**Yarn Numbering (Count) System**

**Md. Ahsan Ul Habib (Saikat) ^{1} & Md. Amir Khasru^{2}**

B.Sc. in Textile Engineering

^{1,2},

Department of Yarn Engineering, SARSTEC

^{1,2}

Email: santo.amirkhasru2002@gmail.com

^{2}

**What is Yarn Count / Linear Density?
**Count is the numerical expression which expresses fineness of yarn. A definition of

**Yarn count**is given by the textile institute,

“Count a number of indicating the mass per unit length or the length per unit mass of Yarn”.

**Types of Yarn Numbering (Count) System:
**Basically two types of yarn counting system –

Direct system: Tex, Denier, lb/spindle

Indirect system: English, Metric, Worsted

**Direct system:**

The direct system of denoting linear density is based on measuring the weight per unit length of yarn. Hair higher the count course are there it is used for silk, jute etc.

Here, higher the count coarser the yarn.

It is used for silk, jute, etc.

Example: Tex, denier, lb/spindle.

N_{d} = (W×l)/(L×w)

**Indirect system:
**The indirect system is based upon the length per unit weight (mass) of yarn.

Here, higher the count finer the yarn.

It is used for cotton, woolen, worsted, linen etc.

Example: English, Metric, Worsted.

N_{id }= (L×w)/(W×l)

Where,

N_{d} = Yarn Count in direct system

N_{id}= Yarn Count in indirect system

W = mass of the sample.

l = Unit length of the system.

L = Sample length.

w = Unit mass of the system

**Count in Different System:**

**Direct:**

Tex = (Weight in gram)/(Length in 1000m or 1 km)

Denier = (Weight in gram)/(Length in 9000m or 9 km)

Pounds per spindle = (Weight in lb)/(Length in 14,400 yds)

Millitex = (Weight in milligram)/(1000m )

Kilotex = (Weight in kilogram)/(1000m )

Wollen = (Weight in grain)/(20 yds)

**Indirect:**

English system, N_{e} = (Length in Hanks of 840 yds)/(Weight in lbs)

Metric, N_{m} = (Length in Hanks of 1000m)/(Weight in kg)

Metric, N_{m} = (Length in 560 yds)/(1 pound )

Wollen = (Weight in grain)/(20 yds)

**Conversion of Count System:**

N_{R} = Count of required system.

N_{k} = Count of Known system.

l_{k} = Unit length of Known system.

l_{R} = Unit length of required system.

W_{k} = Unit Weight of Known system.

W_{R} = Unit Weight of required system.

**Calculation of Conversion Factor:**

**English (Cotton count) to Denier:
**From count conversion system,

Here,

W_{k} = Unit weight of known system = 1 pound = 453.6 gm

l_{k}= Unit length of known system = 840 yds = 840×0.9144 m= 768.096m

W_{R}= Unit weight of required system = 1 gram

l_{R} = Unit length of required system = 9000 meter

Denier = 1/Ne × {(453.6 )/1}/{(768.096)/9000} = 1/Ne ×5315

**Denier to English (Cotton count):**

Here,

W_{k}=Unit weight of known system=1gm

l_{k} = Unit length of known system = 9000 meter

W_{R} = Unit weight of required system = 1 pound = 453.6 gm

l_{R} = Unit length of required system = 840 yds = 840×0.9144 m = 768.096m

N_{e}= 1/D×5315

**Denier to Tex:**

Prove: denier = 9Tex

Here,

W_{k} = 1 gram

l_{k} = 9000 meter

W_{R} = 1 gram

l_{R} = 1000 meter

N_{R }= N_{k}×(1/1)/(9000/1000)

=> Tex = D×1/9

=> D = 9×Tex

**English to Metric:**

Here,

W_{k} = 1 pound =453.6 gm=453.6×10^{-3} kg

l_{k} = 840 yds= 840×0.9144 m = 840×0.9144×10^{-3}km

W_{R} = 1 kg

l_{R} = 1 km

N_{m}=N_{e}×(840×0.9144×10^{-3}/1)/(453.6×10^{-3})/1)

N_{m}=N_{e}×1.69

**Tex to English:**

Here,

W_{k} = 1 gram = 1/(453.6) lb or pound

l_{k} = 1000 meter = 1000 × 1.09 yds

W_{R} = 1 pound

l_{R} = 840 yds

N_{e }= 1/Tex×{(1000×1.09)/840}/[{(1/453.6)}/1]

=> N_{e}=1/Tex×590.5

**Relation between Yarn Dia & Count:
**Pierce approached the problem from consideration of the apparent specific volume of yarns. By experiment an apparent specific volume of 1.1 for cotton yarns was obtained.

So, specific volume of yarn = 1.1 cm^{3}/gm

The weight of 1.1 cm^{3} yarn = 1 gm

Let,

Yarn count = N tex.

∴ Wt of 1000m yarn is N gm

Length of N gm yarn is 1000m

∴ Length of 1 gm yarn is (1000 m)/N or (1000×100 cm)/N

So, Length of 1.1 cm^{3} (Volume) yarn is 10^{5}/N cm

Since, The volume = Cross section ×Length

1.1 cm^{3} = π×(d/2)^{2}×10^{5}/N cm (For Direct system)

1.1 cm^{3} = π×d^{2}/4×10^{5}/N cm

d^{2} =(1.1×4×N)/(π×10^{5} )cm

[Where d = dia of yarn in cm]

d = √{(1.1×4×N)/(π×10^{5} )}

d = 0.375/100 √(N ) cm

d∞√(N_{e})

**Converting cm to Inches and Tex to Cotton Count:**

d (inch) =0.375/100×√(590.5 )/√(N_{e})×1/2.54 (For Indirect system)

d = 3.6/{100√(N_{e} )}

d = 1/{28√(N_{e})}

∴ d ∞1/√(N_{e})

**Count Calculation for Double Yarn:**

For Direct system

N_{r } = N_{1} + N_{2}

For Indirect system

1/N_{r} = 1/N_{1} + 1/N_{2}

Where,

N_{r} = resultant count

N_{1} & N_{2} = component count

**Measurement of Count:**

**Warp Reel**and Balance Method:- Quadrant balance method
- Knowles balance
**Beesley balance**- Sliver, roving count by measuring drum
- Count data system
- Auto sorter by Uster

**References:**

- Handbook of Textile Testing and Quality Control (PDF) – B. Grover, D. S. Hamby
- https://textilelearner.net/yarn-count-numbering-system-and-conversions/
- https://textilestudycenter.com/yarn-numbering-system/
- Textile Testing by Angappan.
- Principles of Textile Testing by Booth
- Physical Testing of Textiles by Saville

**You may also like: **

**Yarn Count Numbering System and Conversions****Yarn Count Measurement System****Count Conversion System for Textile Yarns**

Founder & Editor of Textile Learner. He is a Textile Consultant, Blogger & Entrepreneur. He is working as a textile consultant in several local and international companies. He is also a contributor of Wikipedia.