How to determine number of turns needed to fill spool with known length of wire?

Question

I've been scratching my head on this problem for a while, even bounced it off of some friends who are way better at math than myself with no luck, so I'm hoping someone here can help. I'm trying to find a formula to determine the number of turns or rather complete revolutions a spool will make to fill it with X amount of wire in feet. For example, a spool with a traverse of 20mm, barrel diameter of 70mm, with 100 feet of 1mm wire. The simple estimate is turns = length / circumference. What that doesn't take into account however is that each time the wire completes a layer from flange to flange on the spool, the circumference increases by 2 times the wire diameter, multiplied by pi. Help!

Answer 1

A simple approach which will not be far wrong is to expand the wire to a square or hexagonal cross section. These fill space so you can compute the volume the wire occupies from its length, then find the outer radius of the coil that results in that volume. Take the cross section of the coil and divide by the area your circumscribing shape represents and you have the number of turns. Square is less dense than hexagonal. Depending on how you wind the coil it may be less dense than that, but this gives an answer with basic algebra.

Answer 2

First determine packing factor $f$, fraction of thread occupied in close packing or filling. For tight packing in a rectangular packing it is $ \pi/4 \approx 0.785$ and hexagonal $ \pi \sqrt 3/6 \approx 0.907;$

Let inner, outer radii be $(r_i,r_o)= (30,...)$ respectively and spool width $W (= 20)$mm.

Equate volumes

$$\frac{\pi d^2}{4f}\cdot X = \pi(r_o^2-r_i^2)W;\, (r_o^2-r_i^2)=\frac{Xd^2}{4Wf}=a^2; \tag1$$

Here $d=1$mm, outer radius $r_o$ can be calculated.

Average perimeter

$$ p = \pi ( r_i+r_o) \tag2 $$

Next total length $L =100 $ feet is given, convert it to mm, $ n\cdot p= L $

$$ n= \frac{L}{p} = \dfrac{ L}{\pi( r_i+\sqrt{r_i^2+ a^2} )} \tag3$$

is the required formula.