Let me try to help with this part. The cart reaches a maximum speed, which is determined by friction, the efficiency of the prop, etc. At that maximum speed it is not accelerating, which means that the total force on it is zero. There is a force pushing it up-tread (to the right, let's say), and some forces that total to the same (like friction in the wheel bearings) pushing it to the left.
Now suppose the cart moves a little slower. Several things happen: first, the prop rotates more slowly (which decreases the net force to the right as you said, all else being equal at least). But second, the apparent headwind decreases, changing the efficiency of the prop at a given rotational speed (and I think increasing it, although I'll have to think harder to be sure). Third, friction in the wheel bearings decreases (which increases the net force to the right).
So the net change could be in either direction - one needs to do a more careful analysis and add up all those effects. Video evidence confirms that the maximum speed is stable (i.e. the forces always accelerate the cart "towards" that max speed).
Does that help?
Sol, thanks, yes it does help. One thing I noticed after I asked the question was that I was trying to describe a process concerning the cart on the treadmill moving away from equilibrium, but used the term 'slowing down', to describe the change, as though I were describing the speed wrt the track. You seem to have echoed this, and it does seem natural. I'm just getting used to the idea that nothing is a
correct POV, and all these speeds are relative.
I think I might have underestimated those other two factors, the reduction in efficiency of the prop, caused by the reduction of 'apparent headwind', and the reduction of friction in the machine (at the wheels, but presumably elsewhere too, if all is turning more slowly).
The last bit surprises me. I had imagined that friction remained the same in such things at all speeds, but maybe I'm thinking of a coefficient of friction or something - the 'roughness' rather than the force. So there's another thing I've learned. It's obvious if I think about trying to start a fire with two sticks - fast equals more friction, more energy being dissipated, and ignition. The more important bit, to me, seems to be this reduction in apparent headwind. What is really hard to understand is how, as I imagine the cart 'slowing' (or accelerating wrt the room), I see it moving back down the treadmill so slowly that it is hard to believe that this tiny reduction in 'headwind' is significant enough (and changes in friction, too, maybe) to account for a righting of the equilibrium. It's that that boggles my mind.
I think what's going on is that I imagine moving the cart at that speed, held in my hand, something in the order of a cm a second, say, and I know from experience how doing that won't even turn a prop with usual levels of friction. So the magic ingredient seems to be that this is not analogous, that there is this prop spinning already quite fast, and moving it very slowly backwards makes a big difference to the balance of forces. Maybe.
Anyway, thanks for that.
John, I'll just address two of your points that seem, to me, to put the problem in a nutshell:
No, it indeed goes faster that the "wind" around it: an imp sitting on the cart will feel a headwind. Look again at my little cotton reel: it is really going faster than the piece of paper that is pushing it. Once you have understood that the reason for this is that the piece of paper is always pushing at a specific part of the reel that is going slower than the centre of gravity of the reel, you can transfer this idea to the cart. I know it seems weird to say that a specific part of the cart is always going slower than the the centre of gravity of the cart, but this is what the propeller is effectively doing with respect to the air.
Yes, that makes sense. I think I have mixed up two things, the slowing down of the wind as the prop extracts energy, and the differential you describe here due to the gearing. So although the wind is slowed by the prop, that may only be very slight in comparison. An imp on a mast would have a slightly stronger headwind than one suspended behind the prop?
Think for a while on how a propeller works. I like this explanation: "A propeller moves through the air in a similar manner as a mechanical screw moves forward through a piece of wood". Imagine the propeller burrowing through the air: how far does it move in one complete rotation? If you think of the propeller as a screw, you'll see that it moves the equivalent of the width of the thread of the screw. This is the "pitch" of the propeller.
The gearing of Spork's cart is deigned so that the effective speed of the propeller through the air, defined by its pitch and its speed of rotation, is slower than the speed of the wheel on the ground, defined by the diameter of the wheel and its speed of rotation. This is a fixed ratio, like the fixed ratio of the speed of the point on the cotton reel to the speed of the centre of the reel.
I was just thinking I would never get it, and then you wrote that. I started to realise that what my little brain likes is mechanical stuff like this in terms of actual gears, not sloppy fluids, and I think I've got it, or I'm about to.
Maybe this will help, and others might be able to correct or develop it:
I'm thinking I could have added a drive train and pusher located in the screw instead of just arrows. If one had a pusher device in the screw thread travelling at V, it would push the whole 'vehicle' forward, force the wheel, which would turn my badly drawn cogs, and rotate the screw (I think in the correct sense), which would mean that the whole would progress forward faster than V. Have I got it? Well, of course the fluid is a lot more complicated, and there is the requirement (for steady state maintained indefinitely) in the above of an indefinitely long screw, but that's kinda what a prop is in a fluid.
I think that all the problems of friction here and drag there and energies and relative velocities just confuse me, and this helps to simplify it. There are such massive losses in the above that actually you probably couldn't push the screw at its thread surface without a further wheel there at least. It would also help to redraw it with a sprocket and chain as the surface, because pushing this, it would just slide forward at V.
Again, we can see in this that there's the same relative stuff going on concerning the 'air being pushed backwards'. The pusher would not be going backwards, but as it pushed the thread, the thread would be forced by the ground-chain connection to rotate past it so as to 'slow' it (relatively), which I can imagine as the backward rotation of the screw thread relative to the machine.
There is also a push on the rest of the machine, represented by the lower of the two bold arrows, which in this case is an impossible complication - we can't have two pushing motions at different speeds, but the fluid air averages out such differences.
Equivalence - analogue of the treadmill: given those amendments of a little wheel in the thread fixed to a tracking arm sliding above the contraption, and an unslippable connection at the ground, if the ground is forced to the left, the machine (at rest) will wind the tracking arm such that it overtakes the ground. Slow the tracking arm down a bit to represent the air resistance at the prop, and the machine can only wind itself forward - now moving at a faster rate wrt the ground.
So a wind-powered vehicle can travel downwind faster than the wind? Well, let's just say I'm almost convinced! If I watch the video again, I'll swear it's breaking some law or other. Well done, spork & JB, although I still don't believe you*.
*Besides, I don't know what wine goes with hat.
