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

[zoobyshoe]

The moving brown fence being the ground has enough mass.

Whoops!

How much mass does the ground have here?

 

[schroder]

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

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

It is interesting to see where there's something that's bothering you.

As I told people here already several times, energetic considerations "in the wild" are ERRONEOUS APPLICATIONS of the principle of conservation of energy, so they do not hold if they use energies in different frames.

And so are erroneous applications of reference frames!
I will make another, probably futile waste of time; attempt to explain to you exactly how you are making your error. Galilean relativity and reference frames are incredibly easy to understand. But they can be wrongly applied to a problem and get wrong, even ridiculous solutions. That is what is happening here.
Your claim is that on the treadmill, the velocity of the cart is measured wrt to moving tread. You are using the frame of the cart to come this conclusion and what you have is relative velocity between the cart and the tread, and that relative velocity is faster than the tread, So you claim that is evidence that the cart can go faster than the wind, (at least in principle) Is that correct?
I reject that claim. I claim that when the cart is running on the tread, being powered by the tread, it is working into the air and doing no work on the tread. Therefore, I claim that the proper frame of reference for the cart and the tread is the reference frame of the air, or floor as the air and floor are relatively stationary to each other. Now, based on the usage of that reference, I can measure the speed of the cart relative to appoint on the floor in one direction and the relative speed of the tread measured from the same point in the opposite direction and compare them. It is without any question that the tread is moving much faster than the cart when the velocities are compared. My conclusion from that is that when placed in the wind, the cart will also be moving much slower than the wind. You reject what I am saying.
Ok. Now when the cart is running down wind, how do you measure its velocity? How do you determine how fast the cart is going? Do you measure the velocity of the cart relative to the wind or relative to the ground? I would assume you would measure it relative to the floor Am I right? You would do this because if you measured it relative to the wind, you would get ridiculous answers such as it is moving with a negative velocity a sub wind speeds. And a zero velocity at wind speed and you would only get a positive velocity for the cart if it should ever go over the wind speed. Obviously these numbers are wrong, as you can see the cart moving with a positive velocity the second it starts moving. The reason the result is wrong is because you have chosen the wrong reference frame, the wind, when you should have chosen the ground.
You should have chosen the reference frame of the medium the cart is working into, not the medium that is doing work on the cart!
In the case of the treadmill, the tread is doing the work on the cart and the cart is working into the air. To measure the velocity of the cart you should have chosen the reference frame of the air for your calculations. By choosing the reference frame of the tread you have the ridiculous result that the cart is going faster than the wind which is equally ridiculous to the cart going at a negative velocity on the ground.
You need to choose the correct frame of reference in order to gets results that make any sense. Usually when you get a ridiculous result, you know you did something wrong! In this case, the ridiculous result that the cart is running faster than the tread has lead to all this nonsense which has taken up so much time and trouble.
Now, after reading this and giving it your consideration, rather than rejecting it simply because you don’t like the result, you must agree that you have been using the wrong reference frame and that has led to an inversion of the results which I have been saying since day ONE.
Vanesch, it is time for you to accept that you have been wrong and set the record straight. Will you do it?

 

[vanesch]

Your claim is that on the treadmill, the velocity of the cart is measured wrt to moving tread. You are using the frame of the cart to come this conclusion and what you have is relative velocity between the cart and the tread, and that relative velocity is faster than the tread, So you claim that is evidence that the cart can go faster than the wind, (at least in principle) Is that correct?

yes. Well, I don't know exactly what you understand by those statements yourself, but yes: I could use the frame of the cart to calculate the relative velocity of the cart and the tread, and to calculate the relative velocity of the air and the tread, and conclude that the first is larger than the second. Actually, that calculation is more easily done in the frame of the ground which is the frame of the air of course. But I can go with your proposition here.

I reject that claim. I claim that when the cart is running on the tread, being powered by the tread, it is working into the air and doing no work on the tread.

I don't know exactly what you mean by this, given all the rest you've written. But I don't see in how much this changes the calculation of the relative velocities.

Are you claiming that the relative velocity of the cart wrt the tread, contrary to what I say, is NOT larger than the relative velocity of the air wrt the tread ?

Therefore, I claim that the proper frame of reference for the cart and the tread is the reference frame of the air, or floor as the air and floor are relatively stationary to each other.

What do you mean by "the proper reference frame" ? All reference frames are "proper". That's what galilean relativity means.

Now, based on the usage of that reference, I can measure the speed of the cart relative to appoint on the floor in one direction and the relative speed of the tread measured from the same point in the opposite direction and compare them.

Yes. But that are not the velocities we are talking about of course.

It is without any question that the tread is moving much faster than the cart when the velocities are compared.

Yes. But these are not the velocities we are talking about. We are talking about the relative velocities of
A) The AIR wrt the surface on which the wheel runs (number A)
B) The CART wrt the surface on which the wheel runs (number B).

The claim is that the first number is smaller than the second (that A < B).

You, however, are saying that the velocity of the cart WRT the AIR is smaller than the velocity of the surface wrt the AIR (if you accept that the air is in no motion wrt to the ground).

Well, your first number is (B - A). Your second number is number A.

So you say that we didn't show that A cannot be smaller than B, simply because (B-A) < A.

So you are saying, in effect: from (B-A) < A, we have shown that we cannot have A < B.

Well, that's wrong of course. Take A = 3, B = 5. Yes, (B-A) = 5 - 3 = 2 < A = 3.
Nevertheless, contrary to your "conclusion", 3 < 5.

So you cannot derive that it is wrong that A < B simply because you show that (B-A) < A.

But this is reaching such levels of triviality that I really wonder whether you are trolling or not.

My conclusion from that is that when placed in the wind, the cart will also be moving much slower than the wind. You reject what I am saying.

Of course, because from your premise, your conclusion doesn't follow.

Ok. Now when the cart is running down wind, how do you measure its velocity? How do you determine how fast the cart is going? Do you measure the velocity of the cart relative to the wind or relative to the ground?

You measure with respect to the surface on which the wheel is running, in all these cases.

I would assume you would measure it relative to the floor Am I right? You would do this because if you measured it relative to the wind, you would get ridiculous answers such as it is moving with a negative velocity a sub wind speeds. And a zero velocity at wind speed and you would only get a positive velocity for the cart if it should ever go over the wind speed. Obviously these numbers are wrong, as you can see the cart moving with a positive velocity the second it starts moving. The reason the result is wrong is because you have chosen the wrong reference frame, the wind, when you should have chosen the ground.

I can't make any head or tails from these phrases.

You should have chosen the reference frame of the medium the cart is working into, not the medium that is doing work on the cart!

Galilean relativity tells the following (given that a 6 year old child can understand it according to you, and that I assume that you have been a 6 year old child, you understand this):
- I can make all of my calculations in any frame I like, the frame-independent results will come out all the same.
- relative velocities are frame-independent.

We have of course to calculate in all examples, the same quantities, which are:
- the relative speed of the cart wrt the surface on which the wheel runs
- the relative speed of the air wrt the surface on which the wheel runs

The claim is that the first is larger than the second in these examples.

In the case of the treadmill, the tread is doing the work on the cart and the cart is working into the air.

I explained already several times that what does work on what is dependent on what reference frame one has chosen, and is not a physical quantity by itself. In one frame, the floor is doing work on the cart, and in another frame, the cart is doing work on the floor. "Doing work" is a force (frame independent) times a displacement (frame-dependent).

And, BTW, you still didn't answer my post #378 with the train...

 

[zoobyshoe]

Therefore, I claim that the proper frame of reference for the cart and the tread is the reference frame of the air, or floor as the air and floor are relatively stationary to each other.

In so far as you are working on the side of Good and trying to squelch the notion this thing can work I applaud and support you, but I have to tell you I do not believe there is actually a "proper" frame. Any frame you both agree to analyze it from will give accurate results. The important thing is never to mix frames: don't start out in one frame then suddenly pull in information from another.

 

[cristo]

And, BTW, you still didn't answer my post #378 with the train...

Everyone seems to be ignoring this post, but I think it's quite an interesting post in this discussion. Personally, I'm not too keen on this step of your process of 'transforming' the experiment:

Ok, so now we have a flat train, running at 30 km/h through the fields, with a 100 m track on it, and a cart doing the test. Let us say that instead of having a straight track, we make a circular track and have the train run on a circle of say, 2 km diameter. This won't change much, so can you still accept the test with the flat train running on this track at 30 km/h ?

You're moving from an inertial frame to a non-inertial frame so it's not entirely clear, on first glance, that the two are equivalent. Of course, I've not read the whole thread, so apologies if this has been touched upon (and I've not done any calculations so don't know whether this will make all that much difference).

 

[vanesch]

Everyone seems to be ignoring this post, but I think it's quite an interesting post in this discussion. Personally, I'm not too keen on this step of your process of 'transforming' the experiment:



You're moving from an inertial frame to a non-inertial frame so it's not entirely clear, on first glance, that the two are equivalent. Of course, I've not read the whole thread, so apologies if this has been touched upon (and I've not done any calculations so don't know whether this will make all that much difference).

Whooo ! A real physicist here

You're right of course, this is the part which is disputable. However, the idea is that the effect of the rotation will be smaller when the radius of curvature is bigger. It was indeed the caveat I made in the beginning.

 

[A.T.]

Cart at wind speed means: blue fence(=air) is at rest, but brown fence(=ground) behind it is still moving left, from your perspective. The stick will still accelerate you to the right so you soon see both fences moving left, at different speed. From brown fence(=ground)-perspective you are then moving faster than the blue fence(=air) in the same direction. -> DWFTTW

Whoops!
How much mass does the ground have here?

Where 'here'?
Fence scenario: Both fences are very long and have a huge mass, compared to you+stick.
DWFTTW scenario: Both Earth & atmosphere have a huge mass compared to the cart.

 

[zoobyshoe]

Since DDWFTTW is possible without taking the "greatest advantage', the prop doesn't need to change configurations, as evidenced by the operating fixed pitch prop carts.

I know.

 

[schroder]

In so far as you are working on the side of Good and trying to squelch the notion this thing can work I applaud and support you, but I have to tell you I do not believe there is actually a "proper" frame. Any frame you both agree to analyze it from will give accurate results. The important thing is never to mix frames: don't start out in one frame then suddenly pull in information from another.

It is also important not to invert the way you apply the frames. When you are sitting in your car and the engine is driving the wheels to turn against the road and you want to know how fast your car is going. Do you measure the velocity between the center point of the engine and the centerline of the wheels? Or do you measure the velocity between the centerline of the wheels and a centerline drawn on the road? Of course you measure it between the wheels and the road. If you measured between the wheels and the engine it of course is always constant and you get a wrong result. When the cart is being driven by the wind as the engine and is running on the road, you measure the velocity between the wheel and the road, not between the wheels and the engine (the wind). When the cart is being driven on the tread as the engine and driving into the air with the propeller. You measure between the centerline of the propeller and a centerline in the air. You do not measure between the centerline of the wheels and the centerline of the engine (the tread). You must be consistent in both frames. By using the wheels and the tread to take the measurement on the treadmill you are inverting from the down wind frame and that is giving you the ridiculous result of faster than the tread hence faster than the wind. Be consistant in both frames. Do not invert things!

 

[zoobyshoe]

Where 'here'?
Fence scenario: Both fences are very long and have a huge mass, compared to you+stick.
DWFTTW scenario: Both Earth & atmosphere have a huge mass compared to the cart.

"Here" means in the situation we're discussing: we have a cart going, say, 10 mph over the ground. Wind speed is 0 with respect to the cart. Ground speed is 10 mph with respect to the cart.

What amount of mass should we plug into formula: K=1/2mv^2 to find out how much energy is available to the cart from the ground?

 

[schroder]

Everyone seems to be ignoring this post, but I think it's quite an interesting post in this discussion. Personally, I'm not too keen on this step of your process of 'transforming' the experiment:

I actually read that post and found one question is missing; the first one, which should be “Are you willing to accept that a cart being driven by the wind can go DDWFTTW?” My answer is NO and that makes all the other questions irrelevant! But without that first crucial question, I would have answered Yes to all. It is that first question which I refuse to accept because it is NOT happening on the treadmill once you remove the inversion in the frames which I just pointed out (again)

 

[zoobyshoe]

It is also important not to invert the way you apply the frames. When you are sitting in your car and the engine is driving the wheels to turn against the road and you want to know how fast your car is going. Do you measure the velocity between the center point of the engine and the centerline of the wheels? Or do you measure the velocity between the centerline of the wheels and a centerline drawn on the road? Of course you measure it between the wheels and the road. If you measured between the wheels and the engine it of course is always constant and you get a wrong result. When the cart is being driven by the wind as the engine and is running on the road, you measure the velocity between the wheel and the road, not between the wheels and the engine (the wind). When the cart is being driven on the tread as the engine and driving into the air with the propeller. You measure between the centerline of the propeller and a centerline in the air. You do not measure between the centerline of the wheels and the centerline of the engine (the tread). You must be consistent in both frames. By using the wheels and the tread to take the measurement on the treadmill you are inverting from the down wind frame and that is giving you the ridiculous result of faster than the tread hence faster than the wind. Be consistant in both frames. Do not invert things!

I look at the speedometer.

 

[cristo]

You're right of course, this is the part which is disputable. However, the idea is that the effect of the rotation will be smaller when the radius of curvature is bigger. It was indeed the caveat I made in the beginning.

I must have missed that caveat but, yes, I can accept that as a plausible approximation.

I actually read that post and found one question is missing; the first one, which should be “Are you willing to accept that a cart being driven by the wind can go DDWFTTW?” My answer is NO and that makes all the other questions irrelevant!

Actually, this question is not relevant. What vanesch is showing is that the experiment on a turntable is an equivalent experiment to what he calls the 'true outdoor test.' He's not asking whether or not you think the outdoor test gives a positive result: he's asking that if the outdoor test gives a positive result, then you would accept it as proof of DWFTTW. That must be true, no, since the 'true outdoor test' is the definition of DWFTTW? If you don't agree with this point, then the discussion might as well be over, since if you're not willing to accept the results of an experiment as proof, then you are no longer practicing science.

But without that first crucial question, I would have answered Yes to all.

Good. So you agree that the turntable test is equivalent the true outdoor test? Thus, since the OP shows a video of the turntable test giving a positive result, then you agree the true outdoor test will give a positive result, also? So, we're done!

 

[schroder]

I must have missed that caveat but, yes, I can accept that as a plausible approximation.






Good. So you agree that the turntable test is equivalent the true outdoor test? Thus, since the OP shows a video of the turntable test giving a positive result, then you agree the true outdoor test will give a positive result, also? So, we're done!

Nice Try! Sure they are equivalent and since the TT test clearly shows the cart is moving slower than the TT then I conclude it will also move slower in the wind!

 

[A.T.]

"Here" means in the situation we're discussing: we have a cart going, say, 10 mph over the ground. Wind speed is 0 with respect to the cart. Ground speed is 10 mph with respect to the cart.

What amount of mass should we plug into formula: K=1/2mv^2 to find out how much energy is available to the cart from the ground?

The energy is not available from the ground. It is available from the speed difference between ground and air. The energy received by the cart is frame dependent. It is useless to examine if the cart could accelerate further. You could easilly pick a reference frame where the cart is losing energy by going faster relative to the ground.

What matters for acceleration are forces acting on the cart: Due to transmission, the propeller-thrust-force is higher than the wheel-force needed to drive the propeller, so the cart accelerates further.

 

[zoobyshoe]

The energy is not available from the ground. It is available from the speed difference between ground and air.

You are quite right that the energy is harvested from the difference in the relative velocity of the ground and air! The ground speed and difference between air and ground speed just happen to be exactly the same thing at this point, because the cart and wind are going the same speed in the same direction.

That being the case, what number for mass should we plug into K=1/2mv^2? The ground speed is 10mph, what amount of mass does it represent with respect to the cart?

 

[vanesch]

Nice Try! Sure they are equivalent and since the TT test clearly shows the cart is moving slower than the TT then I conclude it will also move slower in the wind!

You are claiming that the velocity of the cart *wrt the turntable* is smaller than the velocity of the *air* wrt to the turntable ? Or are you claiming that the velocity of the cart *wrt the ground* is smaller (and opposite) to the velocity of the turntable (wrt the ground) ?

Because the equivalence we are after is the FIRST, not the SECOND set of velocities.

This is so terribly trivial that I think you're trolling, honestly.

 

[A.T.]

You are quite right that the energy is harvested from the difference in the relative velocity of the ground and air! The ground speed and difference between air and ground speed just happen to be exactly the same thing at this point, because the cart and wind are going the same speed in the same direction.

That being the case, what number for mass should we plug into K=1/2mv^2? The ground speed is 10mph, what amount of mass does it represent with respect to the cart?

What exactly is K and how is it relevant to examine if the cart could accelerate further? Again: What matters for acceleration are forces, not kinetic energies calculated in some arbitrary reference frame. When you have computed the acceleration of the cart using forces, you can use the formula above to compute the gain in KE of the cart. Then you plug the cart's mass into the formula.

 

[schroder]

You are claiming that the velocity of the cart *wrt the turntable* is smaller than the velocity of the *air* wrt to the turntable ? Or are you claiming that the velocity of the cart *wrt the ground* is smaller (and opposite) to the velocity of the turntable (wrt the ground) ?

Because the equivalence we are after is the FIRST, not the SECOND set of velocities.

This is so terribly trivial that I think you're trolling, honestly.

I don’t know what trolling is. I am seriously trying to establish the truth here about this cart.

If you have two cars passing each other on a two way road and all you wanted to know is the relative velocity of the cars, relative to each other, you could pick either car to be the reference for the frame and get the correct result. But suppose two cars are passing each other and you wanted to know the difference between their velocities. Picking either car as your reference will always give you the sum. To get the difference you need to pick some other reference, in this case the road. Now you can measure the velocity of each car independently with respect to the road and compare them to get the difference. In the down wind frame, you want to know the velocity of the cart with respect to the road, not the wind, so you pick the road as your reference frame. Even though there is also velocity between the wind and the cart in that frame, you ignore this in calculating your velocity because it is obviously not of interest to you. You could pick the wind as your reference but your numbers would not make sense unless you applied a correction factor to convert them into road velocity which is what you want to know. That one is obvious. In the treadmill test, it is not so obvious and you are picking the tread because you see the cart is moving wrt the tread. It is a little harder to see it moving wrt the wind, but that is what you really should be interested in! All you really need to do is look at the nature of the beast. On the tread, it is using wheels for input from the tread, and a propeller for output in the air. It is working into the air and it is moving with respect to that medium. Sure, there is motion between the cart and the tread but that is about as interesting as the motion between the cart and the wind in the wind frame. That is the motion between the cart and what is driving it. That is not of interest. You want to know how fast the cart is going in the medium it is driving into, not the medium which is driving it. You really want the velocity into the wind that is what the cart is driving into. I know this is a bit hard to see but you need to clarify this in your own mind. Once you adopt the wind or floor as the reference frame in the treadmill test, you get believable numbers, but unfortunately they are much less than the tread velocity. But that is exactly in accordance with any machine which has losses between input and output. If you use the tread as your reference you get these fantastic numbers which are greater than tread velocity. Very exciting but not true! I will accept the numbers that make sense even if they are not very impressive. Nature has designed the most efficient thing that will go as fast as possible in the wind, it is a molecule of air! And all it can do is wind velocity. I don’t think we can improve on that by using technology which has been around since ancient times. If you still want to believe in this go ahead, but it will be short lived once someone races tumbleweed against one of these carts. Then you will realize you should have been measuring the velocity as I have told you.

 

[zoobyshoe]

What exactly is K and how is it relevant to examine if the cart could accelerate further? Again: What matters for acceleration are forces, not kinetic energies calculated in some arbitrary reference frame. When you have computed the acceleration of the cart using forces, you can use the formula above to compute the gain in KE of the cart. Then you plug the cart's mass into the formula.

Sorry, that wasn't rigorously expressed. Should have put E_k=1/2mv^2.

As Vanesch will confirm you have to be in a frame, and any frame you pick is OK as long as you stay with that frame. Thus far I have been speaking about the cart's frame. We can analyze from another frame if you want, but whatever frame we pick we have to stick to that frame.

The reason I am asking about energy is because earlier I asserted that the cart is in a situation where it no longer has any energy available to it outside the energy represented by it's own momentum. The cart has a certain momentum, which is its mass times its velocity. We can determine its kinetic energy from the same things, therefore, it's momentum represents a certain amount of energy.

You have asserted that the speed difference between the ground and air represents "plenty" of energy. If that is the case, then we ought to be able to get a ballpark figure pretty easily by determining what mass should be plugged into the formula.

 

[vanesch]

I don’t know what trolling is. I am seriously trying to establish the truth here about this cart.

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

If you have two cars passing each other on a two way road and all you wanted to know is the relative velocity of the cars, relative to each other, you could pick either car to be the reference for the frame and get the correct result.

Yes. You could also take the difference of the velocities, as measured in the ground frame. In fact, what is important is that no matter in what frame you measure the velocities of car 1 and car 2, (and you'll agree with me that v1 and v2 will be dependent on the choice of that frame), the DIFFERENCE, that is v = v1 - v2 will always give the same result, no matter in what frame one calculates it. I guess you agree with that ?

But suppose two cars are passing each other and you wanted to know the difference between their velocities. Picking either car as your reference will always give you the sum.

Are you still talking about the same setup ? Because the relative velocity IS of course exactly this difference.

To get the difference you need to pick some other reference, in this case the road. Now you can measure the velocity of each car independently with respect to the road and compare them to get the difference.

Sorry, I'm lost as what you are trying to say.

Let's say the two cars are driving north on a north-south road, right ? Car A is driving 50 km/h, and car B is driving driving 60 km/h, wrt to the road of course. Well, the relative velocity is of course 60 - 50 = 10 km/h.
But if you measure car B's velocity in car A's frame (with a radar or something), well, you have that car B's velocity is 10 km/h and car A's velocity is 0 (of course), so again we have 10 - 0 = 10 km/h.

You say something different or not ?

Now, imagine that there's also a train track along the road. Train C runs north with a velocity of 100 km/h. What you see is that car A is running at -50 km/h and car B is running at -40 km/h. (in other words, from the train's PoV, the cars have a southward velocity).

Again, the relative velocity is found by: -40 - (-50) = +10 km/h.

Another train is running south on the track at 160 km/h. As seen from the train, car A is speeding north at a velocity of 210 km/h, and car B is speeding north at a velocity of 220 km/h.

But again, the relative velocity: 220 - 210 = 10 km/h.

The relative velocity is the same, no matter in what reference frame one has calculated it. So it is a frame-independent quantity.

You agree with that or not ?

In the down wind frame, you want to know the velocity of the cart with respect to the road, not the wind, so you pick the road as your reference frame.

Yes.

Even though there is also velocity between the wind and the cart in that frame, you ignore this in calculating your velocity because it is obviously not of interest to you. You could pick the wind as your reference but your numbers would not make sense unless you applied a correction factor to convert them into road velocity which is what you want to know.

Huh ?

No matter in what frame I have my velocities (which are of course frame-dependent), any DIFFERENCE gives you frame-independent relative velocities. There are no "correction factors" to be applied.

That one is obvious. In the treadmill test, it is not so obvious and you are picking the tread because you see the cart is moving wrt the tread. It is a little harder to see it moving wrt the wind, but that is what you really should be interested in!

The motion wrt the air is extremely simple in the treadmill test: it is the same as the motion wrt the ground, as the air is still wrt the ground.

All you really need to do is look at the nature of the beast. On the tread, it is using wheels for input from the tread, and a propeller for output in the air. It is working into the air and it is moving with respect to that medium. Sure, there is motion between the cart and the tread but that is about as interesting as the motion between the cart and the wind in the wind frame. That is the motion between the cart and what is driving it. That is not of interest. You want to know how fast the cart is going in the medium it is driving into, not the medium which is driving it.

Tell me, say there is a wind of 50 km/h and if a sailing cart is going at 40 km/h and a car is driving at 40 km/h in the same direction, what is their relative velocity ? Are you going to say that we shouldn't measure the velocity of the sailing cart wrt the ground because the "driving medium" is the air ?? Is the cart going to take over the car or vice versa ?

You really want the velocity into the wind that is what the cart is driving into. I know this is a bit hard to see but you need to clarify this in your own mind. Once you adopt the wind or floor as the reference frame in the treadmill test, you get believable numbers, but unfortunately they are much less than the tread velocity.

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.

Imagine a long ship. Do your test on the ship. Are you going to compare to the ship, or to the water ?

But that is exactly in accordance with any machine which has losses between input and output. If you use the tread as your reference you get these fantastic numbers which are greater than tread velocity.

Sure, and that's what this is about here. Let us consider another case: the "driving of a brick on the road".

Claim: a brick dropped on the road will go at 30 km/h. Proof (according to you): Drop the brick on a treadmill. Of course the brick sticks to the treadmill after some bouncing. You have to look upon its velocity wrt to the ground. It is going at 30 km/h. Hence, proof: the brick is doing 30 km/h when dropped on a road.

Why should I use the reference of the GROUND when I use a treadmill ? In *this* case, you'd be willing to accept that we have to express the velocities in the frame of the treadmill, right ? Now, the velocity of the treadmill is 30 km/h, and the brick is also going at 30 km/h (both measured in the ground frame), so the relative velocity of the brick wrt the treadmill is 0 (30 - 30).

So this experiment on the treadmill shows that a brick we would drop on an outdoor road would not move (wrt to the road). But in order to deduce this on the treadmill we had to work in the frame of the treadmill.

With the cart, it is the same. We work wrt to the treadmill (which will have all velocities wrt to the road in the outdoor test). And wrt the treadmill, the cart is going at 40 km/h (it is doing - 10 km/h in the ground frame, and hence the relative velocity wrt to the treadmill is -10 - 30 km/h = - 40 km/h - like our brick was doing 0 km/h).

In the same way, the air is doing 0 km/h in the ground frame, and doing -30 km/h in the treadmill frame (0 - 30 = -30). So in the treadmill frame, the cart is going faster than the wind (-40 versus -30). As in this experiment, the treadmill frame is the equivalent of the "road" frame (think of the brick!), we have established an equivalent experiment that shows that the cart is going at 40 km/h when the wind is going at 30 km/h, if ever we see the cart move at 10 km/h versus the floor in the opposite direction as a treadmill that runs at 30 km/h.

 

[A.T.]

As Vanesch will confirm you have to be in a frame, and any frame you pick is OK as long as you stay with that frame. Thus far I have been speaking about the cart's frame. We can analyze from another frame if you want, but whatever frame we pick we have to stick to that frame.

The cart's frame is not inertial if the cart's speed varies. So it is easier to consider ground's frame.

The reason I am asking about energy is because earlier I asserted that the cart is in a situation where it no longer has any energy available to it outside the energy represented by it's own momentum.

What do you mean by "the energy the cart has available"? The carts current kinetic energy?

You have asserted that the speed difference between the ground and air represents "plenty" of energy. If that is the case, then we ought to be able to get a ballpark figure pretty easily by determining what mass should be plugged into the formula.

To compute what exactlly? The total kinetic energy in the entire atmosphere? We don't need that much energy. And this quantity is completely irrelevant to the question, which is:

Can the cart accelerate further, when already traveling at wind speed?

acceleration_cart = (force_propeller - wheel force) / mass_cart
therefore:
if force_propeller > wheel force then acceleration_cart > 0 and the transmission makes sure that force_propeller > wheel force
therefore:

Yes, the cart can accelerate further, when already traveling at wind speed!

Once you have the acceleration_cart you can compute the cart's energy consumption, pay your energy bill to the planet and sleep well.

 

[swerdna]

That is something the torpedo can do too. It is just not the standard operating scenario.

Why do you need two cables? The torpedo needs two for steering, but for a minimal design I would recommend using just one.

Wouldn’t that require the torpedo to be used in a river or at least strong sea current?

Yes one cable would be fine. Not even sure why I drew two. Perhaps because the torpedo has two or some intuitive balance maybe.

 

[A.T.]

Wouldn’t that require the torpedo to be used in a river or at least strong sea current?

Yes, I just wanted to point out, that it is exactly the same mechanism.

Yes one cable would be fine. Not even sure why I drew two. Perhaps because the torpedo has two or some intuitive balance maybe.

It could be messy with two cables potentially twisting around each other. One cable coming out at the center rear of the boat scould stabilize it just fine.

And don't forget to attach the end of the cable at the reel, so you can recover your boat.

 

[swerdna]

Yes, I just wanted to point out, that it is exactly the same mechanism.

It could be messy with two cables potentially twisting around each other. One cable coming out at the center rear of the boat scould stabilize it just fine.

Sure I didn’t mean to present it as anything substantially new except instead of moving cables pulling against stationary water, moving water (or air) is pulling against stationary cables.

And don't forget to attach the end of the cable at the reel, so you can recover your boat.

DOH!

 

[rcgldr]

That being the case, what number for mass?

The number for mass is the mass of the air affected by the prop. The force is equal to that the mass of the affected air times the affected air's rate of acceleration (or the integral sum of all the components of affected air).

The carts are designed so when moving downwind at the speed of the wind, the forward force of the air onto the prop is greater than opposing backwards force of the ground onto the driving wheels (plus the backwards force from the ground related to rolling resistance).

The ground can be considered to have a huge amount of mass, any work done on the "ground" results in only a tiny change in velocity.

 

[rcgldr]

Your claim is that on the treadmill, the velocity of the cart is measured wrt to moving tread. You are using the frame of the cart to come this conclusion and what you have is relative velocity between the cart and the tread, and that relative velocity is faster than the tread.

That's not what is claimed. There are no claims about cart speed versus tread speed, only cart speed versus wind speed, or better stated that cart speed relative to the tread is greater than wind speed relative to the tread. I've already stated this in a mathematical form that is independent of frame of reference, and you have yet to respond to this. I'll repeat the claim again:

a DDWFTTW can acheive

|(cart speed) - (ground speed)| > |(wind speed) - (ground speed)|

for some range of wind speed:

(minimum wind speed) <= |(wind speed) - (ground speed)| <= (maximum wind speed)

 

[zoobyshoe]

The number for mass is the mass of the air affected by the prop. The force is equal to that the mass of the affected air times the affected air's rate of acceleration (or the integral sum of all the components of affected air).

Rodger that. The only air that matters is that which directly impinges on the cart, or upon which the cart impinges. Vanesch referred to this as an air column or air section.

The carts are designed so when moving downwind at the speed of the wind, the forward force of the air onto the prop is greater than opposing backwards force of the ground onto the driving wheels (plus the backwards force from the ground related to rolling resistance).

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 ground can be considered to have a huge amount of mass, any work done on the "ground" results in only a tiny change in velocity.

Rodger that. The point I was trying to make is that if you're coasting over the ground in the carts frame looking at the ground as a source of power, that huge amount of mass going by at 10 mph isn't the ginormous power source it looks like at all. The power you could obtain from it by virtue of relative motion is only equal to the power it could obtain from you. Likewise, the force it could exert on you is only equal to the force you could exert on it.

 

[zoobyshoe]

Can the cart accelerate further, when already traveling at wind speed?

acceleration_cart = (force_propeller - wheel force) / mass_cart
therefore:
if force_propeller > wheel force then acceleration_cart > 0 and the transmission makes sure that force_propeller > wheel force
therefore:

Yes, the cart can accelerate further, when already traveling at wind speed!

Unfortunately for the cart when it reaches wind = 0 it becomes inert. There is no force from the wind or ground acting on it other than friction.

The fact the gearing insures that the force exerted on the propeller by the gear train is greater than the force applied at the wheels doesn't insure that the thrust the prop creates is greater than the cart's inertia.

On top of the inertia of the mass as a whole there is the inertia of the gears, wheels, and propeller itself to overcome. The cart will not change speeds unless force is applied and with no force applied to it from the wind or anything else it has no reason to accelerate. The prop won't spin faster unless the cart goes faster and the cart won't go faster unless the prop spins faster.

If the cart and propeller have gathered enough momentum during the downwind acceleration it may continue into the headwind and "coast" at a speed faster than the down wind speed for some length of time, but this is stored energy.

 

[uart]

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

Here's one train of thought that might help convince that the "treadmill in still air" situation is a least feasible.

Imagine for a moment that the rotating a propeller was replaced by a long “cork screw” (or similar). That is, the wheels were coupled through a suitable drive train to turn this long “cork screw”.

Now imagine that the corkscrew is started into a large block of soft foam (representing the air) attached to the front of the treadmill. If the treadmill is run then the wheels turn and the corkscrew turns and the vehicle will move forward as the corkscrew screws into the foam.

Think of the propeller in a similar way as "screwing" itself forward into the stationary air.

Here’s another little thought experiment that might help convince you. In the spirit of de-bunking perpetual motion devices you can usually assume frictionless ideal operation of most components and of course they still fail to achieve "over unity" operation. So in this spirit let's assume that we have an ideal lossless drive train (wheels, belts and gearing) and further that we can adjust the gearing ratio from the wheels to propeller to any desired ratio. Let's just concentrate on non-ideal lift/drag of the propeller.

First we note that the turning of the wheels is driving the prop, so the inevitable blade drag will mean we require constant torque to keep the prop turning at a constant rate and this torque must be provided by the wheels, giving a retarding force on the vehicle.

Second we note that the lift generated by the prop is providing a forward directed force. So we now have two forces in opposition, the lift on the prop giving a forward force and the rotational drag on the prop which, through the drive train, ultimately results in a retarding force at the wheels.

Since at first sight this thing looks like an “over unity” device our first instinct is to think that perhaps the lift force must be less than the retarding force. However the retarding force at the wheels is dependant on the gear ratio, that is, if we gear it so that the prop turns fewer times for each rev of the wheels then the ratio of retarding force at the wheels to blade drag to is also reduced.

So let's play devils advocate and assume that we set this thing up on a treadmill and hold it until it’s at steady state (wheels and prop up to speed) and we find that the retarding force is indeed larger than the propeller lift and our vehicle goes backwards.

No problems, let's just reduce the gear ratio so that the prop turns less times per wheel rev, and this will reduce the retarding force at the wheels. Arh but you say, this will also reduce the prop speed and so reduce it’s lift. Again no problems, just increase the treadmill speed until the prop turns at the same speed as it did before! Now you can't argue with this, the prop is at the same speed so the lift is identical to before, and the drag at the blades (torque required to spin the prop) is also the same as before, but due to the modified gearing the retarding force at the wheels is now lower than before. Can you see that in principle there is no limit to how much we repeat this procedure so eventually it has to work!