chingel said:
But if we did engage it at windspeed, it should also work?
Yes, but the prop makes a lousy sail, so it gets to windspeed quicker if you let the prop do it's job while getting up to wind speed, although the initial start is always slow.
How did you get the numbers that 10 x 20 is how much energy we get?
That's the input power value, 10 lbs 20 mph = 200 lb mph = 5.333 horsepower = 3977 watts. The energy per second would be 2933 lb ft or 3977 Newton meters (joules). The mentioned output power is 1/2 of that (10 x 10).
chingel said:
If we then direct the propeller backwards, the force will push the cart forward, but then the wind isn't pushing on the cart anymore, it is offering a little air resistance because the cart moves faster than the wind, the propeller does work against the air, getting it's energy from the wheels, which should make the cart slow down.
Although the force at the wheels is an opposing (braking) force, the effective gearing multiplies the force at the wheels and divides the speed. In the example above with a 10 mph tailwind, the input force and input speed from the ground to the wheel was 10 lbs, 20 mph. If there were no losses, the output force from the prop to the air could be 20 lbs, at 10 mph, if the tailwind was greater than 10 mph, but with a 10 mph tailwind, as the cart approaches 20 mph, the output force to the prop would decrease to zero, which would also decrease the opposing force on the ground to zero for a zero loss cart (since the wheels are now driving the propeller with zero force).
For the BB, effective gearing from wheels to prop is .8 = 1/1.25, so the force is multiplied by 1.25 and speed divided by 1.25 before losses. A zero loss cart could achieve 5x wind speed with these numbers, but we also know it's actual speed in one of it's runs was "only" 2.8x wind speed. With a 10 mph tailwind, that means 28 mph ground speed. Input power could be = 28 mph x 100 lbs, output power = 22.4 mph x 125 lbs in a no loss case. However there are losses. So output power might be 20 mph x 119 lbs (15% loss). That's 19 lbs of net forward thrust (used to overcome rolling resistance and 18 mph worth of aerodynamic drag), with the tailwind being slowed down by 2 mph, the true energy source from a ground based frame of reference. That 19 lbs of net downwind force is the result of the 2 mph upwind change in airspeed, regardless of the frame of reference. From a ground frame of refernce, the input power from the wind being slowed down is 19 lbs x 10 mph = .507 hp = 378 watts.
chingel said:
I am still wondering how does the fulcrum example work with the ruler cart, where the middle of the wheel is on a shorter section of the fulcrum?
For that cart, the fulcrum effect is due to the ratio of the inner and outer diameters of the spools on the bottom of the cart. In the videos, the inner diameters are 1/2 the outer diametsrs, so it's a cart with an effective gearing of .5. It goes down ruler at 2x ruler speed or upground at -1x ground speed. If the ratio was 2/3, it would go down ruler at 3x ruler speed or upground at -2x ground speed, faster than the difference in speed betwen ruler and ground regardless of which one moves.