How to calculate how many yards of  warp thread to put on each spool when warping in sections.
 
by Tom Vogl, May 2001
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Figuring out how much warp to put on each spool for sectional warping requires a relatively complicated thought process but is, intrinsically, quite simple. Having to go through the thought process all over each time we warp is a not only nuisance but there is always the residual question in our minds whether we did it right this time.  So, to do it once and for all, here is an algorithm for the calculation. What is needed is to put onto each spool the yards of cloth desired plus loom waste times the number of sections that will be used.  An allowance must be made for the waste involved in cutting and tying each section.  The important thing to remember is that the number of 'cloth lengths' on each spool is determined by the number of sections; the number of spools is determined by the number of ends per section; and the cutting waste is determined by the number of times the warp will be cut an retied, that is the number of sections.

So, let
e = end per inch in warp (epi)
i = width of section in inches
s = number of spools required
k = yards of cloth to be woven (k includes loom [but not cutting] waste)
w = width in reed in inches
c = cut waste per section
y = yards of warp on each spool
Thus e x i = s spools will be required.
Each of the s spools must be would with
y = (w/i) x k + (c x [w/i]) yards.

An example:
Want to weave k = 25 yards (including loom waste) of cloth at e = 30 epi, with w = 20 inches wide in the reed.
With a i = 1" sectional beam, will need s = 30 spools each with 20/1 x 25 + (1/2 x 20/1) = y = 510 yards/spool.
With an i = 2" sectional beam, will need s = 60 spools, each with 20/2 x 25 + (1/2 x 20/2) = y = 255 yards per spool.

Note that as a check, the total number of yards (s x y) is the same.  A further check is to calculate e x k x w which is the total length of warp required excluding cutting waste.

In the example, this is 30 x 25 x 20 = 15,000 yards.  This is to be compared to s x y, which for the examples is 30 x 510 = 60 x 255 = 15,300 yards.  The 300 yard difference is the cutting waste allowance.

We deliberately make the cutting waste allowance quite generous, since there is little more frustrating than winding on 19 sections and discovering that you are two yards short on the 20th one.  After all, 300 out of 15,000 is only 2%.
These calculations also point out why it is less trouble (and requires a smaller spool rack) to use one inch rather than two inch sections.

 
 
 
 
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