How Does DDWFTTW Work and What Are Its Key Principles?

[jduffy77]

It would provide thrust in the other direction.

It would not do so using an identical transmission.

This invalidates the argument that all you need for the craft to work is for angled sails to revolve around a "cylindrical earth" (so to speak). Counter-thrusting the props (matched-angled sails ("matched-angled" due to the 180-degree rotation) moving in opposite directions around this "cylindrical earth" (so to speak)) would eliminate the very thing that the Blackbird requires to operate DDTWFTTW: the propeller thrust as advantaged by the interaction between the ground, wheels, transmission, frame, propellers, and the wind.

Note: The notion of a "cylindrical earth" is found in Scenario 2 of the video that has been referenced in this thread several times.

It is not an "argument" to say that angled sails are revolving around a "cylindrical earth". It is a fact. It is called the prop. I do not see how your second prop scenario would invalidate anything even if it were true. You have yet to say exactly what is wrong with the scenario in the video. Do you not believe it could be built?

 

[kmarinas86]

I do not see how an increase in pressure has anything to do with it. The prop is slowing down the wind. There is no increase in potential energy stored in the air at any point. The prop together with the carts drive mechanism is acting as an air brake.

Two winds blowing on each other is just like two like-charges repelling each other, so it is wrong to say that opposing winds cannot store potential energy. Heat does not form immediately. Energy in opposing winds produces pressure (both dynamic pressure and static pressure) before the stored energy can be released as heat.

 

[rcgldr]

A propeller facing the other way through an identical transmission would not function.

The end result would be a directly downwind slower than the wind vehicle. Look at the graph posted by A.T., if the propeller is "facing the wrong way", the vehicle advance ratio is negative, downwind but slower than wind speed. An advance ratio greater than zero but less than 1 resutls in DDWFTTW, an advance ratio greater than 1 results in an upwind vehicle (between 1 and 2, upwind faster than wind on an idealized vehicle (no losses), greater than 2, upwind slower than wind).

I do not see how an increase in pressure has anything to do with it. The prop is slowing down the wind.

The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop. Using a ground based frame of reference, the total energy of the affected air is decreased and consumed by all the losses in a DDWFTTW vehicle, aerodynamic drag (due to relative headwind), drivetrain losses, rolling resistance, ..., and consumed by the increase in kinetic energy of the vehicle during acceleration.

 

[A.T.]

A sail boat whose velocity made good exceeds wind speed uses water as a reaction mass to attain this excess speed (for all one must do with that is to turn the vehicle, after it exceeds wind speed, toward direct downwind),

No. A sailcraft can achieve and hold a steady state downwind VMG > true wind at a constant course in constant true wind. You don't have to turn the vehicle at all.

http://img253.imageshack.us/img253/6694/downwindvectorsen3.png

On top of this, many of your videos (the ones showing the gears) apparently depict the tailwind as providing a steady thrust when actually that tailwind is replaced by an unmistakable headwind.

Propellers can produce thrust in apparent head wind.

The sailboats simply cannot exceed windspeed downwind without changing the angle of the apparent wind...

For a sailboat the angle of the apparent wind changes, even when you accelerate at a constant course in constant true wind. The angle of attack at the propeller blades changes in the same way during acceleration.

...either by turning the boat (requiring a reaction mass) and/or the wind itself.

Nope. Constant course. Constant true wind. Steady state downwind VMG > true wind. See vectors above.

Thus, your analogy with sail boats is flawed.

Once you understand sailboats, you will see that the analogy is perfectly valid.

 

[A.T.]

In both scenarios, the sail (seen here as not jibbing) should only be able to beat the balloons drifting with the wind if the wind is not parallel to the left-to-right axis.

The true wind is perpendicular to the blue stripes on the surface. The balloon goes directly downwind at windspeed. The boat goes at an angle to the true wind, with a downwind VMG > true wind. So the boat outruns the balloon along the true wind direction. This is a physically perfectly valid steady state situation (see vectors in previous post).

Jibbing is irrelevant to achieve this. In fact, changing tacks only slows down the boat as it goes through DDW. The continuous helical tack of the propeller blades eliminates that problem.

 

[jduffy77]

The end result would be a directly downwind slower than the wind vehicle. Look at the graph posted by A.T., if the propeller is "facing the wrong way", the vehicle advance ratio is negative, downwind but slower than wind speed. An advance ratio greater than zero but less than 1 resutls in DDWFTTW, an advance ratio greater than 1 results in an upwind vehicle (between 1 and 2, upwind faster than wind on an idealized vehicle (no losses), greater than 2, upwind slower than wind).

A propeller facing the other way using the same transmission would not function. As I mentioned if you changed the angle of attack to the point where you had a turbine you would be able to generate an upwind force vector. As I mentioned though I do not think this has any bearing at all on the applicability of the tacking prop analogy, least of all invalidating it.

The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop. Using a ground based frame of reference, the total energy of the affected air is decreased and consumed by all the losses in a DDWFTTW vehicle, aerodynamic drag (due to relative headwind), drivetrain losses, rolling resistance, ..., and consumed by the increase in kinetic energy of the vehicle during acceleration.

I agree with all of the above as long as you mean the area of decreased pressure is in front of the prop, downwind relative to the cart. The cart is functioning exactly as a prop plane would except instead of engines the prop cart is powered by the wheels. As long as that is what is meant by the following explanation then ok.

The wheels are spun because the tailwind pushes the frame of the vehicle which holds the axles. So the axles are pushed, dragging the wheels over the ground, causing them to rotate (due to ground friction) which, by connection with the transmission, spins the prop in the direction opposite that the tailwind wants to turn it. In other words, the power of the tailwind that is diverted to turn propeller to generate a "head" wind component that passes from the propeller in the thrustward direction (opposite of the tailwind). Two "winds" blowing at each other (opposite directions toward one another) increases the pressure, and such increases the potential energy stored in the air (at the expense of kinetic energy of the winds). So energy is stored entirely outside the vehicle, if we ignore the phonons traveling through the vehicle's structure. The latter is of little importance as its energy density is very small (acoustic resonance is not an important feature of the Blackbird vehicle).

I do not think it is a very accurate or illuminating explanation of what is happening with the cart. I would be curious to get your opinion on that.

 

[ThinAirDesign]

@kmarinas86:

It appears that you do not believe that traditionally rigged sailing vessels (be they on land, water or ice) do not and cannot achieve VMGs of greater than wind speed while on a steady course heading. This is simply not true. Until you correct this aspect of your physics understanding, you will be fighting a steep uphill battle in understanding the DDWFTTW vehicles.

Perhaps this link would help with your understanding of the topic. It contains all the maths, vector diagrams and real world examples (along with plenty of enlightening exchanges on the discussion pages) that show your position on this to be flawed.

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

Traditionally rigged sailing vessels easily and regularly exceed 1.0 vmg on steady course -- often by a very large margin.

JB

 

[jduffy77]

Two winds blowing on each other is just like two like-charges repelling each other, so it is wrong to say that opposing winds cannot store potential energy. Heat does not form immediately. Energy in opposing winds produces pressure (both dynamic pressure and static pressure) before the stored energy can be released as heat.

Whether or not it makes sense to talk about winds "storing potential energy" I will leave to others more knowledgeable. I will say it does not make make sense to talk about in the context of the cart. No one is saying the cart is 100% efficient. Obviously there is energy being converted to heat at the propeller air interface as well as all over the drive train. What does it have to do with what we are talking about? The cart takes kinetic energy from the air by slowing it with respect to the Earth and uses it to accelerate ddwfttw.

 

[rcgldr]

A propeller facing the other way using the same transmission would not function.

This is how the Bauer DDWFTTW cart started off, it had a variable pitch prop and started off in turbine mode, where the wind drove the propeller which drove the wheels (DDW slower TTW). Once pitch of the propeller is increased from negative to positive but less than 1:1 ratio, the setup is in DDWFTTW mode and the wheels drive the propeller that produces thrust against the wind. The chart that A.T. created on the first page of this thread shows this.

As I mentioned if you changed the angle of attack to the point where you had a turbine you would be able to generate an upwind force vector.

But that angle of attack corresponds to a vehicle advance ratio greater than 1. Note that in the upwind case, both the propeller and the wheels turn the opposite direction as the downwind case, so the transmission direction doesn't change, only the ratio.

The prop slows down the wind by increasing the pressure upwind of the prop. The affected air's pressure and speed of are reduced downwind of the prop.

I agree with all of the above as long as you mean the area of decreased pressure is in front of the prop, downwind relative to the cart.

It's what I meant, although the wording was a bit awkward. The main point was that relative to the ground, the prop extracts energy from the air.

As I mentioned though I do not think this has any bearing at all on the applicability of the tacking prop analogy, least of all invalidating it.

A propeller may be more efficient than other means of generating thrust, but the tacking part of the thrust generation isn't a requrement. Ignoring efficiency issues, any method that uses a force from the wheels to generate thrust (greater force, slower speed) would work for a DDWFTTW cart, and no component of movment perpendicular to the true wind is required for a DDWFTTW cart. Note that for an under the ruler faster than the ruler vehicle, tacking is not used:



The point here is that tacking surfaces aren't required for a DDWFTTW cart, but since a propeller is more efficient at producing thrust than a paddle wheel type device, a propeller is used.

 

[kmarinas86]

@kmarinas86:

It appears that you do not believe that traditionally rigged sailing vessels (be they on land, water or ice) do not and cannot achieve VMGs of greater than wind speed while on a steady course heading. This is simply not true.

It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!). If you are able to travel a straight line and "beat the wind" with a sail, then the wind has to be changing direction over time. In reality, wind does exactly that, but the changing of the wind's direction is supposed to be negligible for the very scenario tested by DDWFTTW Blackbird sand yacht. So for the test, you must rely upon a steady wind of unchanging direction and travel parallel only with respect to that wind.



The blades travel faster than wind speed, but they do not move directly downwind.

Perhaps this link would help with your understanding of the topic. It contains all the maths, vector diagrams and real world examples (along with plenty of enlightening exchanges on the discussion pages) that show your position on this to be flawed.

http://en.wikipedia.org/wiki/Sailing_faster_than_the_wind

Traditionally rigged sailing vessels easily and regularly exceed 1.0 vmg on steady course -- often by a very large margin.

JB

Not so if vmg is parallel to steady wind of unchanging course! I've been to that page several times, and it still doesn't refute my point. Take for example the picture for boat speed downwind:

http://upload.wikimedia.org/wikipedia/en/4/4f/Wiki_sailing_vector_downwind.png

Note that alpha+beta=180-(90-alpha)-(90-beta). So the component of the apparent wind applied to the boat in the direction opposite of the true wind is equal to apparent wind speed*cos(alpha+beta).

Given that alpha and beta are positive angles, when the boat is going downwind faster than true wind (0 degrees < 180-(90-alpha)+(90-beta) < 90 degrees), cos(alpha+beta) > 0. Therefore, as soon as the boat overtakes the true wind, the apparent wind applies a force component on the boat against the direction with respect to the true wind. As alpha+beta approach 0 (parallel condition), the apparent wind speed on the boat against the direction of the true wind approaches boat speed minus true wind speed.

In both scenarios, the sail (seen here as not jibbing) should only be able to beat the balloons drifting with the wind if the wind is not parallel to the left-to-right axis. However, in DDTWFTTW, velocity made good must be parallel to downwind (and anti-parallel to upwind). The video therefore relies upon two contradicting assumptions about the direction of the wind even though it speaks of only one of them.

The above was the further elaboration of my point about this not having to do with tacking.

 

[rcgldr]

alpha ... beta

Can you swap the name alpha and beta in your post and diagram(s)? This would allow beta in your diagram to correspond to the term "beta" as used by the sailing world (where beta is the angle of apparent wind relative to boat's heading).

 

[kmarinas86]

Can you swap the name alpha and beta in your post and diagram(s)? This would allow beta in your diagram to correspond to the term "beta" as used by the sailing world (where beta is the angle of apparent wind relative to boat's heading).

It would be helpful them for sure, but this is not really my diagram, and I have no interest in swapping the alpha and beta terms in the image. One can replace alpha and beta with x and y if they like. They will just have to think a little bit harder when observing the diagram, that's all.

 

[rcgldr]

Therefore, as soon as the boat overtakes the true wind, the apparent wind applies a force component on the boat against the direction with respect to the true wind.

Split up the apparent wind into components perpendicular and parallel to the boat's heading. An apparent headwind applies an aerodynamic drag to the boat, while an apparent crosswind is diverted to produce thrust by the sail. Note that apparent crosswind is independent of the boats speed: apparent crosswind = true wind speed x sin(boat heading - true wind heading). As long as the thrust to drag ratio is high enough, a boat can achieve vmg downwind greater than true wind speed while tacking at some angle (around 30° to 40°, depending on circumstances).

 

[kmarinas86]

Split up the apparent wind into components perpendicular and parallel to the boat's heading. An apparent headwind applies an aerodynamic drag to the boat, while an apparent crosswind is diverted to produce thrust by the sail. Note that apparent crosswind is independent of the boats speed: apparent crosswind = true wind speed x sin(boat heading - true wind heading). As long as the thrust to drag ratio is high enough, a boat can achieve vmg downwind greater than true wind speed while tacking at some angle (around 30° to 40°, depending on circumstances).

The components you mentioned were already in the original picture and had nothing to do with the force directed parallel to the direction of the wind.

Instead of projecting the apparent wind onto axes parallel and perpendicular to the boat speed, I split up the apparent wind into components perpendicular and parallel to the wind's heading in lime green.

Your components:
* x = Boat speed - cos(beta)
* sin(beta)

My components:
* apparent wind speed*cos(alpha+beta)
* apparent wind speed*sin(alpha+beta)

In the second image below, I split up your two components into four, shown in red.

 

[A.T.]

It is possible for them to exceed VMG, but not if VMG is dead downwind (and stays that way!).

They don't "exceed VMG". 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.

 

[kmarinas86]

They don't "exceed VMG". They achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind.

I didn't fix the typo in time. I meant this:

It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!). If you are able to travel a straight line and "beat the wind" with a sail, then the wind has to be changing direction over time.

downwind VMG : boat velocity component pointing directly downwind.

Project the apparent wind speed vector over the boat speed vector component parallel to the true wind speed vector:

That is not steady state. It is obvious that the craft will slow down.

 

[A.T.]

I didn't fix the typo in time. I meant this:

It is possible for VMG to exceed windspeed, but not if VMG is dead downwind (and stays that way!).

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.

Project the apparent wind speed vector over the boat speed vector component parallel to the true wind speed vector:

Why should I do this? It is utter nonsense.

https://www.physicsforums.com/attachment.php?attachmentid=42386&stc=1&d=1325539649

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

That is not steady state. It is obvious that the craft will slow down.

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.

 

[kmarinas86]

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

None, as this was not a force diagram. It is a velocity diagram. The red text I added was not identify the red vectors as forces. The vectors are obviously velocities as none of them are forces. The text "This affects d'' / t' of the boat[,] and it causes d' t' to change." is a description of the consequences of the red vectors in question, not of the red vectors themselves. Two of three red vectors are components of the apparent wind velocity. The third and longest one represents the apparent wind velocity itself. The consequences (not depicted graphically here) are the forces which vary by the square of the vector magnitudes here depicted and the power which varies by the cube of the vector magnitudes here depicted, so taking that into consideration, it is clear that decomposing them in the same way as the velocity vectors would result in a diagram where the angles have no physical relevance to path of the boat other than its derivative with respect to time, and indeed I have not done so with the above modified diagram, as it is only a velocity diagram applicable for an instant of time. The force on the sail is not even balanced here without incorporating friction. Once you do that, then you would have explain how friction would act against the apparent wind, when it is obvious that the apparent wind is able to push as a result of the fact that it is itself a source of friction providing a force on the craft.

 

[A.T.]

None, as this was not a force diagram. It is a velocity diagram.

Then it cannot tell us if the boat will accelerate. Acceleration is determined by forces, not velocities. And the force vectors say it can accelerate:

 

[kmarinas86]

Then it cannot tell us if the boat will accelerate. Acceleration is determined by forces, not velocities. And the force vectors say it can accelerate:

The deflection of the apparent wind off the sail means that the apparent wind does not maintain the same direction when approaching versus leaving the sail. Any deflection at an exact right angle to the sail's momentum direction does not add energy to the boat. It simply redirects the direction of the boat. Any force component at a right angle to a momentum direction only deflects it, leaving its norm unaffected. Does this deflection add to the kinetic energy of the boat? No. It can only deflect whatever kinetic energy is already there. Only forces parallel or anti-parallel to the motion in question will allow for the kinetic energy of the boat to change.

 

[A.T.]

Only forces parallel or anti-parallel to the motion in question will allow for the kinetic energy of the boat to change.

Yes, and the "accelerating sail force" is parallel to the boat velocity. It will accelerate the boat until the hull drag matches it. Here all the vectors for constant velocity (net force = zero):

 

[kmarinas86]

Yes, and the "accelerating sail force" is parallel to the boat velocity. It will accelerate the boat until the hull drag matches it. Here all the vectors for constant velocity (net force = zero):

There is probably more to this system than meets the eye.

The sail drag and lift can be further broken down into components either parallel, anti-parallel, or perpendicular to the sailboat velocity. The part of the sail lift that is parallel doesn't quite make sense though (at first), but I think there might be an account for the energy involved in doing that. I just don't agree that it comes from the wind though.

I will now concede that in this there is a source of energy that would make for the illusion that the apparent wind can accelerate a sail craft in the opposite direction that it is blowing, so such likely has deeper origins in the energy in motion of the atoms and molecules of the craft itself. If this is indeed what is going at a deeper level, then I take back some of the things which I have said. It seems to me that if the matter's energetic motions were somehow deflected internally as a result of external pressure, then that deflection would be sufficient to explain the observational fact (which I have until now have downplayed) that indeed, as stated by A.T., that sailboats can "achieve a downwind VMG greater than true wind, steady state, on constant course, in constant true wind". If so, then some of this phenomenon could be related even to the General Relativistic corrections to Special Relativity (which drop the assumption of "inertial motion only"), as it appears that the fundamental microscopic non-inertial, vibratory/rotational motions involved must some how have changed course to some small degree (even though this is a non-relativistic scenario) as a result of the force interactions involved.

Indeed, General Relativity would predict that the sailboat would undergo an additional "gravitational time dilation" due to the non-inertial motion induced by the deflection of both the apparent wind and the sail, which perhaps could be explain sometime in the future as increasing the effective internal wavelengths that result from "spreading" paths of highly-curvatured motions inside the mass of the sailboat over longer traces of distances with respect to the grid of "spacetime", consequently leading to a decreases in corresponding frequencies and thus decreasing the overall rate of time at the sailboat relative to an external observer.

 

[ThinAirDesign]

If so, then some of this phenomenon could be related even to General Relativity, as it appears that the fundamental microscopic non-inertial, vibratory/rotational motions involved must some how have changed course to some small degree (in this non-relativistic example) as a result of the force interactions involved.

Nothing fancy -- simple straightforward fluid mechanics involved and it's not even anything new, having been done for hundreds (thousands?) of years. A simple flip of the environment (viewed from the perspective of a fish for example) easily shows that the keel of the very first boat to ever tack its way upwind was achieving a downfluid VMG faster than the fluid, absolutely steady state.

JB

 

[A.T.]

I will now concede that in this there is a source of energy that would make for the illusion that the apparent wind can accelerate a sail craft in the opposite direction that it is blowing

It is not "an illusion". It is a well verified empirical fact and in full agreement with Newtonian physics. The "source of energy" is the velocity difference between the air & surface which is always being reduced.

...matter's energetic motions were somehow deflected internally ... some of this phenomenon could be related even to General Relativity...fundamental microscopic non-inertial, vibratory/rotational motions involved ...

LOL. It's just simple mechanics as the vectors diagrams show.

 

[kmarinas86]

Nothing fancy -- simple straightforward fluid mechanics involved and it's not even anything new, having been done for hundreds (thousands?) of years. A simple flip of the environment (viewed from the perspective of a fish for example) easily shows that the keel of the very first boat to ever tack its way upwind was achieving a downfluid VMG faster than the fluid, absolutely steady state.

JB

It is not "an illusion". It is a well verified empirical fact and in full agreement with Newtonian physics. The "source of energy" is the velocity difference between the air & surface which is always being reduced.

LOL. It's just simple mechanics as the vectors diagrams show.

Without positing pre-existing "hidden" momentum inside the mass of the apparent wind and/or the craft itself, I cannot at all see how something that blows at you can pull you forward. It seems to make more sense to imagine that the apparent wind is simply allowing this "hidden" momentum (which we know exists in the form of the \vec{p} in the equation (m_{whole}c^2)^2=E_{whole}^2=(m_{parts}c^2)^2+\left\|\sum \vec{p}\ c \right\|^2 to appear visible to a human observer, than it is to believe that apparent wind would be gaining energy by doing work on the sail. After all, if the apparent wind is moving to the left and pushes the sail to the right, this would mean that the apparent wind would have to accelerate both itself and the sail, which would violate the conservation of energy, if it were not for this hidden momentum. It is no coincidence to me that so many think incorrectly that the DDTFTTW craft is impossible. They see the apparent wind as a source of energy, and they cannot imagine how it would increase speed relative to the craft by being thrusted by the propeller tailwards while at the same time having that power of the propeller being explained by the same incoming headwind. The reason why this is so prevalent is that there is something wrong with apparent wind being able to do that with the cart, if you don't accept that there is a hidden source of energy! Yes, the forces can explain conservation of momentum, but from the inertial frame of the craft at time t, it is very clear that without a hidden source of energy, we cannot explain why the head wind and the craft with increase respect to the frame once t has passed. In this case, it is not hidden because of deception, but rather, it is hidden because the energy is that of atoms and molecules. So the skeptics of DDTFTTW are not entirely wrong in their skepticism. There must be a hidden energy source (It's nature's energy!). :)

 

[A.T.]

I cannot at all see how something that blows at you can pull you forward.

Sailors use this of ages.

...hidden momentum...

Very creative, but there is no need for such obfuscatory nonsense. All the momentum is clearly visible all the time and is being conserved.

 

[ThinAirDesign]

I cannot at all see how something that blows at you can pull you forward.

Before I respond I really must clarify something.

Sailboats sail upwind all the time. They can leave a point downwind and readily arrive at a point directly upwind of where they were by simply sailing towards a point situation to the right (or left) of the upwind goal, and then once halfway there, they turn and sail directly towards that point.

A: If taken literally and without context, your above quoted statement would make it seem as though you don't see how a boat can sail upwind such as the above. I'm pretty sure that's not what you mean but I do want to ask the question.

B: If you believe boats can make upwind progress by sailing at an angle to the wind, but are having difficulty believing that *anything* wind powered can make steady state progress directly into the wind using basic Newtonian physics then I need to know that.

A or B or other?

Thanks

JB

 

[kmarinas86]

Sailors use this of ages.

Very creative, but there is no need for such obfuscatory nonsense. All the momentum is clearly visible all the time and is being conserved.

The net momentum is indeed conserved.

2 + (-2) = (3) + (-3)... etc.

That makes it look like there is no mystery.

What does not make sense to many still-skeptical skeptics is how would the Blackbird DDTWFTTW sand yacht conserve energy. Neither you, nor them, seem to have the explanation.

Note that:

(2)^2 + (-2)^2 is not (3)^2 + (-3)^2... etc.

Would you mind explaining where the energy comes from to allow the wind to do work on the DDWFTTW vehicle (in the time between t and t+\epsilon) at the same time the DDWFTTW vehicle accelerates, with respect to the initial inertial frame of the vehicle at time t? The work is done in opposite directions, conserving momentum even macroscopically, but not the kinetic energies of both (both increase as far as the initial inertial frame is concerned). My "very creative" resolution addresses this problem by bringing up the point about the true and factual existence of the below-macroscopic energy of atoms and molecules as being the entity that accounts for this apparent gap.

My explanation is not that of a hidden net momentum, but a hidden set of vector momenta which sums to zero in the frame being evaluated (i.e. the momentum whose energy is identical to the rest mass of a body, as evaluated from the system frame in question, times the speed of light squared). I have a hunch that somehow this is either the static P-V energy that was already present in the air mass prior to vehicle operation, and/or the vibrational and rotational energy of the vehicle's particle makeup. Probably both.

 

[kmarinas86]

Before I respond I really must clarify something.

Sailboats sail upwind all the time. They can leave a point downwind and readily arrive at a point directly upwind of where they were by simply sailing towards a point situation to the right (or left) of the upwind goal, and then once halfway there, they turn and sail directly towards that point.

A: If taken literally and without context, your above quoted statement would make it seem as though you don't see how a boat can sail upwind such as the above. I'm pretty sure that's not what you mean but I do want to ask the question.

B: If you believe boats can make upwind progress by sailing at an angle to the wind, but are having difficulty believing that *anything* wind powered can make steady state progress directly into the wind using basic Newtonian physics then I need to know that.

A or B or other?

Thanks

JB

Neither. Potential energy must be extracted from the system. Either the static P-V of the wind and/or the energy from the particle make-up of vehicle (and/or even that of the ground, if need be). A correct explanation cannot be found by trying to conserve "kinetic energy+heat" while ignoring potential energy.

The quote (taken out from a sentence after a comma) is taken too far out of context. It doesn't even relate to what I am saying.

 

[ThinAirDesign]

Neither.

<snip>

The quote (taken out from a sentence after a comma) is taken too far out of context. It doesn't even relate to what I am saying.

Ok, got it. I was pretty sure from your previous posts that "A" wasn't what you meant, but didn't want to move forward without confirmation.

Thanks.

It appears that what you are saying the examples I have in that post can and do happen, you just don't believe they can be explained through simple Newtonian physics.

Would that be a fair representation of your position?

JB

 

[rcgldr]

Potential energy must be extracted from the system.

For a fluid or gas, potential energy is used to describe the gravitational potential energy of a gas or fluid. For a wind driven vehicle, the energy extracted from the air affected by the wind driven vehicle corresponds to the reduction in kinetic energy (wrt ground) of the affected air. (Pressure effects are short term and only exist in the immediate vincinity of the propeller.)

 

[kmarinas86]

Ok, got it. I was pretty sure from your previous posts that "A" wasn't what you meant, but didn't want to move forward without confirmation.

Thanks.

It appears that what you are saying the examples I have in that post can and do happen, you just don't believe they can be explained through simple Newtonian physics.

Would that be a fair representation of your position?

JB

I guess that depends on what you call simple, what you call Newtonian, and what you mean by explain. You can explain things using forces without highlighting the apparent non-conservation of "kinetic energy+heat". That is simple (to me), but ignoring where this kinetic energy comes from doesn't do it for me, so I would disagree that it is somehow an adequate explanation. (If a claimed-to-be explanation is inadequate, does it really explain what needs to be explained?) Potential energy latent inside matter isn't exactly something that I would call part of "Newtonian" physics. It's not included in most of the (simple enough to be convincing to most) analyses that have been offered to explain the Blackbird. Certainly it can be explained using classical physics.

The lack of significant mention of potential energy when discussing how tacking can allow sails to move ahead of the wind, in addition to the absolute absence of this point in many of the videos that I have seen that try to explain DDTWFTTW, has (I bet) contributed much confusion for people (including skeptics and naysayers) who wonder where the energy comes from and who, like myself, have for a time not been able to see how tacking would be of any benefit to it.

 

[ThinAirDesign]

The lack of significant mention of potential energy when discussing how tacking can allow sails to move ahead of the wind, in addition to the absolute absence of this point in many of the videos that I have seen that try to explain DDTWFTTW, has (I bet) contributed much confusion for people (including skeptics and naysayers) who wonder where the energy comes from and who, like myself, have for a time not been able to see how tacking would be of any benefit to it.

If by "potential energy" you mean something other than the mass of one fluid moving relative to another (or surface) and the kinetic energy contained thus, then the reason it isn't used in an explanation or video is that it would be flat wrong. There is NO other energy involved in accelerating the craft.

If by "potential energy" you mean the mass of one fluid moving relative to another (or surface) and the kinetic energy contained thus - meaning the power of the wind, I can't imagine how you have missed such explanations.

JB

 

[rcgldr]

where this kinetic energy comes from

You could consider the source of energy for the true wind to be the heat from the sun.

Potential energy latent inside matter isn't exactly something that I would call part of "Newtonian" physics.

I don't recall any mention of potential energy in the descriptions of how wind driven vehicles operates (sail boats, DDWFTTW vehicles, DUW vehicles, ... ). Extracting potential energy within matter involves a chemical or nuclear reaction, which doesn't occur with the wind powered vehicles being discussed here.

The lack of significant mention of potential energy when discussing how tacking can allow sails to move ahead of the wind. ... where the energy comes from

I'm not sure what you mean by potential energy. Wind driven vehicles extract kinetic energy from the wind (using a ground or water based frame of reference).

how tacking would be of any benefit to it.

Tacking isn't required for a DDWFTTW vehicle. A DDWFTTW vehicle could connect the wheels to a treadmill geared so the upper surface of the treadmill moves upwind at some fraction of the vehicles speed, for example 1/2 of the vehicles speed (an advance ratio of .5). The treadmill could pull parachutes along the upper surface and then collapse them (perhaps pull them through a tube) along the lower surface. It wouldn't be as efficient as a propeller, but if the losses could be reduced enough, it would work.

A sail can't generate thrust from an apparent headwind component, so it needs an apparent crosswind component which it diverts to aft of the boat's heading to generate thrust, which is why a sail boat needs to tack in order to achieve vmg downwind greater than true wind.

 

[kmarinas86]

I'm not sure what you mean by potential energy. Wind driven vehicles extract pressure energy and kinetic energy from the wind (using a ground or water based frame of reference).

Pressure energy to me is a form of potential energy, though I tend to look at it from a "molecular" perspective where electric forces reign supreme over gravitational ones.

 

[spork]

Pressure energy to me is a form of potential energy, though I tend to look at it from a "molecular" perspective where electric forces reign supreme over gravitational ones.

Even pressure energy doesn't come into the equation. The vehicle simply slows down the air relative to the ground beneath it. That's all. For pressure energy to benefit the cart it would have to leave a volume of air in its wake that has been expanded to greater volume and lower pressure. It doesn't do this.

 

[rcgldr]

Even pressure energy doesn't come into the equation.

There's a pressure differential in the immediate vincinity of the propeller, but eventually the affected air's pressure returns to ambient and it's velocity is changed. From a DDWFTTW vehicle's frame of reference, the pressure and flow near the propeller corresponds to the description in this NASA article:

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

From a ground frame of reference, the pressure differential at the propeller results in the affected air from the true wind to be slowed down.

 

[spork]

There's a pressure differential in the immediate vincinity of the propeller...

Of course there is. But you said: "Wind driven vehicles extract pressure energy and kinetic energy from the wind"

And this isn't the case. Sure they create a lower pressure in front of the disk and a high pressure behind the disk, but ultimately, they don't take any "pressure energy" from the wind. That would be equivalent to saying that an airplane's exploits the "pressure energy" of the air to stay aloft. The fact that it has a local effect (both positive and negative) on air pressure does not imply that it extracts pressure energy from the wind.

 

[rcgldr]

But you said: "Wind driven vehicles extract pressure energy and kinetic energy from the wind"

Yeah I worded that badly and corrected my previous posts. The pressure effects are short term and only exist in the immediate vicinity of the propeller.

 

[spork]

Yeah I worded that badly. The pressure effects are short term and only exists in the immediate vicinity of the propeller.

I figured it was a simple mis-statement. That's one of the reasons I didn't mention it when you posted it, but only when kmarinas86 talked about it as potential energy. I got the distinct idea he saw it as something it's not.

 

[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.