Split Thread The validity of classical physics (split from: DWFTTW)

[GregLondon]

Still sure he's not a whacko?

I would change one bit of my assessment. When proven wrong in some way that cannot be denied, Humber exhibits malice towards the person proving him wrong.

Rather than acknowledge he used "maximum power theorem" in a way that had nothing to do with the conversation, he got pretty mad.

I do not want to hear any more of this crap.

Why make a fuss? Is it that I should have the expectation that you are infallible?

Those who prevaricate or deliberately try to misunderstand, are taken for short ride down humber lane...or runway.

Considering that you were, and still are, entirely ignorant of this theorem, perhaps you are wrong again.

do you expect that I would wilt under the glare of your piercing interrogation?

Either way it's all about you and hubris!

Assuming all "mistakes" are deception or evasion, should be left at the church door.

It is your lack of understanding and willingness to learn that are the obstacles.

If you want to save face, stick it in the sand.

Talk to the hand.

Inertial frames as applied to the treadmill, is BS.

Please do not tell me what I think, Greg.

Nobel prize sighted.

Field's medal sighted.

I don't know how to teach something to a person who gets mad when you point out a mistake.

 

[humber]

Too funny not to respond to.... gee, I wonder what happens if you take A and add a certain mysterious constant velocity (sometimes known as "wind speed") to it.... jj, could you help poor humber out with that, maybe graphically?

Oh I forgot. Beltspeed is what makes the wind on the belt, so how do you add it
again? No, you don't of course.

There is also the big question:
(1) If it's subtracted that makes the speed zero.
(2) If added that makes it twice winspeed.

Pawned.

 

[humber]

No, I haven't.
That's true (as I said) - and the first link you gave conflated them. Charge is displacement only in the "force-voltage" setup, but the chart the first link gives is (mostly) force-current.

I am afraid you did. All that is needed are force = current and voltage = velocity. You are not reading the chart properly

Gibberish.

I am afraid that won't do. It seems that it's quite popular.
Springer Verlag have a wide range of books.The IEEE have all sorts of publications and seminars on this topic. I wonder if computers may have something to do with it?
http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=973452

Utterly wrong, as always.

No. QED. Nobody does not know that.
http://pubs.acs.org/doi/abs/10.1021/ie50543a027
http://www.springerlink.com/content/f00u470069453746/

Gibberish. Your arrogant stupidity has ceased being amusing.

While you were never amusing. Just a joke.

 

[GregLondon]

You are obnoxious. You are a peculiar sort of slow learner. One who takes a long time to learn nothing. Bye.

You made a mistake humber, that's all. We all make mistakes. But for some reason you can't admit it. No doubt there are some people who use the fact that they understand the cart and otehr people dont to put those other people down. And they've got their own issues going on. But I'm not trying to put you down here. I just don't know how to convey some information to you when you're holding on to information that is wrong and contradicts what I'm trying to tell you.

I'm sorry.

 

[humber]

I would change one bit of my assessment. When proven wrong in some way that cannot be denied, Humber exhibits malice towards the person proving him wrong.

No, just irony.

Rather than acknowledge he used "maximum power theorem" in a way that had nothing to do with the conversation, he got pretty mad.

Nothing wrong with my useage. Just yours. I don't have to justify myself to you.

I don't know how to teach something to a person who gets mad when you point out a mistake.

You have no place teaching anybody, anything. You are making errors of first principles. Walk before running.

 

[jjcote]

It is A. The flow is not observer dependent. If you change it here, you will need to do the same for both ends and in the real model.

Hmm. Not sure what you're saying here. It's clearly the case that the airspeed you will feel will change if your velocity changes (something about "velocity is relative"). So I'm happy to accept A for an answer, but I still need to know if that's what someone sees who is standing on the floor next to the treadmill, or who is on the belt, moving with the belt.

In the real world it's C, but moving the other way. (left to right)
The difference will again appear.

Please explain what you mean by "in the real world". Do you mean in the case of wind over stationary ground, or the treadmill case viewed from another perspective, or something else? And by "C, but moving the other way. (left to right)", do you mean "G"?

 

[humber]

You made a mistake humber....

I can use that principle to good use. You can not. That is your problem.
I exposed your fundamental lack of knowledge and that embarrassed you.
I gave you plenty of time to learn, you chose to take the other path.

Keep your faux apologies where you keep your face. Now that is an insult.

 

[Michael C]

Good Grief. I am forced to agree with Micheal_C. You are obnoxious.
You are a peculiar sort of slow learner. One who takes a long time to learn nothing. Bye.

Humber, if you wish to criticise Greg, that's your affair. Don't drag me into it and imply that I have said anything even remotely similar to your blatant insults (that have nothing to do with "irony"). I haven't.

 

[John Freestone]

Humber, I see a lot of foolish people have been distracting us from finishing our conversation. Forgive them, for they know not what they do. I'm just having difficulty understanding correctly how the BL reverses on the treadmill. I'm sure you're right, and just want your help in clearing up my obvious mistakes. Could you tell me, in the scenario we were discussing, based on your drawings:

Is the land-based 8 m/s to the right above or below the 6 m/s? (above/below)

Does it go at 8 m/s right to left above the treadmill, w.r.t. the still air/room/earth? (Y/N...it goes at).

Thanks again again.
John

 

[GregLondon]

If folks insist on making comparisons to electrical equivalence, then the electrical engineer is going to weigh in on analog circuits.

The thing humber is wrapped up in has nothing to do with the cart. the maximum power theorem only says that if you have a power source with some internal impedance, you can transfer maximum power if the load is the same impedance. the maximum power theorem doesn't say anything about transforming the power or leveraging it, which is what the cart does.

So, if you want to talk about the cart in electrical equivalences, the cart is a block and tackle, you pull teh rope 1 meter and it makes the cart go 2 meters. The electical equivalent for somethign liek that would probably be a transformer. YOu put in an alternating current of 1 amp at 12 volts on one side, and you might get out 2 amps at 6 volts or half an amp at 24 volts.

You could conceivably have a transformer that has an impedance of 4 ohms to match your power source if you wanted to transfer max power from source to the primary coil of the transformer. But nothing says you cant then use a newtonian simple machine like a lever, or a screw, or a block and tackle, or the electical equivalent called a transformer, and change that power into different voltages and currents.

The "Maximum Power Theorem" has nothing to do with newtonian objects. But even if you want to make electrical equivalents for mechanical ideas, the whole thing that makes the cart go faster than the wind is that it has a transformer in it.

You could have a powersource that has a 120 volt ac output. And you could put it through a transformer and get 240 volts ac at half the current.

Just because the wind is moving at 10 mph over the ground doesn't mean you're limited to the speed of the wind. If the cart uses a block and tackle approach, it can transform the power into different speeds.

 

[humber]

Hmm. Not sure what you're saying here. It's clearly the case that the airspeed you will feel will change if your velocity changes (something about "velocity is relative"). So I'm happy to accept A for an answer, but I still need to know if that's what someone sees who is standing on the floor next to the treadmill, or who is on the belt, moving with the belt.

So it's done. You agree the profile is different. The result of the road powering the wind? The model is wrong.

It works for both cases. I must insist that if you acknowledge that, it would mean the "windspeed" observer is the same as a ground observer.
I don't see the problem. It's quite clear.
The cart observer is not in a flow, he is in still air, so all boundary motion, must be right to left for him, as for treadmill side.
The observers in both cases see the same thing.
The cart at windspeed has no significant motion w.r.t the belt.

Conclusion: The belt profile is 'A' moving right to left, inverted profile

Please explain what you mean by "in the real world". Do you mean in the case of wind over stationary ground, or the treadmill case viewed from another perspective, or something else? And by "C, but moving the other way. (left to right)", do you mean "G"?

Real wind over stationary ground.

Conclusion; Real wind profile is 'C' moving left to right. Normal profile.

 

[Dan O.]

It is A
<snip>

It is A.

<snip>

Checkmate.

Wait, was humber saying it was "A" for the belt-speed observer and "A" for the wind-speed observer or was it the other way around??

 

[GregLondon]

Actually, the more I think about it, Humber is arguing that it is impossible to take 120 volts and convert it into 240 volts.

The difference in velocity between wind and ground translates somewhat metaphorically to the difference in voltages between the positive terminal and electrical ground. It is the difference that lets you extract power.

Because the wind is moving at 10 mph over ground, humber argues that the cart cannot go faster than 10 mph over the ground. But in the electrical equivalent version, that means that if the source of your power is 10 volts, humber would be arguing that the output can never exceed 10 volts.

And that just isn't the case.

But more importantly, arguing about the cart in electrical metaphors is only going to be one hell of a source of confusion to try and find the corresponding units for mechanical equivalents. We can talk about the cart in purely mechanical terms and understand how it works. We don't need to invent some electircal equivalent version of physical reality just so Humber can argue that his invocation fo the "maximum power theorem" had some relavence to teh conversation.

Switching to electrical metaphors is more a smokescreen for humber to hide that he doesn't understand it mechanically. The best way to understand the cart is to look at it for what it is, mechanically, rather than metaphorically.

 

[jjcote]

Actually, the more I think about it, Humber is arguing that it is impossible to take 120 volts and convert it into 240 volts.

I don't want to speak for him, but it seems more likely that what he is saying is analogous to maintaining that if you have a circuit and you raise all node voltages by 10V, the circuit will behave differently.

 

[humber]

If folks insist on making comparisons to electrical equivalence, then the electrical engineer is going to weigh in on analog circuits.

I'm bored. Let me have a go.

The thing humber is wrapped up in has nothing to do with the cart. the maximum power theorem only says that if you have a power source with some internal impedance, you can transfer maximum power if the load is the same impedance. the maximum power theorem doesn't say anything about transforming the power or leveraging it, which is what the cart does.

No. It has the same power law. Friction is a direct analog of drag.
It can be modeled to have any characteristic i.e. proportional to velocity or velocity squared...
V= Velocity
I = Force
R = Resistive load
Power = (Voltage)^2/ Resistance (electrical)
Power - (Velocity)^2/ Drag (mechanical)
(also Power= I^2*R)
Power transferred need not be the maximum, it may be less. That value depends upon the ratio and magnitude of the "load resistance" and "source resistance".

So, if you want to talk about the cart in electrical equivalences, the cart is a block and tackle, you pull teh rope 1 meter and it makes the cart go 2 meters. The electical equivalent for somethign liek that would probably be a transformer. YOu put in an alternating current of 1 amp at 12 volts on one side, and you might get out 2 amps at 6 volts or half an amp at 24 volts.

Same drum, new tune. Transformers have no power gain.

You could conceivably have a transformer that has an impedance of 4 ohms to match your power source if you wanted to transfer max power from source to the primary coil of the transformer. But nothing says you cant then use a newtonian simple machine like a lever, or a screw, or a block and tackle, or the electical equivalent called a transformer, and change that power into different voltages and currents.

Then you will not get maximum power. You may get more force or velocity, but not the maximum of both at the same time. Conservation rules.

The "Maximum Power Theorem" has nothing to do with newtonian objects. But even if you want to make electrical equivalents for mechanical ideas, the whole thing that makes the cart go faster than the wind is that it has a transformer in it.

Internally inconsistent. It has nothing to do with it, but...

You could have a powersource that has a 120 volt ac output. And you could put it through a transformer and get 240 volts ac at half the current.

Transformers get warm. That is loss. Similar to gear losses.

Just because the wind is moving at 10 mph over the ground doesn't mean you're limited to the speed of the wind. If the cart uses a block and tackle approach, it can transform the power into different speeds.

Just get a skyhook...

 

[GregLondon]

The bottom two drawing in post 2411 shown what the cart would see in operation, either on the ground with a 10 mph tailwind, or on a treadmill at 10 mph indoors. Either way, both scenarios show the windspeeds measured by the tower (and therefore felt by the cart) are exactly the same.

http://www.internationalskeptics.com/forums/showpost.php?p=4373908&postcount=2411

If we're just talking about windspeed differences between

on the ground, moving at 10 mph, with a 10 mph tailwind

and

on a treadmill at 10 mph

the velocities measured from the tower are identical.

They are equivalent

 

[humber]

Humber, if you wish to criticise Greg, that's your affair. Don't drag me into it and imply that I have said anything even remotely similar to your blatant insults (that have nothing to do with "irony"). I haven't.

OK Micheal_C. Sorry about that.

Perhaps is was not you but Myriad who referred to him as "the obnoxious Greg London"

 

[humber]

Wait, was humber saying it was "A" for the belt-speed observer and "A" for the wind-speed observer or was it the other way around??

See, you've gone and messed your cat litter again.

The profile is 'A' no matter what the observer is doing. The boundary layer is viscously attached to the belt. Difficult for the observer to change that.

 

[GregLondon]

Transformers get warm. That is loss. Similar to gear losses

And yet, somehow, every piece of electronic equipment in your house that uses a 5 volt power plane somehow connects into your wall outlet that has a 120 volt source.

If I pull on a block and tackle with 10 pounds of force, I can lift 20 pounds of weight. Or I could pull on the rope at 15 meters per second and lift a weight at 30 meters per second.

If you insist on the electrical metaphor, then either force is voltage and velocity is current or force is current and velocity is voltage. Either way, you can put your electrical power through a transformer and take an input "velocity" of 10 and have an output velocity of 20.

Just like a block and tackle. Just like a lever.

Push the short end of a lever down an inch, adn the long end could move 2 inches. Push it down at 1 meter per second and teh long end could move at 2 m/s.

Maximum power theorem doesn't say anything about transforming the power you have into different voltages or speeds. conservation of energy doesn't say anything about transforming your power ro energy ito other forms, possibly using simple machines to change your velocity but maintain the same amount of power or energy.

Even IF we accept electrical metaphors to look at the cart, Maximum power theorem has nothing to do with how the cart actually works.

 

[jjcote]

Hmm. Not sure what you're saying here. It's clearly the case that the airspeed you will feel will change if your velocity changes (something about "velocity is relative"). So I'm happy to accept A for an answer, but I still need to know if that's what someone sees who is standing on the floor next to the treadmill, or who is on the belt, moving with the belt.

So it's done. You agree the profile is different. The result of the road powering the wind? The model is wrong.

I haven't agreed to anything. I said I would accept an answer if it were a complete answer. I'll accept A as your answer if you'll say what perspective it's from. I haven't yet said whether I agree that it's the correct answer.

It works for both cases. I must insist that if you acknowledge that, it would mean the "windspeed" observer is the same as a ground observer.
I don't see the problem. It's quite clear.

I agree that in the treadmill case, an observer standing on the floor next to the treadmill is a windspeed observer. Is that what you're saying? An observer moving with the belt is clearly different.

The cart observer is not in a flow, he is in still air, so all boundary motion, must be right to left for him, as for treadmill side.
The observers in both cases see the same thing.
The cart at windspeed has no significant motion w.r.t the belt.

There is no cart in my diagrams that I can see. If there were a cart on the belt, moving at windspeed, an observer on the cart would be the same as an
observer standing on the floor next to the treadmill, and all flow from that perspective would be from right to left, yes.

Conclusion: The belt profile is 'A' moving right to left, inverted profile

Do you recognize that this is not a difference of the road vs. the treadmill, but rather a difference of what the observer is doing? Think about what airspeeds will be observed in a ground/wind case if the observer is in a car moving to the right at 10 m/s.

Please explain what you mean by "in the real world". Do you mean in the case of wind over stationary ground, or the treadmill case viewed from another perspective, or something else? And by "C, but moving the other way. (left to right)", do you mean "G"?

Real wind over stationary ground.
Conclusion; Real wind profile is 'C' moving left to right. Normal profile.

I'll ask again: do you mean "G"? or is "C left to right" something different in the humberverse?