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

[humber]

No, reversal of your reversed way of thinking will make it good. Try this: the method used to move the observer at 2 m/s downwind in the realwind is a very long treadmill. Do the rest of the numbers of the realwind scenario to get the observer matching the speed of the realwind.

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 realwind to belt with the steps in between put in so you can see how it works. Does that help?

No, it is quite obvious that the relationship will hold for all values in real wind.
You have shown that. It is constant in real wind at 4m/s.

The flow in the treadmill is the other way around, so it cannot be constant That is the point.

 

[mender]

I've edited a bit, so you might want to go back and double-check and do any editing that you might want to do. That way we can stay at the same place wrt to each other.

 

[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"
[/FONT="Courier New"]

10 - 6 = 4m/s. Not distance, velocity.
No, Clive. It is the simple subtraction of the two flows. 10m/s - 6m/s = 4m/s.
That applies to standing still as well. The 6m/s is a nominal value, but will change with position within the flow.
This is not so for the treadmill. They are in opposition and can never emulate two parallel flows.


Treadmill Wind/Belt

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

There is no speed difference between the cart and the "windspeed" observer. For both, the flow moves back at 6m/s relative to the still air.

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

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

No, it's 10m/s - 6m/s = 4m/s. You have made the observer in the wind, but he is on the belt. That is not windspeed observation but ground/raod observation.
Simple translation of a true windspeed observer, sees the relative velocities of wind and flow change with motion up the belt. There are errors from both belt and winspeed observers. They must translate by the belt peed, but do not.

 

[Clive]

Humber, let me ask you two quick questions to make sure I understand how you're seeing this stuff in your head:

1. Is this "paddle" something like what a traditional paddle-steamer has on it's sides to drive it?! (There's an axle and two (or maybe more) equal length arms at 180 degrees apart with blades at the ends that in this case can be assumed to be in either the boundary layer or the air above when it comes to determining the direction of rotation.

2. Do you have any objection to viewing the boundary layer as one single layer at some fixed intermediate speed (always relative to the surface below)? That means our simplified model would have ground (or belt), with a boundary layer of some fixed depth moving at (say) 6 m/s relative to the surface, and above that all the air at 10 m/s relative to the surface (which for the treadmill with moving belt means this is at rest).

 

[humber]

I've edited a bit, so you might want to go back and double-check and do any editing that you might want to do. That way we can stay at the same place wrt to each other.

For real wind the difference is quite clearly 10 - 6 =4m/s. If an observer moves through this wind then that velocity affects wind and boundary flow equally, add or subtract the velocity, to both flows in equal amount. A constant.

Doing exactly the same for an observer moving up the belt, the difference will not remain constant, because one value, the belt flow, is negative. It cannot be constant from the observer's point of view. If the flow were the other way, then it would be constant. Both cannot be right.

 

[jjcote]

You have made the observer in the wind, but he is on the belt.

Fine. In that case, exact same diagram from the road example will apply. If you disagree, please provide a diagram like the ones that Clive created showing what you think happens with the puffs of smoke in the treadmill case.

 

[mender]

For real wind the difference is quite clearly 10 - 6 =4m/s. If an observer moves through this wind then that velocity affects wind and boundary flow equally, add or subtract the velocity, to both flows in equal amount. A constant.

Doing exactly the same for an observer moving up the belt, the difference will not remain constant, because one value, the belt flow, is negative. It cannot be constant from the observer's point of view. If the flow were the other way, then it would be constant. Both cannot be right.

But we're not going to move the observer up the belt, because we want to see what is happening when the observer is moving at the same speed as the belt. We're going to go back to the observer being stationary with the ground in a 10 m/s wind. We're going to use the F1/BMW wind tunnel so that we can control the conditions exactly so that we get the right transition from the observer standing on the floor of the wind tunnel with the wind blowing at 10 m/s to the air being at 0 m/s with the floor moving in the opposite direction at 10 m/s. That will keep the difference between the floor and the air at the same speed and direction for each of the steps needed to get to the observer standing on the floor that is moving back at 10 m/s through air at 0 m/s.

I'll let you have the first go at it if you like, 'cause I need to get some sleep. Be careful which way you have the observer facing, you've gotten this mixed up every time so far.

 

[spork]

So here's the latest update (with Sol responding):

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: YES
spacediver: ?
spork: YES
subduction zone: YES

Executive summary: We have exactly one person that tells us every single other member has it wrong. And we have exactly one person not willing to answer a few VERY BASIC physics questions to establish that he is even competent to be a part of this conversation.

So everyone here (except humber) seems to be in complete agreement as to what's going on with this simple frame of reference problem. And that ONE person that is not in agreement, just happens to be the very same person that's unwilling to demonstrate even the most basic competence in the subject matter.

I don't think there's any mystery left here.


But I'm tempted to do the test just to see humb's responses. They're more consistently entertaining than humber anyway.

 

[humber]

Humber, let me ask you two quick questions to make sure I understand how you're seeing this stuff in your head:

1. Is this "paddle" something like what a traditional paddle-steamer has on it's sides to drive it?! (There's an axle and two (or maybe more) equal length arms at 180 degrees apart with blades at the ends that in this case can be assumed to be in either the boundary layer or the air above when it comes to determining the direction of rotation.

This is "it", Clive. The numbers do not matter so much as the fact the the paddle is not consistent, whereas it is in the real world, it is.
You can also see that a pitot tube would yield different results in each case,
Even if the observer is driven up the belt by the wind i.e. he is not attached to the belt, the relative velocities will change with observer velocity. This obviates the confounding mixing of flows at the observer's location ,and shows that it is inconsistent in this way too.

ETA: The wind in the lower left drawing is that generated as the paddle moves through the air, driven by the belt.
The lower right flow is 4m/s slower than the 10m/s wind, so is the same as for the real case.

2. Do you have any objection to viewing the boundary layer as one single layer at some fixed intermediate speed (always relative to the surface below)? That means our simplified model would have ground (or belt), with a boundary layer of some fixed depth moving at (say) 6 m/s relative to the surface, and above that all the air at 10 m/s relative to the surface (which for the treadmill with moving belt means this is at rest).

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.

 

[humber]

So here's the latest update (with Sol responding):

Monty Python again "we are all individuals". There are puzzles enough.

 

[spork]

Monty Python again "we are all individuals". There are puzzles enough.

Horse tank movie towel consideration. Disambiguate inertial credence with one. True. Your claim is denied.

 

[humber]

Fine. In that case, exact same diagram from the road example will apply. If you disagree, please provide a diagram like the ones that Clive created showing what you think happens with the puffs of smoke in the treadmill case.

There are some drawings, I posted to Clive, but the arithmetic is enough. It is quite impossible for two opposing flows to behave like two similar flows. One already has a greater differential that cannot remain constant.

 

[humber]

Horse tank movie towel consideration. Disambiguate inertial credence with one. True. Your claim is denied.

Donkey at windspeed.

 

[spork]

spork: Horse tank movie towel consideration. Disambiguate inertial credence with one. True. Your claim is denied.
Donkey at windspeed.

humber: Donkey at windspeed.


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

 

[mender]

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.

Humber, you need to change your lower left drawing in order to represent equal cases. You have a wind speed of 10 m/s. There is no wind; the observer only feels the wind because they are moving back with the belt at 10 m/s. In your diagram you have a total speed difference between the belt and the air of 20 m/s. That's wrong. That's why you are having trouble understanding this.

Also, your wind diagrams show that the ground changes the speed of the air in the boundary layer by 4 m/s (10m/s wind slowed by 4m/s = 6 m/s for the boundary layer). The treadmill diagrams show a speed difference of 6 m/s, 50% higher speed difference than what you show for the wind diagrams. They should be the same in both cases, which is 4 m/s speed difference between the free air and the boundary layer, caused by the ground or by the belt. The air above the belt in your treadmill diagrams is moving at 4 m/s, not 6 m/s.

All you have to do is correct those mistakes and your diagrams will be right.

 

[spork]

Humber, you need to change your lower left drawing in order to represent equal cases. You have a wind speed of 10 m/s. There is no wind; the observer only feels the wind because they are moving back with the belt at 10 m/s. In your diagram you have a total speed difference between the belt and the air of 20 m/s. That's wrong. That's why you are having trouble understanding this.

Also, the boundary layer is moving 4 m/s differently than the air in your wind diagram. The treadmill numbers show 6 m/s, 50% higher speed difference than what you show for the wind diagrams.

I'm detecting an extreme impedance mismatch. I would anticipate your final RF stage would be seeing an SWR of 35 or greater. The probability of letting the magic smoke out is very high.

All you have to do is correct those mistakes and your diagrams will be right.

So you're saying, I could just cure cancer, and then take lunch. Sounds simple enough.

 

[mender]

At this point you're probably right spork!

 

[humber]

Humber, you need to change your lower left drawing in order to represent equal cases. You have a wind speed of 10 m/s. There is no wind; the observer only feels the wind because they are moving back with the belt at 10 m/s. In your diagram you have a total speed difference between the belt and the air of 20 m/s. That's wrong. That's why you are having trouble understanding this.

The 10m/s wind is that generated by observer motion. So it is OK. You can see that from the numbers. (10m/s - 6m/s = 4m/s)

Also, your wind diagrams show that the ground changes the speed of the air in the boundary layer by 4 m/s (10m/s wind slowed by 4m/s = 6 m/s for the boundary layer). The treadmill diagrams show a speed difference of 6 m/s,
50% higher speed difference than what you show for the wind diagrams. They should be the same in both cases, which is 4 m/s speed difference between the free air and the boundary layer, caused by the ground or by the belt. The air above the belt in your treadmill diagrams is moving at 4 m/s, not 6 m/s.

No, the diagram is correct, Mender. For the cart observer, the belt is moving back at 10m/s, the flow is at at 6/ms: 4m/s slower than the 10m/s wind. That is the same for both cases. The 6m/s is nominal, and if the model surface is a good modle, can be compared no matter how that nominal speed is defined.

All you have to do is correct those mistakes and your diagrams will be right.

They are right, but adjustment would still not explain the variation with observer travel, or cause them to turn in the same direction. Cart and belt speed paddles are always in contra-rotation, but in the real wind not. Applying the same rules, even if somewhat differently than I have done, will also show that inconsistency.

 

[Michael C]

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:?

Yes, bring on the questions.

 

[humber]

I know that at some point you'll tell me that I'm wrong. The purpose is to identify exactly at which point you tell me wrong, and why you think I'm wrong, in order to identify the specific ways in which our understanding of physics differ.

I do not say you are wrong. There is little difference in the physics, but the way the model is derived that is wrong. What happens before you build your explanation.

Just telling us that we're wrong, without explaining to us why we're wrong in a manner in which we can clearly understand you is ultimately futile. It just leads to an endless cycle of "Is not!" "Is too!" posts.

I don't do that without explanation. It is just that it is not accepted.

I'm trying to break that cycle. So please, go through the steps and respond to them clearly for us.

Yes, but that is not easy. I am trying to address the laminar flow for a reason. That is a clear indicator, but even that is not easy.

I can't make much sense from the boundary layer diagram you posted in this thread. How exactly to the results you get from that diagram differ from the ones the stick-man gets in this drawing I made?

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_26573497147e120f0b.png[/qimg]
(Click to Enlarge)​

Note: the readings in the bottom-right square are the same you should get if you were standing next to the treadmill. After all, if there is no wind then remaining stationary is wind speed.

Bottom right should be -7m/s. The wind is 10m/s so the flow is 3m/s less.

(1) For the wind; 3m/s flow-to-wind in both cases
(2) From the belt; 3m/s flow-to-wind with the belt, but 7m/s flow-to-wind at windspeed, but in the other direction.