Understanding DDWFTTW: Exploring Its Principles and Addressing Common Questions

[kmarinas86]

Could you elaborate on what you mean by DIRECT here? It seems to me to be clearly wrong.

If you stand on a bunch of rolling pins (as used for kneading), what happens to the rolling pins when you lift one of your feet and your entire body forward? They move backward relative to your leading foot! The ground moves in the opposite direction. For you to create the imbalance prior to this, you use the mass of your other foot, the rolling pins and the ground as a reacting mass as you stretch forward. The foot moving forward is analogous to the cart, and the foot that was behind is analogous to the wind, and the rolling pins are the equivalent of the wheels.

Analogously, as for the DDWFTTW vehicle:

What direction does the force on the ground propagate if caused by a tire rotating moving in the +x direction? The -x direction!

What direction is the force by the wind on the vehicle in question which causes rotation prior to this? The +x direction! (Despite that the wind moves in the -x direction)! The reaction to this action onto the cart is carried by the mass of the wind thrusted toward the -x direction (in the 30 mph frame)! The same is true with the ground (the -x direction)! And DDWFTTW would still work even if you somehow could make the space from 0 to 2 feet above the ground an airless vacuum.

What makes it possible for the velocity (vector) of the wind respect to the cart and the force (vector) applied by the wind onto the cart in opposite directions with respect to the 30 mph frame from which they both accelerate with respect to? Only one thing: Potential energy. I know spork(33) cannot believe that as he has already made up his mind, for he argues as if he is making the implicit assumption the air was behaving as an incompressible (or non-decompressible) medium. As for the rolling pin analogy, the elastic potential energy stored in muscle fiber tension is the source of this energy.

 

[spork]

What makes it possible for the velocity and the force of the wind to be opposite in directions with respect to the cart? Only one thing: Potential energy. I know spork(33) cannot believe that as he has already made up his mind, for he argues as if he is making the implicit assumption the air was behaving as an incompressible medium.

At the Reynolds numbers involved here the air is behaving as an incompressible fluid.

Potential energy doesn't come into it.

 

[jduffy77]

If you stand on a bunch of rolling pins (as used for kneading), what happens to the rolling pins when you lift one of your feet and your entire body forward? They move backward relative to your leading foot! The ground moves in the opposite direction. For you to create the imbalance prior to this, you use the mass of your other foot, the rolling pins and the ground as a reacting mass as you stretch forward. The foot moving forward is analogous to the cart, and the foot that was behind is analogous to the wind, and the rolling pins are the equivalent of the wheels.

Analogously, as for the DDWFTTW vehicle:

What direction does the force on the ground propagate if caused by a tire rotating moving in the +x direction? The -x direction!

What direction is the force by the wind on the vehicle in question which causes rotation prior to this? The +x direction! (Despite that the wind moves in the -x direction)! In any case, the wind thrusted toward the -x direction (in the 30 mph frame)! The same is true with the ground (the -x direction)! And DDWFTTW would still work even if you somehow could make the space from 0 to 2 feet above the ground an airless vacuum.

This is absolutely wrong. You are needlessly confusing yourself with this flawed analogy. Why not focus on how the cart actually works?

What makes it possible for the velocity and the force of the wind to be opposite in directions with respect to the cart? Only one thing: Potential energy. I know spork(33) cannot believe that as he has already made up his mind, for he argues as if he is making the implicit assumption the air was behaving as an incompressible medium.

It basically is and I am not sure about spork. I know I have not agreed with your concept of potential energy since you introduced it. What makes it possible for the velocity and the force of the wind to be in opposite directions are thrust and leverage.

 

[kmarinas86]

It basically is and I am not sure about spork. I know I have not agreed with your concept of potential energy since you introduced it. What makes it possible for the velocity and the force of the wind to be in opposite directions are thrust and leverage.

Force cannot be multiplied without a lever, ramp, pulley, or some other equivalent. Yes, you can apply a force on one side of the lever to make another force in the other direction. The problem is find what "lever" is in the air as the force propagates from the wind to the sail. The sail itself it would seem at first to be this lever. But the sail does not need to rotate. You just need rotation between the wind with respect to the sail. This can be explained using both the Coandă effect, which is assisted by frictional forces, and Bernoulli's principle, which involves a pressure change. It is absolutely amazing to me how spork(33) can still deny the existence of the latter, after I was shown wrong about the sail force. Not to long ago (just a day ago), I too was thinking of only inertial forces when it came to this problem (thus some of my earlier claims about the force needing to be parallel to the velocity)! It wasn't much long thereafter until I realized that the potential energy in the air was absolutely required in this problem in terms of conservation of energy.

 

[ThinAirDesign]

Force cannot be multiplied without a lever, ramp, pulley, or some other equivalent. Yes, you can apply a force on one side of the lever to make another force in the other direction. The problem is where is the lever in the air as the force propagates from the wind to the sail? The sail itself it would seem at first. But the sail does not need to rotate. You just need rotation between the wind with respect to the sail. This can be explained using both Coandă effect, which is assisted by viscous forces, and Bernoulli's principle, which involves a pressure change. It is absolutely amazing to me how spork(33) can still deny the existence of the latter, after I was shown wrong about the sail force. Not to long ago (just a day), I too was thinking of only inertial forces when it came to this problem! It wasn't much long thereafter until I realized that the potential energy in the air absolutely required in this problem in terms of conservation of energy.

-No.

 

[mender]

It wasn't much long thereafter until I realized that the potential energy in the air was absolutely required in this problem in terms of conservation of energy.

And again, no.

You've been given patient, correct descriptions of how the cart works. You need to stop to a bit, set aside your incorrect theory and reread the replies so far before this becomes more a matter of defending your theory than it is a search for what is really happening.

Ask questions and consider the replies; these gents have been through countless discussions and have a large amount of information and expertise to call on.

The main thing to understand in that they know exactly how the cart works, and arguing with them rather than trying to reach an understanding will only impede your progress.

It's a fascinating concept, one well worth learning about.

Force cannot be multiplied without a lever, ramp, pulley, or some other equivalent. Yes, you can apply a force on one side of the lever to make another force in the other direction. The problem is where is the lever in the air as the force propagates from the wind to the sail?

It's a class three lever, with the cart as the load, the air as the force, and the ground as the fulcrum. It trades the large force and lower speed of the wind to get a high speed for the cart (lower force). Very simple.

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ENG19004

As such, the cart needs to be in direct contact with the ground to work. Your statement here:
"And DDWFTTW would still work even if you somehow could make the space from 0 to 2 feet above the ground an airless vacuum."

Is flat-out wrong. If you think about it and ask questions, you'll learn why.

 

[kmarinas86]

-No.

Sure.

Go ahead and deny it.

I do agree now what all that A.T. said in a previous post here:

Well, that is wrong. They achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind.

downwind VMG = boat velocity component pointing directly downwind.

Why should I do this? It is utter nonsense.

Nope, this makes no sense at all. Acceleration is determined by forces, not velocities. Where are the forces in your diagram?

It is steady state. And it is obvious that the craft will accelerate if the L/D ratio of the sail is good enough. L/D must be at least tan(90°-alpha), but since we have hull drag it must be better than that.

This is where I was corrected. (Yes, I now even agree with A.T. where he replied to one my statement with, "Nope this makes no sense at all.") I realized that lift in sail involves a pressure change; I just didn't apply that knowledge until earlier this morning. Well it turns out that there is potential energy right there. (Although, technically, at a closer inspection, we are dealing with kinetic energy of tiny gaseous particles that is at least to some degree contained by external forces.) Some will like to call it "pressure energy" - fine. In this thread I've already called it "P-V" energy.

 

[A.T.]

Of course the Earth acceleration is negligible but you could run the cart on a floating platform, that will be accelerated in a measurable way. Or you can consider the DDWFTTW boat, where the underwater turbine drag clearly accelerates water forwards, thus slowing it down in the boats frame.

Going from -30 mph to, say, -30 - (10)^(-20) mph, is a speed increase!

You have the sign wrong. The force on the ground points forwards, so the acceleration is positive:
Going from -30 mph to, say, -30 + (10)^(-20) mph, is a speed decrease.

in the 30 mph frame the acceleration of the Earth is negative.

Nope. By our convention where the ground velocity is negative (-30 mph) the acceleration is positive. The force on the ground and thus its acceleration is opposite to its velocity in that frame.

 

[rorix_bw]

Is there a website with blueprints for such craft?

I want to look at how the propeller is linked to the wheels. I found some photos of Blackbird during construction (on their own website) that shows part of the linkage, but not in full detail. It actually looks to me from the photos that the wheels aid in rotating the prop.

I need to first understand the engineering of the design fully before I make further comments. The "they act like keels" comments do not help because not all sail yachts have keels in the sense that people understand them to be, and I don't know how well the guys saying that understand keels or not.

 

[rcgldr]

What direction does the force on the ground propagate if caused by a tire rotating moving in the +x direction?

In the case of a DDWFTTW cart, the tires experience a "braking" torque used drive the propeller against the wind. The result is the tires apply a force to the ground in the +x direction.

 

[ThinAirDesign]

Is there a website with blueprints for such craft?

I want to look at how the propeller is linked to the wheels.

There are no blueprints, but for the Blackbird there are near a thousand construction pics at www.fasterthanthewind.org

It's very simple -- the rear axle is connected to the prop shaft through a long twisted bicycle chain.

On the small models you see on Youtube, a right angle drive connects the axle directly to the prop shaft.

JB

 

[kmarinas86]

You have the sign wrong. The force on the ground points forwards, so the acceleration is positive:
Going from -30 mph to, say, -30 + (10)^(-20) mph, is a speed decrease.

What is the force "on" the ground? There is a force onto the ground, and a force by the ground. The contact forces by themselves are equal and opposite. The force that accelerates the ground has to be in the opposite direction of that of the reaction mass. As far as those are concerned, it should cause acceleration of the ground.

Nope. By our convention where the ground velocity is negative (-30 mph) the acceleration is positive. The force on the ground and thus its acceleration is opposite to its velocity in that frame.

If I step off a skateboard from the front, the skateboard does the goes other way (backward). What logical manipulation is done to make that a positive acceleration? Acceleration is a technically a change in velocity, not speed. So, according to how I understood it, acceleration can be negative while speed^2 increases if the velocity is negative. The force*distance would still cause K.E. to increase because both the displacement and the force onto the skateboard (in this example) are negative, meaning that the work (=force (dot) displacement) is still positive.

Are you guys really saying that the cart is tugging the ground forward somewhat as it moves forward it as opposed to using it as a reaction mass? The only way I can see that happening is if the wheels slip, or if something non-rotating like skis are used. I still insist that the force of static pressure displacement is the primary driving force here, with aerodynamic effects which decompress being a function of relative wind speed, angle of attack, sail geometry and other factors that can affect the pressure.

I've been talking about the (+)30 mph (with respect to the ground) frame, in which the cart initially is moving 30 mph (with respect to the ground), and thus initially stationary at this point, and where the wind and the ground were moving in the (-) direction. This would imply that the center of momentum frame is not this 30 mph frame which I speak of. I cannot see how the wind (20 mph in my example) would be moving in the opposite direction as the ground in this moving frame.

 

[mender]

(Although, technically, at a closer inspection, we are dealing with kinetic energy of tiny gaseous particles that is at least to some degree contained by external forces.)

If by this you mean that the cart uses the kinetic energy of the wind, you are correct. The cart takes KE out of the moving air mass (wind) by slowing it down relative to the ground with the propeller. After the cart passes through the air mass, the air that has been affected by the propeller will be moving slower showing that KE was transferred to the cart.

The KE of the moving air mass is the energy source for the cart.

 

[rcgldr]

Is there a website with blueprints for such craft?

In one of the older threads, I think someone created a parts list for the small treadmill carts, that included the model prop used and differential from some radio control car.

I want to look at how the propeller is linked to the wheels.

In forward motion, the wheels drive the prop against the wind (backwards). In some of the small carts, the differential gear ratio was 1:1, but the pitch of the prop is less than the circumference of the wheels, so the effective gearing in terms of ground speed versus thrust speed is less than 1:1, increasing the force while decreasing the speed. This only works if there's a tailwind relative to the ground (since thrust speed is less than ground speed).

 

[mender]

Are you guys really saying that the cart is tugging somewhat on the ground as it moves forward it as opposed to using it as a reaction mass?

It depends on what you mean by tugging.

As I said, the cart uses the ground as a fulcrum and the moving air mass as the force.

I've been talking about the (+)30 mph (with respect to the ground) frame, in which the cart initially is moving 30 mph (with respect to the ground), and thus initially stationary at this point, and where the wind and the ground were moving in the (-) direction. This would imply that the center of momentum frame is not this 30 mph frame which I speak of. I cannot see how the wind (20 mph in my example) would be moving in the opposite direction as the ground in this moving frame.

If the 30 mph cart is used as the reference for the frame, the ground would be seen moving back at 30 mph while the air mass would be moving back at 10 mph.

Is that how you see it?

 

[rorix_bw]

It's very simple -- the rear axle is connected to the prop shaft through a long twisted bicycle chain.JB

Ya I saw that but you can't see if there are clutches or diffs. Are the rotation rates locked together?

 

[rcgldr]

Are you guys really saying that the cart is tugging the ground forward somewhat as it moves forward it as opposed to using it as a reaction mass?

Yes, the wheels are driving the prop which accelerates the relative headwind. The reaction force from the air results in an opposing torque on the prop which is transferred to the tires, which apply a forwards force onto the ground. The ground applies a backwards force onto the tires, and this force resists the forward motion of the cart. However the air applies a greater forwards force onto the propeller (due to the effective gearing that we've been calling advance ratio), so the cart is propelled forward (as long as there is a tailwind wrt ground).

 

[mender]

It actually looks to me from the photos that the wheels aid in rotating the prop.

That is correct; the wheels drive the prop.

 

[rorix_bw]

In one of the older threads, I think someone created a parts list for the small treadmill carts, that included the model prop used and differential from some radio control car.

Oh cool, will have a tinker. No large flat areas of land near me, but there is water. :-)

EDIT: Looks a lot harder to make it work on water though. Maybe needs a lake. I don't have a lake :-( This is probably doomed to failure!

 

[kmarinas86]

[Well it turns out that there is potential energy right there.] (Although, technically, at a closer inspection, we are dealing with kinetic energy of tiny gaseous particles that is at least to some degree contained by external forces.) [Some will like to call it "pressure energy" - fine. In this thread I've already called it "P-V" energy.]

If by this you mean that the cart uses the kinetic energy of the wind, you are correct. The cart takes KE out of the moving air mass (wind) by slowing it down relative to the ground with the propeller. After the cart passes through the air mass, the air that has been affected by the propeller will be moving slower showing that KE was transferred to the cart.

The KE of the moving air mass is the energy source for the cart.

So it seems that much of the disagreement about the "potential energy" I was invoking to explain the phenomenon was merely a semantic issue.

 

[kmarinas86]

I've been talking about the (+)30 mph (with respect to the ground) frame, in which the cart initially is moving 30 mph (with respect to the ground), and thus initially stationary at this point, and where the wind and the ground were moving in the (-) direction. This would imply that the center of momentum frame is not this 30 mph frame which I speak of. I cannot see how the wind (20 mph in my example) would be moving in the opposite direction as the ground in this moving frame.

If the 30 mph cart is used as the reference for the frame, the ground would be seen moving back at 30 mph while the air mass would be moving back at 10 mph.

Is that how you see it?

Yes.

 

[mender]

So it seems that much of the disagreement about the "potential energy" I was invoking to explain the phenomenon was merely a semantic issue.

Could be; definitions are pretty important!

So we're now on the same page re: the KE of the air mass being used as the energy source? And by your last reply, the frame of reference scenario is clear also.

Moving on; how do you see the cart working at this point? Just want to make sure that is clear as well.

 

[jduffy77]

So it seems that much of the disagreement about the "potential energy" I was invoking to explain the phenomenon was merely a semantic issue.

I am still afraid it was a bit more than semantics:

https://www.physicsforums.com/showpost.php?p=3691985&postcount=33

https://www.physicsforums.com/showpost.php?p=3692118&postcount=45

Could you re-visit the description in post 33? Unless you change that a bit it still seems to me there is some confusion.

 

[ThinAirDesign]

Ya I saw that but you can't see if there are clutches or diffs. Are the rotation rates locked together?

On the small models, direct drive. On the larger Blackbird, there are ratchets which only come into play of you brake quickly. The ratchets keep the mass of the rotating propeller from shearing off the drive if one has to stop quickly in an emergency. Under normal operation is is effectively a direct drive system as the ratchets never click.

JB

 

[ThinAirDesign]

So it seems that much of the disagreement about the "potential energy" I was invoking to explain the phenomenon was merely a semantic issue.

-No.

 

[kmarinas86]

Could be; definitions are pretty important!

So we're now on the same page re: the KE of the air mass being used as the energy source? And by your last reply, the frame of reference scenario is clear as well.

Moving on; how do you see the cart working at this point? Just want to make sure that is clear as well.

Yes.

Outside the context of this problem, I have even considered that the very idea potential energy, in general is an illusion manifested by kinetic energy contained below a certain bound of observation. For this problem, I have seem to adopted the notion of potential energy.

I can now imagine the paths of, say two particles of air, moving in two hyperbolic paths, approaching at a minimum and diverging away (like the "hourglass" figure). The particles change direction, but not necessarily their speed. The result of the latter divergence is to expand the volume occupied by two particles, allowing force to be applied to a surface, which is itself moving in the forward direction (from the point of view of the center of momentum of "two particles+surface") while the center of mass of the other two particles (taken separately from the surface) drifts in this backward direction. Of course, an "hourglass" shape is not required, I'm just using a common visual reference that people can relate to so I don't have draw a picture of random interactions.

Then all I have to do is increase the number of such interactions to extrapolate how the kinetic energy can be absorbed by the sail even though overall mass of air is moving in the opposite direction in the center of momentum frame.

 

[kmarinas86]

-No.

Wow, thanks a lot!

 

[ThinAirDesign]

Wow, thanks a lot!

{blushing}
It's the least I can do ... really, it is.

JB

 

[mender]

Yes.

Outside the context of this problem, I have even considered the very idea potential energy, in general is an illusion manifested by kinetic energy contained below a certain bound of observation. For this problem, I have seem to adopted the notion of potential energy.

I can now imagine the paths of, say two particles of air, moving in two hyperbolic paths, approaching at a minimum and diverging away (like the "hourglass" figure). The particles change direction, but not necessarily their speed. The result of the latter divergence is to expand the volume occupied by two particles, allowing force to be applied to a surface, which is itself moving in the forward direction (from the point of view of the center of momentum of "two particles+surface") while the other two particles whose sum drifts in this backward direction.

Then all I have to do is increase the number of such interactions to extrapolate how the kinetic energy can work on the sail in the opposite direction of the overall drift even though it is moving in the opposite direction.

Interesting theory if I'm understanding what you're saying but I see some issues with it.

It sounds like you're saying that a flat surface that is moving faster than the air mass can somehow generate pressure behind it. Yes?

As I said, if I'm understanding what you're saying.

 

[mender]

{blushing}
It's the least I can do ... really, it is.

JB

 

[kmarinas86]

Interesting theory if I'm understanding what you're saying but I see some issues with it.

It sounds like you're saying that a flat surface that is moving faster than the air mass can somehow generate pressure behind it. Yes?

As I said, if I'm understanding what you're saying.

A non-flat surface to be sure. A glider experiencing a slight downdraft during flight can still have lift fighting against it. But as for a flat surface? No, I'm not trying to say that. Blowing over a piece of paper does reduce the pressure on one side though. I can't imagine how it would increase the pressure on the other though unless some kind of funnel were present. (Okay, now that I think about it, I do think it is possible. The relatively unaffected air mass that is somehow behind the vehicle prop combined with the angle of attack of the blades can create a "funnel cup" to increase the pressure (if the air is compressing that is), or perhaps instead it merely transfers existing force momentum towards more focused "areas" that are projected onto the sail and the reacting mass, such that the average pressure (integrated over a closed surface) doesn't really change. Sorry for my ignorance of the actual technical terms!)

 

[A.T.]

What is the force "on" the ground?

The horizontal force exerted by the wheels on the ground.

Are you guys really saying that the cart is tugging the ground forward

The cart is exerting a forward force on the ground.

The only way I can see that happening is if the wheels slip, or if something non-rotating like skis are used.

Nope. The wheels can be rolling without slippage and still exert horizontal forces on the ground.

 

[kmarinas86]

The horizontal force exerted by the wheels on the ground.

The cart is exerting a forward force on the ground.

Nope. The wheels can be rolling without slippage and still exert horizontal forces on the ground.

So do you think the wind pushes both the cart and the ground even when using wheels?

I can imagine that it would if the wheels were "locked" so that they could not rotate.

I guess I should just go out and say that whether or not the ground experiences a force will depend on what is connected to the wheels and what is driving them.

If the wind can go in the opposite direction (backwards) that it applies a force to something else (forwards), maybe it is also true for a car moving forwards applying a force to something else (backwards).

For a car, I definitely do expect that the ground is a reaction mass to the car's acceleration. But maybe it is different for wind or even propeller thrust in general? What do you think? What would the reason for that be? (I can almost explain it, but I don't know what to say without (perhaps) going into more semantics issues.)

 

[mender]

(Okay, now that I think about it, I do think it is possible. The relatively unaffected air mass that is somehow behind the vehicle prop combined with the angle of attack of the blades can create a "funnel cup" to increase the pressure (if the air is compressing that is), or perhaps instead it merely transfers existing force momentum towards more focused "areas" that are projected onto the sail and the reacting mass, such that the average pressure (integrated over a closed surface) doesn't really change.

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!

 

[mender]

So do you think the wind pushes both the cart and the ground even when using wheels?

I can imagine that it would if the wheels were "locked" so that they could not rotate.

I guess I should just go out and say that whether or not the ground experiences a force will depend on what the wheels do.

If the wind can go in the opposite direction (backwards) that it applies a force to something else (forwards), maybe it is also true for a car moving forwards applying a force to something else (backwards).

For a car, I definitely do expect that the ground is a reaction mass to the car's acceleration. But maybe it is different for wind or even propeller thrust in general? What do you think? What would the reason for that be?

If by reaction mass you mean that the wheels are pushing the cart forward by exerting a force on the ground, you are wrong!

The prop is pushing the cart forward by exerting a force on the air; for the prop to be able to generate that force, the wheels have to exert a force on the ground in the opposite direction (think about levers). The prop generates more force against the air than the wheels generate against the ground, so the net force is pushing the cart forward is greater and the cart accelerates.