of incorporating the two in one. Hence this book. This book being written for and at the instigation of cotton mill men, it is well adapted to their use. While repetition may occur at several places, it is intended to be so, and is believed that it enhances rather than depreciates its value. Only such prin- ciples of mathematics are used as are needed to fully understand the methods used in textile cal- culations. The author wishes to express his thanks to and appreciation of the valuable aid given by officials and employes of the Victor-Monaghan Company in compiling this book. Many of the problems incor- porated were submitted by some of the employes of this Company, and many valuable suggestions were given by others. C. W. McSWAIN. DEDICATION V To those men and boys, employed in the Cotton Mills of the South, who desire to fit themselves to become successful in this, our most important industry, this book is dedicated. — The Author. Numbers A number is an expression showing how many times an object or thing is taken. If we see an object and then another of the same general physical or mechanical construction we say that there are two of these objects,, and if we see still another object of the same make-up we say three and so on for as many times as we conceive in our mind that this thing occurs at a given place or in a given time. The symbols used in writing numbers are com- monly called figures. These figures which we use to denote numbers were brought into use about two thousand years ago. They were brought from India by the Arabs and are still called Arabic or Hindu- Arabic numerals. There are nine of these figures besides zero which is sometimes called naught or cipher. These figures are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. In common practical uses there are three different kinds of numbers all using the same figures in ex- pressing their values. These are : Whole Numbers, Fractional Numbers or Fractions, and Decimal Num- bers or Decimals. WHOLE NUMBERS "When we write figures to show the number of times a certain thing is taken as a whole these figures are called whole numbers. Thus, if we say "two hundred bales of cotton", we write the figures, 200, which means that we are considering this many separate individual bales of cotton, all of which have the same general make-up in regard to their- physical composition as well as to their mechanical construction. In writing large numbers it is found convenient to separate the whole series of figures into groups of three ; thus, 250,615,112. By this means it is easier to read the whole by considering the first group of three to the right as in hundreds place. Thus„ to read this part of the number we say "one hundred and twelve." The second group to the left of this is in thousands place and is read "six hundred and fifteen thousand. ' ' The third group to the left of this one is in millions place and is read "two hundred and fifty million." Thus, the whole of this number is read: "two hundred and fifty million, six hundred and fifteen thousand, one hun- dred and twelve." Exercise "Write in figures : 1. Fifty bales of cotton. 2. One hundred and twelve rolls. 3. Six hundred and one spindles. 4. One thousand and one pounds. 5. Two thousand, one hundred and nine ounces. 6. Two hundred and fifteen thousand, eight hun- dred and twenty-five grains. 7. Forty-five flyers. 8. One hundred and one yards of roving. 9. Two hundred and fifty-three pounds of card strippings. 10. One thousand, six hundred and twelve spin- ning rings. 11. Seven hundred and eighty-five reeds. 12. Ten thousand, eight hundred and six heddles. 13. Six hundred and five thousand, eight hun- dred and two minutes. FRACTIONS When any one whole thing is divided into several parts one of these parts is called a fraction. Thus, if a bale of cotton weighing 400 pounds is divided into two parts each of these parts will be 200 pounds and is called one half of the whole. If this 400 pound bale of cotton is divided into 4 equal parts each of the four parts will weigh 100 pounds and is called one-fourth of the whole. Thus we see that each of the parts when the whole is divided into four parts is exactly one-half of those parts when the division was made in 200 pound lots. There- fore the number one-fourth is one-half as large as the number one-half. The figures used to denote a fraction are placed one above a line and one below, thus, £ (one-half), | (three-fourths), -J (one-eighth), f (three-eighths), y iG (one-sixteenth), %c (three-sixteenths). The figure below the line, called the Denominator, shows into how many parts the whole number has been divided. Thus -J means that the whole was divided into two parts, \ that it was divided into four parts, ■§ into eight parts, and Y 16 into sixteen parts. The figure above the line, called the Numerator, shows how many of these parts are taken. Thus if we divide a number into eight equal parts each of these parts is called one-eighth (|), and if we desire to show that three of these parts are to be taken we write the figure three above the line, thus: f (three-eighths). All fractions are divided into three general classes, known as proper fractions, improper frac- tions, and mixed fractions. PROPER FRACTIONS When the number above the line (numerator) is smaller than the number below the line (denom- inator) the whole is said to be a proper fraction. Thus the following are all proper fractions : •§■ (one- eighth), | (three-eighths), $• (four-fifths), % (three- sixths), f (two-thirds), | (three-fifths), % (six- eighths), % (five-eighths), % 6 (four-sixteenths) r 15 / 1Q (fifteen-sixteenths), 17 / 2 o (seventeen-twen- tieths). IMPROPER FRACTIONS When the number above the line (numerator) is greater than that below the line (denominator) the whole is said to be an improper fraction. Thus, the following are improper fractions : % (five- fourths), % (six-halves), % (seven-thirds), % (nine- fifths), % (six-fourths), xx /-i (eleven-sevenths). MIXED NUMBERS When a whole number precedes or is set before a fraction the whole is said to be a mixed number. Thus, the following are mixed numbers: 5£ (five and one-half), 4| (four and three-fourths), 10f (ten and three-eighths), 25^ (twenty-five and one-fourth), llG 1 ^ (one hundred and sixteen and eleven-six- teenths), HOf (one hundred and ten and five- eighths). 10 Exercise Write in figures: 1. One-half a revolution of a roll. 2. Three-eighths of a bale of cotton. 3. Eleven-sixteenths is the diameter of this pipe. 4. Two and seven-sixteenths is the diameter of this shaft. 5. Eleven-fifths is equal to two and one-fifth. 6. The diameter of this roll is one and seven- eighths inches. 7. Eight and one-half revolutions of the cord cylinder. DECIMALS Decimal notation is a method of expressing the values of numbers which are less than one in a de- creasing ratio of 10. If, for instance, one inch be divided into ten equal parts, one of these parts is called "one-tenth" of one inch. This is expressed decimally by placing a point to the left of the number, thus : .1. If each of these ten divisions be divided in turn into ten more equal parts the total number of divisions in one inch would be 100 and each one of these parts would be called "one- hundredth" and is expressed decimally thus .01. In like manner if we again divide each of these latter divisions into ten more, the total number would be one thousand and each of these divisions would be "one-thousandth" and is written thus : .001. If then we desire to show how many of each of these parts are to be taken we proceed as follows : If we take 2 parts when one inch is divided into 10 parts it is read: "two tenths" and is written .2. If we take 11 3 parts it is read: ''three tenths" and is written .3. If we take four parts it is read "four tenths" and is written .4. If we take five parts or one-half of the total number of parts it is read "five-tenths" and is written .5. When we proceed thus for all ten parts we arrive at the whole or "ten-tenths" and it is written 1.0. Again when the total number of divisions is one hundred,, one of these divisions is called "one hundredth" and is written .01. If two divisions are taken it is read "two one-hundredths " and is written .02, and so on for the total number which brings us back to "one tenth". In like manner do we proceed in writing and reading the parts when the whole is divided into a thousand parts. For instance, if four of such divisions be taken it is read "four one-thousandths" and is written .004. Exercise Write in figures : 1. Three-tenths of a mile. 2. Twenty-five hundredths is equal to £. 3. Seventy-five hundredths is equal to f. 4. Two and five-tenths equals 2\. 5. Three and f ourteen-hundredths multiplied by the diameter equals distance round a circle. 6. Seven and eighty-five hundredths is the cir- cumference of a circle which is 2| inches in diameter. 7. Two-tenths of a day. 8. Eight and seven-tenths revolutions of a cal- endar roll. 9. Six and eighty-seven hundredths yards of, warp. 12 Iii common practice decimals are read by stating first where the point is placed and then giving the figures following in order. Thus .5416 is read: ' ' point, five, four,, one, six ' \ Write in figures : 1. Point, one, two, three, six. 2. Two, point, four, six, eight. 3. Ten, point, six,, eight, two. 4. One hundred and twelve, point, six, five. 13 Addition of Whole Numbers In adding small numbers, such as those which are smaller than 10, it is sometimes customary to place them in a line with the sign of addition, (+, plus), between them. Thus, 2+4+3+5=14, is read, "two plus four, plus three, plus five equals fourteen". When large numbers are added they are usually placed one below the other so that the last figure on the right for each number is directly under the same corresponding figure in the number above. Thus to add 204, 16, 1012 and 12002 we proceed as follows : 204 16 1012 12002 13234 The process above is to add first the row of figures at the right, namely, 2, 2, 6 and 4. This gives 14 and the figure 4 is placed below the line directly below the set of figures thus added and the figure 1 is added into the next line of figures to the left. This is continued until all rows have been thus added and the resulting number below the line is the sum of all the numbers thus added. CARD ROOM PROBLEMS 1. Ten bales of cotton are opened having the following weights : 514 lbs., 476 lbs., 504 lbs., 524 14 lbs., 487 lbs., 497 lbs., 510 lbs., 507 lbs., 494 lbs., and 502 lbs. What is the total weight? 2. Six laps are weighed as follows : 42 lbs., 40 lbs., 43 lbs., 42 lbs., 41 lbs., 40 lbs. What is the total weight? 3. If a man received 18 dollars per week, how much would he have received at the end of four weeks ? "4. A mill has 2 slubbers with 92 spindles each, 2 slubbers with 96 spindles each, and 4 slubbers with 88 spindles each. What is the total number of slubber spindles in mill? SPINNING ROOM PROBLEMS 1. Six doffs are made on spinning frames having following weights : 52 lbs., 51 lbs., 50 lbs., 53 lbs., 51 lbs., 52 lbs. What is the total weight of yarn made on the six frames? 2. A Spinning Room produces 8572 lbs. of 30 's yarn, 10,514 lbs. of 24 's, and 6451 lbs. of 28 's. What is the total weight of yarn made? 3. If a Spinning Room makes 5416 lbs. of 32 's, 11,562 lbs. of 30 's, and 4154 lbs. of 24 's in one week, how many pounds of 32 's should it make in 4 weeks ? (Add 5416 four times.) How many pounds of 30 's should it make in three weeks? (Add 11,562 three times.) How many pounds of 24 's should it make in five weeks? (Add 4154 five times.) 4. If a girl runs 10 sides of 150 spindles each how many spindles does she tend? If each spindle will produce (theoretically) 2 pounds in one week, each side should make 300 pounds. How many pounds should the 10 sides make? (Add 300 10 times. ) 15 WEAVE ROOM PROBLEMS 1. A loom has twelve harness on each of which are the following numbers of heddles : 416,, 420, 424, 418, 82, 82, 106,, 106, 116, 116, 46, 46. What are the total number of heddles on all harnesses? 2. A Weave Room produces the following num- bers of cuts in a week : Monday, 142 ; Tuesday, 264; Wednesday, 250; Thursday, 246; Friday, 232; Saturday, 164. What is the total number of cuts? 3. Eight warper beams are placed behind the slasher weighing respectively 448 lbs., 446 lbs., 450 lbs., 447 lbs., 448 lbs., 446 lbs., 451 lbs. What is the total weight of yarn unsized? 4. If eight warper beams, each having 416 ends, are behind the slasher, what is the total number of ends on each loom beam? (Add 416 eight times.) 1G Subtraction of Whole Numbers Subtraction is the process of finding the differ- ence between two unequal numbers. The sign of sub- traction is made thus : — and is called minus. When this sign is placed between two numbers, it means that the latter one is to be taken, or subtracted, from the former. Subtraction may be considered as an inverted form of addition, because when a smaller number is taken from a larger the resulting number when again added to the smaller will equal the larger. Thus 4—2=2. This is the same as saying that the difference between 4 and 2 is the same as whatever number is required to be added to 2 to make 4. In subtracting large numbers they are placed with the smaller one below the larger the same as in addition. Then by starting at the right side the first figure in the bottom number is subtracted from the one above. If the top figure is smaller than the bottom the number 10 is added to the top, making a temporary number of either 10, 11, 12, 13, 14, 15, 16, 17,. or 18 according as to whether the top number is either 0, 1, 2, 3, 4, 5, 6, 7, or 8. When this is done for a row of figures then to the next row to the left the bottom figure must be increased by "one" before proceeding with the subtraction. Example : Subtract 2168 from 3415. 3415 -2168 1247 ' 17 Ill the above example we take the first two num- bers to the right in bottom and top number, and as 5 is smaller than 8 we mentally add 10 making 16. Then we say "8 from 15= (equals) 7,." and place this number, 7, below the line. Then before subtracting the next two figures add 1 to the bottom figure, 6. This makes 7, and we again proceed as before,, saying "7 from 11 is 4", and place this figure below the line. In like manner Ave proceed subtracting all the figures in the set. In this case the third set of figures contains 4 above and 1 below, hence it is not necessary to call the top number 14. As, however, we added 10 to the top number in the set before this one we must add 1 to the bottom figure. This gives 2 and we say "2 from 4 equals 2" and place the 2 below the line. In the fourth set of figures it is not necessary to add 1 to the bottom number as Ave did not add 10 to the top in the last set. Therefore Ave say "2 from 3 equals 1" and place this beloAv the line. The result is 1247 which is the difference betAveen 3415 and 2168. This can be proved by adding 1247 and 2168 Avhich will give the original number 3415 as follows: 1247 2168 3415 CARD ROOM PROBLEMS 1. If ten bales of cotton each Aveighing respec- tively 487, 515, 520, 492, 498, 522, 508, 494, 498, 522, 508, 494, 496 and 512 pounds has 247 pounds of bagging and ties Avhat is the net (Aveight after deducting bagging and ties) Aveight of cotton? 18 2. Locally ginned cotton has bagging and ties to the amount of 22 pounds. Compressed cotton 24 pounds. If 8 bales of locally ginned cotton weigh respectively 492, 498, 502, 504, 498, 496, 492 and 498 pounds, and 6 bales of compressed cotton -weighing respectively 512, 506, 508,. 498, 502 and 504 pounds are opened, what is the net weight of cotton opened? (Add 22 together 8 times for total weight of bagging and ties on the eight bales of cotton locally ginned. Subtract this weight from total weight of all 8 bales. Then add 24 together six times to find weight of bagging and ties on the 6 bales of com- pressed cotton. Subtract this weight from the total weight of the 6 bales of compressed cotton. Add together the difference in each case and the result in the total number of pounds of cotton opened.) 3. If cotton is opened to the amount of 20,000 lbs. and from this weight 18,642 lbs. of laps are made, how much waste Avould you expect in the Picker Room? Suppose that 1162 lbs. of waste is all that can be accounted for, what has become of the remainder and how much is it? 4. If 4 laps weighing 40 lbs. each are put through the card and 152 lbs. of sliver is delivered, what is the amount of waste? SPINNING ROOM PROBLEMS 1. If a mill has 31,652 spinning spindles and 12,840 are on warp yarn, what is the number of filling spindles? 2. 62,847 pounds is the weekly production of certain spinning room. If 38,117 pounds of this is warp, how much is filling? 19 3. An order is received for 100,000 lbs. of 28 's yarn. 12,450 lbs. are produced the first week, 11,872 lbs. are produced the second week, and 15,117 lbs. the third week. How much remains to be made? 4. A mill has 1492 looms, 216 of these are stand- spindles stopped for one week, how many spindles run? WEAVE ROOM PROBLEMS 1. An order for 50,000 pieces is received and 8752 are made the first week, 7985 the second week and 9415 pieces the third week. How many pieces remain to be made? 2. If 8 warp beams each weighing respectively 432, 434,. 435, 432, 432, 433 and 434 lbs. are run on a set, on slasher and loom beams are made totaling 3683 lbs. together with 9 lbs. of waste, how many pounds of size are put on? 3. A cut of cloth weighs 14 pounds. If it has 8 pounds of warp in it, what is the weight of the filling? 4. A mill has 38,252 spindles. If there are 2560 ing for a week, how many looms are runnings 20 Multiplication of Whole Numbers The process of taking a certain one number for a particular number of times is called multiplica- tion. If we consider the number 2 taken three times the result is 6. Therefore we may consider multiplication as the process of adding together a certain number for a particular number of times. Thus, if we add 3 to itself three times we have 3-f-3+3= (equals) 9. As we have taken this num- ber together three times to make 9, we might say "3 taken three times equals 9," or "3 times 3 equals 9." In like manner if we desire to multiply one number by another it is not necessary for the per- son unskilled in figures to worry himself about trying to remember all of the multiplication tables. Of course as multiplication is simply a short form of addition for each individual number it is more convenient for one who expects to. become proficient in mathematics to learn by practice the result of mul- tiplying all of the numbers from 1 to 9 by any of these same numbers. It is recommended that all who are not familiar with the multiplication tables prac- tice this process until it is thoroughly mastered. Usually when two small numbers are to be multi- plied together they are placed one in front of the other with the sign of multiplication, (X), between them. Thus if we desire to multiply 6 by 4 we place them thus,, 6X4, and say "6 multiplied by 4" or "6 times 4." In arriving at the product or result of multiplying these two numbers together we may proceed in the following manner : If 6 is to be 21 multiplied by 4 it means that 6 is to be added to- gether four times and we might say ' ' 6 plus 6 plus 6 plus 6 equals 24, ' ' which is the same thing as saying that "6 taken 4 times equals 24" or that 6X4= (equals) 24. When multiplying a large number by a smaller one, or two large numbers together, the usual prac- tice is to place one (usually the smaller one) below the other as in addition or subtraction. After this is done start with the first figure to the right in the top number and multiply this by the first figure to the right in the bottom number. Remember always that "zero" or "naught" or the figure "0" Avhen multiplied by any other produces "zero" or "naught" or the figure 0. Suppose we desire to multiply 1463 by 502. In this case we proceed as follows : 1463 502 2926 0000 (usually omitted.) 7315 734426 We take in this example the first figure, 3, in the top number, 1463, and multiply it by the first figure, 2, in the bottom number, 502. This gives 6 and we place it below the line directly below the figure, 2. Then proceed to multiply all of the fig- ures in the top number by this figure, 2. Thus, 2 times 6 equals 12. Place the figure 2 below the line to the left of 6 and add the figure 1 to the next figure 22 produced by multiplying. Thus, 2 times 4 equals 8 and 1 added gives 9. Then 9 is placed below the line to the left of 2 previously obtained. Next the figure 1 is multiplied by 2 and as the previous prod- uct was not greater than 9 no figure is to be added to this one and the result is 2, which is placed to the left of 9. This first step then produces the number 2926 as seen below the line. The next step is to multiply again all of the figures in the top number by the second figure from the right in the bottom number. As, however, this fig- ure is "zero" or the result Avould be zero. Then proceed to multiply all figures in the top number again by the third figure, 5, in the bottom number, 502. First, 5 times 3 equals 15. Place the 5 directly in line and under the figure, 5, by which we are multiplying, and add 1 into the next product. Thus, again, 5 times 6 equals 30 and 1 added gives 31. Place the figure 1 to the left of the 5 previously obtained and add the 3 into the product. Thus 5 times 4 equals 20 and the 3 gives 23. Place the 3 to the left of 1 previously obtained and add 2 into the next product. Thus, 5 times 1 equals 5 and 2 added gives 7. Place this 7 to the left of the figure 3 previously obtained. This completes the process of multiplying. The next step is to add together the numbers produced by each process. In this case we have only the numbers 2926 and 7315 to add, but the number 7315 is placed with its figure 5 three numbers to the left of the first figure, 6, in the num- ber 2926. This happens because the third number which would have been produced by multiplying by the second figure in the bottom number produced zero. If this second figure had been any other fig- 23 ure than zero we would have had a third number in our final adding. Adding then all the vertical rows of figures we find the number 734,426, which is the product of multiplying together 1463 and 502. In multiplying any number by 10 all that is suffi- cient is to annex one cipher or zero or to the num- ber multiplied. .Thus 416X10=4160; 510x10= 5100; 1604X10=16040, etc. In multiplying any number by 100 all that is necessary is to annex two "ciphers" or 0's. Thus, 116X100=11600; 510X100=51,000; 1004X100= 100,400, etc. In multiplying any number by 1000 all that is necessary is to annex three "ciphers" or 0's. Thus, 26X1000=26,000; 41x1000=41,000; 510X1000= 510,000, etc. EXERCISE Multiply 8X9 17X8 16X4 1006X40 25X4 410X20 50X2 120X60 126X10 140X50 115X15 132X66 304X8 201X80 510X16 80X15 In multiplying any numbers by 20, 30, 40, 50, 60, 70, 80, or 90 simply multiply the number by either 2, 3, 4, 5, 6, 7, 8, or 9 and annex one '"naught" to the result. Example: Multiply 1240 by 50. Thus 1240 5 Annexing the ' ' cipher, ' ' 0, we have 62000 6200 24 CARD ROOM PROBLEMS 1. If a lap weighs 14 oz. per yd., how many grains will there be in 1 yard? (1 oz.=437+ grains.) 2. If a 54 grain sliver is being made with a 98 draft in card,, how many grains in one yd. of lap? (Wt. in grain of sliver X draft = wt. in grain of hi p.) 3. If there are 840 yards in 1 lb. of a No. 1 hank roving how many yards would there be in 10 lbs of hank roving? 4. If there are 840 yards in 1 lb. of No. 1 hank roving how many yards would there be in 1 lb. of No. 6 hank roving? 5. If a roll is 4 inches in circumference and runs 116 revolutions per minute (R. P. M.) how many inches of roving would be delivered in 1 minute ? In 1 hour? In 1 day? In 1 week? 6. If the slubber clock registers 8 hanks per spin- dle made in a day, how many hanks would be made on a slubber of 92 spindles ? How many yds. of rov- ing would this be of 840 yards=l hank? SPINNING ROOM PROBLEMS 1. If there are 840 yards of No. 1 yarn in 1 lb., how many yards of No. 1 would there be in 6 lbs. ? 2. If there are 840 yards of No. 1 yarn in 1 lb., how many yards of No. 20 yarn would there be in lib.? 3. If one spindle will produce 6 pounds of No. 8 yarn in one week, how many pounds ought 11640 spindles make in the same time? 4. If the front roll on a spinning frame is 3 inches in circumference (around) and is running 120 25 revolutions per minute, how many inches of yarn will be delivered in 10 hours? 5. The weight of yarn on a full bobbin with 6 inch traverse is about 2 ounces from a If inch ring. How many ounces should a doff weigh from a 32C spindle frame? 6. If a mill has 516 frames with 336 spindles each what is the total number of spindles? WEAVE ROOM PROBLEMS 1. If cloth is to weigh 5 yards per pound, how many yards would there be in 12 pounds? 2. If cloth has 64 warp ends in one inch how many ends would there be in the warp if cloth is 36 in. wide? 3. If a reed has 32 dents on one inch how many dents would be on 38 inches? 4. If a loom makes 160 picks per minute how many picks does it make in one hour? In 10 hours? 5. How many inches of filling would be taken up in 60 picks of shuttle if the reed spread of warp is 38 inches? (One pick would take up 38 inches.) 6. There are 840 yards of No. 1 yarn in one pound. How many yards of No. 1 yarn would there be in 6 pounds? How many yards of No. 20 yarn would there be in one lb. ? 7. If a loom makes 42 yards of cloth in one day, how many yards would 15 looms make? 8. If 8 section beams each with 438 ends are up behind the slasher what is the' total number of ends going on the loom beams? 9. Sixty-four looms are to be put on a style of goods requiring 26 bars in the pattern chain. How many bars would there be on all 64 looms? 26 10. A pattern requires 8 harness with the fol- lowing numbers of heddles on each harness : 1, 432 ; 2, 438 ; 3, 434 ; 4, 438 ; 5, 82 ; 6, 82 ; 7, 64 ; 8, 64. If 24 sets of harness are to be put on, how many hed- dles would be required in all? 27 Division of Whole Numbers Division is the process of splitting up a number into several parts so that when all of these parts are added together the result obtained is the original number. Thus, in the number 2 there are two l's and we say "1 will go into 2 two times." In like manner if we divide 15 into three equal parts each one of these parts will be the number, 5. So there are three fives in 15 and we say "3 into 15 will go 5 times." In all division the number we " divide into" is always equal to the product of multiplying together the number we "divide by" and the num- ber obtained by the division. Thus in dividing 20 by 5 we mean that we desire to divide 20 into 5 equal parts, so we mentally decide what number multiplied by 5 will give 20. This number is 4 and we say "5 into 20 will go 4 times," or "20 divided by 5 equals 4." In dividing small numbers they are usually placed one in front of the other with the sign of division (-=-) between them. Thus if we desire to divide 20 by 5 we have 20-^5= (equals) 4. Also 15-^-3 = (equals) 5; 36-1-6= (equals) 6; 10-^5= (equals) 2; 20^-2= (equals) 10, etc. In some cases a number is indicated to be divided by another by placing the first above a line and the second below that line. Thus if we have 15 it 3 means that 15 is to be divided bv 3. In like man- 28 ner we may have.. 20 15 — = (equals) 4; — = (equals) 3: 5 5 16 — = (equals) 4, etc. 4 Iu dividing a large number by a small number, (usually one less than 10) it is done by a process known as "Short Division." If, for instance, we desire to divide 27612 by 6 the process is as follows : 6127612 4602 In this example Ave first divide 6 into the first figure on the left. If this figure is smaller than the number we are dividing by, we then take the first two figures and mentally divide into' these. In this case 6 would not go into two, so we take 27 and divide 6 into it ; 6 into 27 goes 4 times and 3 over. Place the 4 below the line and putting together the remainder, 3, and the next figure, 6,, we have 36. Then 6 into 36 goes 6 times even, and the 6 is placed below the line. Next 6 will not not go into 1, so Ave place the figure below the line and run 6 into the two figures 1 and 2 making 12. Then 6 goes into 12 tAvo even times and the figure 2 is placed beloAv the line. In dividing one large number by another it is usually done by a process knoAvn as "Long Di- vision." If, for example, Ave desire to divide 28567 by 162 the f olloAving process is used : 29 162|28567|176 ~162 1236 1134 1027 972 55 over In this process the number, 162, is placed to the left, as seen, and is divided into as many figures on the left of large number as is required for it to go into. In this case 162 could not be divided into 2, nor 28, so we take the three figures, 2, 8, and 5, mak- ing 285. Then we try to find out how many times 162 will go into this number. It must be a number which, when 162 is multiplied by it, the result must not be greater than the 285. If we were to try 2, we would find that 162 when multiplied by 2 would give 324. So the first number in the result must be 1. Then 162 is multiplied by this first number, 1, and the result placed under and subtracted from the 285. This remainder is 123. Then annex to this remainder the next figure in the number being divided. In this case it is 6 and we have the num- ber 1236, into which we again divide 162. This we find will go 7 times and we again multiply 162 by 7 and place this result, 1134, under the number 1236 and again subtract. This step gives a remainder of 102 and we draw down the next figure, 7, from the number being divided. This gives 1027, into which we again divide 162 and obtain 6. Again 162 mul- tiplied by 6 gives 972 and this subtracted from the number, 1027, leaves a remainder of 55. Then the 30 result of our division is that 162 goes into 28567, 176 times with 55 over. EXERCISE Solve, 10- ■r-2= 30- r-3= 40- r-10= 16 4 18 2 110 10 Divide 1672 by 2 1480 by 4 1600 by 8 " 21462 by 213 Solve 840 yds. 14 yds. " 1140 inches yds. = yds. 36 inches CARD ROOM PROBLEMS 1. If there are 840 yds. in 1 pound of No. 1 yarn, how many pounds would there be in 11760 yds. of the same size yarn? 2. If a No. 1 hank roving has 840 yds. in 1 pound, what would be the hank roving number if there are 2520 yds. in 1 pound? The actual draft of a card is found by dividing the weight in grains of one yd. of sliver into weight in grain of 1 yd. of lap. 3. If a card is running a 12 oz. lap and 54 grain sliver, what is its draft? Reduce the weight in ozs. of the lap to grains. There are 437+ grains in 1 oz. Then 12x437=5244 grains and 31 54 1 5244 1 97 +dr aft 486 384 378 4. If a 54 grain card sliver, when doubled six on drawing frames produces a 55 grain sliver, what is the draft of drawing ? 54 X 6=324 324^55=draft. 5. What is the weight of sliver produced on a drawing frame doubling 6, 50 grain card slivers into one with a draft of 5? 50X6=300 and 300-f-5= 60 grain sliver. 6. If there are 36 inches in 1 yd.,, how many yds. would there be in 10,188 inches? ,7. If a Coiler Calendar roll is 7 inches in cir- cumference (around), how many yds. of sliver will it deliver in 1 hour if it is turning 168 revolutions per minute? If a roll is seven inches in circumference, when it makes one turn 7 inches of sliver will be delivered. 36 inches 1 yd. 8. If the Coiler Calendar roll in the above ex- ample will deliver 19,600 yds. of a 54 grain sliver in 10 hours, what is the production in pounds ? 7000 grains=l pound. 9. If the circumference of front roll on slubber is 4 inches and it is making 138 revolutions per 32 minute, Iioav many hanks of 840 yds. each will be delivered in 10 hours' running? If the frame has 96 spindles, what is the total number of hanks made? If the roving is 1 hank, it means that there is 1 hank in 1 pound. How many pounds would this be for 96 spindles? SPINNING ROOM PROBLEMS 1. If there are 840 yds. of a No. 1 yarn in one pound, how many pounds would 16,800 yds. weigh of the same number? 2. If a No. 1 yarn has 840 yds. in 1 pound, what would be the number of the yarn if there were 25 ; 200 yds. in 1 pound? 3. If two ends of an 8 hank roving are run be- hind spinning frame, what is the resulting hank roving number? Hank roving -f- doublings = hank roving fed. 4. If a 6 hank roving is doubled on spinning frame and is being spun into a No. 30 yarn, what is the draft in spinning frame? 6^2=3 and 30^-3=10 draft. 5. If a frame has a draft of 8 and is making 24 's yarn, what is the number of hank roving used, if it is being used double ? 24-^8=3, then if double, 2X3=6 hank roving. 6. If 3 pounds of yarn has 51640 yds., what is its number, (840 yds. in 1 pound of number 1) ? 840X3=2520, then 51640-^2520=No. yarn. 7. The number of a yarn is found practically by reeling 120 yds. and weighing in grain. The weight in grain divided into the number 1000, will give the number of the yarn. 33 If 120 yds. of yarn weigh 50 grain, what is its number? If 120 yards of yarn weigh 100 grain, what is its number? If a yarn is No. 20, what should it weigh in grain? 8. If the weekly production of a spindle on 10 's yarn is 2 pounds, how many spindles must be put on to deliver a weekly production of 18500 lbs ? 1 spindle=2 pounds, therefore the number of pounds produced is twice the number of spindles, therefore the number of spindles would be \ the number of pounds, or 18500 — : =9250 spindles. 2 WEAVE ROOM PROBLEMS 1. If a warp has- 2592 ends in it and cloth made from it is 36 inches wide, how many ends per inch has the cloth? 2. If 4 harness are to be used on a warp of 2800 ends with equal number of ends for all harness, how mand heddles should be on each harness? 3. If a warp 2460 ends is drawn 2 ends per dent in reed, how many dents would be taken up in reed? 4. If a warp requires 1230 dents in reed to be drawn and it is to be set 30 inches wide, how many dents per inch should there be in the reed? 5. If five yards of cloth weigh 1 pound, how many pounds should 1 cut of 60 yards weigh? 6. If cloth is to be 36 inches wide, 72 ends per inch and woven with fancy pattern carrying 108 ends to one repeat, how many repeats of pattern would there be across cloth? 7. If cloth is to weigh 4 yds. per pound, what 34 would one yard weigh in ounces? (16 ounces==l pound.) 8. If a loom makes 160 picks per minute weaving cloth with 40 picks per inch, how many inches of cloth would it make for each minute? How many inches in 1 hour? (60 minutes.) How many inches in 10 hours? How many yards w»uld that be in 10 hours? 9. No. 1 yarn means that there are 840 yds. of that size yarn in 1 pound. In No. 2 yarn there are 2X840 yards in 1 pound and so on for any number of yarn taken. If we have 25200 yds. of yarn in 1 pound, what number is it? 25200-^840. 10. If the spread in the reed is 39 inches and the picks per inch 48, how many inches of filling would there be in one inch of cloth? How many yards would this be? How many yards of filling would this give in one yard of cloth? 11. A certain loom makes 40 yards per day. How many cuts of 60 yards will it make in 6 days : 40 yds. = 1 day. In 6 days,, 6X40=240 yds, then 240-^60=4 cuts. 12. A weekly production of 500 cuts of 60 yds. is desired from looms which can produce 40 yds. in 10 hours. How many looms must be put on? 500X60=30,000 yds. required in 60 hours. If 1 loom makes 40 yds. in 10 hours, in 60 hours it will make 40X6=240. 30,000-=-240==number of looms. 35 Fractions REDUCTION OF FRACTIONS For practical methods all proper fractions are reduced to their lowest terms before being used in the process. When a whole number is divided into two parts, (equal), each of these two parts is called one-half, % : and when divided into four parts, (equal), each of these parts is called one-fourth, J. If we compare these two divisions we will see that it will require 2 of the one-fourth divisions to equal one-half of the total number, while it will require only one of the one-half. divisions to make an equal amount. Then we see that the number, %, is the same in value or is equal to the number, %■ I n like manner % is equal to % or to %. Thus we see that multiplying both the top and the bottom figures in a fraction by any number, the value of the fraction is not changed. Also dividing both the top and the bottom figure by a common number does not alter its value. Thus if we multiply % by 2, we have % ; by 3, %; by 4, %; by 5, % ; by 6, 6 / 12 , etc., etc. Also if we divide % 6 , both top and bottom, by 4, we have % ; by 2, %, and if we divide 1 % 2 by 4 Ave have %; by 8, %; by 16, y 2 , etc., etc. Any improper fraction may be reduced to a mixed number by dividing the bottom figure into the top. Thus, iy 5 ==2y 5 , i 2 / 7 =l%, 16 / 5 =3i/ 5 , 22/ 8=2 %,=23/ 4 , etc., etc. 36 Example, 66 4 4 1 66 J 16% ~~4_' 26 24 % *\ mixed number can be reduced to an improper fraction by multiplying the whole number by the number below the line and adding in the number above the line. Use this new number above the line and place the old number below the line. Thus, to reduce 11% to an improper fraction we have 6X11=66, 66+5=71. Then the new number will y 6 . be, ra 37 Addition of Fractions In adding two or more fractions it is necessary to reduce them all to the same term. We cannot add Ys and % because the two numbers are express- ed in different terms and if we add the two in this form, the sum will be neither in "Thirds" or "Fourths." In order then to add two fractions it is necessary to reduce them to similar fractions. This means that all the fractions must be reduced so that the numbers under the line for all of them are the same. Thus it is simple to add Ys, % and % as they are all in the same term or "Eighths." This is done by adding together all the figures above the line and placing the sum above a new line under which must be the same figure which occurs below the line of all numbers thus added. The sum then of Ys and % and %=%=% (by divid- ing top and bottom by 2). If,, however, it becomes necessary to add such fractions as Ys and Y% and Y15 all must be reduced to like terms. Thus, to add Ys> Y% an( l Y15 we mul- tiply both bottom and top figures in each fraction by a number which will produce the same number on the bottom for all. In this case if we reduce % by multiplying bottom and top by 3 we have, % 5 . Also Yz multiplied by 5, top and bottom, gives % 5 . Then this gives us %5+%5-f%5=%5=/ ; 5 (reduced by dividing by 3). When it becomes necessary to add fractions where it is difficult to see how they can all be re- duced to like terms, the following process is used: 38 Example, add % 6 , % and % 5 . These must be reduced to like terms and as it is hard to see by what numbers to multiply, we find a number common to all those below the lines. This number is found by dividing first any one of these by any one other. Thus, 8|16. If the third 2 number is evenly divisible by the result, then do so, but as in this case 2 will not divide evenly into 15 we get our new number by multiplying togeth- er 8, 2 and 15 which gives 240. This number is to be used below the line for all new or reduced frac- tions for each addition. Then to reduce the first fraction, 3 / 1G , divide 240 by 16 which gives 15. Multiply this number, 15 by the number above the line, 3, and, place the result above the line which gives 4 %4o for the first fraction. The second frac- tion, %, is reduced in like manner. Thus 240-^-8 =30, and 30x3=90. Then the reduced fraction will be 9 %4o- The third fraction 2 / 15 is likewise re- duced. Thus 240^15=16 and 16x2=32. Then 3 %40 i s "the third reduced fraction. Thus, *5/ 240+ 90/ 24()+ 32/ 240= 167/ 24() . PROCESSES (a) To add a whole number and a proper frac- tion all that is necessary is to annex the fraction to the whole number. Thus, 2014+% 6 =2014% 6 . (b) To add a whole number to a mixed number, add the two whole numbers and annex the frac- tion. 39 Example, add 456 to llf, 456+11=467, annex t=467|. (c) To add a whole number to an improper frac- tion, first reduce the improper fraction to a mixed number and proceed as above. Example, add 305 and 15 / 4 . 15 / 4 =3f. Then 305+ 3=308. Annex f=308f. (d) To add two mixed numbers together, first add the two whole numbers and then add the two proper fractions as above and annex the two. Example, add together 105f and llf. First, 105+11=116, then f+f= 9 /i 5 + 10 /i5= l9 /io- There- fore, we have 116 1 % 5 . If, as in above, the resulting fraction is an improper one, reduce this to a mixed number and add as above. Then 19 /i 5 =l 4 /i 5 , then 116+lf 15 = H7 4 /l5- When only two fractions of different terms are to be added, the changing of the two in- to like terms is done by multiplying together the two numbers below the line in order to obtain the new bottom number for both fractions. Thus, to add f and f we have, 3X8=24 which is the new number below the line for both fractions. Then to find the new number to use in place of f, divide 24 by 8. Thus, 24-^-8=3. Multiply the old number above the line by this number and we have, 3X3 =9. Then the new number for our reduced frac- tion is % 4 . In like manner reduce the f to a frac- tion having 24 below the line. Thus, 24-^-3=8. Then 2X8=16. This gives for our new fraction X 24. 16 / 2 40 Then % 4 + 1 % 4 = 2 %4=l/<>4 (reduce to mixed number by dividing 25 by 24). CARD ROOM PROBLEMS 1. A roll is 1 % 6 inches in diameter. It is de- sired to increase its diameter £ of an inch, what will be the resulting diameter? 2. A man works 8f hours one day, 1\ the next and 10| the next. What is the total time? 3. The weight of three samples of roving is 56^,. 55f, and 54% grains. What is the total weight ? 4. The circumference of a certain doffer is 28% inches. How many inches of clothing strips will be required to make two wraps round? 5. 6% inches is the circumference of a certain . feed roll on card. How many inches of lap are fed when the roll turns two times? 6. Laps are weighed as follows : 46f pounds, 46^ pounds, 46| pounds and 46f pounds. What is the total weight? SPINNING ROOM PROBLEMS 1. A front roll on spinning frame is 3% inches in circumference. How many inches of yarn will it deliver in 2 turns? 3^+3%= inches. 2. Three weighings of a yarn are made as fol- lows : 32f grains, 31f grains, 32^ grains. What is the total weight? 3. A bobbin is 3^ inches in circumference. How many inches of yarn will be taken up for 3 wraps round ? 41 3|+3^+ 3i= inches. 4. The circumference of a warper cylinder is 37f inches. How many inches of warp is wound when cylinder turns 3 times? 37f+37f+37f= inches. WEAVE ROOM PROBLEMS 1. The front harness of a certain loom lifts 4f inches. If the back harness is to lift % inches high- er, what is its lift if 4%+%= inches. 2. 3 cuts of cloth weigh 14f, 15-J and 14^ pounds. What is the total weight? 3. A man works 8f hours one day, 1\ the next and 8 J the next. What is the total time? 4. The drag roll on a slasher measures 28f inches round. How many inches of warp are de- livered when roll turns twice? 28f-f28f= inches. 42 Subtraction of Fractions In subtracting fractions one from another it is necessary first to reduce to like terms as in addi- tion. When the two fractions are in the same terms all that is necessary is to subtract the small- er number above the line from the larger one and place the remainder above a new line below which must be placed the number occurring below lines in the two fractions subtracted. Thus to subtract % from % we have : % — %=% : =44 (reduced by di- viding top and bottom by 2). Example : Subtract % 5 from |. As the two fractions are in different terms; that is, the two numbers below the line not being the same, it is necessary to reduce both of them to like terms. Thus, 60 is the only number into which both 15 and 4 can be evenly divided so we use this as our new number below the line for both fractions. This number is usually found by simply multiply- ing together the two numbers below the line in both fractions. Then to reduce the first fraction % 5 divide 15 into 60. This gives 4. Multiply this 4 by two, (the number above the line) and we have 8. Then % is the reduced fraction. Next reduce | into a fraction which has 60 below the line. Thus 60-^4=15, and 15X3, (number above) =45. Then 4 % is the reduced fraction. Then, 45 / 6 o-%o= 37 /6o- 43 PROCESSES (a) To subtract a fraction from a whole num- ber, first subtract the fraction from 1 and reduce the first figure to the right in the whole number by 1. Example, subtract -J from 11. As 1 is equal to % (because 8-^-8=1) then% — ^ =%. Reducing the number 11 by 1 we have 10. Then joining or annexing the whole number, 10, and the fraction % we have 10% as our remainder. (b) To subtract two mixed numbers, first sub- tract the two fractions as shown under subtraction of proper fractions above. Then subtract the two whole numbers and join the two remainders to- gether for the result. Example, subtract 2-g- from 12f or 12f — 2-J. First %— %=%=% (reduced by dividing both top and bottom by 4). Then 12— 2=10. Joining the two we have 10y 2 . Example, subtract 3% 5 from 10|. First reduce the two fractions to like terms. Then 15X4 (number below lines) =60, then 60-^- 15=4. And 4X2=8. The first fraction then be- comes % . Then the next fraction f is reduced thus, 60-h1=15 and 15X3 (number above line) =45. Thus the next fraction becomes 4 % . Then 4 %o-%o= 3 %o- Then 10-3=7. Annexing the whole number and the fraction we have 7 3 % as the resulting number. When the fraction in the large number is small- er than that in the smaller number, the figure, 1, must be added to the top fraction and reduced to 44 an improper fraction. Then 1 must be taken from the first right hand top figure before subtracting the two whole numbers. Example, subtract 2f from 13^ or 13^— 2f. As we cannot subtract % from %, we add 1 to % making 1%. 1%=%. Then %-%= %=V 2 (by dividing by 4) . Then as we added this 1 to the frac- tion we must reduce the first number to the right in the large whole number by 1. Then we have 13 — 1=12 and 12 — 2=10. Joining the remaining whole number and the remaining fraction we have ioy 2 . (c)To subtract a mixed number from a whole number. First subtract the fraction from 1 as shown in (a) then reduce the first number to the right in the large number by 1. Then subtract the remaining two whole numbers and annex the remaining frac- tion and the whole number. Example, subtract 3% from 254 or 254—3%. As 1=% then %- %=%. 254-1=253. Then 253- 3=250. Result is 250%. CARD ROOM PROBLEMS 1. A roll is 1 1:l /{q inches in diameter. If it is reduced % of an inch, what is the remaining diam- eter? %=%6- Thenl 1: ^ 6 — % 6 = 1% 6 inches diame- ter. 2. A full bobbin of roving weighs 47f ounces. If the empty bobbin weighs 4^ ounces, what is the weight of roving on bobbin? 45 3. A roll delivers 267f inches of sliver in a min- ute. Another delivers 315^ inches in same time., what is the difference? 4. A card delivers 147% pounds one day and 152% pounds the next, what is the difference? 5. A yard of laps fed to a picker weighs 3| pounds. If this delivers 3-§ pounds, how much waste is made? 6. A lap fed to a card weighs | of a pound. If it delivers 90 yards of sliver weighing f of a pound, how much is lost in waste? SPINNING ROOM PROBLEMS 1. A bobbin of yarn weighs 4| ounces. If the empty bobbin weighs 1^ ounces, what is the weight of yarn on bobbin? 2. A warper beam when full weighs 516| pounds. The empty beam weighs 98^ pounds, what is the weight of yarn on beam? 3. A spindle makes 8,000 turns per minute. If the traveler has to lag behind the spindle 216| turns per minute to wind on the yarn, what is the speed of the traveler? 4. A certain yarn has 22% turns per inch twist. Another has 18% turns per inch, what is the differ- ence? 5. At the beginning of an empty bobbin the traverse is 6 inches. If the taper is 1% inches at both ends, what is the length of traverse when bob- bin is full? 6. A spool of full yarn weighs 18^ ounces. . If empty spool weighs 4| ounces, what is weight of yarn on spool? 46 WEAVE ROOM PROBLEMS 1. The weight of warp in a cut of cloth is 9f pounds. How much filling is in 1 cut if 1 cut weighs 15^ pounds? 2. The reed for a certain warp is 44f inches long. If the spread of warp in reed is only 38^ inches, what space should be left at each end of roed? 3. If a cut of cloth is to weigh 12f pounds and it weighs only 1T§ pounds, how much light is it? 4. If the opening of shed at reed is 2f inches when lay is back, what is the height of shuttle if ■J clearance is allowed? 2f — J= inches, height of shuttle. 47 Multiplication of Fractions When two proper fractions are multiplied to- gether the resulting number is of less value than either of the two fractions. In multiplying whole numbers the resulting number is always greater than either of the two. This difference must be kept in mind in multiplying fractions. For instance when we multiply 2X2 the result is 4 which is greather than either of the other two numbers, but when we multiply i by ^ the result is 5 which is smaller than either of the two fractions multiplied. PROCESSES (a) To multiply two proper fractions, multiply together the two numbers above the line and place the result above a new line. Then multiply together the two numbers below the line and place this under the new line. Example : Multiply % and % or % X V2 or read Y 2 of %. 3X1=3, and 2x4=8. Then the result is %■ In multiplying fractions which have large num- bers above and below the line the result may be ob- tained by using cancellation. Example : Multiply i% 6 by i 6 / 24 or io/ 36 X 16 / 24 . This may be expressed 10 X 16 Then cancel, 36 X24 5 X 2 5 00 X % 3 X 9 27 . 9 48 (b) To multiply a fraction by a whole number, multiply the number above the line by the whole number. Use this result for the new number above the line, and place the old number below the line under this new one. Example : Multiply Hie by 12 or 1:L / 16 X 12. 11X12=132 ; then n/ 16 xl2= 132 / 16 or 8% 6 or 8%. (c) To multiply a whole number by a fraction, multiply the whole number by the number above the line and divide this result by the number below the line. Example : Multiply 204 by f or 204Xt or f of 204. 204 X 3 = 612 ; then 612 -*- 5 = 122%. In multiplying a whole number by a fraction in which the number above the line is 1, it is only nec- essary to divide the whole number by the number below the line in the fraction. Example : Multiply 315 by | or 315 X* or £ of 315. Then 315 -=- 8 = 39f . Examples : How much is > 4 of 100 or 100 X l A ? How much is Y 8 of 56 or 56 X H- How much is Ho of 200 or 200 X Yio * (d) To multiply a whole number by a mixed number, multiply the two whole numbers first. Then multiply the whole number by the fraction and add the two results. Example : Multiply 272 by 4f . First, 272 X 4 = 1088 ; 49 3 272 X 3 816 Then 272 X — = = = 102 8 8 8 and 1088 + 102 = 1190. Examples: Multiply 116 by 8%. Multiply 212 by 3%. Multiply 419 by 7%. (e) To multiply three or more fractions to- gether, place all the numbers above the lines in a row with the sign of multiplication between them. Below these draw a line and place under this all numbers, below the lines of -fraction, with the sign between them. Then to solve proceed as in can- cellation : Example : Multiply together 10 /i 6 , %o, 15 / 3 o, %2 and i%. 8 Then J0 X £ XWXfi X W 1 ifi xwxfflxffix t i6 2^28 This can be more clearly understood when "Can- cellation" is reached. (f) To multiply two mixed numbers together, reduce the two numbers to improper fractions and proceed as in multiplication of proper fractions. Example : Multiply lOf by 4f or 10| X 4f . 10t = 32/ 3and 4 f = 23 /5 . 32 23 736 Then — X— = = 49 1 / 15 . 3 5 15 50 CARD ROOM PROBLEMS 1. If the circumference of a roll is 6| inches, how many inches of sliver will it deliver in turning 5^ times ? 2. If a certain roving has 1-J turns per inch, how many turns would there be in 367f inches? 3. If a roll delivers 450^ inches in a minute and is putting in 1-J turns per inch in roving, what is the spindle speed? As the spindle must make 1^ turns for every inch delivered, it must make 1^ X 450J for this delivery. 4. A certain card produces 147f pounds per day. How much would it make in 6 days ? 5. If a yard of No. 1 roving weighs 8% grains, how many grains would there be in 12 yards of No. 1 roving? SPINNING ROOM PROBLEMS 1. If the cylinder on a warper is 37f inches in circumference, how many inches of w T arp will it wind in turning 12-J turns? 2. If a roll delivers 387-J inches in a minute, how many inches will it deliver in 60 minutes? 3. If a man works lOf hours per day, how much time can he put in in 6£ days ? 4. If 1 yard of No. 1 yarn weighs 8% grains, how many grains would there be in 120 yards of No. 1 roving? WEAVE ROOM PROBLEMS 1. If 52| yards of No. 1 yarn weigh 1 ounce how many yards of No. 1 would there be in 10^ ounces ? 2. The width of spread in reed of certain warp 51 is 38f inches. How many inches of filling are used Avhen shuttle goes across 12 times ? 3. How many ends would there be in warp for cloth 36^ inches wide and 64 ends per inch? 4. If a sand roll on a loom is 14| inches in cir- cumference (around), how many inches of cloth would be made when it makes 3| turns? 5. If the circumference of drag roll on slasher is 28f inches and turns 20^ turns per minute, how many inches of warp are delivered in 1 minute? 52 Division of Fradions Since a fraction is an indicated form of division, 1 the number 1 may thus be shown, — ; because 1 -=- 1 2 * = 1, and 2 may be shown — , etc. If we wish to 1 divide the number 1 into two equal parts, these two 112 1 parts will be ^ and ^ ; because 1 = — = — ■ 2 2 2 1 as we have seen. If then we divide 1 into two parts 1 12 1 and get — , then we can say - — divided by — = — , 2 1 12 12 1 or ; = — . From this we see that the num- 112 ber below the line in the result is found by multi- plying together the top number in the second frac- tion by the bottom number in the first fraction. Also the top number in result is found by multiply- ing together the top number in first fraction by the bottom number in second fraction. 3 4 Example : Divide f by | or f -^ ^ or — 3 1 3X2 6 Then 4 2 1X4 4 6 1 To prove this if we multiply together — and — 4 2 53 the result ought to be | just as in dividing 6 by 3 the result 2, when multiplied by 3 gives 6. 6 16 3 Then — X — = — = — 4 2 8 4 (by dividing top and bottom by 2). In dividing one fraction by another,, always place the fraction to be divided, first ; and proceed as in- dicated above. The following diagram may be useful in dividing fractions : FIG. 1. PROCESS (a) To divide a proper fraction by a whole num- ber, multiply the bottom number, in the fraction to be divided, by the whole number, place this result below a new line and place above this line the same number occurring above the line in the fraction divided. Example : Divide % by 2 or % -^ 2 or — , 4X2 = 8, then placing this number below a new line and the number, 1, above we have -J. 54 3 /l6 Example : Divide % 6 by 4 or % 6 -f- 4 or — , 4 16 X 4 = 64, then the result is % 4 . 6 /l8 Example : Divide % s by 2 or % 8 -f- 2 or - — , 2 6 1 then 18 X 2 = 36. Result = — = — (by reducing). 36 C If the number above the line is evenly divisible by the whole number, the result may be obtainec by dividing this top number by the whole number to produce the new number above the line. The new number below the line in this case would be the same as that below the line in the number divided. 9 /l6 Example : Divide % 6 by 3 or % 6 -^- 3 or , 3 then 9-^3 = 3. Result : % 6 . (b) To divide a whole number by a fraction, multiply the number below the line in fraction by the whole number and place this result above a line in result. The new number below the line in result is same as that above the line in old fraction. 2 Example : Divide 2 by | or 2 -=- i or — . 4X2 = 8. Result : — = 1 55 4 Example : Divide 4 by % 6 or 4 -r- %q or . 64 16 X 4 = 64. Result: — = 21% (by reduction). 3 (e) To divide a mixed number by a whole num- ber,, reduce the mixed number to an improper frac- tion and proceed as under (a). 25| Example : Divide 25| by 16 or 25f -f- 16 or 16 Reduce 25|, 25 X 4 = 100, and 103 100 + 3 = 103 ; then 254 == and 4 103 103 103 =- 16 = = = 18% 4 . 4 4X16 fl (d) To divide a whole number by a mixed num- ber, reduce the mixed number to an improper frac- tion and proceed as in (b). 24 Example : Divide 24 by 5f or 24 -=- 5| or . Reduce 5f, 5 X 8 = 40 and 40 + 3 = 43 ; then 5f = 4 % and 24 ■+- 4 % = 24 X 8 192 = = 420/ 43 43 /43- (e) To divide one mixed number by another, reduce both to improper fractions and proceed as in dividing one fraction by another. 56 25| Example : Divide 25| by 4^ or 25| ^- 4J or ; 25| = 103 /4 and 4£ = 33 /s; then 103 X 8 824 10 % -f- 33 /s =■- - - - 4 X 33 132 6 32 /l32 = 6% 3 . (f ) When two fractions are to be divided which have the same numbers below the lines, or can be reduced to have the same numbers, all that is nec- essary is to divide the two numbers above the lines. 16 /s Example: Divide x % by % or 16 / 8 -j- % or ; % ' then 16 -j- 4 = 4, answer. This is true because in dividing as shown in the 8X16 proceeding we have 1 % -=- % = = 4. The & 8X4 above and below will always cancel. l 3 /l6 Example: Divide 1% 6 by 1 or 1% 6 -=- 1 or ; 1 = i% 6 and 1% 6 = i% 6 , then i% 6 -5" 16 /l6 = 19 /l6 = l 3 /l6 1 Example : Divide 1 by 1% 6 or 1 -i- 1% 6 or ■716 1 = i% 6 and 1% 6 = i% 6 , then i% 6 19 /i6 = 16 /i9. 57 2f Example : Divide 2f by I5 or 2f -f- 1| or n 2f = 1% and li = % = 1%, then ' 19 /s^ 10 /8== 19 /lO = l 9 /lO. CARD ROOM PROBLEMS 1. If a roll delivers 457f inches of roving in a minute and has a circumference of 4^ inches, how many turns is it making in 1 minute ? 457f -=- 4^ == turns per' minute. 4(457) +3 1831 4A = 4 4 2X4+1 9 18 then issi/4 -=- i8/ 4 = 1831 /is — 101 13 /i 8 turns per minute. 2. If a roll delivers 358% inches of roving in a minute and turns 110 times, what is the circum- ference? 358% -r- 110 = circumference. 8X(358)+7 2871 358% = = . 110 = 88%, 2871 880 2871 then 1 = == 3 23 % 80 inches 8 8 880 circumference. 58 3. Circumference of a roll = diameter X 22 A- If a roll is 11%4 inches in circumference, what is the diameter? If 6 = 2 X 3, then 6-^2 = 3 or 6-^3= 2. Then 11%4 -z- 2 % = diameter. 157 22 44 11% 4 = and — = — . 14 7 14 157 44 157 Then : • = = 3^ inches diameter. 14 14 44 4. One yard of No. 1 roving weighs 8% grains, how many yards would there be in 237% grains ? If 1 yard = 8% grains, then 237% grains would 237% have yards. 949 25 237% = and 8% = — . 4 3 949 2847 (multiply by 3) 12 25 100 and — = (multiply by 4) 3 12 2847 100 2847 then i = = 28 4 %oo yards. 12 12 100 5. One card produces 142% pounds of sliver per day. How much time would it be in making 12% pounds 1 59 713 61 142% = and 12% = — , 5 5 61 713 61 then ■- ■ = of a day. 5 5 713 SPINNING ROOM PROBLEMS 1. If one spindle will produce 1^ pounds of 255 warp in a week, how many spindles would be re- quired to make 8,456 pounds of the same number? 8546 -=- If 2. A roll delivers 327f inches of yarn in a min- ute. How many minutes would be required to de- liver 1162% inches? 1162% -v- 327 %. 3. There are 52^ yards of No. 1 yarn in 1 ounce. How many ounces of No. 1 would there be in 1152f yards? 4. Front roll on a certain spinning frame deliv- ers 382f inches of yarn per minute. If the spindle is running 8500 R. P. M., how many turns per inch would be put in? . 8500 -=- 382f. WEAVE ROOM PROBLEMS 1. If the sand roll on a loom is 14f inches in cir- cumference, how many times does it turn in deliv- ering 436 inches of cloth ? 2. The sand roll on a certain loom is 14f inches in circumference. If the loom makes 684 picks while 60 the sand roll turns 1 time, how many picks per inch are put in? 3. One loom will make 37| yards of a certain cloth per day. How many looms must be put on to make 2460 yards in a day? 4. There are 52^ yards of No. 1 yarn in 1 ounce. If we have 2642f yards of yarn and it weighs 2 ounces, what is the number ? 521 X 2 = 105 yards of No. 1 in 2 ounces, then 2462f -*- 105 = No. yarn. 61 Decimals A decimal is a simple form of indicating frac- tions, when they are expressed in "tenths," "hun- dredths," "thousandths," etc. Thus Y 10 is expressed decimally .1. % o is expressed decimally .01. % 000 is expressed decimally .001. Therefore,, when the fraction is in "tenths" it is expressed decimally by having only one figure to the right of the decimal point. Examples : % = .1, 2 / 10 = .2, % = .3, 4 / 10 - .4, 5 /io = -5, 6 /io = -6, 7 /io = -7, 8 /io = -8, % = -^ io/ 10 =l=1.0, ii/ 10 -iy 10 =l.l, i% - iy 10 - 1.2, etc. When the fraction is in "hundredths" it is ex- pressed decimally by having two figures to the right of the decimal point. Examples : Yioo = .01, t4oo = -02, 3 /ioo = .03, 4 /ioo = .04, %00 = .05, 6 /ioo = = -06,. %oo = = .07, 8 /ioc • = .08, 9 /ioo = = .09 , 10 /ioo = .10, iy 100 = • 11, 20 /ioo = .20, 3 %>o = = .30, 40 /ioo = = .40 , 50 /ioo = -5C », 60 /ioo - .60, 7 %oo - -™, so /ioo = -80, 9 %oo - -90, ™% o = 1 == 1.00, ioy 100 = iy 100 = i.oi, io2 /loo = i2 /l00 = i.o2, 110 /ioo = l^/ioo = 1-10, "Koo = l u /ioo = 1.U, etc. When the fraction is in "thousandths it is ex- pressed decimally by having three figures to the right of the decimal point. 62 Examples : % oo = 001, %ooo = -002, % 000 = -003, io/iooo - -010 or i% 000 - Y 10 o = .01, "/iooo = •Oil, 12 /iooo = -012, 2 %ooo = -020, ^/ 1000 = .021, 100 /iooo = -100 or y 10 = .1, 101 /iooo = .101, ^% oo = .110 or ^/iooo^/ioo or .11, H&0-.1U, 120 /iooo= .120 or i.20 /l000 = i2 /l00 = 12 , etc. From the above examples we see that a cipher or the figure, 0, when placed to the right of the last figure in the decimal does not change its value. Hence these ciphers, or noughts, may be added or taken away at will. REDUCING FRACTIONS TO DECIMALS A decimal being a fraction expressed in y > ^4 00 , etc., it is possible to change any fraction to a decimal. If we have the fraction % and desire to change it into its equivalent in % or "tenths," we simply divide the number, % into Yo parts. Then % divided by Y 10 = % -*- % - i0 A = 2% = 2y 2 . Then as 2% is the result of dividing % into % 2.y 2 parts this number is the same as . Now as dec- 10 imals are shown, by placing the point,. ( . ), before the number instead of using 10 below the line, this 2% number may be written .2y 2 . Again, if we de- 10 sire to. change the number ->4 ni to Yioo parts or ""hundredths," we divide it into % 00 parts. Then % -h y 10 o = 10 % = 25. Therefore, if 25 is 63 the result of dividing % into y 100 or "hundredth" part, this number, 25, is the same as 2 % o- There- fore this number may be expressed decimally thus .25. If it is desired to change or reduce the frac- tion % to a decimal and as we do not know whether the resulting decimal will be in one figure, two figures or three figures to the right of the point, it is best to reduce by dividing it into "thousandths" or three places to the right of the point. Then to reduce % we have : % -+- Kooo = 300 % = -750. This is read 75 %ooo> or by reducing, 7 % o- From this we see that to reduce a fraction to a decimal we divide by the fraction %ooo- This is the same as multiplying the number above the line by 1000 and dividing into this result the number be- low the line. In deciding where the decimal point is to come in the result, it is only necessary to de- cide, after dividing, how many of the ciphers or noughts were used in the division. If only one " " is used, then the point would be placed so that one figure is to its right. If two "0's" are used, two figures must be to the right of the point. If three "0's" are used, three figures must be to the right of the point. CARD ROOM PROBLEMS 1. If the licker-in is 9% inches in diameter ex- press it in decimals. As this is a mixed number it is only necessary to reduce the Y 2 to a decimal. 1 X 1-000 Then = .5, then 9% = 9.5. 2 64 2. What is the decimal equivalent of a 2% inch feed roll on card? 3. What is a 1% 6 inch roll equal to in decimals? Al 3 / 16 roll? 4. What is the decimal equivalent of a 1% inch back roll on driving frame? 1% inch front roll? SPINNING ROOM PROBLEMS 1. If whirl on spindle is % inch in diameter ex- press its diameter in decimals. 7 X 1000 = 7000, then 7000 -f- 8 = .875 inches. 2. If whirl is 1% 6 inches, express it in decimal. As this is a mixed number it is only necessary to re- duce the fraction and join the two. Then 1 X 1000 = 1000,. and 1000 -=- 16 = .062+. As this does not come out even in dividing we add another "0" and obtain .0625. Then result = 1.0625. 3. If front roll on spinning frame is V/ 8 inches in diameter, express it in decimals. 4. If a certain bobbin contains 3% ounces of yarn, express it in decimals. WEAVE ROOM PROBLEMS 1. 1 yard of No. 1 yarn weighs 8% grains; ex- press it in decimals. 1 x 1000 = 1000 and 1000 -4- 3 = .333+ then we have 8.333. 2. If a reed has an average of 30% dents per inch, express it in decimals. 3 X 1000 = 3000, and 3000 -h 4 = .750 Then 30% = 30.75. 3. A certain bobbin has 3% ounces of yarn in it ; express it in decimals. 65 4. The spread in reed for a certain cloth is 38% inches ; express it in decimals. UNITED STATES MONEY The money of our Government is expressed in decimals. There are 100 cents in a dollar. Then one cent would be %oo of a dollar and is expressed .01. Twenty cents is 2 %oo of a dollar and is ex- pressed thus, .20. The dollar sign ($) is usually placed before the amount indicated. Thus fifteen cents is written $.15, one dollar and fifty one cents, $1.51, etc. Y 10 of a cent is called a mill and would De /looo °f a dollar and is written or expressed then, $.001. Write in figures the f ollowing : One dollar, fifty-three and one-half cents. $1,535. Twenty dollars, forty-two and one-fourth cents. $20.4225. Fifty-one dollars and fifteen cents. $51.15. 66 Addition of Decimals In finding the sum of two or more decimals, place them one under the other so that the decimal points will come directly under each other and proceed as in addition of whole numbers. Example : Add .125, .364, .7821 and .1234. .125 .364 .7821 .1234 1.3945 Place the point in the result directly under all the points in numbers added. This will always give as many figures to the right of point as there are fig- ures in the longest decimal used. In the above addition the longest decimal contained four figures and we must have four figures to the right of the point in the result. CARD ROOM PROBLEMS 1. If 8 laps weigh as follows : 42% pounds, 41% pounds, 40% pounds, 41% pounds, 42% pounds, 41% pounds,, 42% pounds and 41% pounds, what is the total weight expressed in decimals? % = -5, % = -75, % = .875, % = .25. Then we have 67 42.5 41.75 40.875 41.25 42.75 41.875 42.5 41.75 335.250 pounds = 335% pounds 2. If the overseer makes $31.25 per week, sec- ond hand $21.43 and six fixers $19.38 each, what is the total wages paid? 3. If four samples of roving are weighed as fol- lows, what is the total? 52.4 grains, 53.1 grains, 52.4 grains and 52.6 grains. 4. The circumference of a certain roll is 3.425 inches; how many inches of roving will it deliver in turning 3 times? 3.425 + 3.425 +.3.425 = inches. SPINNING ROOM PROBLEMS 1. A roll is 3.1416 inches in circumference ; how many inches of yarn will it deliver in turning 3 times ? 3.1416 + 3.1416 + 3.1416= inches. 2. The overseer gets $31.42, the second hand $22.60, 4 section hands $18.25 each; what are the total wages paid? 3. A spinning room produces 2624.32 pounds of 28 's, 4681.47 pounds of 30 's, 12822.67 pounds of 40 's and 15461.38 pounds of 22 's in a week; what is the total production? 68 4. 4 samples of the same yarn are taken and weigh 32.4 grains, 31.8 grains, 32.2 grains and 31.9 grains; what is the total weight? WEAVE ROOM PROBLEMS 1. A certain loom produces 38.24 yards in a day ; how many yards would it produce in 4 days? 38.24 + 38.24 + 38.24 + 38.24 = yards. 2. Four samples of cloth weigh 15.27 pounds, 14.84 pounds, 14.92 pounds and 15.18 pounds ; what is the total weight? 3. Overseer receives $32.41, two second hands $22.50 each and 8 section hands $18.75 ; what is the total wage paid? 4. Supplies are received as follows: 2 shuttles $.94 each, 2 reeds $1.14 each, 4 picker sticks at $.09 each ; what is the total cost ? 69 Subtra&ion of Decimals In subtracting one decimal from another,, place the one number under the other as in addition of decimals and proceed as in subtraction of whole numbers. Example : Subtract .1234 from .5216. .5216 .1234 .3982 Example : Subtract 1.923 from 3.134. 3.134 1.923 1.211 ' Example : Subtract .785 from 2. 2.000 .785 1.215 CARD ROOM PROBLEMS 1. A yard of a certain lap weighs 5261.24 grains. If it makes 90 yards of sliver which weigh 5182.36 grains, how many grains are lost? 2. Back roll on drawing frame takes in 110.26 inches of sliver, while the front roll is delivering 624.82 inches. How much more does front roll de- liver than back roll? 70 3. A lap is supposed to weigh 5265.64 grains. If it weighs 5372.72 grains, how much too heavy is it? 4. A full bobbin of roving weighs 47.64 ounces. If the empty bobbin weighs 4.34 ounces, how much roving is on the bobbin? SPINNING ROOM PROBLEMS 1. A full bobbin weighs 5.26 ounces. If the empty bobbin weighs 2.42 ounces, how much yarn is on bobbin? 2. One roll is 3.52 inches in circumference and another is 4.23 inches ; how much more roving will the latter roll deliver than first when both rolls will have turned two times? 3. A spool when full of yarn is 12.56 inches in circumference. When near empty it is 3.75 inches in circumference. How much more yarn will be unwound in 2 turns when full than will be unwound in 2 turns when near empty? 12.56+12.56 = 25.12 inches unwound when full. 3.75-j- 3.75 = 7.50 inches unwound when near [empty. 17.62 inches more. 4. The breaking weight of a 20 's warp is 88.34 pounds, of a 30 's warp it is 66.28 pounds. What is the difference? WEAVE ROOM PROBLEMS 1. A reed is 45.25 inches long, the warp is to be spread 39.6 inches. How many inches must be left on each end of reed? 2. A cloth is to weigh 4.84 yards per pound which makes a cut of 60 yards weigh 12.39 pounds. 71 If a cut weighs 13.25 pounds, how much loss would this be in every cut? 3. A certain cloth weighs 2.85 ounces per yard. If the warp in 1 yard weighs 1.93 ounces, what is the weight of the filling? 4. A sample of sized yarn weighs 26.8 grains, after the size has been washed out and yarn dried it weighs 25.9 grains. How much size was in the yarn? 72 Multiplication of Decimals To multiply one decimal by another place the two numbers as in multiplication of whole numbers re- gardless .of where the two decimal points are. Proceed to multiply as in whole numbers and place the decimal point in the result so that there will be as many figures to its right as are counted to the right of the points in both numbers multiplied. Example : Multiply .3041 by .012. .3041 .012 6082 3041 36492=.0036492 In the bottom number multiplied we count three figures to the right of the point, and in the top num- ber we count four figures. Both together this gives seven. Then we must have seven figures to the right of point in the result. As our multiplication only produced five figures we must make up the seven by prefixing "0's." Therefore we have the number .0036492 as shown to the right above. When two decimals are multiplied together the resulting decimal is smaller than either of the two multiplied. Example : .25 X .25=.0625. When a whole number is multiplied by a decimal 73 the resulting number is smaller than the whole number. Example: 4X .25=1.00. To multiply two mixed, numbers in decimals. Example : Multiply 2.75 by 1.785. 1.785 2.75 8925 12495 3570 4.90875 CARD ROOM PROBLEMS 1. The feed roll on a card is 2.5 inches. If it runs .85 revolutions in a minute, how many inches of lap are fed to licker-in? When the roll makes one turn it will deliver as many inches as there are inches in its circumference (distance around). The circumference is found by multiplying the diameter (distance through) by the standard number, 3.1416. Then 2.5X3.1416=7.854 inches delivered in 1 turn. If it makes .85 in 1 minute, we have, 7. 854X .85=6.6759 inches delivered to licker-in in one minute. 2. The doffer on a certain card is 27% = 27.75 inches diameter through clothing. If it makes 10.3 revolutions per minute, how many inches of sliver will it produce in one minute? 74 Circumference=27.75X 3.1416=87.179 inches ; then 87.179X10.3=879.94 inches delivered in one minute. 3. The diameter of a metallic roll which is 1% inches is figured as 1.83 inches diameter. If front roll on a drawing frame is 1%=1.83 inches in diam- eter and is running 300 revolutions per minute, how many inches of sliver will it deliver in 1 hour? «» Circumference=1.83X 3.1416=5.749 inches then 5.749X300=1724.700 inches delivered in one minute, 1 hour=60 minutes, then 1724.7X60=103482.0 inches in 1 hour. 4. A man works 24% hours at the rate of 2V-/ 2 cents per hour, what is the total amount he should receive ? 24% hours=24.75 2iy 2 cents=$.215 then 24.75 X-215=$5.32. 5. If the back roll on a slubber takes in 275.24 inches of sliver in a minute and the draft is 3.75, how many inches will the front roll deliver in 1 minute ? Draft X inches fed = inches delivered; then 275.24X3.75=1032.15 inches delivered. 6. The twist per inch for roving is found by mul- tiplying the square root of the hank roving to be made by the standard, 1.2. If 4 H. R. is to be made, how many turns per inch should be inserted? V4=2. Then 2x1.2=2.4 turns per inch. 75 Find twists for following numbers of H. R. : .25, V.25=.5, then .5 Xl.2--=.60 turns per inch. .75, V.~75=.S66, then .866X1.2=1.039 turns per inch. .2, Vl*~=1.414, then 1.414 X 1.2=1.696 turns per inch. SPINNING ROOM PROBLEMS 1. The weight of yarn on a bobbin from a 2% ring and 7 inch traverse is about 3.875 ounces. How many ounces would there be in a doff from 300 spindles ? 3.875X300=1162.500 ounces. 2. 1 yard of No. 1 yarn weighs 8.333 grains,, what would 550 yards of a No. 1 weigh? 8.333X550=4583.15 grains. 3. If the ratio of cylinder to whirl is 1 to 7.25, what would be the speed of spindle if cylinder runs 960 revolutions per minute? This ratio means that when cylinder makes 1 turn the whirl and spindle makes 7.25 turns. Then the whirl turns 7.25 times as fast as the cylinder and if the cylinder makes 960 turns the spindle will make, 960X7.25=6960.00 revolutions per minute. 4. The twist per inch for warp yarn is found by multiplying the standard number, 4.75, by the square root of the number of yarn. Find the turns per inch to be put in following warp numbers : 36, V36=6, then 4.75x6=28.50 turns per inch. 28, V28=5.291, then 4.75x5.291=25.13 turns inch. 30, V-30=5.477, then 4.75x5.477 = 26.01 turns inch. 76 per per 5. The twist per inch for filling is found by mul- tiplying the standard number, 3.50 by the square root of the number of yarn. Find the turns per inch to be put into the follow- ing numbers : 38, V38"= 6.164, then 3.50 X 6.164 = 21.57 turns per inch. 40, ViO = 6.324, then 3.50 X 6.324 = 22.13 turns per inch. 44, V44 = 6.648, then 3.50 X 6.648 = 23.26 turns per inch. 6. The diameter of a front roll on a certain spin- ning frame is 1%=1.125 inches, and it is running 112.8 revolutions per minute. How many inches will it deliver in 1 minute? In one turn the roll will deliver as many inches as there are inches in its circumference (distance around). The circumference is found by multiply- ing the diameter by 3.1416. Then 1.125 X 3.1416 = 3.5343 circumference = inches delivered in one revolution. Then if it turns 112.8 times in one min- ute, the number of inches delivered in 1 minute will be 3.5343 X 112.8 = 398.669 inches delivered in 1 minute. 7. If a girl runs 10 sides and is paid 17 cents per day per side, what should be her pay for 6 days' work? 17 cents = $.17. Then 10 X $-17 = $1.70 per day, and $1.70 X 6 = $10.20 pay for 6 days. WEAVE ROOM PROBLEMS 1. If a reed has 28.75 dents on 1 inch, how many dents would there be on 37.84 inches ? 77 28.75 X 37.84 = 1087.9 dents. . 2. A cloth has an average of 84.72 ends in 1 inch. If it is to be 36% inches wide, how many ends should there be in the warp? 36y 2 = 36.5. Then 84.72 X 36.5 = 3092.28 ends ; this would be called 3092 ends. 3. The sand roll on a certain loom is 14.75 inches around. How many inches of cloth will be taken up when the roll turns 5.375 turns? 14.75 X 5.375 = 79.28 inches. 4. If the diameter of drag roll on slasher is 9.452 inches and it is turning 19.52 revolutions per min- ute, how many inches of warp will be delivered in 1 hour? When the roll makes one turn it will deliver as many inches of yarn as there are inches around the roll. The number of inches around the roll (cir- cumference) is found by multiplying the diameter by 3.1416. Then 9.452 X 3.1416 = 29.694 inches delivered in 1 turn. And 29.694 X 19.52 = 579.626 inches delivered in 1 minute. Then 579.626 X 60 (minutes in 1 hour) = 34777.560 inches delivered in 1 hour. 5. If "cloth is to weigh 5.64 yards per pound,, how many yards would there be in 16.75 pounds? 1 pound = 5.64 yards. Then 16.75 pounds would have 16.75 X 5.64 = 94.47 yards. 6. If a weave room prod es in a week, 5236 78 pounds of a cloth which weighs 4.75 yards per pound, how many yards would that be? 5236 X 4.75 = 24871 yards. 7. There are 52.5 yards of No. 1 yarn in 1 ounce. How many yards would there be in 14.62 ounces of No. 1?" 14.6 X 52.5 = 766.5 yards. 8. If there are 52.5 yards in 1 ounce of No. 1 yarn, how many yards would there be in 1 ounce of No. 30 yarn! 52.5 X 30 = 1575.0 yards. 9. The spread in reed of a certain warp is 39.84 inches. How many inches of filling would be put into 1 inch of cloth if there are 64 picks to the inch? 39.84 X 64 = 2549.76 inches. 10. If a weaver receives 24 cents a cut and pro- duces 64 cuts in one week, what should be his pay? 24 cents = $.24. Then 64 X $.24 = $15.36 per week. 79 Division of Decimals In dividing one decimal by another proceed as in division of whole numbers. In placing the decimal point in the result, place it so that there will be as many figures to its right as the difference between the numbers to its right in the two numbers divided. Example : Divide .4562 by .0136. .0136 ) .4562000 ( 33.544 408 482 408 740 680 600 544 560 544 In the above example we place the numbers .0136 and .4562 as shown and divide as in whole numbers disregarding the decimal points and any ciphers or "0's" which occur between the point and the first figure in the number. In this case, then we divide 136 into 456 and it goes 3 times with 48 over as seen. Drawing down the next figure, 2, we have 482 into which we again divide 136. After the figure, 2, is brought down We see that the division is not complete, because we have 74 left over. Then we add "ciphers" or "noughts" to the number di- 80 vided bringing these down in the process as if they were figures. To decide where the point is to come in the result, keep in mind the number of figures to right of the point in number divided by. When we have used up as many figures to the right of point in number divided as there are figures to the right in the number divided by, the point should be placed to the right of the figure in result. In this example, after we have drawn down the figure, 2, we see that we have used as many figures to the right as we have in the number .0136. Then after making the division, "136 into 482 = 3," place the point after this figure, 3. As many "ciphers" or "noughts" may then be added to the number, .4562,. accord- ing as to how many figures Ave desire to the right of point in result. Example : Divide .0231 by .001. A .001) .023 11 (23.1 9 The line, A, may be drawn between two figures to show where the point is to go in the result. We place this line between 3 and 1 thus cutting off three figures to the right of the point in the number .0231, to equal the total number of figures to the right of point in number, .001. Thus after we have brought down the figure, 3, and divided 1 into it to produce 3 we place the point in the result after this figure, 3. Example: Divide 2.0534 by .025. 81 .025 ) 2.053J40 ( 82.136. 2 00 53 50 34 25 90 75 150 150 This division can be proved correct by multiply- ing the result 82.136 by the number divided by, .025.. Thus, 82.136 X -025 = 2.053400, number divided. Example : Divide 1 by .001. .001 ) 1.0001 ( 1000. 000 In this case it was necessary to add as many "noughts" to the figure, 1, as we have in the figure .001. Remember always that when "noughts" are added to a whole number, as in this case, the point must be placed after the whole number. Example : Divide 53.5 by 2.352. 2.352 ) 53.500|00 ( 22.74 47 04 6 460 4 704 17560 16464 10960 9408 Example : Divide .123 by 100. 100. ) |.12300 ( .00123 100 230 200 300 300 CARD ROOM PROBLEMS 1. A lap weighs 40.75 pounds and is 56 yards long; how many ounces does 1 yard weigh? 1 pound = 16 ounces ; therefore, 40.75 X 16 = 652 ounces, wt, of 56 yards ; then 652 -4- 56 = 11.6+ ounces per yard. 2. Laps are being made weighing 40.75 pounds. If 30 bales of cotton averaging 475 pounds per bale are opened, how many laps will it make? 30 X 475 == 14250 pounds of cotton run ; then 14250 --=- 40.75 = 349.6 laps made. 3. A lap 56 yards long weighs 41.25 pounds ; how many yards of sliver will be made weighing 54.5 grains per yard? 1 pound = 7000 grains; then 41.25 pounds = 41.25 X 7000 = 288750 grains ; each yard of sliver weighs 54.5 grains,, 288750 then = 5298.1 yards of sliver. 54.5 4. A feed roll on a card is 2% inches in diameter ; how many times will it turn in taking in 1 yard of lap? 2.5 X 3.1416 = 7.854" circumference; 1 yard = 36 inches ; then 36 -f- 7.854" = 4.583 turns. 5. The cylinder on a card is 50" in diameter and is 45 inches wide. How many feet of card fillet will it take to cover it, if fillet is 2" wide ? 3.1416 X 50 = 157.08 inches circumference ; so it will require 157.08 inches of fillet for each round. The taper will be 2" in each 157.08 inches. 45 -=- 2 = 22% rounds of fillet to cover 45 inches; then 157.08 X 22% = 3534.3 inches of fillet to cover cylinder. Or 3534.3 -r- 12 = 294.5 lineal feet of fillet. 6. A man draws $12.75 for running frames at 8 cents per hank. How many hanks did he run? 7. Fine frame spindles are running 1200 R.P.M. and front roll is delivering 629.42 inches per minute ; how many turns per inch are being put in? SPINNING ROOM PROBLEMS 1. Spool when full is 12.56 inches in circumfer- ence. When near empty it is 3.75 inches in circum- 84 ference. How many more turns must the latter size spool make to deliver the same amount of yarn as the full spool? In 1 turn of the full spool it will deliver 12.56 inches of yarn, and in 1 turn of the empty spool it will deliver 3.75 inches; therefore, 12.56 -=- 3.75 = 3.35 times as fast. Example : If the spool above when full turns 100 R.P.M. to deliver a certain amount of yarn it will have to turn 100 X 3.35 = 335 R.P.M. when near empty to deliver the same amount. Why is the cone drive put on a warper? 2. If spindle is running 8400 R.P.M. and front roll is delivering 437.48 inches per minute, how many turns per inch are being put in? 8400.00 -f- 437.48 = 19.2. turns per inch. 3. If spindle speed is 8600 R.P.M. on spinning frame and yarn No. is 36 warp, what is the front roll delivery in yards in 1 hour? V 36 = 6 and 4.75 X 6 = 28.5 turns per inch, then 8600^-28.5 = 301.75 inches delivery in 1 minute and 301.75 X 60 = 18105 inches delivery in 1 hour, then 18105 ~ 36 (inches per yd.) = 502.91 yards delivery in 1 hour. 4. If the front roll is 1 inch in diameter, how many turns per minute is it making for the deliv- ery in above example? It is delivering 301.75 inches per minute ; in 1 turn it will deliver 1 X 3.1416 = 3.1416 inches (be- cause its circumference is 3.1416")- Then 301.75 -r- 3.1416 = 96.4 revolutions per minute. 85 WEAVE ROOM PROBLEMS 1. If cloth weighs 5.64 yards per pound, how many ounces would 1 yard weigh? 5.64 yards = 1 pound = 16 ounces ; then 16 -=- 5.64 = 2.83 ounces weight of 1 yard. 2. If cloth weighs 4.75 yards per pound, what would be the weight of a 60-yard cut? 60 -f- 4.75 = 12.63 pounds. 3. 6 repeats of a fancy pattern measure 7.25 inches. Each repeat has 96 ends. What is the "over-all" construction or average ends per inch? 96 ends X 6 repeats = 576 ends on 7.25 inches ; then 576 -f- 7.25 = 79.44 average ends per inch. 4. If a sand roll on loom is 14.35 inches in cir- cumference, how many times will it turn in deliv- ering 1 yard of cloth? 1 yard = 36 inches; then 36 -=- 14.35 = 2.5 turns. 86 Ratio and Proportion The relation of one quantity to another of the same kind is called the ratio of the one to the ifcher. Thus if Ave have two pulleys one running 10€ turns per minute and the other 200 turns in the same time the first one will make 1 turn while the second is making 2 and we say that the ratio of the first to the second is as 1 to 2. Also if we have a pulley or drum which in turning one time drives by a band or belt another one which makes 6 turns in the same time, the ratio of speed between the two is 1 to 6. The ratio between numbers is usually in- dicated by the sign, :, and is read "is to." Thus in the latter example above Ave may express the ratio thus, 1 : 6. This is read "1 is to 6." In some cases the ratio betAveen numbers is indi- cated by placing one above a. line and the other be- low. Thus in shoAving the ratio betAveen tAvo num- bers say 1 and 5 Ave may use the 1 above the line 1 and the 5 beloAv, thus — . This is read also "1 is to 5. " 5 When Ave have two ratios which are equal to each other this is said to be a proportion. Thus if Ave have tAvo pulleys one running 100 turns per minute and the other 200 turns and compare these speeds to tAvo more pulleys one of Avhich makes 300 Avhile the other makes 600 turns, these tAvo ratios are said to be in proportion. This is true because the ratio of the first tAvo pulleys is 1 : 2 Avhile the second two 87 also have the ratio 1:2. Then these two relative speeds are in proportion and we may say "100 : 200 = (equals) 300 : 600." In writing proportions it is sometimes customary to use, instead of the equal- ity sign (=), the sign, : :, which is read "as." Thus in the above proportion it may be written thus : 100 : 200 : : 300 : 600, and is read "100 is to 200 as 300 is to 600." In the above proportion we can readily see that the second term (200) in the first ratio when mul- tiplied by the first term, 300,. in the second ratio is equal to the number obtained by multiplying the other two terms together. Thus 200 X 300 = 60000, and 100 X 600 = 60000. This must be true in all proportions. Therefore in any proportion the two inside terms multiplied together are equal to the two outside terms multiplied together. From this law it is plain that if three terms are given the fourth or remaining term can be found by multi- plying together the two inside or outside terms as the case may be, and dividing this result by the third term given. Thus if we again use the proportion 100 : 200 : : 300 : 600, any one of the terms may be found, if not given, if the other three are known. The term which is not given is usually represented by the letter X. Thus if we leave out the third term, 300, and insert the letter, X, we will have, 100 : 200 : : 100 X 600 X : 600. Then to solve we have, = 300, 200 which is the unknown number, X, or the third term, 300. In like manner any one of the other terms can be found if the other three are given. Thus, 200 X 300 X : 200 : : 300 : 600. Then = 100 or X, 600 or the first term. 100 X 600 Also, 100 : X : : 300 : 600. Then = 300 200 or X or the second term. And 100 : 200 : : 200 X 300 300 : X. Then = 600 or X or the fourth 100 term. This method is commonly called the "Rule of Three." Proportions are usually stated in con- sidering the relative values of cause and effect. Thus if a man in 6 days makes 12 dollars he will make in 12 days 24 dollars. In this case he works 6 days and makes 12 dollars. The ratio of time to money or work to pay is 6 : 12 or 1 : 2. As he works at the same rate of pay for the whole 12 days and receives 24 dollars the ratio is again 12 : 24 or 1 : 2 and these ratios being equal are in proportion and may be stated, 6 : 12 : : 12 : 24. In this we see that, time : pay : : time : pay, provided the rate of pay per day has remained the same. We also see that the two inside terms, 12 and 12, when multi- plied together will give 144; and the two outside terms when multiplied together will give 144. Again if it is desired to find any one of these terms when the other three are given it is only necessary to mul- tiply either the two inside or outside terms together and divide this by the remaining term. 89 Example : If a man works 6 days and makes 18 dollars, how many days will he have to work at the same rate of pay to make 90 dollars ? Proportion 6 : 18 :: X :,90. 6 X 90 Then = 30 days. 18 Example : If a man works 6 days and makes 18 dollars, how much will he make in 60 days \ Proportion 6 : 18 : : 60 : X. 18 X 60 Then = 180 dollars. 6 In like manner the other two terms may be found. CARD ROOM PROBLEMS The draught gear on a card is directly propor- tioned to the weight in grains per yard of sliver made, provided the weight of the lap has not been changed. Thus if an 18 teeth gear makes a 54 grain sliver, what gear will be required to make a 60 grain sliver from same lap? The proportion is, 18 : 54 : : X : 60 ; because if an 18 teeth gear produces a 54 grain sliver, X No. of teeth will be required to produce a 60 grain sliver. To solve multiply together the two outside terms, 18 and 60, and divide this by the remaining inside term 54. The result is the unknown number of teeth or the unknown quantity, X. 18 X 60 Thus = 20 — teeth draught gear required. 54 90 1. If a 20 teeth draught gear is making a 60 grain sliver from a certain lap, an 18 teeth draught gear Avill produce what weight sliver from same lap? Proportion 20 : 60 : : 18 : X. Solve : 2. The production gear on the card is also direct- ly proportional to the pounds production. Thus, if a 25 teeth production gear is making 150 pounds per day of a certain weight sliver, what gear will be in- quired to make 126 pounds per day of same sliver 1 If a 25 teeth gear makes 150 pounds then X num- ber of teeth will make 126 pounds and avc have : 25 : 150 :: X : 126 ; 126 X 25 and = X = 21 — teeth production gear. 150 3. If a 25 teeth production gear on card makes 150 pounds of sliver in a day, what size gear would produce 144 pounds in a day of the same weight sliver ? State proportion and solve. SPINNING ROOM PROBLEMS 1. If an 8 inch cylinder is driving a 1 inch whirl.. how many times would the whirl turn while the cylinder turns one time? Answer 8. What is the ratio of cylinder to whirl? 1 : 8. 2. If ratio of cylinder to whirl is 1 : 8, how many turns per minute must the cylinder make to give a spindle speed of 6400 turns per minute? 91 Proportion 1 : 8 : : X : 6400 ; 6400 then X = = 800 turns per minute of cylinder. 8 3. If it takes 4 spindles to produce 1 pound of yarn per day, how many pounds would be made on 30000 spindles? Proportion 4:1:: 30000 : X ; then 4 X X = 1 X 30000 ; 30000 then X = = 7500 pounds. 4 4. If an average number of sides per girl is 14, how many girls would be required to run 280 sides? Proportion 1 : 14 : : X : 280 ; 14 X X = 1 X 280 ; 280 X = = 20 girls. 14 WEAVE ROOM PROBLEMS 1. If a 32 teeth gear gives 64 picks per inch, what pick gear will give 100 picks ? Proportion 32 : 64 : : X : 100 ; 64 X X = 32 X 100 ; 32 X 100 and X = = 50 teeth gear. 64 2. 8 looms produce 20 cuts in a week. How many looms would it take to make 250 cuts per week? 92 Proportion 8 : 20 : : X : 250. 20 X X = 8 X 250 ; 8X250 X = = 100 looms. 20 3. If crank shaft on loom makes 2 turns while the cam shaft makes 1 turn, what is the ratio of speed of cam to crank shaft ? Answer 1 : 2. If the crank shaft has a gear of 50 teeth, how many teeth on cam shaft gear? 1 : 2 :: 50 : X. 1 X X = 2 X 50; X = 100 teeth. 4. If 4 yards of cloth weigh 1 pound, how many yards of cloth would there be in 15 pounds ? 4 : 1 : : X : 15. 1 X X = 15 X 4; X = 60 yards. 5. If 4 yards of cloth weigh 1 pound, how many pounds would there be in 60 yards? 4 : 1 : : 60 : X. 4 X X = 1 X 60; 60 X = — = 15 pounds. 4 93 6. If a spread in reed of 38 inches makes 36 inch cloth, what must be the spread to make 72 inch cloth of the same construction? 38 : 36 : : X : 72 36 X X = 72 X 38 72 X 38 X = = 76 spread in reed. 36 94 Square and Square Root When one number is multiplied by itself the re- sulting number is said to be the square of the num- ber thus multiplied. If we multiply 2 by itself the result is 4 and the number, 4, is said to be the square of 2. Thus to square any number all that is neces- sary is to multiply that number by itself. EXERCISE What is the square of 4? 4 X 4 = 16. What is the square of 5 •? 5 X 5 = 25. What is the square of 6? 6 X 6 = 36. What is the square of 10? 10 X 10 = 100. What is the square of 20 ? 20 X 20 = 400. What is the square of 25? 25 X 25 == 625. When we find the number which was used in multiplying by itself to produce a certain other number this number is said to be the square root of the one thus obtained. For instance, when we mul- tiply 2 by itself to produce its square, 4, then the number 2 is said to be the square root of 4. Thus, 4 is the square root of 16 because 4 multiplied by it- self produces 16. In like manner 5 is the square root of 25, 6 the square root of 36, 8 the square root of 64, 10 the square root of 100 and so on. When- ever we extract the square root of any number this square root when multiplied by itself must produce the number from which we started. In finding the square root of numbers of large 95 value,, or those in which the square root is not a whole number, the following process is used : Find the square root of 14884. 1,48,84 ( 122 1 20 48 22 44 240 4 84 242 4 84 The first thing to do is to mark off, as shown, the number into groups of 2 figures starting at the right of the number. Then select a number which when multiplied by itself is not greater than the first group on the left of number. In this case only one figure is in this last group on the left. This is the figure 1, and we select 1 as our first figure in the result because 1 multiplied by itself produces 1 which is not greater than our first group. This fig- ure is then placed to right of line as shown and its square which is also 1 is placed under the first group to the left. This is then subtracted from the first group and the next group of two figures brought down. In this case 1 from 1 leaves noth- ing and our only remaining number is 48. Then multiply the figure, 1, in the result by figure, 2, and annex to it the .figure, 0. This gives 20 as shown to the left of 48. Then divide this number 20, into 48. This gives 2, and we place this next to 1 in the result. This figure 2 must next be added to 20 giving the number 22 which is placed directly under 20. Then multiply 22 by 2 in result and place this result under 48 and subtract as in long division. 96 Subtracting then we obtain 4 and then drawing- down the next group, 84, we have 484. Again mul- tiply the result, which is now 12, by 2 and add to this the figure, 0. This gives 240 and we divide this into 484 and find that will give a result of 2. This is placed to right of 12 and is also added to 240 making 242 which is placed below 240. Then mul- tiply this 242 by the figure last obtained in result, namely 2„ and we obtain 484 which is placed below the last remainder 484. The square root then of 14884 is the whole number, 122. If it becomes nec- essary to find the square root of a number which is not a whole number, the following method is given. This is the same as the one just given ex- cept that the figures, 0, in groups of two are placed to the right of the whole number after the decimal point has been inserted. Example : Extract the square root of 30. 30.00,00 ( 5.47 25 100 5 00 104 4 16 JL080 84 00 1087 76 09 The only thing to note in such examples is that the whole number is put to the left of the decimal point, and must be divided into groups of two fig- ures from right to left and the ciphers or figure 0, are put to the right of the decimal point and must be divided into groups of two from left to right. The following numbers are given showing how to 97 divide into groups so as to produce two figures to the right of the decimal in the result. 1,33.00,00. .20.00,00. 2,63.00,00. 21,53.00,00. 1,42,53.00,00. 2.00,00. EXAMPLES FOR CARD ROOM The twist per inch to be inserted into any size or number of hank roving is found by multiplying the square root of the number of roving by a certain given standard number. Run the following to two figures to the right of decimal point. Find the square root of No. 2 hank roving. Find the square root of No. 3 hank roving. The sign of square root is made thus, ^J and when placed before a number it means that this num- ber is to have its square root found. Find V4 hank roving. Find V 5 hank roving. Find V 6 hank roving. Find V7 bank roving. Find V8 hank roving. SPINNING ROOM PROBLEMS The number of ends that will lie side by side in one inch is found by extracting the square root of the number of yards of that size yarn in one pound. A No. 28 's yarn has 28 X 840 = 23520 yards in one pound. How many ends will lie side by side in one inch? V23520 98 A No. 30 's yarn has 30 X 840 = 25200 yards in one pound. How many ends will lie side by side in one inch? V25200 Find the square root of the following numbers of yarns : VslT slW, a/22~, Vl2~, V36^ V25~, a/32~, V^T WEAVE ROOM PROBLEMS (Square Root) 1. How many ends will lie side by side of a 30 's yarn V30 X 840 = 2. How many ends will lie side by side of a 36 's warp? V36 X 840 = 3. How many ends will lie side by side of a 22 's warp? V22 X 840 = 4. The diameter of a yarn is found by dividing •034 by Vn^ Find the diameter of a 36 yarn. .034 1034 — — == — — =• .0056 inches. V36 6 99 Equations When we wish to show that several values when taken collectively,, either expressed by addition, subtraction, multiplication or division, are equiva- lent or equal to certain other indicated values this is shown in a form known as an equation. Thus, 8 + 2 + 1 = (equals) 7 -f 2 + 2, or 12 + 2 — 3 = 6 + 3 + 2, or 3 X 6 X 10 = 5 X 6 X 6 or 10 X 16 4 X 20 = are all equations. As both sides of an equation are the same, dividing or multiplying both sides through by a common number does not alter its value. Thus, if we have 16 X40 the equation, 16 X 10 = , and divide both 4 sides through by 16 the relative values of both sides have not been changed, and the remaining equation 40 would be 10 = . 4 When one of the numbers in an equation is un- known it is usually indicated by the letter, X. Thus, if we have 10 X X = 15 X 10 we can find the value of X by dividing both sides through by 10. Then this would give us the equation X = 15. 100 EXAMPLES The surface speed of two pulleys which are driven by a belt must be equal. This surface speed, or speed of a point on the outer rim of a pulley is found by multiplying the diameter, or distance across, by the number, 3.1416, which gives the cir- cumference,, or distance round. This in turn mul- tiplied by the revolutions per minute of the pulley gives the surface speed. Example : A 24 inch pulley is running 200 revo- lutions per minute and is driving, by a belt, another pulley which is 10 inches in diameter. Find the speed of the 10 inch pulley. As the surface speeds of the two pulleys must be equal (unless belt slips), and as surface speed of each pulley is found as given ' above we have : (a) 24 X 3.1416 X 200 = surface speed of 24 inch pulley, and (b) 10 X 3.1416 X (X unknown) = sur- face speed of 10 inch pulley. Then as these two surface speeds are equal we have : 24 X 3.1416 X 200 = 10 X 3.1416 X (X unknown). Then as dividing both sides of an equation does not alter its value, we can divide both sides of this one through by 3.1416 and our remaining equation is 24 X 200 = 10 X (X unknown speed). Thus we see that in all cases of this kind the num- ber, 3.1416, will be left out and we have the general principle that the diameter (distance across) of a pulley, multiplied by its speed always equals the 101 diameter of the other pulley multiplied by its speed. In the above remaining problem, then if both sides of the equation are divided by 10 the value is not changed and we have 24 X 200 = X (unknown speed of 10" pulley). 10 Then the speed of the 10" pulley would be 480. In like manner the diameter of either the driver or driven pulleys may be found as well as the speeds. Example : A 12 inch pulley is to be run 160 rev- olutions per minute from a shaft which is running 192; what size pulley must be put on the driving shaft? 12 X 160 == 192 X (X unknown pulley). 12 X 160 Then — = X = 10 inch pulley. 192 Example : Driving shaft runs 212 and has a 16 inch pulley driving a 12 inch pulley; what is speed of 12 inch pulley? 212 X 16 = 12 X (X unknown speed) ; 212 X 16 Then X = . 12 When one gear is driving another the relative speeds of the two are determined from the number of teeth in each gear. Thus if a gear having 50 teeth is driving an- other of 50 teeth the driver in turning one time 102 will take up the 50 teeth in the driven gear, thus causing- it to make one turn at the same time. If the driver makes 100 turns in one minute, then the driven will also make 100 turns in the same time. If, however, we have 50 teeth in the driver and only 25 in the driven, the driver making one turn must take up 50 teeth in the driven gear. But as this gear only has 25 teeth we readily see that it will have to turn two times while the driver of 50 teeth is turning one time. Then the relative speeds of two gears is found by dividing the num- ber of teeth in larger by the number of teeth in the smaller. Thus if we have a 160 teeth gear driving a 40 teeth gear the relative speeds of the two is found by dividing 40 into 160, which goes 4. This means that while the 160 teeth gear is making one turn the 40 teeth gear will make 4 turns. Again we see that if this 160 teeth gear makes 10 turns in a minute the total number of teeth passing by a cer- tain point in one minute will be 10 X 160 = 1600. If then this gear meshes into the 40 teeth gear it will cause this same number of teeth from this gear to pass the same point. Then as this 40 teeth gear must deliver 1600 teeth to this point in the same time it must turn 40 divided into 1600 = 40 turns in one minute. From the above then we can see that when two gears are meshing, the number of teeth in one gear multiplied by its speed must equal to the number of teeth in the other gear multiplied by its speed. Suppose we have a gear of 40 teeth running 150 revolutions per minute meshing into another gear of 30 teeth and wish to know how fast the 30 teeth gear is turning. Then we have the 103 equation, 40 X 150 = 30 X (X unknown speed). 40 X 150 Therefore, by solving we have = X, the 30 speed of 30 teeth gear = 200 speed required. In like manner we can find the speed of either gear thus meshing or the size of either gear, to pro- duce a certain speed. Example: A gear has 40 teeth and is running 116 turns per minute; what size gear must be used with it to produce 145 turns per minute ? Equation, 40 X 116 = 145 X (X unknown gear) ; 40 X H6 then, = X = 32 gear required. 145 In solving equations, long processes are avoided by substituting equal values for certain numbers. Example : If the back roll in a machine takes in 20 inches in a minute and front roll delivers 200 inches in a minute, what is the draft and what the revolutions per minute of front 1" roll ? As front roll delivers 200 inches while back roll is taking in 20 inches, the front roll is delivering 200 -f- 20 = 10 times as much as the back roll. This is draft, or the draft is 10. Surface speed of front roll Then, draft = . Surface speed of back roll Now as draft in this case is 200 200 Surface speed of front roll we have = : . 20 20 Surface speed of back roll 104 As surface speed of back roll is % that of the front roll, we have, 200 Surface speed of front roll 20 /4oX (Surface speed of front roll) Now as surface speed of a roll = circumference X revolutions or turns we have, 200 circumference X revolutions 20 Yiq X (surface speed of front roll) And as the surface speed of' front roll is 200 inches per minute, we have 200 circumference X revolutions 20 y 10 X (200) Also if circumference = diameter X 3.1416, and as 200 the diameter of front roll is 1 inch we have, = 20 lX3.1416Xrevolutions lX3.1416Xrevolutions Mo X (200) 20 Because Y 10 of 200 or y 10 X 200 = 20, 200 IX 3.1416 X revolutions then, = . 20 20 If both sides of this equation is multiplied by 20 the value is not changed and we have, 20 X 200 20 X 1 X 3.1416 X revolutions 20 20 105 This will do away with the number 20 on both sides because the 20 below the line on left goes into the 20 above the line 1 time, and 1 time 200 1 = 200. Also the 20 below line on right hand side of the equation will go 1 time into the 20 above the line. Then we have, 200 = 1 X 3.1416 X revolutions. Again dividing both sides by 1 X 3.1416 will not change the value and we have,. 200 1 X 3.1416 X revolutions 1 X 3.1416 3.1416 And as the 3.1416 above and below the line will can- cel each other we have, 200 = revolutions, or 1 X 3.1416 200 revolutions = = 63.6 revolutions. 1 X 3.1416 From this as a general rule we have, R.P.M., (revo- surface speed per minute lutions per minute) = dia. X 3.1416 And R.P.M. X diameter X 3.1416 = surface speed. FORMULAS A formula is an equation showing in a general way how a problem is to be solved. Thus from the preceding we have the following formulas : 106 To find the speed of pulleys when diameters are given or visa versa. Formula: SpeedXdiameter = speed X diameter. ( one pulley ) ( other pulley ) To find the speed of gears when the number of teeth are given or visa versa. Formula: Speed X No, teeth = speed X No. teeth, (one gear) (other gear) CARD ROOM PROBLEMS 1. If the driving pulley of a card is 20 inches in diameter and it is to run 165 revolutions per min- ute, what size pulley must be on the line shaft which runs 200 revolutions per minute? Formula: SpeedXdiameter = speed X diameter: Then 165 X 20 = 200 X diameter (pulley on shaft). Then solving the equation we have by dividing both sides by 200 : 165 X 20 200 = X diameter. 200 200 And as the 200 above and below the line on right will cancel we have : 165 X 20 = dia. of pulley on shaft=16.5=16^ in. 200 2. If the doffer is running 10 revolutions per minute and doffer gear has 192 teeth, how many revolutions per minute is the doffer change gear of 20 teeth making ? Formula: Speed X number teeth = speed X 107 number teeth; then substituting we have, 10 X 192 = speed X 20. And, speed = 10 X 192 = 96 Rev. per minute of change gear. 20 SPINNING ROOM PROBLEMS 1. If the cylinder on spinning frame is to run 800 revolutions per minute and has a 12 inch pul- ley, what size pulley must be on line shaft which has a speed of 300 revolutions per minute? Formula : Speed X diameter = speed X di- ameter; then substituting we have, 800 X 12 = 300 X diameter ; then diameter = 800 X 12 = 32 inch pulley on shaft. 300 2. If the cylinder is turning 800 revolutions per minute and has a 26 teeth gear, how fast is the stud gear of 132 teeth turning? Formula : Speed X number teeth = speed X number teeth; then substituting we have, 800 X 26 = speed X 132, then speed = 800 X 26 = 157.5 Rev. per minute of stud gear. 132 WEAVE ROOM PROBLEMS 1. If a loom has a 12 inch pulley and is to run 160 picks per minute, what size pulley must be on line shaft which is running 200 revolutions per minute ? 108 Formula: Speed X diameter = speed X diame- ter ; then substituting we have,, 160 X 12 = 200 X diameter, and diameter = 160 X 12 . = 9.6 inches Dia. of pulley on shaft. 200 2. If an auxiliary harness cam shaft is to be used mi a loom to make 1 turn for 6 picks and it has a 60 teeth gear on it, what size gear must be on the cam shaft of loom to drive it? As the cam shaft of loom makes 1 turn for every 2 picks of loom this auxiliary shaft will make 1 turn for every 3 turns of cam shaft in order to equal 6 turns of the crank shaft. Then we have the for- mula : speed X number teeth = speed X number teeth and by substituting,, 1 X 60 = 3 X number teeth on cam shaft. Then number teeth = 1X60 = 20 teeth gear on cam shaft. 109 Cancellation Cancellation is a short form of division in which both the number divided and the number by which we are dividing are reduced by dividing both by a common number. Thus, if we have 36 to be divided 36 by 6 and indicate it thus : — , we can divide both the 6 top number and the bottom number through by 2. This will give 18 on top and 3 below. This is done in practice as follows : 18' ^0 18. 3 3 This reduces the top and bottom number to lower terms in which it is simple to divide the one into the other. Example : Use cancellation and divide the product of 20 X 16 X 30 X 4 by the product of 10 X 8 X 24. This is shown thus : 2o >< iy> x -0 x 4. -=20 ;P X ? X 24 2 6 By selecting any number either above or beloAv the line which we see is easily divided the one into the other we thus divide, striking out, or cancell- 110 ing, these two numbers and placing the result of this division above or below the number thus divided. In the above example we readily see that 4 will go into 24 evenly, giving the result, 6. Then we place 6 below 24 and draw a line through both 4 and 24. Again we see that this 6 will go 5 times into 30 above the line and we place 5 above 30,, striking out both 6 and 30. Again this 5 will go 2 times into 10 and we strike out the 5 and 10 and place 2 below 10. Also we see that this 2 will go 8 times into 16 and we place 8 above 16 and strike out 2 and 16. Again this 8 will go one time into the 8 below the line and these two numbers are cancelled, leaving as a result of this division 20 above the line and 1 below. As 1 goes into 20, twenty times the result of our division is 20. CARD ROOM PROBLEMS 1. The doubling on a drawing frame is 6 ends of a 50 grain sliver. The Aveight of one yard going in back must be 6 X 50. If the draught is 5 what is the weight of sliver produced in front? 10 ffl X 6 = 10 X 6 = 60 grains. 2. The doubling on drawing frame is 6 ends of a 50 grain sliver. The weight of one yard going in must be 6 X 50. If the weight per yard in front is desired to be 60 grains what draft must be used? 5 xm draught = 5 draught. 00 ) ' 111 3. (a) If jack shaft on the slubber makes 290 revolutions per minute and jack shaft gear has 52 teeth driving to spindles while gear on spindle shaft has 58 teeth, what is the speed of spindle shaft? As seen under equations, 290 X 52 = 58 X (X, unknown speed). Then, 290 X 52 = X, speed of spindle shaft. 58 (b) If spindle shaft has gear of 50 teeth driving spindle gear of 26 teeth, what is speed of spindle? As above, (a), speed of spindle shaft X its gear = speed of spindle (X, unknown) X spindle gear. Then as speed of spindle shaft is represented by 290 X 52 the unsolved part of equation, , if we place 58 this number in to represent this speed we have from (a) : 290 X 52 , (speed of spindle shaft), X 50, 58 (spindle shaft gear), = speed of spindle,. (X, unknown), X 26, (spindle gear). This gives 290 X 52 X 50 = 26 X X, unknown spindle speed. 58 Then to solve for-X as in equations we have (di- viding both sides through by 26) 290 X 52 X 50 = spindle speed. 58X26 Then to solve by cancellation we have : 112 2 10 M 25 200 X&2X £0 58 X 2(> 10 X 2 X 25 = 500 revolutions per 20 18 minute spindle speed. SPINNING ROOM PROBLEMS 1. How many pounds of No. 20 's warp yarn per spindle per 10 hours will be produced by a frame with 1 inch front roll running 100 E. P. M. ? 5 5 J0 1X3.1416XWX00XJ0 5X5X3.1416 = = .28 pounds. 36X840X20 36X7 1 2. If 3.1416 is the circumference of front roll, front roll gear 100 teeth, stud gear 132 teeth, cylin- der gear 26 teeth, change gear 20, and ratio of whirl to cylinder 7.8, what is the twist per inch? 66 m X WX7.8 66X7.8 = = 63.02 turns per inch. 3.1416X20X2$ 2.6X3.1416 ft 2.6 WEAVE ROOM PROBLEMS 1. A loom runs 160 picks per minute, on goods with 40 picks per inch ; how many yards will it make in 10 hours ? i jmx 20X10 20X10 = = 66.66 yards. MxW s 3 113 2. A cloth has 72 ends per inch of 30 's warp and is 36 inches wide. What is the weight of warp in 840 yards allowing 6% take up? 1.2 72X£0XW 72X1.2 = 91.9 pounds. .94, (1— .O6)XWX00 .94 114 Percentage As most all items in cotton manufacturing, such as production, waste, machines stopped and run, and dividends made, are expressed as being a certain "per cent," it is necessary for us to first clearly understand what "per cent" or "percentage" means. The word "per cent" comes from the Latin words "per" meaning "by" or "by the" and "centum" meaning "one hundred." Thus, the word "per cent" means "by the hundred," and when Ave say "ten per cent" we mean "ten by the hundred" or "ten parts of a hundred." Per cent then means the relation of a certain number to the number, 100. Thus if we say "ninety-eight per cent, ' ' we mean that in every one hundred parts of a certain number we are considering 98 parts. If Ave run through the mill 100 pounds of cotton and pro- duce 88 pounds of cloth Ave sell 88 per cent of the cotton bought or put in. If we run 500 pounds of cotton and get 88 pounds of cloth from every 100 pounds, then we will produce 5 X 88 = 440 pounds of cloth. Thus we see that multiplying the number of pounds made from each 100 by the number of hundreds put in will produce the number of pounds made. As per cent means a certain part of 100, when Ave say "5 per cent" it can be expressed thus: 5 / 100 of 100 or % 00 X 100. Then % 00 X 100 = ™% 00 = 5. Percentage being thus always expressed in hundredths it can be shoAvn as a decimal 115 and is almost always so shown. Thus 4 per cent = /loo = -04 ; 6 per cent = % o = -06 ; 10 per cent = io/ 100 = .10; 15 per cent = 15 / 100 = .15; 25 per cent = 2 %oo = -25, etc. The sign of "percentage" is made thus, % ; and when used after a number it is read "per cent." Thus 6% = six per cent = .06 ; 8% = eight per cent = .08; 10% = ten per cent = .10; 50% = fifty per cent = .50; 100% = one hundred per cent = 1.00; 116% = one hundred and sixteen per cent = 1.16; 13%% = thirteen and one-half per cent = .135; 8%% = 8.75% = eight and three- fourths per cent = .0875; 2%% = 2.25% = two and one-fourth per cent = .0225; %% = .5% = one-half per cent = .005 ; %% = .25% = one-fourth per cent = .0025; 2%% = 2.5% = two and one- half per cent = .025. Thus we see that to reduce a number expressed as % to its decimal equivalent we divide by 100. Examples: Reduce .253% to decimal equivalent. 100. ) .25300 ( .00253 decimal equivalent. 200 530 500 300 300 Reduce .052% to decimal equivalent. .052 -=- 100 = .00052. Reduce .125% to decimal equivalent. .125 -=- 100 = .00125. . 116 Reduce 2%% to decimal equivalent. 2%% = 2.375%; then 2.375 -=- 100 = .02375. Prom these we see that to reduce a number ex- pressed in % to its decimal equivalent all that is necessary is to move the decimal point two figures to the left. PROCESSES (a) To find the "per cent" (%) of a given num- ber, multiply the number by the number of per cent expressed in decimals. Examples : 1. How much is 20% of 75? 20% = .20; then 75 X .20 = 15.00 (fifteen) ; then 15 is 20% of 75. 2. How much is 22%% of 82.5? 22y 2 % = 22.5% = .225; then 82.5 X -225 = 18.5625 ; then 18.5625 is 22%% of 82.5. 3. Find 8%% of 1012. 83/ 4 % = 8.75% = .0875; then 1012 X -0875 = 88.55 ; then 88.55 is 8%% of 1012. 4. If a mill makes 12% waste and uses 120,000 pounds of cotton in a month, how many pounds of waste does it make ? 12% = .12 and 120,000 X -12 = 14,400 pounds waste. (b) To find what "per cent" one number, A, is of another, B, multiply the first number, A, by 100 and divide by the second number, B. 117 Examples: What % is 2 of 100? 2 X 100 = 200, and 200 -r- 100 = 2% = .02. 5 is what % of 75 ? 5 x 100 = 500, and 500 -=- 75 = 6.666% = .06666: What % of 116 is 52? 52 X 100 = 5200, and 5200 -+- 116 = 44.82% = .4482. If a mill opens 252,147 pounds of cotton and 25,622 pounds of waste are made, what is the % of waste ? 25622 X 100 = 10.16% = .1016. 252147 (c) When a number is given which is a certain % of another number, to find this number, multiply the given number by 100 and divide by the % (not in decimals). Examples : If 50 is 20% of a number, what is the number? 50 X 100 = 5000, and 5000 -f- 20 = 250 the number. This can be proved as follows : If 250 is the number then what number is 20% of it? 250 X -20 (20%) = 50, or if 250 is the number, 50 is what % of it? 50 X 100 = 5000, and 5000 ~ 250 = 20%. 118 2. What is the number if 116 is 5y 2 % of it? 116 X 100 = 11600, and 11600 -^- 5.5 (5y 2 %) = 2109.09. 3. 30% of a certain number is 416. What is the number? 416 X 100 = 41600, and 41600 -4- 30 = 1386.6. 4. A mill produces in one month 187,.516 pounds of cloth, which is 96.7% of what it could produce. How much could it produce? 187516 X 100 = 193915.1 pounds. 96.7 (d) When a number, A, is given which is the result after allowing a certain % loss on another number, B; to find, the number, B. Subtract the % (in decimals) from 1.00, and divide the result into the number, A. Example: If 81 is the result after allowing 10% loss, what is the number? 1.00 — .10 = .90. Then 81 -f- .90 = 90. Then 81 -^ .90 = 90. Proof : If 90 is the number then 10% of 90 = 90 X -10 = 9, and 90 — 9 = 81. Example : If 10% waste is made how much cotton was used in making 18000 pounds of cloth? 1.00 — .10 = .90. 18000 -f- .90= 20000 pounds. 119 (e) When a number, A, is given which is the result after allowing a certain % gain on another number, B ; to find the number, B. Add the % (in decimals) to 1.00 and divide the result into the number, A. Eample : If 88 is the result of a number's gaining 10%, what is the number? 1.00 + .10. = 1.10. 88 ~ 1.10 = 80. Proof: If 80 is the number then 10% of 80 = 80 X .10 = 8, and 80 + 8 = 88. Example : If a loom beam has 109 pounds of sized warp on it, what was the weight of unsized yarn if 10% is put on? 1.00 + .10 = 1.10 109 -f- 1.10 = 99.09 pounds. CARD ROOM PROBLEMS 1. If a picker makes 4% waste, how many pounds of Avaste would you expect from running 22,412 pounds of cotton? 22412 X -04 = 896.48 pounds of waste. (See a.) 2. In above example, if only 710 pounds of waste is collected, what is the % of "invisible" waste? 896.48 — 710 = 186.48 pounds of invisible waste; 186.48 X 100 then (see b) = .83% invisible waste. 22412 The "actual" waste being 710 pounds the % is 710 X 100 (see b) = 3.16% actual waste. 22412 120 What is "invisible" waste? 3. A man receives $.85 which is a 3% premium on what he makes in a week. What is his actual wage ? .85 X 100 (see c) = $28.33 actual wage. 3 4. A card room has 2516 speeder spindles and 412 are stopped for a week. What is the % of spindles run? 2516 — 412 = 2104 spindles run ; 2104 X 100 = 83.6% spindles run. 2516 SPINNING ROOM PROBLEMS 1. A mill has 42,168 spindles and 5% are stopped for a week; how many spindles stopped? 42168 X -05 (see a) = 2108.40 spindles stopped. How many spindles run? 42168 — 2108 = 40060 spindles run. What is the % of spindles run? 40060 X 100 (see b) = 95% spindles run. 42168 2. If a frame is figured to put in 25 turns per inch in a certain yarn and the contraction in length due to twisting is 3%, what is the actual turns per inch put in? 25 X .03 = .75 turns extra on account of twist ; then 25 -f- .75 = 25.75 actual turns per inch. 121 3. The yarn from a frame numbers 20 's. What is the % of contraction if a 3 hank roving is used with a 6.87 draft? The number of yarn ought to be 6.87 X3 = 20.61. As,, however, it actually num- bers 20 's, which is heavier than 20.61 's, there must be some contraction in twisting. 20 X 100 Then = 97%, 20.61 which means that the resulting number is 97% as long as the untwisted strand would be. Then the % of shrinkage is found by subtracting .97 from 1 which gives 3% contraction. 4. If a frame is calculated to make 52 pounds per day after allowing 2% off for doffing, oiling, etc., what is the figured production? If 52 is the result after deducting 2%, then 52 pounds is 98% of the whole. 52 X 100 Then = 53.06 figured production. 98 WEAVE ROOM PROBLEMS 1. A warp is set 39.5 inches in reed and makes 37.5 inch cloth; what is the % of shrinkage? 39.5 - — 37.5 = 2 inches shrinkage ; 2 X 100 then = 5.06% shrinkage. 39.5 2. The cut marks on a warp are 68 yards apart. Cloth made from this produces 64 yard cuts. What is the % "take up?" 122 68 — 64 = 4 yards take-up ; 4 X 100 then = 5.88% take-up. 68 3. If cloth cuts are to be 60 yards and 6% is al- lowed for the take-up, how far apart must the cut marks be placed? If 60 yards is 6% less than the warp length then it would be 94% of the warp length. 60 X 100 Then = 63.83 yards between cut marks. 94 60 or = 63.83. 1.00 — .06 4. A cloth is to be made 36.5 inches wide. If 7% shrinkage is allowed, how wide must the warp be in reed? If 36.5 inches is 7% less than the width in reed, it is 93% of this width. Then we have, 36.5 X 100 = 39.24" Avidth in reed. 93 36.5 or = 39.24" 1.00 — .07 5. Cut of cloth weighs 12.4 pounds and has 7.46 pounds of warp in it. What is the % of warp and filling in the cloth? 6. If a mill produces 146,218 pounds of cloth which has 68% of warp, how many pounds of warp and filling are in this amount of cloth? 123 Weights and Measures There are three principal standard measures used in cotton manufacturing. These are weights, lengths and time. WEIGHTS The system of weights used is that of the stand- ard pound in which there are 16 ounces. 1 pound (lb) =16 ounces (ozs.). For weighing small amounts of yarn, cloth or cotton, the weights are usually obtained in grains, and for fractional parts of grains the decimal sys- tem is used. Thus, we may have 136.25 grains of yarn or cloth, 18.47 grains of yarn, etc. In order to use smaller numbers it sometimes becomes neces sary to change from one system of weights to an- other ; i.e., from grains to pounds or ounces, or from pounds to ounces or ounces to grains. To do this we must bear in mind the following relation : 1 pound (lb) =16 ounces (ozs.) =7000 grains (grs.) If there are 7000 grains in 1 pound then in 2 pounds there would be 2 X 7000 = 14000 grains. Then to reduce pounds to grains we have, No. pounds X 7000 = grains. Examples : How many grains in y 2 pound ? y 2 X 7000 = 7000 / 2 = 3500 grains. How many grains in % pounds? % X 7000 = 49000 / 8 = 6125 grains. 124 How many grains in % pounds? % X 7000 = 2100 % = 5250 grains. How many grains in 1.37 pounds? 1.37 X 7000 = 9590 grains. If 7000 grains = 1 pound = 16 ounces, then the number of grains in 1 ounce can be found by divid- ing 7000 by 16. Thus, 7000 -j- 16 = 437.5 grains in one ounce. Then No. ounces X 437.5 = grains. Examples : How many grains in 2 ounces ? 437.5 X 2 = 875.0 grains. How many grains in 12 ounces ? 437.5 X 12 = 5250 grains. How r many grains in 3.45 ounces? 3.45 X 437.5 = 1509.375 grains. Again, if 1 pound = 7000 grains, then in 14,000 grains there would be 14000 ■ — = 2 pounds. 7000 Then to bring grains to pounds we have, No. of grains = No. pounds. 7000 Examples : How many pounds in 15,640 grains ? 15640 -f- 7000 = 2.234+ pounds. How many pounds in 4860 grains? 4860 U- 7000 == .6942 pounds. 125 Also if 1 ounce = 437.5 grains, then in 875 grains there would be 875 = 2 ounces. 437.5 Then to reduce grains to ounces we have No. of grains = No. ounces. 437.5 Examples : How many ounces in 1575 grains ? 1575 -=- 437.5 = 3.6 ounces. Hoav many ounces in 4562 grains? 4562 -f- 437.5 = LENGTHS In measuring lengths the standard of 1 yard is taken in which there are 36 inches. When Ave have yards to change to inches all that is necessary is to multiply the number of yards by 36. No. of yards X 36 = No. inches. Examples : Hoav many inches in 2 yards ? 2 x 36 = 72 inches. How many inches in 3.75 yards? 3.75 X 36 = 135 inches. When Ave have inches to change to yards we divide the number of inches by 36. No. of inches = No. of yards. 36 126 Examples: How many yards in 416.84 inches? 416.84 -=- 36 = 11.578 yards. How many yards in 5624.92 inches? 5624.92 -r- 36 = 156.24 yards. CIRCULAR MEASURE , The distance around a circle, such as a roll or cylinder, is found by multiplying the diameter (distance through) by 3.1416. Examples : A roll measures 1% inches in diame- ter; how many inches around (circumference)? 1% inches = 1.25 inches ; then 1.25 X 3.1416 = 3.927 inches circumference. If a cylinder measures 27% inches in diameter, what is its circumference? 27% = 27.25 ; then 27.25 X 3.1416 = 85.6086 inches circumference. TIME The measure of time is based on the standard of hours in which there are 60 minutes. Examples : How many minutes in 60 hours ? 60 X 60 = 3600 minutes. How many hours in 672 minutes ? 672 -=- 60 = 11.2 = liy 5 hours. CARD ROOM PROBLEMS 1. If a roll delivers 637.42 inches in 1 minute, how many yards will it deliver in 10 hours? 127 637.42 -=- 36 = 17.7 yards delivered in 1 minute. 1 hour = 60 minutes; then 10 hours = 10 X 60 = 600 minutes. 17.7 X 600 = 10620.0 yards delivered in 10 hours. 2. If a roll delivers 436.42 inches of sliver weigh- ing 56 grains per yard in 1 minute, how many pounds A\ r ill it deliver in 10 hours? 436.42 -=- 36 = 12.122 yards delivered in 1 minute ; then 12.122 X 56 = 678.832 grains delivered in 1 minute. 10 X 60 = 600 minutes in 10 hours. Then 678:832 X 600 = 403699.200 grains delivered in 10 hours. And, 403699.2 -r- 7000 = 57.67 pounds delivered in 10 hours. 3. If a roll 1% inches in diameter is running 200 revolutions per minute, how many pounds of a 60 grain sliver will it deliver in 1 hour? 1% inches = 1.375 inches ; 1.375 X 3.1416 = 4.3197" circumference. If it turns 200 times in 1 minute it will deliver 4.3197 X 200 = 863.9400 inches per minute; 863.94 -T- 36 = 23.97 yards delivered per minute. 1 hour = 60 minutes, then 23.97 X 60 = 1438.20 yards delivered in 1 hour, then 1438.2 X 60 = 86292 grains delivered in 1 hour, and 86292 -f- 7000 = 12.327 pounds delivered in 1 hour. 4. If a lap weighs 12 ounces, how many grains would it weigh? 128 437.5 grains in 1 ounce; then 437.5 X 12 = 5250 grains. 5. If 100 yards of sliver weigh 52 grains per yard, what would it be in ounces? 100 X 52 = 5200 grains in 100 yards ; then 5200 -f- 437.5 = 11.88 ounces. 6. If a roving can containing 56 a grain sliver weighs 12 pounds, how many yards of sliver in it? 12 X 7000 = 84000 grains weight of sliver. Then 84000 -=- 56 = 1500 yards. SPINNING ROOM PROBLEMS 1. If a roll is 3.14 inches in circumference and is. turning 100 r.p.m., how many yards will it deliver in 1 hour? 3.14 X 100 = 314 inches delivered in 1 minute ; then 314 X 60 = 18840 inches delivered in 1 hour,. 18840 and = 523.3 yards delivered in 1 hour. 36 2. If 120 yards of yarn weigh 60 grains, how many pounds would there be in 22,860 yards? 120 : 60 : : 22860 : X 22860 X = 60 X = H430 grains 120 weight of 22860 yards. Then, 11430 -f- 7000 = 1.63 pounds. 129 3. If bobbins are 1% = 1.25 inches in diameter, how many would lie side by side on 6 feet on the floo.- ? 6 feet = 6 X 12 = 72 inches, and 72 -=- 1.25 = 57.6 or almost 58. 4. If 1 bobbin contains 967.5 grains of yarn, how many pounds would 330 bobbins weigh? 967.5 X 330 = 319275 grains in 330 bobbins ; then 319275 -=- 7000 = 45.6 pounds. WEAVE ROOM PROBLEMS 1. If 6 inches of 34" cloth weigh 142 grains, how many yards would there be in 1 pound? 6 : 142 : : 36 : X 142 X 36 X = = 852 grains 6 weight of 36" or 1 yard of 34" cloth ; then if 1 pound = 7000 grains, 7000 -H- 852 = 8.21 yards per pound. 2. If a shuttle travels 460 inches in a second, how many feet would it go in 1 minute ? 60 seconds = 1 minute ; then 460 X 60 = 27600 inches in 1 minute, and 27600 -=- 12 = 2600 feet in 1 minute. 3. A certain loom makes 2.5 inches of cloth per minute ; how many yards would it make in 60 hours? 130 2.5 X 60 = 150.0 inches made in 1 hour ; 150 X 60 = 9000 inches made in 60 hours, and 9000 -=- 36 = 250 yards in 60 hours. 4. If a cloth weighs 3.72 ounces per yard, how many pounds ought 60 yards to weigh? 3.72 X 60 = 223.20 ounces, and 223.20 -=- 16 = 13.95 pounds. 131 Mechanical Calculations AREA A square inch is a surface area which is, or is equal to, a surface which measures one inch on all four sides of a square. To find the number of square inches in figures like those shown in Fig. 2 multiply together the length of two sides. If the figure is like (b) a short side, A, and a long side, B, is multiplied together. Example : If a square measures 14 inches on all sides, how many square inches does it contain? 14 X 14 = 196 square inches. A square yard is a surface area equal to the product of multiplying 36 X 36 = 1296 square inches. To change square yards to square inches multiply by 1296. Example : How many square inches in 2.5 square yards? 1296 X 2.5 = 3240 square inches. To change square inches to square yards divide by 1296. 132 Example : How many square yards in 5642 square inches? 5642 -f- 1296 = 4.35 square yards. A square foot is obtained by multiplying 12 X 12 = 144 square inches. To change square feet to square inches multiply by 144. To change square inches to square feet divide by 144. To find the area (square inches) in a figure like (b) Fig. 2, multiply a short side by a long side, or, A X B. Example : How many square inches in a piece of cloth 36 inches long and 28 inches wide? 36 X 28 = 1008 square inches. © © FIG. 3. v -^ To find the area of a figure like (a) Fig. 3. XXY = area (square inches if X and Y are in inches). To find the circumference (distance round) from A back to A again in (b) Fig. 3, multiply the di- ameter, X, by 3.1416. Example: What is the circumference of a roll 4 inches in diameter? 3.1416 X 4 = 12.5664 inches. 133 To find the area of a disk or the square inches contained in a circle like (c) Fig. 3. Square the diameter, X, (multiply by itself) and then multiply by .7854 (% of 3.1416). (Because (X) 2 X 2 area = x 3.1416,. X 3.1416 = X 2 X-7854). (2) 2 4 Example : How many square inches in a spool head 5 inches across? 5 X 5 = 25 (5 squared or 5 2 ), then 25 X .7854 = 19.635 square inches. VOLUME When we have a square block which measures 1 inch on all edges it is called a cubic inch. If the block is 2 inches on all edges it will contain 2X2x2 = 8 cubic inches. This is true because the area of a square 2 inches on all sides will be 2X2 = 4 square inches. If we build this area up to 1 inch in height, we have a volume of 4 cubic inches, and if we build again so as to make the whole 2 inches deep we double the volume and have a block which measures 2 by 2, by 2 inches and a volume of 8 cubic inches. Square Tanks Then to find the volume of a square shaped vessel we multiply the two sides together, then multiply by the height. 134 Example : Find the volume in cubic inches of tank like (a) Fig. 4, in which the two sides, X and X, measure 22" each, and the height, or depth, Y, is 52 inches. 22 X 22 = 484, and 484 X 52 = 25,168 cubic inches. A standard gallon contains 231 cubic inches. How many gallons would the above tank hold? 25168 -i- 231 = 108.9 gallons. One cubic foot = 12 X 12 X 12 = 1728 cubic inches; how many cubic feet in the above tank? 25168 -r- 1728 = 14.56 cubic feet. Example : Find the volume in cubic inches of a block or tank like (b) Fig. 4, when the two sides X and X measure 22 and 46 inches respectively and the height or depth, Y, is 10 inches. 22X46 = 1012, and 1012X10 = 10120 cubic inches. How many cubic feet will it contain ? 10120 -v- 1728 = 5.85 cubic feet. How many gallons will it hold? 10120 -r- 231 = 43.8 gallons. If 1 cubic foot = 1728 cubic inches and there are 231 cubic inches in a gallon then one cubic foot 135 = 1728 -4- 231 = 7.48 gallons. One gallon of water weighs 8.35 pounds; then one cubic foot of water will weigh 7.48 X 8.35 = 62.5 pounds. Example: If a tank like .(b) Fig. 4 is 3 feet by 8 feet and 3 feet deep, how many cubic feet will it contain ? 3 X 8 = 24 and 24 X 3 = 72 cubic feet. How many gallons will it hold? 72 X 7.48 = 538.56 gallons. How many pounds of water will it hold? 538.56 X 8.35 = 4496.976 pounds. Circular Tanks When we have a tank or cylinder like (c) Fig. 4, the volume is found by first finding the area of the top or the bottom, X, and multiplying this by the depth or height, Y. Example : Find the volume of a circular tank which is 54 inches deep and is 24 inches in diameter. Area of bottom = (24) 2 X -7854 = 576 square inches. Then 576 X 54 = 3104 cubic inches. How many gallons will this tank hold? 231 cubic inches = 1 gallon ; then 31104 -f- 231 = 134.6 gallons. What will be the weight of water? 134.6 X 8.35 = 1123.91 pounds. To find the volume of tanks like (d) Fig. 4, in which the top X is larger than the bottom X, and 136 viee versa. Find the average diameter by subtract- ing the smaller from the larger and adding one-half of this difference to the smaller. Then find the area of this average diameter and multiply by the depth of tank. Example : If top of tank is 34 inches in diameter and bottom is 28 inches in diameter what is its volume if depth is 30 inches ? 34 _ 28 = 6, and % = 3. Then average diameter is 28 -j- 3 = 31 inches. Then (31) 2 X -7854 = (31X31) X -7854 = 754.7694 square inches, and 754.7694 X 30 = 22643.0820 cubic inches. How many gallons will it hold? 22643.082 -r- 231 = 98 gallons. SPEED OR VELOCITY Speed = distance X time. There are two kinds of speeds as generally con- sidered although they are closely related. These are rotary speed and surface speed. Rotary speed is spoken of as "revolutions per minute" (r. p. m.), and surface speed as "feet per second," "feet per minute," "inches per minute," or "yards per min- ute,, second," etc. Thus we say a roll or pulley has 210 r. p. m., meaning it turns 210 ! times in 1 minute. Also the speed of a belt, say, is 2100 inches per minute, meaning that a point on the belt is travel- ing 2100 inches in 1 minute. The speed or velocity of a creek or river can be found by throwing a cork or light article on its surface and noting how far it will travel in a certain time. Rotary speed 137 can be converted to surface speed, as in the case of two pulleys connected by a belt. Also surface speed can be converted to rotary speed, as in the case of an engine piston driving a crank. Rotary Speeds When two pulleys are connected by a belt, as this changes rotary to surface speed the rotary speeds of both pulleys must accommodate themselves to the resulting surface speed of the belt. If they do not, one or the other must slip. Then the surface speed produced by both pulleys, no matter what their diameter, must be equal. If the circumference of a pulley is 36 inches and if we roll it on the floor for one complete turn it will pass over 36 inches on the floor. If we roll it this distance in 1 second, then it passes over 36 inches in 1 second, which is its surface speed. If we suspend this pulley and pass around it a belt and turn the pulley 1 turn in one second it will take up 36 inches of belt in 1 second, which is its surface speed. Then to change rotary speed to surface speed we multiply the rotary speed by the circumference of the roll or pulley. (The circumference = 3.1416 X diameter.) Example : What is the surface speed of a pulley 12 inches in diameter and running 160 r. p. m. ? 12 X 3.1416 X 160 = 6031.872 inches per minute; 6031.872 -~ 12 = 502.65 feet per minute ; 502.65 X 60 = feet per second. If in the above example this 12 inch pulley is con- nected by a belt to a 14 inch pulley we find the speed (rotary) of this pulley as follows : 138 As the surface speeds of both pulleys must be equal; then 12 X 3.1416 X 160 = 14 X 3.1416 X (X unknown). Dividing both sides by 3.1416 we always eliminate this number 3.1416 and we have 12 X 160 = 14 X (speed unknown). Always, then, we have this fact : Speed of a pul- , ley X its diameter = speed of other pulley X its diameter. Then from equations we see that speed of 14 inch 12 X 160 pulley = = 137.1 r. p. m. 14 Using this equation we can find the sizes or speeds or any pulley desired. Example : A 9 inch pulley is running 200 r. p. m. and is required to drive another which is to run 160 r. p. m. What must be the size of other pulley? 9 x 200 = (X) X 160 9 X 200 then (X) = = 11.25 = 11% inches 160 diameter of other pulley. Surface Speeds Surface speeds can be converted to rotary speed, as in the case of an engine piston driving an engine shaft by means of a crank. If an engine fly wheel is turning 200 r. p. m.„ the piston having a stroke of 2 feet, the piston will make two strokes for each turn of the crank and fly wheel. Then the piston will have a surface speed of 2 X 2 X 200 = 800 feet per minute. 139 About 3500 feet per minute is a good average speed for main driving belts. Then if we have a motor running 1200 r. p. m., what size pulley should it have so as not to give a greater surface speed to the belt than 3500 feet per minute? Surface speed = diameter X 3.1416 X r. p. m. Then 3500 = diameter X 3.1416 X 1200. Dividing both sides by (3.1416 X 1200) we have 3500 = diameter in feet =.928 ft. diameter 3.1416 X 1200 and .928 X 12 = 11.1 inches diameter of pulley. Example : If a machine has a 16 inch driving pulley which is to be run 320 r. p. m., what surface speed will the belt have ? 16 X 3.1416 X 320 = 16084.99 inches per minute, and 16084.99 -f- 12 = 1340.4 feet per minute. Is this too high? POWER Power is the energy required to do a certain amount of work in a given time. Work is the result of moving a certain weight for a certain distance. Then power takes into consideration, weight, dis- tance and time. If we raise a block weighing 1 pound a distance of 1 foot the work done is ex- pressed in feet and pounds and this is said to be 1 foot-pound. If we raise the same block a distance of 2 feet, the result is 2 foot-pounds ; or if we raise a weight of 2 pounds a distance of 1 foot, it is said to be 2 foot-pounds. If we raise a weight of 2 140 pounds a distance of 2 feet the result is 4 foot- pounds. Thus we see that work or foot-pounds = weight X distance. As speed = distance in a given time, power = weight X speed in a given time. Example : If it be found to require a pull of 50 -pounds to start up a machine and the speed of driv- ing belt is 1500 feet per minute, how much power is required to run it? Power = weight X speed in a given time, then power = 50 X 1500 = 75000 foot-pounds per minute. The standard for horse power (H. P.) is, 1 H. P. = 33000 foot-pounds per minute or 33000 1 H. P. = = 550 foot-pounds per second. 60 Then in the above example if it requires 75,000 foot-pounds to move the machine with a belt speed of 1500 1 feet per minute, the horse power will be 75000 -=- 33000 = 2.27 horse power. From the above the power required to drive any machine may be determined. The following is given as a simple method of finding the horse power required to run a machine : Pass a cord around the driving pulley of machine and give it several turns so that when the cord is pulled the machine will be started. Then attach to the end of this cord a re- liable spring balance. A pull on this spring with sufficient force to turn the machine will show the weight on balance. Then multiply this weight by 141 the working speed of belt in feet per minute and divide by 33000. Example : A machine has a 12 inch pulley which is to run 160 r. p. m. The tension as shown on the spring is 22 pounds; how much H. P. will be re- quired to run it? The surface speed of driving belt will be, 12 X 3.1416 X 160 = 6031.872 inches per minute, or 6031.872 -+- 12 = 502.656 feet per minute ; 502.656 X 22 then = .335- horse power, or about 33000 y 3 H. P. Steam Power The pressure of steam is expressed in pounds per square inch. Thus if steam is admitted to an engine cylinder with an average pressure of 30 pounds per square inch and the piston head has an area of 122 square inches, the total pounds pressure exerted against the piston is 122 X 30 = 3660 pounds. An average coal consumption is about 23 pounds per hour for every square foot of grate surface. Thus if four boilers have a grate surface of 18 square feet, the coal consumption will be near 18 X 23 = 414 pounds per hour, and 414 X 10 = 4140 pounds per day of 10 hours. One pound of coal will evaporate about 8% pounds of water. Hence in the above where 4140 pounds of coal are used in 1 day it will require about 8% X 4140 = 35190 pounds of water and as water weighs near 8y 2 pounds per gallon this will be 4140 gallons of water consumed in a day. An average water con- sumption is 20 pounds for each indicated horse 142 power (1 H. P.). Then if the above boiler consumes 35,190 pounds of water, the H. P. produced should be 35190 -r- 20 = 1709.5. To find the indicated horse power of a steam en- gine the f ollowing formula may be used : 2RASP H. P. = in which 33000 H. P. = Indicated horse power. E. = Revolutions per minute of shaft or fly wheel. A = Area of piston in square inches. S = Stroke in feet. P = Average steam pressure in pounds per square inch. It must be remembered that the indicated horse power is always more than the actual horse power on account of friction in the engine itself. Thus the "efficiency" of an engine is the ratio of actual horse power to indicated horse power. Thus Ave have, A. H. P. I. H. P. in which A. H. P. means actual horse power. For instance, if the indicated horse power of an engine is 250 and 235 the actual horse power, then 23% B0 = .94 = 94^, efficiency. The average steam pressure being used in an engine is found by the use of an instrument known as an "indicator." The application of this instru- ment to the engine is familiar to most all engineers. After the tracing is obtained on an indicator card, the pressure is found as shown by following: 143 If the area of card is 6 square inches and it is 5 inches long, while the scale of indicator spring is Y 3 q the steam pressure would be 6 X30 = 36 pounds per square inch. 5 Example : Find the indicated horse power of an engine which has an 18 inch cylinder (diameter), a stroke of 32 inches and is running 80 r. p. m., if the steam pressure is 36 pounds per square inch. Allow for a 2" piston. Using the above given for- mula we have, 11=80 revolutions per minute. A= (18X18) X -7854=254.469 square inches. As this area is less on the piston rod side, we find the area of piston rod and take half of this area from the piston head area to give the effective area. Then area of piston = (2 X 2) X -7854 = 1.5708, 2 then 254.469 — 1.5708 = 252.8982 square inches. S = 32 inches = 32 -f- 12 = 2.666 feet. P = 36 steam pressure. Then 8 12 2X^X252.89X2.666X^ H. P. = = 117.6 H. P. 550 144 Water Power The pressure -exerted by a head of water on a given area is equal to the weight of a volume of water Avhose area is equal to the area considered, multiplied by the height of this column from the given area. Thus the weight of a volume of water whose area is equal to 1 square inch would be the weight of 1 cubic inch of water, which is 62.5 -4- 144 (as 1 cu. ft. = 62.5 lbs. 1 cu. ft. = 144 cu. inches) = .434 pounds. Then if the height of this water (head) above a certain point is 40 feet the pounds per square inch would be .434 X 40 = 17.360 pounds per square inch. It makes no difference what the total volume of water at the head is, if we multiply the distance below a tank or reservoir by .434 the result is always (shown by example) pounds per square inch. The weight of a cubic foot of water being 62% pounds, if we know the volume of water passing a certain point in a stream in a given time we can find the foot-pounds and hence the horse power devel- oped. Example, if a stream passes 2162 cubic feet of water in a minute at a 20 foot fall we will have 2162 X 62.5 X 20 = 2702500 foot-pounds per minute of power developed. Then as 33000 foot-pounds per minute = 1 H. P. we have, 2702500 -4- 33000 = 81.8 H. P. developed. As, however, water turbines do not give more than 75% mechanical efficiency, then 81.8 X -75 = 61.35 effective horse power. PUMPS The number of gallons of water or other liquid delivered by a pump is found by obtaining the vol- ume of the cylinder in cubic inches and dividing by 145 231. This gives the number of gallons for each stroke. Then the number of strokes per minute multiplied by this will give the gallons capacity of the pump in 1 minute. If a pump has a 4 inch cylin- der and a 6 inch stroke, and is running 40 strokes per minute, how many gallons will be pumped in 1 minute ? 4 X 4 = 16, then 16 X -7854 = 12.5664 square inches area of piston or cylinder. 12.5664 X 6 (stroke) = 75.3984 cubic inches per each stroke. Then 75.3984 X 40 = 3015.936 cubic inches pumped in 1 minute, and 3015.936 -h 231 (eu. in. per gal.) = 13.05 gallons pumped in 1 minute. To find the power required to run a pump hav- ing a certain capacity and lifting the water a cer- tain height multiply the number of gallons pumped per minute by 8% (lbs. of 1 gallon) and multiply this by the height pumped and divide this by 33000. This will give the theoretical horse power. To this must be added an allowance for pipe friction, loss in pumps, gears, belts, etc. Example : What horse power will be required to run a pump with a capacity of 500 gallons per min- ute pumped to a height of 50 feet ? 500 X 8^4 = 4250 pounds per minute. 4250 X 50 = 212500 foot-pounds per minute. And 212500 -=- 33000 = 6.43 H. P. Theoretical. ELECTRIC POWER AND LIGHTING The terms used in electricity correspond to those used in mechanical engineering and when thus com- pared are more easily understood. . 146 Ampere Means the amount of current in a wire, light or motor and is equivalent to weight as commonly used in mechanics. Volts or Voltage Means the rate of flow of an electric current in a „wire, light or motor and is equivalent to velocity, speed or "feet per minute" in mechanics. Ohm Is the unit of resistance as set up against a cur- rent by its conductor and is equivalent to loss due to friction in mechanics. In mechanical power the work performed is found by multiplying the weight by the speed in a certain time. So in electric power the work performed is found by multiplying the amperes by the volts. This product is called watts, which represents the unit of electric power. Then we have, Electricity : Amperes X volts = watts. Mechanics : Weight X velocity = power. One electrical horse power is equal to 746 watts. One kilowatt is equal to 1000 watts. So one kilo- watt would be equal to 100 %46 = 1-34 horse power. And one horse power would equal .746 kilowatt or 746 watts. One watt-hour means one watt ap- plied for one hour and is equal to % 46 horse power applied for one hour. One kilowatt-hour means one kilowatt applied for one hour and is equal to 1.34 horse power applied for one hour. 147 One horse power-hour means one horse power applied for one hour and is eual to .746 kilowatt- hours. Examples : Find watts generated by a dynamo having a capacity of 60 amperes and 120 volts. 120 X 60 = 7200 watts, and 7200 ^- 1000 = 7.2 kilowatts. As 1 K. W. (kilowatt) = 1.34 H. P. this motor would generate 7.2 X 1-34 = 9.648 H. P. What current is required for 40, 16-c. p. 3.5 watt lamps at 110 volts? A lamp at 3.5 watts means that it requires 3.5 watts for each candle power. Then for a 16 c. p. 3.5 watt lamp it would require 16 x 3.5 = 56 watts. Then 56 X 40 = 2440 total watts for all lamps = 3.27 H. P. 2440 watts And = 22.18 amperes. 110 volts How many horse power would be required to run 100, 40 watt lamps? Sometimes the candle power of lamps is not given, the term 30, 40 or 50 watts, etc., being given. Prom this we can find the candle power if we know the usual worth required for each. Thus a Mazda lamp requires 105 to 115 volts and will produce one candle power for about every .95 to 1.03 watts. So to solve the above problem we have, 100 X 40 = 4000 total watts used, 4000 and = 5.36 horse power. , 764 (watts per H. P.) 148 How many 16 c. p. 3.1- watts (per candle power) could be run from a 25 K. W. (kilowatt) dynamo? IK. W. = 1000 watts and 25 X 1000 == 25000 watts, then 16 X 3.1 = 49.6 watts for each lamp, and 25000 -f- 49.6 = 504 lamps at dynamo. To find size of wire to use on two wire service, the following is given as a formula: CXDX21 M = P In this M = circular mils of wire which is found by squaring the diameter of any wire expressed in mils (1 mil = .001"). C = Current amperes, D = Average distance one way of wire, P = Per cent loss allowed, 21 = Constant. Example : Find size of wire required to carry current for 150, 16 c. p. 3.5 watt lamps at 120 volts from a dynamo a distance of 500 feet at a loss of 5%. 120 x -05 = 6 volts loss ; so, 120 — 6 = 114 volts applied to lamps ; then 150 X 16 X 3 - 5 = 84 00 watts required, and 8400 -f- 114 volts =73.6 amperes current. Then using formula as given above for size of wire we have : 73.6 (current) X 500 (distance) X 21 (constant) 5 (per cent loss) 154560 mils size of wire. Looking at the table we find this area to be closest to a No. 000 wire. 149 GEARS In finding the relative speeds of two gears when in mesh, no matter whether they be straight, spur or bevel gears, we proceed as follows : number teeth X speed = number teeth X speed, (one gear) (other gear) From this we can find the number of teeth or the speed of either gear when the other data is given. Examples : If a 30 teeth gear is running 210 r. p. m. and is meshing with a gear of 50 teeth, what speed is the latter gear making? 30 X 210 = 50 X (X) ; then (X) speed of 50 t gear = 30 X 210 = 126 r. p. m. 50 A 40 teeth gear is running 200 r. p. m. and is do- sired to run another gear at 250 r. p. m. ; how many teeth must the latter gear have? 40 X 200 = (X) X 250 ; then, (X), No. teeth in gear = 40 X 200 = 32 teeth. 250 When two gears are meshing the speed of one gear is opposite to that of the other. 150 Thus in Fig. 5 the two gears A and B are running .toward each other, and the two gears,. C and D, are running away from each other. When two gears are connected by a carrier gear as shown in Fig. 6 the direction of speed of the two gears considered are in the same direction. Also the relative speeds of the two gears are not affected. J A ® As shown in (a) Fig. 6 the directions of A and C are both the same when connected by the carrier gear, B, neither is the relative speeds of the two gears, A and C, affected by this carrier gear. Sup- pose A in (b) is on a shaft making 100 r. p. m. and has 40 teeth, it drives through the carrier gear, B, to C which has 60 teeth, what is the speed of this roll? 40 X 100 = 60 X (X) ; then (X) speed of roll = 40 X 100 60 = 66.6 r. p. m. 151 Yarn Calculation As all calculations in the mill are based on yarn and hank roving numbers, it is important to have a thorough knowledge of the methods used to determ- ine these numbers. As a general rule yarn numbers are in an inverse ratio to the sizes or diameter of yarns. Thus a number 1 yarn has a greater diam- eter than a number 2 yarn. In like manner a num- ber 16 yarn is larger in diameter than a number 30 yarn. r-The number of a yarn is based on length and weight. If we know the weight, either in pounds, ounces or grains, of a certain length of yarn we can then determine its number. The standard used in finding yarn numbers is based on the method which has come into constant universal use. This standard first takes 840 yards as a length and is called one "hank." It also takes 1 pound as the weight of a hank if it be number 1 yarn. Then for one hank of 840 yards if the weight is reduced the yarn becomes smaller and the number is increased. Thus if a hank (840 yards) weighs only % pound the number would be increased twice and we call it a. number 2 yarn. If the weight be reduced to % pound the number would be increased to 3 and so on. Again,, if we increase the length of a certain weight the yarn will become smaller in diameter and the number then increased. For instance, if we in- crease the length of 1 hank (840 yards) so as to make 2 hanks (1680 yards) and if it still weighs 1. ,152 pound it. is called. a.. number 2 hank or a number 2 yarn. From this we see then that as the number of hanks in 1 pound is increased, the number is in- creased. If a pound of yarn contains 6 hanks (= 840 X 6 = 5040 yards) it is a number 6 yarn or hank roving. If a pound contains 10 hanks (= 840 X 10 = 8400 yards) it is a number 10 hank •roving or yarn. In like manner in one pound of yarn we have the following : 1 hank (840 yards) =No.l hank, roving or yarn, 2 hanks (840X2=1680) =No.2 hank, roving or yarn, 3 hanks (840x3=2520)=No.3 hank, roving or yarn, 4 hanks (840X4=3360) =No.4 hank, roving or yarn, and so on. The number of hanks., then, in 1 pound is the number of the roving or yarn. Examples: 1. How many hanks in 1 lb. of No. 30- yarn? 2. How many hanks in 2 lbs. of No. 40 yarn? 3. If a yarn is No. 18.25, how many hanks in 1 lb. ? 4. If 1 lb. of roving has 2.75 hanks, what is the No.? ■ 5. How many yards in 1 lb. of No. 3 hank roving? 840 = 1 hank; then 840 X 3 = 2520 yds. in 1 lb. 6. How many yards in 2 lbs. of No. 2 hank roving? 1 lb. of a No. 1 hank = 840 yds. ; 2 lbs. = 2 X 840 — 1680 yds. in 1 lb. of No. 1 hank; then 1680 X 2 = 3360 yds. in 2 lbs. of No. 2 hank roving. ms 7. How many yds. in 1 lb. of No. 20 yarn? No. 1 yarn = 840 yds. ; then No. 20 = 840 X 20 = 16800 yds. 8. How many yds. in 1 lb. of No. 30 yarn? No. 1 yarn = 840 yds. ; then No. 30 = 840 X 30 = 25200 yds. 9. If a No. 1 hank roving has 840 yds., what is its weight ? 1 lb. = 840 yds. = No. 1, then 840 yds. of No. 1 hank roving weighs 1 pound. 10. How many pounds in 1680 yds. of No. 1 hank roving ? 1 lb. = 840 yds. ; then 1680 yds. = 1680 / 840 = 2 lbs. 11. How many pounds in 16800 yards of No. 20 yarn ? 1 lb. of No. 20 = 840 X 20 = 16800 yards ; then 16800 yds. = 16800 /i 6 800 = 1 pound. 12. How many pounds in 33600 yards of No. 10 yarn? 1 lb. No. 10 = 840 X 10 = 8400 ; then 3360 %4oo = 4 pounds. 13. If 2520 yards of roving weigh 1 pound, what is its No. ? 1 lb. of No. 1 = 840 yds. ; then 1 lb. of (X) yarn = 252 % 40 = No. 3 roving. 154 14. If 840 yards of roving weigh 2 pounds, what is its No. 1 1 lb. of No. 1 = 840 yds. ; 2 lbs. of No. 1 = 840 X 2 = 1680 yds. ; then if 2 lbs. of No. 1 = 1680 yds., 2 pounds of (X) yarn = ^% 680 = No.. % = .5 H. R. . From the above examples we see that if we have the pounds and yards given we can find the num- ber. Thus if we have 1680 yards weighing 1 pound, the number is found by dividing 1680 by the No. yards of No. 1 in 1 pound which is 840. Then 1680 -r- 840 X 1 = No. 2. Again if we have 25200 yards weighing 3 pounds we find the number thus, 25200 = No. 10. 840X3 (a) Then if we have weight in pounds and yards given we have, number yards (1) = No. 840 X pounds Also if we have the number and the number of yards given we have, number yards (2) = Pounds. 840 X No. Again if we have pounds and No. of yarn or rov- ing given, we can find the number of yards thus, (3) 840 X No. X pounds = number of yards. 155 (b) If the length is given in yards and the weight in ounces, it is only necessary to find, first, the num- ber of yards in 1 ounce of No. 1 hank. If there are 840 yards in 1 lb. = 16 ounces of No. 1, then there would be 840 -=- 16 = 52.5 yards of No. 1 in 1 ounce. Then as in (a) we will have, number yards (1) = No. ' 52.5 X ounces number yards (2) = ounces. 52.5 X No. (3) 52.5 X No. X ounces = number yards. (c) If the length is given in yards and the weight in grains we must likewise find the number of yards of No. 1 in 1 grain. If there are 840 yards of No. 1 in 1 lb. = 7000 grains, then there would be 840 -=- 7000 = 21 / 175 = .12 yards of No. 1 in 1 grain. Then as in (b) we have,, number yards (1) = No. . .12 X grains number yards (2) = grains. .12 X No. (3) .12 X No. X grains = number yards. CARD ROOM PROBLEMS In finding the H. K. No. (hank roving No.) of caret or drawing slivers the; usual practice is to weigh 1 yard in grains. If it is desired to find th3 156 H. R. No. this weight in grains is multiplied by .12 and divided into the number of yards, 1. As 1 yard is always taken a constant is found which will give the H. R. No. by simply dividing the weight in grains into this constant. This constant is as fol- lows : number yards If = No. it is the same as .12 X grains number yards 1_ grains. .12 As 1 yard is always taken then we have 1 1 1 12 '— grains, now = '— ■ — — because .12 .12 1 100 .12 = i2/ 10 o, and 1 = ft Then 1 / 1 ~ 12 Aoo = 100 /i2 = 8% ; then we have, 8i/ 3 (= 8.333) = No. if 1 yard is taken. grains In finding the H. R. No. of roving 12 yards are usually taken and weighed in grains. We also find a constant which by dividing the weight in grains into it the H. R. is found. This constant is found as follows : number yards If = No., it is the same as .12 X grains number yards -f- grains. .12 157 12 And as = 100 we have, .12 100 = No. if 12 yards are taken, grains 100 or = grains. No. Examples : 1. If 12 yards of roving weigh 25 grains, what is the H. R. No. ? 100 -=- 25 = 4 H. R. 2. 12 yards of roving weigh 35 grains ; what is the H. R. No. ? 100 -=- 35 = 2.857 H. R. 3. How many grains ought 12 yards of 1.75 H. R. to weigh? 100 -^ 1.75 = 57.14 grains. 4. Four 12-yard lengths of roving are taken weighing 52.5, 51.75, 52 and 52.25 grains ; what is the average H. R. No.? An average weight is found by adding all weights and dividing by the number of weighings made. Thus, 52.5 51.75 52. 52.25 4 ) 208.50 52.12 average weight. 100 = 1.91 H. R. 52.12 158 5. What H. R. is a 56 grain drawing sliver? 1 yard = 56 grains, 12 yards = 12 X 56 = 672 grains ; then 100 -f- 672 = .1488 H. R. ; or 8.3333 -=- 56 = .1488 H. R. 6. If a 48 grain drawing sliver is fed to the slub- ber and 1 yard of sliver makes 3 yards of roving, uwhat is the H. R. being made on slubber? 1 yard roving would weigh 48 -~ 3 = 16 grains ; then 12 yards would weigh 12 X 16 = 192 grains, and 100 / 192 = .52 H. R. 7. If hank clock on a frame registers 8 hanks run in a day and 4 H. R. is being made, what is the pounds production per spindle per day? 8 hanks = 840 X 8 = 6720 yards per day. From 2 under (a). 6720 2 pounds, 840 X4 or as 6720 = 840 X 8 we have, X8 8 = — = 2 pounds per spindle per day. X4 4 From this we get the rule for pounds production. No. hanks from clock X No. spindles = pounds H. R. production. SPINNING ROOM PROBLEMS 1. If 12 yards of H. R. weigh 40 grains, what is its No.? 159 From (1) in (c) we have 12 100 = = 2.5 H. R. No. .12 X 40 40 2. If 12 yards of 2.5 H. R. are fed into spinning frame, one "end up," and 120 yards are delivered, what is the yarn No.? From (1) under (c) we have 12 100 = = 40 grains, .12 X 2.5 2.5 weight of 12 yds. of 2.5 H.R. Then if 120 yards of yarn are made from this it will weigh approximately 40 grains. Then we have, 120 1000 = 25 yarn No. .12 X 40 40 The usual practice in finding the number of yarn is to reel off 120 yards and weigh in grains. The weight of this 120 yards in grains divided into the constant will give the number of yarn. If 120 yards are taken Ave have, 120 120 -r- grains, from (1). under (c), and — — = 1000 .12 .12 which is our constant. Then we always have, 1000 = yarn No. wt. in grains of 120 yds. 3. If 120 yards of yarn weigh 25 grains, what is the number? 160 1000 = 40 yarn No. 25 4. If a 1 inch front roll is running 136 revolutions per minute on 30 's yarn, what is the pounds pro- duction in a day per spindle ? Circumference of roll = 1 X 3.1416 = 3.1416 m inches, and 3.1416 X 136 = 427.2576 inches delivered in 1 minute ; 427.2576 -=- 36 = 11.86 yards delivered in 1 minute ; 60 X 10 hours = 600 minutes in a day ; then 11.86 X 600 = 7116.00 yards per day of 10 hours, and from (2) under (a) 7116 7116 = .282 lbs. per spindle 840 X 30 25200 per day. 5. If a spool has 14 ounces of No. 28 's yarn on it, what is the length? From (3) under (b) we have, 52.5 X 28 X 14 = 14700 yards. 6. Spools are to be made with enough 30 's yarn to make 18000 yard warp on warper ; what should be weight in ounces of yarn on each spool? From (2) under (b) we have, 18000 = 11.4 ounces or 12. 52.5 X 30 7. If a warper beam has 436 ends of No. 30 's and is 16000 yards long, what should it weigh in pounds ? 436 X 16000 = 6,976,000 yards of yarn on beam ; then from (2) under (a) we have, 161 6,976,000 = 276.8 pounds. 840 X 30 8. A warper beam weighs 400 pounds and has 342 ends of 28 's on it; what is the length of warp? 400 -=- 342 = 1.169 pounds, weight of each end on beam; then from (3) under (a) we have, 840 X 1-169 X 28 = 27494 yards. WEAVE ROOM PROBLEMS 1. 8% size is put on a warp of 30 's which has 2040 ends. If the beam has 15 cuts, each 64 yards in length, what is the weight of yarn on beam? 15 X 64 = 960 yards, length of warp ; then from (2) under (a) we have, 960 X 2040 = 77.71 pounds without size. 840 X- 30 Now there are two ways of figuring the % of the size. First, it may be figured based on the result- ing weight of finished warp; or, second, it may be figured on the weight of yarn before sizing. To illustrate the first case,, if we have 2000 pounds of yarn going to slasher and the finished yarn on beams weighs 2200 pounds, we have, 2200 — 2000 = 200 pounds of size put on, and 200 X 100 = 9.09%. 2200 To illustrate the second case we can take the same example and we have, 162 2200 — 2000 = 200 pounds of size put on, and 200 X 100 = 10%. 2000 It seems that as the size is put on yarn and sold as part of the finished cloth the resulting weight of ^yarn after sizing should be the basis from which to figure the % of size. Then if we use the first method of figuring % in example 1, 77.71 pounds would be 1.00 — .08 = 92% of the weight of yarn on beam. Then if 1 pound of sized yarn has .92 pounds of cotton in it, (X) pounds would have 77.17. This can be solved by the fol- lowing proportion : 1 : .92 : : X : 77.17; then X = 77.17 = 83.8 pounds. .92 2. A cloth is to be 36 inches wide with 68 ends per inch and to be made into cuts of 60 yards. If we allow 7% take-up in weaving, how many pounds of warp would there be in a cut of cloth using 28 warp with 8% size? 36 X 68 = 2448 ends in warp, 1.00 — .07 = .93 ; then 60 yard cut of cloth is 93% of the length in a cut of warp. Hence we have, 60 -r- .93 = 64.5 yards in 1 cut of warp or 64. Then from (2) under (a) we have, 64 X 2448 = 6.66 pounds of unsized yarn. 840 X 28 163 As under example 1 we have, 6.66 -f- .92 (1.00 — .08) = 7.239 pounds of sized yarn in 1 cut of cloth. 3. If the cloth in example 2 has 64 picks per inch, how many wards of filling would there be in a cut of cloth, if the warp is 38 inches at the reed? 38X64X^0 = 2432 yards of filling in 1 fifi yard of cloth. then 60 X 2432 = yards of filling in 1 cut. If the cloth is to weigh 4 yards per pound, what must be the filling No. ? 4 yards per pound = 6 % = 15 pounds per cut. From example 2 we see that there are 7.239 pounds of warp in 1 cut. Then 15 — 7.239 = 7.761 pounds of filling in 1 cut. From (1) under (a) we have, 60 X 2432 = 22.38 filling No. 840 X 7.761 PLY YARNS When two single yarns are twisted together to form a strand, the resulting yarn is said to be "two ply yarn." When 3 single yarns are twisted to- gether the result is said to be "three ply yarn,"- and so on for as many single strands as are twisted together. Twisted yarns are shown in figures by placing the number which represents the single yarn above a line and the figure Avhich shows how 164 many of such strands were used below this line. Thus, a 4 % yarn means that 2 ends of single 40 's are twisted together, 3 % two ends of 30 's, and so on. When two ends of single yarn are thus twisted to- gether the resulting hank number is practically one-half of the single yarn. It will be a little heavier owing to the contraction in twist. Thus a 3 % yarn = 15 counts, a 4 % = 20 counts, a 3 % = 10 *counts, and so on. When two yarns of different sizes are twisted together the resulting number is found by multiply- ing the two single numbers together and dividing this product by the sum of the two numbers. Example : If a 30 and 40 are twisted together what is the resulting number? 30 X 40 1200 = = 17.14. 30 -f 40 70 Proof : 120 yds. 30 's = 1000 -=- 30 = 33% grains. : 120 yds. 40 's = 1000 -*- 40 = 25 grains. 120 yds. twisted yarn 58% grains. Then 1000 -^ 58.33 = 17.14. WEIGHT OF YARN ON BEAMS, SPOOLS, ETC. Sometimes it becomes necessary to find the weight of yarn contained on a beam or spool of certain size ! To do this 60 cubic inches of yarn are allowed to 1 pound. Example : A warper beam is 54 inches between heads and has a 9 inch barrel and 24 inch heads; how many pounds of yarn will it hold? First, multiply the result of adding the diameters 165 of head and barrel by the result of subtracting bar- rel from head. 24 + 9 = 33 and 24 — 9 = 15 ; then 33 X 15 = 495. Second, multiply the above result by .7854, then .7854 X 495 = 388.773 square inches. Third, multiply this area by distance between heads, then 388.773 X 54 = 20993.742 cubic inches capacity. Fourth, as 60 cubic inches = 1 pound, then 20993.742 -=- 60 = 349.8 pounds. 166 Draft Calculations The process of drawing cotton sliver and roving by means of rolls was invented by John Wyatt in 1737. This process was afterwards improved upon J)y James Hargeaves, 1764, Arkwright in 1767 and Samuel Compton in 1779. Arkwright is usually given the credit for this improvement in cotton spinning. The drawing out of slivers or roving is done by feeding into a roller with a constant speed a certain length of sliver or roving. The cotton then passes to another roller which has a faster speed than the previous one, thus drawing out the strand of cotton between the two. This drawing out of strands com- posed of cotton fibers can be done to a greater ex- tent from fine strands than from heavier ones. For this reason we see greater drawing as the strand gets finer in the mill. For instance, starting with the first process in drawing we gradually increase this process to the last machine. This drawing out of a strand of cotton is com- monly called Draft. Draft then means the amount of attenuation or drawing out that this strand re- ceives in a certain machine. If a yard of sliver is fed into, a machine and four yards delivered the re- sulting strand is one-fourth as large as it was at the beginning,, and draft is said to be 4. If 2 yards are fed into a set of rolls and 8 yards are delivered the ratio of the fed strand to delivered strand is 1 to 4 and the draft is 4. From this we find the fol- lowing : 167 EFFECT OF DRAFT ON LENGTH If 2 yards fed = 8 yards delivered, draft is 4. If 4 yards fed = 20 yards delivered, draft is 5. If 400 inches fed = 2400 delivered draft is 6. From this we see that, (a) length delivered -=- length fed = draft. Also if 3 yards are fed and draft is 4 = 12 yards delivered. If 4 yards are fed and draft is 5 = 20 yards de- livered. If 20 inches are fed and draft is 4 = 80 inches delivered. From this we see that, (b) length fed X draft = length delivered. Also if 36 yards are delivered and draft is 6 = 6 yards fed. If 100 yards are delivered and draft is 10 = 10 yards fed. If 3600 inches are delivered and draft is 12 = 300 inches fed. From this we again see that, (c) length delivered -4- draft = length fed. EFFECT OF DRAFT ON WEIGHT In considering the weights of certain sliver or rovings the effect of draft is as" follows: If one yard weighing 5400 grains is fed, and 1 yard delivered weighs 50 grains, the draft is 90. If 12 yards weighing 25 grains are fed and 12 yards delivered weigh 10 grains- the draft = 2.5. 168 (a) From this we see that, grains in 1 yard fed -r- grains in 1 yard delivered = draft. Or grains in 12 yards fed -=- grains in 12 yards delivered = draft. Also if 1 yard weighing 50 grains is delivered and draft is 100 the weight of 1 yard fed is 5000 grains. If 12 yards weigh 25 grains delivered and draft is 4 the weight of 12 yards fed is 100 grains. (b) From this we see that, grains in 1 yard de- livered X draft = grains in 1 yard fed. Or grains in 12 yards delivered X draft = grains in 12 yards fed. Also if 1 yard weighing 4500 grains is fed and draft is 90, the weight of 1 yard delivered is 50 grains. If 12 .yards weighing 100 grains is fed and draft- is 4, the weight of 12 yards delivered is 25 grains'. (c) From this we see that, grains in 1 yard fed -r- draft = grains in 1 yard delivered. Or grains in 12 yards fed -=- draft = grains in 12 yards delivered. EFFECT OF DRAFT ON WEIGHT WHEN DOUBLING If two or more laps, sliver or rovings are fed to the rolls it is only necessary to find the resulting weight of a certain length after this doubling. Thus if 4 laps are fed into a finisher picker and each lap weighs 14 ounces to the yard the resulting weight of 1 yard being fed is 4 X 14 = 54 ounces. If a lap weighing 14 ounces to the yard is delivered from this the draft is 4. From this we have, No. of doublings X weight of each (a) = draft. weight delivered 169 and No. of doublings X weight of each (b) = weight draft delivered. or weight delivered X draft (c) = weight fed. No. doublings EFFECT OF DRAFT ON H. R. NOS. AND YARN NOS. In considering the effect of draft on numbers of yarns or hank roving it is only necessary to remem- ber that the size of a yarn or roving is reduced as the number is increased, or that the yardage is in- creased as the number is increased. Thus a No. 1 yarn or hank roving has 840 yards in 1 lb.„ a No. 2 yarn or hank roving has 1680 yards in 1 lb., and so on. Then as draft deals directly with relative lengths fed and delivered it has a similar effect on numbers of yarns and hank roving fed and delivered. If a 1 hank roving is fed and = 2 H. E. delivered, draft is 2. If a 2 hank roving is fed and = 8 H. B. delivered, draft is 4. If a 6 hank roving is fed and = 36 yarn delivered, draft is 6. EFFECT OF DRAFT ON NOS. WHEN DOUBLING If the H. E. is doubled the resulting H. R. fed is found by dividing the No. H. R. by the No. doublings. 170 Thus if a 2 H. E. is being fed two ends into one, the resulting H. R. is 1. If a 4 hank roving is being fed two ends into one, the resulting H. R. is 2, etc. Then, If a 2 H. R. is fed doubled 2, and = 2 H. R. de- livered, draft = 2. If a 4 H. R. is fed doubled 2, and = 8 H. R. de- livered, draft = 4. If a 6 H. R. is fed doubled 2, and = 30 yarn de- livered, draft = 10. From this we see that, H.R. or yarn delivered X No. doublings (a) = H.R. fed draft Also, if 2 H. R. is fed doubled 2, and draft is 2 = 2 H. R. delivered. If 4 H. R. is fed doubled 2, and draft is 4 = 8 H. R. delivered. If 6 H. R. is fed doubled 2, and draft is 10 = 30 yarn delivered. From this we see that, H.R. fed X draft (b) = No. H.R. or yarn No. doublings delivered. When Doubling" Is 2 If 2 H. R, is delivered and draft is 2 = 2 H. R. fed (single). If 8 H. R. is delivered and draft is 4 = 4 H. R. fed (single). If 30 yarn is delivered and draft is 10 = 6 H. R. fed (single). From this we see that, 171 H.R. or yarn No. delivered X No. doubling (c) : = draft H. R, (single) fed. DRAFT OF MACHINE The draft in a machine is figured from the rel- ative length of sliver or roving being received by one roll and that delivered by another. For instance,, if back roll takes in 300 inches of sliver or roving in one minute and front roll de- livers 900 inches in the same time, the front roll will deliver 3 inches while the back roll is taking in one inch. Then the draft is said to be 3. Example : If back roll is 1%" in diameter and is making 100 revolutions per minute, and front roll is 1%" in diameter and makes 200 revolutions per minute, what is the draft? The number of inches per minute taken in by the back roll Avill be the same as the surface speed of this roll expressed in inches per minute. The sur- face speed is found by multiplying together the di- ameter, iy 8 " (= 1.125"), 3.1416 and its speed, 100. Then, 1.125 X 3.1416 X 100 = 353.43 inches taken in by back roll in 1 minute. Also, 1%" (= 1.375"), X 3.1416 X 200 = 863.94 inches delivered by front roll in one minute. Then the draft is found by dividing 863.94 by 353.43 as seen in (a) under "Effect of Draft on Lengths. ' ' Then 863.94 -=- 353.43 = 2.44 draft. 863.94 If (1) = draft, 353.43 172 1.375 X 3.1416 X 200 then (2) = draft, 1.125 X 3.1416 X 100 because (1) and (2) are equal or (1) is the same as (2). By cancelling the 3.1416 on the top and bottom we have left, 2 1.375 (front roll dia.) X 200 (speed front roll) (a) ^ = 1.125 (back roll dia.) X W (speed back roll) 2.750 = 2.44 draft, or Ave may show this in the 1.125 following manner : 1.375 X 200 = 2.44 draft. 1.125 X 100 200 If we consider the above, , as finding the rel- 100 ative speeds of the two rolls, as it is, it is not nec- essary to know the speed of either roll to determine the draft. All we want to know is to find out ratio of speed between the two,, or to find when the back roll makes 1 turn how many turns the front roll will make. This can be found out by considering the gearing between the two. If we consider the following sketch as the gear- ing between front roll and back the total draft be- tween the two is found as follows : 173 I '/a" BACK ROLL _/~ 24 B A-70 l^-fRONT ROLL FIG. 7. C-96 D-30 If we consider the back roll as making 1 turn then the gear, A, will make 1 turn and the gear, B, will make 1X70 24 turns = 2.916 -j- turns. Then, if B makes 2.916 turns the gear, C, on same stud will make 2.916 turns and the gear, D, will make 2.916 X 96 = 9.331 turns. 30 From equations we saw that if we use 2.916 in the last process we could substitute for it the value 1X70 24 and bring all of the calculation under one head. 174 1 X 70 96 70 X 96 Then (which is 2.916) X — = 24 . 30 24 X 30 and using cancellation we have, 4 W X H 28 = — = 9.333 turns of front roll while 4fy£ X fift 3 back roll turns 1 time. 3 ft If then we consider the gears B (= 24 teeth) and D {== 30 teeth) as the drivers and the gears C (= 96 teeth) and A (= 70 s teeth) as driven gears, then the relative speed of the back to the front roll is found by multiplying the driven gears together and dividing this product by the driving gears mul- tiplied together. Now, as we saw under (a) that, diam. of front roll X speed of front roll = draft. diam. of back roll X speed of back roll And as in this example the front roll is 1% inches in diameter = 1.375, and back roll 1% inches in diameter = 1.125 we have, 1.375 X 9.333 12.832 (b) = = 11.4 draft. 1.125 X 1 1.125 Again, if in the above equation, (b) 9.333 70 X 96 1 24 X 30 we can substitute and have the following 175 1.375 X 70 X 96 (c) = 11.4 draft. 1.125 X 24 X 30 RULE TO FIND DRAFT In order then to find the draft of any machine, start with the diameter of back roll, or first roll into which cotton is fed. Place this diameter under the line as seen above and place the diameter of front roll,, or last roll through which cotton passes, on top of the line. Then place all driven gears above line with the sign, X, between them. Then place all driver gears below the line with the sign, X, between them and solve by cancellation. CHANGE GEARS In Fig. 7 if the gear, B, is a change gear the draft is affected by changing the number of teeth in this gear. If the number of teeth is increased the back roll is run faster and the draft is reduced. Also if the number of teeth is decreased the back roll runs slower and the draft is increased. From this we see that a larger gear produces less draft and a smaller gear produces more draft. Then the draft is inversely proportional to the number of teeth in change gear. Thus if a 24 teeth gear pro- duces a draft of 11, what gear will produce a draft of 6 ? If we use a direct proportion to. find this we have, 24 : 11 : : X : 6 6X24- then = 13 teeth gear, 11 176 which being a smaller gear than the 24 teeth ought to produce more draft. Hence this proportion is not true, because we want less draft than that produced with the 24 teeth gear. Then if we turn the proportion completely around and say 24 teeth is to 6 draft as X teeth is to 11 draft w T e have an inverse proportion as follows : 24 : 6 : : X : 11 ; 11 X24 then = 44 teeth change gear. 6 The above proportion is correct and may be used to advantage. DRAFT CONSTANTS As the gear, B, in Fig. 7 can be changed to pro- duce different drafts, the number, 24, could be left out in the equation (c), and the resulting number thus obtained when divided by whatever number draft gear is to be used will give the draft. Thus the draft constant in (c) will be, 7 32 1.375 X JT0 X H = 273.77 constant. 1.125 X (X) X^ Now as the number 24 was left out below the line if this number be divided into 273.77 the draft will be obtained. Thus, 273.77 -=- 24 = 11.4 draft. Such constants are usually obtained for all ma- chines and when it is desired to change the draft it is quickly obtained. 177 If the constant 273.77,, when divided by the gear, 24, gives the draft, 11.4, then the draft, 11.4, divided into the same constant, 273.77, will produce the gear, 24, as follows : 273.77 -^ 24 gear = 11.4, and 273.77 -=- 11.4 draft = 24 gear. Then when we have a draft constant we can ob- tain the draft gear to use for any draft, and also the draft obtained by using any draft gear. constant Thus, = draft. change gear constant = change gear. draft required 17P DRAFT OF A PICKER v "c \ S* X f ^t M *\ \/ P )q vrA /~Sr* V <\) <*> its yiM * Si ^ ^'V Jx-X -■c\ , s Hk i L-Z X " 'M^ no Li \ a 1 > CI c 5 U u c o (J o U 179 Using the above sketch the draft of a finisher or intermediate picker is found as follows : cal. roll ACBGI KMO 9 X32X22X90X12X28 X20X15X13 = 3.8 draft. 2.5X40X 9X1 X24X45 X52X72X54 feed B D F H J L N P roll change If the change gear, J, is left out Ave have the constant number, 171. If this gear, 45, be divided into 171 we get 3.8 draft. constant Then = draft. change gear constant And = change gear. draft Examples : If 4 laps weighing 14 ounces to the yard each are fed to this machine using a 45 teeth gear, what will each yard of lap weigh in front? constant 171 = 3.8 draft. change gear 45 Then as number of doublings X wt. of each draft = wt. in front. From (b) under "Effect of Draft on Weight When Doubling." 180 4X14 "We have = 14.73 ounces, weight of each 3.8 yard delivered. As there is some of this lost in going through in the form of waste this allowance must be made. Then allowing 2% for waste we have 14.73 ounces as 100% of weight fed in. Then, 14.73 X -02 = .29 ounces lost as waste. Then, finished lap should weigh 14.73 — .29 = 14.44 ounces. To find the % of waste made. If 4 laps each weighing 14 ounces to the yard are fed and 1 yard of finished lap weighs 14% ounces and draft is 3.8, the finished lap should weigh 4X14 — = 14.73 ounces. 3.8 But if it weighs only 14% ounces the loss is 14.73 — 14.5 = .23 ounces. Then the % of waste would be .23 X 100 = 1.56% waste. 14.73 Example : If 4 laps each weighing 15 ounces are fed to machine and the lap to be delivered is to weigh 14% ounces, what draft gear must be used, allowing 2% for waste? 4 X 15 = 60 ounces, weight of 1 yard fed to machine. * 181 As 14.75 ounces is to be the finished lap after al- lowing for waste, the theoretical weight would be found as follows : 14.75 is (1.00 — .02) 98% of the theoretical weight; then 14.75 -=- .98 = 15.05, weight of 1 yard if no waste were made. Then 60 -^ 15.05 = 4 draft practically. Using the constant, 171, as found above we have, 171 constant = 42.7 or 43 draft gear. 4 draft Example : If a 14 ounce lap is required and draft in machine is 4, what must be the weight per yard of each of the 4 laps if 2% waste is allowed? 14 ounces is (1.00 — .02) 98% of the theoretical weight allowing for waste. Then 14 -=- .98 = 14.28 ounces lap if no waste were made. Then if draft is 4, 14.28 X 4 = 57.12 ounces, weight of 1 yard of cotton going in feed roll. As there are 4 laps going in, 57.12 -f- 4= 14.28 weight of each lap behind. 182 DRAFT OF A CARD FLOOR PLAN OF 40-INCH CARD WITH 27-INCH DOFFER FIG. 9 183 Using the above sketch the draft of a card is found as follows. coiler roll 2 X 120 X 40 X 214 X 27 = 76.73 draft. 2.25 X 20 X 45 X 21 X 17 feed bevel roll change If the change gear, 20, is left out we will get a result of 1534.6 which is the draft constant. If 1534.6 is divided by 20 the draft, 76.73, is found thus, 1534.6 constant = 76.73 draft. 20 change gear Or, 1534.6 constant = 20 change gear. 76.73 draft Example : If a 12 ounce lap is fed to the card using an 18 teeth change gear, what will be weight of sliver delivered? constant 1534.6 = 85.2 draft with 18 change gear 18 teeth gear. A 12 ounce lap = 12 X 437.5 = 5280 grains per yard; then, 5280 = 61.9 grain sliver delivered. 85.2 draft 184 As there is some waste the resulting sliver will not weigh quite this much. If 4% is allowed for Avaste this weight would be reduced as follows : 61.9 = 100% stock fed; so 61.9 .X -04 ===== 2.476 grains lost in waste ; and 61.9 — 2.476 = 59.424 grains actual weight delivered. * To find actual % of waste in card. Example : If the weight of lap is 12 ounces == 5280 grains and card has a draft of 90 and is de- livering a 56 grain sliver we have, 5280 -f- 90 1 ===== 58.66 grain sliver, theoretical. But as it only weighs 56 grains the loss would be 58.66 — 56 = 2.66 grains ; then 2.66 X 100 = 4.5+% waste. 58.66 Example : If 12 ounce laps are used and a 54 grain sliver is to be produced, what draft gear must be used? Allow 4% waste. A 54 grain sliver would be (1.00 — .04) 96% of total weight that should be delivered if there were no waste. Then, 54 -f- .96 = 56.25 theoretical weight of sliver. A 12 ounce lap = 5280 grains; then 5280 -f- 56.25 = 93.86 draft required for 54 sliver ; and, constant 1534.6 = 16-}- draft gear. draft 93.86 185 Example : If a 52 grain sliver is desired and the card has a draft of 95, what must the lap weigh in ounces. Allow 4% waste. 52 -+- .96 (1.00 — .04) = 54.16 grain sliver if no waste is made ; then 54.16 X 95 draft = 5145 grain lap, and 5145 -i- 437.5 grains in 1 ounce = 11.76 or 11% ounces lap. What draft gear will give 95 draft? constant 1534.6 = 16 gear. draft 95 DRAFT OF DRAWING FRAME BACK ROLL MtGEJ 2" CAL ROLL As most all drawing is now done with metallic rolls it is necessary for us to remember the follow- ing in figuring the draft : Rolls With 16 Pitch V/g" diameter figured as 1.66" 1%" diameter figured as 1.83" 1%" diameter figured as 2.00" 1%" diameter figured as 2.13" 186 Rolls With 24 Pitch 1%" diameter figured as 1.50" 1%" diameter figured as 1.66" 1%" diameter figured as 1.83" V/2" diameter figured as 2.00" Using the above sketch the draft is as follows : As 1%" back roll = 2" for metallic rolls 8i 2 % X 42 X m X W = 7.77 draft. % X %i X 45 X00 ^ change gear If the change gear, 30, is left out of the calcula- tion we have the constant,, 7.77 X 30 = 233.10. Then if change gear is divided into constant we find the draft. Thus, and, constant 233.10 change gear 30 constant 233.10 7.77 draft; = change gear, 30 draft 7.77 or, draft, 7.77, X change gear, 30, = constant, 233.10. Example : If a 54 grain sliver is fed to. drawing with doublings of 6, what draft gear would be re- quired to make a 52 grain sliver? 54 X 6 = 324 grains per yard fed in behind ; and, 324 -^ 52 = 6.23 draft required. 187 constant 233.10 Then = 37+ teeth change gear. draft 6.23 Example : A 54 grain sliver is required to be de- livered with a draft gear of 36. What must be the weight per yard of card sliver if the doubling is 6? constant 233.10 6.47 draft. draft gear 36 Then 54 X 6.47 = 349.38 weight of 1 yard fed with 6 doublings. Therefore, 349.38 -f- 6 = 58.23 grain card sliver. Example : If a 52 grain sliver is used behind drawing with a 38 teeth change gear and doubling 6, what will be the weight in grains of sliver de- livered ? constant 233.10 6.1 draft. draft gear 38 52 X 6.1 = 317.2 weight of 1 yard fed to machine; and 317.2 ~=- 6 = 52.8 grain sliver. DRAFT OF A SLUBBER BACK ROLL I" DIA. 30- CMAN6E FRONT ROLL iVs'DIA- 1.1875 FIG. 11. 188 From the above sketch the draft of a slubber is found as follows : 10 1.1875 X 56 X 10 = 6.717 draft. 1 X 00 X 33 3 Leaving out the change gear, 30, Ave have the constant number, 30 X 6.717 = 201.51, which can be used for finding the draft if gear is given, or for finding the gear if draft is given. Thus,. constant = draft ; change gear constant and = change gear. draft Example : If a 52 grain sliver is used behind slubber with a 42 teeth draft gear, what will be H. R. No. in front? constant 201.51 = 4.79 draft. change gear 42 Then if a 52 grain sliver is used behind, the weight of 1 yard in front of slubber would be, 52 -r- 4.79 = 10.85 grains per yard. Then 12 yards would weigh 12 X 10.85 = 130.2 grains ; then from yarn calcu- lations 100 = .76 H. E. No. 130.2 DRAFT OF AN INTERMEDIATE FRAME BACK ROLL I" DIA. 30— f CHANGE FRONT ROLL l^'DIA* 1. 1875 FIG. 11. From the above sketch the draft of the interme- diate is figured as follows : 1% 6 ". = 1.1875" = 6.71 draft. .3958 20 ;j^7ox56x;w> 1 x w x w 3.3 $ change gear If we leave off the change gear, 30, we will have 6.71 X 30 = 201.3 constant. Then we have, and, constant 201.3 change gear 30 constant 201.3 = 6.71 draft ; draft 6.71 = 30 change gear. Example: If a .62 H. R, comes from slubber and a 36 teeth change gear is used on intermediate, what will be the H. R. No. in front of intermediate? 190 constant 201.3 = 5.6 draft. change gear 36 As the roving is doubled behind we have from fty X 30 yarn No. warp in 840 yds. 58.6 = 95.6 pounds of sized warp in .92 (1.00 — .08) 840 yards. If cloth is to weigh 4.75 yards per pound 840 yards will weigh 840 -^ 4.75 = 176.84 pounds. Then, 176.84 — 95.6 = 81.24 pounds of filling in 840 yards, and $40 X 2322 yards filling : = 28.5 filling No. jB£0 X 81.24 246 Pra&ice and Cost Finding By practice is meant the method of finding the total cost of materials when rated at so much a pound, yard, etc. Cost finding is the reverse of prac- tice in that it is the method of determining the cost per pound, yards, etc., after including all materials going into the manufacture of that article. Examples : 1. A mill buys 400 bales of cotton weighing on an average of 510 pounds each. The price paid is 25 cents per pound. What is the total cost? 400 X 510 = 204000 pounds of cotton ; then 204000 X $0.25 = $51000 cost. If the bagging and tires weigh 24 pounds per oale, what is the amount paid ? Note, these are paid for at the same rate as the cotton. 24 X 400 = 9600 pounds of bagging and tires ; and 9600 X $0.25 = $2400 paid for bagging and tires which is included in $51000 above. If, when sold, they bring $2400, the actual cost of the cotton would be $51000 — $2400 = $48600.00. But if they bring only 12 cents per pound,, we would get only 9600 X -12 = $1152.00' for them. Then they would cost the mill $2400 — $1152 = $1248.00. Then the actual cost of the resulting weight of cotton, 204000 — 9600 = 194400 pounds, would be $51000.00 + $1248.00 = $52248.00. 247 To find the cost per pound in cents, multiply the total cost by 100 and then divide by the pounds. Thus, to find the actual cost per pound of the 194400 pounds of cotton which cost a total of $52248.00, as we have seen, we have, 52248.00 X 100 = 26.87 cents per pound. 194400 2. A mill runs in a week, 500,000 pounds of raw cotton and makes 12% waste. How much cloth should be made? 500000 X .12 = 60000 pounds waste. Then it should make 500000 — 60000 = 440000 pounds of cloth. If the raw cotton cost 26 cents per pound and the waste is sold at an average of 12 cents, what is the cost per pound of cotton going into cloth? 60000 pounds of waste which was paid for at 26 cents per pound would cost 60000 X -26 = $15600. If sold at 12 cents we receive back 60000 X -12 = $7200. Then we must add the difference, $15600 — $7200 = $8400, to the cost of the cotton put into cloth. Thus, 440000 X -26 = $114400.00 ; and 114400 + 8400 = $122800.00 cost of 440000 pounds of cotton. Then, 122800 X 100 = 27.9 cents per pound. 440000 3. What would be the cost of 216 looms at $185 each? 216 X $185 = $39960. 248 4. At $40 per 1000 (usually indicated by letter, M), what would be the cost of 50000 wire heddles? If price is by the thousand, cut off three figures to the right of the number indicating the amount and then multiply. Thus, 50.000 X 40 = $2000. 5. At $15 per M (1000) what would be the cost of 11650 travelers? 11.650 X $15 = $174.75. At $12 per hundred what would be the cost of 2652 picker sticks? When the price is by the hundred cut off two fig- ures by decimal point to. the right of number show- ing the amount. Then, 26.52 X 12 = $318.24. 6. If cloth is sold at 12 cents per yard and it Weighs 4.48 yards per pound,, what is the selling price per pound ? 4.48 X -12 = 53.76 cents per pound. 7. A bale of cloth weighing 520' pounds contains cloth weighing 5.25 yards per pound and is worth 15 cents per yard. What is the bale worth? 520 X 5.25 = 2730 yards in the bale; and 2730 X -12 = $327.60. 8. The price paid for 1 horse power per year is around $20. If a mill has 840 looms requiring 1 H. P. for every 3 looms, what should be the weekly cost for power? 840 -i- 3 = 280 horse power required. This would be 280 X 20 = $5600 per year, or $5600 = $107.69 per week for 840- 52 weeks in a year looms. 249 If these looms produce 42000 pounds per week, what is the power cost per pound? 107.69 X 100 = .256 cents. 42000 The above, .256 cents, is a fraction of a cent or about % of a cent. In showing cost per pound this is the best method to use. It is sometimes shown with the dollar mark. Thus, in the above, if it is to be shown expressed in dollars we would not have used the number 100, to multiply the number $107.69. When we use 100, as there are 100 cents in a dollar, we reduce the dollars to cents and the re- sult is in cents or fractions thereof. 9. A card room produces 46824 pounds of rov ing per week. The weekly payroll is $513.85. What is the cost per pound of labor? 513.85 X 100 = 1.09 cents per pound. 46824 10. A man runs two roving frames and makes 162 hanks in a week at 8 cents per hank. What is his pay? 162 X .08 = $12.96. 11. A weaver running 16 looms makes 62 cuts per week at 21 cents per cut. What is his pay? 62 X -21 = $13.02. 12. A cloth is to be made which will give 3% cuts per week on a loom, allowing an average of 16 looms per weaver. How much should be paid per cut so that weavers can make $16.75 per week? 16 looms X 3% cuts = 56 -cuts per week. 250 16.75 X 100 Then = 29.9 cents or 30 cents per 56 cut to be paid. 13. A cut of cloth has 8.46 pounds of warp valued at 42 cents per pound, 6.57 pounds of filling at 39 cents per pound. What is the cost of yarn in 1 cut? 8.46 X -42 =. $3.5532 cost of warp. 6.57 X -39 = $2.5623 cost of filling. $6.1155 total cost of warp and filling. 14. If a cloth weighs 4.30 yards per pound and has 65% warp in it valued at 50 cents per pound, what is the cost per yard if filling is worth 42 cents and cloth has 60 yards per cut? 60 -f- 4.30 = 13.953 pounds, weight of 1 cut. 13.953 X -65 = 9.069 pounds of warp in 1 cut. 9.069 X -50 = $4.5345 cost of warp in 1 cut. 13.953 — 9.069 = 4.884 pounds of filling in 1 cut. 4.884 X -42 = $1.8512 cost of filling in 1 cut. 4.5345 + 1.8512 == $6.3857 cost of 1 cut of 60 yards. 6.3857 X 100 Then = 10.64 cents per yard. 60 15. A speeder tender runs two frames of 160 spindles each and produces 98.8 hanks of No. 7 H. R. at 8 cents a hank. What is the cost per pound for labor ? 1 spindle on each frame will have made 98.8 = 49.4 hanks. 2 Then 49.4 -f- 7 H. R. = 7.057 pounds per spindle. Then 160 X 2 = 320 total spindles. 251 And 320 X 7.057 = 2258.24 pounds production on 320 spindles. 98.8 X -08 ■— $7,904 paid for producing 2258.24 pounds. Then 7.904 X 100 = .35 cents per pound. 2258.24 This can also be found as follows : 98.8 X -08 X 100 X 2 X 7 = .35 cents per pound. 320 X 98.8 Then to find the cost per pound for any price paid per hank regardless of the amount of production we have, price paid in cents X 100 X H.R. No. = cost per number spindles per frame pound. ill h flfflraffl] Sal Wm SB Pil HHH I B I iiSii : H ill h »m ■Hi mm mm- fBUm