Re: toroid shaped coil capacitance
From: Mark S. Rzeszotarski, Ph.D.[SMTP:msr7-at-po.cwru.edu]
Sent: Sunday, January 04, 1998 8:50 AM
To: Tesla List
Subject: Re: toroid shaped coil capacitance
At 06:57 PM 1/2/98 -0600, you wrote:>
>From: D.C. Cox[SMTP:DR.RESONANCE-at-next-wave-dot-net]
>Do your reference books contain any equations for the "suface area" of both
>a sphere and torus? I'm working on an article for the TCBA and have the
>equations for capacitance but not surface area. Thanks in advance.
For a sphere of radius R, the surface area is 4 pi R^2, where pi=3.14159...
More typically, you have a piece of a sphere, because the bottom and/or top
or both have been flattened. In this case assume the upper radius is b and
the lower radius is a, where a and b are the radii of the circles which
define the upper and lower edge of the cut-away of the sphere. If only one
is present, a or b goes to zero. In this case the surface area is simply 2
pi R h, where R is the largest sphere radius, and h is the perpendicular
distance between the circles of radii a and b. Note that in the limit where
a and b both go to zero, h becomes 2R, and the formula becomes that of a
For a torus the surface area is 4 pi^2 R r, where R is the radius from the
center of the torus to the center of the circle of the donut, and r is the
radius of the donut. Hence, the i.d. of the torus is 2(R-r) and the outer
dimension is 2(R+r). You might want to include the surface area of the
circle [ pi (R-r)^2 ] which connects things together, although
electrostatically this seems to be rather unimportant in a firing tesla coil.
Isotropic capacitance of an isolated sphere or toroid should follow a
proportionality with surface area, anthough in the case of a toroid, it may
be more closely related to one half of the surface area, since charge
distributes itself primarily on the outside surface of the object.
Mark S. Rzeszotarski, Ph.D.