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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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      Cars have this neat thing called fuel they burn to produce work. They run out, and then need more fuel put in.
      It's pretty mindblowing stuff.

      EDIT: It doesn't really need saying, but in case: a car will always output total work equivalent to less than the amount of energy embedded in the fuel by the time the fuel is fully expended.

      It has been explained numerous times why this won't work. Your refusal to accept basic physics doesn't negate that.

      The numbers don't say anything about input, and very little about output. My previous statement stands with no modification
       
      Last edited: Jan 14, 2022
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    3. Justcurioustwo

      Justcurioustwo Well-Known Member

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      The numbers show the input and output.
      I am still waiting for you or anyone to point put why it does not work.
      s.weinberg, why does it not work?
      Please be specific, generalities just doesn’t cut it.


      :)-
       

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    4. Justcurioustwo

      Justcurioustwo Well-Known Member

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    5. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      There is literally nothing in your little schematic about either input or output. Like several other terms in this conversation, I'm beginning to suspect that you don't actually know what they mean.

      The bits of specifics about why you think this should work have been disproven in detail. There have also been general explanations as to why the concept is flawed from inception.
      Both have been done probably a dozen times by now. I don't hold hope that doing it again will yield any positive result, so I'll just refer you back to the numerous pages already doing this.
       
    6. Justcurioustwo

      Justcurioustwo Well-Known Member

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      It is clear to me that you did not look at the seaengine diagram that clearly states the input and output. Having said that it is also clear to me that you have no idea what the term “torque” means. Now if you can set your pride aside, look up the term “torque”, try to understand what it means and then we will continue this discussion; setting aside whether you graduated from high school or not.
       
    7. Erich

      Erich Well-Known Member

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      Water weighs 64 pounds per CUBIC foot not square foot.

      Your design needs air supplied to the bottom of the stack so that it can be put into a balloon and allowed to expand.
      The air starts at the surface at 1 atm pressure.
      What is the source of energy you use to compress that air so that it will force its way to the bottom of the stack and be available to expand?
      Can you calculate how much energy it will take to accomplish that task?
      Now compare that number with the energy released as the balloons rise.
       
    8. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      Ok, pretty well confirmed that you don't know what "input" or "output" actually mean.
      So I'll help.
      Input = what you are putting into the system - e.g. in this system, are you putting in compressed air? Are you putting in electrical power? Anything else?
      Output = what you are taking out of the system - e.g. in this system, do you simply watch it run? Do you try to extract energy or materials? If so, what and how?

      For your edification, torque is instantaneous rotational force, equivalent to Force*Distance
       
    9. s.weinberg

      s.weinberg Well-Known Member EngineeringClicks Expert

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      And, for the billionth time, I'll humour you. It's hard to say why the machine, "won't work," because you have started dodging specifying what exactly you claim the machine should do, after finally admitting way back that it won't do what you originally claimed.

      However, in a general sense, why it, "won't work," is because there's nothing clever about the machine. You imagine you are multiplying torque or energy or power or whatever it is that you fancy on a particular day by looking at a particular moment in time, and noting that there is more energy embedded in the system than you are instantaneously adding.
      Which is not notable.

      You straight up refuse to consider the time/effort/energy already expended to add said energy into the system.

      It is exactly the same as carrying a rock up a cliff, dropping it, and marveling that it falls even though you are not currently pushing it!

      Not that I expect this reiteration of the obvious to get through to you...
       
    10. Justcurioustwo

      Justcurioustwo Well-Known Member

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      If you drop one (1) five (5) pound weight, when it hits the ground it will apply a five pound force onto the ground.
      If you drop ten (10) five (5) pound weights at the same time you are applying fifty pounds of force upon the ground.
      Which is greater, 5 pounds or 50 pounds?

      :)-
       
    11. Erich

      Erich Well-Known Member

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      Actually, no. A 5 pound weight sitting on the ground applies a 5 pound force on the ground. A falling weight applies a far bigger force. Its why hammers work. But hammers stop doing their thing the second you stop swinging.

      But what energy did you use to get the 1 or 10 weights up in the air so that they could be dropped? Did they do anything useful when they hit the ground that was worth spending the energy to lift them?
       

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