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    Discussion in 'The main mechanical design forum' started by Justcurioustwo, May 11, 2021.

    1. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      If you carry one (1) five (5) pound weight up a cliff, you will have wasted some amount of energy, even if you harness energy when you drop that rock.
      If you carry (10) five (5) pound weights up a cliff, you'll have wasted even more energy! Tremendous.

      Same thing with bubbles underwater. Your 'machine' will probably do an excellent job at wasting energy. If that's your actual goal, bravo! Can't say it won't work.

      If you post one thousand (1000) posts of zero (0) value, refuse to accept responses of actual value, but continue to post the original post/diagram, as if it held some value, have you accomplished anything?
       
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    3. Justcurioustwo

      Justcurioustwo Well-Known Member

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      If you tie a rope to a tree branch---

      you attach a 5 lb. weight to the bottom of the rope
      you attach a 10 lb. weight 2 feet above the lower one
      you attach a 15 lb. weight just above the one below
      you attach a 9 lb. weight just above the one below

      How much weight is now pulling down on the tree branch-?
      The answer is 39 pounds, not 5 or 10 or even 15 but 39.

      I am beginning to understand where you are coming from, but I do not believe you understand where I am coming from.

      If the force I needed to snap a telephone pole in half was 300 lbs. of force and all I had was a force of 5 pounds the pole would not break.

      If I tie together 60, 5-pound weights together and then apply this force to the pole it would snap into.
      What am I missing?
      Remember the goal is to break the pole.
      :)-
       
    4. Justcurioustwo

      Justcurioustwo Well-Known Member

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      Let’s look at this from an energy out and energy in point of view.

      My example showed a pulling force of 118,428 pounds at three feet per second.
      How much electric power can I get from this-? Say it is (X)

      How much electric power is being expended to get the air down to the bottom? Say it is (Y)

      If (X) is greater than (Y) then I have a positive return. I am getting more out than expended at any one moment in time.
       

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    5. Justcurioustwo

      Justcurioustwo Well-Known Member

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      You also need to keep in mind that these air bubbles are expanding as they rise. This will create an acceleration force that needs to be included in the equation.
       
    6. Justcurioustwo

      Justcurioustwo Well-Known Member

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      I sincerely apricate your efforts to enlighten me on my misguided view. In one respect my efforts in this idea have given me something to contemplate which is better than drinking until I fall asleep. In the end I will find something else to ponder once I put this defunct idea to bed.

      Thank you, s.weinberg for your efforts.
       
    7. Justcurioustwo

      Justcurioustwo Well-Known Member

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      I got some responses from you !!
      Still, no one has dug into the numbers that would disprove the design.

      That would require some real thinking, maybe even checking the ”numbers” instead of making broad generalities like “it just does not work” or “it will not work”.

      At least my diagram has real numbers
       
      Last edited: Feb 23, 2022
    8. Justcurioustwo

      Justcurioustwo Well-Known Member

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      My last dying set of questions---

      I need to convert the mechanical energy needed to power the system; into electrical power, i.e., power(X).

      I need to convert the energy required to get the compressed air to the bottom; into electric power, i.e. power (Y)

      Do I consider voltage output?
      Amps?
      Current?
      I.E.; electrical properties?

      Torque; i.e. mechanical

      In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Torque is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically {\displaystyle {\boldsymbol {\tau }}}{\boldsymbol {\tau }}, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.


      In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector[2] connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:


      https://en.wikipedia.org/wiki/Torque

      https://en.wikipedia.org/wiki/File:Torque_animation.gif
       
      Last edited: Feb 23, 2022
    9. Erich

      Erich Well-Known Member

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      Electric power (actually mechanical power as well) is measured in Watts. Watts=Volts * Amps
      Please do work through your calculations, when you do you will arrive where Steve and I have been all along.
      It will take more energy to get the compressed air to the bottom of the ocean, than you get back letting the balloons expand and rise. I base that on my understanding of the Second Law of Thermodynamics.
       
    10. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      To paraphrase Erich's last line, so you understand better how it directly affects your case.
      You talk about calculating various energy forms, and you are welcome to do that to check your numbers, but it's not really necessary to evaluate your concept.

      You can convert from electrical energy to magnetic energy to torque to pressure to flow to potential energy of a balloon underwater to kinetic energy as it rises, to torque at your pivot to magnetic flux in a generator, back to electrical energy...but each of these has a formula that converts to energy. Each conversion will have the same value, minus some waste due to inefficiencies in conversion.
      You will never, at any stage, make a conversion that yields more energy after the conversion.
       
    11. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      To go back to the rock up a cliff example. You incorrectly stated that a 5 lb weight would exert 5 lbf on the ground when dropped. It would actually be far higher, due to impact, as Erich noted.
      You've taken a 5lb weight and multiplied the force, possibly many times over, simply by dropping it! How does this not mean that your output exceeds your input?!

      The reason is that Force is nearly irrelevant to what we are discussing. The reason the force is high is because the energy is being converted over a much smaller distance (the distance it takes the rock to stop upon hitting the ground) than the distance it took to carry that weight up the cliff. The energy is the same.

      There is a place for mechanical advantage. But it doesn't do what you imagine it does.
       

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