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Textile Calculation Questions with Worked Answers: Spinning, Weaving and Knitting

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22 exam-style textile calculation questions with worked answers, from yarn count and twist to loom and knitting production, each checked in a calculator.

These textile calculation questions cover the numericals that B.Tech, diploma and GATE textile papers set most often: yarn count conversion, twist and draft, spinning production, winding, warping and sizing, loom production, reed count, fabric GSM, cover factor, crimp, and knitting stitch length, GSM and production. There are 22 questions, all with our own numbers. Each gives the answer first, then the formula and the working, a link to the formula page to revise from, and a link that opens the site’s calculator with the question’s figures already entered, so you can check the answer yourself. Every answer below was computed in that calculator on 18 October 2026.

Try each question before reading the working. The formulas themselves, with the constants explained, are on the yarn spinning formulas and weaving calculations pages; this page is for practice. Revision notes (PDF) for the formula pages and the other study pages are gathered on the study page.

Calculation questions at a glance

Topic Formula Questions Calculator
Count conversion tex = 590.5 ÷ Ne; denier = 9 × tex; Nm = 1000 ÷ tex 1, 2, 3 Count converter
Folded (ply) yarn tex adds; 1 ÷ R = 1 ÷ N1 + 1 ÷ N2 in Ne 4 Count converter
Twist TPI = TM × √Ne; TPM = TPI × 39.37 5, 6 Twist multiplier
Draft draft = hank delivered ÷ hank fed × doublings 7, 8 Ring frame draft, Drawframe draft
Card and drawframe production kg/h = m/min × 60 × ktex × efficiency ÷ 1000 8, 9 Card production
Ring frame production lb per spindle-hour = rpm × 60 ÷ (TPI × 36 × 840 × Ne) 10 Ring frame production
Winding kg/h = m/min × 60 × tex ÷ 10⁶, less stops 11 Winding production
Warping and sizing beam kg = ends × m × tex ÷ 10⁶; pick-up = size ÷ dry yarn 12, 13, 14 Warping, Size pick-up, Warp requirement
Loom production m/h = rpm × 60 × efficiency ÷ (PPI × 39.37) 15 Loom production
Reed count dents per inch = ends ÷ (reed width × ends per dent) 16 Reed count
Fabric GSM and weft GSM = 23.25 × [EPI(1 + c1) ÷ Ne1 + PPI(1 + c2) ÷ Ne2] 17 Fabric GSM
Cover factor K = threads per inch ÷ √Ne; Kc = K1 + K2 − K1K2 ÷ 28 18 Cover factor
Crimp crimp = (yarn − fabric) ÷ fabric; take-up = (yarn − fabric) ÷ yarn 19 Crimp
Knitting stitch length = course length ÷ needles; GSM = loops/m² × l × tex ÷ 10⁶ 20, 21, 22 Stitch length, Knitted GSM, Knitting production

Conventions used throughout: Ne is the English cotton count (840-yard hanks per pound), tex is grams per 1,000 metres, and tex = 590.5 ÷ Ne, the constant every calculator on the site uses. Efficiency is written as a percentage in the question and used as a fraction in the working. The calculators print two decimals, so the answers do too.

Yarn count and conversion questions

Q1. Convert 32s Ne to tex, Nm and denier

Question: a ring-spun cotton yarn is 32s Ne. Express its linear density in tex, metric count (Nm) and denier.

Answer: 18.45 tex, 54.19 Nm and 166.08 denier.

Formula: tex = 590.5 ÷ Ne; Nm = 1000 ÷ tex; denier = 9 × tex.

  1. tex = 590.5 ÷ 32 = 18.4531.
  2. Nm = 1000 ÷ 18.4531 = 54.19 (or Nm = 1.6935 × Ne).
  3. denier = 9 × 18.4531 = 166.08.

Revise: yarn count and linear density. Check: open the calculator with these inputs entered.

Q2. A 150 denier filament in tex, decitex and Ne

Question: a polyester filament yarn is 150 denier. Find its tex, decitex and the English cotton count of a yarn of the same linear density.

Answer: 16.67 tex, 166.67 dtex, 35.43 Ne.

Formula: tex = denier ÷ 9; dtex = 10 × tex; Ne = 590.5 ÷ tex.

  1. tex = 150 ÷ 9 = 16.6667.
  2. dtex = 10 × 16.6667 = 166.67.
  3. Ne = 590.5 ÷ 16.6667 = 35.43. A direct count (denier, tex) rises as the yarn gets coarser; an indirect count (Ne, Nm) falls.

Revise: yarn count and linear density. Check: open the calculator with these inputs entered.

Q3. Card sliver hank to ktex

Question: a card delivers 0.12 hank sliver (0.12 Ne). What is its linear density in ktex, that is, grams per metre?

Answer: 4,920.83 tex, which is 4.92 ktex or 4.92 g per metre.

Formula: tex = 590.5 ÷ hank; ktex = tex ÷ 1000.

  1. tex = 590.5 ÷ 0.12 = 4,920.83.
  2. ktex = 4,920.83 ÷ 1000 = 4.92. Sliver and roving hanks are Ne counts like any other, only very coarse.

Revise: yarn count and linear density. Check: open the calculator with these inputs entered.

Q4. Resultant count of a 30s and a 40s folded together

Question: a 30s Ne and a 40s Ne yarn are twisted together into a two-ply yarn. Ignoring twist contraction, find the resultant count in tex and in Ne.

Answer: 34.45 tex, or 17.14s Ne.

Formula: in a direct system the linear densities add, so tex(ply) = tex1 + tex2. In Ne this becomes 1 ÷ R = 1 ÷ N1 + 1 ÷ N2.

  1. tex of 30s = 590.5 ÷ 30 = 19.6833; tex of 40s = 590.5 ÷ 40 = 14.7625.
  2. tex(ply) = 19.6833 + 14.7625 = 34.45.
  3. Ne = 590.5 ÷ 34.45 = 17.14. The reciprocal rule gives the same: 1 ÷ (1/30 + 1/40) = 17.14.

Two equal yarns, such as 2/60s, simply halve the count (30s). Twist contraction makes the real yarn a little coarser than either figure. The site’s resultant count calculator takes equal plies only, so check this one by entering the ply’s tex in the count converter.

Revise: yarn count and linear density. Check: open the calculator with these inputs entered.

Twist and draft questions

Q5. TPI and TPM from the twist multiplier

Question: a 30s Ne warp yarn is spun with a twist multiplier (TM) of 4.2. Find the twist in turns per inch and turns per metre, and the tex twist factor (TPM × √tex).

Answer: 23.00 TPI, 905.68 TPM, tex twist factor 4,018.14.

Formula: TPI = TM × √Ne (NPTEL, Yarn Manufacture II, ring spun yarns); TPM = TPI × 39.37.

  1. √30 = 5.4772; TPI = 4.2 × 5.4772 = 23.0043.
  2. TPM = 23.0043 ÷ 0.0254 = 905.68.
  3. Tex twist factor = TPM × √tex = 4,018.14. Because tex = 590.5 ÷ Ne, this always equals 956.7 × TM, whatever the count.

Revise: twist. Check: open the calculator with these inputs entered.

Q6. Find the twist multiplier from TPI

Question: a 25s Ne hosiery yarn has 19 turns per inch. What is its twist multiplier, and its twist in turns per metre?

Answer: TM = 3.80; 748.03 TPM.

Formula: TM = TPI ÷ √Ne, the same relation as Q5 turned round.

  1. √25 = 5; TM = 19 ÷ 5 = 3.80.
  2. TPM = 19 × 39.37 = 748.03.
  3. Check by entering TM 3.80 and 25s: the calculator returns 19.00 TPI.

Revise: twist. Check: open the calculator with these inputs entered.

Q7. Total and main draft on the ring frame

Question: 40s Ne yarn is spun from 1.1 hank roving. The break draft is 1.25. Find the total draft and the main (front zone) draft.

Answer: total draft 36.36; main draft 29.09.

Formula: total draft = count delivered ÷ count fed (in Ne, or tex fed ÷ tex delivered); main draft = total draft ÷ break draft.

  1. Total draft = 40 ÷ 1.1 = 36.36. In tex: roving 590.5 ÷ 1.1 = 536.82 tex, divided by yarn 590.5 ÷ 40 = 14.7625 tex, gives the same.
  2. Main draft = 36.36 ÷ 1.25 = 29.09.

Revise: draft. Check: open the calculator with these inputs entered.

Q8. Drawframe draft and production

Question: eight card slivers of 4.8 ktex are fed to a drawframe, which delivers one sliver of 4.2 ktex at 500 m/min with 80% efficiency. Find the draft and the production in kg per hour.

Answer: draft 9.14; production 100.80 kg/h.

Formula: draft = doublings × ktex fed ÷ ktex delivered; kg/h = delivery m/min × 60 × ktex × efficiency ÷ 1000.

  1. Draft = 8 × 4.8 ÷ 4.2 = 9.14.
  2. kg/h = 500 × 60 × 4.2 × 0.80 ÷ 1000 = 100.80. ktex is grams per metre, so m/min × g/m gives g/min.

Revise: draft and card, drawframe and winder production. Check: open the calculator with these inputs entered.

Spinning production questions

Q9. Card production, gross and net

Question: a card delivers 4.5 ktex sliver at 200 m/min with 90% efficiency from a chute feed of 450 ktex. Flat strips are 1.6% and licker-in droppings 1.2%. Find the draft, the gross and net production per hour, and the net output in an 8-hour shift.

Answer: draft 100.00; gross 48.60 kg/h; net 47.24 kg/h; 377.91 kg per shift.

Formula: kg/h = m/min × efficiency × 60 × ktex ÷ 1000; net = gross × (1 − waste fraction).

  1. Draft = 450 ÷ 4.5 = 100.00.
  2. Effective speed = 200 × 0.90 = 180.00 m/min; gross = 180.00 × 60 × 4.5 ÷ 1000 = 48.60 kg/h.
  3. Waste = 1.6 + 1.2 = 2.8%; net = 48.60 × 0.972 = 47.24 kg/h; × 8 h = 377.91 kg.

Revise: card, drawframe and winder production. Check: open the calculator with these inputs entered.

Q10. Ring frame production per spindle and per frame

Question: a ring frame of 1,008 spindles spins 36s Ne at 18,000 rpm spindle speed and 22.5 TPI, with 92% efficiency. Find the front roller delivery, the production per spindle-hour in grams, and the frame’s production per hour and per 8-hour shift.

Answer: delivery 20.32 m/min; 18.40 g per spindle-hour; 18.55 kg/h; 148.38 kg per shift.

Formula: delivery (inches/min) = spindle rpm ÷ TPI; lb per spindle-hour = rpm × 60 ÷ (TPI × 36 × 840 × Ne); 1 lb = 0.45359237 kg (NIST SP 811).

  1. Delivery = 18,000 ÷ 22.5 = 800 inches/min = 800 × 0.0254 = 20.32 m/min.
  2. lb per spindle-hour at 100% = 18,000 × 60 ÷ (22.5 × 36 × 840 × 36) = 0.0441.
  3. g per spindle-hour = 0.0441 × 453.59 × 0.92 = 18.40.
  4. Frame: 18.3997 × 1,008 ÷ 1000 = 18.5469 kg/h, which rounds to 18.55; × 8 = 148.38 kg per shift.

The formula ignores traveller slip and twist contraction, as exam answers normally do. Mills that quote ounces per spindle per shift (OPS) multiply the pounds by 8 hours and by 16: here 5.64 oz at 100% and 5.19 oz at 92%.

Revise: ring frame production. Check: open the calculator with these inputs entered.

Winding, warping and sizing questions

Q11. Drum speed and net winding production

Question: (a) a drum-driven winder has a 90 mm drum running at 4,250 rpm. What is the winding speed? (b) Taking 1,200 m/min, a 60-drum autoconer winds 20 tex yarn. The clearer makes 20 cuts per 100 km, each splice stops the drum 7 s, packages are 2 kg and each doff takes 25 s; other losses leave 90% efficiency. Find the gross and net production per drum and the machine’s net production per hour.

Answer: (a) 1,201.66 m/min. (b) gross 1.44 kg/h per drum; net 1.25 kg/h per drum; machine 75.19 kg/h, an effective efficiency of 87.03%.

Formula: on a drum-driven winder the package is driven by surface contact, so yarn speed is the drum’s surface speed, π × d × n. Production kg/h = m/min × 60 × tex ÷ 10⁶, less the time lost to splices and doffs.

  1. (a) π × 0.090 m × 4,250 = 1,201.66 m/min.
  2. (b) Length wound = 1,200 × 60 = 72 km/h; gross = 72 × 20 ÷ 1000 = 1.44 kg/h.
  3. Cuts = 72 ÷ 100 × 20 = 14.40 per hour; splicing loses 14.40 × 7 s, which is 2.80% of the hour.
  4. Doffs = 1.44 ÷ 2 = 0.72 per hour; × 25 s = 0.50% of the hour.
  5. Net = 1.44 × (1 − 0.028 − 0.005) × 0.90 = 1.2532 kg/h, which rounds to 1.25; × 60 drums = 75.19 kg/h.

Revise: card, drawframe and winder production. Check: open the calculator with these inputs entered.

Q12. Warping: run time and beam weight

Question: a direct warper runs at 900 m/min with 70% efficiency. A beam carries 640 ends of 30s Ne, 18,000 m long. Find the effective production in metres per hour, the time to build one beam and the yarn weight on it.

Answer: 37,800.00 m/h; 0.48 h (about 28.6 minutes) per beam; 226.75 kg of yarn.

Formula: m/h = m/min × 60 × efficiency; beam kg = ends × length (m) × tex ÷ 10⁶.

  1. Production = 900 × 60 × 0.70 = 37,800.00 m/h.
  2. Run time = 18,000 ÷ 37,800.00 = 0.48 h.
  3. tex = 590.5 ÷ 30 = 19.6833; beam kg = 640 × 18,000 × 19.6833 ÷ 10⁶ = 226.75.

Revise: warping and sizing. Check: open the calculator with these inputs entered.

Q13. Size pick-up

Question: a warp sample weighs 120 g oven-dry before sizing and 134.4 g oven-dry after. Find the size pick-up, and the dry size carried by a set of 1,800 kg of unsized warp at that pick-up.

Answer: pick-up 12.00%; 216.00 kg of size; sized warp 2,016.00 kg.

Formula: size pick-up % = (sized dry mass − unsized dry mass) ÷ unsized dry mass × 100.

  1. Size added = 134.4 − 120 = 14.4 g; pick-up = 14.4 ÷ 120 × 100 = 12.00%.
  2. Size on the set = 1,800 × 0.12 = 216.00 kg; sized warp = 2,016.00 kg.

The base is the unsized yarn, not the sized yarn. Dividing by 134.4 gives 10.71%, a common wrong answer.

Revise: warping and sizing. Check: open the calculator with these inputs entered.

Q14. Warp yarn to order for a fabric

Question: an order needs 2,000 m of fabric with 5,400 ends of 40s Ne. Warp crimp is 8% and preparation waste 2%. How much warp yarn should be ordered?

Answer: 175.63 kg.

Formula: warp kg = ends × fabric length (m) × tex ÷ 10⁶ × (1 + crimp) × (1 + waste).

  1. tex = 590.5 ÷ 40 = 14.7625; net yarn = 5,400 × 2,000 × 14.7625 ÷ 10⁶ = 159.4350 kg.
  2. Crimp allowance = 159.4350 × 0.08 = 12.7548 kg.
  3. Waste allowance = (159.4350 + 12.7548) × 0.02 = 3.44 kg; total 175.63 kg.

Revise: yarn requirement. Check: open the calculator with these inputs entered.

Weaving questions

Q15. Loom production

Question: an air-jet loom runs at 750 rpm with 88% efficiency, weaving 72 PPI cloth 67 inches wide. Find the production in metres per hour, metres per 8-hour shift and square metres per hour.

Answer: 13.97 m/h; 111.76 m per shift; 23.77 m²/h.

Formula: a single-width loom inserts one pick per revolution, so metres per hour = rpm × 60 × efficiency ÷ PPI × 0.0254.

  1. Picks per hour = 750 × 60 × 0.88 = 39,600.00.
  2. Inches woven = 39,600.00 ÷ 72 = 550; metres = 550 × 0.0254 = 13.97 per hour.
  3. Per shift = 13.97 × 8 = 111.76 m; area = 13.97 × (67 × 0.0254) = 23.77 m²/h.

Revise: loom production. Check: open the calculator with these inputs entered.

Q16. Reed count

Question: a cloth 58 inches wide has 4,800 ends. Weft crimp is 5% and the warp is drawn 2 ends per dent. Find the reed width, the reed count in dents per inch and in the Stockport system, and the EPI in the reed and in the cloth.

Answer: reed width 60.90 inches; 39.41 dents per inch, or 78.82 Stockport; 78.82 EPI in the reed and 82.76 in the cloth.

Formula: reed width = cloth width × (1 + weft crimp); dents per inch = ends ÷ (reed width × ends per dent); Stockport count = dents in 2 inches.

  1. Reed width = 58 × 1.05 = 60.90 inches.
  2. Dents per inch = 4,800 ÷ (60.90 × 2) = 39.41; Stockport = 2 × 39.4089 = 78.82.
  3. EPI in reed = 4,800 ÷ 60.90 = 78.82; in cloth = 4,800 ÷ 58 = 82.76. The cloth is denser than the reed because the weft draws it in.

Revise: ends, picks and the reed. Check: open the calculator with these inputs entered.

Q17. Fabric GSM and yarn for 100 metres

Question: a grey cloth is 132 × 72, 40s × 40s, 63 inches wide, with 7% warp crimp and 5% weft crimp. Find its weight in g/m², and the warp and weft needed for 100 m of cloth with 2% process waste.

Answer: 126.04 g/m²; warp 13.40 kg and weft 7.17 kg per 100 m.

Formula: GSM = 23.25 × [EPI × (1 + warp crimp) ÷ warp Ne + PPI × (1 + weft crimp) ÷ weft Ne]. The 23.25 is 39.37 threads-per-inch to per-metre × 590.5 ÷ 1000.

  1. Warp term = 132 × 1.07 ÷ 40 = 3.5310; weft term = 72 × 1.05 ÷ 40 = 1.8900.
  2. GSM = 23.25 × (3.5310 + 1.8900) = 126.04.
  3. Warp: 8,316 ends × 100 m × 1.07 × 1.02 × 14.7625 tex ÷ 10⁶ = 13.40 kg. Weft: 72 picks per inch × 63 inches × 100 m × 1.05 × 1.02 × 14.7625 ÷ 10⁶ = 7.17 kg.

Selvedge ends and the extra width in the reed are left out, as in most exam answers; add them when they are given.

Revise: fabric weight from the construction and yarn requirement. Check: open the calculator with these inputs entered.

Q18. Cover factor

Question: for the 132 × 72, 40s × 40s cloth in Q17, find the warp, weft and cloth cover factors, and the fraction of the area covered.

Answer: warp 20.87, weft 11.38, cloth 23.77; 84.89% of the area covered.

Formula: K = threads per inch ÷ √Ne; cloth cover Kc = K1 + K2 − K1 × K2 ÷ 28, where 28 is full cover, because Peirce took the diameter of a cotton yarn as 1 ÷ (28√Ne) inch (Peirce, “The geometry of cloth structure”, 1937).

  1. √40 = 6.3246; K1 = 132 ÷ 6.3246 = 20.87; K2 = 72 ÷ 6.3246 = 11.38.
  2. Kc = 20.87 + 11.38 − (20.87 × 11.38 ÷ 28) = 23.77.
  3. Area covered = 23.77 ÷ 28 = 84.89%.

Revise: cover factor. Check: open the calculator with these inputs entered.

Q19. Crimp and take-up

Question: a 20 cm × 20 cm fabric sample is unravelled. A warp thread straightened under tension measures 21.3 cm and a weft thread 20.9 cm. Find the warp and weft crimp and the corresponding take-up.

Answer: warp crimp 6.50%, warp take-up 6.10%; weft crimp 4.50%, weft take-up 4.31%.

Formula: crimp % = (yarn length − fabric length) ÷ fabric length × 100; take-up % = (yarn length − fabric length) ÷ yarn length × 100. Crimp measurement is standardised in ISO 7211-3.

  1. Warp crimp = (21.3 − 20) ÷ 20 × 100 = 6.50%; take-up = 1.3 ÷ 21.3 × 100 = 6.10%.
  2. Weft crimp = 0.9 ÷ 20 × 100 = 4.50%; take-up = 0.9 ÷ 20.9 × 100 = 4.31%.

Revise: crimp and take-up. Check: open the calculator with these inputs entered.

Knitting questions

Q20. Stitch length and tightness factor

Question: one course unravelled from a single jersey fabric knitted on 2,640 needles measures 7,656 mm. The yarn is 30s Ne. Find the stitch length and the tightness factor.

Answer: stitch length 2.90 mm; tightness factor 1.53 with the loop in millimetres (15.3 with the loop in centimetres).

Formula: stitch length = course length ÷ needles; tightness factor = √tex ÷ loop length. NPTEL’s knitting course works it with the loop in centimetres and gives 14.7 as the commercial value for plain single jersey (NPTEL, Knitting Technology, module 5); the site’s calculator uses millimetres, so its figures are a tenth as large.

  1. Stitch length = 7,656 ÷ 2,640 = 2.90 mm.
  2. tex = 590.5 ÷ 30 = 19.6833; √tex = 4.4366.
  3. TF = 4.4366 ÷ 2.90 = 1.53 (per mm), or 4.4366 ÷ 0.290 = 15.3 (per cm): a little tighter than 14.7.

Revise: the formula section of the stitch length calculator. Check: open the calculator with these inputs entered.

Q21. Knitted GSM from courses, wales and loop length

Question: a single jersey fabric has 44 courses and 34 wales per inch, a stitch length of 2.75 mm, and is knitted from 26s Ne. Find the stitch density per cm² and the fabric weight in g/m².

Answer: 231.88 loops per cm²; 144.82 g/m².

Formula: GSM = loops per m² × loop length (m) × tex ÷ 1000, the relation NPTEL’s knitting course uses (module 5).

  1. Loops per in² = 44 × 34 = 1,496; per cm² = 1,496 ÷ 6.4516 = 231.88.
  2. Loops per m² = (44 × 39.37) × (34 × 39.37) = 2,318,804.64.
  3. Yarn per m² = 2,318,804.64 × 2.75 ÷ 1000 = 6,376.71 m.
  4. tex = 590.5 ÷ 26 = 22.7115; GSM = 6,376.71 × 22.7115 ÷ 1000 = 144.82.

Revise: the formula section of the knitted GSM calculator. Check: open the calculator with these inputs entered.

Q22. Circular knitting production

Question: a 30 inch, 24 gauge single jersey machine with 90 feeders runs at 28 rpm and 85% efficiency, knitting 30s Ne at 2.8 mm stitch length. Find the number of needles, and the production in kg per hour and per 8-hour shift.

Answer: 2,262 needles; 16.02 kg/h; 128.18 kg per shift.

Formula: needles = π × diameter × gauge; courses per minute = rpm × feeders; kg/h = courses/min × needles × stitch length × 60 × tex × efficiency ÷ 10⁹ (stitch length in mm).

  1. Needles = π × 30 × 24 = 2,261.95, so 2,262.
  2. Courses per minute = 28 × 90 = 2,520.00; loops per minute = 2,520.00 × 2,262 = 5,700,240.00.
  3. Yarn per hour = 5,700,240.00 × 2.8 mm × 60 ÷ 1000 = 957,640.32 m.
  4. kg/h = 957,640.32 × 19.6833 ÷ 10⁶ × 0.85 = 16.0221, which rounds to 16.02; × 8 = 128.18 kg.

Revise: the formula section of the circular knitting production calculator. Check: open the calculator with these inputs entered.

Mistakes that cost marks

  • Treating Ne as a direct count. A higher Ne is a finer, lighter yarn; divide by it, as in Q1, Q4 and every production formula.
  • Leaving efficiency as a percentage. 92% enters the arithmetic as 0.92.
  • Mixing inches and metres. TPI, EPI, PPI and loom width are per inch; 1 inch = 0.0254 m exactly, so 39.37 inches make a metre (NIST SP 811).
  • Confusing crimp with take-up (Q19), and dividing size pick-up by the sized weight (Q13).
  • Giving the reed count without its system: dents per inch, Stockport (dents per 2 inches) or metric (dents per 10 cm) (Q16).
  • Comparing a tightness factor on the wrong scale: per millimetre and per centimetre differ by ten (Q20).

Questions students ask

Is there a spinning or weaving calculation book in PDF?

The site’s two formula pages each carry free revision notes in PDF: Yarn spinning formulas (count, draft, twist, and card, drawframe, ring frame and winder production) and Weaving calculations (reed, GSM, cover factor, crimp, yarn requirement, warping, sizing and loom production). This question page is not a PDF, but it prints cleanly from the browser.

What is the ring frame production formula, and what is “ops”?

Production per spindle in pounds per hour is spindle rpm × 60 ÷ (TPI × 36 × 840 × Ne): rpm ÷ TPI is the delivery in inches per minute, 36 converts inches to yards and 840 yards make one hank. The query “ops = (spindle speed × 60 × 8) ÷ (yarn count × 840 × 36 × tpi) × 16” is the same formula for an 8-hour shift, with × 16 turning pounds into ounces per spindle per shift. Q10 works it through.

How do you convert TPI to TPM, and TM to TPI?

TPM = TPI × 39.37 and TPI = TPM ÷ 39.37, because a metre is 39.37 inches. TPI = TM × √Ne, so TM = TPI ÷ √Ne. Q5 and Q6 show both directions.

Winding speed depends on what?

On a drum-driven (random) winder, where the package rests on a grooved drum or cylinder, the yarn speed is the drum’s surface speed, π × drum diameter × drum rpm, and it stays constant as the package grows. On a spindle-driven winder the speed is π × package diameter × spindle rpm, so it rises as the package builds unless the spindle slows. Q11 computes the drum case.

What is the loom production formula?

Metres per hour = loom rpm × 60 × efficiency ÷ (PPI × 39.37). Picks inserted per hour divided by picks per inch gives the inches of cloth woven; dividing by 39.37 gives metres. Q15 works an example.

Where are the knitting calculation formulas?

Q20 to Q22 cover the three that exams set most: stitch length from course length, GSM from courses, wales and loop length, and circular machine production. Each calculator page linked from those questions sets out its formula and a worked example.

Sources

How this page is kept

Last verified: 18 October 2026. The questions and their numbers were written for this page; none is copied from a past paper. On that date every answer was computed with the formula of the site’s calculator named in the question, and each “check” link was opened to confirm the calculator shows the figures quoted. The few figures no calculator shows (Q4’s folded count, Q10’s ounces, Q11’s drum speed, Q16’s Stockport count, Q20’s per-centimetre factor and Q22’s needle count) were computed from the formula cited beside them. If a calculator changes, the answers are re-checked against it. Corrections are welcome through the contact page: name the question and the source that disagrees.

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