General > General Technical Chat
Mess with your minds: A wind powered craft going faster than a tail wind speed.
electrodacus:
--- Quote from: Kleinstein on December 16, 2021, 07:07:50 pm --- So using the full wind speed for the power transferred to the vehicle is obviously wrong.
--- End quote ---
I do not use the full wind power since I use (wind speed - vehicle speed) for a direct downwind vehicle unlike the wrong (vehicle speed - wind speed) Derek used in his video.
You just seems to not understand what air is. Like in my small house at this moment I have over 200kg of air.
electrodacus:
--- Quote from: IanB on December 16, 2021, 07:13:41 pm ---
That's irrelevant. A sailboat, with no turbine, and no energy storage, can sail 20 miles downwind in less than an hour with a 20 mph following wind. It could do this indefinitely.
--- End quote ---
If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.
bdunham7:
--- Quote from: electrodacus on December 16, 2021, 07:34:27 pm ---If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.
--- End quote ---
How? The boat could go 20 miles downwind and 5 to the side without changing direction, or it could make one turn halfway through the course. In that turn, which only takes a few seconds out of that hour, its speed before the turn and after the turn are the same, so what kinetic energy?
electrodacus:
--- Quote from: bdunham7 on December 16, 2021, 07:48:36 pm ---
--- Quote from: electrodacus on December 16, 2021, 07:34:27 pm ---If you are referring to zigzag (not direct down wind travel like blackbird) then you can use the kinetic energy as an energy storage device.
--- End quote ---
How? The boat could go 20 miles downwind and 5 to the side without changing direction, or it could make one turn halfway through the course. In that turn, which only takes a few seconds out of that hour, its speed before the turn and after the turn are the same, so what kinetic energy?
--- End quote ---
That will not be directly downwind it will be at an angle and it will get to another location compared to someone that went direct down wind.
As for kinetic energy the formula is this 0.5 * mass * (vehicle speed)^2
So you can see the relation between speed and stored kinetic energy
Kinetic energy will not be helpful for a sail type vehicle driving directly down wind but as soon as it drives in other direction than down wind the kinetic energy storage can be used.
Kleinstein:
--- Quote from: electrodacus on December 16, 2021, 07:32:55 pm ---
--- Quote from: Kleinstein on December 16, 2021, 07:07:50 pm --- So using the full wind speed for the power transferred to the vehicle is obviously wrong.
--- End quote ---
I do not use the full wind power since I use (wind speed - vehicle speed) for a direct downwind vehicle unlike the wrong (vehicle speed - wind speed) Derek used in his video.
--- End quote ---
At a vehicle speed much smaller than the air speed it does not matter if one subtracts the vehicle speed. At zero vehicle speed your (wrong) formular gives the full power theoretical power in the wind, and this is not the power actually captured. A sail is not at all 100% efficient in converting the energy ! Going against the wind actually needs extra power, while the could provide power if used in a different way. At zero speed the efficency is zero.
When the vehicle moves at the speed of the wind it does not matter if w-v or v-w is used, both would be zero. So even if the sign is wrong it would not make a difference at that point. When using the correct formular for the power, there is (w-v)² * v and in the square the sign makes no difference.
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