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

[Clive]

No, it was the number that jjcote used, so I adopted his version and use.
It is a nominal value, but of course it will vary within the flow.

Thanks for the diagram showing the "paddle" Humber. The really great thing is that you've shown other details also and so (as already noted by mender) you've exposed another error that I suspected you were making (and which was also the main motivation for my second question). Anyway, I had a long post mostly written but can throw some of that away.

10 - 6 = 4m/s. Not distance, velocity.

You should note that I still very little idea what you're really trying to say here Humber. What distance are you talking about? (I'm not sure I really want to know actually.) Remember, I am not inside your head. I don't know precisely what you were looking at or thinking about when you wrote this. You seem to have very "different" ways of thinking about things and you've also stated that yourself. That probably means that when explaining something that you think is wrong or whatever, that you also need to spend a bit more time "filling in the gaps" as it were. Instead your comments often seem to add confusion rather than shedding light on the matter. To much like a stream of consciousness perhaps? No?, 6? Yes. I think so, no... m/s maybe. Or a distance. Do you think so? Who really knows?

Anyway, those diagrams of mine used the values that jjcote used in the original set-up. All the velocities I have shown out to the right are as "you" (Humber, standing still on Earth) would measure them. I thought that would be fairly obvious given that one part was always labelled as "still" but I guess I should have stayed with the "diagrams only" approach to minimise any possible new sources of "confusion".

So please just look at the diagrams again and see if you can at least follow those. They're just three rectangular "photographs" from a short movie really, "clipped" on the left and right. You can think of the asterisk and plus sign as puffs of different coloured smoke. There is really nothing tricky here at all. They don't prove the cart works or doesn't. They're about as simple as you could get short of actually filming the scenes (complete with smoke and a reference mark on the ground/belt) with a non-moving camera on a tripod and then posting three images each taken a second later than the previous one. In the first three we see air moving to the right over non-moving ground. In the second set of three we see the belt moving left while the air high above doesn't move. That is what the numbers and the diagrams both show and say. Please give me a cheap thrill by telling me you can understand them now!

Back to one of your main errors (and as already noted by mender). The "nominal" boundary layer's velocity must be some constant value relative to the surface in both the road and treadmill scenarios (because we're assuming those are essentially the same kind of surface in terms of drag with an air mass moving past at some particular speed). It's the result of friction with the surface below. The belt "drags" on the air above it. The road "drags" on the air above it. It's the interface between the surface below and the air mass above, and the speed (relative to the surface) builds up from 0 at the bottom at the same rate in both "real wind over ground" and "still air over moving belt" scenarios. You can't just take one "nominal value" in one frame of reference and then plug the same value into a different frame of reference.

Consider a layer that is just below the very top of the boundary region. That is where the "boundary layer" effect will be least. In other words that is where the boundary layer air is moving at very close to "wind speed". Relative to the road (with "real wind" above at 10 m/s) that might be moving at 9.9 m/s (for example). But relative to the "cart observer" (room) for the treadmill that same point in the boundary layer would be moving at 0.1 m/s. It's obviously not 9.9 m/s again! Also note that the two correct values add to wind speed and this must always be the case if they are being measured relative to ground and "cart observer" as you are doing.

So, instead of doing it correctly, you have often (always?) been using the same value (such as 6 m/s) in both cases. Mixed frames in, garbage out..

 

[John Freestone]

spork: Are you willing to answer some questions from a basic physics text?
humber: Woof!

Classic - even this I can't get an answer to. In my business "woof!" means "affirmative" or "go" or "next". But in the humberverse I assume it means "woof".

Have we discounted the possibility that humber is a highly intelligent dog?

 

[H'ethetheth]

I'm in for the questions, but it's not half as much fun if humber won't.

 

[Dan O.]

GUYS! I think I've almost broken the code!!! At least I've figured out humber's baud rate and stop bits.

When the answer is that he has negative stop bits, you can never be sure when you are in sync. Multiple readings of the same transmission could produce different interpretations depending on the starting phase.

 

[John Freestone]

Hi jjcote - further notes on our anthropological expeditions into humberverse:

OK, from the front of the treadmill toward the back is minus. That's fine.


So far, so good. You're first examining this as a person standing on the floor (i.e. at windspeed).

I thought that was from a person on the cart at windspeed. There was a line, which may have been a heading, that said 'Cart windspeed', IIRC, and he got 0 wind (correct) and -6 (nothing like +4, which is what I get).

I can't be bothered to try to decipher the rest. He gives the potentially useful heads-up that when he says "Cart" that means he's measuring from the frame of the cart, and "Observer" means from the frame of the belt. He then, twice, puts what again look like they might be headings, though who knows, "Cart Observer". Need I say more? He's on the cart and watching from the belt at the same time. No wonder his answers are right with a frequency no better than chance would predict.

My intention now is to clarify the absolutely most simple version I can of humberphysics on this issue, and see if that leads to some kind of break-...er...-through?/down?...

It seems to go like this: I travel backwards on a driven surface (treadmill) in 'still' air in a room. Some lower portion of the air is moved with the belt, 'backwards'. From my perspective as I am moved back (facing forward) at belt-speed, I feel something of the order of these winds: 10 m/s from the back, on the back of my head; 6 m/s from the front, on my knees or ankles, or wherever.

Humber, if you're reading this, check that if you like. It seems to fit with your version of events. Be careful. You might want to start talking French or Double-Dutch or the stronger Humberese dialect about now, because when you've agreed to that, I plan to utterly destroy your physics right before your very eyes. You'll like it later. You can look back and say it was the day you caught up with Enlightenment thinking.

It will pay you back for calling my hard work, in which I was trying to explain my view as clearly and fully as possible, 'guff', and offering instead 'a little 'maths', which you got wrong again.

So, I'll post a diagram if you like. You say, if I understood, 10 on the back of head, 6 on the front of your knees, as you travel with the belt. Don't move those feet now will you 'til I get back?

HINT: JJcote chose 6 for the boundary layer windspeed (the forward speed the wind over the ground is going, since it is slowed by the ground); an arbitrary choice depending on height. It's 10 at the top. Very slow near the ground/belt. So it's 9 at his neck, say. Now, try reversing that as you did so casually with the 6! It's not so easy when the numbers aren't close to half the speed of the maximum. Head feels 10 on the back. Neck - what are you going to say now? 9 m/s from where? Back or front? Then chest - say that's 8. Front or back? See where this is going? Where does the wind suddenly change direction from hitting his back to going almost the same speed and hitting his front? Don't move those feet. Don't change the frame of reference again. This is the windspeed you feel as you are moved backwards by the belt.

ETAs above are bolded

 

[John Freestone]

Here's another quick representation of it: For this thought experiment, you need one piece of thought-card with a thought-nail through it.

We hold out said card with nail stuck through it. We blow across it, having practised long and hard so we can blow a steady mass of air just like 'real wind'. The nail feels 10 (ignore units) at its top end, 6 near it's middle, both air flows from the same side, near our mouth.

We stop blowing. Next we wave the card and nail though the air in the room. Now the nail feels 10 at its top again, on the 'headwind' side, and 6 nearer the card, FROM THE DIRECTION OF THE POSITION IT HAS JUST LEFT, THE DOWNWIND SIDE!

What will humber's secret rabbit hole be now? Is there any difference in these scenarios? What will it be?

If I blow, is that not 'real wind'?
If I push card and nail through the air, is that not equivalent to a treadmill driven with a motor?
Is a nail driven through a piece of card nothing like a person standing on a treadmill going at treadmill speed?
Will all this guff be irrelevant, and a little wrong maths put me right?
Or will I just have misunderstood what he said before, and of course the nail will still feel the air from the same direction if we wave it through the air?

Who knows? Any side bets? I think he'll go with waving_nail_&_card DOES NOT EQUAL treadmill.

 

[John Freestone]

ANOTHER PRECISE ANALOGY:

A fly is sitting on my car roof. My car is going North on a still day, driven by its engine. The fly feels some strength of wind, being right up against the car roof and thus in its 'flow' (boundary layer). The wind it feels is from the back end of the car, a Southerly, which blows it off the front of the car. I see it shoot off down the road ahead of me. But then I get pulled over and busted. My drivers licence doesn't apply in this particular universe.

 

[humber]

Thanks for the diagram showing the "paddle" Humber. The really great thing is that you've shown other details also and so (as already noted by mender) you've exposed another error that I suspected you were making (and which was also the main motivation for my second question). Anyway, I had a long post mostly written but can throw some of that away.
You should note that I still very little idea what you're really trying to say here Humber. What distance are you talking about? (I'm not sure I really want to know actually.) Remember, I am not inside your head. I don't know precisely what you were looking at or thinking about when you wrote this. You seem to have very "different" ways of thinking about things and you've also stated that yourself.

But the paddle says it all, and the simple arithmetic confirms it. You seem to be taking the cart's motion into account. I mean, that you are confusing the apparent motion of the cart, with that of the real relative velocities of the flows. I think

That probably means that when explaining something that you think is wrong or whatever, that you also need to spend a bit more time "filling in the gaps" as it were. Instead your comments often seem to add confusion rather than shedding light on the matter. To much like a stream of consciousness perhaps? No?, 6? Yes. I think so, no... m/s maybe. Or a distance. Do you think so? Who really knows?

OK, but all of the responses have been flawed. I have used four methods to describe the same result. They are consistent, so what is wrong with them?
The first "dropped ball" drawings, the smoke, direct arithmetic and the paddle.
All show a difference.

Anyway, those diagrams of mine used the values that jjcote used in the original set-up. All the velocities I have shown out to the right are as "you" (Humber, standing still on Earth) would measure them. I thought that would be fairly obvious given that one part was always labelled as "still" but I guess I should have stayed with the "diagrams only" approach to minimise any possible new sources of "confusion".

That was not the problem. It is not so clear with how you derived the position of the asterisk. jjcote's version is not correct. I said so the first time he posted it.

So please just look at the diagrams again and see if you can at least follow those. They're just three rectangular "photographs" from a short movie really, "clipped" on the left and right. You can think of the asterisk and plus sign as puffs of different coloured smoke. There is really nothing tricky here at all. They don't prove the cart works or doesn't. They're about as simple as you could get short of actually filming the scenes (complete with smoke and a reference mark on the ground/belt) with a non-moving camera on a tripod and then posting three images each taken a second later than the previous one. In the first three we see air moving to the right over non-moving ground. In the second set of three we see the belt moving left while the air high above doesn't move. That is what the numbers and the diagrams both show and say. Please give me a cheap thrill by telling me you can understand them now!

OK. I will look, but jjcote says his is the same as mine , but they are not.

There is no need to even think about it, Clive. One is going one way, and one the other. There are no such winds in the real world, only one boundary flow.
The paddle does not change direction as it does on the belt.
There is no doubt that the flow at windspeed will be faster and in the opposite direction. It is no possible to manipulate the numbers to correct for that, so any consistently applied method will perhaps cure one, but create another.

Back to one of your main errors (and as already noted by mender). The "nominal" boundary layer's velocity must be some constant value relative to the surface in both the road and treadmill scenarios (because we're assuming those are essentially the same kind of surface in terms of drag with an air mass moving past at some particular speed). It's the result of friction with the surface below. The belt "drags" on the air above it. The road "drags" on the air above it. It's the interface between the surface below and the air mass above, and the speed (relative to the surface) builds up from 0 at the bottom at the same rate in both "real wind over ground" and "still air over moving belt" scenarios. You can't just take one "nominal value" in one frame of reference and then plug the same value into a different frame of reference.

They are said to be equivalent, so a cut and paste value is good for both. If applied consistently that is a valid approach. The application and results must be consistent.

Consider a layer that is just below the very top of the boundary region. That is where the "boundary layer" effect will be least. In other words that is where the boundary layer air is moving at very close to "wind speed". Relative to the road (with "real wind" above at 10 m/s) that might be moving at 9.9 m/s (for example). But relative to the "cart observer" (room) for the treadmill that same point in the boundary layer would be moving at 0.1 m/s. It's obviously not 9.9 m/s again! Also note that the two correct values add to wind speed and this must always be the case if they are being measured relative to ground and "cart observer" as you are doing.

The cart observer is the same as the speed of the wind. So the difference is that of the speed of the laminar and air. All stratifications. The 6m/s is a nominal "mid layer" value. The problem is nothing more than it is going the wrong way.

So, instead of doing it correctly, you have often (always?) been using the same value (such as 6 m/s) in both cases. Mixed frames in, garbage out..

It is not 6m/s, it is 4m/s less than the beltspeed of 10m/s. That is consistently applied.
It is so simple. One has a minus sign, no matter how the value is defined. It is x less than the wind in each case.

 

[humber]

When the answer is that he has negative stop bits, you can never be sure when you are in sync. Multiple readings of the same transmission could produce different interpretations depending on the starting phase.

Bored rating

 

[humber]

Can you follow this humber? (This diagram has been updated to match jjcote's set-up.)

"Real wind"/"Real Ground"

Time=0
--------------------*----------------------------- 10 m/s ->
--------------------+----------------------------- 6 m/s ->
01234567890123456789=12345678901234567890123456789 "still"

Time=1
------------------------------*------------------- 10 m/s ->
--------------------------+----------------------- 6 m/s ->
01234567890123456789=12345678901234567890123456789 "still"

Time=2
----------------------------------------*--------- 10 m/s ->
--------------------------------+----------------- 6 m/s ->
01234567890123456789=12345678901234567890123456789 "still"

No, Clive, you using the observer as a reference and taking the perspective from there. That is what I meant by distance. The paddle and direct calculation remove this component.
The difference is (wind - laminar flow) = 10 - 6 = 4m/s .That does not change with the cart's velocity. The difference in actual velocities is always 4m/s regardless of he cart's velocity.
If broken down to remove the observer's component the answer will be the same, anyway.

Treadmill Wind/Belt

Time=0
--------------------*----------------------------- "still"
--------------------+----------------------------- 4 m/s <-
01234567890123456789=12345678901234567890123456789 10 m/s <-

Time=1
--------------------*----------------------------- "still"
----------------+--------------------------------- 4 m/s <-
0123456789=123456789012345678901234567890123456789 10 m/s <-

Time=2
--------------------*----------------------------- "still"
------------+------------------------------------- 4 m/s <-
=1234567890123456789012345678901234567890123456789 10 m/s <-

The same problem here. The smoke test was not intended to be analyzed like this, but a simple way of seeing that there is a difference of some unknown magnitude. The paddle and other tests quantify that.

 

[CORed]

I beg your pardon? Can we hear that again?
"So let's forget the sign and say that the belt flow is 6m/s in the opposite direction."
You want to ignore the sign, and simply assert that it's going the other way? Yeah, you can "prove" a lot of strange things if you're allowed to do that. It must be late, and I must be tired, because now I'm seeing the techniques of humberithmetic, and they boggle the mind.

In Humbermath, any sign can be reversed at will so:

2 + 2 = 4
or
2 + (-2) = 0
or
(-2) + 2 = 0
or
(-2) + (-2) = -4.

In Humbermath 2 + 2
can be 4, 0 or -4. depending on which answer best proves that the treadmill is wrong and backwards.

 

[CORed]

Hey guys - I had a thought (don't look so surprised)....

Speak up if you're willing to answer some questions from a basic physics text. I'll go first. I am willing. Who else is game? (and who is not?)

Willing to answer some basic physics questions:

spork: yes
Clive: ?
Christian: ?
humber: ?
Brian-M: ?
John Freestone: ?
JB: ?
fredriks: ?
spacediver: ?
subduction zone: ?
jjcote: ?
sol invictus: ?
H'ethetheth: ?
CORed: ?
Mender: ?

others:?

I'm game, but it would be much better if Humber participated. He won't because he knows how miserably he would fail.

 

[humber]

In Humbermath, any sign can be reversed at will so:

2 + 2 = 4
or
2 + (-2) = 0
or
(-2) + 2 = 0
or
(-2) + (-2) = -4.

In Humbermath 2 + 2
can be 4, 0 or -4. depending on which answer best proves that the treadmill is wrong and backwards.

No, jjcote is wrong. The belt is already going at beltspeed, so that requires tht it is negated. You can do the same adding ten to both,and taking it from the observer view. Try again

 

[John Freestone]

humber, I know you might not have had time to get to my recent posts, but here's an accompanying sketch. This seems to be something like what you have given us so far. Clearly, my gradient is only approximate. It isn't going to be linear, but I didn't quite get the arrow spacing right. It doesn't really matter if it were linear or not. Maybe you can help me fill in the areas where there are question-marks. If the observer on the belt feels 10 m/s wind on the back of his head, and 6 on the front of his knees, can you explain where it changes from one to the other? ETA: Please note, the right hand diagram represents the person just letting himself move backwards at belt-speed. ETA again: Just to be absolutely clear, the treadmill is in a room with 'still' air, other than for the boundary layer gradient caused by the belt. It should also be noted that in reality the gradient would tail off much closer to the belt - let's just say this is a rather small person!

I don't hold much hope again, after I saw one of your earlier criticisms of someone's analysis - that they were talking about the winds from the point of view of the observer! That's what the experiment is about. That is what we're interested in, what a person, or indeed a cart on the treadmill will experience. Please acknowledge this point too.

ETA: Whoops, .... ok, never mind, I fixed the diagram now. I missed a 6 earlier.

Off you go.

 

[CORed]

No, jjcote is wrong. The belt is already going at beltspeed, so that requires tht it is negated. You can do the same adding ten to both,and taking it from the observer view. Try again

Humber, it doesn't matter how many times I try it. The only reason you keep coming up with the bizarre notion that the boundary layer wind on the belt is moving the opposite way that it is in the real wind is that you simply can't stop switching frames of reference in the middle of your "analysis". Either frame is valid, but you can't figure some velocities relative to the belt and some relative to the ground. If you keep doing that, you're going to keep flipping signs and coming up with the wrong answer.

 

[spork]

Have we discounted the possibility that humber is a highly intelligent dog?

Well... I haven't discounted the possiblity that he's a dog. But a highly intelligent dog - not likely.


So here's the latest update (with more members reporting):

Members willing to answer some questions from a basic physics text:

Brian-M: YES
Christian: YES
Clive: YES
CORed: YES
fredriks: ?
H'ethetheth: YES
humb: HELL YEAH
humber: NO
JB: ?
jjcote: YES
John Freestone: YES
Mender: YES
Michael C: YES
sol invictus: YES
spacediver: ?
spork: YES
subduction zone: YES

So it still looks like the ONLY guy that understands physics on this forum happens to be the one guy that refuses to answer some basic physics questions whose answers are not subject to bizarre opinions or gibberish.

 

[H'ethetheth]

...So here's the latest update (with more members reporting):...

By the way, I missed where humber said no. Did he PM you?

 

[H'ethetheth]

Also, I'm still hoping to be aced by humber's engineering prowess and credentials.

 

[spork]

By the way, I missed where humber said no. Did he PM you?

Previous page:

spork: So it's pretty much down to you humber. Give me the go ahead and I'll do the test.
humber: Sorry, not interested.

Also, I'm still hoping to be aced by humber's engineering prowess and credentials.

These questions are not intended to be a competition. I expect most everyone to get most of them right without much difficulty. I don't expect humber to answer them - even if he agrees to. I don't think he understands what a responsive answer to a question really is.

 

[H'ethetheth]

These questions are not intended to be a competition. I expect most everyone to get most of them right without much difficulty.

I know, the second post was about my rant friday after being told I don't understand wind tunnels. Seeing how, in my workplace, I'm pretty much surrounded by wind tunnels, I told him that, after which he told me he wasn't impressed and could easily "ace" my argument from authority, if he so chose, which he didn't, in case you were wondering.

ETA: Thanks for the reference.