Re: Conical primary length formula

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

Original poster: "Richard Modistach" <hambone-at-dodo-dot-com.au> 

wouldn't it be easier to use tesla map or some other program.
all the specs.are plastered right in front of you.
just punch in the parameters.

regards
richard
aus



----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Monday, May 10, 2004 10:20 AM
Subject: RE: Conical primary length formula


 > 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
 >
 >
 >
 >
 >
 >
 >
 >