| General > General Technical Chat |
| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
| << < (144/285) > >> |
| IanB:
--- Quote from: electrodacus on December 14, 2021, 11:59:27 pm ---Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1. You think you can manage that with 60mW ? (ideal vehicle no friction). --- End quote --- Yes, of course. The force acting on the sail is 60 N. The vehicle speed required is 0.001 m/s. Therefore the power required from the motor is 60 N x 0.001 m/s = 60 mW. |
| electrodacus:
--- Quote from: bdunham7 on December 15, 2021, 12:09:53 am --- --- Quote from: electrodacus on December 14, 2021, 11:59:27 pm ---You need to use brakes and sufficient large contact with ground to transfer that static force to ground so then ground deals with that. Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1. You think you can manage that with 60mW ? (ideal vehicle no friction). --- End quote --- I think perhaps you don't realize that "the ground deals with that" in all of the situations, not just the zero speed one? Yes, if the sail is producing about 60N force, then 60mW will get me upwind at 1mm/s, provided there are no losses. --- End quote --- You forgot about the 10m/s headwind. |
| electrodacus:
--- Quote from: IanB on December 15, 2021, 12:11:14 am --- --- Quote from: electrodacus on December 14, 2021, 11:59:27 pm ---Let's take this in reverse and say you have an onboard battery and electric motor and want to drive upwind at 0.001m/s in to a 10m/s wind with a 1m^2 frontal area and COD (coefficient of drag) of 1. You think you can manage that with 60mW ? (ideal vehicle no friction). --- End quote --- Yes, of course. The force acting on the sail is 60 N. The vehicle speed required is 0.001 m/s. Therefore the power required from the motor is 60 N x 0.001 m/s = 60 mW. --- End quote --- Where is the 10m/s headwind in all this ? |
| IanB:
--- Quote from: electrodacus on December 15, 2021, 12:22:05 am ---Where is the 10m/s headwind in all this ? --- End quote --- It produces a force on the sail of 60 N. |
| electrodacus:
--- Quote from: IanB on December 15, 2021, 12:23:08 am --- --- Quote from: electrodacus on December 15, 2021, 12:22:05 am ---Where is the 10m/s headwind in all this ? --- End quote --- It produces a force on the sail of 60 N. --- End quote --- You need to consider your speed relative to air not relative to ground so 9.999m/s not 0.001m/s which is just the speed relative to the ground. |
| Navigation |
| Message Index |
| Next page |
| Previous page |