The Wind Powered Sail-less Boat

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

  1. Guest625101138

    Guest625101138 Previous Member

    Yeah; but they have gone away either not prepared to say they were completely wrong or simply not wishing to continue digging holes and embarrassing themselves any more. I take their silence as acceptance of the feasibility of sailing DDWFTTW.

    There is also the pilot on the DDWFTTW thread who has gone away and not returned. At least he offered concrete calculations to show his reasoning that enabled the misunderstandings to be specifically identified.

    It would be nice to get a thank you for taking the time to correct their misguided opinions but that is difficult for some when they have strongly defended an ill-thought opinion and later realised they were totally wrong.

    If you had not revived the thread they would have lived in blissful ignorance and not realised the joke until it becomes well known, accepted knowledge. So they have you to thank for now being better informed on this curious slice of sailing technology.

    I look forward to your post with a clip slowing a full-scale buggy easily exceeding wind speed directly downwind.

    Rick W
     
  2. sailor2
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    sailor2 Senior Member

    That's the flawed assumption which leads incorrect conclusion by you that principle of equivalent IRF:s doesn't work.
    In reality it works just fine without flawed assumptions. When you use it correctly the airspeed relative to the reference frame is the same in all cases by definition. An assumption not in agreement to that is flawed by definition.
     
  3. sailor2
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    sailor2 Senior Member

    Assuming power from shaft being 12.5 Watts :
    1) I agree.
    3) The sum is indeed 150 W, not 125W as initially claimed.
    2) the rate of work being done by the air on the vehicle is what the air loses, nothing to do with the sum including other sources than air.
    4) If something was alreade allowed for and you didn't take that into account then the error would have been 12.5W not 25W as it was. (150 instead of 125)
    Don't understant this 4th statement anyway.

    In both cases:
    If prop is 100% efficient, then there are no swirl losses, no tangential force between blades & air, and no torque due to that. Therefore no power from shaft. Assumed 12.5W from shaft and 100% efficiency are in cotnradiction.
     
  4. Guest625101138

    Guest625101138 Previous Member

    The power out of the prop is 12.5W. The is the power required to accelerate the air.

    Mechanical efficiency is defined as Power Out/ Power In. With 100% efficiency both power In and Out are 12.5W. The prop has to do this work. 100% efficiency does not mean it is doing no work. It just means it has no conversion losses. So the shaft power to the prop is 12.5W and converts all this to accelerating the airflow through the disc. There is no swirl, viscous drag or induced drag. JUst perfect conversion from rotation to thrust by accelerating air through the disc.

    Rick W
     
  5. sailor2
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    sailor2 Senior Member

    Accelerating air in the direction of initial flow means losing energy into air as induced losses by definition. See PDF Marc Drela posted to see that.
    If you want 100% propulsion efficiency, no amount of air can be accelerated linearly at all, since that would mean induced losses. You would have to have no slip at all. Therefore I initially assumed 100% efficiency meant thermodynamic efficiency instead. Perhaps you didn't mean that either. So what efficiency did you mean ?!?!?!?!?

    "With 100% efficiency both power In and Out are 12.5W." And depends how power out is defined. In ground IRF we are using the power out is 150W if power in is 12.5W. Does that mean the efficiency is 1200% ? I wouldn't define it that way since the power out comes from air by the prop, it's not converted from rotation to linear.

    "JUst perfect conversion from rotation to thrust by accelerating air through the disc." does not mean 100% propulsion efficiency as induced losses are there. It means more than just 100% thermo eff. as swirl losses are zero. And why assume that both powers are 12.5 W why not both as zero Watts ? That would avoid some of the contradictions, so some logic would still be there.
    If that's what you mean then you need a new definition for efficiency instead of just saying 100% and you also assume conservation of rotation energy is invalid. What's the point of analysing by math if laws of physics are assumed to not hold ?
     
  6. Guest625101138

    Guest625101138 Previous Member

    The power put in by the prop can be determined by the mass flow through the disc by the velocity change.

    In either case this is:
    Power = 0.5 x 100kg/s x 0.5m/s^2
    = 12.5W

    This formula is identical to the original analysis that Mark Drela provided.

    I am using the prop mechanical efficiency as 100% as defined in any text book you care to look at. That means 12.5W applied to the air will require 12.5W applied to the prop shaft. This obviously cannot be achieved but again is the same assumption that Mark Drela offered before he applied the efficiencies for the various components.

    In case 2 the power flows are 150W applied by the air. 12.5W of this is contributed by the prop leaving 137.5W given up by the air. That power goes into mechanical losses in the power transfer system, overcoming vehicle drag and accelerating the vehicle if any is left over.

    The whole aim of the exercise was to show that once there is relative movement of the air over the ground a small power input from the prop can result in a large power reduction in the air flow. It makes the point that the energy comes out of thin air. There is no perpetual motion involved.

    Rick W.
     
  7. sailor2
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    sailor2 Senior Member

    O.K. I did find only 6 hits on the google and one of them is from a book:
    http://books.google.fi/books?id=Ns5...&hl=fi&sa=X&oi=book_result&resnum=1&ct=result
    Engineering analyses of flight vehicles by Holt Ashley
    It seems to define prop mechanical efficiency exactly the same as propulsive efficiency in one of the problems, not mentioned in anywhere else in the book. That means that 100% is acheaved only if there is no slipstream and no swirl. Assuming 0.5 m/s slipsream therefore contradicts 100% efficiency assumption,
    since some energy goes into accelerating air with slipstream reducing that efficiency from 100%.
    I can not read every book in this planet to find out if any of them uses some other definition.
    Definition of thrust power = The power usefully expended on thrust, equal to the thrust (or net thrust) times airspeed.
    Airspeed is defined as relative speed of free stream air to the plane.
     
  8. clmanges
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    clmanges Senior Member

    I'm starting to think that this reference frame thing may have driven some people to insanity . . .
    As I stated, "Air speed relative to ground is zero in both cases."

    This is not a "flawed assumption." It is a condition of the experiment. It is the environment in which the activity takes place.
    It's quite easy to set up such a thing in real life; all you need is a big, empty building with all the doors and windows closed and the ventilation system turned off. Treadmill off in a corner, and a little track for the runner to use when he's not on the treadmill.

    Man on treadmill: the man and his tethered airplane experience no wind, despite their groundspeed (relative to the treadmill belt). In this case, the only thing experiencing windspeed is the belt on the treadmill, since it's the only thing in motion relative to the air. The treadmill belt feels a tailwind. Plane gets no headwind, thus no lift.

    Man off treadmill, running on ground: man with plane moves relative to ground, while the air does not. Man with plane see headwind, plane flies.

    What frame of reference do you use that would change the above outcomes?
     
  9. ThinAirDesigns
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    ThinAirDesigns Senior Member

    Me too.

    I'm not sure what your getting at clmanges, you're correct in your above described outcomes, but in your scenarios, you haven't change frames, you've changes entire scenarios. In your comparison, you haven't take the same story and looked at it through another 'frame', you've started telling a different story.

    A: If I were outdoors in a 6mph wind and were to run straight downwind at 6mph towing the plane, the plane would most certainly get no wind and thus no lift.

    B: I use your scenario with the man running 6mph on the treadmill in the enclosed room -- the plane would most certainly get no wind and thus no lift.

    A and B are equivalent frames -- both men are running 6mph and neither feels any wind.

    Your own two comparisons fail as equivalent because on the treadmill, the runner has a tailwind that matches his groundspeed (both 6mph). When the runner runs 6mph across the floor of the large enclosed building, he has no 6mph tailwind to cancel out the runners motion and thus sees a 6mph headwind.

    Again, not exactly sure what point your attempting to make, but just wanted to point out that even though your 'outcomes' were correct, if you were attempting to use those outcomes as results from 'equivalent frames', your conclusion is flawed.

    JB
     
  10. Guest625101138

    Guest625101138 Previous Member

    Just Google mechanical efficiency. Here are a couple of examples:
    http://en.wikipedia.org/wiki/Mechanical_efficiency
    http://www.britannica.com/EBchecked/topic/371842/mechanical-efficiency
    http://encyclopedia2.thefreedictionary.com/Mechanical efficiency
    Hope this helps.

    Rick W
     
  11. spork

    spork Previous Member

    It seems clear to me that you two are now down to debating the definition of "efficiency" or "mechanical efficiency" (or whatever). The "mechanical efficiency" you're using Rick is a much broader one. Both definitions have a place (and are useful for given applications). But arguing definitions is no fun. I recommend that we agree on the meaning of what you're trying to say (which I think most of us understand and accept), and not worry about what to call it. Just a suggestion.
     
  12. Guest625101138

    Guest625101138 Previous Member

    I never intended to debate the meaning of mechanical efficiency.

    My intention was to show in the simplest possible way the propensity for wind over the ground to give up energy simply by applying a little extra energy from a propeller. The aim was to answer the question - where does the energy come from? The answer in this case is "from thin air" of course.

    Rick W
     
  13. sailor2
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    sailor2 Senior Member

    My point was and still is that although mechanical efficiency is always defined as power out / power input, but despite of that power out can have more than one reasonable definitions for a prop. As a result just saying 100% efficiency is too fuzzy expression to use if one wants to teach somebody something.
    It's unfortinate that Rick refuse to realize that, because his teaching methods would otherwise be one of the best. Sad really.

    Ps. I wasn't debating what the definition is or should be, just pointing out it wasn't ( and still isn't ) properly defined like it should have been.
    Hope Rick now looks closely the bold part of your statement quoted above to understand what it was all about since you seem to agree more than one definition exists for props. Thanks for pointing that out.
     
  14. sailor2
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    sailor2 Senior Member

    It's not a condition of an experiment in 2 different frames. It's 2 different conditions of 2 different experiments in either the same or different frames of reference.

    Of course no IRF would change the outcomes of your 2 different scenarious.
    JB alredy correctly responded for the same issues, perhaps you could read it carefully.
     

  15. clmanges
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    clmanges Senior Member

    I did read JB's post carefully; almost replied to it, but decided to let my brain cool off a little first.

    I'll take the blame for the misunderstanding here; somewhere along the way I failed to properly comprehend "equivalent reference frames." Silly me, I thought that a reference frame was the same thing as "point of view." I need to work on that a bit more, I guess.

    Maybe I need a "Relativity for Dummies" book.
     
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