Yes, it's at 4:15 on this video.
"There, I've run rings around you logically."
"Ohhh... Intercourse the penguin."
Thanks Brian. We got there already, but you can't know that without trailing further on. I love the bit: "Penguins don't come from next door, they come from the Antarctic.
BURMA! ....
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.... (long pause) ....
....
....
Why did you say Burma?
I panicked."
How about this guys,
Yes Humber, you are right about EVERYTHING.
Thankyou, and goodbye!
Yes, except the thing I've been hanging out for is a clear contradiction, such that his humberphysics regarding the treadmill cannot be true as well as its contradiction, although I am weary enough of his ways now to know that I will probably only approach this position closely enough to satisfy me (and maybe others...read on and enjoy...); the humberverse may include a humberlogic different from ours, such that two contradictory statements can both be true as long as they were both uttered by humber.
That's why I was so persistent in asking my questions regarding his diagram of the boundary layer. I think jjcote noticed, and maybe RossFW.
Statement 1.
If the layer is say, 100mm and has a linear profile, what moves at 10mm above the ground, moves at that same speed 90mm above the belt, but in the opposite direction.
This, to orthodox science, is effectively correct. I posted a table of the full boundary layer according to that scheme:
Code:
HEIGHT OUTSIDE INSIDE
120 10 > 0
110 10 > 0
100 10 > 0 ____theoretical 'beginning' of boundary layer
90 9 > < 1
80 8 > < 2
70 7 > < 3
60 6 > < 4 ----------- Case in your diagram
50 5 > < 5
40 4 > < 6
30 3 > < 7
20 2 > < 8
10 1 > < 9
_0_ 0 <10
Interlude
This I am posting again, just because I want to enjoy the soothing lineament of it after banging my head against this immovable wall for so long (I don't actually remember who he was addressing at the time):
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.
Ah. Lovely.
Statement 2. (I'll leave the full post, but have emphasised the important bit if you want to cut down on your drivel intake and get to the meat of the humberism - with especial thanks to jjcote for joining in this pincer movement on the enemy position! It was in fact jjcote's original scenario, I think.)
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).
So, the fast bits of air near the top in 'real wind' become the same speed fast bits of air at the bottom over the treadmill, but also the 6 m/s 'real wind' boundary layer becomes the same speed in the other direction over the belt, it's hard to imagine how to have both of these at the same time. Could humber have been wrong about one of them, or could they somehow both be true? Could the gradient have squashed up somewhere, so that the 60 --> is still opposite the <-- 60, but the 90s and the 10s have reversed position? He hasn't specifically acknowledged the correctness of my table. Could air suddenly have changed its characteristics by being stopped and having a moving belt put under it?
Of course, we all know that his diagram was wrong, and that Statement 1 is his own refutation of the paddle. It now remains only to see whether he will choose one or the other, try to make them fit somehow (probably by saying that I've misinterpreted the statements), ignore this blatant self-contradiction, waffle about the well-known antialiasing effect at windspeed and different aerodynamic aspect ratios...?
Finally, let me just make this point utterly clear. Ages ago, some of us noted that it was unfortunate that 6 m/s was chosen arbitrarily as the nominal velocity of "the boundary layer" (there of course being all velocities between the full windspeed and zero), because in this case it was near to half the windspeed, and made its incorrect reversal easy to miss or misunderstand. Now, had we chosen 9 m/s as a different example of boundary layer velocity, humber would now have to choose, on changing to the treadmill scenario: should the speed near the top now become the same speed lower down, with the direction reversed, and hence it can't be 9, and must be 1 m/s at this level; or should "the boundary layer" windspeed maintain the same difference between itself and "the fastest component", and hence be 9 m/s?
And as we argued endlessly, the +6 should have become a -4, not a -6. This corrects the flaw in the paddle's differential to, as we stated 0--4 = 4 (w.r.t. the ground). There, I've run rings round you logically.
However, I see that since then he is still "refusing to understand" how to calculate velocities from a different frame. I think he needs a different kind of boost first.
QED
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