capacitance of horned toroid

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

Hello Antonio

refer to http://www.pupman-dot-com/listarchives/2004/February/msg01419.html and
http://www.pupman-dot-com/listarchives/2004/March/msg00164.html.
I have not yet heard back from NASA, so I went ahead with
my own developments.

With the above, the capacitance can be written as C =

8(Pi)(permittivity)d times the below

Sum[a(n)^(-1)BesselJ[1,a(n)]^(-2)(-1+Integrate[BesselJ[0,t],{t,0,a(n)}]),{n,
1,Infinity}].

Borrowing an asymptotic expansion for the Struve function H(z)
and some properties of Bessel functions, I have derived
an asymptotic expansion for the terms of the series above
where n is large. I'll send the details to your University
address after I get them written up in logical order. The
expansion is a finite sum of p terms. By increasing p,
the accuracy of the expansion increases. But p = 1 gives the
dominate term. For p > 1, the terms are very small and would
only become valuable if one were going for many significant
figures. Using the dominate term only,

BesselJ[1,a(n)}^(-1)Integrate[E^(-a(n)Sinh[t]),{t,0,Infinity}]

is approximated by (-1)^(n+1) 2^(1/2) (4n-1)^(-1/2).

For n = 1000, Mathematica gives -0.0223635 for the integral
expression. The asymptotic expression gives -0.0223635. This
is real good. So my idea is to compute with Mathematica Pi
times the sum of the first 3000 terms of the series at the
top of the page. Then sum, using asymptotic terms, the rest
of the terms from n = 3001 to Infinity. The asymptotic terms
sum to Zeta functions, which Mathematica handles quickly.

Using machine precision only, Mathematica gave the sum of the
first 3000 terms as 1.3474. The contribution from the asymptotic
terms came to 0.020278. The sum of the two is 1.367678. As in
the posts above, the best I could get using the Legendre stuff
was 1.36765, which was still growing. Why did I not get a better
value with Mathematica---with the machine precision increased to
20 significant figures for the formation of big tables of numbers,
I lose that precision when I perform operations on the tables. I
don't see how to get around this yet. I'm not an artist with
Mathematica. I'll figure out some way so I can give a better
value of 1.3077 below. Maybe out of the blue, NASA will send the
number.

So the best so far in the MKS system of units is

C = 8(permittivity)(1.3677)d.

Your previous calculation with d = 0.4999, gave C = 48.43867pF.
My calculation gives C = 48.42980pF.

BTW, Bert Pool's formula gives 38.00029pF. It surely breaks
down when (D-d)/d is close to 1.

Godfrey Loudner