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Author Topic: Recover energy from gravity  (Read 72118 times)

rc4

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Re: Recover energy from gravity
« Reply #30 on: February 25, 2014, 09:10:05 AM »
Other side view. Cable has a F force to up (air inside cable). Cone stop F force from its axis, this don't cost nothing. The cable slide on cone 1, turn like a helix inside water and slide on cone 2. The container has more weight when it move down and the potential energy is always the same.
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rc4

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Re: Recover energy from gravity
« Reply #31 on: February 25, 2014, 11:41:12 AM »
Or maybe with 4 pulleys like that. 2 up pulleys are conic like that they take the additional length of the cable
« Last Edit: February 25, 2014, 02:14:56 PM by rc4 »
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rc4

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Re: Recover energy from gravity
« Reply #32 on: February 25, 2014, 07:17:10 PM »
Maybe with 2 densities ? The black stem is fixed. Force F give and extra force to the left. 2 containers have always the same position.
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rc4

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Re: Recover energy from gravity
« Reply #33 on: February 26, 2014, 09:38:10 AM »
The stem can move like the system because of force is perpendicular to the surface. I think it must be exist one position for the stem where the lost of surface of the container 1 is not equal to the lost of liquid at a slope giving.
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rc4

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Re: Recover energy from gravity
« Reply #34 on: February 27, 2014, 09:46:30 AM »
Maybe like that ?

Triangle has weight and it is stable in water without black stem (it's like a boat). There are fixed black stems inside water. Triangle move from right to left. It's possible to see there is always a left force on triangle except at final where this need very few energy.
« Last Edit: February 27, 2014, 02:10:38 PM by rc4 »
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Gabriele

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Re: Recover energy from gravity
« Reply #35 on: February 27, 2014, 02:31:15 PM »
Seems logic. But how you return to initial situation? you have to give the same energy you have gained... or not?  Perhaps if the black hose is weightless you can re-put inside the triangle gaining energy
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rc4

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Re: Recover energy from gravity
« Reply #36 on: February 27, 2014, 02:35:35 PM »
Imagine the container like a disk (top view), triangle turn in the container. All black stems are fixed. The left position is like right position (return to initial position). Triangle is always inside water. Triangle pass through stems from right to left.

At final step, it's possible to move down grey rectangle because there is less up force, when all rectangle is there, move to the left and move up and repeat cycle.
« Last Edit: February 27, 2014, 05:12:39 PM by rc4 »
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Gabriele

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Re: Recover energy from gravity
« Reply #37 on: February 27, 2014, 06:47:01 PM »
Doesn't work.. Cause right covered surface is bigger than left...
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Gabriele

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Re: Recover energy from gravity
« Reply #38 on: February 27, 2014, 06:59:11 PM »
in the case of a triangle and a rectangle too...
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Gabriele

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Re: Recover energy from gravity
« Reply #39 on: February 27, 2014, 07:02:06 PM »
And the last shape too..  >:(
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rc4

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Re: Recover energy from gravity
« Reply #40 on: March 05, 2014, 12:32:00 AM »
An idea with half torus inside water. All the system is under gravity. I compute forces F1 and F2 and they don't depend of radius (R), only of height (h) in the integrale. So this would say I can suppress bottom surfaces (grey color) so the weight is reduced too. Normally, surfaces must be compensated by volume of air, but here it's lower volume. I can move up or down all the system (water+air) the energy is not the weight by distance. Like F1=F2 all the time I can let at bottom air volume without force, and this suppress surfaces at bottom.

The result is always 2h, for any R.
« Last Edit: March 05, 2014, 08:43:22 AM by rc4 »
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rc4

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Re: Recover energy from gravity
« Reply #41 on: March 05, 2014, 01:34:15 PM »
small error, it's more that: 2h/pi, don't depend of R
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Gabriele

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Re: Recover energy from gravity
« Reply #42 on: March 05, 2014, 02:09:43 PM »
So you are saying F1=F2? Are you sure of that?
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rc4

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Re: Recover energy from gravity
« Reply #43 on: March 05, 2014, 02:27:58 PM »
arf, this don't work with homogenous liquid, maybe if density change
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rc4

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Re: Recover energy from gravity
« Reply #44 on: March 28, 2014, 11:46:49 PM »
New idea with container of water with an object with air inside. It's like a weighing scale, but here I use the difference of distance for change the torque. If container can be like drawings, the air object give a difference of torque. It's possible to change the position of air object because container of water is always at horizontal. Red line is a rope that attach the air object to the stem.
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