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| Mess with your minds: A wind powered craft going faster than a tail wind speed. |
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| electrodacus:
--- Quote from: bdunham7 on December 14, 2021, 11:06:37 pm --- --- Quote from: electrodacus on December 14, 2021, 11:01:40 pm ---Yes in theory if you want to keep the vehicle at 1mm/s brakes will need to dissipate 600W --- End quote --- One of us has a fundamental misunderstanding of a very basic principle of physics. I think the brakes will dissipate approximately 60mW under those conditions. --- End quote --- If you apply only 60W of breaking power will get in about 1 second to around half the wind speed so 5m/s maybe just slightly more I will need to calculate exactly. We can test this and see that a real test will exactly be predicted by my theory (is not my theory but the one I presented to you). |
| bdunham7:
--- Quote from: electrodacus on December 14, 2021, 11:12:14 pm ---If you apply only 60W of breaking power --- End quote --- You need just under 60N of braking force, not 'braking power'. Then the power would be 60N X 0.001m/s = 60mW. Power is force x speed. |
| IanB:
--- Quote from: electrodacus on December 14, 2021, 11:12:14 pm ---If you apply only 60W of breaking power will get in about 1 second to around half the wind speed so 5m/s maybe just slightly more I will need to calculate exactly. We can test this and see that a real test will exactly be predicted by my theory (is not my theory but the one I presented to you). --- End quote --- Work = Force x Distance (for example: joules = Newtons x meters) Power = Force x Speed (for example: watts = Newtons x meters / second) If the vehicle speed is zero the power is zero because Power = Force x zero. If the vehicle speed is nearly zero the power is nearly zero because Power = Force x (very small number). If you calculate power vs speed for a simple sail driven vehicle, you should get power = zero when the speed is zero, and power = zero when the vehicle is travelling at the same speed as the wind (if the vehicle speed and the wind speed are the same, no power can be transferred to the sail). When you plot this on a graph, you should find that the maximum power transfer occurs somewhere in the middle, when the vehicle speed is approximately half the wind speed. This has some similarity with the power transfer theorem in electronics, where maximum power transfer occurs when source and sink impedances are equal. |
| electrodacus:
--- Quote from: bdunham7 on December 14, 2021, 11:33:02 pm --- --- Quote from: electrodacus on December 14, 2021, 11:12:14 pm ---If you apply only 60W of breaking power --- End quote --- You need just under 60N of braking force, not 'braking power'. Then the power would be 60N X 0.001m/s = 60mW. Power is force x speed. --- End quote --- Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ? The mistake you make is using the 0.001m/s instead of (10m/s-0.001m/s) So you are thinking about slowing down 10m/s wind on a 1m^2 sail to 9.999m/s |
| IanB:
--- Quote from: electrodacus on December 14, 2021, 11:44:09 pm ---Not sure if you are familiar with power (it is an electronics forum) but do you find that a 10m/s (22mph) wind pushing on 1m^2 (10.7sqft) sail can be slowed down to 0.001m/s by 60mW ? --- End quote --- Yes! |
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