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Weaving Calculations · Lecture 06

Sett Calculations – Beams and Warping Production

How many warper’s beams an order needs, how many ends go on each, and how long the warper will take.
Creel capacityBeamsWarping productionOrder time

From total ends to beams

A sized weaver’s beam is made from several warper’s beams run together on the sizing machine. The warper’s creel holds a limited number of cones (here 840), so the total ends — including selvedge ends — are divided over an even number of beams (beams are paired on the sizing creel).

beams = round-up-to-even( total ends ÷ creel capacity )   ·   ends per beam = total ends ÷ beams
Number of beams
ABC
1Fabric AFabric B
2Ends per inch
3Width (in)
4Extra selvedge ends
5Creel capacity (cones)
6Total ends incl. selvedge3,8208,104
7Beams calculated4.559.65
8Beams used (round up to even)610
9Ends per beam (average)636.67810.40
10… in practice2 beams × 636 + 4 beams × 6376 beams × 810 + 4 beams × 811
blue = input (edit me)grey = formula (click to see it)

Warping production and time for an order

Warping production (fabric A)
AB
1Warper
2Warping speed (yd/min)
3Efficiency
4Shift (hours)
5Ends per beam
6Warp count Ne
7Yards warped per shift268,800
8Weight warped per shift (lb)10,187
9Weight warped per hour (lb)1,273
10Order
11Fabric ordered (m)
12Warp contraction
13Total ends in fabric
14Fabric (yd)21,872
15Warp length required (yd)24,059
16Warp weight required (lb)5,413
17Warping time (hours)4.3
blue = input (edit me)grey = formula (click to see it)
A 20,000 m order of fabric A needs 5,413 lb of warp and keeps the warper busy for only about 4.3 hours — warping is rarely the bottleneck; sizing and loom capacity are (Lecture 10).

Original lecture material

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