Down wind faster than the wind

[Michael C]

If someone such as sol thinks humber has a test that's either useful or entertaining, I'll certainly consider doing it to their specs.

Oh, please do the one where Charles Platt (or Humber) runs in circles after the cart with an electric fan. I'm sure it will be very entertaining.

 

[humber]

Oh, please do the one where Charles Platt (or Humber) runs in circles after the cart with an electric fan. I'm sure it will be very entertaining.

Or we can use a monkey.

 

[humber]

It sounds like you still maintain the absurd position you took at the "beginning" of the thread:

Ahh, the imperious 'absurd'. I shall expect a 'nonsense' soon. Same position, it has only changed, when others have changed it for me.
The propeller and wheel shaft as are in series, and in contra-rotation as any two, mated shafts must be. The propeller produces a torque in opposition to that of the wheel. The cart is the fulcrum of these two forces, so the net force upon the cart is zero in the limiting case.
A flywheel might work, but I am not sure there would be enough resistance in the system for the cart to remain stable. I know that it can support the propeller load, so can have at least have that much. If I use that level, I avoid changing the cart in other ways. It's called 'controlling the variables'.

If you still believe that there is an extremely easy test spork and co. could do: replace the propeller with a brake, or just orient the propeller the other way. Hell, just put the cart on the treadmill facing backwards.

Your suit of irony needs a polish. Those modifications will increase the load, or reduce it as appropriate. The balance remains, as explained above. The wheels will turn and drive the propeller, until the load becomes greater than the friction to the belt can reliably support.
The cart has been developed to run forward, so reversing it makes it a different model, but based upon the limited information available, I would expect a similar result. The difference being that the video is evidence that the cart in the forward orientation, can remain in place. However, failure would add nothing to your argument, because the forces do oppose each other so as to drive them towards zero, whereas in the wind, you would hope that to be a maximum.

So you think if I aim the propeller straight up the cart will remain motionless on the belt. I suggest you think again, and if you find that difficult, imagine the propeller is encased in a container full of thick viscous oil. If there is any resistance in the propeller - which there always will be in still air unless it's not rotating - there is a torque on the wheels. It's precisely like braking a car.

Propeller up, reason as before. There are two limiting cases; not enough drag to oppose the belt, and too much drag for the friction between the belt and wheel to sustain. Also, an upturned propeller may cause a downward thrust, so increasing the friction. Please control your analogies.
How would you suspend the oil tank? If held, the cart may move in either direction until hitting the sides. Set up correctly to avoid that situation, and the load is kept in working range, it should work. The experimental science so far displayed is so bad, I doubt that would be an appropriate task.

In humber's world none of that would have any effect on the cart's behavior. In the real world it will quickly fall of the back end of the treadmill.

And what would that demonstrate? That I do indeed need just a little bit of thrust to make my argument stick? I can only be wrong by that tiny amount. The cart is stationary, so the forces are in balance, whether you like it or not. There seems to be such a tiny amount of thrust generated by such fast moving belt. We can at least question the efficiency of such a machine.
That is not important though, because your equivalent model is wrong, and so is your implementation. The balance is just an artifact of those facts.

So - if you haven't flipflopped and still believe that utter nonsense, say so and maybe they'll oblige by proving you wrong yet again.

Yeah.

 

[sol invictus]

Just a friendly correction: Putting the cart on backwards will also reverse the wheel and prop direction, thus still producing thrust against the treadmill/road direction. It will be less efficient as most props are asymetrical and work poorly in reverse, so it may not work at the available treadmill speeds.

// CyCrow

Agreed - my point was that humber thinks the cart will somehow always be "balanced", regardless of prop orientation, and one potentially very simple way to disprove that is to reverse direction where (if the prop isn't symmetric) it will behave differently. But it would be better to aim it up, or just put some friction on it.

 

[sol invictus]

Your suit of irony needs a polish. Those modifications will increase the load, or reduce it as appropriate. The balance remains, as explained above. The wheels will turn and drive the propeller, until the load becomes greater than the friction to the belt can reliably support.

So in humberland pointing the prop another way increases the load felt by the belt? That's backwards, but regardless it's just too stupid to bother with.

 

[humber]

I'm having trouble with your wording here, but it appears you're just rephrasing the statement that the treadmill and the road are not equivalent.

They are certainly not equivalent in this case. There are at least two interpretations of that word. I mean that the propeller on the treadmill does not have an equivalent on the real cart.
<snip>

If you reply to this post, please limit your reply to the numbered statements, and tell me why and how they are incorrect.

I am afraid that I can't answer your questions H'ethetheth, and keep to your last request. The assumption of windpeeed is not valid, because the treadmill does not show that it is. There is no point continuing unless we can agree on that.

Some time ago I mentioned that the logic behind the treadmill is flawed, because windspeed is equated with 0dV cart/wind speed. I argued that this can well be an indicator that the cart was actually (nearly) still in still air, also a case of 0dV cart/wind velocity, rather than being at windspeed. The treadmill's error is wished away by the use of equivalency, but 'treadmills' are equivalent models, not equivalent views derived from that principle. However, if correctly applied that principle can be used to freely translate from one perspective to the other. OK,I will do that.

(A) is a cart in real wind
(B) is the translation to the treadmill model

In (B), I have allowed the cart to move back with the belt, and applied force to accelerate it to windspeed. Now, you see that in the translations, the velocities and the indicator 0dV are correct, but not the kinetic energy. You can see that driving the cart with force, actually reduces its kinetic energy! How can that be? From a standing start, the kinetic energy is zero, so it must be that it rises to some maximum and then falls to zero at windspeed? Really?

The balance mechanism that I have described, is not connected with this error, but another. However, both serve to drive the cart's kinetic energy to a minimum, as I have claimed. The balance problem is a failure to actually make a good model, regardless of equivalence. That is the propeller loading issue.

It is not possible to construct an accurate model without wind. The depiction in the treadmill has no real-world equivalent. Try and devise, by any means other than the treadmill approach, a way in which there can be 0dV cart/wind velocity, that does require the cart to move at windspeed. The treadmill cannot be the only model can it? The treadmill does, because it is logically flawed. Part of the equivalency of (A) dV0, has been equated with the same at (B), but not the kinetic energy.
The correct conclusion is that the final velocity is the origin of the graphs; a still cart in still air, just as it looks to be. This says nothing about the real cart, just the treadmill model.
Now that I hope you agree, then the whole supporting hypothesis looks less plausible. The remainder, we can perhaps later discuss.

 

[humber]

So in humberland pointing the prop another way increases the load felt by the belt? That's backwards, but regardless it's just too stupid to bother with.

Well supported. Even though I tell you that no thrust is required, you still think the propeller direction to be important.
You cannot refute the underlying balance argument, so you select trivial matters to hide behind.
The treadmill is wrong, and that balance proves it.

 

[sol invictus]

In (B), I have allowed the cart to move back with the belt, and applied force to accelerate it to windspeed. Now, you see that in the translations, the velocities and the indicator 0dV are correct, but not the kinetic energy.

The KE is different, yes.

You can see that driving the cart with force, actually reduces its kinetic energy!

Yes.

How can that be?

Basic Newtonian mechanics.

From a standing start,

It didn't have one, according to your own diagram.

the kinetic energy is zero,

Wrong.

so it must be that it rises to some maximum and then falls to zero at windspeed?

No.

Enough.

 

[John Freestone]

Although the treadmill is initially a little confusing for people not used to these sorts of things, it is educational and it has been shown to be equivalent to all intents and purposes to my satisfaction. However, further land demos would be great and, since some are less persuaded by galilean relativity after 300 years or so than others, might help the cause.

I asked a while ago about places where winds are more reliably steady, but I don't know about any, and there have been no responses. I was thinking salt flats, etc.

I'm a little dubious about shielding a cart as it does tests, but again maybe from ignorance of such matters. However, I would have thought that with suitable tweaking to get as much speed as possible, and reasonably careful checks of windspeed and direction, there should be little room for doubt. In this I'm taking the pedantic stance needed to object that a treadmill isn't the same thing, and turning it round: people are asking the impossible if the cart has to be proven never to stray by a few degrees from the downwind direction, to avoid them raising objections that it is using some other method. The same goes for other objections too. Other than really changeable conditions, I would think most reasonable people would accept success even if it included moments of higher windspeed, as long as the cart maintained a significant speed above the average. If it only manages it by jumping, skipping or slipping (another Humberism), well, it's doing it anyway.

 

[humber]

double post

 

[humber]

The KE is different, yes.



Yes.

So you must agree, that at windpeed on the belt, an equivalent of 10kph, it has no kinetic energy. This also means that the velocity of the cart w.r.t the ground is only a little more than zero.

Basic Newtonian mechanics.

Basically wrong. The cart has more KE at, say 1/2 windspeed than it does at windspeed?

It didn't have one, according to your own diagram.

Both cases are standing start. If the cart is moving back withe the belt st -10kph, and has the kinetic energy of that velocity, though w.r.t to the belt, that is motionless. That is correct and equivalent.

You are losing. The treadmill drives the kinetic energy to zero. So in Sol_world, the kinetic energy of the cart is o at 10kph?

You still have not refuted the balance argument, because that is basic Newtonian mechanics, and correct to boot.

 

[H'ethetheth]

They are certainly not equivalent in this case. There are at least two interpretations of that word. I mean that the propeller on the treadmill does not have an equivalent on the real cart.
<snip>

I am afraid that I can't answer your questions H'ethetheth, and keep to your last request. The assumption of windpeeed is not valid, because the treadmill does not show that it is. There is no point continuing unless we can agree on that.

Some time ago I mentioned that the logic behind the treadmill is flawed, because windspeed is equated with 0dV cart/wind speed. I argued that this can well be an indicator that the cart was actually (nearly) still in still air, also a case of 0dV cart/wind velocity, rather than being at windspeed. The treadmill's error is wished away by the use of equivalency, but 'treadmills' are equivalent models, not equivalent views derived from that principle. However, if correctly applied that principle can be used to freely translate from one perspective to the other. OK,I will do that.
[qimg]http://www.internationalskeptics.com/forums/attachment.php?attachmentid=12462&stc=1&d=1228912500[/qimg]
(A) is a cart in real wind
(B) is the translation to the treadmill model

In (B), I have allowed the cart to move back with the belt, and applied force to accelerate it to windspeed. Now, you see that in the translations, the velocities and the indicator 0dV are correct, but not the kinetic energy. You can see that driving the cart with force, actually reduces its kinetic energy! How can that be? From a standing start, the kinetic energy is zero, so it must be that it rises to some maximum and then falls to zero at windspeed? Really?

The balance mechanism that I have described, is not connected with this error, but another. However, both serve to drive the cart's kinetic energy to a minimum, as I have claimed. The balance problem is a failure to actually make a good model, regardless of equivalence. That is the propeller loading issue.

It is not possible to construct an accurate model without wind. The depiction in the treadmill has no real-world equivalent. Try and devise, by any means other than the treadmill approach, a way in which there can be 0dV cart/wind velocity, that does require the cart to move at windspeed. The treadmill cannot be the only model can it? The treadmill does, because it is logically flawed. Part of the equivalency of (A) dV0, has been equated with the same at (B), but not the kinetic energy.
The correct conclusion is that the final velocity is the origin of the graphs; a still cart in still air, just as it looks to be. This says nothing about the real cart, just the treadmill model.
Now that I hope you agree, then the whole supporting hypothesis looks less plausible. The remainder, we can perhaps later discuss.

I cannot agree, because it's wrong: Kinetic energy is not an absolute quantity! In both cases the change in kinetic energy is identical, namely 1/2mV². That is, the total energy uptake of the cart is the same because the forces acting on it and the work they do are the same.
If you do not believe that kinetic energy is relative, then please try to calculate the kinetic energy of the earth.
Additionally, this supposed flaw in the treadmill representation would apply exactly the same to wind tunnel testing, yet wind tunnel testing is still being used by engineers all over the world. Why do you think this is?

 

[humber]

I cannot agree, because it's wrong: Kinetic energy is not an absolute quantity! In both cases the change in kinetic energy is identical, namely 1/2mV². That is, the total energy uptake of the cart is the same because the forces acting on it and the work they do are the same.
If you do not believe that kinetic energy is relative, then please try to calculate the kinetic energy of the earth.
Additionally, this supposed flaw in the treadmill representation would apply exactly the same to wind tunnel testing, yet wind tunnel testing is still being used by engineers all over the world. Why do you think this is?

No it is not. The cart is said to be at windspeed. How can it have zero kinetic energy? When moving with the belt, then is does have kinetic energy, that is correct, but it cannot be that it gets less as it accelerates. It looks wrong, not because "equivalence" is wrong, but that there is no actual equivalence. It is not moving, the model is wrong!
Do you expect that if you pick up a cart at 10kph, it will have no kinetic energy? Yet, that is obviously so, it shows no reaction at all. It is not travel ling any faster than it appears to be.

Wind tunnels have wind. The test vehicle is often stationary, so its kinetic energy is zero. That is correct. The wind is moving and has kinetic energy and that is also correct. No calculations are done to account for the velocity of the earth in either case because they are in the same frame and environment. That is common to all objects.
I can turn it around. If wind is not necessary, why do they go to all that expense?
Equivalency guarantees that this is so, all views are equivalent, objects will have the same kinetic energy from any view, otherwise, there is no equivalency.

ETA:
When at the roadside, a cart traveling at windspeed certainly has kinetic energy from your point of view, the observer on the ground.
At the treadmill, you are in the same reference, the cart is said to be moving at windpeed, but has no kinetic energy. How can that be? The wind is necessary. Can you calculate work or power knowing only the velocity of an object? No.
If the cart is by the treadmill, you must agree that it has no kinetic energy over a stationary object. Why would putting it on a belt change that, if the relative velocities are only slightly different in each case? The treadmill does not model the wind. Only the boundary layer, very close to the belt is actually driven by it. That never gets near the propeller, so it is not coupled to the air in the way that it is to the wind. The model is incomplete, and that results in some odd behavior, when taken from the perspective that it is. Viewed as you see it, the correct reference, it all adds up. No velocity, no kinetic energy. The beltspeed is irrelevant.

 

[Christian Klippel]

Cart at windspeed: What is the kinetic energy of the cart, relative to the wind? What is the kinetic energy of the road, relative to the cart, and what would it be if the earth had 4 times it weight?

Cart on treadmill: What is the kinetic energy of the cart, relative to the wind? What is the kinetic energy of the belt, relative to the cart, and what would it be if the belt had 4 times its weight?

Is it really that hard to understand that the cart on the treadmill is indeed equivalent to a cart on the street at windspeed?

Edit: And what would be the kinetic energy of the cart in the wind, relative to the wind, if the earth had 4 times it's weight? Or if the belt had 4 times it's weight?

 

[H'ethetheth]

No it is not. The cart is said to be at windspeed. How can it have zero kinetic energy? When moving with the belt, then is does have kinetic energy, that is correct, but it cannot be that it gets less as it accelerates. It looks wrong, not because "equivalence" is wrong, but that there is no actual equivalence. It is not moving, the model is wrong!
Do you expect that if you pick up a cart at 10kph, it will have no kinetic energy? Yet, that is obviously so, it shows no reaction at all. It is not travel ling any faster than it appears to be.

Wind tunnels have wind. The test vehicle is often stationary, so its kinetic energy is zero. That is correct. The wind is moving and has kinetic energy and that is also correct. No calculations are done to account for the velocity of the earth in either case because they are in the same frame and environment. That is common to all objects.
I can turn it around. If wind is not necessary, why do they go to all that expense?
Equivalency guarantees that this is so, all views are equivalent, objects will have the same kinetic energy from any view, otherwise, there is no equivalency.

ETA:
When at the roadside, a cart traveling at windspeed certainly has kinetic energy from your point of view, the observer on the ground.
At the treadmill, you are in the same reference, the cart is said to be moving at windpeed, but has no kinetic energy. How can that be? The wind is necessary. Can you calculate work or power knowing only the velocity of an object? No.
If the cart is by the treadmill, you must agree that it has no kinetic energy over a stationary object. Why would putting it on a belt change that, if the relative velocities are only slightly different in each case? The treadmill does not model the wind. Only the boundary layer, very close to the belt is actually driven by it. That never gets near the propeller, so it is not coupled to the air in the way that it is to the wind. The model is incomplete, and that results in some odd behavior, when taken from the perspective that it is. Viewed as you see it, the correct reference, it all adds up. No velocity, no kinetic energy. The beltspeed is irrelevant.

Tell me: From a frame of reference travelling with the air (or standing still with it), what is the kinetic energy of the cart in both cases?
And from a frame of reference fixed to the road or treadmill surface?
The wind is indeed necessary, but only insofar that it creates a difference in speed between the surface in contact with the wheels and the air in contact with the propeller.

 

[Christian Klippel]

The debate with humber seems to be the true proof of circular perpetual motion.

Fixed it for you

Greetings,

Chris

 

[Christian Klippel]

It's really funny. Let's consider that humber had the idea that the cart uses stored kinetic energy to move faster than the wind, maybe coming from gusts, making the cart slip. Then he goes on and says that the cart on the treadmill has no kinetic energy. But somehow he fails to connect the dots: If the cart would be driven as he thinks, and if the treadmill was all so different as he thinks, it should _not_ work the way it does on the treadmill because it doesn't have that kinetic energy available then.

But, it does work, as we all can see. So, it can not be any kind of stored kinetic energy that moves the cart, so the treadmill is indeed equivalent to the situation on the street.

Oh well ...

 

[Dan O.]

If anyone is going to discuss Kinetic Energy^WP here, I suggest they get on the same page and pay particular attention to "Frame of Reference".

 

[humber]

Tell me: From a frame of reference travelling with the air (or standing still with it), what is the kinetic energy of the cart in both cases?

In the real world 1/2mv2, where V is relative to the ground, in both cases, because the treadmill is also in the real world. It is going as fast
as it appears to be.

And from a frame of reference fixed to the road or treadmill surface?
The wind is indeed necessary, but only insofar that it creates a difference in speed between the surface in contact with the wheels and the air in contact with the propeller.

Only the ground is relevant. It is common to the roadside and treadmill side observer. You could do it from the belt, but that will not mean any effective change in any parameter. If you find a real moving road then perhaps it may be useful to know from that vantage point, but why consider it?
Velocity in that last case cannot be used to calculate power for example.
It means nothing without knowing the force as well. Low friction in this case, means low force, and it is clearly low

There is a word game here. Equivalency means for example, that if you view on object from another, they remain in the same relative locations. Also equivalency means, that unless velocities are very high, all objects are unchanged relative to each other by their velocities. There are no Einstein effects. Two parked cars may have a lot of KE from traveling around the sun, but they, like all objects on this planet have it too, so the relaative difference is zero. That is our reference frame. There is nothing to be gained from making the observation that one can be viewed equally from the other. So what?
This has nothing to do with the problems of the treadmill. It is a distraction.

Imagine a cart in a 20kph wind but on a belt. When the beltspeed is zero, that is like a fixed road. The only way that the cart can be at 0V relative to a ground observer, is if the belt and wind are equal and opposite. That is, the belt is going back as in the treadmill, at 10kph. Two opposing velocities mean that the cart is stationary.
This is also true for the observer in the car. In the real world, that is the only case, save for no motion at all, where these two views are the same

How is it that the treadmill can model a situation, that of being at windspeed, but stationary to a ground observer and the driver?
There is no relative velocity between you and the driver. When that happens it is not possible to be moving. The model is wrong

Only the boundary layer is modeled, and this is insufficient to couple the cart to the belt in a useful way. in fact, it tends to lift the wheels away, which is why the cart appears to float. What should happen, is the cart should go with the belt, dragged back by the missing wind. The cart would need to drive against that to achieve windspeed.
The entire model is wrong, because there is no real world equivalent that it could model.
You are seeing an artifact that results in the balance I have mentioned, that lends the illusion that the cart is at windspeed. Without that, you would not give it a second thought.

 

[Marcus]

You lost me at that point Marcus. There's nothing more "definitive" about a non-treadmill test. In fact, the treadmill test is the *most* definitive as it allows for the tight control and repeatability unrivaled in the great outdoors.

Having said that, an outdoor test is rather easy to do but hard to do well -- that is covering all the bases for documentation purposes is absolutely non-trivial.

JB

The problem is that you can't show the cart going faster than the wind more than a short distance, the length of the treadmill. A test showing the cart going faster than the wind a substantial distance is need to rule out other explanations.