Re: OT Here is an example of pseudo science.

[ThinAirDesigns]

dennis@home

So you agree that a propeller *can* slow the real wind (you said "B" was true), but you don't believe that *our* propeller is slowing the true wind.

Is that correct?

[dennis]

Ok.

So they are travelling at 2.9 times the speed of the wind, no prop They need to slow the wind as currently they are speeding it up through drag. Start prop.. now they have to accelerate a mass of air to greater than 1.9 times the real wind speed before the real wind starts to be slowed. Does it take more energy to accelerate it to >1.9 times the wind speed than they get by slowing the real wind down by less than 1?

The claim does not make sense even before you add in friction, turbulence, etc.

Does WD40 do magic too?

Anyway as far as I am concerned this thread id dead, other people can believe it if they want, nothing in the videos explains it.

[Clive George]

I'm not doing any deflecting. I'm taking you through it, slowly and carefully.

Can you stick with my argument, rather than starting your own? You're introducing confusion into the discussion which isn't helping.

No, it's a step along the way to explaining it to you.

Can you answer my questions?

[ThinAirDesigns]

dennis@home

Well there's you problem -- no such thing has to happen.

Let's to back to the skateboard and use your numbers.

Sidewalk traffic = 1ms Skateboarder = 2.9ms

To start to slow the pedestrian down he need not throw them back at

1.9ms per your assertion. If he DID throw them backwards at 1.9ms, he would be reversing their direction and they would now be traveling the OPPOSITE direction on the sidewalk at 0.9ms (1ms - 1.9ms =

-0.9ms).

Now, the skateboards *arms* definitely need to travel 1.9ms backwards (2.9ms - 1.0ms) before those arms can begin to contact and slow the pedestrian, but your claim that the weight of the pedestrian must be accelerated as such is simply false.

Similarly, the propeller need not accelerate the mass of the air rearward at your claimed 1.9ms before it begins to slow down. If the wind is merely accelerated rearward by 0.1ms, its speed has been reduced by 10% (1ms - 0.1ms = 0.9ms) and that energy was harvested by the mechanism.

[Ronald Raygun]

I apologise for my earlier misunderstanding, which came from being fixated on the boat example being driven by a water propeller using power harvested from an air turbine. I see that in the case of the land based car you're discussing it is not using the air screw as a source of power with which to drive the wheels, but the wheels as a source of power with which to power the air propeller.

I don't doubt that this is possible, but I believe your analysis below is wrong. I offer a corrected version below.

Fine.

This is wrong. See below.

Fine.

Fine.

This is wrong. See below.

This would be fine if the prop did in fact only need 1/2hp. Actually it needs a bit more than that. See below.

No, this is wrong. See below.

But look, when there's no wind, and the car is not moving, you can still claim to be travelling at twice the wind speed. There is no problem, since two times zero is zero! There will be no losses either! :-)

Now let's get to what's wrong with B and part 2 of D above:

You're using the simple formula that Power = Force * Speed. This is OK for the wheel situation, but it doesn't apply to the prop, where the air is being accelerated. More below.

The other problem is that you seem to be arbitrarily using the input air speed for the prop calc, i.e. the speed of the headwind (aka the apparent wind, i.e. wind speed minus car speed), which is wrong. To see why it's wrong, do the calcs for the situation where the car it going at exactly air speed:

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. At

0hp the prop won't be turning, and would be producing no thrust.

It would be more correct to use the output air speed. But when I say "more correct" that just means "less obviously wrong", it's not actually correct yet. To see why it's not right, let's redo the calcs for your original example with car speed 55ft/s and air speed 27.5ft/s.

I'm persuaded by your comment elsewhere, regarding the energy balance, that the optimum acceleration the prop must impart to the wind is such as to reduce its ground speed to zero. So the prop needs to reduce the wind's speed over the ground by 27.5 ft/s from 27.5 ft/s to 0 ft/s, and this is equivalent to changing the wind speed as seen by the car from 27.5 ft/s coming in to 55 ft/s going out. The air is leaving the back of the prop at

55 ft/s, and if we were to apply your simple formula, the prop would require a full 1hp to deliver 10lb of thrust. Nothing left for losses, so if the darned thing works in real life, the formula must be wrong.

How do we fix this? We need to go back to your excellent skateboarder.

He's skating along at 2m/s, going 1m/s faster than the pedestrians around him who are moving at 1m/s. Suppose he grabs one pedestrian each second, decelerating them from 1m/s to 0m/s relative to the ground, or accelerating them from 1m/s to 2m/s relative to himself. The acceleration required is 1m/s/s. If each pedestrian has a mass of 60kg, the skater needs to apply a force of 60N to achieve that acceleration. So far, so good. How much power does the skater need to keep up this rate of work? Well, if we were to apply the simple power formula of force times speed, we'd get either 60 W or 120 W depending on whether we used the 1m/s starting speed or the 2m/s final speed of the pedestrian. Both are wrong. To get the correct answer we need to get a bit more basic.

Energy = Force * distance. How far does the pedestrian travel during the second he is being accelerated? It turns out to be 0.5m relative to the ground, or 1.5m relative to the skater, and so the amount of work the skater is doing on the pedestrian is 60N * 1.5m = 90J, and if this is done over the course of one second, the power is 90W. Notice how 90 is equal neither to

60 nor to 120.

How do we apply this to the car? The prop needs to accelerate a certain quantity of air each second by 27.5ft/s in order to get 10lbf of thrust. How much air? Ah, curse your antediluvian non-metric units! Having taken great care to distinguish lb force from lb mass, the simple formula Force = Mass * Acceleration gives mass = 10 lbf / 27.5 ft/s/s which comes to about 11.7 lbs of air. Air weighs about 1.7 lb/yd3, so the prop needs to "pump" about 6.9 yd3/s, and since this needs to leave the back of the prop at 55ft/s, its effective cross sectional area needs to be 0.38 yd2. But I digress.

Although the prop accelerates the air continuously within its own thickness, we could instead think of it as a set of arms grabbing a boxful of air each second, each box containing 11.7lb of air. How far would each box travel during the second? Turns out to be 1.5 times 27.5ft = 41.25ft. So each second the prop does work equal to 412.5ftlbf.

That's 0.75hp. What luck, eh? Still comfortably below 1hp.

[Clive George]

Dennis, if you want to continue looking dim, you're welcome to. But if you stick with this you might actually learn something. It does require that you stop assuming it's impossible, and instead look at the actual physics involved.

[Rick Cavallaro]

I applaud your effort, but I've been through this with too many Dennis's. I assure you that assistance is futile.

That's not how this game works. Dennis will NEVER late you take him anywhere near the answer. When he sees any sign of it, he'll just make an immediate left-turn and start down some other pointless alley.

I have one squillion dollars that says no. But it's entertaining to watch.

[Rick Cavallaro]

Well, you're making progress. I suspect if you'll continue a rational discussion with JB he'll be able to convince you that your "corrected" analysis is in fact wrong.

[The Natural Philosopher]

Hey, it was just a way to get the idea across.

[The Natural Philosopher]

Dennis doesn't do actual physics.

[ThinAirDesigns]

@Ronald Raygun

Hey Ron -- thanks for your detailed response. I'll put together a reply in a bit -- I've got some important stuff at work I'm into at the moment.

Later.

[Ronald Raygun]

I'll approach what he has to say with an open mind.

I would point out, though, that my result (of the prop requiring 3/4 hp) means that 1/4 hp is available to be dissipated on losses and on the car's air resistance etc. Since we are considering the car as being in steady state motion, no power is needed to accelerate it.

This spare 1/4 hp has to come from somewhere, and it is from the wind being slowed down, its kinetic energy being transferred into those losses. Right?

It is no coincidence that the kinetic energy of 11.7 lb of air at

27.5 ft/s is exactly 1/4 hp s.

[ThinAirDesigns]

@Ronald Raygun

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.

If the vehicle is traveling at exactly windspeed, the propeller is working in what's called 'static conditions' and according to standard propeller efficiency terms is doing no work (different "efficiency terms" are used for fans).

In static conditions, a 100% efficient propeller would be infinitely large and would be turning infinitely slow but would still be producing 10lbs of thrust at 0hp. Its not unlike leaning a 2x4 against the wall =96 it can happily exert 10lbs of force on the wall all day long while needing 0hp to do it. Same with a lossless propeller in static conditions.

[Ronald Raygun]

That's a poor excuse and I don't accept it. We've disregarded losses only because allowing for them quantitatively makes the calculations tedious and because we'd need to understand the finer points of aerodynamics (you might - I admit I don't, but I don't think I should need to in order to understand the basic energy budgeting here). We just allow for them qualitatively by observing that the power harvested by the wheels must exceed that consumed by the prop by a margin comfortable enough to feed any losses.

The exactly-at-windspeed scenario is a bit of a crazy special case in that there is no air resistance; if the wheels are also frictionless, then the car doesn't actually need any power to remain at its speed. Therefore the presumption that the wheels must generate 1hp and the prop must therefore provide 10lb of thrust is a bit silly, since clearly the prop really doesn't need to provide any thrust, and so it's perfectly all right to run it at 0hp, and our generator will happily supply 0hp using 0lb of wheel drag.

It's a bit like the skateboarder trundling along at the same speed as the pedestrians. He still has some thrust requirement to meet his losses, and he can still accelerate the pedestrians backwards, but if they are starting out as stationary relative to him, he's going to need ever longer arms to grab more, they're no longer going to be coming at him all by themselves, as they did when he was going faster than they were.

The propeller must have a similar problem. Suppose we introduce an artificial loss into an otherwise lossless system, for example by inclining the road slightly uphill. The car will then require nonzero power to climb the hill at windspeed. The car will still see a zero headwind, but the prop has to deliver nonzero quantities of air out of the back of it each second, and at nonzero speed, and the air has to come from somewhere. Like the skateboarder who uses up all the pedestrians he can reach, the fan, if it's to keep working, must suck the air from somewhere, so there isn't really any point at which its input air speed can be considered zero despite zero headwind.

[Clive George]

Yes, but the freewheeling case isn't interesting. Introducing the wheel driven prop is what makes it interesting.

You've introduced the assumption that the car must be at steady state - there's no reason for that to be the case. Drop that assumption, work with the forces and see where you get.

[ThinAirDesigns]

@Ronald Raygun

Ron, math doesn't care *why* we've disregarded the losses. Present all the arguments you wish against my calculations, but arguing that *why* we ignore losses impacts the math is just silly.

Fact is, at that point in the analysis it's clear were employing a

100% efficient propeller and a 100% efficient propeller will in static conditions produce 10lbs of force from 0hp just as a 2x4 against the wall will.

JB

[Clive George]

I know. I played a similar game a few years back with a rather simpler problem - I persuaded a few, but there was at least one person who disappointed me, since I thought he ought to have the ability to understand it - he stuck to saying I was wrong rather than explaining why.

> Can you stick with my argument, rather than starting your own?

Dennis would rather carry on looking stupid than admit he's wrong, so your squillion dollars is safe.

[ThinAirDesigns]

@Ronald Raygun

You did notice that it was *you* who brought up the "crazy special case" of exactly at wind speed. :-)

It is your own special case that has brought into play the seemingly odd (but true) case of the 100% efficient propeller which can produce

10lbs of force on 0hp.

The way the term "efficiency" is applied in any given situation can be a bit arbitrary. As I previously mentioned, propellers and fans use the term very differently because they have very different applications and thus very different definitions of "work". What is most important is that the the term and it's relevent formula be used consistently throughout any given. We can't use one definition at 2x wind speed and another at exactly wind speed. I have been consistent in my application.

It just so happens that the standard propeller efficiency formula turns up looking a bit odd (but still valid) in static conditions.

[Andy Champ]

Actually I brought it up a few days ago... but I'm not sure anyone noticed...

Ron,

imagine a case where in order to generate that 10lbs of thrust at zero relative windspeed the prop needs half a horsepower. Real losses, quite achievable - in fact I think TAD did better in real life.

The cart has that 10lbs of thrust, and it's going downwind at exactly windspeed, and that 10lbs through the wheels at that speed makes a whole HP. You have half a HP left over to accelerate with!

Andy

[Grimly Curmudgeon]

We were somewhere around Barstow, on the edge of the desert, when the drugs began to take hold. I remember ThinAirDesigns saying something like:

No need to get sarky. That's my job.

It might never have occurred to you that the post might be missing?