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).