On Sat, 6 Sep 1997 19:40:42 -0000 Stephen H Alsop
<s.ssystems@EASYNET.CO.UK> writes:
>Dear Harold
>
>Thanks for your prompt posting to the PIClist following
>my request for help.  It appears that you have done what
>I am trying to do (except I am not controlling relative brightness
>but 6,000 amps NDT equipment)

If the "NDT equipment" is effectively a resistor, then the solution for
RMS power delivered is just to integrate sin^2(x) over ranges up to a
complete half cycle:
      > x=pi
RMS = | sin^2(x) dx
      < x=f

      ^ (ASCII integral sign)

where f is the firing angle from 0 (earliest firing, maximum power) to pi
(latest firing, minimum power).  As a check, the result for f=0 should be
sqrt(2)/2 (maybe half that since it is a half cycle)

I've been away from this long enough I won't attempt to integrate it but
it is something that could be readily looked up in a math handbook.

In order to generate the table it would be nice to have f as a function
of RMS; this is not obvious how to formulate.  Either a little C program
or a general-purpose math package could be used to search the solutions
of the integral above for values of f that satisfy evenly-spaced RMS
values.