Naked Science Forum
On the Lighter Side => New Theories => Topic started by: LB7 on 29/08/2018 11:39:56
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In the attached file
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You say "I study the sum of the energy during a deformation of a device, the device doesn’t return to the start
position. ".
The simplest example I can think of is a spring compressed and then tied with a string.
I can connect it to some apparatus and cut the string so the expanding spring drives a generator and get power from it.
But that's not "free energy" it's just using the energy that was put into the spring by compressing it.
It's nothing new, or very useful unless you can mine compressed springs or find a plant that grows them.
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But that's not "free energy" it's just using the energy that was put into the spring by compressing it.
I counted the energy stored in the springs at start and at final:
- potential energy at start: -√2/2+(√2-1-d)S
- potential energy at final: +√2/2-√2/2*δ-((√2-1-d)S - (Sδ-Sdδ) )
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Calculations: with the depth of the device = 1 m, ‘S’ the surface of the white shape, ‘δ’ an infitisimal angle of rotation for the lateral walls, the black arm and the orientation of the springs:
Work to rotate the lateral walls:
-√2/2*δ
Work from the black arm:
+Sdδ
Work recovered to move out the white parts and move in the blue spheres:
The energy lost by the move out/in of the blue spheres is -δS-δdS*√S√2, the second term is very small in comparaison to the first. Note the move out/in in the direction of the attraction don't give/need energy what I lost in difference of pressure I win in difference of the length of the springs, the move out/in perpendicularly to the attraction have no difference of pressure but I need to increase the length of the springs
Potential energy lost by springs:
The difference of potential energy of the springs is
- potential energy at start: -√2/2+(√2-1-d)S
- potential energy at final: +√2/2-√2/2*δ-((√2-1-d)S – (√2Sδ-2Sdδ) )
- the sum of potential energy of the springs is: -√2/2*δ+(√2Sδ-2dSδ)
The work won by the green line:
√2*δ-√2δS with S the surface of the white shape
The laterals walls rotate, so I need to move out/in :
I need to move out/in the volume (1+d+δ/2)*δ*√2*√2*√S
The mean of difference of pressure is √S/2
The energy recovered is (1+d+δ/2/√2)*δ*S the term ‘δ/2/√2*δ*S’ is very small compared to the others
Sum of energy:
The sum is 0
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the device doesn’t return to the start
position. ".
That's the bit that matters.
If it doesn't return to the original condition then you haven't got a closed cycle from which you can repeatedly take energy.
It's like hydroelectric power- it only works because of rain.
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the device doesn’t return to the start
position. ".
That's the bit that matters.
If it doesn't return to the original condition then you haven't got a closed cycle from which you can repeatedly take energy.
It's like hydroelectric power- it only works because of rain.
Even during a deformation, the sum of energy is conserved. I counted the potential energy stored at start and the potential energy recovered at final