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

[spork]

So here's the update:

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

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

As to when the test is - who knows. Unless humber agrees there's hardly a point to it at all. I figured he may think we're being condescending to ask him to answer some basic questions. But it seems pretty clear most of the rest of us aren't afraid to do it. So if he's in, it's a go. If he's not in, but you guys really want to do it, I'll get my physics text, post the questions, and I'll play along as well (it's been a LONG time, but I presume there are plenty of questions without answers in the back - or you can just trust me not to peek).

That would be "5 by 9". A strenght of 5 would be mediocre though I have given out 2's on the county hunter nets.

My recollection is that there are three parameters, and the full set would be "5 x 5 x 9", but it seems most leave off the 9 (at least in my experience).

Wikipedia claims: Five by five is the best of 25 possible subjective responses used to describe the quality of communications...Five by five therefore means a signal that has excellent strength and perfect clarity — the most understandable signal possible http://en.wikipedia.org/wiki/Five_by_five

I think the 3rd parameter (1 to 9) is tone. But at least on hams I never hear it used.


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".

So it's pretty much down to you humber. Give me the go ahead and I'll do the test. But after tomorrow I'll be away from my physics book for 2 weeks.

 

[jjcote]

What a lot of guff, when a little arithmetic shoots it down.

I am going to define going with the belt as minus.

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

Cart means view from the cart
Observer means the view of an observer fixed to the belt

Windspeed Cart
Wind = 0
Flow = -6m/s (CW)

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

The observer is traveling with the belt, and so at -10m/s, w.r.t the cart,
so we can subtract 10m/s from the Windspeed Cart results to get the Observer's view.

Cart Observer
Wind =( 0) -10 = -10
Flow = (-6) -(-10) = +4m/s (CCW)

(Sigh.) So, the correct calculation is:

Our observer is now moving with a speed of -10. So we subtract that from each airspeed:
Cart Observer
Wind =( 0) - (-10) = +10
Flow = (-6) -(-10) = +4m/s
and the difference is still that "flow" is less than "wind" by 6 m/s. You have to do the same thing to each part. You can't subtract -10 in one place and add -10 in the other. He now feels a 10 m/s wind on the back of his head, and a 4 m/s wind on the back of his ankles. Both from the back of the treadmill toward the front, so both positive. He feels no headwind when he's standing on the belt.

QED

I think the term you were looking for is "FAIL".

 

[humber]

So here's the update:

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

Spork's willing or able to answer questions; zero

Unless humber agrees there's hardly a point to it at all. I figured he may think we're being condescending to ask him to answer some basic questions.

Indeed

But it seems pretty clear most of the rest of us aren't afraid to do it.

Fearful expectation over few high school questions?

So it's pretty much down to you humber. Give me the go ahead and I'll do the test. But after tomorrow I'll be away from my physics book for 2 weeks.

Sorry, not interested.

 

[Clive]

I am going to define going with the belt as minus.
Cart means view from the cart
Observer means the view of an observer fixed to the belt

Windspeed Cart
Wind = 0
Flow = -6m/s (CW)

The observer is traveling with the belt, and so at -10m/s, w.r.t the cart,
so we can subtract 10m/s from the Windspeed Cart results to get the Observer's view.

Cart Observer
Wind =( 0) -10 = -10
Flow = (-6) -(-10) = +4m/s (CCW)

Well, I see that jjcote has beaten me to it. But I think there's another glitch that slipped through the net. In the original description the boundary layer was 4 m/s slower than windspeed (moving across the ground at 6 m/s). So I think the corrected version should be should be more like this:

Windspeed Cart
Wind = 0
Flow = -4m/s (CW)

The observer is traveling with the belt, and so at -10m/s, w.r.t the cart,
so we can subtract -10m/s from the Windspeed Cart results to get the Observer's view.

Cart Observer
Wind =( 0) -(-10) = 10Flow = (-4) -(-10) = 6m/s (CW)

ETA: The original set-up.

Okay, we have a smoke generator sitting on the ground. There is a 10 m/s wind blowing out of the south, and we are approaching the smoke generator at 10 m/s, northbound (i.e. at windspeed) in some sort of conveyance (not necessarily wind powered). When we are 10 m from the generator, it emits two puffs of smoke, one at eye level, the other at ankle level. What do we see? The puff that is at eye level travels northward at 10 m/s, and remains 10 m in front of us, while the one at ankle level, in the boundary layer, moves forward at only 6 m/s, so we close the distance to it at 4 m/s, and we pass by it in 2.5 seconds.

Now we have a smoke generator attached to a (sufficiently large) treadmill belt that is moving from north to south at 10 m/s. It is in a room with still air, and we are maintaining our position with respect to that air (and to the "floor") by moving relative to the belt at 10 m/s in some sort of conveyance, so the smoke generator is coming toward us at 10 m/s. When the smoke generator is 10 m in front of us, it emits two puffs of smoke, one at eye level, the other at ankle level. What do we see? The puff that is at eye level stays put relative to the air (and to the "floor"), and remains 10 m in front of us., while the one at ankle level, in the boundary layer, is dragged back by the belt at 4 m/s, and we pass by it (or it passes by us) in 2.5 seconds.

Exactly the same.

 

[mender]

As I'm sure everyone expected, humbre won't take the chance of being shown up. Not surprising. He'd rather blather on and on, repeating the same mistakes over and over. I imagine it's a lot easier do that than having to come up with more bizarre misinterpretations of the laws of physics, especially when the answers are known and are beyond "debate".

Probably a good thing; he can't even keep his own explanations straight, as shown by the need to correct his simple diagrams, which somehow didn't show what the whole issue of the cart is! Besides, I doubt that his answers could be deciphered.

 

[humber]

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).


(Sigh.) So, the correct calculation is:

Our observer is now moving with a speed of -10. So we subtract that from each airspeed:
Cart Observer
Wind =( 0) - (-10) = +10
Flow = (-6) -(-10) = +4m/s
and the difference is still that "flow" is less than "wind" by 6 m/s. You have to do the same thing to each part. You can't subtract -10 in one place and add -10 in the other. He now feels a 10 m/s wind on the back of his head, and a 4 m/s wind on the back of his ankles. Both from the back of the treadmill toward the front, so both positive. He feels no headwind when he's standing on the belt.


I think the term you were looking for is "FAIL".

No, that is not correct, jjcote, but by your own calculation, you can see that there must be a point where they will be equal and opposite.

It is simple. The rear wind can be said to +10 stricking the rear of the observer. So let's forget the sign and say that the belt flow is 6m/s in the opposite direction. The resultant flow will be 10 - 6 = 4m/s, whereas at windspeed, it is the other way a 6m/s.

Forget the observer, and simply make the "belt wind" real, so there is a +10 flow from left to right, with the belt-flow under that, at 6m/s in the other direction. The difference is 4m/s viewed from the belt. Only as a "windspeed" observer do you again see -6m/s. An observer moving up the belt as "windborne", and at 2m/s, will see the flows equal and in opposition,

10 - 2 = 8 Difference between wind and observer at 2m/s
-6 -2 = -8 Difference between flow and observer at 2m/s

ETA;

Our observer is now moving with a speed of -10. So we subtract that from each airspeed:
Cart Observer
Wind =( 0) - (-10) = +10
Flow = (-6) -(-10) = +4m/s
and the difference is still that "flow" is less than "wind" by 6 m/s. You have to do the same thing to each part. You can't subtract -10 in one place and add -10 in the other. He now feels a 10 m/s wind on the back of his head, and a 4 m/s wind on the back of his ankles. Both from the back of the treadmill toward the front, so both positive. He feels no headwind when he's standing on the belt.

No, the reason for the change in sign of the 10m/s in the belt case, RossFW, is that the belt is already going at -10m/s. (in order to get the -6m/s reading).

 

[humber]

Well, I see that jjcote has beaten me to it. But I think there's another glitch that slipped through the net. In the original description the boundary layer was 4 m/s slower than windspeed (moving across the ground at 6 m/s). So I think the corrected version should be should be more like this:

Windspeed Cart
Wind = 0
Flow = -4m/s (CW)

The observer is traveling with the belt, and so at -10m/s, w.r.t the cart,
so we can subtract -10m/s from the Windspeed Cart results to get the Observer's view.

Cart Observer
Wind =( 0) -(-10) = 10Flow = (-4) -(-10) = 6m/s (CW)

No, the original is what I meant. You seem to be counting the belt twice. See my answer to jjcote, Clive.

 

[humber]

As I'm sure everyone expected, humbre won't take the chance of being shown up. Not surprising. He'd rather blather on and on, repeating the same mistakes over and over. I imagine it's a lot easier do that than having to come up with more bizarre misinterpretations of the laws of physics, especially when the answers are known and are beyond "debate".

Probably a good thing; he can't even keep his own explanations straight, as shown by the need to correct his simple diagrams, which somehow didn't show what the whole issue of the cart is! Besides, I doubt that his answers could be deciphered.

No, the criticisms are incorrect, Mender. There is a definite difference, and it changes as the observer moves up the belt. There is no doubt, because it is going the wrong way. There is no way that the flow coming from the other way can be corrected it for all possible relative motions.

 

[mender]

No, the criticisms are incorrect, Mender. There is a definite difference, and it changes as the observer moves up the belt. There is no doubt, because it is going the wrong way. There is no way that the flow coming from the other way can be corrected it for all possible relative motions.

See? Your smoke test was also supposed to show that the flow is going the wrong way too. You didn't read why your own smoke test proved you wrong then either.

C'mon, I'm just a car mechanic and I'm willing to answer physics questions publicly. Surely you don't think that you would get shown up by l'il ol' me?

I offered to do your tests for you and report the results no matter what those results were. Return the favour and join in on the testing. At the very worst you might learn something new.

 

[Clive]

No, the original is what I meant. You seem to be counting the belt twice. See my answer to jjcote, Clive.

No no, Humber! You did slightly better (I think) with this "problem" earlier on but now you seem to be getting more and more mixed up.

Have you looked at the "diagrams" I created earlier? Perhaps I should tweak those to use the same wind, belt, and boundary layer speeds as jjcote's set-up?

The point is that there is a fixed difference in velocity between the boundary layer and the air above that regardless of whether you're on the ground in the "real wind" or on the treadmill. The air above moves 4 m/s faster than the boundary layer below (and to the north/right). If you put some kind of "paddle" in between so the two layers are driving it (that is what we're talking about with the paddle isn't it?) then it will always be turned clockwise as viewed from the eastern side (your "conventional" viewpoint), and that doesn't change if the paddle is moving north or south either.

 

[humber]

See? Your smoke test was also supposed to show that the flow is going the wrong way too. You didn't read why your own smoke test proved you wrong then either.

C'mon, I'm just a car mechanic and I'm willing to answer physics questions publicly. Surely you don't think that you would get shown up by l'il ol' me? I offered to do your tests for you and report the results no matter what those results were.

The smoke tests are fine. The other tests too. It's not a matter of wanting to or not wanting to show anything. I simply do not want to. It does not interest me.

 

[Dan O.]

For online physics problems, we could have one questioner propose a problem and the other participants would return their answer by PM. These answers would be randomized and printed by the questioner where they would then be critiqued by the participants. Then the questioner would release the names of the participant that submitted each answer and a link to a video of a 5th grader providing the correct answer.

For scoring, each participant receives 1 point for each correct answer and 5 bonus points for being judged the best answer by a majority of the participants. The answer judged most likely to be humber would of course earn -5 points for the participant that submitted it.

The participant with the highest score at the end wins. Humb wins if he submits more posts judged most likely to be humber than humber.

 

[mender]

The smoke tests are fine. The other tests too. It's not a matter of wanting to or not wanting to show anything. I simply do not want to. It does not interest me.

I guess I can understand that. Too bad; most of us have been participating in your game for quite a while.

 

[Dan O.]

My recollection is that there are three parameters, and the full set would be "5 x 5 x 9", but it seems most leave off the 9 (at least in my experience).

RST Code^WP

I think the 3rd parameter (1 to 9) is tone. But at least on hams I never hear it used.

Tone is only used for digital modes like CW, RTTY or PK31.

 

[humber]

No no, Humber! You did slightly better (I think) with this "problem" earlier on but now you seem to be getting more and more mixed up.

Have you looked at the "diagrams" I created earlier? Perhaps I should tweak those to use the same wind, belt, and boundary layer speeds as jjcote's set-up?

The point is that there is a constant difference velocity difference between the boundary layer and the air above that regardless of whether you're on the ground in the "real wind" or on the treadmill. The air above moves 4 m/s faster than the boundary layer below (and to the north/right). If you put some kind of "paddle" in between so the two layers are driving it (that is what we're talking about with the paddle isn't it?) then it will always be turned clockwise as viewed from the eastern side (your "conventional" viewpoint), and that doesn't change if the paddle is moving north or south either.

No they are not the same. It is quite simple.
For an observer traveling with the wind in each case.
From the the belt shows the same problem.

Realwind
10m/s (+2m/s) = 8m/s Difference in wind to observer at 2m/s downwind
6m/s (+2m/s ) = 4m/s Difference in flow to observer at 2m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

Belt
10m/s (+2m/s) = 8m/s Difference in wind to observer at 2m/s from L to R
-6m/s (+2m/s) = -8m/s Difference in flow to observer at 2m/s from R to L
Wind to flow +16m/s.
Dependent upon observer velocity. It is going the wrong way. Reversal of flow makes it good.

 

[jjcote]

It is simple. The rear wind can be said to +10 stricking the rear of the observer. So let's forget the sign and say that the belt flow is 6m/s in the opposite direction.

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.

 

[Clive]

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"


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 <-

 

[sol invictus]

So here's the update:

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

Sure - I'm in.

 

[mender]

No they are not the same. It is quite simple.
For an observer traveling with the wind in each case.
From the the belt shows the same problem.

Realwind
10m/s (+2m/s) = 8m/s Difference in wind to observer at 2m/s downwind
6m/s (+2m/s ) = 4m/s Difference in flow to observer at 2m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

Belt
10m/s (+2m/s) = 8m/s Difference in wind to observer at 2m/s from L to R
-6m/s (+2m/s) = -8m/s Difference in flow to observer at 2m/s from R to L
Wind to flow +16m/s.
Dependent upon observer velocity. It is going the wrong way. Reversal of flow makes it good.

No, reversal of your reversed way of thinking will make it good. Try this: do the rest of the numbers of the realwind scenario to get the observer matching the speed of the realwind. And it'll be easier if you use the proper signs as you go,

Like this:
10m/s (-4m/s) = 6m/s Difference in wind to observer at 4m/s downwind
6m/s (-4m/s ) = 2m/s Difference in flow to observer at 4m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

10m/s (-6m/s) = 4m/s Difference in wind to observer at 6m/s downwind
6m/s (-6m/s ) = 0m/s Difference in flow to observer at 6m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

10m/s (-8m/s) = 2m/s Difference in wind to observer at 8m/s downwind
6m/s (-8m/s ) = -2m/s Difference in flow to observer at 8m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

10m/s (-10m/s) = 0m/s Difference in wind to observer at 10m/s downwind
6m/s (-10m/s ) = -4m/s Difference in flow to observer at 10m/s downwind
Wind to flow = 4m/s
Constant difference between wind and flow of 4m/s, independent of observer velocity

There. We've made the transition from standing on the ground in a wind to moving at the same speed as the air with the steps in between put in so you can see how it works. Does that help? Note that the direction of the wind to flow difference hasn't changed either.

If you agree with this then we'll go to the next step so you can have something to disagree with.

 

[humber]

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.

No. that is the same as adding a minus. It makes it easier to see the reason for the difference. But it is the same. The remainder is correct.
The difference is quite clear, the numbers are only an indicator that it is going the wrong way. This can simply never be like parallel flows moving in the same direction. Seems obvious to me.