I DDWFTTW: Looking for the least confusing explanation

[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.

 

[A.T.]

The cable is moving relative to both ends ...

The question asks about the rest frame of the straight part of the cable. In that frame the material of the straight part of the cable does not move, and cannot transfer any mechanical energy. Invoking motion relative to some points, which are not at rest in the chosen frame, simply fails to address the question.

If the cable were to be stretched, that would constitute work / lost energy.

For the purpose of the question you can assume the cable is not being stretched.

Have you missed the hint I gave you in my previous post?

 

[jbriggs444]

That could apply to any rigid part of any mechanism, though.

Not exactly.

A rigid, non-rotating, non-accelerating part cannot be a net source of work. Nor can it be a net sink of work. If that rigid part is moving, it can be a source of work in one place and a sink of work in another. But the net will still come out to zero.

A rigid, non-rotating, non-accelerating part that is at rest cannot be a source or sink of work at all. Not only is the net zero, the work done by/on the part at every interface must be zero (*).

The latter point is, perhaps, the one you meant to make. It is certainly true. I hear no one saying anything different.

(*) If there is relative motion at an interface, work may be done on the target object despite our part not being in motion. For instance, friction can drain kinetic energy. Or force-at-a-distance interactions such as gravity or electrostatic repulsion/attraction can add or remove kinetic energy. One would normally attribute such gains or losses to the interaction rather than to our motionless part.

Tension in the cable is the same value all along it. The cable is moving relative to both ends

As @A.T. points out, this is relevant to the work done or energy transferred in other frames of reference. In those frames, the cable does zero net work but does act as a conduit through which mechanical energy can flow.

One way to trace mechanical energy flow is to draw a cross-sectional boundary through an object across which you suspect mechanical energy might be flowing. Consider the stress tensor for the object at a point on this boundary. Multiply it by an incremental directed area along the cross section at that point. This will give you a force vector. Multiply by the local [frame-relative] velocity. This will give you a rate of power transfer at that point. Take the surface integral of this quantity across the cross section. That will give you the total power being carried by the object across the cross section.

For a cable at rest you trivially get zero power transferred.

 

[sophiecentaur]

For a cable at rest you trivially get zero power transferred.

So you seem to be saying that work is done at one end and at the other end but no work is done at a point on the wire? Is that counter-intuitive? It's something that I could come to terms with. If you look in another frame then work will be done because the cable is not stationary.

 

[jbriggs444]

So you seem to be saying that work is done at one end and at the other end but no work is done at a point on the wire?

No. No one is saying that.

For a wire at rest, no work is done at the one end. No work is done at the other end. No energy is transferred through the wire.

The fact that in some other frame, positive work is done on one end, negative work on the other and energy is transferred through the wire is irrelevant. Work is frame variant. Power is frame variant.

One can find invariants if one knows where to look. The net mechanical energy created by the wire is invariant and is zero. The net mechanical energy created by the motor is invariant and is non-zero.

 

[sophiecentaur]

“For a wire at rest” a wire is always at rest in its own frame, isn’t it?

 

[jbriggs444]

“For a wire at rest” a wire is always at rest in its own frame, isn’t it?

Yes. And in motion in other frames. So?

You realize that we can contemplate a wire without adopting its rest frame, right?

 

[sophiecentaur]

The net mechanical energy created by the wire is invariant and is zero.

The word 'net' seems to be the relevant thing to resolve an apparent problem.

No. No one is saying that.

For a wire at rest, no work is done at the one end. No work is done at the other end. No energy is transferred through the wire.I think you must be leaving out . . . .

The fact that in some other frame, positive work is done on one end, negative work on the other and energy is transferred through the wire is irrelevant.

You say that no work is done at the ends but energy is transferred on to and off the wire because the motor's energy is spinning the propellor. To resolve an apparent paradox, there needs to be a more rigorous definition of work . The winch causes tension on the wire and there is motion of the drum along the wire. That implies (is?) work.
It is pointless to think that nothing is transferred along the wire. The energy involved is known so how is it just "irellevant"?

 

[Ibix]

I think the point is that the motor is not only connected to the wire. In the frame where the ground is moving and the wire is not only one of the third law pair of forces is doing work, and it is not the one on the wire.

 

[sophiecentaur]

I think the point is that the motor is not only connected to the wire. In the frame where the ground is moving and the wire is not only one of the third law pair of forces is doing work, nd it is not the one on the wire.

I think that is the most useful (and brief) contribution towards clearing up the problem. The ground moves and so does the wire on the drum.

 

[A.T.]

.... because the motor's energy is spinning the propellor.

Wrong. We are using the rest frame of the wire, were the wire cannot transfer any mechanical energy. In that frame all the energy the torpedo gets comes from the moving water.

The winch causes tension on the wire and there is motion of the drum along the wire. That implies (is?) work.

No. If the wire material doesn't move, no force is doing work on it.

It is pointless to think that nothing is transferred along the wire.

It is correct in the rest frame of the wire. You just fail to stick to that frame.

 

[A.T.]

I think that is the most useful (and brief) contribution towards clearing up the problem. The ground moves and so does the wire on the drum.

The wire on the drum is not where you will find the solution. Think about the forces exerted by the whole motor assembly on other objects. Aside of the wire, what is the motor assembly exerting forces on? Are those forces doing work?

 

[jbriggs444]

You say that no work is done at the ends but energy is transferred on to and off the wire because the motor's energy is spinning the propellor.

In the rest frame of the cable, no work is done at the ends.
In the rest frame of the cable, the motor's energy is not spinning the propellor. As I recall, you've been asked where the motor's energy is going from the perspective of this frame and have been given some pretty broad hints. But you've never answered.
In the rest frame of the cable, energy from the flowing water is spinning the propellor and providing the propulsive thrust for the torpedo.

In the rest frame of the ground, work is being done by the motor on one end of the cable.
In the rest frame of the ground, power is being transmitted through the cable.
In the rest frame of the ground, work is being done by the cable on the torpedo's propellor.

Pick a frame. Any frame. But stick to it. Don't frame jump without fair warning and appropriate transformation rules. Your statement quoted above involves a frame jump.

A claim that "no energy is transferred" is frame-relative.
A claim that the motor's energy is spinning the propellor is frame-relative.
Those two claims are each true for one frame and false for another. There is no single frame where both are correct. The contradiction you wish to claim does not exist.

 

[sophiecentaur]

In the rest frame of the cable, no work is done at the ends.

A claim that the motor's energy is spinning the propellor is frame-relative.

That is not what I said (or at least meant to say). Work is done but not 'on' the ends if the 'ends' of the wire are taken to be where the wire enters or exits the moving drums. At each end of the wire there is force and there is motion (but not motion of the string). This is just the same as when a car drives up hill or accelerates. Earth stays still but the wheels turn and work is done. You have to allow work to be done somewhere or energy can't be transferred.
This justifies my worry that the 'on or by' adverbs are often a needless confusion in this sort of scenario.

But you've never answered.

Not broad enough, I'm afraid.

Don't frame jump without fair warning

I appreciate that but I have only allowed the drums to move relative to the wire so how am I frame jumping?

A claim that "no energy is transferred" is frame-relative.

I could be struggling with this because there is a causal chain involved with coal being burned on shore and the torpedo is being pulled along through the water. The on shore drum is moving along the wire and the torpedo is moving along the wire. Merely stating a fact about use of frames can hardly change that - can it?

 

[jbriggs444]

That is not what I said (or at least meant to say). Work is done but not 'on' the ends if the 'ends' of the wire are taken to be where the wire enters or exits the moving drums. At each end of the wire there is force and there is motion (but not motion of the string).

This much is not quite incorrect. Just misguided.

This is just the same as when a car drives up hill or accelerates. Earth stays still but the wheels turn and work is done.

This part is flatly wrong.

In the rest frame of the Earth (or, equivalently, the rest frame of the contact patch on the tires) zero work is done by the Earth on the tires. And zero work is done by the tires on the Earth.

The wheels turn, yes. But that does not affect the work done across the interface between tires and road. The work done across that interface is zero.

If you want to look for work done, you will need to look more closely. The wheel is exerting forward force on the frame of the car. Work is being done there. Meanwhile, the axle is exerting torque on the wheel. Work is being done there. Ideally, the net work done on the wheel is zero. In the real world, rolling resistance causes the wheel to dissipate mechanical energy internally, so really, the wheel absorbs positive net work. That is an invariant.

We could chase the energy flow through the drive train to the engine. The engine exerts torque on the drive shaft doing work across that interface. The countering torque is from the motor mounts which do not rotate. So no work is done there. The motor is a net energy source. This is an invariant.

You have to allow work to be done somewhere or energy can't be transferred.

But in the rest frame of the cable, energy is not transferred. Nor is work done on or by the cable.

Work is done somewhere in this frame. I know where. @A.T. knows where. We want you to figure it out.

This justifies my worry that the 'on or by' adverbs are often a needless confusion in this sort of scenario.

The supposed justification is lacking.

I will grant you that those adverbs are often unnecessary and are eliminated for brevity. For instance, if a team of horses is pulling a plow through a field and the student is asked for the work is done over the length of a furrow, we are clearly asking for the work done by team on plow.

It would not be amiss if the student multiplied the furrow length by its cross-section, multiplied by the height to which the soil had been raised above its prior resting place and then multiplied by the local acceleration of gravity. That would give the work done by plow on soil and would miss the energy dissipated by friction between plowshare and earth.

I appreciate that but I have only allowed the drums to move relative to the wire so how am I frame jumping?

Do you understand how work is defined? It appears that you do not.

Work is the vector dot product of the force across an interface and the displacement of the material of the target object at the interface.

In the case of work done by Earth on tires, the tire material at the contact patch has zero displacement parallel to the road. Zero work is done by road on tires.

In the case of work done by tires on Earth, the road material at the contact footprint has zero displacement parallel to the road. Zero work is done by tires on road.

In the case of work done by drum on wire, the wire material at the contact point(s) has zero displacement parallel to the wire. Zero work is done by drum on wire.

In the case of work done by wire on drum, the drum material at the contact point(s) has zero displacement parallel to the circumference of the drum. Zero work is done by wire on drum.

In the rest frame of the wire it is a simple fact that the drum does zero work on the wire.

I could be struggling with this because there is a causal chain involved with coal being burned on shore and the torpedo is being pulled along through the water. The on shore drum is moving along the wire and the torpedo is moving along the wire. Merely stating a fact about use of frames can hardly change that - can it?

Nothing that anyone has said here denies the causal chain.

We are looking at energy flows, not causation chains or information flows. They are not at all the same thing.

Energy flows are frame relative. We could talk about momentum flows, angular momentum flows or information flows instead. But we should be clear on which one we are analyzing.

 

[sophiecentaur]

Work is the vector dot product of the force across an interface and the displacement of the material of the target object at the interface.

Thanks for your time; it must have been frustrating.It's all down to having a consistent definition of work. I won't waste more of your time looking for 'loopholes'. It makes more and more sense now.