# DDWFTTW - Directly Downwind Faster Than The Wind

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

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### keroseneSenior Member

Friction where? In the mechanism of the cart?

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### zawyJunior Member

Friction anywhere. They are both penalized by any friction, but to different degrees based on a difference in the pressure profile across the prop. There is not much if any difference in total pressure drop (v=v1+v2). But some is on the downwind side of the prop (v2), so it changes the net power calculation by Dr Drela's Jan 2, 2009 derivation (eq 3) where his dW/2 is Dr Bauer's v.

In Dr Bauer's eq 13, there is an s=+/-1 for the "windmill" (sail) and propeller modes (basically two different equations). With the correction, it becomes a continuous equation for a smooth transition from Vc=0 to Vc>>Vw. The treadmill derivations are not valid for Vc<Vw, but my correction combines the derivations. Dr Drela's work, as he states, is only good for Vc>Vw. To correct Dr Bauer's eq 13 for a smooth equation at all speeds:

Change
[L(n-1)+L^2)]
to
[L(|n-1|)+(v1^2-v2^2)/Vw^2]

and take out the s. v1=0 for sail mode Vc<Vw-v2. I believe it is at about Vc=Vw-v2 that v1 starts to increase and v2 starts to decrease.

There may also be an implied divide by zero in Dr Bauer's work at the transition point that this fixes.

I am not happy that the || is still in there as a carry-over from Dr Bauer's s=+/-1. I would like to make sure I am starting the physics at Vc=0 and correctly deriving Vc>Vw from there without assuming endpoint conditions. Declaring a change in pressure on only one side of the prop is the historical error.

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### WindmasterSenior Member

Why don't you convert all that into something that normal people can understand?
Why is it always doubters or nearly non-believers that generate all this gobbledygook? What planet are they on? How many angels can dance on the point of a needle?

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### zawyJunior Member

My first post has the easy intuitive explanation at the bottom. I am typing only for those who are serious about about getting the perfect prop and prop speed.

I would like to see people stop saying the energy comes from the ground. We don't say wind farms get their energy from the ground. It's terminology that obscures how it works. It's possible to derive it from the viewpoint of the power leaving the ground, but people unknowingly use the air mass entering the prop as their stationary frame of reference rather than the air mass on the sides of the car. This is the source of the error. The thermodynamic problem I mentioned before can be resolved if the kinetic energy decrease downwind of the prop is accounted for. But it's hard to see that decrease when you think the ground is accelerating the vehicle into the air mass. This is in regard to derivations based on Earth. In regard to real treadmills, there is a tire friction that the treadmill motor overcomes, but it doesn't cause a pressure drop downwind of the prop. In a wind machine, energy is taken out of the air and a pressure drop occurs across the prop. There is also no air resistance on the treadmill, and yet this causes more pressure drop in real air.

If you don't like my assertion that it proves the treadmill is not exact, then you can still use the equation and assert that the vehicle is 100% a sail at slow speeds and 100% a prop at high speeds You can do this by letting v1 completely replace v2 at high speeds and vice versa at low speeds. Then you just have Bauer's 2 equations. But the equation allows it to be part propeller and part sail at anytime.

I did not look at Bauer and try to decide how to modify his equations to fit my beliefs. I had a belief and I used Drela's derivation to see where it would take me. I plugged the result into Bauer and it came out exactly equal to Brauer for the endpoints (all sail or all propeller). Therefore I probably did not make a math error and it adds support to my initial belief. My initial belief was that all the energy comes from the wind. After first seeing the video on Sunday and realizing it was not a hoax, i developed a model in my mind of how it should work. Luckily I did not see the treadmill example, and when I did it gave me fits as to how to explain the treadmill was the "same".

If *half* the pressure drop across the blade is due to friction losses (v2=v1), then the net force available for acceleration and overcoming the friction is about 30% less for the wind-machine verses traditional derivations (L=0.4, eff=90% for n=2 to 3.) To clarify, I am not saying this drop is due to more friction or that there is a greater pressure drop, but that when the same bulk data is plugged into the different equations, there is a noticeable difference by using the more precise equation.

The importance for anyone serious about this stuff is that if affects the choice of prop. Firstly, the intuitive view that led to this equation and that I expressed in my first post indicates that the prop should have a higher angle of attack closer to the axis in proportion to radius. This might be obvious to prop guys. Secondly, and more directly the result of the equation, is that the prop should be designed with an efficiency that matches the expected friction. In other words, selecting the prop should include the knowledge that v2 is not zero in order to get the highest efficiency out of it. Thirdly it must be realized that the prop is deriving all energy by acting like a sail at all velocities. Fourthly and most importantly the proper speed of the prop throughout its length should be such that no matter how fast the car is moving, that if there is zero wind relative to ground, then the prop should be turn that it provides no drag or thrust. In steady state this also means the wheels do not experience any force. When you push it in dead wind, it should go easily without resistance while the prop is turning.

Most of this is already pretty accurately included in the methods of Bauer. But the equation needs to be changed if you want to be precise, and the intuitive view I expressed before may be invaluable.

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### ThinAirDesignsSenior Member

Again, no.

JB

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### zawyJunior Member

You have not pointed out an error in my comments except to express your opinion. Do you have any writings on the subject I can critique for accuracy? The only thing I see on your website appears to be the discussion below. It didn't go into detail so it didn't make an error. My discussion applies when there is enough detail to derive the acceleration force at all speeds based on prop specifics.

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### ThinAirDesignsSenior Member

You're arguing with Galileo, Newton and Einstein. We decided not to cover their work on our site -- it's old news.

High school physics text books are what you need.

JB

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### zawyJunior Member

How did you guys get 2.8 in 10 mph when Dr Bauer and Dr Drela's power analysis indicates only 18 pounds of thrust? I used the data below. Your estimated resistance at only 20 mph was 36 pounds.

17.5' prop
Electric generator efficiency: 85%
Electric motor efficiency: 85%
Propeller efficiency: 85%
Coefficient of rolling resistance: 0.02
Vehicle gross weight: 650 lbs
Coefficient of aerodynamic drag: 0.3
Projected frontal area: 20 sq-ft

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### keroseneSenior Member

electric generator? I thought blackbird was geared with a twisted chain - did I miss a major update?

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### ThinAirDesignsSenior Member

Replace the 72% efficiency of your above listed power transmission method with one that is well in to the 90s% and see what happens with your numbers.

There is no generator or motor on the Blackbird.

JB

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### zawyJunior Member

Now I see that I included a penalty of 20 pounds retarding force from the wheels that I shouldn't have. You should take a closer look at my equation. My predicted max force for your system came out to be exactly equal to your rolling resistance, 13 pounds. The method presented by Dr Bauer and Dr Drela predicted 18 pounds was the max.

To calculate theoretical max speed, you plug in prop eff (eff), wind speed (Vw), prop radius (R), and then keep plugging in "n" (Vc/Vw) and L (0.1->0.9) until you find the force that equals your total resistance. L is v/(2*Vw) where v is how much the prop sped up the air. This comes mostly from Bauer's paper that Wired I believed linked to, which uses the same power analysis as Dr Drela. It's also derivable from the eff calc on your web site.

pounds force = Q*[1-(n-1+L)/(n*eff)]*(L*n-L+L^2)
Q=2*rho*(pi*R^2)*Vw^2 = 0.0146*(R*Vw)^2

Change the last +L^2 to -L^2 to get my max speed correction to Dr Bauers et al work. I used rho units so that things are in ft, sec, lbs force. But the equation is good for S.I. units.

So Bauer gets 17.8 pounds and my max is 13 pounds for 2.8*Vw with 10 mph. The eq is different for slower speeds.

Bauer's paper:
http://projects.m-qp-m.us/donkeypus...aster-Than-The-Wind-The-Ancient-Interface.pdf

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### zawyJunior Member

Can I get paid for this research? Your current set up should be able to get you up to 4x if you can get your L down to 0.25 in a 15 mph wind. Unless my equation is wrong, you should have no trouble getting 3.5 in a 13 mph wind. alternatively, you would have to increase prop diameter to 20. 13 mph and 20' is another way to get 4x. 5x is possible in 15 mph if you can get eff up to 0.88. 5.5 if you also increase prop to 20'. Efficiency is the biggest bang for your buck, but I think your equation already tells you that. What is the theoretical max for all designs? If you get a 30' prop in 20 mph with 90% efficiency then you can get 7x. Anything more than that seems difficult.

At 85% eff, a 5' prop on an airplane going downwind with a really small battery in 5 mph wind should go 15 mph with no problem. It seems like it should be a fruitful area.

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### zawyJunior Member

Dr Drela and Dr Bauer predicted blackbird's max speed was 3.4x. My correction to the equation gave the measured 2.8x. I derived it from insisting that all the energy comes from the wind. I had an initial insight that prevented me from taking prior derivations at face value.

Here is a repeat of understanding how this phenomena works. You do a wheel-to-prop gearing ratio that causes zero air resistance and zero wheel resistance when you push it in zero wind conditions. The only resistance is gear losses and acceleration of the prop which is gained back when the cart slows down. It's a wonderful little cart with little resistance. Push it at 20 mph. Everything's great, but it starts slowing down a little from normal rolling, gear, and prop efficiency friction. Now add a 10 mph tail wind. The wind thinks it is a stationary sail, so you first look at it like a SAIL no matter how fast the cart is moving. Whatever SAIL energy is lost to ANY friction is translated to a pressure drop in front of the prop (downwind side). But the *acceleration* force (non-friction) from this SAIL is transferred to the wheels which causes the prop to *immediately* accelerate which causes a pressure drop in the back. A pressure drop in the back means it has converted sail energy into prop energy. So you use sail equations for the friction losses and prop equations for the acceleration, but it initially was all SAIL energy. It's a sail boat at it's core. It's beautiful.

The primary purpose of the wheels are to tell the prop how fast to turn (or the attack angle) to achieve the above, not to provide energy. This could be done with an airplane facing downwind with a really small battery and tracking ground movement to achieve highest lift efficiency. The prop has a reverse windmill effect from the sail-action that causes a certain amount of force on the wheels. This is what prevents the speed from getting really high.

Here's another paper that treats max speed as if it were a propeller when max speed returns to being a being a fully sail case (friction losses are equal to pressure drop on sail side, not a pressure increase on the propeller side). The object is to design a sail that can act like a prop until it reaches full speed where you want it to be 100% sail again.
http://www.ewec2009proceedings.info/allfiles2/517_EWEC2009presentation.pdf

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### WindmasterSenior Member

Please set me straight on this. You seem to be saying that SAIL power is different than ROAD power. However, common sense would say that the power is the difference between the two mediums and it doesn't matter what power you call it - its the same power.

Can you therefore explain the difference between a sailboat sailing in still water in a 5mph wind, and the same boat sailing in water moving at 5mph in still air? It doesn't matter if you call it SAIL power or WATER power, isn't it the same thing?

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### zawyJunior Member

Case 1: A sail boat moving forward relative to the water at 2.5 mph with a 5 mph wind is taking energy from the air.

Case 2: A sail boat (pointing the same direction) moving backwards relative to the air at 2.5 mph is gaining energy from the water.

Notice the direction of energy flow depends on a velocity direction. If you want to say the velocity was in the same direction, then one of them has to be considered running backward in time. The direction of time determines direction of energy flow. So they are the same thing, but you have to be careful when you try to derive an equation from it. I think we should stick with the convention that the energy comes out of the softer medium so that the velocity is relative to the harder medium and *forward* motion in *forward* time.

Historically the equations for DDWFTTW have been derived for backwards time flow but they did not realize it. Said another way, they used the wrong frame of reference (downwind of the prop instead of *exactly* with the cart). Said another way, they did not realize there is still a sail component on the downwind side of the prop. This last error has its source in the belief that the energy is coming out of the ground in forward time. It's not, it's coming out of the ground only if you go backwards in time, or let the cart slow down (reverse acceleration).

At the max speed, if you place a positive pressure upwind of the prop, then you are declaring it to be a prop instead of a sail. This is the mistake researchers have been making. It's really 100% sail (v2=v, v1=0) at max speed with a negative pressure on the downwind side. So when someone declares it is a prop, his frame of reference quietly shifted from the cart to the downwind side of the prop which is moving at a slightly slower speed than the cart. So he thinks the cart can go faster than it can. Below max speed, you model the sail force as a mixture of upwind sail vacuum (v2) and downwind prop pressure (v1).

To clarify what happens at max speed: when there is no longer an acceleration, all the friction forces are canceling the sail forces. There is no longer a prop or gearing resistance force at the wheels. That force drops to zero as they get near max speed. Blackbird might be able to confirm this observation. At the same time, the pressure upwind of the prop shifts to become a negative pressure on the downwind side. This shift is what makes prior derivations less accurate.

L=0.28 seems ideal for a lot of velocities and eff=0.85 seems accurate, so I can simplify the above equation to determine max speed:
Rolling friction + air friction = Q*(0.30-0.05*Vc/Vw-0.30*Vw/Vc)

and solve for Vc. Or plug in a Vc and see how much extra sail force you have for acceleration.

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