Indirect count system and relation between unit length and unit mass in indirect system

This article is about Indirect count system and about the  relation between unit length and unit mass in indirect system. 

•  In this system, The length of a fixed weight of yarn is measured.
• In the other, The length per unit weight is the yarn count.
• So overall, The common features of all indirect count systems are the weight of yarn is fixed and the length of yarn varies according to its fineness.

The below formula  is use calculate the yarn count:

N= (L×w)/(l×W)

Where, N= yarn count of numbering system
              W= Weight of the sample
              L= Length of the sample 
              w= Unit of weight (sample)
               l= Unit of the sample

Relation between unit length and unit mass in indirect system

Numbering system	Unit of length (l)	Unit of mass (w)
English cotton count, NeC (Ne

B) 840 yards 1 pound (1b) Metric count, N_m 1000 meters/1km 1 kg Woolen count (ysw) 256 yards 1 pound (1b) Woolen count (Dewsbury) 1 yard 1 ounce (oz.) Worsted count, NeK 560 yards 1 pound (1b) Linen count, NeL 300 yards 1 pound (1b).

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  In brief definition of the above system is as follows:

  
  

  English cotton count,NeC	No of 840 yards length per pound
  Metric count,Nm	No. of kilometers per kilograms
  Woolen count(ysw)	No of 256 yd length per pound
  Woolen count(Dewsbury)	No. of yds. lengths oz
  Worsted count,NeK	No. of 560 yds. per pound
  Linen count,NeL	No. of 300 yds. lengths per pound

  
  
  The Nₑ indicates how many hanks of 840 yards length weigh one English pound.
• So that 32Nₑ means 32 hanks of 840 yards i.e. 32×840 yards length weigh one pound.
• The Nₘ indicates how many hanks of 1000 meters length weigh one kg.
• So that 50 Nₘ means 50 hanks of 1000 meters i.e. 50×1000 meters length weigh one kg and 100 Nₘ means 100 hanks of 1000 meters i.e. 100×1000 meters length weigh one kg.
• From above discussion it is concluded that, higher the yarn number (count) finer the yarn and lower the number coarser the yarn. 

In practice, 3 problems have to be solved in yarn calculations:

• To find Count, length and weight have to know. 
• To find Weight, count and length have to know.
• To find Length, count and weight have to know.

Example 1: On a cone, there are 9800m yarn which weight 490gm. What is the Nₑ, Nₘ Tex and Denier of the yarn?

Solution: 

For Nₑ

             We know that, Nₑ = (L×w)/ (l×W)

                                                           Here, L= 9800m 

                                                     W = 490gm

                                   w = 1 lb = 453.6

l = 840yds= 840× 0.90

                     Nₑ = (9800×453.6)/(840×0.91×490)

                     Nₑ= 11.87

                     Nₑ= 12

 For Nₘ
We know that, Nₘ= ( L×w)/ ( l×W)

Here, L= 9800m

W= 490gm

w= 1 kg = 1000gm

                          Nₘ = (9800×1000)/ (1000×490)

                         Nₘ = 20

For Tex

We know that, Tex = (W×l)/ L

                               Here, L = 9800m

                                             W = 490gm

                                            l = 1000m

                              Tex = (490× 1000)/ 9800

                               Tex = 50

 For Denier

We know that, Denier= (W×l)/L

          Here, L = 9800m

                        W = 490gm 

                      l =  9000m

                               Denier= (490×9000)/9800

                                Denier= 450