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

[vanesch]

Nothing can possibly surprise me more than your refuasl to accept the facts when you see tham in front of your eyes. I have nothing more to say here. Have a nice day!

Well, you have shown to have some strange understandings of basic Newtonian mechanics, to say the least. I still insist on you answering (y/n) my post on the train business (post 378 in this thread). It would indicate me exactly where your (first) misunderstanding resides.

 

[zoobyshoe]

The fences have plenty of energy, just like the wind. They can accelerate any mass in the opposite direction of their movement, if the stick and the fences are strong enough.

Then the stick and fences analogy does not represent the downwind cart situation. In the downwind cart there comes a point where there is no longer plenty of energy.

First you have to remind yourself that "speed" is not, by itself, the whole picture. There is no such thing as "pure speed". Speed represents energy only so long as we ascribe that speed to something with mass. k = 1/2mv^2 requires that we know the mass, not just its speed.

Then, to analyze the cart's situation you have to account for the point where it reaches downwind speed, and the wind contribution plummets to 0. A cart going at the same speed as the wind sees the wind as standing still.

Here's how I put it eariler:

"If we keep our focus on this vehicle as one which harvests power from relative motion of surrounding media then we have also to contend with the fact that the relative motion of the surrounding media has two separate and distinct configurations to deal with. In the first the cart sees the wind as a tailwind and the ground as a head wind (provided the cart's in motion at all.) In the second the cart sees both the wind and the ground as headwinds. These two headwinds have energy to harvest, in principle, by virtue of the fact they are moving relative to each other at different speeds (according to MGrandin's description of the situation). However, they are still both headwinds, and will require an engineering solution specific to that situation, which must be different than the solution to the tailwind verses headwind situation.

I think it would be very useful to begin analyzing each design according to how it solves for two separate situations 1.) the Tailwind-Headwind (TH) and 2.) the Headwind-Headwind (HH). This would greatly clarify design strategies and discussions.

Personally I think that designing for the HH situation is the critical problem to solve. I would design for that and let the TH solution be direct blowing (DB) i.e. accept and allow for the startup to consist of the wind simply blowing the whole thing physically downwind. Otherwise you have to design transmissions or rotors that change configuration somehow."

 

[rcgldr]

Before it was ice boats, now it is torpedos.

Ice boats operate on a different principle, but the Brennan torpedo operates on the same principle as DWFTTW cart.

It consists of a single axle with four wheels. Two wheels will sit on the tread, and extract power from the tread. Two larger diameter wheels sit on the floor and do work on the floor. If you do actually build this be careful as it will take off like a bat out of hell in the opposite direction

It would move backwards, in the same direction as the treadmill. It's the equivalent of pulling a string on a yo-yo as the string winds onto the yo-yo's axle from below the axle. The advance ratio is > 0, < 1, and the speed of the single axle 4 wheel device on the tread would be tread speed / (1 - ar), where ar = advance ratio = (smaller wheel diameter) / (larger wheel diameter). The wiki Brennan torpedo link refers to a similar example: Brennan was inspired to create his torpedo's unique propulsion system in 1874, when he noticed that a cotton reel, if the thread is pulled toward the operator from underneath, moves forward rather than backward.

http://en.wikipedia.org/wiki/Brennan_Torpedo

no outdoor test which can demonstrate or has demonstrated, DDWFTTW with this cart or any other cart

Jack Goodman's video is of a outdoor test:

http://www.youtube.com/watch?v=aJpdWHFqHm0&fmt=18

 

[vanesch]

Then the stick and fences analogy does not represent the downwind cart situation. In the downwind cart there comes a point where there is no longer plenty of energy.

First you have to remind yourself that "speed" is not, by itself, the whole picture. There is no such thing as "pure speed". Speed represents energy only so long as we ascribe that speed to something with mass. k = 1/2mv^2 requires that we know the mass, not just its speed.

Yes, and on top of that, it depends in which frame you describe it, because "v" depends on it.

It is true that when the cart has wind velocity, the wind is (of course) at rest wrt the cart. But now the floor isn't. In this reference frame, the floor is "flowing" in the same way as a river is flowing wrt a watermill. You can tap into that.

The error resides (as I said already a few times) in thinking that it is "the wind which brings in energy, and not the steady floor" ; which is a correct, but frame-dependent statement. It is correct in the frame of the floor.

Imagine, out there in space, an iron windtunnel, a cart with magnetized wheels (to stick to the surface - there's no gravity!). Then it is not so simple to say that it is only the wind that "can provide for energy". The energy comes in fact from the velocity difference between the floor and the air, but how much you attribute to each is dependent on the frame in which you are working. In the frame of the floor, yes, your only source of energy is the moving wind. But in the frame of the wind, the only source is the moving floor. And in another frame, it depends.

Then, to analyze the cart's situation you have to account for the point where it reaches downwind speed, and the wind contribution plummets to 0. A cart going at the same speed as the wind sees the wind as standing still.

Yup, but in that frame, you have a moving surface from which you can tap.

"If we keep our focus on this vehicle as one which harvests power from relative motion of surrounding media then we have also to contend with the fact that the relative motion of the surrounding media has two separate and distinct configurations to deal with. In the first the cart sees the wind as a tailwind and the ground as a head wind (provided the cart's in motion at all.) In the second the cart sees both the wind and the ground as headwinds. These two headwinds have energy to harvest, in principle, by virtue of the fact they are moving relative to each other at different speeds (according to MGrandin's description of the situation). However, they are still both headwinds, and will require an engineering solution specific to that situation, which must be different than the solution to the tailwind verses headwind situation.

I see what you mean. You would like to say that somehow one needs to "change gears" exactly at the point of overtaking the wind. Probably this comes about because the faster the cart is moving wrt the ground, the larger the force that is applied by the propeller in the forward direction. In other words, the propeller doesn't have a fixed gearing ratio wrt displacements or velocities (as in the case of a ruler), but only as a force.

So at low velocities (wrt to the floor) the propeller "doesn't do much" and acts at best as a sail. The larger the velocity, the harder it starts to turn, and the larger the force is it exerts on the air (by having a larger and larger momentum transfer).

 

[rcgldr]

This cart clearly has two wheels which sit on the tread and extract energy from the tread. It also has two more wheels which sit on the tread and do basically nothing but balance the weight as a symmetrical object so it does not fall over. It also has a propeller which interfaces with THE AIR! As the tread does work on this cart, the cart does work on the AIR. It moves in the opposite direction to the tread, but it does NO work against the tread!

The cart driving wheels exerts a forward force against the tread. This force is exerted over a distance on the tread over a period of time, and the work done in this period of time is force times distance. The motor driving the treadmill at a specific speed has to use more power to drive the treadmill with the cart on the treadmill, than it does to drive the treadmill without the cart on the treadmill.

If you want to measure the velocity of this infamous little cart, you could try to mark a spot in the air and measure against that.

We could mark a spot in the air with a bubble, indoors or outdoors.

the little cart is moving rather slowly with reference to the floor, much more slowly than the tread which is driving it.

The goal is to achieve speed faster than wind, not the tread. Restated the goal of a DDWFTTW cart is to achieve:

|cart speed - ground speed| > |wind speed - ground speed|

Note that this inequality is independent of frame of reference.

 

[rcgldr]

Then, to analyze the cart's situation you have to account for the point where it reaches downwind speed, and the wind contribution plummets to 0.

The cart's prop generates upwind thrust. The wind's contribution is to apply a forwards force onto the air acclerated upwind by the propeller. The affect air transfers the forward thrust from the wind onto the prop. When the cart is as wind speed, the thrust from the propeller is still well below wind speed, so the wind has the air from the prop wash to interact with.

TH versus HH

Spork and swerdna's carts are self starting (T0 situation) with a fixed pitch prop and achieve HH speed, so they are good enough. A variable pitch prop would allow for a better acceleration rate (with a low or negative pitcth) while in TH mode, and then switch increasing positive pitch for HH mode to maximize speed. Since the goal here is any speed above wind speed as opposed to maximum speed, the fixed pitch prop carts are good enough.

 

[zoobyshoe]

If you think that theory is more robust evidence than an actual, practical physical demonstration then I respectfully suggest that perhaps you should get out into the real world more often.

As I said earlier, stored energy was my main concern before I did testing for over 10 minutes on my turntable 10 minutes is a lot longer than the one or two you suggest. I also did a turntable test to show how the kinetic energy of a marble lasts a long time on a turntable - http://nz.youtube.com/watch?v=kc88SrMG5fA

I repeat A.T’s question - “How did the Brennan Torpedo maintain itself in this situation for 2000 yards? Stored momentum? Under water?”

And ask you the same as I asked Schroder - “If myself or someone else built a cart and provided a video of it clearly beating bubbles floating in an outside wind would you accept this as proof? If not, what you accept as proof?”

If it does not work on paper, but works in real life then two possible situations might exist:

1.) There was a problem with the math. The wrong formula was used, someone put the wrong number into the formula.
or:
2.) The demonstrations aren't what they look like.

None of you has disputed my formula, because its the formula you taught me to apply. But when that leads to the realization that the cart comes to a point in the directly downwind scenario where it has only its stored energy, you all abandon the need for it to work on paper and take refuge in demonstrations that "seem" to work.

I am more inclined to suppose those demonstrations aren't what they look like.

If you think that theory is more robust evidence than an actual, practical physical demonstration then I respectfully suggest that perhaps you should get out into the real world more often.

It often happens that things are claimed to be operating by a certain interaction of forces, when it turns out they are operating by a different interaction of forces. Demonstrations demonstrate something, yes, but it isn't always what has been claimed they demonstrate. Crooke's Radiometer:

http://en.wikipedia.org/wiki/Crookes_radiometer

Called the "Light Mill". Reputed to work by pressure exerted directly on the vanes by photons:

They come in various forms, such as the one pictured, and are often used in science museums to illustrate "radiation pressure" – a scientific principle that they do not in fact demonstrate.

 

[zoobyshoe]

The cart's prop generates upwind thrust. The wind's contribution is to apply a forwards force onto the air acclerated upwind by the propeller. The affect air transfers the forward thrust from the wind onto the prop. When the cart is as wind speed, the thrust from the propeller is still well below wind speed, so the wind has the air from the prop wash to interact with.

The cart is running off of energy stored in the cart's system here.

Spork and swerdna's carts are self starting (T0 situation) with a fixed pitch prop and achieve HH speed, so they are good enough. A variable pitch prop would allow for a better acceleration rate (with a low or negative pitcth) while in TH mode, and then switch increasing positive pitch for HH mode to maximize speed. Since the goal here is any speed above wind speed as opposed to maximum speed, the fixed pitch prop carts are good enough.

I am persuaded there's an excellent possibility the cart will make it into the HH situation under these circumstances, but only on stored energy. It cannot sustain this. Once it expends enough of its stored energy it will slow back down to where energy is being added by the tailwind.

 

[schroder]

Well, you have shown to have some strange understandings of basic Newtonian mechanics, to say the least. I still insist on you answering (y/n) my post on the train business (post 378 in this thread). It would indicate me exactly where your (first) misunderstanding resides.

At this point in the discussion, your post #378 is irrelevant. It is clear to me that you are religiously clinging to a theory of DWFTTW which I have disproven using the Fundamentals of Physics. If a non-staff member of this forum engaged in this practice, he would be banned as a crackpot. I do not accuse you of crack pottery, but by clinging to a belief when the fundamentals of physics have been shown to go overwhelmingly against that belief, is just unacceptable. I am sorry this has come down to a personal duel. My only objective here is the truth of the matter. You cannot reasonably claim the cart is working against the tread without also claiming it is over unity because it is getting all of its energy from the tread. That is an inescapable fact which you refuse to acknowledge. Once you acknowledge that the cart is in fact working against the air, you must also acknowledge that it is moving in the reference frame of the air. And once you do that, you see that the velocity is measured wrt the air (or floor which is easier) and certainly not wrt the tread. Then it is clear it is NOT moving faster than the tread and by the proper application of Galilean relativity it is not moving faster than the wind. This entire matter could have been cleared up by a simple outdoor test with the most primitive of equipment but no one sees fit to do it because the outcome will destroy this myth which at this point is a hoax. Release the cart downwind and after a head start release one of Nature’s own tumbleweeds and see what happens! The tumbleweed will pass that cart faster than you can say “Faster than the Wind”. And, tumbleweeds do Not exceed wind velocity. I do not wish to damage your reputation or your ego, but you are wrong about this and I just wonder if you will be man enough to admit it. If I am wrong I would expect to be thrown off this forum as a crackpot. But if I am right, and you ban me for being right, that would be a very serious crack in the prestige of this forum.

 

[vanesch]

At this point in the discussion, your post #378 is irrelevant.

It isn't. But even if it were, why do you refuse to consider it ?

It is clear to me that you are religiously clinging to a theory of DWFTTW which I have disproven using the Fundamentals of Physics.

You have not demonstrated any thing of the kind. You have stated a large number of erroneous statements concerning basic classical mechanics, reference frames and things as simple as relative velocity. As I said already a few times before, without at least a sound basic understanding of these concepts, and the principle of galilean relativity, it is indeed impossible to understand the relevance of the demonstration of the treadmill or turntable experiments, and without its understanding, it is of course also impossible to accept it as a proof of the matter.

There could have been valid ways to attack these demonstrations. For instance on the turntable, there is the fact that the reference frame is not 100% inertial, and that there is also an extra force on the cart due to the arm. You could have considered these things. On the treadmill experiment, you could argue that steady state has not been reached, or something else. All these would be interesting points. But you miss the basic point, which is a transformation of coordinates.

If a non-staff member of this forum engaged in this practice, he would be banned as a crackpot. I do not accuse you of crack pottery, but by clinging to a belief when the fundamentals of physics have been shown to go overwhelmingly against that belief, is just unacceptable.

Except that no such thing is happening, and that what you consider as a proof against it is an incoherent babble full of elementary mistakes. It is not because I write that 100 - 200 is greater than zero that I have demonstrated in any way that negative numbers don't exist. Your "proof" amounts logically to a similar set of statements.

I am sorry this has come down to a personal duel. My only objective here is the truth of the matter. You cannot reasonably claim the cart is working against the tread without also claiming it is over unity because it is getting all of its energy from the tread.

I explained already at least 5 times why this reasoning is false (once again, because "getting energy from" is a frame-dependent concept).

That is an inescapable fact which you refuse to acknowledge. Once you acknowledge that the cart is in fact working against the air, you must also acknowledge that it is moving in the reference frame of the air. And once you do that, you see that the velocity is measured wrt the air (or floor which is easier) and certainly not wrt the tread. Then it is clear it is NOT moving faster than the tread and by the proper application of Galilean relativity it is not moving faster than the wind.

This is the kind of nonsensical derivation I pointed to, which you consider as "proof".

This entire matter could have been cleared up by a simple outdoor test with the most primitive of equipment but no one sees fit to do it because the outcome will destroy this myth which at this point is a hoax. Release the cart downwind and after a head start release one of Nature’s own tumbleweeds and see what happens!

Sure, but that's not the matter. If this happens, there is a serious problem in the application of galilean relativity. It would be almost as surprising as finding out that the test works in the Mid-West, but not in London, because of the distance to Big Ben.

If I am wrong I would expect to be thrown off this forum as a crackpot. But if I am right, and you ban me for being right, that would be a very serious crack in the prestige of this forum.

It is not a matter of crackpottery, it is a matter of not understanding the basic concepts of Newtonian mechanics. I'm not interested in DDWFTTW per se, I'm interested in the nice demonstration this is using transformations of frames in Galilean relativity. It could very well be that it is not possible for some or other technical reason to make a DDWFTTW cart, but given the demonstrations, the practical proof is there if you understand how frame transformations work. It is also true that DWFTTW doesn't violate any basic principle, such as "over unity" or the conservation of energy of momentum - that's easily established too.

 

[zoobyshoe]

Yes, and on top of that, it depends in which frame you describe it, because "v" depends on it.

It is true that when the cart has wind velocity, the wind is (of course) at rest wrt the cart. But now the floor isn't. In this reference frame, the floor is "flowing" in the same way as a river is flowing wrt a watermill. You can tap into that.

The error resides (as I said already a few times) in thinking that it is "the wind which brings in energy, and not the steady floor" ; which is a correct, but frame-dependent statement. It is correct in the frame of the floor.

Imagine, out there in space, an iron windtunnel, a cart with magnetized wheels (to stick to the surface - there's no gravity!). Then it is not so simple to say that it is only the wind that "can provide for energy". The energy comes in fact from the velocity difference between the floor and the air, but how much you attribute to each is dependent on the frame in which you are working. In the frame of the floor, yes, your only source of energy is the moving wind. But in the frame of the wind, the only source is the moving floor. And in another frame, it depends.

Yes, I completely agree with this. It is essential to know what frame you are analyzing.

Yup, but in that frame, you have a moving surface from which you can tap.

Indeed you do! Just because the wind velocity is now 0 and is not contributing does not mean the cart is without an energy source, because in the cart's frame, the ground is moving! The ground is moving backward underneath the cart, and represents an energy source.

I see what you mean. You would like to say that somehow one needs to "change gears" exactly at the point of overtaking the wind. Probably this comes about because the faster the cart is moving wrt the ground, the larger the force that is applied by the propeller in the forward direction. In other words, the propeller doesn't have a fixed gearing ratio wrt displacements or velocities (as in the case of a ruler), but only as a force.

So at low velocities (wrt to the floor) the propeller "doesn't do much" and acts at best as a sail. The larger the velocity, the harder it starts to turn, and the larger the force is it exerts on the air (by having a larger and larger momentum transfer).

I think you are following my reasoning: among other considerations there is a purely mechanical, engineering problem to solve, which arises from the shift in direction of the motion of the two surrounding media. The propeller must somehow change configurations when the TH switches to HH to take the greatest advantage of both the TH situation and the HH situation.

 

[A.T.]

I’ve designed a “Brennan Boat” that uses the same principal as his torpedo except it uses the flow of water as the power source instead of pulling the cables.

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

[PLAIN]http://www.accommodationz.co.nz/images/brennanboat.bmp[/QUOTE][/URL]
Why do you need two cables? The torpedo needs two for steering, but for a minimal design I would recommend using just one.

 

[vanesch]

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.

 

[A.T.]

Then, to analyze the cart's situation you have to account for the point where it reaches downwind speed, and the wind contribution plummets to 0. A cart going at the same speed as the wind sees the wind as standing still.

This 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

Sorry, I don't see any problem extracting energy from the fences when just one of them is at rest to you. This is exactly the situation when the cart travels at wind speed.

 

[vanesch]

g
I think you are following my reasoning: among other considerations there is a purely mechanical, engineering problem to solve, which arises from the shift in direction of the motion of the two surrounding media. The propeller must somehow change configurations when the TH switches to HH to take the greatest advantage of both the TH situation and the HH situation.

I'm not entirely sure about this, but as I said, I think it comes down to the fact that we have, in the case of a propeller, a force which is a function of velocity (which isn't the case in the ruler-based system). Mathematically, this gives you an "extra integration" to perform.

Let us analyze this in more detail. I'm discovering this myself as I'm typing...

The force a propeller generates is by the change in momentum it causes to the air mass. That is to say, if a propeller has an effective "outgoing" section S (which might be different to its geometrical section, and is equivalent to an idealized situation where a "cylinder of air" is uniformly accelerated by the propeller, and outside of it, nothing is done, which is clearly not what happens in reality, where there will be a gradient in velocities).

We work in the frame of the cart.
Now, in a time dt, the propeller will give to a volume V_out = S x dt x v_out, which has initially a velocity v_in, a final velocity v_out.

Assuming a non-compressible flow, the "in" surface has to be larger than the "out" surface (to keep the volume right) and will depend on the incoming velocity.
The momentum gained is the mass of the air times the velocity change:
dp = V_out x rho_air x (v_out - v_in). The momentum transfer per unit of time is dp/dt is the force:

F = rho_air x S x v_out x (v_out - v_in)

I suppose we can say that v_out is entirely fixed by the properties of the propeller and its rotation speed (but maybe there's an influence of the incoming velocity, I don't know). So we take it that v_out = K x v_cart (K includes the wheel and the gearing ratio and all that).

So we get:

F = rho_air x S x K x v_cart x (K x v_cart - v_in)

This is the force which the propeller exerts on the air.
Let us think about the signs. I took the positive direction in the direction of the tail of the cart. That means: v_in positive is "headwind", v_in negative is "tailwind" (which is the starting condition). v_cart however, is taken in the positive direction of the motion of the cart wrt the floor. So from the moment the cart is moving downwind, v_cart is positive.
F is positive if the propeller pulls the cart in the forward direction.

Through the gearing mechanims, F_wheels is the force exerted on the wheels.

The energy balance is the following. The energy given to the air by the propeller is given by (in the cart ref. frame):

rho_air x V_out x 1/2 x (v_out^2 - v_in^2)

(gain in kinetic energy = 1/2 mass x v^2)

This can be written as:
rho_air x S x K x v_cart x (K^2 x v_cart^2 - v_in^2)

The energy taken from the wheels is v_cart x F_wheels. In the lossless case, we have that:

rho_air x V_out x 1/2 x (v_out^2 - v_in^2) = v_cart x F_wheels

From this, F_wheels can be calculated, and hence the total force on the cart (without drag):

F = F - F_wheels.

...

 

[zoobyshoe]

This 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

Sorry, I don't see any problem extracting energy from the fences when just one of them is at rest to you.

I completely agree with you. There is no problem extracting energy from the fences, but only so long as we assume the moving fence has more than enough speed and mass to move you.

This is exactly the situation when the cart travels at wind speed.

There is an important difference, and this difference is the problem.

In the gound/air/cart scenario, what represents the fence that is still moving when the wind speed = 0 with respect to the cart?

 

[vanesch]

I'm not entirely sure about this, but as I said, I think it comes down to the fact that we have, in the case of a propeller, a force which is a function of velocity (which isn't the case in the ruler-based system). Mathematically, this gives you an "extra integration" to perform.

Let us analyze this in more detail. I'm discovering this myself as I'm typing...

The force a propeller generates is by the change in momentum it causes to the air mass. That is to say, if a propeller has an effective "outgoing" section S (which might be different to its geometrical section, and is equivalent to an idealized situation where a "cylinder of air" is uniformly accelerated by the propeller, and outside of it, nothing is done, which is clearly not what happens in reality, where there will be a gradient in velocities).

We work in the frame of the cart.
Now, in a time dt, the propeller will give to a volume V_out = S x dt x v_out, which has initially a velocity v_in, a final velocity v_out.

Assuming a non-compressible flow, the "in" surface has to be larger than the "out" surface (to keep the volume right) and will depend on the incoming velocity.
The momentum gained is the mass of the air times the velocity change:
dp = V_out x rho_air x (v_out - v_in). The momentum transfer per unit of time is dp/dt is the force:

F = rho_air x S x v_out x (v_out - v_in)

I suppose we can say that v_out is entirely fixed by the properties of the propeller and its rotation speed (but maybe there's an influence of the incoming velocity, I don't know). So we take it that v_out = K x v_cart (K includes the wheel and the gearing ratio and all that).

So we get:

F = rho_air x S x K x v_cart x (K x v_cart - v_in)

This is the force which the propeller exerts on the air.
Let us think about the signs. I took the positive direction in the direction of the tail of the cart. That means: v_in positive is "headwind", v_in negative is "tailwind" (which is the starting condition). v_cart however, is taken in the positive direction of the motion of the cart wrt the floor. So from the moment the cart is moving downwind, v_cart is positive.
F is positive if the propeller pulls the cart in the forward direction.

Through the gearing mechanims, F_wheels is the force exerted on the wheels.

The energy balance is the following. The energy given to the air by the propeller is given by (in the cart ref. frame):

rho_air x V_out x 1/2 x (v_out^2 - v_in^2)

(gain in kinetic energy = 1/2 mass x v^2)

This can be written as:
rho_air x S x K x v_cart x (K^2 x v_cart^2 - v_in^2)

The energy taken from the wheels is v_cart x F_wheels. In the lossless case, we have that:

rho_air x V_out x 1/2 x (v_out^2 - v_in^2) = v_cart x F_wheels

From this, F_wheels can be calculated, and hence the total force on the cart (without drag):

F = F - F_wheels.

...

Now, in the above, v_in is of course (v_wind - v_cart).

So what happens at low speeds ? (v_cart << v_wind)

F is negative:

F = rho_air x S x K x v_cart x (K x v_cart - v_in)

Indeed, for small v_cart, we have that the term in parentheses reduces to:
( (K+1) x v_cart - v_wind ) and if v_wind is larger than (K+1) x v_cart, we see that the propeller actually acts as a sail.

At the point where there is an equality, v_wind = (K+1) x v_cart, we see that there is no force on the propeller: the propeller smoothly "follows" the air without any force.
But the cart is still running at a velocity smaller than that of the wind. From the moment that v_cart becomes larger than v_wind / (K+1), we have that the propeller starts to act as a propeller, and the "power flow" reverses in the cart.

BTW the "turning over" point (K+1) x v_cart = v_wind is actually that when the "out" velocity of the air equals exactly the apparent incoming wind velocity: so indeed, the velocity of the air doesn't change when going through the propeller. This happens when the cart is running at a lower-than wind speed.PS: I'm a bit puzzled by K+1. I expected K-1. I will look for sign errors...

EDIT: indeed, I made a sign error. See post #477.

 

[atyy]

I suppose we can say that v_out is entirely fixed by the properties of the propeller and its rotation speed (but maybe there's an influence of the incoming velocity, I don't know). So we take it that v_out = K x v_cart (K includes the wheel and the gearing ratio and all that).

I've actually given up thinking about this for the moment, because I'm the sort of person to whom even a wheel rolling without slipping on flat ground is counterintuitive. But just to explain what I hope to study when I get the time. Not bothering me: yes, DDWFTTW is possible in principle because of the simple models; yes; there is no obvious problem with the energy and momentum conservation laws; yes a straightline treadmill test is in principle good; yes, the models do seem to instantiate DDWFTTW. What's bothering me: the tests are not perfect, even in a straight line test, how do we know the air is really stationary? how do we know the velocity of the treadmill is really constant? These caveats are even worse in the turntable test. Some theory is needed to interpret what we observe. Nonviolation of conservation laws doesn't mean that dynamics exist to actually instantiate a particular phenomenon. So a clear dynamical model, like the one you are trying here (and before in #214, but Jeff Reid seemed to have some problem with it), for the non-simplified cases that have actually been built is what would really help those of us without mechanical intuition.

 

[A.T.]

This 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

There is no problem extracting energy from the fences, but only so long as we assume the moving fence has more than enough speed and mass to move you.

We can assume that. The moving brown fence being the ground has enough mass. And "enough speed" is just a matter of leverage: total stick length vs. distance between the fences.

In the gound/air/cart scenario, what represents the fence that is still moving when the wind speed = 0 with respect to the cart?

The ground. Read what I wrote in the brackets.

 

[vanesch]

I've actually given up thinking about this for the moment, because I'm the sort of person to whom even a wheel rolling without slipping on flat ground is counterintuitive. But just to explain what I hope to study when I get the time. Not bothering me: yes, DDWFTTW is possible in principle because of the simple models; yes; there is no obvious problem with the energy and momentum conservation laws; yes a straightline treadmill test is in principle good; yes, the models do seem to instantiate DDWFTTW. What's bothering me: the tests are not perfect, even in a straight line test, how do we know the air is really stationary? how do we know the velocity of the treadmill is really constant? These caveats are even worse in the turntable test. Some theory is needed to interpret what we observe. Nonviolation of conservation laws doesn't mean that dynamics exist to actually instantiate a particular phenomenon. So a clear dynamical model, like the one you are trying here (and before in #214, but Jeff Reid seemed to have some problem with it), for the non-simplified cases that have actually been built is what would really help those of us without mechanical intuition.

Yes, you are right (actually my first attempt at a model seems to be flawed as there is clearly no first-order term present, and only a second-order term in the velocities). This is why some discussions here are a bit sad, in that they miss the essence. There *are* points to be risen in the presented tests, as you point out. So first having a crude model in which one can trace out all one's elementary objections is a good start.

 

[rcgldr]

The propeller must somehow change configurations when the TH switches to HH to take the greatest advantage of both the TH situation and the HH situation.

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.

velocity of the air doesn't change when going through the propeller.

This never happens at below wind speed. The velocity of the air is always slowed down (accelerated upwind) by the prop, regardless of how much the prop is acting as a blunt body (sail) or acting as a thrust generating device (prop) at below wind speed.

The cart speed at which the velocity of the air doesn't change when going through the prop occurs when (cart speed) - (prop pitch speed (relative to cart)) = (wind speed), or restated in terms of advance ratio, when (cart speed) = (wind speed)/(1 - ar), where ar = advance ratio = (prop pitch speed (relative to cart)) / (cart speed). Using the ratios from swerdna's cart, and a 10 mph tailwind, this occurs when cart speed = 23.33 mph. Cart speed = 23.33 mph, prop pitch speed = (6 /10.5) x 23.33 mph = 13.33 mph. Prop pitch speed is 13.33 mph upwind of the cart moving downwind at 23.33 mph, so zero change in the velocity of a 10 mph tailwind.

 

[vanesch]

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.

This never happens at below wind speed. The velocity of the air is always slowed down (accelerated upwind) by the prop, regardless of how much the prop is acting as a blunt body (sail) or acting as a thrust generating device (prop) at below wind speed.

Ok, where's my error ? I feel a bit unsecure about the signs of some contributions. I should check it.

EDIT: right, I found it. I took v_in as the incoming wind direction as seen from the propeller, and then this is given by v_cart - v_wind and not vice versa.

So I have to take back what I wrote in post #472, and correct it:Now, in the above, v_in is of course (- v_wind + v_cart).

So what happens at low speeds ? (v_cart << v_wind)

F is negative:

F = rho_air x S x K x v_cart x (K x v_cart - v_in)

Indeed, for small v_cart, we have that the term in parentheses reduces to:
( (K-1) x v_cart + v_wind ) and we see that if K > 1, this never acts as a sail.

And there is no turning-over point, as Jeff pointed out.

 

[rcgldr]

Ok, where's my error?

I suppose we can say that v_out is entirely fixed by the properties of the propeller and its rotation speed (but maybe there's an influence of the incoming velocity).

The incoming velocity does effect the thrust. Compare the thrust when apparent headwind = 0 to when the apparent headwind = prop pitch speed (thrust is zero in this second case). Even if thrust isn't greater if the apparent headwind is negative, the thrust is still non-zero. I'm not sure what the thrust to apparent heawind curve looks like, other than it's approaches infinity as apparent headwind approaches negative infinity, and approaches zero as apparent headwind approaches prop pitch speed, transitioning to negative thrust when apparent headwind is greater than prop pitch speed.

I think the sub-wind case you're considering is when the true wind speed is slowed to 0 mph by the propeller, which does occur when the cart speed is sub-wind.

And there is no turning-over point, as Jeff pointed out.

There is a turning over point, but this occurs when cart speed is well above wind speed (v_cart >> v_wind). I did the math in my last post.

 

[rcgldr]

Now, in the above, v_in is of course (- v_wind + v_cart).
F = rho_air x S x K x v_cart x (K x v_cart - v_in)

Regardless of v_cart, can't this be simplied as you mentioned:

F = rho_air x S x K x v_cart x (K x v_cart - (vcart-v_wind))
F = rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind)

now I'm having trouble with this equation, since F = 0 at some (v_cart >> v_wind).

update - just realized that K > 0 but < 1. Relative to cart, v_out = K x v_cart. K is what I've been calling advance ratio. Except now we need to fix the equation so F is postive at slow speeds (assuming you want F to be positive for downwind acceleration of the cart). Setting the last term = 0

(K-1) x v_cart + v_wind = 0
(K-1) x v_cart = -v_wind
(1-K) x v_cart = v_wind
v_cart = v_wind / (1 - K)

which matches my equation for when F = 0.

 

[vanesch]

The incoming velocity does effect the thrust. Compare the thrust when apparent headwind = 0 to when the apparent headwind = prop pitch speed (thrust is zero in this second case).

The incoming velocity has an effect on the thrust, but does it have a big effect on the outgoing velocity ? I would think that pitch x turning ratio determines more or less the outgoing velocity (yes, it will be affected a bit: even at zero turning speed, some air will "blow through" the propeller ; this would be much less the case with a turbine for instance, which would essentially "shield" the wind).

The thrust is affected by the incoming velocity because it is given by the amount of displaced mass times the velocity difference per unit of time.

I'm still trying to figure out the meaning of the K-value. I first thought it had to be larger than 1, but I think you're right, and that it is between 0 and 1.

What is interesting is that at v_cart = 0, we have according to my
F = rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind)

that F = 0.

So initially (and that's trivial!) when the cart is at rest (wrt the floor) the propeller is not giving any thrust either.
So here we need a little bit of drag by the wind on the rest of the cart (or even on the propeller, but this has not been taken into account in the model).

So we need to add a term F_drag = W x (v_wind - v_cart)

Now, this will probably even be: F_drag = W x (v_wind - v_cart)^3.

Note that this term flips sign when the cart overtakes the wind. Initially it works as a sail, and afterwards, it works as an aerobrake.

So the total aerodynamic force is rather:

F_aero = F + F_drag = rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind) + W x (v_wind - v_cart)^3

We have on the other hand the reaction force, which we can estimate from the power balance:

From rho_air x V_out x 1/2 x (v_out^2 - v_in^2) = v_cart x F_wheels

we have: F_wheels = rho_air x S x K x v_cart x (K^2 x v_cart^2 - (v_cart - v_wind)^2) / v_cart

or: F_wheels = rho_air x S x K x (K^2 x v_cart^2 - v_cart^2 + 2 x v_cart x v_wind - v_wind^2)

We now have the total force on the cart:

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)

Special cases:

v_cart = 0, we have:

F_tot = W x v_wind^3 + rho_air x S x K x v_wind^2

So this is the force that sets the cart in motion, and indeed in the forward direction.

v_cart = v_wind, we have:

F_tot = rho_air x S x K^2 x v_cart^2 -

... rho_air x S x K x ( (K^2) x v_cart^2)

= rho_air x S x K^2 x (1-K) x v_cart^2

Now, if K < 1, indeed, we see that we have a forward force, while if K > 1, we have a negative force (You were right, Jeff).

 

[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?