The Wind Powered Sail-less Boat

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

  1. Joakim
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    Joakim Senior Member

    Boston

    You agree that an ice racer can reach the downwind mark faster than the wind. Thus it is not perpetual motion going faster than the wind dead downwind. This physical fact does not change wether one goes directly from A to B or by jibing.

    Why doesn't the sledge towed by a ice racer suffice as going dead downwind faster than wind? The sledge can be steered dead downwin and it only uses wind with a help of a ice racer. The mechanism an ice racer uses could also be build into the sledge, but very likely the ice racer would be more efficient and thus faster.

    It is obvious that using a normal sail powered vessel it is impossible to go dead downwind faster than the wind, just as it is obvious that it is impossible to go dead headwind. Both can be accomplished by tacking or with a propeller turbine combination. Why do you believe in going dead headwind, but not in going dead downwind faster than the wind?

    The design of a turbine propeller vessel and the physics behind it is already shown on the 3rd page of this thread. And the videos on treadmills shows that in practice for anyone having basic understanding of physics.

    The gearing calculations are far from silly, since a propeller is a gear that "connects" fluid to the vessel. There can be slip (=gear ratio not constant) and efficiency is not very good, but it still is a gear.

    Accelerating from zero velocity to faster than the wind dead downwind, the water turbine is turning the air propeller all the time. The handness of the air propeller is such that it would turn the wrong way, if wind would turn it. So at very low speed it doesn't turn at all and it acts as a sail. With sufficient water velocity the propeller starts to turn the turbine and nothing special happens when the wind speed is reached, just as nothing special happens when a ice racer reached wind speed as VMG.

    Joakim
     
  2. TeddyDiver
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    TeddyDiver Gollywobbler

    Actually in some of the videos the air propeller did turn wrong way a bit, maybe half a turn, before the water turbine overcame, so B made a point there.
    Later, after gaining some speed, it seemed like smth similar to happen, which actually is an effect of frame speed of the video camera and turbine speed..
     
  3. Windmaster
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    Windmaster Senior Member

    Here is the latest video on youtube.
    It explains everything.

    http://www.youtube.com:80/watch?v=_fBDcchw5nw

    Let's face it, it is hardly worth arguing with "flat earthers".

    What is amazing, is that some who claim to be scientifically trained, cannot grasp a simple concept of moving interfaces.
     
  4. Tiny Turnip
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    Tiny Turnip Senior Member

    Some of the images that helped me 'get' the principle of a turbine vessel going DDWFTTW:

    1. Imagine the whole vessel but with no water propeller being blown down wind, moving at, or close to the same speed as the wind, in still water. The air blades will not be rotating, or rotating very slowly, as there is little difference between vessel velocity and air velocity.
    There is considerable relative difference between vessel velocity and the stationary water however.

    2. then consider putting the water propeller on the end of the Shaft. The 'considerable' velocity of the stationary water relative to the vessel will spin the water prop, and drive the Air blades. The air blades will spin, clawing their way forward through the air which is relatively stationary with respect to the vessel, and thus pull the vessel forward faster than the wind.

    3. Of course there are many complex losses in the system, and my rusty physics prevents me from readily tacKling the detail, but the fact that energy is lost *does not matter* to the *principle*, it just affects the efficiency.

    The vessel is a machine allowing work to be applied in a different way. The gears on a bicycle allow us to apply the work we do pedalling to go much faster, further than we can walking, or even pedalling round a wheel of the same diameter as the pedal cranks, like a stunted penny farthing. (high wheeler?) The bicycle can do this *even *though *the *gearing *means *additional *energy *losses *in *the system*.
    (I'm sure the losses in the bicycle system are small compared with the turbine vessel, but the principle remains)
     
  5. Guest625101138

    Guest625101138 Previous Member

    Day two and still no rep points awarded.

    Once again I will post the question sheets.

    As is convention, velocity provided is with reference to the fixed ground surface and the answer has to be given with respect to the ground reference.

    Happy to have anyone provide the correct answers. First correct answers get rep points.

    Rick W
     

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  6. Guillermo
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    Guillermo Ingeniero Naval

    No sailboat can travel faster than the wind if going dead-downwind. No windmill can either sail faster than the wind, be it moving a prop, wheel or whatever. That's a fake.

    The amount of power transferred to a wind turbine is directly proportional to the area swept out by the rotor, to the density of the air, and the cube of the wind speed.

    The power P in the wind is given by:

    P = 1/2 π α ρ r^2 v^2

    where P = power in watts., α = an efficiency factor determined by the design of the turbine, ρ = mass density of air in kilograms per cubic meter, r = radius of the wind turbine in meters, and v = velocity of the air in meters per second.

    As the wind turbine extracts energy from the air flow, the air is slowed down which causes it to spread out. Albert Betz, a German physicist, determined in 1919 that a wind turbine can extract at most 59% of the energy that would otherwise flow through the turbine's cross section, that is α can never be higher than 0.59 in the above equation. The Betz limit applies regardless of the design of the turbine.

    This equation shows the effects of the mass rate of flow of air traveling through the turbine, and the energy of each unit mass of air flow caused by its velocity. As an example, on a cool 15 °C (59 °F) day at sea level, air density is 1.225 kilograms per cubic metre. An 8 m/s (28.8 km/h or 18 mi/h) breeze blowing through a 100 meter diameter rotor would move almost 77,000 kilograms of air per second through the swept area. The total power of the example breeze through a 100 meter diameter rotor would be about 2.5 megawatts. Betz' law states that no more than 1.5 megawatts could be extracted.


    Cheers
     

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  7. Joakim
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    Joakim Senior Member

    You should have read page #3! I'm sure you know enough physics to understand this. It's not a windmill, it's an air propeller.

    Use your equation for the water turbine to calculate how much power is available there and then calculate how much is needed for the air propeller.

    Joakim
     
  8. Guest625101138

    Guest625101138 Previous Member

    Guillermo
    You are correct with the above statement. But this is not what we are debating.

    We have a WATER TURBINE extracting power from the water and delivering it to the AIR PROPELLER producing thrust.

    If you have a go at answering the questions I have posed in post #110 above you will likely gain the necessary understanding to realise that sailing DDWFTTW is indeed possible and those various land vehicles are no trick.

    You will do yourself a disservice if you do not try to understand what is going on. It simply requires an open mind and some reasoning power, roughly that of a fifth grader.

    Rick W
     
  9. Guillermo
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    Guillermo Ingeniero Naval

    That makes no sense to me. You're asuming a boat is sailing faster downwind than the wind to demonstrate it can sail faster downwind than the wind. That's circular. And you're saying it's possible to extract energy from nothing. The only physics where it's possible to extract energy from nothing is on quantum mechanics, not in everydays life.

    The cart thing, which video I posted (as well as several others) before, is neatly a fake. Bringing its reasoning to the extreme it is like saying that if in absence of wind we push the cart fast enough it will create its own wind and run forever. If that's not perpetual movement, let God come here and see it. I'm with Boston in this one.

    Cheers.
     
  10. Guillermo
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    Guillermo Ingeniero Naval

    Unluckily I'm not a fifth grader but a humble Naval Architect.
    Let me ask you: Could it be possible you're doing a disservice to yourself?

    Cheers.
     
  11. Guest625101138

    Guest625101138 Previous Member

    Guillermo
    No. I have taken the time to understand it.

    The original demonstration was done by Mr Goodman as Windmaster has pointed out. There are published papers on his work. I applaud his efforts to demonstrate it and it is encouging to see others making similar vehicles. It is only a matter of time befor it is demonstrated on the water if it has not already been done. It is not that hard to do. There is no magic. There is no need for perpetual motion.

    I have posed a very simple problem that will help you understand what is going on. If you cannot work out the answers then find a fifth grader and ask them to give it a go. Someone with an open mind and not prejudiced about what is and is not possible.

    Rick W
     
  12. philSweet
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    philSweet Senior Member

    Rick.

    case 1. yes, to the right, at 0.5m/s
    case 2 yes, to the right, at 1.5 m/s
     
  13. Guillermo
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    Guillermo Ingeniero Naval

    OK. Let's see how your fifth grade mind works. Could you please set out the energy balance of the system for me to understand? Perhaps I'm missing something.

    Cheers.
     
  14. philSweet
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    philSweet Senior Member

    Rick,
    first post requested workings.

    case 1 (2m/s -v) = 1/3v
    case 2 (2m/s -v) = 3v

    sorry for the distraction, I caught the second post and have just caught up with the discussion from the first one.
     
    Last edited: Nov 16, 2008

  15. Jim_Hbar
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    Jim_Hbar Junior Member

    Phil:

    I agree with your first post, where;

    But in the follow up post,
    I believe you have inadvertently reversed your equations.

    My explanation -

    Consider the "buggy" as a free body - If the input velocity is called Vin, and the output velocity Vout, and the ratio of Vin/Vout = N

    One can write the following equations:
    Vin + Vout = 2m/s
    Vin = N x Vout

    Substituting Vin results in (1+N)Vout = 2 m/s

    Now, in Case 1, N = 3 and in Case 2, N= 1/3

    So, Case 1, 4Vout = 2 m/s or Vout = .5 m/s

    and case 2, 4/3Vout =2 m/s or Vout = 1.5 m/s

    BTW Vout = the speed of the buggy


    This whole discussion reminds me of the high-school science experiment, where one pulls on the thread wrapped around a spool... And most do not understand when the spool wraps up the thread, and moves faster than the thread is pulled..
     
    Last edited: Nov 16, 2008
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