Author Topic: free energy circuit setup (Read 49683 times)

NerzhDishual · « **Reply #30 on:** April 15, 2006, 10:17:51 PM »

I wish I could be able to insert an image in a post.

NerzhDishual · « **Reply #31 on:** April 15, 2006, 10:24:51 PM »

Inserting image in a post could avoid wasting everybody time....

pg46 · « **Reply #32 on:** April 16, 2006, 06:11:56 PM »

Hi NerzhDishual -

Nice work. Too bad though that you didn't prove any gains in actual power

I just did some crude experiment myself today with this set up and found about 50% gain in time running the small motor without a load. Not much good is it without it being able to do any work?? Still, I find it odd that it will turn the little DC motor 50% longer...hmmmm... What does it mean do you think and why does it do it?
Also, what about the lifting test the original experimenter did? He claims he gained around 30% more in real lifting work.
It's really cool that you got at the experiment so fast. I'm sure everyone appreciates it as I did. I am going to try some more testing myself and will report back in if I find out anything interesting.

hartiberlin · « **Reply #33 on:** April 16, 2006, 06:17:31 PM »

Hi NerzhDishual
please try with a DC motor that has a strong sparking commutator
without a smoothing capacitor at it.
There you should get much more running time.
There will be the Newman effect at work,
recharging the caps via the spark !
Please let us know. Thanks.

pg46 · « **Reply #34 on:** April 17, 2006, 02:56:45 PM »

Some questions-

Maybe you brighter minds can help me out here with a couple of questions?
I was mucking about with the charging and decharging of capicitors through the small DC motor when I found that when I charged the C1 to 13.5 Volts and then decharged through the motor to C2 I would end up with up with 7.6 V in C1 and then 6.5V in C2. I was first wondering why they are not the same and then maybe more importantly why they total to more than 14V which is more than I started with? I repeated this several times with the same results.
A question regarding the DC/DC converter - This converting of energy isn't 100% efficient, right? The wikipedia says the most common converters are from something like 85% to 95% efficient. So, wouldn't using capicitors and DC/DC converters in this experiment eventually lose power and soon just come to a complete stop?
Unless of course more energy is being produced than you are losing.

just wondering..

NerzhDishual · « **Reply #35 on:** April 20, 2006, 02:02:06 PM »

Hi everbody

Actually, without a load I also got about 50% gain (just counting up the seconds white the motor is running).

BTW: I had not checked the voltage of the *two* capacitors as pg46 did.

Weigh lifting:
I have done a couple of quick experiments with an Igarashi motor (20G-380).

(http://jeanyves.hervouet.free.fr/4overun/Igarashi_Motor_20G_380.gif)

It runs at 25 turns/min.
I lifted a small 10 grams bolts at about 22 cm high.
I got *no* "indisputable" gain.
I will do more experiments and swap the caps.

(http://jeanyves.hervouet.free.fr/4overun/motor_2_1.jpg)
(http://jeanyves.hervouet.free.fr/4overun/Motor_2_2.jpg)

I will also try another motor (witout a small smoothing capacitor).
I have a very sparky motor but it draws too much amp...

I will aslo try this:
(http://jeanyves.hervouet.free.fr/4overun/JLN_exp.gif)

And this small DC-DC converer, which is nore convenient that the big one I was previously talking about.

)
(http://jeanyves.hervouet.free.fr/4overun/Max_732.gif)

Best

pg46 · « **Reply #36 on:** April 20, 2006, 05:57:14 PM »

Hi All-

NerzhDishual, I sure like the way you get at the projects in a "no nonsense" manner and report back quick with your results. I for one really appreciate your input and the work that you are doing, as I am sure many others do as well. I like the photos too.
I got stalled with my own project since I am having difficulty with a faulty DC/DC converter I got and haven't replaced it yet.
I'll look forward to hearing from you regarding whether you are able to do a complete circle with the original charge. If so that would really be something because even if you didn't do any "more work under a load" then back in step #1, you would still have the same energy charge that you began with. There would be your energy gain because you'll have done the same work(or nearly so) without losing the entire energy as you undoubtably do when you just dump the 13V charge out of a single capacitor through the motor back in step #1

Elvis Oswald · « **Reply #37 on:** April 20, 2006, 09:46:45 PM »

I'm not trying to bust anyone's bubble. Everyone wants to find overunity. But I believe this experiment has been duplicated now enough to know that the original poster was right.

The next logical step is to evaluate the findings. My evaluation is that running a dc motor off one cap yields a certain amount of time. If you put another cap on the other end of the motor... and end up with a charge in that cap - you are only proving that energy was lost using the one cap. Probably due to the inability of the motor to utilize the energy at the rate of discharge.
Catching that energy may be "efficient", and a circuit that would do this and run the power through the motor again would certainly mean that you have a super efficient circuit for a dc motor... but it's not overunity.

pg46 · « **Reply #38 on:** April 21, 2006, 01:23:24 PM »

Hi All-

If one is moving the same amount of energy from capacitor to capacitor while turning a motor(s) without losing the energy then you may have "perpetual motion" which is pretty cool on its own really. While its not over unity as such it might be depending on what those motors are doing for work while they are spinning .. like maybe turning a generator??

NerzhDishual · « **Reply #39 on:** April 21, 2006, 08:33:39 PM »

Hi folks!

I am setting up new experiments:
- another motor without load, just a very light plastic 'needle' glued on the shaft to see the motion and possibly visually count up the turns;
- the JLN and the DC-DC ones.

According to Hartiberlin, you could recharge a cap via the motor sparks (Newman effect).

Just my crackpot(ed) thought: what if a motor were less sparky when loaded?

Best

NerzhDishual · « **Reply #40 on:** April 22, 2006, 12:12:43 AM »

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.

Quote

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 · « **Reply #41 on:** April 22, 2006, 04:17:50 AM »

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 · « **Reply #42 on:** April 22, 2006, 04:24:21 AM »

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 · « **Reply #43 on:** April 25, 2006, 02:46:32 PM »

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 · « **Reply #44 on:** April 25, 2006, 03:17:08 PM »

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

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