Re: Charge distribution on a Toroid (was spheres vs toroids)

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

Original poster: Paul Nicholson <paul-at-abelian.demon.co.uk> 

Antonio wrote:
 > Size      Bela    Tssp (C  Max field   Breakout) Inca (C   Breakout)
 > 12 x 3:   302 kV  14.01pF, 9.87 V/m/V,  304kV    13.11 pF  250.6 kV

All the tssp C values for the toroids are all wrong, as in the example
above .... hmm I seem to have messed up the conversion of the input
dimensions, confusing the radius of the central hole with the tube
radius!   Must have been a Friday!

Here are the corrected figures...

Size    Bela  Tssp________________   Inca_________
         kV    C pf   V/m/V    kV      C pF     kV
12x3    302   13.10  11.97   250.6   13.11   250.6
16x4    386   17.47   8.97   334.4   17.48   334.2
20x5    458   21.84   7.18   417.8   21.84   417.7
26x6    532   28.02   5.75   521.7   28.03   522.0
34x8.5  795   37.12   4.22   710.9   37.14   710.1
48x12  1353   52.41   2.99  1003     52.43  1003.
12 sph  850   33.89   3.33   901     33.91   914.4

Now we're back on track. I think this demonstrates that for toroids
and spheres the decomposition into tubular rings with 1/pi spacing
ratio is working at least as well as decomposition into flat tape
rings, and offers the added advantage that the potential coefficients
are easier and quicker to calculate.

Antonio, could you summarise for us now the formulas you're using for
the Pij and the Pii coefficients?

The Bela breakout voltages now seem to be on the high side.

 > I can try an open hemisphere, that has also a simple expression for
 > capacitance...

I don't have an open hemisphere in tcap, I'd better put one in...
...ok 'tis done, here's what we get:

 > Hemisphere with 1 meter of diameter:
 > Exact:     45.5246270366 pF
             Inca               Tssp (tcap)
10 rings:  44.9288700205 pF    44.53 pF
20 rings:  45.2259355350 pF    45.00 pF
40 rings:  45.3749442201 pF    45.26 pF
80 rings:  45.4496825615 pF    45.39 pF
200 rings: 45.4946218478 pF    45.48 pF

The tubular rings seem to be converging better than the flat tapes do.

Can we have a look at a thin circular disk?
Say 1 metre diam, C = 4 * e0 * diam = 35.41675 pF

rings     tssp
  10       34.297 pF
  20       34.833 pF
  40       35.114 pF
  80       35.263 pF
200       35.405 pF

Godfrey wrote:
 > I can't pull up the document.

These are GNU 'gzip' compressed postscript files.  Your web browser
should know how to uncompress gz files on-the-fly.  Can 'ms-word'
read postscript - I'm not sure?  Maybe you'll have to install
a tool such as ghostview.   Postscript is a very good, very well
established standard that's been around for decades. Everything
speaks postscript - except in windows land!

While we're on the subject of postscript files, here's another one.
Pages 2 and 3 give a concise account of the boundary element method
and show how the equivalent charge formulation can be fitted in to
the overall scheme to deal with dielectrics:

  http://faculty.smu.edu/tausch/Papers/mtt1.ps.gz

Gerry wrote:
 > tell me more about gcc and does this work in windows or DOS.

gcc is the GNU C compiler, which is available for almost all systems,
except windows and DOS.  There are variants of gcc available for
these (djb, cygwin?) I think.  It will compile/cross-compile for
any (32 bits or more) machine from a PC to a cray - you just tell it
the target machine and OS type, and will compile C, C++, and assembler
code.  It is now a very strong standard in the C programming world.
And like all the best quality software, it doesn't cost anything!
It's entirely open source so you can inspect and modify it.
The trouble with DOS is it's only 16 bits, and windows I think is
just too peculiar :)

Anyway, back to capacitances...

Gerry wrote:
 > I assumed a charge and need to determine the voltage the charge
 > corresponds to in order to scale.

 > The E field will need to be computed from Coulombs Law for each
 > point on the path of E.dl,

Ok, so the steps are:
a) Assume an arbitrary total charge Q on the topload, with an
    arbitrary initial distribution.
b) By some process, redistribute the charge so that the potential is
    the same at all points on the surface.
c) Compute the surrounding E-field field, as a function of Q.
d) Integrate E along a path from infinity to the topload to get
    the topvolts as a function of Q, and thus the capacitance.

I think that steps (c) and (d) are the reverse of one another, and
that you can expect the topvolts potential to pop out at step (b).

Let's say that you have arrived in step (b) at a charge distribution
that gives a uniform potential to all the tiles of the topload.

So you now have a list of fractional tile charges, ie against each
tile 'j' you know the what fraction q[j] (range 0 to 1) of the total
charge it holds.

In other words, you have the charge distribution as a function of the
total charge.  Now just consider any one of the tile charges and
note that the potential at its location is the sum of the Coulomb
potentials due to all the other tile charges, plus its own self-
potential.  This tile is at the same potential as the rest of the top-
load, so we can go on to say, where 'i' labels any 'test' charge:

  V_topload = self_potential_coefficient[i] * q[i] * Q +
              sum{ for all j != i; q[j] * coulomb_coefficient[i,j]} * Q

Thus, since you know all the coefficients and all the q[i],
you can immediately calculate the constant P in the equation

    V_topload = P * Q

and so the total topload capacitance is then just 1/P.

Thus, really, all the work will be done by your charge re-distribution
algorithm in step (b).   Then you'll run step (c), not to determine
topvolts, but maybe to assess the likely streamer paths.
--
Paul Nicholson
--