The Newman comedy company

kmarinas86 · October 16, 2010, 06:14:50 PM

Quote from: kmarinas86 on July 04, 2010, 07:50:29 PM
There are few demonstrations by Newman that defy this pattern. Most of them are demonstrations his Newman's small motor generator:

Older videos from 2008:
http://video.google.com/videoplay?docid=-5399697456012877363 (He prepares to start motor hooked to a generator at 9:50, and prepares to measures amps and volts at 11:40 (17 watts) and after that, the motor speed is doubled and eventually tripled. This could mean that the power input increased by a factor of 4x then (9/4)x (which is 68 watts and 153 watts respectively). 153 watts input fails to explain how the meter picks up (which occurs at 7.5 horsepower according to the Grainger catalog for this generator).)
http://video.google.com/videoplay?docid=-3747078809628665374 (This is the previous video. He does not show power input of the motor (only amps), but if you compare it to 10:40 in the second video above (10 volts and 1 amp) and have your speakers on, you will see that is it faster than that. In this setup he used 18 nine-volt batteries. Obviously 9V is the maximum voltage. It is going about 1/4th as fast the speed where the meter picks up beyond 7.5 horsepower. 5595 watts / 4^2 = 350 watts, which is more than 98 volts * 3.5 amps. If the internal resistance of each battery is 2 ohms, then that is already a loss of 7 volts per 9V battery. If you had 3.5 amps, the voltage of all eighteen 9V's combined under load would be 22 volts.)

Something is wrong about the last few lines apparently, but proper calculation may speak even more in favor that the device actually operates as claimed.

18 batteries * 9 volts/battery = 162 volts

http://video.google.com/videoplay?docid=-3747078809628665374

Power used to drive the motor = (162 Volts - Battery Resistance*(Current))^2 / Coil Resistance

Newman say his coil weighs 200 pounds. The resistance of the coil will depend on the wire gauge chosen.

0 AWG:   0.1 ohms
1 AWG:   0.1 ohms
2 AWG:   0.2 ohms
3 AWG:   0.2 ohms
4 AWG:   0.4 ohms
5 AWG:   0.6 ohms
6 AWG:   1.0 ohms
7 AWG:   1.6 ohms
8 AWG:   2.5 ohms
9 AWG:   4.0 ohms
10 AWG:   6.4 ohms
11 AWG:   10 ohms
12 AWG:   16 ohms
13 AWG:   26 ohms
14 AWG:   41 ohms
15 AWG:   65 ohms
16 AWG:   103 ohms
17 AWG:   163 ohms
18 AWG:   260 ohms
19 AWG:   413 ohms
20 AWG:   657 ohms
21 AWG:   1044 ohms
22 AWG:   1660 ohms
23 AWG:   2640 ohms
24 AWG:   4197 ohms
25 AWG:   6674 ohms
26 AWG:   10612 ohms
27 AWG:   16874 ohms
28 AWG:   26830 ohms
29 AWG:   42662 ohms
30 AWG:   67836 ohms
31 AWG:   107864 ohms
32 AWG:   171510 ohms
33 AWG:   272720 ohms
34 AWG:   433620 ohms
35 AWG:   689500 ohms
36 AWG:   1096340 ohms
37 AWG:   1743240 ohms
38 AWG:   2771800 ohms
39 AWG:   4407400 ohms
40 AWG:   7008000 ohms

If the wire gauge is greater than 12 AWG, we can safely ignore the batteries' internal resistance as it will be insignficant in comparison to the coil resistance.

If the output of the motor is 350 watts, and if the input is 162V * current, then:

Efficiency = [350 watts] divided by [(162 Volts)^2 / Coil Resistance]

AWG
14 AWG:   54.17%   4.0 amps
15 AWG:   86.15%   2.5 amps
16 AWG:   136.99%   1.6 amps

Wires lower than 14 AWG would draw more amps while wires above 15 AWG would require efficiency greater than 100%.

If it turns out that current is less than "Volts / Coil Resistance" (possible due to the massive coil inductance), then the efficiency values in the table just above further OVERestimate the actual current and further UNDERestimate the actual efficiency. In that case, to maintain a status of less than 100% efficiency, lower gauges wires would be necessary, or you would need to connect coils in parallel.

Nevertheless, the current must be above 2.16 amps if the efficiency is less than 100%. In one minute, 2.16 amps translates into 36 mAh (36=2160/60). If it were drawing that much power, then the 9Vs are gone in two minutes. The video at:

http://video.google.com/videoplay?docid=-3747078809628665374

...is cut at several places. However, more than once is the motor shown running longer than 2 minutes. If the motor runs any longer than that, then it is already showing overunity.

Where the 350 watts output estimate came from:

Quote from: kmarinas86http://video.google.com/videoplay?docid=-3747078809628665374 (This is the previous video. He does not show power input of the motor (only amps), but if you compare it to 10:40 in the second video above (10 volts and 1 amp) and have your speakers on, you will see that is it faster than that. In this setup he used 18 nine-volt batteries. Obviously 9V is the maximum voltage. It is going about 1/4th as fast the speed where the meter picks up beyond 7.5 horsepower. 5595 watts / 4^2 = 350 watts.

Obviously, if the batteries are already down 10%, such decline means the denominator of the efficiency calculation is even smaller than assumed above, which causes the answer to rise.

The result of the other video is clear:

Quote from: kmarinas86http://video.google.com/videoplay?docid=-5399697456012877363 (He prepares to start motor hooked to a generator at 9:50, and prepares to measures amps and volts at 11:40 (17 watts) and after that, the motor speed is doubled and eventually tripled. This could mean that the power input increased by a factor of 4x then (9/4)x (which is 68 watts and 153 watts respectively). 153 watts input fails to explain how the meter picks up (which occurs at 7.5 horsepower according to the Grainger catalog for this generator).)

Even if you took 17 watts and multiplied by 3*3*3, you would get 459 watts, which cannot account for even one horsepower. You could also multiply it by 3 a fourth time and still not get 7.5 horsepower! Multiplying 17 by 4*4*4*4 doesn't give you the answer either! Regardless, required power should increase with the square of the speed, or perhaps even the cube of the speed (if the generator has a lot of friction), but NEVER to the fourth power of the speed.