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, 10:03:56 pm ---Some ( a small fraction when at low speed) of the power would be used if going with the wind, none of this power would be used when moving against the wind. It is wrong to assume that whole available power would be transferred to the vehicle when moving with the wind - only a small fration ( vehicle speed / wind speed) is.
--- End quote ---
When you are start moving slowly in to wind direction you need to apply brakes so some small part of the power will be used to increase the vehicle kinetic energy while acceleration and the difference will be burned by the friction brakes (assuming that is what you use to slow down the vehicle).
So if you use no brakes then vehicle will use all that wind power to accelerate the vehicle (increase kinetic energy).
Thus as soon as the vehicle moves the power is either all used to increase vehicle kinetic energy thus super fast get to a large fraction of the wind speed or if you want to maintain that low speed constantly then you need to do something with all that wind power available either convert to heat using friction brakes or generate electricity and store that in some sort of energy storage.
I think this is important to understand so I will try to insist on this point.
Say wind speed is 10m/s and vehicle is 100kg and has a 1m^2 sail driving directly down wind.
When you are stationary no work is done and we all agree with that.
Now say you want the vehicle to get to 1m/s and maintain that speed after it got there.
Kinetic energy for that vehicle at 1m/s will be 0.5 * 100kg * 1^2 = 50Ws
So with 600W available at the start and 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W available when vehicle gets to 1m/s the vehicle need about 100ms to get from 0m/s to 1m/s
And if you want to stop accelerating at that point you will need to setup a generator at the wheel that can take 437.4W and do whatever it wants with them you can heat water or air or store that in a battery but as long as you take out those 437.4W the vehicle can maintain 1m/s and not accelerate or decelerate.
That 437.4W of wind power will be available to vehicle as long as wind matains 10m/s
If you understand this example and explanation you should be able to understand my entire explanation about how blackbird works.
I wanted to answer the other questions but I feel we need to concentrate on this and once this part is understood the rest should be simple.
bdunham7:
--- Quote from: electrodacus on December 16, 2021, 10:40:15 pm ---That 437.4W of wind power will be available to vehicle as long as wind matains 10m/s
If you understand this example and explanation you should be able to understand my entire explanation about how blackbird works.
I wanted to answer the other questions but I feel we need to concentrate on this and once this part is understood the rest should be simple.
--- End quote ---
Try doing your math with the units!
electrodacus:
--- Quote from: bdunham7 on December 16, 2021, 10:58:26 pm ---
Try doing your math with the units!
--- End quote ---
Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.
bdunham7:
--- Quote from: electrodacus on December 16, 2021, 11:02:01 pm ---Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.
--- End quote ---
Take the expression " 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W" and put the actual units in everywhere the number refers to units.
So the 0.5 refers to area of the sail, so 0.5m2, I don't know offhand what the 1.2 or 1 refer to, the (10-1)3 refers to speed, so (9m/s)3 becomes 729m3s-3 and so on, then your result will be some monstrosity of units until you simplify it. If it doesn't simplify down to a usable unit, then your formula or math are wrong.
electrodacus:
--- Quote from: bdunham7 on December 16, 2021, 11:16:21 pm ---
--- Quote from: electrodacus on December 16, 2021, 11:02:01 pm ---Can you explain what you mean ? I think I used units everywhere in my example.
If I miss one somewhere please point that out.
--- End quote ---
Take the expression " 0.5 * 1.2 * 1 * (10-1)^3 = 437.4W" and put the actual units in everywhere the number refers to units.
So the 0.5 refers to area of the sail, so 0.5m2, I don't know offhand what the 1.2 or 1 refer to, the (10-1)3 refers to speed, so (9m/s)3 becomes 729m3s-3 and so on, then your result will be some monstrosity of units until you simplify it. If it doesn't simplify down to a usable unit, then your formula or math are wrong.
--- End quote ---
Sorry OK I see what you mean. Here is the equation with units.
0.5 * 1.2kg/m^3 * 1m^2 * (10m/s-1m/s)^3 = 437.4W
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