spiral length

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

Original poster: "Godfrey Loudner by way of Terry Fritz <twftesla-at-qwest-dot-net>" <ggreen-at-gwtc-dot-net>

Hi All

Because I was wondering about the accuracy of the approximate formula for
the length of the spiral of Archimedes, I did a bit of analysis. The spiral
is assumed to be a filament. Let d = the distance for the center of the
spiral to the beginning of the first turn, G = the spacing between the
turns, and n = the number of turns. Of course d > 0 and G > 0. Unfortunately
I must use tt and x^2 as symbols for Pi and x squared respectively. The
symbols ln{x} and Sq[x] denote the natural logarithm of x and the square
root of x  respectively. The exact formula for the length of a spiral is
given by the expression A + B where

A = (tt/G)(d + nG)Sq[(d + nG)^2 + G^2/(4tt^2)] - (ttd/G)Sq[d^2 + G^2/(4tt^2]
and

B = (G/4tt)ln{(d + nG + Sq[(d + nG)^2 + G^2/(4tt^2)])/(d + Sq[d^2 +
G^2/(4tt^2)])}.

The traditional approximation formula used by coilers is ttn(2d + nG). It
turns out that

0 < A - ttn(2d + nG) < G/(8tt) for n > 0. There is even a finer inequality

0 < A - ttn(2d + nG) < G/(8tt) - ttd(Sq[d^2/G^2 + 1/(4tt^2)] - d/G) <
G/(8tt) for n > 0.

It also turns out that

0 < B < (G/4tt)ln{4ttn + 1} for n > 0. Finer inequalities are possible, but
this one is good enough

for our purposes. This inequality says that for tesla coiling applications,
B is so small that it can be neglected. For example if
G = 1" and n = 25, 0 < B < 0.46".

The above inequalities imply that for the range of parameters used by
coilers, ttn(2d + nG) is a excellent approximation
for the length of the spiral of Archimedes. I give a table below to
strengthen the point with d = 11", G = 1/4".

                turns     Exact length                    Approx. length

                1          69.9009"                        69.9004"
                5          365.2123"                      365.2101"
                10        769.6943"                      769.6902"
                15        1213.4460"                    1213.4402"
                20        1696.4675"                    1696.4600"
                25        2218.7588"                    2218.7498"
                30        2780.3198"                    2780.3095"
                100      14,765.5091"                 14,765.4855"
                1000    854,513.2648"               854,513.2018"

I do see a problem in using the above exact or approx. formulas to estimate
the length of copper tubing
needed for given d, G, and n. Suppose we are using 1/4" outside diameter
copper tubing, a spacing
between turns of 1/4", a distance of 11" from the center to the inside of
the beginning of the first
turn, and n = 20 turns. Here I use the exact formula. Coming from d = 11", G
= 1/2", and n = 20 turns, the length of the inside filament of the copper
spiral is 2010.65". Coming from d = 11.25", G = 1/2", and n = 20 turns, the
length of the
outside filament of the copper spiral is 2042.06". Hence the outside length
is 31.4105" more than the inside length of the copper tube. If a straight
length of copper tubing is bent, the copper will surely experience
stretching and compression. Considering that copper tubing is sold in
spools, the copper has already experienced the majority of the stretching
and compression. Perhaps I would use the outside filament length to estimate
the length of copper tubing needed to fashion a
flat spiral primary.

The length of a filament spiral has a clear meaning, but the length of a
copper tube spiral is ambiguous.

The approximate formula ttn(2d + nG) used by coilers for the length of a
filament spiral can be interpreted as
the sum over k = 1, 2, ... , n of the circumferences of circles of radii d +
(2k - 1)G/2.

Godfrey Loudner