free energy circuit setup

[NerzhDishual]

Hello chaps

I was just asking myseft about some calculations....

Is there any guy that could do some maths?

Back to Basic and to Back-EMF.
I am refering at the initial post from FreeEnergy.
"Back EMF from the Heaviside Layer"
http://energy.spiritworld.info/free_energy_systems.html?back-emf.html

It is claimed to use 4 (four) 10 000 uF caps hooked in parallel = 40 000 uf.

This was just a repetition of the last setup except for measuring centimeters
of weight lifting instead of motor running time.

A thread was fixed to the shaft of the motor allowing it to wind up a 74 gr. weight.

The starting voltage was 10 volts.
Calibrating in step 1 gave us 71 cm lifting.

So, he can lift a mass of 74 gr at 71 cm height with 10 volts and a 40 000 uf capacitor.

For my part, I can only lift a 10 grams bolt at about 22 cm heigh with 12.8 volts and a
0.1 farad cap (claimed precision -10+30%, I have no meter to measure such capacity).

These results are not matching at all.

My question is :

If I give you:
- a "C" farad capacitor;
- a V volts battery;
- a m kilogramme mass;
- a 100% efficient motor;
- idealistic conditions: no losses at all.

You charge the capacitor with the battery.
At what height can you theoretically elevate this mass by discharging the cap into the motor?

I have already done some maths but I wish I could have it checked because my result sounds weird.

Best

[hartiberlin]

The formular is:
0.5 x C x V^2 = m x g x h

C= Capacitance in Farad
V Voltage in Volts
m = mass in kg
g= earth acceleration constant  9.81
h = height in meter

You get the height when you put this formular this way:
h= ( 0.5 x C x V^2 ) / (m x g)

Remember these small motors will only have about 50 % efficiency...

[hartiberlin]

hmm, I am getting 83.5 Meters lift for just a 10 gramms weight with a 0.1 Farad cap charged
to 12.8 Volts... Hmm, you seem to have a very inefficient motor...

[NerzhDishual]

Thanks for the maths..

Actually, I did these calculations and got the same results as yours. ) (83.5 meters)...

It sounded too strange to me, I thought I was wrong.

83.5 metres vs 22 cm is 0.22/83.5 *100 = 0.26% efficency.

With 40 000 uf, 10 volts and 74 grams, the theorical heigh is: h=0.5*0.04*100/(0.074*9.81)= 2.75 metres.
Efficiency = 0.71/2.75*100 =/=26%

This efficiency is 100 times as big...

My 2 caps and my motor are new.
I do not understand.

[hartiberlin]

Maybe you should try a bigger weight !
Also these small motors have a very low efficiency....