Overunity.com Archives

Gerard Morin's  8) Replication with Eleman Magnet Motor Test 1. Well this is an interesting PROJECT! This is my version of Gerrard Morrin... Not with drain pump, but with my own ELEMAN motor. I think it's really interesting, not just for the results...but because it forces us to think differently :o ... www.youtube.com/watch?v=jAcbZ6CvDTc&list=PLT7Zi_DRtP7fpdEHDlFIfaI_bOwZo5zyF&index=21

If the 0.38A @ 24V is the input, then the ~9.5W that constitutes seems a whole lot greater than the apparent light from the fluorescent tube.

No offense but you can light small tube that size up a lot brighter using a "dead" AA battery and a Joule Thief modified flash camera circuit.  Your input is 240 volts?  Try .5 volts and much better light.  (Also, a lot lower amp draw.)

Bill

Quote from: Pirate88179 on March 07, 2015, 07:15:33 PM
No offense but you can light small tube that size up a lot brighter using a "dead" AA battery and a Joule Thief modified flash camera circuit.  Your input is 240 volts?  Try .5 volts and much better light.  (Also, a lot lower amp draw.)

Bill

It's 24.0V.  The decimal point on the power supply is a little hard to see.  Still, from the picture the light output seems very anemic for almost 10W in.

Quote from: MarkE on March 07, 2015, 08:04:11 PM
It's 24.0V.  The decimal point on the power supply is a little hard to see.  Still, from the picture the light output seems very anemic for almost 10W in.

Thanks for pointing that out.  My mistake.  Still, that is much more input than from an AA battery.  Unless I am missing something else here?

Bill

what can you say about this picture ? The first picture has effects on camera....BUT LOOK THE VIDEO www.youtube.com/watch?v=jAcbZ6CvDTc&list=PLT7Zi_DRtP7fpdEHDlFIfaI_bOwZo5zyF&index=21

I can say the white levels are different in each.  In each case the fluorescent bulb appears much dimmer to me than I would expect from ~10Watts of input power.

Personally, I dislike the use of light bulbs of any kind as a proxy for power.  But if you are intent on using a light bulb in front of a camera to try and indicate power, then I suggest that you run two bulbs side by side where one is a reference driven by a power supply adjusted such that the light output of both lamps is as close to equal as possible.  The much better alternative is to perform proper power measurements of both the input and the output.

Your measuring the output voltage on the wrong side of the transformer.

You Ampere meter is on the input side of the transformer and
your Volt meter must also be on that side. So the transformer
is a 230 / 9 ratio = 25,6. 125 Volt / 25,6 = 4,9 Volt. So the
output = 4,9 * 200mA = 0,98 Watt. Input = 24 * 0,38 = 9,1 Watt.
COP = 0,98 / 9,1 = 0,1.

GL.

I agree that the voltage and current should be measured on the same side of the transformer, preferably the output side.  I agree that the transformer turns ratio means that the output current is no more than 9/230 * 200mA <= 7.83mA, which at 125V <= 0.98W into the fluorescent bulb, or alternately that the transformer input voltage is: ~9/230 * 125V ~= 4.89V which at 200mA is also 0.98W, under the favorable assumption of a unity power factor.  And I agree that the input power is:  0.38A * 24.0V =  9.1W making the power to the fluorescent tube less than 11% of what is drawn from the power supply.  The less than 1W power accounts for the low tube brightness.  How much less than 1W / 11% the fluorescent tube power is we do not know because we do not have any phase information.