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

[humber]

Well, that's easy. Because you didn't actually build one. If you had, you would have seen that it doesn't behave the way that you seem to think that it will. When you brought it up, several people pointed out the errors in your description and explained what it will actually do. But since you've asserted that it will behave in a certain way, I guess that's how it will behave in the humberverse. Meanwhile, the ball is on your court to tell us why the ladder with toilet paper hanging from each rung is wrong. (It's hard to imagine how it could be wrong, since none of us has said what it will do, I don't think, but we want to know what you believe it will do.) Planted vertically in the ground in a wind, or mounted vertically on a treadmill in still air, what direction will each piece of tissue get blown?

No. I posted the paddle before that. Get in the queue/line. A smart guy like you does not need to see to work it out, otherwise I will have to ask you to conduct your ladder test.
There have been no useful responses to the paddle, let alone refutation.
So, Mr Aero.You tell me why it's wrong. Use any method you like, but be prepared to defend it to the end. A solution one way or the other.

 

[humber]

WOW - you do one day of real work
and this thread grows by three pages. Sadly I had to put everyone on fast scroll and simply search on spork to see if there are questions or complaints to address (I take the compliments and adoration as read )

Working in the hall, no doubt.

Originally Posted by humber View Post
Yes. Spork's silence is golden

Yes, I can certainly see how reason is your enemy.

A vacuum makes no sound

 

[humber]

LOL. Note, NOT a physics textbook!!!

A non sequitur, as usual, Captain. Over.

 

[humber]

I seem to recall humber making some imagined claim involving a paddle wheel but he never clarified what he was saying.
posted.

Taking a cat-nap?

If he would actually state clearly what he thinks the paddle wheel will do and how it will be different between real wind and the effective wind of a treadmill reference, we could produce a video showing once again that humber is wrong.

By all means, make that video!

I presume humber is still following this thread...

pussy hunting.

...so I'll present a simple scenario using the paddle wheel and no numbers (since we already know humber can't do math).
In the unlikely event that humber responds with a clear claim, I expect it will be quoted (with loud gasps) so I'm not going to even bother taking humber off ignore.

Nobody bothers to ignore you. In accordance with your skills, you need only address the top left and bottom right. Assume that the force on the paddle is the sum of the flow in but as a nominal '6m/s' in each case.

In the first scenario, we are flying in a hot air balloon high enough above the flat ground (this must be Kansas) to be in free flowing air.

Lyrical

The paddle wheel is suspended by its axle on two strings so that it can rotate freely about the axle and kept parallel to the ground and perpendicular to our direction of motion over the ground. As we lower the paddle wheel starting in the free flowing air down to just above the ground level, describe the action of the paddle wheel.

Done

In the second scenario, there is no wind and we are suspended (possibly in a hot air balloon) high enough above a large moving treadmill to be in still air. The paddle wheel is again suspended by its axle on two strings so that it can rotate freely about the axle and kept parallel to the treadmill surface and perpendicular to the direction of motion of the treadmill surface. As we lower the paddle wheel starting in the still air down to just above the treadmill surface, describe the action of the paddle wheel.

Done. All that because you don't see that the flow must go with the belt. Patently obvious.

 

[humber]

Oh Good! More big words that have nothing to do with the subject!!

Kindly supply one example of the application of "Negative drag' to aerodynamics???

And, of course, "Cavitation in gas" is LONG gone!!

Missed the runway again, Captain.
"Big words"...normal vocabulary for me.

http://www.path.berkeley.edu/PATH/Publications/PDF/PRR/98/PRR-98-05.pdf
There are many more, and then there is other class with negative incremental (dynamic) coefficients and hysteric.
As "things" operate the under the same laws, they apply to air too, and so does cavitation. Sometimes the nomenclature is not so specific.
(The tires of your aircraft are hysteric, but fortunately for your passengers, you don't need to know.)
Over.

ETA:
That's today's trash dealt with. Serious comments welcome.

 

[John Freestone]

Taking a cat-nap?

<snip>
Done. All that because you don't see that the flow must go with the belt. Patently obvious.

But humber, with respect, I have tried several times to clarify this point about the boundary layer, and you keep evading the question. For those of us who believe in the equivalence of these situations, it is like this:

Yes, the boundary layer goes with the belt - it has to and that is obvious. But, because we are discussing velocities w.r.t. the belt, that velocity of the b.l flow is slower than the surface itself. Hence, for a molecule of the actual treamill belt, looking up, all it sees is molecules of air that it's trying to grab and push along with it, but because it's not a solid, they are slipping behind. Thus, from the frame of reference of the belt, the air is either "going with it" completely (i.e. would be measured as stationary in that frame), or is going left to right w.r.t. it.
[ETA: had to reverse that myself then!]

This then means that anything going at the same speed as the belt, like your rabbit, does not see or feel the boundary layer going with the belt faster than the belt. To go back to my scenario, the person on the belt facing backwards, feels the air down their back at all heights, and the speed of that wind on their backs is exactly the same as it would be if they stood still on an equivalent surface (or the very same surface if you like) in "real wind".

Look. You wanted to solve this BL problem a long time ago, and that's what I was trying to do. It's quite simple. Answer these questions:

1. Do you mean that the boundary layer goes in the direction of the belt faster than the belt? Y/N

2. If the answer to 1. is No, then, when you're on the belt going with the belt, at belt speed, is the BL going to overtake you?

3. If the answer to 1. is Yes, how does a belt accelerate still air to go faster than it?

These depend on you being able to visualise things from different speeds, i.e. frames of reference, or be able to do the maths for that correctly, or be willing to test the hypothesis empirically, none of which seem to pertain in your case at any kind of reliable level. If you don't like the above examples, there is the other I suggested. Stick a "ladder" to a large piece of card, with very light pieces of tissue paper stuck to the rungs. Wave it through the air so that the card slices the air like the surface of a treadmill, and the tissue catches the still air in the room. Is there any level from top to bottom where the paper swings forwards? If not, what does it mean to say that the direction of the BL is reversed? They all blow backwards. If you stop that and put it down and blow it with a hairdryer, they blow 'backwards' (downwind) as well.

Or answer my question about how the air over a car in a windtunnel (still surface, "real wind") should all go in the same direction, but when the car drives off down the street in still air, the BL goes the wrong way, according to what you seem to be saying?

And I'm sorry to burst your bubble, but the diagram above is not "clarity". You don't specify what the figures are measured in relation to, which is precisely the point. And if I interpret them as best I can, there are glaring errors...not only are things from different frames of reference, but the BL shown going R-L over the treadmill (which is the right direction relative to the experimenter's ground, the room) should be 4, not 6, as has been explained to you by several people, including me. If the BL we chose to measure in the real wind was slowed to 1 m/s (closer to the ground than the layer going at 6), then in the treadmill situation that would have to be going 9 m/s right to left w.r.t. the room. How could it be going 1 m/s, with the belt, dragged by the belt, yet air above dragged by the belt with the belt faster than that? The belt drags the air right next to it to 10 m/s, not zero.

And just look at it - in the bottom left diagram, you've got "10 m/s ---->" written three times down the left hand side. What the hell are they? It's anybody's guess.

 

[jjcote]

In the interest of keeping people from going down a blind alley, criticism of humber for misuse of the term "negative drag" is probably misplaced. As far as I know, he did not introduce this into the discussion, I did. He had said that removing the prop from the cart would remove its drag, and I sarcastically replied that it's a funny kind of drag because it moves the cart forward rather than holding it back. In addition, there is a concept of "negative drag", where for example one component may exhibit this behavior when part of a system that has overall positive drag. An example is a cowling that has an increasing forward force as airspeed increases, but only when in the altered airflow pattern of the rest of the aircraft. This has nothing to do with the matter at hand, of course, and there are plenty of other weird things that humber has said to taunt him about.

Carry on.

 

[John Freestone]

Besides the small matter of the boundary layer, humber, you don't seem to be clear on whether a moving surface in still air is a suitable equivalent platform for modelling something at rest with wind blowing over it. It seems from earlier arguments about people on treadmills, that you still dispute that as well as the boundary layer direction.

So here's another little brainteaser for you. We have a large - very large - flat trolley on wheels, with a motor to drive the wheels. It is placed in the middle of a wide flat desert. On it stands a man, facing West (I'm just using West because it more naturally fits with left), and a wind turbine attached to a generator and ammeter. He's also got his mobile phone charging in parallel so he can phone home and ask a friend to get him the hell out of there if the engine cuts out.

The vehicle is still, and a westerly is blowing at 10 m/s. The man feels a nice cooling breeze on his face, and the ammeter reads 432.1 mA. However, the breeze drops a little, to 9 m/s, and the man feels just a little bit warm. The ammeter reading drops. He's a bit worried about his mobile battery and his chances of survival. He starts the motor and drives the vehicle westwards at 1 m/s. His motion through the air is now back up to 10 m/s, and the ammeter's too, so it now reads 432.1 mA again.

Let's imagine the wind keeps dropping by 1 m/s, and each time, he adds 1 m/s to the speed of the vehicle. Then we have the following results for W (windspeed from the west m/s), V (vehicle speed to the west m/s), R (relative windspeed felt by man, from west, m/s), C (current reading on ammeter, mA):

 W      V      R       C
10      0     10    432.1
 9      1     10    432.1
 8      2     10    432.1
 7      3     10    432.1
 6      4     10    432.1
 5      5     10    432.1
 4      6     10    432.1
 3      7     10    432.1
 2      8     10    432.1
 1      9     10    432.1
 0     10     10    432.1

and if the wind even reversed and blew from the east, he could continue accelerating beyond 10 m/s, and we would get:

-1     11     10    432.1
-2     12     10    432.1
-3     13     10    432.1
...etc.

Anything you disagree with there? See how absolutely nothing of any relevance within that changes? The wind could be racing round the planet at a squillion megapardres per fortnight for all he cares. If he maintains that 10 m/s difference, everything, including his face, is cool.

Stationary air is real wind. It's only 'stationary' w.r.t. things going at the same velocity. Because

velocity is relative

 

[jjcote]

I went ahead and created a corrected version of the paddle diagram. Please, everyone take a look, and see if I've made any errors (it's certainly possible). That includes you, humber. Note that the paddle is doing the same thing in all cases, that is, indicating that the higher air is moving to the right relative to the lower air.

 

[spork]

take a look, and see if I've made any errors

I haven't looked at any of the numbers or arrows, but I see a problem right off. You have the paddle turning the same direction in all cases.

Perhaps humb or humber can check it out in detail and find the source of the error.

 

[John Freestone]

I went ahead and created a corrected version of the paddle diagram. Please, everyone take a look, and see if I've made any errors (it's certainly possible). That includes you, humber. Note that the paddle is doing the same thing in all cases, that is, indicating that the higher air is moving to the right relative to the lower air.

Yep, jjcote, I agree with that. It seems there are two easy bits for someone to misunderstand or misinterpret from there:

1) They might not see where that 4 m/s comes from. Hence my other argument about the relationship between the fast air and slower air dragged by the belt - that argument should demonstrate that the BL outside must be subtracted from the windspeed maximum when considered relative to the belt.

2) They might ignore the direction of those 4 m/s arrows in the "at windspeed" diagrams on the right.

If they get those, then the maths is as I think I put it for the paddle result, which is the difference in windspeeds:
Outside at rest = 10-6
Outside with wind = 0--4
Inside with belt = 10-6
Inside with wind = 10--4

10-6 = 4
0--4 = 4

Result, paddle differential windspeed is 4 m/s. What's more, you could drive it through the air at that level in either direction at any speed you like, and the answer would still be 4.

Hint: that's the answer to the one spork set too, BTW.

Hey, jjcote, I think we're working up to a rock solid proof here. Pity Newton isn't around to thank us.

 

[mender]

But where's the bunny, jj? We have to know the location of the bunny!

JJ, thank you for posting those diagrams. I agree that they are correct.

See, humber? It isn't all that hard to get things right but you have to apply the correct physics, logic and math. Have a good look at jj's diagrams and figure out where you made your mistakes.

Try not to rationalize your way out of doing at least that much.

 

[John Freestone]

Ok, just so there aren't any bits of jigsaw left on the playroom floor, how do we get that 4 m/s (to the right, i.e. -4 m/s)? There's a BL going at 6 to the right, when the wind is going at 10. We want to go at the windspeed and see what the BL is doing. We set off downwind. As we get up to 1 m/s, we're catching up with the BL at 1 m/s, while it's running away from us at 6. The result, it's still on our backs, at 6-1=5. We go faster, and it's
6-2 = 4
6-3 = 3 ...
6-6 = 0 ...
6-7 = -1 ... we're now running past it at 1 m/s... we go faster still, cause we want to get up to the 10 m/s windspeed by hook or by crook...
6-8 = -2 ...
6-10= -4

Ok for you, humber? If you run at 10 m/s "downwind" in a 6 m/s wind, it's reversed, and blows on your face at 4 m/s. Hence if you run at 10 m/s downwind in a 10 m/s wind that has a boundary layer going 6 m/s (4 m/s slower than it), you feel no wind at the top level, but you experience a BL of 4 m/s right to left. Yeah? You had it going at 6 right to left.

ETA: No you didn't, sorry. You had it going 6 right to left on the treadmill. That's why the treadmill isn't a valid equivalent model. I give up. You're right after all. All treadmills are false. Smash the treadmills!

 

[humber]

I went ahead and created a corrected version of the paddle diagram. Please, everyone take a look, and see if I've made any errors (it's certainly possible). That includes you, humber. Note that the paddle is doing the same thing in all cases, that is, indicating that the higher air is moving to the right relative to the lower air.

[qimg]http://www.internationalskeptics.com/forums/imagehosting/thum_29182497b2d315dd09.jpg[/qimg]

It is corrected, though you may be it may just be possible that you are wrong? One small step for jjcote, one giant step for this thread.

Both the real wind cases are incorrect. The difference between the flows of 10m/s and 6m/s, is 4m/s at standstill. This must remain the case at windspeed because it is a simple translation;
10ms + 10ms = 20m/s
6m/s + 10m/s =16m/s
Difference = 4m/s
This is analogous to two parallel belts of 10m/s and 6m/s; the difference being 4m/s. The motion of the cart observer w.r.t the ground is also 10m/s at windspeed, so that is the same as an observer walking on the ground parallel to the belts at that same speed. That will not change ratio of the belt speeds for the observer, so once again, the answer is 4m/s. The absolute value is not all that important, but that difference will be constant for all observer velocities. This is is not the case on the treadmill. (This problem can be solved directly using superposition.)

That is enough, but the treadmill windspeed is certainly wrong. The belt-flow is with the belt at 10m/s. The fastest moving air is at the belt surface, and slowest at the still air, which is the opposite of real wind. (The belt surface is the road surface.)
The speed of the belt-flow is therefore ( 10m/s - 4m/s) and like the wind 4m/s slower than the fastest component, and so an effective 6m/s in the direction of the belt. Referenced to still air that is 6m/s back with the belt.

The paddle averages the force of the entire flow, as stated, so from that point of view, it is also 6m/s in the direction of the belt.
Pitot tubes will confirm the above result (and the reversal of the flow's profile).

 

[humber]

Ok, just so there aren't any bits of jigsaw left on ....

All the information is in the diagrams. No new scenarios, please.

 

[RossFW]

Missed the runway again, Captain.
"Big words"...normal vocabulary for me.

http://www.path.berkeley.edu/PATH/Publications/PDF/PRR/98/PRR-98-05.pdf
There are many more, and then there is other class with negative incremental (dynamic) coefficients and hysteric.
As "things" operate the under the same laws, they apply to air too, and so does cavitation. Sometimes the nomenclature is not so specific.
(The tires of your aircraft are hysteric, but fortunately for your passengers, you don't need to know.)
Over.

ETA:
That's today's trash dealt with. Serious comments welcome.

Did you even read your own link? There was LOTS or talk about drag, especially the way it is produced (Form, parasite, induced etc.) there was lots of use of Cd in order to make calculations. They all refered to drag in it's excepted definition, a sum of the forces in the opposirte direction of motion.

THERE WAS NOT ONE MENTION of the term "Negative drag". It is not used in Aerodynamics,and once again you have simply blustered to try and sound like you weren't in error.

I am aware that my tyres are part of a system that may be in a number of different states, as can my altimeter.

Who are you tryin to kid?

I also notice, that after posting several links that DON'T show that Negative Drad exisits as a term in aerodynamics, you don't even try to show anyone desribing Cavitation in liquids.

Full of it...

 

[humber]

But where's the bunny, jj? We have to know the location of the bunny!

JJ, thank you for posting those diagrams. I agree that they are correct.

See, humber? It isn't all that hard to get things right but you have to apply the correct physics, logic and math. Have a good look at jj's diagrams and figure out where you made your mistakes.

Try not to rationalize your way out of doing at least that much.

You are a bunny. Try again. I have corrected jjcote's mistakes.

 

[humber]

Did you even read your own link? There was LOTS or talk about drag, especially the way it is produced (Form, parasite, induced etc.) there was lots of use of Cd in order to make calculations. They all refered to drag in it's excepted definition, a sum of the forces in the opposirte direction of motion.
..

blah, blah, blah. Find your own. Don't get hysteresis-terical.
You have NO IDEA what you are talking about, and are trying to bluff, Captain.
I know quite a lot about rheology, so your failure is 100% guaranteed. Over.

 

[John Freestone]

It is corrected, though you may be it may just be possible that you are wrong? One small step for jjcote, one giant step for this thread.

Both the real wind cases are incorrect. The difference between the flows of 10m/s and 6m/s, is 4m/s at standstill. This must remain the case at windspeed because it is a simple translation;
10ms + 10ms = 20m/s
6m/s + 10m/s =16m/s
Difference = 4m/s
This is analogous to two parallel belts of 10m/s and 6m/s; the difference being 4m/s. The motion of the cart observer w.r.t the ground is also 10m/s at windspeed, so that is the same as an observer walking on the ground parallel to the belts at that same speed. That will not change ratio of the belt speeds for the observer, so once again, the answer is 4m/s. The absolute value is not all that important, but that difference will be constant for all observer velocities. This is is not the case on the treadmill. (This problem can be solved directly using superposition.)

That is enough, but the treadmill windspeed is certainly wrong. The belt-flow is with the belt at 10m/s. The fastest moving air is at the belt surface, and slowest at the still air, which is the opposite of real wind. (The belt surface is the road surface.)
The speed of the belt-flow is therefore ( 10m/s - 4m/s) and like the wind 4m/s slower than the fastest component, and so an effective 6m/s in the direction of the belt. Referenced to still air that is 6m/s back with the belt.

The paddle averages the force of the entire flow, as stated, so from that point of view, it is also 6m/s in the direction of the belt.
Pitot tubes will confirm the above result (and the reversal of the flow's profile).

Ok, no new scenario. Can you tell me this. At the level where the real wind is blowing at 8 m/s, how does its height compare with the 6 m/s and the 10 m/s, and what velocity does it have down belt w.r.t. room/still air? See, if I apply your method, and keep it 2 m/s different from the full speed, as you said the 6 m/s flow should be 4 m/s from full speed, then it's going R to L at 8 m/s.

But the 8 m/s BL in real wind must be higher up, away from the belt, so we have the 8 m/s reversed BL above the 6 m/s BL. Do we apply the rule throughout, and have a 0 m/s wind on the belt and 10 m/s reverse BL (going right to left) just below the still air? Doesn't that create an awful sheer force, where the opposite flows of air pass each other? ETA;...I mean, where the 9.999999 m/s BL going right to left, dragged by the belt, passes the still air in the room? Like I asked before, if you just filled in the blanks on my earlier diagram, we'd already have solved these little problems.

 

[Dan O.]

Is that really humber's paddle wheel argument? Why does he use a different relative speed of the boundary layer over the surface when the surface is a belt instead of the ground? Is this one of those "lets solve a different problem and get a different answer and claim that the problems are different" sort of things?

And why does he screw up so badly in the lower left panel? Are there to many vectors for him to keep track of?