# Textile Calculation

Problems and solutions of textile

## Yarn Wrapping Calculation

Calculation of Probable Weight Due to Yarn Wrapping
Amit Kumar Das
Founder of "টেক্সটাইল ম্যানিয়া" Blog
Pabna Textile engineering college, Pabna, Bangladesh
Cell: 01770 456240
Email: amitptec6th@gmail.com

Yarn Wrapping:
The object of taking wrapping in spinning is to ascertain the actual hank or count of the product which is expected to be not significantly different from what is desired or predetermined.
Therefore it becomes much necessary to lay down certain expected upper and lower limits for the data, to decide a wheel change is to be made or not. Some suggested formula for the determination of these probable limits are discussing bellows. Hope you would be benefited by it a lot.

Probable limits for the average weight,
= Nominal Weight  ± 2 Nominal Wt / 100 x [ √ { (CVB)² ÷ b } + {(CVW)² ÷(b x 1) } ]

Where,
CVB = CV% between cards, deliveries or bobbins
CVW = CV% within cards, deliveries or bobbins
b = Number of cards, deliveries or bobbins from which wrapping are taken
I = The number of wrapping or leas tested from one card, delivery or bobbin.

To calculate the limits they have also given different values of CV% (as given below) that can be considered satisfactory.

 Material Length in yards CV% between CVB CV% within (CVB) Blow room lap 2 0.75 1.5 Card sliver 5 1.5 7.0 Finisher Drawing sliver 5 2.0 1.0 Inter frame Yarn upto 60s Yarn upto 60s 15 120 129 2.5 3.0 2.5 2.5 2.5 2.0
Mathematical Problems & Solutions:

Problem: 01
What will be the limits of card sliver whose nominal weight is 300 grains per 5 yards? If one wrapping is taken from each of four cards then the limits for the average of 4 wrappings?

Solution:

As we know,

Probable limits for the average weight,
= Nominal Weight ± 2 Nom. Wt / 100 x [ √ { (CVB)² ÷ b } + { (CVW) ² ÷(b x 1) } ]

Limit of card sliver will be = 300 ± 300 / 100 x [ √ [{ 15² ÷ 4} + { 7 ² ÷ 4 } ] grains
= 300 ± 21.5 grains (ANS)

Problem: 02
If 2 wrappings are taken from each of 8 bobbins and if the nominal weight of lea is 30 grains then what will be the limit for average of wrapping 16?

Solution:

As we know,

Limit of card sliver will be = 30 ± 2 x 30 / 100 x [ √ [{ 3² ÷ 8} + { 2.5 ² ÷ 16 } ] grains
= 300 ± 0.74 grains (ANS)