If it is required to know the number of grains which 120 yards of any count should weigh, the method of procedure is the reverse of the foregoing.

When testing the counts of cops, it is usual to wrap two, three, or four cops, in order to arrive at a more satisfactory test.

If two leas, or two-sevenths of a hank, are weighed, the counts can be obtained by dividing the weight into 2000 grains, or two-sevenths of 1 lb. If three leas, or 360 yards, are weighed, divide the weight into 3000 grains, and the result is the counts. If 480 yards are weighed, the dividend is 4000; if 600 yards, or five leas, are weighed, the dividend will be 5000; if six leas, or 720 yards, are weighed, the dividend is 6000; and when seven leas, or one hank, is weighed, the dividend will be 7000 grains, or 1 lb.

As it takes a considerable time to take 120 yards of weft out of a piece, a shorter length is often weighed and the counts found therefrom. A balance is extensively used which registers the counts when twenty yards of yarn are put upon the pointer. This is a very useful, though not always accurate, method.

When any odd length of yarn is weighed, the counts may be obtained by proportion, thus—

If 34 yards of yarn have been found to weigh 8 grains, what count is it?

The yards in 1 lb. can first be found as follows:—

and this divided by 840 will give the counts, thus:—

29750840 = 35·41 counts.

From this we get the formula:—

7000 × yards weighed840 × counts = counts.

This is a very useful formula, as when only a small piece of cloth is available to be tested it is necessary to get as near as possible to the counts from weighing sometimes only 10 or 15 yards, or any odd length.

A calculation may occur in the following form:—

How many grains should 16 yards of 20’s cotton weigh?

There are 840 × 20 = 16,800 yards of 20’s in 1 lb., or 7000 grains.

Then if 16,800 yards weigh 7000 grains, how many grains will 16 yards weigh?

This may be stated in a formula as follows:—

7000 × yards weighed840 × counts = weight in grains.

Staub’s Yarn Balance is a small balance which is made to test the counts of very small quantities of yarn. A template is given with the balance, and the yarn is cut into lengths the size of the template, about two inches. One end of the balance is slightly heavier than the other, and the number of threads the size of the template which are required to draw the balance indicate the counts of the yarn. If twenty threads or about 40 inches balance the small weight, the count of the yarn is 20’s, and so on.

The principle is the same as if a 1 lb. weight were put on one end of a balance, in which case the number of hanks required to draw the weight would indicate the counts, because if 20 hanks = 1 lb. the counts are 20’s, and if 21 hanks = 1 lb. the counts are 21’s. The balance may be made to weigh any length, according to the weight on one end of the balance.

The form in which it is usually made makes it specially suitable for testing the counts in small patterns of a few inches.

The test is, of course, only approximate, as could only be expected from weighing so short a length.

If the foregoing examples are thoroughly understood, the following will not be found difficult.

If a warp has 2000 ends, and is 500 yards long, and weighs 60 lbs., what counts is it?

The ends multiplied by the length will give the total length of yarn in the warp, and this divided by 840 will give the hanks. If the hanks are divided by the weight, the result will be the counts. The result may be obtained at once as follows:—

2000 × 500840 × 60 = 19·84 counts.

If a beam has 2200 ends, the counts being 40’s, and the weight 50 lbs., find the length.

By multiplying 40 by 840 the yards in 1 lb. are obtained, and multiplying this by 50, the yards of yarn on the beam are arrived at. If this is divided by the ends in the warp, the result will be the length of warp thus:—

40 × 840 × 502200 = 763·6 yards.

A simple method of mentally calculating the number of hanks in a piece is as follows:—

A warp 84 yards long will contain just one-tenth as many hanks as ends. Thus a warp of 2000 ends, 84 yards long, contains 200 hanks. This can be proved as follows:—

2000 × 84840 = 200 hanks.

The number of hanks in a warp 84 yards long can thus be seen at once, and it is a very simple matter to mentally calculate the difference for any other length.

The hanks of weft can also be calculated mentally in a similar manner.

If the piece is 84 yards, the counts multiplied by the width and divided by 10 will give the number of hanks required for 84 yards. Thus, find the hanks of weft in a piece 34 inches wide, 84 yards long, 60 picks per inch.

60 × 3410 = 204 hanks.

The calculation is really simpler than it looks in the above form, as the dividing by 10 can be done by simply pointing off the last figure in the product of the picks and width. The formula may be proved correct by working out fully as follows:—

34 × 84 × 60840 = 204 hanks.

This system of mentally calculating the hanks is very useful, as it serves as a check upon a full calculation.

The Firmness of Cloth.—The number of ends and picks per inch which can advantageously be put into a fabric depends upon the number of intersections per inch in the pattern or weave, and on the counts or diameters of the yarns used. In a plain cloth woven with 32’s twist and 32’s weft, the number of threads per inch which could be put into the cloth without undue compression would be a little more than one-half the number which could be laid side by side touching each other. The reason for this is that the warp and weft threads interlace with each other every pick, and therefore, supposing that 156 threads of 32’s occupy one inch when laid side by side, one-half of these threads would have to be left out to allow of the intersection of the weft between every end.

In a “two and two” twill the weft intersects once for every two ends, or twice in the pattern; therefore there are four threads and two intersections in the pattern. It is obvious, therefore, that to keep the same firmness in the twill as in the plain cloth with the same yarns, a larger number of threads per inch both in warp and weft will be required.

To keep the same “firmness” the threads must be kept as close together in one cloth as in the other, and as in a plain cloth one-half the threads which occupy one inch are dropped out, so in a twill with two intersections for four ends there must be one-third of the ends occupying one inch left out. Thus with 32’s yarn, of which the diameter is 1/156 of an inch, there will require to be about 102 threads per inch in a “two and two” twill.

A perfectly balanced plain cloth may be defined as a cloth in which the warp and weft yarns are equal in diameter, and the spaces between the threads are equal to the diameter of the yarn.

If the diameters of yarns of various counts are known, it is an easy matter to find the number of threads per inch which will produce the desired firmness in any simple weave.

The diameters of yarns of cotton, woollen, worsted, and other threads are given by the late Mr. T. R. Ashenhurst in an excellent little work on “Textile Calculations and the Structure of Fabrics,” which has done much to promote this branch of the art of weaving.

Mr. Ashenhurst estimates the diameter of a 32’s cotton yarn at the 1148th part of an inch; but this is probably somewhat under the mark, and in the following table I have taken 1156th inch as the diameter of 32’s.

The variation in the thickness of any yarn, and the fact that they are not strictly cylindrical, renders measurements of little avail, but taken in conjunction with an examination of a range of woven cloths, the approximate or practical diameter can be estimated.

TABLE OF DIAMETERS OF COTTON YARNS.

The preceding is a table of the diameters of cotton yarns from 1’s counts to 200’s. The number given as the diameter is the number of threads which occupy the space of one inch when laid as close together as possible without compression.

A perfectly balanced plain cloth will require one-half this number of threads per inch, plus, perhaps, 5 per cent. for the threads being forced somewhat out of the same plane in weaving.

Relative Diameters of Yarns.—The “counts” of yarns indicate the number of hanks in 1 lb., and therefore a given length of 30’s is twice as heavy as the same length of 60’s; but the diameter of the 30’s will not be twice that of the 60’s, as the yarns are cylindrical, and the diameters will vary as the square roots of the areas, which in this case are as 1: 2.

If one thread is four times as heavy as another, and if it is of the same density—which in these calculations is assumed, although it is not strictly correct—the diameters of the two threads will be as 2: 1. For example, looking at the tables, the diameter of a 60’s is seen to be the 1/213 of an inch, whilst the diameter of a thread four times the weight, viz. 15’s, is seen to be 1/106½ of an inch, or exactly twice the diameter of the 60’s thread.

The diameter of one yarn being known, the diameter of any other may be obtained by the following rule:—

Rule.—As the square root of one count is to the square root of another count, so is the diameter of one to the diameter of the other.

In this form the calculation necessitates the extraction of two square roots, and with most numbers would require the use of two fractions in the calculation. By squaring all the three terms the calculation is much simpler, as in the following example:—

As the diameters of yarns vary as the square root of their counts, it follows that the diameters will always bear a certain relation to the yards in 1 lb. If this relation is once obtained, it becomes easy to calculate the diameter of any yarn on this principle.

Taking the diameter of a 32’s yarn from the table, viz. 156, it will be found that this is equal to the square root of the yards in 1 lb., less 5 per cent.

The number of ends and picks per inch required to make plain cloths of equal firmness from different counts may be at once seen from the table of diameters, as one-half the number given as the diameter is required.

Thus if a plain cloth with 78 threads per inch of 32’s is taken as the standard, and it is required to make a cloth of equal firmness, with 60’s yarns, the number of threads per inch required would be 106½. In 20’s yarns about 62 threads would be required. In 16’s yarns 55 threads per inch, and so on.

In twills, or other regular weaves, the following rule will give the number of threads per inch required of any count:—

Rule.—As the sum of the ends and intersections in the pattern is to the ends, so is the diameter to the number of threads required.

One of the most useful purposes to which a knowledge of this principle can be put is in changing the weave of a fabric, to find the threads per inch of a given count of yarn required to keep the same firmness as in a sample cloth.

It must be remembered that the word “firmness” is here used as implying that the space between the threads bears the same relation to the diameters of the threads in both cases, or, if the given cloth is perfect, the proposed one will also be perfect.

Suppose it is desired to make a “two and two” twill of the same “firmness” as a plain cloth made with 103 threads per inch.

The yarns being the same, the number of threads per inch required will be as the ends plus intersections in a given number of ends in both patterns.

In the above question the given cloth is plain, with 103 threads per inch, and the proposed cloth is a “two and two” twill. Taking the same number of threads in each case, we get—

It must not be forgotten that it is necessary to take an equal number of ends of each pattern in this class of calculation. In more complex patterns it is often advisable to take the number of ends which is the L.C.M. of the ends in the two patterns in order to get a complete number of intersections in each case.

If it is required to make a cloth with the same number of threads as a sample cloth, and to change the pattern and keep the same firmness, it is necessary to change the counts on the following principle:—

Rule.—As the sum of the ends and intersections in the sample cloth is to the sum of the ends and intersections in the proposed cloth, so is the square root of the counts in the sample to the square root of the counts in the proposed cloth.

This may be proved correct by referring to the table of diameters on page 335, where it will be seen that a plain cloth with 82½ threads per inch of 36’s is “perfect,” and a “two and two” twill with 82½ threads of 20¼’s counts is equally perfect.

To change the Counts, the pattern and threads per inch remaining the same.

If a sample cloth has 78 threads per inch of 32’s yarn, and it is proposed to make a cloth of the same weave with 55 threads per inch, what counts of yarn are required to keep the same “firmness”?

This is simple enough. The diameters of yarns vary as the square root of their counts, and therefore as the threads in one cloth are to the threads in another, so will the square root of the counts in one be to the square root of the counts in the other.

On referring to the table of diameters (p. 335), it will be found that a plain cloth with 78 threads of 32’s is “perfect,” and that a plain cloth with 55 threads of 16’s is also perfect. Therefore the above calculation is correct.

To change the Threads per Inch, the counts and pattern remaining the same.

If a sample has 78 threads per inch of 32’s, and it is proposed to weave a cloth of the same pattern, but with 60’s yarns, find the number of threads per inch required to keep the same firmness.

This is simply a continuation of the previous statement.

If the two counts are known, the number of threads will vary as the square roots of the counts; thus—