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RE: Conical primary length formula



Original poster: "Daniel Kline" <daniel_kline-at-med.unc.edu> 



 > -----Original Message-----
 > From: Tesla list [mailto:tesla-at-pupman-dot-com]
 > Sent: Thursday, April 29, 2004 6:50 PM
 > To: tesla-at-pupman-dot-com
 > Subject: Conical primary length formula
 >
 >
 > Original poster: pepperman-at-softhome-dot-net
 >
 > 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

For a straight helix:

N * (sqrt( ( (pi*D)^2) + ( (d)^2) ) )

where:

N = Number-of-turns
D = diameter of helix
d = pitch, i.e., the spacing between the turns

Basically, it's like this (just in case I didn't get my parenthesis
right ;)

1) Find the Diameter of the cylinder around which the helix is wrapped
(either real or imaginary),
multiply times pi, square the total.

2) Measure the center-to-center spacing between the turns of the helix
and square that amount.

3) Add the totals from 1) and 2) together, and take the square root of
the new total.

4) Multiply the result from 3) by the total number of turns to get the
length of the helix.
(Multiply by 1 to get the length of a single turn)

Visualization:

Picture a right-triangle where the length of one 90-degree leg is the
circumference of your form,
and the length of the other 90-degree leg is the spacing between the
turns.
The hypoteneuse is the actual length of a single turn.
Apply Pythagorian Theorem to calculate the length, and multiply by
number of turns :)

Hope this helps,
Dan K.
P.S. This works for space-wound secondaries too.