• Welcome to engineeringclicks.com
  • What follows is an off-the-wall idea

    Discussion in 'The Leisure Lounge' started by Justcurioustwo, Nov 9, 2019.

    1. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      Erich said it short and sweet, and given how this thread's gone, I don't expect you to stop repeating the same idea, no matter how many times we explain why it won't work, but once more:

      1. foot-lbs are torque, not force. You don't have ft-lbs.
      2. Your 39,312x64 lbs is basically made up math. Those numbers don't correlate to anything meaningful. The energy in each balloon is maximum at the lowest point, and less as it goes up. At the surface, it's whatever energy is embedded in the rubber required to stretch the balloon to its current volume. The volume at any point isn't particularly relevant.
      3. You are spending the full lifting force generated filling up each balloon at maximum pressure. No additional energy is created at any point.
      In a perfect world, you will get motion equivalent to the full amount of energy put in, minus. In reality, you'll get less. Either way, there will be no energy left to extract from the machine.
       
    2.  
    3. Justcurioustwo

      Justcurioustwo Well-Known Member

      Joined:
      Nov 2019
      Posts:
      71
      Likes Received:
      0
      Of course the drawing is based on real numbers.

      The numbers are based calculated psychical math using known variables.

      You are just wrong. As the air bubble rises it expands displacing more water giving it more liftin force.

      The volume of displaced water is inversely proportional to the lifting force

      s.weinberg, I appreciate your contribution here and having said that, I strongly suggest you re-visit the machine schematic and do the math yourself and/or consult a friend that has an engineering background.
       
    4. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      With respect, your first 2 bits don't make any sense - not that you're wrong, there's literally no statement of any sense being made - so I'll respond to 3 and 4.

      3. Nope, time for you to review basic hydrostatics. The REASON the balloon expands is that the same air, at lower pressure, due to less depth, takes up more space. No additional force is created. Certainly no energy - there's less and less energy the closer you get to the surface.

      4. There is no displaced water at the surface. At any other depth, you're ignoring pressure, which is kind of a critical part.

      The 'machine schematic' is a paper-napkin sketch of a non-useful device. There is nothing interesting about it, and I've reviewed and discussed all salient bits.

      Done the math myself, and I'll consult a friend who does engineering professionally right now:
      "Hello self, does Justcurioustwo have a point?"
      "No."
      "Like, at all?"
      "Still no."

      Well, there ya have it!
       
    5. Dana

      Dana Well-Known Member

      Joined:
      Sep 2010
      Posts:
      407
      Likes Received:
      3
      @Justcurioustwo, you need to study the basic physics relationships between force, torque, energy (or work), and power. Hint: They're not the same thing.

      You are correct that the lifting force of each balloon will increase as it rises and expands. However, the energy remaining to be extracted is continuously decreasing as the distance remaining decreases.

      The trap you're falling into is the same one that gets most people who think they've figured out how to make a perpetual motion (or "free energy") machine: A complex system of gears and levers and wheels (and/or inflated balloons) with enough conversions of one form of energy into another, it's easy to focus on one side of the mechanism where energy seems to be increasing, with the complexity of the system masking the fact that the energy had to be put into the system elsewhere (in this case, the energy required to compress the air).
       
    6. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      You're right. Because water is noncompressible, the pressure doesn't affect buoyancy.

      I take back that bit. Rest still applies
       
    7. Justcurioustwo

      Justcurioustwo Well-Known Member

      Joined:
      Nov 2019
      Posts:
      71
      Likes Received:
      0
      The lifting force of an enclosed container, (bubble); is equal to the volume of water being displaced. To push one cubic foot of enclosed air under water takes a force of 64 pounds at one ATM. If you push that same cubic foot of air down to 2 ATM’s the volume will be compressed to half its volume.

      This is the equation that determines the volume of air at different depths.

      Formula used (ATM/V1) X V = bubble size

      V1 is the volume at 1 ATM. Assuming the volume at 1 ATM is 100 cubic feet; at 2 ATM’s the volume is compressed to half its size or 50 cubic feet.

      At 18 ATM’s that 100 cubic feet is compressed down to (18/100) X 100 = 18 cubic feet
       
    8. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      The volume decrease is due to pressure difference. The increase on the way up is exactly the same.

      Pressure in water is caused by the full weight of the water above.

      Which is why to fill a balloon at depth uses the full energy required to lift all of the water above that point. You don't get more energy as the balloon expands at lower pressure. You just get less energy density in more space (pressure*volume is energy).

      Nothing else matters, as far as your idea is concerned. You've already spent all the energy you'd be able to extract when you filled your balloon - in a perfect world. In reality, you'll end up with even less.
       
    9. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      Think of it this way:

      Buoyancy is caused by the heavier water being pulled down by gravity, displacing the balloon.

      When you fill the balloon, you've essentially lifted up (pushed away might be more accurate, but lifted helps for the illustration, and is functionally identical) a bunch of water. When the balloon rises, all that happens is that the water you lifted falls back down, pushing the balloon ahead of it as it goes.

      No more water falls than you lifted. So there's nowhere that energy could possible be drawn from.
       
    10. Justcurioustwo

      Justcurioustwo Well-Known Member

      Joined:
      Nov 2019
      Posts:
      71
      Likes Received:
      0
      You are still missing the point. What I am proposing is the same principle that keeps a boat afloat.

      If a boat weighs 100 pounds and floats it is because the weight of the boat displaces 100 pounds of water. Water weighs 64 pounds per square foot. The volume of water being displaced is therefore 100/64 or 1.56 cubic feet of water.

      So far you have not successfully proven that the principles of the machine does not work.

      Having said that I do applaud your efforts.

      :)-
       
    11. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

      Joined:
      Nov 2012
      Posts:
      315
      Likes Received:
      0
      I'm not sure what you think the point of that last post was.

      If you think what I described is different than how a boat floats, you need to go review buoyancy.

      There is no principle to how your device works.

      You expend energy to fill some balloons underwater.
      They rise, using up that energy.
      That's it.
       

    Share This Page

    1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies.
      Dismiss Notice