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Old April 28th 07, 11:23 AM posted to rec.radio.amateur.antenna
Keith Dysart Keith Dysart is offline
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First recorded activity by RadioBanter: Mar 2007
Posts: 124
Default Rotational speed

On Apr 27, 10:32 pm, Cecil Moore wrote:
Keith Dysart wrote:
On Apr 27, 9:22 pm, Cecil Moore wrote:
Owen Duffy wrote:
Isn't hopping onto the rotor (assuming synchronous speed) to make your
observations called moving from the time domain to the frequency domain,
and all the mathematical shortcuts are only valid if all quantities share
the same angular velocity (or frequency), implying sinusoidal waveform.
Ever wonder which direction, clockwise or counter-clockwise,
a standing-wave phasor is rotating?


Clockwise, of course, by convention. Always look at the end of the
machine that lets you draw your phasor diagram clockwise.


The equation for a standing wave,

E(x,t) = 2*Eot*sin(kx)*cos(wt)

would have an identical value if it were written,

E(x,t) = 2*Eot*sin(kx)*cos(-wt)

Thus, a standing wave phasor can be considered to be
rotating either clockwise or counter-clockwise.


I suggest that the solution to this ambiguity is to do the
same analysis for the current, which should be found to
be 90 degrees shifted from the voltage. The real current
is either leading or lagging the voltage. Rotate the frame
of reference in the direction that will cause a lagging
current to appear counter-clockwise from the voltage
and a leading current to appear clockwise from the
voltage.

This is all convention, of course. You can rotate the
frame of reference in either direction, you just need
to remember which direction on the graph represents
earlier time and which represents later.

Physically, it is just a question of which end of the
rotor you climb on to.

....Keith