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Author Topic: Recover energy from gravity  (Read 71798 times)

Gabriele

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Re: Recover energy from gravity
« Reply #45 on: March 29, 2014, 12:49:11 AM »
doesn't work. The air object must not t be linked with the lever.But to the container.And perhaps it will not work

rc4

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Re: Recover energy from gravity
« Reply #46 on: March 29, 2014, 06:05:04 PM »
doesn't work

Why, could you explain please ?

Like water is at ALL the bottom surface, the container is in equilibrium position, with or without air container. If I attach the air container to the lever, the distance is not the same and the torque too, no ?

Gabriele

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Re: Recover energy from gravity
« Reply #47 on: March 29, 2014, 06:43:28 PM »
You are right,the torque is not the same. But when each air container rises or fall,lose potential energy respect the water tank

rc4

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Re: Recover energy from gravity
« Reply #48 on: March 29, 2014, 09:30:31 PM »
If there is a torque, energy can be receover when lever basculate, no ? For move right to left (or left to right) air container I don't need to give energy (without count friction), no ?

Gabriele

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Re: Recover energy from gravity
« Reply #49 on: March 30, 2014, 04:34:13 AM »
No,no,no... i'm interested insted in the drawing you made about a circular spring that compressed change his shape

rc4

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Re: Recover energy from gravity
« Reply #50 on: March 30, 2014, 09:27:16 AM »
ok, but for the #44 post ? air container can be attached directly to the water container, I think it's the same.
« Last Edit: March 30, 2014, 04:30:14 PM by rc4 »

Gabriele

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Re: Recover energy from gravity
« Reply #51 on: March 30, 2014, 04:31:18 PM »
Each cycle you move the lever the air container lose his potential energy untill he is completly out of the water

rc4

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Re: Recover energy from gravity
« Reply #52 on: March 30, 2014, 05:25:01 PM »
if I use only one container and turn it around like drawing shows, in half part I don't have more energy ?

rc4

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Re: Recover energy from gravity
« Reply #53 on: March 31, 2014, 06:24:59 PM »
Maybe with electric like that ? V can be multiplied with external iron. At secondary, reluctance is divided by 2 because there are 2 iron in parallel. If reluctance is divided by 2, there are 2 fluxes created by I, so this multiplied voltage by 2, no ?

With number of turns at primary = number of turns at secondary:

Input: give V*I
Output: recover 2V*I
« Last Edit: April 01, 2014, 12:23:23 AM by rc4 »

rc4

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Re: Recover energy from gravity
« Reply #54 on: April 01, 2014, 09:45:25 AM »
V is divided by 2, so at primary I need VI/2 and at secondary I recover VI/4 ?

So if I place source in the middle with 2 charges, I give VI/2 and recover VI ?
« Last Edit: April 01, 2014, 12:38:14 PM by rc4 »

rc4

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Re: Recover energy from gravity
« Reply #55 on: April 02, 2014, 03:51:28 PM »
This case destroy energy ? Same number of turns N for each coil. Reluctance is the same in each iron circuit.

rc4

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Re: Recover energy from gravity
« Reply #56 on: April 04, 2014, 04:18:09 PM »
a test with NI SIM : there is a little difference of power

first and second : win energy

third: lost energy
« Last Edit: April 04, 2014, 07:37:51 PM by rc4 »

rc4

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Re: Recover energy from gravity
« Reply #57 on: April 04, 2014, 08:17:29 PM »
more output than input
« Last Edit: April 04, 2014, 10:51:04 PM by rc4 »

rc4

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Re: Recover energy from gravity
« Reply #58 on: April 05, 2014, 11:04:23 AM »
I changed value of Kx, because it must lower than sqrt(Lx*Ly), but the system is very stable like that (even voltages are high !)

if K < 0.95 the system don't give energy.

works with Kx=0.9 but lower power

« Last Edit: April 05, 2014, 04:09:54 PM by rc4 »

rc4

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Re: Recover energy from gravity
« Reply #59 on: April 06, 2014, 05:21:29 PM »
with possible values of L and K. You can test, NI sim is free for 15 days. This circuit need superconductivity material for coils.