Gravity powered devices => Gravity powered devices => Topic started by: EOW on April 27, 2018, 08:56:04 AM

Title: Deformation of a device
Post by: EOW on April 27, 2018, 08:56:04 AM
I adapted my theoretical device and instead of use small spheres and springs I use water and gravity. I choose the device like I drew but maybe it is not like that because, that I drew is not the same device I study without mass and gravity. I choose that one because like that the center of gravity is constant and it is easier to think like that.

No friction to simplify the calculations. All volumes are constant. Gravity is always vertical. At start, the device has near 0% of water, just thin layer between walls. At final, the device has 29% of water inside. At final, there is 29% of the white parts of the rectangles outside the device. The white rectangle is "nothing", just air or polystyrene for example (something with a small density).

1/ The device without the white rectangles keep constant the sum of energy, like that I don't need to calculate the deformation of the 4 walls.

2/ The device with the white rectangles: I move out the bottom of the white rectangles FOREVER and I move in water between the white rectangles. The external stock of water is at start at the altitude of the green dot like that the water never win/lost a potential energy. I need to move down the white rectangles when the device is deformed. I need to move down each white rectangle and I move out from the bottom the white part. I move down 50% and I move out 50%, to keep constant the center of gravity (the green dot). The energy needed to move down the first 50% is recover by the move out of the white parts. The second 50% that I move out win more energy because I move out from the bottom and I move in water in mean at the middle: the difference of energy it there.

It is not a cycle. I study from start to end. At start it is a parallelogram and at final it is a square. I need to move out 29% of white rectangles and move in 29% of water.

Images:

1/ the device at start
2/ the device 10° after (angle of the black arm)
3/ The device at start and after in the same image
4/ The device at start and at final in the same image
5/ The device with details
6/ How I move out the white parts, and move in water.
Title: Re: Deformation of a device
Post by: EOW on April 27, 2018, 04:52:08 PM
I drew the device at start and at final (at final by transparency).

I study the energy just at the middle, at left and at right.

At the middle, I need to move down the rectangle of a length of X1 but I move out the white part at the bottom, so what I need to X1 I recover at the bottom. I need to move out the length of X2, that length is at the pressure P and I move in water at a mean of P/2.

At left, I move out the white part of a length of X5 at a pressure P' and I move in at a mean pressure P'/2. I need to move up the rectangle of the length X3, so I recover an energy.

At right, the energy won by X3 is lost by the length X4. I move down the rectangle of X6 but I move the white part in the same time, so I lost nothing.

It is the same logic for all rectangles.

The potential energy from start to end is at the green dot. The water I move in is at the altitude of the green dot.

I don't count the energy to modify the external walls because it must be at 0, in the contrary the device without the white rectangles create/destroy the energy.

Title: Re: Deformation of a device
Post by: Low-Q on April 27, 2018, 11:21:32 PM
Maybe you should learn how to use SketchUp Make for the 3D effect. Easier to see the principle of operation when in 3D. Hard (at least for me) to just read the text, looking at 2D drawings, and understand how it works :-)

Vidar
Title: Re: Deformation of a device
Post by: EOW on April 27, 2018, 11:28:21 PM
Thanks, but I drew only in 2D (front view of the device), the arrows to show the move out/in are perpendicularly to the screen, the image show the side view:

Maybe some images are difficult to read because I drew the device in 2 positions in the same image, to show how is the device at start and at final, or at start and another position. Take the 2 first images, you will have the device at start and after a rotation. Just look at the green dot: it is fixed on the ground.

The second image here shows the device at the final position

Title: Re: Deformation of a device
Post by: EOW on April 28, 2018, 11:09:43 AM
In my calculations of the sum of energy I didn't count the energy needed to move in water because water is at the green dot at start (water is outside the device at start), so the mean energy to move in is 0. Because the water (outside the device) is at the green dot, when I move in the water, I need an energy if I move in more bottom but I win the same energy because I move in water more upper.

This is a good news the device works with theoretical spheres+springs or with water (mass) and gravity. That device is easy to build in reality, I mean it is not special technology, just water and a deformation of a device. Sure I need a lot of pumps to "move in" the water near it must be, to prevent flow bad friction.

The animation: https://imgur.com/a/cFXyeuF

That forum doesn't accept gif.
Title: Re: Deformation of a device
Post by: EOW on April 30, 2018, 01:46:40 PM
No, it doesn't work with gravity just with my theoretical device. Maybe with a far attraction, I drew 3 cases:

1/ z1 & z2 : the energy is conserved
2/ z3 & z4 : the energy is conserved, move out and move in the white parts
3/ z5 & z6 : the energy is not conserved, move out the white parts and move in the water
Title: Re: Deformation of a device
Post by: EOW on May 01, 2018, 08:45:55 AM
Or like that
Title: Re: Deformation of a device
Post by: EOW on May 02, 2018, 10:59:39 AM
Like my theoretical device ! with gravity, water and polystyrene.