# DDWFTTW - Directly Downwind Faster Than The Wind

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

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### zawyJunior Member

Sail has much higher efficiency, according to Bauer. At slow speeds the blades are turning, so are they not acting like a true prop (at least a little) as well as the strong effect of being a sail at that time? I am suggesting the reverse effect is occurring at high speeds and I want the equation for it. So far I have treated the small vacuum in front of the prop as if it was affecting a prop, not as if the was a true sail, despite my emphasis on calling it all sail-derived energy. If I treat it like a true sail, then it is going to increase the "prop" efficiency. Instead I have been using the same efficiency factor for sail component (v2) and prop component (v1).

If what I'm saying wrong and the simpler equations of Bauer and Drela are correct, then if prop efficiency times chain efficiency is 0.93*0.98 = 91%, then blackbird should have been experiencing 30 pounds of air and rolling resistance (10.1 mph and 2.77x). If JB thinks 91% is accurate and says "no way it was that much" to 30 pounds of resistance, then I have support for my modification. My modification says 20 pounds resistance at max speed. JB has the closest thing available to resolve which theory might be closest.

I think the objective of the blackbird should be to make the wind speed in its wake equal to zero. That would indicate it took 100% of the wind energy (relative to ground) out of the wind.

I agree. If the frame of reference is the cart, then energy is flowing in from the wind at Vground<Vw and from the wheels when Vground>Vw. If the frame of reference is the cart, the Vc is always zero.

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### ThinAirDesignsSenior Member

Zawy, I couldn't find the answer to the above question in your last couple of posts.

If your still considering the question, that's cool -- just didn't want it to get lost in the shuffle.

JB

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### zawyJunior Member

JB, I didn't know how to answer the question. Let me try. As my response shows, I do not consider the treadmill and wind cases to be significantly different, if they are different. In both cases, stating with tread mill (aka wind) starting out zero and slowly increase: air resistance causes a push on car to move (sail effect), which causes tires to move, which causes prop to turn (prop effect). Notice that no prop energy was available without the air resistance. (calling it air resistance instead of "wind" makes it applicable to both wind and treadmill). In addition to this, there is a windmill effect on the prop which tries to turn it against the wheels. It is small compared to the sail effect. The prop effect is always the result of the air resistance (sail effect) in both treadmill and wind. This is how it starts at slows speeds and I do not think the physics changes for high speeds.

So I think my answer is no. The wheels are turned exclusively by the sail effect at all speeds (forces through body of cart). You can view the sail effect as caused by wind running into the prop, or as the treadmill pulling the prop back into the air resistance.

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### ThinAirDesignsSenior Member

Thanks, that is correct. We have a 'tight side' and a 'loose side' on our chain drive. The slack takeup device is of course on the loose side. At no time and in no phase of operation does the 'loose side' come under tension from the rotor attempting to drive the wheels.

What I am aiming for with these questions is to find out what of your theory can be measured and how it would be measured.

If there is ANY difference between the treadmill and the natural wind scenario, one should be able to measure it. Could you describe a way to instrument and demonstrate this difference that you claim exists?

JB

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### zawyJunior Member

Yes, air velocity meters (or pressure meters) on both sides of prop, relative to side of cart or wheels (not relative to each other). If treadmill is different like I theorize, then "downwind" side of prop (what I call front of prop) will have higher velocity than in the case of wind. You would have to set the blackbird up on "tire spinners" in relatively still air.

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### zawyJunior Member

How was the wind measured during the trial? Based on the equations below for wind above a lake, the wind at the bottom of the propeller may have been 8.7 mph, 10.1 at the axis, and 10.7 at the top. The wind speed over the area of the prop is not linear. 10.1 mph at the axis may have averaged 9.5 over the area. A treadmill and especially "tire spinners" will have an advantage because of this, unless the size of the cart is small compared to the length and width of the treadmill. The "infinite" plane of the flats slows down the wind close to ground more than the finite plane of a treadmill. It seems like the bottom of the prop would need to be 5 meters above ground for the effect to be small.

http://sts.bwk.tue.nl/drivingrain/fjrvanmook2002/node7.htm#SECTION00311000000000000000

[edit: the 10.7 to 8.7 difference prevents the ability to set set the optimum attack angle of the blades (or the rotation speed if that is adjustable). ]

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### ThinAirDesignsSenior Member

Nalsa rules require the wind to be measured at hub height. As you note from your link, this location places the measurement at a point above the point where the speed would be the average across the propeller.

The goal of course is to race the wind that you are using for propulsion and while the average speed of said wind would occur just a bit lower than the center of the propeller, we agreed that the disadvantage to the Blackbird would be slight to measure it at hub height and that point was easy to describe and explain and so it was.

You will note however that previously and on non-record runs we have placed streamers a full 6ft above the top of the propeller, racing wind that is WAY above and faster than the wind powering us and had no problem running multiples of this wind.

JB

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### zawyJunior Member

The streamers fell almost exactly as the wind equation predicted. The bottom one fell at 13.3 and the other two at about 15. If the prop were an extra 2 feet above the ground it, your 2.8x run would have been about 3.2x.

Are you adjusting the attack angle of the blades at higher speeds, or do you have the ability to change rotation ratio with the wheel?

If speed at 12 feet is 10 mph, here are the theoretical wind speeds from 1 foot to 22 feet:

1) 7.5
2) 8.2
3) 8.6
4) 8.9
5) 9.1
6) 9.3
7) 9.5
8) 9.6
9) 9.7
10) 9.8
11) 9.9
12) 10.0
13) 10.1
14) 10.2
15) 10.2
16) 10.3
17) 10.4
18) 10.4
19) 10.5
20) 10.5
21) 10.6
22) 10.6

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### zawyJunior Member

Sorry guys, I think I'm wrong about the treadmill. I think it's the same. That means my equation modification is wrong. Which means air, rolling, and chain resistance in the blackbird during the 10.1 mph (9.5 mph average) run for 2.8x was 26 pounds.

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### ThinAirDesignsSenior Member

Yep.

JB

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### Nick.KSenior Member

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### keroseneSenior Member

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### rushiNew Member

conservation of energy problem solved.

the kinetic energy of wind is converted to kinetic energy of vehicle, some portion of the vehicle's kinetic energy is converted into rotational energy of propeller. slowly the vehicle's speed reaches the speed of wind and by that time as the propeller is aerofoil that can generate lift and it again converts rotational energy to kinetic energy. this delay makes vehicle go faster than air but only for sometime. as it goes faster than wind, the drag of head wind slows the vehicle and when speed drops below the wind speed, same process repeats. basically it's oscillation. so instantaneous speed can be faster than wind speed but long term average speed will not be more than wind speed.

there's even better way that can beat current world record. use the flywheel with gear box.
have static sail that drags the vehicle in wind direction, connect wheels of cart to flywheel with gear, change gears to keep speed below wind speed so that push of wind keeps increasing flywheel energy. once you have required flywheel energy, disengage flywheel, let sail drag vehicle to near wind speed. then connect flywheel in reverse gear and lower the sail so it won't cause drag as vehicle goes faster than wind.

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### keroseneSenior Member

Nope not how it works.

conservation of energy is solved by thinking of power generated by the wheels and power consumed by output of the propeller. Formula is P=FV (power equals force times speed)

and example with made up numbers but hopefully illustrates the point.

5m/s wind, vessel has reached 6m/s speed. The ground to vessel speed is that 6m/s. Propeller operates in almost still air, the air to prop speed difference is inly 1m/s.

Lets say we take torque from wheels (to run the prop) that result in 100 Newton breaking force. Using P=FV we get power of 600 watts.
P = 100N * 6m/s = 600W

now what kind of Force can we generate from 600W in 1 m/s speed?
From P=FV we get F=P/V, so 600W/1 m/s = 600N.

so 100 N braking force can be geared to 600N pushing force thanks to different relative speeds in the different interfaces (ground-to-vessel and aur-to-vessel). Plenty room for operational losses.

this explains why the record vessel did 2.8 (if I recall right) wind speed gradually accelerating to that speed with no oscillation and in a sanctioned setting.

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### keroseneSenior Member

I have summarized it as:

the relative speed differences in two interfaces
([vessel to ground] and [vessel to air])
allow for leveraging the higher speed difference of [vessel to ground] into higher force in the lower speed difference of [vessel to air].

Last edited: Jun 4, 2021
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