Class T S 1 K ^ \ ' Rnok Y-^ Copyright }1^ COPYRIGHT DEPOSIT. A PRACTICAL TREATISE YARN AND CLOTH CALCULATIONS COTTON FABRICS THOMAS YATES Instructor in W^eaving, IVarp Preparation and Calculations New Bedford Textile School. NEW BEDFORD, MASS. 1904. -^-^ LIBRARY of 0ON6RESS TVmOopiesRwetved SEP 7 1904 «tAS8 CL xxe. Na OePYB Entered according to Act of Congress, in the year 1904, by Thomas Yates, In the oflSce of the Librarian of Congress, at Washington, D. C. Preface to Second Edition. The demand for the first edition of ''A Practml Treatise on Yarn and Cloth Calculations for Cotton Fabrics'' has been such that the author has been encouraged to present a second edition. This second edition has been enlarged by the addition oi new calculations and explanations of the rules given for their solution. That this volume in its new form may meet with the same approval as the first edition is the hope of the author. TEOMAS YATES. New Bedford^ 1904. YARN CALCULATIONS. In dealing with Yarn Calculations it is necessary to have two tal>les or standards, one of leno^th and one of weight. The standards of length for different materials vary, l)ut the calculations are all based on a given length that is contained in a given weight. Tables of lengths and weights will be given, with rules for calculations under the specified materials. The table of lengths for Cotton Yarn is as follows : 11. yds. = 1 thread or cir. of wrap reel, 120 " = 80 threads or 1 skein or lea, 840 " — 560 " =7 skeins " " =: 1 hank The term "hank" in the table does not mean that the yarn is in hank form, as it may be in any of the usual forms, such as cops, Ijobbins, warps, skeins, or hanks, etc., but it is used to indicate a definite length of yarn, i. e. 840 yards. The table of weights for Cotton Yarn is as follows : 437i grains ^ 1 oz. avoirdupois, 7000 " = 16 oz. = 1 lb. The above table is the standard for weight used for for all materials except raw silk. The table of lengths given is the standard for cotton in the United States and England. Cotton Yarns are indicated by numbers or counts, and the number of hanks of 840 yards that weigh 1 lb. is the count of the varn. If it should be found that 2520 yards of a coiton yarn weiglied 1 lb. there would be 2520 ^ 840 = 3 hanks, and the counts of the yarn would be 3s. Therefore the following rule may be used : To find the counts of a Cotton Yarn when the length and weight are known : Rule. Divide the length by the standard, 840, re- sult equals hanks : divide hanks by weight in lbs., equals counts. Examples. If 16800 yards of Cotton Yarn weigh 1 lb., find the counts. 16800 ^ 840 = 20 hanks -^ 1 lb. = 20s counts. If 33600 yards of Cotton Yarn weigh 5 lbs., what are the counts ? 33600 -^ 840 = 40 hanks -i- 5 lbs. = 8s counts. If 80 bundles of yarn each contain 2100 yds. and weigh 20 lbs.- in all, find the cotton counts. 2100 X 80 = 168000 yds. 168000 -^ 840 = 200 hanks. 200 -^ 20 = 10s counts. If a warper beam contains 420 ends, 1'8000 yards long, and weighs 600 lbs. net, what are the cotton counts ? 18000 X 420 = 7560000 yds. 7560000 -f- 840 = 9000 hanks. 9000 ^ 600 z= 15s counts. If a warper beam contains 8 wraps of .500 ends and weighs 700 lbs. net, what are the cotton counts? (A standard wrap is equal to 3000 yards). 500 X 3000 X 8 840 X 700 20.40s counts. If a set of 8 section beams of 7 wraps and 450 ends each, weigh 750 lbs. in all, find the cotton counts. 450 X 8 X 3000 X 7 ^^^ := 120s counts. 840 X 750 If a loom beam of 2400 ends and 20 cuts of 50 yds. each, with 10% size, weighs 45 lbs., what are the cot- ton counts of the yarn ? 45_10% = 40.50 lbs. net. 20 X 50 X 2400 840 X 40.50 70.54s counts. The preceding examples show that yarn in any form may be reduced to yards, and by dividing by 840 be changed to hanks ; by dividing the hanks by the weight the counts of the yarn is obtained. REELING YARN. The counts of yarn on bobbins, cops, or spools, etc., may be found by reeling 120 yards on a wrap reel and dividing its weight in grains into 1000. Example. If 120 yards or 1 skein of Cotton Yarn weighs 20 grains, what are the counts ? 1000 ~ 20 = 50s counts. A full hank is too much to reel, so 1 skein, or one- seventh of a hank is taken, and its weight divided into 1000, which is one-seventh of a lb. of 7000 grains. If 2 skeins, or 240 yards, are reeled, 2000 must be used for a dividend, as it is the same proportion of 7000 as 240 yards is of a hank. If 4 skeins or 480 yards are weighed, then 4000 is tlie dividend, etc. It is the practice in mills to test the yarn directly from the mules or spinning frames at least once a day, to see if the yarn is of the correct counts. These tests 8 are made by taking 4 cops or bobbins of each count of yarn being spun and winding 120 yards, or 1 skein, from each on a wrap reel, weighing each skein sepa- rately, and dividing its weight in grains into 1000. The result is the counts of the yarn in each skein, and shows the variation in the counts of the yarn. These separate counts may then be added together and the result divided by 4 to find the average counts, or the 4 skeins may be weighed together and their weight in grains divided into 4000, the result l)eing the counts, as follows : 20 = 50s counts. 1000 1 skein 20 grains, 1000 1 u 20.5 1 u 19.5 1 u 19 20.5= 48.78 1000 ~ 19,5 =r 51.28 1000 -^ 19 = 52.63 202.69 202.69 -f- 4= 50.67 average counts. Examples. If 4 skeins are reeled and together they weigh 200 grains, what are the counts ? 4000 -f- 200 = 20s counts. If 1 skein of cotton yarn weighs 20 grains, find the counts. 1000 -^ 20 = 50s counts. If 2 skeins of cotton yarn weigh 35 grains, what are the counts ? 2000 ^ 35 = 57.14 counts. If 4 skeins are wrapped and together weigh 60 grains, find the counts. 4000 ^ 60 = 66.66 counts. To find the Cotton counts from irregular or short lengths of yarn: Rule. Multlyly the numher of yards weighed by 8i and diride the result by their weight in grains. The constant 8^ is the weight in grains of 1 yard of number 1 cotton yarn, and is found l)y dividing the standard for weight, 7000 grains, by the standard for length, 840 yards. 7000 ^ 840 = 8i grains, weight of 1 yard of num- ber 1 cotton yarn. Examples. If 96 yards of cotton yarn weigh 20 grains, what are the counts ? As 1 yard of number 1 yarn weighs 8i grains, 96 yards will weigh 96 X 81 = 800 grains, and as the 96 yards in the example weigh 20 grains, it will be 800 -f- 20 = 40 times lighter than a number 1 or 40s yarn. If 66 yards of cotton weigh 12 grains, what are the counts ? 66 X 8^ = 550 grains. 550 -^ 12 = 45.83s counts. If 135 yards of cotton yarn weigh 20 grains, what are the counts ? ^ -^ = 56.25s counts. 20 If 240 yards of cotton yarn weigh 50 grains, what are the counts ? 240 yards = 2 skeins. 2000 -^ 50 = 40s counts, or 240 X 81 ^ ^ 40s counts. 50 10 Another rule that will be found useful when only- short lengths of thread are available is as follows: Rule. TJie number of inches of cotton yarn that weigh 1 grain, multiplied by .2314, equals the counts. Example. If 63 threads, 3 inches long, weigh 1 grain, what are the cotton counts ? 63 X 3 = 189 inches of yarn that weigh 1 grain. 189 X .2314 =r 43.73s counts. The constant .2314 is the weight ii> grains of 1 inch of number 1 cotton yarn ; hence, any number of inches that weigh one grain will be .2314 times finer than number 1 yarn. 840 X 36 = 30240 inches in one hank of cotton yarn. 7000 grains -^ 30240 = .2314 grains, weight of 1 inch of number 1 yarn. This same problem may be worked out by reducing the inches to yards and using the constant 8i as follows : 63 X 3 = 189 inches. ' 189 -f- 36 = 5.25 yards. 5.25 X 81 ^ m 43.73s counts. 1 To find the length in yards when the counts and weight are known : Rule. Multiply the standard length, 840 yards, by the counts, result equals yards in 1 /6., multiply by weight in lbs., equals length. Examples. Find the length of yarn in 10 lbs. of 20s cotton yarn. 840 X 20 = 16800 yards in 1 lb. of 20s yarn. 16800 X 10 = 168000 yards in 10 lbs. 11 What is the length contained in 25 lbs. of 40s cot- ton yarn ? 840 X 40 X 25 = 840000 yards. If a bobbin or spool filled with yarn weighs 1 lb. 4 oz., tlie empty bobbin or spool 6 oz,, and the counts of the yarn 36s, what length of yarn is contained on the bobbin or spool ? 1 lb. 4 oz. rrr 20 OZ. 20 — 6 = 14 oz., weight of yarn. 14 X 437.5 = 6125 grains weight of yarn. 6125 H- 7000 == .875 lbs. 840 X 36 X .875 = 26460 yds. on bobbin or spool. If a warper beam contains 420 ends of 15s yarn, and weighs 600 lbs. net, what is the length of yarn on the beam ? 840 X 15 X 600 = 7560000 yards of continuous thread. 7560000 ^ 420 ends ^ 18000 yards. If 32 cops of 55s cotton yarn w^eigh 1 lb. and each cop contains an equal amount of yarn, what is the length contained on each cop ? 840 X 55 = 46200 yards in 1 lb. of 55s yarn. 46200 -^ 32 cops = 1443.75 yards on each cop. How many yards in a i lb. of 60s cotton yarn ? 840 X 60 = 50400 yards in 1 lb. of 60s. 50400 ^ 2 = 25200 yards iu i lb. To find short lengths when counts and weight in grains are known : Rule. MuUijjIy the weight in grains by the counts and divide the result bij 8^. For convenience, the decimal 8.33 may be used instead of the common fraction 8^ ; although 8.33 is 12 not the exact figure it is near enough for practical purposes. Example. How many yards of 40s cotton yarn will weigh 36 grains ? 36 X 40 _ ^ = 172.8 yards, or ^ili-°= 172.86 yards. 8.33 ^ To find weight when length and counts are known: Rule. Divide the length of yarn in yards by the sta7idard 840, result equals hanks; divide hanks by counts, equals weight in lbs. Examples. Find the weight of 168000 yards of 20s cotton yarn. 168000 ^ 840 = 200 hanks. 200 -^ 20s r= 10 lbs. What is the weight of 18480 yards of 44s cotton yarn? 18480 ^ 840= 22 hanks. 22 -^ 44 = .5 or i lb. If a warper beam contains 500 ends of 32s cotton yarn and is 24000 yards long, what is the weight of the yarn ? 24000 X 500 = 12000000 yards. 12000000 -^ 840 = 14285.71 hanks. 14285.71 -f- 32 = 446.42 lbs. If you were running 410 ends of 36s yarn on your warpers, what should 6 wraps weigh ? 410 X 3000 X 6 840 X 36 = 244.04 lbs. 13 Find the weight of 27720 yards of 44s cotton yarn, = . <5 or 4 II). 840 X 44 To find the weight of short lengths of Cotton Yarn: Rule. Multiply the length by 8^, result equals the weight in grains of that length of namher 1 Cotton Yarn; divide by any other counts, equals the weight in grains of that counts. Examples. Find the weight of 144 yards of 50s Cotton Yarn. 144 X 81 = 1200 grains, weight of 144 yards of 50s Cotton Yarn, and as 50s is 50 times lighter than number 1, 1200 -^ 50 = 24 grains. Find the weight of 75 yards of 60s Cotton Yarn. 75 X 81 60 = 10.41 srrains. What is the weight of 96 yards of a 24s Cotton Yarn ? ^^-^^ = 33.33 grains. SPUN SILK. Spun silk in single yarn is numbered by the same method as Cotton Yarn, the same table of lengths and weights being used. Thus the rules given for calcu- lating single Cotton Yarn will apply to spun silk. Ply yarn, however, is indicated in a different manner. Cotton ply yarn is indicated by placing the number of the threads twisted before the numbers of the single 14 yarn ; a two ply 40s would be composed of two single ends of 40s and would be indicated thus, 2 /40s, and would be equal in weight to a single 20s. In spun silk, the method is to place the actual size or counts of the ply thread first, and the number of threads twisted after it. Thus a two ply thread of spun silk equal to. a 20s would be composed of two threads of 40s and would be indicated 20s/2 ; a 30s/3 would be composed of three threads of 90s. Thus the count of a ply thread in spun silk, multi- plied by the number of threads twisted, equals the counts of the single threads. Examples. If 302400 yards of spun silk weigh 9 lbs., what are the counts ? 302400 — z=i 40s counts. 840x9 If 115 yards of spun silk weigh 20 grains, find the counts. 3 _ 47 90s counts. 20 Find the length of 25 lbs. of 40s spun silk. 840 X 40 X 25 = 840000 yards. What counts of spun silk are equal to a 60s cotton ? The same counts, 60s. What counts of single yarn would be used to make a 20s/^3 spun silk thread ? 20 X 3 = 60s counts. How many yards of ply yarn are contained in 10 lbs. of 20s/2 spun silk ? 840 X 20 X 10 = 168000 yards. 15 How many yards of single yarn would be required to make 10 lbs. of 20s/2 spun silk? 20s/2 is made by doubling single 40s. 840 X 40 X 10 = 336000 yards of single yarn required. How many yards of ply yarn are contained in 20 lbs. of 30s/3 spun silk ? 840 X 30 X 20 = 504000 yards. How many yards of single yarn would ])e contained in 20 lbs. of 30s/3 spun silk? 30s/3 would be made of 90s single. 840 X 90 X 20 = 1512000 yards. RAW SILK. Silk is the product of the silk worm, which, at a certain stage of its existence, discharges the silk in a semi-fluid condition from two glands near its head. These filaments unite as they are discharged and form one thread which hardens immediately on exposure to the air. The worm winds the thread around itself, so that when it has finished it is entirely enclosed in what is called a cocoon, which contains on a average about 400 yards of thread. In about three weeks, after it has finished the cocoon, the worm changes to a moth which forces its way out of the cocoon, cutting or breaTcing some of the threads which spoils the cocoon. To prevent this, every cocoon, not intended for breeding purposes, is placed in a steam heater to stifle the chrysalis. The silk may then be reeled at any future time. Four or more of these cocoons (according to the size of thread desired) are placed in a basin of warm 16 water to soften. Then the threads are slightly twisted around themselves, run on a reel and made into skeins. The skeins weigh from one to several ounces and are packed into bundles called books, weighing from five to ten pounds. The silk is then made up into bales weighing from one hundred to one hundred and sixty pounds weight and is called thrown silk. When preparing thrown silk for weaving purposes, it is sometimes necessary to twist several threads together. The threads to be used as warp are twisted harder than those for filling and will now be known as organzine, while those for filling will be known as tram. As in the case of Cotton Yarn, the warp threads are generally coarser than the filling threads and this is obtained by twisting more threads together for the warp than for the filling. A 4 thread organzine means that 4 threads of thrown silk, (raw silk) have been twisted sufficiently hard for warp, while if 3 threads were twisted with a less number of turns per inch for filling it would be called tram. Raw silk is numbered on an entirely diff'erent basis from either spun silk, cotton, worsted, or woolen yarn. In the United States and in England, the length of the hank used is 1000 yards, and the number of drams that such a hank weighs is the counts of the silk. Thus, in raw silk, the coarser the silk the higher the counts. For calculating raw silk in the United States and English system the following table of weights is used : 1 dram = 27.34 grains. 16 " =1 oz. = 437.5 256 " = 16 '^ z= 7000 " = 1 lb. 17 The table of length is : 1000 yards equals 1 liaiik. Thus, if 1000 yards weigh 1 dram, it would be a 1 dram silk, and as there are 256 drams in 1 lb., there will be 1000 X 256 := 256000 yards in 1 lb. of number 1 dram silk. To find the dram silk counts when length and weight in lbs. are known : Rule. Multiply 256 ^y/ 1000, result, equals yards of 1 dram silk in 1 lb.; multiply by iveight in lbs., equah total length; divide by yards iveighed, equals counts. Examples. If 64000 3^ards of raw silk weigh 2 lbs., find the dram silk counts. In 2 lbs. there are 512 drams and 512000 yards of 1 dram silk, and, as the 64000 yards in the example weigh 2 lbs. or 512 drams, they are: 512000 ^ 64000 = 8 times heavier or 8 dram silk. If. 128000 yards of raw silk weigh 3 lbs., what are the dram silk counts ? 256 X 1000 X 3 ^ ^ —- 6 dram silk. 128000 If 512000 yards of raw silk weigh 10 lbs., what are the dram silk counts ? 256 X 1000 X 10 _ = 5 dram silk. 512000 To find length when counts and weight are known: Rule. Midtiply 25Q by 1000, result equals yards of 1 dra?n silk in 1 lb.; midtiply by weight in lbs., equals total length ; divide by the counts, equals length of that count of silk. 18 Examples. Find the length in 5 lbs. of 4 dram silk. 256 X 1000 X 5 = 1280000 yards in 5 lbs. of 1 dram silk. A 4 dram silk is 4 times heavier than a 1 dram silk, consequently there will be 4 times less length in the same weight of a 1 dram silk. 1280000 ^ 4 = 320000 yards. Find the length in 10 lbs. of 8 dram silk. 256 X 1000 X 10 = 320000 yards. 8 ^ Find the length in 5 J lbs. of 6 dram silk. 256 X 1000 X ^5.5 ^_^^ _ = 234666.66 yards. 6 ^ To find the weight when the length and counts are known : Rule. Multiply 25Q bij 1000, result equals yards of 1 dram silk in 1 lb.; divide by the counts, equals yards in 1 lb. of that counts; divide total yards by the yards jier lb., equals weight in lbs. Examples. Find the weight of 160000 yards of 8 dram silk. 256000 yards of 1 dram silk weigh 1 lb., and as an 8 dram silk is 8 times heavier than a number 1, there will be 256000 -^ 8 = 32000 yards of 8 dram silk in 1 lb., and 32000 is contained in 160000 5 times, which equals 5 lbs. Find the weight of 234667 yards of 6 dram silk. 256 X 1000 ^ 42666.66 yards of 6 dram silk in 1 lb. 6 ^ 234667 -^ 42666.66 = 5.5 lbs. 19 The same result may be obtained by another method as follows : Rule. Multiyly the yards hy the counts, result equals the length in 1 dram co2mts, divide by 256 X 1000. equals the weight in lbs. The last example by this rule is as follows : 234667 X 6 256 X 1000 5.5 lbs. To find counts when length and weight in ounces are known : Rule. Multiply freight in ounces by 16, result equals iveight in drams ; divide length by drams, equals yards jmr dram; divide 1000 by yards per dram, equals counts. Example. If 2400 yards of raw silk weigh 3 ounces, what are the dram counts ? 16 X 3 = 48 drams weight. 2400 ^ 48 = 50 yards per dram. 1000 ^ 50 = 20 dram silk. To find length when counts and weight in ounces are known : Rule. Multiply the iveight in ounces by 16, result equals weight in drams; divide 1000 by counts, equals yards jier dram ; multiply iveight in drams by yards per dram, equals length. Example. Find the length of raw silk in 7 ounces of 10 dram silk. 16 X 7 = 112 drams weight. 1000 -f- 10 =100 yards per dram. 112 X 100= 11200 yards. 20 To find weight in ounces when length and counts are known : Rule. Divide loot) bij the counts, result eqidcils yards per dram ; divide total yards by yards per dram, ecjuals weight in drams ; divide iveight in drams by IQ, e(pials iveight in ounces. Example. What is the weidit in ounces of 2000 yards of 12 dram silk? 1000 2000 24 12 ^ 83.33 yards per dram. 83.33 = 24 drams weight. 16 := 1.5 ounces weidit. ■■o' To find the dram silk counts when length and weight in grains are known : Rule. Divide weight in grains by 27.34, result equals weight in drams ; divide length by iveight in drams, equals yards per dram ; divide 1000 by yards per dram, equals counts. Example. If 1280 yards of raw silk weigh 70 arrains, what are the dram silk counts ? 70 -^ 27.34 = 2.56 drams weight. 1280 ^ 2.5.6 = 500 yards per dram 1000 ^ 500 = 2 dram silk. To find length when counts and weight in grains are known : Rule. Divide weight in grains by 27.34, residt equals weight in drams; divide 1000 bij counts, equals yards per dram; multiply iveight in drams by yards per dram, equals length. Example. Find the length of raw silk in 100 grains weight of 4 dram silk. 100 -^ 27.34 = 3.65 drams weight. 1000 ^ 4 =: 250 yards per dram. 3.65 X 250 = 912.50 yards. 21 To find weight in grains when length and counts are known: Rule. D'tride lOfKJ hy the coaTtn. -^ti'^c t^ju/j'S yardi per dram : diride UAal yardt hy yard* per dram. eqvaU i«aVA/ in dramjs ; multiply 27-M by wfight in dram*. (^piaU tceighi in grains. Example. Find weight in grains of 1500 yards of 12 dram silk- 1000 -^12 — >t?j.?jZ yard* per dram. 1500 ^ 83.33 = 1^ drams weiglit. 27-34 X 1« = 492.12 grains weight In TT2inQe and other coantrie* of Europe, the standard length of the hank is 476 metres, or 520 yards- and the siandard weight is 1 denier. The nam1>er of deniers that 1 hank of 520 yards weighs is the counts of the silk. The table of weights is as follows : 533i deniers == 1 oz. aroirdupois- 8533' - r= 16 ^ =1 lb. Raw sflk varies in weight to such an extent that in numbering it in the denier system it is the practice to allow a variation of two numbers, and instead of specifying the counts Ijy a single number, two are given- In the case of a 15 denier silk, the counts would be marked 14/16. meaning 14 to 16. allowing a variation of one number each way and the silk would be known as 14/16 denier silk: other counts would be known as 18/20. 20/22 etc. If a calculation were made on a 14/16 denier silk the number between the two specified, or 15. would be used: on an 18/20. 19 would ]>e used etc. 22 To find the denier counts of raw silk, when length and weight in lbs. are known : Rule. Multiplij 8533 hij 520, result equals yards of 1 denier silk in 1 lb. ; multiply by weight in lbs., equals total length; divide by yards weighed, equals counts. Examples. If 400000 yards of raw silk weigh 2 lbs., what are the denier counts ? 8533 X 520 X 2 = 8874320 yards in 2 lbs. of 1 denier silk, and, as the 400000 yards in the example weigh 2 lbs:, they are 8874320^400000 = 22.18 times heavier or 22.18 denier counts. If 320000 yards of raw silk weigh li lbs., what are the denier counts ? 8533 X 520 X 1.5 — = 20.78 denier counts. 320000 To find length when weight in lbs. and counts are known : Rule. Multiply 8533 by 520, result eq^ials yards oj 1 denier silk in 1 lb.; multij^ly by weight in lbs., equals total length in that weight of I denier silk; divide by the counts, equals length. Examples. How many yards are contained in 4 lbs. of 13/15 denier silk ? 8533 X 520 X 4= 17748640 yards of 1 denier silk in 4 lbs. weight, and, as a 14 denier silk is 14 times heavier than a 1 denier, there will be 17748640 ^ 14 = 1267760 yards. 23 How many yards are contained in 7 lbs. of 20 / 22 denier silk ? 8533 X 520 X 7 1^ = 1479053 yards. 21 ^ Find the length in 10 lbs. of 14/16 denier silk. 8533 X 520 X 10 = 2070674.66 yards. 15 ^ To find weight when length and counts are known : Rule. Multiply 8533 by 520, resiik equals yards of 1 denier silk in 1 lb.; divide by the counts, equals yards in 1 lb. of that counts; divide total yards by yards per lb., equals weight in lbs. Example. Find the weight of 450000 yards of 15/17 denier silk. 8533X 520 ~ =277322 yards of 16 denier silk in 1 lb. 16 450000 -^ 277322 = 1.62 lbs. weight of the 450000 yards. The same result may be obtained by another method as follows : Rule. Multiply the yards by the counts, result equals the length in 1 denier counts, divide by 8533 X 520, equals the weight in lbs. The last example by this rule is as follows. 450000 X 16 = 1.62 lbs. 8533 X 520 Practically all the raw silk used in the United States is imported and is numbered in the denier system. 24 For^'convenience in calculating according to English and United States methods, the denier counts may l)e changed to dram silk counts as follows : To find the equivalent counts of denier silk in dram silk counts: Rule. Multiply 8533 by 520, result equals yards of 1 denier silk in 1 lb. ; divide by the denier counts, equals yards in 1 lb. of that counts; divide by 25^000, .the yards of 1 dram silk in 1 lb.; and ansiver is dram silk coimts. Example. Find the equivalent of a 11/13 denier silk in dram silk counts. 8533X520 — z= 369763 yards of 12 denier silk in 1 lb. 12 369763 -^ 256000 = 1.44 dram silk. Or in one operation as follows 8533 X 520 =r: 1.44 dram silk. 12 X 256000 Thus a 11/13 denier and a 1.44 dram silk are equal. To find the equivalent counts of dram silk in cotton counts : Rule. Multiply 256 by 1000, result equals yards of 1 dram silk in 1 lb.; divide by the dram silk counts, equals the yards iii 1 lb. of that counts ; divide by 840, equals the counts in the cotton system. Example. Find the equivalent counts of a 4 dram silk in cotton counts. 256 X 1000 = 64000 yards of 4 dram silk in 1 lb. 4 ^ 64000 ^ 840 = 76.19 cotton counts, or 256 X 1000 = 76.19 cotton counts. 4 X 840 25 Thus a 4 dram silk and a 70 cotton yarn are equal. To find the equivalent counts of a denier silk in cotton counts : Rule. Multiply 8533 hy 520, result e(pials yards of 1 denier silk in 1 lb.; divide by the dewier silk counts, equals the yards in 1 lb. of that counts; divide by 840, equals the counts in the cotton system. Example. Find the equivalent of a 21/23 denier silk in cotton counts. ^ = 201689 yards of 21/23 denier silk 22 in 1 lb. 201689 4- 840 = 240.1 cotton counts, or 8533 X 520 22 X 840 240.1 cotton counts. Thus a 21/23 denier silk and ^ 240 cotton yarn are equal. WORSTED YARN. The table of lengths for worsted yarn is as follows : 1 yard = 1 thread or cir. of worsted reel. 80 '' = 80 " =1 lea or knot. 560 '^ = 560 " = 7 " " " == 1 hank. The table of weights is the same as that for Cotton Yarn. The number of hanks of 560 yards contained in 1 lb. indicates the counts. If 560 yards instead of 840 are used, the rules given for Cotton Yarn will apply equally well for worsted, except those given for dealing with short lengths. 26 Examples. If 252000 yards of worsted yarn weigh 15 lbs., what are the counts ? 252000 -^ 560 = 450 hanks. 450 -f- 15 = 30s counts. How many yards are contained in 25 lbs. of 20s worsted yarn ? 560 X 20 X 25 == 280000 yards. What is the weight of 268800 yards of 40s worsted yarn ? 268800 -f- 560 = 480 hanks. 480 -^ 40s = 12 lbs. If a beam that contains 1568 ends, 300 yards long, weighs 30 lbs. net, what are the counts ? 1568 X 300 = 470400 yards. 470400 ^ 560 = 840 hanks. 840 -^ 30 = 28s counts. To find the worsted counts from irregular or short lengths : Rule. Multiply the number of yards weighed by 12.5 and divide the result by their tveight in grains. The constant 12.5 is found by dividing 7000 grains by 560 yards, length of a worsted hank. 7000 ^ 560 = 12.5 grains, weight of 1 yard of number 1 worsted yarn. Examples. If 56 yards of worsted yarn weigh 35 grains, what are the counts ? 56 X 12.5 = 20s counts. 35 27 As 1 yard of number 1 yarn weighs 12.5 grains, 56 yards will weigh 56 X 12.5 = 700 grains; and, as the 56 yards in the example weigh 35 grains, it will be 700 -i- 35 = 20 times finer than a number 1 or 20s counts. If 63 yards of worsted yarn weigh 40 grains, what are the counts ? 63 X 12.5 z= 19.68s counts. 40 If 110 yards of worsted yarn weigh 50 grains, what are the counts ? 110 X 12.5 =^ 27.5s counts. 50 WOOLEN YARN. There are two standards for length used in the calculations for woolen yarn ; viz. the run and the cut systems. In the cut system, 300 yards is the standard for lengtli, and the number of cuts of 300 yards contained in I lb. is the counts of the yarn. In the run system, 1600 yards is the standard for length, and the number of runs of 1600 yards that weigh 1 lb. is the counts of the yarn. The full table of woolen lengths is as follows : 1 yard =^ 1 thread or cir. of the woolen reel. 80 " = 1 knot. 300 " = 3| ^' = 1 cut. 1600 '^ =20 ^' = 51 '^ = 1 run. 28 To find counts, length, or weight, apply rules given for cotton and worsted yarns (except those for short lengths) and use the standard length in the system desired. WOOLEN CUT YARN. Examples. What are the cut counts of a woolen yarn when 1200 yards weigh 2 lbs.? 1200 ^ 300 r=r 4 cuts. 4 -f- 2 = 2s cut counts. If 45 bundles of woolen yarn each contain 600 yards and weigh 30 lbs. in all, what are the cut counts ? 600 X 45 = 27000 yards. 27000 ^ 300 = 90 cuts. 90 -^ 30 = 3s cut counts. Find the length in 100 lbs. of 3 cut yarn. 300 X 3 X 100 = 90000 yards. If 40 spools of 5 cut yarn weigh 6 lbs. net and each spool contains an equal amount of yarn, find t?he yards on each spool. 300 X 5 X 6 = 9000 total yards. 9000 -^ 40 = 225 yards on each spool. Find the weight of 60000 yards of 4 cut yarn. 60000 -^ 300 = .200 cuts. 200 ^ 4 = 50 lbs. If a loom beam contains 1400 ends of 6 cut yarn and is 500 yards long, find the weight of the yarn. 1400 X 500 =3 700000 total yards. 700000 ^ 300 = 2333.33 cuts. 2333.33 ^ 6 = 388.88 lbs. weight of yarn. 29 To find the cut counts from short lengths : Rule. Multlpbj the jjards weighed by 23^ and divide result by weight in grains. Example. If 45 yards of woolen yarn weigh 240 grains, wliat are the cut counts ? 45 X 23.33 = 4.3 i cut counts. 240 The constant 23^ is the weight in grains of 1 yard of a 1 cut yarn and holds the same relation to the 300 yard cut that 8i does to the 840 yard hank, and 12.5 does to the worsted hank of 560 yards. WOOLEN RUN YARN. Examples. If 100000 yards of woolen yarn weigh 7^ lbs., what are the run counts ? 100000 -^ 1600 =rr 62.5 runs. 62.5 ^ 7.5 ^ 8.33 run yarn. Find the length of 75 lbs. of 2 run woolen yarn. 1600 X 2 X 75 = 240000 yards. What is the weight of 38400 yards of 3 run woolen yarn ? 38400 ^ 1600 = 24 runs. 24 ^ 3 = 8 lbs. To find «ounts from short lengths: Rule. Multiply the yards weighed by 4.375 and divide result by weight in grains. If 60 yards of woolen yarn weigh 20 grains, find the run counts. 60X4.875 ,„,„ =r Id. 12 run counts. 20 30 In the run system, the counts equal the number ot hundred yards in an ounce of the yarn. 1600 yards in 1 run divided by 16 oz. per Ib.=:100 yards. Thus, to find the counts when the weight is known in ounces : Rule. Divide the length by 100 and divide the result by the number of ounces the yarn weighs. Examples. If 13200 yards of woolen yarn weigh 33 ounces, what are the run counts ? 13200 ^ 100 =z 132. 132 -f- 33 ^ 4 run yarn. If 430 yards of woolen yarn weigh 1^ ounces, what are the run counts ? 430 -f- 100 = 4.30. 4.30 -^ 1.5 = 2.86 run yarn. To find the weight in ounces when the length and run counts are known : Rule. Divide the length by 100 and divide the result by the counts of the yarn. Examples. Find the weight in ounces of 6600 yards of a 4 run yarn. 6600 -f- 100 =: 66. • 66 -i- 4 = 16.5 ounces weight. Find the weight in ounces of 320 yards of 3 run yarn. 320 ^ 100 = 3.2. 3.2 ^ 3 = 1.06 ounces weight. 31 To find the length when counts and weight in ounces are known : ' Rule. Multiply the iceight in ounces by 100 and multiply this result by the counts of the yarn. Example. Find the length of 27 ounces of 6 run woolen yarn. 27 X 100 X 6 = 16200 yards. LINEN YARN. The length of the linen hank is 300 yards, and the standard of weight is 1 ll>. The number of hanks that weigh 1 lb. is the count of the yarn. The rules given for calculating woolen yarn in the cut system apply to linen yarn also. Cotton, in process of manufacture, is worked tlirough the pickers, cards, and drawing frames, witliout Iiaving twist put into it, but from the slubber to the spinning frame or mule the amount ot twist put in increases with every process. Twist is put in roving and yarn to give it strength, and is calculated by the number of turns put in one inch. The turns per inch vary according to the quality ot the cotton used and the class of yarns being made, and is determined by multiplying the square root of the counts by a constant. The constant for warp yarn in the U. S. is 4.75. " " " filling " " 3.25. " " '' hosiery" " 2.50. To find the turns of twist per inch : Rule. Multiply the square root of the counts by one of the constants. 32 Examples. Find the turns per inch for a 25s warp yarn, using 4.75 for a constant. V25 = 5, the square root of 25 is 5. 5 X 4.75 = 23.75 turns per inch. Find the turns per inch for a 20s hosiery yarn. \/20~= 4.47. 4.47 X 2.50 r= 11.17 turns per inch req. Warp yarns are tested for strength by breaking one skein on a testing machine that registers the breaking weight in lbs. To find the standard breaking weight of warp yarn : Rule. Divide 1760 by the counts of the yarn; result is the St an do rd hrcakinor weiorht in lbs. b o Example. What is the standard breaking weight of a 40s cotton warp yarn ? 1760 -^ 40s = 44 lbs. Although 1760 is accepted as the standard constant for breaking weight, it will be found in cloth mills, where medium and finer counts are woven, that a higher constant must be used to get satisfactory results in the weave room, while in mills on coarser counts, a lower constant may be used and satisfactory results follow. The following table of constants has been compiled partly from Draper's table of brCakinsi; weights and from the writer's personal experience in the weaving of fine counts. 33 Counts. Constants, 20 to 29 1760 30 - 39 1780 40 '• 49 1825 50 '^ 59 1880 60 -' 69 1930 70 '' 79 1960 80 ^' 89 1980 90 " 99 2040 100 '• 109 2150 110 '' 119 2200 Constant divided by warp counts equals the lbs. that one skein of cotton yarn should register on the breaking machine. EQUIVALENT COUNTS. When dealing with the numbering of diiferent yarns, it frequently becomes necessary to know what the counts of a certain yarn would be if numbered accord- ing to a different standard. This is known as converting tlie counts of yarn in one system into equivalent counts of another system. To find the equivalent counts of a yarn in any other system (except raw silk) : Rule. Mtdtijfhj the given, counts hij its own standard'' length, result equals yards in 1 lb., divide by the standard/ yards in the system desired, equals the counts. Examples. What are the equivalent counts of a 15s cotton in woolen run counts ? 840 X 15 = 12600 yards in 1 lb. of 15s cotton yarn. 34 12600 ^ 1600 yards in 1 run woolen = 7.87 run counts. Find the equivalent counts of a 50s worsted in cotton counts. 560 X 50 840 33.33 cotton counts. Find the equivalent of a 40s/ 2 spun silk in worsted counts. 840 X 40 560 60s worsted. Find the equivalent of an 8 cut woolen in run counts. 300 X 8 ^ ^ =1.5 run counts. 1600 PLY YARNS. Ply yarns are made by twisting two or more threads of the same counts of single yarn, and are then numbered according to the counts of the single yarn with the number of threads twisted put betore it (except in spun silk). Thus, if two threads of 40s were twisted, it would be numbered 2/^40s, and would equal a 20s single ; if three threads of 35s were twisted it would be numbered 3/ 35s, and would equal a 11.66s single etc. Finding the counts of the yarn after they have been twisted is known as finding the re- sultant counts. RESULTANT COUNTS. It sometimes occurs in fancy yarns that threads of unequal counts are twisted together, and in this instance, the resultant count must be found in another way than when the counts of the threads are equal. 35 To find the resultant counts of a ply thread when two or more threads of different counts are twisted together : Rule. Dir'ule the highest counts by \t self and by each of the other counts; add the results, and- dirlde into the highest counts. Examples. Suppose one thread of 60s and one of 20s worsted yarn are twisted together ; what is the resultant counts of the twisted thread ? Assuming there is no variation in the contraction ot either thread during the twisting, the same length of each counts will be required. Suppose that 60 hanks or 1 11). of the 60s were used, then 60 hanks of the 20s would also be used, and when these have been twisted together there will be 60 hanks in length of the combined threads. But as 60 hanks of 60s weigh I lb., 60 hanks of 20s will weigh 3 lbs.; consequently the 60 hanks of twisted thread will weigh 4 lbs. and 60 -i- 4 = 15 hanks per lb. or 15s resultant counts. This may be stated as follows : 60 hanks of 60s = 1 lb. 60 " '^ 20s = 3 lbs. 4 lbs. total. 60 lianks of twisted thread now weigh 4 lbs., and 60 -f- 4 = 15s resultant counts. This same method can be used when more than two threads are twisted. 36 Example. Find the resultant counts when a 40t 20s, and 10s cotton threads are twisted together. 40 -^ 40 = 1 lb. 40 -^ 20 rr= 2 - 40 -^ 10 = 4 " 7 lbs. total. 40 hanks of twisted thread now weigh 7 lbs. and 40 ^ 7 = 5.71 resultant counts. When ply yarns are composed of difterent materials it is first necessary to reduce all the yarns to one standard, (see equivalent counts) and then proceed as- by last rule. Example. If a ply thread is composed of one thread of 24s worsted, one thread of 16s worsted, and one thread of 32s spun silk, what are the resultant counts in worsted ? 840 X 32 ^^ m 4bs worsted. 560 Tlie equivalent of a 32s spun silk in worsted counts is 48s ; thus we have one thread of 48, one of 24, and one of 16s worsted; then by rule, 48 -^ 48 = 1 lb. 48 ^ 24 = 2 " • 48 ^ 16 =^ 3 " 48 hanks weigh 6 lbs. 48 hanks -^ 6 lbs. = 8 hanks per lb., or 8s re- sultant counts. To find the cotton counts of yarn required to twist with a known count of single yarn, to produce a given count of ply yarn : Rule. Divide 840 by the difference in weight of the known single and ply yarns. 37 Example. What counts of cotton yarn is required to twist with a 40s cotton yarn to make a ply yarn equal to a 15s ? 840 -^ 15s = 56 lbs. weight of 840 hanks of 15s. 840 ^ 40s =21 '^ " '• ^' '' '• 4:0s. 35 lbs. diflference. 840 hanks 4- 35 lbs. = 24s counts required. It has previously been explained that the same length of each counts of single yarn will be required to make a ply yarn but that the weights will be different. In this case, 840 hanks is taken so as to have the weights show in lbs., any other length could be used and tlie answer obtained the same way ; as for instance, if 1000 yards were used: 1000 ^ 15s = 66.66 represents total weight. 1000 — 40s ~ 25 " weight of 40s. 41.66 represents difference in wgt. 1000 ^ 41.66 = 24s counts required. To find the weight of each counts of yarn to obtain a given weight of ply yarn, when the threads twisted are of unequal counts : Rule. First find the counts of the yhj yarn, then multiply by the iveight of ply yarn required, divide the result by the counts of either of the single yam, and the result equals the iveight of that yarn ; deduct this iveight from the total iveight and the remainder is the iveight of tlie other counts of single yarn. Example. What weight of 40s cotton yarn must be twisted with a 32s cotton yarn to produce 100 lbs. of ply yarn ? 38 40 -f- 40 = 1 lb. 40 -^ 32 = 1.25 '' 40 hanks weigh 2.25 lbs. 40 -i- 2.25 = 17.77 counts of the ply yarn. 17.77 X 100 lbs. = 1777 hanks in 100 lbs. 1777 ^ 40s = 44.44 lbs. of 40s required. 100 lbs. — 44.44 = 55.56 lbs. of 32s. It will be plain that the length being the same, the weight of the co irser yarn will be greater than the finer counts; 17.77 counts of the ply yarn means that 17.77 hanks weigh 1 lb., and when multiplied by 100 equals the hanks in 100 lbs. ; as the lengths of the 40s and 32s that compose the ply thread are equal, the hanks divided by either of the single counts will give its weight in that length etc. The rule applies when more than two threads are twisted. Example. Find weight of each of 120s, 80s, and 40s required in making up 100 lbs of ply yarn. 120 -f- 120 = 1 lb. 120 -f- 80 = 1.5 ^' 120 -^ 40 =z 3 ^' 120 hanks weigh 5.5 lbs. 120 ^ 5.5 = 21.82 counts of the ply yarn. 21.82 X 100 = 2182 hanks of ply yarn in 100 \ht weight. 2182 -^ 120s = 18.18 lbs. of 120s. 2182 -^ 80s = 27.27 " • -' 80s. 2182 ^ 40s = 54.55 '^ " 40s. 100.00 " Or multiply the total weight of yarn by the relative weight of each count, and divide the result by the sum of the relative weights of all tlie yarns. 39 The last example by this rule will be 100 lbs. X 1 5.5 100 X 1.5 5^5 ~ 100 X 3 5^5 = 18.18 lbs. of 120s. = 27.27 lbs. of 80s. ^ 54.55 lbs. of 40s. 100.00 ^' COST OF PLY YARNS. To find the cost of a ply yarn when the threads are of equal counts, material, and value, is simply to multiply the lbs. of ply yarn by the cost of the single yarn plus the cost for twisting. In the manufacture of fancy yarns, the counts, ma- terials, and cost per lb. of the yarns twisted are often different. In this case, the price of the finished yarn must be found in some other way. To find the cost per lb. of a ply yarn composed of threads of different counts and values: Rule. Divide the highest counts by itself, and then by each of the other counts; the result in each case will repre- sent the relative weight of each thread in lbs. ; multiply the relative weight (f each thread by its own cost ; then divide the total cost by the sum of the relative weights; the answer equals the cost per lb. of the ply thread. Examples. If a 10s cotton thread at 15c. and a 40s cotton thread at 34c. per lb. are twisted" together, find the cost per lb. of the ply thread. 40 -^ - 40 = 1 lb. of 40s X 34c. 40 -^ :- 10 = 4 " '' 10s X 15c, 40 34c. _ 60c. 5 lbs. of ply yarn cost 94c. 94c. -^ 5 lbs. = 18.8c. per lb. If the threads are of diflPerent materials and the counts calculated on a different standard, reduce them to the same standard (see equivalent counts) and then proceed as in the last example. Find the cost per lb. of a twisted thread composed of 1 thread of 40s worsted at 75c. per 11). and 1 thread of 60s/2 spun silk at $2.30 per lb. * First reduce the worsted to its equivalent counts in spun silk ; 560 X 40 =^ 26.66 equivalent count of the worsted ^^^ in spun silk counts. We shall now have yarns as follows : A 26.66s at 75c. and a 60s at $2.30 per lb. ; then by the last rule : 60 -^ 60 = 1.00 lb. of 60s at $2.30 = $2.30 60 ^ 26.66 = 2.25 " " 26.66s at 75c. = 1.69 3.25 lbs. of ply yarn cost $3.99 $3.99 -f- 3.25 lbs. = $1.22 cost per lb. of tlie ply yarn. If three or more threads are twisted together, first find the cost of any two when twisted, as by last rule ; then use this ply thread and its cost as though it were a single thread, and find the cost of this thread and the third when twisted together. Find the cost of a ply thread composed of 1 thread of 50s worsted at 80c. per lb., 1 thread of 28s cotton at 30c., and 1 thread of 20 run woolen yarn at 56c. per lb. 41 First reduce the worsted to cotton counts and lind the cost of the worsted and cotton threads twisted. = 33.33 equivalent of 50s worsted in ^40 cotton counts, then 33.33 -^ 33.33 = 1.00 lb. of 33.33s at 80c. = 80c. 33.33^28 =1.19 '• ^' 28s at 30c. = 35c. 2.19 lbs. of ply yarn cost $1.15 $1.15 ^ 2.19 lbs. = 52c. cost per lb. of the worsted and cotton threads twisted. Next find the resultant counts of the two threads, 33.33 ^ 33.33 = 1.00 lb. of 33.33s 33.33 ^28 = L19 •' " 28s 2.19 lbs. of ply yarn. 33.33 ^ 2.19 =: 15.21 resultant counts. Thus we have a yarn equivalent to a 15.21s cotton, costing- 52c. per lb., to be twisted with a 20 run woolen at 56c. per lb. We must now reduce the woolen yarn to cotton counts : ^^^^ ^ ^^ = 38.09 equivalent counts of the 20 840 run woolen in cotton counts at 56c. per lb., then to follow the rule : 38.09 -^ 38.09 = 1.00 lb. of 38.09s at 56c. = $ .56 38.09 ^ 15.21 = 2.50 '• '^ 15.2 Is at 52c. = OO 3.50 lbs. of ply yarn cost $1.86 $1.86 ^ 3.50 lbs. == 53c. per lb. cost of the three ply yarn. 42 EXAMPLES FOR PRACTICE. Find the length in 3 lbs. of 40s cotton yarn. 100800 yards. What is the length in 8 lbs. of 60s worsted yarn ? 268800 yards. If 40 cops of 60s cotton yarn weigh 1 lb., and each cop contains an equal length, what is the length on each cop? 1260 yards. Find the length in 10 lbs. of 6 cut woolen yarn. 18000 yards. Find the length of 15 lbs. of 8 run woolen yarn. 192000 yards. If 224000 yards of worsted yarn weigh 10 lbs., what are the counts ? 40s counts. What is the length in 6 lbs. of 40s/ 2 spun silk ? 201600 yards. If 56 yards of worsted yarn weigh 30 grains, what are the counts ? 23.33 counts. Find the equivalent of a 4 run woolen yarn in cut counts. 21.33 counts. Find the number of yards in 9 lbs. of 5 dram silk. 460800 yards. If a loom beam with 3000 ends and 14 cuts of 50 yards each, with 8% size, weighs 60 lbs., what are the cotton counts? 45.29 counts. Find the equivalent of a 36s cotton yarn in worsted counts. 54s counts. How many lbs. of 60s cotton yarn will be required to make the following set of section beams — 2 beams 43 of 450 ends, 3 beams of 400 ends, and 1 beam with 375 ends; each beam to contain 6 wraps? 883.92 lbs. Find the equivalent of a 1 dram silk in spun silk counts. 304.76 spun silk. If 60 yards of cotton yarn weigh 5 grains, what are tlie counts? 100s counts. Find the weight of yarn on a section beam that contains 450 ends of 32s cotton yarn and is 10 wraps of 3000 yards long. 502.23 lbs. If a section beam contains 8 wraps of 3000 yards, 500 ends, and weighs 700 lbs. net, what are the cotton counts ? 20.40s counts. If 80 yards of worsted yarn weigh 50 grains, find the counts. 20s counts. Find the length in 8 lbs. of 2/ 30s worsted yarn. 67200 yards. Find the weight in grains of 112 yards of 40s cotton yarn. 23.33 grains. If 75 lbs. of 2/40s worsted yarn is made into a warp of 1400 ends, how long will the warp be? 600 yards. If 128000 yards of raw silk weigh 1 lb., what is the dram silk counts ? 2 dram silk. Find the yards contained in 7 lbs. of 6 dram silk. 298666.66 yards. If 72 threads, each 2 inches long, weigh 2 grains, what are the cotton counts? 16.66 counts. If 120 yards are wrapped from each of 3 cops, and together weigh 50 grains, what are the cotton counts ? 60s counts. 44 If 132740 yards of raw silk weigh 3i lbs., what are the dram silk counts ? 6.75 dram silk. How many yards of 50s cotton yarn will weigh 40 grains ? 240 yards. Find the equivalent counts of a 40s worsted in cotton counts. 26.66s cotton. How many yards are contained in 6 lbs. of 16/18 denier silk? " 1566056.47 yards. What is the equivalent of a 60s cotton yarn in spun silk counts ? The same, 60s. If 400000 yards of raw silk weigh 2 lbs., what are the denier counts ? 22.18 or 21/23 denier counts. If 2400 yards of woolen yarn weigh 2 lbs., what are the cut counts ? 4 cut yarn. Find the weight of 65 yards of 70s cotton yarn. 8.88 grains. Find the equivalent counts of a 10/12 denier silk in dram counts. 1.57 dram counts. Find the equivalent counts of a 20/22 denier silk in cotton counts. 251.54s cotton counts. How many yards of single yarn are contained in 7 lbs. of 60s/2 spun silk?" * 705600 yards. Find the resultant counts of a thread when a 60s, 40s, and 20s cotton threads are twisted together. 10'.90s resultant counts. How many yards of single yarn will be required to make 8 lbs. of 2/ 20s worsted yarn ? 89.600 yards. Find the equivalent counts of a 2 dram silk in denier counts. 34.66 denier or 34/36. CLOTp CRLCULflTIONS CLOTH CALCULATIONS, In dealing with cloth calculations, it may be best to explain the necessity for the rules that are given later. As is well known, cloth is made by interlacing warp and filling threads. When the threads cross each other they are bent more or less out of a straight line, a fact that causes the cloth to contract both in width and length. Thus the rate per cent of contraction will vary on different cloth constructions. For this reason, reeds of a lower count than the required count of the cloth must be used, making the cloth wider at the reed than on the roller. The length of a cut must be made longer on the slasher than the required length of the cut when woven. The threads of the warp must be sized before they can be woven, and this adds to the weight so that size and contraction are factors to be considered in all cloth calculations. The principal particulars of a plain cloth are the number of threads of warp and filling in one inch, the width of the cloth when woven, and the weight, which is designated as so many yards per pound. The number of warp threads per inch is known as the sley, and the filling threads as the picks, and when speaking of a cloth construction, the sley is invariably put first. A cloth that contains 56 warp threads and 60 filling threads per inch, would be classed as a 56 X 60 cloth. Warpthreads are commonly known 48 as ends, and this term will be used in all the calcu- lations. To find the number of ends required in a warp when the sley and width are g-iven: Rule. Mtiltiphj the sley hy the width and add extra ends for the selvages . Selvages are made by drawing some of the ends on the sides double, the number so doubled depending on the width of selvage desired. Extra ends must be added to keep the cloth at the required width. On heavy cloths, ply yarn is sometimes used for the selvage ends. Examples. IIow^ many ends are required to weave a cloth 56 sley, 36 inches wide, with 24 ends extra for selvages? 56 X 36 = '2016 + 24= 2040 ends. How many ends will be required to weave a cloth 76 sley, 30 inches wide, allowing 36 ends for selvages ? 76 X 30 = 2280 + 36 = 2316 ends. In the process of weaving, the warp and tilling threads are bent more or less out of a straight line, causing the' cloth to contract in both length and width. When the sley and picks are equal and the w^arp and tilling of the same counts, the contraction will be nearly equal in length and width. When the sley is higher than the picks the contraction in width will be less than if the picks were higher than the sley. If coarse tilling is used, the contraction in width will be less than if tine tilling is used, more especially if the coarse tilling is slack twisted. A plain cloth of ordinary construction about 50 sley and pick will contract about 7 per cent, while a cloth 49 100 sley and pick will contract about G per cent. Wlien warp stop motions are used, the warp has to be held at a greater tension on account of the drop wires and this extra tension causes the cloth to contract from 1 to 2 per cent more, or from 8 to 9 per cent. There are two methods used for finding the dents per inch in a reed for a given sley cloth. One is to estimate how much a cloth will contract and make it that much wider in the reed, and from the sley re- quired and the estimated width at the reed find the dents per inch required in the reed. Another method is to make a calculation from the sley of the cloth required and use a rule that gives a sliding rate of contraction which decreases as the sley increases. By the first method, long practice and experience is absolutely necessary to estimate correctly the amount to allow for contraction, while the second method may be used by those less experienced. To illustrate the first method for finding the count of the reed, — suppose a cloth 100 sley, 40 inches wide is required, it would take 100 X 40 = 4000 ends ^ 2 = 2000 dente. The next thing is to estimate the amount of con^ traction, which, under ordinary conditions, would be- about 2i inches, making the ends 42i inches in the reed •2000 dents ^ 42.50 inches = 47.05 dents per inch or a 47 dent reed. If this cloth were woven on a loom equipped with a warp stop motion, the width at the reed might be one inch more or 43^ inches, then, 2000 dents ~ 43.50 = 45.97 dents per inch or a 46 dent reed. 50 The width of cotton cloth is seldom changed in the process of finishing, while worsted and woolens which pass through fulling and washing processes may change in width considerably. Thus the finishing of the cloth, as well as the weave, must be taken into consideration in finding the reed for worsteds or woolen cloths. For these reasons, the above method is well suited to the woolen trade. The second method for finding the reed is to find the dents per inch from the sley of the cloth by rules that give a sliding rate per cent, of contraction which decreases as the sley increases. One of these rules is as follows : To find the dents per inch for any sley cloth: Rule. Subtract \ from the sley, from the result sub- tract b per cent. J divide this result by the Jiumber of ends per dent and aiiswer is dents per inch required. Examples. Find the dents per inch in a reed to weave a 48 sley cloth, the ends to be drawn 2 in a dent. 48 — 1 = 47 X .95 = 44.65. 44.65 -f- 2 = 22.32 dents per inch required. To subtract 5 per cent., multiply by .95. If no contraction took place, a reed for a 48 sley cloth would require 24 dents per inch ; the rate per cent, for contraction allowed by the rule given is 7 per cent. 24 — 22.32 = 1.68 dents less. 1.68 ^ 24 = .07 or 7%. Find the dents per inch in a reed to weave a 96 sley cloth, ends drawn 2 in a dent. 96 — 1 = 95 X .95 = 90.25. 90.25 -f- 2 = 45.12 dents per inch. 51 If no contraction took place, 48 dents would be necessary, but by the rule 45.12 dents, which is 6 per cent, less, is required. 48 _ 45.12 == 2.88 dents less. 2.88 -f- 48= .06 or 6%. The explanation of this rule is as follows : One deducted from any given number gives a higher per cent, reduction than the same amount taken from any higher number. This is where the varying rate of contraction is obtained by the rule. Five per cent, is then deducted in every instance, as per rule. To find the sley cloth a reed will weave when length and total dents in the reed are known : Rule. Divide the total dents by the length of the reed inside the reed bars ; the result is dents per inch; multiply dents per inch by ends per dent ; divide the result by .95 and to this result add 1. Examples. What sley cloth will a reed weave that contains 1030 dents on 36 inches, the ends drawn 2 in a dent? 1080 -f- 36 = 30 dents per inch. 30 X 2 = 60 ends per inch in reed. 60 ^ .95 =: 63.15 + 1 = 64 sley cloth. If a reed contains 23.12 dents per inch, what sley cloth will it weave if the ends are drawn 3 in a dent ? 23.12 X 3 = 69.36 ends per inch in the reed. 69.36 -^ .95 = 73.01 + 1 = 74 sley cloth. Bars a quarter of an inch wide are put in the ends ot the reed to protect them from damage ; they also serve to have marked on them the number of dents, the length of the reed, and the sley of the cloth it would weave with the ends drawn 2 in a dent. 52 Looms are made in different widths, and are gen- erally known by the Ml width of cloth that can be woven on them. A loom that will weave cloth 40 inches wide is styled a 40-inch loom, while a loom that will weave a cloth only 30 inches wide is styled a 30- inch loom, etc. Allowance is made for contraction in the width of the cloth, so that a 40-inch loom is made with about 44 inches of reed space, while a 30-inch loom would hold a reed about 34 inches long. Sometimes the orders for narrow cloths exceed the possible production of the looms of the correct width, so that it is necessary to use the wider looms to weave it. When a cloth is woven that requires the full width of the loom, the reed is made to fill the reed space on the loom. When narrow cloth is woven on a wide loom, the reed is usually made about an inch longer than is actually required for the warp ends, and the reed space on the loom filled in with short pieces of old reeds. Selvage ends sometimes spring the dents in the reed, causing the ends to break. If some extra length is allowed on the reed, the selvage ends may be drawn in some other dents when a new warp is drawn, whereas, if the reed is only the exact length a new reed would be required. A method for finding the number of dents and the length of the reed, also the number ot harness eyes on each harness shaft will be explained by an example. Examples. Find the number of dents, the length ot the reed, and the number of eyes per shade on the harness to weave a 64 sley cloth in a loom, with 40 inches reed space. 53 64 — 1 = 63 X .95 = 59.85 4- 2 = 29.92 dents per inch in the reed and 29.92 eyes per inch on each shade of the harness, 2 shades per set. 40 inches reed space less a half inch for reed bars equals 39.50 inches. 39.50 X 29.92 = 1171 dents on 39.50 inches and 1171 eyes on each shade of the harness : the reed will measure 40 inches with the reed bars. If the harness is made double shade, the number of eyes per shade would be one-half the number of dents in the reed. Find the number of dents in the reed, and harness eyes required on each shaft to weave a 5 end warp sateen, 114 sley, 40 inches wide, allowing a half inch extra width on harness and reed. 114 X 40 = 4560 ends -f- 5 ends per dent =912 dents required. 114— 1 = 113 X .95= 107.35 -f- 5 = 21.47 dents per inch in the reed. 912 ^ 21.47 = 42.47 inches space that 912 dents will occupy in the reed. 42.47 4- -50 for extra width = 42.97 inches length of reed inside reed bars. 42.97 X 21.47 = 923 dents on 42.97 inches; add a half inch for reed bars and the reed will measure 43.47 inches over all. The harness will require 923 eyes per shade on 42.97 inches, 5 shades in the set. When making a calculation to find the weight of filling in a given length of cloth, the width of the warp ends in the reed must be taken as the length of the filling pick, and not the actual width of the cloth. To find the width of the warp ends in the reed: Rule. Divide the total number of ends in the warp by the number of ends in a dent, equals the dents required if 54 all the ends were drawn 2 in a dent; subtract as many dents as are to contain double ends in the selvages, equals dents that actually contain ends ; divide by the dents -per inch in the reed, ecpials the width at the reed. Example. Find the width in the reed of a 64 sley cloth, 36 inches wide when woven, the ends to be drawn 2 in a dent with 48 extra ends in 12 dents for selvages. 64 X 36 + 48 = 2352 ends. 2352 ^ 2 == 1176 dents required if all dents contained 2 ends. 48 of the 2352 are to be used 4 in a dent, thus there will be 1176 — 12 =1164 dents that contain warp ends. 64 — 1 = 63 X .95 = 59.85 ^ 2 = 29.92 dents per inch in a 64 sley reed. 1164 dents used -^ 29.92 = 38.90 inches wide at the reed. To find the weight of filling in a given length of cloth : Rule. Multiply the width at the reed by the picks per inch, result equals inches of filling in 1 inch of cloth, it also equals the yards of filling in 1 yard of cloth ; multi- ptly by length of cloth in yards, equals total yards of filling required; divide by 840. equals hanJcs; divide hanks by counts, equals weight in lbs. (The width at the reed multiplied by the picks per inch equals the inches of filling in 1 inch of cloth, which, divided by 36, equals the yards of filling in 1 inch of the woven cloth. To obtain the yards of filling in 1 yard of cloth, multiply by 36 and the result is the same in yards of tilling in 1 yard of cloth as the inches in 1 inch of cloth, thus they will cancel and are omitted in the rule.) 55 Examples. Find the weight of filling in 50 yards of cloth woven with 72 picks of 60s filling and 32.50 inches wide at the reed. 32.50 X 12 = 2340 inches of filling in 1 inch of cloth, also the yards in 1 yard of cloth. 2340 X 50 = 117000 yards of filling in 50 yards of cloth. 117000 ^ 840 = 139.28 hanks. 139.28 -^ 60s = 2.32 lbs. of filling. If a cotton cloth is woven 64 X 68 — 36 inches wide, with 48 ends extra in 12 dents for selvages, what weight of 45s filling would be required to weave 100 yards of cloth ? 64 X 36 + 48 = 2352 ends. 2352 -^ 2 =: 1176 — 12 = 1164 dents used. 64 — 1 = 63 X .95 := 59.85 -^ 2 = 29.92 dents per inch. 1164 -^ 29.92 = 38.90 inches wide at reed or length of each pick of filling, 38.90 X 68 X 100 = 264520 yards of filling in 100 yards of cloth. 264520 -^ 840 = 314.90 hanks. 314.90 -f- 45s filling = 6.99 lbs. filling req. To find the weight of filling of each color in a checked gingham when the picks of each color in the pattern, picks per inch, width at reed, and counts of filling are known : Rule. Multiply the width at reed by the picks per inch and this result by the cloth length, result equals total yards of filling required ; divide by 840, equals hanks ; divide hanks by counts, equals lbs. Divide the picks of each color in the pattern by the total pi^ks per pattern and 56 result equals the per cent, of each color of filling ; multi- ply the total weight of filling in the cloth length bij the per cent, of each color , ecpials the weight required of each color. Example. If a checked gingham is woven 30 inches at the reed, with 68 picks of 35s filling, what weight of each color of filling will it require to weave 100 yards of cloth, with the following pattern ? 40 picks white 12 " blue 8 " black 6 " yellow 66 " total picks in the pattern. 40 -^ 66 = .61 or 61% white filling. 12 -^ 66 = .18 " 18 '' blue " 8 -^ 66 = .12 ^^ 12 " black " 6 -^ 66 == .09 " 9 " yellow " 30'/ X 68 X 100 ^ ^^ „ ^ . , =6.93 lbs. total weidit. 840 X 35 "" 6.93 X 61% = 4.23 lbs. white. 6.93 X 18^' =: 1.25 " blue. 6.93 X 12 " = 0.83 " black. 6.93 X 9 " = 0.62 " yellow. 6^93 " To find the weight of filling required for a given length of cloth when two different counts are used : Rule. Multij)ly the width at the reed by the picks per iiich and the result by the cloth length, result equals the total yards of fdliiig required ; divide by 840, equals total hanks. Divide the number of picks of each counts of filling in one repeat of the pattern by the total picks per pattern , result equals the per cent, of length of each counts. Multiply the total hanks by the per cent, equals hayiks of each counts, divide hanks by counts equals the required weight of each counts. 57 Example. Find the weight of each counts of filling required to weave 100 yards of cloth (filling welt pattern) woven 30 inches at the reed with 12 picks of 60s filling for face, 4 picks of 10s filling for wadding per pattern, and 124 picks per inch. 30^^X 124 X 100 ..^o^ , , ,, , =: 442.85 total hanks. 840 12 picks face -^16 picks per patt. =.75 or 75% in length of the 60s. 4 picks wadding -f- 16 picks per patt. =.25 or 25% in length of the 10s. 442.85 hanks X 75% = 332.13 hanks of 60s. 332.13 ^ 60 = 5.53 lbs. of 60s. 442.85 hanks X 25% = 110.71 hanks of 10s. 110.71 -^ 10 = 11.07 lbs. of 10s. WARP CONTRACTION. The contraction that takes place in the length of the warp is caused by its being bent out of a straight line in passing around the filling. It will be plain that as the picks per inch increase, the rate of contraction will also increase, and that coarse filling will require a longer warp length than fine filling, because the warp ends will be bent more out of a straight line in passing around it. The sley will also have an effect on the rate of contraction, as for instance, a warp satin stripe with a plain ground may be woven from one beam, because the warp ends in the stripe are drawn four to six in a dent, and on account of their crowded condition are prevented from laying as flat as they do in a plain weave. The weave, sley, picks, counts of warp and filling and twist of yarns each have an effect on the con- traction of the cloth, and on account of the varied 58 constructions used in the manufacture of cotton cloths it is impossible to give a rule for finding the per cent, of contraction that will apply to all cloths. Long practice and familiarity with different cloths, and comparison of the cloth to be made with others of a smilar construction are the surest means for obtaining the desired results. For plain cloths of ordinary construction the following rule may be used to find the approximate per cent, of contraction from warp to cloth : Rule. Multiply the picks per inch by 3.5 and divide the result by the counts of the filling. Example. Find the slasher length of a cut with 68 picks of 32s filling, the cut to be 50 yards when woven. 68 X 3.5 = 238. 238 -f- 32 = 7.4 per cent contraction. 50 X T.4% = 3.70 yards. 50 + 3.70 = 53.70 " slasher length. When a cloth is made where the construction is such that the contraction would be more than ordinary, 4 may be used as a multiplier, or if there is reason to expect a less rate of contraction, 3 may be used, etc. On fancy cloths, yarns of different counts are often used ; ply yarns are also used either for backing, as in bedford cords, or for figured work, as on lenos and lappets. When making new patterns. on such cloths, there is no rule by which the length of yarn required to weave a given length of cloth can be obtained, so that samples must be woven and the lengths and weights obtained in that way. When it is desired to reproduce a cloth from a small sample, a piece from three to six inches, more or less, 59 as may be obtainable, is taken and a warp end removed from each distinct weave, straiglitened out, and compared with the length of the cloth ; from the differ- ence in the lengths of the end and the cloth the per cent to allow for contraction on the cloth length may be obtained. To find the per cent to allow for contraction from a cloth sample : Rule. Pull a warp end out of the cloth sample^ straighten it out and measure it ; divide the difference in the lengths of the end and the cloth by the length of the cloth, the result equals the per cent, to allow for contraction on the cloth length. Examples. If an end from a piece of cloth 3 inches long measures 3.25 inches, what is the per cent, to allow for contraction on the cloth ? 3.25 — 3 == .25 difference in length of the end and the cloth. .25 ^ 3 = .083 or 8.3% contraction. If an end from a piece of leno cloth 4 inches long, measures 6 inches, what is the per cent, of contraction to allow on the cloth ? 6 — 4^2 inches difference. 2 ^ 4 = .50 or 50% contraction. It will be clear that if 4 inches of cloth require 6 inches of yarn, the difference in the lengths of the yarn and the cloth is half the length of the cloth or 50% . If an end from a piece of cloth 4 inches long, measures 4.50 inches, what length of yai-n must be run on the slasher to obtain a cut of cloth 60 yards long ? 60 4.50 — 4 = .50 difference in length of the end and the cloth. .50 ^ 4 = .125 or 12.5% contraction. 60 yards X 12.5% = 7.5 yards. 60 -\- 7.5 = 67.5 yards of yarn required on slasher to weave a 60 yard cut. To find weight of warp yarn in a given length of cloth : Rule. Mnltiply the ends of each counts by the slasher le7igth, result equals yards of ivarj); divide by 840, equals hanks; divide hanks by warp counts, equals weight. If the tveight when sized is desired, add from 5 to 8%, ac- cording to the amount of size to be put 07i the yarn. Ply yarns, except when mercerized, do not require size. Examples. If a cotton cloth is constructed 68 sley, 36 inches wide, with 40 ends extra added for selvages, what weight of 36s warp yarn would be contained in 50 yards of cloth, allowing 5 per cent, for contraction and 7 per cent, for size ? 68 X 36 + 40 == 2488 ends. 50 rds X 5% = 2.50 yards. 50 " -{- 2.50 = 52.50 yards slasher length. 2488 X 52.50 = 130620 total yards warp. 130620 -^ 840 = 155.50 hanks. 155.50 -^ 36s = 4.31 lbs. without size. 4.31 lbs. X 7% size = .30 lbs. 4.31 -\- .30 = 4.61 lbs. weight of warp with size added. When the cloth contains ply yarns, the counts of the single yarn is considered; as for example, 100 ends ot 2/40s would be calculated as 200 ends of 40s. When the ends in one part of the cloth vary from another part, either in counts or in the rate of con- traction, they must be put on separate beams. 61 If a cloth is woven witli 2100 ends of 40s with 5 per cent, contraction, and 7 per cent, size, 200 ends of 2/ 15s, with 15 per cent, contraction, and 350 ends of 3/20s, with 25 per cent, contraction, what weight of warp yarn will be required to weave 100 yards of cloth ? 2100 X 105 = 220500 yards -^- 840 = 262.50 lianks. 262.50 -^ 40s. = 6.56 lbs. + 7% size = 7.01 lbs. weight of the 40s. yarn. 200 X 2 X 115 = 46000 yards -^ 840 = 54.76 hanks. 54.76 -^ 15s = 3.65 lbs. of 15s yarn. 350 X 3 X 125 = 131250 yards -^ 840 = 156.25 hanks. 156.25 -^ 20s = 7.81 lbs. of 20s yarn. 7.01 -f 3.65 H- 7.81 = 18.47 lbs. total weight. AVERAGE NUMBERS. One of the principal particulars in a plain cloth is the weight, which is designated in yards per lb., with a given sley, picks, and width. When the numbers of the warp and filling are not given, they may be found by first finding the average numbers, i. e., a number, which, if used for warp, and filling, would give the weight desired ; then to assume a number to be used for the warp that is coarser than the average numbers in the cloth and one that is perhaps already being produced, then by a calculation find the numbers of the filling required to give the yards per lb. On the coarser grades of cloth, the difi*erencc in the numbers of the warp and filling will be slight, but on the finer grades the difference will be greater. 62 The reason for this is that warp yarns are subjected to greater strain than filling yarns and a stronger yarn is necessary. To find average numbers when sley, picks, width, and yards per lb. are known : Rule, First find number of eiids, then multiply by estimated slasher length, result equals yards of tear p ; divide by 840, equals hanks; add a certain per cent to represent size, and consider it as yarn. Multiply ividth at reed, by picks per inch, result equals iiiches of filling in one inch of cloth, also the yards of filling in one yard of cloth ; midtiply by length of cloth, result equals yards of filling; divide by 840, equals hanks. Divide length of cloth by yards per lb., resiilt equals weight of cloth. Add hanks of warp and filling and divide by weight of cloth, equals average numbers. In cloth calculatiolis where the length is not specified, 100 yards will be found the most convenient length to use on account of the percentages for contraction and from the fact that it is an easier sum to use than any part of 100. Example. Find average numbers on a cloth 48x 52, 36 inches wide, and weighing 6 yards per lb. 48 X 36 + 40 == 1768 ends. 1768 X 106 yards (6% added for contraction) = 187408 yards. 187408 ^ 840 = 223.10 hanks. 223.10 + 7% for size = 238.71 hanks as re- presented with the addition of size. 1768 -^ 2 = 884—12 = 872 dents occupied with yarn. 48 — 1 = 47 X .95 = 44.65 -i- 2 = 22.32 dents per inch. 63 872 total dents ^ 22.32 = 39.06 inclies wide at reed. 39.06 X 52 X 100 = 203112 yards of filling in 100 yards of cloth. 203112 4- 840 — 241.80 hanks filling. 238.71 hanks warp + 241.80 hanks filling=:480.51 total hanks. 100 yards of cloth -^ 6 yards per lb. = 16.66 lbs. of cloth. 480.51 4- 16.66 = 28.84 average numbers. It will be seen from this problem, as in yarn calcu- lations, that having found the amount of yarn in hanks it is divided by the weight to find the counts. To find the average numbers when sley, picks, and counts of warp and filling are known: Rule. Divide the sley by the icarp counts, result equals relative weight of ivarj) yarn ; divide the picks by the filling counts, ecpials relative iveight of filling yarn; divide the sum of the sley and picks by the sum of the relative weights, e(pials the average counts. Example. If a cloth is woven 68 X 72 with 34s warp and 50s filling, what is the average numbers ? 68 ^ 34 = 2 relative weight of warp. 72 -f- 50 = 1.44 ^' """ " filling. 140 3^ 140, the sum of the quantities -^ 3.44, the sum of the relative weights = 40.69 average numbers ; or, The sley and picks, being quantity of yarn in a given space, it may be considered as hanks, and 68 hanks of warp 4- 34s counts = 2 lbs. of warp. 72 '^ ^' filling -^- 50s " =1.44" "filling. 140 " 3.44 " total. 140 hanks -^ 3.44 lbs. = 40.69 average counts. 64 On fancy cloths, the rate of contraction on the different weaves that are combined in the cloth, is often unequal, which makes it necessary to use more than one loom beam. Sometimes ply yarns are used, and in this case, it will require separate beams ; therefore it follows that the different yarns must be calculated separately. To find average numbers when the cloth contains more than one counts of warp, and the number of ends of each counts is known, with picks per inch, filling counts, and width of cloth at reed : Rule. Multiply each member of ends by the slasher length, result equals yards of warp ; divide length of ivarji by 840, equals hanks ; divide hanks by counts, equals iveight. Multiply width at reed, by picks per inch, equals yards of filling i7i one yard of cloth; imiltiply by length of cloth or cut length, equals total yards of filling; divide by 840, equals hanks ; divide hanks by counts, equals iveight. Add all the weights and divide into the total number of hanks of warp and, filling and result is average number. Example. If a cloth is 30 inches wide at reed and contains 76 picks per inch of 70s filling and requires the following number of ends, what is the average number ? 2000 ends 60s with 4% contraction. 300 " 2/15s " 15 '^ ^' 600 " 3/20s " 25 " " ^^^iAj^* = 247.61 hanks- 60s = 4.12 lbs. 840 30^2^^)015^ 82.14 " ^15s= 5.42 " 840 65 600 X 3 X 125 267.85 '^ -^ 20s =13.36 lbs. 271.42 '• -f-70s= 3.87 " 840 30^^ X 76 X 100 ^^^ 869.02 '^ 26.77 " 869.02 -^ 26.77 = 32.46 average numbers. To find the average number when the number of ends and picks per inch, width at reed, and yards per lb. are known : Rule. Maltiply each numher of ends by its ow7i slasher length, add a given per cent, for size to the single yarn, and consider the increase as yards ; multiply the width at the reed by picks per inch, equals the yards oj filling in one yard, of cloth ; multiply by length of cloth, equals total yards ofjilling ; add the yards of warp and filling and divide by 840, equals hanks of yarn ; divide hanks by weight in lbs., equals counts, which, in this case, will be the average counts of the warp and filling. The length of the cloth, on which the calculation is made, divided by the yards per lb., equals the weight of the cloth. Example. If a cloth contains 2200 ends of 50s which contract 5 % and carry 5 % size, 400 ends of 2/30s which contract 25%, 200 ends of 3/15s which contract 20%, and 76 picks per inch, is 34 inches at the reed, and weighs 5 yards per lb., what is the average counts of the warp and filling ? As the single yarn contracts 5% and carries 5% size it equals 10% of an increase in its weight, and if the calculation is made on 100 yards of cloth it equals 10 yards; — thus, 66 2200 X 110 = 242000 yards of 50s 400 X 2 X 125 = 100000 " " 30s 200 X 3 X 120 = 72000 ^' '' 15s 34^' X 76 X 100 = 258400 " " filling. 672400 yards of yarn. 672400 -f- 840 = 800.47 hanks. 100 yards of cloth -i- 5 yards per lb. = 20 lbs. of cloth. 800.47 hanks -f- 20 lbs. = 40.02 average counts. It has been explained how the warp and filling yarns contract in the process of weaving, and that the con- traction and the size which is added to the single warp yarn increase the weight. In making a calculation on cloth, contraction is considered by adding to the cloth length an amount which varies on difi'erent cloth constructions ; cloth made from coarse yarns, however, will contract more, and will also absorb more size than finer yarns. If a thread is removed from a piece of woven cloth it will be found to have a wavy appearance, caused by its bending around other threads; consequently, in this condition, it will weigh more than a thread that is perfectly straight. The standard lengtli of the cotton hank is 840 yards, but if this length is put into cloth it reduces its length which will vary according to the construction of the cloth. In previous examples, the contraction has been added to the cloth length or width and the warp and filling calculated separately. Shorter methods are obtained by adding the amount of warp and filling as it appears in the cloth and making the allowance for contraction and size on the yards in the hank. 67 A list of lengths or constants is given that may be used in place of 840 yards, also the yarns on which they may be used. It will necessarily require some experience to determine the most suitable number to use on any given class of goods, but when once ob- tained, their use will be found to be practical. Use 745 yards on counts from 18 to 24 '^ 750 ' ^' ^' '' '^ 25 '^ 35 u 755 u u u u 36 u 45 u 760 u u u u 46 u 60 " 765 " " " ^^61 '^ 80 u 770 '.' u u u gi i, 100 The counts of yarns given are the average of warp and filling. To find average number on plain cloths when sley, picks, width and yards per lb. are known : Rule. Add sley and jjicks, result equals inches of yarn in one inch of cloth ; multiply by the width, equals inches of yarn in one inch of cloth of that width; it also equals the yards of yarn in one yard of cloth; multiply by yards per lb., equals the yards of yarn in one lb. of the cloth ; divide by the most suitable constant or length in place of 840, equals counts, which will be the average of waip and, filling. Examples. If a cloth 52 sley, 48 picks, 36 inches wide, weighs 4 yards per lb., what is the average numbers ? { Use 745 for constant). (52 +48) X 36 X 4 745 constant = 19.32 average numbers. If a cloth 84 X 88 — 32 inches wide, weighs 11 yards per lb., find the average numbers. (Use 760 for a constant). 68 (84+ 88) X 32 X 11 760 constant = 79.66 average numbers. To find the average counts of warp yarns when two or more different counts are used : Rule. Divide each number of ends by its own counts, equals relative iveight ; add the weights and divide into the total number of ends. Examples. If a cloth contains 2800 ends of 70s and 400 ends of 40s, what is the average counts of the warp ? 2800 -f- 70 = 40 relative weight. 400 -^ 40 = 10 '^ 3200 total ends 50 total 3200 -f- 50 = 64 average counts of warp. If a cloth contains 2200 ends of 60s, 300 ends of 3/ 15s and 200 ends of 2/40s, what are the average warp counts ? 60 = 36.66 rel. wgt. 15 = 60.00 40 = 10.00 2200 ends of 60s ^ 2200 300 >^ 3/15s =: 900 200 " 2/40s = 400 3500 total 106.66 3500 ^ 106.66 = 32.81 average warp counts. To find the yards per lb., when sley, picks, width and counts of warp and filling are known : Rule. Multiply the ends by slasher length, result equals yards of warp ; divide by 840, equals hanks ; di- vide hanks by warp counts, equals weight ; add a certain per cent, for size, equals total weight of warp yarn. Mul- tipty width at the reed by picks per inch, equals yards of filling in one yard of cloth ; multiply by cloth length, equals total yards of filling ; divide by 840, equals hanks; 69 divide hanks hy JiUlncr roiints, rqimls ireight of filling xjarn ; add weight of warp and filling and divide into cloth length, equals yards per lb. Examples. Find the yards per 11). of a cloth with the following particulars : . 56 X 60 — 36 inches wide, 20s warp and 24s filling. 60 X 3.5 = 210 -^ 24 = 8.75% for warp con- traction. 56 X 36 + 40 for selvages = 2056 ends. 2056 X 108.75 yards slasher len,o:th =: 223590 yards of warp. 223590 -f- 840 = 266.17 hanks. 266.17 ~ 20s = 13.35 lbs. + 7% size = 14.2S lbs. weight of warp sized. 2056 -f- 2 =: 1028—12 ~ 1016 dents that contain yarn. 56—1 = ^5 X .95 = 52.25 ^ 2 =z 26.12 dents per inch. 1016 -f- 26.12 =: 38.89 inches wide at reed. 38.89 X 60 X 100 = 233340 yards of filling in 100 yards of cloth. 233340 ^ 840=: 277.78 hanks. 277.78 -^ 24s = 11.57 lbs. filling. 14.28 + 11.57 = 25.85 lbs. warp and filling. 100 yards -^ 25.85 = 3.86 yards per lb. If a piece of cloth contains 1800 ends of 40s, with 5% contraction and 6% size, 200 ends 2/35s with 10% contraction, 500 ends 3/45s with 17% contrac- tion, 72 picks of 60s filling, 30 inches wide at reed, what are the yards per lb. ? 70 5.62 lbs. of yarn+6% size=5.95 lbs. = 1.49 " = 4.64 ^' =z 4.28 " 1800 X 105 840 X 40 200 X 2 X 110 _ 840 X 35 ~ 500 X 3 X 117 _ 840 X 45 ~ 30^^X 72 X 100 _ 840 X 60 ~~ Total weidit, 16.36 lbs. 100 yards of cloth -^ 16.36 lbs. = 6.11 yds. per lb. By making use of the constants as explained under average numbers, a shorter method for finding the yards per lb. is obtained. To find the yards per lb. when the sley, picks, width, and counts of warp and filling are known : Rule. First find the average numbers by short method as previously explained, then add the sley and picks, re- sult e(pials inches of yarn in one square inch of cloth ; mul- tiply by width of the cloth, equals inches of yarn in the cloth width and one inch long ; it also equals the yards of yarn in one yard, of the cloth. Multiply the average num- bers by the most suitable constant, equals yards of yarn in one lb. of that average number with allowance made for contraction and, size. Divide the yards of yarn in one lb. of the average number by the yards of yarn in one yard of the cloth, equals the yards of cloth in one lb. Example. If a cloth is woven 56x60, with 20s warp and 24s filling, and 36 inches wide, what are the yards per lb. ? 56 -f- 20 = 2.80 relative weight of warp. 60 -^ 24 == 2.50 " '' ^' filling. 116 5.30 total weight of yarns. 71 116 ^ 5.30 = 21.88 average numbers. 21.88 X 745 = 16300 yards in 1 lb. with allow- ance for contraction and size. (56 + 60) X 36 = 4176 inches of warp and filling in a strip of cloth one inch long, 36 inches wide ; it also equals the yards of yarn in one full yard of the cloth. 16300 ^ 4176 = 3.90 yards of cloth in one lb. or 21.88 X 745 ,^^ , = 3.90 yards per lb. (56 + 60) X 36 To find the yards per lb. from short lengths of cloth : Rule: Multiply 7000 grains by the number of square inches of cloth weighed, divide the result by the weight of the cloth in grains multiplied by the width of cloth desired, and by SQ inches per yard. Example. If a piece of cloth 3 inches square weighs 1 1 grains, what would be the yards per lb. if it were woven 30 inches wide ? 3 X 3 X 7000 .oc, A 11 =5.30 yards per lb. 11 X 30 X 36 The problem is worked by proportion. The square inches in the piece of cloth weighed is to its weight as the square inches in one yard is to the grains in one lb. 30 X 36 ^ 1080 square inches of cloth in one yard of cloth 30 inches wide ; then if 9 square inches weigh 11 grains, 1080 square inches will weigh 1080 X 11 = 1320 grains, weight of 1 yard. V 7000 grains in one lb. -h 1320 grains, weight of one yard =:r 5.30 yards per lb. 72 More accurate results are obtained by weighing larger pieces, and when possible a full yard should be weighed and its weight divided into 7000 grains. If any fractional part of a yard is weighed, its weight in grains must be divided into the same fractional part of 7000. Examples. If a yard of cloth weighs 650 grains, how many yards will weigh 1 lb. ? 7000 ^ 650 = 10.85 yards per lb. If a quarter yard of cloth should weigh 250 grains, how many yards will weigh 1 lb. ? 7000 -^ 4 = 1750 grains in a quarter lb. 1750 -f- 250 = 7 yards per lb. In mills where a large variety of cloths are made, it is the practice to spin but three or four different counts of warp yarn, and to regulate the weights of the cloth by varying the counts of the filling. It is economy to manufacture cloth l)y this method, on account of the various processes warp yarn must pass through, while filling may be taken to the loom direct from the spinning frame or mule. To find the counts of filling required to govern the yards per lb., when sley, picks, counts of warp, width and yards per lb: are known : Rule. MnltipJy ends by slasher length, equals yards of warp ; divide by 840, equals hanlts ; divide hanks by counts, equals weight ; add a certain per cent, for size ivhen required J equals the weight of the warj) ivith size added. When more than one counts of warj) yarfi are 2ised in the cloth each count must be calculated, separately and their weights added for the total weight of . the warp yarn. Multiply luidth at the reed by jncks per inch, 73 apials ijards ofJiUing in one yard of cloth ; midtiphj by doth length, equals total yanh of Jilling ; divide by 840, equals hanks of filling. Divide length of cloth by yards yer lb., equals weight of the cloth. Subtract iveight of warj), and the balance is the weight of the filling ; divide the hanks of Jilling by the weight of filling, equals the counts of filling required. Examples. Finds the counts of filling required on a cloth made with the following particulars : 88 X 92 with 60s warp, 32 inches wide and 11 yards per lb ; 4% warp contraction and 6 % size. 88 X 32 + 40 = 2856 ends. 2856 X 104 = 297024 yards of warp. 297024 -f 840 = 353.60 hanks ^ 60 = 5.89 lbs. 5.89 + 6% for size = 6.24 lbs. weight of warp with size added. 100 yards of cloth ^ 11 yards per lb. — 9.09 lbs. of cloth. 9.09 — 6.24 lbs. of warp = 2.85 lbs. of filling. 2856 ^ 2 = 1428 — 12 = 1416 dents filled with yarn. 88—1 == 87 X .95 = 82.65 ^ 2 — 41.32 dents per inch. 1416 _i_ 41.32 = 34.26 inches wide at the reed. 34.26 X 92 X 100 — 315192 yards of filling. 315192 ^ 840 = 375.22 hanks of filling. 375.22 -^ 2.85 lbs. = 131.65 counts of filling re- quired. If a cloth is woven 30 inches wide at reed, with 76 picks per inch, and contains 2000 ends of 60s warp with 5% contraction and 6% size,. 300 ends of 2/15s with 15% contraction, 200 ends of 2 /40s with 25% contraction, the cloth to weigh 6.50 yards per lb., what counts of filling is required ? 7.4 2000 X 105 = 4.16 lbs. warp + 6% size == 4.41 lbs. = 5.47 " = 1.48 " 840X60 300X2X115 840X15 200X2X125 840X40 Total lbs. warp 11.36 30^^ X 76 X 100 = 271.42 hanks filling. ■840 * 100 yards cloth -^ 6.50 yards per lb. = 15.38 lbs. of cloth. 15.38 — 11.36 lbs. warp = 4.02 lbs. of tilling. 271.42 hanks of filling -^ 4.02 lbs. = 67.51 counts of fillina: required. If the constants 745/770 are used, the process for finding the filling required is made considerably shorter. To find the filling required to govern the yards per lb. (on plain cloth) when the sley, picks, width and yards per lb. are known : Rule. First find the average numbers, (see short method for finding average numbers) the7i add the sley and picJcSy result equals quantity of yarn in one square inch ; divide by average numbers, equals a sum that re- 'presents the weight of ivarj) and filling ; divide the sley by the counts of the ivarp, equals the relative weight of the warp ; deduct the relative weight of the warp from the total, the balance represents the relative weight of the filling; divide the picks by the iveight of the filling, equals the counts of filling required. Example. If a cloth is made 88 x 92, 32 inches wide, with 60s warp, and is required to weigh 11 yards per lb., what counts of filling are required ? 75 (88 + 92) X 32 X 11 ^ ^^.82 average number. 765 88 + 92 = 170 ^ 82.82 = 2.17 total weight. 88 -^ 60s warp — 1.46 relative weight of warp. 2.17 — 1.46 r= .71 relative wei^^ht of filling. 92 -^ .71 = 129.57 counts of filling required. It has been previously explained that quantity divided by counts equals weight. In this case, sley + picks equals the quantity and may be considered as so many hanks; i.e., 88 hanks of warp and 92 hanks of filling, 88 + 92 = 170 total hanks. 170 hanks ^ 82.82 counts = 2.17 lbs. 88 hanks of warp -^ 60s counts = 1.46 lbs. of warp. 2.17 — 1.46 — .71 lbs. of filling. 92 hanks of filling ^ .71 lbs. = 129.57 counts of filling required. The approximate counts of warp and filling from a small sample of ordinary plain cloth may be found as follows : Example. If a piece of cloth 4x5 inches, weighs 14 grains and counts 76 X 80, what counts of warp and filling could be used if it were woven 30 inches wide? First, find yards per lb. by short method. 4 X 5 X 7000 ^ ^_ , ,, z= 9.26 yards per lb. 14 X 30 X 36 Second, find average number by short method. (76 + 80) X 30 X 9.26 ^^^^ , ^- ^ i = 57.02 average number. 760 Use a 50 warp and find filling required. 76 76 + 80 ^ 57.02 = 2.73 total weight. 76 sley ^ 50 warp = 1.52 weight of warp. 2.73 — 1.52 = 1.21 weight of filling. 80 picks -f- 1.21 weight of filling = 66.11 counts of filling. To find the per cent, of warp and filling in a piece of cloth, is to find the weights of the different yarns used in its construction and divide the weight of each by the weight of the whole. Example. If a cloth is constructed 88 X 92, 32 inches wide, with 60 warp, and 125 filling, allowing 5 per cent, for warp contraction and 5 per cent, for size, what is the per cent, of warp and filling ? Rules have been previously given for the following work : 88 X 32 + 30 = 2846 ends. 2846 X 105 = 5.92 lbs. of warp + 5% ^o^* size = 840 X 60 6 21 lbs. of warp and size. 2846 -f- 2 = 1423 — 12 = 1411 dents required. 88 — 1 =r 87 X .95 =r 82.65 ^ 2 = 41.32 dents per inch in the reed. 1411 -^ 41.32 =z 34.14 inches wide at the reed. 34.14 X 92 X 100 r^ 2.99 lbs. of filling. 840 X 125 6.21 lbs. warp + 2.99 lbs. filling = 9.20 lbs. total. 6.21 ^ 9.20 = .67 or 67% warp. 100% — 67% =33% filling. A shorter method is to find the relative weights of warp and filling, and divide the relative weight of warp by the sum of the relative weights ; answer is the per cent, of warp ; then deduct the per cent, of warp from 100%, result equals the per cent, of filling. 77 The last example worked by this method would be as follows : 88 sley ^ 60 warp = 1.46 relative wgt. of warp. 92 picks -f- 125 filling = ^ -' " " filling. 2.19 total weight. 1.46 -^ 2.19 = .66 or 66% warp. 100% — 66% = 34% filling. To find per cent, of warp and filling when ends, picks, and counts of warp and filling are known : Rule. Divide ends by warp counts, result equals rclatlvx weight of warp ; multiply picks by ividth and divide result by Jilling counts, equals relative weight of filling ; divide iveight of warp by weight of warp and filling, equals per cent, of warp ; deduct per cent, of warp from 100%, equals per cent, of filling. Example. If a cloth 30 inches wide contains 2560 ends of 50s warp and 72 picks per inch of 60s filling, what is the per cent, of warp and filling? 2560^50 ^ 51.2 relative weight of warp. '^ ^^ ^^ = 36.0 - - " filling. bO 87.2 weight of warp and filling. 51.2 -^ 87.2 = .58 or 58% warp. 100% —58% =42% filling. In one yard of the cloth in the last example, 2560 ends equal the yards of warp yarn in one yard of cloth ; the picks per inch multiplied by the width equals the yards of filling in one yard of cloth; thus, if the amount of warp and the amount of filling is each divided by its own counts the result will equal the proportionate weight of each as per rule. 78 To find the per cent, of warp and filling in a cloth when sley, picks, average counts, and warp counts are known : Rule. Divide the sum of sley and jncJcs by average counts, ecpials weight of u^arp and Jilling ; divide sley by warp coujits, equals relative weight of warp; divide weight of warp by weight of warp and filling, ecpials per cent, of warp; deduct per cent, of warp from 100%, equals per cent, of filling. Example. If a cloth is made 8S X 84, with 45s warp and an average number of 50, what is the per cent, of warp and filling ? 88 4- 84 z= 3.44 weiorht of warp and fillinp^. 50 ^ V ^ 88 -f- 45 = 1.95 relative weight of warp. 1.95 -^ 3.44 = .56 or 56% warp. 100% — 56%= 44% filling. To find the per cent, of warp and filling when the cloth contains more than one counts of warp, and the counts of warp and filling and width at reed are known : Rule. Multiply each iiumber of ends by its slasher length ; divide by 840, equals hanks ; divide hanks by coimts, equals iveight ; add all the weights, equals weight of warp yarns. Multiply width at reed by picks per inch and restdt by cloth length, equals yards of filling ; divide by 840, equals hanks ; divide hanks by counts, equals weight of filling ; add weight ,of vmrp and filling and divide into the weight of ivarp, equals per cent, of warp ; deduct per cent, of warp from iOOfo , equals the per cent, of filling. Example. If a cloth is woven 30 inches at reed with 76 picks of 70s filling, and contains the following warp ends, what is the per cent, of warp and filling ? 79 2000 ends of 60s witli 5% contraction. 300 U ii 2/15s a 15% u 200 u u 2/40s u 25% u 2000 X 105 4.16 lbs. of 60s. 840 X 60 300X 2 X 115 _ 840 X 15 5.47 " " 15s. 200 X 2 X 125 _ 840 X 40 1.48 11.11 " " 40s. ^' warp. 30^^X76X100 _ 3.87 lbs. of 70s fillins-. 840 X 70 11.11 + 3.87 = 14.98 lbs. of warp and filling. 11.11 -^ 14.98 = .74 or 74% warp. 100% — 74% =26% fillincr. When the per cent, of each counts of warp is required, divide^ the weight of each counts by the total weight of the warp and filling. In the last example there are 4.16 lbs. of 60s; 5.47 lbs, of 15s; 1.48 lbs. of 40s; and 3.87 lbs. of filling, making a total of 14.98 lbs. of warp and filling. What is the per cent, of each counts of warp ? 14.98 = .28 or 28% of 60s warp. 14.98 = .36 or 36% " 15s " 4.16 5.47 1.48 -^ 14.98 = .099 or 9.9%" 40s '^ 73.9% warp. 100% — 73.9% = 26.1% filling. On many fancy cloths, the ends are not drawn in the reed in regular order, and some dents contain more ends than others ; therefore other methods than those previously given must be used to find the number of ends in a warp, also the count of the reed to use, etc. 80 On plain cloths, tlie sley means the number of ends in each inch of cloth ; on fancy cloth, the sley is under- stood to be the average number of ends in one inch, which is found by dividing the total ends by the width of the cloth. Example. If a cloth, 30 inches wide, contains 2000 ends of 60s, 250 ends of 2/ 15s, and 100 ends of 3/ 20s, what is the average sley of the cloth ? 2000 X I = 2000 ends of 60s. 250 X 2 =: 500 '^ " 15s. 100 X 3 rr= 300 " " 20s. 2800 total ends. 2800 ~ 30^^ = 93 ends per inch or average sley. If this cloth were woven with an 80 sley reed and 84 pick per inch, the particulars would be specified as follows : 80 X 84 93 showing that an 80 sley reed is used and that the cloth averages 93 sley and 84 picks per inch. The term sley reed is understood to be a reed that will weave a given sley cloth with the ends drawn two in each dent, but when more or less ends are drawn in a dent and every dent does not contain the same number, the sley given is the average number of ends per inch. To find the number of ends in a cloth with an ir- regular reed draft, when the ends in one pattern, the sley reed, and the width of the cloth is given: Rule. Divide the sley of the reed by tiro, equals the dents per inch in the cloth after contraction ; multiply by the width of cloth required, equals total dents required ; deduct as many dents as may be required for selvageSf 81 equals dents to be used for the body of the cloth; divide by the dents in one pattern^ equals the number of patterns ; multiply the ends in one iiattern by the number of patterns, equals ends required ; add extra, ends to he used, for selvages, equals total ends. Example. If a cloth is made with the following reed draft on a 76 sley reed, how many ends are required to make it 28 inches wide? Dents. Ends. 14 28 5 20 6 12 J_ 20 30 80 76 -^ 2 = 38 dents per inch as represented in the cloth after contraction. 38X28^^ = 1064 — 12 for selvages = 1052 dents required for the body of the cloth. 1052 -^ 30 dents per patt ==35 patts and 2 dents over. If this warp were drawn as per draft, the cloth would contain 14 dents, 2 ends in a dent, next to the selvage on one side, and 5 dents, 4 ends in a dent, on the other. This would not look just right and it is the practice in designing a cloth to have a few dents plain weave next to the selvages when it is possible. In this case, the 2 dents over the 35 patterns would be utilized by adding them to the 14 dents plain, making 16 dents, and the first time the pattern was drawn, 8 of these dents would be drawn instead of 14, leaving 8 dents to be drawn after the draft has been drawn 35 times, thus making both sides of the cloth with 8 dents of plain weave next to the selvages. The ends may now be found as follows : 80 ends per patt X 35 patts = 2800 ends. 82 2800 + 4 for the 2 dents over + 48 for the 12 dents, 4 in a dent, for selvages = 2852 total ends required. When the cloth contains two or more different counts of warp yarns, the reed draft would be made out as in the last example ; except that the ends in each dent that were of different counts would be placed in a separate column as follows : Dents Ends. B.B. M.B. T.B. IC u u u The ends in each column would be multiplied by the number of patterns, adding selvage ends to that column that contained the single yarn, or the ends from which the selvage ends would be taken, which is generally the bottom beam. Mill managers often receive cloth samples which they are asked to reproduce and give prices on, etc. When this is the case, the reed to use, the picks per inch, the counts of the warp and filling yarns must be obtained from the sample of cloth furnished. A method for finding the counts of yarns by comparison will be found on later pages of this book. To illustrate a method for finding the ends required from a cloth sample, reference will be made to the cloth shown in Fig. 1, which is a satin stripe with a plain ground. As this cloth contains a wide plain stripe, the sley reed may be found by the use of a pick glass ; then having the sley reed, the dents and ends per dent for the satin stripe may be found as follows : Suppose the plain stripe counts by glass 64 ends per inch, or 64 sley ; then use a thin steel rule that is graded in lOOths and measure the satin stripe, which, for an example, will be supposed to measure 13 83 hundredths of an inch. The number of dents for the stripe may be found by the following rule : To find the number of dents in a given space, when the sley reed is known : Rule. Divide the sley reed by 2, residl erjuals the dents jjer inch as represented in the cloth ; multiply by the space the dents occiqnj in the cloth measured in lOOths, ecpmls the number of dents required. Fig. 1. The dents required for the satin stripe as found by this rule are : 64 sley -^ 2 = 32 dents as represented in 100 hundredths or 1 inch of cloth, and 32 X .13 = 4.16 or 4 dents required. There are 16 ends in the stripe and 16 ends -^ 4 dents = 4 ends per dent. There are 11 dents in the plain stripe, so that one complete pattern contains 15 dents and 38 ends. Dents. Ends. 11 22 15 38 84 If the cloth is required to measure 30 inches, the ends would be found as follows : 64 -^ 2 = 32 dents per inch in the cloth, 32 X 30 inches =960 total dents required. 960 — 12 for selvages == 948 dents for body of cloth. 948 -f- 15 dents per pattern = 63 patterns and 3 dents over. These 3 dents may be used as previously explained by adding them to the 11 dents of plain weave in the pattern and using half or 7 dents the first time the pattern is drawn, leaving 7 dents to be used at the finish of the 63 patterns, which gives 7 dents of plain on each side next to the selvages. The ends are then found by multiplying the ends per pattern by the number of patterns, and adding the ends for the dents left over and also the selvage ends. 38 ends per patt. X 63 patts.= 2394+ 6 +48 = 2448 total ends required. When the cloth contains a narrow plain stripe that is less than, a quarter of an inch wide, it will be difficult to obtain the sley reed by using a pick glass, so the following rule is given : To find the sley reed when the number of dents and the space they occupy in the cloth, measured in lOOths, is known : Rule. Divide the number of dents measured by the space they occupy in the cloth measured in lOOths, result equals the dents per inch a^ rey resented iri the cloth ; multiply by 2, equals the sley reed. Example. If the space in a cloth that requires 6 dents measures 15 hundredths of an inch, w^hat is the sley of the reed ? 6 ^ .15 = 40. 40 X 2 = 80 sley reed. s^ It must not he understood thnt tlie dents per inch in the cloth and dents per incli in the reed mean the same thing, for they do not ; for, as the measurements have been taken from a ch»th sample, contraction has taken i)lace and the dents per inch in the reed will be found by a rule ])reviously explained. Fig. 2. On some cloths, where the sley is very high, and especially if ply yarns are used either for backing or decorative pui-poses, it is necessary to put several ends in the same dent. Sometimes a ply end is drawn in the same dent with single yarn, and if the dents are very close the ply ends will break the single yarn, making it impractical to weave it, etc. When it is 86 desired to make a cloth of this construction, it will be necessary to make out a reed draft for one complete pattern, the desi.s^ner placing the ends in the dents in such order as he thinks will be satisfactory. The width of the pattern is measured in lOOths of an inch and the sley reed obtained as explained when the dents and the space they occupy are known. If the sley reed found is one that would be considered too fine or too coarse for the cloth, a new reed draft is made, and more or less ends put in a dent, thus changing the number of dents per pattern. The calculation is now made again to see if the new draft has brought the sley reed to what is considered a more practical reed to use on that cloth. This idea will be better understood by referring to Fig. 2, which shows both sides of a bedford cord cloth sample. A bedford cord, on its face, appears to be complete in one rib, when in fact it requires two, as will be understood if the under side or back is examined. The sample. Fig. 2, contains 10 face ends and 2 ply ends in each rib, and to illustrate the method for find- ing the sley reed, two different reed drafts are given. Dents. Ends. C 1 4 1 rib. ^1 2 2 ply ends for wadding. 1 rib. } 6 20 4 Supposing two ribs as per draft, or one complete pattern, measure 22 hundredths of an inch, the sley reed may be found as per rule. 6 X 100 =: 600. 600 -^ 22 = 27.27 X 2 = 54.54 or 54 sley reed. 4 2 2p 4 4 2 2 4 87 Tliis reed may be considered as being too coarse or open, so another reed draft is given with less ends per dent and two more dents per pattern. Ends. r 1 rib. I I 1 ril). -{ 3 2 1 2 1 3 3 2 1 2 I 3 8. 20 4 Fig. 3. 8 X 100 =:: 800. 800 -^ 22 = 36.36 X 2 = 72.72 or 72 sley reed. A 72 sley reed would be the best to use as it permits of the ply ends being drawn in separate dents. To further illustrate this method for finding the ends required from a cloth sample, reference will be made to Fig. 3, a cloth tiiat combines plain, satin, and leno weaves. This cloth has a plain section that is wide enough to permit the sley reed to be found with a pick glass, but 88 for practice in finding the sley reed by measurements, the reed will be found by the latter method. The plain sections of the cloth contain 1 1 dents and 22 ends each, and, assuming that each measures 27 hundredths of an inch 11 X 100 = 1100. 1100 -f- 27 = 40.74. 40.74 X 2 = 81.48 or 82 sley reed. Having found the sley reed, it will be necessary to find the dents and ends required for the satin and leno stripes. Assuming that the satin stripe measures 15 hund- redths of an inch, the dents required are 82 4- 2 = 41. 41 X .15 = 6.15 or 6 dents, and, as there are 30 ends in the stripe, 30 -^ 6 = 5 ends per dent. Assuming that the leno stripe measures 27 hund- redths of an inch 82 -f- 2 = 41. 41 X .27= 11.07 or 11 dents. In this stripe are 3 dents that contain 2 double ends in each dent, therefore there will be 8 empty dents, which, divided into 2 spaces, allow 4 dents in each space. Starting the pattern at a plain stripe, the reed draft is as follows : Dents. Ends. 11 6 11 1 22 30 22 4 4 empty 1 4 4 empty 1 4 39 86 39 dents and 86 ends in one pattern. 89 If the cloth were required to he made 27 inches wide, the ends would be found as follows : 82 -^ 2 = 41 dents per inch in the cloth. 41 X 27 = 1107 — 12 for selvages equals 1095 dents required for the body of the cloth. 1095 -^ 39 dents per pattern = 28 patterns and 3 dents over. These dents may be used by adding them to the 1 1 dents of plain weave and drawing 7 dents the first time the pattern is drawn, leaving 7 dents to be drawn after the pattern has been drawn 28 times. 86 ends per patt. X 28 = 2408 ends. 2408 + 6 for the 3 dents + 48 for selvages = 2462 total ends required. Most leno cloths require two or more beams, either because the leno ends are of different counts than tlie ends in the other weaves in the cloth, or because the rate of contraction is not the same. In the cloth, Fig. 3, the leno ends are the same counts as the ends in the plain and satin weaves, and as the leno ends in this cloth cross but once in six picks, they will not contract any more than the plain weave. If the ends in the satin stripe were drawn two in a dent the contraction would be less on the stripe than on the plain cloth, but to obtain the warp satin effect it is necessary to crowd more ends in a dent, thus increasing the sley in the stripe. This crowding of the ends increases the .con- traction, making it possible to weave such clotlis from one beam. In the reproduction of cloth from samples it is some- times required that the cloth be reproduced on a different sley reed than that on which it was woven. The object may be to reproduce the cloth in a lighter weight without changing the yarns, or it may be to make it heavier ; in both cases, however, it is required that the width of the pattern l)e the same as is in the cloth sample. 90 When it is required to reproduce a cloth on a differ- ent sley reed, and at the same time, preserve the same size of pattern and general appearance of the cloth, it will be necessary to measure each section or stripe in the pattern and obtain the number of dents that would be required in the new reed in each of the different parts of the pattern. For an illustration. Fig. 1, which was made in a 64 sley reed and contains 1 1 dents plain and 4 dents satin weave will be reproduced on an 80 sley reed. Assuming that the plain stripe measures 34 hund- redths of an inch, the dents required in an 80 sley reed would be 80 -^ 2 = 40. 40 X .34 = 13.60 or 14 dents and 28 ends for the plain stripe ; then taking the measurement of the satin stripe as 13 hundredths of an inch, there would be 80 -^ 2 r= 40. 40 X .13 = 5.20 or 5 dents 4 ends in a dent, or 20 ends for the satin stripe ] one complete pattern would contain Dents. Ends. 14 28 5 20 19 48 A dimity cloth is one that has cords that run warp way in the cloth, and are formed by drawing several ends in one eye of the harness. These ends are run on a separate beam, on account of the difference in the contraction on them and other ends in the cloth. The pattern may be formed by having the cords drawn in regular order across the cloth or by having them drawn in the reed in groups, etc. Sometimes patterns of this kind are required that must count a given average sley in the cloth when 91 woven, and in that case, it is necessary to find the sley reed from the reed draft and the average sley required. To find the sley reed to use on a dimity pattern when the number of dents and ends in one pattern and the average sley cloth required is known : Rule. Multiply the average sley by the dents i)cr ■pattern and divide the result by the ends per pattern; multiply this result by two, ecpials the sley reed required. Examples. If a dimity cloth that contains 10 dents and 23 ends per pattern is required to be 116 sley when woven, what sley reed is required ? ^^^ ^ ^^ = 50.43 X 2 = 100.86 or 100 sley reed. 23 Find the sley reed and number of ends required to make a cloth 30 inches wide and 112 average sley, with the folio wins: reed draft. ;nts. Ends. B.B. T.B 6 12 3 2 3 2 3 11 16 9 ^^^ ^ ^^ = 49.28 X 2 = 98.56 or 98 sley reed. 25 Now find ends by rule previously given. 98 -f- 2 = 49 dents per inch as represented in the cloth. 49 X 30^^ — 1470 total dents requihed. 92 1470 — 12 for selvages = 1458 dents for the body of the cloth. 1458 -f- 11 dents per patt. = 132 patts. and 6 dents over, which may be used after the pattern has been drawn 132 times. 16 ends per patt. X 132 patts. = 2112 -f 12 for the 6 dents + 48 for selvages = 2172 ends in bottom beam. 9 ends per patt. X 132 patts. = 1188 ends in top beam. 2172 + 1188 r=: 3360 total ends. 3360 -f- 30^^ cloth = 112 average sley. The cords in a dimity cloth have a tendency to pre- vent contraction in width, and to obtain the correct width of the cloth about 8 dents less than the calcu- lated number should be used. In the last example, 1470 — 8 =: 1462 dents would be required. COMPARISON OF YARNS FROM CLOTH SAMPLES. When a cloth has to be reproduced from a small sample, the counts of the yarns in the cloth may be found by comparison with yarn of which the counts are known. It is the practice to pull out 6 or 8 ends from the cloth and to estimate the counts as near as possible ; then to take the same number of ends from a bobbin or cop which is of the same counts estimated for the ends from the cloth and pass one set of ends around the other, twisting each group of ends to form a cord of 12 or 16 ends. If the cords appear to be of equal thickness it is fair to assume that the counts of the yarn from tlie cloth and those from the bobbin or cop are the same. If the cord formed bv the ends of the known 93 counts is smaller tliai) the cord formed of those taken from the cloth, try a coarser counts, if it is larger, try a finer counts, until both cords appear to be the same size. In check patterns, where the fillinij; stripe is required to be the same width and have the same appearance as to weight, etc., of the warp stripe, it is necessary to put in more picks in tlie filling stripe than there are ends in the warp stripe, on account of the filling being finer than the warp. To find the number of picks required in a filling stripe to equal the warp stripe in weight : Rule. MultlpUj the number of ends in the warp stripe by the filling cmmts, and divide the result by the warp counts, result equals the number of picks reipiired in the filling stripe. Example. If a warp satin stripe contains 64 ends of 40s warp, how many picks of 60s filling will be required to weave a filling satin stripe to equal the warp stripe ? 64 X 60 =z 96 picks of 60s. 40 ^ PROOF. 7000 X 64 — — — z=z 13.33 grains weight of the 64 ends of 840 X 40 40s warp. 7000 X 96 -— — — - = 13.33 grains weight of the 96 picks of 840 X 60 (50, i^iiing. To find the average counts of filling when two or more different counts are used : Rule. Divide the number of picks of each counts by its own counts, add the results, and divide into the total jncks per pattern. 94 Example. If a cloth contains 38 picks of 15s and 360 picks of 60s filling in a pattern, what are the average counts of the filling ? 38 -^ 15 = 2.53 relative weight of 15s. 360 -^ 60 =: 6.00 ^' '^ " 60s. 398 8^ 398 4- 8.53 = 46.65 average counts of filling. It is the rule in mills, whenever it is possible, to pay the operatives by the piece. Weavers are paid by the cut, as a general rule, although in mills engaged in making very heavy cloths they are sometimes paid by the pound. A cut may contain almost any length, although 50 yards is about the average. The price paid for weaving will vary with the length of the cut, picks per inch, counts of the yarns, speed ol the loom, and number of looms it is possible for a weaver to run. The speed and number of looms will be governed by the counts of the yarns and width of the cloth, or, if it is a fancy cloth, the number of harnesses and difficulty of the pattern must be taken into consideration. Loss in time for mending threads and for repairs must also be allowed for, which amount will vary with the character of the cloth. On plain cloth, the loss in time will vary from 8 to 20 per cent., while on fancies, the loss may be as high as 35 per cent. The weekly rate or amount of wages it is desired that the weaver should earn is' then fixed, and the price per cut may be calculated as follows : First find the production of one loom in yards per week, multiply by the looms per set, result equals total yards ; divide by yards per cut, equals total cuts per week. Now divide the weekly rate by the cuts per week, equals the weaving price per cut. 95 To find the production of a loom : Rule. Mult'qybj the speed of the loom by the minutes per hour, result equals incks per hour ; multiply by hours per week, eipiah jyicks yer week ; multiply by jjer cent, of production, eijuals picks per iveek with loss deducted. Divide by picks per inch, equals inches per tveek ; divide by SQ, equals yards per week ; divide by yards per cut, equals cuts ivoven. Examples. How many 50 yard cuts will a loom weave in 58 hours with a loss in time of 15 per cent., the loom running 160 picks per minute on cloth containing 64 picks per inch ? 160 X 60 X 58 X 85% ,,r,, , . ^^ = 4.10 cuts per loom per 64 X 36 X 50 ^eek. Find the weaving price per cut from the following particulars : 76 picks per inch. 1 80 picks per minute. 85 per cent production. 8 looms per set. 50 yards per cut. 58 hours per week. $10.00 weekly rate. 180 X 60 X 58 X 8 X 85% ^ ^^^^ ^^^^^ ^^^, ^^^^ 76 X 36 X 50 on 8 looms. $10.00 ^ 31.13 = 32.12c. weaving price per cut. COST OF PRODUCTION. A large portion of the product of cotton mills is subject to a variation in the selling price which is fixed by the condition of the market. The prices on other 96 cloths are fixed by contracts, i. e., agreements to make cloths with given constructions at stated prices per yard. In both instances it is necessary to know the exact cost of producing a yard of cloth. In the first case this is necessary so that the margin between the market price and the actual cost may be known, while in the second case it is necessary to know what price to l)id on a new cloth when a contract is made, Contracts are sometimes made on standard goods, but on the finer grades of plain and fancies it is the rule. It is the custom in every cotton mill to keep an ac- count of the cost of production in every department ; the cost is obtained by dividing the production by the amount ot the pay roll and is called the labor cost. The cost of certain kinds of supplies can very readily be charged to tlie departments to which they belong, but it may be difficult to apportion a proper share of the general expense of running the mill to any one department, so for convenience the cost of supplies and general expense for the whole mill are added together and charged to the weaving room, — the whole expense divided by the number of looms, equals the cost per loom for supplies and general expense. This amount will vary in different mills and will average about $1.35 per loom per week on plain to $1.80 on fancies. Weavers, as a general rule, are paid by the piece, but the overseer, second hand, loom fixers, filling men, etc., are paid by the day; the amount thus paid is divided by the number of looms in the mill and the result is the amount paid for oversight, and varies from about 30c. to 36c. on plain and 40c. to 46c. per loom per week on fancies. 97 The cost of the cotton and the lal)Oi- cost on the yarns being known, the cost per lb. of a cloth may be found by addinjx the additional cost of weavinu;, oversight, and general expen.^c. The cost per yard may be found by di- viding the cost per lb. l)y the yards of the cloth in one lb. The above method for finding cost will be illustrated by an example as follows : Find the cost of production of a cloth from the following particulars. 84 X 1)2. 32 inch cloth, 34 ipxhes at reed ) 60s warp 4% contraction \ estimated. 80s filling. 6 looms per set. 160 picks per minute speed. 85% production. $11.00 weekly rate of wages. .16 cotton. .07 labor cost on warp. .08 " '^ " filling. $1.50 general expense per loom per week. .40 oversight '' " " i.' 58 hours per week. 84 X 32 + 48 = 2736 ends. 2736 X 104 ———--—- = 5.64 lbs. of warp in 100 yards of o^u X bu Q^^^Yi. ^4c" X 92 X 100 l^Af^Zro^ == ^-^^ ^^'^- «f filing in 100 yards 840 X 80 of cloth. 5.64 + 4.65 = 10.29 lbs. weight of yarns in 100 yds. of cloth. 100 yards -^ 10.29 lbs. = 9.71 yards per lb. 5.64 lbs. of warp -f- 10.29 lbs. total weight = b^% warp. 4.65 lbs. of filling -^ 10.29 lbs. total weight == 45% filling. 98 16c. cotton X 55%=.0880 cost per lo. of stock for warp 16c. " x45%=.0720 '' " " " " '' filling .1600 total cost per lb. of stock. 7c. labor cost on warpX 55% =.0385 labor cost per lb. on warp. 8c. labor cost on fillingX 45% =.0360 labor cost per lb. on fillins:. .0745 total labor cost on stock. 160 X 60 X 58 X 85% ^ .^ .. , ^ , — ^-^ = 142.89 yards on 1 loom 92 X 36 pgj. ^veek. 142.89 X 6 = 857.34 yards on 6 looms. 111.00 ^ 857.34 = .0128 weaving cost per yard. 40c. -^ 142.89 = .0028 oversight " - .0156 total cost of weaving and oversight per yard. .0156 X 9.71 yards per lb. = .1514 cost of weav- ing and oversight per lb. $1.50 ^ 142.88 yards per week on 1 loom = .0105 cost of general expense per yard. .0105 X 9.71 yards per lb. = .1019 cost of general expense per lb. Summary of Costs : Stock .1600 Labor cost on stock .0745 Weavipg and oversight .1514 General expense '.1019 .4878 cost per lb. .4878 cost per lb. -^ 9.71 yards per lb. = .0502 cost per yard. 48.78c. per lb. 5.02c. " yard. 99 EXAMPLES FOR PRACTICE. Find the ends required for a cloth 56 sley, 36 inches wide, allowing 36 extra ends for selvages. 2016 ends. Find the dents per inch required in a reed for a 72 sley cloth, with the ends drawn 2 in a dent. 33.72 dents. Find the dents per inch required in a reed to weave a cloth 114 sley, with the ends drawn 4 in a dent. 26.84 dents. If a reed contains 1077 dents and is 36 inches long inside the reed bars, what sley cloth will it weave with the ends drawn 2 in a dent ? 64 sley. If a cloth is 76 sley and 30 inches wide when woven, with 48 ends in 12 dents, extra for selvages, how wide will it be in the reed ? 32.34 inches. Find the hanks of filling in 100 yards of cloth woven 68 X 72, — 30 inches wide. 277.97 hanks. Find the lianks of warp in 100 yards of cloth woven 48 sley, 30 inches wide, 48 ends extra for selvages, and 7 per cent, contraction. 189.54 hanks. If a cloth contains 3000 ends of 60s warp, which contract 6 per cent, in weaving, what weight of yarn is required to weave 100 yards of cloth ? 6.31 lbs. How many lbs. of 36s filling is required to weave 100 yards of cloth, 36 inches wide in the reed, and 60 picks per inch? 7.14 lbs. If a cloth contains 2000 ends of 60s warp, with 6 per cent, contraction and 7 per cent, sizing, 250 ends of 2 / 20s warp, with 12 per cent, contraction, what weight of warp yarn will be required to weave 50 yards of cloth? 3.90 lbs. l.ofC. 100 If a cloth is woven 68 X 72 — 45s warp — 60s filling, find the average number by short method. 51.66 average number. If a cloth is woven 36 inches wide at the reed and contains 68 picks per inch of 45s fiUincr, and the following ends, what is the average number ? 1800 ends of 40s with 5% contraction. 250 " ''2/35s " 12% " 340 " ^•3/15s " 20% " 30.75 average number. Find the average number on a cloth woven 64 X 64 — 28 inches wide — 7 yards per lb. (Use short method.) 33.44 average number. Find the average number in the previous example by long method, and allow 7 per cent, for contraction on warp and 6 per cent, for size, 30 ends extra for selvages and 12 dents. 33.25 average number. Find the yards per lb. from the following particu- lars : (By long method.) 76 X 80 — 30 inches wide — 60s warp — 95s filling — 5% warp contraction and 5% size. 12.01 yards. Find the yards per lb. in the previous example. (By short method.) 12.09 yards. Find the yards per lb. from the following particu- lars : 1466 ends of 40s with 6% contraction and 5% size. 100 " "2/20s " 23% 100 '' ^'2/20s '^ 15%" 30 inches wide at reed with 68 picks of 50s filling. 7.98 yards. If a piece of cloth, 3x3 inches, weighs 9 grains, find the yards per lb. if woven 30 inches wide. 6.48 yards. 101 If the cloth in the previous example was woven 56 X 60, find the average number. (By short method.) 30.06 average number. If a yard of cloth weighs 896 grains, how many yards will weigh 1 lb.? 7.81 yards. If a quarter yard of cloth weighs 250 grains, how many yards will weigh 1 lb. ? 7 yards. Find the counts of fillinsr required to weave a cloth to weigh 11 yards per 11)., constructed 76 X 84, 30 inches wide, 60s warp witli ^\% warp contraction and 6% size. (By long method.)" 81.05 filling. If a cloth is woven 30 inches wide in the reed, with 76 picks per inch, and requires the following ends, what counts of filling is required to make the cloth weigh 6i yards per lb. ? 1950 ends of 50s with 5% contraction. 300 ^' - 2/20s '• 12% 200 " '^2 /40s '' 25% ^• 57.26 filling. If a piece of cloth, 4x4 inches, weighs 20 grains and counts 64 X 72, make the cloth 30 inches wide, using a warp 4 numbers coarser than the average numbers, and find the counts of filling required. (Use short methods.) ' 33.18 filling. Find the weaving price per cut from the following particulars : Speed of loom 180 picks per minute, 72 picks per inch, 7 looms per set, 90% production, 50 yards per cut, 58 hour week and $10.00 weekly rate. 32.84c. per cut. Find the per cent, of warp and filling in a cloth woven 72 X 78, 60s warp and 100s filling. 60% warp. 40% filling. 102 Find the per cent, of warp and filling in a cloth woven 30 inches wide at the reed, with 76 picks of 70s filling and the following number of ends : 1850 ends of 60s with 5% contraction. 250 - ^^ 2/ 15s '' 15% 300 " -^ 2 /40s " 25% '^ 73% warp. 27% filling. MISCELLANEOUS. To find the speed of a loom : Rule. MuUiphj the speed of the driving shaft by the diameter of the driving pulley ^ and divide the result by the diameter of the loom pulley. Example. If a loom has a 14 inch pulley and is driven by an 8 inch pulley on the driving shaft which makes 280 revolutions per minute, what is the speed of the loom ? 280 X 8 = 160 picks per minute. 14 To find the size of loom pulley : Rule. Multiply the speed of driving shaft by the diameter of driving pulley, a?id divide the result by the speed required. Example. If the speed of the driving shaft is 280 R. P. M. and the diameter of the driving pulley is 8 103 inches, what size loom pulley is required to liave the loom make 160 picks per minute? 280 X 8 ^14 inch pulley required. To find the size of driving pulley : Rule. Multiply the speed required by the diameter of loom pulley and divide the result by the speed, of the drivinor shaft. Example. What size driving^ pulley is required on the driving shaft which makes 280 R. P. M. to drive a loom 160 picks per minute, the loom having a 14 inch pulley ? := 8 inch pulley. 280 To find the production of a loom : Rule. Multiply the speed of the loom by the number of minutes it has run, divide the result by picks per inch, and this result by 36 inches per yard. Example. How many yards will a loom weave in 10 hours, running 160 picks per minute, on a cloth that contains 60 picks per inch ? 160 X 10 X 60 ^^ ^^ ^ = 44.44 yards. 60 X 36 *" It is necessary to make an allowance for stoppages on a loom for loss in time, in piecing up ends, changing shuttles, etc. If the loom in the last example were stopped 10 per cent, of the time, the actual production would be as follows : 160X10X60X.90 ^^ ^ =z 40 yards. 60 X 36 ^ 104 When the production per week is desired, the method may be shortened by cancellation. The hours and minutes per week are placed on the top line and the inches per yard under it, and, as they will be present in every calculation, they may be cancelled as follows. (Mass. has a 58 hour week.) 96.66 ^12^ = 96.66. So that the method with a 58 hour would be Speed of loomX 96.66 X % production ^ =yds. per week. picks per inch. For a 60 hour week the cancellation would be 100 0X11 = 100. Use 100 in place of 96.66 as in the case of a 58 hour week. To find the number of teeth required in the pick gear on a loom : The simplest way is to find the constant, i. e., the number of teeth required in the pick gear to put in one pick per inch ; then use the constant to find the teeth required for any other number of picks. To find the constant : Rule. Multiphj all the driven gears togetlter and divide the result by the product of the drivers and cir- aimference of sand roll in indies. Answer equals the constant. Leave the pick gear (and carrier gears if there are any) out of the calculation. 105 When the pick gear is a driver, the constant will be a dividend ; when it is a driven gear, tlie constant will be a divisor. When the constant is a dividend, divide the constant by the picks required, and answer equals teeth required in pick gear. When the constant is a divisor, divide the picks per inch required by the constant, and answer equals the teeth required in the pick gear. Ratchet gears are always driven gears and the sand roll is always a driver. When the take-up motion is driven from tiie bottom shaft, multiply the product of the driven gears by two before dividing by the product of the drivers. To set the harnesses (on a plain loom) for time and position : Turn the loom crank to bottom center, loosen the cams and turn theni until the harness treadles are level ; then turn the loom crank to top center and make adjustments on the harness straps to bring the shed, that is down, to a point almost touching the race board ; turn the loom over and adjust the other slied the same way. When cover is required, set the harnesses a little earlier, i. e., have them level with tlie loom crank a little back of tlie bottom center, also raise the wliip-i'oll so that it wili be a little above the level of the harness eyes when they are level. On twills or sateens, have the loom crank on the bottom center, then loosen the cams and turn tliem until any two of the treadles that are changing become level ; make adjustments to obtain the correct position of the shed as directed for a plain loom. 106 To set the picking motion : Have the loom crank on the top center, loosen the picking cam and bring it up against the picking ball or cone, and make it fast, turn the loom until the picking stick is brought forward, adjust the lug straps to bring the stick within one and a half inches of the bunter. (This distance will vary on different makes of looms.) Try the loom and get more or less pick by moving the . lug strap up or down on the stick, taking up or letting out the lug straps, or moving the picking cam in or out on the shaft. To set the protector rod : Bring the lay forward until the dagger strikes the bunter, set the fingers on the protector rod against the shuttle box binders and make them fast; adjust the spring on the protector rod to give the tension on the binders ; put the shuttle in the box and adjust the binder (or box front) until the shuttle opens the binder sufficient to have the dagger clear the bunter a quarter of an inch ; set both sides the same way and set the check on the protector rod to hold the shuttle firmly in the box. To set the filling stop motion : Turn the loom to the front center and have the shuttle on the filling fork side of the loom ; loosen the cam that operates the stop motion finger ; and set it to raise the lever and move the finger forward, then turn the loom until the lever rests on the smallest diameter of the cam; adjust the finger so that there will be a quarter of an inch between the end of the finger and the hook on the fork, when the lay is forward see that the tines of the fork do not touch the grate bars. To time the harnesses on a dobby : Have the loom crank on the bottom center, loosen the dobby crank and place it on either back or front 107 center; adjust the connecting rod until the rocker arms are vertical ; turn the loom until the bottom knile is at its extreme inward position and set the knife a quarter of an inch back of the hooks; turn the loom over and set the top knife in the same manner. If the throw of the knives were changed to obtain either a larirer or smaller shed, they must be readjusted to bring them a quarter of an inch l)ack of the hooks. To time the pattern chain cylinder on a dobby, with a worm drive : Turn loom until bottom knife is at its extreme inward position ; loosen the cylinder and turn until the pegs lower the hooks on the knife ; set the worm with the straight part of the tooth passing through the cylinder gear ; then tighten all the parts. To time the pattern chain cylinder on a dobby with a pawl and ratchet drive, when the pawl pulls the cylinder : Have the bottom knife at its extreme inward po- sition, loosen the cylinder and turn until the pegs lower the hooks on the knife ; loosen the ratchet gear and place it in contact with the pawl ; set the stop or check wheel to hold cylinder stationary while the pawl reaches for another tooth ; tighten all the parts. When the pawl reaches for a new tooth, have throw enough on the pawl to pass a quarter of an inch beyond the tooth so that the top knife will catch tlie hooks before the cylinder is moved. On box looms, the boxes should commence to move when the loom crank is at the bottom center and the shuttles are at the box end of the loom, and should become stationary when the loom crank reaches the top center where the loom commences to pick. INDEX. Yarn Calculations o To find the counts of a cotton yarn when the length and weight are known 6 Reelin(; Yarn 7 To tind the cotton counts from irregular or short lengths of yarn 8 To find the length in yards when the counts and weight are known 10 To find short lengths when counts and weight in grains are known 11 To find weight when length and counts are known .... 12 To find the weight of short lengths of cotton yarn . . . lo Spun Silk I'-i Raw Silk 15 To find the dram silk counts when length and weight in lbs. are known 17 To find length when counts and weight are known .... 17 To find the weight when the length and counts are known IS To find counts when length and weight in ounces are known li> To find length when counts and weight in ounces are known 1-^ To find weight in ounces when length and counts are known ^0 To find the dram silk counts when length and weight in grains are known 20 To find length when counts and weight in grains are known -0 To find weight in grains when length and counts are known 21 To find the denier counts of raw silk, when length and weight in lbs. are known 22 To find length when weight in lbs. and counts are known . 22 To find weight when length and counts are known .... 28 To find the equivalent counts of denier silk in dram silk counts 24 To find the equivalent counts of dram silk in cotton counts 24 To find the equivalent counts of a denier silk in cotton counts 25 Worsted Yarn 25 To find the worsted counts from irregular or short lengths 26 Woolen Yarn 27 To find counts, length, or weight, apply rules given for cotton and worsted yarns (except those for short lengths) and use the standard length in the system desired ... 28 Woolen Cut Yarn 28 To find the cut counts from short lengths 29 Woolen Run Yarn 29 To find counts from short lengths 29 To find the counts when the weight is known in ounces . 30 To find the weight in ounces when the length and run counts are known 30 To find the length when counts and weight in ounces are known 31 Linen Yarn 31 To find the turns of twist per inch 31 To find the standard breaking weight of warp yarn . ... 32 Equivalent Counts 33 To find the equivalent counts of a yarn in any other system (except raw silk) 33 Ply Yarns 34 Resultant Counts 34 To find the resultant counts of a ply thread when two or more threads of different counts are twisted together . 35 To find the cotton counts of yarn required to twist with a known count of single yarn, to produce a given count of ply yarn 36 To find the weight of each counts of yarn to obtain a given weight of ply yarn, when the threads twisted are of un- equal counts 37 Cost of Ply Yarns 39 To find the cost per lb. of a ply yarn composed of threads of different counts and values 39 Examples for Practice 42 Cloth Calculations 47 To find the number of ends required in a warp when the sley and width are given 48 To find the dents per inch for any sley cloth 50 To find the sley cloth a reed will weave when length and total dents in the reed are known 'A To find the width of the warp ends in the reed ')■] To find the weight of filling in a given length of cloth . . 54 To find the weight of filling of each color in a checked gingham when the picks of each color in the pattern, picks per inch, width at reed, and counts of filling are known 55 To find the weight of filling required for a given length of cloth when two different counts are used 5() Warp Contraction 57 To find the approximate per cent, of contraction from warp to cloth 58 To find the per cent, to allow for contraction from a cloth sample 51) To find weight of warp yarn in a given length of cloth . . (iO Average Numbers (51 To find average numbers when sley, picks, width, and yards per lb. are known (VI To find the average numbers when sley, picks, and counts of warp and filling are known (58 To find average numbers when the cloth contains more than one counts of warp, and the number of ends of each counts is known, with picks per inch, filling counts, and width of cloth at reed (54 To find the average number when the number of ends and picks per inch, width at reed, and yards per lb. are known (55 To find average number on plain cloths when sley, picks, width and yards per lb. are known <57 To find the average counts of warp yarns when two or more different counts are used (i8 To find the yards per lb., when sley, picks, width and counts of warp and filling are known . . . • (58 To find the yards per lb. when the sley, picks, width, and counts of warp and filling are known 70 To find the yards per lb. from short lengths of cloth ... 71 To find the counts of filling required to govern the yards per lb., when sley, picks, counts of warp, width and yards per lb. are known 72 To find the filling required to govern the yards per lb. (on plain cloth) when the sley, picks, width and yards per lb. are known 74 To find the approximate counts of warp and filling from a small sample of ordinary plain cloth 75 To find the per cent, of warp and filling in a piece of cloth, is to find the weights of the different yarns used in its construction and divide the weight of each by the weight of the whole 70 To find per cent, of warp and filling when ends, picks, and counts of warp and filling are known 77 To find the per cent, of warp and filling in a cloth when sley, picks, average counts, and warp counts are known . 78 To find the per cent, of warp and filling when the cloth contains more than one counts of warp, and the counts of warp and filling and width at reed are known .... 78 To find the number of ends in a cloth with an irregular reed draft, when the ends in one pattern, the sley reed, and the width of the cloth is given 80 To find the number of dents in a given space, when the sley reed is known 88 To find the sley reed when the number of dents "and the space they occupy in the cloth, measured in lOOths, is known 84 To find the sley reed to use on a dimity pattern w4ien the number of dents and ends in one pattern and the average sley cloth required is known 91 Comparison of Yarns from Cloth Samples .... 92 To find the number of picks required in a filling stripe to equal the warp stripe in vyeight 93 To find the average counts of filling when two or more different counts are used . 93 To find the production of a loom 95 Cost of Production 95 Examples for Practice 99 Miscellaneous 102 To find the speed of a loom 102 To find the size of loom pulley 102 To find the size of driving pulley 103 To find the production of a loom 103 To find the number of teeth required in the pick gear on a loom 104 To find the constant 104 To set the harnesses (on a plain loom) for time and position 105 To set the picking motion 106 To set the protector rod . . . . • 106 To set the filling stop motion 106 To time the harnesses on a dobby 106 To time the pattern chain cylinder on a dobby, with a worm drive .'.... 107 To time the pattern chain cylinder on a dobby with a pawl and ratchet drive, when the pawl pulls the cylinder . . 107 SfP 7 1904