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

[humber]

I don't think our alter-abled humber understands the difference. I think we can assume his answer (A) is for the "windspeed" observer, i.e. someone standing next to the treadmill and measuring the airspeed at different heights above it.

I must say I preferred 'the humber one'

Which raises the question: what would someone moving along with the belt measure? What happens if we add the windspeed vector to picture A?

An interesting conundrum, professor. I think the simplest solution is likely to be the best. An observer going at beltspeed, is stationary ,and not at windspeed.

 

[humber]

Humber,

Regarding:

the scenario I drew in 2411,
the two scenarios in the bottom half of that picture,
described in text in post 2390,

question for you:

How would the truck/tower know which of the two scenarios it is in if all of its sensor data measures teh exact same values? Spedometer says 10 mph, wind speed indicators on the top of the tower say zero.

How is it the truck/tower can measure the exact same inputs and yet you insist that the treadmill is not a legitimate equivalent to running on teh ground?

How would the truck/tower even know there is a difference?

Fuel bills.

 

[GregLondon]

That is partially correct. MPT does not imply the absolute level of transfer but the amount of that transferred. It does say that the maximum will occur when the load and source match. This is why terminal velocity, happens.

MPT has nothing to do with anything. It's an electrical engineering term. It only applies to physical motion if you play make believe. If MPT is a valid outcome of your make believe analogy, then there ought to be a physical law equivalent for the electrical engineering Maximum Power Theorem. If there isn't an equivalent physical law, then you're blowing smoke and the MPT doesn't apply to this discussion.

There is no need to talk in electrical metaphors other than to avoid talking about how the cart really works.

What is the physical law that is equivalent to the electrical max power theorem?

If there is such a law, talk about the cart in terms of that law. If there is no such law, you're blowing smoke.

 

[GregLondon]

How would the truck/tower even know there is a difference?

Fuel bills.

In both situations, the truck speedometer says its moving at 10 mph, and the windspeed indicators on teh tower say it has a 10 mph tailwind. In both cases, the truck would consume fuel at the same rate.


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

 

[Clive]

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:

http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b6095446c0.jpg

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).
http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b62962bc21.jpg
http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b65e557836.jpg

Devil's Advocate: "Clearly all these diagram must be showing all velocities w.r.t. the floor. If you attempt to interpret any of them as showing velocity w.r.t. the belt, then you are hoisted on your own petard immediately (ouch!), for how can the belt move to the left at 10 m/s w.r.t. itself as is clearly shown? (Similar paradoxes arise in the "corrected" paddle diagrams also of course.) Surely you are not mixing frames of reference in order to sew confusion jjcote?"

 

[Clive]

Hey Humber, let me just quickly say that I think you've produced some really vintage stuff in the last few pages - just like the "good old days". I had a few really good belly laughs. So, even if you don't know why (or do) thanks!

PS: try not to "learn" too fast - you may strain your brain.

 

[jjcote]

Does the observed belt speed change depending on what the observer is doing? In other words, doesn't an observer on the floor see it moving at 10 m/s, but an observer moving with the belt sees it as stationary (0 m/s)?

Don't see how. Both observers see the belt going at the same speed

Well, I guess this has now degenerated to a very simple utter lack of understanding. Seems to me that this is tantamount to 2 + 2 = 2, would you concur, spork?

 

[humber]

MPT has nothing to do with anything. It's an electrical engineering term. It only applies to physical motion if you play make believe. If MPT is a valid outcome of your make believe analogy, then there ought to be a physical law equivalent for the electrical engineering Maximum Power Theorem. If there isn't an equivalent physical law, then you're blowing smoke and the MPT doesn't apply to this discussion.

There is no need to talk in electrical metaphors other than to avoid talking about how the cart really works.

What is the physical law that is equivalent to the electrical max power theorem?

If there is such a law, talk about the cart in terms of that law. If there is no such law, you're blowing smoke.

No it does not. It's universal. The mechanical analogues fit that model.
It was developed from;
http://mysite.du.edu/~jcalvert/tech/jacobi.htm

It's fundamental. When you store energy in that pendulum you mentioned, the losses are also very low. Small bearing friction.
But, if the energy transferred to it from a mechanically 'real' source is 100J, then 200J must come from the source. 50% is lost in the source as perhaps heat. When the source and load are both real, i.e. dissipative, then 50% goes in the load and 50% is lost in the load when maximum power transfer occurs. A law of nature.

Too lazy to support your own argument.

 

[mender]

Stay away from the wight!

 

[John Freestone]

Is it not clear now, John? I mean after the subsequent posts and drawings?

No, it isn't. You see, you have said that the 6 m/s left to right BL velocity in real wind becomes a 6 m/s left, on the treadmill, and that makes me wonder if it's because it's always the same number, but in the opposite direction, for a particular elevation, do you see?

So in order to understand, I need to ask some questions. There are only two of them. Forgive me if I have missed the information. It can only take a moment of your time to say "Above. Yes" or "Below. No, it's 12". I think I'll be fine from there on my own.

Q1. Is the land-based 8 m/s left to right BL above or below the 6 m/s (obviously measured from the ground's frame of reference)? (Please answer 'above' or 'below', it can only be one or the other.)

Q2. Does that same elevation of air over the treadmill move at 8 m/s right to left, (real velocity, i.e. w.r.t. the still air/room/earth)? (Please answer 'yes' or 'no'. Again, it can only be one or the other. If no, you may want to tell me what velocity it does move at).

Thanks again again again.
John

 

[humber]

No, it isn't. You see, you have said that the 6 m/s left to right BL velocity in real wind becomes a 6 m/s left, on the treadmill, and that makes me wonder if it's because it's always the same number, but in the opposite direction, for a particular elevation, do you see?

It means that the belt flow is in the opposite direction of the real wind boundary layer. So apart from a few coincidental cases, will always be different. It's going the wrong way, and the profile is inverted.

So in order to understand, I need to ask some questions. There are only two of them. Forgive me if I have missed the information. It can only take a moment of your time to say "Above. Yes" or "Below. No, it's 12". I think I'll be fine from there on my own.

Q1. Is the land-based 8 m/s left to right BL above or below the 6 m/s (obviously measured from the ground's frame of reference)? (Please answer 'above' or 'below', it can only be one or the other.)

above

Q2. Does that same elevation of air over the treadmill move at 8 m/s right to left, (real velocity, i.e. w.r.t. the still air/room/earth)? (Please answer 'yes' or 'no'. Again, it can only be one or the other. If no, you may want to tell me
what velocity it does move at).

no
If the layer is say, 100mm and has a linear profile, what moves at 10mm above the ground, moves at that same speed 90mm above the belt, but in the opposite direction.

Thanks again again again.
John

 

[GregLondon]

No it does not. It's universal. The mechanical analogues fit that model.
It was developed from;
http://mysite.du.edu/~jcalvert/tech/jacobi.htm
It's fundamental. .

Forget the electrical metaphor, humber. Give me a mechanical law that is equivalent to the max power theorem or drop it.

When you store energy in that pendulum you mentioned, the losses are also very low. Small bearing friction..

Yeah, sure, fine.

But, if the energy transferred to it from a mechanically 'real' source is 100J, then 200J must come from the source

woah woah woah.

What do you mean a "real" power source? Are you telling me if I have a weight lifted up vertically with potential energy of 50 joules, I'll only be able to extract 25 joules from that weight, because your maximum power theorem for electricity says so?

50% is lost in the source as perhaps heat.

The maximum efficiency I can convert energy is 50%? Is that what your electrical metaphor is saying?

When the source and load are both real, i.e. dissipative, then 50% goes in the load and 50% is lost in the load when maximum power transfer occurs. A law of nature.

Wait, wait, wait.

You're telling me, that if I have some source of power, that's "real" power, what "real" power means in your vocabulary I have no idea, but if I have a source of this so called "real" power, then only half of that power can be converted into doing work? The rest must be lost as heat?

So, if I have something like, say, a hydroelectric dam that spins a turbine connected to a generator, if I calculate the energy I get from lowering a certain mass of water from some height, I can only get, at most, HALF of that energy converted into some mechanical energy like spinning a turbine, because your maximum power theorem from electrical engineering says so?

Say I have 1 kilogram of water, and it's 1 meter above the turbine outlet of the dam. Then that water would have m*g*h = 9.8 joules of potential energy, right? And you're saying that if I drop that 1 kilogram of water by 1 meter, teh most I can extract from that would be 4.9 joules?

You're saying if I have a mass of air, say 1 kilogram, and it's moving at a velocity of 1 meter per second over the ground, then the kinetic energy of that air would be 0.5 *m*v*v = 0.5 joules, and you're saying the most energy I could extract from that air would be 0.25 joules becaue the maximum power theorem, a theorem for calculating the best electrical impedance for an analog electrical circuit, says so?

Too lazy to support your own argument.

The cart works off of the difference between air and ground. Some carts float and work off the difference between air and water. so, if you want to support your max power theorem argument, I have one question for you: How do these cart-related sources of power (earth, air, water), these "real" sources of power, behave such that when we look at the power transfer, we find that "50% goes in the load and 50% is lost in the load when maximum power transfer occurs. A law of nature"

I can support my argument that you're totally wrong with a few simple examples:

a pendulum converts potential energy to kinetic back to potential energy back to kinetic energy without any loss of power during the conversion.

A newton's cradle transfers energy from one sphere to another with almost no loss of energy.

Billiard balls collide, transferring almost all the energy from one ball to another. head on shots can have the cue ball come to a complete halt and whatever ball it hit take on all the speed.

Of all the physics tests that I've seen that deal with collisions, I've never seen one say "reduce the kinetic energy of the incoming object by 50% to account for the max power theorem, that will be the maximum energy that can be transferred to the object it is about to collide with".

Again, humber, please tell me what physical law of motion is equivalent to the electrical Maximum Power Theorem.

I've got conservation of momentum, force = mass * acceleration, for every action there's an opposite action, conservation of energy. Newton's three laws plus conservation of energy.

I have NO IDEA what the maximum power theorem law would be for the motion of physical objects or the transfer of energy or power between physical objects.

And since this propeller cart is strictly a mechanical device, the law must have a mechanical equivalent that would affect physical motion. Not electrical.

 

[Clive]

If the layer is say, 100mm and has a linear profile, what moves at 10mm above the ground, moves at that same speed 90mm above the belt, but in the opposite direction.

Yay!

All of what you've said above is relative to your beloved earth, right Humber?

So.... RELATIVE TO THE BELT... the air flow at any given height above it will be moving in exactly the same direction and speed (relative to the ground) as what we see at the same height in the "real wind".

In other words, all the directions and speeds are the same if they are measured from the ground under the "real wind" in that case, and from the belt in the case of the treadmill.

PS: Of course I'm merely playing with you Humber. This can't possibly be correct for various reasons that I look forward to hearing from you.

 

[humber]

Forget the electrical metaphor, humber. Give me a mechanical law that is equivalent to the max power theorem or drop it.
Yeah, sure, fine.
woah woah woah.
What do you mean a "real" power source? Are you telling me if I have a weight lifted up vertically with potential energy of 50 joules, I'll only be able to extract 25 joules from that weight, because your maximum power theorem for electricity says so?
The maximum efficiency I can convert energy is 50%? Is that what your electrical metaphor is saying?
ait, wait, wait.
You're telling me, that if I have some source of power, that's "real" power, what "real" power means in your vocabulary I have no idea, but if I have a source of this so called "real" power, then only half of that power can be converted into doing work? The rest must be lost as heat?
So, if I have something like, say, a hydroelectric dam that spins a turbine connected to a generator, if I calculate the energy I get from lowering a certain mass of water from some height, I can only get, at most, HALF of that energy converted into some mechanical energy like spinning a turbine, because your maximum power theorem from electrical engineering says so?

Say I have 1 kilogram of water, and it's 1 meter above the turbine outlet of the dam. Then that water would have m*g*h = 9.8 joules of potential energy, right? And you're saying that if I drop that 1 kilogram of water by 1 meter, teh most I can extract from that would be 4.9 joules?

You're saying if I have a mass of air, say 1 kilogram, and it's moving at a velocity of 1 meter per second over the ground, then the kinetic energy of that air would be 0.5 *m*v*v = 0.5 joules, and you're saying the most energy I could extract from that air would be 0.25 joules becaue the maximum power theorem, a theorem for calculating the best electrical impedance for an analog electrical circuit, says so?

As far as the entire USA is concerned,oOnly 13% of all the energy available from the source, ever does useful work.

The cart works off of the difference between air and ground. Some carts float and work off the difference between air and water. so, if you want to support your max power theorem argument, I have one question for you: How do these cart-related sources of power (earth, air, water), these "real" sources of power, behave such that when we look at the power transfer, we find that "50% goes in the load and 50% is lost in the load when maximum power transfer occurs. A law of nature"
I can support my argument that you're totally wrong with a few simple examples: a pendulum converts potential energy to kinetic back to potential energy back to kinetic energy without any loss of power during the conversion.A newton's cradle transfers energy from one sphere to another with almost no loss of energy.
Billiard balls collide, transferring almost all the energy from one ball to another. head on shots can have the cue ball come to a complete halt and whatever ball it hit take on all the speed. Of all the physics tests that I've seen that deal with collisions, I've never seen one say "reduce the kinetic energy of the incoming object by 50% to account for the max power theorem, that will be the maximum energy that can be transferred to the object it is about to collide with".

Despite my making the distinction of 'real' and 'dissipative', my crystal billiard ball told me that you would raise this point.
When low-loss items such as billiard ball collide, a great deal of the energy is transferred. However, there is little work done. Idealised, none.That's the difference. Carts and parachutes, are real also in the sense that they do work. Now look that up in the physics book you did not consult before posting.

Your windmill analogy is not correct. A windmill on a treadmill, is not like the real thing. The aerodynamics are different. When real wind blows over a propeller, there is drag on the other side. This consumes power, and limits that power. There is a balance, a feedback system.
A treadmill's power is not limited, and will simply 'power' the loss and overcome the feedback control by brute force. You may get more power that way, amusing the efficiency is perhaps the 60% you cited.
"Equivalency" does allow you to swap relative velocities, but not the medium, or power source. That is why a car on the road is not like a car on belt, and why it cannot move w.r.t the belt unless it's engine is running, and so providing power. Like the real thing, in fact. The belt can change only the relative velocity. As far as the car is concerned, 'it can't tell' that the belt is moving. So, guess what? It can't.

 

[sol invictus]

"Equivalency" does allow you to swap relative velocities

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b62962bc21.jpg[/qimg]

So: what happens when you "swap velocities" from the frame of an observer standing next to the belt - which the humber tells us is option A - to that of an observer moving with the belt?

To do so, one subtracts the beltspeed vector <----- from every velocity vector in image A. As a hint for the mentally challenged poster in this thread, that's the same as adding ----->. The result of that operation appears here:

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b6095446c0.jpg[/qimg]

Remind you of anything?

 

[Dan O.]

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.

I agree. Before switching metaphors we would have to agree that all the properties are equivalent between the metaphors. Humber won't even agree to the simple properties of mechanical systems that have been known for over 300 years. If he can't see the equivalence of a simple translation in velocity, how is he going to know what rules from a electrical perspective will apply to the mechanical system.

 

[John Freestone]

Thank you for answering my questions simply. You don't often do that, and it's refreshing when you do.

It means that the belt flow is in the opposite direction of the real wind boundary layer. So apart from a few coincidental cases, will always be different. It's going the wrong way, and the profile is inverted.

I think that's sort of right, if you have a particular meaning of 'profile' and 'inverted'. Your details below explain that perfectly:

above

Yes, that's fairly easy isn't it - in the outside conditions, obviously the ground slows the air more lower down, so the 8 ---> is higher up than the 6 --->.

no
If the layer is say, 100mm and has a linear profile, what moves at 10mm above the ground, moves at that same speed 90mm above the belt, but in the opposite direction.

Again, this is what I say. You seem to be suggesting this situation (all velocities measured from the ground / room):

HEIGHT      OUTSIDE       INSIDE
120           10 >          0
110           10 >          0
100           10 >          0  ____theoretical 'beginning' of boundary layer
 90            9 >        < 1
 80            8 >        < 2
 70            7 >        < 3
 60            6 >        < 4  ----------- Case in your diagram
 50            5 >        < 5
 40            4 >        < 6
 30            3 >        < 7
 20            2 >        < 8
 10            1 >        < 9
 _0_           0          <10

So, do you see the point yet? The 6 m/s 'real wind' is above the middle, and that same layer of air is going at 4 m/s downbelt, not 6 m/s, as you have it in your diagram.

This may also solve the other argument we had, because if you imagine someone moving with the belt, you have to take "< 10" off each of those figures, or add "10 >". If you do that, you find that you get the same as in the first windspeed column, and nowhere does someone moving with the belt feel a wind from the up-side of the belt; it all hits the back of my man going backwards down the belt. It hits the back of his head and the back of his ankles. I didn't get clear about whether you agreed with that before or disputed it, but there were strong suggestions that you disputed it; you certainly failed to fill in the blanks or answer my direct questions that time.

Furthermore, you should be able to see that your own description of the gradient will cause balloons to travel towards your rabbit on the ground in 'real wind' exactly in the same fashion as s/he travels towards them on a treadmill in still air, and their separation is exactly the same too.

Indeed...well, I hardly need to go on, do I? You seem to have demolished your earlier argument...or have I missed something?

 

[semper]

Well, I guess this has now degenerated to a very simple utter lack of understanding. Seems to me that this is tantamount to 2 + 2 = 2, would you concur, spork?

Yeah, give it up Humber, this kind of complete lack of observation and reasoning capability is beyond even the most visceral excuse here. We are saying that the belt is both moving and not moving, simultaneously. This is an utter paradox for you. However we understand that it is only our perception which has lead to our feeling of paradox.

Logically consistent mathematical theory has been developed which yields accurate description of physical systems and can safely predict, predictable outcomes, and show where they are not predictable. For most people here who are participating in a consensus reality, that is enough. The theory is quite easy to learn and use and becomes a natural and mostly intuitive way of understanding motion of physical objects. It won't solve all problems, but your approach would seem to solve very few if any.

You might have any number of fascinating ideas, valuable insights etc, but they are wasted if you can't form any consensus with other people about some fundamentals of reality. You dispute the "validity of classical physics" as applied to mechanical problems of motion, thats about all you seem to do.

Sometimes things come along which seem to confront the conventional notions. DDWFTTW is a classic and wonderful case. When forms of non-intuitive energy transfer and practical use are observed, they are often met with skepticism. This can be quite healthy, usually necessary, and hopefully progressive. Spork's foray into ensuring healthy skepticism and progressive analysis has been brilliant, most people involved have learn't really valuable truth. Just as valuable is to allow a clear picture of how some people resist the flow of learning, we can't say why, but we see clearly how, and you are making a steady contribution.

 

[semper]

Does the observed belt speed change depending on what the observer is doing? In other words, doesn't an observer on the floor see it moving at 10 m/s, but an observer moving with the belt sees it as stationary (0 m/s)?

Don't see how. Both observers see the belt going at the same speed, so it would be odd that the viscously attached layer would not be seen doing the same thing.

Well, I guess this has now degenerated to a very simple utter lack of understanding. Seems to me that this is tantamount to 2 + 2 = 2, would you concur, spork?

Q.E.D.

 

[humber]

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b62962bc21.jpg[/qimg]

So: what happens when you "swap velocities" from the frame of an observer standing next to the belt - which the humber tells us is option A - to that of an observer moving with the belt?

That is not the point. All view are equivalent. If instead of being a dedicated observer, I watch as if it were a movie, and in the real world, I would perhaps see the sequence from the top, (the wind) 10...9,8,4,0,. left to right. From the treadmill side, movie mode, I would see from the top 0...4,8,9,10, but moving right to left. Later moving the observer, cannot change that.relationship.
The difference is, that an observer attached to the belt, generates a wind due to motion through the air, that is not present for the windspeed observer. The net result is different form the real wind, and for different velocities up the belt. This is the result of the cart going the wrong way (it should be powered, and so go with the belt like the car), and the wind being powered by the belt,and not an external source.

To do so, one subtracts the beltspeed vector <----- from every velocity vector in image A. As a hint for the mentally challenged poster in this thread, that's the same as adding ----->. The result of that operation appears here:

But not if the power comes from the road. That vector is OK, but all other relationships must be preserved.
The real boundary is not powered by the road, but the wind. The belt is 'dumb' power source, the wind is a dynamic, interactive power source It controls the process, and cannot be a 'slave' if it is to be correctly modeled. The cart is powered by the wind via the propeller, not via the wheels.

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b6095446c0.jpg[/qimg]

Remind you of anything?

Yes. http://www.burj-al-arab.com/