# DDWFTTW - Directly Downwind Faster Than The Wind

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

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### WindmasterSenior Member

If you were in the sailboat, how would you know which medium you were gaining the energy from?

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### zawyJunior Member

That's a good question because my answer seems to sum it all up best:

If you consider yourself to be traveling forward, the wind. Backward, the water. (for the particular set up I described)

In other words, the frame of reference you consider stationary determines where energy is coming from. Choosing the harder medium as the stationary one makes the most common sense.

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### WindmasterSenior Member

Why do you have to choose? Why not admit there is just a difference in velocities. You seem to be making a whole lot of confusion for yourself by insisting on looking at it from some arbitrary fixed point of reference. Energy doesn't "come" from anywhere, it is just a difference in velocities between two mediums. It's really simple leverage.

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### zawyJunior Member

I'm not confused about anything. Yes, a leverage viewpoint might be OK. I haven't thought or read about that perspective. When I do the math, I use the car as the frame of reference. But when I write down car velocity, it is forward in time and relative to the ground. That requires energy to flow out of the wind.

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### WindmasterSenior Member

Your answer seems to infer that the wind is some kind of reservoir of energy of itself.
But in fact, apart from the heat energy contained within it (molecular movement) it only has energy in relation to something that is moving at a different velocity. It has no energy OF ITSELF - only in relation to something else.
The very term "wind" means moving air. If it is not moving it is not "wind" only "air".
This is why the treadmill test is EXACTLY the same as the downwind demonstration. It doesn't matter whether the ground is stationary or the air is stationary - that's only to an outside observer.

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### zawyJunior Member

"it only has energy in relation to something that is moving at a different velocity"
That's what I said. I want to view the ground as stationary because I want to derive the velocity and acceleration of the car relative to ground. Requiring this viewpoint forces the view that energy comes from the wind.

There is a subtle difference in the treadmill that I explained before. The energy input from the electrical motor to the wheel is above and beyond the actual velocity difference between the treadmill and the air. That extra energy prevents a small pressure drop in front of the prop and it is 100% acting like a prop. A real wind machine can only operate off the difference between the ground and air speed which causes a small pressure drop proportional to the friction in the wheels, air resistance, and gearing. It does not have an extra energy that is above and beyond the velocity differences. It's a small but real effect.

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### ThinAirDesignsSenior Member

zawy -- did I miss the answer to this question?

JB

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### TeddyDiverGollywobbler

Why you wan't to do that? You force yourself to do the observation on a spinning object. Air is 50% of the time more stationary than earth

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### zawyJunior Member

Yes, my answer was in post 390. Direction of energy/power flow depends on chosen frame of reference combined with chosen direction for the resulting velocity/acceleration. If I switch either reference frame or redefine direction of velocity, then direction of energy/power flow reverses. An exception is heat energy (such as friction in a wind machine) which always goes towards higher entropy.

My equation above says the blackbird should have been in 13 mph winds when you reached 3.5*Vw. Was that the observed wind speed?

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### ThinAirDesignsSenior Member

I don't believe that answers the question he was asking, but perhaps I misunderstood his question. As clarification, I'll ask a related question:

Two sailboats:

A: One on a lake, sailing into a 10knot wind coming from the south

B: One sailing down a river (calm wind day) with the river flowing south at 10knots.

How does the boat behave differently and how can you tell?

The Blackbird reaches such multiples at many different wind speeds, but other than the HP limits of our drivetrain, the higher the wind speed, the higher the multiple that can be achieved.

JB

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### zawyJunior Member

I can't think of a difference.

Can the blackbird get 3.5x in less than 11.50 mph? My equation says no. I'm checking it for accuracy.

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### ThinAirDesignsSenior Member

It's a question I can't answer as we haven't tested down in the single digits much. As you know it would simply depend on all the system efficiencies.

I believe our prop is also well above the 85% that you quoted. That is a typical number for a GenAv prop and ours in comparison is turning far slower and accelerating the air far less dramatically. This will give us a nice increase in efficiency.

From our performance numbers we believe we are into the 90s% with both our prop and our transmission.

JB

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### ThinAirDesignsSenior Member

Zawy, another quick question so I can make sure I totally understand your position:

When comparing the treadmill test and the tests in the natural wind, are you insisting that in one or other of the scenarios, the spinning rotor is providing the force to turn the wheels?

JB

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### zawyJunior Member

If I am correct in saying there is a difference between the treadmill and the wind, then the difference in acceleration is not easy to explain with words. In math it is (K-L^2)/K where K is large and L is small, so the effect is small.

The treadmill is a continuing source of energy outside of the speed difference between the treadmill and wind. This is different than taking energy out of the wind. The difference I'm proposing is based on the conservation of energy. However, to the extent the wind is also a continuing source of energy, they are identical. Obviously to a very large extent the wind *is* a continuing source of energy to the cart, so I am saying they are almost identical. The difference I'm proposing is that there is a *local* effect (slight vacuum) at the front of the prop as a result of removing energy out of the wind *locally*, which does not happen when you apply an electrical motor to the wheels in still air.

The question of if the treadmill and wind have any difference is not important to me. I just want to get the right equation for acceleration at all car and wind speeds in a real situation. If it turns out this is exactly the same as on a treadmill, then great. But I don't want to start a derivation based on an electrical motor applied to the wheels in still air because I want to make sure I'm not violating the conservation of energy even at a local level (in front of the prop).

Bauer says chain efficiency is 98% to 99%. He also says sail efficiency is about 1/0.85 = 1.17%, which I don't understand. If I'm right in saying there is a sail effect at all speeds by creating a small vacuum in front of the prop, then it could cause a prop to get higher efficiency than expected.

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### keroseneSenior Member

would you care to explain what is the fundamental difference between prop and sail (besides the spinning). If you ask me they act the same way here. before the apparent wind swings around they act like bluff bodies (parachute) but as the apparent wind starts shifting there is airflow over the blade/sail and it creates lift.