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Mess with your minds: A wind powered craft going faster than a tail wind speed.
electrodacus:
--- Quote from: Kleinstein on December 27, 2021, 07:25:42 pm ---There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.
The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part. It would still need a good explaination if the 2 parts together are actually the maximum avialable power.
So both formulas can be correct, but are made for different problems. For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.
--- End quote ---
We are now talking about direct upwind not direct down wind.
For direct down wind you have w-v as wind power available to vehicle
For direct upwind you have w+v as power needed by the vehicle to overcome drag.
Both a wind turbine and a sail will have available the same wind power but the wind turbine is limited by Betz limit around 59% while a sail has no such limitation as wind speed behind a sail can be zero thus a sail is much more efficient in theory 100%
So while stationary a vehicle with a wind turbine can at most have 59% of that ideal Wind power equation while a sail can have 100% in ideal case.
Of course if vehicle is not moving (brakes applied) then none of that potential power is used by the vehicle.
Thus if you wanted to move that sail vehicle against wind (same as a cyclist is a sail) it requires that potential wind power + some extra to move.
My earlier example is perfectly correct and for a bicycle to move at 1km/h against a 230km/h head wind it will require a minimum of 66kW and gears can not help with anything as a cyclist may be able to provide 300W for minutes and maybe peak around 1kW it will never be able to provide 66kW and so never be able to move against a 230km/h head wind let a lone the possibility that 300W is enough to do that as the wrong formula and that calculator will imply.
IanB:
--- Quote from: electrodacus on December 27, 2021, 08:13:05 pm ---
--- Quote from: Kleinstein on December 27, 2021, 07:25:42 pm ---There is absolute need to have only the vehicle speed in there as one factor:
Consider not driving the bicycle directly, but use a rope or pully to pull it. No doubt that with some extra gearing from the pully the force is lower and thus less power needed to pull the rope with a constant speed. So half the speed needs half the power.
If you have the (V_w-v_v) factor instead the power needed would be essentially constant, and the pully would magically need to increase the power. So this is obvious nonsense and thus the (w-v)³ form is wrong for calculating the power needed to move the vehicle against or with the wind.
The (w-v)³ type formula is "correct" (an approximation ignoring the Betz limit) for a wind turbine on the vehicle to produce power there (e.g. to power the light). This is a different thing from the power needed to drive against a head wind or a sail uses to drive a vehicle with head wind. It is more like the 2 parts together make up the wind power. The sail uses force times speed and the wind turbine used the (w-v)³ part. It would still need a good explaination if the 2 parts together are actually the maximum avialable power.
So both formulas can be correct, but are made for different problems. For resistors you also know formulas for 2 parallel and 2 series resistors - different formulas for different problems.
--- End quote ---
Both a wind turbine and a sail will have available the same wind power but the wind turbine is limited by Betz limit around 59% while a sail has no such limitation as wind speed behind a sail can be zero thus a sail is much more efficient in theory 100%
So while stationary a vehicle with a wind turbine can at most have 59% of that ideal Wind power equation while a sail can have 100% in ideal case.
--- End quote ---
The direct downwind sail analysis is here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830
Such a sail can be at most 15% efficient.
If a sail is traveling directly downwind at the speed of the wind, then (w-v) is zero, (w-v)^2 is zero, (w-v)^3 is zero. Any way you look at it, if the sail is moving at the same speed as the wind, then the power will be zero.
gnuarm:
--- Quote from: electrodacus on December 27, 2021, 06:53:55 pm ---Here again the wrong formula https://en.wikipedia.org/wiki/Drag_(physics)
They add vehicle speed and wind speed for force part of the equation but not for the power part ?
--- End quote ---
This will be very easy to explain if you apply the correct formula correctly.
The drag equation is Fd = 1/2 · Cd · A · p · v^2
The v is relative to the medium so Fd = 1/2 · Cd · A · p · (vw + vo)^2
That's the force created by the vehicle moving through the air.
The power to move the vehicle is measured at the point of the wheels in contact with the ground. There the force is equal to the drag force, so the power is P = Fd · vo
You don't include the wind velocity because the wheels don't see the wind velocity only the drag force.
Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero. In that case vo has approximated zero and so the power required to move the car has approximated zero.
--- Quote ---How will that make any sense unless people just understand force but have no clue what power is.
That last therm also need to be vo+vw then result will be correct.
This can be tested relatively easy why is this not done at universities ? If they did this test they will realize their formula is just wrong and use the correct one.
Check this https://www.omnicalculator.com/sports/cycling-wattage use 50kg cyclist weight 20kg bike weight 1km/h (0.277m/s) bike speed and 230km/h (63.88m/s) headwind.
The Cd*A for tops is 0.408
So 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2 * 0.277 = 284.9 less than they get 300W with no elevation but they add the other less significant parts as riling resistance
So yes you think 300W is enough to deal with 230km/h head wind ? If so you never experienced strong winds (I'm sure nobody here experienced 230km/h).
The most I have ever experienced is 110km/h gusts (not even continues wind speed) and this is 8x less damaging than 230km/h
Correct answer for that is 0.5 * 0.408 * 1.225 * (63.88 + 0.277)2*(63.88 + 0.277) = 66kW
Unless you are a super hero with super human strength you will not going to pedal against a 230km/h wind at any speed not even 1km/h
--- End quote ---
How much power does that formula say is required for a zero bike speed? The correct answer is zero. I think anyone who understand work and power understand there is no work if there is no motion, so no power also. Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into. We could make windmills without any movement! That would be a big plus!!!
electrodacus:
--- Quote from: IanB on December 27, 2021, 08:26:15 pm ---
The direct downwind sail analysis is here:
https://www.eevblog.com/forum/chat/mess-with-your-minds-a-wind-powered-craft-going-faster-than-a-tail-wind-speed/msg3888830/#msg3888830
Such a sail can be at most 15% efficient.
If a sail is traveling directly downwind at the speed of the wind, then (w-v) is zero, (w-v)^2 is zero, (w-v)^3 is zero. Any way you look at it, if the sail is moving at the same speed as the wind, then the power will be zero.
--- End quote ---
No vehicle can move directly upwind powered only by wind without energy storage (Fact).
Yes that is correct a sail can not exceed wind speed directly down wind and the equation is correctly predicting that.
electrodacus:
--- Quote from: gnuarm on December 27, 2021, 08:30:58 pm ---
Consider the worst case where the speed of the vehicle has dropped infinitesimally toward zero. In that case vo has approximated zero and so the power required to move the car has approximated zero.
How much power does that formula say is required for a zero bike speed? The correct answer is zero. I think anyone who understand work and power understand there is no work if there is no motion, so no power also. Otherwise there are a lot of buildings doing a lot of work which we should be able to tap into. We could make windmills without any movement! That would be a big plus!!!
--- End quote ---
That is not the correct answer. You assume there is brake. Remove the brake and think again what will be the power needed to maintain around zero speed while you have a headwind.
You will need to remove the brakes in order to move so at very low speed you have no brakes to anchor you to the ground.
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