**What is 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.

Warping is an important process in weaving that involves preparing the warp yarns that are held under tension on the loom. The warp yarns run vertically or longitudinally in a woven fabric and are interlaced with the weft yarns, which run horizontally across the fabric. 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]