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Re: Sigh. Can't we get some better mudslinging? (Score: 1)
by Archer on Tuesday, April 27, 2004 @ 09:23:57 EDT
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Well, John "the Pickle" Lichtenstein:

You are such a bonehead I don't know why I bother, but I'll try explaining this issue to you one last time.

First, it's false to say or even imply that I do not believe Lenz's law applies to disk induction dynamos - especially when I posted an enormous and highly-detailed webpage some time ago that clearly shows the magnitude of the generated magnetic back-torque in such devices and how it's properly calculated!
Tesla himself believed (with good reason) that by craft it is possible to substantially reduce the magnitude of the back-torque created by magnetic induction in the DC disk machines. I've also clearly shown on our new over-unity Faraday Dynamo page just why he was correct and how to demonstrate the rather simple principle involved. But it's obvious that you haven't bothered looking at this rigorous engineering study's evidence and conclusions, which I pointed out to you in my own defense against your condescending crap once before, nor will you bother in all likelihood to do so when I'm done with this latest attempt at "full and fair disclosure".

Lenz losses are, by definition, magnetic in nature. The EDF Generator uses electrostatic induction to avoid creating such losses in the first place! Although I can't really afford to take the time, I'm going to show you how the magntitude of the 'electric binding force' upon an EDF Generator's rotor is properly calculated, as extracted directly from pg. 243 of my book, and you'll see that it is negligibly tiny by comparison to 'normal' Lenz magnetic retarding force.

" . . . First, we must calculate the electric field intensity across the Primary Arrays, and in [a] . . . 4ft. Thermal Unit this intensity's peak value is equal to 574,730 v/m [using data from WorkSheet VIII(c)]. For capacitive parallel plates, this intensity is also equal to q/εA where q is the charge on one plate of area A. Each primary cathode's area is 14.09 cm2, so each Array's total A equals 0.057 m2, and q = 2.58 x 10-7 coul. Here, the binding force F caused by each Array is thus equal to Eq = 0.1483 N and the total force equals 2Eq = 0.2966 N . . ."

The effective Primary Array torque radius in the 4ft. device is ~ 14", or 0.35 m. So, T = Fr = (0.2966)(0.35) = 0.104 N-m = 0.077 ft-lbs. Obviously, this is entirely trivial. Although we could go on to do similar calculations for each of the rotor's 4 induction ring pairs, the result (I can assure you) is quite similar - and the total electrostatic retarding torque is still thoroughly insignificant by comparison to the magnetic back-torque produced in a conventional rotary induction device of comparable output capability.

True, we as yet have no extant prototype, since the projected cost of its development approaches a quarter million dollars. Do you have it? I don't; I don't owe it to you people to be rich as well as gifted. The point of the new corporation is to enable us to complete this important work without the help of those like you who are too snide, cynical, sarcastic, stupid, etc. to be of any use at all. And, no, we will not (as many of 'you' seem to think) need to prey on gullible regular folks for their hard-earned dollars. Not everyone out there is as foolish and trivial as you.

Mark R. Tomion
Archer Energy Systems, Inc.

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