General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
Alex Eisenhut:
https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
IanB:
--- Quote from: Alex Eisenhut on December 28, 2021, 10:31:05 pm ---https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
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Good find. Thank you!
That will take some time to digest.
electrodacus:
--- Quote from: IanB on December 28, 2021, 11:10:01 pm ---
--- Quote from: Alex Eisenhut on December 28, 2021, 10:31:05 pm ---https://projects.m-qp-m.us/donkeypuss/wp-content/uploads/2009/06/Bauer-Faster-Than-The-Wind-The-Ancient-Interface.pdf
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Good find. Thank you!
That will take some time to digest.
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Do not waste your time it is the same rubbish from another person that has no understanding of power and energy.
gnuarm:
--- Quote from: fourfathom on December 28, 2021, 09:20:30 pm ---
--- Quote from: gnuarm on December 28, 2021, 08:51:51 pm ---
--- Quote from: fourfathom on December 28, 2021, 08:40:21 pm ---
--- Quote from: gnuarm on December 28, 2021, 08:19:33 pm ---So the full equation for the power applied to the vehicle through the wheels is
0.5 * air density * area * (wind speed - vehicle speed)² * vehicle speed
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Are you sure about that? When wind speed = vehicle speed, that equation still reduces to zero.
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Yes, that is correct. When the wind speed and the vehicle speed are the same, there is no wind force on the vehicle and so no force required to maintain the vehicle speed... other than friction which we are not factoring in here. This is just about the wind forces.
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OK. I will shut up and wait until friction is re-introduced. I want to see the situation when the vehicle speed = windspeed, and force has to be found to overcome friction. I know this works, but how the math applies to the actual real-world blades, gears, and wheels still isn't clear to me.
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I assume you are talking downwind? That's not complex, but figuring out the math is not so simple as it depends on a lot of details regarding not just the gearing but the propeller design and adjustment. The details really aren't important. When the vehicle is moving the same speed as the wind, the wheels are being turned by the motion of the vehicle. The wheels in turn work through the gearing to turn the prop. The prop pushes against the wind, in essence adding to the wind speed.
I'm not going to try to figure out the details of how much force or power is happening where. This is remotely reminiscent of an old regenerative radio I built as a kid. It used feedback to amplify the signal. This is analogous to that. The wind blows pushing the car initially, which in turn spins the prop. The prop makes its own wind blowing against the oncoming wind making the "virtual wind" higher. By the time the vehicle is moving at the speed of the wind this virtual wind is significantly higher. The only limitation is from the losses, friction and propeller drag, ect. It takes work to spin the prop and work the machinery, etc. That has to come from the wind itself. So when all the losses add up to the power from the wind, that's the top speed.
Earlier someone drew an imaginary vehicle that used sails like oars to move the vehicle. Clearly that vehicle could move faster than the wind as the sails moved backwards when the vehicle moved forward. The prop is the same thing.
gnuarm:
--- Quote from: IanB on December 28, 2021, 09:24:02 pm ---
--- Quote from: gnuarm on December 28, 2021, 08:19:33 pm ---The equation 0.5 * air density * area * (wind speed - vehicle speed)3 may be the right equation, but for the wrong purpose.
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There is an equation, 0.5 x (air density) x (area) x (wind speed)^3
This equation gives the flow of wind kinetic energy through the swept area of a turbine.
It breaks down into three parts:
1) kinetic energy of wind per unit mass of air
2) mass flow of air per unit area
3) swept area of turbine
Hence:
(flow of kinetic energy) = (energy per unit mass of air) x (mass flow per unit area) x (area)
If any one of those terms is zero, the available energy will be zero.
If you try to apply this to a sail, you find that the mass flow of air though the sail is zero (because the sail is a wind barrier). So you cannot use this equation in this form for sails.
When I brought up this point with ED and asked him to explain it, he quickly changed the subject and avoided giving an answer.
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I'm not going to get into all the details of how the propeller or a sail interacts with the wind. That is very complex and I expect the equation above is actually a limit or something else fundamental rather than a practical equation for a given propeller.
Doesn't matter. Either way it is for the power in the wind. The issue is how much power does it take to move the vehicle?? That is set by the force applied to the vehicle by the wheels which are propelling it against the wind. That force is the same force as generated by the wind, but the speed is not the relative wind speed (the difference in wind and vehicle speed) but rather is just the vehicle speed because that's what the wheels see pushing against the ground.
So the power applied to move the vehicle against the wind is the force the wind exerts (which the wheels must push against) times the speed the wheels are turning against, not the relative speed of the vehicle and the wind.
The equation with the cubed factor is incorrect because it is the wrong power being calculated. As you say, it is the kinetic power in the wind rather than the power applied to the car to maintain a speed.
Is that clear?
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