For the instantaneous acceleration only the instantaneous forces are relevant. Which force exactly is acting on the cart with a true tailwind, that doesn’t act on it without the true tailwind. Keep in mind that the relative wind over the cart can be identical in both cases:
a)
true tailwind = 0 m/s
cart speed = 5 m/s
relative wind = -5 m/s
b)
true tailwind = 5 m/s
cart speed = 10 m/s
relative wind = -5 m/s
You aren’t taking into account the fact that the cart’s wheels are touching the ground, and the speed at which the wheels turn determines how fast the propeller turns.
Assuming for this exercise that the cart is geared such that it will move at twice windspeed, and neglecting friction:
Speed of the ground relative to the cart: -5m/s.
Wheels on the cart turn propeller backwards to push air backwards at a speed of 2.5 m/s. Actual relative wind is 5 m/s backwards. The propeller isn’t turning fast enough for the relative windspeed, so the cart slows down. As the cart slows down, the propeller slows down, and is still not going fast enough. Eventually the cart stops.
Speed of the ground relative to the cart: -10m/s.
Wheels on the cart turn propeller backwards to push air backwards at a speed of 5 m/s. Actual relative wind is 5 m/s backwards, so the forces balance out. The cart stays at the same speed. (Neglecting friction and inefficiencies, of course.)
The propeller is just a sail. When it’s rotating, it’s effectively a sail that’s continually moving upwind.
Take a look at any of the mechanical analogues that have been made for this device (one of them is the walkway cart linked in this thread).
If this vehicle can’t exceed the speed of the wind powering it, then a bicycle cannot exceed the speed of the cyclist’s feet.
Force isn’t measured in metres per second, and you appear to be trying to apply a linear formula. The cart moves at a multiple of wind speed. Zero multiplied by anything is zero.
I asked a very specific question about the difference in forces acting in two situations. None of the above addresses it directly.
Who said it is?
I’m not applying anything. I just gave two situations at which the cart is released, and asked about the differences in forces.
Nobody, directly. It just seemed strange to me that you were very strictly focusing on force (not energy, power, or speed), then you suddenly started expressing yourself in terms of velocity.
Fair enough (and I was wrong to use the term linear, also), but, cards on the table please - are you of the opinion that this device is impossible?
In your first situation (zero wind speed, cart given a push) the force on the propeller acts to slow the cart down, because it isn’t turning fast enough to keep up with the movement of the cart through the air.
In your second situation (non-zero wind speed, cart moving faster than the wind) the propeller is turning just fact enough to match the relative wind flowing backwards over the cart. The force on the propeller from the wind is just enough to overcome the frictional losses that would slow the cart down.
It is trivial to adjust the transmission ratios such that the propeller spins at the same rate in both scenarios. Since the airspeed is also the same, the thrust of the propeller will be the same. So where is the difference in forces coming from?
You are assuming, what you are supposed to show. I gave you a different configuration: Choose the pitch/transmission such, that the propeller wants to advance 10m/s through air, when the wheels roll 5m/s over the ground. So when I push the cart at 5m/s in still air, the propeller wants to go faster and produces a forward force.
The difference between the two situations (wind blowing vs an initial push) is that there is no difference in velocity between the ground and the air.
The Blackbird DDFTTW extracts energy from the difference in those two media (via the linking gear) to drive forward - the greater that difference, the more energy available. As the cart moves forward it leaves slower moving air behind the blades of the propellor.
You can push the cart when the wind is not blowing as fast as you like, but without the speed differential between air and ground, the frictional forces will stop your cart in short order once you stop.
Si
How does this affect the forces acting in the cart? Keep in mind that:
That’s not how the cart is set up to work at all. If you’re going to argue against the DDWFTW cart, you need to actually argue against how it was actually built, not some strawman.
The cart is geared such that the propeller wants to push air backwards at a speed which is some fraction of the speed at which the cart is moving forwards relative to the ground. For example, in the 2:1 ratio cart, if the cart is moving forwards at 10 m/s, the the propeller will be attempting to push air backwards at 5 m/s relative to the cart. If the ambient wind is 5 m/s relative to the ground, it will be blowing backwards at 5 m/s relative to the cart. That’s the same speed at which the propeller is pushing backwards, so the forces are balanced.
Say this cart slows down, and is going for example at 9 m/s with a 5 m/s tailwind. The cart will feel a wind flowing backwards at 4 m/s. The propeller will be trying to blow wind backwards at 4.5 m/s relative to the cart. The cart will speed up, because the force of the ground turning the wheels is greater than that required to turn the propeller, because there is a mechanical advantage in the gearing between the wheels and the propeller.
This has by the way been demonstrated to work with actual physical hardware multiple times.
Now, we can consider the completely hypothetical case of a different cart where the propeller pushes backwards at twice the ground speed. Give it a push at 5 m/s in still air. The propeller is trying to push air backwards at 10 m/s relative to the cart, or 15 m/s relative to the ground. Here you have a mechanical disadvantage, and the force required to turn the propeller will exceed the force generated by pushing against the air. The cart will slow down and eventually stop.
Yes, that’s the key issue. To achieve the same spin rate in the windless case, one would need twice as much force at the wheels, in order to counter the same prop reaction torque. If you try it via the pitch instead, the prop reaction torque will be greater in the windless case, so again more braking at the wheels.
Right. Essentially, there’s a torque fight between the wheels and the prop. The cart is geared so that the torque it takes to turn the prop is less than the torque it would take to slow the cart down, so the cart keeps going (so long as there’s a tailwind). It does this by having the propeller push the air back more slowly than the speed of the ground relative to the cart, and the tailwind makes up the difference. If you geared it the other way, the torque that it takes to turn the propeller would be greater than the torque to slow down the cart, and the cart would slow to a stop.
There’s no point talking about a rotating propellor as if it’s a stationary object - it’s an inclined plane, so its rotation effectively gives the working surfaces forward motion along the axis of rotation, even if the propellor itself, as a whole physical object, isn’t going anywhere.
Just to a stop?
Well, that explains a lot.
What if the wheels are frictionless, but have a non-negligible moment of inertia. Can you then prevent the plane from taking off?
Here’s a video of the Lego car I made to demonstrate the way the Blackbird works:
In this video, the toothed rack takes the place of the wind, and therefore the coupling between power source and vehicle is more efficient than a propellor-as-sail, but in all other respects, it’s analogous.
In fact, there’s only one real difference between the Blackbird and any powered vehicle that uses gearing - such as a bicycle or car - and that difference is that the power source isn’t onboard. And it’s a trivial difference. The overall motive power of a vehicle can’t exceed that of its engine, but the speed can (and commonly does).
Yes and no. Yes, if you don’t just run the conveyor belt at some high speed, but continually accelerate it. But that still won’t work for long: The acceleration needed is pretty high, so the belt will quickly reach a ludicrous speed, and entrain air to move along with it. And that air moving over the plane’s wings will then lift it.
Thanks, that was a great video. It really made it clear.