How Does DDWFTTW Work and What Are Its Key Principles?

[kmarinas86]

So let's get this straight.

If we have wind blowing at 20 mph relative to ground and the cart moving at 30 mph and still accelerating, at the 30 mph frame, the wind would appear to move at -10 mph and the cart (not) move at 0 mph.

But if we wait just a split second, the cart will start moving at greater than 0 mph in the 30 mph frame and the wind will start moving at less than -10 mph.

So in the 30 mph frame, both the wind and the cart will gain relative velocity (relative to the 30 mph frame that is). Obviously the mass will remain the same for each, so they both gain kinetic energy. Is there any reason to believe that they don't both gain kinetic energy in the 30 mph frame? It sure looks like that to me!

Why? Why would this be possible if there is no potential energy involved?

As far as wheel friction is concerned, the ground would be slightly nudged by the cart rearwards as well, although the change in speed is almost non-existent (the ground -the Earth- is very massive) so its basically just pressure wave being sent to the ground that you can't see. So, for being thrusted by the cart in the same direction as the wind is thrusted by the cart, the ground's reaction to the cart doubles my suspicion for the 30 mph frame.

Due to the wheels, there is no net force of the cart crosswise here, so there is no expectation that there should be any net cross-wise force on the "wind+ground". So as far as net forces are concerned, they exist only along the line that the DDTFTTW cart travels. Like was shown in a previous diagram, the lateral hull force and the lateral sail force cancel each other out! Also note the that transition from compression to rarefaction in a pressure wave is the conversion of potential energy into kinetic energy, and the transition from rarefaction to compression is the conversion of kinetic energy into potential energy. Buffeting anyone?

 

[rcgldr]

wind blowing at 20 mph relative to ground and the cart moving at 30 mph and still accelerating, at the 30 mph frame, the wind would appear to move at -10 mph and the cart (not) move at 0 mph.

But if we wait just a split second, the cart will start moving at greater than 0 mph in the 30 mph frame and the wind will start moving at less than -10 mph.

So in the 30 mph frame, both the wind and the cart will gain relative velocity.

And by the way, as far as wheel friction is concerned, the ground would be slightly nudged by the cart rearwards as well, although the change in speed is almost non-existent (the ground -the Earth- is very massive) so its basically just pressure wave being sent to the ground that you can't see.

In the 30 mph frame, the surface of the Earth is moving at -30 mph, and the Earth's's KE is huge. The forward force at the wheels decreases the magnitude of the Earth's velocity (wrt 30 mph frame) by a tiny amount, but the decrease in KE of the Earth will be enough to account for the overall increase in KE of the affected air and the cart, plus the losses that end up as heat.

 

[kmarinas86]

In the 30 mph frame, the surface of the Earth is moving at -30 mph, and the Earth's's KE is huge. The forward force at the wheels decreases the magnitude of the Earth's velocity (wrt 30 mph frame) by a tiny amount, but the decrease in KE of the Earth will be enough to account for the overall increase in KE of the affected air and the cart, plus the losses that end up as heat.

That's wrong. The cart can work as long as there is tailwind. So this can cause the Earth to increase or decrease rotational speed (or in a special case, not at all) depending on the direction. What happens, say, if the cart is going westward, accelerating (not decelerating) the Earth's rotation. Does the cart cease to function? Obviously not!

In the 30 mph frame, the true wind remains constant at -10 mph regardless of the cart's speed.

The true wind is a separate "parcel" from the affected wind. There is absolutely no way that the affected wind could have the same speed relative to the 30 mph frame after interacting with the blades of the prop.

The prop thrust speed (wrt 30 mph frame) will decrease as the cart speed increases, since the prop thrust speed is a fraction of the cart's ground speed (about 4/5 in the case of the blackbird).

The thrust consists of a series of air parcels that change speed. While the thrust speed decreases as cart accelerates, this thrust speed is of different air parcels all of which are already moving in the opposite direction of the cart in the 30 mph frame. To have them being thrown the other way (implying a change of speed as being the basis for thrust) can only accelerate the cart forwards, and it can only make the thrusted air even more negative in their speed relative to the 30 mph frame!

As far as changes in kinetic energy of the wind is concerned, knowing that it does (indeed) provide some input to the cart, the kinetic energy of the wind, which was already moving at -10 mph from the cart frame speeds up in the 30 mph frame and slows down in the ground frame. Note that going from -10 mph, to say, -11 mph in the 30 mph frame is actually an increase of kinetic energy as seen in that frame by a factor of (11/10)^2 given constant mass. The reason for that is that energy is not a vector!

 

[rcgldr]

In the 30 mph frame, the surface of the Earth is moving at -30 mph, and the Earth's's KE is huge. The forward force at the wheels decreases the magnitude of the Earth's velocity (wrt 30 mph frame) by a tiny amount, but the decrease in KE of the Earth will be enough to account for the overall increase in KE of the affected air and the cart, plus the losses that end up as heat.

So this can cause the Earth to increase or decrease rotational speed (or in a special case, not at all) depending on the direction.

The direction doesn't matter because you defined the frame of reference to be in the same direction as the cart wrt Earth's surface, at +30 mph wrt Earth's surface.

The true wind is a separate "parcel" from the affected wind. There is absolutely no way that the affected wind could have the same speed relative to the 30 mph frame after interacting with the blades of the prop.

Assuming the advance ratio is 4/5 then from the cart's frame of reference, a ground speed of -30 mph translates into prop thrust speed of -24 mph, and a ground speed of -35 mph translates into prop thrust speed of -28 mph.

In the 30 mph frame of reference, at a cart speed of +0 mph, prop thrust speed is -24 mph, and the affected air is accelerated from -10 mph to -24 mph, increasing its KE. At a cart speed of +5 mph, prop thrust speed is (5 - 28 =) -23 mph, and the affected air is accelerated from -10 mph to -23 mph, increasing its KE.

In the 30 mph frame of reference, KE is extracted from the Earth and added to the air and cart.

 

[A.T.]

In the 30 mph frame, the surface of the Earth is moving at -30 mph, and the Earth's's KE is huge.

... Earth's rotation...

It has nothing to do with the Earth's rotation. Even a non rotating Earth would have KE in a reference frame, which moves relative to the Earth at 30mph. KE is frame dependent. In the frame of the cart the wheels are doing negative work on the ground. So they are harvesting energy from it in that frame.

 

[kmarinas86]

It has nothing to do with the Earth's rotation. Even a non rotating Earth would have KE in a reference frame, which moves relative to the Earth at 30mph. KE is frame dependent. In the frame of the cart the wheels are doing negative work on the ground. So they are harvesting energy from it in that frame.

In the 30 mph frame of reference, KE is extracted from the Earth and added to the air and cart.

A.T. is right as far as the fact that it has nothing do to with the speed of Earth's rotation. We are talking about the norm of the derivative of Earth's speed (not really the "rate of rotation", which is angular velocity). It's about the derivative because are now talking about forces (which cause changes in speed), and it's about the norm because the kinetic energy is based on the square of the speed, which removes the sign!

And we are talking primarily about one frame at the moment, not so much the others.

And to both of you, why is it so hard to believe that potential energy is involved? It is clearly required from the 30 mph frame. Not the cart, not the wind, and not the Earth -none of them- would lose KE in this frame, they all gain it, just as you would expect for two negative charges in proximity to one another. Negative velocities in this frame become larger negative velocities (those of the wind (-10 mph) and the Earth (-30 mph)), and positive velocities become larger positive velocities (those of the propeller and the cart (+0.0000... mph onward)).

And as we should already know, "positive" and "negative" are arbitrary here, being assigned to forward and backward relative to the vehicle's direction. They have nothing to do with some non-physical explanation that equates a growing negative velocity (i.e. from -10 mph to -11 mph) as somehow corresponding to a loss of kinetic energy supposedly extracted by the cart. Kinetic energy is not a vector! Going from -10 mph to -11 mph is not an increase in kinetic energy, it corresponds to it!

 

[A.T.]

Why is it so hard to believe that potential energy is involved? It is clearly required from the 30 mph frame. Not the cart, not the wind, and not the Earth -none of them- would lose KE in this frame,

The Earth is loosing KE in an inertial frame that initially was moving 30mph wrt to the Earth. In that frame the cart exerts a force on the ground opposite to the velocity of the ground.

If the huge mass difference confuses you, consider a DDWFTTW boat with a turbine in the water, instead of the wheels. It affects only some of the water locally, just like the propeller does with air. In the rest frame of the airmass (outside of the prop influence) you see some of the air initially at rest, being accelerated back by the prop (gaining KE), and some of the water initially moving back at -30mph being pushed forward by the turbine and slowed down (loosing KE).

 

[kmarinas86]

The Earth is loosing KE in an inertial frame that initially was moving 30mph wrt to the Earth. In that frame the cart exerts a force on the ground opposite to the velocity of the ground.

That's impossible. Kinetic energy is not a vector. Kinetic energy is NOT a vector! The Earth already has a velocity of -30 mph in the 30 mph frame. Are you able to conceive of that? I'm not so sure that you can. For it to that to change to -29 mph requires that the Earth accelerates the same direction as the cart, so by the law of non-contradiction it does not do so. (-30)^2 is 900 and is greater than (-20)^2 (=400). The Earth has more kinetic energy going -30 mph in the 30 mph frame than it does going at -29 mph. But acceleration of the Earth in this linear case is negative, so the speed must be going from -30 mph to a little bit larger in the negative direction (perhaps, say (-30 - 10^(-20)) mph). So in this 30 mph frame, the kinetic energy of the Earth increases!

 

[A.T.]

The Earth is loosing KE in an inertial frame that initially was moving 30mph wrt to the Earth. In that frame the cart exerts a force on the ground opposite to the velocity of the ground.

requires that the Earth accelerates the same direction as the cart,

That is correct in the inertial frame that initially was moving 30mph wrt the Earth. The acceleration vectors for cart and Earth point in the same direction. But due to different velocities the cart speed increases, while the Earth speed decreases.

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.

 

[spork]

Kinetic energy is not a vector. Kinetic energy is NOT a vector! ... Are you able to conceive of that? I'm not so sure that you can.

Chill out Francis. Everyone but you seems to understand this just fine.

 

[kmarinas86]

That is correct in the inertial frame that initially was moving 30mph wrt the Earth. The acceleration vectors for cart and Earth point in the same direction. But due to different velocities the cart speed increases, while the Earth speed decreases.

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

Going from 10^(-20) mph to 0 mph is a speed decrease!

How would you express speed of a laterally moving body? Not by using positive and negative of course! The + and - directions are not special. They are not responsible for causing "opposite" values for speed, nor for kinetic energy. They do represent opposite tendencies for momentum and velocity for a one-dimensional case. Momentum and velocities are vectors! Speed and kinetic energy are scalars!

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.

Yes, and in the 30 mph frame the acceleration of the Earth is negative. The DDWFTTW vehicle can work just fine without any DIRECT friction coupling between the wind and the ground. The coupling between friction and the ground is insignificant in contrast to that of between the wind and the cart as well as that of between the cart and the ground, so the wind and the ground are accelerated in the same direction (negative)! A negative acceleration causes a negative velocity to become a greater negative, and a positive velocity becomes lesser positive (either that, or becomes negative). If the cart is moving faster than 30 mph relative to ground, and the Earth is the ground and thus less than 30 mph relative to the ground, then it stands to the reason that the velocity of the Earth relative to the 30 mph frame is negative, as contrasted with the cart's velocity that is positive relative to the same 30 mph frame!

 

[rcgldr]

In the 30 mph frame of reference, the tires on the cart exert a force on the Earth in the positive direction, changing it's velocity from -30 mph to -29.99999999999... mph, so KE of the Earth is decreased (extracted).

 

[kmarinas86]

In the 30 mph frame of reference, the tires on the cart exert a force on the Earth in the positive direction, changing it's velocity from -30 mph to -29.99999999999... mph, so KE of the Earth is decreased (extracted).

I already refuted this:

The DDWFTTW vehicle can work just fine without any DIRECT friction coupling between the wind and the ground. The coupling between friction and the ground is insignificant in contrast to that of between the wind and the cart as well as that of between the cart and the ground, so the wind and the ground are accelerated in the same direction (negative)!

 

[spork]

I already refuted this:

Yes, but you were completely wrong.

 

[jduffy77]

The DDWFTTW vehicle can work just fine without any DIRECT friction coupling between the wind and the ground.

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

 

[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?