**Formula and Examples of Warping Calculation**

**Warping:**

The parallel

**winding**of warp ends from many winding packages (cone or cheese) on to a common package (warp beam) is called warping. Main purpose of warping is to arrange a convenient number of warp yarns of related length so that they can be collect on a single warper’s beam, as a continuous sheet of yarns which can be used for sizing or next process.

**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)**