How to determine Yarn Count

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How to determine Yarn Count


Calculation of  Yarn Count

Simplified Yarn Calculation:
By definition one’s count mean a bundle of 840 yards which weights one pound. Similarly 10s count mean 10 bundles (each of 840 yards length) in one pound. It should be borne in mind that,

1 pound = 7000 grains
1 penny weight (dwt.) = 24 grains
1 lea =120 yards
7 leas or 840 yards =1 hank

From the above a lea of 1s count weight is 1000 grains.

If 1000 grains is the lea weight the count is 1s

And If 50 grains is the lea weight the count is 1000/50 = 30s

And If 25 grains is the lea weight the count is 1000/25 = 40s

And so on.

Thus the constant of 1000 becomes very useful for finding out yarn counts by obtaining the lea weights.

Note: If two or three leas weight is given ,to find count either find out one lea weight and divide 1000 by one
lea weight; or multiply 1000 by the given number of leas and then divide the result by total given leas weight.



Mathematical Problem: 01
Given the weight of 5 leas equal to 100 grains, Find the yarn count?

Solution:

By first method weight of one lea = 100/5 = 20grains

And count =1000/20 = 50s

By second method 1000 x 5 leas = 5000

And count 5000/100 = 50s (ANS).

The same can be conveniently used even to find the count of a given length of yarn shorter than a lea.

Mathematical Problem: 02
If a 24 yards length yarn piece weights4 grains what is the count ?

Solution:

24 yards weight is 4 grains

120 yards weight is (4 x 120)/20 = 20 grains

And count = 1000/20 = 50s (ANS).

Conversely, If count is given the lea weight can be quickly determined by dividing the constant 1000 by the given count.

Mathematical Problem: 03
What is the lea weight of 30 s yarn ?

Solution:

Weight is =1000/30 = 33 1/3 grains (ANS).

Even short lengths weight can be calculated if the count is known.

Mathematical Problem :04
What the weight of 15 yards of 40 s yarn?
 

Solution:
 
Weight of 1 lea (120 yards) = 1000/40 = 25 grains
 
So 15 yards weight = (25 x 25)/120 = 1 7/3 grains. (ANS).
 

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