Re: OT Here is an example of pseudo science.

[Ronald Raygun]

I'm sure that's what it was meant to do.

But at first it doesn't, because his analysis made the explicit assumpti ## Suppose car speed and wind speed are *both* 55ft/s. The ## wheels are still delivering 1hp. The headwind is zero, so you ## would calculate the power needed by the prop to produce 10lb ## of thrust as 0hp. Can't be right.

His reply was:

# But it is right, because at that point in the calculations we are # still describing the scenario using theoretical, lossless components. # We've defined the transmission method as lossless and the propeller # also as 100% efficient.

That's a good approach, but the prop efficiency stuff ought to have gone into the engineering part, whereas he seems to have it initially in the physics part. It seems to me that the physics analysis would be more illuminating if it allowed us to work with finite values of air throughput.

I have done, and he didn't. See above.

What actually got me going was that he stated the prop's power output would equal thrust times the incoming airspeed. I thought, by considering the work it had to do on the air, that it should be more than that. something like thrust times a speed value somewhere (probably halfway) between the incoming and outgoing air speeds.

If that is wrong, I'd like to know why it's wrong. I'm sure you could just tell me, but please don't just yet, as it'll be better for my soul if I can work it out myself.

Indeed.

It was a minor correction; I call that a quibble. If you understand a quibble to be more of a dispute, that's not what I meant.

OK. Watch this space.

[dennis]

For once I can agree with rick (shock + horror), use the same frame of reference and do the sums.

try this example..

assume a wind of -10 m/s at the cart assume the ground is doing -20 m/s

What does this mean you ask..

well its a cart going at twice the wind speed down wind. Its the same as the wind being 10 m/s and the cart travelling down wind at

20 m/s which I will use for the example as it makes the maths easier. You can use any other figures you like, the conclusions are the same.

This gives a kinetic energy to the relative wind passing the cart of 100 units (-10 x -10) but do notice it is going backwards and is being generated by ricks prop.

Ah look at the energy needed, we are expending energy to make the wind go backwards.

However we are not slowing the real wind at all (it is still -10 m/s) so it is not losing any energy to anywhere

This is impossible. There is no energy input.

Now rick will say the energy comes from the wheels and drives the prop..

so in this steady state the wheels must be providing the 100 units of energy but its not slowing the real wind so there can't be any energy coming from the wind. In actual fact the energy is going into the prop and none is coming back.

Opps that's impossible too you can't just create energy from nowhere ( not even using wheels and props).

so lets increase the speed of the prop to throw more air back and slow the wind to extract energy as rick says he does..

So we increase the prop wind to -11 m/s , the energy used has gone up from

100 to 121 = 21 extra units (using the energy = constant x mass x velocity squared formula and remembering two negatives multiplied are a positive).

So what has happened to the real wind? Yes it has gone from -10 m/s to -11 m/s and its energy has changed by 121-100=21 extra units.

Remember we are using a single frame of reference, the cart.

Do I see a problem here? is 21 greater than 21? no so we have no extra energy extracted from the wind by the cart.

What does it say..

well even with perfect conditions there are no circumstances using any device (even a 100% efficient prop) where you can extract more energy by slowing the wind when the device is moving faster than the wind in the direction of the wind than the energy it takes to do it. (This isn't a surprise to a physicist as its just conservation of energy.)

Well is this what really happens? probably not, after all this is the perfect case, lets look at some of the problems rick appears to avoid in the real world..

To stop the physical impossibility of the air piling up behind the the cart would have to increase speed to maintain the steady state of the wind speed from the prop being the same as the wind, if it doesn't turbulence will occur even with a 100% efficient prop. (The excess air has to go somewhere and the prop has no control over it) there is no energy being input so this can't happen.

To actually overcome losses the prop would have to accelerate the air to a higher speed than in the steady state this results in the pressure build up behind the prop and that causes turbulence as the air trys to even its self out, this loses energy so more energy is needed by the prop, its impossible to get this energy back as it was shown above. there are friction losses, etc.

Are there any values of speed where it works and you can actually extract energy? well yes, if the cart is going slower than the wind you can get energy from the wind, this is not really surprising.

[Doctor Drivel]

A lighter than air balloon is bowing to gravity rather than defying it. Gravity is actually forcing it UP.

[Ronald Raygun]

Do we? It seems obvious enough.

To what wind? There is -10 m/s wind all around the cart, but we're only interested in that part of it which is going through the prop. We need to know how much that is. Suppose you put 6kg of air through it every second. Then the kinetic energy of those 6kg of air before they go through the prop will be (1/2)x(6kg)x(-10m/s)^2 which is 300 kg m^2/s^2, or 300 J.

No, it's there already, it's the headwind resulting from the cart's ground speed being 10 m/s more than the wind's ground speed. What the prop does is make the 10 m/s backwards relative wind go even faster backwards.

Yes, of course we are expending energy to make the relative wind go faster backwards, but how much energy? To what speed are you wanting the prop to accelerate the wind? Suppose you want to accelerate that 6kg of air from -10 m/s to -20 m/s.

This would increase its kinetic energy to 1200 J, which means adding 900 J to the 300 J it already had.

We are slowing both the relative wind ("slowing" in the sense of making its speed more negative, i.e. from -10 m/s to -20 m/s), and also slowing the real wind (from +10 m/s to 0 m/s). But beware, you are switching frames when changing from relative to real wind. Both winds experience a reduction in speed by 10 m/s, but in the two frames, the kinetic energy changes which accompany the speed changes are not the same. In the moving frame it increases by 900 J (from 300 J to 1200 J), in the stopped ground frame it decreases by 300 J (from 300 J to 0 J).

Momentum change rules tell us that if you accelerate 6kg by 10 m/s each second, this will give 60 N of thrust. If you didn't have the propeller and instead had a mule pulling the cart using a string with 60 N tension, without accelerating the cart, and if the cart is moving at 20 m/s, then the mule would be doing 1200 J of work each second (i.e. delivering

1200 W of power). Obviously if you're applying a force to the cart and it isn't accelerating, then the force must be doing something else, and in this case it is operating a generator driven by the wheels. The generator's 1200 W power output could be used to boil an on-board kettle. Or we could divert 900 W of this power to run a motor to drive the prop to provide the thrust to make the mule redundant. The spare 300 W is available for losses and air drag.

Is it a coincidence that the spare 300 J is the same energy as the real wind is giving up?

[ThinAirDesigns]

Doctor Drivel

Of course the lighter than air balloon is going up; it has no choice

-- gravity is forcing it UP.

Lighter than air craft ... slaves to gravity.

[Rick Cavallaro]

That's exactly what it does. *At first* it considers a 100% efficient device - as you suggest. He then looks at the excess energy available and answers the engineering question - "is this enough energy to make it work in the real world?"

You're getting stuck on the notion that a 100% efficient propeller would have to be infinite in size. Why? 100% efficient ball bearings and zero rolling resistance tires can no more easily exist in the real world than a prop of infinite diameter. Exactly how large a prop would you allow in the first part of the proof when he is simply trying to see if a DDWFTTW cart would violate the laws of physics?

In other words, he did precisely what you suggest. He starts with lossless components, determines that no laws of physics would be violated, and then addresses the real-world questions. In the very beginning of the analysis he explains that he's starting with lossless components, but that losses will be dealt with later. There's nothing hidden or tricky about this.

100% efficient components don't exist in the real world. I think he assumes that everyone but Dennis would recognize that.

Yes, and his reply is correct.

The imperfect prop is introduced in the engineering part. The perfect prop is assumed in the physics part. There's no other way to follow the very approach you suggest.

If you're asking whether the very concept of DDWFTTW violates the laws of physics it would be wrong to limit yourself to ANY specific engineering dimensions.

I don't know how we can be seeing different things here - but indeed he does. So I guess we're stuck now at "does too.... does not"

Yes, you made that somewhat arbitrary assumption and stated quite definitively that his analysis was wrong. That's when I made the mistake of thinking that you were channeling Dennis.

In any event, if the ultimate goal is to reach a common understanding, and you feel JB's explanation isn't the best for you, then we've probably spent more than enough time on it for now. Let's move forward with your skateboard approach. But I think we should do it initially with the idea of understanding how this works - not whether JB's analysis was right or appropriate - agreed?

[Ronald Raygun]

I guess.

Yes indeed. And thank you for your patience.

Let us begin with a preliminary simple case without power harvesting, just to see if my analyis of that passes muster.

Here we have a population of skateboarders all of whom weigh (I mean have a mass of) 60kg each, including their boards.

The boards' wheel bearings are lossless. Oh my God, what am I saying? I mean they do have losses but we'll disregard them for now. :-)

They're all moving at 1 m/s in the same direction, except for one (whom we'll call our hero, the others we'll call plebs) whose speed is 2 m/s instead, in the same direction.

Our hero decides he wants to go faster, at 3 m/s, and intends to steal some momentum by grabbing one pleb and pushing him back with a constant force of 60 N for as long as it takes.

Because both objects (hero and pleb) are 60kg in mass each, a force of

60 N will accelerate each of them at a rate of 1 m/s^2, positive for the hero and negative for the pleb, assuming our frame of reference is the fixed ground, relative to which the speeds are as stated above.

It will therefore take one second for the acceleration to achieve the hero's desired speed increase from 2 to 3 m/s.

I'll use M to denote the mass of 60kg, and bare unsuffixed A to denote the constant value 1 m/s^2.

I'll use suffixed A, V, S, and E to denote accelerations, speeds, distances travelled, and kinetic energies of hero or pleb, depending on the suffix h or p. Extra suffixes 0 or 1 denote values at times t=0s or t=1s. All pretty obvious.

We start the experiment at t=0s when the hero grabs the pleb and starts pushing. We end the experiment at t=1s.

So we have:

Ah = A Vh = A t + Vh0 Sh = 1/2 A t^2 + Vh0 t + Sh0 Eh = 1/2 M Vh^2

Ap = -A Vp = -A t + Vp0 Sp = -1/2 A t^2 + Vp0 t + Sp0 Ep = 1/2 M Vp^2

Starting conditions: Sh0 = 0 m, Sp0 = 0 m, Vh0 = 2 m/s, Vp0 = 1 m/s.

We calculate: Vh1 = 3 m/s, Vp1 = 0 m/s, Sh1 = 2.5 m, Sp1 = 0.5 m, Eh0 = 120 J, Eh1 = 270 J, Ep0 = 30 J, Ep1 = 0 J.

In particular, therefore, the sums of host+pleb kinetic energies are

150 J at the beginning and 270 J at the end. The difference of 120 J is accounted for by the work which the hero's arms have done in the pushing. He has applied a constant force of 60 N, and has done so over a distance of 2 m, because that's the difference Sh1-Sp1.

So that's the energies balanced in the ground frame.

I can also do it in a moving frame, for example in that moving at the hero's initial speed, and get the same answer. I can even do it in a frame which accelerates with the hero, and still get the same answer, but it isn't really very interesting, and dennis wouldn't understand the inertial forces adjustment.

Before I move to the next step, are you happy with the analysis so far?

[Rick Cavallaro]

On a quick read-through, yes. I reserve the right to scrutinize it more closely if I find we missed something.

[Ronald Raygun]

OK, then. Let's add deliberate energy-wasting drag.

The hero still exerts 60 N of force on a passing pleb, but there is also drag of 60 N on the skateboard, because he is towing a suitable wheel-less box along the ground behind him which at 2 m/s generates just the right amount of drag.

Alternatively he has installed an electric generator on his board, coupled to the wheels, and this powers a lightbulb. If this is to generate 60 N of drag, while the road is moving at 2 m/s, it must be consuming 120 W of power, or 120 J of energy each second. Enough for two 60 W lightbulbs.

The hero's board now accelerates at zero rate, because the forces of thrust and drag are in balance. His speed stays at 2 m/s.

As before, the pleb is accelerated backwards for 1 second and, relative to the ground, is brought to a stop in a distance of 0.5m.

As before, the pleb loses 30 J of kinetic energy, but the hero's kinetic energy does not change.

This time, the hero's arms are exerting 60 N of force over a distance of only 1.5 m, so they are doing 90 J of work.

The energy balances: 90 J stored energy plus 30 J kinetic energy from the pleb give us the 120 J for the hero's lightbulbs.

Agreed?

[dennis]

No we don't.

Suppose you put ten times that mass it doesn't matter. Its a perfect prop the mass on either side is the same, it doesn't change the mass in the flow so you can ignore it. This is an exercise under perfect conditions just to prove there is no energy gain from slowing the wind if you are travelling faster than the wind which is what it shows.

You just keep switching frames of reference like rick does. Its easy to get free energy doing that sort of thing.

So where is that energy from then if not the prop.

No the real wind is unchanged so you can't get any energy from there. I specifically chose values where the real wind is unchanged for the first part and you omitted the bit where I changed them. It is a steady state and it is there to prove that you get no energy at all from slowing the wind even though you are travelling twice as fast as the wind. We know that this particular case is valid because rick has stated they can go faster than that.

I'm not switching frames at all, that is the point, it is all relative to the cart.

No I am not the speeds are all relative to the cart, there is no switching of frames in what I said.

There is no change in momentum, look at the speeds, they do not change. Not yet anyway they change in the second half you have omitted.

Its the wrong way around even in your description the cart is losing that energy and not having it spare.

Is it a coincidence that you have changed what I said to something that isn't using the same frame of reference just as rick likes to do?

[dennis]

.

After seeing your reply to my post where you changed frames of reference and said I was changing them when I clearly did not I don't think you understand it at all.

You shouldn't be, not until you understand what the same frame of reference means. It might help to understand what physical effects a perfect propeller means too.

[Doctor Drivel]

So if defies gravity.

[Ronald Raygun]

Something doesn't defy gravity just by going up.

Suppose you have a jar half full of syrup. The syrup occupies the bottom half. There is air in the top half.

Now turn the jar upside down. It's OK, the lid is closed. What heppens next is that gravity forces the syrup down and the air up. So they both *obey* gravity. The air does not defy it.

A helium filled balloon goes up because gravity is pulling on the air around and above it more strongly than on the balloon itself.

Put the balloon in a car, and a football on the back seat. Drive round a corner quickly, and the football will shoot off to one side, and the balloon to the other. Both items are obeying, not gravity this time, but a similar phenomenon, inertial acceleration.

[Ronald Raygun]

Don't we? I thought you said we were *giving* kinetic energy to the wind, expending energy. That means we'd have to make it go faster. We can only make air go faster if we push it, and if we make 2kg of air go faster, this will take twice as much energy as making only

1 kg of air go similarly faster. So we do need to know how much air we're making faster.

I assumed, by the way, that you're referring to air which the propeller is making go faster. If, on the other hand you mean the air which the structure of the car is pushing out of its way (kind of like the bow wave of a boat), then that's different. That's just ordinary air drag which we are for the moment teating as neglible.

No you can't. Each kg of air has kinetic energy of 50 J if it's moving at 10 m/s. If you speed it up to 12 m/s, it will have 72 J of energy. So if you're putting 1 kg through per second, you'll be imparting 22 W of power, and if you're putting through 2 kg, it's 44 W, so you need to know both input speed and output speed and also quantity per second.

The prop is not *making* energy, it is *converting* (mechanical) energy from the electric motor which drives it, into kinetic energy which it is adding to the wind's. The electric motor in turn derives its electric energy from the generator, which converts mechanical energy from the wheels into electricity. The mechanical energy which drives the wheels essentially comes from the thrust generated by the propeller, and it is this fact which initially makes people think that something is being got for nothing, that the *energy* for creating the thrust comes from the thrust itself.

That's why it's probably easier to work with the skateboarder.

[Ronald Raygun]

I didn't change frames. I did the analysis in just the one frame, with all measurements relative to the ground. I showed the workings, which confirmed that the energies at the beginning add up to the same as at the end.

Then I also did the whole analysis of the same situation in a different frame, and then in a third one too, and got the energies to balance there too, but I merely reported this without showing the workings.

Did you see my subsequent post in which the hero is able to power a 120 W lightbulb by using only 90 W of muscle power, plus the spare 30W harvested from slowing down the plebs? What did you think of it?

I hope I'm not speaking too soon, as that post has not yet passed Rick's inspection.

[Ronald Raygun]

Assuming Rick approves this, the next step is trivial. We replace the hero's arm with an electric robot arm programmed to push the pleb. We plug this robot arm into the generator instead of one of the two

60W lightbulbs, and replace the other with a 30W bulb. That way the arm gets the same 90W as the hero's arm had.

Now, how does this all work if we want to look at it, as dennis seems to, entirely from a frame of reference moving with the hero.

In the hero's frame, obviously Vh and Sh and Eh are zero.

Ap = -A Vp = -A t + Vp0 Sp = -1/2 A t^2 + Vp0*t Ep = 1/2 M Vp^2

Initial condition: Vp0 = -1 m/s

Reminder: A = 1 m/s^2 M = 60kg

Calculate: Vp1 = -2 m/s Sp1 = -1.5 m Ep0 = 30 J Ep1 = 120 J Arm work = 90 J

Now we have a problem getting the energies to balance. The hero has done

90 J work, but now instead of gaining 30 J from the pleb, we're losing 90 J to him. Dear oh dear oh dear, our energy budget has been completely depleted to zero and there's nothing left for the lightbulb.

There's something wrong!

No there isn't.

It's easy to forget what we took for granted when our reference frame was the ground. Notice how once we moved the frame to the hero, his own acceleration, speed, distance, and kinetic energy are all zero, as were those of the ground before we moved. Well, now that the frame is moving with the hero, the ground's motions are of interest.

The wheel drag is a force which the ground is exerting on the hero's wheels, but of course the same force is pushing back against the ground, moving it forward, i.e. tending to reduce the relative speed between the two. The ground is hugely massive, of course, but we do slow it down ever so slightly, so we need to take into account the way its kinetic energy changes.

The Earth is initially moving at speed Ve0 (equal to -2 m/s which we'll denote by unsuffixed V) in the hero's frame. We put a nominal figure on the earth's mass (which for convenience we model as flat and thin so we don't need to worry about rotational effects). Let's say its mass is H (for huge). Its kinetic energy is therefore

Ee = 1/2 H Ve^2

We are slowing down the Earth using a puny 60 N of force, giving an acceleration of -60N/H. Another way of expressing this is (M/H)*A.

Ve = M/H A t + V

We calculate: Ve1 = (M/H m/s + V) m/s Ee0 = 1/2 H V^2 Ee1 = 1/2 H (M/H m/s + V)^2

The energy difference Ee1-Ee0 = 1/2 H ((M/H m/s)^2 + 2 M/H m/s V + V^2 - V^2) = 1/2 M (M/H m/s + 2 V m/s)

Because H is huge compared to M, we can ignore the M/H term since it'll be vanishingly small compared to 2V. The earth is losing M V m/s = 120 J of kinetic energy.

Now it balances again: The hero puts in 90 J of work, the Earth contributes

120 J, the pleb takes away 90 J, and so we have 120 J left over for the lightbulb.

[Gib Bogle]

I can't believe Drivel is as stupid as he appears. Therefore I think he's pulling everyone's legs.

[dennis]

Read what I said in the last post you replied to. I just showed that you never get any energy back from slowing the wind when you are travelling down wind even with perfect conditions and a perfect prop. It doesn't matter at all what size the prop is (that is air mass) which you appear to think it does.

In the absence of getting any energy back it doesn't matter in the slightest about the mechanics including the mechanics extracting the energy from the wheels and giving it to the prop. All that happens is that you add in the various energy loses like turbulence, friction, the braking done by taking energy form the wheels and the whole thing just slows down.

I am not going to go through all your points one by one as you fail to grasp even the simplest perfect case of just a perfect prop. If that can't extract any energy from the wind the rest doesn't matter, it is the *only* *source* of energy that you have.

The energy from the wheels is just stored energy you have already extracted from somewhere else, it is not a *source* of energy.

You can bullshit like rick if you want, I will just put you on the bullshit list.

[dennis]

You changed them from my example and got them wrong, there was no slowing of the wind in my example and you just slowed it by using a different frame of reference. Check it if you want, I know that there was no slowing of the wind. Which is why there was no energy being extracted and that was the absolute best case for you to get energy from the wind.

[dennis]

Does that make him go faster than pulling their arms? We do need to maximise this free energy in this thread.