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

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No, you still don't have a clue, and are trying to make much of a definition. Got ya Google. Take a look for any of those terms. You can even find cavitation for air, if you look. It won't mean anything to you of course, Captain. There is a cretin certain amount of "negative credibility" when a pilot writes about the stubbornness of other pilots, only in order to soil himself.
Over.


Really? Not any of the several times I've looked. If that's the case, why did you produce four links that DON'T use the term "Negative drag" in conection with aerodynamics, and NONE that talk about cavitation in a gas?

AH! Got it!!

You said Google, but meant Humoogle!!
 
No, you still don't have a clue, and are trying to make much of a definition. Got ya Google. Take a look for any of those terms. You can even find cavitation for air, if you look. It won't mean anything to you of course, Captain. There is a cretin certain amount of "negative credibility" when a pilot writes about the stubbornness of other pilots, only in order to soil himself.
Over.

But of course you're not going to link to any of them, because then people would follow the links and point out that they don't even come close to showing what you claim they show. Your act is wearing very thin, Humber.
 
(1) Do your own work, Greg, or accept mine
(2) Electrical maximum power theorem is one manifestation of that theorem.
Conversion if necessary, can be made using in the standard list of analogues.
http://www.eas.asu.edu/~holbert/analogy.html
.

There is nothing in that link that talks about maximum power theorem (MPT). MPT simply means that if your amplifier has an internal impedance of 4 ohms, then you'll get maximum power transfered to your speaker if your speaker has an impedance of 4 ohms.

There is no work for me to do. THere is nothing to accept. Maximum Power Theorem has nothing to do with anything we're talking about here.

And there's nothing in that link that says if you extract power from the difference beween wind and groudn that you can't use that power to go faster than the wind. Matching speaker impedances doesn't mean you can't go faster than teh wind.



Is it that I should have the expectation that you are infallible?

I pointed out several places were I admit I was wrong.

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

If you google "Maximum power theorem", you will find numerous websites with that exact phrase in it describign exactly what I'm describing. You prove me wrong by providing a link that doesn't even talk about "maximum power theorem".

One page is on wikipedia, it describes exactly what I'm talking about.

http://en.wikipedia.org/wiki/Maximum_power_theorem



Let me help you out. I was wrong, power transfer is organized and controlled by elves. Happy now?

You used a specific term to mean something it didn't. That's all.


You are being illogical; why ask a liar to tell the truth? If I were to be such a liar, do you expect that I would wilt under the glare of your piercing interrogation?

If you said 2+2=5, I'd expect you to admit it was a mistake.




Only if all parameters are the same. For example, to run the belt-generator, requires that the windmill first be accelerated to windspeed. That is KE that is not in the stationary generator. There will be some loss that is not recoverable when the motor is tuned off and the KE returned.
"Steady State" is misused to cover these sort of inconvenient facts.

So, the only inertial frame is one that doesn't require acceleration to get to?

That doesn't even make sense. An inertial frame is any frame moving at fixed velocity. You could have a frame movign at 0 mph over the ground, or a frame moving at 10 mph over the ground. Both are inertial frames. To get from one frame to the other, you would have to accelerate, but that doesn't mean they're not inertial frames.

teh standard example of inertial frame is to calculate teh trajectory of an artillery piece fired from a stationary position, then put the gun on a train, have the train move at a constant velocity, and fire the gun from the train. If the gun has radar to track its own shell, it will track the same trajectory on the ground or on the train.

On the ground with no wind, or on a train moving at 10 mph wtih a 10 mph tailwind are exactly the same for the gun.

on the ground with a 10 mph tailwind or moving at 10 mph with no wind are exactly the same for the gun.


Energy cannot come from the ground. See your own remark (2)
Why is there no energy when there is no wind?

YOu can only extract power if there is a difference, humber. YOu cannot extract power from 500 degree air if your machine adn all its components are also 500 degrees and you have no access to anything but 500 degree air.

you cannot extract power from teh wind if you're floating on the wind in a balloon at the same speed as the wind.

You can only extract power from the wind by tying it to something that isn't moving with the wind, liek the ground. The cart does that by gearing the prop to the wheel. As long as the air over the prop is different than the ground under the wheel, the cart can extract power, regardless of how fast the cart is moving.

And yes, you can extract power from the ground, if you're in a car moving at 50 mph and you need to slow down.

Why do you not see wind-generators on tracks?

wait. you just changed the subject here.

You said the treadmill in still air isn't equivalent to being on the ground with a tailwind. That the cart on the treadmill doesn't prove that the cart works on teh ground.

The reason you don't see windmills on tracks is because it takes power to move a track. from newtonian mechanics point of view, inertial frames are equivalent. But now you're talking about whats the best place to generate power, not whether the inertial frames are equivalent.
 
Any thoughts? I'm not interested in Greg's new scenario, like you asked me not to post any new scenarios. I want you to answer these simple questions about the one we were discussing.

Sorry, Greg, I'm sure it's fine when I get round to reading it.

Whivh videos? I have commented on the geared cart video. They have notinh to do with the boundary. I will not comment anything other than my drawings.
 
There is nothing in that link that talks about maximum power theorem (MPT).

Trust humber to pick a source full of errors. If q is displacement, current is not force - it's velocity. The author has confused two different analogues.

MPT simply means that if your amplifier has an internal impedance of 4 ohms, then you'll get maximum power transfered to your speaker if your speaker has an impedance of 4 ohms.

There is no work for me to do. THere is nothing to accept. Maximum Power Theorem has nothing to do with anything we're talking about here.

You're correct of course, but nevertheless I think there are mechanical analogues. The MPT follows from some properties of the relevant differential equations. I'm sure one can construct mechanical systems that also obey those equations. As an example, I'll bet one can transfer maximum power between two coupled damped mechanical oscillators when the frictions are equal.

Nothing to do with the treadmill, of course.
 
(1) The wind boundary is fastest adjacent to the wind, but slowest at the ground.
(2) The belt-wind is fastest at the belt, and slowest adjacent to the "wind"

For equal profiles, the speed of 6m/s or 8m/s or some other value, will be at different heights w.r.t the surface. Because the belt surface is the road surface, that is the reference point not the "wind", meaning that the profile is also wrong, being inverted.
Whoa, whoa, stop the presses. I think I finally understand what humber is trying to say when he says that "the boundary layer is going the wrong way". I think we can kill two birds with one stone here by also resolving the ladder/tissue question, by means of multiple choice.

First, here's a diagram of what the airspeed is at various levels abouve the ground on the "outdoors/wind" situation. I hope that everyone will agree that this is the accepted situation for laminar flow conditions:



Now, which of the following is the correct diagram for the boundary layer flow over a treadmill belt? We will make the assumption that the belt has the same surface characteristics as the ground (we can just cover the ground with belt material in the outdoor case to ensure that), and also assume that the treadmill is large enough to minimize edge effects.. I'll note that there are two possible answers, depending on whether the wind is being measured by someone standing on the belt (beltspeed observer), or on the floor next to the treadmill (windspeed observer). Either answer is fine, just specify which case you are considering, and either pick one of A through H, or say that none are correct (in which case you'll need to describe what would be correct).


[ETA: changed last diagram to make it less confusing.]
 
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And of course he has it wrong. The first automatic wind turbine was built on tracks so it could point itself into the wind. http://en.wikipedia.org/wiki/File:Wind_turbine_1888_Charles_Brush.jpg

Any gain, should it not be possible to incorporate it in a static design, will not be because of the motion, but better use of the wind.
(1) Yes. That is what that means DOH!
(2) 1888? Perhaps your next car will be a Stanley or Cleveland Steamer
Perhaps modern generators are more efficient, and do not warrant that effort. DOH!
(3) You contradict your mentor, Spork DOH!
(4) Even wind vanes automatically point inot the wind. DOH!
(5) In Australia, you can see fixed windpumps that do the same DOH!
 
And of course he has it wrong. The first automatic wind turbine was built on tracks so it could point itself into the wind. http://en.wikipedia.org/wiki/File:Wind_turbine_1888_Charles_Brush.jpg

humber wrote;
Any gain, should it not be possible to incorporate it in a static design, will not be because of the motion, but better use of the wind.
(1) Yes. That means what you wrote. DOH!
(2) 1888? Perhaps your next car will be a Stanley or Cleveland Steamer
Modern generators are more efficient and do not warrant that effort. DOH!
(3) You contradict your mentor, Spork DOH!
(4) Even wind vanes automatically point inot the wind. DOH!
(5) In Australia, you can see fixed windpumps that do the same DOH!
 
OK, here's my wind measurements, put onto a really quickly drawn sketch

picture.php


I think the main thing to note is that the bottom two images show that the tower moving at 10 mph over the ground with a 10 mph tailwind measures exactly the same windspeeds as the tower moving on a treadmill indoors.

So the two are exactly equivalent.

Edited to add: The windspeed measurements along the right side of the tower are all windspeed measurements as measured by the windspeed devices on teh tower. They show teh speeds that the cart would see.
 
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First, here's a diagram of what the airspeed is at various levels abouve the ground on the "outdoors/wind" situation. I hope that everyone will agree that this is the accepted situation for laminar flow conditions:


Oh, crap. THat's a lot more legible than my scribblings
 
Trust humber to pick a source full of errors. If q is displacement, current is not force - it's velocity. The author has confused two different analogues.

No you have. There are different models.
http://www.swarthmore.edu/NatSci/echeeve1/Ref/Analogs/ElectricalMechanicalAnalogs.html
That would not be the only mistake either. Mass = inductance.


You're correct of course, but nevertheless I think there are mechanical analogues. The MPT follows from some properties of the relevant differential equations. I'm sure one can construct mechanical systems that also obey those equations. As an example, I'll bet one can transfer maximum power between two coupled damped mechanical oscillators when the frictions are equal.
You think? de facto, mate.
You cannot have ever being employed in science, if you don't know that.

Nothing to do with the treadmill, of course.
No, that would be to make the losses perhaps the same, but not the power. The oscillators are reactive.
Everything to do with it, and Greg's yo-yo.
 
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Any thoughts? I'm not interested in Greg's new scenario, like you asked me not to post any new scenarios. I want you to answer these simple questions about the one we were discussing.

Sorry, Greg, I'm sure it's fine when I get round to reading it.

Whivh videos? I have commented on the geared cart video. They have notinh to do with the boundary. I will not comment anything other than my drawings.
Exactly, I'm not asking you to. I was asking you to ignore michael's videos and concentrate on your drawing and where we had got to in discussing them. I said:
Humber, I realise there's something new to distract you - or rather, quite old now - in the form of Michael's videos, but if you wouldn't mind answering the questions, please, I'm sure we'll have the misunderstanding cleared up in no time, and you will, of course, be proved right. I just want to get to the bottom of it.

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.
Would you please answer those questions. Last time you just said that the direction of the gradient depends on.... Just answer the specific questions with one of the options in brackets after them:

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.
 
Whoa, whoa, stop the presses. I think I finally understand what humber is trying to say when he says that "the boundary layer is going the wrong way". I think we can kill two birds with one stone here by also resolving the ladder/tissue question, by means of multiple choice.

No, he knows that both the profile and direction are wrong

First, here's a diagram of what the airspeed is at various levels abouve the ground on the "outdoors/wind" situation. I hope that everyone will agree that this is the accepted situation for laminar flow conditions:
[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b6095446c0.jpg[/qimg]

Similar to mine, posted months ago

Now, which of the following is the correct diagram for the boundary layer flow over a treadmill belt?
It is A
(1) Like my drawing of months ago
(2) Like the toilet paper of video #7. End game

We will make the assumption that the belt has the same surface characteristics as the ground (we can just cover the ground with belt material in the outdoor case to ensure that), and also assume that the treadmill is large enough to minimize edge effects..
Assume away.

I'll note that there are two possible answers, depending on whether the wind is being measured by someone standing on the belt (beltspeed observer), or on the floor next to the treadmill (windspeed observer). Either answer is fine, just specify which case you are considering, and either pick one of A through H, or say that none are correct (in which case you'll need to describe what would be correct).
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.
In the real world it's C, but moving the other way. (left to right)
The difference will again appear.


Checkmate.
 
Greg,

Still sure he's not a whacko?

Keep your hand on your joystick, Captain
Over.

Originally Posted by humber View Post
No, you still don't have a clue, and are trying to make much of a definition. Got ya Google. Take a look for any of those terms. You can even find cavitation for air, if you look. It won't mean anything to you of course, Captain. There is a cretin certain amount of "negative credibility" when a pilot writes about the stubbornness of other pilots, only in order to soil himself.
Over.

Really? Not any of the several times I've looked. If that's the case, why did you produce four links that DON'T use the term "Negative drag" in conection with aerodynamics, and NONE that talk about cavitation in a gas?

AH! Got it!
I should explain. It's not the sort of cavity search you are used to.
 
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No you have.

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.

You think? de facto, mate.

Gibberish.

You cannot have ever being employed in science, if you don't know that.

Utterly wrong, as always.

No, that would be to make the losses perhaps the same, but not the power. The oscillators are reactive.
Everything to do with it, and Greg's yo-yo.

Gibberish. Your arrogant stupidity has ceased being amusing.
 
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But of course you're not going to link to any of them, because then people would follow the links and point out that they don't even come close to showing what you claim they show. Your act is wearing very thin, Humber.

I don't know. I am having more trouble making the insults veiled.
Perhaps you can help?

Anyway, negative resistance, is but one of many such phenomenon in all sorts of materials and systems. It would do you no good to know of them.
 
It is A
(1) Like my drawing of months ago
(2) Like the toilet paper of video #7. End game

<snip>

It is A.

<snip>

Checkmate.

:dl:

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?
 
There is nothing in that link that talks about maximum power theorem (MPT). MPT simply means that if your amplifier has an internal impedance of 4 ohms, then you'll get maximum power transfered to your speaker if your speaker has an impedance of 4 ohms.

There is no work for me to do. THere is nothing to accept. Maximum Power Theorem has nothing to do with anything we're talking about here.

And there's nothing in that link that says if you extract power from the difference beween wind and groudn that you can't use that power to go faster than the wind. Matching speaker impedances doesn't mean you can't go faster than teh wind.





I pointed out several places were I admit I was wrong.



If you google "Maximum power theorem", you will find numerous websites with that exact phrase in it describign exactly what I'm describing. You prove me wrong by providing a link that doesn't even talk about "maximum power theorem".

One page is on wikipedia, it describes exactly what I'm talking about.

http://en.wikipedia.org/wiki/Maximum_power_theorem





You used a specific term to mean something it didn't. That's all.




If you said 2+2=5, I'd expect you to admit it was a mistake.






So, the only inertial frame is one that doesn't require acceleration to get to?

That doesn't even make sense. An inertial frame is any frame moving at fixed velocity. You could have a frame movign at 0 mph over the ground, or a frame moving at 10 mph over the ground. Both are inertial frames. To get from one frame to the other, you would have to accelerate, but that doesn't mean they're not inertial frames.

teh standard example of inertial frame is to calculate teh trajectory of an artillery piece fired from a stationary position, then put the gun on a train, have the train move at a constant velocity, and fire the gun from the train. If the gun has radar to track its own shell, it will track the same trajectory on the ground or on the train.

On the ground with no wind, or on a train moving at 10 mph wtih a 10 mph tailwind are exactly the same for the gun.

on the ground with a 10 mph tailwind or moving at 10 mph with no wind are exactly the same for the gun.




YOu can only extract power if there is a difference, humber. YOu cannot extract power from 500 degree air if your machine adn all its components are also 500 degrees and you have no access to anything but 500 degree air.

you cannot extract power from teh wind if you're floating on the wind in a balloon at the same speed as the wind.

You can only extract power from the wind by tying it to something that isn't moving with the wind, liek the ground. The cart does that by gearing the prop to the wheel. As long as the air over the prop is different than the ground under the wheel, the cart can extract power, regardless of how fast the cart is moving.

And yes, you can extract power from the ground, if you're in a car moving at 50 mph and you need to slow down.



wait. you just changed the subject here.

You said the treadmill in still air isn't equivalent to being on the ground with a tailwind. That the cart on the treadmill doesn't prove that the cart works on teh ground.

The reason you don't see windmills on tracks is because it takes power to move a track. from newtonian mechanics point of view, inertial frames are equivalent. But now you're talking about whats the best place to generate power, not whether the inertial frames are equivalent.

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.
 
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