Jerel

From 1998: http://mathforum.org/library/drmath/view/53155.html *** 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! ***