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Down wind faster than the wind

spork: hmm... yes and no. Look at the first picture where the sailboat is tacking upwind (same as iceboat tacking downwind). The direction of the force from the sail is more sideways to our direction than forward. With the cart it is vice versa. The iceboat is NOT going DIRECTLY downwind, its zigzagging. Also no part of the iceboat sails drag transfers into force directed downwind. I would say the idea and final function are the same, but the situation is not the same. Anyway, I get your point. And I am able to go reverse with my boat in real life too:)

Is your propellor symmetrical? If not, you could try reverse both the gearing and the prop. It should work better one of the ways, I think.

edit: ofcours not reversing the gearing...

edit2: spork, consider this situation. Both the cart and iceboat are running steady state and the wind suddenly stops and the "sails" are backing. Are the situations equivalent?
 
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Look at the first picture where the sailboat is tacking upwind (same as iceboat tacking downwind). The direction of the force from the sail is more sideways to our direction than forward. With the cart it is vice versa.

Nope. The tips of our prop are exactly the same as the ice boat on a downwind tack.

The iceboat is NOT going DIRECTLY downwind, its zigzagging.

Gosh - thanks for that. I think that must be what I was missing all along when I came up with this prop cart concept from the tacking ice boat, built it, and demonstrated it doing precisely what I said it would.

Is your propellor symmetrical? If not, you could try reverse both the gearing and the prop. It should work better one of the ways, I think.

In this design the propeller functions as a propeller (not a turbine). There is only one correct way to mount it.

spork, consider this situation. Both the cart and iceboat are running steady state and the wind suddenly stops and the "sails" are backing. Are the situations equivalent?

Yes.
 
Doesn’t that negate the claim that the cart (which includes the propellers) is going directly downwind?

Nope. If that were the case it wouldn't even be legal for the wheels to turn. The frame is going directly downwind. The C.G. is going directly downwind. And every single component is going directly downwind. Some components just happen to be spinning. As such, things like the prop tips carve a continuous downwind tack.
 
Nope. The tips of our prop are exactly the same as the ice boat on a downwind tack.

Yes, ofcourse. I'm sorry, I'm not native english speanking and maybe I was't clear on what I ment. What I mean is that as the sail and tip of prop are exactly the same (same airflow, same direction of movement, etc.), the cart "frame" and the iceboat are not moving in the same direction. (This is also obvious ofcourse).

Gosh - thanks for that. I think that must be what I was missing all along when I came up with this prop cart concept from the tacking ice boat, built it, and demonstrated it doing precisely what I said it would.

Oh for f..cks sake. I said the cart works. I just wonder about some details trying to understand better. Why do you count me as humber?

In this design the propeller functions as a propeller (not a turbine). There is only one correct way to mount it.

Yes, ofcourse. But I didn't know if your propeller was originally designed to be a turbine or a propeller, what is the shape of the airfoil. If it was a turbine that you use as a propellor, it should be installed "the wrong way". I was thinking maybe more about the other guy who's cart is not yet working.

Hmm, here I need to do some thinking myself.

... about 10 minutes ...

I still stick to my opinion for a while and hope you will explain.

Another picture attached.

A assume that we agree that
1) looking at the apparent wind at the tip of the propeller and at the sail they are identical and so are the forces produced
2) the propeller tip and the sail are moving in the same speed and same direction relative to the ground
3) the cart as a whole and the sailboat are moving in different directions and different speeds, the downwind component of speed is the same, however

In the following I assume, that after the wind stops, the resulting force from the sail is backward and left (on stb tack) compared to the direction the iceboat is moving. If I look at this same force on the cart, the force is pointing *forward* and left compared to the direction *the cart* is moving. This should be quite possible, although I still only "scetch" the pictures without doing any calculus.

On the iceboat, the skates take care of the sideways force (bigger) and all we have left is the braking force (smaller). On the cart there is no direct geared coupling between sideways force from the skate and forward-back forces from the sail. Effectively there might be a "gearing" as you explained with the iceboat starting windward -example and I just fail to see it.

On the cart, however, forces from the propeller change magnitude and direction because of the gearing. We can divide the force from the prop to a forward component (smaller), relative to the cart's direction, not the sails movement) and a sideways component (bigger) which is also "CCW component" for the prop. The forward component is transferred via gearing to form a CW force and a braking force on the wheels. The CCW component from the prop is transferred via gearing to a accelerating force on the wheels.
Are these situations identical? Maybe. Maybe all the forces from sail/skate and prop/gear/cart sum up to nothing. But the answer definately is not obvious to my mind.

What is obvious though, is that the cart will not accelerate after the wind stops, so the lenghts and directions of arrows in my drawing are definately wrong.

But still, intuitively, I would say that the cart both speeds up and slows down "slower" compared to an imaginary iceboat because of the gearing. Maybe the kinetic energy of the propeller's spinning is somehow related to this? The iceboat has no moving parts in this example, the cart has.

On the other hand, intuitively, I am wrong.:o

huh
 

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There should be a law that prohibits thinking while tired.

The angle of the keel/iceboat defines the braking force. The gearing of the cart defines the braking force. There you go. It is just a bit more difficult equation to solve graphically.

That leaves us only with the spinning, does it play any role in acceleration? The total mass of the cart is moving "slow" and the prop is spinning. The total mass of the iceboat is moving "fast" (as fast as the propeller tip) but there are no moving parts. Assuming the total masses (cart/iceboat) the same and frictions the same, is the situation the same?
 
Oh for f..cks sake. I said the cart works. I just wonder about some details trying to understand better. Why do you count me as humber?

Sorry. I sometimes have a hair trigger. Far too often we have people that come around only to say "now I get it - it's just that you explained it all wrong - here's how it really works..." Followed be an explanation that isn't correct.

I'll have a look at your analysis shortly.
 
In the following I assume, that after the wind stops, the resulting force from the sail is backward and left (on stb tack) compared to the direction the iceboat is moving.

I started to look over your analysis, but got stuck on this point. Once the wind stops the sail should luff straight back and all forces on the boat (drag from both aero and ice) should directly oppose the boat's motion. With no wind we simply have an ice boat coasting in still air until it comes to rest - right?

I suspect there's a key point you're making that I'm completely missing.

With regard to the rate of acceleration of each we'd have to look at the moment of inertia of the prop, prop-shaft, gears, and axle as compared with the mass of the ice boat and sail. We'd also have to compare the efficiency of the rotating prop as compared to the sail. None of that has been within the scope of the point I've tried to bring across up to this point. But I strongly suspect you could make it go either way based on design decisions.
 
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Doesn’t that negate the claim that the cart (which includes the propellers) is going directly downwind?

No one claims that a Porsche isn't going directly south just because the pistons are bobbing to the east and west inside the engine.

No one claims that a Cessna isn't going directly south just because the tips of the prop are zig-zagging east and west.

No one claims that a sailboat isn't going directly south just because someone is grinding a winch on deck.

Components will move around on any mechanism, but it's the mechanism as a whole that determines it's direction.

JB
 
On a completely unrelated note, I spoke to Bauer's wife yesterday and today spoke to a colleague he co-authored a book with. Both were incredibly friendly and interesting to talk with. Unfortunately Bauer is now in an assisted living facility, and no longer able to pursue his life's work. In speaking with his colleague it became clear just how sharp a guy Bauer was during his career. These guys are the kind of engineers I aspire to be. In the next few days I hope to speak to a couple of colleagues that worked with Bauer when he first developed his DDWFTTW cart.

Interestingly, this whole thing came about when one of the senior engineers at McDonnell Douglas posed a challenge at lunch one day - can you make a wind powered vehicle that goes directly downwind faster than the wind?. A few days later Bauer proposed his idea, and not long after (as a side project) he demonstrated his cart doing what he designed it to do. I'm told that all witnesses were very well satisfied that it did in fact beat the wind on a straight downwind heading.

Bauer demonstrated his cart several times over the years, most recently to the Aerovironment group headed by Paul MacReady. They were all impressed and convinced as well. As most of you probably know, MacReady passed away recently. He was truly one of the great aviation pioneers. I highly recommend this talk MacReady gave at a TED conference in 2003: http://www.ted.com/index.php/talks/paul_maccready_flies_on_solar_wings.html

The colleague of Bauer that I spoke with today has had an amazing career in engineering as well. Among other things, he designed and built the first animatronics (Lincoln in the Hall of Presidents) for Walt Disney (who he came to know during the project). He and Bauer did some of the most comprehensive (and radical) work on propeller design, and co-authored a book on the subject.

Yesterday JB and I also met with a Stanford Aero professor that happens to be a friend of a friend. We brought our little cart with us of course. He was well familiar with the concept and had even presented it as a problem for his PhD candidates. But I think what was most eye opening was to watch the most definitely surprised reactions of his employees (other Aero PhD's) at the company he founded.
 
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Spork, Do you want a place where you can go and leave all the non-believers behind?

Propose a simple math physics problem involving the cart. For the conditions with the cart moving directly down wind in at 5 m/s in a 5 m/s wind and given all the necessary measurements of the cart, compute the acceleration of the cart. Use that number to form the URL to your secret bulletin board.
 
Spork, Do you want a place where you can go and leave all the non-believers behind?


Not at all. Skeptics I like. People that know for sure 100% that it can't be done, and call us fools, I could do without.
 
I started to look over your analysis, but got stuck on this point. Once the wind stops the sail should luff straight back and all forces on the boat (drag from both aero and ice) should directly oppose the boat's motion. With no wind we simply have an ice boat coasting in still air until it comes to rest - right?

I suspect there's a key point you're making that I'm completely missing.

Yes. It is a key point in understanding my misunderstanding:boggled:.

Once the wind stops a real boat would behave like you said and the sails would be flapping ofcourse. I was assuming that the shape of the airfoil was fixed (like a sail made of some hard material, like the propeller blade). Then it would still produce some force sideways because of its shape, right? Anyway, somewhere between the wind completely stopping and just a lull, there is a situation were the force is directed the way it is in my example. I wanted to study this situation and just used the wind stopping example for reasoning that this situation is possible.

What I was doing in my mind is that I was just looking at the forces using two differently aligned coordinate systems (one with the axel parallel to the iceboats movement and another one with the axel parallel to carts movement). With the iceboat I came to the conclusion that all resultant forces slow down the boat. With the cart there was this coupling with the gears that produced also an accelerating force. This messed my head. I thought that there was no equivalent thing with the iceboat, but there is. The angle of the keel not only defines the direction of the resultant force (=slowing down) but also the magnitude of the force (they are coupled just like in the gearing). Putting it this way it seems so bloody obvious...:o By messing with the coordinated I just managed to hide this coupling.

A stupid analogy: with the iceboat, I found the decelerating force to be say A. With the cart I found the force to be C + D and wondered if A = C + D. I should have thought the iceaboat as A + B where B=0 because of the coordinate system I chose. Therefore I failed to see the obvious equation "1 + 0 = 2 - 1".

With regard to the rate of acceleration I would like to consider an ideal model of the situation just as a brainteaser. So the propeller blade producing the forces would be differantially small "dot" with a certain mass at the tip of the prop etc. In this context I think my last question in my previous post makes sence. I could be more precice by asking "Assuming the total masses (cart/iceboat) the same and frictions the same, is the situation the same, REGARDLESS OF THE MASS DISTRIBUTION BETWEEN CART FRAME / PROP TIP?
 
And yet in spite of your empty assertions, energy is extracted from the relative motion between two mediums every moment of every day the world over.

JB

Yes, there must be two. Try extracting the difference between something and itself.

Media = plural of medium (materials)
Mediums = plural medium (psychics)
 
The only harmonic wheel I'm familiar with is music related and to that the answer would be no.
JB

Harmonic, as in harmonic motion. Often rotating mass damper or counterweight

Spork, Do you want a place where you can go and leave all the non-believers behind?

Propose a simple math physics problem involving the cart. For the conditions with the cart moving directly down wind in at 5 m/s in a 5 m/s wind and given all the necessary measurements of the cart, compute the acceleration of the cart. Use that number to form the URL to your secret bulletin board.

Oh, nice. I wonder if there isn't just a little more hubris there than is warranted.
 
Yes, there must be two. Try extracting the difference between something and itself.

Media = plural of medium (materials)
Mediums = plural medium (psychics)

Harmonic, as in harmonic motion. Often rotating mass damper or counterweight

Oh, nice. I wonder if there isn't just a little more hubris there than is warranted.
I see you have no more arguments on topic. Does this mean you understand now?
 
But the cart is motionless in still air on the treadmill. You really are losing me here.

Re:above post, no.

It is too simplistic to conclude that no differential wind = windspeed.
It is not logical to conclude, that because there is no differential wind, the cart is at windspeed.

There are reasons for this: It has been found to work, and more importantly, it can be reasoned to be true, and observed to be true.
I'm sorry, I'm not going to say it again. You seem to have had a very odd physics teacher.

The principle itself is not in dispute.
In what sense does it work? As a tool of convenience to allow simpler calculation, or a mental aid to gain a perspective, yes, but as you say, that changes nothing about the relationship between the relative elements. 'Equivalence' is ubiquitous. It is part of a common background, as are time and gravity. I do not see how invoking this principle changes one fact about the relationship between the cart and the wind or anything else.

No! If you sit on a missile you can know your velocity, but only if you define a frame of reference! Velocity is meaningless without a frame of reference, that's why I asked you how fast the earth is travelling. You gave the correct answer, but for some reason you think that this is not true for things on earth.
No, it is not redundant. It can be enlightening, because it shows you that you don't have to measure flight in the air only; you can use a wind tunnel.

You are comparing the general to the specific. How is it enlightening to view the cart from the belt? Do you need to invoke a recondite theorem to justify concluding that the wind moves reactive to the ground? Well, it just does, doesn't it?
The Earth's velocity is not significant because the motion is between the cart and the ground; the wind is the energy conveyor. Yes, the entire solar system is in motion and the galaxies are moving apart. And? There is such a thing as 'locality' to be considered.

In all the math examples I have seen are simplistic. The wind is replaced by a singly-dimensioned vector 'V'.

Statics and Dynamics are equated. Equivalence does not allow for that!.

The wind is not only a dynamic force, it is a system. The air gains energy and momentum from the sun, which is returned as it blows over the road, and other objects. The "equivalence" is much more than the just the relative velocity between the surface of road and the wind, because that is also (one of) the means by which the energy is circulated.

The next step it seems, is to take that vector, and make it manifest by the use of a treadmill. There is no other supporting mathematics or logic that allows for any more equivalence than one superficial similarity. It moves.

It is "observer based" only. Dynamic systems cannot be treated as if they were static, just by means of a new perspective.

It is also an astounding failure of reason. All things may be relative, but that does not mean they are all equal, nor that the absolute magnitude of each becomes unimportant. To 'prove' this failure in a van....

No. As an aerospace engineer I can assure you that this is a failure of your understanding.
To be fair, H'ethetheth, you have made more effort than those you have cited, but I do get tired of the general appeal to authority, especially when the arbiter is that self-same person. There are good engineers and bad engineers. I don't know you. Do you think automatically that you can match me in my fields of expertise?

I do not like to see education used as a weapon, if only because there are those that deserve an explanation despite not having the necessary formal skills. They do not need the pseudo-intellectual posturing of those I refer to as the 'mice'. That's my 2 cents on that topic.

As an aerospace engineer you should be aware that all design and development tools, from software simulations to environmental chambers, undergo a process of verification by comparison with information derived independently of the test object. In use, they are calibrated against standards. The list of failures in this regard are to numerous to mention.

The only force the cart needs to overcome when it is riding at wind speed is the friction in the mechanism. The energy to do this comes from the fact that, in this situation, the ground is passing by at wind speed.

What a happy coincidence it is that we can write;

Required force at windspeed = Stray friction in treadmill.

You are thinking only of the incremental force. In reality, of course, the cart will be constantly changing its velocity as it meets varying circumstances, so making almost continuous demands for significant incremental force. That not withstanding, there must be a static force behind it sufficient to have driven the cart too windspeed in the first place.

The reason that the cart has no 'go' is simple. There is no force at the propeller.
For that to happen, there must a force capable of developing as differential pressure across the propeller. Obviously, in static air, this is not the case.
The small amount of energy that trickles up from the motor to power the drag of the propeller as it makes a local disturbance, is orders of magnitude below what is actually required.

I have earlier described the treadmills act, but perhaps this is clearer.

Replace the propeller with a tangential impeller, (like a waterwheel
with end-plates, such that it cannot generate lateral forces).
If you spin such a device in the hand, you will feel the opposing drag of the air.

Put this on the cart, and spin the wheel clockwise. The 'prop' will rotate
clockwise, and you will feel the opposing drag.
Turn the wheel CCW, and you will also feel opposing drag.
The cart is trapped in this force 'well'. I cannot so eadily move forward or backwards, so its natural tendency when on the belt, is to stay where it is.
The prop and the drive shaft oppose each other, resulting in an absolute torque that is the drag of the prop at that particular angular velocity, and a differential (driving) force of next to zero.
( Assuming there is adequate friction with the belt to transmit that absolute torque)

The real propeller is not symmetrical with respect to drag, though it is turned against its natural bias, so that drag will be dominant. The 'thrust' component will be smaller, as the propeller is not designed to spin in that direction, so the absolute 'drag' will be the result of the difference of those two.
Same process, different 'well' depth. The motion forward I have also previously described, but given that the basic principle is so flawed, I don't feel that it requires more elaboration than has been given

ETA:
To make things clear,H'ethethet, I do not say that vehicles cannot travel faster than the wind, but:
1. A direct propulsion device like this cart, no
2. I have not seem proof that it can, therefore I stick to 1.
3. The treadmill says nothing about the carts capacities, other than the wheels spin
 
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All things may be relative, but that does not mean they are all equal, nor that the absolute magnitude of each becomes unimportant. To 'prove' this failure in a van....

The wind is blowing at a speed of 10 kph. The cart is moving downwind along the road at a speed of 12 kph relative to the road. What is the "absolute magnitude" of the cart's velocity?
 
Yes. It is a key point in understanding my misunderstanding...

Thanks for the explanation. I see what you were up to now.

"Assuming the total masses (cart/iceboat) the same and frictions the same, is the situation the same, REGARDLESS OF THE MASS DISTRIBUTION BETWEEN CART FRAME / PROP TIP?


Interesting question. I'm nearly sure the answer is no. Imagine all the mass is in the frame. In this case we only have to accelerate in in the linear mode.

Now assume some or all of that mass is in the prop tips. We still have to accelerate it all linearly, PLUS we now have to get the prop spinning, with it's additional M*R^2 term. Or looking at it slightly differently, if we put all the mass at the prop tips we would have to accelerate it to a greater velocity for the same cart velocity.
 

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