DDWFTTW: Looking for the least confusing explanation

[A.T.]

Derek has posted the entire discussion with Kusenko that preceded the wager. I'm linking to the timepoint, where Derek asks him about connecting two boats with downwind velocity component > windspeed (which Kusenko accepts as achievable by a boat), and thus creating a craft that goes DDWFTTW. Kusenko just waves his hands, and claims this has no relation to the propeller vehicle, because there is no wheel in the sailboat analogy.



Here is an animation that explains how the coupling to the wheels forces each section of the propeller blade to move across the wind:



Just like the keel forces the sail to move across the wind when downwind velocity component > windspeed:



I posted the vectors for both situations here:
https://www.physicsforums.com/threa...ast-confusing-explanation.896869/post-6499006

 

[rcgldr]

I've been searching in vane for the parts list people were using to build the treadmill models. Was it included in one of the treadmill test videos, or was it included here?

 

[A.T.]

I've been searching in vane for the parts list people were using to build the treadmill models. Was it included in one of the treadmill test videos, or was it included here?

There is a part list in the video description here:

 

[pbuk]

I've been searching in vane

Freudian slip?

 

[rcgldr]

There is a part list in the video description here.

I meant the old ones from 10 years ago. No 3d printing required, and they were lighter. Several of these were made by different people here at Physics Forums, and a parts list posted somewhere.



"searching in vane" - this pun was caught a bit sooner that I thought it would be.

 

[A.T.]

I meant the old ones from 10 years ago. No 3d printing required, and they were lighter.

Here

 

[anorlunda]

I just read those threads from 2008. Wow! There were a lot of posts back then. Most of the posts in this thread repeat points made 2 or 3 times before.

 

[rcgldr]

Here

Thanks.

 

[OmCheeto]

No, I think we're all good and in agreement. Device works, Propeller explanation is good.

Well, I'm not in agreement.
The device does not work.
Even Kusenko said so.

 

[hutchphd]

All we have is definitive theoretical and experimental evidence. Seems pretty weak to me.

 

[Merlin3189]

[Moderator Note -- sub-thread merged into this main thread]

Having seen some discussion of FTTWDDW in the past on PF, I thought people might be interested in Veritasium's latest videos:
Original (land car) experiment
Enhanced "proof" to settle a $10k bet with physics prof
No doubt it's a bit light for PF, but the second one certainly convinced me he could be right.

Edit: Apologies for this post. I did search for this thread, but did not find it, perhaps because I used the wrong acronym. But now that I've looked at recent posts, I can see you've already discussed it and some of my searches should have found those posts?

Thanks to Mods for moving it, but they are welcome to delete it since it adds nothing to what's gone before.

 

[.Scott]

You could use this to power an airplane - until it left the ground.

 

[DaveE]

Plus this one: How to build your own.

 

[Ivan Seeking]

It seems to me that the reason the math blows up in the solutions discussed in the Veritasium videos as well as the physics papers referenced is that everyone ignores the increased velocity of the air due to the fan. They use the speed of the air through the fan as V-W and therefore the fan power as F(V-W), where V is the speed of the car relative to the ground and W is the speed of the wind relative to the ground. But they ignore the delta V induced by the fan motion. It should be V + delta V - W. Add this and you no longer have the divide by zero problem without any need for complex fan efficiency equations. And you have the correct power when the wind speed matches the vehicle speed.

The increased velocity of the air molecules then manifests as increased pressure behind the fan.

Right?

 

[anorlunda]

It seems to me that the reason the math blows up in the solutions discussed in the Veritasium videos

Have you read this entire thread? Have you watched this video?

 

[Ivan Seeking]

Have you read this entire thread? Have you watched this video?

Yes I was quoting their equation that fails.

Derek justifies it by referencing a more complex equation needed for the prop efficiency but that doesn't seem to be necessary to make sense of things. It seems to me that they have missed a key term - the energy added to the air by the fan.

 

[jbriggs444]

Yes I was quoting their equation that fails.

Derek justifies it by referencing a more complex equation needed for the prop efficiency but that doesn't seem to be necessary to make sense of things. It seems to me that they have missed a key term - the energy added to the air by the fan.

If one hypothesizes (as the simplistic equation requires) an arbitrarily large propeller rotating at an arbitrarily small rotation rate with an arbitrarily large torque from a transmission with an arbitrarily high gear ratio then an arbitrarily large thrust can be obtained, even without 100% efficiency.

The larger you make the prop, the larger the deflected air mass becomes, the lower the imparted delta V needs to be and the problem with energy loss due to induced drag can be made as small as one pleases.

[Obviously, there are practical limitations with material rigidity and the finite depth of the atmosphere]

For any fixed finite propeller size, infinite torque means zero rotation rate and you get zero propulsion from an infinite gear ratio.

 

[rcgldr]

It should be V + delta V - W. Add this and you no longer have the divide by zero problem without any need for complex fan efficiency equations.

That just shifts the divide by zero problem to some speed slightly less than wind speed, where V + delta V - W = 0.

 

[A.T.]

It seems to me that the reason the math blows up in the solutions discussed in the Veritasium videos as well as the physics papers referenced is that everyone ignores the increased velocity of the air due to the fan. They use the speed of the air through the fan as V-W and therefore the fan power as F(V-W), where V is the speed of the car relative to the ground and W is the speed of the wind relative to the ground. But they ignore the delta V induced by the fan motion. It should be V + delta V - W. Add this and you no longer have the divide by zero problem without any need for complex fan efficiency equations.

As @rcgldr notes, it doesn't help at all, because you just have another point where the airspeed you use is zero. And additionally you now need to know delta V. The math doesn't blow up at this point, when you include the propeller efficiency, which also goes to zero, when the airspeed does.

But all this is not needed to show that steady state above wind-speed is possible, regardless how it got there. For this the simple idealized formula that the video shows is enough, since it shows excess thrust, which can account for the loses.

The static thrust estimation is only needed to predict the acceleration phase. But if you want to do that, then you also have to consider the propeller working in reversed flow below wind speed, basically with negative efficiency.

 

[Ivan Seeking]

But all this is not needed to show that steady state above wind-speed is possible, regardless how it got there. For this the simple idealized formula that the video shows is enough, since it shows excess thrust, which can account for the loses.

The static thrust estimation is only needed to predict the acceleration phase. But if you want to do that, then you also have to consider the propeller working in reversed flow below wind speed, basically with negative efficiency.

But the equation is wrong. Where does it show the energy added to the air moving through the prop area?

 

[A.T.]

But the equation is wrong. Where does it show the energy added to the air moving through the prop area?

See the post by @jbriggs444. It's an ideal (limiting case) where Δv is negligible. The rate of work done on the air then approaches: F * v.

 

[Keith_McClary]

Link

 

[A.T.]

If there is actual wind relative to the ground, the above would work. Aside of the directly downwind record of 2.8 x windspeed, the Blackbird also established a record for going directly upwind, at 2.1 x windspeed.

 

[Swamp Thing]

The OP was looking for the easiest way to demystify this concept. Here's my attempt... It's a bit long, but each step is hopefully easy and convincing.

Take something like the old-time Dutch windmills, with long, narrow cloth sails making up the vanes. Mount it on a wheeled platform, with the windmill free to turn. Park it facing downwind with the brakes on. The wind will turn the blades, let's say, clockwise.

Now add a ratchet mechanism that allows only anticlockwise rotation. So the blades are locked as far as the wind is concerned. They now act as plain old sails, and try to push the cart forwards. Take off the brakes, and let it catch up with the wind (which it will do, absent friction).

Now park somewhere and add a generator to the wheels, and use it to charge a battery. Again let the cart run with the wind. It won't quite catch up with the wind, but it could get pretty close if the gear ratio between the wheels and generator is high enough, such that the wheels are hardly loaded.

Note that, without friction, the cart can get arbitrarily close to the wind velocity at arbitrarily high wind speeds -- it all depends on the gear ratio. So it can generate arbitrarily high power as well, in theory, without friction.

Next step, replace the battery with a motor. Again let the cart get very close to wind velocity (as in the previous paragraph) and use the motor to turn the windmill anticlockwise, which the ratchet will freely allow. Now the windmill will push back against the air and cause the cart to speed up. Given a proper blade angle and/or gear ratio between the motor and windmill, this can more than make up the lag caused by the generator loading the wheels. If you find this isn't the case, just lighten that load by further increasing the wheel-to-generator ratio.

Finally, replace the electrical energy conversions with equivalent mechanical ones.

It remains to convince ourselves that a small amount of friction won't break the above scheme. This is reasonable because, as stated, we can generate plenty of energy given a high enough wind speed. Remember that the Dutch style vanes make pretty good sails, delivering plenty of power per unit of relative wind speed, by forcing the wheels to turn the generator.

 

[A.T.]

The OP was looking for the easiest way to demystify this concept. Here's my attempt... It's a bit long, but each step is hopefully easy and convincing.

I think your core point can be made much simpler, and is demonstrated in the video below.

with the windmill free to turn.

This is shown in the video around 4:00.

Now add a ratchet mechanism

We can skip that and go directly to the rotor linked to the wheels such that they spin it opposite to a free spin windmill mode, when the cart rolls forward. This is shown in the video around 6:00.



Once you realize, that with the right gearing, the cart can be pushed downwind, while the rotor is spun by the wheels against the aerodynamic torque, it is clear that it can still produce thrust at windspeed, in zero relative wind. So passing windspeed is just a a matter of efficiency.

 

[Crispin Miller]

I've avoided these discussions thus far - I find them confusing force and energy, and the use of a boating analogy to people who have never sailed (I know that's not you, @anorlunda ) isn't helpful. I also think treadmill analogies are not helpful either.

I maintain that thinking of this in terms of wind energy is unhelpful, because with an infinite volume of air, there's an infinite amount of energy that can be extracted from it. (In real life, replace "infinite" with "very large")

1. Consider a cart with a sail, going exactly as fast as the wind. Do the wheels have to be frictionless? The answer is no, the requirement is that the thrust from the wind is equal to the drag from air plus from the wheels. If I have more wheel friction, I need a larger sail, to be sure. Agree?

If you think you understand, answer this: You are riding in this cart at the speed of the wind, and the brakes are partially engaged. What happens when you release the brakes? The answer is - nothing.

2. Consider a cart with a sail, going exactly as fast as the wind. A battery-powered fan is mounted on it, to move air from the front to the rear, but avoiding the sail. The fan is switched on. What happens to the cart? It accelerates forward. Agree?

Our cart is now going downwind faster than the wind. No problem here - it has its own power source.

3. Now we put the two together. The wheel brakes are taken from a Toyota Prius, and so generate electricity. I remove the battery and use this electricity to power my electric fan. And there we go.

Of course we can then replace the electrical system with a purely mechanical one.

I agree with your thought experiment in (2) and (3), so I recommend abandoning scenario (1) because it's unnecessary and no, I don't agree that the sail in it can overcome wheel drag. A sail moving directly downwind can function only as a (horizontal) parachute (i.e., not as an airfoil), and the only way it can generate thrust is to exert drag on air which is outrunning it. If it's moving at the speed of the wind, it's hanging slack and you get no thrust from it.

 

[rcgldr]

A simple explanation. Assume cart is moving at the same speed or slightly faster than the true wind. Use the cart's current velocity as the frame of reference. There is a newton third law pair of forces at the wheels, the earth exerts a backwards force onto the wheels, coexistent with the wheels exerting a forwards force onto the earth, slowing down the earth's relative speed by a tiny amount, extracting energy from the earth. There is another Newton third pair law of forces at the propeller, the propeller exerts a backwards force onto the air, adding energy to the air (but less than the energy extracted from the earth), coexistent with the air exerting a forwards force onto the propeller. There is reduction gearing that decreases speed and increases force at the propeller, so the force at the propeller is greater than the force at the wheels, allowing the cart to accelerate until the forces equalize at some speed.

The key factor here is with a tailwind, from the cart's perspective, the relative air speed is less than the earth speed, allowing for reduction gearing, since the propeller interacts with the relatively slower moving air, while the wheels interact with the relative faster moving earth.

 

[Crispin Miller]

A simple explanation. Assume cart is moving at the same speed or slightly faster than the true wind. Use the cart's current velocity as the frame of reference. There is a newton third law pair of forces at the wheels, the earth exerts a backwards force onto the wheels, coexistent with the wheels exerting a forwards force onto the earth, slowing down the earth's relative speed by a tiny amount, extracting energy from the earth. There is another Newton third pair law of forces at the propeller, the propeller exerts a backwards force onto the air, coexistent with the air exerting a forwards force onto the propeller. There is reduction gearing that decreases speed and increase force at the propeller, so the force at the propeller is greater than the force at the wheels, allowing the cart to accelerate until the forces equalize at some speed.

The key factor here is with a tailwind, from the cart's perspective, the relative air speed is less than the earth speed, allowing for reduction gearing, since the propeller interacts with the relatively slower moving air, while the wheels interact with the relative faster moving earth.

Sure -- and I've expounded more or less the same thing elsewhere to friends by email. Wherein the fundamental point is that the propellor is driven by the wheels and is never a turbine at any stage of the run. For the less-than-wind-speed runup, its pitch faces the wrong direction. And for FTTW operation, Kusenko's claim that it can function simultaneously as a propellor and turbine is simply hocus-pocus and a failure to examine the force balance on the blades. You can't consume shaft work and collect it both at once. Also a failure to consider the blade's angle of attack, which is opposite to what turbine operation would require.

I think Vanadium50's points 2 and 3 are just an electrically-mediated (therefore a tad cumbersome) scenario for the same wheels-to-prop power transfer you describe, and I agreed with those two points. My quibble was just with the unnecessary and fallacious initial scenario that posited a sail moving (DDW) at full wind speed in spite of a drag load on it.

 

[ridgerunner]

I've written a paper that applies to an idealized DDWFTTW vehicle traveling at wind speed, i.e. the special case where a toy cart is hovering in a stationary position on a treadmill. This analysis assumes that the cart has a 100% efficient transmission connecting the rear wheels to the propeller such that the power coming in at the wheels is equal to the power going out at the propeller. Because the treadmill belt under the wheels is moving faster than the air passing through the propeller, the forward thrust force acting on the propeller can be greater than the retarding force acting on the rear wheels. For a real world cart, this unbalanced net forward force is used to overcome inefficiencies and to accelerate the cart faster than the wind. However, for this analysis, a steady state condition is assumed, where the forces acting on the cart are in equilibrium, and the cart travels at a constant speed equal to the speed of the wind.

To achieve force equilibrium and steady speed, an idealized regenerative braking system is attached to the front wheels, where a balancing rearwards force is being applied, and this force is used to run a 100% efficient generator that harvests useful power from the "wind". As it turns out, the maximum net power that can be harvested from such a vehicle is the same as that for a stationary turbine, and this number is equal to the Betz limit, i.e. 16/27. In honor of the designer of the Blackbird, Rick Cavallaro, aka "Spork", I christened this upper bound limit for a hovering treadmill cart power harvesting machine as: "The Spork Limit."

Note that the derivation provided in this paper very closely parallels the derivation of the Betz limit. To illustrate the close parallels, I have written a companion paper that derives the Betz limit. Both papers are written in a "GIVEN:/FIND:/SOLUTION:" engineering text example problem format. Here are links to the PDF files hosted on my server:

BETZ LIMIT DERIVATION
SPORK LIMIT DERIVATION

Note that, like the Betz limit, this "Spork Limit" number represents a theoretical upper bound limit to the amount of power that could be extracted from an idealized wind power harvesting propeller cart traveling at wind speed; a real world device could never actually achieve this performance.

I've also been working on a more generalized paper, which analyzes an idealized Blackbird vehicle traveling both upwind, (using turbine blades,) and downwind, (using propeller blades,) at any speed, but this paper is not yet finished. (I may eventually submit this to a peer reviewed journal.) Here is a summary of the general equations from that paper that I claim apply to an idealized Blackbird:

JMR EQUATIONS

Interestingly, for the propeller cart, the math indicates that the optimal ΔV change in air velocity caused by the propeller, (to maximize the net forward force,) is greater than the wind speed, (so that to an observer on the ground, the wake behind the vehicle is moving upwind.)

Cheers,
Jeff Roberson

 

[A.T.]

Interestingly, for the propeller cart, the math indicates that the optimal ΔV change in air velocity caused by the propeller, (to maximize the net forward force,) is greater than the wind speed, (so that to an observer on the ground, the wake behind the vehicle is moving upwind.)

This is indeed a nice counter-intuitive result, because it means that (for the propeller downwind cart) to maximize acceleration (and wind energy extraction rate), you would not only stop the air relative to the ground (extract all energy from true wind), but actually then make it move in the opposite direction, upwind (thus put energy back into true wind).

I was skeptical about this result at first, but then it was correctly pointed out to me, that this allows to process more air per time. And in terms of energy per time, this outweighs that we extract less energy per air mass.

However, this derivation has some caveats, when applied to the real world: A real propeller cart will have two types of losses:
1) Losses that do not depend on ΔV (e.g. hull drag, rolling resistance)
2) Losses that depend on ΔV (e.g. propeller swirl losses, transmission losses)
How close the optimal ΔV derived here is to reality depends on the ratio of type 2 to type 1 losses.

It should also be noted, that this approach doesn't give us the maximal achievable speed, which is only limited by losses. Therefore the idealization employed here doesn't have an upper speed limit, but it gives us the ΔV for maximal acceleration or maximal energy harvesting at constant speed, under idealized assumptions.

 

[ridgerunner]

A.T.,
Thank you for the response. I greatly respect your opinion. Indeed, you were the first person to respond to my Spork limit derivation when I first presented it, (in rough, handwritten form,) nearly three years ago. In your response at that time, you stated: "I have not checked the math. ..." I now have a couple questions for you:

1. Have you taken the time to read my recent, more formalized paper?
2. If so, did you find any errors?

Thanks again.

 

[ridgerunner]

For the sake of further discussion on this topic, allow me to present a working example problem which illustrates the principles at play. We'll use nice small integer values for all relevant parameters to simplify the needed calculations. The following example represents a DDW propeller cart operating at precisely wind speed, just like a hovering treadmill cart:

Example 1: (ΔV = 4/3 Vwind)

Given:
An idealized, optimized DDW cart is pushed up to wind speed on a flat, level surface, with:
ρ = 1 kg/m^3 = density of air,
S = 18 m^2 = swept area of rotor disk,
Vwind = 3 m/s = velocity of the wind,
Vcart = Vwind = 3 m/s = velocity of the cart, and
ΔV = 4/3 Vwind = 4 m/s = optimal change in air velocity caused by propeller (for n = 1).

Find:
Immediately after release, compute (from the cart reference frame):
Ft = force of air acting on prop (forward direction),
Pout = Pra = power transferred from the rotor into the air,
Pin = Pgw = power transferred from the ground into the wheel,
Fg = force of ground acting on wheel (reverse direction), and
Fnet = Ft - Fg = net forward force acting on cart.

Solution:
V1 = 0 m/s = air velocity far upstream
V2 = V1 + ΔV = 4 m/s = air velocity far downstream
V = 1/2 * (V1 + V2) = 1/2 * (0 + 4) = 2 m/s= air velocity at the rotor disk
ṁ = ρ*S*V = 1 * 18 * 2 = 36 kg/s
Ft = ṁ * ΔV = 36 * 4 = 144 N
Pout = Pra = Ft * V = 144 * 2 = 288 W
Pin = Pgw = Fg * Vcart
Pin = Pout
Fg * Vcart = Pra
Fg = Pra / Vcart = 288 / 3 = 96 N
Fnet = Ft - Fg = 144 - 96 = 48 N

Extra credit (from the ground reference frame):
Pnet = Fnet * Vcart = 48 * 3 = 144 W
Pwind = 1/2 * ρ * S * Vwind^3 = 1/2 * 1 * 18 * 3^3 = 243 W
Pnet = Pwind * CPnet
CPnet = Pnet / Pwind = 144 / 243 = 16/27 = Spork Limit!

Notes:
To an observer on the ground, the flow behind the cart is indeed reversed and is blowing in the upwind direction at 1 m/s, but so what? To an observer on the cart, the propeller is simply accelerating still air up to 4 m/s; is there anything wrong with that? And from the cart frame, the velocity of the air passing through the rotor disk, 2 m/s, is less than the velocity of the ground passing under the cart, 3 m/s, so the thrust force of the air on the rotor, 144 N, can be greater than the force of the ground on the wheels, 96 N, resulting in a net forward force of 48 N. As far as I can tell, no laws of physics were being harmed in the making of this DDW example.

Furthermore, if one runs this example problem using a value of ΔV other than 4/3 Vwind, the net force and power extracted will both have lesser values. Some might argue that slowing the air down to a stop is the best that can be done, i.e. ΔV = Vwind. Let's go ahead and run the numbers for that case:


Example 2: (ΔV = Vwind)

Given: (ΔV = Vwind)
An idealized DDW cart is pushed up to wind speed on a flat, level surface, with:
ρ = 1 kg/m^3 = density of air,
S = 18 m^2 = swept area of rotor disk,
Vwind = 3 m/s = velocity of the wind,
Vcart = Vwind = 3 m/s = velocity of the cart, and
ΔV = Vwind = 3 m/s = non-optimal change in air velocity caused by propeller (for n = 1).

Find: (ΔV = Vwind)
Immediately after release, compute (from the cart reference frame):
Ft = force of air acting on prop (forward direction),
Pout = Pra = power transferred from the rotor into the air,
Pin = Pgw = power transferred from the ground into the wheel,
Fg = force of ground acting on wheel (reverse direction), and
Fnet = Ft - Fg = net forward force acting on cart.

Solution: (ΔV = Vwind)
V1 = 0 m/s = air velocity far upstream
V2 = V1 + ΔV = 3 m/s = air velocity far downstream
V = 1/2 * (V1 + V2) = 1/2 * (0 + 3) = 1.5 m/s = air velocity at the rotor disk
ṁ = ρ*S*V = 1 * 18 * 1.5 = 27 kg/s
Ft = ṁ * ΔV = 27 * 3 = 81 N
Pout = Pra = Ft * V = 81 * 1.5 = 121.5 W
Pin = Pgw = Fg * Vcart
Pin = Pout
Fg * Vcart = Pra
Fg = Pra / Vcart = 121.5 / 3 = 40.5 N
Fnet = Ft - Fg = 81 - 40.5 = 40.5 N

Extra credit (from the ground reference frame):
Pnet = Fnet * Vcart = 40.5 * 3 = 121.5 W
Pwind = 1/2 * ρ * S * Vwind^3 = 1/2 * 1 * 18 * 3^3 = 243 W
Pnet = Pwind * CPnet
CPnet = Pnet / Pwind = 121.5 / 243 = 1/2 < Spork Limit!

Notes:
So it turns out that to extract the maximum power from the wind, (and achieve a maximum net forward force,) the DDW propeller cart must accelerate the air by an amount that is greater than the speed of the wind!

Comments? Corrections?

 

[A.T.]

Steve Mould made a great video on the Brennan torpedo, the probably first practical guided missile, which used a propulsion system based on the same principle as the DDWFTTW carts.



If you would put the Brennan torpedo into a fast river (aimed downstream), and just fixed the cable to the ground (no motor), it would go downstream faster than the stream, powered only by the stream (see the attached diagram). Just like the Blackbird, it is a great example to explain the frame dependence of kinetic energy and mechanical work.

And here is a question to ponder: In case A2 (rest frame of the cable), since the motor cannot power the torpedo via a static cable, where does the energy from the motor go to?

 

[A.T.]

The ground is moving (in the cable frame).

Yes, very good.

Work is done by the motor on the cable because it is 'pulled'.

No. You cannot do work by applying a force to a static object. In the rest frame of the cable no energy is going into or out of the cable.

Work is done on the torpedo due to the reaction of the water on the prop.

Work is done on the torpedo by the moving water (in the cable frame). But that doesn't explain where the energy from the motor goes to. It cannot go to the torpedo via the static cable for the reason above.

Maybe you should think more about the part I quoted first.

 

[sophiecentaur]

No. You cannot do work by applying a force to a static object.

That could apply to any rigid part of any mechanism, though. Tension in the cable is the same value all along it. The cable is moving relative to both ends so you can identify Work at each ends as the torpedo and the drive motor winch rotates. The work in at one end is equal to the work out at the other. If the cable were to be stretched, that would constitute work / lost energy.