First off, you can number me among the skeptics - I don't think it's possible to produce a free energy device.
Second, I had an idea for one anyway, and I'm curious as to why it wouldn't work.
The idea just uses gravity and buoyancy to attempt to keep a ping pong ball in motion, and a valve that probably doesn't exist, so treat that as a hypothetical for now.
Fill the tube with water and leave it open at the top. At the bottom of the tube, place the hypothetical valve such that the ping-pong ball can be pushed into the bottom of the tube without letting the water out.
You probably get the idea now - let the ball float up to the top of the tube, and use a device at the top to kick it out of the tube (depending on how fast it goes up, a clever design might be able to bounce it out just by using the upward motion of the ball if it's moving up fast enough when it breaks the surface of the water.)
So now the ball falls down next to the tube (and here is where someone could have it interact with devices to produce the energy needed to run a mechanism at the bottom to push the ball back into the tube).
Given that the tube can be quite tall, there would be room for several such devices between the top of the tube and the bottom - hopefully enough to power the device at the bottom to push the ball back into the tube.
Thoughts?
to push the ball in throughout the valve, u will need to displace the volume of water equal to the ping pong ball volume. By pushing the ball in, the water level will rise, so will the pressure. Not to mention the pressure that the valve will have to hold and if to push anything inside the tube, this pressure will have to be overcome.
no go.
I can understand that, thanks for the explanation.
Also, I found this (thanks to another thread on the forum) that goes into far more detail.
http://www.lhup.edu/~dsimanek/museum/buoy4.htm
Has anyone seen this?
http://www.frank.germano.com/oceanpower.htm
@Creativity
Quote:
"to push the ball in throughout the valve, u will need to displace the volume of water equal to the ping pong ball volume. By pushing the ball in, the water level will rise, so will the pressure. Not to mention the pressure that the valve will have to hold and if to push anything inside the tube, this pressure will have to be overcome.
no go."
I have found that there are people who throw their hands in the air and declare "that is impossible" and there are people who use deductive reasoning to simplify and solve what would seem to be impossible problems. The only problem is that usually the solution is so simple it is embarrassing for all involved.
Consider the following---
You have a thin walled hollow tube extending downward to the bottom of a tank of water, the tube has neutral boyancy and there is a valve on the bottom of the tube so no water can enter. Next a cylindrical hollow weight with an O-ring seal having a slight positive boyancy in water is dropped down the tube and work is extracted from this drop until it reaches the bottom of the tube. Now everyone tells me I cannot remove the hollow weight from the tube without displacing an equal volume of water, equal to the hollow weight volume---it is impossible. In fact it is easy, I open the valve, hold the hollow weight in place at the bottom of the tank and lift the tube, a portion of the hollow tube is now above the tank water level a distance equal to the length of the hollow weight. The weight is now in the tank and basically no water has been displaced, the valve is closed and the weight is given a little kick to the side and it floats to the surface. To reinsert the weight into the tube a set of arms attached to the tube grabs the weight, as the volume of the weight is lifted "out" of the water the tube gets heavier and falls "in" displacing an amount of water equal to that lost by the weight rising----net change in displacement of water = zero. I imagine someone will state the obvious, the weight cannot fall in the tube due to air pressure below it, we will place an air channel through the weight with a valve in it.
I am not claiming anything here gains anything, all I am saying is that all the experts stating that water must be displaced when a ball or hollow weight enters the tank equal to the ball or weight volume are wrong. In this example there is basically no displacement of the water because every action is countered with an opposite reaction, weight enters as tube leaves(rises)--weight leaves as tube enters(falls). I imagine some will say I am cheating or bending the rules, to this I would say ---people who succeed do not follow the rules they make their own.
PS-- I found this simple solution within five minutes of reading your post:)
Regards
AC
Actually, you are indeed displacing water equal to the volume of the ball - you're just doing it at a different time - specifically, between your last and second to last pictures, when you take the hollow tube while it is raised and put it back down into the water. Just because you do it when there is equivalent negative displacement elsewhere in the tank doesn't mean you spend no effort on displacing the water.
@comassion
Quote:
"Actually, you are indeed displacing water equal to the volume of the ball - you're just doing it at a different time - specifically, between your last and second to last pictures, when you take the hollow tube while it is raised and put it back down into the water. Just because you do it when there is equivalent negative displacement elsewhere in the tank doesn't mean you spend no effort on displacing the water."
Let's look at where any displacement could occur?
In the third tube from the left the tube rises up from around the weight therefore any displacement in the tank must be equal to the difference in diameters of the tube and weight. We will say the tube is microns thick so the change in the level of the tank is equal to a few drops of water---- essentially zero.
Next consider the picture where the weight rises from the water and the tube is pushed into the water----at exactly the same rate. Any real change in displacement would raise or lower the tank level. If the weight rises incrementally from the tank at the same rate as the tube falls incrementally into the tank then the tank level would not change thus there is no real displacement. The volume of tube entering the tank is equal to the volume of the weight leaving the tank on a microsecond to microsecond basis, the tube enters the water at the same speed as the weight leaves the water-- the tank level cannot change thus no real displacement of water in the context of the tank.
Quote:
"Just because you do it when there is equivalent negative displacement elsewhere in the tank doesn't mean you spend no effort on displacing the water."
Ah, now you are adding variables that have no context in my statements, I did not mention anything in regards to effort or work as it is not relevant to this topic. As I said in my last post I make no reference to gains (or losses) of any kind, I am stating the weight can enter the bottom of the tank with little or no displacement of the water in the tank--period. I did this in response to many people stating it is impossible--apparently it is quite easy. Maybe someone should send this to that Donald Simanek skeptic site, he seems to think this is impossible.
This machine does have one issue which I thought was obvious but----maybe it is not obvious to everyone.
Regards
AC
@allcanadian
Well, what will u do when there is no pipe in the water as in first from the left case? Obviously pushing a pipe into the tank all the way in, is a required and necessary precondition for placing the weight under the water according to ur scheme. Even when not taking any forces into consideration to get into the first stage of your experiment for the first time u basically displace the volume of the weight upfront of placing it under water. With no weight rising to compensate ur water tank level rises => displacement happens.
I do agree with the rest of ur scheme with compensation, just to start the scheme u need to displace :)
@Creativity
Quote:
"Well, what will u do when there is no pipe in the water as in first from the left case? Obviously pushing a pipe into the tank all the way in, is a required and necessary precondition for placing the weight under the water according to ur scheme. Even when not taking any forces into consideration to get into the first stage of your experiment u basically displace the volume of the weight upfront of placing it under water. With no weight rising to compensate ur water tank level rises => displacement happens.
I do agree with the rest of ur scheme with compensation, just to start the scheme u need to displace"
You make some good points, again I want to state I am not claiming anything, I like to look at complex problems and find creative solutions. In my first post I stated the tube has neutral boyancy and this means the tube must have a weight equal to the water it displaces. In this case if the tube displaces 10 liters of water it must weigh 10 kg, in the context of this thought experiment I would use a super thin tube of unobtanium. If the weight of the tube is 10kg and it displaces 10 liters then the tube neither sinks nor floats when fully submerged in the tank. It should also be noted that when the movable weight is in the water tank (outside the tube) that the tube is out of the water thus a 10kg tube would have 1kg of length not in the water. As well the movable weight "MUST" rest on the bottom of the tank or external pressure from below would force it back into the tube, the external water pressure acts in all directions as a pressure gradient. There is something very strange which happens in this device that nobody has mentioned yet?
Regards
AC
@allcan
will not the air inside the tube make it extremely buoyant?
the tube might be neutral, but that's only if its filled with water
Crap post, needed deleting:)