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Mess with your minds: A wind powered craft going faster than a tail wind speed.
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bdunham7:

--- Quote from: electrodacus on December 16, 2021, 11:18:27 pm ---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

--- End quote ---

OK, now if you could just explain what each of the terms is, actually just the 1.2kg/m3 term--I see the 1m2 is area and the others are the speed of the wind and the sail.  Then instead of magically arriving at 'Watts' at the end, do all the operations with the units and see if what you get simplifies to watts or not.
electrodacus:

--- Quote from: bdunham7 on December 16, 2021, 11:27:16 pm ---
--- Quote from: electrodacus on December 16, 2021, 11:18:27 pm ---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

--- End quote ---

OK, now if you could just explain what each of the terms is, actually just the 1.2kg/m3 term--I see the 1m2 is area and the others are the speed of the wind and the sail.  Then instead of magically arriving at 'Watts' at the end, do all the operations with the units and see if what you get simplifies to watts or not.

--- End quote ---

The 0.5 is a constant has no unit
1.2kg/m^3 is the air density
1m^2 is the equivalent are of the sail
then there is the wind speed and vehicle speed in m/s
I use all international system of unit so I do not need to be as careful as if using all sort of strange unrelated units.
And yes that result is in Watt's
bdunham7:

--- Quote from: electrodacus on December 16, 2021, 11:33:59 pm ---The 0.5 is a constant has no unit
1.2kg/m^3 is the air density
1m^2 is the equivalent are of the sail
then there is the wind speed and vehicle speed in m/s
I use all international system of unit so I do not need to be as careful as if using all sort of strange unrelated units.
And yes that result is in Watt's

--- End quote ---

Where are you getting that the result is in watts?  And are you saying that the force on a sail is just the mass of the air x its speed x the area of the sail?  Where do you get that?
Kleinstein:

--- Quote from: electrodacus on December 16, 2021, 10:40:15 pm ---
--- 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 fraction ( 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.

--- End quote ---
If all the power or only a small farction is actually used makes a big difference. The problem with the assumption all the theoretically possible power would be added to the kinetic energy is that it is rather hard (essentially impossible) to do this.

With a 100 kg vehicle to add the first 0.5 Ws to the kinetic energy it would need a speed of 0.1 m/s.  If the sail would provide a full 500 W to add to the kineatic energy, this would be only 1 ms to reach 0.1 m/s. It is only a low speed, but 0.1 m/s   / 1 ms is still 100 m/s² and thus a bit more than 10 times gravity for the acceleration needed (on averge). Things get even more carzy when looking at lower speed / shorter time.   So there must be an error in the calculation.

The error in this example is in the idea that a "sail" would be 100 % energy efficient in converting to kinetic energy. Energy is still conserved, but most is converted to heat and not to the kinetic energy of the vehicle.

I somehow have the feeling the concept of force is not really understood. Using a power for the drag is an indication.  Drag is a force and not a power.
Except from this error I have not seen a problem with the units. The desire to do the calculations with examples instead of the letters is often found with beginners who try to use intuation instead of math. However even beginners are usually way better in realizing that they may be wrong.
bdunham7:

--- Quote from: Kleinstein on December 16, 2021, 11:40:25 pm ---Except from this error I have not seen a problem with the units.

--- End quote ---

Well, sure, it's not the only error, but you have to start somewhere.  This example appears to come out as kg*m2/s3, does that come out to watts somehow? 

A newton is 1 kg*m/s2, so then we have N*m/s, so I'll be damned, it is watts.  Oh well.  Maybe he's right.   :)
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