Incidentally, the downwind vehicle others have linked doesn’t actually get propulsion from its wheels, but from the big propeller. The wheels drive the propeller, not the other way around.
And what drives the wheels?
The movement of the car over the ground drives the wheels. The wind pushes on the prop, which pushes the entire car along. The ground turns the wheels, which turn the prop to push against the wind.
So the wheels drive the prop, the prop drives the entire car forward, and that drives the wheels again…?
The wind pushes the car. The wheels and prop act in such a way to create an effective surface that the wind is pushing against that is moving at a fraction of the speed of the car itself. It’s no different than sailing faster than the wind by tacking at an angle to the wind, except here the angle is generated by the prop blades turning around the prop hub at a speed that’s a fixed ratio to the ground speed of the car.
So why do you need the wind? Why can’t you just push it, so that the wheels turn the prop, which creates thrust and pushes the cart forward, which drives the wheels again?
The effect of the prop and wheels is to cause the cart to move at a speed that is a fixed multiple of the speed of the wind. If the speed of the wind is zero, then the speed of the cart will be zero. The cart isn’t powering itself, it’s drawing power from the difference in speed between the wind and the ground, and just using a clever mechanism to ensure that the cart itself is moving at a speed that is some multiple of the windspeed instead of at windspeed.
Not if I push it initially. The propeller is spinning and creating thrust. So what *exactly *prevents it from accelerating on release?
Conservation of energy.
Acceleration is determined by forces. What force prevents it from accelerating?
Wrong question. It should be “what force is causing it to accelerate?” After you push it, there is no more force acting on it and there is no wind to gain energy from so it doesn’t accelerate (actually it does, just in a negative sense.)
Rolling friction with the wheels, and drag as the propeller will be hitting the still air on the front side of its blades rather than the rear side.
Don’t think of the propeller as something which simply pushes air. Think of it as a screw which wants to advance a certain distance through the air with each revolution.
The cart’s wheels and propeller are geared such that the cart will assume a speed that’s about three times the windspeed. If the cart is going below this speed, it will speed up. If it is going above this speed, it will slow down. When the windspeed is zero and you give the cart a push, it will be going above the speed it’s geared to go, and will slow down to zero.
I “believe” (and will try to explain) why the plane will not take off.
First, let us consider the scenario where the conveyor belt isn’t turning around so that it opposes the forward motion of the plane, but rather supplements it. In that case, the speed of the airflow over the plane’s wings, ceteris paribus, will be increased specifically by the amount equal to the conveyor belt’s turning speed. Let’s assume that this plane has been manufactured so that it takes off when its “runway speed” is 200 mph.
Then, if the plane’s speed relative to the conveyor belt (which is parallel to the runway) is 100 mph, and the speed of the conveyor belt relative to the runway is 100 mph, again, both of these speeds’ corresponding velocity vector directionalities being in the same direction, the plane will surely be on the threshold of taking off. In other words, I think it’s “fairly obvious” to say that the conveyor belt “helps” the plane take off at a lower speed (100 mph) relative to whatever it is moving along before takeoff than it would had the conveyor belt not existed (200 mph).
The faster the conveyor belt moves forward in the same direction as the plane on it, the “more” the belt “helps” the plane take off at lower speeds and lower takeoff fuel consumptions. If the conveyor belt’s speed exceeds 200 mph, then the plane can take off all on its own without “pulling its own weight” so to speak. This is true because the velocity of the plane relative to the air is what matters in the finality, not the velocity of the plane relative to anything else. It does not matter what the velocities of the plane relative to the conveyer belt and that of the conveyor belt with respect to the surrounding air in an ultimate sense, but rather only to the extent that they are among the determining factors of what does ultimately matter, the velocity of the plane relative to the air (or the air relative to the plane, equivalently), the necessary and sufficient condition that must be satisfied being that the velocity of the plane relative to the air must exceed, in this case, 200 mph.
In the case that the plane moves forward along the conveyor belt at 200+ mph but that the conveyor belt moves backward at 200+ mph, their motions being antiparallel rather than parallel in my example discussed in the previous two paragraphs, the necessary and sufficient condition that the speed of the plane with respect to the air exceeds 200 mph is not met, for the velocity that does exceed 200 mph is that of the plane relative to the conveyor belt, rather than the air. As long as these two opposing velocities are exactly and instantaneously equal, no matter how large in magnitude they may be, they sum to be 0 => zero lift.
So I guess I’d like to conclude…that a conveyor belt ALONE (meaning the plane’s engine is not even turned off) can cause a plane to take off, but that the case illustrated by your post and the topic at hand does not satisfy the requirements (imo) that have to be met. I checked out a Mythbusters video, but the plane still moved forward with respect to the air; I suppose if it remained stationary with respect to the air at every instant of the plane’s forward motion (w.r.t. the conveyor belt), so that there would be no net displacement of the plane relative to the air (provided the air was ‘perfectly still’, i.e. a calm day) throughout the pertinent time interval, that the lack of airflow over the plane’s wings due to the net velocity vectors canceling out would result in zero/negligible lift generation and hence a failure to takeoff.
-supery00n
Just wow. It’s a shame that you’re wasting your ability to spin bullshit into fine gold on discussions of aerodynamics on the Internet. You should be in politics.
If your assumption is that the plane’s brakes are on, then you can make a pretty good case that a plane not on a treadmill can’t take off either.
That is not different from the the case with wind. So it can’t be the reason why it needs wind.
Why should the still air hit the front side the blades? I can adjust the prop pitch such that the propeller will produce positive thrust when pushed in still air.
And that certain distance can be freely chosen. I can choose the pitch such, that the propeller wants to advance 2m/s through air, when the wheels roll 1m/s over the ground. So when I push the cart at 1m/s in still air, the propeller wants to go faster and produces a forward force.
So again, which force (that is not present in the case with wind) prevents further acceleration on a windless day.
Friction, drag, entropy. Same as any other vehicle given a single push, then left alone.
The wind is nothing more than the power source for the DDFTTW vehicle. It moves faster than the wind only because it exploits mechanical advantage, like a yacht, or, for that matter, like a cyclist on a bike with gears.
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So again, which force (that is not present in the case with wind) prevents further acceleration on a windless day.
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Entropy is not a force. Friction and drag exist too when is goes faster than the direct tailwind. So what is the difference** in terms of forces** to a windless day?
Yes, whatever.
Yes.
The difference is: when the wind blows there is a sustained force acting upon the system.
Perhaps it would be easier if you say why it is you think it shouldn’t be able to work? (assuming I am correct in supposing that is the thrust of your argument)