On Sun, 7 Sep 1997 20:16:42 +1000 "Paul B. Webster" snarled: >Mike Keitz wrote: > >> If the "NDT equipment" is effectively a resistor, then the solution >> for RMS power delivered > > I know I am being picky, but can we please refrain from this fetish >of >referring to the *meaningless* term "RMS power"? Ok I menat to say RMS voltage, which relates to average power dissipated by a resistive load for periodic waveforms. The power dissipated during a complete repetition of a waveform's period (or integral number of repetitions) will be V^2/R regardless of the waveshape of the voltage as long as V is expressed as RMS equivalent. It has become popular to use the term "RMS" where it doesn't belong when talking about equivalent or average powers. For example, manufacturers of stereo amplifiers are (were) required to publish a power rating in the form of "35 Watts RMS into 8 ohm load (with less than such and such THD)". This means that the voltage and current capabilities of the amplifier's output are sufficient to output a sine wave which will dissipate 35W in an 8 ohm resistor. The "RMS" is actually talking about voltage but the FTC requires the specification to be written this way. so it is easy to get an impression that RMS is a sort of power. It would be equivalent to write the specification "16.73 V RMS or 2.09 A RMS into resistive load" The hapless consumer would have to calculate that this amplifier could deliver 35 W into a 8 ohm load optimally, or 17.5W to 16 ohms (reaching the voltage limit) or 17.5W to a 4 ohm load (reaching the current limit and possibly causing damage). Use of this term >indicates an essential lack of grasp of the situation in question! Remember you get what you pay for for advice received via Internet mail. >> ... just to integrate sin^2(x) over ranges up to a complete half >> cycle: >> > x=pi >POWER = | sin^2(x) dx >> < x=f Paul is right (if not very polite), after integrating the square root of the result should be taken and scaled by 1/pi if the RMS voltage is to be obtained (This should be the formula Harold posted from the GE book). In the present problem, to control the firing angle to get a linear increase in power dissipated in the load, it is useful to leave it as shown. The GE formula could be converted to a percentage of power formula by squaring it and rescaling. >> 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) > > No, it is sine SQUARED, so no square root. What I was trying to say here is that since the integral assumes the peak voltage is 1, then applying whole cycles (f=0) should produce the familiar result that the RMS voltage of a 1 volt peak sine wave is sqrt(2)/2 volts. This won't happen without modifying the formula for true RMS voltage by taking the square root and multplying by 1/pi. The GE formula (see Harold's posting) reduces to Eo = E/sqrt(2) for a=0.