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

In summary, this turntable and cart seem to be able to move faster than the wind, but it's not conclusive proof of DDWFTTW. There are some possible explanations for the effect, including lift.
  • #386
zoobyshoe said:
Here is a rough explanation of how this is accomplished in a conventional sailboat:


"Lets say a boat is trying to go directly downwind and has a hull design that will allow it to plane and take maximum advantage of the wind it experiences. Plus the sails / crew are all optimized...

Starting at a beam reach the boat starts to accellerate and generate more apparent wind. Which will be a vector of the true wind speed and the forward direction of the boat. This new apparent wind will be coming from further forwards that the true wind speed and at a faster speed.

From here the boat turns downwind to keep the apparent wind speed on the beam of the boat. As it does so the boat continues to speed up, and the now new apparent wind continues to accellerate while moving further forward.

The boat responds by turning downwind another five degrees and the cycle repeats over and over, until the boat could be sailing completely on its' own apparent wind actually beating into a headwind that exists completely in its' own mind.

The boats that do this the best are ice boats that are always sailing upwind to their apparent wind regardless of their point of sail on the compass, and can reach rediculous SOG due to the effectively zero drag.

The problem with maintaining this type of thing in the real world is that if a boat slowes down and looses the apparent wind, or is effected by an event that robs is of velocity (running into the back of a wave for instance) the boat now has to start the cycle over again. This leads to a practical limit on this, but not a theoretical one."

That was written by "stumble" in this thread on a sailing forum:

http://www.boatdesign.net/forums/sailboats/tacking-downwind-faster-than-wind-24761.html

Quite fascinating!

What people need to do then is scrap the current designs and make a cart with a self tacking iceboat sail sticking out in front of it on a pivoted arm that can swing from side to side. As this little "servo" iceboat zigzags down wind it should drag the cart directly down wind faster than the wind.

Why would they need to scrap their prop cart? It works quite elegantly on an equivalent of a continuous beam reach tack. That is what the propellers are actually doing. The fact that we are used to thinking of propellers only as driven by a motor is prejudicing our perception.
 
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  • #387
Subductionzon said:
Why would they need to scrap their prop cart? It works quite elegantly on an equivalent of a continuous beam reach tack. That is what the propellers are actually doing. The fact that we are used to thinking of propellers only as driven by a motor is prejudicing our perception.
The trouble, as I see it, is that at windspeed the DDWFTTW cart the propellers are not in any wind. Once the cart reaches windspeed, there is no wind. How does the blade get a beam reach out of 0 wind?
 
  • #388
zoobyshoe said:
The trouble, as I see it, is that at windspeed the DDWFTTW cart the propellers are not in any wind. Once the cart reaches windspeed, there is no wind. How does the blade get a beam reach out of 0 wind?

A prop that's spinning in no wind creates a wind.
 
  • #389
swerdna said:
A prop that's spinning in no wind creates a wind.

Yes, which is what is happening in a fan. When you shut the motor off, the wind stops, though.

When the cart reaches windspeed its prop has only momentum left to create the wind. That energy will soon be expended.
 
  • #390
Wow. All I can say is that it's been an interesting read.

:biggrin:
 
  • #391
zoobyshoe said:
Yes, which is what is happening in a fan. When you shut the motor off, the wind stops, though.

When the cart reaches windspeed its prop has only momentum left to create the wind. That energy will soon be expended.


Then how do you explain the advance of the various carts on treadmills and turntables against the apparent wind? Or do you disagree with the frame of reference equivalents of the treadmill and the turntable.
 
  • #392
zoobyshoe said:
Yes, which is what is happening in a fan. When you shut the motor off, the wind stops, though.

When the cart reaches windspeed its prop has only momentum left to create the wind. That energy will soon be expended.
The wind blows, that moves the cart, that turns the wheel, that spins the prop, that thrusts against the wind that blows, that moves the cart when the cart is at the speed of the wind and beyond (that lives in the house that Jack built ;-).

Not a technically correct explanation but the way I visualise it is that when the cart is at wind speed and beyond the thrust of the prop is moving backwards relative to the cart and therefore is traveling slower than both the cart and the wind in the direction of the wind. Therefore the wind still “pushes” on the thrust of the prop (that moves the cart, that moves . . . ). The prop thrust is always essentially a sail in the wind.
 
  • #393
Jeff Reid said:
You're right, the Brennan torpedo is a DWFTTW (W = water) device.
A device patented in 1877 and used by the Royal Navy for years. Fortunately for Brennan he lived in the 19th century, and did not have to defend his invention against internet "physicists". :wink:
 
  • #394
zoobyshoe said:
When the cart reaches windspeed its prop has only momentum left to create the wind. That energy will soon be expended.
The only momentum that matters, is that of the cart, which can only decrease if the net force on the cart points backward. At windspeed there are two forces (gear friction aside):
- propeller thrust (forward)
- breaking force of the wheels (back)
Gear translation makes sure that the breaking force is smaller that the propeller thrust. So the net force is facing forward and accelerates the cart further beyond windspeed.
 
  • #395
Subductionzon said:
Then how do you explain the advance of the various carts on treadmills and turntables against the apparent wind? Or do you disagree with the frame of reference equivalents of the treadmill and the turntable.
Rather than taking a demonstration that apparently works and then fitting some logic or reasoning to explain it, I would rather first determine if it can possibly work. At this point I am attempting to follow the train of energy: at each given point a certain amount of energy is needed. Each time I look at it I find that when the cart reaches windspeed it has no energy to continue except that amount represented by its own momentum. It can't go much farther.
To counter this by saying that iceboats, or demonstration carts, or anything, seem to be able to do it, doesn't address the problem. If we keep looking and looking and still can't find the energy, then we have to start considering that what "seems" to be happening with ice boats and demonstration carts isn't what it looks like.
 
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  • #396
A.T. said:
The only momentum that matters, is that of the cart, which can only decrease if the net force on the cart points backward. At windspeed there are two forces (gear friction aside):
- propeller thrust (forward)
- breaking force of the wheels (back)
Gear translation makes sure that the breaking force is smaller that the propeller thrust. So the net force is facing forward and accelerates the cart further beyond windspeed.
How does the gear translation make sure the breaking is smaller than the thrust?
 
  • #397
Personally I think this debate is over and thank everyone for participating and the very helpful information provided. I think any intelligent (self-honest) person has to conclude that DDWFTTW is possible and there is a practical working example in the Brennan Torpedo. Thanks for the link to it A.T. (post #349)

If anyone DOESN’T accept that the Brennan Torpedo is an actual, practical working example of DDWFTTW (no treadmills or turntables required), that is essentially the same as the carts in the videos, I would appreciate an explanation why it isn‘t.
 
  • #398
Zooby if you want to know how it works then you will have to talk with some aero majors. You would have to know how a sail works like an airfoil. Then realize that the spinning propeller is working the same way. The spinning activity of the propeller allow the props to act like a sail on a beam reach. One more point you should know what your BRAKES are or else you might BREAK something:biggrin:
 
  • #399
swerdna said:
The wind blows, that moves the cart, that turns the wheel, that spins the prop, that thrusts against the wind that blows, that moves the cart when the cart is at the speed of the wind and beyond (that lives in the house that Jack built ;-).
I have a remarkable new source of energy and I am going to offer it to you in a special, limited time offer. It's called a brick. Now watch: when I hold the brick here, about a yard off the ground, then release it, it suddenly goes slamming down into the ground, as if by magic! Look here, I show you again: I bring the brick to this position, about a yard off the ground, and simply take my hand away, and it slams into the ground with astonishing force! I did not throw it at the ground! You saw me: I simply released it, and it slammed into the ground all by itself!

It works by quantum physics: it is packed full of quanta which are then put through a special and proprietary scintering process requiring thousands of degrees of heat. It is cooled and now it is ready to take advantage of the amazing forces of General Relativity as explained by Einstein, inventor of the bomb. Who understood energy better than Einstein, eh?

Here's the amazing part: the brick works indefinitely, and needs no batteries!

I am going to offer this amazing product to you for only 5 easy payments of $19.95! plus 21.95 shipping and handling Can you believe it! Unlimited energy for so little!

Act quickly. Offer ends soon.
 
  • #400
zoobyshoe said:
I have a remarkable new source of energy and I am going to offer it to you in a special, limited time offer. It's called a brick. Now watch: when I hold the brick here, about a yard off the ground, then release it, it suddenly goes slamming down into the ground, as if by magic! Look here, I show you again: I bring the brick to this position, about a yard off the ground, and simply take my hand away, and it slams into the ground with astonishing force! I did not throw it at the ground! You saw me: I simply released it, and it slammed into the ground all by itself!

It works by quantum physics: it is packed full of quanta which are then put through a special and proprietary scintering process requiring thousands of degrees of heat. It is cooled and now it is ready to take advantage of the amazing forces of General Relativity as explained by Einstein, inventor of the bomb. Who understood energy better than Einstein, eh?

Here's the amazing part: the brick works indefinitely, and needs no batteries!

I am going to offer this amazing product to you for only 5 easy payments of $19.95! plus 21.95 shipping and handling Can you believe it! Unlimited energy for so little!

Act quickly. Offer ends soon.
Thanks for your amazing and generous offer but I already have purchased such a brick from a street seller. (I traded a cow for it).

However, If you have a brick that can move a yard from the ground to my hand, that works indefinitely with no batteries required, I would definitely be interested in purchasing several of those.
 
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  • #401
Subductionzon said:
Zooby if you want to know how it works then you will have to talk with some aero majors. You would have to know how a sail works like an airfoil. Then realize that the spinning propeller is working the same way. The spinning activity of the propeller allow the props to act like a sail on a beam reach. One more point you should know what your BRAKES are or else you might BREAK something:biggrin:
Yes, aero majors would be nice. I understand the sail on a beam reach: Bernoulli on the one side, Newton on the other. I don't understand how a fan can keep running indefinitely after you shut it off.

Spellcheck knows no homonyms. (And it is now telling me I spelled "spellcheck" wrong.)
 
  • #402
swerdna said:
However, If you have a brick that can move a yard from the ground to my hand, that works indefinitely with no batteries required, I would definitely be interested in purchasing several of those.
We have bricks that can do that. They are enhanced with a remarkable technology called a spring. You simply release the brick as before and it will return to your hand. You'll love them!

I've set you down for a dozen. Your paypal account has been billed and your order will be shipped within 24 hours.

Thank you for your order!

Zoobonic Technologies Inc.
 
  • #403
zoobyshoe said:
How does the gear translation make sure the breaking is smaller than the thrust?

By using the http://en.wikipedia.org/wiki/Mechanical_advantage" : In short: small force at one end, large force at the other.

Do you understand why http://de.youtube.com/watch?v=k-trDF8Yldc" is moving in the same direction as the ruler? If it had no transmission (cylinders instead of the two reels) it would not move at all (sliding aside), because the thrust-force from the ruler would be transmitted 1:1 to the ground as breaking-force. But due to the transmission by the reels you get a breaking-force smaller than the thrust-force.
 
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  • #404
zoobyshoe said:
We have bricks that can do that. They are enhanced with a remarkable technology called a spring. You simply release the brick as before and it will return to your hand. You'll love them!

I've set you down for a dozen. Your paypal account has been billed and your order will be shipped within 24 hours.

Thank you for your order!

Zoobonic Technologies Inc.
GREAT!. Next year’s Xmas gifts problem solved!
 
  • #405
A.T. said:
By using the http://en.wikipedia.org/wiki/Mechanical_advantage" : In short: small force at one end, large force at the other.
I believe I understand the lever, yes. It's more complex than small force at one end, large force at the other. It's more accurately: small force at one end applied over a greater distance, large force at the other end applied over a smaller distance.

Do you undestand why http://de.youtube.com/watch?v=k-trDF8Yldc" is moving in the same direction as the ruler?
Yes
If it had no transmission (cylinders instead of the two reels) it would not move at all (sliding aside), because the thrust-force from the ruler would be transmitted 1:1 to the ground as breaking-force. But due to the transmission by the reels you get a breaking-force smaller than the thrust-force.
What I understand about the wheels is that they impart forward motion to the cart at some number times the speed of the ruler in the same direction as the ruler, but they have less torque per unit distance traveled than the ruler. The ruler is pressing with great force on the short end of the lever which causes the other end, the cart to move much faster,over a greater distance, but with less torque per unit distance moved than the ruler.

It seems from the context you are referring to the ground's friction as "braking force", but I'm not sure. You'd better clarify "braking force" for me.
 
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  • #406
zoobyshoe said:
Here is a rough explanation of how this is accomplished in a conventional sailboat ... boat could be sailing completely on its' own apparent wind actually beating into a headwind that exists completely in its' own mind.
Most of those apparent wind explanations, including this one, read like an over unity situation. For sailcraft there are two key factors:

1. The apparent crosswind is independent of the sailcrafts forward speed. The apparent crosswind is perpendicular to the sailcraft's path, so this is a given. The apparent crosswind = (true wind speed) x sin (angle between true wind direction and sail craft direction). For a given crosswind, some sailcraft can accomplish a forward speed so that the (apparent crosswind) / (apparent headwind) is < 1. This ratio is normally expressed as an angle, called Beta, = atan(apparent crosswind / apparent headwind). For iceboats, I've read the Beta can be as low as 8 degrees under ideal conditions (true wind speed near 10 mph.)

2. The sailcraft obtains it's power by slowing down the true wind. This requires that even though the sailcraft's downwind component of speed is greater than the wind, the sail has to divert a sufficient amount of the apparent headwind to true upwind, so that the air flow off the sail slows down the true wind. Slowing down the true wind is the source of power for the sailcraft (or any wind powered vehicle).

Starting at a beam reach the boat starts to accellerate and generate more apparent wind.
This isn't required, the apparent crosswind is independent of the boat's forward speed. The main reason for starting near perpendicular to the true wind is that the apparent crosswind will be higher (sin() is maximum when angle is 90 degrees), resulting in more acceleration.

zoobyshoe said:
Each time I look at it I find that when the cart reaches windspeed it has no energy to continue
Although the cart reaces wind speed, the air flow through the prop doesn't. The true wind is still being slowed down by the thrust from the prop, and as long as the true wind is slowed down, power can be extracted from the wind, as with any wind powered device.

zoobyshoe said:
How does the gear translation make sure the braking is smaller than the thrust?
It doesn't. What determines if the braking force is smaller than thrust depends on the propeller parameters, diameter, pitch, and efficiency. The power input to the propeller is torque time angular velocity. The power output is thrust times air speed through the propeller. The forward force from the prop is thrust, the opposing "braking" force from the ground is the prop input torque divided by the radius (times any gearing factor) of the driving wheels. In addition, rolling resistance, aerodynamic drag, and internal losses opposed the forward thrust from the prop, and the cart reaches a terminal velocity when these forces cancel.

The cart requires that forward force due to thrust be greater than the opposing force from the ground to compensate for the other loss factors. Since the cart isn't an over unity or unity device, power output < power input, so the prop + air speed has to be less than the ground + wheel speed by more than the difference between forces. I've been using the term "advance ratio" to describe the ratio (prop pitch speed) / (vehicle speed). A tailwind allows the prop + air speed to be lower and still generate thrust.
 
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  • #407
zoobyshoe said:
I believe I understand the lever, yes. It's more complex than small force at one end, large force at the other. It's more accurately: small force at one end applied over a greater distance, large force at the other end applied over a smaller distance.
Exactly. That's why it only works if air and ground have different speeds:
- The ground puts in a small backwards force over a long distance.
- The air puts in a large forward force over a small distance.
-> The net force points forward
zoobyshoe said:
You'd better clarify "braking force" for me.
The large forward-force applied to the top gear by the ruler (thrust force) does two things:
a) Push the cart forward
b) Create a torque on the gear which is transmitted to the ground, where it causes a small backwards force (=breaking force).

Here is another simple analogy of DDWFTTW:
Imagine you are facing a blue picket fence which is moving very slowly to the left. Close behind it there is a brown picket fence moving also to the left but much faster than the blue one. If you now hold a stick and put it trough both fences simultaneously it will accelerate you to the right.

So you are using two things which both are moving left to accelerate yourself to the right. Thats what the DDWFTTW-cart also does with the ground (brown fence) and air (blue fence):
 
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  • #408
zoobyshoe said:
How does the gear translation make sure the breaking is smaller than the thrust?

Jeff Reid said:
It doesn't. What determines if the braking force is smaller than thrust depends on the propeller parameters, diameter, pitch, and efficiency.

Just to clear up: We were talking about the cart under the ruler, with the reels (which I called gears) accomplishing transmission.

In the propeller cart the gears are not important for the transmission. It is done by wheel size vs. the propeller parameters you mention above.
 
  • #409
zoobyshoe said:
Rather than taking a demonstration that apparently works and then fitting some logic or reasoning to explain it, I would rather first determine if it can possibly work. At this point I am attempting to follow the train of energy: at each given point a certain amount of energy is needed. Each time I look at it I find that when the cart reaches windspeed it has no energy to continue except that amount represented by its own momentum. It can't go much farther.

You always make the same error: you calculate energies in two different reference frames to make the energy balance, which you shouldn't. When the cart is up to windspeed, indeed, in the reference frame of the cart, the wind is "still" and has no available energy to tap from. But *in this frame* the ground is moving, and you can tap from that movement.
You come to the conclusion that the ground has no energy to tap from because it is "not moving" but that is *in another frame*, namely the frame in which the *ground* is standing still (but in that frame, the wind is not standing still).

Consider, instead of the ground as being a flat surface, that it is water (and the wheel is changed into a paddle wheel). In a frame where the water is moving, you surely can tap energy with a paddle wheel from the running water, right ? That's how a water mill works. If you have a small river flowing under a paddle wheel, it turns the paddle wheel and delivers energy to the mechanics attached to the paddle wheel.

Well, in the frame where the wind isn't moving, the water (or the ground) is moving, and the paddle wheel (the wheel) extracts energy from that moving medium.

Now, this comes at a price, of course: if you want to extract power P from a moving medium, then you will have to exert a force on it, F, such that P = v x F. Whatever extracts the power P from that medium will undergo (at least) a force F - in the sense of being dragged along with the medium. This is the "braking force".
It is the force that a water mill undergoes (and is compensated by an identical but opposite force by the floor on the building). But the point is that power IS available in this frame.

And the trick is to use that power to power the propeller. If that propeller, with that power P, can exert a force in the other direction which is bigger than F, say F', then the overall force on the cart is in the direction of the wind. And it is possible to have F' larger than F, because the wind is motionless, and giving it a small speed v', smaller than v, is good enough: P = F' x v' and because v' is smaller than v, F' can be bigger than F.

This wouldn't work if the wind was not in relative motion to the ground or the water, because you couldn't then have v' smaller than v (it already comes in at v and you need to accelerate it a bit to have F' in the right direction). This is why this doesn't work if you have a car driving on a road (with no wind), and try the same: in the frame of the car, yes, you can extract energy from the moving road medium. But you won't be able to create with that available power anything which gives you a forward overall force, and hence your car will slow down.
 
  • #410
Here is the answer, in a nutshell. From Vanesch post #377:

Well, I guess it is semantics, but the frame is not limited, it extends to all of space, and you can describe all objects in them. It is: as long as their observations are limited to *the system at hand and their relevant boundary conditions*.

Now all we need to do is go back to the original question posed on the treadmill: Is the cart going faster than the treadmill? The proponents who claim that the cart is going faster are using a limited reference frame, that of the moving cart and the moving tread only. There are no boundary conditions. It is actually impossible to say which is going faster because the movement between them is relative. Based on this frame you cannot even determine which is moving. All that needs to be done is to enlarge the frame to include more relevant information, to include a boundary condition. I choose to include the floor and make it my boundary condition, my reference for the new enlarged frame. I am not claiming the floor is an absolute reference. It is not, but any thinking, logical person is aware that the floor is not moving relative to its own frame. It can therefore be used to act as the reference for both the cart and the tread. In this frame, it is very easy to determine which is going faster. From the videos, and watching the motion of the cart and the tread it is obvious to even the most dubious observer that the tread is traveling much faster than the cart, relative to the floor and relative to the new reference frame. Since the tread is the source of energy and it is clearly moving faster than the cart, it is easy to say that when the wind is the source of energy, it is also moving faster than the cart. This entire problem was artificially created by the proponents who have an agenda to prove that the cart is going faster than the tread and faster than the wind. I consider that claim to be false and I believe I have just proved it is false.
 
  • #411
schroder said:
From the videos, and watching the motion of the cart and the tread it is obvious to even the most dubious observer that the tread is traveling much faster than the cart, relative to the floor and relative to the new reference frame. Since the tread is the source of energy and it is clearly moving faster than the cart, it is easy to say that when the wind is the source of energy, it is also moving faster than the cart.
Since the floor moves at the same speed as the air, then let's switch this to the air as the frame of reference, which is a frame of reference that works equally well on a treadmill or outdoors. The treads moves the fastest, while the cart moves the next fastest in the opposite direction, and the air moves the slowest (zero). No one claimed the cart's speed would be faster than the tread, just that the cart's speed relative to the ground is faster than the air's speed relative to the ground. Since this is about the difference in speed, (cart - ground) versus (air - ground), the frame of reference doesn't matter.
 
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  • #412
Jeff Reid said:
Since the floor moves at the same speed as the air, then let's switch this to the air as the frame of reference, which is a frame of reference that works equally well on a treadmill or outdoors. The treads moves the fastest, while the cart moves the next fastest in the opposite direction, and the air moves the slowest (zero). No one claimed the cart's speed would be faster than the tread, just that the cart's speed relative to the ground is faster than the air's speed relative to the ground. Since this is about the difference in speed, (cart - ground) versus (air - ground), the frame of reference doesn't matter.

The air does not work equally well both indoors and outdoors. Indoors I have included the floor in the frame because it is “relatively” stationary to both the cart and the tread. I could have chosen the air but did not because it will NOT serve as a relatively stationary reference outdoors. And in the floor’s frame of reference the tread is moving faster than the cart. By the way, I believe you are the first person to acknowledge that. Others have claimed the cart is outrunning the tread!
Outdoors the air is moving in the form of wind. It serves no purpose to choose the air as the reference for the frame. We can once again use the floor to be consistent with the indoor frame as once again it is self evident that the ground is not relatively moving in its own frame of reference. And by using the exact same frame of reference, the only logical conclusion I can come to is that once again the driving force will be moving faster than the cart. In other words, the wind is moving faster than the cart. In all the video evidence I have seen, I have never seen any evidence to indicate that the cart is going faster than the wind, but I have seen plenty of evidence, in the form of debris being blown past the cart, that the wind is going faster than the cart.

The reference frame doe not matter? I see this as the entire issue. Once you can discount the reference frame “proof” there is nothing at all to support the claim of DDWFTTW. It is the heart of the matter.
 
  • #413
schroder said:
Others have claimed the cart is outrunning the tread!
I don't recall any comparason to tread speed versus cart speed with except that the advance ratio make's it very clear that prop pitch speed is less than tread speed, relative to the cart.

Outdoors the air is moving in the form of wind.
Not if the observer is in a hot air balloon or in hovering helicopter. The air is a perfectly valid frame of reference.

It (the air) serves no purpose to choose the air as the reference for the frame.
It provides a frame of reference where tread speed appears the same indoors or outdoors.

We can once again use the floor to be consistent with the indoor frame
What if I choose the "floor" to be a flatbed train moving at wind speed to be consistent with the indoor frame?

(floor indoors, ground outdoors ...) exact same frame of reference
It's not the same frame of reference. The tread moves with respect to the floor indoors, but the ground doesn't move with respect to itself as you pointed out.

I can come to is that once again the driving force will be moving faster than the cart.
The driving force is (prop thrust) - (opposing force from the ground), there's no speed factor here, forces and accelerations don't have a speed component, so the forces and accelerations are independent of the frame of reference,

The reference frame does not matter?
The reference frame only affects velocity related aspects to the situation, such as work done, change in energy, or power.

Once you can discount the reference frame ... there is nothing at all to support the claim of DDWFTTW.
DWFTTW implies a ground based reference, so let's restate the claim to be:

|cart speed - ground speed| > |wind speed - ground speed|

which is a way of describing the "DWFTTW" situation independent of frame of reference (|...| meaning magnitude).
 
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  • #414
schroder said:
Here is the answer, in a nutshell. From Vanesch post #377:

Well, I guess it is semantics, but the frame is not limited, it extends to all of space, and you can describe all objects in them. It is: as long as their observations are limited to *the system at hand and their relevant boundary conditions*.

Now all we need to do is go back to the original question posed on the treadmill: Is the cart going faster than the treadmill? The proponents who claim that the cart is going faster are using a limited reference frame, that of the moving cart and the moving tread only. There are no boundary conditions. It is actually impossible to say which is going faster because the movement between them is relative.

In as much as you are right concerning *absolute* velocities (which are just quantities which are frame-dependent, and hence have no intrinsic physical meaning), there is no arbitrariness concerning *relative* velocities. The velocity of the cart wrt the floor is a physically significant and frame-independent quantity: that means that I will find the *same* quantity, no matter what reference frame I use to calculate it. And so you CAN compare relative velocities, and say which one is larger than another one.

Now, the claim of a DWFTTW is that it is possible to make a device that:
1) is mechanically coupled to a flat surface and an air mass (and to nothing else) ; meaning, exchanges momentum, or has forces due to, or interacts with the flat surface and an air mass ;
2) doesn't have any internal source of mechanical energy (motor or something of the kind)
3) is put in a situation where the relative velocity of the device wrt the flat surface (relative velocity, so a frame-independent vector) is in the same direction and larger than the velocity of the air mass wrt to the flat surface (relative velocity, so again frame-independent vector), as long as that relative velocity of the air mass wrt the flat surface is given within some finite limits (say, between 15 km/h and 20 km/h or something).
Based on this frame you cannot even determine which is moving. All that needs to be done is to enlarge the frame to include more relevant information, to include a boundary condition.

Moving has, per Galilean relativity, no absolute meaning, but it does have a relative meaning. You can say that the cart is moving wrt the surface. And that does have a physical meaning, independent of the frame in which you express it.

Now the whole thing is that we can write frame-independent quantities as a function of other frame-independent quantities. If we can write such a relationship (a mathematical function), then we have "solved the problem". And Galilean relativity tells us that this function is independent on the frame in which we perform the calculations.

What interests us here is the function: v_cart-wrt-surface-steady-state = F(v_air-wrt-surface, v_cart-wrt-surface-initial). Note that the argument as well as the result are frame-independent quantities (they are the same number independent of the frame in which they are seen). So if we calculate the exact mathematical form of F, which is given by the forces of interaction between the cart and the surface, and the cart and the air, and the internal construction of the cart, then we have solved this problem. We can calculate this function F in any reference frame we like, Galilean relativity tells us that its mathematical form is independent of the choice of the reference frame in which we prefer to do the calculation (and a good physicist is one that intuitively picks the frame in which the solution is most easily obtained).

I choose to include the floor and make it my boundary condition, my reference for the new enlarged frame. I am not claiming the floor is an absolute reference. It is not, but any thinking, logical person is aware that the floor is not moving relative to its own frame. It can therefore be used to act as the reference for both the cart and the tread. In this frame, it is very easy to determine which is going faster. From the videos, and watching the motion of the cart and the tread it is obvious to even the most dubious observer that the tread is traveling much faster than the cart, relative to the floor and relative to the new reference frame.

I don't exactly understand what you are talking about. On a treadmill, a DWFTTW demonstration would be to see the cart move against the treadmill in the ground frame. At any speed. From the moment it moves against the treadmill (call it the positive speed v_cart if it goes against the mill, say, 3 km/h), it goes DWFTTW, because the RELATIVE velocity of the cart and the surface is v_cart + v_mill (say, v_mill = 7 km/h) while the relative velocity of the air and the surface is v_mill. In other words, we get that F(7 km/h) = 10 km/h. So the first number is larger than the second, and that is what needed to be demonstrated, because that's what comes out of the mathematical function F above. As this function is supposed to be independent of the reference frame in which it is calculated (Galilean relativity), this would then result in exactly the same number if the floor were now a "real floor". If the wind were blowing at 7 km/h, we would have to put 7 km/h in this function, and find 10 km/h, all the same.

What you are claiming is that 3 km/h is smaller than 7 km/h, in other words, that F doesn't DOUBLE its argument, but only adds 3 to it. Yes, but that was not the claim. Your requirement comes down to Down Wind more than twice faster than the wind.

Since the tread is the source of energy and it is clearly moving faster than the cart, it is easy to say that when the wind is the source of energy, it is also moving faster than the cart.

What is "source of energy" is frame-dependent. If a fighter plane flying at Mach 2 would shoot a rocket backwards at Mach 3, then in the process, from the frame of the air or the ground, the rocket has LOST energy (it went from Mach 2 to Mach 1). In the frame of the fighter, the rocket WON energy (it went from 0 to Mach 3).


This entire problem was artificially created by the proponents who have an agenda to prove that the cart is going faster than the tread and faster than the wind. I consider that claim to be false and I believe I have just proved it is false.

You have not done so at all. You switch to some very ad hoc claims about "source of energy" to jump to the conclusion you wanted to. You really have demonstrated a very poor understanding of what reference frames and galilean relativity are about. As I said, the proof is ununderstandable to someone who doesn't accept, or doesn't understand, galilean relativity because it is essential in these demonstrations. But that doesn't invalidate the proof. It is not because one doesn't understand induction on the natural numbers, that the proof that there are an infinitude of prime numbers is not valid.
 
  • #415
schroder said:
The reference frame doe not matter? I see this as the entire issue. Once you can discount the reference frame “proof” there is nothing at all to support the claim of DDWFTTW. It is the heart of the matter.

BTW, you've still not answered my post concerning the train. It would be interesting to see where you switch from "yes" to "no".

See: https://www.physicsforums.com/showpost.php?p=2036877&postcount=378

It will allow me to pinpoint exactly where you are having a misconception.
 
  • #416
zoobyshoe said:
Yes, aero majors would be nice. I understand the sail on a beam reach: Bernoulli on the one side, Newton on the other.
It's Newton all over the place. If fluid dynamics doesn't follow from Newton, something's wrong it--but of course it does.
 
  • #417
Jeff Reid said:
2. The sailcraft obtains it's power by slowing down the true wind.

Hi Jeff. One small point. Not to distract from your expertise in this. I haven't got all the elements lined up, so I can't be critical of your overal analysis. I would't bother, but I've heard you say something similar to this before.
An airfoil would work perfectly well in an ideal inviscous fluid--one without energy loss from drag of any sort, and where it has a bound vortex. (You might look into bound vorteces later--rather an interesting model of an idealized airfoil--if you haven't already.) Looking at this in an wind tunnel, one would see the airsteam redirected as it leaves the foil. The airstream is redirected slighly, but the velocity of the airsteam remains the same after it interacts with the foil. The kinetic energy of the trailing steam is the same.
As the velocity must remain the same where there is no energy lost from viscous drag or otherwise, the downsteam component of velocity is reduced, but taken-up with a velocity component normal to the free steam velocity of the air. For a propeller the airflow would develops a helical flow. The force on the airfoil from this redirection is the induced drag. There's no energy loss with induced drag. Without going into details, the viscous drag is one of the components of force that would tend to hinder progress DDW.
 
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  • #418
Phrak said:
Not to distract from your expertise in this.
I'm not an expert, but I have been following the DDWFTTW threads for a while now.

Jeff Reid said:
The sailcraft obtains it's power by slowing down the true wind.

The viscous drag is one of the components of force that would tend to hinder progress DDW.
Which it does, the cart itself experiences an apparent headwind and the associated drag from the apparent headwind hinders progress DDW. However slowing down the true tailwind means that air is accelerated upwind, and the affected air reacts with a downwind force. This force times the speed of the ground (relative to the cart) is the power input.

The rest of this post is a bit off topic:

An airfoil would work perfectly well in an ideal inviscid fluid ... zero drag ...
zero drag only if the fluid is inviscid and incompressable at constant velocity based the math from D'Almebert's paradox:

http://en.wikipedia.org/wiki/D'Alembert's_paradox

My issue with this is the idea of an incompressable fluid. This would require that the molecules have some magical geometric shape that leave no gaps, but allows frictionless movement. Since the molecules are incompressable, then there is no relationship between pressure and deformation of the molecules. If such a fluid were in a cylinder, also made of uncompressable material, then what's the pressure of the fluid? What if this cylinder was in a perfect vacuum, and an attempt was made to expand the cylinder (for example withdrawing a perfectly sealed pistion), what would stop the expansion of the cylinder?

For a propeller, the airflow develops a helical flow. The force on the airfoil from this redirection is the induced drag. There's no energy loss with induced drag.
I've seen multiple definitions for induced drag, so I'm not sure on what you mean by induced drag. I've always read that the helical flow off a propeller is related to total drag, not just induced drag.
 
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  • #419
vanesch said:
What interests us here is the function: v_cart-wrt-surface-steady-state = F(v_air-wrt-surface, v_cart-wrt-surface-initial). Note that the argument as well as the result are frame-independent quantities (they are the same number independent of the frame in which they are seen). So if we calculate the exact mathematical form of F, which is given by the forces of interaction between the cart and the surface, and the cart and the air, and the internal construction of the cart, then we have solved this problem. We can calculate this function F in any reference frame we like, Galilean relativity tells us that its mathematical form is independent of the choice of the reference frame in which we prefer to do the calculation (and a good physicist is one that intuitively picks the frame in which the solution is most easily obtained).


That is a major part of the problem. An honest and straight forward evaluation of this cart does not require a “physicist” who deals in abstractions. I have the utmost respect for physicists, I am only saying this problem is not complicated enough to require the services of one. It only requires a qualified test engineer with a great deal of practical experience in this sort of thing. That’s where I come in. I am now going to run a test for you and demonstrate that this cart has never run faster than the wind, cannot run faster than the wind and never will run faster than the wind.
As test director, I choose the reference frame and I choose it to be the most logical frame which can act as a reference for both the treadmill test and the outdoor test. I choose the ground as my frame of reference. It is not an “absolute” frame, so it does not violate Galilean relativity but it is easily shown to be stationary “relative” to both the moving treadmill, the cart on the treadmill, the wind which is blowing, and the cart which is moving with the wind. It is a relatively stationary reference to all the moving components in the test. I am setting up this test so the treadmill test and the outdoor test can be run simultaneously and observed by me or any other witness by standing in one position on the common ground reference. The test is conducted outdoors on a large flat surface (the “ground”). The treadmill is sitting on the ground and is enclosed by Plexiglas so as to shield it from the wind. The wind is cooperating by blowing at a nice steady velocity and is constant at 10 m/sec and is being constantly monitored by calibrated anemometers. The treadmill has a setting which has been carefully calibrated to run at exactly 10 m/sec exactly the same as the wind and both velocities referenced to the ground reference. The re is no need to read the cart’s actual velocity on the treadmill, only to demonstrate that it is either greater or lesser than the tread velocity all velocities referenced to the ground. By video analysis, it is clearly demonstrated that the cart is moving at much less than the tread velocity. I have estimated it to be at about 30% of tread velocity at most. The actual number is not important. As for the cart running down wind, for some reasons nobody has provided me with a radar gun so I can measure it directly. I don’t know why my request has been turned down. I find this very strange but I will make do. All I need is to see if there is any evidence at all if the cart is going faster than the wind or slower, all velocities measured in respect to the ground reference. As the cart is running down wind a video is recorded and on playback it clearly shows that some debris which is also being blown by the wind passes the cart at a much higher velocity than the cart is moving. All comparisons are made with respect to the ground reference. From the evidence I conclude: 1) The cart on the treadmill, being driven by the treadmill, is moving much slower than the treadmill. 2) The cart in the wind, being driven by the wind is moving much slower than the wind. All of my comparisons are with respect to one common reference and can be considered as one reference frame. I can only conclude that the evidence is overwhelmingly against the claim of DDWFTTW. In fact, I now consider the claim to be so ridiculous that I will not participate in any more discussion or debate on this matter. My findings are here presented and open to inspection to all. I consider the matter as closed.
 
  • #420
Jeff Reid said:
1. The apparent crosswind is independent of the sailcrafts forward speed. The apparent crosswind is perpendicular to the sailcraft's path, so this is a given. The apparent crosswind = (true wind speed) x sin (angle between true wind direction and sail craft direction). For a given crosswind, some sailcraft can accomplish a forward speed so that the (apparent crosswind) / (apparent headwind) is < 1. This ratio is normally expressed as an angle, called Beta, = atan(apparent crosswind / apparent headwind). For iceboats, I've read the Beta can be as low as 8 degrees under ideal conditions (true wind speed near 10 mph.)
"Apparent crosswind"? Can you link me to some site that explains "apparent crosswind"? I googled but I found no definitions, just mostly posts by you here and on other forums. I don't understand why you aren't speaking of everything in terms of plain "apparent wind". Why "apparent crosswind" and also "apparent headwind? You'll have to bring me up to speed on the advantage of analyzing this way.
2. The sailcraft obtains it's power by slowing down the true wind. This requires that even though the sailcraft's downwind component of speed is greater than the wind, the sail has to divert a sufficient amount of the apparent headwind to true upwind, so that the air flow off the sail slows down the true wind. Slowing down the true wind is the source of power for the sailcraft (or any wind powered vehicle).
This search for slowed wind sounds very idiosyncratic and seems misleadingly incomplete. It suggests that the only operative force here is Newton's Third Law. It strikes me as essential to include Bernoulli. We aren't only applying force to the back of the sail we are also removing opposition to it from the front. The relative vacuum in front of the sail explains a lot more to me than just the Newton III behind it does.
This isn't required, the apparent crosswind is independent of the boat's forward speed. The main reason for starting near perpendicular to the true wind is that the apparent crosswind will be higher (sin() is maximum when angle is 90 degrees), resulting in more acceleration.
Why this isn't required may become apparent after you fill me in on "apparent crosswind." In the meantime everything I find says that apparent wind is correlated to boat speed.


It doesn't. What determines if the braking force is smaller than thrust depends on the propeller parameters, diameter, pitch, and efficiency.
Rodger that.
The power input to the propeller is torque time angular velocity.
OK.
The power output is thrust times air speed through the propeller.
The propeller can put out this much power. I am not sure of the significance of this. This power is being given to the air?

The forward force from the prop is thrust, the opposing "braking" force from the ground is the prop input torque divided by the radius (times any gearing factor) of the driving wheels. In addition, rolling resistance, aerodynamic drag, and internal losses opposed the forward thrust from the prop, and the cart reaches a terminal velocity when these forces cancel.
OK.

The cart requires that forward force due to thrust be greater than the opposing force from the ground to compensate for the other loss factors. Since the cart isn't an over unity or unity device, power output < power input, so the prop + air speed has to be less than the ground + wheel speed by more than the difference between forces. I've been using the term "advance ratio" to describe the ratio (prop pitch speed) / (vehicle speed). A tailwind allows the prop + air speed to be lower and still generate thrust.
So, do you see the cart ever being able to sustain FTTWDDW? It seems you are adding things up to the conclusion that if it makes it into the HH it will only last as long as the energy represented by the cart's total momentum lasts.
--------------------------------------------------------------------------------------
Because he thought to affix a tell tale, this video is the most apparently impressive I've seen:

http://www.youtube.com/watch?v=aJpdWHFqHm0&fmt=18

However, if you go to the linked address for the expanded look at the cart he specifically says it's designed to be brought up to speed as a windmill: the fan is configured to receive power from the wind from behind and transmit it to the wheels, in which case the fan blades are pitched to resist acceleration into the HH, not encourage it. Very weird.
 

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