The Wind Powered Sail-less Boat

Discussion in 'Boat Design' started by DuncanRox, Oct 20, 2008.

  1. clmanges
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    clmanges Senior Member

    SamSam wrote:
    I understand this argument. The man running on the treadmill experiences zero wind velocity, regardless of his groundspeed, so the sailplane (which should be tethered to the man and not the treadmill, btw) also experiences zero windspeed. Take him off the treadmill, and he will run at the same groundspeed, but will now have windspeed equal to that, so the plane will fly.

    The problem is that a sailplane is not the same device as the prop-driven cart or boat; the sailplane is not capable of generating its own wind.
     
  2. spork

    spork Previous Member

    This is a complete red herring. The equivalent frame is that of the belt, not the structure of the treadmill. Tether the sailplane to the treadmill's belt and it will fly just fine. The belt IS an equivalent frame to the road surface with a wind blowing over it - thus it will behave the same in all respects and give the same result for any experiment.
     
  3. sailor2
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    sailor2 Senior Member

    No, replace word source with cause and I would agree.
    Source is still the relative speed between air & ground (or water)
    Extra source isn't necessarily even needed, but for any given time intervall, moving laterally, like most iceboats do, increases the source volume and with given density also source mass and thus energysource is greater as well.
    Apparent wind is not related to source of energy only the cause of more being available.
     
  4. sailor2
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    sailor2 Senior Member

    I disagree. With your assumed 100% efficient prop, you get 137.5 watts of power, not just 125W. Off course there is no 100% efficient prop in reality, but if you assume it in the first case, why not in the second case ?
    Substracting something out from 137.5W suggests violation of conservation of energy with 100% efficient prop. All the energy extracted from air has to go somewhere, if you claim 125W is available, what happens to the rest of it ?
     
  5. sailor2
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    sailor2 Senior Member

    Guillermo, that's the best post by far you have posted in this thread.
    Now apply same logical thinking to the rest of it, and you'll see how ddwfttw is possible.
     
  6. sailor2
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    sailor2 Senior Member

    1)The man running on the treadmill experiences zero apparent wind velocity, regardless of his groundspeed, so the sailplane (which should be tethered to the man and not the treadmill, btw) also experiences zero apparentwind speed and the plane will not fly.

    2)Take him off the treadmill, and he will run at the same groundspeed downwind as the true wind, but will have zero apparentwindspeed, so the plane will not fly in either case.

    Just as principle of equivalent reference frames requires to happen, proving once more, how great and correct and useful that principle is.
     
  7. kerosene
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    kerosene Senior Member

    sailor2 beat me -
    it is exactly the same case.

    Traveling on treadmill is same as traveling downwind at wind speed = apparent wind zero. Exactly same rules apply to the model plane as the DDW-cart
     
  8. Guest625101138

    Guest625101138 Previous Member

    The only way to get the air to lose 137.5W under the stated conditions is to add 12.5W from the propeller. The conditions for the propeller do not change in either case. This leaves 125W to overcome vehicle drag and transmission losses in transferring said 12.5W power from the wheels back to the air via the propeller. If there is excess power then this will will result in excess thrust accelerating the cart.

    Rick W
     
  9. clmanges
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    clmanges Senior Member

    Sailor2 wrote:
    Sorry for the confusion; I should have been more specific about the conditions.
    In the case I set up, there is no wind velocity; treadmill with man on it are in still air. Then, man running in still air off the treadmill experiences apparent wind equal and opposite to his groundspeed. This relates to the treadmill videos, because there was no ambient wind in those examples, either.

    Sailplane tethered to the treadmill's belt -- well, I hadn't thought of it that way, but yes, it will fly -- just long enough to get pulled into the machinery, even though it has zero groundspeed relative to the belt. The belt and the plane both experience a headwind due to their motion relative to the still air, while the man running on the treadmill would not.

    Better?
     
  10. Guest625101138

    Guest625101138 Previous Member

    What can I say to that - There are slow learners and poor educators. Maybe we have a better mix here.

    However it does raise the question on progress with full-scale demonstration and indisputable evidence.

    Rick W
     
  11. sailor2
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    sailor2 Senior Member

    No, under stated conditions 100% efficient prop needs zero watts from the shaft to extract 137.5W from the air. If you add 12.5W from the shaft to keep conditions of the prop the same & 137.5W from the air that's 150W total which have to go somewhere, not 125W !!! That's a scalar addition and both are positive !

    The prop doesn't have energy from which to lose at 12.5 Watt rate.
    The shaft can deliver that rate if you want, but it's then added to the amount from the air, not substracted. Just like in the first scenario where power delivered from air is negative and again addition of -12.5W & 12.5W = zero total.
    The difference of the cases is still 150 Watts.

    Read this part from your original post again :
    Case1 - case2 = 150 Watts. The difference of the cases is 150 W
     
  12. sailor2
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    sailor2 Senior Member

    Yes, and that's exactly the case I wrote about and which you just replyed to.
    In both cases no apparent wind. No difference if outside or on treadmill.
     
  13. Guest625101138

    Guest625101138 Previous Member

    The calculation of the work is unrelated to the difference between case 1 and case 2. It is simply the rate of energy either added to or taken from the air stream as calculated from the velocity changes and the mass flow rate in either case.

    In the second case the air certainly gives up 137.5W.

    However the rate of work being done by the air on the vehicle IS 150W. This is the sum of what the air loses and what the prop contributes. So you are correct in that the 12.5W is already allowed for in the 137.5W.

    Rick W
     
  14. clmanges
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    clmanges Senior Member

    Let me try this again. Air speed relative to ground is zero in both cases. On the treadmill, the man -- and the plane he's pulling -- both experience zero wind velocity; even though they both have some speed relative to the treadmill belt, neither they, nor the surrounding air, are moving relative to the fixed ground.
    Off the treadmill, the man is then moving at some speed relative to the ground, but the air is not, so the man sees a headwind, and so does the plane.

    Try using the frame of reference of the surrounding air or the fixed ground, since neither is moving relative to the other.
     

  15. ThinAirDesigns
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    ThinAirDesigns Senior Member

    LOL Rick -- that's a bit optimistic as 3 of the primary critics when you gave up on this thread and when I restarted this tread are still critics:

    Guillermo
    Boston
    3dYachts


    JB
     
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