Here in the US we believe strongly in equal rights. :-)
No, not once -- since the way I specifically knew that I mentioned it in the first sentence of the first paragraph of my first post was I went back in the thread until I found it ... right in the order I posted it. I just now again went back and sure enough it's still right there.
There's no crime in you personally missing that post, but if you tell me you wish I had said something sooner that I opened with, I get to correct you.
Before you and Rick slink off to repeat this argument elsewhere, I'd like to say thanks for popping in and giving us your explanation.
You could probably save yourselves some hassle by revising the introductory paragraph on your explanation page, on the grounds that it didn't seem to help anyone here to "get it", adding a few fancy animations and the skateboarder analogy would help too.
Indeed. Guilty as charged. But I did it merely in order to illustrate why I felt it wrong of you to apply the "Power = Force * Speed" formula using the speed of the input air. I still feel that's so, and moreover that it's wrong to apply that simple formula at all to cases where the force in question is applied to objects being accelerated.
Well, I find that hard to take, because surely a prop can only give thrust as a result of imparting impulse to air, which necessarily involves accelerating (doing work on) that air, which in turn necessarily consumes power. If not, why not? It seems to me that a prop giving thrust must always consume power even if the thrust itself is not going to do useful work, and even if the fluid's initial speed is zero. Think of a boat with an outboard engine. Its water prop consumes power not just when it's pushing the boat through the water, but also when it's pushing it against an immobile dock. An air prop is in principle just the same as a water prop, isn't it?
Let me get away from any crazy prop effects by returning to the skater.
Suppose he needs thrust of 60N (that's quite a lot, but I guess his wheels are very poorly lubricated) to maintain his speed of 2m/s. He applies this constant and continuous force of 60N to a steady stream of 60kg pedestrians coming at him at 1m/s, and he grabs hold of each pedestrian just long enough to accelerate him to 2m/s. The instant he lets go of one guy, he grabs the next, with a seamless changeover so that his thrust is constant. With an acceleration of 60N/60kg=1m/s/s, he pushes each pedestrian for 1 second.
Now if you were to apply your simple formula "Power = Force * Speed" to calculate the power requirement of the skater's arms, in the same way as were doing to the car's propeller, would you use the force of 60N and the *arrival* speed of 1m/s? If so, you would get a power of 60W.
This, as I pointed out earlier, is incorrect because if you do the energy budget, 90J of work is done on each pedestrian, because the pedestrian travels 1.5m during the second he is being accelerated. Therefore the mean power needed is not 60W but 90W.
The problem is that "Power = Force * Speed" is no more universally true than is "Distance = Speed * Time". They only hold where the speeds are constant. Power is the gradient, with respect to time, of energy, and energy is the integral, with respect to distance, of force.
Where force is constant, you can say that "Energy = Force * Distance". When, in addition, distance is linear with time, then energy is also linear with time, and *then* you can apply the simple formula, but not generally otherwise.
In the situation of our pedestrians, force is constant but speed is not, and neither is power. For each pedestrian, as his speed increases from 1 to 2 m/s, the instantaneous power increases from 60W to 120W, and its mean value is 90W.
We can corroborate the 90W figure by looking at the gain in kinetic energy of each pedestrian. It starts at 1/2*60kg*(1m/s)^2 when he's grabbed (that's 30J) and finishes at 1/2*60kg*(2m/s)^2 when he's let go (that's 120J), so the gain is 90J per pedestrian, corresponding to
No. Static thrust does no work any more than a labrador sitting on your foot and imparting thrust, is doing work.
Thrust is very misleading in fluid dynamics. There are so many ways in which thrust is easy to measure, but the actual air mass accelerated both in size and velocity change is not.
In the limit an infinite mass suffering infinitesimal acceleration generates any thrust you want. For zero power
It is far easier to look overall at the drag profile if the machine, its speed, and the wind speed and see whether you can using the prop extract enough work from the air mass to overcome the power to move the vehicle AND spin the prop to blow air backwards to extract the wind.
It seems to me this concept is about creating a 'dynamic sail' consisting of a slipstream of air being blown backwards
In a boat, you could have a set of sails on a caterpillar track that were driven backwards by a screw underwater, and then furled and led forward to the bow of the ship. Thought experiment only.
Or another way to look at it is that the whole thing represents no more than wind overdrive. A gearing system.
If I was to pose the question in this form: "A rope runs through a tunnel at 1m/s. It is driven by a very powerful motor and serious loads can be applied and it will remain at 1 m/s. Devise a mechanism that will utilise this rope to drive a cart faster than 1m/s in the direction the rope is moving'
I think a very simple geared system with one gear being driven by the rope and the other by the ground wheels, meshed together would end up with a very high speed of the axles' centres..more than 1m/s.
ISTM this is the same principle.
By having the wheels or the waterscrew you can establish a zero velocity frame with respect to which the windspeed can be offset to extract power. The fact that the vehicle is not tied to this frame doesn't matter, as long as conceptually your 'sail' is.
To my mind it cuts two ways. Theory says it should work, my gut says it cant,. If the tests are verified tho, it sees to say it can and does work.
It would be a fairly trivial exercise to build a RC model though and try it out.
If not, why not? It seems to me that a prop giving
No, props need not consume any power to make thrust In the theoretical limit.
That they do is down to beating and vortex and sound generation. And practical size limitations, a very large very thin prop turning very slowly can generate enormous thrust. Until you start to move, then thrust times speed dominates and that's the power. A variable pitched prop in very fine pitch is a lot less draggy and power hungry at slow speeds an accelerates a plane smartly..as speed builds up, pitch coarsens to keep up with incoming airflow, and the prop load increases on the engine. Thrust times speed is here.
Yes it takes power to hover a helicopter, but less than you think. They have oddly, very long very thin blades, spinning very slowly..if they were longer, thinner and slower they would take even less power, but then the aircraft would be infeasibly large.
I agree with everything in your post, with one slight clarification that I expect you'll agree with. Static thrust does no work on the craft. It does do work on the air however (except of course in the limit as you approach a 100% efficient prop).
What The Natural Philosopher tells you above is the simple truth. Do the math.
Momentum = MV Energy = 1/2 MV^2
You have tried to to show that my force/energy analysis is wrong because those calcs show that in a perfect world 0hp is required at exactly windspeed. Well, the analysis and equations are correct and would be wrong if they *didn't* show this.
force * time =3D impulse or momentum =3D mass * velocity
So: force (or thrust) =3D velocity * (mass/time)
This means we can get exactly the same thrust if we double the mass flow rate and accelerate the mass half as much.
Looking at the other equation:
energy =3D mass * velocity^2 / 2
Power =3D energy / time.
So power =3D (mass/time) * velocity^2 / 2
Clearly if you double the (mass/time) term and halve the velocity term, power is reduced. You can repeat this as many times as you like
- and thus make the required power as small as you'd like without reducing thrust.
Helicopters do in fact use this principle to hover with far less power than would be required if they used an airplane propeller. A Cessna and an R22 have similar powerplants, but a Cessna couldn't dream of hanging from its prop.
Yup - we were going to go with one of those on the Blackbird, but JB brought up all these silly problems about the height of the pylons, cost of fiberglass - blah, blah, blah.... : )
So what have you guys done with dennis? Did he just slink away quietly because he figured out he's probably wrong - but still doesn't quite understand it.
Oops, pressed wrong button and the message went out before I'd written the rest.
The thrust you get is equal to the rate of change of impulse, and this corresponds to the air speed difference (output air speed minus input air speed) multiplied by the rate at which mass is put through the prop.
This means that to get a particular thrust value, you can either put through one particular quantity of air per second at one particular speed difference, or alternatively you can put through k times that quantity at 1/k times the air speed difference, for any real positive k.
This impinges on the power requirement, which is equal to the difference in the input and output airs' kinetic power.
I concede that this shows that in principle the power needed to produce a given amount of thrust can be made as small as we want by making the output air speed as small as we like. Unfortunately the mass throughput required per unit of time is now equal to thrust/v2, which means that if we wanted to reduce the power to zero, we'd need to put through an infinite amount of mass each second.
This is clearly unachievable, and so I must reject your analogy with the two-by-four plank leaning against the wall, until such time as you can produce a 2X4 of infinite length and zero density. :-)
Depends what you mean by perfect. You merely stipulated that the components (propeller, motor, generator) should initially be 100% efficient. A propeller is 100% efficient if the mechanical power you put into it from the motor matches the power which comes out, which is represented by the difference between the before and after kinetic energies of the air being put through per unit of time. I did not expect it would necessarily have to have infinite size. If input and output power are zero, its efficiency is 0/0 which is undefined. How can you say it's 100%? Hardly a perfect world!
Anyway, let's leave the silly static special case and return to the one in which I originally took exception to one aspect of your analysis. The case with the car travelling at 55ft/s in a 27.5ft/s wind, and 10lb of thrust.
If you're wanting to get 10lb thrust in those circumstances using only
1/2hp, then (if my calculations are right) this is possible only if you play the same silly game as you've done with the static case, i.e. you would end up with prop air output speed the same as air input speed, i.e. zero speed difference and hence a need for infinite air throughput.
I had thought that the purpose of your discussion which included the analysis was to try to persuade unbelievers (who no doubt think the whole DDWFTTW idea is impossible, and that your documented demonstrations of it nevertheless actually working must therefore be elaborate hoaxes). I would comment that an analysis which only works with infinitely large propellers is hardly going to be terribly persuasive.
You'll kindly notice that it was YOUR analysis that required an infinitely large propeller. My analysis works just fine as written and involved no such thing. It's a fact.
Had I picked exactly wind speed to analyize (as YOU did), I would have picked a different (but equally as accurate) method. Not all things are perfect for illustrating all situations.
I might add here that you have yet to produce an *accurate* analysis, let alone a persuasive one, so you might want to go a bit easy on the high horse.
OK, sorry if I came across that way. I have no expertise in aerodynamics or propellerology and am trying to understand your analysis at the level of basic laws of motion using nothing more advanced than high school level physics, addled by the passage of nearer 4 than 3 decades. I'd be grateful if you'd care to point out any mistakes in the following.
Let V1 be the air input speed to the prop (27.5ft/s in the example). Let V2 be the air output speed from the prop (you left this value open in your example, and I picked 55ft/s for reasons I stated earlier, but let's leave it open for now).
For convenience, I'm not going to use V2 directly but will instead use the air speed difference across the propeller: Let DV be V2-V1. For further convenience I shall express DV as a multiple of V1: Let k = DV/V1, or DV = k*V1, which just means V2 = V1*(k+1). Let F be the thrust required from the prop (10lb in the example). Let M be the rate at which (air) mass is being put through the prop.
We know that M = F/DV = F/(k*V1).
[For example, if I were to pick V2 in the way I did before, then we would have k=1 and M=11.7 lb/s]
Let me define kinetic power in the obvious way: If a mass m of air moving at speed V has kinetic *energy* (1/2)*m*V^2, then a stream of air passing through a particular area at rate M [lb/s] has kinetic *power* (1/2)*M*V^2.
The kinetic power of the air stream entering the prop is therefore
P1 = (1/2)*M*V1^2
and that of the exiting stream is
P2 = (1/2)*M*V1^2 * (k+1)^2
The kinetic power *gain* of the air stream, as a result of passing through the prop, in other words the propeller output power, is
DP = P2-P1 = (1/2)*M*V1^2*k*(k+2)
Since M = F/(k*V1), this simplifies to
DP = F*V1*(k+2)/2
If I pick V2=2*V1, as I did earlier, this corresponds to k=1, which gives
DP = (3/2)*F*V1
You will recall that with F=10lb and V1=27.5ft/s, F*V1 = 1/2hp, and so in the case k=1, we get DP = 3/4hp, the result I reported earlier.
I'm now interested to see what values of k, and hence V2 and M, are implied by YOUR result for DP, which was 1/2hp, in the same circumstances, i.e. where F and V1, and therefore F*V1, are given as 10lb, 27.5ft/s, and 1/2hp respectively.
We have DP = F*V1*(k+2)/2. On the LHS we substitute 1/2hp for DP, and on the RHS we substitute 1/2hp for F*V1, and we get
1/2hp = 1/2hp * (k+2)/2
and therefore
(k+2)/2 = 1
This means k=0, DV=0, V2=V1, and M=infinite.
Looks like your analysis also requires an infinite propeller.
My apologies. Perhaps I read too much into your explaining confidently that JB's analysis was wrong (rather than suggesting that maybe it was wrong, or asking for clarification). I've been through this thing a whole bunch of times, and that's almost always a dead giveaway.
Fair enough. The mistake is not in your math, but in your assumptions (or perhaps a simple miscommunication). In your above post you say "Looks like your analysis also requires an infinite propeller". Yes - that's right. JB starts with a lossless prop (as he states). I'd have to go back and look at his analysis, but I suspect he shows that there's lots of excess power available with that assumption. This lets us distinguish the difference between an engineering problem and a problem of violating physical law. If we have excess power in the no-losses situation, we know it's just down to an engineering problem.
And yes, as you suggest (and as we also have suggested) a lossless prop will always be of infinite diameter and will have zero delta- velocity across the disk. An 85% efficient prop however can be significantly smaller than that (and will have a non-zero delta-V).
One thing to keep in mind is that prop efficiency is typically defined relative to "V-infinity". This is the free-stream velocity. In the case of an airplane, it's the plane's airspeed. For this definition eff =3D power_out / power_in
Relative to the aircraft this is the definition that makes sense. The pilot really doesn't care how much breeze the prop makes. He just cares how much force it creates at a given airspeed. If you're designing house fans you wouldn't use that definition for efficiency. In this case you DO care how much breeze your prop creates.
Of course choosing different definitions for efficiency cannot change the outcome of an experiment, but it can lead you to an incorrect result if you use a different definition than what is assumed by the person that did the analysis.
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