Understanding DDWFTTW: Exploring Its Principles and Addressing Common Questions

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In summary, the propeller can apply more force at faster-than-wind speeds because it does not travel as far through the air. This allows the cart to extract more power from the wind.
  • #246
Tom_K said:
These two statements (from the test question) look contradictory to me:
1) All of the power used to move the car comes from the wind.
2) Power should always be produced by the force corresponding to the larger relative velocity.
In case you haven't already figured it out on your own:

1) Is a general statement about the energy source: velocity difference between air and ground (true wind).

2) Is explained best by Figure 1 in Gaunaa's paper (right side labels):

ddwfttw_energy_2.png


Since P = F * v you can generate lots of power with a small braking force at the fast moving surface with the wheels. And because the air moves slower relative to the cart than the surface, you need less power to generate equal or greater thrust, than the braking force at the wheels.
 
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  • #247
Tom_K said:
These two statements (from the test question) look contradictory to me:
1) All of the power used to move the car comes from the wind.
2) Power should always be produced by the force corresponding to the larger relative velocity.
The difference is the frame of reference.

1) - From a ground based frame of reference, all of the power comes from reducing the speed (kinetic energy) of a portion of the wind.

2) - From the vehicles frame of reference, the source of power comes from the faster moving medium (surface or air), which is then effectively geared down to produce more force at a lower speed upon the slower moving medium (air or surface), and at a lower rate of power output (due to losses) than power input.

Both 1) and 2) have to be true in order for these vehicles to work. For example, in the case of the DDWFTTW vehicle, the thrust from the propeller has to slow down a portion of the wind wrt (with respect to) ground. Say the wind speed is 10 mph wrt ground, and the DDWFTTW vehicle is traveling at 30 mph wrt ground, then the thrust speed wrt vehicle has to be greater than 20 mph in order to slow down a portion of the wind wrt ground.
 
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  • #248
That doesn’t sound right. Generally speaking, a propeller operating in a tailwind is less efficient than when operating in still air or into a relative headwind. At cruise speeds an airplane propeller produces very little thrust so is very inefficient and it has the benefit of an internal combustion engine.
You seem to be saying the propeller on the car is producing thrust above wind speed yet all of the power to produce thrust is coming from the wind. Or are you saying the surface moving under the wheels is somehow transferring energy to the car, like a belt drive turning the wheels?
I don’t see the logic here at all!
By my reasoning, once the car is traveling at the same speed as the wind there is no longer any wind force acting on the car since there is no longer any Δ Velocity. Force = mass x acceleration which is mass x Δ Velocity. No Δ Velocity, no Force.
I can see you saying there is Δ Velocity at the propeller because it is spinning but that is a circular argument. The car is moving so the propeller is spinning and the propeller is spinning so the car is moving! Every perpetual motion aficionado will believe you, but I don’t!
 
  • #249
Tom_K said:
Generally speaking, a propeller operating in a tailwind is less efficient than when operating in still air or into a relative headwind.
In DDWFTTW the propeller is operating in a relative headwind. But the relative headwind is less than it would be without the true tailwind. This improves the achievable propeller performance. You need less power to generate the same force on something that moves slower relative to you. This follows directly from:

P = F * v

Tom_K said:
You seem to be saying the propeller on the car is producing thrust above wind speed
Yes

Tom_K said:
yet all of the power to produce thrust is coming from the wind.
Yes (wind = velocity difference between surface and air, which is being reduced by the car)

Tom_K said:
Or are you saying the surface moving under the wheels is somehow transferring energy to the car, like a belt drive turning the wheels?
Yes, but note that some energy statements are frame dependent.

Tom_K said:
By my reasoning, once the car is traveling at the same speed as the wind there is no longer any wind force acting on the car since there is no longer any Δ Velocity.
Propellers can produce thrust in this condition, so your reasoning fails right here.

Tom_K said:
I can see you saying there is Δ Velocity at the propeller because it is spinning but that is a circular argument. The car is moving so the propeller is spinning and the propeller is spinning so the car is moving!
Feedback loops are circular, but they do work and can amplify certain quantities (here velocity). This is where your logic fails again: You cannot analyze feedback loops using a linear cause-effect chain.

Tom_K said:
Every perpetual motion aficionado will believe you, but I don’t!
PM ideas often involve isolated feedback loops, that work without any energy input. This one is not isolated at all, as it continuously removes kinetic energy from the surface-air system, by reducing the velocity difference between the two.
 
  • #250
Or are you saying the surface moving under the wheels is somehow transferring energy to the car, like a belt drive turning the wheels?

A.T. said:
Yes, but note that some energy statements are frame dependent..

You really are saying that energy is transferred from the ground to the wheels? Fascinating! In what frame of reference is that happening? What is the source of this amazing energy from the ground? Energy is frame dependent, but forces are not. Please describe the forces involved in this supposed energy transfer from the ground to the wheels. Can I roll a wheel on the ground and extract this energy?

A.T. said:
Propellers can produce thrust in this condition, so your reasoning fails right here.


Feedback loops are circular, but they do work and can amplify certain quantities (here velocity). This is where your logic fails again: You cannot analyze feedback loops using a linear cause-effect chain.


No, nothing has failed here except your strawman arguments against something I did not say.
 
  • #251
Tom_K said:
Or are you saying the surface moving under the wheels is somehow transferring energy to the car
The ground applies a force to the wheels, generating a torque around their axles. The rest follows as described here:
http://en.wikipedia.org/wiki/Torque#Relationship_between_torque.2C_power.2C_and_energy

Tom_K said:
You really are saying that energy is transferred from the ground to the wheels? Fascinating! In what frame of reference is that happening?
In the frame of the car, for example.

Tom_K said:
What is the source of this amazing energy from the ground?
Kinetic energy of an object (here the ground) is simply a consequence of choice of the frame of reference.

Tom_K said:
Please describe the forces involved in this supposed energy transfer from the ground to the wheels.
See post #246.
 
  • #252
A.T. said:
The ground applies a force to the wheels, generating a torque around their axles.

No it doesn't. The rolling wheel deforms on contact with the ground and the normal forces are redirected against the direction of motion and oppose that motion. No torque is applied to keep the wheel rolling.

rollres3.gif


If the gound could exert such a torque, a rolling wheel would roll forever!


A belt drive is different. In that case, the belt must deform around the contact patch with the wheel in order to provide a force difference across the contact patch and develop a torque to turn the wheel.

440px-Illustration_of_creepage_for_a_railway_wheel.png



If you are trying to make a Galilean transform between a belt driven wheel and a wheel rolling on the ground, you are wrong.

The forces are not the same in the two cases and the transform is bogus.
 
  • #253
Tom_K said:
The rolling wheel deforms on contact with the ground and the normal forces are redirected against the direction of motion and oppose that motion.
You are confusing rolling resistance (due to deformation) with traction (static friction). Rolling resistance is a dissipative force, like sliding friction. The force of static friction applied at the contact patch generates a torque around the axle, which represents a power transmission according to:

##P=\tau \cdot \omega##
 
  • #254
No, I am not confusing anything. You are claiming there is a net force at the bottom of a rolling wheel that produces a torque in the direction to keep the wheel rolling.
That is just not true. I have shown you the forces formed when the wheel deforms oppose the direction of rolling and in fact apply a torque that slows the wheel down. That force is greater than any supposed net force that would keep the wheel rolling.
You can also examine this in terms of energy. The wheel deforms and flexes continuously as it rolls on the ground losing energy in the form of heat. The ground can transfer no energy to a rolling wheel in any frame of reference.
I have also shown you that a belt drive cannot develop a force at the bottom of a wheel unless the belt deforms around the bottom, at least a bit, and develops a force differential there; that is a torque to turn the wheel. That is why drive belts are flexible. Without the belt deformation it cannot exert a torque force and the wheel deformation would try to spin the wheel the other way. Have you ever heard a belt squeal because it has been adjusted too tightly?
All claims of a Galilean transform of a belt driven wheel to a wheel rolling on the ground are false. The forces must be the same and they simply aren’t.
 
  • #255
Tom_K said:
No, I am not confusing anything.
Yes, you are confusing traction and rolling resistance. The force relevant to power generation in DDWFTTW is static friction. Rolling resistance due to wheel deformation is just one of the losses.
 
  • #256
Once again, I am not confusing anything. Static friction is present, indeed it is necessary for the wheel to roll and not slip. But that friction forces does no work on the wheel, either to slow it down or speed it up. The work done on the wheel to slow it is the rolling resistance which is a result of the wheel deformation changing the angle of the normal force.
Read this:

Friction is also present in this scenario. In the case where the "flat" surface is deformed by the weight of the cylinder wheel, the cylinder is continually rolling slightly "uphill" up the front portion of the depression. To do this without slipping, surface friction is necessary. This friction has a horizontal component that acts upward and forward on the cylinder, preventing slipping. But, it is still a zero-work force since it produces no displacement of the cylinder.
But the compression and decompression process that are responsible for rolling resistance do involve forces that displace portions of the bodies, and therefore they are doing work on the cylinder and the flat surface.
 
  • #257
First note that the DDWFTTW vehicles actually work, videos have been posted in previous threads. Here's an older video before the aerodynamic frame was added to the "blackbird", reaching 37.5 mph in about a 15 mph wind. Note the change in direction of the streamers attached to the poles on the vehicle. Intially the propeller acts as a bluff body, but as speed picks up it then acts as a true propeller.



Tom_K said:
propeller operating in a tailwind
As posted previously, the propeller operates wtih a relative headwind. The example I used was a 10 mph tail wind and a DDWFTTW vehicle moving at 30 mph tail wind. From the frame of reference of the vehicle, the propeller operates with a 20 mph headwind, and the ground moves backwards under the vehicle at 30 mph. (Note the actual vehicle with aerodynamic frame achieved about 35 mph with a 10 mph tailwind).

Tom_K said:
You seem to be saying the propeller on the car is producing thrust above wind speed yet all of the power to produce thrust is coming from the wind. Or are you saying the surface moving under the wheels is somehow transferring energy to the car, like a belt drive turning the wheels?
The axle for the wheels is connected via a chain drive to drive the propeller to produce thrust, and the propeller generates an opposing torque eventually onto the wheels, which in turn apply a forwards force onto the ground, part of a Newton third law pair where the ground applies a backwards force onto the wheels. As mentioned before, there is an effective reduction gearing of the propeller, so it generates more thrust than the opposing backwards force from the ground, but at a lower speed than the 30 mph that the ground is moving under the vehicle, somewhere between 20 mph and 30 mph, most likely around 23 mph since it's a big prop and a 3 mph relative acceleration of the air (wrt vehicle) will generate enough thrust.

From the ground frame of reference, the propeller slows the affected wind down by 3 mph, which is it's source of energy.

From the vehicle's frame of reference, the forwards force of the wheels on the ground slows down the surface of the Earth (wrt vehicle) by a very tiny amount, since the Earth is massive, but the decrease in kinetic energy of the Earth (wrt vehicle) is a bit larger in magnitude than the increase of kinetic energy of the air accelerated (by about 3 mph in my theorectical example here) by the propeller (wrt vehicle). It is in this frame of reference that energy is extracted from the ground.

Tom_K said:
By my reasoning, once the car is traveling at the same speed as the wind there is no longer any wind force acting on the car since there is no longer any Δ Velocity.
When the vehicle is traveling at wind speed, the wind force acting on the vehicle is zero, but the propeller is producing thrust. The Newton third law pair of forces are the forces related to the deceleration of wind (wrt ground) by the propeller, the propeller exerts a backwards force onto the air, and the air exerts a forwards force onto the propeller.

Again note that this only works when there is a wind relative to the ground. In my example here, at 30 mph ground speed, due to effective gearing, the propeller generates 23 mph thrust speed. Without a tail wind greater than 7 mph, the propeller would not be slowing the wind.

Looking at the external forces, there's a forwards force exerted by the air onto the propeller, which is greater in magnitude than the backwards force exerted by the ground onto the wheels, resulting in a net forwards force that is opposed by aerodynamic drag of the vehice moving against a relative head wind, rolling resistance in the wheels, losses in the drive train, ... . Eventually the vehicle reaches some maximum speed where the net force ends up zero.
 
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  • #258
Tom_K said:
But that friction forces does no work on the wheel...
Work is frame dependent. Here a remainder which frame we were talking about:
Tom_K said:
You really are saying that energy is transferred from the ground to the wheels? Fascinating! In what frame of reference is that happening?
A.T. said:
In the frame of the car, for example.
In the frame of the car static friction is doing work on the wheel.
 
  • #259
rcgldr said:
First note that the DDWFTTW vehicles actually work, videos have been posted in previous threads. Here's an older video before the aerodynamic frame was added to the "blackbird", reaching 37.5 mph in about a 15 mph wind. Note the change in direction of the streamers attached to the poles on the vehicle. Intially the propeller acts as a bluff body, but as speed picks up it then acts as a true propeller.

Why should I believe they work? That is what I would like to see some real evidence of. I have already asked to see some peer reviewed analysis that demonstrates this is possible. You are showing me YouTube videos?
Physics Forums has a policy of valuing quality:
PF values quality• Topics based on science published in real scientific journals or textbooks
Reference https://www.physicsforums.com/
I feel justified then, in asking for that quality.
rcgldr said:
From the vehicle's frame of reference, the forwards force of the wheels on the ground slows down the surface of the Earth (wrt vehicle) by a very tiny amount, since the Earth is massive, but the decrease in kinetic energy of the Earth (wrt vehicle) is a bit larger in magnitude than the increase of kinetic energy of the air accelerated (by about 3 mph in my theorectical example here) by the propeller (wrt vehicle). It is in this frame of reference that energy is extracted from the ground..

I have just shown that no energy is extracted from the ground, in any frame. That is impossible.


rcgldr said:
When the vehicle is traveling at wind speed, the wind force acting on the vehicle is zero, but the propeller is producing thrust. The Newton third law pair of forces are the forces related to the deceleration of wind (wrt ground) by the propeller, the propeller exerts a backwards force onto the air, and the air exerts a forwards force onto the propeller..

Below wind speed the propeller acts as a bluff body, a sail basically but not a very good one! A standard flat sail of similar diameter would be much more efficient.
But even allowing the car should reach wind speed (which I doubt) the only way it can accelerate from that point is for a net force to be acting on it. No net force from the wind can be acting on the body of the car as the velocity is the same as the wind. Force = mass x Δ Velocity and the Δ Velocity is 0. No wind force on the body of the car.
Is there a wind force on the propeller? If the propeller is turning, then yes. That begs the question of what energy is being used to turn the propeller to produce thrust? It isn’t some mysterious ground energy, as my previous post has shown. Can it be wind energy? If so, the claim now must be that wind energy is used to turn the propeller and then the turning propeller extracts still more wind energy. It all sounds very shaky to me! If wind turbines could do the same thing they could produce unlimited power, but they don’t. The idea is silly.
Can you prove the energy/power needed to turn the propeller and produce thrust is less than the energy the propeller can take out of the wind? Do you see the problem here? This machine, by virtue of its own motion, is able to produce energy that not only can sustain its own motion, but go even faster!
If this is as well established as you claim, then by now surely someone has formulated the differential equation showing all of the forces involved and the solution shows a net forwards force. Surely then, you can show me this equation? That would be more convincing than you just saying it works.
 
  • #260
A.T. said:
In the frame of the car static friction is doing work on the wheel.

No it doesn't. Have you read the reference I linked to?

The wheel is instantaneously at rest with respect to the ground. Static friction does no work on the wheel in any frame of reference. The deformation forces (normal force redirected) is what does the work on the wheel to take energy out of the wheel and slow it down.

Your claim just does not hold up to what the physics is telling me.
 
  • #261
A.T. said:
In the frame of the car static friction is doing work on the wheel.
Tom_K said:
The wheel is instantaneously at rest with respect to the ground.
But I'm not talking about the frame of the ground.
Tom_K said:
Static friction does no work on the wheel in any frame of reference.
Wrong, work is frame dependent.
 
  • #262
A.T. said:
Wrong, work is frame dependent.

It doesn't matter what sort of reference frame game you want to play, static friction does no work on the rolling wheel in any frame.

Read up on it.

Rolling Friction

A rolling wheel requires a certain amount of friction so that the point of contact of the wheel with the surface will not slip. The amount of traction which can be obtained for an auto tire is determined by the coefficient of static friction between the tire and the road. If the wheel is locked and sliding, the force of friction is determined by the coefficient of kinetic friction and is usually significantly less.

Assuming that a wheel is rolling without slipping, the surface friction does no work against the motion of the wheel and no energy is lost at that point. However, there is some loss of energy and some deceleration from friction for any real wheel, and this is sometimes referred to as rolling friction. It is partly friction at the axle and can be partly due to flexing of the wheel which will dissipate some energy. Figures of 0.02 to 0.06 have been reported as effective coefficients of rolling friction for automobile tires, compared to about 0.8 for the maximum static friction coefficient between the tire and the road.
 
  • #263
Tom_K said:
...static friction does no work on the rolling wheel in any frame. Read up on it.
I don't need to read pop-sci articles, which tacitly assume the ground frame for simplicity. I can apply the definition of work in the frame of the car myself: The contact patch is not static in that frame, so the static friction force on it is doing work.
 
  • #264
Tom_K said:
I have just shown that no energy is extracted from the ground, in any frame.
Again from the vehicles frame of reference, say the force at the wheels is 400 Newtons (a forward force that the wheels exert onto the ground, and a backwards force that the ground exerts onto the wheels, which is used to drive the propeller), and the speed is 14 meters / second, so power would be 400 x 14 = 5600 watts. This is power extracted from the Earth's angular kinetic energy, with respect to the vehicles frame of reference.

Simplified example:

mass of Earth = 5.972e+24 kg
radius of Earth = 6.371e+6 meters
angular intertia of Earth = I = 2/5 m r^2 ~= 9.696e+37 kg meters^2
initial angular velocity (wrt vehicle) = ω = (14 m/s) / 6.371e+6 m ~= 2.197457e-6 radians / second
torque exerted by wheels to slow the Earth (wrt vehicle) = -f r = -400 Newtons x 6.371e+6 m ~= -2.5484e+9 Newton meters
angular acceleration = torque / inertia ~= -2.6283e-29 radians / second^2
change in angular velocity each second = Δω = -2.6283e-29 radians / second
change in angular energy each second = 1/2 I ((ω + Δω)^2 - ω^2) ~= I ω Δω ~= -5600 joules
power extracted from Earth = 5600 watts
 
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  • #265
The misconceptions you have about kinetic energy in different reference frames have been addressed in this old thread:

https://www.physicsforums.com/threads/ke-of-system-different-reference-frames-question.564246/

Please read the answers given to Humber by DaleSpam and others there. And if you still have questions left, ask the mods to reopen that old thread or start a new one. Your confusion has nothing to do with DDWFTTW as such, but more with basic issues of KE frame dependence and wheel dynamics. So please stop derailing this thread.
 
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  • #266
It's time to close this thread, as TomK's misunderstanding of how ##W=Fd## transforms from frame to frame has been pointed out several times, and we're starting to repeat ourselves.

[EDIT: the thread has been re-opened, and the most recent re-iteration of the misunderstanding is deleted. I hope that it can continue more constructively]
 
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  • #267
Tom_K said:
It doesn't matter what sort of reference frame game you want to play, static friction does no work on the rolling wheel in any frame.

Read up on it.
Note, this reference does NOT support the claim made by Tom_K. Specifically, the reference shows only that static friction does no work in some frame (specifically the frame where the contact patch is at rest), but the reference does not make any statement or claim whatsoever regarding how the work done in that frame transforms to work done in other frames.
 
  • #268
DaleSpam said:
Static friction does no work in some frame (specifically the frame where the contact patch is at rest).
Its my understanding that in a rolling object situation, the velocity of the contact patch or the average velocity of a point on the surface of a rolling object, is the same as the velocity of the axis of the rolling object, with respect to any inertial frame. In the case of a vehicle with wheels, if the ground is used as a frame of reference, then the contact patches move at the same velocity as the vehicle (wrt ground).

In case it was missed earlier, there is a wiki link for the full scale DDWFTTW vehicle:

http://en.wikipedia.org/wiki/Blackbird_(land_yacht)
 
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  • #269
rcgldr said:
Its my understanding that in a rolling object situation, the velocity of the contact patch or the average velocity of a point on the surface of a rolling object, is the same as the velocity of the axis of the rolling object, with respect to any inertial frame. In the case of a vehicle with wheels, if the ground is used as a frame of reference, then the contact patches move at the same velocity as the vehicle (wrt ground).
In rolling without slipping the velocity of the contact patch is the same as the velocity of the "ground". In the frame of the ground, that is 0. The contact patch is always at rest with respect to the ground, even in a frame where the ground is moving.

That is not usually the same as the average velocity of a point on the rim nor the same as the velocity of the axle.

I think that you are thinking of the velocity that you might assign to the contact patch if it were a single physical object instead of just a label for the part of the wheel touching the ground at any moment. I don't know a name for that.
 
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  • #270
DaleSpam said:
In rolling without slipping the velocity of the contact patch is the same as the velocity of the "ground". In the frame of the ground, that is 0.
In articles about vehicle dynamics (try a web search for "vehicle dynamics contact patch"), contact patch is often used as a dynamic term, and in this context, the contact patch moves with the vehicle (it has the same velocity wrt ground). There are also statements made about how the tread surface in a tire deforms as it flows through the contact patch, yet another dynamic usage of the term contact patch.

Regardless of the terminology, coexistent with a non zero torque, the static friction force between tire and road can perform work. The amount of work performed and what that work is performed on on depends on the frame of reference (ground, vehicle, air, ... ), or the relative velocity between vehicle and ground can be used to eliminate frame of reference issues. In the case of a DDWFTTW vehicle, as previously mentioned, the backwards static force exerted by the ground onto the tires times the relative velocity between the vehicle and the ground (minus losses in the drive train) equals the power that drives the propeller.
 
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  • #271
rcgldr said:
In articles about vehicle dynamics (try a web search for "vehicle dynamics contact patch"), contact patch is often used as a dynamic term, and in this context, the contact patch moves with the vehicle (it has the same velocity wrt ground). There are also statements made about how the tread surface in a tire deforms as it flows through the contact patch, yet another dynamic usage of the term contact patch.
OK, that makes the terminology confusing, but that happens sometimes. The important velocity for determining work is the velocity of the material in the contact patch (which is equal to the velocity of the ground assuming rolling without slipping), not the velocity of the "vehicle dynamics contact patch" (which is equal to the velocity of the vehicle under standard assumptions).
rcgldr said:
the static friction force between tire and road can perform work
Yes, in a frame where the ground is moving.
 
  • #272
I think that it is time to close the thread permanently. I tried to keep it open to support our good members, but it seems to be full of physics misunderstandings which I think are too muddled now to continue.
 
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