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

[OmCheeto]

Jeff's and mines for instance

Especially #378.

Each step in post #378 seems to founded on the fact that a legitimate DDWFTTW device exists, which I've already said I don't believe. Until you can show that one does exist, the whole post is invalid. Sorry.

And the force model.

https://www.physicsforums.com/showpost.php?p=2032570&postcount=214"
As you said, it was a crude model, and I don't see how it proves anything.
Or are you referring to another model?

There was a specific reason for my question, it was not something I just threw out.
So could you answer it (the London and Mid West thing)

Yes. Yes. No.

and argue your answer ?

I have no answer. How can I argue that which I do not have.

I am simply not convinced.

 

[vanesch]

Each step in post #378 seems to founded on the fact that a legitimate DDWFTTW device exists, which I've already said I don't believe. Until you can show that one does exist, the whole post is invalid. Sorry.

Then you misunderstood the post - maybe I expressed myself badly.

The point is not: "given that a DWFTTW exists, blah blah...", but rather:

"*if* this test gives a positive result, would you consider that a DWFTTW device has been demonstrated ? "

For instance, if I say:

Suppose I have a cart that goes at 20 km/h downwind on a road when the wind is blowing at 10 km/h, and I show this to you. Would you consider that this demonstrates a DWFTTW device ?

I don't see how your answer can be anything but "yes", because having a cart that goes downwind faster than the wind (we assume that 20 > 10 is an accepted mathematical expression which doesn't need argumentation) is exactly what it means to have a DWFTTW demonstration. If it happens in front of your eyes, you can hardly deny it, if we are to have any scientific debate at all, can we ?

Mind you, I don't say that there *IS* such a demonstration. I'm asking you that *IF THERE WERE* such a demonstration, would you consider it a proof ? (and have to reconsider your position?)

So I'm asking you to place yourself in the hypothetical mode that such a demonstration is, against all your expectations, given, and your reaction to that.

Your PoV can of course be that such demonstration is not possible, given that no DWFTTW device exists (for whatever reason you might have convinced yourself).

What I'm doing, in that post, is to transform the original demonstration, step by step, into equivalent demonstrations.

So your PoV can be: well, given that such a cart doesn't exist, the first demonstration is of course not possible. So then you should read my transformations, and tell me at what point you DO think that a demonstration is possible. Or in what way the fact that the first demonstration is (according to you) not possible implies that the next one in the row is also not going to be possible.

As you said, it was a crude model, and I don't see how it proves anything.
Or are you referring to another model?

I'm referring to posts #517, #516, #480, #477 and #472 and #471 although there was a sign error in these two last posts (but they do contain the whole calculation and argument).

Yes. Yes. No.

Ok, on what theoretical basis do you conclude that the cart will work in the Mid West (even though there has been no video and no demonstration given of this actually happening in the Mid West) when you've seen a demonstration in London ?

Similar question: on what theoretical basis do you conclude that the test will work out in London next week when you have seen the test last week in London ?

These are not, contrary to what you might be thinking, questions to make fun of you. They are very fundamental, and will contain the key to the further discussion.

Because you can say: DWFTTW is not possible next week. Or, DWFTTW is not possible in the Mid West.
Why would such a statement (which is less strong than your statement DWFTTW is simply not possible) loose all its credibility when the demonstration were given last week in London ?

 

[vanesch]

Is it wrong not to want to be an idiot?

I think the question is rather:
Is it wrong not to find out that one IS actually an idiot ?

For one's ego, it surely is a good idea not to. But in as much as knowledge is power, it is maybe strategically not a bad idea to find out that one has been an idiot all the time (maybe in order to avoid any such error in the future).

 

[zoobyshoe]

I think the question is rather:
Is it wrong not to find out that one IS actually an idiot ?

For one's ego, it surely is a good idea not to. But in as much as knowledge is power, it is maybe strategically not a bad idea to find out that one has been an idiot all the time (maybe in order to avoid any such error in the future).

The point is: I can find out I'm wrong for free. Only an idiot would pay to find it out.

 

[vanesch]

The point is: I can find out I'm an idiot for free. Only an idiot would pay to find it out.

 

[zoobyshoe]

Did you happen to read that article about the Johnson Magnet motor I linked you to in the Magnet Motor thread?

 

[vanesch]

Did you happen to read that article about the Johnson Magnet motor I linked you to in the Magnet Motor thread?

Yes, but I fail to see the point ?

 

[Subductionzon]

This seems to be one of the longer threads on this topic in this forum that I have seen. Maybe it has not been locked because the powers that be have recognized the validity of spork and JB's work. At least no one is getting overly excited yet, though shcroder seems a bit trollish to me and he can be a bit provocative.

OmCheeeto, do you not believe that the treadmill or turntable is the equivalent of a cart going directly downwind at a speed greater than the wind? It is a fairly obvious equivalent frame of inertia. Or do you think that the various videos were fraudulent? If not then you have observed DWFTTW. And the question becomes not can it be done, but how are they doing it.

 

[rcgldr]

In case this one got lost:

I restated your multiple choice for a horiztonal tread since that is what we mostly have in the videos:

If the (horizontal) tread is running at -10 m/s and the cart moves horizontally at +2 m/ sec , what is the cart’s true velocity on the surface of the table?
a) 10 m/sec (Idiot)
b) +12 m/sec (Vanesch)
c) +2 m/sec (Schroder)
d) Impossible to determine. (Jeff)

c) + 2 m / sec.
Note the wind speed is 0 m / sec. Therefore:

|v_cart - v_tread| = |2 m/sec - (-10 m/sec)| = 12 m/sec
|v_wind - v_tread| = |0 m/sec - (-10 m/sec)| = 10 m/sec

and my claim:

|v_cart - v_tread| > |v_wind - v_tread|

holds true since 12 m /sec > 10 m / sec.

 

[vanesch]

OmCheeeto, do you not believe that the treadmill or turntable is the equivalent of a cart going directly downwind at a speed greater than the wind? It is a fairly obvious equivalent frame of inertia. Or do you think that the various videos were fraudulent? If not then you have observed DWFTTW. And the question becomes not can it be done, but how are they doing it.

The problem is that many people here seem never to have grasped what is the principle of Galilean relativity (to my great surprise). Schroder is apparently a hopeless case, as he doesn't even get vector addition of velocities.

It is true that the demonstrations given only have a meaning when one sees the link through a galilean transformation to the actual case, and this (a priori elementary) exercise seems to be lost on different persons, so we should work this out a bit more.

The thing I'm trying to do now with OmCheeto is to make him first see that he already uses a theoretical transformation when doing the London - MidWest and the last week - next week demonstrations: these are the translations and rotations in space and translations in time. He already uses part of the Galilean group. I hope to pick on from there to make him see that a galilean boost is just another member of this symmetry group of nature (in its classical formulation) and that "generalisations" through this symmetry group are all right as demonstrations.

 

[rcgldr]

Galilean relativity

For the others here, why not just call it Newtonian physics and frame of reference? Personally I'm not sure what Schroeders issue is, but it doesn't appear to be a Newtonian physics based frame of reference.

I keep posting using the magnitudes about the speed differences between cart and ground being greater than wind and ground, which should settle that issue, but so far no response.

 

[vanesch]

I keep posting using the magnitudes about the speed differences between cart and ground being greater than wind and ground, which should settle that issue, but so far no response.

I think the problem some have is much more elementary than that: they don't see why this speed difference corresponds to the "real speeds" in a "real demonstration" outdoor. I think they see it just as some number magic which is unrelated to a genuine outdoor test, like in "why would you consider the *speed difference* in this particular case, and take the genuine *ground speed* in the true test ? What's their relationship ? In this way, I can find any number I like, but that doesn't prove anything, does it ?

So what is needed is to show, step by step, what corresponds to what, and why. I know, this is truly elementary. But if I read that Schroder even thinks that a car going north and a car going south have relative speeds which are the sum, and that you get something else when you calculate the difference, we are dealing with a very low level of understanding and we have to work up from there.

I'm going to give it a try, if there are any responses.

 

[rcgldr]

But if I read that Schroder even thinks that a car going north and a car going south have relative speeds which are the sum.

Which is I resorted to using mathematical forms like

|v_cart - v_ground| > |v_wind - v_ground|

since it eliminates the frame of reference and similar abstract concepts with a simple to calculate relationship. Then just it's a case of being able to grasp how that mathematically inequality applies to both indoor and outdoor situations.

 

[schroder]

In case this one got lost:

I restated your multiple choice for a horiztonal tread since that is what we mostly have in the videos:

c) + 2 m / sec.
Note the wind speed is 0 m / sec. Therefore:

|v_cart - v_tread| = |2 m/sec - (-10 m/sec)| = 12 m/sec
|v_wind - v_tread| = |0 m/sec - (-10 m/sec)| = 10 m/sec

and my claim:

|v_cart - v_tread| > |v_wind - v_tread|

holds true since 12 m /sec > 10 m / sec.

You know what my response is: It is exactly what it always has been from day ONE.
The cart is going slower than the tread, no faster. You and Vanesch have now, finally agreed to that by picking 2 m/sec in my example. The reason you picked that number is by referencing to the stationary table (please do not tell me about absolute or preferred references. The table is “relatively” stationary to the cart and to the tread and is a perfectly valid reference. Even if you decide to use the tread, if you do the proper vector sum you still will get 2 m/sec. Notice I said proper vector sum!
I think we all agree that 2 are LESS then 10. And as these are Velocities, 2 is SLOWER than 10.
Now, you have not demonstrated that the PHYSICS is significantly different with the cart on the Horizontal tread than the cart being driven by the Vertical tread.
Vanesch, in one of his typical brilliant analyses, has simply waved his hand and said that the Horizontal tread is a DDWFTTW situation and the Vertical tread is a “motorized cart” But since the PHYSICS is the same, how can Vanesch’s claim be taken seriously? Is that all you have? An opinion?
I have much more than that. Since the Physics is the same, and we get a much lower cart velocity than the tread velocity, then we must get a much lower cart velocity in the Horizontal tread as well.
You believe the cart is DDW because of the orientation of the tread, even though the Physics is the same? Then you are letting your eyes fool you and you are being extremely childish in insisting on adding the relative velocities any different. You are adding up to 12 m/sec in the Hor frame and 2 m/sec in the VER frame with the same Physics in BOTH!
Once you realize what your outlandish error is you must agree that in the wind, the cart velocity is much less than Wind velocity.
I have spelled it out so even an IDIOT can understand it. Isn’t there anyone monitoring this Forum! Where is ADMIN, where is Chroot? Where is Fred Garvin? Halls of IVY.. All people who I respect. Why are the people who understand this staying so silent when this Forum’s integrity is at stake?

 

[vanesch]

Galilean relativity (but that name only stuck after Einstein introduced HIS relativity, so to many the name might not say anything in fact) is the following principle:

Consider a frame of reference (that is, an origin, and 3 orthogonal axes X Y and Z).

Consider a mechanical system under study, and all the relevant elements that exert forces on it. Call that ensemble "the system under study".

The positions (and orientations) of all these elements of the system under study can, at a given instant t0, be expressed in the frame of reference. Points can be given 3 coordinates x0, y0 and z0 in this coordinate system, rigid bodies can be given 3 coordinates (of a reference point on the body) and an orientation with 3 angles (Euler angles for instance) ...
All this is geometry in space.

These positions and so on CHANGE from one time t0 to another time t1, and this change is called *motion*. That means that for a given point, we have a certain value of the x-value of the coordinates as a function of time t, which we can write mathematically as x(t).

This is a mathematical function, and hence we can take the *derivative*. The derivative of a position coordinate wrt time is called a *velocity component*. The 3 velocity components of a point (x, y, z) form a 3-some which we can call a VECTOR.

The combination of geometry and time is called kinematics. It is the description of the geometrical setup as a function of time.

Now, the thing is that we can consider another reference frame, O', X', Y', Z'. Of course that will lead, at instant t0, to *other* coordinates (x',y',z') and so on for the *same* objects.

Now, the trick is that if we know O', and X', Y' and Z' as described in the frame OX,Y,Z, then we can CALCULATE what will be the coordinates (x',y',z') as a function of the coordinates (x,y,z). This is what is called a geometrical TRANSFORMATION. It is studied in 3-dimensional euclidean geometry.

And now comes the crux: It can be that the description of O',X',Y',Z' in the frame OXYZ is different for different times. If that is the case, we say that the frame O'X'Y'Z' *is in relative motion* wrt the frame OXYZ. The description of O'X'Y'Z' is then a function of time in OXYZ. At any moment, we can still calculate x'(t), y'(t) and z'(t) from this and the functions x(t), y(t) and z(t).

This is what is called a "change of reference frame".

And now comes something somewhat more difficult: the *velocities* (derivatives wrt time) calculated in the frame O'X'Y'Z' can also be calculated from the *positions* and the *velocities* in OXYZ of a given point. This is done, in all generality, by using the chain rule.

All this is still simply kinematics. It is the description of the variations of 3-dimensional geometry as a function of a parameter, called time, and their derivatives.

It turns out that, in Newtonian mechanics, there exists at least one reference frame OXYZ in which Newton's law is valid: F = m a. a is the acceleration, which is the derivative of the velocities as a function of time. This law is not valid in just any reference frame, but at least in one. The reference frames in which this law holds are called "inertial frames".
Applying Newton's law all components in a system under study gives us a set of differential equations, which are called the "equations of motion". Their solution gives us the functions x(t), y(t), z(t) for all the elements in the setup ; in other words, their solution gives us a complete description of the motion. In order for the solution to be unique, we need to specify an "initial position" and "an initial velocity" for each component.

It turns out that if there exists one single inertial frame, then there exist a whole family of them and they are all linked between them by what is called a galilean transformation.

A galilean transformation consists of an arbitrary combination of the following:
1) a geometrical translation (X' is parallel to X, Y' is parallel to Y, Z' is parallel to Z, but O' is different from O).
2) a geometrical rotation (O is the same as O', but X'Y'Z' is rotated wrt XYZ)
3) a translation in time (t' = t + t0)
4) a galilean boost:
that is the same as a geometrical translation, except that now, the coordinates of O' are a linear function of time. The derivatives of these linear functions are a constant of course, and these 3 constants form the velocity vector of the frame O' wrt the frame O.

From kinematical considerations, it turns out that to each transformation corresponds a very specific way of transforming coordinates (that's pure geometry) and velocities.

The remarkable thing is that velocities transform easily. For the geometrical transformations 1) and 2), velocities transform simply as position vectors. For a boost, 4), velocities simply transform by subtracting the velocity vector that specifies O' wrt O.
*it is from this property that follow all the velocity transformations we need*

[ edit: oops, I made an error. For transformation 1), velocities remain the same. For transformations 2) velocities transform as position vectors ]

And now we come to the principle of Galilean relativity:

If we have a setup of a system under study where all the relevant forces are taken into account, described in an inertial frame OXYZ, then this will give us a set of equations of motion, with a certain set of solutions given a certain set of initial conditions. Looking upon this system into another inertial frame of reference, *we obtain the same equations of motion*.

That's the principle of galilean relativity: the equations of motion remain the same under a transformation from one inertial frame to another. They don't change form. If we have a set of equations of motion expressed in x,y and z (from the frame Oxyz), then we can simply replace them by x', y' and z', and we now have the equations of motion in the frame Ox'y'z'.

This is not kinematics. It could be that the laws of nature don't give us equations of motion which are so. But they are. And that property is called Galilean relativity.

Now, this is important, because it means that, if we have a setup in one frame, and we have another system in another (inertial) frame, but we can find a Galilean transformation which maps every element (position, velocity) of the first system onto the second one, then we know that they have *identical equations of motion*. That is, that they will behave, each in their frame, in an identical way.

 

[schroder]

Get you head out of the clouds and look at what is happening!

 

[zoobyshoe]

Yes, but I fail to see the point ?

That's because I haven't tried to make a point about it yet. I was first going to find out if you'd read it.

 

[rcgldr]

|v_cart - v_tread| = |2 m/sec - (-10 m/sec)| = 12 m/sec
|v_wind - v_tread| = |0 m/sec - (-10 m/sec)| = 10 m/sec

and my claim:

|v_cart - v_tread| > |v_wind - v_tread|

holds true since 12 m /sec > 10 m / sec.

The cart is going slower than the tread, not faster.

The claim states nothing about cart speed versus tread speed, only about cart speed versus wind speed, which is 2 mph versus 0 mph in the case of the treadmill.

If outdoors, and switching the frame of referenece to the ground, I get the same results using my mathematical statement.

|v_cart - v_ground| = |12 m/sec - (0 m/sec)| = 12 m/sec
|v_wind - v_ground| = |10 m/sec - (0 m/sec)| = 10 m/sec

My claim includes nothing about cart speed versus tread or ground speed, so why do you keep bringing up this comparason? No one is claiming the cart can go downwind at over twice the speed of the wind, just faster than the wind. Cart speed > 1x wind sped, but not > 2x wind speed.

 

[zoobyshoe]

Galilean relativity ...

I don't have any problem with this. The reason I did not want to sort out the list of different frames is because you stipulated in the first one that the device worked as claimed and the events in the frame constituted proof. Therefore going through the list and deciding which other frames also proved it worked would have required me to carry a what I take to be fiction from one frame to the rest, which seemed sort of absurd.

 

[vanesch]

I don't have any problem with this. The reason I did not want to sort out the list of different frames is because you stipulated in the first one that the device worked as claimed and the events in the frame constituted proof. Therefore going through the list and deciding which other frames also proved it worked would have required me to carry a what I take to be fiction from one frame to the rest, which seemed sort of absurd.

Galilean relativity is not just about looking at one and the same thing in different frames. That's pure kinematics. The principle in Galilean relativity resides in that two *different* setups have the *same equations of motion* if they are linked by a galilean transformation.

Having the equations of motion means: knowing how it will behave, mechanically. Knowing what are the functions of time of the coordinates of the different elements of the system under study.

That means that if you know how one setup behaves, there is no doubt as how the second one will behave. In other words, the principle of galilean relativity makes actual predictions about the behavior of a new system (which might not even be built) by knowing how another system behaves. As such, it means that we can be sure about how the not-yet-built system will behave, without even have to build it.

Now, the hardest part is with the galilean boost. People accept this for the other galilean transformations, such as translations and rotations in space, or shifts in time. That was what I wanted to show in the reasoning with Omcheeto. Visibly he already applies the principle of galilean relativity for space transformations (rotations and translations), when he accepts that if it is demonstrated in London, then it will also work in the Mid West. He already applies it for time translations: if it was demonstrated last week, it will also work next week. What some here don't accept, is the boost. It is however part of the same transformation group (the galilean symmetry group).

Let us compare the "genuine" setup in London, and consider the (hypothetical) setup in the Midwest. The claim is that the behavior of the setup in London "will be the same" as the behavior of the setup in the Midwest.

How do we do this ? First, we specify the relevant elements that will exert a force on the cart. These are the floor and the air. Big Ben doesn't matter. The Tower of London doesn't matter. Now, it can be that the wind during the experiment was blowing 10 km/h from east to west in London (call this direction the X-axis). We work in a coordinate frame in which:
-the ground was having velocity 0
-the air was having velocity (10km/h,0,0)
-the cart was having velocity (v_cart,0,0)
The axis Y is going north-south, and the Z-axis is going into the sky.

We want to deduce from this a behavior of the cart in the Midwest.
If the wind is blowing north-south in the midwest, this would have a velocity component in our original frame of (0, 10km/h,0), along the Y-axis. However, the X-axis in london is now almost going into the sky in the midwest, and the Z-axis is going east-west.
This would be true if we were on the equator, but actually the rotation of frames is more complicated. We also take our "origin" at a different point.
We take the "equivalent midwest frame" to be the frame O'X'Y'Z'.
So we used a geometrical rotation, and a geometrical translation to map the positions and directions of the London situation to the "midwest situations".
What was position x,y,z of the cart in the London frame of the London setup, we take it to be the position x',y',z' of the cart in the Midwest frame in the Midwest setup.
We do the same for the air, and the ground, and everything, and we see that we find a 1-1 mapping between an element of the setup in London in coordinates x,y,z in the Oxyz frame, and the corresponding element of the hypothetical setup in the Midwest in coordinates x',y',z' in the O'x'y'z' frame.
The principle of galilean relativity tells us then that what was the result x(t),y(t),z(t) in london will be the same function, but written as x'(t), y'(t),z'(t) in the Midwest.
So all the derivatives will be the same numbers too, hence all the velocities and everything.

Now, a specific property of those derivatives in London, say v = 13 km/h, will be identical in the Midwest frame.

So we don't have to do the experiment in the Midwest. We know how it will behave. Galilean relativity + the result of the London test gives us that.

 

[vanesch]

...

And now we come to the crux of the business.

Consider the hypothetical outdoor test, in the frame Oxyz. In this frame, the ground is at rest, the wind is blowing in the X direction with velocity v_wind, and the cart is also running in the X direction, with velocity v_cart.

Now, consider the treadmill situation. We consider a O'X'Y'Z' in which the treadmill is at rest, in which the cart is moving in the X' direction (so X' is pointing in the opposite direction as the motion of the treadmill in the frame of the floor on which the treadmill machine is standing), and in which the air is moving in the X' direction.

We see that there is a 1-1 mapping between all the relevant elements and their position in the hypothetical outdoor test in the frame OXYZ, and the corresponding elements and their positions in the real treadmill situation in the frame O'X'Y'Z'.
We also see that there is a galilean transformation linking both (maybe a rotation and a translation just as in the midwest case, but moreover a boost, where the frame O'X'Y'Z' is moving at constant velocity wrt the frame OXYZ).

So we know that all the results we find for the treadmill test *expressed in the frame O'X'Y'Z'* can be taken over in the frame OXYZ for the hypothetical outdoor test. Specifically, if v_cart (in the frame O'X'Y'Z') is larger than v_wind (in the frame O'X'Y'Z') in the treadmill test, then it will be larger too in the hypothetical outdoor test.

That is what constitutes the demonstration.

 

[vanesch]

The cart is going slower than the tread, no faster. You and Vanesch have now, finally agreed to that by picking 2 m/sec in my example.

Yes. But I never said that the cart was going faster than the mill in the frame of the ground. I said that the cart *in the frame of the mill* is going faster than the air *in the frame of the mill*.

The reason you picked that number is by referencing to the stationary table (please do not tell me about absolute or preferred references. The table is “relatively” stationary to the cart and to the tread and is a perfectly valid reference.

Yes, of course it is.

Even if you decide to use the tread, if you do the proper vector sum you still will get 2 m/sec. Notice I said proper vector sum!

Could you work that out ?

Consider the original setup, all horizontal.

In the ground frame, treadmill is going 10 m/s to the right. Cart is going 2 m/s to the left.
Air is stationary.

Question to schoder:
in the frame of the mill, what is:
-- the velocity of the cart ?
-- the velocity of the air ?
-- the velocity of the ground ?

in the frame of the cart, what is:
-- the velocity of the ground ?
-- the velocity of the mill ?
-- the velocity of the air ?

Can you answer those ?

 

[vanesch]

Now, you have not demonstrated that the PHYSICS is significantly different with the cart on the Horizontal tread than the cart being driven by the Vertical tread.

In the "outdoor test" there is no vertical surface rubbing on the wheel, which would correspond to this, and the horizontal force balance is hence different. Now, the fun thing is actually that if you would do that, the cart with your vertical surface would actually go much faster than in the case of the horizontal motion driving tread, simply because the horizontal reaction force that was exercised on the cart by the treadmill and which had to be compensated by the propeller, is now absent.

In other words, if, in the horizontal drive case, the propeller was pulling 12 N on the cart, and the treadmill was pulling 12 N BACK on it (hence total horizontal force 0, and steady state), this last 12 N is absent in your vertical treadmill example, and so now the cart is going to accelerate forward. Because now the 12 N on the wheel are applied VERTICALLY (and hopefully compensated by the weight of the cart, or it will lift in the air), but there's no horizontal component anymore. The total horizontal force on the cart is hence 12 N forward.You haven't demonstrated that you are in steady state in your example. The forces are different.

Vanesch, in one of his typical brilliant analyses, has simply waved his hand and said that the Horizontal tread is a DDWFTTW situation and the Vertical tread is a “motorized cart” But since the PHYSICS is the same, how can Vanesch’s claim be taken seriously? Is that all you have? An opinion?

vanesch just did so.

 

[schroder]

Consider the original setup, all horizontal.

In the ground frame, treadmill is going 10 m/s to the right. Cart is going 2 m/s to the left.
Air is stationary.

Question to schoder:
in the frame of the mill, what is:
-- the velocity of the cart ?
-- the velocity of the air ?
-- the velocity of the ground ?

in the frame of the cart, what is:
-- the velocity of the ground ?
-- the velocity of the mill ?
-- the velocity of the air ?

Can you answer those ?

Yes I can and none of it is relevant because the cart and the tread are moving in opposite directions and they are not working against each other. These velocities are similar to calculating the relative velocities of two cars passing each other in opposite directions on the same road. Since they are doing no work against each other it is a SUM of their individual velocities with respect to the road. We have been all through this before. If you want to know the velocity of either car with respect to the road, use the road as your reference!

In the frame of the floor, what is:
The velocity of the air? 0 m/sec
The velocity of the tread? 10 m/sec
The velocity of the cart? 2 m/sec
And since the cart is slower than the tread, by Galilean Relativity (your favorite subject) it will also be 2 m/sec going down wind in a 10 m/sec wind.
You are saying you can pick any reference you want, so why not pick the floor? If a jet was flying by at 600 mph you could use that as your reference and claim the cart will do 600+ mph in the wind. You can pick any reference you want, as long as it makes sense in relation to the problem you are trying to solve! I have shown you explicitly, by way of rotating the tread Vertically, that the tread does not represent a reference which makes any sense to the problem at hand. Forgetting about the mechanical interference with the propeller, I can rotate the tread in a complete circle around its interface point with the wheel and the dynamics of the cart being driven by the tread at no time represent a Down wind situation It is ALWAYS just a “motorized cart” which travels at a much slower speed than the tread. And in the Wind, it will be also be a motorized cart with the wind providing the force, and it will always travel at a much slower speed than the wind.
I have ONE BIG FACTOR on my side: Plain old fashioned common sense!

 

[vanesch]

Yes I can and none of it is relevant

Show me that you can. I'm not convinced. The point was not whether they are relevant, but whether you understand the concept of velocity transformation between one frame and another.

 

[schroder]

Show me that you can. I'm not convinced. The point was not whether they are relevant, but whether you understand the concept of velocity transformation between one frame and another.

I am not convinced that you understand the importance of picking a reference which is relevant to the problem you are trying to solve. You want me to add some numbers together that have no relevance? The best mathematics in the world means nothing if it is not properly applied. Thomas Edison had a very good understanding of OHM’s Law and DC circuits. However, he did not understand AC theory at all, so he applied Ohm’s Law for DC and did all the math correctly and came up with numbers which made no sense. It is the application of the proper math to the problem here.
You want your meaningless numbers?
Question to schoder:
in the frame of the mill, what is: (by “mill” I assume you mean the moving tread and my whole point is that this is not a meaningful reference to this problew)
-- the velocity of the cart ? 12 m/sec
-- the velocity of the air ? 10 m/sec
-- the velocity of the ground ? 10 m/sec

in the frame of the cart, what is:
-- the velocity of the ground ? 2 m/sec
-- the velocity of the mill ? 12 m/sec
-- the velocity of the air ? 2 m/sec

 

[vanesch]

in the frame of the mill, what is: (by “mill” I assume you mean the moving tread and my whole point is that this is not a meaningful reference to this problew)
-- the velocity of the cart ? 12 m/sec
-- the velocity of the air ? 10 m/sec
-- the velocity of the ground ? 10 m/sec

in the frame of the cart, what is:
-- the velocity of the ground ? 2 m/sec
-- the velocity of the mill ? 12 m/sec
-- the velocity of the air ? 2 m/sec

Wow, I'm amazed. All correct. There is more hope than I thought.

 

[vanesch]

Now, what are the elements which exert a force onto the cart in the treadmill experiment ?

The air ?
The surface of the treadmill ?
The ground ?
Big Ben ?

edit: next: what are the equivalent elements which exert the same forces on the cart in the outdoor experiment ?

The air ?
The treadmill surface ?
The ground ?
Big Ben ?

 

[schroder]

Wow, I'm amazed. All correct. There is more hope than I thought.

Now if you can show me that you know how to pick a meaningful reference for two cars passing in opposite directions in order to solve for the velocity of either car with respect to the medium the car is driving on, I might consider that there is some hope for you as well!

 

[vanesch]

Now if you can show me that you know how to pick a meaningful reference for two cars passing in opposite directions in order to solve for the velocity of either car with respect to the medium the car is driving on, I might consider that there is some hope for you as well!

I showed you that this can be done in any frame.

Look, car 1 is going 50 mph north (on the road), car 2 is going 60 mph south.

A train is going 100 mph north. Question to schroder:
what is the velocity (+-sign is north, - - sign is south) of car 1 in the train frame ?
what is the velocity of car 2 in the train frame ?
what is the velocity of the ground in the train frame ? (be careful of the sign).

Now, can we, or can't we, calculate, with these data, calculate:
- the relative velocity of car 1 and the ground ?
- the relative velocity of car 2 and the ground ?
- the relative velocity of car 1 in car 2's frame ?

Tell me how you obtain those from the velocities in the train frame.