Okay, you seem to understand frames of reference. But I don't get why you think that an observer standing on the ground beside the treadmill is analogous to a person observing an outside test. It has been pointed out numerous times just on this page that you would have to be standing on the belt, which is moving relative to the cart. The other part of your post, credit, debts, books, ect. makes no sense. Frames are frames, history and nature are irrelevant.
Marcus, your comments would be correct, and we would largely agree, if that is what the treadmill actually did, or the observer played any part. As I have said many times, the treadmill does nothing to deny or support windspeed claims. It is simply irrelevant. I say that the observer is seeing the correct and literal view, and not one relative to winspeed.
Please take a look at my drawings.
They are from the view of an observer on the ground. The first is a cart in real wind. The belt speed is zero, and so it is like a fixed road. The belt speed is then increased until the current treadmill situation is reached of zero local wind. Note that the co-ordinates of the graphs are translated according to the belt speed. When at windspeed, the wind relative to the cart is said to be zero (so justifying the still air argument). However, does that not suggest that even if successful, the final destination is (0,0) with respect to the observer? If the real one is correct, surely the last is too, or do I just add something else to it later, perhaps? You can observe your luggage from the airport carousel, but why? Is that worthwhile if you need to be on the belt to gain any benefit?
Okay. Let's accept that the "correct view" is is from the belt. When placed on the belt, the cart does not move back with the belt, but is already at windspeed. Is this because the cart is capable of immediate acceleration to windspeed? It displays none of the power that one would expect of a cart capable of driving itself against a powerful belt at any speed. No, it is put there by the operator, by external influence. What you are seeing is an illusion, supported by the misuse of equivalency, which is NOT required to explain this treadmill, nor real windtunnels. The latter do share a superficial similarity to the mathematical idea of equivalency, but a windtunnel is a model; a physically "equivalent" system. Windtunnel wind is not 'like' the wind it IS a wind.
How would it get from moving back with the belt as the graph suggests, to being instantaneously at windspeed? There is an additional translation, an external unseen 'vector' to (0,0) and not (20,20) as is suggested. That's how.
The cart does look like is conforming to TAD's reasoning because it does stay on the belt, but that is simply the result of the cart balancing itself against friction with the belt. It is a secondary, spurious effect that completes the illusion. (When I say illusion, I do not mean a trick, just what is said to be happening)
Take a book and tie a thread to it. Place it on the belt. You will feel considerable force as you try to hold it against the belt. Now, cover the belt and book in teflon, and try again. Much easier. It can be moved up the belt with relative ease. (On an incline, the only work required would be against gravity). The cart behaves just like this. It is barely in contact with the belt, and may be pushed around with a plastic fork. The belt has no means by which it can 'grip' the cart so as to carry it back as it should. The only force that the cart needs to overcome, is that tiny amount of residual friction. If friction is increased, by using soft rubber wheels or some weight over the drive wheel, then the cart will go backwards (Unless is can indeed generate the required thrust)
Many attempts have been made to deny my balance argument, and they all seem to rely upon the fact that the cart will then go back with the belt. Yes, too much load and the balance fails, as I have said, but that is not a failure of my argument, but a failure of theirs. Too much friction exposes the illusion. Easy to test, I think. This is already understood I suspect, but the "it's already at windspeed" argument is used to deny it.
The treadmill contains many mistakes, but "equivalency" plays no part in its success or failure. All failures result in the same thing; back down the belt. The cart and treadmill are interdependent, it is a system that cannot be independently verified, and serves only to 'demonstrate' its initial premise, because the test system is also based upon that same premise. That is not science.
"Equivalency" as used in this thread, is a metaphysical idea. I will try to describe the error, but it does not influence the conclusion, because it has
no influence on anything, but the subject always returns, silly as it is.
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A car hitting a person at 60kph, is the same as a person traveling at 60kph hitting a static car. No.
What equivalency says, if it can actually be applied here, is that the transfer of energy and the changes of velocities, will be the same from each point of view. Each 'observer' sees the same thing from their perspective, which is correct. (The observer: that's another canard). The car "feels" what the person "feels" and vice-verse. All vectors and forces are the same. A direct invocation not of "equivalence" but of Newton's laws of motion!
The equivalent of (A hits B), is NOT (B hits A), but perhaps (-B hits -A). (B hits A) is another
event.
A car traveling at 60kph, hits a person. (A hits B). So what can (B hits A) mean for this event? It means nothing, they are the same. There are
one and the same collision. There is another option. The car is
labeled as being stationary, and the earth 'rotated' beneath the wheels, so that the person is driven into the car, but that would still have to be a the same as the original event, but can never be realised in a model.
Try to model a golfer. We are told the mass of the club, and the ball and its velocity of 30m/s. That is said to be the same as the ball traveling at 30m/S hitting the club. (A hits B) = (B hits A). So, the ball is projected at the club at that speed. It hits the club, bounces off, and goes out through the window. It seems we have forgotten about the golfer. Okay, we build a model golfer that absorbs the ball's energy so as to bring it to a stop at the tee. But how ? Don't we need to know in advance how to set up the golfer for each shot, or look into the future? To be accurate, would that not mean producing a time-reversed but carbon copy of the same event? There is no (B hits A) for the same event, save for that spurious time-reversed event. Time had an arrow, the collision is over.
A person traveling at 60kph, hitting a stationary car (B hits A) is another
event. Totally different collision, and not a mirror of a 60kph car hitting a person. The kinetic transfer will be different. Timescales will be different. Impulse collisions are time defendant, and reversing the velocity vectors won't make them the same. The list is endless. A battery heating a resistor, is the same as heating a resistor connected to a battery? As I said, this makes no difference to the treadmill, either way. Good, that's done.
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A propeller in the wind is not always like a propeller in still air. They are not the same thing in this particular case, and that is simply a physical fact, and not of how the world is viewed, but "equivalency" is used to say that they are the same.
The force of the wind drives the propeller to drive the cart, where as on the treadmill, the belt drives the propeller but without driving the cart. There is no wind to drive the propeller in opposition to the force from the belt, so there is no force to drive the cart.
However, if the cart is dragged back by the belt, the propeller then meets moving wind, which will drive the propeller in opposition to the force from the wheels. Now, that is certainly a more accurate model, though the cart would be driven forward, but only if the force generated by the propeller is greater than from the belt, but that is obviously not possible, which is why it causes so much fuss. How can the forward thrust be greater than the force dragging it back?
I have been told that I must "understand" that the cart is already at windspeed, so no force is required. We have a cart moving at windspeed, yet has no applied force, and no kinetic energy that indicates a history of acceleration? That would be unknown to science. Gearboxes cannot provide power gain, so arguments to that effect are not valid.
So how does it stay where it is? The friction is too low as I have said, but there is another consideration. Because the propeller and wheels are connected by gears, there is an analogy with the rear differential drive of a car. If one of the drive wheels is off the ground, the car cannot be moved by turning that wheel, no matter how much force or speed is applied to it. If put on a jacks such that both wheels are off the ground, and a wheel turned, the other contra-rotates. That is the situation with the wheel and propeller shafts of the cart. They spin in contra-rotation only to react against each other, because the friction is so low and the propeller so lightly coupled to its air mass. Without moving wind, there is insufficient effective air mass to do any work.
The cart is as if it were on floating jacks, like a disembodied differential gearbox. This is likely to be the result of laminar airflow lifting the wheels. The only downward force is gravity, and the cart is light. The motion up the belt is an artifact, that is the consequence of those facts.
My remarks about an object's history, are so, but not really relevant. I used them as a foil against the prevailing frames idea, that seems to consider only disembodied and isolated interactions. All objects on this planet are influenced by the earth's motion around the sun. It can be argued that the influence does not affect all objects equally, but the effects are trivial, and the local differences between objects even more so - were it even possible to take advantage of them. That is called 'locality'. Sometimes, in some experiments, it is necessary to acknowledge these effects, but they do not apply to this cart. The differences in local gravity will be of more significance, as would the local gravitational influence of nearby buildings. Differences in air temperature alone would swamp any effects. What about the difference is treadmill belts? Are they "equivalent" As you say, frames are frames. It makes no difference, other than a change in perspective. All else remains the same, so how can it help?
Do scientists agree? You can see in this thread, that a 'good' professor is one who agrees with the claimant. There is no value in this vernacular information. Radar engineers are masters of vectored velocities, but the aircraft don't seem to care. I have tried not to use outside references, because I think that it is not necessary in this case, as the errors are so fundamental, and can be otherwise explained.