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

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This is for science, not spork.

Hey, I just happen to work at the "Hall of Science" (at least that's what the guys in the field call it). In fact, I'm the "Chief Scientist" at "The Hall of Science". How much more sciency can you get than that?



Let's welcome RossFW and Spacediver to our little list.

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: YES
jjcote: YES
John Freestone: YES
Mender: YES
Michael C: YES
RossFW: Yes
Sol invictus: YES
Spacediver: YES
spork: YES
subduction zone: YES


So, it seems there's a pattern beginning to emerge. The only one that cannot be tested, is he who knows all. How bizarre.
 
Humber, I have a few serious questions for you

Please answer yes or no.

1) Do you think you have a good grasp of basic physics?

2) Do you acknowledge that there are many people who think you do not have a good grasp of basic physics?

3) If you answered yes to the above two questions, do you think it's worth spending a few minutes answering three questions from a high school physics exam to prove that you do indeed know your stuff?
 
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.



[qimg]http://www.internationalskeptics.com/forums/imagehosting/22697497348179d162.bmp[/qimg]

The observer does not matter, John. It's not a matter of a tree falling in the forest without a witness, or the sound of one hand clapping.

The flows are there if you are aware of them or not. Pitot tubes, paddles, anemometers, or some other measurement device, will contradict your conclusion. Also, an observer of the kind you suggest, but on the ground or somewhere else, will also contradict your cart observer, whereas all will agree with the direct measurements and the calculations. My solution is correct, if only because it is the same for all observers, and quite clearly not like real wind.
 
Sorry, I've been waiting so long to read about humber recanting I just had to see it before I died. I sleep now.

Did I get anyone else?:)


Sort of. It surprised me too much to either accept or reject until after I'd clicked on the link to check the original quote.
 
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.

No COREed. there are no frames of refernce, just the datum from which you take the measurements. Do you need to be at windspeed to take them in the real wind.? No. You don't ask such a stupid question, yet on the treadmill, you do. All views, the point from which you take the measurements will give you the same result. You can test my idea from any and all frames, in both real and treadmill cases, and come to the same conclusion.
 
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.

No I did not I denied the quiz, because I an not interested.
It seems to me though, H'ethetheth, that you need your colleagues to support you.
Post any of the windtunnels that you think support your case, from seeds to cars, and I will tell you why they are not like the treadmill. Agreed?

An equal appeal to both our "authorities".
 
What a lot of guff, when a little arithmetic shoots it down.

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)


I think I'm following you so far.

less clumsily, from the "wind's" perspective
Cart Observer
Wind = +10 + 0 = +10
Flow = +10 -6 = +4m/s (CCW)


Wait a sec... you started off by saying



Are you claiming that the wind is travelling at +10 m/s faster than itself?

It's logically impossible for something to be travelling faster than itself from it's own perspective. Everything travels at 0 from it's own perspective. So to see things from the perspective of any given object, you just have to subtract that object's velocity from all velocities.

Since you started off by stating that Wind = 0, then you started from the wind's perspective.

The fact that you are incapable of switching to another perspective and back again without getting completely different figures from what you started with is a very strong indicator that you have no idea what you are doing.

This is how you should have done it.

You begin with:

Wind = 0
Flow = -6

To get the observer's perspective, you subtract the observer's velocity (-10) from everything:

Wind = 0 - (-10) = 10
Flow = -6 - (-10) = 4

To get back to the wind's perspective, you subtract the wind's velocity (10) from everything:

Wind = 10 - 10 = 0
Flow = 4 - 10 = -6

Easy!
 
I'd like to see humber fill in the blanks in John's diagram...

[qimg]http://www.internationalskeptics.com/forums/imagehosting/22697497348179d162.bmp[/qimg]

You do that,Sol. I have posted my solution, so I don't see why I should post yours, and we can compare. Ir seem that others are making an effort.
 
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.


Wait a minute!

Negative indicates velocity in the opposite direction.
You have the "flow" moving the opposite direction to the belt.

Since the flow boundary layer is the air that is being dragged along with the belt then it must move in the SAME direction.

So to correct your mistake by fixing your negatives...

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) = 4m/s Difference in flow to observer at 2m/s from R to L


ETA: When you do the math right you find that there is no difference.
 
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I denied the quiz

You cannot "deny" the quiz. You can "decline" to participate. Only in the bizarre world of humber do you get to deny things such as this. This is similar to the way you tell us "the treadmill is wrong". Do you not realize that in our language such a statement has no meaning?

...because I an not interested.

That makes perfect sense. You've been arguing with us for 50 pages about how you understand physics, and we don't. But you're "not interested" in spending 15 minutes to show us that you're right and we're wrong.

Nope, I just don't buy it. If you spend another 15 minutes on this thread we'll all know that "not interested" is pure B.S. "not willing" is clearly more accurate. I'd say that reflects your fear that you'd do poorly - but that gives you too much credit. You know from the start that you're just plain unable to answer a single question of physics, no matter how basic.
 
It's logically impossible for something to be travelling faster than itself from it's own perspective.
Brian, I used to think like you. But now I have a small plaque with a message that I read often. It was a gift to me, and at first I thought it was just one of those "feel good" things. But now I see that it is the new way... the master's way... and it says:

If you can dream it, you can do it.​

Commit this to memory and then your path to the dark side will flow as freely as a boundary layer with no datum...
 
Bottom right should be -7m/s. The wind is 10m/s so the flow is 3m/s less.


But it's being measured from a balloon hanging motionless on a windless day. There is no wind, so wind-speed is 0m/s, not 10m/s. And last I checked 0m/s - 3m/s equals -3m/s.
 
Humber,
I've been away with work for a couple of days (Hamburg, beutiful but bloody cold!!), and, given time to reflect, I realise you have been right about everythiing.
Yes, it has been cold. Hamburg is indeed splendid. Snow makes it luminously so.

As such, I now have a problem I hope our newly melded outlook can fix:-
It about Kites.

Most men have trouble withe their 'kite' now and again. How may I help you?

I have a kite that flys just fine if I'm standing still and the wind is 10M/s.
The wind powers the kite.

Thing is, it ALSO flys if there is NO wind and I RUN at 10M/s.
You power the kite.

I'm also PRETTY SURE that if I had a long treadmill that went at 10m/s and stood on it, towing the kite, it would also fly.
The belt powers the kite.

There must be a problem with my perception, as these things obviously can't be true.
As I said, 'kite' problems are so often in the mind. 'Kite performance anxiety'
is quite common. Nothing to be ashamed of.

For one, in the first instance it's a REAL wind, where the others aren't, so there's no way I could model the kites performance in 10m/s REAL wind by pulling it through still air at 10m/s, right?

So relative motion between a kite an its surrounding medium, may be REAL or what?

The last one can't be true, because I'm not even running on a REAL road, but on a treadmill, and that's NOT a road!!
Maybe Hamburg has been moved. Perhaps it's the motion that is important, and not the nature of the 'road'

Thirdly, and most importantly, there is the issue of KE.

The first kite has no KE in any frame, as it is not moving wrt the earths surface. In the second two, it does.
Glad you brought that up. The first kite has no KE because it is restrained by you. There you go, no motion no KE, just like the cart. Glad to see that you have also dismissed that nonsense of "relative to the moon".
Yes, the other two are in motion w.r.t the ground. It seems quite difficult to avoid KE when that happens.

Now we established (or at least you SAID it many times) that an object with no KE can't model on that has some, so I assume the reverse is also true?
Trouble is, the damned kite still insists on flying.
I know what I SAID, RossFW, you appear not to know it.
First statement generally correct. Second is conditional. You are fond of false syllogisms.

But you agree that your first example the kite has no KE because it has no motion w.r.t the ground.
The cart is restrained in the sense that it is not driven to motion, so you agrees that it has no KE.

Can you tell me where I've gone wrong?

Done.
 
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