DDWFTTW - Directly Downwind Faster Than The Wind

Discussion in 'Propulsion' started by Guest625101138, Jan 4, 2009.

  1. InetRoadkill
    Joined: Jan 2009
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    InetRoadkill Junior Member

    If the prop is producing 9N at 4m/s, it's energy demands are 36W, not 15W. If you allow for a prop efficiency of 80%, the demands are closer to 45W.
     
  2. Guest625101138

    Guest625101138 Previous Member

    The prop only has an airflow of 1m/s. The same as it did when there was no tail wind. At 9N the power output is 9W which results in a battery drain of 15W allowing for the losses.

    If you apply your logic to an airplane with 100kts airspeed in a tail wind of 100kts it would require to double its power to achieve 200kts over the ground.

    The propeller on the cart is not reacting against the ground. It is reacting against the air. This is exactly the same condition as an aircraft flying at 10,000m. It does not matter if it is in the air or on the ground. Increasing ground speed with a tail wind does not increase the power required by the prop.

    So if you can prove that an aircraft increases power to maintain its airspeed in a tail wind you may be able to mount a plausible argument against DDWFTTW.
     
  3. InetRoadkill
    Joined: Jan 2009
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    InetRoadkill Junior Member

    Your diagram clearly shows that 9N are required to push the cart 4 m/s. Therefore, 36W are required. It doesn't matter what the airspeed thru the prop is. You indicated that there is 9N of drag. So 36W of power are required to overcome it at 4 m/s.

    FYI: I happen to be a pilot and I do understand indicated airspeed vs. ground speed. I've actually flown a plane backwards in level flight according to the GPS.
     
  4. sailor2
    Joined: Jan 2009
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    sailor2 Senior Member

    In ground IRF cart needs 36W to go 4m/s with 9N drag force.
    That fact has nothing to do with how much electrical power is needed to rotate the prop. You are mixing too very different things. In craft IRF the craft doesn't move at all meanig speed is zero, and power is 9N*0m/s = 0 Watts.
    That doesn't tell anything about needed shaft power.
    If you do, why do you use any other than airspeed to determine shaft power needed by the prop ?
    Or do you claim your power consumption was negative while going backwards related to the ground ? If not, why don't you just say so.
     
  5. Guest625101138

    Guest625101138 Previous Member

    The 36W is correct. 9W comes from the electric motor and 27W from the wind.

    If I was to replace the propeller with a spinnaker in 3m/s tail wind I could get maybe 2m/s over the ground. The power input from the wind in this case is 18W.

    My cart is capable of generating 9N of thrust with an airspeed of 1m/s. If I give it a tail wind of 3m/s then it will generate 9N with an airspeed of 1m/s and a ground speed of 4m/s. The 9N is enough to maintain the ground speed of 4m/s.

    Go back to your aeroplane. If it takes 100HP for level flight at 100kts you do not double horsepower to achieve 200kts over the ground in a 100kt tail wind. You only need the same 100HP. Exactly the same with my cart. The conditions for the prop have not changed. It sees a headwind of 1m/s. It is not bothered by what it happening at ground level. The prop operates in air.
     
  6. InetRoadkill
    Joined: Jan 2009
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    InetRoadkill Junior Member

    You are pushing against the ground and surrounding environment with 9N of force at 4 m/s. That's 36W. You can't escape that by arbitrarily juggling the frame of reference until you find one that makes your energy demands go away.

    As far as the airplane goes, its frame of reference is the free air stream. Ground speed is meaningless in flight. And before you say that cart's frame of reference should be the free airstream like the airplane, let me remind you that the cart's wheels are still attached to the ground. Even if we allow for zero aerodynamic drag in spite the fact that the cart has a headwind, you are still in need of 20W of power just to turn the wheels against the ground which has busted your budget of 15W.
     
  7. Guest625101138

    Guest625101138 Previous Member

    My propeller is the only source of propulsion. The wheels carry the weight and have constant rolling drag. The propeller is accelerating the incoming 1m/s airstream to something a bit faster relative to the vehicle to generate that thrust. That is all it sees - air incoming at 1m/s. Why do you think it needs to increase power when it is already producing the 9N of thrust required to keep the vehicle rolling at any speed? I have said the rolling drag does not change with ground speed so that 9N is enough to maintain 1m/s or 4m/s across the ground.

    When it is operating in a tail wind of 3m/s and doing 4m/s across the ground the wind provides the extra power. I have not busted the budget because the power demanded by the prop has not changed. It sees 1m/s wind and is producing 9N. The wind provides the rest of the power.

    You have not responded to how a spinnaker could actually provide energy to the cart doing 2m/s in a 3m/s tail wind. How much power is the spinnaker generating. By your analysis you are saying the spinnaker needs to produce at least 18W. What is the energy input to the spinnaker. Where is its battery connected?????
     
  8. spork

    spork Previous Member

    Thanks for the warning. We'll try not to use any complicated aero terms or formulas.


    That's why you're a pilot and not an aero engineer. The plane's frame of reference is whatever inertial frame we choose to perform the analysis in (or even a non-inertial frame if we're particularly sadistic).

    Something must be wrong with all the carts I built then - because they can all achieve DDWFTTW steady state.
     
  9. InetRoadkill
    Joined: Jan 2009
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    InetRoadkill Junior Member

    Fine. they have constant rolling resistance. But the power demands increase linearly with velocity.

    That's ok too. You've exceeded the wind speed and now you have a headwind. No problem.

    I'll come back to this.

    But power needed to overcome drag does increase with velocity. This is an important point.

    Houston, have a problem. You're saying that the prop is drawing energy from the wind and then turning around to say the prop is providing thrust (expending energy) against the wind. Umm, that doesn't make sense. Which is it?

    As long as the air is behind the spinnaker, it can power the boat. I don't think I've seen any spinnakers deployed with a headwind.

    You're twisting my point. Ground speed may be important to the pilot as a matter of convenience by saving some time, but it has no bearing on the mechanics of flight. A tail wind may save you time on your trip, but it doesn't change the efficiency of the aircraft.

    Wait, weren't the wheels providing power to the prop? How can the wheel drag go to zero then?
     
  10. spork

    spork Previous Member

    We're not talking about rolling resistance. We're talking about the torque that must be provided by the road's force on the wheel to turn the axle, and thus the prop-shaft.

    This is a brain-teaser; it's not supposed to be trivial. Questions of torque and force will be frame invariant. Questions of energy (such as "is this vehicle powered by the wind or the ground?") will depend on the frame chosen for the analysis.

    Yes, and as you may be aware, a traditional sailboat can never achieve steady-state DDWFTTW - not even with a really big spinnaker.

    True.

    I guess that depends on how you choose to define efficiency. It does if you're interested in miles/gallon (along the ground).
     
  11. InetRoadkill
    Joined: Jan 2009
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    InetRoadkill Junior Member

    Sorry, but that torque has to come from somewhere and it's not free. Since the wheels are turning against the torque, power is being expended. The force needed to produce the torque shows up as rolling resistance against the ground.

    You dodged my point. What is driving the prop? Where is the energy coming from? Is the prop driving the air to produce the thrust needed to exceed the wind speed?
     
  12. spork

    spork Previous Member

    No need to apologize. We know you're confused. The torque comes from the tangential force of the road against the rolling wheels. And no one said it's free. It's always the sceptics that like to talk about perpetual motion. We like to explain where the energy is coming from, and the sceptics like to ignore us.

    In the frame of the road there is no work being done by the wheels on the road or by the road on the wheels. From the frame of the wind, the road does work on the wheels.

    I recommend you avoid terms you're not familiar with.

    You failing to understand an answer does not constitue me dodging your point. If you want to actually understand this thing rather than tell me I'm flat wrong about an object I've built and tested many times over, we can chill and explain it carefully and thoroughly. Your call.

    What's driving the prop? The prop shaft. What's driving the prop shaft? The axle (through a 90 degree bevel gear set). What's driving the axle? The wheels. What's driving the wheels? Well, that depends on how you look at it. We can say it's the road (which isn't the most pleasing answer) or the propeller thrust (which SEEMS like it would constitute perpetual motion, but it does not for reasons we can easily explain).

    Yes, the prop is producing the thrust needed to exceed the speed of the wind. The energy is extracted from the wind/ground interface. The cart leaves a wake of slowed wind in its path.
     
  13. InetRoadkill
    Joined: Jan 2009
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    Location: San Antonio, Texas

    InetRoadkill Junior Member


    I notice that you're mixing rolling resistance with rolling power. Slide 1 shows the energy required at this point is 1.75W and that you have a surplus of power available from a battery. Cool.

    Slide 2 shows that the power needed to roll the cart is now 3W. So far so good.
     
  14. Guest625101138

    Guest625101138 Previous Member

    In Slide 3 I have ditched the motor and propeller and fitted a generator to my buggy. I am using another vehicle to push it.

    Do you agree with my analysis here?

    Rick W.
     

  15. InetRoadkill
    Joined: Jan 2009
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    Location: San Antonio, Texas

    InetRoadkill Junior Member

    Your power demands in slide 3 are now 33W.

    Windage power; 1N * 1 m/s = 1W
    Rolling power; ((2N + 5N) + 1) * 4 m/s = 32W;

    Total power = windage + rolling = 32W + 1W = 33W
     
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