DDWFTTW Turntable Test: 5 Min Video - Is It Conclusive?

[schroder]

Ok, then there are some *very elementary* misunderstandings to be cleared up before we can go any further.

Yes and the first elementary misunderstanding is: I said two-way road! The cars are passing each other going in opposite directions! You are not going to sell me your old Camarro because you clocked it doing 250 mph wrt to a Maserati that was passing it going the opposite way! For all I know, your Camarro was parked in the driveway. And you are certainly not going to sell me DDWFTTW because you clocked the cart wrt to the tread which was going in the opposite direction. Let me ask you this; If you wanted to know the velocity of the tread would you place a cart on it and measure wrt the moving cart? And if you cannot measure the velocity of the tread wrt a moving cart, you cannot measure the velocity of the cart wrt the moving tread. You can forget all about Galileo, prop pitch even Newton and reduce this down to a simple sum and difference problem and apply good old fashioned common sense.
You measure the velocity of something wrt the medium that something is working against. I have shown you the cart cannot possibly be working against the tread. If it were, and it is advancing, then by definition it is over unity. But since it IS advancing, and it is NOT over unity, it is NOT working against the tread. That is pure logic and you CANNOT argue with that.
And since it is NOT working against the tread, the tread is NOT the medium to use to measure the cart’s velocity. It IS working against the AIR which is painfully obvious if you look at that propeller spinning around. Since it IS working against the air, you measure the cart’s velocity wrt the AIR.
What you are doing is SUMMING the velocity of the tread with the velocity of the cart and of course you get a number higher than tread velocity! I suppose you can do the same on the down wind side; SUM up the velocity of the wind with the velocity of the cart and shout EUREKA! Faster than the wind! But the men in white coats would come to take you away!
All you have to do here is THINK.

 

[vanesch]

Yes and the first elementary misunderstanding is: I said two-way road! The cars are passing each other going in opposite directions!

Yes, that was ambiguous: if there are two lanes, you can use them in the same direction, or in opposite directions, that's why I made a choice as it wasn't clear which case you wanted.

But nothing changes here, except a few numbers. So let's go through this exercise again, this time the Camarro at 20 mph north wrt the road, the Maserati 230 mph south (wrt to the road). So from the Camarro, we see the Camarro at 0 mph, and the Maserati at 250 mph south. The relative velocity of the Camarro wrt the Maserati is 0 - (- 250 mph) = +250 mph (we take + to be north).

From the Maserati, the Camarro is doing + 250 mph north, and the Maserati is doing 0. So we have 250 - 0 = 250 mph again.

If we have a train going north at 130 mph, it sees the Camarro going 110 mph south and it sees the Maserati going 360 mph south. So the relative velocity of the Camarro wrt the Maserati is -110 - (-360) = +250 mph again.

I could continue. I can calculate the relative velocity between two objects by using their velocities as seen in just any frame: their difference is the relative velocity.

I can do something else:
For instance, from the frame of the train, I can see that the ground is moving 130 mph south, and the Camarro going 110 mph south. As such, the velocity of the Camarro wrt the ground is (-110) - (-130) = 20 mph. As calculated from the velocities in the train frame.

You are not going to sell me your old Camarro because you clocked it doing 250 mph wrt to a Maserati that was passing it going the opposite way! For all I know, your Camarro was parked in the driveway.

That is because I'm not interested in the relative velocity between the Camarro and the Maserati, but rather between the Camarro and the ground. I can calculate it using the velocities in any frame. It gives me always the same, correct, result.

And you are certainly not going to sell me DDWFTTW because you clocked the cart wrt to the tread which was going in the opposite direction. Let me ask you this; If you wanted to know the velocity of the tread would you place a cart on it and measure wrt the moving cart?

Depends wrt I want to know the velocity of the tread. If I wanted to know the velocity of the tread wrt the cart, yes, why not ?

And if you cannot measure the velocity of the tread wrt a moving cart, you cannot measure the velocity of the cart wrt the moving tread.

If I know the velocity of the tread wrt a frame, and I know the velocity of the cart wrt that same frame, I can calculate the relative velocity between them, which would be the one that an observer sitting on the cart would see.

Look at our train: because I had the velocity of the Camarro wrt the train (110 mph south) and I had the velocity of the ground wrt the train (130 mph south), I could calculate the velocity of the Camaro wrt the ground (20 mph which was -110 - (-130) = 20).
And that was indeed what the speedometer in the Camaro (which measures directly the speed wrt the ground) indicated.

Fun, no ?

You can forget all about Galileo, prop pitch even Newton and reduce this down to a simple sum and difference problem and apply good old fashioned common sense.
You measure the velocity of something wrt the medium that something is working against. I have shown you the cart cannot possibly be working against the tread. If it were, and it is advancing, then by definition it is over unity. But since it IS advancing, and it is NOT over unity, it is NOT working against the tread. That is pure logic and you CANNOT argue with that.

No, indeed, I cannot argue with that because there's no logic there. "Working against" is not something that has an intrinsic meaning, and from there on, the rest fails. But what is working against what doesn't matter when we see velocities, no ?

And since it is NOT working against the tread, the tread is NOT the medium to use to measure the cart’s velocity. It IS working against the AIR which is painfully obvious if you look at that propeller spinning around. Since it IS working against the air, you measure the cart’s velocity wrt the AIR.

Listen, the Maserati is not doing work against the train either, but there's nothing wrong in expressing the velocity of the Maserati wrt the train. There's a frame-independent physical meaning to "the velocity of the Maserati wrt the train" independent of what are the interactions between trains and Maseratis.

What you are doing is SUMMING the velocity of the tread with the velocity of the cart and of course you get a number higher than tread velocity!

No, I'm taking their ALGEBRAIC DIFFERENCE. Because the two directions are opposite, the *signs* of the velocities are opposite (one + the other -) and the difference of a positive and a negative number is the same as the sum of their absolute values.

Look: 3 - (-5) = 3 + 5 = 3 + |-5| = 8.
Fun, no ?

But the men in white coats would come to take you away!
All you have to do here is THINK.

Fun, no ?

 

[schroder]

We want to know the velocity with respect to the floor on which the cart is driving ! In the outdoor test, this is the floor, and in the treadmill test, this is the treadmill. It is the thing that is touched with the wheels of the cart.

Ah Ha! There it is! There is your mistake which is preventing you from understanding what is going on here.
It is Not Running on the treadmill at all! You are looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road. You are completely forgetting about that propeller that is spinning in the air. That is where all the work is being done By the cart; where all the force is being exerted BY the cart. That is where the cart is doing all it’s Running! It is running on or in the air! The wheels on the tread represent the Source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted ON the cart. The cart is no more running against the tread than a sail being blown by the wind is running against the wind. The running is between the hull and the water for the sailboat. The running is between the propeller and the air, for the cart on the treadmill. You do not measure the velocity of a sailboat by measuring how fast it is going wrt the wind that is pushing on the sails. You do not measure the velocity of the cart on the treadmill by measuring how fast it is going wrt the tread that is turning the wheels. Look elsewhere! Look where the moving vehicle is exerting force against something. The sailboat exerts force against the water so you measure its velocity wrt the water. The cart is exerting force against the wind, so you measure its velocity wrt the wind. I have been saying this since day ONE.

 

[zoobyshoe]

Hi Zooby. Here are some simple explanations of what's happening from the first thread on DWFTTW posted previously. Please read.

Hi, uart.

Both these gedankens seem to demonstrate a cart merely holding place given a certain motorized treadmill and air speed.

As AT pointed out, the cart has to do more than hold its own, it has to accelerate. (The more gearing you have the more elements have to be accelerated.) When it reaches wind speed there are no more outside forces adding power. The cart is feeding stored power to the surrounding media. It may be able to hold its own and even accelerate at first if it has enough stored power, but this will be used up soon, or eventually, and the cart will slow to wind speed.

If you were to put the cart in this condition without any stored energy, it would never work: suppose you drove down wind in a car till you reached wind speed, then picked the cart up from the seat next to you and placed the cart on the road. It's obvious it would never accelerate. It would instantly start to lose forward momentum to the road. Therefore, I suspect that non-motorized demonstrations that seem to work are most likely working off of stored energy.

 

[vanesch]

Ah Ha! There it is! There is your mistake which is preventing you from understanding what is going on here.
It is Not Running on the treadmill at all! You are looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road. You are completely forgetting about that propeller that is spinning in the air. That is where all the work is being done By the cart; where all the force is being exerted BY the cart. That is where the cart is doing all it’s Running! It is running on or in the air! The wheels on the tread represent the Source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted ON the cart. The cart is no more running against the tread than a sail being blown by the wind is running against the wind.

But what force is exerted on what doesn't matter to express *velocities*.

Wrt to what are you expressing the velocity in the outdoor test ? Wrt the by-going train ? Wrt the observer running next to it ? Wrt the air ? Wrt the moon ? Or with respect to the thing the wheel is touching, namely the floor ?

If you do the test on the train in a wind tunnel, would you compare to the floor of the windtunnel (touched by the wheels of the cart), or by the guy walking in the corridor of the train, or wrt the tracks, or wrt the cow running through the field next to the train ?

Again, unless you object, I suppose we can agree that it will be the floor of the windtunnel.

So what we call "the speed of the cart" is the speed of the cart wrt the thing the wheels are touching.
What we call "the speed of the wind" is the speed of the air wrt the thing the wheels are touching. Note that both velocities are expressed in the same frame, so these are relative velocities, they have a physical meaning. It is concerning THESE velocities that there is a claim. So there's no point calculating OTHERS. That can be done, but there's no claim regarding OTHER velocities.
The claim "DWFTTW" is the claim that the speed of the cart wrt the thing the wheels are touching is a bigger number in absolute value than the speed of the air wrt that same thing in absolute value.

If I say that the speed of a train is 150 mph on its track, and you claim that the velocity of the train wrt the glass on a table in the train is 0, then your statement doesn't disprove mine.

If I say that 5 > 3, then you claiming that 4 < 6 doesn't disprove mine. Now, you can say all you want about that it is absolutely necessary to compare 4 to 6, it doesn't alter anything concerning the comparison of 5 and 3.

The cart is exerting force against the wind, so you measure its velocity wrt the wind. I have been saying this since day ONE.

But my friend, this is most basic nonsense. There is no relationship between what is exerting a force on what, and any obligation to only consider relative velocities between them.

Tell me, is the moon exerting a force on the ocean water ?
So, is the only sensible way to express the velocity of the ocean water, the velocity in the frame of the moon ?

A personal question. I had the impression, along this thread, that you were an engineer. Are you ?

 

[vanesch]

As AT pointed out, the cart has to do more than hold its own, it has to accelerate. (The more gearing you have the more elements have to be accelerated.) When it reaches wind speed there are no more outside forces adding power. The cart is feeding stored power to the surrounding media. It may be able to hold its own and even accelerate at first if it has enough stored power, but this will be used up soon, or eventually, and the cart will slow to wind speed.

Look at the total force balance I worked out in post #480 and which gave the following total force:

F_tot = rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind) + W x (v_wind - v_cart)^3 - ...

... rho_air x S x K x ( (K^2-1) x v_cart^2 + 2 x v_cart x v_wind - v_wind^2)

Here, rho_air is the specific mass of the air, S was the "effective outgoing surface" of the propeller flow, K was the constant giving us the "gearing ratio" which was determined by v_out = K v_cart where v_out is the outgoing air from the propeller in the cart's frame, and v_cart was the velocity of the cart wrt to the ground (the velocity with which the wheel is spinning). W was a constant describing the drag on the structure (we can assume it small).

v_wind and v_cart are taken positive both in the "sense of the wind" and are both expressed in the "floor" frame.

The first term rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind)

represents the thrust by the propeller. It is obtained by making the momentum balance between the mass of air that is accelerated from "incoming velocity" to "outgoing velocity" by the propeller (of course, in the frame of the propeller).

If it is positive, it means a *forward* force, which *accelerates* the cart.

The second term: W x (v_wind - v_cart)^3

represents in general, the drag on the cart by the wind (including any drag by the propeller).
It works in the sense of the velocity difference between the wind and the cart of course. For a back wind ("wind faster than cart") it pushes forward, and for a head wind, it pushes backward (negative force).

The third term (note the minus sign): - rho_air x S x K x ( (K^2-1) x v_cart^2 + 2 x v_cart x v_wind - v_wind^2)

is the force that the ground exerts on the cart. It is obtained from the power balance calculated in the cart's frame.
The propeller doesn't only exchange momentum with the air column, it also exchanges energy. Indeed, the air wins or looses kinetic energy by changing from v_in to v_out and this energy is provided by (or to) the propeller. Per unit of time, there is hence a power flow from the wheel to the propeller (or vice versa).
Assuming perfect mechanical energy transfer between the wheel and the propeller, this power is obtained by driving the wheel with a certain force: the force exerted by the ground on the wheel (and hence on the cart). It works to the back direction if the wheel delivers energy to the propeller (that is, if the air is accelerated by the propeller, and hence wins kinetic energy, which is provided by the propeller, and hence must receive it from the ground).

You can find the total force on the cart when the cart is driving at the same velocity as the wind (v_cart = v_wind), and then we find:

F_tot = rho_air x S x K x v_wind x ((K-1) x v_wind + v_wind) + W x (v_wind - v_wind)^3 - ...

... rho_air x S x K x ( (K^2-1) x v_wind^2 + 2 x v_wind x v_wind - v_wind^2)


Or after some algebra:
1) the first term becomes: rho_air x S x K x v_wind x K x v_wind = rho_air x S x K^2 x v_wind^2

Note that the propeller is giving a forward force, so "wants to accelerate" the cart.

2) the second term is 0 of course: there's no drag in a wind-still situation

3) the third term is: - rho_air x S x K x ( (K^2-1) x v_wind^2 + 2 x v_wind x v_wind - v_wind^2) = - rho_air x S x K x ( K^2 x v_wind^2) = - rho_air x S x K^3 x v_wind^2

Note two things: the third term is NEGATIVE (wants to brake the cart, which is logical: we take energy from the wheels to provide for the energy of the propeller)

It looks a lot like the first term, except for an extra factor of K.

So the final force is: F_tot = rho_air x S x K^2 x v_wind^2 - rho_air x S x K^3 x v_wind^2

or: F_tot = rho_air x S x K^2 x (1 - K) x v_wind^2

Note that if K is between 0 and 1, this is positive, so there is a total force on the cart that wants to accelerate it.

In all of this, we didn't need any "stored kinetic energy of the cart".

 

[vanesch]

For instance, when using some dimensionless quantities and having numerically:

rho_air x S = 0.5

w = 0.1

K = 0.3

v_wind = 1.0

we have the plot in attachment as the force as a function of v_cart. Note that the force is positive beyond the point v_cart = 1, which means that there is a forward force (accelerating the cart) until a point around v_cart ~ 1.4 or so.

 

[rcgldr]

the transmission makes sure that force_propeller > wheel force

There gear ratio is 1:1. The force is greater due to the properties of the propeller, diameter, pitch, and efficiency. The effective prop pitch needs to be less than the circumference of the wheels so that prop to air speed is slower than wheel to ground speed, so that the prop + air interface involves greater force but at less power than wheel + ground.

The only air that matters is that which directly impinges on the cart, or upon which the cart impinges.

Unfortunately for the cart when it reaches wind = 0 it becomes inert.

No just as you mentioned, the air that matters includes the air that the cart impinges on. The thrust from the prop provides an upwind acceleration and speed of the air, which interacts with the wind. The combined effect of the upwind wash and the downwind flow increase the pressure aft of the prop, increasing the forward force applied through the air to the prop. The net result is that the wind speed of the air is slowed down as the cart passes through, and the reduction in wind speed is the source of power. The cart is outrunning the wind, but not the air flow from the prop wash. The air flow aft of the prop is slower than the wind, even when the cart is moving faster than the wind.

This is the prop as "bluff body" you mentioned before, right? It's very apparent in that video of the large cart with the tell tale. The prop is quite different than the usual airplane prop.

The prop is different because it has a low pitch versus diameter, a typical "slow flyer" prop. The carts need a prop with a relatively small pitch for the reasons stated previously. The prop only acts as a bluff body only at start up. During acceleration the prop acts as both bluff body as well as thrust generator. You could consider the air itself from the prop wash to be a bluff body moving upwind with respect to the cart.

seem to demonstrate a cart merely holding place given a certain motorized treadmill and air speed.

Both swerdna's and spork's videos demonstrate a cart advancing on the treadmill, not just holding place.

 

[vanesch]

Both swerdna's and spork's videos demonstrate a cart advancing on the treadmill, not just holding place.

If I understood it well, the trick was to incline the treadmill so that the cart had to move "uphill". This means that on top of the wheel and air forces, it had a backward working force, namely a component of its weight. The fact that it was holding still in this situation means that the air+wheel force balance was in the forward direction, given that it compensated the backward working force of gravity. This then means that at the point of being at wind velocity, the cart still had a forward force working on it by the wheels + propeller, and if there hadn't been the gravity "holding it in place", it would have accelerated forward.

It was again a nice experimental trick to make the treadmill experiment last so that the argument of "steady state is not reached yet" is refuted.

 

[rcgldr]

looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road.

or wheels running on a treadmill:

For this one, how would the experience on the treadmill be different for the rider if the rider and bicycle were outdoors riding downwind at the same speed as the wind?

http://www.youtube.com/watch?v=pS_tIfbfCLM&fmt=18

How about a treadmill on a bicycle?

http://www.youtube.com/watch?v=Sg-KpT9RNXE&fmt=18

propeller that is spinning in the air. where all the force is being exerted by the cart. ... wheels on the tread represent the source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted on the cart.

Newton's 3rd law, forces only exist in equal and opposing pairs. The force the prop applies to the air coexists with an equal and opposing force that the air applies to the prop. The force that the tread applies to the wheels coexists with the equal and opposing force the wheels apply to the tread.

The cart is no more running against the tread than a sail being blown by the wind is running against the wind. The running is between the hull and the water for the sailboat.

So why isn't the cart case the running between the wheels and the ground?

The running is between the propeller and the air, for the cart on the treadmill.

The cart involves two interactions. 1 - between prop and air. 2 - between wheel and ground.

You do not measure the velocity of a sailboat by measuring how fast it is going wrt the wind that is pushing on the sails.

Yet aircraft measure speed relative to the air they travel in. There's no reason sail craft can't use instruments to measure apparent wind and direction (and some do).

You do not measure the velocity of the cart on the treadmill by measuring how fast it is going wrt the tread that is turning the wheels.

Why not? It's a perfectly reasonable frame of reference.

Look where the moving vehicle is exerting force against something.

The cart applies an upwind force to the air, and a downwind force onto the tread, so either could be a reasonable choice as a frame of reference. The cart doesn't exert a force on the floor, so it would seem that the floor would be a bad choice.

You have yet to commnet about my mathematically described claim about these carts:

|v_cart - v_ground| > |v_wind - v_ground|

 

[A.T.]

Unfortunately for the cart when it reaches wind = 0 it becomes inert.

No, it doesn't become inert. The propeller is producing more forward force than is needed to drive the propeller at the wheels pointing back. The net force is not zero -> acceleration occurs.

the transmission makes sure that force_propeller > wheel force

There gear ratio is 1:1.

With transmission I mean the entire transmission chain from wheel to propeller, that you described.

 

[rcgldr]

If I understood it well, the trick was to incline the treadmill so that the cart had to move "uphill". It was again a nice experimental trick to make the treadmill experiment last so that the argument of "steady state is not reached yet" is refuted.

I was referring to sporks videos where the cart advances. Swerdna's turntable allowed the cart to advance indefinately to give a good idea of the steady state situation.

I captured swerdna's video so I could time it and calculate the actual results to show some real numbers:

Turntable speed = 13.1 mph
Cart speed = 5.6 mph.

This is equivalent to v_wind = 13.1 mph, and v_cart = 18.7 mph.

Cart speed = 1.43 times wind speed.

For the given paramters, prop pitch 6 inches, wheel circumference 10.5 inches, the theoretical limit would be

v_cart_max = v_wind / (1 - ar)
v_cart_max = v_wind / (1 - 6/10.5)
v_cart_max = 2.33 v_wind

So the cart is achieving about 61% of the theoretical (no loss) limit.

 

[schroder]

Fun, no ?



No, indeed, I cannot argue with that because there's no logic there. "Working against" is not something that has an intrinsic meaning, and from there on, the rest fails. But what is working against what doesn't matter when we see velocities, no ?

Now right here you are doing something that is unscientific. You may consider it to be “fun” and I agree there is an element of that, but there is also a serious side of this. This ridiculous claim needs to be rebuked by the physics community. You are dismissing what I wrote as illogical without showing where the logic fails. That is like saying “You are wrong because my name is Vanesch, and I say so.” I would expect better from you.
And of course it matters what is working against what! The velocity of the wind working against the sail is not used as a reference to measure the velocity of the boat because the boat is not working against the wind, it is working against the water. How can you honestly say it does not matter what is working against what when you measure velocities?
So I am going to walk you through the logic one more time, and this time You need to show me where the logic fails. If you cannot do so, you must accept the conclusion that the logic leads you to. I am going to be very fair to you and give you advance notice that I am now placing you in checkmate. You cannot simply knock over the chess board and say this is nonsense. You need to show how and why you are not in checkmate and then show me that you can get out of it.
Just follow the logic:
1) A machine that can do more work than the work that is done on it is by definition an over unity machine.
2) There are no over unity machines.
3) A machine that can do more work on a tread than the tread does on it, is an over unity machine.
4) There are no over unity machines.
5) From 1 - 4 : a machine that has work done on it by a tread cannot do more than that amount of work on the tread.
6) From 1 - 5 : A machine that has work done on it by a tread, but is clearly seen to be advancing on that tread in the opposite direction to the tread cannot possibly be an over unity machine because there are no over unity machines.
7) From 1 - 6 : The cart is being powered by the tread. It is also advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine.
8). From 1 – 7 : Since the cart is not an over unity machine, but it is advancing on the tread in the opposite direction to the tread and it is being powered by the tread it cannot possibly be working against the tread. Because if it were working against the tread , due to being being worked on by the tread , it would be doing more work on the tread than the tread is doing on it which means it is an over unity machine and they do not exist!
First definitive conclusion: From 1 – 8 : The cart is being powered by the tread. It is advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine. The cart is not working against the tread.
If you accept (and you have no choice) that the cart is not working against the tread, then you need to justify why you insist on measuring the velocity with respect to the tread. If you can justify that, you would also be saying that the velocity of a sailboat can be measured with respect to the wind which is pushing it. You would get some very low numbers because the boat is moving in the same direction as the wind. And the reason you are getting high numbers for the cart velocity is because the tread and the cart are moving in opposite directions and you are measuring the velocity of the cart with respect to the tread.
In both cases, sailboat and cart, the only CORRECT approach is to measure the velocity of the vehicle with respect to the interface it is moving against. The sailboat moves against the water so you measure the velocity at the interface between the hull and the water. The cart moves against the wind (air) so you measure the velocity of the cart at the propeller and air interface.

Since you have displayed a disturbing tendency, when cornered, to simply dismiss even the most logical presentation as mere gibberish, I now openly challenge you to go through the above presentation and show where you feel it is wrong. You like chess?

 

[A.T.]

For the given paramters, prop pitch 6 inches, wheel circumference 10.5 inches,

Have you taken into account, that the propeller is further away from the turntable axis than the wheel?

 

[cristo]

1) A machine that can do more work than the work that is done on it is by definition an over unity machine.
2) There are no over unity machines.
3) A machine that can do more work on a tread than the tread does on it, is an over unity machine.

You can't just pick a part of the system that you want energy conservation to hold in, and ignore the rest! 1 and 3 are not equivalent statements!

 

[rcgldr]

Have you taken into account, that the propeller is further away from the turntable axis than the wheel?

I ignored that factor. Guestimating that prop is 8 inches further out than wheel:

Turntable speed at wheel = 13.1 mph
Turntable speed under prop = 17.5 mph
Prop advance speed = 7.5 mph

 

[Subductionzon]

Schroder, your last argument had one huge logical fallacy to it, no one is claiming that the cart is an over unity device except for you. It runs off of the difference between the speed of the wind and the speed of the surface. Outdoors that would be the speed that the wind goes over the ground. Indoors on the treadmill it would be the speed of the treadmill. The cart cannot tell the difference between the two cases.

Here is a "real life" device that can go faster than the wind, an ice boat. It also extracts energy from the difference between the speed of the wind and the speed of the ground. They have been observed going four to five times the speed of the wind on a broad reach, that is sailing at an angle 45 degrees to the wind. So if the wind was blowing straight North the ice boater would tack back and forth with the wind going northeast and northwest. Let's say he is an accomplished sailor with a good boat and the wind is 10 mph. He would be able to maintain a 50mph speed on those tacks with a resultant straight north speed of 35 mph. Now if he sailed due north the most he could go is that, 10 mph, but by tacking he can outrace the wind by 25 mph. That is what the props on the propeller are doing, they are acting as a sail and driving the boat.

 

[rcgldr]

The velocity of the wind working against the sail is not used as a reference to measure the velocity of the boat because the boat is not working against the wind, it is working against the water.

The boat is working against both the wind and the water. Both the air and the water are accelerated by the boat, even when the boat itself is not accelerating. The cart also interacts with air and ground, but the effect on the ground is tiny because the mass of the ground versus cart is huge.

It is a closed system, and momentum is being conserved. If the wind speed decreases, then the cart and/or ground speed increase in order to conserve momentum. In this case, most of the increase in speed occurs with the cart (as oppose to with the ground), because of the difference in mass of ground versus cart, and because the cart has wheels that allow it to advance with respect to the ground.

 

[rcgldr]

8). Since the cart is not an over unity machine, but it is advancing on the tread in the opposite direction to the tread and it is being powered by the tread it cannot possibly be working against the tread. Because if it were working against the tread , due to being being worked on by the tread , it would be doing more work on the tread than the tread is doing on it which means it is an over unity machine and they do not exist!

Applying Newton's 3rd law, any force exerted by the tread onto the cart is opposed by an equal and opposite force by the cart onto the tread. The work done by the tread onto the cart and the work done by cart onto the tread are equal and opposite. Energy isn't being created here, just exchanged. The same logic applies to the prop + air interface.

 

[Subductionzon]

I also like to use the phrase "difference between the speed of the wind and the speed of the surface" because it helps get you from thinking of the wind blowing directly on the cart, or with a sailboat on the sail. If you think of the wind blowing on the sail it would seem impossible for the boat to go faster than wind speed when sailing on a beam reach, I mean its outrunning its power source isn't it? But no it isn't and this has been observed countless times, it is still "within" the wind. It is running off of the difference in speed between the wind and the ice. If you are really nuts you could put the treadmill on the back of a flatbed truck on a day with a known windspeed. For example say it was another 10mph wind day and the truck was going with the wind at 10mph. Net result no wind, now we start up the treadmill like we did before and run it at 10 mph. Just like the indoors test the cart will advance up the treadmill. But how do you measure its speed now schroder? With respect to the treadmill and the ground it is going 10 mph plus whatever its advance speed up the treadmill is. With respect to the flatbed truck and the wind it is just going up the treadmill at its advance speed up the treadmill. In either case it is going faster than the wind.

 

[vanesch]

Now right here you are doing something that is unscientific. You may consider it to be “fun” and I agree there is an element of that, but there is also a serious side of this. This ridiculous claim needs to be rebuked by the physics community. You are dismissing what I wrote as illogical without showing where the logic fails. That is like saying “You are wrong because my name is Vanesch, and I say so.” I would expect better from you.

I never used any argument of authority as far as I remember. I do make "authoritative statements" concerning certain properties of Newtonian mechanics, because I think that I know the theory well enough to make them. Like I could make the authoritative statement that Newton's equation is F = m.a without arguing it, because I think that the "physics community" won't dispute it.
If I tell you (or if I say in general) some statements like "you can do the calculations in any frame you like", then this is not on my authority, but simply because I know it is an entirely accepted fact, and that I don't think that there is any physics professor out there who will dispute it.

However, I have taught university level courses, and I can tell you that any student of mine who would display such ignorance of basic material together with such a refusal to reconsider, would have been in serious trouble. I consider this discussion on my side as an exercise in the Art of Zen

And of course it matters what is working against what! The velocity of the wind working against the sail is not used as a reference to measure the velocity of the boat because the boat is not working against the wind, it is working against the water.

You see, that phrase, by itself, doesn't have any meaning when you use standard terminology. The only meaning one could give to it is that no physically meaningful quantity can ever be calculated using the velocity of the wind. That's of course glaring nonsense. After all, the totally valid question "what is the velocity of the boat wrt to the wind" would need to use the velocity of the wind AND the velocity of the boat, in *any randomly chosen frame*, to be able to calculate the answer to the question.

So if you want to know where your logic fails, it is already right here.

How can you honestly say it does not matter what is working against what when you measure velocities?

Because it is true. The kinematical description of a situation (that is, the positions and velocities and so on) are independent of the dynamical description of the situation (the interactions, the forces...). That is what is explained in about the first chapter of every book on classical mechanics.
For instance, in my engineering education, I started out with 45 hours of kinematics, without even mentioning the word "force". It was only in the next course, 90 hours of "dynamics" that force was introduced. I'm not saying that this is the only or best way to teach mechanics, but at least it demonstrates that the kinematical description has nothing to do with what thing is exercising what force on what other thing.

So I am going to walk you through the logic one more time, and this time You need to show me where the logic fails.

Well, it failed already, but let us continue (in my practice of the Art of Zen...: I want at least black belt 6th dan in it...)

If you cannot do so, you must accept the conclusion that the logic leads you to. I am going to be very fair to you and give you advance notice that I am now placing you in checkmate.

Whoo...

You cannot simply knock over the chess board and say this is nonsense. You need to show how and why you are not in checkmate and then show me that you can get out of it.
Just follow the logic:
1) A machine that can do more work than the work that is done on it is by definition an over unity machine.

Yes. At least in steady state. Granted. However, be very careful what is "doing work on" and "work done on it". These are frame-dependent quantities. You can of course have a machine that "does work" on something in one frame, "on which work is done" in another frame, and nevertheless find out that it does more work as per the first quantity than it receives as per the second quantity, without it being an over-unity machine. So, granted on the condition that both amounts of work are calculated in the same frame.

2) There are no over unity machines.

Granted.

3) A machine that can do more work on a tread than the tread does on it, is an over unity machine.

No, not necessarily. You have to find out if there are other contributions. Otherwise, the motor driving the treadmill is an over unity device. But the electric grid does work on the motor, so this is not a problem in this case.

4) There are no over unity machines.

Granted (second time).

5) From 1 - 4 : a machine that has work done on it by a tread cannot do more than that amount of work on the tread.

I don't know exactly what you mean by this. If you mean: *in a specific reference frame, the force exerted by the treadmill on the system, multiplied by the displacement of the treadmill in that frame, is the power provided by the treadmill*, yes, that's the power provided by the treadmill in that reference frame. And this power should not be larger than any work the device is doing on anything else, such as the air, *in that same reference frame*.

6) From 1 - 5 : A machine that has work done on it by a tread, but is clearly seen to be advancing on that tread in the opposite direction to the tread cannot possibly be an over unity machine because there are no over unity machines.

Well, you can simply say: the machine cannot be an over unity device because there aren't any. And the first part of your sentence is not related to this, nor to 1-5.

7) From 1 - 6 : The cart is being powered by the tread. It is also advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine.

Well, whether the cart is powered by the tread or not is dependent on from which frame you look upon it. In the frame of the tread, it isn't, of course. Because "powered by" means: has a positive value of "the product of force exerted by and displacement by". But as in the frame of the tread, the tread is not moving, the displacement is 0, and hence it is not doing any work. In the frame of the ground, this might be the case, if the force exerted by the treadmill on the cart is in the same direction as the movement of the treadmill.

8). From 1 – 7 : Since the cart is not an over unity machine, but it is advancing on the tread in the opposite direction to the tread and it is being powered by the tread it cannot possibly be working against the tread.

Not at all. This is wrong. The treadmill can exert a force on the cart in the same direction as it moves, and that is then the power given by the treadmill to the cart *in this frame*. And this number depends on what frame one uses.

The state of motion (position and velocity) of the cart wrt the treadmill is independent, a priori, of the direction and magnitude of the force that one can exert on the other. Moreover, the statement "moves in the opposite direction as the treadmill" is also a frame dependent quantity. In the airplane that flies by, both move in the same direction.

Because if it were working against the tread , due to being being worked on by the tread , it would be doing more work on the tread than the tread is doing on it which means it is an over unity machine and they do not exist!

No, that's wrong, for several reasons, but the main reason is this: the exchange of power is not only with the treadmill, and, as I said before what exchanges power with what is frame dependent. You can do the balance in any frame, and as long as you stick to it, things will come out all right. But when you switch frames during the calculation, you will make errors.

In the frame of the tread, the tread is NOT doing any work on the cart, and the air is doing all the work. That means: you calculate the total force exerted on a certain mass of air, and the displacement that goes with it, you multiply and that gives you the amount of energy (in the given time lapse) that the air has given to the system. As the treadmill didn't move in this frame, it can't give any energy.

In the frame of the air (= frame of outside ground), the air is not moving and not doing any work, and the treadmill is doing all the work. Now, in this frame, the treadmill is exerting a certain force on the cart, and undergoes a certain displacement in doing so, hence delivers an amount of work to the system. That's the power that is available. It doesn't matter at what velocity the cart is moving, the power that is extracted from the treadmill is the velocity of the treadmill here (in the ground/air frame) times the force it exerts on the system. It is this amount of power that the motor of the treadmill will have to deliver.

In the frame of the cart, the treadmill is moving (faster than in the previous frame btw). So in this frame, the cart receives energy from the treadmill (a larger number than the number calculated in the previous case), but has to spend also energy with the propeller on the air.

Now, the erroneous objection might be that in *this frame* (the cart frame) the cart receives more energy from the treadmill than the treadmill delivered *in the ground frame*. So did there go *more energy* in the cart than the treadmill actually delivered ?
No, the error here is that we are using energies calculated in two different frames, and we shouldn't use them in the same balance. It is an error that is easy to make. It is the one I pointed out already several times.
It is even more confusing. Let us place ourselves in the ground/air frame. Now, there's something weird. The treadmill is going at velocity v1 and is undergoing a force - F by the wheel (v1 and F positive numbers). So it is "receiving" power from the cart v1 x (-F), or in other words, the treadmill is DELIVERING the power v1 x F.
The air is still and is exerting a force F on the cart (as the cart is in steady state, there's no net force on it). However, the air being steady, there is no power delivered by the air to the cart.
The cart is going at a velocity v2 in the opposite direction as the mill (v2 positive number).

So the cart is delivering a power F x (v2) to the treadmill and is receiving power F x v2 from the air.

Now, note that it is extremely confusing that what the treadmill is giving to the cart, is not what the cart is receiving from the treadmill, and that what the air is giving to the treadmill (nothing) is not what the cart is receiving from the air.

That is because we used an abuse of language. It is not because we have a force, and its reaction, that the displacements along them are equal, in a random reference frame. So a force F and its reaction -F do not correspond to individual power balances because the v1 of the point where F acts, and the v2 of the point on which -F acts, doesn't have to move at equal velocities ; hence F x v1 will in general not be equal and opposite to - F x v2.

So the abuse of language has been to take couples of "action / reaction" and to call the associated powers "the power given to ... " and the "power received from".

Newtonian mechanics doesn't require a detailed power balance for every interaction individually, but it does require a power balance overall.

And that's ok here: power balance of the treadmill: lost v1 x F.
power balance of air: 0
power balance of cart: delivered F x (v2), received v2 x F.

The power balance of the cart is actually trivial in the steady state, as the total force on it is 0, so it is normal that the total amount of mechanical power received by it is 0. If the system were not steady yet, there would be an effect: the acceleration of the cart would make for a positive power balance which is going into the kinetic energy of the cart.

So we have that the overall mechanical power received by the system is - v1 x F + 0 + v2 x F - F x (v2) = -v1 x F < 0 as it should be. Energy balance is ok: in the steady state, the entire energy delivered by the treadmill (the only external source of power) is dissipated.

If the total balance would have been > 0, we would have had an over-unity device.
So far from being an over-unity device, this thing is just dissipating the power of the treadmill, which is obvious, because once steady state is reached, no energy is stored anymore in the system, no power is done by the system on anything, and the treadmill is delivering power to the system (in the frame of the ground). So whatever is pumped into it, is dissipated.

Now, this amount of dissipated energy IS a frame-independent quantity. However, the *source* of it is dependent on the frame. Let's redo the exercise in the frame of the treadmill.

Here, the cart is moving at (v1 + v2), and the air is moving at v1, both in the opposite directions as the treadmill was moving in the ground frame. Let us call "forward direction" this direction of the motion of the air and the cart. (in this frame, they move in the same direction, v1 and v2 are positive numbers).

Now, in *this* frame, the air is exerting a force F on the cart in the positive direction, the air is hence undergoing a force -F, and the work done on the air is v1 x (-F) ; in other words the air is DELIVERING power to the system here, and that amount of power is v1 x F.

The cart undergoes a force +F by the air, and is undergoing a force -F by the treadmill. Hence the cart receives a power (v1 + v2) x F from the air and delivers a power (v1 + v2) x F to the treadmill (but we know again that this "to this" or "to that" is just indicating where the force is coming from, and isn't a power balance). Again, we could also have said that because the total force on the cart is 0 (steady state) that the cart doesn't receive or do any work overall.

The treadmill undergoes a force +F by the cart. As it is not moving in this frame, however, it is not delivering any work.

Overall balance: power received by all components: - v1 x F + (v1+v2) x F - (v1 + v2 ) x F + 0 = - v1 x F.

Total balance negative, this is what is dissipated. It is again equal to v1 x F.

Third way of calculating: the frame of the cart.

Here, the air is moving at velocity v2 and the treadmill is moving at velocity (v1 + v2), both in the sense of the motion of the treadmill in the frame of the ground. Let us call this direction positive.

The air is undergoing a force F in the positive direction. It is hence RECEIVING the power v2 x F.
The cart will again give a 0 balance, this time for two reasons: total force = 0, and on top of that, velocity is 0.
The treadmill is undergoing a force F in the negative direction. So it is DELIVERING the power
(v1 + v2) x F.

Total balance of received power: v2 x F - (v1 + v2) x F = - v1 x F.

Again, this indicates that the total power dissipated is v1 x F.

You see that no matter what frame one uses, we come out all the same. However, we see that the individual contributions of treadmill and air are different according to the frame in which we do the calculation.This comes about because in steady state, if there were no dissipation, the force F would actually be 0 when steady state would be reached. This would be (in the frame of the cart) when the propeller is not delivering any work anymore to the air (that is, when v_out would be equal to the apparent headwind velocity), and if dissipation-less, would hence not require any power from the wheel, which would then be rolling totally freely over the surface.

First definitive conclusion: From 1 – 8 : The cart is being powered by the tread. It is advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine. The cart is not working against the tread.

I think I explained you why it is not correct.

If you accept (and you have no choice) that the cart is not working against the tread, then you need to justify why you insist on measuring the velocity with respect to the tread.

First of all, it is not right that the cart is not "working against the tread" in any frame. But second, I can measure a velocity wrt to anything I like. So your statement hasn't head or tails (let alone follows from any logical deduction).

If you can justify that, you would also be saying that the velocity of a sailboat can be measured with respect to the wind which is pushing it.

Of course it can. That's actually what is most easily done with an anemometer on the boat.

You would get some very low numbers because the boat is moving in the same direction as the wind.

yes.

And the reason you are getting high numbers for the cart velocity is because the tread and the cart are moving in opposite directions and you are measuring the velocity of the cart with respect to the tread.

yes.

In both cases, sailboat and cart, the only CORRECT approach is to measure the velocity of the vehicle with respect to the interface it is moving against.

This is a nonsensical statement.

Since you have displayed a disturbing tendency, when cornered, to simply dismiss even the most logical presentation as mere gibberish, I now openly challenge you to go through the above presentation and show where you feel it is wrong. You like chess?

I do.

 

[vanesch]

I also like to use the phrase "difference between the speed of the wind and the speed of the surface" because it helps get you from thinking of the wind blowing directly on the cart, or with a sailboat on the sail. If you think of the wind blowing on the sail it would seem impossible for the boat to go faster than wind speed when sailing on a beam reach, I mean its outrunning its power source isn't it? But no it isn't and this has been observed countless times, it is still "within" the wind. It is running off of the difference in speed between the wind and the ice. If you are really nuts you could put the treadmill on the back of a flatbed truck on a day with a known windspeed. For example say it was another 10mph wind day and the truck was going with the wind at 10mph. Net result no wind, now we start up the treadmill like we did before and run it at 10 mph. Just like the indoors test the cart will advance up the treadmill. But how do you measure its speed now schroder? With respect to the treadmill and the ground it is going 10 mph plus whatever its advance speed up the treadmill is. With respect to the flatbed truck and the wind it is just going up the treadmill at its advance speed up the treadmill. In either case it is going faster than the wind.

Ha, we're back to my train experiment

 

[schroder]

I never used any argument of authority as far as I remember.




Not at all. This is wrong.





This is a nonsensical statement.

I have laid it all out in front of you in black and white. I have given you and Physics Forum the opportunity to redeem your prestige.

False pride is all that stands in your way. You are very skillful at dancing around the issue.

But, you know, at some point that old tumbleweed will go past that ridiculous little cart as if it was standing still. ( Relative Velocity and all that)

What will all your dancing numbers mean then?

What will your professional credibility and the credibilty of this forum mean then?

I am done playing this silly game.

I will wait and see how this plays out.

Meanwhile, the people who are running this scam are guilty of fraud. Selling these carts through the mail while making fraudulent claims. That is mail fraud, A Federal offense. This is no game.

I can take this to the Academy but they already consider it to be nonsense.

I tried to do you a favor.

It is now too late.

Oh, and you need not bother with the ban, or the infractions That does not affect me.

 

[vanesch]

False pride is all that stands in your way. You are very skillful at dancing around the issue.

It is amazing how applicable your advice is to your own viewpoint.

I am done playing this silly game.

That must be the 10th time you say you will quit.

The bare bones fact is that you haven't gotten a clue of the most basic concepts of mechanics, and that you've smeared this out along this thread.

You take offense at anyone pointing out the myriads of errors in what you call your reasoning, which is nothing else but a succession of demonstrations that you never opened a book on theoretical mechanics or had a decent course in it - or if you had, that you never understood anything there, or have forgotten all of it.
There's no problem in being wrong. But there's a problem in keeping to one's opinion despite careful explanation, while not once trying to understand what people try to explain to you. That comes because you have convinced yourself of the end result of the reasoning, and as such, that you think you don't even have to consider the other person's reasoning, given that he arrives at the conclusion you think is impossible, he must hence be wrong.

There's nothing difficult in the presented mechanics. It is elementary, first year mechanics in any physics or engineering curriculum. True, it is at first somewhat counter intuitive. If you would have asked me 2 weeks ago whether such a cart would work, from my bones I would probably have said, no.
Then there was this video demonstration, and then one had to sit down a few minutes to "understand the trick", by working it out in a bit more detail. And that's all there is to it.

But given your total lack of the most basic understanding of mechanics, you can't do that, and hence you keep with your intuition.

One single statement of yours illustrates this: the fact that you are requiring "proper" frames which are related to what you call "the medium the thing works against". That is such an obvious glaring nonsensical statement that it doesn't need any further comments.

Meanwhile, the people who are running this scam are guilty of fraud. Selling these carts through the mail while making fraudulent claims. That is mail fraud, A Federal offense. This is no game.

Unless, of course, you haven't gotten a single clue in the whole discussion.

I can take this to the Academy but they already consider it to be nonsense.

I tried to do you a favor.

It is now too late.

Oh dear. Well, I tried to do you a favor, by going all out of my way to teach a very elementary course in mechanics. You are visibly not interested in learning anything here.

BTW, telling you that you are wrong is not a statement of authority. I *explain* to you why you are wrong. If I write: 3 + 4 = 12, and someone tells me that this is wrong because 3 + 4 = 7, is he making an authoritative statement ?

EDIT: one more thing. You are visibly convinced that a DWFTTW cart is an over-unity device, right ?

If I give you one, how would you go about in making unlimited amounts of electricity from it (which is what an over unity device can do) ?

 

[uart]

I have laid it all out in front of you in black and white. I have given you and Physics Forum the opportunity to redeem your prestige.

False pride is all that stands in your way. You are very skillful at dancing around the issue.

But, you know, at some point that old tumbleweed will go past that ridiculous little cart as if it was standing still. ( Relative Velocity and all that)

What will all your dancing numbers mean then?

What will your professional credibility and the credibilty of this forum mean then?

I am done playing this silly game.

I will wait and see how this plays out.

Meanwhile, the people who are running this scam are guilty of fraud. Selling these carts through the mail while making fraudulent claims. That is mail fraud, A Federal offense. This is no game.

I can take this to the Academy but they already consider it to be nonsense.

I tried to do you a favor.

It is now too late.

Oh, and you need not bother with the ban, or the infractions That does not affect me.

This is just mind boggling. I can't believe the patience that vanesch and Jeff (along with others) have displayed in trying to explain to you the operation of this device. I gave up trying to explain anything to you in the last DWFTTW thread and absolutely won't waste anymore time on it. Your attitude is completely un-scientific in the way you have simply assumed that this device violates some physical law (which it doesn't) and on the basis of that assumption you’ve basically refused to even try to understand anything anyone tells you. That's sure how it looks to me anyway.

 

[zoobyshoe]

The first term rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind)

represents the thrust by the propeller. It is obtained by making the momentum balance between the mass of air that is accelerated from "incoming velocity" to "outgoing velocity" by the propeller (of course, in the frame of the propeller).

If it is positive, it means a *forward* force, which *accelerates* the cart.

In all of this, we didn't need any "stored kinetic energy of the cart".

In order to understand the non-stored forces at work on the cart at wind=0 we have to imagine a "virgin" situation where the propeller is not in motion at all. Not rotating.

Wind speed = 0

Ground speed = say, 10mph but contrary to the direction we want the cart to accelerate into

Prop rotation speed = 0

Is there a total force on the cart here in the right direction that will cause acceleration in the right direction?

 

[zoobyshoe]

Yes if the prop speed is zero then it can't work, but how did you get this far into the discussion without working out the the prop is driven by (directly coupled) to the wheels, so it won't (can't) be at zero speed when the cart is moving at wind speed. I'm not sure if you really didn't understand this or if you just wanted to change things so it couldn't work? (Like Schroder, he wanted Swerder to connect the prop to run in the wrong direction and re-test it for whatever that would prove).

I understood this from the start. The counterargument, though, was that there is still relative velocity between air and ground, therefore available power. I have been waiting for someone to show me a way for the cart to take advantage of that.

 

[OmCheeto]

... Selling these carts through the mail ...

Has swerdna taken my advice and put them on the market? That would be awesome. I want one to amaze my friends with. And to do my own testing on of course. I suppose I could build one myself. I've most of the parts on the back porch. I'm only missing an efficient wheel to prop interface device.

I was actually thinking about this problem the other day and came up with a 4th DDWFTTW device. Only this time, the power comes from an on-board motor. Exact same setup though, with the wheels and prop linked up. I was curious at which point the wheels stopped being the primary source of power and the propeller took over.

Sorry if this has already been mentioned. I've not read the middle 30 or so pages in the thread.

And btw:

Zoob, could you maybe consider my post #378 and answer the y/n questions ?

https://www.physicsforums.com/showpost.php?p=2036877&postcount=378
...

Did anyone ever answer the questions to post #378? I thought I might have missed it amongst all the shopping carts, windmill boats, sail boats, ice sail vehicles, trains, fences, fence posts, torpedo's, cart #1, cart #2, cart #3, etc. etc.

 

[uart]

I understood this from the start. The counterargument, though, was that there is still relative velocity between air and ground, therefore available power. I have been waiting for someone to show me a way for the cart to take advantage of that.

Ok I think I misunderstood what you were trying to get at here. You want the wind at zero and the cart is moving in the negative direction correct. Well you might have to modify the design but you could make it take power from the wheels to help it more rapidly reach the wind speed (that is to stop). It's already been established that this thing can't go faster than the wind if the wind speed is too low. No it can't sustain faster than wind speed in zero wind, nobody ever claimed it could.

 

[rcgldr]

Wind speed = 0, Ground speed = say, 10mph but contrary to the direction we want the cart to accelerate into, Prop rotation speed = 0. Is there a total force on the cart here in the right direction that will cause acceleration in the right direction?

This is virtually the same as the treadmill test, except the prop has been disconnected from the wheels and is fixed in place. From start up, the prop and cart still act as a bluff body, so the cart accelerates in the "right" direction, but at terminal speed, the cart retreats instead of advacing on the treadmill, because it's speed with a non-spinning prop will be slower than the wind.