Wax Actuator watts?

The RMS voltage is halved - the voltage profile stays the same for one half-cycle (hence same RMS for this portion) and is zero for the next half-cycle, so the average over a full cycle is halved (ignoring diode voltage drop). Nonetheless, the confusion on this point accidentally gets the correct answer that the half-cycle version of 240v is effectively equivalent to 170v in terms of power, but that?s because for a resistive loud the power consumption goes as v^2 (ignoring any temperature dependence of heater resistance)

Peak voltage is 339V for both full and half-wave rectified sinewaves. The equations for RMS voltage differ in each case. A quick google will show you them and they are as I quoted above. You have taken the RMS (not peak) voltage of a half-wave rectified sinewave and multiplied it by the factor for converting a full-wave sinewave's peak voltage into an RMS one!

SteveW

Unlike RMS power, which is a seriously confused concept, RMS voltage of any repetitive waveform, regardless of shape, remains a precisely defined quantity and, importantly, continues to goven the average power delivered to a resistive load over that repetitive waveform. The relationship to peak voltage is, of course, arbitrary in the general case. But a rectified sine wave still allows simple, accurate calculation. But not by me, at least at this time in the morning.

Its all a matter of definition. RMS volts is, strictly I guess, heating power, but in a weird waveform it doesnt make much sense to think in those terms.

Easiest is to simply halve the power of a full wave as it only is there half the time and take the root.

You can confirm your result by simple inspection. The mean square voltage of a half wave rectified sine wave must be half that of a full wave sine wave because half the readings are replaced by 0^2. But for RMS we take the square root of mean square voltage. Square root of a half is 1/1.414 which is 0.707. So half wave rectification halves the power into a resistive load. Any other result would be surprising because it is applying the same waveform for half the time, ignoring polarity which we can for a resistive load.

half-cycle (hence same RMS for this portion) and is zero for the next half-cycle, so the average over a full cycle is halved (ignoring diode voltage drop). Nonetheless, the confusion on this point accidentally gets the correct answer that the half-cycle version of 240v is effectively equivalent to 170v in terms of power, but that's because for a resistive loud the power consumption goes as v^2 (ignoring any temperature dependence of heater resistance)

Sorry (a passive aggressive sorry) but if you halved the RMS voltage then you'd get a quarter of the power. This is intuitively nonsense because the load is getting the same power for half the time. So it is not surprising that the actual calculation shows that the RMS voltage of a half wave rectified sine wave is actually 0.707 of the RMS voltage of a full sine wave. The RMS voltage is not linearly related to the mean voltage.

Agreed!

Lol.

Half power yes, which equates to 240v x 0.707 rms volts, as I said. Wakey w akey.

NT

I don't think Harry has a clue what he's done.

Different ways of getting to the same result, unlike Harry's oddball guess. Yours of course does have the advantage of showing how it gets to the result, whereas mine just applies the different equations.

SteveW