By Ft = Fc/2 I meant that the force with which the the cart advances would be half the amount being applied to it by the conveyor belt, which is the same thing as saying Fc = 2Ft, or the force being applied by the conveyor belt is twice the force whith which the cart is advancing, so I
think we're in agreement here.
So, the cart advances to twice the speed of the board pushing it, with twice the... (drag? backward force?)... of a cart that advances to the same speed as the board.
Now what if, instead of being pushed by a board, the top wheel was being pushed by the wind? You wouldn't have to worry about running out of length on the board. Of course, you'd have to redisign it a bit, replacing the top wheel with a propeller to get enough "traction", and it would be highly inefficient, advancing with even less force than it would with the board, but it would still work on the same principle.
So, the cart advances to twice the speed of the wind pushing it, with twice the... (drag? backward force?)... of a cart that advances to the same speed as the wind.
It all makes perfect sense... eventually.
Brian-M,
A real object's velocity is governed by the power that is available from the source. The gears cannot offer a power gain. Perhaps there will be some change the acceleration, but not the terminal velocity, which will be the same for the same object without gears. The same for an equivalent propeller driven object. Considering that is a major difference of opinion within this thread, we can leave it at that for the real case.
Some confusion. There is your geared cart and the "skateboard" device. They have different limitations because in the latter case, the velocity increases in the direction of the hand, whereas yours is opposite.
On your device, if you pull at 1m/s backwards, then the cart moves 2m/s forward. Correct?
The graph with the sine waves shows that the average difference will be zero, meaning that your velocity gain, is only
incrementally 2. That means that to get the gain, the cart must
constantly accelerate.
You know what that means for the input power and the cart's final velocity.
To keep the track at a constant velocity (-V) and the cart at (2V), is not
directly possible. To solve this, the average force must controlled. The cart is allowed to accelerates a bit above its average velocity, and then allowed to slow below it by the same amount. (which can be incrementally small).
When calculated using the integral rather than the instantaneous values, you will see that there is no gain over a standard device. If you trade, v for f to gain an improvement, that will also be seen in a standard cart.
Try it.
To get the gain, the cart must constantly accelerate.
If the bottom wheel does two revolutions, then the top wheel will turn once. So at that point, you are already at 2V. And from there?
Give the cart some mass, and then apply a constant force or constant velocity.
ETA:
There is nothing wrong with you kinematic collision calculations of an ealrier post . The momentum exchange is not the point. It relates to the way "equivalence" is being misapplied, and the consequences for the cart and treadmill. I am not sure that re-stating it would help