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

[spacediver]

Two batteries connected in series opposition A +[ V ]--[ V ]+ B
The voltage between A and B = 0V. I say that has no meaning.

It has meaning. It is coherent and intelligible, and conveys something about the world.

The following propositions are meaningless:

The train was moving at a mass of three acres.
I am telephone my toe is
The kinetic energy of the pencil is negative 20 joules with respect to the desk.

The following propositions are meaningful:

The voltage between A and B is 5 volts
The voltage between A and B is 0 volts

10 minus 3 is 7
6 minus 6 is 0

ETA: just realized I've been using Newtons as a unit of energy this whole time - DOH!

 

[humber]

No not a flippant question - just a stab in the dark. Your thought processes seem a bit accelerated and jumbled at times, though lately you've been a bit more focused. No offense intended - just a wild stab in the dark as I said.

I know what you meant SD, and no offense has been taken, but do you think that perhaps I am trying to explain things that are so very fuzzy, yet it is assumed that all of the 'confusion' is my doing alone?

 

[humber]

I keep thinking of that line from the Monty Python film The Holy Grail, "On second thought, let us not go to Camelot; it is a silly place."

I was thinking of the Castle Anthrax from the same film. You know the one, where all the girls want a spanking?

ETA:
"can analyze the complex"
You don't think my description of the cart's operation is not remarkably similar to that of the Russian puzzle? How did I work that out, John?

 

[CORed]

I was thinking of the Castle Anthrax from the same film. You know the one, where all the girls want a spanking?

I'm thinking of the scene where they prove the woman is a witch. Lots of humber-style "logic" in that scene.

 

[humber]

And yet
He once again conclusively demonstrates that he has no idea what the ideas of physics actually are. Only in the Humberverse is the ground magic.

Yes, CORed. I am suggesting that the treadmill violates many accepted scientific principles and so making necessary such matters as;
(1) motion without displacement,(?!)
(2) a vehicle at rest can be said to be moving.
(3) Real motion over the ground without KE?

Now it may well be true that kinetic energy is relative, but that does not mean that you can make it so at will. For example, by what mechanism can the implied KE of the cart be transferred to the belt, so as make the numbers add up?
Now you don't have to take "beltworld" literally, to conclude that is strange that it is 'possible', do you?

He once again conclusively demonstrates that he has no idea what the ideas of physics actually are. Only in the Humberverse is the ground magic.

In CORedworld, all objects are like quarrelsome neighbors always tying to defy each other "to tell" where the fence line is.

 

[Subduction Zone]

(1) motion without displacement,(?!)
(2) a vehicle at rest can be said to be moving.
(3) Real motion over the ground without KE?

Three statements by humber that prove he has no idea what a frame of reference is.

 

[jjcote]

In the Japanese baseball-gun-on-a-truck video, we see that the baseball drops to the ground.
1) Does it have any lateral motion when they fire it out of the gun?
2) If you were on a trailer being pulled by the truck, directly in the line of fire, would the baseball hit you?
3) How would it feel?
4) Would it feel different than if the truck was parked when they fired the gun?
5) What about if instead of firing the baseball out of a gun, they just dangled it on a string and let you come along in the 100 kph trailer and hit it. Would that feel the same, or different?
6) What's moving in these scenarios?
7) What objects have displacement?
8) How much KE do the objects have?
9) From a physics point of view, are they all the same?
10) If you were standing on the track when the baseball was fired, could you just reach out and grab it barehanded?

You will be graded on your answers.

 

[spork]

do you think that perhaps I am trying to explain things that are so very fuzzy, yet it is assumed that all of the 'confusion' is my doing alone?

Nothing "fuzzy" except your understanding of this puzzle. And I'm pretty sure that's intentional.

In the Japanese baseball-gun-on-a-truck video, we see that the baseball drops to the ground.
1) Does it have any lateral motion when they fire it out of the gun?
2) If you were on a trailer being pulled by the truck, directly in the line of fire, would the baseball hit you?
3) How would it feel?
4) Would it feel different than if the truck was parked when they fired the gun?
5) What about if instead of firing the baseball out of a gun, they just dangled it on a string and let you come along in the 100 kph trailer and hit it. Would that feel the same, or different?
6) What's moving in these scenarios?
7) What objects have displacement?
8) How much KE do the objects have?
9) From a physics point of view, are they all the same?
10) If you were standing on the track when the baseball was fired, could you just reach out and grab it barehanded?

You will be graded on your answers.

Wait - I know this one! Is it "ten questions that humber won't answer"?

 

[Clive]

Humber,

Seeing that our earlier exchanges didn't seem to lead anywhere useful, here's another thought experiment for you to consider. You might see shadows of the hobo on the train and so on, but I'm hoping you won't just brush me off.

Say we have two large ocean going liners, sister ships, essentially identical (at "macroscopic level at least). There is a special "Classical Physics" room on each ship reserved for passengers interested in that kind of thing. As such these rooms have various apparatus and measuring equipment and so forth so the lucky occupant can perform experiments involving macroscopic objects, balls, blocks, springs, etc. There is also a green light above the entrance door that is lit when the ship is moving at constant velocity (including not moving at all) relative to the surface of the Earth. We'll assume nice smooth seas also so even when at sea and moving (and when the green light is on) there will be no significant tilting or vibrations in the room in terms of the experiments we might want to conduct. Assume no external forces have any significant effect on the experiments that we do inside the room except for gravity. You can have a window. And of course you may have a glass of your favourite beverage on the table with no worries about spillage or sloshing (in my opinion at least) - because the ship moves very smoothly!

Okay, so let's say you are in the special "Classical Physics" room on one of these ships and it is berthed in some port. I.e. It is not "moving". I am in the other ship out at sea somewhere, travelling (close enough) at some constant speed and direction. We note that our green lights are on and conduct various experiments as agreed in advance and record the results.

Assuming these experiments are not sensitive enough to be affected by things like the "Coriolis Force" and as already noted it's really only gravity that has any significant influence on our experiments from the "outside", do you expect there to be any differences in any of the results that we record (beyond what could reasonably be expected from the limited precision of our measuring tools and so on)? I'm not specifying the experiments precisely but this isn't meant to be a "trick". Just think of the typical range of things people might be able to do during classes at high school level physics if that makes sense. Dropping objects from a height, rolling things down slopes, measuring speeds, swinging pendulums, etc, etc.

I expect your answer to be something like "yes, the results will be the same", but I just want to be sure before continuing. But if it is no, then please feel free to give examples and explain further.

 

[Subduction Zone]

Nothing "fuzzy" except your understanding of this puzzle. And I'm pretty sure that's intentional.




Wait - I know this one! Is it "ten questions that humber won't answer"?

No, no you're wrong!! The answer is the truck is not an inertial frame of reference.

 

[CORed]

Yes, CORed. I am suggesting that the treadmill violates many accepted scientific principles and so making necessary such matters as;
(1) motion without displacement,(?!)

The cart is displaced, relative to the belt, or any speck of dust that happens to be stuck to it.

(2) a vehicle at rest can be said to be moving.

Vehicle moves relative to the belt.

(3) Real motion over the ground without KE?

Of course something moving relative to the ground has kinetic energy of mv^2/2, with respect to the ground, or any object stationary relative to the ground. It has 0 KE relative to anything that has 0 velocity relative to it.

Now it may well be true that kinetic energy is relative, but that does not mean that you can make it so at will. For example, by what mechanism can the implied KE of the cart be transferred to the belt, so as make the numbers add up?
Now you don't have to take "beltworld" literally, to conclude that is strange that it is 'possible', do you?

The fundamental error you continue to make, humber is to regard kinetic energy as an intrinsic property of an object. Kinetic energy is simply a measurement of the energy required to accelerate an object to a velocity relative to your frame of reference, or the energy that will be transferred if you decelerate it to a velocity of zero relative to your frame of feference Certainly you can calculate kinetic energy relative to any frame of reference, but it is only meaningful if the object interacts with something at rest in that frame of reference.

Say I am in a car with a mass of 1000kg, traveling at a speed of 30 m/s, relative to the ground. That car has 450,000 Joules of kinetic energy, relative to the ground. If I run the car into a concrete wall, that energy will be expended crumpling the body, frame and other parts of the car (and very likely crushing my body), and ultimately converted to heat. If, instead of running into the wall, I wisely brake the car to a stop (again, relative to the ground), the brakes will generate 450,000 Joules of heat. Also, if I want to accelerate back to 30 m/s (In a direction that won't run me into that wall, one would hope), I will have to expend another 450,000 Joules of energy by burning gasoline (and waste quite a bit more as heat) to do so.

OTOH, if I run into a big truck going 29 m/s (I'm making a simplifying assumption hear that the truck is so massive that its change in velocity is negligible), 500 Joules of kinetic energy will be expended denting my front bumper, and again, ultimately converted to heat. If , again, I take the more sensible course, and tap my brakes to match my speed with the truck, my brakes will convert 500 Joules of kinetic energy to heat. Where did the other 449,500 Joules go? Nowhere, of course. If I want to stop, I'm still going to have to convert it to heat. But, if I want to figure out how much damage I'm going to do to my car running into that truck going 1 m/s slower than me, or how much heat my brakes are going to have to dissipate to avoid running into it, 500 Joules is the meaningful figure, not 450,000 Joules.

I could, if I wanted to, calculate the kinetic energy of my car relative to the sun. I'm too lazy to look the velocity up, and crunch the numbers, but let's just say it's going to be a really big number. However, that number is only meaningful if somebody somehow manages to park a truck, stationary relative to the sun, in front of me. In that case, my brakes aren't going to save me. I'm going to need some really big, powerful rockets (powered by antimatter, maybe). But since that hardly ever happens, I don't worry about it too much. So, when I calculate my kinetic energy relative to the ground, where does all that kinetic go? Nowhere: it's still going to smash me and my car to atoms, just as soon as somebody manages to park that truck, stationary relative to the sun, in front of me.

 

[humber]

O.K. let's expose humber for the 11,397th time this week:

What question? You talk about my inability to answer a simple question. Give me the question. You haven't formed one. Hint: it should be a complete sentence with a subject and everything - and it should end with a question mark. Point me to that post.

Yes. I would have thought the colored smoke test to be obvious, but it is perhaps clearer when diagrammatically expressed.

Clearly the boundary layers are not same in each case. The cart is not moving relative w.r.t the apparent driving wind or the boundary layer, suggesting that the cart is not in motion, yet in direction, opposite to expectation. It would seem that lack of any real wind excludes correction of this situation.

 

[humber]

The cart is displaced, relative to the belt, or any speck of dust that happens to be stuck to it.

No, the belt moves away from the cart. (Heresy!)

vehicle moves relative to the belt.

The cart is not in motion, CORed

Of course something moving relative to the ground has kinetic energy of of mv^2/2, with respect to the ground,

Yes. But "of course"? How can you say that?

or any object stationary relative to the ground. It has 0 KE relative to anything that has 0 velocity relative to it.

Agreed. Stationary objects are said to have zero of both. "W.r.t the ground" is redundant, considering you have little or no choice in the matter.

Now it may well be true that kinetic energy is relative, but that does not mean that you can make it so at will. For example, by what mechanism can the implied KE of the cart be transferred to the belt, so as make the numbers add up?
Now you don't have to take "beltworld" literally, to conclude that is strange that it is 'possible', do you?

Begets

The fundamental error you continue to make, humber is to regard kinetic energy as an intrinsic property of an object. Kinetic energy is simply a measurement of the energy required to accelerate an object to a velocity relative to your frame of reference, or the energy that will be transferred if you decelerate it to a velocity of zero relative to your frame of feference Certainly you can calculate kinetic energy relative to any frame of reference, but it is only meaningful if the object interacts with something at rest in that frame of reference.

Simply? Too simple, CORed! How does the cart get the KE it is said to have? Your remarks suggest that what happens to that object over time - its history - may be important.

Say I am in a car...

In a tuxedo, not as a hobo. But, yes.

with a mass of 1000kg, traveling at a speed of 30 m/s, relative to the ground. That car has 450,000 Joules of kinetic energy, relative to the ground. If I run the car into a concrete wall, that energy will be expended crumpling the body, frame and other parts of the car (and very likely crushing my body), and ultimately converted to heat.

Yes. "Conservation of energy".

If, instead of running into the wall, I wisely brake the car to a stop
(again, relative to the ground), the brakes will generate 450,000 Joules of heat.

Yes, but there is the definite physical mechanism of friction present to transfer the car's KE to the brakes, and so to heat.

Also, if I want to accelerate back to 30 m/s (In a direction that won't run me into that wall, one would hope), I will have to expend another 450,000 Joules of energy by burning gasoline (and waste quite a bit more as heat) to do so.

Yes. But it would be a waste of fuel and your time to demonstrate that.

OTOH, if I run into a big truck going 29 m/s (I'm making a simplifying assumption hear that t he truck is so massive that its change in velocity is negligible), 500 Joules of kinetic energy will be expended denting my front bumper, and again, ultimately converted to heat. If , again, I take the more sensible course, and tap my brakes to match my speed with the truck, my brakes will convert 500 Joules of kinetic energy to heat. Where did the other 449,500 joules go? Nowhere, of course. If I want to stop, I'm still going to have to convert it to heat. But, if I want to figure out how much damage I'm going to do to my car running into that truck going 1 m/s slower than me, or how much heat my brakes are going to have to dissipate to avoid running into it, 500 Joules is the meaningful figure, not 450,000 Joules.

Yes, again. All present and correct. All quantities are "meaningful" because they are real.

I could, if I wanted to, calculate the kinetic energy of my car relative to the sun. I'm too lazy to look the velocity up, and crunch the numbers, but let's just say it's going to be a really big number.

Not lazy, just sensible. It is so big, that it is vast, and the Universe to which the Sun is relative, vast with a capital V. Are you respecting another even bigger 'reference'?

However, that number is only meaningful if somebody somehow manages to park a truck, stationary relative to the sun, in front of me. In that case, my brakes aren't going to save me. I'm going to need some really big, powerful rockets (powered by antimatter, maybe). But since that hardly ever happens, I don't worry about it too much. So, when I calculate my kinetic energy relative to the ground, where does all that kinetic go? Nowhere: it's still going to smash me and my car to atoms, just as soon as somebody manages to park that truck, stationary relative to the sun, in front of me.

Again, you are employing real mechanisms to make your metaphorical examples, real.

How do you do that for the cart and treadmill belt, CORed ?

 

[humber]

I'm thinking of the scene where they prove the woman is a witch. Lots of humber-style "logic" in that scene.

Including "that is how we know the world to be banana-shaped"

 

[spork]

Yes. I would have thought the colored smoke test to be obvious, but it is perhaps clearer when diagrammatically expressed.

I don't see a diagram.

Clearly the boundary layers are not same in each case.

Like I said, I don't see a diagram. But if the diagram is done properly the boundary layers will be identical.

 

[mender]

I won't post this as a brain teaser but if you are floating in the basket of a balloon at the exact same speed as the air just above the ground over a flat snow covered field (featureless landscape) when it is snowing reasonably hard, could you tell if the wind is blowing or not?

That is one of the benefits of living in the land of the ice and snow. I don't need a smoke machine to see what is happening in a wind. The snow shows that very nicely.

For example, most Canadians know that if you're driving along in a blizzard blowing directly at you and you can hardly see the road, all you need to do is find a place to turn around, and when you go the other way the blizzard turns into a gentle snowfall. When you match the speed of the wind exactly, the snow looks like it is falling straight down - except just above ground level, where the snow looks like it is suddenly moving towards the car. That's what happens when the snow enters that slower moving boundary layer.

Guess what? The direction of the boundary layer is just like what a smoke test on the treadmill would show if a video was taken while riding on the cart! Amazing!

 

[humber]

That is because of the way you use them. Here's an example:
"The torque delivered to the propeller is in opposition to the wheel, so there can be little load, and therefore no axle motion."
It sounds like you are saying "there is no load across a voltage, so there will be no current flow", which does not equate to a mechanical system.

I was not making an analogue there, Mender, but I will. In the simplifying case, toruqe is a linearly related to current, and so to force. The gears reverse or 'reflect' these currents so as to be in opposition, and much of the expected heat (work) produced would all be within the cart. There is no significant equivalent voltage (velocity) being developed, suggesting that the work being done is very low, or at least inefficiently employed.

There is also the matter of reactive (complex) loads, where energy is stored. Inductance needs no voltage to sustain a current. The mechanical analogues can include mass and other storage mechanisms, such a springs and propeller torsion tubes.

I've noticed this quite a few times. Mechanical systems are not the same as electrical systems, and trying to apply the same logic to energy flows will result in bizarre term useage, statements that don't make sense, flawed reasoning, etc. and will prevent you from ever understanding what the cart is really doing while on the road and on the treadmill.

No, they arevery close, Mender. Perhaps only "ground" is the exception, but that is notional anyway.
Both mechanical and electrical devices can be described by a set of differential equations, so it should be possible to convert from one to the other. Because electricity is abstract as far as we humans are concerned, a system of notation was developed to describe it. It need not only apply to electrical devices. It is extremely useful.

As part of one project, I helped develop a low voltage DC/DC converter to take 1v and 12 amps to a more usable 13.8v and .8 amps. It started at about 80% efficiency and ended up at 91%, with a little room for more but was good enough. I initially had to come up with mechanical analogues to understand how it worked so I could think of ways to improve the performance. It was different enough from my normal way of thinking to stump me at first but with the help of a professor of EE (the designer of the circuit), I was able to get it and improve it.

Yes, I got the idea that you worked in that field when you mentioned regenerative braking, but that is what I am suggesting. If you can work one way, why not the other ? Yes, 80/85% is typical for a good design, and 91% is an improvement!

You have a number of people here who would like very much to help you get this, but you have to listen to what they say rather than argue. You've come close a few times but then you take off again on a rant.

I know, Mender, I understand and appreciate that, but what you are all doing, is mixing the figurative withe real. If I demonstrate real physical errors, that seems to have little effect, because the figurative always wins.

 

[humber]

I don't see a diagram.



Like I said, I don't see a diagram. But if the diagram is done properly the boundary layers will be identical.

You are right. That is yet another description of the same treadmill error. So now that is cleared up, perhaps you would like to
(a) How me the errors within those diagrams. (Without reference to ice-carts please)
(b) Present diagrams with your interpretation of events.

Otherwise, I will assume that you can't do either.

 

[humber]

I won't post this as a brain teaser but if you are floating in the basket of a balloon at the exact same speed as the air just above the ground over a flat snow covered field (featureless landscape) when it is snowing reasonably hard, could you tell if the wind is blowing or not?

That is one of the benefits of living in the land of the ice and snow. I don't need a smoke machine to see what is happening in a wind. The snow shows that very nicely.

For example, most Canadians know that if you're driving along in a blizzard blowing directly at you and you can hardly see the road, all you need to do is find a place to turn around, and when you go the other way the blizzard turns into a gentle snowfall. When you match the speed of the wind exactly, the snow looks like it is falling straight down - except just above ground level, where the snow looks like it is suddenly moving towards the car. That's what happens when the snow enters that slower moving boundary layer.

Guess what? The direction of the boundary layer is just like what a smoke test on the treadmill would show if a video was taken while riding on the cart! Amazing!

I didn't realise that Canadians had visceral experience of riding on a cart on a treadmill, Mender. Perhaps you can show me the errors in my diagrams?

 

[humber]

I don't see a diagram.



Like I said, I don't see a diagram. But if the diagram is done properly the boundary layers will be identical.

ETA:
The diagrams appear on my screen, but just in case, here is the attachment.