New type of gravity power plant offers chance to be landmark use of fusion

Discussion in 'Pseudoscience' started by trevor borocz johnson, Nov 13, 2015.

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  1. DaveC426913 Valued Senior Member

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    And how much energy would you expend excavating, lifting, transporting and decanting all this rock and dirt? Then moving your powerplant to a new location once that spot gets full?
     
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  3. billvon Valued Senior Member

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    OK. How much would such a device cost? How much energy would it require to construct? How much energy would it generate?
     
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  5. trevor borocz johnson Registered Senior Member

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    You don't move the power plant, you move the platform for lowering weight to different spots in the crater. Damn that's a question my patent lawyer asked me four years ago, we could be on the moon right now living it up by some crater front property.
    It's not to make money. Its to save the trouble of sending fuel to a moon colony.
     
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  7. trevor borocz johnson Registered Senior Member

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  8. DaveC426913 Valued Senior Member

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    Lowering the weights provides energy. How do you collect this energy? You have a turbine generator of some sort. This will have to be moved every time you want to change drop points.

    Again: how much energy do you expend excavating all the rock/regolith, loading it, transporting it and decanting it? I'll give you a hint: orders of magnitude more than you'd gain back by dropping it a few hundred metres into a crater.

    Do you wonder why they don't have hydro dams in the middle of the Sahara desert? Because it would take a lot of energy to physically carry the water there. Much, much more than you would gain from the energy of a paltry 100 metre fall. And that's using water - which will flow of its own accord. Your moon rocks will have to be carried.
     
  9. trevor borocz johnson Registered Senior Member

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    Well. you could have many buckets full of rock being lowered operating the generator at once. The energy from all that weight could be used to operate a device for digging, cutting, or drilling, then to just lower the weight from the top lip of the crater to the bottom of it. Pulling up say a ten foot cube of moon soil would be a lot less energy then what would be gained if you lowered it 30 times that distance. Since your converting mechanical to mechanical with the turbine operating your digging or drills or whatever, you're getting a good energy conversion efficiency as well. If you were using the energy to operate say a moon dwelling you could have a self operational system that lowers the weight down at different speeds based on the amount of energy that was needed at any given moment.
     
  10. DaveC426913 Valued Senior Member

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    You're figuring that cutting that ten foot cube of moon soil, lifting it, moving it and putting it in your bucket won't use up any energy.

    I suggest you try this at home on a small scale. See how much energy you expend, and see how much you gain. Then see what happens when you have to move the operation elsewhere when you've run out.
     
  11. trevor borocz johnson Registered Senior Member

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    I don't really wonder that. Try this, tying two of the same plastic cups together with a piece of string. then drop one cup off a table or ledge to see how far you pull the other cup along the surface of the table. There is more energy in the distance of the gravity then that of the same distance on the table.
     
  12. DaveC426913 Valued Senior Member

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    You should.

    Don't forget to count the energy required to get the cup, take it to the a bucket of standing water, fill the cup with water, bring it to the kitchen and tie it to the other cup. It will be significantly more than what you gain by a cup of water falling three feet.

    But it's a great example, since it scales down well to a kitchen experiment. If what you say were true, why don't you literally power your house that way? If you can extract energy from that 2-cup system continually, then you really could power your house.
     
  13. trevor borocz johnson Registered Senior Member

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    Now you just don't make sense Dave. win for me
     
  14. Kristoffer Giant Hyrax Valued Senior Member

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    No wins for you, trev.

    Tell us how expending more energy than what we'll get out of it equals a profit?

    Your model makes as much sense as the South Park underpants gnomes.
     
  15. DaveC426913 Valued Senior Member

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    You not being able to make sense of basic physics has been the problem since post 1.
     
  16. exchemist Valued Senior Member

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  17. trevor borocz johnson Registered Senior Member

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  18. DaveC426913 Valued Senior Member

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    Trevor, we get what you're saying about extracting energy from dropping stuff into craters. Really. We get it. It's not a complex concept.

    Your attempts to beat on the initial unpolished concept (such as the above image) are ways of evading the followup questions that are raised by the concept.

    What you don't get is that you can't extract that energy in any meaningful way without expending energy in the process. And you will expend more energy than you gain.
     
  19. billvon Valued Senior Member

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    OK. How much would it cost (and weigh) to send all that machinery and nuclear explosives to the Moon? Now how many solar panels could you buy (and lift) with the same money? And which system would be more hazardous to colonists on the Moon?

    Remember, on the Moon, solar panels produce more power from the get-go - and if you site them right, they produce power 24 hours a day, 7 days a week, 365 days a year.
     
  20. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Hey, the greater the head, the more energy you can produce!
     
  21. trevor borocz johnson Registered Senior Member

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    how so?
     
  22. DaveC426913 Valued Senior Member

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    There's 137 posts that will explain it.
    We'll wait here while you catch up..
     
  23. trevor borocz johnson Registered Senior Member

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    Oh yeah well I ask anyone if they have any useful information about fusion lasers then. I'm tired of explaining in circles.
     
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