RE: Conical primary length formula

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

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

Hello Michael

Imagine that the copper tubing is wound around a cone with
bottom radius r. Take

P = angle of the cone edge with the horizontal,

d = the diameter of the copper tubing,

G = the distance between the turns, and

n = the number of turns.

The coil must start at the bottom of the cone. Take

R = r + d(1 + Cos[P])/(2Sin[P]),

A = R + (G + d)nCos[P], and

B = (G + d)/2Pi.

The formula for the length is Pi/((G + d)Cos[P]) times the below

A(A^2+B^2)^(1/2) - R(R^2 + B^2)^(1/2) +

((G + d)/(2Pi))^2Log[(A+(A^2+B^2)^(1/2))/(R+(R^2+B^2)^(1/2))].

Note that x^2 is x squared, x^(1/2) is the square root of x,
and Log[x] is the natural logarithm of x.

Note that the length formula is the distance along the center
of the copper tube. This should be close enough. I suppose one
could define the length to be an average of all the lengths
through the tube, but is this really worth the effort?

The helix formula is gotten from the above with P = 90 degrees.

Good luck with this mess.

Godfrey Loudner

Does anyone know a formula to find the length of tubing required for
a conical primary sloped at a given angle?  I found one for a flat
spiral and made a guess at a derivation for a straight helix, but
couldn't figure this one out.  Also, if you have one for a straight
helix, would you mind posting it so I can compare what I got?

Thanks!
Michael Johnson