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

[zoobyshoe]

You cannot have a thing X which goes slower WRT to a thing Y, than the thing Y ITSELF is going WRT to itself, can you ?

Because a thing doesn't move wrt itself.

Maybe the thing Y is having an Out-Of-Body-Experience.

 

[schroder]

Where don't you believe me anymore ?

Here is where I don't believe you, or more correctly, I do not believe DDWFTTW because it is all based upon a misintrepretation of what is happening on the treadmill.

I AM saying that the cart with respect to the tread, is going SLOWER than the tread with respect to the air! Is that clear enough for you to understand?
Is anyone here familiar with a heterodyne? It would help if you are, because that is what is happening with the cart on the treadmill. It may not be immediately apparent on the treadmill, but it is very apparent when looking at Swerdna’s turntable. The table and the cart are clearly in a heterodyne. The solution to a heterodyne is very simple, even simpler than the RMS solution I attempted earlier!
But first let us look more closely at the interface between the wheel and the tread. As the tread moves from Right to Left, the wheel is turning CCW. At the point where the tread and the circumference of the wheel meet, the two are moving in the same direction. Now, I want you to imagine a wheel which is turning so that the linear velocity of the circumference is exactly the same as the velocity of the tread. Since they are moving in the same direction, and at the same velocity, their relative velocity is Zero. The wheel will have no translational motion to the right or to the left, but will hold it’s position on the tread. It is rotating, but it is not translating. A length of tread is passing under the wheel equal to the length of circumference of the wheel that is passing the point of contact. Let that sink in.
Now, I want you to imagine a wheel which is turning on the tread but also has translational motion in the direction opposite to that of the tread. If the tread is moving from left to right, the wheel rotating on the tread is seen to be moving from right to left. It is “advancing against the tread” to use the popular interpretation. Now, in order for this to happen, MORE tread must be passing under the point of contact than circumference of wheel! Read that again, and think it over. The wheel is translating to the Left, so more tread has passed the contact point than in the stationary condition. For more tread to pass by than circumference of wheel, the linear velocity of the tread is greater than the linear velocity of the wheel. In other words, the wheel is moving SLOWER than the tread, not faster !
This is a classic heterodyne, where two rotational motions result in a translational motion (the heterodyne) which is the Difference between the two velocities of the rotating objects. If the tread is running at 10 m/s and the wheel is turning with a linear velocity at the circumference, 8 m/sec the resulting translational motion will be 2 m/sec and it will be in the opposite direction to the tread, as what we see on the treadmill and on the turntable. The translational motion in the opposite direction of the tread is PROOF that the linear velocity of the wheel is LESS than the linear velocity of the tread. In other words, the cart is moving with respect to the tread SLOWER than the tread with respect to the air.
The reason this happens is that the propeller is being made to do work which in turn results in the wheel turning slower than it was before the propeller started working. The slower wheel then translates in a direction opposite to the tread or turntable.
The cart in no case runs faster than the tread and in no case will it run faster than the wind.

 

[vanesch]

Here is where I don't believe you, or more correctly, I do not believe DDWFTTW because it is all based upon a misintrepretation of what is happening on the treadmill.

No, indicate me in the story with the truck where is the step that you think won't work.

Is it when the treadmill is lifted on the truck ? Is it when the truck starts moving ?
Is it when the cart is going off the treadmill ? Is it when we remove the cover ?
Tell me.

Because the first step, you accept: the cart goes 2 mph forward, while the treadmill goes 10 mph backward.

The last step is that we have our little cart running at 12 mph on the road when the wind is blowing 10 mph.

So somewhere in between there must be a step you think won't work.

Tell me which one.

 

[vanesch]

I AM saying that the cart with respect to the tread, is going SLOWER than the tread with respect to the air! Is that clear enough for you to understand?

As seen from the observer on the ground, the cart is going 2 mph to the left, the treadmill is going 10 mph to the right. Do you dispute that the velocity of the cart wrt the treadmill is 12 mph ?
Do you dispute that the air is doing 10 mph wrt the treadmill ?

 

[schroder]

As seen from the observer on the ground, the cart is going 2 mph to the left, the treadmill is going 10 mph to the right. Do you dispute that the velocity of the cart wrt the treadmill is 12 mph ?
Do you dispute that the air is doing 10 mph wrt the treadmill ?

I DO dispute that "the velocity of the cart wrt the treadmill is 12 mph". I dispute that loud and clear! You are making a linear addition when this is clearly a heterodyne problem. Can you not recognize a heterodyne when you see it? The heterodyne which is the 2 mph is the Difference between the velocity of the tread and the velocity of the cart. The velocity of the cart is 8 mph much less than the tread and much less than the wind. Do you dispute that this is a heterodyne?

 

[vanesch]

But first let us look more closely at the interface between the wheel and the tread. As the tread moves from Right to Left, the wheel is turning CCW. At the point where the tread and the circumference of the wheel meet, the two are moving in the same direction. Now, I want you to imagine a wheel which is turning so that the linear velocity of the circumference is exactly the same as the velocity of the tread. Since they are moving in the same direction, and at the same velocity, their relative velocity is Zero. The wheel will have no translational motion to the right or to the left, but will hold it’s position on the tread.

Yes. That's pretty evident.

It is rotating, but it is not translating. A length of tread is passing under the wheel equal to the length of circumference of the wheel that is passing the point of contact. Let that sink in.

Yes. Obviously.

Now, I want you to imagine a wheel which is turning on the tread but also has translational motion in the direction opposite to that of the tread. If the tread is moving from left to right, the wheel rotating on the tread is seen to be moving from right to left. It is “advancing against the tread” to use the popular interpretation. Now, in order for this to happen, MORE tread must be passing under the point of contact than circumference of wheel!

No, of course not, the wheel will be spinning faster of course! Although I'm not sure I understand your wordings. In a same amount of time, the wheel will have spun over a larger angle in the second case than in the first. The "strip of contact" will be longer on the tread surface than in the first case.

Read that again, and think it over. The wheel is translating to the Left, so more tread has passed the contact point than in the stationary condition. For more tread to pass by than circumference of wheel, the linear velocity of the tread is greater than the linear velocity of the wheel. In other words, the wheel is moving SLOWER than the tread, not faster !

Here you've lost me. Do you talk about the velocity of the AXLE of the wheel, or of the contact point of the wheel ?

The velocity of the point of contact at the rim of the wheel is of course exactly the same as the velocity of the tread (however, this is a quantity that is dependent on the frame in which one expresses it).

This is a classic heterodyne, where two rotational motions result in a translational motion (the heterodyne) which is the Difference between the two velocities of the rotating objects.

To me, heterodyne is mixing frequencies to displace a portion of the spectrum on the frequency axis, like mapping HF onto an IF in a radio receiver. This has nothing to do with it here, or at least I don't see any link. Heterodyne effects come about because of the development of sin(a) x sin(b) into a component with sin(a+b) and a component with
sin(a-b).

We are simply adding vectors here.

If the tread is running at 10 m/s and the wheel is turning with a linear velocity at the circumference, 8 m/sec the resulting translational motion will be 2 m/sec and it will be in the opposite direction to the tread, as what we see on the treadmill and on the turntable.

No, of course not, you are making a sign error. If the tread is running at 10 m/s and the cart is running at 2 m/s in the other direction (that is, the axis of the cart wheel is moving at 2 m/s) then the wheel will be spinning with a velocity that corresponds to 12 m/s at its rim.

Obviously. This is so trivial that I have difficulties believing you don't see it.

Imagine some rope wound up on the wheel circumference, that you are unwinding. Assume the end of the rope glued to the treadmill surface. Imagine that you let the thing run for 1 second. How much rope has been unrolled from the wheel ?

 

[vanesch]

I DO dispute that "the velocity of the cart wrt the treadmill is 12 mph". I dispute that loud and clear!

Then some people were right that you are not capable of vector addition.

If thing A goes at 10 mph to the right, and thing B goes at 2 mph to the left, then the velocity of A wrt B is 12 mph.

Disputing that is beyond any hope.EDIT: maybe it is because you don't understand negative numbers.

If person A runs at 5 mph to the north, and I ALSO run at 5 mph to the north, we're actually running together, right ? So my relative velocity wrt A is 0. We remain at the same distance. That's what it means, to have relative velocity: that the DISTANCE BETWEEN BOTH is growing or shrinking by so much per second.

The mathematical explanation is that the relative velocity is the VECTOR DIFFERENCE of both our velocities: A goes at + 5mph, I go at +5mph, relative velocity: 5 mph - 5 mph = 0.

We are both running at 5 mph, and our relative velocity is 0.

Ok, now I run SOUTH, while A still runs north. What happens ? The VECTOR indicating my velocity now flipped sign: my velocity is actually - 5 mph, while person A's velocity is still 5 mph. Of course, in ABSOLUTE VALUE I still run at 5 mph, but my vector velocity has now a different sign.

AGAIN, or relative velocity is: 5 mph - (-5 mph) = 5 mph + 5 mph = 10 mph. Our relative velocity is 10 mph.

Nevertheless, we are both running (in absolute value) at 5 mph.

Tricky, isn't it ?

In the last case, if we had a rope between the two of us, its length would increase by 10 mph.
In the first case, if we had a rope between us, the rope would not change length.

Now, back to our treadmill and cart:

If we glued a rope to the treadmill surface, and have a reel on the cart, the rope would have to unwind with a velocity of 12 mph. Not 8 mph. If you had 12 miles of rope on your reel, after just one hour, the reel would be unwound. And not one hour and a half. (assuming a very long treadmill of course).

The reason is that 10 mph to the right is +10 mph, and 2 mph TO THE LEFT is vectorially, - 2 mph.

Their difference is 10 - (-2) = 12.

 

[vanesch]

Maybe the thing Y is having an Out-Of-Body-Experience.

Yup, that must be it.

 

[A.T.]

As seen from the observer on the ground, the cart is going 2 mph to the left, the treadmill is going 10 mph to the right. Do you dispute that the velocity of the cart wrt the treadmill is 12 mph ?

I DO dispute that "the velocity of the cart wrt the treadmill is 12 mph". I dispute that loud and clear!

I mentioned his name and he reappeared. Sorry guys. I shall burn in hell for that!

 

[schroder]

No, of course not, the wheel will be spinning faster of course! Although I'm not sure I understand your wordings. In a same amount of time, the wheel will have spun over a larger angle in the second case than in the first. The "strip of contact" will be longer on the tread surface than in the first case.



Here you've lost me. Do you talk about the velocity of the AXLE of the wheel, or of the contact point of the wheel ?

The velocity of the point of contact at the rim of the wheel is of course exactly the same as the velocity of the tread (however, this is a quantity that is dependent on the frame in which one expresses it).

Here, it is obvious that you do not understand a simple heterodyne problem. The velocity at the point of contact with the wheel is NOT “of course exactly the same as the velocity of the tread”. If it were always exactly the same as the velocity of the tread, the wheel could neither advance or retard. The velocity must change! The tread velocity is constant, so only the velocity of the rim of the wheel changes; the rpm of the wheel changes! For a translation of the axle in the direction opposite to the direction of the tread, the rpm of the wheel must slow down! The rpm slows, the linear velocity at the wheel rim slows, and more tread passes the point under the axle than circumference of wheel does. That results in a translation to the left, in the opposite direction to the tread. This is basic mechanics, Vanesch and you cannot deny it any more! When the propeller starts working on the air, the wheel slows down, resulting in the translation you see on the treadmill. It seems most everyone has been wrong about this from the beginning, with the exception of myself and a few others. What this shows, is that this cart can be pushed faster downwind, as a bluff body, without a propeller than with one! The propeller working in the air slows the cart down! Of course, you would need to give the cart a broad sail, to compensate for the loss of propeller. A sailboat can do something over 90% of wind velocity sailing directly down wind. This cart with that ridiculous propeller churning away may do 80% but no more than that.
All of this can be easily proved, if anyone cares to do honest experiments and a proper interpretation of the treadmill and turntable evidence. Push the cart to a known velocity while observing the force required to do this. Do that test with and without a propeller. I predict it will be easier to push without the propeller, as the propeller slows the cart down!

 

[ThinAirDesign]

All of this can be easily proved, if anyone cares to do honest experiments and a proper interpretation of the treadmill and turntable evidence. Push the cart to a known velocity while observing the force required to do this. Do that test with and without a propeller. I predict it will be easier to push without the propeller, as the propeller slows the cart down!

Schroder, I truly am intersted in perform test(s) related to this. You know the tool I have (standard treadmill seen in videos), and I also can come up with ways to measure force etc.

I do have to say that at this point I don't understand your test:

You say "push the cart to a known velocity". Unless I'm misunderstanding you (and that seems quite likely), this is what we do every time we 'start' the device on the treadmill -- that is we set the tread speed to say 8mph, and then hold(push) the cart until it catches up with the tread.

Please be more detailed as to the test you are looking for.

JB

 

[vanesch]

If it were always exactly the same as the velocity of the tread, the wheel could neither advance or retard. The velocity must change! The tread velocity is constant, so only the velocity of the rim of the wheel changes;

I'm always doubting that you are in fact trolling, as the things you write are so elementary and so terribly wrong that I'm torn between "this guy is just trying to tickle my b**ls" and "this is the most funny cocksure ignoramus I've ever met".

What is the instantaneous velocity of a point on the rim of a wheel, when we know the velocity of its center, and we know its rotation velocity ?

Let us consider the wheel in an XY frame (X horizontally to the right, Y upward), with the origin at the center of the wheel. Let us first consider the case that the wheel center is not moving in this frame. The rim points are on a circle with radius R and with center (0,0). They can be described by an angle theta, which indicates a point as (R cos(theta) , R sin(theta) ).
This corresponds to the standard goniometric convention: angle 0 corresponds to the point (R,0) on the positive X-axis (to the right), angle pi/2 (90 degrees) corresponds to the point (0,R) on the point on the positive Y-axis (to the top), etc...
So the angle gives us the position in counterclock wise fashion, starting at the right.

If the wheel is turning with an angular velocity of w (w rad/second), that means that it is running counter clock wise for w positive: for a material point on the rim, the value of theta is increasing at a rate of w per second.

Now, this means that the point that was at the right at (R,0) is moving instantaneously UPWARD with a velocity of... w R. The velocity vector can be written (0, wR) (no X component, and a positive y component, upward). So the point with theta = 0 has a velocity component (0, w R). The point ON TOP (at (0, R) ) is having a velocity to the LEFT of w R, so its velocity vector is written ( - w R, 0).
Note the minus sign, it means "to the left" (towards the NEGATIVE X-axis).
It was the point at theta = pi/2. The point to the left, at (-R, 0), has a downward velocity of w R, so its velocity vector equals (0, - w R). Note again, the minus sign: it is directed downward, towards the NEGATIVE Y axis.

The point at the bottom, for theta = 3/2 pi, has a velocity of (w R, 0). It goes in the positive X direction.

We could continue that way. It is easy (well, all is relative, of course!) to establish that the point at angle theta is having a momentary velocity vector given by

(- w R sin(theta) , w R cos(theta))

Check it: theta = 0 gives us (0, w R) ok
theta = pi/2 gives us (- w R, 0 ) ok
theta = pi gives us ( 0 , - w R) ok
theta = 3/2 pi gives us (+ w R, 0 ) ok.

Looks ok.

EDIT: see attachment for a picture.

Note that the lowest point of the wheel has a velocity in the direction of the positive X axis of w R.

NEXT, let us add a translation movement to the wheel. We give the center of the wheel, on top of its rotation movement, a translation movement with a velocity (vx, vy).

That simply means that we ADD this velocity to all the points of the rim, which move of course also with that center. So the velocity of the points on the rim of a wheel, turning with a rotation speed of w, and a center undergoing a velocity (vx, vy), is given by:

(- w R sin(theta) , w R cos(theta)) + (vx,vy)
= ( - w R sin(theta) + vx ; w R cos(theta) + vy)

Now, look at the velocity of the lowest point (theta = 3/2 pi) again:

Hell, it becomes ( + w R + vx, vy) !

Assume now that the wheel center is moving to the left. That means that vx is - vl where vl is the absolute value of the velocity of the wheel center (vx is then a negative number). We assume no vertical velocity of the wheel center.

We see that the velocity of the bottom point of the wheel is given by (w R - vl, 0).

Application: wheel on a steady road. If the wheel turns counterclockwise (w positive), and the bottom point of the wheel touches the road and is hence momentarily AT REST (in non-slipping contact) with the road, we have that the bottom velocity is 0:

w R - vl = 0

Or: w = vl / R...

Hurray ! We found the rotation speed of the wheel ! A wheel ROLLING to the left on a steady road will rotate COUNTERCLOCKWISE with an angular velocity of vl / R, which is a positive number!

Second application:
Now, if the wheel is not on a road, but on a treadmill that GOES TO THE RIGHT with a velocity v_tread (positive number: the velocity vector of a point on the treadmill is (v_tread,0) and this is a vector oriented to the positive X-axis, so to the right), then, if the wheel is making a NON SLIPPING CONTACT, we see that the point at the bottom of the wheel is having the same velocity as the tread (as it isn't slipping there and in contact), so we have equality of the two velocity vectors:

( + w R + vx, vy) = (v_tread,0)

from which:
w R + vx = v_tread and vy = 0

In other words: w = (v_tread - vx) / R.
and no vertical motion.

The velocities here were all positive TO THE RIGHT (positive X-axis).

Special cases:
a) If the tread is not moving (v_tread = 0) and the wheel center is not moving (vx = 0), then w = 0: the wheel doesn't rotate ! That's true, isn't it ?

b) if the tread is not moving (v_tread = 0) and the wheel rolls with a velocity vx > 0, then we have that w = - vx / R is negative. The wheel is rotating CLOCKWISE. It is moving to the right. Yes, that's right!

c) if the tread is moving (v_tread > 0) and the wheel is kept in place, then w = v_tread/R, the wheel is rotating COUNTER CLOCKWISE. Hell, right again !

d) tricky! If the tread is moving and the wheel is MOVING WITH IT, we have vx = v_tread.
We find... w = 0 ! If the wheel is moving with the tread, it is NOT ROTATING! Damn, that's correct! If you glue the wheel to the tread, it doesn't rotate and moves with it. Right again...Taaaa tadam !
e) if the wheel... is moving to the LEFT (vx < 0) and the tread is moving to the right, then...

w = (v_tread - vx) / R is a bigger number than when the wheel were not moving (vx = 0).

It is spinning FASTER counterclockwise than if it were stationary...

the rpm of the wheel changes! For a translation of the axle in the direction opposite to the direction of the tread, the rpm of the wheel must slow down!

Nope. It accelerates.

BTW, this is so trivially silly. If you are running a bike on a treadmill, and you pedal FASTER than needed to remain in place so that you advance on it (while the treadmill runs backwards of course), do you have you bike wheels spin SLOWER ??

This is basic mechanics, Vanesch and you cannot deny it any more!

Tell me, are you tickling them (my b**ls) or are you really that misguided ?

 

[The Dagda]

I think schroder is working in 11 dimensions? Either that or he's missing something fundamental here?

48 pages for this experiment, it just goes to show you never can tell?

 

[rcgldr]

The path of a point on the tread surface of the wheel is called a cycloid.

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

Such a point on the wheel cycles between 0 speed relative to the ground to double the axis speed relative to the ground, with an average horizontal speed = the axis speed relative to the ground.

 

[vanesch]

The path of a point on the tread surface of the wheel is called a cycloid.

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

Such a point on the wheel cycles between 0 speed relative to the ground to double the axis speed relative to the ground, with an average horizontal speed = the axis speed relative to the ground.

Whoops. I was just trying to get him understand the circle...

 

[rcgldr]

The path of a point on the tread surface of the wheel is called a cycloid. http://en.wikipedia.org/wiki/Cycloid

I was just trying to get him understand the circle.

but the animated diagram of a wheel rolling at a constant rate looks so cool.

 

[ThinAirDesign]

BTW, this is so trivially silly. If you are running a bike on a treadmill, and you pedal FASTER than needed to remain in place so that you advance on it (while the treadmill runs backwards of course), do you have you bike wheels spin SLOWER ??

Yeah vanesch, I'm certainly a bit mystified by schroder's position here.

I keep thinking there is some little piece he's missing that will explain the disconnect ... like he thinks the treadmill belt is moving the opposite way it actually is, or he believes that the wheels are slipping on the belt to some relevant degree or perhaps something else.

He writes as if he has some education relating to physics, but his position here is at such odds with the most basic principles of physics that I just can't figure it out.

The 'if the bike holds steady on the treadmill and then begins to advance, are the wheels then spinning slower?' question you raise above is just perfect and I can't imagine how anyone can answer anything other than "no, they're then spinning faster than before".

An example of a 'missing piece": We had an extended exchange with a physicist who also happens to be an acquaintance of mine -- many, many rounds of "you're doing all your math wrong", etc. Just like here with schroder. Suddenly out of the blue he says "Whoah ... you mean that prop isn't free spinning ... it's geared to the wheels?". LOL

I keep thinking there will be some "Eureka" moment with schroder where he goes the equivalent of the above with some tidbit.

JB

 

[ThinAirDesign]

Schroder, after thinking about your rough test description from post #760, I think perhaps I understand what you are looking for. Tell me if I'm correct.

If I were to put the cart in a large warehouse or gym, tow it across the floor with a string to say 5mph, and measure the towing force (tension) on the string with and then again without the prop engaged on the shaft. Would the readout of those two test forces be what you are looking for?

Thanks

JB

PS to others ... yes, of course I know the answer as to the relative numbers on the two tests.

 

[zoobyshoe]

Schroder, after thinking about your rough test description from post #760, I think perhaps I understand what you are looking for. Tell me if I'm correct.

If I were to put the cart in a large warehouse or gym, tow it across the floor with a string to say 5mph, and measure the towing force (tension) on the string with and then again without the prop engaged on the shaft. Would the readout of those two test forces be what you are looking for?

Thanks

JB

PS to others ... yes, of course I know the answer as to the relative numbers on the two tests.

In my estimation it also all boils down to the propeller. If you'll allow me to erroneously refer to the force the ground exerts on the wheel linked to the propeller as "thrust", then: can the propeller generate more thrust on the cart from its interaction with the air than the thrust it takes to turn the wheel that turns the propeller? The possibility exists in my mind that it could for the reason that the way it interacts with air is different than the way the road surface interacts with the wheel.

However, I must ask you where you stand in the Bernoulli vs Newton debate on the force most responsible for lift on airplane wings.

 

[swerdna]

In my estimation it also all boils down to the propeller. If you'll allow me to erroneously refer to the force the ground exerts on the wheel linked to the propeller as "thrust", then: can the propeller generate more thrust on the cart from its interaction with the air than the thrust it takes to turn the wheel that turns the propeller? The possibility exists in my mind that it could for the reason that the way it interacts with air is different than the way the road surface interacts with the wheel.

However, I must ask you where you stand in the Bernoulli vs Newton debate on the force most responsible for lift on airplane wings.

The thrust of the prop ALONE can never exceed the “thrust” (rolling resistance) of the wheel that gives the prop the thrust energy to begin with. If it could and did you could simly push start the cart in calm conditions and it would travel forever (free energy). The “thrust” of the wheel is overcome by a combination of the thrust of the prop and the “push” of the wind that the prop thrusts against. The prop thrust is motion in the opposite direction to the motion of the cart and is therefore slower than the cart. So when the cart is traveling at the speed of the wind and beyond the prop thrust isn’t and the wind still “pushes“ on it.

 

[ThinAirDesign]

The thrust of the prop alone can never exceed the “thrust” (rolling resistance) of the wheel that gives the prop the thrust energy to begin with.

I disagree -- the thrust of the prop ALONE can and obviously does "exceed the rolling resistance of the wheels that gives the prop the thrust to begin with"

Now in truth, swernd and I likely don't disagree all that much -- perhaps it's even just a matter of semantics: The prop is able to do this because it's operating in such a favorable environment -- a tailwind.

Let's take a well known type certified combo (Cessna 172, Lycoming engine, Sensenich prop) and add only one variable ... tailwind. Tie the plane down and instrument with load cells for thrust and you'll find that of course with a tailwind there's a lot more thrust for a given horsepower than in still air. No mystery there.

Same with the cart prop -- it's working in a very favorable environment and thus doesn't have the same horsepower requirement to generate the needed thrust to move the cart forward as it would in still air.

Again, I'm pretty sure swerdna and I don't disagree on this point and are just wording it differently.

JB

 

[ThinAirDesign]

However, I must ask you where you stand in the Bernoulli vs Newton debate on the force most responsible for lift on airplane wings.

Oh, I'm all for it.

JB

 

[OmCheeto]

Here is where I don't believe you, or more correctly, I do not believe DDWFTTW because it is all based upon a misintrepretation of what is happening on the treadmill.

I AM saying that the cart with respect to the tread, is going SLOWER than the tread with respect to the air! Is that clear enough for you to understand?

Whilst looking at this post, my acquaintance mentioned the term "luftmensch".
I'd never heard the term, so I queried him for a definition.
Either he is not a good communicator, or I am not a good interpreter.
We spent some time laughing after doing some surfing the net on the origin of the word.

Anyways, which video are we referring to at the moment? There are many treadmill videos.

 

[rcgldr]

In my estimation it also all boils down to the propeller. If you'll allow me to erroneously refer to the force the ground exerts on the wheel linked to the propeller as "thrust", then: can the propeller generate more thrust on the cart from its interaction with the air than the thrust it takes to turn the wheel that turns the propeller?

Yes, the propeller generates more thrust, but at a much lower speed. Since power = force x speed, the force is higher, but the power is less because of the slower speed speed of the air through the prop versus the speed of the ground at the wheels.

Similarly, the torque at the driven tires of a car is greater than the torque at the engine's crankshaft because of gearing, but the power is reduced because of drivetrain losses.

However, I must ask you where you stand in the Bernoulli vs Newton debate on the force most responsible for lift on airplane wings.

Start another thread for this one please.

 

[rcgldr]

Anyways, which video are we referring to at the moment?

swerdna started this thread with this video:

http://www.youtube.com/watch?v=MCB1Jczysrk&fmt=18

 

[OmCheeto]

Yes, the propeller generates more thrust, but at a much lower speed. Since power = force x speed, the force is higher, but the power is less because of the slower speed speed of the air through the prop versus the speed of the ground at the wheels.

Similarly, the torque at the driven tires of a car is greater than the torque at the engine's crankshaft because of gearing, but the power is reduced because of drivetrain losses.

Start another thread for this one please.

Hmmmm... More thrust at a lower speed...

= more push at less than the speed of the wind...

Where's that lever that makes the "cha-ching" noise?

Actually. I'm interested in reducing swerdna's device to a purely mathematical model based on observational data. swerdna, can you recreat your original video, with the camera placed directly above the platform, with the video starting from nothing moving, and ending with the DDWFTTW sprite moving at maximum speed? Thanks!

 

[swerdna]

Hmmmm... More thrust at a lower speed...

= more push at less than the speed of the wind...

Where's that lever that makes the "cha-ching" noise?

Actually. I'm interested in reducing swerdna's device to a purely mathematical model based on observational data. swerdna, can you recreat your original video, with the camera placed directly above the platform, with the video starting from nothing moving, and ending with the DDWFTTW sprite moving at maximum speed? Thanks!

Sorry but I canabalised the turntable cart to use some parts to build a small cart for outdoor wind testing. Unfortunately there hasn’t been a decent wind where I live for about three weeks (it‘s the “calm“ season) so I haven‘t been able to test it yet. Also I don’t have much spare time at present. I have a DDStreamFTTStream device I want to build and test as well. Besides, what information you would get from an overhead movie that you can’t get from the existing one?

 

[schroder]

I'm always doubting that you are in fact trolling, as the things you write are so elementary and so terribly wrong that I'm torn between "this guy is just trying to tickle my b**ls" and "this is the most funny cocksure ignoramus I've ever met".


NEXT, let us add a translation movement to the wheel. We give the center of the wheel, on top of its rotation movement, a translation movement with a velocity (vx, vy).


We see that the velocity of the bottom point of the wheel is given by (w R - vl, 0).




( + w R + vx, vy) = (v_tread,0)

from which:
w R + vx = v_tread and vy = 0




Taaaa tadam !
e) if the wheel... is moving to the LEFT (vx < 0) and the tread is moving to the right, then...

w = (v_tread - vx) / R is a bigger number than when the wheel were not moving (vx = 0).

It is spinning FASTER counterclockwise than if it were stationary...




Nope. It accelerates.




Tell me, are you tickling them (my b**ls) or are you really that misguided ?

Your calculation is incorrect because you are treating this the same as a wheel rolling on the road and ignoring the fact that this is clearly a heterodyne.
You are saying w R – vx = v tread That is Wrong!
The Correct expression is : w R – v tread = vx In a heterodyne the Difference is ALWAYS produced!
Look, see if you can follow along here:
The wheel is rotating CCW and stationary on the tread such that the linear velocity at the rim of the wheel is 10 m/sec. That is from Left to Right at the base of the wheel.
The tread is moving from Left to Right at 10 m/sec so there is no relative motion between the tread and the wheel. The wheel appears to be stationary on the tread.
NOW use your powers of visualization! Magnify the point where the wheel is in contact with the tread until you can see tow horizontal surfaces which are in contact and both are moving to the Right at 10 msec.
Can you do that? You see the two surfaces. Place a dot on each one, one over the other so that you can draw a vertical line between the two dots. This shows they are not moving with respect to one another.
OK. Now let the top dot, which is on the wheel (the bottom of the wheel) translate to the LEFT.
What does it take to allow that to happen? Only two ways that can happen: Either the lower dot speeds up to the Right which represents an accelerating tread, OR THE UPPER DOT SLOWS DOWN AND MOVES TO THE LEFT!
In order for the wheel to “advance” in the opposite direction the tread is moving, it must SLOW down!
As I said before, this makes perfect sense. The wheel is initially operating idly on the tread, without working into a load, so it goes to the RIGHT, and it is turning Faster than the tread initially. As the propeller starts working, the wheel is now working into a load and it SLOWS DOWN which results in it moving to the LEFT.
You do not even need to understand a heterodyne to see that what I am saying is correct.
Think it over.
Now, about your comments: I have been called a (somewhat) eccentric genius, but nobody has ever publicly called me a “cocksure ignoramus” and I will not stand for it now! I expect an apology or you should be removed from your position as a “PF mentor”. I am also not “tickling your b**ls” but would instead prefer to administer a swift kick there! After you realize how wrong you are about this cart, you will no doubt feel as if such a kick has indeed been delivered.

 

[vanesch]

Y
You are saying w R – vx = v tread That is Wrong!
The Correct expression is : w R – v tread = vx

That's the same expression !

But it is not the correct one, in its two forms. See my other post for a detailed explanation. I can't get more explicit than that.

Look, see if you can follow along here:
The wheel is rotating CCW and stationary on the tread such that the linear velocity at the rim of the wheel is 10 m/sec. That is from Left to Right at the base of the wheel.

Yes.

The tread is moving from Left to Right at 10 m/sec so there is no relative motion between the tread and the wheel. The wheel appears to be stationary on the tread.

That's what it means: rolling without slipping, indeed. (I mean, the fact that the point of contact is stationary, not that "the wheel" - by which we mean of course the axle of the wheel - is stationary of course)

NOW use your powers of visualization! Magnify the point where the wheel is in contact with the tread until you can see tow horizontal surfaces which are in contact and both are moving to the Right at 10 msec.

Yes.

Can you do that? You see the two surfaces. Place a dot on each one, one over the other so that you can draw a vertical line between the two dots. This shows they are not moving with respect to one another.

Very, very good. That's what it is called: rolling without slipping.

OK. Now let the top dot, which is on the wheel (the bottom of the wheel) translate to the LEFT.

Ok, but it would be slipping if that were the case.

What does it take to allow that to happen? Only two ways that can happen: Either the lower dot speeds up to the Right which represents an accelerating tread, OR THE UPPER DOT SLOWS DOWN AND MOVES TO THE LEFT!

Sure. But first of all, note that when you do that, the wheel axle itself doesn't move. What you are describing is the stationary wheel, which is now SLIPPING on the tread.

But that's not what we are going to do. We are going to leave those dots IN CONTACT (non-slipping), and we are going to move THE AXLE of the wheel to the left. As such, the angular velocity of the line linking the axle of the wheel with the dot will increase of course.

As I said before, this makes perfect sense. The wheel is initially operating idly on the tread, without working into a load, so it goes to the RIGHT, and it is turning Faster than the tread initially. As the propeller starts working, the wheel is now working into a load and it SLOWS DOWN which results in it moving to the LEFT.

We were talking about the situation where the cart with the wheel was moving to the left at 2 m/s (or mph take your pick) and the tread was moving at 10 m/s to the right. The wheel was rolling without slipping on the tread. Both velocities are expressed as seen from an observer on the firm ground, right. The cart is soon going to fall off the tread on its left side.

You were claiming that
1) the velocity of the cart wrt to the tread was not 12 m/s but rather 8 m/s
2) that the rotation velocity of the wheel moving to the left was slower than the rotation velocity of a wheel stationary with the ground.

These are, honestly, totally ludicrous statements, that I think even most serious high school students would recognize as erroneous (if not, one must stop teaching physics in high school because it is time lost).

Now, about your comments: I have been called a (somewhat) eccentric genius, but nobody has ever publicly called me a “cocksure ignoramus” and I will not stand for it now! I expect an apology or you should be removed from your position as a “PF mentor”. I am also not “tickling your b**ls” but would instead prefer to administer a swift kick there! After you realize how wrong you are about this cart, you will no doubt feel as if such a kick has indeed been delivered.

There's a first time for everything

Usually I do not use such language here, but honestly, the words there are really what I'm thinking. You leave me perplex. Come on, you can't be THAT ... well. So you just want to play, right ? You can't really mean what you write, right ? Or do you ?

Seriously, or we are talking about two totally different setups, or you do not have the slightest bit of understanding of kinematics - at a level that I've never seen before. Given your algebra demonstration in the beginning of this post...

 

[rcgldr]

The wheel is rotating CCW and stationary on the tread such that the linear velocity at the rim of the wheel is 10 m/sec. That is from Left to Right at the base of the wheel.

All of the videos I've seen so far have the tread moving from right to left with the cart wheels rotating CW, but it doesn't matter.

The tread is moving from Left to Right at 10 m/sec so there is no relative motion between the tread and the wheel.

Only at the contact patch.

Magnify the point where the wheel is in contact with the tread until you can see two horizontal surfaces which are in contact and both are moving to the right at 10 msec. You see the two surfaces. Place a dot on each one, one over the other so that you can draw a vertical line between the two dots. This shows they are not moving with respect to one another. Now let the top dot, which is on the wheel (the bottom of the wheel) translate to the left.

This isn't what happens. Using the floor as a frame of reference, when the wheel speed is zero relative to the floor, the wheel dot's horizontal component of speed = tread speed x cos(angle between vertical line down from axis and line to the dot on the wheel). At 0 degrees, the dot is in contact with the tread and moving left to right at 10 m / sec. At 90 and 270 degrees, the dot is moving vertically, with zero horizontal speed. At 180 degrees, the dot is at the top of the wheel moving right to left at 10 m / sec, ... The average speed of the dot is 0 m / sec if the cart is not moving with respect to the floor.

If the cart is moving right to left at 2 m / sec, then at 0 degress wheel dot speed = 10 m / s to the right. At 90 and 270 degrees,the horizontal dot speed = 2 m / sec to the left. At 180 degrees witht the dot at the top, speed = 14 m / sec to the left, ... for an average speed of 2 m / sec to the left, same as the axis. The wheel is rotating CCW and relative to the axis of the wheel the wheel surface is moving at 12 m /sec.

Another way to look at this case is to imagine a second horizontal surface in contact with the top of the wheel. The speed of the cart is the average speed of the two surfaces, 1/2 (upper surface speed + lower surface speed). If lower surface speed is -10 m / sec and upper surface speed is +10 m / sec, then wheel axis speed is 0 m / sec. If lower surface speed is -10 m /sec and upper surface speed is +14 m / sec, then wheel axis speed is +2 m / sec. Note that wheel surface speed = (upper surface speed) - (wheel axis speed) = (wheel axis speed) - (lower surface speed).

 

[A.T.]

Yes, the propeller generates more thrust, but at a much lower speed. Since power = force x speed, the force is higher, but the power is less because of the slower speed speed of the air through the prop versus the speed of the ground at the wheels.

Hmmmm... More thrust at a lower speed...

= more push at less than the speed of the wind...

No. The lower speed of the propeller is relative to the air (while the higher speed of the wheels is relative to the ground). You have to add this lower speed to windspeed in order to get the cart's speed relative to the ground. And this gives you more than the speed of the wind.

 

[schroder]

That's the same expression !



Ok, but it would be slipping if that were the case.



Usually I do not use such language here, but honestly, the words there are really what I'm thinking. You leave me perplex.

Yeah, yeah, I made a typo, there should have been a plus sign as in:
You are saying w R + vx = v tread That is Wrong
Only someone like you would jump on that to try and discredit the rest of what I am saying.
Translating IS rolling without slipping, but I now seriously doubt that you are capable of understanding that. I also seriously doubt that you want to understand what is happening with this cart on the treadmill. You are a typical member of academia who sees this as some sort of joke. You can afford to be wrong and have a laugh about it. I am a professional engineer and cannot accept being wrong, even when it is others who are wrong. Maybe someday you will grow up and understand that.
Swerdna could easily put a tachometer on the wheel and see it slowing down as it translates to the left.
TAD and Co could easily race the propeller cart against a cart with a sail and four freely rolling wheels.
This whole thing stinks to heaven for the lack of verification and the reason is it is not true.
And you, sir are a MORON!

!

 

[ThinAirDesign]

I am a professional engineer and cannot accept being wrong, ...

TAD and Co could easily race the propeller cart against a cart with a sail and four freely rolling wheels.

Really schroder? -- that's all you need to accept that you are wrong is a race between a prop cart and a sail cart?

I'm asking seriously schroder -- you haven't answered any of my previous questions regarding what sort of test you were referring to above so it makes me think that when it comes to the details of any test you are avoiding an exchange with me.

If a test makes sense to me I'll perform it (for all other tests ask swerdna). If you want to see a prop cart vs sail cart race, I'm happy to arrange one -- all I need to know is what excuse you're going to make when the sail cart gets left in the proverbial dust.

??

And you, sir are a MORON!

Tell you what ... I will include "I sir am a MORON" and a link to this thread, at the end of every one of my PF posts if the sail cart wins. You will do the same if the prop cart wins. DEAL?

JB

 

[rcgldr]

Swerdna could easily put a tachometer on the wheel and see it slowing down as it translates to the left.

Unless you're looking at a mirror image, swerdna's turntable is moving to the left (the side closest to the viewer), while the cart eventually translates to the right. The speed of the wheel and prop are constantly increasing until the cart reaches it's terminal speed.

http://www.youtube.com/watch?v=MCB1Jczysrk&fmt=18

 

[vanesch]

Yeah, yeah, I made a typo, there should have been a plus sign as in:
You are saying w R + vx = v tread That is Wrong

It is correct, if you understand the smallest bit of mechanics, if you know what a vector in 2 dimensions is, you see that it is right, and that's it. Not being able to understand the velocity field of a rotating circle, especially when the explanation is written in front of you, and when you have a picture drawn, but even disputing someone who explains it to you means that you are genuinely beyond the point of hope of ever understanding the most basic elements in mechanics - coupled with an attitude that precludes any improvement. This discussion has no sense anymore, I could just as well trying to have my cat do trigonometry.

Translating IS rolling without slipping, but I now seriously doubt that you are capable of understanding that. I also seriously doubt that you want to understand what is happening with this cart on the treadmill. You are a typical member of academia who sees this as some sort of joke. You can afford to be wrong and have a laugh about it. I am a professional engineer and cannot accept being wrong, even when it is others who are wrong. Maybe someday you will grow up and understand that.

I'm also a professional engineer, btw. I have 2 masters, one in engineering and one in physics. I have doubts you are one, but then, surprises are not excluded. If, as an engineer, you are not capable of seeing that if thing A is going 2 m/s to the left, and thing B is going 10 m/s to the right in a frame of reference, then thing A is going 12 m/s wrt to thing B, I don't know how you got your degree, or you must have forgotten all of it, or... I don't know. As I said, I've never met anybody so clueless as to the most elementary concepts of mechanics.

And you, sir are a MORON!

As I said, I've never met anybody at the same time so clueless as to the most basic notions in a field, and at the same time so terribly cocksure about his totally erroneous claims.

There's really no point anymore pursuing this discussion with you.

 

[ThinAirDesign]

There's really no point anymore pursuing this discussion with you.

Correct -- all that is left is the race between a sail cart and a prop cart.

You know how with big fights they always have some tag line to promote -- per his comment to you and my "I Sir am a MORON!" offer to schroder seen above, I proposed we call it the following:

"The race to be the MORON!"

Don't dilly dally schroder. You suggested the race. Let's work out the details and GET IT ON!

JB

 

[rcgldr]

More thrust at a lower speed ... = more push at less than the speed of the wind ...

The magnitude of the thrust from the prop is less than the wind speed (at least for the current DDWFTTW carts), but the direction is opposite. Swerdna's cart goes about 1.43 times the wind speed. If the wheel was directly under the prop, it would advance 14.07 inches per revolution, while the prop pitch advances 6 inches per revolution, for an advance ratio of .426. I guestimated the prop path radius to be about 31.6 inches (8 inches from wheel). The turntable rotates backwards at about 1.58 times per second, and the prop circles forward at about .69 times per second. Translating this into an outdoor equivalent:

wind speed +17.8 mph
cart speed +25.6 mph (1.43 wind speed)

apparent wind (relative to cart) = (wind speed - cart speed) = -7.8 mph
prop thrust speed based on prop pitch (relative to cart) = -10.9 mph

The prop attempts to accelerate an apparent headwind of 7.8 mph to 10.9 mph upwind, a 4.1 mph acceleration of air upwind. The actual speed will be less due to slip, but a sufficient amount of air is accelerated upwind against the tailwind to provide the thrust required to propel the cart.

From the ground frame of refererence, the prop is attempting to slow the 17.8 mph tailwind down to 13.7 mph with the same 4.1 mph acceleration of air upwind, even though the prop itself and the cart are moving downwind at 25.6 mph, 7.8 mph faster than the wind.

 

[swerdna]

Swerdna could easily put a tachometer on the wheel and see it slowing down as it translates to the left.

Assuming we are talking about this video - http://nz.youtube.com/watch?v=MCB1Jczysrk&fmt=18

When you say “put a tachometer on the wheel” I’m guessing that you don’t mean to fit the tachometer to the axle of the wheel of the cart to measure the revolutions of the wheel as it’s clearly obvious that the wheel speeds up from being stationary to a constantly sustainable speed. The revolutions of the wheel never slow down. I’m guessing you mean to fit the tachometer to the stationary axle of the turntable to measure the revolutions of the cart and tether around that axle. Is my guess correct?

If so then a tachometer is not required to see what occurs. The cart starts being stationary wrt the turntable axle. It then starts to move (clockwise) wrt the turntable axle and for a period this movement at first speeds up then slows down and briefly stops. It then starts to move again wrt the turntable axle (anti-clockwise) and progressively speeds up to constantly sustainable speed. The reason the cart slows and briefly stops moving clockwise to then move anti-clockwise is because the increasing thrust of the prop makes it do so. The net effect of the cart wrt the turntable axle is that it speeds up from being stationary to a constantly sustainable speed. It only slows down and briefly stops to change direction.

ETA - The period when the cart moves clockwise has nothing to do with the DDWFTTW claim. The claim isn’t that a cart can go from stationary to DDWFTTW, It’s that a cart can sustain DDWFTTW regardles of how it gets there.

 

[OmCheeto]

Sorry but I canabalised the turntable cart to use some parts to build a small cart for outdoor wind testing. Unfortunately there hasn’t been a decent wind where I live for about three weeks (it‘s the “calm“ season) so I haven‘t been able to test it yet. Also I don’t have much spare time at present. I have a DDStreamFTTStream device I want to build and test as well. Besides, what information you would get from an overhead movie that you can’t get from the existing one?

I don't know how to do frame by frame scrolling in these video's, and there is no clock, so my guestimates on velocity wouldn't be very accurate. But that's ok. I've nearly everything needed to build one on my back porch. Might be as soon as this weekend, and I'll have my very own DDWSTTW device built.

Btw, I loved your http://www.youtube.com/watch?v=VgaXpHOxtQg&feature=related" video.

hmmm... a new design just popped into my head: a giant hamster cage DDWFTTW device.

at least there'd be no gyroscopic torsional forces, which I believe is what pitted vanesch and myself against each other.

 

[zoobyshoe]

Oh, I'm all for it.

I said "in" not "on".

 

[swerdna]

I don't know how to do frame by frame scrolling in these video's, and there is no clock, so my guestimates on velocity wouldn't be very accurate. But that's ok. I've nearly everything needed to build one on my back porch. Might be as soon as this weekend, and I'll have my very own DDWSTTW device built.

Btw, I loved your http://www.youtube.com/watch?v=VgaXpHOxtQg&feature=related" video.

hmmm... a new design just popped into my head: a giant hamster cage DDWFTTW device.

at least there'd be no gyroscopic torsional forces, which I believe is what pitted vanesch and myself against each other.

Good luck with building your turntable and cart. Can’t wait to see the movie, read the book, wear the tee shirt . . .

 

[ThinAirDesign]

OmCheeto:

I'll have my very own DDWSTTW device built.

Well, there certainly are many ways to do exactly that -- the wrong ratios are a really great place to start.

JB

 

[ThinAirDesign]

I said "in" not "on".

Humor zoo. Humor.

JB

 

[OmCheeto]

Good luck with building your turntable and cart. Can’t wait to see the movie, read the book, wear the tee shirt . . .

Someone is selling T-shirts? I'll take 10! xxx Large.

Sewn together, they might make a good sail for my device, in the "I am a Moron" race.

 

[zoobyshoe]

Humor zoo. Humor.

JB

Answer thi. Answer.

Also: I think you can stop including your initials at the end of each post; no one gives a hoot, so you're just contributing to the heat death of the universe.

 

[ThinAirDesign]

Answer thi. Answer.

Off topic. New thread.

Also: I think you can stop including your initials at the end of each post; no one gives a hoot, so you're just contributing to the heat death of the universe.

Sure thing.

JB

 

[zoobyshoe]

Off topic. New thread.

No. Goes to the question of how the prop creates thrust.

Sure thing.

JB

Why not include a list of your memberships and college degrees?

 

[schroder]

When you say “put a tachometer on the wheel” I’m guessing that you don’t mean to fit the tachometer to the axle of the wheel of the cart to measure the revolutions of the wheel as it’s clearly obvious that the wheel speeds up from being stationary to a constantly sustainable speed. The revolutions of the wheel never slow down. I’m guessing you mean to fit the tachometer to the stationary axle of the turntable to measure the revolutions of the cart and tether around that axle. Is my guess correct?

If you are serious about doing this test, then this is my suggestion:

Fix a flexible cable to the unused end of the wheel’s axle (one end is presently used to power the propeller) Mount a small digital tachometer on top of the cross arm which supports the wheel and connect the new flexi shaft to the tach. You really should have another tach on the turntable, to ensure that it is not accelerating during the test as that will cause a false reading on the wheel tachometer.

When the test starts, everything is at rest. When the turntable starts running CW, and the cart initially moves along with it and the wheel is turning CCW. There may even be a very short spike when the cart is pushed ahead of the TT by the shock of turning it on, but that is a short transient response and may be impossible to capture. It is of no significance anyway. During the time that the cart is moving along with the TT in the CW direction, the linear velocity of the edge of the wheel rim is exactly the same as the linear velocity of the TT and they are both accelerating in the same direction at the same time. Similar to when they are both standing still 0 = 0 in the stationary case. In the run up case maybe 5 = 5 whatever the velocities are, they need to be either measured or computed.

The next stage is when the air resistance to the prop and the crossarm causes the CW motion of the cart to slow down and then it will reverse to CCW. Corresponding to this slowing down in the CW direction, there will be a slowing down in the CCW rpm of the wheel. It is THIS that you should look for with the tachometer! You WILL see the rpm reduce as the cart starts to translate in the CCW direction, opposite to the direction the TT is going. I GUARANTEE that you will see this reduction in RPM as it is the ONLY explanation which is within the laws of physics and mechanics.

The reason for the reduction in RPM is simple, the propeller is starting to work against the air and it is extracting energy from the wheel which slows the wheel down. The loss in rpm of the wheel is compensated for by the translational movement of the cart, in accordance with the conservation of energy. If you have an electric fan handy, turn it on and let it get up to speed. Next, place a plastic bag over it so it has little or no air to work against. Amazingly, the fan speeds up! This is like the wheel which has nothing to work against. Lift off the bag and the fan slows down, this is like the wheel which is now driving the propeller to do work against the air.

Everyone here assumes the wheel is speeding up, because that is what your eyes and your mind conditioning is telling you from everyday experience of watching wheels rolling on a road. But that is NOT the case here and if you do this test you will prove it beyond all doubt.

Finally, Swerdna, once you have this demo perfected, with tachs on the TT and the wheel and a nice readout on a PC, go and get a patent on it FAST! It does not demonstrated DDWFTTW (it debunks it) but it serves as an excellent teaching aid for rolling and translating motion. In particular, it teaches the heterodyning of two rotating wheels and how that produces a translational motion. Every mechanical engineering school and physics department should have one of these. This is the silver lining which will come out of this DDWFTTW FARCE!

 

[vanesch]

When the test starts, everything is at rest. When the turntable starts running CW, and the cart initially moves along with it and the wheel is turning CCW.

What if the cart is GLUED to the TT ? Is the wheel then also turning CCW ?? Is it turning at all ?
Do you realize what you are saying here ? What if the wheel is a square wheel ? Not a wheel at all ? A bolt ?

The next stage is when the air resistance to the prop and the crossarm causes the CW motion of the cart to slow down and then it will reverse to CCW.

What you describe would be correct if the wheel were touching a stationary surface (a ring outside of the turntable, that is fixed to the ground or something). But it is running on the turntable itself, you are aware of that, no ? (otherwise - as I suggested earlier - we are in fact talking about two totally different setups). The wheel is ON THE TURNTABLE, not on a fixed support next to it.

 

[schroder]

What if the cart is GLUED to the TT ? Is the wheel then also turning CCW ?? Is it turning at all ?
Do you realize what you are saying here ? What if the wheel is a square wheel ? Not a wheel at all ? A bolt ?




What you describe would be correct if the wheel were touching a stationary surface (a ring outside of the turntable, that is fixed to the ground or something). But it is running on the turntable itself, you are aware of that, no ? (otherwise - as I suggested earlier - we are in fact talking about two totally different setups). The wheel is ON THE TURNTABLE, not on a fixed support next to it.

The wheel is NOT glued, square or bolted. It is a wheel and it is turning. Care to argue that point?
What I describe IS correct. Are you afraid of the test? Why not just let the test be done and keep your opinions to yourself in the meanwhile?