Hi George,
Not wanting to waste anymore bandwidth on this
device than is necessary, if it is not OU, I sent the following
message to intereseted parties, which I hope gets to JNL or to Nicholas Moller
very soon - as they are difficult to contact directly. COPY of
posting:
A serious question has been raised about the
accuracy of the MAHG
power measurement, i.e. the P-in measurement, and I hope that if
Sterling Allan, or anyone else reading this, has ongoing contact
with J.L. Naudin, that we can get a prompt answer. Better yet, if
Nicholas Moller could address the question, and also the issue of
his expectations concerning replication of the experiment, it
would be very helpful.
As George Holz has suggested, the problem is not just that the
duty cycle is incorporated into the voltage calculation in order
to give the low average voltage BUT that it also appears to have
been incorporated in both the voltage calculation and the current
calculation at the same time - in effect, Naudin seems to be
underestimating the actual power by a factor of 20 (the reciprocal
of the duty cycle).
Both I and V were marked as average readings in earlier versions
of the graphs on JLN's site. When it is incorporated in both
readings by averaging it multiplies the resulting input power by
(.05)^2 rather than the correct single (.05). To get the right
answer we must average the peak power by applying the duty cycle
factor only once. The average power in a pulsed system is not equal to V(avg)*I(avg). Using this incorrect method shows that a resistor is 20 times overunity at .05 duty factor.
An easy resolution to this problem is IF he will take a reading at
the battery itself, preferably with an analog ammeter, during the
tube operation, and IF he is getting about 1/3+ amp, average
current being drawn at the battery - then most would agree that
the claimed ~5 watt input is accurate. Otherwise, we are left with
lingering doubts.
Actually, as George mentions a 166 A (peak)* 12 V(peak) * .05
(duty factor) gives about 100 watts input at an average current of
about 8 A and an average voltage of .6 V. This would be the same as 100
watts in and 100 watts out, which is what would be expected in the
case where there is zero excess energy.
Unfortunately, although George did mention this earlier, he did
not make this crystal clear. He says: "Perhaps I failed to make my
earlier observations on the MAHG list clear enough. I have done
pulse power measurements in so many prior experiments and the
possible error was so clear to me that I thought it would be
obvious to everyone."
We are all hoping that some prompt resolution can be made of this
apparent problem.
Jones Beene
power measurement, i.e. the P-in measurement, and I hope that if
Sterling Allan, or anyone else reading this, has ongoing contact
with J.L. Naudin, that we can get a prompt answer. Better yet, if
Nicholas Moller could address the question, and also the issue of
his expectations concerning replication of the experiment, it
would be very helpful.
As George Holz has suggested, the problem is not just that the
duty cycle is incorporated into the voltage calculation in order
to give the low average voltage BUT that it also appears to have
been incorporated in both the voltage calculation and the current
calculation at the same time - in effect, Naudin seems to be
underestimating the actual power by a factor of 20 (the reciprocal
of the duty cycle).
Both I and V were marked as average readings in earlier versions
of the graphs on JLN's site. When it is incorporated in both
readings by averaging it multiplies the resulting input power by
(.05)^2 rather than the correct single (.05). To get the right
answer we must average the peak power by applying the duty cycle
factor only once. The average power in a pulsed system is not equal to V(avg)*I(avg). Using this incorrect method shows that a resistor is 20 times overunity at .05 duty factor.
An easy resolution to this problem is IF he will take a reading at
the battery itself, preferably with an analog ammeter, during the
tube operation, and IF he is getting about 1/3+ amp, average
current being drawn at the battery - then most would agree that
the claimed ~5 watt input is accurate. Otherwise, we are left with
lingering doubts.
Actually, as George mentions a 166 A (peak)* 12 V(peak) * .05
(duty factor) gives about 100 watts input at an average current of
about 8 A and an average voltage of .6 V. This would be the same as 100
watts in and 100 watts out, which is what would be expected in the
case where there is zero excess energy.
Unfortunately, although George did mention this earlier, he did
not make this crystal clear. He says: "Perhaps I failed to make my
earlier observations on the MAHG list clear enough. I have done
pulse power measurements in so many prior experiments and the
possible error was so clear to me that I thought it would be
obvious to everyone."
We are all hoping that some prompt resolution can be made of this
apparent problem.
Jones Beene
