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

[rcgldr]

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

The cart only moves along with the TT when it's forward motion was prevented by a block. In this case the cart wheel isn't turning at all.

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.

Also inertia during the time the TT is accelerating.

Corresponding to this slowing down in the CW direction, there will be a slowing down in the CCW rpm of the wheel.

The wheel on the cart is turning CW.

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 prop exerts an CCW torque on the wheel, but the force between tread and wheel result in a greater CW torque on the wheel, so the wheel accelerates in the CW direction until the cart reaches it's terminal speed.

 

[swerdna]

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!

Sorry but I can’t agree with this bit - “When the turntable starts running CW, and the cart initially moves along with it and the wheel is turning CCW.”

The turntable moves CW and the cart initially also moves CW at an increasingly slower speed, but the wheel always turns (revolves) CW (never CCW). To revolve CCW the wheel would have to travel faster then the motion of the turntable in the same direction and this never happens. The direction of the motion of the wheel as a whole changes but the direction of the wheel revolving doesn't. Either you are wrong or I’m completely misunderstanding you. Please explain where and when the wheel ever revolves CCW. Here’s a pic to help explain what I mean.

http://www.accommodationz.co.nz/images/directions.bmp

ETA - The shorter arrow from the axle indicates it's traveling slower than the TT.

 

[schroder]

Sorry but I can’t agree with this bit - “When the turntable starts running CW, and the cart initially moves along with it and the wheel is turning CCW.”

The turntable moves CW and the cart initially also moves CW at an increasingly slower speed, but the wheel always turns (revolves) CW (never CCW). To revolve CCW the wheel would have to travel faster then the motion of the turntable in the same direction and this never happens. The direction of the motion of the wheel as a whole changes but the direction of the wheel revolving doesn't. Either you are wrong or I’m completely misunderstanding you. Please explain where and when the wheel ever revolves CCW. Here’s a pic to help explain what I mean.

http://www.accommodationz.co.nz/images/directions.bmp

ETA - The shorter arrow from the axle indicates it's traveling slower than the TT.

Since the TT is a circular device, I suppose there is room for confusion. Your drawing is correct in a frontal view of the TT and the cart is running on the front edge closest to the viewer. In the video I saw, the cart began translating while it was running on the back edge, farthest from the viewer. In that frame of reference, the TT is turning CW, the edge of the TT is moving from Left to Right, the cart is moving from Right to Left and the wheel is turning CCW. Correct?

 

[rcgldr]

Since the TT is a circular device, I suppose there is room for confusion. Your drawing is correct in a frontal view of the TT and the cart is running on the front edge closest to the viewer. Left to Right, the cart is moving from Right to Left and the wheel is turning CCW. Correct?

Yes, but for consitency in this thread, let's keep the view of the cart as seen from outside the TT looking in, where the wheel turns CW. Most of the straight line cart videos also have the tread moving right to left and/or the cart moving left to right with the cart wheels turning CW as viewed from the camera.

I'm still confused why you don't think that the carts speed relative to the tread isn't 12 mph ("left to right") when the tread is going "right to left" at 10 mph and the cart is going "left to right" at 2 mph relative to the floor.

 

[vanesch]

The wheel is NOT glued, square or bolted. It is a wheel and it is turning. Care to argue that point?

Yes, but IF it were bolted or glued, it would move all the same with the TT, no ? The motion of the cart would be exactly the same as initially in the actual test, no ? And then the wheel wouldn't turn, would it ? So how come you think it is turning when the cart is moving all the same ?

What I describe IS correct.

I don't know anymore what you describe. But it is certainly not the setup on the video of the turntable.

Are you afraid of the test? Why not just let the test be done and keep your opinions to yourself in the meanwhile?

Hell no I'm not "afraid of the test". Your test comes down to:
"the bike's wheels, of a bike placed upon a truck, are spinning when the truck advances, look at the speedometer of the bike". No need to look at the speedometer of the bike to see that it reads 0 mph, and that the wheels are not turning, even if the truck is moving.
What I'm saying is: "the wheels of the bike are of course not turning when the truck advances, because imagine that the bike is glued or bolted to the truck". And you tell me that I'm wrong, because the bike is not glued on it, but just *placed* on it.

I'll ask you this in a different way: imagine that the cart had a remote-controlled brake on the wheel. Imagine that you apply the brake when the TT starts up. The cart would move with the TT, as in fact it does in the beginning (and we're talking about that phase, right) ? Imagine those brakes ultra-strong. Still convinced that the wheels are turning ?

 

[ThinAirDesign]

Are you afraid of the test?

Apparently you are the one afraid of the sail cart vs prop cart test schroder.

Why? Got a bit of doubt going there, or is it like you said previously ...

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

One thing for sure, no one is going to be calling you an "eccentric genius" after the test ... just plain old "eccentric" as the sail cart get's its butt kicked.

C'mon schroder ... the test was your suggestion. Let's work out the protocol together.

JB

 

[ThinAirDesign]

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

Ok, fair enough. Convince me that the outcome of a "Bernoulli vs Newton" argument somehow alters the outcome of a DDWFTTW argument and I will engage.

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

Would that somehow make you more comfortable?

JB

 

[vanesch]

Just once more, an illustration of what I wrote up in post 762 ( https://www.physicsforums.com/showpost.php?p=2052782&postcount=762 )

of the basic kinematics of a wheel rolling without slipping on a surface.

I realize that my orientations are the opposite of all videos here, sorry about that. The principle should remain, though.

In as much I can find references on the web, here is one:
http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html

(although it only treats rolling without slipping on a stationary surface, not a moving one).

There is an educational paper here:
http://www.iop.org/EJ/article/0143-0807/24/6/001/ejp3_6_001.pdf?request-id=4c8428fa-8110-4545-9d19-1b1284d162f3

(although it centers more on the study of acceleration)

Another reference is this, where you see the superposition of rotation and translation:
http://cnx.org/content/m14311/latest/

(note especially equation (3) )In any case, these sources indicate that the motion of individual points of a rigid body (such as a wheel) is given by the vectorial superposition of the translational motion of its center, and the rotational motion around that center.

I try to illustrate that in the picture (it was the calculation done in post 762).

From that, we see that the point of contact (wheel side) has the velocity of w . R + v_x which must - rolling without slipping - have exactly the same velocity as the point of contact (surface side), which is the linear velocity v_tread.

So we have that R.w + v_x = v_tread.

From which: w = (v_tread - v_x)/R

Note, btw, that the TOP point of the wheel has a velocity of v_x - R . w.

Depending on whether v_x is larger or smaller than R.w, this point moves forward or backwards.

 

[ThinAirDesign]

schroder, when viewed from the outside of the TT/cart combo, other than when the device is at rest or held to the TT through non-standard means, swernda's drawing is perfectly correct 100% of the time:

http://www.accommodationz.co.nz/images/directions.bmp

ETA - The shorter arrow from the axle indicates it's traveling slower than the TT.

The TT is *always* moving from right to left relative to the cart wheel, thus the cart wheel is *always* turning CW.

His drawing is correct during both startup phase and operation phase. Initally, as the TT is accelerated, the cart wheel moves right relative to the TT simply because if its inertial resistance and a bit of bluff drag. Once in steady state, it moves right relative to the TT because of the thrust of the prop.

JB

 

[schroder]

Yes, but for consitency in this thread, let's keep the view of the cart as seen from outside the TT looking in, where the wheel turns CW. Most of the straight line cart videos also have the tread moving right to left and/or the cart moving left to right with the cart wheels turning CW as viewed from the camera.

I'm still confused why you don't think that the carts speed relative to the tread isn't 12 mph ("left to right") when the tread is going "right to left" at 10 mph and the cart is going "left to right" at 2 mph relative to the floor.

OK. I agree we need some consistency here to avoid confusion and unnecessary argumentation. From now on, we use the front view of the TT and consider the cart to be moving on the front edge, closest to the viewer.
The TT is turning CW, so the surface is moving Right to Left.
The wheel is turning CW.
The direction of the cart is dependent on whether we are looking at startup, beginning of translation, or steady state. But in all these cases, the wheel is rotating CW.

During startup, the TT, moving from Right to Left gives an initial transient shock to the cart which was sitting stationary. This may actually cause the cart to be momentarily pushed ahead of the TT for an instant. It is of little or no consequence, unless you are studying transient response (an interesting field btw).
After the initial transient, the cart will move along with the TT in the same direction for a brief time. It begins by moving at exactly the same speed as the TT and the linear velocity on the edge of the wheel is the same as the linear velocity on the surface of the tread. Inertia, and rolling resistance are in play here but this condition does not last long. Important to note here that the wheel is rolling on the TT surface, it is NOT glued down or flat or any other silly supposition as Vanesch has offered up!
Air resistance to the cart, mainly the propeller as well as the crossarm will cause the cart to slow down in the direction of the TT. This is the CRITICAL point! As the cart’s motion from Right to Left slows, it’s RPM on the TT slows also! Some of the rotational motion is being exchanged for translational motion! This is the classic heterodyne.
Now, we enter the final stage, where the cart has slowed down enough to a steady RPM which is less than it had when it was moving Right to Left. The cart has slowed enough that the translational motion causes it to move from Left to Right. This is where everyone believes the cart has outrun the TT and you want to add the velocities as relative velocities to arrive at a cart velocity which is greater than the TT velocity. This is precisely where you are all going wrong! This is a heterodyne, the velocities are mixing and the translation velocity is the DIFFERENCE between the cart velocity and the TT velocity. If the TT is running at 10 m/sec and the translation is 2 m/sec the cart’s velocity (linear velocity at the edge of the wheel) is 8 m/sec. It is that simple and it is also undeniable and verifiable!

Put a tach on the wheel and a tach on the TT and do your tests!

I wish I knew how to do computer animations and I do not really have the time to learn right now, but a computer animation of a heterodyne would come in handy right now. I will try to make some drawings to demonstrate what I am saying. I am hoping that someone reading this and understands a heterodyne will chime in and help me with this explanation.

 

[vanesch]

After the initial transient, the cart will move along with the TT in the same direction for a brief time.

Now, do you agree, or not, that at that point (when the cart is moving (almost) simultaneously as the TT), the wheel is (almost) not turning, and that if the cart were running completely in sinc with the TT, that the wheel would not turn at all ?

It begins by moving at exactly the same speed as the TT and the linear velocity on the edge of the wheel is the same as the linear velocity on the surface of the tread.

The linear velocity of the (bottom of) the wheel will ALWAYS be the same as the linear velocity of the surface of the TT at that point as it isn't slipping.

Important to note here that the wheel is rolling on the TT surface, it is NOT glued down or flat or any other silly supposition as Vanesch has offered up!

In the beginning it is rolling *very slowly* and that is because it is not following *completely* the TT. If it were, the wheel wouldn't turn. As if it were glued.
As if it were a bicycle standing on a truck, while the truck is driving. The wheels of the bicycle don't turn (or do they ?).

Air resistance to the cart, mainly the propeller as well as the crossarm will cause the cart to slow down in the direction of the TT.

Yes, and as the difference in velocity is increasing, the wheel is rotating faster and faster.

This is the CRITICAL point! As the cart’s motion from Right to Left slows, it’s RPM on the TT slows also!

You are talking about the RPMs of the wheel, right ?

It is of course increasing, from almost 0 (and would be perfectly 0 if the cart were following perfectly the TT), to a rotation rate which is exactly that of a wheel held fixedly by a bystander and letting it roll on the TT when the direction of the cart turns over, that is, when, in the ground frame, it flips from going in the direction of the TT, to the other direction.

Some of the rotational motion is being exchanged for translational motion! This is the classic heterodyne.

You are confusing again kinematics and dynamics. And btw, no need to bring in a heterodyne ; Lagrange or Euler would have been totally capable of solving the mechanics of this thing, and had never heard of a heterodyne. So leave that out here. You don't need the concept of a heterodyne (and I don't see how it can be useful but that doesn't even matter) to do mechanics.

Now, we enter the final stage, where the cart has slowed down enough to a steady RPM which is less than it had when it was moving Right to Left. The cart has slowed enough that the translational motion causes it to move from Left to Right. This is where everyone believes the cart has outrun the TT and you want to add the velocities as relative velocities to arrive at a cart velocity which is greater than the TT velocity.

Yes, because in order to do so, you have to run even faster backwards.

Imagine that you are going to a childrens' manège (ride in english ?). You put your kid on a horse (bolted) on the manège while it is still not turning. Your other child is standing still, outside. While you are still on the manege with your first kid,it is starting up, and it starts turning CW. In the beginning, you remain steady with your kid on the horse. 1) ARE YOU WALKING NOW ?
Next, you want to keep up with the kid outside. 2) ARE YOU SLOWING DOWN OR SPEEDING UP to keep level with the kid outside as compared when you were with the kid on the horse ?
The kid outside wants to play a trick on you, and runs CCW AGAINST the CW direction of the manege. You want to keep at his side. 3) SHOULD YOU RUN FASTER OR SLOWER TO KEEP UP WITH HIM than when the kid was still standing still ?

Answer these 3 questions in upper case, please.

We can go further. Imagine that the outside kid wants to keep up with his little brother on the horse (bolted on the manege). Should the kid:
a) stand still
b) run CW around the manege which is turning CW
c) run CCW around the manege which is turning CCW ?

to keep at the same place has his little brother ?

 

[schroder]

Yes, and as the difference in velocity is increasing, the wheel is rotating faster and faster.






You are confusing again kinematics and dynamics. And btw, no need to bring in a heterodyne ; Lagrange or Euler would have been totally capable of solving the mechanics of this thing, and had never heard of a heterodyne. So leave that out here.

Imagine that you are going to a childrens' manège (ride in english ?). You put your kid on a horse (bolted) on the manège while it is still not turning. Your other child is standing still, outside. While you are still on the manege with your first kid,it is starting up, and it starts turning CW. In the beginning, you remain steady with your kid on the horse. 1) ARE YOU WALKING NOW ?
Next, you want to keep up with the kid outside. 2) ARE YOU SLOWING DOWN OR SPEEDING UP to keep level with the kid outside as compared when you were with the kid on the horse ?
The kid outside wants to play a trick on you, and runs CCW AGAINST the CW direction of the manege. You want to keep at his side. 3) SHOULD YOU RUN FASTER OR SLOWER TO KEEP UP WITH HIM than when the kid was still standing still ?

Answer these 3 questions in upper case, please.

The wheel is rotating slower and slower.

I will not leave the heterodyne out of here because this is a heterodyne problem!

Your example has nothing to do with a heterodyne, it is an example of relative velocities which has nothing to do with this problem. NOTHING to do with this problem!

I suggest you let Swerdna do his test and I do suggest you study up on heterodynes as you clearly do not understand anything I am saying. Sure, LaGrange and Fourier could have solved this, but you are neither! Fourier understood heterodynes very well btw! Ever heard of Fourier transforms and Bessel functions?

 

[rcgldr]

It begins by moving at exactly the same speed as the TT and the linear velocity on the edge of the wheel is the same as the linear velocity on the surface of the tread.

Ok, back to the floor as a frame of reference.

Air resistance to the cart, mainly the propeller as well as the crossarm will cause the cart to slow down in the direction of the TT.

Again using the floor as the frame of reference.

As the cart’s motion from right to left slows, it’s rpm on the TT slows also!

As the difference in speed between cart and TT increases, so does the rpm of wheel and prop. I've been considering right to left as negative and left to right as positive, and CW angular speed as positive. The angular speed of the wheel in radians / unit time = (cart speed - TT surface speed at wheel) / (wheel radius). Some example states using swerdna's cart, wheel radius = 4.25 cm:

cart speed = -10 m / s
tt speed = -10 m / s
w = ((-10) - (-10)) / (4.25) = (0 / 4.25) radians / sec

cart speed = -5 m / s
tt speed = -10 m / s
w = ((-5) - (-10)) / (4.25) = (+5 / 4.25) radians / sec

cart speed = 0 m / s
tt speed = -10 m / s
w = ((-0) - (-10)) / (4.25) = (+10 / 4.25) radians / sec

cart speed = 2 m / s
tt speed = -10 m / s
w = ((+2) - (-10)) / (4.25) = (+12 / 4.25) radians / sec

 

[vanesch]

Your example has nothing to do with a heterodyne, it is an example of relative velocities which has nothing to do with this problem. NOTHING to do with this problem!

We ARE talking about relative velocities. You were the one saying that the relative velocity between both is 8, and not 12 m/s (or mph whatever). It is ALL about relative velocities.

I suggest you let Swerdna do his test

Sure. We'll have a good laugh at you then.

and I do suggest you study up on heterodynes as you clearly do not understand anything I am saying.

I don't understand much of what you are saying, that is correct. The things I understand are so full of elementary errors that I can't make head or tails of it.

Sure, LaGrange and Fourier could have solved this, but you are neither! Fourier understood heterodynes very well btw! Ever heard of Fourier transforms and Bessel functions?

I wasn't talking about Fourier, I was talking about Euler. Yes, I've "heard about Fourier transforms and Bessel functions"

I will tell you where your "equivalence" with frequency mixing goes wrong: there are no negative frequencies (well, there are, of course, in a Fourier analysis, but when you're doing a frequency transformation in a radio receiver, only positive frequencies count). This is why, when you try to do vector addition in the frequency domain the way you are trying to do an analogy here, you mix up all your signs.

The idea of frequency shifting (heterodyning) is that you apply the sum of two sine functions to a non-linear element which includes a quadratic term:

s1(t) = sin(a t)
s2(t) = sin (b t)

f(t) = (s1(t) + s2(t))^2 = s1^2 + s2^2 + 2 s1 s2

The trick is that as such, the product between s1 and s2 appears.

Now, we have the identity sin(u) sin(v) = 1/2 cos(u-v) - 1/2 cos(u+v)

That means that the "sum frequency" and the "difference frequency" appears somewhere.

Now, imagine that we have a frequency of 10 MHz, and we mix it with a 2 MHz signal: we will obtain, in our output, an 8 MHz signal and a 12 MHz signal of course.

However, this is the trick: there's no such thing as "mixing with a -2 MHz signal"... This is just a 2 MHz signal with eventually a phase shift. So you can't "mix with a -2 MHz signal.

You don't see the difference between mixing 10 MHz and 10 MHz, or 10 MHz and "-10 MHz".

In other words, heterodyning is not a good analogy to learn about the behaviour of the addition of negative and positive numbers, and that is what you want to do here.

This thing has absolutely nothing to do with heterodyning, or with frequency mixing, or whatever. It is a relative velocities problem. You pulled in the concept of heterodyning. I could pull in Feynman diagrams, or transactional analysis, or any other unrelated thing. That doesn't change anything about this basic problem.

So could you answer the 3 questions on the children's manege ?

 

[Subductionzon]

shroder your problem seems to be that you do not understand the concept of inertial frame of reference. As Jeff Reid pointed out you were comparing velocity relative to the floor when you should have been measuring the velocity of the cart on the turntable.

In the beginning the velocity of the cart is zero relative to the turntable. That is when you observe the cart moving CW with the turntable. At this point the wheels are not rotating at all. When the propeller starts to turn the cart moves forward relative to the turntable, it is still going slower than the turntable so from above we can still see it moving clockwise, but it moves slower than the turntable.

It eventually is moving just as fast as the turntable, at this point it is stationary with respect to the ground, but the frame of reference that matters here is the one relative to the turntable.

Finally the cart is moving faster than the treadmill, from above it is seen to be moving backwards. Again you have to be aware what frame of reference in which you are measuring the carts speed. If you had a speedometer on the carts wheels you would see that it had a higher speed than the speed of the treadmill when measured from the ground. It might be a little tricky to see what is being modeled here because you are looking at different frames of reference to understand what is happening, but is not really all that difficult.

Lastly the only references I can find for heterodynes refers to radio waves, and has nothing to do with mechanical motion at all.

 

[Subductionzon]

vanesch, you were describing a "merry go round". Big ones at fairs are powered by a motor of some sort and the horses go up and down, the small park ones are hand pushed and may have horses or just a platform to sit or stand on with some bars for gripping or pusing.

 

[rcgldr]

Finally the cart is moving faster than the treadmill.

The direction of treadmill opposes the direction of the cart. The cart is never "faster" than the treadmill if it's direction opposes that of the treadmill. If you're referring to magnitudes of speed, you need some frame of reference. Using the floor as a frame of reference (schroeder's choice), the cart's magnitude of speed never exceeds that of the treadmill with the current carts. The cart advances 1 revolution for each time the turntable retreats 2.33 revolutions. Relative to the treadmill or to the floor, the cart speed is greater than the wind speed. I've previous restated the claim as:

|cart_speed - treadmill_speed| > |wind_speed - treadmill_speed|

in order to eliminate any frame of reference issues.

 

[Subductionzon]

The direction of treadmill opposes the direction of the cart. The cart is never "faster" than the treadmill if it's direction opposes that of the treadmill. If you're referring to magnitudes of speed, you need some frame of reference. Using the floor as a frame of reference (schroeder's choice), the cart's magnitude of speed never exceeds that of the treadmill with the current carts. The cart advances 1 revolution for each time the turntable retreats 2.33 revolutions. Relative to the treadmill or to the floor, the cart speed is greater than the wind speed. I've previous restated the claim as:

|cart_speed - treadmill_speed| > |wind_speed - treadmill_speed|

in order to eliminate any frame of reference issues.

Yes, I could have worded that a bit better. Let's just say that the speed of the cart measured from its wheels is faster than the speed of the turntable with respect to the ground. It can be a bit tricky when you are comparing various frame of reference. And of course at this point the cart is faster than the modeled "wind".

 

[swerdna]

Since the TT is a circular device, I suppose there is room for confusion. Your drawing is correct in a frontal view of the TT and the cart is running on the front edge closest to the viewer. In the video I saw, the cart began translating while it was running on the back edge, farthest from the viewer. In that frame of reference, the TT is turning CW, the edge of the TT is moving from Left to Right, the cart is moving from Right to Left and the wheel is turning CCW. Correct?

That’s a disappointing answer but let’s move on.

To keep this as simple as possible let’s forget about directions and the prop and just consider what’s happening between the TT surface and wheel. The TT has one motion in one direction and this motion gradually increases until it reaches a terminal speed. The wheel has two motions - the motion of the wheel as a whole with or against the motion of the TT (motion of the axle) and the motion of the wheel around the axis of the axle (revolving).

(1) If the axle of the wheel is traveling at the same speed and in the same direction as the TT, then the wheel doesn’t revolve.

(2) If the axle of the wheel travels at a different speed than the TT (faster or slower), then the wheel revolves.

(3) The greater the difference in speed between the axle and the TT, then the greater the speed that the wheel revolves.

(4) In the video the difference in speed between the axle and the TT surface gradually and constantly increases until the axle reaches a terminal speed in the opposite direction from the TT.

(5) As the difference in speed between the TT surface and the axle constantly increases (see 4) then it follows that the speed of the wheel’s revolution also constantly increases (see 3).

If “rotating“ means the wheel revolving about it’s axis your claim that “The wheel is rotating slower and slower” is obviously incorrect. Even if by “rotating” you mean the motion of the axle then you are still incorrect. For the wheel to rotate (revolve about it’s axis) slower and slower then the difference in speed between the TT and the axle has to become less and less. Please explain where and when this happens in the video - http://nz.youtube.com/watch?v=MCB1Jczysrk.

 

[vanesch]

vanesch, you were describing a "merry go round".

Yes, thanks, that was the word I was looking for. I wanted an abstract merry go around in which there are no superfluous movements of horses going up and down..., but powered by a motor of course...

 

[ThinAirDesign]

Prop car vs sail cart rules schroder. ??

Looks like the tag line might have to change to "the race that schroder suggested and then dropped because he's afraid of the results".

JB

 

[zoobyshoe]

Ok, fair enough. Convince me that the outcome of a "Bernoulli vs Newton" argument somehow alters the outcome of a DDWFTTW argument and I will engage.

You and Jeff never address the low pressure (favorable to the cart) that is created in front of the prop. Without trying to sort out if that is a Bernoulli, Venturi, or Coanda effect, it remains low pressure, and can only be contributing to the force the prop exerts on the cart.

Jeff outright dismisses the importance of this contribution , claiming it all ends up as "slowed air" in the end, and you don't address it at all, focusing instead exclusively on the air that is compacted behind the prop as the only operative force on the prop:

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.

I've done a little googling and find that the faster, lower pressure air on the front surface of the props blades does not slow down upon meeting the slower air behind the blades, it reacts by forming a vortex with that slower air, which is shed as the blade tip vortex. The lower pressure at the front surface of the blades, therefore, represents a net forward pull on the prop which is automatically applied to the cart body. This forward pull when taken into account gives the cart more momentum than when you only consider a Newton III push from behind.

I do not understand why you guys ignore something so favorable to your argument. It is only when I consider the potential contribution of the forward pull that the cart seems like it might work as claimed.

 

[ThinAirDesign]

You and Jeff never address the low pressure (favorable to the cart) that is created in front of the prop.

If I have not addressed it (and I'll take your word for that), it's because it's SOP relating to basic prop operation. I know for a fact I have stated (but perhaps on other threads) that the definitive characteristic of a prop (vs turbine) is that it moves air from a low pressure region to a high pressure region.

So, in short a prop functions not just by creating a high pressure area to the rear, but also an associated low pressure area to the front. It's the combo that does the trick.

Without trying to sort out if that is a Bernoulli, Venturi, or Coanda effect, it remains low pressure, and can only be contributing to the force the prop exerts on the cart.

Absolutely agree and would be surprised if you could find something that implies I feel differently.

Jeff outright dismisses the importance of this contribution , claiming it all ends up as "slowed air" in the end, ...

Can't speak for Jeff, but if that is his position, I disagree. The prop behaves no differently when installed on the cart than it does when installed on the RC Cessna for which it was designed. I would be very surprised if Jeff feels that there isn't a low pressure in front of a Cessna's prop influencing the motion of the plane.

... and you don't address it at all, focusing instead exclusively on the air that is compacted behind the prop as the only operative force on the prop:

Please show me where I have even *implied* that the "air that is compacted behind the prop [is] the only operative force on the prop".

I do not understand why you guys ignore something so favorable to your argument. It is only when I consider the potential contribution of the forward pull that the cart seems like it might work as claimed.

Again, I don't think you can produce the evidence to support your assertion that I focus on the area behind the prop and ignore the area in front of. You certainly haven't produced a quote to imply that yet at least, and I'll be happily waiting.

JB

 

[zoobyshoe]

If I have not addressed it (and I'll take your word for that), it's because it's SOP relating to basic prop operation. I know for a fact I have stated (but perhaps on other threads) that the definitive characteristic of a prop (vs turbine) is that it moves air from a low pressure region to a high pressure region.

So, in short a prop functions not just by creating a high pressure area to the rear, but also an associated low pressure area to the front. It's the combo that does the trick.



Absolutely agree and would be surprised if you could find something that implies I feel differently.



Can't speak for Jeff, but if that is his position, I disagree. The prop behaves no differently when installed on the cart than it does when installed on the RC Cessna for which it was designed. I would be very surprised if Jeff feels that there isn't a low pressure in front of a Cessna's prop influencing the motion of the plane.



Please show me where I have even *implied* that the "air that is compacted behind the prop [is] the only operative force on the prop".




Again, I don't think you can produce the evidence to support your assertion that I focus on the area behind the prop and ignore the area in front of. You certainly haven't produced a quote to imply that yet at least, and I'll be happily waiting.

OK. This is all I wanted. Any implication I saw was simply the result of your not mentioning it assuming it was understood. I made a spurious connection between your stand and what I take Jeff's to be. Conceptually, then, I can see your cart working, while Jeff's, I cannot.

 

[tsig]

Of course it is -- no one questions that, but we don't argue about whether it's the same wind turbine as the one up on the hill do we? We don't argue about whether it works exactly the same way do we?

What *creates* the wind is irrelevant to a wind powered device -- it's doesn't know nor does it care. Solar wind (outdoors), nuke wind (aircraft carrier), elecric wind (wind tunnel, treadmill), gravity (river), all the same to a turbine, sailboat, DDWFTTW cart.

One more time, a wind turbine, sail and DDWFTTW cart are all examples of wind powered devices -- arguments about what powers the wind that powers the devices are not relevant to the details of how a wind powered device works.

JB

In your example the wind loosely couples the windmill to the steam turbine. You could have done the same thing with a belt and been more efficient.

You have to look at the entire system.

 

[tsig]

Well just imagine than an outdoor wind is being generated by steam powered wind turbines. Does it matter what the power source is that creates the relative wind?

Not to the windmill but it does to anyone trying to get energy out of the system (windmill + steam turbines)

Maybe we can use the electricity from the windmill to heat the water that powers the turbines.
Would that work?

 

[A.T.]

The lower pressure at the front surface of the blades, therefore, represents a net forward pull on the prop which is automatically applied to the cart body.

How can the lower pressure in front of the blade pull the blade? It is still pressure - it doesn't pull, it pushes. But it pushes less than the higher pressure behind the blade.

This forward pull when taken into account gives the cart more momentum than when you only consider a Newton III push from behind.

The cart gets momentum from the difference between the push from behind and the push from the front (pressure difference). There is no pull, only pushing. How should air molecules actually pull something?

 

[A.T.]

You have to look at the entire system.

To see what?

 

[Subductionzon]

Not to the windmill but it does to anyone trying to get energy out of the system (windmill + steam turbines)

Maybe we can use the electricity from the windmill to heat the water that powers the turbines.
Would that work?

No, you can't. Now that would take an over unit device. His point about the cart working is that it does not care where it gets its energy from. There is a lot of energy in the wind that could be harvested to power the cart. There is also a lot of energy from the treadmill producing a relative wind that can power the cart. A lot less energy is harvested by the cart than there is in the actual system.

JB, I know a way that a sail cart could beat the prop cart, we just don't limit the cart to directly downwind. Of course if it is a directly downwind race the cart would win hands down.

 

[zoobyshoe]

How can the lower pressure in front of the blade pull the blade? It is still pressure - it doesn't pull, it pushes. But it pushes less than the higher pressure behind the blade.

The cart gets momentum from the difference between the push from behind and the push from the front (pressure difference). There is no pull, only pushing. How should air molecules actually pull something?

It's a casual manner of speaking to make a point clear. The low pressure "pulls" in the same way a vacuum cleaner "sucks".

 

[ThinAirDesign]

Me:

A wind turbine generates just as much electricity bolted to the deck of an aircraft carrier steaming at 15 knots on a still air day as does one mounted on the ridge in a 15 knot wind. Wind turbine can't know. Wind turbine doesn't care.

Tsig responds:

In your example the wind loosely couples the windmill to the steam turbine.

It matters not how "loosely coupled" the other relationships are -- that's the entire point of the example; there's a solid relationship between the deck and the turbine. The deck moves and the turbine spins.

It doesn't matter that the sun shone and the photosynthesis caused plants to grow which were eaten by herbivores which were then eaten by the carnivores which died and became crude which was refined and burned to make the steam to drive the turbine which moved the freakin' deck. The only relevant point is ... power the deck however you wish and in the end as long as the deck moves, the turbine can't tell the difference.

You have to look at the entire system.

No you don't. To analyze the wind turbine you need not look any farther than what the turbine can see -- the wind. It matters not what creates it.

To analyze the DDWFTTW cart, you need not look any farther than what the turbine can see -- the wind. It matters not what creates it.

JB

 

[schroder]

To keep this as simple as possible let’s forget about directions and the prop and just consider what’s happening between the TT surface and wheel. The TT has one motion in one direction and this motion gradually increases until it reaches a terminal speed. The wheel has two motions - the motion of the wheel as a whole with or against the motion of the TT (motion of the axle) and the motion of the wheel around the axis of the axle (revolving).

(1) If the axle of the wheel is traveling at the same speed and in the same direction as the TT, then the wheel doesn’t revolve.

(2) If the axle of the wheel travels at a different speed than the TT (faster or slower), then the wheel revolves.

(3) The greater the difference in speed between the axle and the TT, then the greater the speed that the wheel revolves.

(4) In the video the difference in speed between the axle and the TT surface gradually and constantly increases until the axle reaches a terminal speed in the opposite direction from the TT.

(5) As the difference in speed between the TT surface and the axle constantly increases (see 4) then it follows that the speed of the wheel’s revolution also constantly increases (see 3).

If “rotating“ means the wheel revolving about it’s axis your claim that “The wheel is rotating slower and slower” is obviously incorrect. Even if by “rotating” you mean the motion of the axle then you are still incorrect. For the wheel to rotate (revolve about it’s axis) slower and slower then the difference in speed between the TT and the axle has to become less and less. Please explain where and when this happens in the video - http://nz.youtube.com/watch?v=MCB1Jczysrk.

OK here is my explanation of your points with regard to the video:
(1). In your video, this does not quite happen. The closest the cart comes to this is right at the very beginning, T = 7 seconds, when the TT starts up and the cart momentarily moves at nearly the same speed, both moving CW. The wheel is obviously turning because the propeller is turning. They are linked by a flexishaft, remember?
(2). Yes. At T = 8 seconds both the cart and the TT are still accelerating and moving in the same direction, CW and the TT linear velocity is greater than the cart’s linear velocity on the wheel circumference.
(3) This is true for as long as they are going in the same direction only! At T = 12 seconds, the cart is moving at it’s Max speed because it is still moving CW with the TT. If you had a tach on the wheel, HERE is where you would record the highest RPM.
(4). Yes the Difference increases because the cart starts to slow down! The TT does not accelerate continuously does it? At T = 13 seconds, the cart is seen to be stationary for a moment on the TT. At this point, the linear velocity on the edge of the wheel is exactly the same as the linear velocity on the surface of the TT. This is a critical point.
(5) NO. Again, the Difference increases because the cart is slowing down and the TT is no longer speeding up. At T = 17 seconds the cart has reached its terminal linear velocity, which is less than at T = 12 seconds. From this point onwards, the TT and the cart are in a heterodyne. The translation of the cart in the CCW direction (measured in m/sec) is the Difference between the linear velocity of the TT surface in m/sec MINUS the linear velocity at the wheel in m/sec.

If you can give me very accurate dimensions of the TT diameter, wheel diameter, and a very accurate measurement from center of TT to center of the track of the wheel, and the RPM of the TT, I can calculate the ballpark figures for the RPM of the wheel at T = 12 seconds (High RPM) and T = 13 seconds (RPM where wheel linear velocity equals TT linear velocity) and at T = 17 seconds (when the wheel linear velocity is LESS then the TT linear velocity and it is translating) But it is much preferable that you put a tach on the wheel and measure this directly.
I can give you a calibration method if and when you are ready to do this test. Thanks!

 

[Subductionzon]

schroder, when are you going to see that the important velocity is the velocity of the cart relative to the turntable. If the velocity of the turntable at the rim is 10 m/s that is modeling a wind speed of 10 m/s. If the cart is stationary w.r.t. the ground that means its wheels are rotating at a speed that would translate as 10m/s. Once the cart starts moving counterclockwise w.r.t. the ground it indicates that the wheel speed of the cart is faster than the "wind" that is modeled by the turntable.

 

[ThinAirDesign]

----- Jeff Reid:

Well just imagine than an outdoor wind is being generated by steam powered wind turbines. Does it matter what the power source is that creates the relative wind?

Not to the windmill but it does to anyone trying to get energy out of the system (windmill + steam turbines)

Yes, and lucky for us we're not trying to get energy out of the system. The cart's a user, not a producer. We don't care how much "wind" we use to go DDWFTTW, nor do we care how much sun/electricity/petrol it took to power that wind.

We don't claim that it's cheap to go DDWFTTW --- just that it's easy.

Maybe we can use the electricity from the windmill to heat the water that powers the turbines. Would that work?

No better than us claiming that we could put a generator on the cart to power the treadmill. That's a flat impossible claim.

Just as it takes an *outside* source of power to turn a wind turbine, it takes an *outside* source of power to go DDWFTTW. It's just that neither we nor the turbine nor the cart give a flyin' flip what that source is.

JB

 

[rcgldr]

never address the low pressure (favorable to the cart) that is created in front of the prop.

I thought I covered this in an earlier post.
The prop accelerates the air upwind, and it doesn't matter how much of this acceleration takes place fore (via reduction in pressure) or aft of the prop (via increase in pressure), just the total acceleration. (more on this below).

claiming it all ends up as "slowed air" in the end

The upwind acceleration of the air, ends up slowing down the speed of the air in the vicinity of the prop with respect to the ground, and it's reduction of wind speed versus ground speed that powers any wind powered (wind speed versus ground speed) device.

I've done a little googling and find that the faster, lower pressure air on the front surface of the props blades does not slow down upon meeting the slower air behind the blades.

For a typical situation, the air speed across the prop disk changes little, while the pressure increases significantly, and this is called pressure jump. The thrust from a prop equals this pressure jump times the cross sectional area swept by the prop disk. The pressure jump increases the air pressure above ambient, and the air then continues to accelerate aft of the prop as it's pressure returns to ambient. There is a term called "exit velocity" which is the speed of the air when it's pressure returns to ambient. Link to NASA prop analysis which covers this fairly well including your concern about Bernoulli (it applies before and after but not during the time when air crosses through the propeller "disk", because the propeller peforms work on the air):

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

 

[rcgldr]

When you look at the whole system you see that the windmill on the carrier is being powered by the carrier's steam turbines.

Well just imagine than an outdoor wind is being generated by steam powered wind turbines. Does it matter what the power source is that creates the relative wind?

Not to the windmill but it does to anyone trying to get energy out of the system (windmill + steam turbines). Maybe we can use the electricity from the windmill to heat the water that powers the turbines.

I stated steam powered wind turbines, ones that input mechanical (thermal) energy and output wind, not windmills. My point was to compare a steam powered propeller or turbine used to generate a wind to drive the cart to the steam powered spools that retracted the wires to drive a Brennan torpedo.

 

[ThinAirDesign]

JB, I know a way that a sail cart could beat the prop cart, we just don't limit the cart to directly downwind. Of course if it is a directly downwind race the cart would win hands down.

Yep, and this is why I'm likely not doing any test until he specifies and we agree on what he's asking for. So far he's being a wuss and won't specify.

Clearly (to those who utilize the normal laws of physics) if we place a sail cart and a prop cart next to each other and aim them both DDW, the prop cart will destroy the sail cart.

Schroder(who subscribes to a rather abnormal set of physics laws) has stated he believes otherwise but apparently has come into some doubts recently.

JB

 

[A.T.]

It's a casual manner of speaking to make a point clear. The low pressure "pulls" in the same way a vacuum cleaner "sucks".

Yes, in casual manner you can say that the lower pressure is pulling, instead that the higher pressure is pushing more. But this are only two ways to describe the very same thing: net force from relative pressure difference.

You seem to argue (and I rejected that) that these are two different effects that add up, and both need to be taken into account / considered as separate contributions:

This forward pull when taken into account gives the cart more momentum than when you only consider a Newton III push from behind.

It is only when I consider the potential contribution of the forward pull that the cart seems like it might work as claimed.

 

[rcgldr]

It's a casual manner of speaking to make a point clear. The low pressure "pulls" in the same way a vacuum cleaner "sucks".

Yes, in casual manner you can say that the lower pressure is pulling, instead that the higher pressure is pushing more. But this are only two ways to describe the very same thing: net force from relative pressure difference. You seem to argue (and I rejected that) that these are two different effects that add up, and both need to be taken into account / considered as separate contributions.

Well they could be considered separate aspects of a propeller. How the propeller reduces the air pressure fore of the propeller doesn't have to be related to how the propeller increases pressure at the propeller, and the two effects combined contribute to acceleration of air. One major difference is that Bernoulli applies fore of the propeller, but not at the propeller where the total energy of the air is increased (relative to the prop), mostly the pressure. Aft of the propeller disk, Bernoulli applies again as the air continues to accelerate and decrease in pressure back to ambient.

 

[A.T.]

How the propeller reduces the air pressure fore of the propeller doesn't have to be related to how the propeller increases pressure at the propeller,

My point was that it is the pressure difference that matters, and not how it is achieved: reducing pressure fore of the propeller, increasing pressure aft of the propeller or both. I do not understand why zoobyshoe thinks that DDWFTTW could only work if pressure fore of the propeller is reduced:

It is only when I consider the potential contribution of the forward pull that the cart seems like it might work as claimed.

This "forward pull by lower pressure" is just a casual description of the "bigger push by the higher pressure against smaller push by lower pressure". These two don't add up. They are the same thing.

 

[rcgldr]

I do not understand why zoobyshoe thinks that DDWFTTW could only work if pressure fore of the propeller is reduced.

Probably an efficiency issue. The propeller has to draw air from somewhere to accelerate the air behind the prop. In order to draw air the propeller has to reduce pressure. If the source of the air came from a diverted flow perpendicular to the prop, the net thrust would be less even though the pressure differential at the prop would be the same, because there'd be a second intake diverting surface in front of the prop with low pressure on the prop side and ambient pressure on the forward side, opposing the thrust. Ultimately the thrust is related to the acceleration of air affected by a prop times the mass of the affected air (the intergral sum of the component parts). The Newton 3rd law pair here is the force exerted by the prop onto the air and the coexistant force exerted by the air onto the prop.

 

[zoobyshoe]

You seem to argue (and I rejected that) that these are two different effects that add up, and both need to be taken into account / considered as separate contributions:

I'm not sure what you're objecting to. Are you objecting to that fact there are two separate effects (which there are) or are you objecting to an implied need to add them separately? I am aware that if you know the pressure difference, that's all you need. My complaint is that Jeff doesn't ever seem to figure in the low pressure created simply by virtue of the air passing over the front surface of the prop, air which is accelerated, not slowed, and which is shed as vortices at the wing tips, as opposed to being compacted behind the propeller. This effect increases the pressure difference. It's favorable to the cart. The "pull" it creates does, indeed, make the "push" far more effective. Jeff explains thrust from the prop on the cart according to how much air is slowed: his own, idiosyncratic way of apprehending the situation.

 

[swerdna]

(1). In your video, this does not quite happen. The closest the cart comes to this is right at the very beginning, T = 7 seconds, when the TT starts up and the cart momentarily moves at nearly the same speed, both moving CW. The wheel is obviously turning because the propeller is turning. They are linked by a flexishaft, remember?

I didn’t say it was, I said IF it was. The TT and the cart have to move at different speeds for the wheel to revolve. I hope we agree on that.

(2). Yes. At T = 8 seconds both the cart and the TT are still accelerating and moving in the same direction, CW and the TT linear velocity is greater than the cart’s linear velocity on the wheel circumference.

So we agree that it’s the TT and cart moving at different speeds which causes the wheel to revolve.

(3) This is true for as long as they are going in the same direction only! At T = 12 seconds, the cart is moving at it’s Max speed because it is still moving CW with the TT. If you had a tach on the wheel, HERE is where you would record the highest RPM.

I disagree. The TT and cart are traveling at different speeds when they are moving in the same direction, when the cart “stops”, and when it travels against the motion of the TT. During this whole process the difference in speed between the TT and cart is steadily increasing and the speed of the revolutions of the wheel are also increasing. The highest RPM would be recorded when the cart reaches terminal speed against the motion of the TT.

(4). Yes the Difference increases because the cart starts to slow down! The TT does not accelerate continuously does it? At T = 13 seconds, the cart is seen to be stationary for a moment on the TT. At this point, the linear velocity on the edge of the wheel is exactly the same as the linear velocity on the surface of the TT. This is a critical point.

Do you agree that the differential speed increase caused as the cart slows down means that the revolutions of the wheel speed up?

(5) NO. Again, the Difference increases because the cart is slowing down and the TT is no longer speeding up. At T = 17 seconds the cart has reached its terminal linear velocity, which is less than at T = 12 seconds. From this point onwards, the TT and the cart are in a heterodyne. The translation of the cart in the CCW direction (measured in m/sec) is the Difference between the linear velocity of the TT surface in m/sec MINUS the linear velocity at the wheel in m/sec.

What you call “slowing down” could equally (and probably more correctly) be described as “moving in that direction less“. When the cart is moving against the motion of the TT it is “moving in that direction less” every bit as much as it is when it is “slowing down“ (more so in fact).

If you can give me very accurate dimensions of the TT diameter, wheel diameter, and a very accurate measurement from center of TT to center of the track of the wheel, and the RPM of the TT, I can calculate the ballpark figures for the RPM of the wheel at T = 12 seconds (High RPM) and T = 13 seconds (RPM where wheel linear velocity equals TT linear velocity) and at T = 17 seconds (when the wheel linear velocity is LESS then the TT linear velocity and it is translating) But it is much preferable that you put a tach on the wheel and measure this directly.
I can give you a calibration method if and when you are ready to do this test. Thanks!

As I explained to another member earlier I no longer have an operating cart for my turntable as I’ve used some pieces for another cart. To do your experiment I would have to reconstruct the cart and also purchase and fit a tachometer. I’m not prepared to spend the time and money on a test that I can’t see any sense in or need for. I don’t need to fit a tachometer to the cart to know that the wheel increases it’s revolutions until it reaches it’s terminal speed. That you can’t see this I find really amazing, especially for an engineer. I will continue to give serious consideration to any argument you wish to present that you’re right and I’m wrong.

 

[vanesch]

My complaint is that Jeff doesn't ever seem to figure in the low pressure created simply by virtue of the air passing over the front surface of the prop, air which is accelerated, not slowed, and which is shed as vortices at the wing tips, as opposed to being compacted behind the propeller.

You can try to figure all this out, but it is extremely complicated. It is much easier to trust Newton's 3rd law, which says that action = reaction, together with Newton's second law (F = m a), but here in the form: F = d p / d t.

In other words, the trick is first to calculate the entire force that is exerted (by the propeller) on the air, and to find that out, we make the momentum balance: we see what was the momentum of the air "before" and then the momentum of the air "after".

Remember that momentum of something = mass of something x velocity.

Now, here we make an approximation: we consider that there is air that is left "untouched" by the propeller, and an amount of air that is "uniformly displaced" by the propeller. In reality, there will be different categories of air, those that are untouched, there are some that are a bit displaced, there are others that are more displaced etc... so we should sum all these different contributions, but we don't, we simply say that some amount of air is affected in the same way by the propeller, and the rest not at all.

The momentum balance of the unaffected air will of course be 0: it will have the same momentum before and after. So we don't need to take it into account ; what counts is only what is displaced by the propeller. Per unit of time, a certain mass of air is affected by the propeller, and we call that "m". That means that the mass in an amount of time dt is M = m dt. That amount of mass had a velocity v_in before it was affected by the propeller, and is then accelerated and ends up at a velocity v_out (it is the uniformity of v_out which is the approximation we make: not all bits of air are accelerated to the same velocity).

That means that the momentum "after" is M x v_out while the momentum "before" is M x v_in. (for the non-affected air, the momentum in was MMM v_vin and the momentum out is also MMM v_in, where MMM is the big mass of the unaffected air ; so the balance doesn't matter).

So the momentum gained, by the air, in a time dt is given by M x v_out - M x v_in.
That momentum was gained in an amount of time dt, so the force (Newton's first law) that must have been exerted on that air must have been: F = ( M x v_out - M x v_in ) / dt.

Or: F = m x (v_out - v_in). (as M / dt = m).

So, no matter what or how, this must be the force exerted on the air. And as it is the propeller that does this, the opposite force must be the one exerted on the propeller.

So, if we know that a certain amount of air changes velocity, all the rest doesn't actually matter: it is this momentum balance which counts, and it will be the final sum of all the little bits of pressure that work on all the pieces of surface of the propeller. We don't have to go into that detailed description: we have the end result of the balance.

 

[vanesch]

If you can give me very accurate dimensions of the TT diameter, wheel diameter, and a very accurate measurement from center of TT to center of the track of the wheel, and the RPM of the TT, I can calculate the ballpark figures for the RPM of the wheel at T = 12 seconds (High RPM) and T = 13 seconds (RPM where wheel linear velocity equals TT linear velocity) and at T = 17 seconds (when the wheel linear velocity is LESS then the TT linear velocity and it is translating) But it is much preferable that you put a tach on the wheel and measure this directly.
I can give you a calibration method if and when you are ready to do this test. Thanks!

Ok, let us take the example: radius of the turntable 1 m (take it that this is the track of the wheel, and that the table itself is 1.2 m), angular rotation 6.28 rad/second of the table, diameter of the little wheel 5 cm.
So, show us your calculation for the angular rotation of the axle of the wheel
1) when the cart is perfectly in sync with the TT
2) when the cart is momentaneously stationary wrt the outside ground (when it flips from going CW to CCW on the TT)
3) at the end when it is moving CCW. BTW, it would be fun if you could calculate the axle rotation velocity in the hypothetical case that the arm of the cart were now turning CCW at also 6.28 rad / s.
Don't forget to tune your heterodyne

 

[swerdna]

I’ve built an air cart version of the Brennan Torpedo but I’m having a problem with it. The string is being pulled from a spool on the prop shaft and the cart is being held against a backstop until it gets up to thrust speed. It works fine except when the cart leaves the stop it surges slightly which makes the spool overrun and that tangles the string up. It was worst when I tried it with nylon fishing line. That horrible stuff is just designed to tangle. If I put enough friction on the spool to stop it overrunning it puts too much friction on the system and it doesn’t work. Any ideas? I may just concentrate on a water model as I think it would be more constant and stable.

 

[schroder]

I disagree. The TT and cart are traveling at different speeds when they are moving in the same direction, when the cart “stops”, and when it travels against the motion of the TT. During this whole process the difference in speed between the TT and cart is steadily increasing and the speed of the revolutions of the wheel are also increasing. The highest RPM would be recorded when the cart reaches terminal speed against the motion of the TT.





What you call “slowing down” could equally (and probably more correctly) be described as “moving in that direction less“. When the cart is moving against the motion of the TT it is “moving in that direction less” every bit as much as it is when it is “slowing down“ (more so in fact).



As I explained to another member earlier I no longer have an operating cart for my turntable as I’ve used some pieces for another cart. To do your experiment I would have to reconstruct the cart and also purchase and fit a tachometer. I’m not prepared to spend the time and money on a test that I can’t see any sense in or need for. I don’t need to fit a tachometer to the cart to know that the wheel increases it’s revolutions until it reaches it’s terminal speed. That you can’t see this I find really amazing, especially for an engineer. I will continue to give serious consideration to any argument you wish to present that you’re right and I’m wrong.

If you put the TT and cart back together, with a tachometer on the TT and a tachometer on the wheel, you will have built a very valuable teaching aid for rotational and translational motion. In particular, a very nice demonstration of a mechanical heterodyne which every mechanical engineering department and physics department will be happy to have. That is where you will get the most benefit out of this, not by chasing after some DDWFTTW myth. You have something valuable there if only you recognize it! You find it amazing that I believe the cart slows down to reach the steady state. I find it amazing that you believe in DDWFTTW. One of us is wrong! The best way to determine that is to do the test with the tachometer and I guarantee you will be amazed to see the cart is actually slowing down. There is nothing magical or mystifying about that. Translational motion is replacing rotational motion while energy is being conserved. What I am explaining is according to the laws of physics while DDWFTTW is not! I find it amazing that you choose to believe the stranger of the two claims rather than believe in a heterodyne. I will try to find some more examples of heterodynes in mechanical machinery. They are rare, which is why your TT and cart is a valuable asset. I believe heterodynes also happen on large rollers in paper mills and in some other machinery used in the lumber industry, as well as inside gas turbines. Once you understand the principle, the cart on the TT makes perfect sense. I think you are wasting your time chasing the wind when your TT is far more valuable and certainly worth the cost of at least one digital tachometer and preferably two. But that is entirely up to you. I will try to make some drawings to explain how translation takes the place of rotation, but I am not good at computer animations so it may be a bit crude. Give what I say some consideration as you will benefit if I am right.

 

[Subductionzon]

schroder, I am pretty sure which one of us is wrong! No matter how many times you use the word I am pretty sure that everone but you agrees that there are no radio frequencies involved here. So unless you have some unique definition for the word "heterodyne" I would appreciate it if you would use mechanical terms here. You still have not answered my claim that the relative velocity that you should be looking at is the velocity of the cart relative to the turntable. Two questions. First what are you measuring velocity relative to? Second regardless of what you are measuring velocity relative to, what is the velocity of the cart measured to relative to the turntable? ie, if you put a speedometer on the wheels of the cart what value would you get? And if you measured the speed of the outer edge of the turntable in m/s what value would you get?

 

[vanesch]

If you put the TT and cart back together, with a tachometer on the TT and a tachometer on the wheel, you will have built a very valuable teaching aid for rotational and translational motion.

THAT's for sure !

You find it amazing that I believe the cart slows down to reach the steady state. I find it amazing that you believe in DDWFTTW. One of us is wrong!

THAT's for sure too !

Translational motion is replacing rotational motion while energy is being conserved. What I am explaining is according to the laws of physics while DDWFTTW is not! I find it amazing that you choose to believe the stranger of the two claims rather than believe in a heterodyne.

You have clearly not understood what is the mathematics behind a heterodyne, nor what is the mathematics behind the concept of velocities, nor the most basic notions of kinematics. However, you are somehow *convinced* that you know something about it, you even think you have some exceptional expertise on it.

I've seen other people self-deluded like that, but I have never seen people who do this on an elementary level like this - that's what fascinates me with you. As I said, even without a formal training, I don't know many people who would make the claims about the rolling wheel you are making (especially that the wheel is turning when the cart goes in sync with the table). For most people this would even intuitively be obvious - even though they don't have the formal tools to confirm it. You have distorted your view entirely by thinking that you have a superior understanding of the thing (which is nothing else but a complete confusion between two totally unrelated phenomena, which are frequency mixing in a non-linear component of a scalar signal and vector addition in a velocity field) to a point which is downright fascinating.

But that is entirely up to you. I will try to make some drawings to explain how translation takes the place of rotation, but I am not good at computer animations so it may be a bit crude. Give what I say some consideration as you will benefit if I am right.

Will it benefit to you if you finally will see that you are wrong on this ? If so, swerdna's test might have some use. But the outcome is evident.

 

[vanesch]

You still have not answered my claim that the relative velocity that you should be looking at is the velocity of the cart relative to the turntable.

The problem is that even the notion "relative velocity" is distorted by schroder. Some posts back, when I asked him if he disputed the claim that if a cart was going 2 m/s to the left, and a tread was going 10 m/s to the right, the velocity of the cart wrt the tread was 12 m/s, and for sure he disputed that (see post 755). He claimed that the relative velocity was 8 m/s. (ok, it might have been mph instead of m/s). At that point, I still thought that there might be a misunderstanding about the actual setup - or that he was just trolling.

But it becomes more and more clear that it was not a misunderstanding concerning the setup.

Schroder doesn't understand what it means "relative velocity", or how to obtain it when you have the individual velocities, as he confuses this with mixing frequencies of 2 and 10 (in other words, he gives himself the liberty to alter signs at will in the velocity composition).

I have never met anyone who has such a profound confusion. It is a remarkable phenomenon, and a true challenge to find out how to tackle it, pedagogically. Have you EVER met anybody who disputed such an elementary claim with such vehemence ?
(especially somebody who claims to be an engineer ?)