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

[schroder]

Everyone seems to be ignoring this post, but I think it's quite an interesting post in this discussion. Personally, I'm not too keen on this step of your process of 'transforming' the experiment:

I actually read that post and found one question is missing; the first one, which should be “Are you willing to accept that a cart being driven by the wind can go DDWFTTW?” My answer is NO and that makes all the other questions irrelevant! But without that first crucial question, I would have answered Yes to all. It is that first question which I refuse to accept because it is NOT happening on the treadmill once you remove the inversion in the frames which I just pointed out (again)

 

[zoobyshoe]

It is also important not to invert the way you apply the frames. When you are sitting in your car and the engine is driving the wheels to turn against the road and you want to know how fast your car is going. Do you measure the velocity between the center point of the engine and the centerline of the wheels? Or do you measure the velocity between the centerline of the wheels and a centerline drawn on the road? Of course you measure it between the wheels and the road. If you measured between the wheels and the engine it of course is always constant and you get a wrong result. When the cart is being driven by the wind as the engine and is running on the road, you measure the velocity between the wheel and the road, not between the wheels and the engine (the wind). When the cart is being driven on the tread as the engine and driving into the air with the propeller. You measure between the centerline of the propeller and a centerline in the air. You do not measure between the centerline of the wheels and the centerline of the engine (the tread). You must be consistent in both frames. By using the wheels and the tread to take the measurement on the treadmill you are inverting from the down wind frame and that is giving you the ridiculous result of faster than the tread hence faster than the wind. Be consistant in both frames. Do not invert things!

I look at the speedometer.

 

[cristo]

You're right of course, this is the part which is disputable. However, the idea is that the effect of the rotation will be smaller when the radius of curvature is bigger. It was indeed the caveat I made in the beginning.

I must have missed that caveat but, yes, I can accept that as a plausible approximation.

I actually read that post and found one question is missing; the first one, which should be “Are you willing to accept that a cart being driven by the wind can go DDWFTTW?” My answer is NO and that makes all the other questions irrelevant!

Actually, this question is not relevant. What vanesch is showing is that the experiment on a turntable is an equivalent experiment to what he calls the 'true outdoor test.' He's not asking whether or not you think the outdoor test gives a positive result: he's asking that if the outdoor test gives a positive result, then you would accept it as proof of DWFTTW. That must be true, no, since the 'true outdoor test' is the definition of DWFTTW? If you don't agree with this point, then the discussion might as well be over, since if you're not willing to accept the results of an experiment as proof, then you are no longer practicing science.

But without that first crucial question, I would have answered Yes to all.

Good. So you agree that the turntable test is equivalent the true outdoor test? Thus, since the OP shows a video of the turntable test giving a positive result, then you agree the true outdoor test will give a positive result, also? So, we're done!

 

[schroder]

I must have missed that caveat but, yes, I can accept that as a plausible approximation.






Good. So you agree that the turntable test is equivalent the true outdoor test? Thus, since the OP shows a video of the turntable test giving a positive result, then you agree the true outdoor test will give a positive result, also? So, we're done!

Nice Try! Sure they are equivalent and since the TT test clearly shows the cart is moving slower than the TT then I conclude it will also move slower in the wind!

 

[A.T.]

"Here" means in the situation we're discussing: we have a cart going, say, 10 mph over the ground. Wind speed is 0 with respect to the cart. Ground speed is 10 mph with respect to the cart.

What amount of mass should we plug into formula: K=1/2mv^2 to find out how much energy is available to the cart from the ground?

The energy is not available from the ground. It is available from the speed difference between ground and air. The energy received by the cart is frame dependent. It is useless to examine if the cart could accelerate further. You could easilly pick a reference frame where the cart is losing energy by going faster relative to the ground.

What matters for acceleration are forces acting on the cart: Due to transmission, the propeller-thrust-force is higher than the wheel-force needed to drive the propeller, so the cart accelerates further.

 

[zoobyshoe]

The energy is not available from the ground. It is available from the speed difference between ground and air.

You are quite right that the energy is harvested from the difference in the relative velocity of the ground and air! The ground speed and difference between air and ground speed just happen to be exactly the same thing at this point, because the cart and wind are going the same speed in the same direction.

That being the case, what number for mass should we plug into K=1/2mv^2? The ground speed is 10mph, what amount of mass does it represent with respect to the cart?

 

[vanesch]

Nice Try! Sure they are equivalent and since the TT test clearly shows the cart is moving slower than the TT then I conclude it will also move slower in the wind!

You are claiming that the velocity of the cart *wrt the turntable* is smaller than the velocity of the *air* wrt to the turntable ? Or are you claiming that the velocity of the cart *wrt the ground* is smaller (and opposite) to the velocity of the turntable (wrt the ground) ?

Because the equivalence we are after is the FIRST, not the SECOND set of velocities.

This is so terribly trivial that I think you're trolling, honestly.

 

[A.T.]

You are quite right that the energy is harvested from the difference in the relative velocity of the ground and air! The ground speed and difference between air and ground speed just happen to be exactly the same thing at this point, because the cart and wind are going the same speed in the same direction.

That being the case, what number for mass should we plug into K=1/2mv^2? The ground speed is 10mph, what amount of mass does it represent with respect to the cart?

What exactly is K and how is it relevant to examine if the cart could accelerate further? Again: What matters for acceleration are forces, not kinetic energies calculated in some arbitrary reference frame. When you have computed the acceleration of the cart using forces, you can use the formula above to compute the gain in KE of the cart. Then you plug the cart's mass into the formula.

 

[schroder]

You are claiming that the velocity of the cart *wrt the turntable* is smaller than the velocity of the *air* wrt to the turntable ? Or are you claiming that the velocity of the cart *wrt the ground* is smaller (and opposite) to the velocity of the turntable (wrt the ground) ?

Because the equivalence we are after is the FIRST, not the SECOND set of velocities.

This is so terribly trivial that I think you're trolling, honestly.

I don’t know what trolling is. I am seriously trying to establish the truth here about this cart.

If you have two cars passing each other on a two way road and all you wanted to know is the relative velocity of the cars, relative to each other, you could pick either car to be the reference for the frame and get the correct result. But suppose two cars are passing each other and you wanted to know the difference between their velocities. Picking either car as your reference will always give you the sum. To get the difference you need to pick some other reference, in this case the road. Now you can measure the velocity of each car independently with respect to the road and compare them to get the difference. In the down wind frame, you want to know the velocity of the cart with respect to the road, not the wind, so you pick the road as your reference frame. Even though there is also velocity between the wind and the cart in that frame, you ignore this in calculating your velocity because it is obviously not of interest to you. You could pick the wind as your reference but your numbers would not make sense unless you applied a correction factor to convert them into road velocity which is what you want to know. That one is obvious. In the treadmill test, it is not so obvious and you are picking the tread because you see the cart is moving wrt the tread. It is a little harder to see it moving wrt the wind, but that is what you really should be interested in! All you really need to do is look at the nature of the beast. On the tread, it is using wheels for input from the tread, and a propeller for output in the air. It is working into the air and it is moving with respect to that medium. Sure, there is motion between the cart and the tread but that is about as interesting as the motion between the cart and the wind in the wind frame. That is the motion between the cart and what is driving it. That is not of interest. You want to know how fast the cart is going in the medium it is driving into, not the medium which is driving it. You really want the velocity into the wind that is what the cart is driving into. I know this is a bit hard to see but you need to clarify this in your own mind. Once you adopt the wind or floor as the reference frame in the treadmill test, you get believable numbers, but unfortunately they are much less than the tread velocity. But that is exactly in accordance with any machine which has losses between input and output. If you use the tread as your reference you get these fantastic numbers which are greater than tread velocity. Very exciting but not true! I will accept the numbers that make sense even if they are not very impressive. Nature has designed the most efficient thing that will go as fast as possible in the wind, it is a molecule of air! And all it can do is wind velocity. I don’t think we can improve on that by using technology which has been around since ancient times. If you still want to believe in this go ahead, but it will be short lived once someone races tumbleweed against one of these carts. Then you will realize you should have been measuring the velocity as I have told you.

 

[zoobyshoe]

What exactly is K and how is it relevant to examine if the cart could accelerate further? Again: What matters for acceleration are forces, not kinetic energies calculated in some arbitrary reference frame. When you have computed the acceleration of the cart using forces, you can use the formula above to compute the gain in KE of the cart. Then you plug the cart's mass into the formula.

Sorry, that wasn't rigorously expressed. Should have put E_k=1/2mv^2.

As Vanesch will confirm you have to be in a frame, and any frame you pick is OK as long as you stay with that frame. Thus far I have been speaking about the cart's frame. We can analyze from another frame if you want, but whatever frame we pick we have to stick to that frame.

The reason I am asking about energy is because earlier I asserted that the cart is in a situation where it no longer has any energy available to it outside the energy represented by it's own momentum. The cart has a certain momentum, which is its mass times its velocity. We can determine its kinetic energy from the same things, therefore, it's momentum represents a certain amount of energy.

You have asserted that the speed difference between the ground and air represents "plenty" of energy. If that is the case, then we ought to be able to get a ballpark figure pretty easily by determining what mass should be plugged into the formula.

 

[vanesch]

I don’t know what trolling is. I am seriously trying to establish the truth here about this cart.

Ok, then there are some *very elementary* misunderstandings to be cleared up before we can go any further.

If you have two cars passing each other on a two way road and all you wanted to know is the relative velocity of the cars, relative to each other, you could pick either car to be the reference for the frame and get the correct result.

Yes. You could also take the difference of the velocities, as measured in the ground frame. In fact, what is important is that no matter in what frame you measure the velocities of car 1 and car 2, (and you'll agree with me that v1 and v2 will be dependent on the choice of that frame), the DIFFERENCE, that is v = v1 - v2 will always give the same result, no matter in what frame one calculates it. I guess you agree with that ?

But suppose two cars are passing each other and you wanted to know the difference between their velocities. Picking either car as your reference will always give you the sum.

Are you still talking about the same setup ? Because the relative velocity IS of course exactly this difference.

To get the difference you need to pick some other reference, in this case the road. Now you can measure the velocity of each car independently with respect to the road and compare them to get the difference.

Sorry, I'm lost as what you are trying to say.

Let's say the two cars are driving north on a north-south road, right ? Car A is driving 50 km/h, and car B is driving driving 60 km/h, wrt to the road of course. Well, the relative velocity is of course 60 - 50 = 10 km/h.
But if you measure car B's velocity in car A's frame (with a radar or something), well, you have that car B's velocity is 10 km/h and car A's velocity is 0 (of course), so again we have 10 - 0 = 10 km/h.

You say something different or not ?

Now, imagine that there's also a train track along the road. Train C runs north with a velocity of 100 km/h. What you see is that car A is running at -50 km/h and car B is running at -40 km/h. (in other words, from the train's PoV, the cars have a southward velocity).

Again, the relative velocity is found by: -40 - (-50) = +10 km/h.

Another train is running south on the track at 160 km/h. As seen from the train, car A is speeding north at a velocity of 210 km/h, and car B is speeding north at a velocity of 220 km/h.

But again, the relative velocity: 220 - 210 = 10 km/h.

The relative velocity is the same, no matter in what reference frame one has calculated it. So it is a frame-independent quantity.

You agree with that or not ?

In the down wind frame, you want to know the velocity of the cart with respect to the road, not the wind, so you pick the road as your reference frame.

Yes.

Even though there is also velocity between the wind and the cart in that frame, you ignore this in calculating your velocity because it is obviously not of interest to you. You could pick the wind as your reference but your numbers would not make sense unless you applied a correction factor to convert them into road velocity which is what you want to know.

Huh ?

No matter in what frame I have my velocities (which are of course frame-dependent), any DIFFERENCE gives you frame-independent relative velocities. There are no "correction factors" to be applied.

That one is obvious. In the treadmill test, it is not so obvious and you are picking the tread because you see the cart is moving wrt the tread. It is a little harder to see it moving wrt the wind, but that is what you really should be interested in!

The motion wrt the air is extremely simple in the treadmill test: it is the same as the motion wrt the ground, as the air is still wrt the ground.

All you really need to do is look at the nature of the beast. On the tread, it is using wheels for input from the tread, and a propeller for output in the air. It is working into the air and it is moving with respect to that medium. Sure, there is motion between the cart and the tread but that is about as interesting as the motion between the cart and the wind in the wind frame. That is the motion between the cart and what is driving it. That is not of interest. You want to know how fast the cart is going in the medium it is driving into, not the medium which is driving it.

Tell me, say there is a wind of 50 km/h and if a sailing cart is going at 40 km/h and a car is driving at 40 km/h in the same direction, what is their relative velocity ? Are you going to say that we shouldn't measure the velocity of the sailing cart wrt the ground because the "driving medium" is the air ?? Is the cart going to take over the car or vice versa ?

You really want the velocity into the wind that is what the cart is driving into. I know this is a bit hard to see but you need to clarify this in your own mind. Once you adopt the wind or floor as the reference frame in the treadmill test, you get believable numbers, but unfortunately they are much less than the tread velocity.

We want to know the velocity with respect to the floor on which the cart is driving ! In the outdoor test, this is the floor, and in the treadmill test, this is the treadmill. It is the thing that is touched with the wheels of the cart.

Imagine a long ship. Do your test on the ship. Are you going to compare to the ship, or to the water ?

But that is exactly in accordance with any machine which has losses between input and output. If you use the tread as your reference you get these fantastic numbers which are greater than tread velocity.

Sure, and that's what this is about here. Let us consider another case: the "driving of a brick on the road".

Claim: a brick dropped on the road will go at 30 km/h. Proof (according to you): Drop the brick on a treadmill. Of course the brick sticks to the treadmill after some bouncing. You have to look upon its velocity wrt to the ground. It is going at 30 km/h. Hence, proof: the brick is doing 30 km/h when dropped on a road.

Why should I use the reference of the GROUND when I use a treadmill ? In *this* case, you'd be willing to accept that we have to express the velocities in the frame of the treadmill, right ? Now, the velocity of the treadmill is 30 km/h, and the brick is also going at 30 km/h (both measured in the ground frame), so the relative velocity of the brick wrt the treadmill is 0 (30 - 30).

So this experiment on the treadmill shows that a brick we would drop on an outdoor road would not move (wrt to the road). But in order to deduce this on the treadmill we had to work in the frame of the treadmill.

With the cart, it is the same. We work wrt to the treadmill (which will have all velocities wrt to the road in the outdoor test). And wrt the treadmill, the cart is going at 40 km/h (it is doing - 10 km/h in the ground frame, and hence the relative velocity wrt to the treadmill is -10 - 30 km/h = - 40 km/h - like our brick was doing 0 km/h).

In the same way, the air is doing 0 km/h in the ground frame, and doing -30 km/h in the treadmill frame (0 - 30 = -30). So in the treadmill frame, the cart is going faster than the wind (-40 versus -30). As in this experiment, the treadmill frame is the equivalent of the "road" frame (think of the brick!), we have established an equivalent experiment that shows that the cart is going at 40 km/h when the wind is going at 30 km/h, if ever we see the cart move at 10 km/h versus the floor in the opposite direction as a treadmill that runs at 30 km/h.

 

[A.T.]

As Vanesch will confirm you have to be in a frame, and any frame you pick is OK as long as you stay with that frame. Thus far I have been speaking about the cart's frame. We can analyze from another frame if you want, but whatever frame we pick we have to stick to that frame.

The cart's frame is not inertial if the cart's speed varies. So it is easier to consider ground's frame.

The reason I am asking about energy is because earlier I asserted that the cart is in a situation where it no longer has any energy available to it outside the energy represented by it's own momentum.

What do you mean by "the energy the cart has available"? The carts current kinetic energy?

You have asserted that the speed difference between the ground and air represents "plenty" of energy. If that is the case, then we ought to be able to get a ballpark figure pretty easily by determining what mass should be plugged into the formula.

To compute what exactlly? The total kinetic energy in the entire atmosphere? We don't need that much energy. And this quantity is completely irrelevant to the question, which is:

Can the cart accelerate further, when already traveling at wind speed?

acceleration_cart = (force_propeller - wheel force) / mass_cart
therefore:
if force_propeller > wheel force then acceleration_cart > 0 and the transmission makes sure that force_propeller > wheel force
therefore:

Yes, the cart can accelerate further, when already traveling at wind speed!

Once you have the acceleration_cart you can compute the cart's energy consumption, pay your energy bill to the planet and sleep well.

 

[swerdna]

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

Why do you need two cables? The torpedo needs two for steering, but for a minimal design I would recommend using just one.

Wouldn’t that require the torpedo to be used in a river or at least strong sea current?

Yes one cable would be fine. Not even sure why I drew two. Perhaps because the torpedo has two or some intuitive balance maybe.

 

[A.T.]

Wouldn’t that require the torpedo to be used in a river or at least strong sea current?

Yes, I just wanted to point out, that it is exactly the same mechanism.

Yes one cable would be fine. Not even sure why I drew two. Perhaps because the torpedo has two or some intuitive balance maybe.

It could be messy with two cables potentially twisting around each other. One cable coming out at the center rear of the boat scould stabilize it just fine.

And don't forget to attach the end of the cable at the reel, so you can recover your boat.

 

[swerdna]

Yes, I just wanted to point out, that it is exactly the same mechanism.

It could be messy with two cables potentially twisting around each other. One cable coming out at the center rear of the boat scould stabilize it just fine.

Sure I didn’t mean to present it as anything substantially new except instead of moving cables pulling against stationary water, moving water (or air) is pulling against stationary cables.

And don't forget to attach the end of the cable at the reel, so you can recover your boat.

DOH!

 

[rcgldr]

That being the case, what number for mass?

The number for mass is the mass of the air affected by the prop. The force is equal to that the mass of the affected air times the affected air's rate of acceleration (or the integral sum of all the components of affected air).

The carts are designed so when moving downwind at the speed of the wind, the forward force of the air onto the prop is greater than opposing backwards force of the ground onto the driving wheels (plus the backwards force from the ground related to rolling resistance).

The ground can be considered to have a huge amount of mass, any work done on the "ground" results in only a tiny change in velocity.

 

[rcgldr]

Your claim is that on the treadmill, the velocity of the cart is measured wrt to moving tread. You are using the frame of the cart to come this conclusion and what you have is relative velocity between the cart and the tread, and that relative velocity is faster than the tread.

That's not what is claimed. There are no claims about cart speed versus tread speed, only cart speed versus wind speed, or better stated that cart speed relative to the tread is greater than wind speed relative to the tread. I've already stated this in a mathematical form that is independent of frame of reference, and you have yet to respond to this. I'll repeat the claim again:

a DDWFTTW can acheive

|(cart speed) - (ground speed)| > |(wind speed) - (ground speed)|

for some range of wind speed:

(minimum wind speed) <= |(wind speed) - (ground speed)| <= (maximum wind speed)

 

[zoobyshoe]

The number for mass is the mass of the air affected by the prop. The force is equal to that the mass of the affected air times the affected air's rate of acceleration (or the integral sum of all the components of affected air).

Rodger that. The only air that matters is that which directly impinges on the cart, or upon which the cart impinges. Vanesch referred to this as an air column or air section.

The carts are designed so when moving downwind at the speed of the wind, the forward force of the air onto the prop is greater than opposing backwards force of the ground onto the driving wheels (plus the backwards force from the ground related to rolling resistance).

This is the prop as "bluff body" you mentioned before, right? It's very apparent in that video of the large cart with the tell tale. The prop is quite different than the usual airplane prop.

The ground can be considered to have a huge amount of mass, any work done on the "ground" results in only a tiny change in velocity.

Rodger that. The point I was trying to make is that if you're coasting over the ground in the carts frame looking at the ground as a source of power, that huge amount of mass going by at 10 mph isn't the ginormous power source it looks like at all. The power you could obtain from it by virtue of relative motion is only equal to the power it could obtain from you. Likewise, the force it could exert on you is only equal to the force you could exert on it.

 

[zoobyshoe]

Can the cart accelerate further, when already traveling at wind speed?

acceleration_cart = (force_propeller - wheel force) / mass_cart
therefore:
if force_propeller > wheel force then acceleration_cart > 0 and the transmission makes sure that force_propeller > wheel force
therefore:

Yes, the cart can accelerate further, when already traveling at wind speed!

Unfortunately for the cart when it reaches wind = 0 it becomes inert. There is no force from the wind or ground acting on it other than friction.

The fact the gearing insures that the force exerted on the propeller by the gear train is greater than the force applied at the wheels doesn't insure that the thrust the prop creates is greater than the cart's inertia.

On top of the inertia of the mass as a whole there is the inertia of the gears, wheels, and propeller itself to overcome. The cart will not change speeds unless force is applied and with no force applied to it from the wind or anything else it has no reason to accelerate. The prop won't spin faster unless the cart goes faster and the cart won't go faster unless the prop spins faster.

If the cart and propeller have gathered enough momentum during the downwind acceleration it may continue into the headwind and "coast" at a speed faster than the down wind speed for some length of time, but this is stored energy.

 

[uart]

Hi Zooby. Here are some simple explanations of what's happening from the first thread on DWFTTW posted previously. Please read.

Here's one train of thought that might help convince that the "treadmill in still air" situation is a least feasible.

Imagine for a moment that the rotating a propeller was replaced by a long “cork screw” (or similar). That is, the wheels were coupled through a suitable drive train to turn this long “cork screw”.

Now imagine that the corkscrew is started into a large block of soft foam (representing the air) attached to the front of the treadmill. If the treadmill is run then the wheels turn and the corkscrew turns and the vehicle will move forward as the corkscrew screws into the foam.

Think of the propeller in a similar way as "screwing" itself forward into the stationary air.

Here’s another little thought experiment that might help convince you. In the spirit of de-bunking perpetual motion devices you can usually assume frictionless ideal operation of most components and of course they still fail to achieve "over unity" operation. So in this spirit let's assume that we have an ideal lossless drive train (wheels, belts and gearing) and further that we can adjust the gearing ratio from the wheels to propeller to any desired ratio. Let's just concentrate on non-ideal lift/drag of the propeller.

First we note that the turning of the wheels is driving the prop, so the inevitable blade drag will mean we require constant torque to keep the prop turning at a constant rate and this torque must be provided by the wheels, giving a retarding force on the vehicle.

Second we note that the lift generated by the prop is providing a forward directed force. So we now have two forces in opposition, the lift on the prop giving a forward force and the rotational drag on the prop which, through the drive train, ultimately results in a retarding force at the wheels.

Since at first sight this thing looks like an “over unity” device our first instinct is to think that perhaps the lift force must be less than the retarding force. However the retarding force at the wheels is dependant on the gear ratio, that is, if we gear it so that the prop turns fewer times for each rev of the wheels then the ratio of retarding force at the wheels to blade drag to is also reduced.

So let's play devils advocate and assume that we set this thing up on a treadmill and hold it until it’s at steady state (wheels and prop up to speed) and we find that the retarding force is indeed larger than the propeller lift and our vehicle goes backwards.

No problems, let's just reduce the gear ratio so that the prop turns less times per wheel rev, and this will reduce the retarding force at the wheels. Arh but you say, this will also reduce the prop speed and so reduce it’s lift. Again no problems, just increase the treadmill speed until the prop turns at the same speed as it did before! Now you can't argue with this, the prop is at the same speed so the lift is identical to before, and the drag at the blades (torque required to spin the prop) is also the same as before, but due to the modified gearing the retarding force at the wheels is now lower than before. Can you see that in principle there is no limit to how much we repeat this procedure so eventually it has to work!

 

[schroder]

Ok, then there are some *very elementary* misunderstandings to be cleared up before we can go any further.

Yes and the first elementary misunderstanding is: I said two-way road! The cars are passing each other going in opposite directions! You are not going to sell me your old Camarro because you clocked it doing 250 mph wrt to a Maserati that was passing it going the opposite way! For all I know, your Camarro was parked in the driveway. And you are certainly not going to sell me DDWFTTW because you clocked the cart wrt to the tread which was going in the opposite direction. Let me ask you this; If you wanted to know the velocity of the tread would you place a cart on it and measure wrt the moving cart? And if you cannot measure the velocity of the tread wrt a moving cart, you cannot measure the velocity of the cart wrt the moving tread. You can forget all about Galileo, prop pitch even Newton and reduce this down to a simple sum and difference problem and apply good old fashioned common sense.
You measure the velocity of something wrt the medium that something is working against. I have shown you the cart cannot possibly be working against the tread. If it were, and it is advancing, then by definition it is over unity. But since it IS advancing, and it is NOT over unity, it is NOT working against the tread. That is pure logic and you CANNOT argue with that.
And since it is NOT working against the tread, the tread is NOT the medium to use to measure the cart’s velocity. It IS working against the AIR which is painfully obvious if you look at that propeller spinning around. Since it IS working against the air, you measure the cart’s velocity wrt the AIR.
What you are doing is SUMMING the velocity of the tread with the velocity of the cart and of course you get a number higher than tread velocity! I suppose you can do the same on the down wind side; SUM up the velocity of the wind with the velocity of the cart and shout EUREKA! Faster than the wind! But the men in white coats would come to take you away!
All you have to do here is THINK.

 

[vanesch]

Yes and the first elementary misunderstanding is: I said two-way road! The cars are passing each other going in opposite directions!

Yes, that was ambiguous: if there are two lanes, you can use them in the same direction, or in opposite directions, that's why I made a choice as it wasn't clear which case you wanted.

But nothing changes here, except a few numbers. So let's go through this exercise again, this time the Camarro at 20 mph north wrt the road, the Maserati 230 mph south (wrt to the road). So from the Camarro, we see the Camarro at 0 mph, and the Maserati at 250 mph south. The relative velocity of the Camarro wrt the Maserati is 0 - (- 250 mph) = +250 mph (we take + to be north).

From the Maserati, the Camarro is doing + 250 mph north, and the Maserati is doing 0. So we have 250 - 0 = 250 mph again.

If we have a train going north at 130 mph, it sees the Camarro going 110 mph south and it sees the Maserati going 360 mph south. So the relative velocity of the Camarro wrt the Maserati is -110 - (-360) = +250 mph again.

I could continue. I can calculate the relative velocity between two objects by using their velocities as seen in just any frame: their difference is the relative velocity.

I can do something else:
For instance, from the frame of the train, I can see that the ground is moving 130 mph south, and the Camarro going 110 mph south. As such, the velocity of the Camarro wrt the ground is (-110) - (-130) = 20 mph. As calculated from the velocities in the train frame.

You are not going to sell me your old Camarro because you clocked it doing 250 mph wrt to a Maserati that was passing it going the opposite way! For all I know, your Camarro was parked in the driveway.

That is because I'm not interested in the relative velocity between the Camarro and the Maserati, but rather between the Camarro and the ground. I can calculate it using the velocities in any frame. It gives me always the same, correct, result.

And you are certainly not going to sell me DDWFTTW because you clocked the cart wrt to the tread which was going in the opposite direction. Let me ask you this; If you wanted to know the velocity of the tread would you place a cart on it and measure wrt the moving cart?

Depends wrt I want to know the velocity of the tread. If I wanted to know the velocity of the tread wrt the cart, yes, why not ?

And if you cannot measure the velocity of the tread wrt a moving cart, you cannot measure the velocity of the cart wrt the moving tread.

If I know the velocity of the tread wrt a frame, and I know the velocity of the cart wrt that same frame, I can calculate the relative velocity between them, which would be the one that an observer sitting on the cart would see.

Look at our train: because I had the velocity of the Camarro wrt the train (110 mph south) and I had the velocity of the ground wrt the train (130 mph south), I could calculate the velocity of the Camaro wrt the ground (20 mph which was -110 - (-130) = 20).
And that was indeed what the speedometer in the Camaro (which measures directly the speed wrt the ground) indicated.

Fun, no ?

You can forget all about Galileo, prop pitch even Newton and reduce this down to a simple sum and difference problem and apply good old fashioned common sense.
You measure the velocity of something wrt the medium that something is working against. I have shown you the cart cannot possibly be working against the tread. If it were, and it is advancing, then by definition it is over unity. But since it IS advancing, and it is NOT over unity, it is NOT working against the tread. That is pure logic and you CANNOT argue with that.

No, indeed, I cannot argue with that because there's no logic there. "Working against" is not something that has an intrinsic meaning, and from there on, the rest fails. But what is working against what doesn't matter when we see velocities, no ?

And since it is NOT working against the tread, the tread is NOT the medium to use to measure the cart’s velocity. It IS working against the AIR which is painfully obvious if you look at that propeller spinning around. Since it IS working against the air, you measure the cart’s velocity wrt the AIR.

Listen, the Maserati is not doing work against the train either, but there's nothing wrong in expressing the velocity of the Maserati wrt the train. There's a frame-independent physical meaning to "the velocity of the Maserati wrt the train" independent of what are the interactions between trains and Maseratis.

What you are doing is SUMMING the velocity of the tread with the velocity of the cart and of course you get a number higher than tread velocity!

No, I'm taking their ALGEBRAIC DIFFERENCE. Because the two directions are opposite, the *signs* of the velocities are opposite (one + the other -) and the difference of a positive and a negative number is the same as the sum of their absolute values.

Look: 3 - (-5) = 3 + 5 = 3 + |-5| = 8.
Fun, no ?

But the men in white coats would come to take you away!
All you have to do here is THINK.

Fun, no ?

 

[schroder]

We want to know the velocity with respect to the floor on which the cart is driving ! In the outdoor test, this is the floor, and in the treadmill test, this is the treadmill. It is the thing that is touched with the wheels of the cart.

Ah Ha! There it is! There is your mistake which is preventing you from understanding what is going on here.
It is Not Running on the treadmill at all! You are looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road. You are completely forgetting about that propeller that is spinning in the air. That is where all the work is being done By the cart; where all the force is being exerted BY the cart. That is where the cart is doing all it’s Running! It is running on or in the air! The wheels on the tread represent the Source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted ON the cart. The cart is no more running against the tread than a sail being blown by the wind is running against the wind. The running is between the hull and the water for the sailboat. The running is between the propeller and the air, for the cart on the treadmill. You do not measure the velocity of a sailboat by measuring how fast it is going wrt the wind that is pushing on the sails. You do not measure the velocity of the cart on the treadmill by measuring how fast it is going wrt the tread that is turning the wheels. Look elsewhere! Look where the moving vehicle is exerting force against something. The sailboat exerts force against the water so you measure its velocity wrt the water. The cart is exerting force against the wind, so you measure its velocity wrt the wind. I have been saying this since day ONE.

 

[zoobyshoe]

Hi Zooby. Here are some simple explanations of what's happening from the first thread on DWFTTW posted previously. Please read.

Hi, uart.

Both these gedankens seem to demonstrate a cart merely holding place given a certain motorized treadmill and air speed.

As AT pointed out, the cart has to do more than hold its own, it has to accelerate. (The more gearing you have the more elements have to be accelerated.) When it reaches wind speed there are no more outside forces adding power. The cart is feeding stored power to the surrounding media. It may be able to hold its own and even accelerate at first if it has enough stored power, but this will be used up soon, or eventually, and the cart will slow to wind speed.

If you were to put the cart in this condition without any stored energy, it would never work: suppose you drove down wind in a car till you reached wind speed, then picked the cart up from the seat next to you and placed the cart on the road. It's obvious it would never accelerate. It would instantly start to lose forward momentum to the road. Therefore, I suspect that non-motorized demonstrations that seem to work are most likely working off of stored energy.

 

[vanesch]

Ah Ha! There it is! There is your mistake which is preventing you from understanding what is going on here.
It is Not Running on the treadmill at all! You are looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road. You are completely forgetting about that propeller that is spinning in the air. That is where all the work is being done By the cart; where all the force is being exerted BY the cart. That is where the cart is doing all it’s Running! It is running on or in the air! The wheels on the tread represent the Source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted ON the cart. The cart is no more running against the tread than a sail being blown by the wind is running against the wind.

But what force is exerted on what doesn't matter to express *velocities*.

Wrt to what are you expressing the velocity in the outdoor test ? Wrt the by-going train ? Wrt the observer running next to it ? Wrt the air ? Wrt the moon ? Or with respect to the thing the wheel is touching, namely the floor ?

If you do the test on the train in a wind tunnel, would you compare to the floor of the windtunnel (touched by the wheels of the cart), or by the guy walking in the corridor of the train, or wrt the tracks, or wrt the cow running through the field next to the train ?

Again, unless you object, I suppose we can agree that it will be the floor of the windtunnel.

So what we call "the speed of the cart" is the speed of the cart wrt the thing the wheels are touching.
What we call "the speed of the wind" is the speed of the air wrt the thing the wheels are touching. Note that both velocities are expressed in the same frame, so these are relative velocities, they have a physical meaning. It is concerning THESE velocities that there is a claim. So there's no point calculating OTHERS. That can be done, but there's no claim regarding OTHER velocities.
The claim "DWFTTW" is the claim that the speed of the cart wrt the thing the wheels are touching is a bigger number in absolute value than the speed of the air wrt that same thing in absolute value.

If I say that the speed of a train is 150 mph on its track, and you claim that the velocity of the train wrt the glass on a table in the train is 0, then your statement doesn't disprove mine.

If I say that 5 > 3, then you claiming that 4 < 6 doesn't disprove mine. Now, you can say all you want about that it is absolutely necessary to compare 4 to 6, it doesn't alter anything concerning the comparison of 5 and 3.

The cart is exerting force against the wind, so you measure its velocity wrt the wind. I have been saying this since day ONE.

But my friend, this is most basic nonsense. There is no relationship between what is exerting a force on what, and any obligation to only consider relative velocities between them.

Tell me, is the moon exerting a force on the ocean water ?
So, is the only sensible way to express the velocity of the ocean water, the velocity in the frame of the moon ?

A personal question. I had the impression, along this thread, that you were an engineer. Are you ?

 

[vanesch]

As AT pointed out, the cart has to do more than hold its own, it has to accelerate. (The more gearing you have the more elements have to be accelerated.) When it reaches wind speed there are no more outside forces adding power. The cart is feeding stored power to the surrounding media. It may be able to hold its own and even accelerate at first if it has enough stored power, but this will be used up soon, or eventually, and the cart will slow to wind speed.

Look at the total force balance I worked out in post #480 and which gave the following total force:

F_tot = rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind) + W x (v_wind - v_cart)^3 - ...

... rho_air x S x K x ( (K^2-1) x v_cart^2 + 2 x v_cart x v_wind - v_wind^2)

Here, rho_air is the specific mass of the air, S was the "effective outgoing surface" of the propeller flow, K was the constant giving us the "gearing ratio" which was determined by v_out = K v_cart where v_out is the outgoing air from the propeller in the cart's frame, and v_cart was the velocity of the cart wrt to the ground (the velocity with which the wheel is spinning). W was a constant describing the drag on the structure (we can assume it small).

v_wind and v_cart are taken positive both in the "sense of the wind" and are both expressed in the "floor" frame.

The first term rho_air x S x K x v_cart x ((K-1) x v_cart + v_wind)

represents the thrust by the propeller. It is obtained by making the momentum balance between the mass of air that is accelerated from "incoming velocity" to "outgoing velocity" by the propeller (of course, in the frame of the propeller).

If it is positive, it means a *forward* force, which *accelerates* the cart.

The second term: W x (v_wind - v_cart)^3

represents in general, the drag on the cart by the wind (including any drag by the propeller).
It works in the sense of the velocity difference between the wind and the cart of course. For a back wind ("wind faster than cart") it pushes forward, and for a head wind, it pushes backward (negative force).

The third term (note the minus sign): - rho_air x S x K x ( (K^2-1) x v_cart^2 + 2 x v_cart x v_wind - v_wind^2)

is the force that the ground exerts on the cart. It is obtained from the power balance calculated in the cart's frame.
The propeller doesn't only exchange momentum with the air column, it also exchanges energy. Indeed, the air wins or looses kinetic energy by changing from v_in to v_out and this energy is provided by (or to) the propeller. Per unit of time, there is hence a power flow from the wheel to the propeller (or vice versa).
Assuming perfect mechanical energy transfer between the wheel and the propeller, this power is obtained by driving the wheel with a certain force: the force exerted by the ground on the wheel (and hence on the cart). It works to the back direction if the wheel delivers energy to the propeller (that is, if the air is accelerated by the propeller, and hence wins kinetic energy, which is provided by the propeller, and hence must receive it from the ground).

You can find the total force on the cart when the cart is driving at the same velocity as the wind (v_cart = v_wind), and then we find:

F_tot = rho_air x S x K x v_wind x ((K-1) x v_wind + v_wind) + W x (v_wind - v_wind)^3 - ...

... rho_air x S x K x ( (K^2-1) x v_wind^2 + 2 x v_wind x v_wind - v_wind^2)


Or after some algebra:
1) the first term becomes: rho_air x S x K x v_wind x K x v_wind = rho_air x S x K^2 x v_wind^2

Note that the propeller is giving a forward force, so "wants to accelerate" the cart.

2) the second term is 0 of course: there's no drag in a wind-still situation

3) the third term is: - rho_air x S x K x ( (K^2-1) x v_wind^2 + 2 x v_wind x v_wind - v_wind^2) = - rho_air x S x K x ( K^2 x v_wind^2) = - rho_air x S x K^3 x v_wind^2

Note two things: the third term is NEGATIVE (wants to brake the cart, which is logical: we take energy from the wheels to provide for the energy of the propeller)

It looks a lot like the first term, except for an extra factor of K.

So the final force is: F_tot = rho_air x S x K^2 x v_wind^2 - rho_air x S x K^3 x v_wind^2

or: F_tot = rho_air x S x K^2 x (1 - K) x v_wind^2

Note that if K is between 0 and 1, this is positive, so there is a total force on the cart that wants to accelerate it.

In all of this, we didn't need any "stored kinetic energy of the cart".

 

[vanesch]

For instance, when using some dimensionless quantities and having numerically:

rho_air x S = 0.5

w = 0.1

K = 0.3

v_wind = 1.0

we have the plot in attachment as the force as a function of v_cart. Note that the force is positive beyond the point v_cart = 1, which means that there is a forward force (accelerating the cart) until a point around v_cart ~ 1.4 or so.

 

[rcgldr]

the transmission makes sure that force_propeller > wheel force

There gear ratio is 1:1. The force is greater due to the properties of the propeller, diameter, pitch, and efficiency. The effective prop pitch needs to be less than the circumference of the wheels so that prop to air speed is slower than wheel to ground speed, so that the prop + air interface involves greater force but at less power than wheel + ground.

The only air that matters is that which directly impinges on the cart, or upon which the cart impinges.

Unfortunately for the cart when it reaches wind = 0 it becomes inert.

No just as you mentioned, the air that matters includes the air that the cart impinges on. The thrust from the prop provides an upwind acceleration and speed of the air, which interacts with the wind. The combined effect of the upwind wash and the downwind flow increase the pressure aft of the prop, increasing the forward force applied through the air to the prop. The net result is that the wind speed of the air is slowed down as the cart passes through, and the reduction in wind speed is the source of power. The cart is outrunning the wind, but not the air flow from the prop wash. The air flow aft of the prop is slower than the wind, even when the cart is moving faster than the wind.

This is the prop as "bluff body" you mentioned before, right? It's very apparent in that video of the large cart with the tell tale. The prop is quite different than the usual airplane prop.

The prop is different because it has a low pitch versus diameter, a typical "slow flyer" prop. The carts need a prop with a relatively small pitch for the reasons stated previously. The prop only acts as a bluff body only at start up. During acceleration the prop acts as both bluff body as well as thrust generator. You could consider the air itself from the prop wash to be a bluff body moving upwind with respect to the cart.

seem to demonstrate a cart merely holding place given a certain motorized treadmill and air speed.

Both swerdna's and spork's videos demonstrate a cart advancing on the treadmill, not just holding place.

 

[vanesch]

Both swerdna's and spork's videos demonstrate a cart advancing on the treadmill, not just holding place.

If I understood it well, the trick was to incline the treadmill so that the cart had to move "uphill". This means that on top of the wheel and air forces, it had a backward working force, namely a component of its weight. The fact that it was holding still in this situation means that the air+wheel force balance was in the forward direction, given that it compensated the backward working force of gravity. This then means that at the point of being at wind velocity, the cart still had a forward force working on it by the wheels + propeller, and if there hadn't been the gravity "holding it in place", it would have accelerated forward.

It was again a nice experimental trick to make the treadmill experiment last so that the argument of "steady state is not reached yet" is refuted.

 

[rcgldr]

looking at those wheels turning and your mind is conditioned by every day experience to interpret that as wheels running on a road.

or wheels running on a treadmill:

For this one, how would the experience on the treadmill be different for the rider if the rider and bicycle were outdoors riding downwind at the same speed as the wind?

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

How about a treadmill on a bicycle?

http://www.youtube.com/watch?v=Sg-KpT9RNXE&fmt=18

propeller that is spinning in the air. where all the force is being exerted by the cart. ... wheels on the tread represent the source of all the power the cart is receiving from the tread... That is the place where all the force is being exerted on the cart.

Newton's 3rd law, forces only exist in equal and opposing pairs. The force the prop applies to the air coexists with an equal and opposing force that the air applies to the prop. The force that the tread applies to the wheels coexists with the equal and opposing force the wheels apply to the tread.

The cart is no more running against the tread than a sail being blown by the wind is running against the wind. The running is between the hull and the water for the sailboat.

So why isn't the cart case the running between the wheels and the ground?

The running is between the propeller and the air, for the cart on the treadmill.

The cart involves two interactions. 1 - between prop and air. 2 - between wheel and ground.

You do not measure the velocity of a sailboat by measuring how fast it is going wrt the wind that is pushing on the sails.

Yet aircraft measure speed relative to the air they travel in. There's no reason sail craft can't use instruments to measure apparent wind and direction (and some do).

You do not measure the velocity of the cart on the treadmill by measuring how fast it is going wrt the tread that is turning the wheels.

Why not? It's a perfectly reasonable frame of reference.

Look where the moving vehicle is exerting force against something.

The cart applies an upwind force to the air, and a downwind force onto the tread, so either could be a reasonable choice as a frame of reference. The cart doesn't exert a force on the floor, so it would seem that the floor would be a bad choice.

You have yet to commnet about my mathematically described claim about these carts:

|v_cart - v_ground| > |v_wind - v_ground|

 

[A.T.]

Unfortunately for the cart when it reaches wind = 0 it becomes inert.

No, it doesn't become inert. The propeller is producing more forward force than is needed to drive the propeller at the wheels pointing back. The net force is not zero -> acceleration occurs.

the transmission makes sure that force_propeller > wheel force

There gear ratio is 1:1.

With transmission I mean the entire transmission chain from wheel to propeller, that you described.

 

[rcgldr]

If I understood it well, the trick was to incline the treadmill so that the cart had to move "uphill". It was again a nice experimental trick to make the treadmill experiment last so that the argument of "steady state is not reached yet" is refuted.

I was referring to sporks videos where the cart advances. Swerdna's turntable allowed the cart to advance indefinately to give a good idea of the steady state situation.

I captured swerdna's video so I could time it and calculate the actual results to show some real numbers:

Turntable speed = 13.1 mph
Cart speed = 5.6 mph.

This is equivalent to v_wind = 13.1 mph, and v_cart = 18.7 mph.

Cart speed = 1.43 times wind speed.

For the given paramters, prop pitch 6 inches, wheel circumference 10.5 inches, the theoretical limit would be

v_cart_max = v_wind / (1 - ar)
v_cart_max = v_wind / (1 - 6/10.5)
v_cart_max = 2.33 v_wind

So the cart is achieving about 61% of the theoretical (no loss) limit.

 

[schroder]

Fun, no ?



No, indeed, I cannot argue with that because there's no logic there. "Working against" is not something that has an intrinsic meaning, and from there on, the rest fails. But what is working against what doesn't matter when we see velocities, no ?

Now right here you are doing something that is unscientific. You may consider it to be “fun” and I agree there is an element of that, but there is also a serious side of this. This ridiculous claim needs to be rebuked by the physics community. You are dismissing what I wrote as illogical without showing where the logic fails. That is like saying “You are wrong because my name is Vanesch, and I say so.” I would expect better from you.
And of course it matters what is working against what! The velocity of the wind working against the sail is not used as a reference to measure the velocity of the boat because the boat is not working against the wind, it is working against the water. How can you honestly say it does not matter what is working against what when you measure velocities?
So I am going to walk you through the logic one more time, and this time You need to show me where the logic fails. If you cannot do so, you must accept the conclusion that the logic leads you to. I am going to be very fair to you and give you advance notice that I am now placing you in checkmate. You cannot simply knock over the chess board and say this is nonsense. You need to show how and why you are not in checkmate and then show me that you can get out of it.
Just follow the logic:
1) A machine that can do more work than the work that is done on it is by definition an over unity machine.
2) There are no over unity machines.
3) A machine that can do more work on a tread than the tread does on it, is an over unity machine.
4) There are no over unity machines.
5) From 1 - 4 : a machine that has work done on it by a tread cannot do more than that amount of work on the tread.
6) From 1 - 5 : A machine that has work done on it by a tread, but is clearly seen to be advancing on that tread in the opposite direction to the tread cannot possibly be an over unity machine because there are no over unity machines.
7) From 1 - 6 : The cart is being powered by the tread. It is also advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine.
8). From 1 – 7 : Since the cart is not an over unity machine, but it is advancing on the tread in the opposite direction to the tread and it is being powered by the tread it cannot possibly be working against the tread. Because if it were working against the tread , due to being being worked on by the tread , it would be doing more work on the tread than the tread is doing on it which means it is an over unity machine and they do not exist!
First definitive conclusion: From 1 – 8 : The cart is being powered by the tread. It is advancing on the tread in the opposite direction to the tread. The cart is not an over unity machine. The cart is not working against the tread.
If you accept (and you have no choice) that the cart is not working against the tread, then you need to justify why you insist on measuring the velocity with respect to the tread. If you can justify that, you would also be saying that the velocity of a sailboat can be measured with respect to the wind which is pushing it. You would get some very low numbers because the boat is moving in the same direction as the wind. And the reason you are getting high numbers for the cart velocity is because the tread and the cart are moving in opposite directions and you are measuring the velocity of the cart with respect to the tread.
In both cases, sailboat and cart, the only CORRECT approach is to measure the velocity of the vehicle with respect to the interface it is moving against. The sailboat moves against the water so you measure the velocity at the interface between the hull and the water. The cart moves against the wind (air) so you measure the velocity of the cart at the propeller and air interface.

Since you have displayed a disturbing tendency, when cornered, to simply dismiss even the most logical presentation as mere gibberish, I now openly challenge you to go through the above presentation and show where you feel it is wrong. You like chess?

 

[A.T.]

For the given paramters, prop pitch 6 inches, wheel circumference 10.5 inches,

Have you taken into account, that the propeller is further away from the turntable axis than the wheel?

 

[cristo]

1) A machine that can do more work than the work that is done on it is by definition an over unity machine.
2) There are no over unity machines.
3) A machine that can do more work on a tread than the tread does on it, is an over unity machine.

You can't just pick a part of the system that you want energy conservation to hold in, and ignore the rest! 1 and 3 are not equivalent statements!