But the force available from the parachute is proportional to velocity squared, where velocity is the difference between the air and the parachute. Power of the parachute is velocity cubed.
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Split Thread The validity of classical physics (split from: DWFTTW)
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But the force available from the parachute is proportional to velocity squared, where velocity is the difference between the air and the parachute. Power of the parachute is velocity cubed.
I'm glad you agree. For the sake of understanding then, it would make the most sense to stay in the mechanical realm to avoid losing meaning when translating in and out. Thank you for that, it will help keep things simple.
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John Freestone
Graduate Poster
Easier to get the right answer, that's the problem with that.
Electrical METAPHORS aren't nearly taxing enough. I think we should be dealing with this in cookery terms, or Freudian psychodynamics. I'm sure we could be motivated enough to overcome the resistance, and get the souffle - I mean, kite - to rise.
Dan O.
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If you take a one farad capacitor with one coulomb of charge measuring one volt and transfer the charge to a second one farad capacitor initially uncharged measuring 0 volts by using a pair of 1 henry inductors to make the connections, in about 6 seconds the first capacitor will be fully discharged and the second capacitor will have acquired most of the one coulomb charge or about 1 volt.
This has nothing to do with MPT.
This has nothing to do with MPT.
Mmm...I don't need to convince you.
I bet yours is a $10 Rolex. No, it's genuine, see, it's got 'Rolex' on it.
I would take it but...
Yes, that is correct. Then it goes badly.
There are idealised velocity sources ( velocity independent of force/velocity)
There are idealised force sources ( force independent of load/velocity )
Actual sources are not ideal, they have a power limit, that is the product of the two.
Force = Current
Velocity = Voltage
Resistance = V/I
The resistance is the ratio of velocity to force; the slope of the load line of the power source, either mechanical or electrical.
dv/df = source "impedance'. It's actually more complicated than that, of course. All sorts of parametric variables.
It is obvious, The power will be a parabola, and the vertex the maximum power available.
Take a sphere and attach a wire to it,and tow it through a trough of oil using a powerful motor. Plot the force/velocity graph. Let's say that it is square law.
Perhaps you think that putting the same sphere in moving oil will be equivalent? No.
When towing the sphere, try to stop it by hand. If the motor is powerful enough, then the sphere will continue undisturbed. There is essentially unlimited power (available), that is directly coupled to the sphere (load).
The same is not true for the oil bath. The power is limited by that from the oil, and it is coupled by drag (friction). The oil and sphere are loosely coupled, like a pair of hooters under a sweater.
That is why a generator on a treadmill, is not the same as one in the wind.
I mentioned that before. There is power to overcome losses and natural impediments, that is not available to the wind. That means different aerodynamics.
Equivalence is but a perspective. Relative velocities may be changed, but all else must be preserved. The two have been conflated and abused, to make "frame" nonsense.
Well done. You have made the entire process reactive.
Note that time is now involved, and the voltage and current are no longer in phase.
That's the difference between situations where both charge and energy are conserved, or momentum and kinetic energy.
That was the point. (In reality some will be lost because the components are real, both inductors and capacitors have resistance.)
A little knowledge is a DANgerOus thing.
That is correct. It is lost in the source. Newton's cradle is different. Reactive to reacticve, with small loss.
I would try the Salvation Army. Nicer people.
Ooops. I think a good ducking should do the job, though.
Over-cooked, but at least an idea. The recipe analogy is not bad. Follow the instructions A - B = 0. Easy.
Freud need to be put in the past. Terrible ideas.
Actually, the issue has never been about how much power a wind has, or what the horsepower rating of a treadmill is. Both are assumed to have more than enough power to maintain the speed difference between the air and the ground. The issue is the amount of energy that the cart can exchange between the air and the ground. Different problem altogether from what your example illustrates.
When trying to stop the sphere with your hand, you are only testing the strength of the tether and/or the power of the electric motor. If you try to stop the flow of oil out of the pump that is pushing the oil, you are testing the power of the pump.
The test that is applicable is to test how much force is acting on the sphere when it is moving at a specific speed through oil, or when the oil is moving past the sphere at a specific speed. Two ways of saying the same thing. The force will be the same for the same speed. Either test can be used to accurately study what is happening.
No. The force is approximately linearly proportional to the relative velocity of chute and wind. This is essentially how Raleigh's equation works, seen again in the meteorological balloon book, and ance again in Dela's paper on carts.
Enough?
Mechanical resistance and conductance are not widely used, and rather specialized. Electrical analogues cross interdisciplinary boundaries.
Most information is not so freely available, but here is another reference for power.
http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel5/10559/33412/01583122.pdf?temp=x
Exactly the same. The cart cannot exchange energy with the air, unless it is dragged to motion by the belt.
No, the force being tested is the driving force. The motor is essentially unlimited in power or force.
In oil, only the oil flowing immediately over the sphere powers it. That force is coupled by drag between the oil and ball, and therefore a limited means of transfer. The oil grips the sphere, but tenuously.
Increasing the force demand, causes increased difference in velocity of oil and sphere. Like current through a resistor develops voltage across it.
This is like friction to the belt. No friction, no power to the object.
True also for wind, but wind and road are not alike in other ways. The surface of the road, is not like the momentum exchange of real moving wind.
No, they are perhaps equivalent relative velocities, but they are not symmetrically coupled.
Different. That is derived from fan laws. The turbine is stationary.
And exploits the difference between the speed of the air and the turbine. Same laws apply, humber.
As a result, a turbine mounted on a treadmill belt moving at 10 m/s in still air will produce exactly the same power as a turbine outside mounted on the ground in a 10 m/s wind. That is equivalency. True, and very useful as a result. The formula doesn't specify whether the air is moving or the turbine is moving because it doesn't matter. The speed difference between the air and the turbine is specified because it does matter.
If the cases were not interchangeable, a glider towed by a winch at 50 m/s through still air wouldn't have the same lift as a glider tethered to the ground (kite, perhaps?) in a 50 m/s wind. That to me is an extraordinary claim, and would definitely require pretty solid real world proof before it can be used as a argument.
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I thought this version of the lift equation
http://www.grc.nasa.gov/WWW/K-12/airplane/lifteq.html
and the drag
http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html
was pretty well known and correct but I realize now that it is wrong. Humbers argument (he said it was false with a reference to a linearized differential equation) convinced me. I should never trust what is written in any aerodynamics book any more.
http://www.grc.nasa.gov/WWW/K-12/airplane/lifteq.html
and the drag
http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html
was pretty well known and correct but I realize now that it is wrong. Humbers argument (he said it was false with a reference to a linearized differential equation) convinced me. I should never trust what is written in any aerodynamics book any more.
GregLondon
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Except that this new metaphor has nothing to do with the maximum power theorem, which leavest the fact that MPT has nothing to do with treadmills and objects in motion, and you have no idea what you're talking about.
time wasn't involved before???? HA!
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GregLondon
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Don't bother humber with facts. They get in the way of his metaphors.
GregLondon
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You're talking in complete gibberish, Humber. There is no "real" source of power versus "non real". There is no 50% hit when you transfer power from one object to another. Absolutely everything you have said has been wrong. Complete and total rubbish. A con. A lie.
sol invictus
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We discussed this extensively upthread. How shocking that humber learned nothing.
The drag force is generally proportional to velocity squared, for the simple reason that both the number of molecules encountered and the momentum transferred to each is proportional to velocity.
The drag force is generally proportional to velocity squared, for the simple reason that both the number of molecules encountered and the momentum transferred to each is proportional to velocity.
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