"I've thought about that post often.  And this was back when he thought the device was in the cellar of a suspicious stranger... Due diligence. Uh huh. It's difficult to conclude anything other than there was never any interest in going, just interest in tagging the project as secretive and resistant to scrutiny.  It didn't work out that way."


"As a dog returns to his own vomit, so a fool repeats his folly." - PROVERBS 26:11

AMEN

Nobody seems to have posted any theoretical treatment of this so far so I thought I'd have a go at it.

We can model the experimental setup in the following equation:

e.P_in + f.P_gen  = P_out + P_gen 

where:

e -- the basic efficiency of the unmodified motor (prime mover) ? usually in the order of 90%+ for a standard modern induction motor

f -- the proportion of the generator power that is fed back to enhance motor performance ? it is zero with the brass coupling and only comes into play when the steel coupling is used.

P_in -- the electrical power input delivered to the motor (prime mover).

P_gen -- the power used to drive the generator

P_out -- the mechanical power output from the system ? presently zero because the system is not driving anything but will be non-zero when installed to drive the buggy, for example.

Here?s what it will look like when used in practice to drive a mechanical load:

1. Without the generator in place, the equation simplifies to:

P_in  = P_out /e

This is the situation where the motor is used in the conventional way to directly drive the mechanical load.

2. With the generator in place but with the coupling of brass instead of steel, the feedback is zero and the equation becomes:

P_in  =  (P_out + P_gen)/e

The power input has to be higher than in the first case because extra power must be supplied to drive the generator. This of course is not useful in practice because it just wastes power in the generator - but it serves for comparison with the third case, below.

3. With the feedback enabled by inserting the steel coupling, the equation gives:

P_in  =  (P_out + P_gen.(1-f))/e

The power input in this case may be lower (better) or higher (worse) than in case 1 depending on the value of f. With f<100%, it is worse; with f>100% it is better.

Conventional theory states that f must always be <100% since the generator cannot produce more power than it consumes. OU theory would allow >100%. The measured value for f is presently about 20% (using Thane?s numbers). It is hard to predict what value of f may eventually be achievable.

PB

Pgen.(1-f)

The above seems to me to suggest that we should expect either a) an decrease in generator output when coupled or b) an increase in draw to maintain the same generator output when coupled

Yes? No?

If yes, I don't think either are in evidence?

Quote from: JustMe on March 22, 2008, 02:01:27 PM
Pgen.(1-f)

The above seems to me to suggest that we should expect either a) an decrease in generator output when coupled or b) an increase in draw to maintain the same generator output when coupled

I assume when you use the term "generator output" you mean the power dissipated in the load resistor. But P_gen is not that: it's the power used to drive the generator. P_gen has three components:

1. Power dissipated as heat in the coils and the cores.
2. Power fed back magnetically to the motor along the shaft (in the coupled case only)
3. Power dissipated as heat in the load resistor (what you're calling "output")

The total amount of power for P_gen was reported by Thane to be about 500W - 600W so #3 is quite tiny compared to that total and it would be hard to detect its influence on the other components. Mostly it looks like the power is going 80:20 to #1 and #2 respectively.