Maybe the subject is misleading, but I have tried different approaches to generators that does not need input energy to generate power out.
Here in this thread I want to display an idea that might seem like the obvious solution, but I do not have knowledge to say if this claims below is true or not. So please come with inputs and comments.
If you have a wheel with many containers filled with air - looks like a ferris wheel (See picture below). If this wheel is under weater, the water will provide more pressure at the bottom than the top. This will force the air inside the containers to compress more at the bottom than the top, so the volume will shrink as the container goes deeper and deeper. It is obvius that work has been carried out on one side from the top to the bottom. On the other hand, the opposite work has been carried out on the other side, so no net energy has been carried out. That means it takes no extended energy to rotate the wheel except fighting against the water - the wheel can be designed to easily rotate in water, so we do not take this loss into account.
The containers can be modified to have a flexible membrane on one side so the volume can change easily. This means that the membrane will move in and out for every revolution of the wheel. We still do not apply energy to rotate the wheel.
So then we attach a dynamo on each membrane so the dynamo can run and generate electricity as the membrane push and pulls a magnet through a coil - the dynamo.
Even if we load the coils with a light bulb, the membrane will still move in and out to pull the magnet through the coils as the wheel rotate - and we still do not apply energy to rotate the wheel. Or do we?
I have thought that if we take energy out of all coils at the same time, which we must, I believe the voltage must be rectified first so the energy doesnt cancels out due to the +30 degree phase shift between each next coil in this case. Left side versus right side will without being rectified be 180 degrees delayed which means energy cancellation.
So what happens when we load all the coils?
First, the membranes will not move as much, but they will still move due to the pressure difference. The energy we can take out is the difference in force the water does on the membranes at the top and the bottom, times the distance the membranes are traveling in and out.
Say we have 12 membranes as in the picture. Each membrane are 10cm^2 and they travel 10cm p-p. The wheel is 1m in diameter.
The energy for one membrane to produce per revolution:
100N x 0,1m/2 x 0,7071 = 3,53J
Energy for all 12 is therfor 3,53J x 12 = 42,4J or 42,4Ws
Then we have to convert this energy into electric energy. We have loss as resistance in the coils. In fact we can short circuit the coils and generate heat without doing any extra work on the wheel!!
How absurd does that sound - heating up water without energy input? But it still seems so obvious it can work this way. Because the wheel would spin just as easy with or without change in the membrane excursion - under no load!
Any comments are welcome. I cannot see the reason why this wheel isn't OU. Can you?