| General > General Technical Chat |
| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
| << < (238/285) > >> |
| electrodacus:
--- Quote from: IanB on December 27, 2021, 06:20:19 am ---But every physicist in the world, every engineer in the world, every textbook in the world, agrees with that formula in Wikipedia. So are you going to tell the whole world they have been getting it wrong for the past 200 years and re-write all the physics textbooks? --- End quote --- That is your opinion and not a fact. Anyone that designs any sort of vehicle will know that second formula is not correct. Also I made the mathematical proof that it is not correct since it will not provide the same result as the first equation and they should provide the same result if both where correct equation calculating the same thing. --- Quote from: IanB on December 27, 2021, 06:20:19 am ---This is interesting. That formula that appears in Wikipedia is not derived from logic. It is derived from experiment. Many experiments. Over the years, scientists and experimenters measured the amount of power required in different situations, and they found that all of their experiments match the formula given by Wikipedia. What is more, you can repeat those experiments yourself, and if you do so, you will also obtain results matching that formula. --- End quote --- Witch one ? The first one is correct while the second one is not as I already demonstrated. Double the wind speed and drag power increases 8x (that is a fact and anyone can test to see that is the case). Why do you think vehicle traveling trough air is any diff rent in therms of drag than air traveling at the same speed (headwind). As far as drag surface is concerned it will not be able to differentiate. Air is a fluid so imagine the drag on a boat traveling on a lake at some speed v and then same boat in a river that flows at speed v (there will be the same drag in both cases). Air is no different from water as is still a fluid the only difference is lower density and important for blackbird it is a compressible fluid. |
| bdunham7:
--- Quote from: electrodacus on December 27, 2021, 05:14:27 pm --- Anyone that designs any sort of vehicle will know that second formula is not correct. --- End quote --- Can you cite any actual person that 'designs any kind of vehicle? Is not the designer of the Blackbird such a person? Do you think he agrees with you? --- Quote ---therms of drag --- End quote --- |O :-DD Why not 'pints of drag'? Or perhaps 'acres of drag'. |
| electrodacus:
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 ? 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 |
| Kleinstein:
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. |
| gnuarm:
--- Quote from: bdunham7 on December 27, 2021, 05:27:39 pm --- --- Quote from: electrodacus on December 27, 2021, 05:14:27 pm --- Anyone that designs any sort of vehicle will know that second formula is not correct. --- End quote --- Can you cite any actual person that 'designs any kind of vehicle? Is not the designer of the Blackbird such a person? Do you think he agrees with you? --- Quote ---therms of drag --- End quote --- |O :-DD Why not 'pints of drag'? Or perhaps 'acres of drag'. --- End quote --- You need to get your units correct. Drag is furlongs squared per fortnight • firkins. I believe the term for a drag unit is the RuPaul. ^-^ |
| Navigation |
| Message Index |
| Next page |
| Previous page |