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

[Christian Klippel]

Oh, and by the way, humber.

I take your evasion of explaining where and what in my diagrams was false as an acknowledgment that in fact it was correct and that you are unable to spot an error. And even more, i take it as an acknowledgment that you are unable to admit that you were wrong.

Anyone else taking it this way?

 

[John Freestone]

Oh, and by the way, humber.

I take your evasion of explaining where and what in my diagrams was false as an acknowledgment that in fact it was correct and that you are unable to spot an error. And even more, i take it as an acknowledgment that you are unable to admit that you were wrong.

Anyone else taking it this way?

I am, and I don't think I've seen your diagrams yet.

 

[Clive]

(1)In nature, the boundary layer flows in the direction of the wind, but slower.

(2)On the treadmill, the tailwind is said to flow up the belt, to drive the rear ofthe cart. The boundary layer, moves with the belt, and so in the opposite direction of the tailwind.

(3) It is not possible to reconcile this situation in a manner that makes them equivalent.

(4) The "layers" do not matter, because of that eclipsing error.

See below for a version with highlighted clarifications and explanations that make what you've written less ambiguous and more accurate.

(1)In nature, relative to the surface below the wind, the boundary layer flows in the direction of the wind, but slower.

(2)On the treadmill, the tailwind is said to flow up the belt, to drive the rear ofthe cart. The boundary layer, moves with the belt, and so in the opposite direction of the tailwind, but slower. Note that by talking about the "direction of the tailwind" being "up the belt" we're implying use of something like "relative to the belt", but then to get "opposite direction" when talking about the boundary layer we have to be using a reference frame that is moving reasonably quickly in the same direction as the "tailwind" but not too fast otherwise there is room for confusion about what direction the "tailwind" is going in now! That kind of forces us to jump around between different viewpoints because of the awkwardly worded original text. It would be much easier to simply say something like this: "On the treadmill, relative to the belt, the boundary layer flows in the direction of the induced wind, but slower". Look at how similar that is point 1.

(3) If you are easily confused or willfully stupid, it is not possible to reconcile this situation in a manner that makes them equivalent.

(4) The "layers" do not matter, because of that eclipsing error, unless there is an even bigger eclipsing error otherwise known as "humber".

Humber, do you need someone to explain to you how "directions" can change delending on the relative movements and orientations between "observer" and the other objects in "the world"?

Say we have two objects constrained to a straight line (through space or on some surface, it doesn't really matter). The distance between those two objects (measured along the line) is steadily increasing (say at 4 m/s). You are observing them from some other fixed point that is not on the same line as the objects. To keeps things reasonably simple let's also say that your head (and hence eyes also) are pointing towards the general vicinity of the two objects, and the line as a whole is what you would called "horizontal". (You are looking at the objects!)

Now, what further information do you need in order to determine unambiguously if the objects appear to to be both moving in the same direction along the line (two choices there), or moving in opposite directions? All of these are possible.

 

[John Freestone]

Sorry Christian, but you know very little. That very reference shows that it is an impedance, a complex frequency dependent impedance, (Z)
These effects are present in flows. Standing waves in water, air?
Also acoustics, again. How do you think a Helmholtz resonator works?

Shame on you. That's a cruel way to catch fish, with a Helmholtz resonator.

 

[Christian Klippel]

I am, and I don't think I've seen your diagrams yet.

Hey John,

here they are, some crude ASCII art.

Don't forget to read humb's refutation, because, well, it's just prue comedy gold, imho.

Greetings,

Chris

 

[Christian Klippel]

Say we have two objects constrained to a straight line (through space or on some surface, it doesn't really matter)......

You mean, as if someone was driving on the highway at 20 km/h and you would be overrunning him at 40 km/h? Yes, in our universe, we would see that 20 km/h car going backwards, but in the humberverse, it would just shooting by ahead of you.

Greetings,

Chris

 

[John Freestone]

humber,
Outdoors, "real" wind coming from behind you. You are traveling at the speed of that wind (e.g. 10 m/s wind from the south, and you are going north at 10 m/s). If you reach down to ankle level, you will feel the air in the boundary layer coming toward you, from in front of you.

Agree or disagree?

That can't be. Two winds going in opposite directions!?

 

[humber]

Then explain where it is wrong. Feel free to use your own numbers. Draw a simple diagram showing the airflow and the flow of the slower boundary layer at different observer speeds.

Can you do that, or do you evade that simple request as well?

I don't see your point, but OK, Christian.

Wind = 10km/hr
Boundary = 2km/hr
Difference = 8km/hr.
Yes.

Observer moving at 5km/hr (10km wind / 2km boundary)
-----------Subtract------Add
Wind = 5km/hr or 15km/hr
Boundary = -3km/hr or 7km/hr
Diff = 8km/hr or 8km/hr
Yes.

In real wind the difference is independent of observer travel, so would expect that to be the case at windspeed too; 8km/hr.
But as you see, the observer has no influence on the flows, they are both at 10km/hr and 2km/hr. They don't go slower or move backwards. There are changes form the observer's point of view, but so what?

OK, we agree for real wind, but that is not the case with the treadmill.
On the treadmill, there are definite changes in the winds themselves, and that is not the case for real wind.

On the belt, it is not so clear. If the beltspeed is the windspeed, (10km/hr right to left) then the boundary is 8km/hr slower. So that must be 2km/hr (right to left)
If windspeed 'air' is zero w.rt the observer, then that flow is -2km/hr (the cart speed is positive left to right).

So for the cart observer
Wind = 0 (windspeed)
Boundary = -2km/hr.

Now, the observer on the belt is moving backwards at 10km/hr or -10km/hr w.r.t the cart observer. To get that observer's speed subtract 10km/hr

Wind = ( 0km/hr) - (10km/hr) = -10km/hr
Boundary = (-2km/hr) -(10km/hr) = -12km/hr
The difference is (-10)- (-12) = 2km/hr
That is the opposite direction of the windspeed condition. of -2km/hr

If you argue that the boundary flow should not be changed then;
Wind = ( 0km/hr) - (10km/hr) = -10km/hr
Boundary = (-2km/hr) = - 2km/hr
The difference is (-10)- (-2) = -8km/hr
That is the samedirection of the windspeed condition, but -8km/hr

Either way , it is not the same.

 

[John Freestone]

To simplify, then, humber, when wind is blown over a surface, the b.l. is in the same direction, but if that same (or similar) surface is moved across (below, through?) an air mass, the boundary layer is reversed.

Could you now help me understand what physical limits or conditions apply to this law - the kinds of surfaces and fluid masses it applies to. I mean, was I right in my prediction that it means that an aeroplane's wing has a boundary layer travelling backwards - tsk, look at me - forwards - over it (not in a windtunnel, of course, where the wind is powered, just when the wing is powered, via the engine, through stationary air)? Ta eversomuchly.

 

[Clive]

On the belt, it is not so clear. If the beltspeed is the windspeed, (10km/hr right to left) then the boundary is 8km/hr slower. So that must be 2km/hr (right to left)
If windspeed 'air' is zero w.rt the observer, then that flow is -2km/hr (the cart speed is positive left to right).

So for the cart observer
Wind = 0 (windspeed)
Boundary = -2km/hr.

You've screwed it up again, almost immediately.

Choose a reference frame, say the "still air" above the treadmill. ("Cart observer" in your terminology.) The "wind" (air) in that frame is moving a 0 km/hr. You don't need all that other waffle. Now, the boundary is 8km/hr slower (as determined by your set-up and calcs with the "real wind").

So.... watch carefully... nothing up my sleeve....

0-8 = -8 km/hr.

Excellent!

Correct version:

So for the cart observer
Wind = 0 (windspeed)
Boundary = -8km/hr.
​

Now proceed with the rest of your calculations...

 

[humber]

You must not be a very imaginative person; I would think that sound generation in crickets and the dynamics of dishwashing are much more inappropriate.

I imagine you would. So according to you, a self-powered fish, in motion, in water, is like a cart driven from the wheels, but with no motion, in still air? Perfect match.

I told you above. You cannot introduce vertical velocities when modelling a boundary layer, which is necessarily a horizontal phenomenon. Therefore, you are changing the relative velocities of everything involved and your model becomes inherently inaccurate. If you want an accurate boundary layer, use a bigger belt. The flow will become nice and turbulent automatically and the boundary layer thickness will increase until it is no longer different from a wind induced boundary layer.

That is not the point and arguable. I can control the flow as I wish. That is only a matter of technology. I could use other many other methods. An array of MMIC's perhaps.
The headwind from the fan would combine as expected. In fact, there is no reason why I could not pre-condition that as well.

Belts induce harmonic motion into the flow, so they do have side effects.
Motion of the belt is not necessary to the creation of the boundary conditions under the car, so there can be no implied motion for the car. That is the point. The belt is used to imply motion on the treadmill. That is false.

Also, you are mistaken. Though perhaps "religiously blunt", this was not the reaction of someone presented with an idea that goes beyond them. This was the exasperated reaction of someone faced with a wall of smug ingorance.

Self-contradictory. Not unlike yourself.
I mentioned your personal appeal to authority on our first encounter. You have done it again, and with no justification.
I do not see that you have met your other claim that your knowledge of relative velocities or equivalence affords you expert status. In fact, the offering is about as far wrong as it is possible to be, and not original.
-----------------
Now we have that out of the way, perhaps you would like to tell me how Bainbridge's device is comparable with the treadmill, as you claimed you could in #1738. Two similarities, with exemptions do not make for "equivalency".
You may chose another device if you like.

 

[humber]

You've screwed it up again, almost immediately.

That's a sure sign you have.

Choose a reference frame, say the "still air" above the treadmill. ("Cart observer" in your terminology.) The "wind" (air) in that frame is moving a 0 km/hr. You don't need all that other waffle. Now, the boundary is 8km/hr slower (as determined by your set-up and calcs with the "real wind").

So.... watch carefully... nothing up my sleeve....

Nor anywhere else.

0-8 = -8 km/hr.

Mistake number one. -8km is backward w.r.t the cart. The wind result is +8km, so in the direction of the cart. Slower is not backwards, or do you think a slower car in the next lane on the highway, is going backwards?
If so, truly backwards cars must make you go faster.

No need for the rest. Keep trying, you can't do it consistently and achieve the correct result. One is opposite to the other, so failure is guaranteed.

 

[Dan O.]

What evidence has anyone shown that there is a boundary layer over a running treadmill? If the treadmill surface is moving right to left at 5 m/s in normally still air, what is the direction and maximum velocity of the boundary layer? Solve the Navier-Stokes equations for the air over the treadmill and compare the results with the measured air flow over the ground on a windy day.

 

[CORed]

Unfinished post, see below for complete post

 

[Clive]

Mistake number one. -8km is backward w.r.t the cart. The wind result is +8km, so in the direction of the cart.

Your "wind result" is actually +8km/hr and that is the difference between the speed of the wind (w.r.t. your chosen reference frame) and the speed of the boundary layer in the same frame. So, let's do the same for the treadmill, continuing to use the same reference frame that I used to get the -8 km/hr for the boundary layer.

Difference
= Wind - boundary_layer
= (0 km/hr) - (-8 km/hr)
= + 8 km/hr
yes!

Any more "problems" humber? By the way, that's not really a question. I know there will be... no doubt you'll "forget" which reference frame I'm using (hint: "Cart Observer"!) and try to use 10 km/hr or dig up a -10 km/hr from somewhere or just ignore me...

 

[humber]

What evidence has anyone shown that there is a boundary layer over a running treadmill? If the treadmill surface is moving right to left at 5 m/s in normally still air, what is the direction and maximum velocity of the boundary layer? Solve the Navier-Stokes equations for the air over the treadmill and compare the results with the measured air flow over the ground on a windy day.

The belt is said to mimic the road's surface, so the belt flow should be the same. or at least similar The paper test in #7, shows that the flow is definitely there. That flow is opposite to the direction of the cart, for all speeds of the cart.
A cart starting from traveling with the belt, will experience relative boundary flow of 0 to -beltspeed. The flow from the rear is actually generated by the cart as it moves with the belt, and that diminishes as the cart approaches "windspeed". This means that a constant differential between wind and boundary flow is impossible, and will change in magnitude and apparent direction as the cart progresses up the belt.

 

[CORed]

I don't see your point, but OK, Christian.

Wind = 10km/hr
Boundary = 2km/hr
Difference = 8km/hr.
Yes.

Observer moving at 5km/hr (10km wind / 2km boundary)
-----------Subtract------Add
Wind = 5km/hr or 15km/hr
Boundary = -3km/hr or 7km/hr
Diff = 8km/hr or 8km/hr
Yes.

In real wind the difference is independent of observer travel, so would expect that to be the case at windspeed too; 8km/hr.
But as you see, the observer has no influence on the flows, they are both at 10km/hr and 2km/hr. They don't go slower or move backwards. There are changes form the observer's point of view, but so what?

OK, we agree for real wind, but that is not the case with the treadmill.
On the treadmill, there are definite changes in the winds themselves, and that is not the case for real wind.

On the belt, it is not so clear. If the beltspeed is the windspeed, (10km/hr right to left) then the boundary is 8km/hr slower. So that must be 2km/hr (right to left)
If windspeed 'air' is zero w.rt the observer, then that flow is -2km/hr (the cart speed is positive left to right).
So for the cart observer
Wind = 0 (windspeed)
Boundary = -2km/hr.

Oops!
From the frame of reference of the room containing the treadmill, aka "the ground", the boundary layer is being dragged by the belt, in the same direction (negative) as the belt, at a speed 2km/hr slower than the belt. In the real wind, we are defining the boundary layer speed relative to the ground. Here, we have to start by defining it relative to the belt. Otherwise we are comparing apples to oranges (stuck on the belt, probably).

Boundary = (-10km/hr) - (-2km/hr) = (-8km/hr) relative to the room.

Now, the observer on the belt is moving backwards at 10km/hr or -10km/hr w.r.t the cart observer. To get that observer's speed subtract 10km/hr

You defined right to left as the negative direction. When you convert from the frame of reference of the room to the frame of reference of the observer on the treadmill belt, you are changing his velocity from - 10km/hr to 0km/hr. That is adding 10. So you add 10 to everything else. So, one frame of reference error, and one flipped sign, and you end up with nonsense.

Wind = ( 0km/hr) - (10km/hr) = -10km/hr
Boundary = (-2km/hr) -(10km/hr) = -12km/hr
The difference is (-10)- (-12) = 2km/hr
That is the opposite direction of the windspeed condition. of -2km/hr

If you argue that the boundary flow should not be changed then;
Wind = ( 0km/hr) - (10km/hr) = -10km/hr
Boundary = (-2km/hr) = - 2km/hr
The difference is (-10)- (-2) = -8km/hr
That is the samedirection of the windspeed condition, but -8km/hr

Either way , it is not the same.

Here it is, correctly.

Observer on belt (-10km/hr + 10 km/hr) = 0 km/hr
Wind = ( 0km/hr) + (10km/hr) = 10km/hr
Boundary = (-8km/hr) + (10 km/hr) = 2km/hr
The difference is (10km/hr) - (2km/hr) = 8 km/hr

The same as in the "real wind"

 

[Clive]

Occam's Razor^WP: "All other things being equal, the simplest solution is the best." (One English version.)

My question: what is the simplest explanation for the fact that thousands of posts seem to have barely narrowed the gap between humber's view of the validity of the treadmill (in particular) but physics in general, and the more or less common view of most or all others posting here? Part 2: what is the best way to proceed from here?

In no particular order, some possible candidates (where "we" means more or less all recent posters here except humber, I hope that's not presuming too much, and sorry humb, you're in a category all on your own again):

1. We are all wrong. That is, we're not deliberately colluding but somehow have all "independently" come to the same incorrect understanding (or slight variations on it). Possibly there are rogue textbooks out there and we're all part a group that has been infected with wrong ideas while Humber has managed to avoid that fate.

2. Humber genuinely has great difficulty with concepts like "relativity" and/or "directions", positive versus negative... I don't know exactly what it seems to be something of that general nature. So he's genuine in his beliefs but just terribly misguided in some way. Perhaps he had a good understanding once but that has mostly gone now.

3. Humber knows a lot more than he lets on, but for some reason is playing the role of... well... the person we all "see". In other words it's all basically a big game to him. He understands, but pretends not to in some critical areas, and deliberately evades answering questions that would expose him or end the game before he's good and ready.

4. We (all except Humber) are colluding in some fashion and trying to confuse Humber with our essentially common (but incorrect) story. We arrange to make the same mistakes, or follow the lead of whoever has the best story going, etc.

5. There are at least two different "games" being played. We're essentially using Newtonian mechanics and arguing in terms of the usual idealised "models" found in textbooks and so on, while humber might believe he is "more correct" when he "denies" things (usually without explanation) by internally justifying himself on the basis of any sloppiness or ambiguity he find's in the questions or scenarios put to him and whatever other ideas or knowledge he may be working from. So we're just basically not talking about the same things...

Any other ideas?

If the truth is #3 or #4 from my list then this may continue for as long as someone is alive and not bored to tears by the whole thing. #5 is almost as bad if true. Of course I "know" that #1 and #4 aren't really true but I'm trying to be even-handed here and perhaps Humber really thinks one of those is the explanation. #2 is presumably the reason that many of the "we" are still here, although if you see #3 as true then I guess you could still be here for entertainment or other purposes...

 

[RossFW]

Lets see how Humber refutes this analogy (I'll even take him off ignore, temporarily):-

Humber,

Lets say you are sitting sideways in a car, facing it's RIGHT side. The general traffic flow is from your RIGHT to your LEFT wrt the ground.

You overtake a slower vehicle to it's LEFT (your RIGHT) side. From your point of view, the car start out on YOUR LEFT, passes in FRONT of you, then ends up on YOUR RIGHT.

SO, even though the car you passed is travelling RIGHT to LEFT wrt the surface, it is travelling LEFT to RIGHT wrt to you.


SIMILARLY, a boudary layer on the treadmill which is travelling RIGHT to LEFT wrt to the ground, but slower than the belt (ALL of which you admit is the case) is travelling LEFT to RIGHT wrt an observer travelling on the belt.

I know this is wrong because of my ignorence and lack of intuition, but please tell me why??

 

[humber]

Your "wind result" is actually +8km/hr and that is the difference between the speed of the wind (w.r.t. your chosen reference frame) and the speed of the boundary layer in the same frame.

That is correct, Clive, but it is independent of veloicty. At windpeed, I can say that the wind velocity of w,r,t wind is "windspeed" that is 0.
This means that the boundary is still moving forward, bit not at windspeed, so that is 2km/hr in "treadmill" terms. However, the actual differential of wind and flow, remains at 8km/hr.

So, let's do the same for the treadmill, continuing to use the same reference frame that I used to get the -8 km/hr for the boundary layer.

Difference
= Wind - boundary_layer
= (0 km/hr) - (-8 km/hr)
= + 8 km/hr
yes!

No. Only by swapping methods. The equivalent for the real wind is +2km/hr. It still moves forward, but slower. It does not move away from the cart as it does on the belt.

Any more "problems" humber? By the way, that's not really a question. I know there will be... no doubt you'll "forget" which reference frame I'm using (hint: "Cart Observer"!) and try to use 10 km/hr or dig up a -10 km/hr from somewhere or just ignore me...

Like I said, a little arithmetic proves that your claim is impossible.