**Introduction**

Winding is the first step of woven fabric manufacturing (**Weaving Mill**). It is the process of transferring yarns from ring, bobbin, hank etc. into a suitable package is called **winding**. There are three types of package – Parallel Wound Package, Near Parallel Wound Package and Cross Wound Package. This process may be electrical or mechanical. The functions of winding are control the yarn path, minimize yarn vibration and reduce chance of **balloon formation**. Also, there are three auxiliary functions – Creeling, Piecing and Doffing. So, calculations play an important role in Winding machine. Some important production calculation (Actual production, Calculated production, Time required, No. of drums required, No. of spindles, Winding Efficiency, Weight of yarn, Length of Yarn etc.) of **Winding machine** are discussed below.

**Some Important Conversions for Winding 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);
- Tex x Ne = 590.5

…………Yards Hank

lb = ——————–; lb = ————–; Hank = lb x Count

……Count x 840 Count

**Winding Production Calculation with Formula in Weaving**

**…………………………………Quality of Yarn in lb to be Wound**

**1. Time Required = ————————————————————————–**

**…………………………….Actual Production in lbs/hr x No of Drum**

………………….Quantity of Yarn in Hank

= ——————————————————————————–

….Actual Production in hanks/hr x No of Drums

……..Quantity of Yarn in lbs x Yarn Count

= —————————————————————————-

….Actual Production in Hank x No of Drums

…………….Total Length of Warp in Yards

= —————————————————————————-

….Actual Production in yds/hr x No of Machine

**………………………………..Calculated Yards**

**2. Actual Production = ———————————– x Efficiency**

**………………………………………..840**

….Calculated Hank

= ———————————– x Efficiency

……..Yarn Count

= Calculated Yards x Efficiency

**Example 01:** How much time will be required to wind 2244 lbs of 20 Ne cotton on 40 drums of a super speed cone winder, if the calculated of winding is 1300 yards per min and the efficiency is 85%.

**Solution:**

Here given,

Calculated yards = 1300

Efficiency = 85% = 0.85

Quantity of Yarn to wind = 2244 lbs

**Yarn Count** = 20 Ne

No of Drums = 40

**Actual Production per drum per hour** = Calculated Yards x Efficiency

= 1300 x 0.85 x 60 yards/hour

….1300

= ————- x 0.85 x 60 Hank

…..840

= 78.93 Hank

…………………………………Quantity of Yarn in lbs x Yarn Count

**Time Required** = ————————————————————————-

……………………………Actual Production in Hank x No of Drums

….2244 x 20

= —————————— Hours

….78.93 x 40

= 14.22 Hours **(Ans)**

**Example 02:** The actual production per spindle per min of an automatic super speed winder is 652 yards of 12’s **cotton yarn**. Calculate the time that will be required to wind 1140 lbs of yarn on 40 spindles.

**Solution:**

Here given,

Actual production per spindle = 652 yards/min

Quantity of yarn to be wound = 1140 lbs

No of spindles = 40

Yarn Count = 12’s

…………………………………………………………………………652 x 60

**Actual production per spindle per hour** = ———————– = 3.88 lbs

…………………………………………………………………………840 x 12

………………………………..Quantity of Yarn in lb to be Wound

**Time Required** = ———————————————————————

……………………………Actual Production in lbs/hr x No of Drum

……..1140

= ———————- = 7.35 hours **(Ans)**

….3.88 x 40

**Example 03:** Special bobbins each containing 1.2 lb of cotton yarn are used on super speed automatic spooler type of the counts is 20’s and if the calculated hanks/min is 58 & the efficiency is 80%. Calculate how long of time required to take for a speed to be exhausted.

**Solution:**

Here given,

Calculated Hank = 58

Yarn count = 20’s

Length of yarn = 1.2 lb

Efficiency = 80% = 0.8

………………………………………………..Calculated Hank

**Actual Production per hr** = ——————————— x Efficiency

…………………………………………………..Yarn Count

….58

= ———- x 0.8 = 2.32 lbs

….20

……………………………….Length of Yarn in lb

**Time Required** = —————————————————

……………………………Actual Production per hr

…..1.2

= ———– = 0.52 hours **(Ans)**

….2.32

**3. Actual Production= Calculated Production x No of Drums x Efficiency**

**………………………………………………Actual Production**

**4. Production in Weight/shift = —————————————– x 60 x 8**

**………………………………………………..Yarn Count (Ne)**

**Example:** Calculated the actual production per shift of a winding m/c of 150 drums. If the calculated production per min per drum is 1400 m/min. Take the efficiency as 85%. Calculate also the weight of yarn actually wound, if the yarn count is 20 Tex.

**Solution:**

Given that,

The yarn count = 20 Tex

We know, Ne x Tex = 590.5

So, Ne = 590.5/Tex = 590.5/20 = 29.5’s

Also Given,

Calculated Production = 1400 m/min

No of drum = 150

Efficiency = 85% = 0.85

**Actual Production** = Calculated Production x No of Drums x Efficiency

….1400 x 1.09

= ——————- x 150 x 0.85

………840

= 231.625 lbs/min

…………………………………………………Actual Production

**Production in Weight/shift** = ———————————– x 60 x 8

…………………………………………………Yarn Count (Ne)

….231.625

= ————- x 60 x 8 lbs/shift

……29.5

= 3768.6 lbs/shift. **(Ans)**

**……………………………Target Production in kg**

**5. Time Required = —————————————————**

**………………………………..Production Rate**

**Example:** Calculate the time that would be needed to wind 2000 kg of 40 Tex yarn. The number of drums available is 120 and the actual production per drum per min is 1080 meter.

**Solution:**

Here given,

The yarn count = 40 Tex

We Know, Ne x Tex = 590.5

So, Ne = 590.5/Tex = 590.5/40 = 14.76

Also given,

Actual Production = 1080 m/min/drum

No of drums = 120

Target Production = 2000 kg

…………………………………………………………………1080 x 1.09 x 60 x 100

Now, the actual production of 100 drums is = ——————————————- kg/hr

……………………………………………………………………840 x 14.76 x 2.21

= 309.33 kg/hr

= Production Rate

…………………………….Target Production

**Time Required** = ————————————–

……………………………..Production Rate

…..2000

= —————— = 6.47 hours** (Ans)**

….309.33

**…………………………………………………..Quantity of Yarn to Wound**

**6. No of Drums Required = ——————————————————————————–**

**……………………………………..Time in hrs x Actual Production per hr per drum**

**7. Quantity of Yarn to be Wound = Quantity of Yarn in Cloth – Waste%**

**Example 01:** Calculate the number of drums required for winding 750 kg of 20 Ne cotton yarn in a day of 8 hrs. If the actual production per drum per min is 1040 meters.

**Solution:**

Here Given,

Actual production per drum per min = 1040 meters

Yarn Count = 20 Ne

Quantity of yarn to wound = 750 kg

Time = 8 hrs.

…………………………………………………………………1040 x 1.09 x 60

Actual Production in kg per drum per hour = ——————————— kg/hr

………………………………………………………………….840 x 20 x 2.21

= 1.83 kg/hr

…………………………………………………………Quantity of Yarn to Wound

**No of Drums Required** = —————————————————————————–

………………………………………..Time in hrs x Actual Production per hr per drum

…….750

= —————- = 51.23 ≈ 52 **(Ans)**

….8 x 1.83

**Example 02:** The rate of winding of modern high speed cone winding machine is 890 yards per min. Calculate the no of drums required to wind 388 lbs of 30’s from ring bobbin in 8 hrs. If the efficiency is 88%. Allow 1% for waste and left on the bobbins.

**Solution:**

Here given,

Calculated production per min = 890 yards

Quantity of yarn to be wound = 388 lbs

Waste = 1%

Efficiency = 88%

Yarn Count = 30’s

Time = 8 hrs

………………………………………………………..800 x 60 x 88

Actual production per drum per hour = ———————– Hanks

………………………………………………………….840 x 100

= 50.28 Hanks

Quantity of yarn to be wound = 388 – 1% of 388 lbs

= 384 lbs

= 384 x 30 Hanks

= 11520 Hanks

……………………………………………………Quantity of yarn to be wound

**No of Drums Required** = ——————————————————————–

………………………………………..Actual Production per drum per hour x Time

…….11520

= ——————-

….50.28 x 8

= 28.64 ≈ 29 **(Ans)**

**8. Length of Yarn = Production of warping m/c per 8 hrs+ Waste%**

**Example:** Calculate the no of winding drums per m/c that will be required for feeding 5 high speed warping m/c. The actual production of the warping m/c is 74600 yds per 8 hrs and the average no of ends warped on a back beam on an average production on this machine is 420. The winding m/c is supposed to have on an actual production of 306 hanks per 8 hrs.

**Solution:**

Here given,

Actual production per winding drum per 8 hrs = 306 hanks

Actual production of warping m/c per 8 hrs = 74600 yards

No of warping m/c = 5

No of ends = 420

………………………………………………74600 x 420 x 5

So total length of yarn warped = ——————————– Hanks

……………………………………………………..840

= 186500 Hanks

= Production of warping m/c per 8 hrs

Waste% = 1%

**Length of Yarn** = Production of warping m/c per 8 hrs+ Waste%

= 186500 + 1% of 186500

= 188365 Hanks

= Production of warping m/c per 8 hrs.

…………………………..Length of Yarn in Hank

**No of Drums** = ————————————————–

………………………..Actual Production in Hank

….188365

= —————– = 615.57 ≈ 616 **(Ans)**

…….306

**…………………………Total Quantity of Weft in Hank per hour x No of Loom**

**9. No of Spindles = —————————————————————————————**

**………………………………..Actual Production in Hanks/Spindle/hour**

**Example:** Calculate the no of spindles required to supply weft to 110 looms. The cloth produced on a loom required on an average 15.32 hanks of weft per hour excluding waste. The actual production per spindle per hour of the super speed automatic pirn winder is 50 hanks. Allow 0.5% waste.

**Solution:**

Here given,

Required weft per hour = 15.32 hanks (Excluding waste)

Actual Production per spindle per hour = 50 hanks

No of looms = 110

Waste = 0.5%

Total Quantity of weft = 15.32 + (15.32 x 0.5%) hanks/hour

= 15.3966 Hanks/hour

……………………………Total Quantity of Weft in Hank per hour x No of Loom

**No of Spindles** = —————————————————————————————-

…………………………………….Actual Production in Hanks/Spindle/hour

….15.3966 x 110

= ———————

………..50

= 33.87 ≈ 34 **(Ans**)

**…………………………….Actual Production**

**10. Efficiency% = —————————————— x 100**

**………………………..Calculated Production**

**Example:** Calculate of modern high speed cone winding machine if its actual production per 8 hrs is found to be 394 hanks. The rate of winding is 780 yards per min.

**Solution:**

Here,

Calculated Production = 780 yards/min

= 780 x 60 x 8 yards/8 hrs

Actual Production = 394 hanks/8 hrs

= 394 x 840 yards/8 hrs

……………………….Actual Production

**Efficiency%** = ———————————– x 100

…………………….Calculated Production

……394 x 840

= ——————– x 100

….780 x 60 x 8

= 88.4% **(Ans)**

**……………………………………………………………………….Length of Yarn in yard**

**11. Weight of Yarn to be Wound on each bobbin = ———————————————**

**……………………………………………………………………….840 x Yarn Count (Ne)**

**Example:** An ordinary slow speed warping machine is working with 40 Ne. The production of the machine is 22000 yards per day of 9 hours. If it is required that creeling is to be wounded on the supply package, allowing 4.6% for wastage and material left on the bobbins.

**Solution: **Here, production of warping machine per 9 hrs = 22000 yards

Waste = 4.6%

Yarn Count = 40 Ne

Length of yarn = 22000 + (22000 x 4.6%) yards = 23012 yards

……………………………………………………………………………..Length of Yarn in yards

**Weight of Yarn to be Wound on Each Bobbin** = ———————————————

………………………………………………………………………………840 x Yarn Count (Ne)

…….23012

= —————- lbs

….840 x 40

= 0.685 lbs/9 hrs.** (Ans)**

**12. Length of Yarn on Each Bobbin = Wt. of Yarn on Each Bobbin x Count (Ne)**

Example: A set of 5 warping beam each containing 89 lbs of yarn is made from 480 bobbins. If the count of yarn is 40’s cotton, what should be the length and weight of the yarn on each bobbin? Allow 4.5% wastage and material left on the bobbin.

**Solution:**

Here, Weight of yarn per beam = 89 lbs

So, Total weight of yarn on 5 beams = 89 x 5

= 445 lbs

Waste = 4.5%

Total weight of yarn on bobbin = 445 + 445 x 4.5% lbs = 465.025 lbs

No of bobbins = 480

Now. Weight of yarn on each beam = 465.025/480 lb = 0.9688 lb

If, Yarn Count = 40 Ne

Therefore,

**Length of yarn on each bobbin** = Wt. of yarn on each beam x Count (Ne)

= 0.9688 x 40 hanks

= 38.752 hanks **(Ans)**

**Conclusion**

I have tried to explain/show all the production calculation of winding 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.

**Author of this Article:**

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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It’s me, a Textile Consultant, Blogger & Entrepreneur. I’m working as a textile consultant in several local and international companies. I’m also a contributor of Wikipedia.