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non-displacement boyancy device

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  • non-displacement boyancy device

    @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
    I made this post over at OU.com and thought you may find it interesting as I have never seen this before. As well the skeptic Mr.simanek at this site---http://www.lhup.edu/~dsimanek/museum/buoy4.htm seems to think this is impossible as well.
    There also seems to be some confusion as to what must happen in the fifth frame of the image below. The weight rises above the water line at the same speed that the tube falls to the water line thus there is no displacement of water in the tank. The weight is then pushed horizontally to the start position and the cycle repeats.
    AC
    Attached Files
    Last edited by Allcanadian; 11-04-2009, 04:58 AM.

  • #2
    About the air pressure, instead of having the o ring on the weight, the valve at the bottom of the tube could be made to allow the weight out without letting water in, then you'd have a small gap around the weight equalizing the air pressure.
    If you forget about the arm lifting the the weight out of the water, it looks like 2 different cycles. one with the hollow tube moving up and down which should be completely neutral because you could recapture the energy it took to lift it out of the water. And the other the weight which should create a gain. Gain (should be) roughly equal to (the weight falling the distance of the length of the tube) - (the small bit of the tube sticking out of the top of the water) - (lifting the weight out of the water [about half of the weight falling it's own length])
    The only problem I see is that the weight is smaller than the tube, so there will be some change in the water level when you raise and lower the tube.
    Your logic seems completely sound.

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