**Warping:**

The parallel winding of warp ends from many winding packages (cone or cheese) on to a common package (warp beam) is called warping. The main purpose of warping is to arrange a convenient number of warp yarns of related length so that they can be collected on a single warper’s beam, as a continuous sheet of yarns that can be used for sizing or the next process. In this article, we will discuss the warping calculation formula and examples.

**Warping Calculation Formula and Examples**

**Formula of Warping Calculation:**

1. Production = Surface speed of drum × π × Dia. of drum × Creel Capacity

……………………………………Weight in lbs

2. The length of warp = ……………………….

………………………………………No. of ends

……Weight of warp in lb X Count X 840

= …………………………………………………………. (Yards)

……………………..No. of ends

……….Weight of warp in lb X Count

= …………………………………………………. (Hanks)

……………………….No. of ends

3. The total length of yarn in the warp,

……..Length of warp X No. of ends

= ………………………………………………. (Hank)

……………………….840

………………………………………….Length of warp in yds X No. of ends

4. Weight of warp in-lbs = …………………………………………………………….

………………………………………………………….840 X Count

…………………………………………….Count X Weight X 840

5. No of ends in the warp = …………………………………………….

……………………………………………Length of warp in yds

………Count X Weight in lbs

= …………………………………………..

…….Length of warp in hanks

……………………………………………………Length of warp in yds X No of ends

6. Count of warp or Beam count = ………………………………………………………

………………………………………………………………840 X Weight in lbs

……………………………….Total length of warp to be produced in yds

7. Time required = …………………………………………………………………… hrs

……………………………Actual production in yds per hr X No of m/cs

8. Total length of warp to be produced in yds = Length of warp in yds required per beam X No of beams per set X No of ends

**Mathematical Problem of Warping:**

A super speed beam warpers with a warping speed of 840 yds per min is preparing a standard warp of 525 ends. If the yarns count 30’S and overall efficiency is 84%. Calculate the following the length of warp on each beam is required to be 44352 yds. Ignore waste.

- Total length of warp produced per day of 8 hrs
- No of beam produced per day of 8 hrs
- Total weight of yarn in lbs warped per drum
- Weight of yarn on a beam

1. Total length of warp produced per day of 8 hrs = Calculated production X Efficiency X 60

……………………….84

= 880 X 60 X ….……… X 8 yds

………………………100

= 354816 yds (ANS)

2. No of beam produced per day of 8 hrs

……..Total length of warp produced per day of 8 hrs in yds

= …………………………………………………………………………………….

…………………Length of warp on a beam in yds

……….354816

= ……………………

……….44352

= 8 beams (ANS)

3. Total weight of yarn in lbs warped per drum

……..Total length of warp yarn in yds X No of ends

= ……………………………………………………………………….

…………………….840 X Yarn count

……354816 X 525

= …………………………

……….840 X 30

= 7392 lbs. (ANS)

4. Weight of yarn on a beam

…..Total weight of yarn in lb

= ………………………………………

……..No of beam produced

………7342

= ………………

…………8

= 924 lbs (ANS)

Author of this Article:S. M. Hossen UzzalB.Sc. in Textile Technology Monnu Fabrics Ltd. Manikgonj Email: [email protected]