Understanding DDWFTTW: Exploring Its Principles and Addressing Common Questions

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In summary, the propeller can apply more force at faster-than-wind speeds because it does not travel as far through the air. This allows the cart to extract more power from the wind.
  • #141
kmarinas86 said:
So do you think the wind pushes both the cart and the ground even when using wheels?
Yes. The air pushes the ground downwind via the cart.
kmarinas86 said:
I can imagine that it would if the wheels were "locked" so that they could not rotate.
No, rolling wheels can also exert horizontal forces.
kmarinas86 said:
For a car, I definitely do expect that the ground is a reaction mass to the car's acceleration.
When you gently use the brakes in a car the wheels are not sliding. And yet the car slows down. So there must be a horizontal force backward on the wheels, and an equal but opposite force forward on the ground.
 
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  • #142
kmarinas86 said:
So do you think the wind pushes both the cart and the ground even when using wheels?
Yes.

This is very clear when the cart is stopped (wheels locked with brakes for example). If the brakes are released slightly so that the cart rolls, the wind is exerting a force on the cart which still exerts a force on the ground, meaning that the wind indirectly is pushing the ground. Good so far?

Duplication of posts, and I have to get some work done! I'll check in later but remember you're in good hands here!
 
  • #143
mender said:
Or more simply, the angle of the attack of the prop plus its rotational speed allows the prop to push the air back as the prop moves through the air mass (by definition moving faster than the air), pushing the cart forward.

I think you're making things a little too complicated!

It's just hard for me to think of this wind as being a perfectly homogeneous entity while trying to explain why this is possible.

The problem was of course whether or not the prop had to do additional work on the particles of air to make the air push against the prop, combined with the fact that we are trying to explain the acceleration of the craft purely using wind. Well I know that average speed of air particles is always faster than the wind they make up (for it to equal it would be like having a condition of absolute zero for the wind, which is impossible). A simple deflection of a momentum doesn't require kinetic work (because its changing the direction only). So a few particles here and there can actually move faster than the craft temporarily, pushing the cart, while particles kicked back are moving slower.
 
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  • #144
kmarinas86 said:
It's just hard for me to think of this wind as being a perfectly homogenous entity while trying to explain why this is possible.

The problem was of course whether or not the prop had to do additional work on the particles of air to make the air push against the prop, combined with the fact that we are trying to explain the acceleration of the craft purely using wind. Well I know that average speed of air particles is always faster than the wind they make up (for it to equal it would be like having a condition of absolute zero for the wind, which is impossible). A simple deflection of a momentum doesn't require kinetic work (because its changing the direction only). So a few particles here and there can actually move faster than the craft temporarily, pushing the car, while particles kicked back are moving slower.
Yup, too complicated!

Go back to thinking about a homogeneous air mass moving at a uniform speed rather than the individual particles; macro rather than micro! The way the cart works is explainable without invoking Brownian movement!
kmarinas86 said:
It's just hard for me to think of this wind as being a perfectly homogenous entity while trying to explain why this is possible.
That's 'cause you don't have the right explanation yet!:wink: As I said, it's much simpler than this, so pull out another sheet of paper and have another go!
 
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  • #145
mender said:
Yup, too complicated!

Go back to thinking about a homogeneous air mass moving at a uniform speed rather than the individual particles; macro rather than micro! The way the cart works is explainable without invoking Brownian movement!

If a train moves at 120 mph and I throw a curve ball at (initially) -60 mph, if the ball does not reach close enough to the train at the rear of one the train cars to hit the train, then it does not help the train to accelerate. Instead it will just pass by. (The exception is if it was a really good curve ball and could return to the back!)

Let's pretend for a moment that this ball is the "homogeneous air mass". How can the ball hit the back of a train car after that kind of throw? I can imagine it hitting the edge of the face of the back of one the train cars. But onto the face of it? No, I cannot imagine that. The ball is supposed to be the donor of energy, not the train. What could the train do to the ball to allow the ball to contribute net kinetic energy to the train? Bounce between train cars? There's no net gain for the train to be found there. A net loss of kinetic energy for the train is to be had if the ball settles down on the train itself, or if it bounces forward faster than the train as it leaves.

That's why wind to me has got to be inhomogeneous in order to explain how this DDTWFTTW stuff works.
 
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  • #146
mender said:
The way the cart works is explainable without invoking Brownian movement!
And sailboats can be explained without General Relativity and bending of space time. Believe us kmarinas.
 
  • #147
kmarinas86 said:
That's why wind to me has got to be inhomogeneous in order to explain how this DDTWFTTW stuff works.
That's 'cause you don't have the right explanation yet!:wink: As I said, it's much simpler than this, so let's pull out another sheet of paper and have another go!

Okay. When an airplane flies through the air (homogeneous air mass), does the prop on the plane move faster than the air around it? Of course it does. Is the prop able to exert a force on the air while it is moving faster than the air? Of course it is; that's what a prop is designed to do and it does it quite well.

Do you agree that a prop can exert a force on an air mass as it is moving through (and therefore faster than) the air mass?
 
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  • #148
kmarinas86 said:
Let's pretend for a moment that this ball is the "homogeneous air mass". How does the ball hit the back of the train"? I can imagine it hitting the edge of the face of the back of one the train cars. But onto the face of it? No, I cannot imagine that.

This animation shows how the "ball" (air particle) hits the back of the propeller blade:

https://www.youtube.com/watch?v=FqJOVHHf6mQ

Here the version without lateral movement:

https://www.youtube.com/watch?v=Ufk6HVWdSzE
 
  • #149
A.T. said:
This animation shows how the "ball" (air particle) hits the back of the propeller blade:

https://www.youtube.com/watch?v=FqJOVHHf6mQ

Here the version without lateral movement:

https://www.youtube.com/watch?v=Ufk6HVWdSzE

Okay. So the slant of the propeller blades combined with its rotation makes it look like the propeller is moving slower than it actually is.

Well, okay then, that does make sense.
 
  • #150
kmarinas86 said:
Okay. So the slant of the propeller combined with its rotation makes it look like the propeller is moving slower than it actually is.
Yes. The intersection point of the blade surface with the path of the particle is moving slower than the particle. Even if the cart is moving faster than the particle.

In the paddle wheel example it is more obvious that the lower paddles move slower than the air, even if the center of the wheel is moving faster.
 
  • #151
A.T. said:
Yes. The intersection point of the blade surface with the path of the particle is moving slower than the particle.

Settled! :smile:
 
  • #152
kmarinas86 said:
Settled! :smile:

Hey do you think that would work for Humber? Just kidding.

Seriously, what do you think about the sail to prop animation now marinas. I think you probably see it differently, I am just asking because I always thought it was a great visualization of what you just discovered.
 
  • #153
jduffy77 said:
Hey do you think that would work for Humber? Just kidding.

Seriously, what do you think about the sail to prop animation now marinas. I think you probably see it differently, I am just asking because I always thought it was a great visualization of what you just discovered.

It's all good to me now.

Some of the other models I was talking about (I think) do exist, except that now I think the mechanisms I described in those models have no meaningful degree of presence for this particular problem. I think they might exist in some very extreme situations - just not at these slow speeds and accelerations.
 
  • #154
kmarinas86 said:
It's all good to me now.
Cool, huh?:smile:

A.T.'s animations are great!
 
  • #155
Yes.
 
  • #156
A.T. said:
This animation shows how the "ball" (air particle) hits the back of the propeller blade:

https://www.youtube.com/watch?v=FqJOVHHf6mQ

A.T. said:
The intersection point of the blade surface with the path of the particle is moving slower than the particle. Even if the cart is moving faster than the particle.

Now I'm no longer disputing the calculation of input and output energy. I'm also not disputing the forces and their orientation in the video. I think the video sufficiently shows that... but:

Concerning frame of reference of the intersection point (seen as the corner made by the blue line in the video), the air does not change speed, just the direction.

In other words, something possessed by the particles of the air, the momentum per air mass, the norm of which gives us the air speed (scalar), does not appear to change here. How then does this cause the kinetic energy of the combined mass of the cart and the ground to increase relative to the intersection point? Maybe it does not!

At 0:52 and 0:53 we see the vectors "force on air" and "force on airfoil", which are both diagonal. The air bounces off the surface of the airfoil at the intersection point. I imagine that the combined mass of the cart and the ground reacts to this air in an equal but opposite manner, as shown by the blue and maroon arrows that I added to the diagram, which are oriented CCW and CW respectively.

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


So it does not even seem to me that the kinetic energy of the wind even needs to be extracted by the cart in order for this to work. All there needs to be is a deflection of existing energy. After all, if we discount the hypothesis that potential energy is somehow relevant in this system, while at the same time we assume the conservation of energy, then it would stand to reason that "kinetic energy+heat" is conserved both before and after the force interactions. If the air does not lose kinetic energy in the reference frame of the intersection, then neither could the rest of system gain it.

Finally, I would also like to point out that at from 1:07 to 1:22 in the video, the amount that the energy in and the amount of energy out is calculated to be different, even though the friction was not modeled! Clearly the video is not taking into to account the work done tangentially. It turns out that the video as a whole takes into account the work done along the path of the wind and the cart (horizontally that is), but not the tangential work. For example, consider how much faster the prop cuts up through the air than the air itself does. If they took into account the work done by the propeller onto the air tangentially to the wind, it turns out that net work that the wind does on the rest of the system, and vice versa, combining both horizontal and tangential components, is zero from the frame of reference of the intersection between the air and the propeller.
 

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  • #157
kmarinas86 said:
Now I'm no longer disputing the calculation of input and output energy. I'm also not disputing the forces and their orientation in the video. I think the video sufficiently shows that... but:

Concerning frame of reference of the intersection point (seen as the corner made by the blue line in the video), the air does not change speed, just the direction.

In other words, something possessed by the particles of the air, the momentum per air mass, the norm of which gives us the air speed (scalar), does not appear to change here. How then does this cause the kinetic energy of the combined mass of the cart and the ground to increase relative to the intersection point? Maybe it does not!

But it does change. I would suggest you research propeller thrust in order to understand what is happening.

kmarinas86 said:
At 0:52 and 0:53 we see the vectors "force on air" and "force on airfoil", which are both diagonal. The air bounces off the surface of the airfoil at the intersection point. I imagine that the combined mass of the cart and the ground reacts to this air in an equal but opposite manner, as shown by the blue and maroon arrows that I added to the diagram, which are oriented CCW and CW respectively.

So it does not even seem to me that the kinetic energy of the wind even needs to be extracted by the cart in order for this to work. All there needs to be is a deflection of existing energy.

Could you elaborate on what you mean by this? It does not make sense to me.

kmarinas86 said:
After all, if we discount the hypothesis that potential energy is somehow relevant in this system, while at the same time we assume the conservation of energy, then it would stand to reason that "kinetic energy+heat" is conserved both before and after the force interactions.

This does not "stand to reason" for me in the sense that it constitutes an argument for the cart using the "deflection of existing energy".

kmarinas86 said:
If the air does not lose kinetic energy in the reference frame of the intersection, then neither could the rest of system gain it.

But it does lose energy.
 
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  • #158
jduffy77 said:
But it does change. I would suggest you research propeller thrust in order to understand what is happening.

The change of speed does depend on the frame of reference. In one reference frame the velocity of a particle can flip 90 degrees, 180 degrees, or whatever degrees without changing speed. An example of this is light bouncing off a mirror. The velocity flips across an axis normal to a mirror's surface. The speed of light did not change.

In any other frame than that in which the lines of the air flow in the video were traces, the speed will undoubtedly change. This includes the ground frame, the cart frame, and the wind frame.

jduffy77 said:
Could you elaborate on what you mean by this? It does not make sense to me.

This does not "stand to reason" for me in the sense that it constitutes an argument for the cart using the "deflection of existing energy".

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


I don't know how this could be clearer. Do I need to show how two balls can bounce off each other from an angle without either gaining more kinetic energy from the other than the other is gaining from it?

jduffy77 said:
But it does lose energy.

That depends on the frame of reference!

I'm just saying that there is a way to look at it (a particular frame of reference) where it does not involve the transfer of net energy.
 
  • #159
kmarinas86 said:
The change of speed does depend on the frame of reference. In one reference frame the velocity of a particle can flip 90 degrees, 180 degrees, or whatever degrees without changing speed. An example of this is light bouncing off a mirror. The velocity flips across an axis normal to a mirror's surface. The speed of light did not change.

In any other frame than that in which the lines of the air flow in the video were traces, the speed will undoubtedly change. This includes the ground frame, the cart frame, and the wind frame.

You are confusing yourself again. None of this has any relevance to the cart. In the ground frame the prop is slowing down the air. It does not need energy to do this. The air is doing the work.

kmarinas86 said:
I don't know how this could be clearer. Do I need to show how two balls can bounce off each other from an angle without either gaining more kinetic energy from the other than the other is gaining from it?

A. No. B. This is nothing to do with the cart.

kmarinas86 said:
That depends on the frame of reference!

I'm just saying that there is a way to look at it (a particular frame of reference) where it does not involve the transfer of net energy.

The air loses kinetic energy with respect to the ground. If by net energy you mean that COE is not violated you are correct but I don't expect that's what you are getting at.
 
  • #160
jduffy77 said:
You are confusing yourself again. None of this has any relevance to the cart. In the ground frame the prop is slowing down the air. It does not need energy to do this. The air is doing the work.

The air is doing the work with respect to the ground frame and other frames. Not in all inertial frames. That is relativity, you know, the theory invented by Albert Einstein?

jduffy77 said:
A. No. B. This is nothing to do with the cart.

It has something to do with the statement I was responding to:

jduffy77 said:
This does not "stand to reason" for me in the sense that it constitutes an argument for the cart using the "deflection of existing energy".

Saying this means you deny that the cart involves the deflection of existing energy. That's like saying that no matter can be changing direction in this system. That has everything to do with blue and maroon arrows in the diagram below:

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


https://www.physicsforums.com/attachment.php?attachmentid=42518&stc=1&d=1325884076

Edit: You added:

jduffy77 said:
The air loses kinetic energy with respect to the ground. If by net energy you mean that COE is not violated you are correct but I don't expect that's what you are getting at.

By net energy, I mean there is a reference frame in which the net transfer of energy across the intersection between the depicted air stream and the propeller is zero.
 
  • #161
kmarinas86 said:
The air is doing the work with respect to the ground frame and other frames. Not in all inertial frames. That is relativity, you know, the theory invented by Albert Einstein?

I do indeed. You seem to be having some problems with it at the moment though. You have made a correct statement here which should lead to an enhanced understanding of the cart mechanism, but instead it is leading you to incorrect conclusions.

kmarinas86 said:
It has something to do with the statement I was responding to:

Saying this means you deny that the cart involves the deflection of existing energy. That's like saying that no matter can be changing direction in this system. That has everything to do with blue and maroon arrows in the diagram below:

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


https://www.physicsforums.com/attachment.php?attachmentid=42518&stc=1&d=1325884076

Your diagram is incorrect. That is what I am trying to help you with. I think talking "deflection of existing energy" in the context of the cart is nonsensical.
 
  • #162
kmarinas86 said:
The air is doing the work with respect to the ground frame and other frames. Not in all inertial frames. That is relativity, you know, the theory invented by Albert Einstein?



It has something to do with the statement I was responding to:



Saying this means you deny that the cart involves the deflection of existing energy. That's like saying that no matter can be changing direction in this system. That has everything to do with blue and maroon arrows in the diagram below:

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


https://www.physicsforums.com/attachment.php?attachmentid=42518&stc=1&d=1325884076

Edit: You added:



By net energy, I mean there is a reference frame in which the net transfer of energy across the intersection between the depicted air stream and the propeller is zero.

-No.
 
  • #163
kmarinas86 said:
Edit: You added:

By net energy, I mean there is a reference frame in which the net transfer of energy across the intersection between the depicted air stream and the propeller is zero.

I wish you would explain how you can think this.
 
  • #164
jduffy77 said:
I do indeed. You seem to be having some problems with it at the moment though. You have made a correct statement here which should lead to an enhanced understanding of the cart mechanism, but instead it is leading you to incorrect conclusions.
Your diagram is incorrect. That is what I am trying to help you with. I think talking "deflection of existing energy" in the context of the cart is nonsensical.

Doesn't the angle of deflection of existing energy matter?

A.T. said:
The best way to avoid confusion is to be precise:

- Make clear which reference frame you are analyzing (power/kinetic energy are frame dependent quantities)
- Distinguish between "air" with "wind" (movement of air relative to something)
- Distinguish between true wind (relative to ground) with relative wind (relative to cart)
- Distinguish between work done by the cart chassis on the air, with work done by the propeller on the air.

Being precise in formulating the questions, often makes the answer obvious.That is not necessarily true. Depending on the propeller pitch the acceleration can be not maximal at WS but rather above it. So the increase in KE (seen from the ground frame) is maximal there. But the power transmitted though the vehicle always increases with speed. Here I posted some simulated values:

https://www.physicsforums.com/showthread.php?p=3352297From the ground frame: Some of the air is doing negative work on the cart chassis. But there is more positive work done on the propeller blades by the air.

The propeller always slows down air relative to the ground. The faster you go, the more volume of air you encounter, that you can draw KE from. But that increase is linear, while chassis drag and transmission inefficiency increase non linearly with speed.Have a look at the table below that shows different settings for a variable blade pitch propeller, coupled to the ground via wheels. What you describe above is starting out in CASE A (that gives you maximal initial acceleration) and then at some point below 1WS switching to CASE C that allows you to go faster than wind.

2gv0kew.png


Note that the Blackbird didn't have that ability (even when it had variable pitch later). They didn't want the ability to turn the wheels with the prop, to avoid confusion about using stored energy. They used CASE C only.

But Andrew Bauer was using his propeller as a turbine below windspeed. Here is video where you can see him starting in "windmill mode" and change the blade pitch later.
http://www.fasterthanthewind.org/2010/09/sad-news-in-world-of-ddwfttw.html
See also the graphs on page 15 in Bauer's paper.
http://projects.m-qp-m.us/donkeypus...aster-Than-The-Wind-The-Ancient-Interface.pdf

"Make clear which reference frame you are analyzing (power/kinetic energy are frame dependent quantities)"

I probably wasn't clear enough as my analyses were often switching from one reference frame to the next. Most people on this forum can only think about one reference frame at a time. It's easy for me to switch them, but I guess that hard it's to notice the reference frame switching from my dense way of writing.

BTW: I am from the point of view that potential energy is itself a frame-dependent quantity that balances exactly the changes in kinetic energy (dKE/dPE = -1), such that things like inertia and their effect on gravity are frame-invariant. But that is a totally different subject anyway.
 
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  • #165
kmarinas86 said:
Saying this means you deny that the cart involves the deflection of existing energy. That's like saying that no matter can be changing direction in this system.

Does burning a piece of wood or hitting a baseball involve "deflection of existing energy"? If that is the sense you mean it in then ok. It seems a strange way of putting things.
 
  • #166
kmarinas86 said:
Doesn't the angle of deflection of existing energy matter?

The energy for the ddwfttw cart comes from slowing down the wind with respect to the ground. Is that the "deflection of existing energy" that you are talking about?
 
  • #167
jduffy77 said:
The energy for the ddwfttw cart comes from slowing down the wind with respect to the ground. Is that the "deflection of existing energy" that you are talking about?

By deflection I mean an angular change of the direction with respect to some surface. If by deflection I meant the actual thing deflected, then I should really call this "deflected matter". So what you are taking about is "deflected matter" as seen from the ground point of view.
 
  • #168
kmarinas86 said:
By deflection I mean an angular change of the direction with respect to some surface. If by deflection I meant the actual thing deflected, then I should really call this "deflected matter". So what you are taking about is "deflected matter" as seen from the ground point of view.

Air is certainly being deflected by the prop. It is being shoved out the back which gives an equal and opposite reaction. That's pretty much the definition of thrust.

This does not help your first post today make any sense however:

kmarinas86 said:
So it does not even seem to me that the kinetic energy of the wind even needs to be extracted by the cart in order for this to work. All there needs to be is a deflection of existing energy. After all, if we discount the hypothesis that potential energy is somehow relevant in this system, while at the same time we assume the conservation of energy, then it would stand to reason that "kinetic energy+heat" is conserved both before and after the force interactions. If the air does not lose kinetic energy in the reference frame of the intersection, then neither could the rest of system gain it.

Finally, I would also like to point out that at from 1:07 to 1:22 in the video, the amount that the energy in and the amount of energy out is calculated to be different, even though the friction was not modeled! Clearly the video is not taking into to account the work done tangentially. It turns out that the video as a whole takes into account the work done along the path of the wind and the cart (horizontally that is), but not the tangential work. For example, consider how much faster the prop moves cuts up through the air than the air itself does. If they took into account the work done by the propeller onto the air tangentially to the wind, it turns out that net work that the wind does on the rest of the system, and vice versa, combining both horizontal and tangential components, is zero from the frame of reference of the intersection between the air propeller.
 
  • #169
kmarinas86 said:
How then does this cause the kinetic energy of the combined mass of the cart and the ground to increase relative to the intersection point?
In the frame where the air's KE doesn't change, the carts KE increases while the grounds KE decreases.
kmarinas86 said:
The air bounces off the surface of the airfoil at the intersection point.
Keep in mind that this is strongly simplified picture. The real airflow is quite different, but the result in terms of force on the blade is the same.
kmarinas86 said:
If the air does not lose kinetic energy in the reference frame of the intersection, then neither could the rest of system gain it.
See first comment.
kmarinas86 said:
Clearly the video is not taking into to account the work done tangentially.
Work is done only along the direction of movement, at the rate P = F dot v (the dot product cancels the tangential force component). It is an idealized model that assumes just a short interaction with the blade. The inefficiencies of a real world propeller (like swirling the air tangentially) constitute the losses.
 
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  • #170
A.T. said:
If the frame where the air's KE doesn't change, the carts KE increases while the grounds KE decreases.

Keep in mind that this is strongly simplified picture. The real airflow is quite different, but the result in terms of force on the blade is the same.

See first comment.

Work is done only along the direction of movement, at the rate P = F dot v (the dot product cancels the tangential force component). It is an idealized model that assumes just a short interaction with the blade. The inefficiencies of a real world propeller (like swirling the air tangentially) constitute the losses.

Thank you for spending the time to answer.
 
  • #171
I did the animation of spiral track sail to prop.

2 things I think will help people to understand what's going on.

1. Sail carts can frequently "beat a balloon" downwind with a significant factor. Downwind velocity component being 2-2.5 x wind speed. This is well documented and can be studied in many places. People follow their intuition and think that the dw component (for some reason) cannot exceed wind speed. If you argue against this fact then read about it 1st.

2. The wind cart works by using "gearing" and the relative speed differences of the two different interfaces. You can can turn a resisting force to a larger pushing force as long as the speed differences support this.

The breaking power at wheels (ground to wheel) could be for example 10N at 10m/s =100w. Same 100w at 5 m/s apparent wind speed can generate 20N thrust.

Obviously there are efficiencies at play in real application but by using above thinking you can throw different numbers to check what is going on at different wind and cart speeds. You will see that there is plenty of "excess" to be wasted - and thus the cart does work.
 
  • #172
kerosene said:
People follow their intuition and think that the dw component (for some reason) cannot exceed wind speed.

The one and only thing that first got me to accept that this is possible was to think of the effective velocity between the air mass and the blade. The difference between the wind velocity vector and the effective velocity vector can have a sign opposite of that between the wind velocity vector and the craft velocity vector. Because of this reason, both DDWFTTW and DUWFTTW are possible.

Evidence is now easy to come by: youtube.com/results?search_query=Dynamic+Soaring

Thinking about tacking wasn't helping that much. As evident on this thread, I and many others wasted a lot of time on that issue. I suggest that next time someone doubts that this is possible, one can skip the whole discussion about tacking and go directly towards talking about the effective velocity between the air mass and the rotating, pitched blade in the sense of vectors (velocities), not scalars (speeds). It's the only thing that should really matter here.
 
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  • #173
kmarinas86 said:
Thinking about tacking wasn't helping that much. As evident on this thread, I and many others wasted a lot of time on that issue. I suggest that next time someone doubts that this is possible, one can skip the whole discussion about tacking and go directly towards talking about the effective velocity between the air mass and the rotating, pitched blade in the sense of vectors (velocities), not scalars (speeds). It's the only thing that should really matter here.

We have found that there is no single explanation that works for everyone. Whenever someone finally "gets it" they usually ask why we wasted their time with all the other useless or even wrong explanations. But the fact is, we have a whole bunch of different explanations. They're all accurate, and each person seems to respond to a different one - and think the rest are nonsense.
 
  • #174
spork said:
We have found that there is no single explanation that works for everyone. Whenever someone finally "gets it" they usually ask why we wasted their time with all the other useless or even wrong explanations. But the fact is, we have a whole bunch of different explanations. They're all accurate, and each person seems to respond to a different one - and think the rest are nonsense.

Ohhhh welll...
 
  • #175


kmarinas86 said:
Saying this means you deny that the cart involves the deflection of existing energy. That's like saying that no matter can be changing direction in this system. That has everything to do with blue and maroon arrows in the diagram below:

attachment.php?attachmentid=42518&stc=1&d=1325884076.gif


https://www.physicsforums.com/attachment.php?attachmentid=42518&stc=1&d=1325884076

physics.aps.org/assets/1e41c2ebe02d4468 [Broken]

Focus: Getting an Extra Bounce said:
Focus: Getting an Extra Bounce
Published October 4, 2004 | Phys. Rev. Focus 14, 14 (2004) | DOI: 10.1103/PhysRevFocus.14.14

Computer simulations and experiments show that a ball can rebound from a surface with more vertical speed than it had initially.
Anomalous Behavior of the Coefficient of Normal Restitution in Oblique Impact

Hiroto Kuninaka and Hisao Hayakawa
Phys. Rev. Lett. 93, 154301 (2004)
Published October 5, 2004

Figure 1
NASA-JPL

http://physics.aps.org/assets/1e41c2ebe02d4468 [Broken]

Quick sand. Computer simulations agree with experiments suggesting that a disk can hit a surface and rebound with a surprisingly vertical trajectory. Research on such impacts can help improve models of the flow of granular materials–such as these Martian sand dunes (shown in false color).

Figure 2
M. Louge/Cornell Univ.

http://physics.aps.org/assets/389adf7d129904e2 [Broken]

Pop-up. In previous experiments [2] the trajectory of a ceramic ball hitting an elastic surface made a larger angle with the surface after impact than before. (See computer simulation video below.)
Animation courtesy of H. Kuninaka, Kyoto University.

http://physics.aps.org/assets/0d26d47f89100be8/video-v1.ogg [Broken]

High ball. This two-dimensional simulation shows how a ball can deform an elastic surface when it bounces. The surface becomes in effect a ski jump, which redirects the ball’s velocity skyward.

Like a gymnast who runs toward a vaulting horse and then hurls herself skyward, a ball can, under certain conditions, rebound from a glancing impact with a surprisingly vertical trajectory. It’s a phenomenon that’s been observed but never fully explained–and at times even doubted. But now researchers report in the 8 October PRL that they have developed a theory that explains the phenomenon and have tested it with computer simulations. Their explanation–which hinges on the ball’s impact deforming the surface it hits–could help refine models of the flow of granular materials such as sand dunes, cement, and soil.

The coefficient of normal restitution compares the vertical component of the velocity of an object before and after it has bounced. Conventional wisdom says it’s less than one–that is, a ball can’t leave the ground moving faster than when it arrived, because that would require extra energy (in the case of the gymnast, her body creates energy). But since the early 1990s several research groups have reported experiments with oblique impacts in which they found what seemed to be absurd results: A hockey-puck-like disk glanced against a wall and then appeared to pop away with an increased perpendicular velocity [1] and a ceramic sphere rebounded off a softer surface with a noticeably more vertical trajectory [2].

“They first thought I was crazy!” says Michel Louge of Cornell University, who performed the sphere-bouncing experiment with student Michael Adams. But after carefully ruling out experimental error, he concluded that the ball must deform the surface in such a way that it changes the trajectory of the ball. In that way, he thought, some of the horizontal component of the velocity could be transferred to the vertical component. He wasn’t sure exactly how this would happen, although it was clear that the effect was limited to special situations. “My conjecture in the [experimental] paper was just that–a conjecture,” he says.

Now Hiroto Kuninaka and Hisao Hayakawa of Kyoto University in Japan report that they have simulated the small-scale interactions between a disk and an elastic surface that can lead to a greater-than-one coefficient of normal restitution. Their computer simulation calculates a coefficient of 1.3 when the disk strikes the surface at an angle of about 11 degrees; at that angle, their simulated ball rebounds at about 15 degrees. The simulation results resemble Louge’s experimental data, according to the authors.

The simulation allowed the team to see the virtual disk denting the surface when it hit at oblique angles. Bill Stronge of Cambridge University in England describes the indentation as a kind of ski jump, which redirects the sphere’s velocity skyward. Because of this phenomenon, the coefficient of normal restitution can be greater than one without breaking the laws of physics. “The point is that the target material is softer than the ball,” says Kuninaka.

But Stronge doubts that the impact could make as vertical a ski jump as the researchers’ model suggests. It may have some effect, he says, “but I think that it’s certainly not nearly as dramatic as they have portrayed.” Computer models of granular materials such as cement and soil must account accurately for the collisions between grains, which can be treated like collisions with walls. So the work could contribute to practical advances in industries that manage and transport these materials, says Louge–and add to the understanding of the physics of ball sports.

–Chelsea Wald

References

J. Calsamiglia, S. W. Kennedy, A. Chatterjee, A. Ruina, and J. T. Jenkins, “Anomalous Frictional Behavior in Collisions of Thin Disks,” J. Appl. Mech.66, 146 (1999).
Michel Y. Louge and Michael E. Adams, “Anomalous behavior of normal kinematic restitution in the oblique impacts of a hard sphere on an elastoplastic plate,” Phys. Rev. E 65, 021303 (2002).

From this, we can see that "DDWFTTW" phenomenon isn't limited to sails or propellers:

Anomalous behavior of normal kinematic restitution in the oblique impacts of a hard sphere on an elastoplastic plate
http://masters.donntu.edu.ua/2010/fimm/kutnyashenko/library/nem_1/nem_1.pdf [Broken]
 
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<h2>1. What is DDWFTTW and what does it stand for?</h2><p>DDWFTTW stands for "Don't Do Work For The Work". It is a concept that challenges traditional ideas about work and productivity, suggesting that by not forcing ourselves to work, we can actually be more productive and efficient.</p><h2>2. How does DDWFTTW work?</h2><p>The principle behind DDWFTTW is that by allowing ourselves to take breaks and not forcing ourselves to work, we can tap into our natural rhythms and energy levels. This can lead to increased focus, creativity, and overall productivity.</p><h2>3. Is DDWFTTW supported by scientific research?</h2><p>While there is not a lot of research specifically on DDWFTTW, there is evidence to support the idea that taking breaks and allowing for rest and relaxation can improve productivity and overall well-being. Studies have shown that breaks can improve focus, creativity, and problem-solving abilities.</p><h2>4. Can DDWFTTW be applied to all types of work?</h2><p>DDWFTTW can be applied to most types of work, but it may not be suitable for all individuals or industries. Some jobs may have strict deadlines or require constant attention, making it difficult to implement this approach. It is important to consider individual needs and job requirements when applying DDWFTTW principles.</p><h2>5. Are there any potential downsides to using DDWFTTW?</h2><p>While DDWFTTW can be effective for some individuals, it may not work for everyone. Some people may feel guilty or unproductive when taking breaks, and it may not be feasible for certain jobs or industries. Additionally, it is important to find a balance and not use DDWFTTW as an excuse to avoid work altogether.</p>

1. What is DDWFTTW and what does it stand for?

DDWFTTW stands for "Don't Do Work For The Work". It is a concept that challenges traditional ideas about work and productivity, suggesting that by not forcing ourselves to work, we can actually be more productive and efficient.

2. How does DDWFTTW work?

The principle behind DDWFTTW is that by allowing ourselves to take breaks and not forcing ourselves to work, we can tap into our natural rhythms and energy levels. This can lead to increased focus, creativity, and overall productivity.

3. Is DDWFTTW supported by scientific research?

While there is not a lot of research specifically on DDWFTTW, there is evidence to support the idea that taking breaks and allowing for rest and relaxation can improve productivity and overall well-being. Studies have shown that breaks can improve focus, creativity, and problem-solving abilities.

4. Can DDWFTTW be applied to all types of work?

DDWFTTW can be applied to most types of work, but it may not be suitable for all individuals or industries. Some jobs may have strict deadlines or require constant attention, making it difficult to implement this approach. It is important to consider individual needs and job requirements when applying DDWFTTW principles.

5. Are there any potential downsides to using DDWFTTW?

While DDWFTTW can be effective for some individuals, it may not work for everyone. Some people may feel guilty or unproductive when taking breaks, and it may not be feasible for certain jobs or industries. Additionally, it is important to find a balance and not use DDWFTTW as an excuse to avoid work altogether.

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