From 1998:
http://mathforum.org/library/drmath/view/53155.html
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Date: 11/16/98 at 12:19:55
From: Doctor Peterson
Subject: Re: Calculating the length of two twisted copper wires
Hi, Roland. If we picture one of your wires making a helix of radius R,
with one turn taking H inches, it will look like this:
|<--R-->|
| |
******|****** |
** **
| ******o****** |----
| o| | ^
| o | | |
|. o | | |
ooo. | | |
| . | | |
| .| | |
| |. | |
| | . | |
| | . | H
| | .| |
| | | |
| | | |
| | o |
| | o |
| | o| |
| | o | |
| ******|****o* | |
** | o ** v
******o******------
The length of the wire can be found by unwrapping it from the cylinder
to form a right triangle:
o
o |
o |
o |
o |
o |
o |
o |
o |
L o | H
o |
o |
o |
o |
o |
o |
o |
o |
o |
*********************************************************
2 pi R
We'll get:
L = sqrt(H^2 + (2 pi R)^2)
as the length of one turn of wire, so the ratio of the length of the
twisted pair to the length of the wire will be:
L / H = sqrt(1 + 4 pi^2 (R/H)^2)
The radius of the cylinder about which the center of the wire will
spiral will be about the same as the radius of the wire itself:
******** oooooooo ********
*** ooo *ooo ***
** oo ** ** oo **
* o * * o *
* o * R o *
* o *----------o *
* o * o *
* o * * o *
* o * * o *
**** oooo* **oooo ****
******** oooooooo ********
So in your example, with R = .020 in and H = 1 in, we get:
L / H = sqrt(1 + 4 pi^2 .020^2) = 1.0079
and for a 100 ft pair, each wire is 1.0079 * 100 ft = 100.79 ft.
That doesn't sound like much difference. It will increase if the wires
don't touch tightly all the way around. You may have to adjust the
radius to correspond to reality!
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