[R-390] Dead horse foaling,. Monkey wrench thrown!

[twleiper@juno.com]

On Mon, 21 Jan 2002 14:02:43 -0600 "Bill Hawkins" <bill@iaxs.net> writes:
> Gotta love that subject line! Joe Foley does have a way with
> words. But it's time to pour a can of Aircraft Stripper on
> this to remove the gum from the works.

[ Stripper most effective, false arguments cleansed ]

Half the waveform, half the power...but we want a third the
power...how are you going to get there? Besides, half of those
arguments about "variability" induced by 30 hz instead of
60 hz are silly, because the frequency does NOT change
when you put a diode in...you are still getting half the wave
at a 60hz RATE. And, as I said before, if you REMOVE the
diode, I fail to see any argument one can make that would
cause the power to be TRIPLED instead of doubled.

Just to pound a little more, see below:

> [Bottom line: a diode in series with a resistive load in an AC
> circuit cuts the power in half. Since the resistance stays
> pretty much constant in a heated filament emitting the same
> amount of light, E and I have a constant ratio. The correct
> equation for half power is P/2 = E/(sqrt 2) * I/(sqrt 2) or
> P/2 = (E * I) / 2. The equation for quarter power is
> P/4 = E/2 * I/2. But we have half power, not quarter power,
> so the voltage cannot possibly be E/2. You get the same result
> using P = (E * E) / R. The value of volts for half power must
> be E/(sqrt 2).]

Tom