# Warping Production Calculation Formula with Example

Introduction
Warping is the second step of the weaving mill, the parallel winding of warp ends from many winding packages like cone or cheese onto a common package warp beam is called warping. The main tasks of warping are to arrange a convenient number of warp yarns of related length, make a parallel package and overall to prepare continuous warp sheet beam according to given pattern or required length, width and no of ends. So, production calculation play an important role in warping machine. Some important calculation (Actual production, Calculated production, Time required, No. of drums required, No. of spindles, Efficiency, Weight of yarn, Length of Yarn etc.) of Warping machine are discussed below.

Some Important Conversions for Warping Production Calculation

• 36 inch = 3 Feet = 1 Yard;
• 1 Meter = 1.0936 Yards;
• 1 Pound (lb) = 453.6 Gram = 16 Ounce (Oz);
• 1 Lea = 120 Yards;
• 840 Yard = 7 Lea = 1 Hank;
• 1 Pound (lb) = 7000 Grains;
• 1 Meter = 39.37 Inch;
• 1 inch = 2.54 cm;
• 1 cm = 10 mm;
• 1 Kg = 2.205 Pound (lb)

…………Yards                          Hank
lb = ———————-; lb = ——————-; Hank = lb x Count
…….Count x 840                  Count

### Warping Production Calculation Formula with Example

1. Total Length of Warp Produced per Day of 8 hrs = Calculated Production x Efficiency x 60 x 8

……………………………………………………..Total Length of Warp Produced/Day of 8 hrs in yds
2. No of Beam produced/Day of 8 hrs = ——————————————————————-
…………………………………………………………………..Length of Warp on a Beam in yds

……………………………………………………Total Length of Warp Yarn in yds x No of ends
3. Total Weight of Warp in lbs = —————————————————————————-
………………………………………………………………………….840 x Yarn Count

………………………………………………………Total Wt. of Yarn in lb
4. Weight of Yarn on a Beam = ————————————————–
……………………………………………………….No of Beam Produced

Example: A super speed beam warpers with a warping speed of 890 yds per min is preparing a standard warp of 525 ends. If the yarns count 40’s and overall efficiency is 86%. Calculate the following, where the length of warp on each beam is required to be 43955 yds. Ignore waste.

1. Total Length of Warp Produced per Day of 8 hrs
2. No of Beam produced per Day of 8 hrs
3. Total Weight of Warp in lbs
4. Weight of Yarn on a Beam

Solution:

Here,
Calculated Production = 890 yds/min
No of Warp ends = 525
Yarn Count = 40’s
Efficiency = 86% = 0.86
Length of Warp on a Beam in yds = 43955

a) Total Length of Warp Produced per Day of 8 hrs = Calculated Production x Efficiency x 60 x 8
= 890 x 0.86 x 60 x 8 yds/8 hrs
= 367392 yds/8hrs (Ans)

…………………………………………………Total Length of Warp Produced per Day of 8 hrs in yds
b) No of Beam produced per Day of 8 hrs = ——————————————————————–
…………………………………………………………………..Length of Warp on a Beam in yds

367392
= —————- beams
43955

= 8.36 ≈ 9 beams (Ans)

……………………………………………..Total Length of Warp Yarn in yds x No of ends
c) Total Weight of Warp in lbs = ———————————————————————–
…………………………………………………………………840 x Yarn Count

367392 x 525
= ——————— lbs
…….840 x 40

= 5740.5 lbs (Ans)

……………………………………………………Total Wt. of Yarn in lb
d) Weight of Yarn on each Beam = ————————————————
…………………………………………………..No of Beam Produced

5740.5
= ————— = 637.83 lbs (Ans)
……9

……………………………………….Count x Wt. of warp in lbs x 840
5. No of ends in the Warp = ———————————————————–
…………………………………………….Length of Warp in yds

Count x Wt. of warp in lbs
= ————————————————–
….Length of Warp in Hanks

Example: Calculate the total no of ends on back beam which weights 253 lbs. The weight of the empty beam as indicated from the marking on its flanges are 64 lbs. The count of the yarn is 30’s cotton and the length of warp on the beam is 13940 yds.

Solution:

Here,
Wt. of warp beam = 253 lbs
Wt. of flanges = 64 lbs
Weight of warp yarn = 253 – 64 = 189 lbs
Yarn Count = 30’s
Length of warp = 13940 yds

…………………………………..Count x Wt. of warp in lbs x 840
No of ends in the Warp = —————————————————
…………………………………………Length of Warp in yds

….30 x 189 x 840
= ———————–
……..13940

= 341.66 ≈ 342 (Ans)

………………………………………………….Length of Warp in yds x No of ends
6. Count of Warp or Beam Count = —————————————————————
……………………………………………………………840 x Wt. of Warp in lbs

….Total Length of Yarn on the Beam in yds
= ————————————————————————-
………………840 x Wt. of Warp in lbs

Example: Calculate the count of yarn on a back beam which weights 306 lbs and has 12400 yds on it. The total no of ends in the warp is 455 and the weight of empty beam is 64 lbs.

Solution:

Here,
Length of Warp = 12400 yds
No of Warp ends = 455
Weight of Beam with Warp yarns = 306 lbs
Empty Beam Weight = 64 lbs

Weight of Yarn = 306 – 64 = 242 lbs

……………………………..Length of Warp in yds x No of ends
Count of Warp Yarn = ————————————————————
………………………………………840 x Wt. of Warp in lbs

12400 x 455
= ——————- = 27.75 ≈ 28’s (Ans)
……840 x 242

…………………………………Wt. of warp in lb x Count x 840
7. The Length of Warp = ————————————————————– (Yard)
………………………………………………No of ends

……………………………………………………..Length of Yarn x No of ends
8. Total Length of Yarn in the Warp = ————————————————- Hanks
………………………………………………………………………..840

Example: A beam made on a high speed warping  machine weights 374 lbs with the warp, the weight of the empty beam 65 lbs. The no of ends on the beam are 430 and count 30’s, find out-

• The length of warp
• The total length of yarn in the warp
• Amount the warper is to be paid if the rate of warping is 7 praises per 1000 yds of warp.

Solution:

Here given,
No of warp ends = 430
Yarn Count = 30’s
Weight of warp beam = 374 lbs
Weight of empty beam = 65 lbs
Weight of warp yarn = 374 – 65 = 309 lbs

………………………………..Wt. of warp in lb x Count x 840
The Length of Warp = —————————————————- (Yard)
………………………………………………No of ends

….309 x 30 x 840
= ——————————- yds
………….430

= 18108.84 yds (Ans)

…………………………………………………..Length of Yarn x No of ends
Total Length of Yarn in the Warp = ——————————————————-
……………………………………………………………………840

….18108.84 x 430
= ———————————-
………….840

= 9270 Hanks (Ans)

………………………Total No of ends
9. No of Beam = ————————————-
……………………….No of Bobbins

Example: Calculate the no of warper beams and the length of warp that can be made from 445 bobbins, each of which contains 0.5 lb of 50’s cotton. The no of ends of warp required is 2280. Allow 4% waste and yarn left on the bobbins.

Solution:

Here given,
No of bobbins = 445
Weight of yarn per bobbin = 0.5 lb
Yarn Count = 50’s
No of ends = 2280

Total weight of yarn = (445 x 0.5) – 4% = 214.1 lbs

………………………..Wt. of warp in lb x Count x 840
Length of Warp = ——————————————————
……………………………………No of ends

….214.1 x 50 x 840
= ————————— yds
………..2280

= 3943.95 yds (Ans)

……………………..Total no of ends
No of Beams = ——————————
………………………No of Bobbins

….2280
= ———— = 5.12 ≈ 5 (Ans)
….445

10. Total Length of yarn for the Set = Length of Warp in yds Required per Beam x No of ends per Beam x No of Beams

…………………………………………………………….Total Length of Yarn in yds
11. Wt. of Yarn in lbs Required for the Set = ————————————————
………………………………………………………………………Count x 840

Example: Calculate the quantity of yarn in lbs which will be required for a set of 6 back beams to be produced on a modern high speed beam warper. The length of warp on each back beam is 24000 yds and there are 460 ends on each beam count 40’s. Allow 1.5% for waste during warping.

Solution:

Here given,
Length of warp on each beam = 24000 yds
No of ends on each beam = 460
Yarn Count = 40’s
No of Beams = 6

Wt. of Warp Required Per Beam = 24000 – 1.5% = 23640 yds

Total Length of yarn for the Set
= Length of Warp in yds Required per Beam x No of ends per Beam x No of Beams
= 23640 x 460 x 6
= 65246400 yds

………………………………………………………….Total Length of Yarn in yds
Wt. of Yarn in lbs Required for the Set = —————————————————-
…………………………………………………………………..Count x 840

65246400
= ———————— lbs
….40 x 840

= 1941.86 lbs (Ans)

12. 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 Sets

……………………………..Total Length of Warp to be Produced in yds
13. Time Required = —————————————————————————-
………………………………Actual Production in yds/hr x No of m/c

Example: 8 sets of 6 warpers beam each is to be produced on 4 warping machines. The length of warp on each beam is required to be 14000 yds. If the actual production of each machine is 2800 yds per hr. how long will it take to complete the sets?

Solution:

Here given,
Length of warp in yds required per beam = 14000 yds
No of beams per set = 6
No of sets = 8
Actual production = 2800 yds/hr
No of machines = 4

Total Length of warp in yds = Length of Warp in yds Required per Beam x No of beams per Set x No of Sets
= 14000 x 6 x 8 yds
= 672000 yds

……………………….Total Length of Warp to be Produced in yds
Time required = ———————————————————————-
………………………..Actual Production in yds/hr x No of m/c

……672000
= ———————– hrs.
….2800 x 4

= 60 hrs. (Ans)

………………………..Actual Production
14. Efficiency = ——————————————- x 100
…………………….Calculated Production

Example: A modern high speed beam warping machine produces 8 beams each containing 222460 yds of warp per day of 8 hrs. If the calculated warping speed of the warper is 590 yds per min. Calculate efficiency.

Solution:

Actual Production per 8 hrs = 222460 yds

Actual Production per hr = 222460/8 = 27807 yds

Calculated Production per min = 590 yds

Calculated Production per hr = 590 x 60 = 35400 yds

…………………….Actual Production
Efficiency = ——————————————- x 100
…………………Calculated Production

…..27807
= ————- x 100
….35400

= 78.55% (Ans)

…………………………………….Total no of ends
15. Number of Sections = ————————————-
…………………………………….Creel Capacity

Example: The total number of ends in a warp is 2700. Calculate the number of sections to be made, if the capacity of creel is 500 bobbins.

Solution:

Here,
No of ends = 2700
Creel Capacity = 500

………………………………….Total no of ends
Number of Sections = ————————————-
………………………………….Creel Capacity

2700
= ————- = 5.4 ≈ 6 (Ans)
….500

……………………………….Width of Warp on Loom Beam
16. Width of Section = ———————————————————–
……………………………………….Number of Sections

Example: If 6 sections are to be made for a warp, which is 29 inches wide in the reed, calculate the width of each section.

Solution:

Here, Total width of warp = 29”

Number of Sections = 6

……………………………………29
Width of each section = ————— = 4.83” (Ans)
……………………………………6

Example: A stripe warp of the following particulars is to be made on a sectional warping machine-

• Total no of ends = 2240
• Number of ends per Pattern = 32
• Number of Extra Pattern ends,
• At Both ends Near Selvedges = 24
• Selvedge ends at Each Side = 20

If the creel capacity is 480 bobbins and the width of warp in the reed is 35”, calculate the following-

1. Number of complete patterns in the warp
2. Number of sections to be made
3. Number of ends per section
4. Width of warp on the weaver’s beam
5. Width of a section.

Solution:

……………………………………………2240 – (24 + 20×2)
a) Total number of patterns = ————————————
…………………………………………………….35

= 68 (Ans)

……………………………………Total no of ends
b) Number of sections = —————————————
…………………………………….Creel Capacity

….2240
= ———— = 4.67≈ 5 (Ans)
…..480

…………………………………………………………Total no of ends
c) Number of ends on each section = ———————————–
………………………………………………………….No of section

….2240
= ———— = 448 (Ans)
…….5

3 center sections each contains 14 patterns i.e., 448 ends

2 ends sections each contains 13 patterns + 12 extra pattern ends + 20 selvedge ends = 448 ends.

d) Width of warp on a weavers beam = 35 + 2.5 = 37.5” (Ans)

………………………………..Width of warp on loom beam
e) Width of a section = ———————————————————-
……………………………………….Number of sections

….37.5
= ————- = 7.5” (Ans)
…..5

Conclusion
I have tried to explain/show all the production calculation of warping machine in weaving mill with so many examples through this article. Hope this article will be helpful for those who are new to the weaving mill or those who are students.

You can see all process shortly in this video:

Md. Imran Hossain
B.Sc. in Textile Engineering
Shahid Abdur Rab Serniabat Textile Engineering College, Barisal.
Email: mdimranhossain.te@gmail.com

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