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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: fourfathom on December 21, 2021, 07:27:42 pm ---
--- Quote from: electrodacus on December 21, 2021, 07:03:38 pm ---If you install the wind turbine on a vehicle and then drive direct down wind then equation of the power output for that wind turbine will again contain (wind speed - vehicle speed) since the wind speed relative to that moving wind turbine will be lower the faster the vehicle moves.
--- End quote ---

This is exactly correct, at least up to (vehicle speed = wind speed), for a turbine/generator bolted onto a moving vehicle.  However, that is not what the rest of us are discussing; the Blackbird and related DDWFTTW vehicles.

--- End quote ---

Wind power available for any direct down wind vehicle no matter how it is designed is always 0.5 * air density * area * (wind speed - vehicle speed)^3 and yes this includes DDWFTTW vehicles.
This if you understand is sufficient proof that without energy storage or an external energy source no vehicle can exceed wind speed.

Kleinstein:

--- Quote from: electrodacus on December 21, 2021, 07:36:15 pm ---
--- Quote from: Kleinstein on December 21, 2021, 07:17:31 pm ---
P_sail =  0.5 * air density * area * (wind speed - vehicle speed)^2* vehicle speed

With the active driven prop there can be even more power available, as the air may slow down even more. This is especially true for a speed close to or higher than the speed of the wind. This is because this way the wind can be slowed down to a speed below the vehicle speed.

--- End quote ---

Say wind speed is 10m/s and vehicle speed is half that 5m/s directly down wind
What sail will see is a 10 - 5 = 5m/s wind relative to it and vehicle thus correct equation for a 1m^2 sail will be
 0.5 * 1.2 * 1 * (10-5)^3 = 75W

Using your equation 0.5 * 1.2 * 1 * ((10-5)^2 * 5) = 75W

Seems to provide the same value for half wind speed but
Vehicle speed of 2m/s

0.5 * 1.2 * 1 * (10-2)^3 = 307.2W

0.5 * 1.2 * 1 * ((10-2)^2 * 2) = 76.8W  no longer correct


--- End quote ---
In the comparison you forgot one important point - the point with vehicle speed zero.
In this case the  (w-v)³ type gives 0.5 * 1.2 * 1 * (10)^3 = 600W.
For sail driven vehicle at zero speed there is only force and not power transfered to the vehicle. So this equation is obviously wrong at that point.  :horse:
It gets even worse when you drive agains the wind  :-DD

IanB:

--- Quote from: bdunham7 on December 21, 2021, 05:07:35 pm ---As I pointed out much earlier, starting with reply #20 to this thread, you can pretty much explain it without much more than the principles of Archimedes.

--- End quote ---

Indeed, and for reference, here is an analysis of the cart/wheel/belt system.

Refer to the diagram attached below.



Assume that it is an ideal system, no friction losses, and both wheels are able to spin freely on their bearings. No slip occurs between the wheels and the belts.

There are five variables: \$v_W\$, \$v_R\$, \$v_M\$, \$v_G\$ and \$v\$

\$v_M\$ and \$v_G\$ are the rim speeds of the wheels, in m/s.

\$v_W\$ and \$v_R\$ are given, leaving three degrees of freedom.

Two equations can be written relating the belt velocities, the rim speeds of the wheels, and the cart speed:
$$v=v_M-v_W$$ $$v=v_G-v_R$$
These two equations take up two degrees of freedom, leaving one degree of freedom remaining.

With no additional constraints this system has an infinite number of solutions. The cart velocity, \$v\$ can have any value we choose.

This makes sense, intuitively, since the wheels are free running, and without friction losses any initial cart velocity will remain the same forever.

To remove the remaining degree of freedom in the system, the wheels can be linked with a connecting gear. If we let the gear ratio be \$\alpha\$, then we can write a linking equation:
$$v_G=\alpha\,v_M$$
For each revolution of wheel M, wheel G makes \$\alpha\$ revolutions.

(Note that the gearing could be mechanical, or it could be electrical. Mechanical is simpler.)

By making a suitable choice of \$\alpha\$, we can arrange for a desired cart speed, \$v\$.

For instance, suppose we wish the cart to move at 1 m/s to the right.

We set \$v\$ = 1 m/s, which gives \$v_M=v+v_W\$ = 5 m/s, and \$v_G=v+v_R\$ = 11 m/s.

From this, we have: \$\alpha = v_G / v_M\$ = 11 / 5 = 2.2

Therefore, with a gear ratio of 2.2, the cart will move to right at a speed of 1 m/s.

electrodacus:

--- Quote from: Kleinstein on December 21, 2021, 07:49:01 pm ---
In the comparison you forgot one important point - the point with vehicle speed zero.
In this case the  (w-v)³ type gives 0.5 * 1.2 * 1 * (10)^3 = 600W.
For sail driven vehicle at zero speed there is only force and not power transfered to the vehicle. So this equation is obviously wrong at that point.  :horse:
It gets even worse when you drive agains the wind  :-DD

--- End quote ---

Yes that calculation for vehicle speed zero is correct. There is a potential 600W available but with no movement there is no power available to vehicle. The vehicle is stuck to earth so those 600W push the earth trough the vehicle that is attached/anchored with a brake to earth.
If you remove the brakes those 600W will be available to vehicle instead of earth.
 

bdunham7:

--- Quote from: electrodacus on December 21, 2021, 08:08:34 pm ---If you remove the brakes those 600W will be available to vehicle instead of earth.

--- End quote ---

And if you release the brakes, but not all the way, so that the vehicle creeps, isn't most of your 600W still 'available to the earth'?  The vehicle is still pushing on the earth...

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