RE: Conical primary length formula

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Godfrey Loudner" <ggreen-at-gwtc-dot-net> 

Hello Michael

In my previous posting of a conical primary length
formula, I wrote that you could take P = 90 degrees
to get the formula for your straight helix, but this
would lead to the indeterminate form 0/0.
http://www.pupman-dot-com/listarchives/2004/May/msg00023.html
Sorry I did not notice this before. Rather, you would
have to take the limit as P -> 90 degrees, using
l'Hospital rule. This is too much work. Its easier
to go back and do the calculus for your straight
helix. The copper tube starts against the imaginary
helix and the plane of the base of the helix.

For the length of the filament along the center of
the copper tube,

L = n[4Pi^2(r+d/2)^2 + (G+d)^2]^(1/2).

For the length of the filament where the copper tube is
in contact with the imaginary helix,

L =  n[4Pi^2r^2 + (G+d)^2]^(1/2).

Either one of these is a good estimator for the length of
the copper tube. But for me there is a problem. If one
deforms a straight copper tube into the shape of a ring, the
copper metal would experience compression and stretching.
It seems like the copper on the outside radius of the ring
would be stretched, while the copper on the inside radius
would be compressed. I really don't know anything about
the physics of the deformation of soft metals. I don't
think its a problem worth working on for coiling.

Godfrey Loudner