RV Camping and Outdoors Videos

livelounge106 (September 23rd, 2008 @ 11:42 am)

This appears a similar system to how a car ignition coil works, by stopping the flow of current, a hige back spike is produced by the coil in the transformer. have you tried putting the (unrectified) output from this into another transformer to reduce the voltage and step up the current, then put the rectifier on the low voltage output, to get dc

Nonetub (August 9th, 2008 @ 4:38 pm)

Of course... Finally this experiment is claimed as Overunity or not?

dallasgoldbug (July 6th, 2008 @ 5:12 am)

You might want to use something other than the alkaline 9v batteries. They tend to resist giving up their current from time to time. Anyone out there experience this?

massoonmann (June 20th, 2008 @ 12:18 am)

yer not even close dude...

Paz4254 (June 4th, 2008 @ 12:58 am)

As other posters have pointed out, the volt/ammeter is absolutely the wrong instrument. In every video claiming overunity you see someone using a VOM as proof. You are not measuring rectified DC, but a waveform. It probably looks a lot like some sort of saw tooth waveform. You need an oscilloscope. You need the amplitude and period of the voltage and current to calculate the actual power. I'm guessing it's less than the input.

jerktrucker (May 22nd, 2008 @ 12:23 am)

If you put a maximum load directly to the batteries or just short out the batteries with an ammeter and measure the current then calculate the power and you will see more power than your getting with all that other junk connected. Now were talking free energy and you can easily show that there is nothing plugged into the wall!!!!!

mammamca (May 19th, 2008 @ 10:42 pm)

eugneo1111 is half right. From the information (if correct) you can find the Thevenin parameters... test1 (open circuit) v = 500V and I = 0A test2 (short circuit) v = 0 and I = 0.250A there for internal resistance is: 250V/0.125A = 2kohm and maximum power is: 250V*0.125A = 31,25W and of course the presented system never do any work. If you add a load of 2Kohm (that can support 31.25W) it will be at peak efficiency, and your battery will be drained in no time

eugeneo1111 (May 19th, 2008 @ 7:43 am)

You are almost shorting the output so P=IV where I=250ma and V=almost 0 So P= almost 0

overunitydotcom (May 8th, 2008 @ 9:10 pm)

Please try to hook up an 8 Watts fluorescent tube directly to your output ( In parallel with your blue cap) and you could probably light it up pretty nicely... Good luck. Regards, Stefan.

billwcali (April 13th, 2008 @ 3:45 am)

unfortunately open circuit voltage doesnt equal useable power - as others say you need to measure under load use any load like may be a large (high wattage) resistor, maybe 200 ohms - then measure the voltage across the resistor and the current through it - you can buy a coupel of resitors for low cost at radio shack

jordyx (March 24th, 2008 @ 6:29 pm)

Let say you obtained about 500V x 0.25A = 125W You should use 5 x 30W/110vac incandescent light bulbs connected in series as a load. If the bulbs does not light up you don't have 0.25A. Try to measure the output voltage with the light bulbs connected. Add the ampermeter in series with the bulbs to measure the current and update us with your findings. Good Luck.

WisdomVendor (March 13th, 2008 @ 4:15 pm)

In any circuit that contains any signal other than pure dc, you can't get an accurate volt or amp reading with a multimeter. You have to have an oscilloscope to see the actual signal to be able to find the duty cycle of that signal. Then you can calculate the power involved.