A little more patience, and you might get one puzzle right this year.
Ok here goes for the paddle question. I don't know what the 'paddle' is really about, but it seems some device for measuring two air velocities, from which we can calculate the difference.
Good, just like the conclusion.
First of all, it seems right up to '4 m/s', for someone 'standing in real wind', if that means someone standing on the ground. Yes. The top layer is going 10, the particular layer lower down that we're measuring the other velocity of, jj said was going slower, at 6. Difference = 4. Now, I don't get why you've put CW. You're obsessed with clocks.
Yes 4m/s CW. Stand in front of the treadmill as is conventional. Belt moves right to left.
All I can imagine is that if I stood to the East of the 'paddle', the top layer would be going 4 m/s faster than the lower layer, and we can imagine that would spin a 'paddle device' clockwise as we look at it. Yes? I think jjcote's wind direction was from the south, but I can't be sure now.
That is correct. The tailwind that pushing the cart (the real world analogue) comes from the left (South in jjcote's case)
I'll say that was it anyway for this. As I've pointed out before to you, CW and CCW don't mean diddly unless you give enough information.
The information is there, John. CW/CCW are correct. The motion is relative, so make a choice if you don't like that one, but be consistent.
If we stood to the West of the device, the southerly wind would still be faster at the top layer and slower in the lower layer, but now it would move the paddle CCW. It hasn't changed direction. You just looked at it from the other side. Ok so far. 4 CW looking from the east side.
But now you say same again at windspeed. If you mean we are now travelling with the wind holding the device, you must surely realise that it can only measure the relative windspeed (by now, surely, can't you, by now, surely, please?). So it's top measuring device gets 0 m/s and its lower one gets
Held when stationary w.r.t the ground, yields the same when held moving from a cart. (real wind)
Wind - flow = constant, so (Wind + x) - (flow + x) = constant.
(the speed of the wind jj said it had relative to the ground) minus (whatever speed we are going at)
= 6 - 10
= -4
i.e. the bottom measurer is getting a backwards wind (it's going slower than us at our 10 m/s) of 4 m/s. The top one is stationary. So your bit in brackets should read (0 - - 4 = 4) not (10 - 6 = 4).
The difference in velocity of the flows is constant, so the paddle wheel will be constant.
So when we are travelling with the wind at windspeed, we get: 4 CW looking from the east side. Gosh, that's the same as above. How amazing. A difference in two velocities is the same even when we measure it relative to a different 'ground-zero'. [ETA: I removed one particularly bitchy bit here, because you did get the same answer, and I was criticising the wrong thing, perhaps. The error seems just that you have a magical paddle device that can measure the windspeed relative to the ground when you're travelling at windspeed, rather than reading 0, and the boundary layer w.r.t. the ground (6) instead of w.r.t. the device (-4), but 10-6 does = 0--4 anyway).]
You do not appear to understand your own ideas, here. I choose the belt to meet that expectation. Not even smoke!
It works from all "frames" of course, because all frames are equivalent. There is, nor can there be, a difference.
The difference in velocity of the real flows is constant, so the paddle wheel will be constant.
Superposition says so.
No. The top one is right, 0. The bottom one is wrong. Unfortunately, I'm finding it difficult to think of how to walk you through how to get it right without just employing the relative velocities trick normal people use all the time when doing this kind of thing.
Guaranteed to lead to error, time after time. And again. This question I raised months ago. He's wrong (shrug), No, you are all wrong, again.
Here goes: JJ said that the boundary layer wind was going at 6 wrt the ground. Now, to any of us, it is obvious that that means that the equivalent layer must also be going at 6 w.r.t. the belt
He can mean what he likes. The belt moves with the belt, so the flow moves with the belt,
from speeds of 0 to 10m/s dependant on height. 6m/s matches jjcote's example.
(or we just haven't got a correct setup and need a wider belt, or the air pressure has changed or something - this depends on such laws of physics - ok?). Therefore, if the belt is going backwards (i.e. negative) 10 and the boundary layer is going +6 with respect to it, then the boundary layer is going -10 +6 with respect to us.
No need to walk, stand fixed to the belt. Attach the paddle by a spindle to the belt to make it easier to see.
= 6 - 10 = -4
So, to correct you above, the paddle will turn (0 - - 4) = 4 CW looking from the east side. Ring any bells?
Only a clanger.
It moves 4m/s CCW. (10 - 6. ) If both flows are 10m/s, no motion. If the bottom flow were
slightly faster, which way? CW. If slower = CCW. Simple.
No. Going with the belt, your back is pushed into a wind. That is what you called a positive wind, so you are right to say the top one gets 10. The bottom one is right as well.
Yes. When both lower and upper are in the same velocity flow, no motion.
But where did you get CCW from? The top is still faster than the bottom, and from the rear, just as before. It should therefore be 4 CW looking from the east side. Hmmm.
?
No. JJ said that the boundary layer he was using was going 6 m/s over the ground. In the equivalent scenario, it is also going over the belt at 6 m/s, although from our usual earthbound perspective we might call that the belt going past the wind at -6 m/s.
The belt is the road.
See above calculation. 4 CW looking from the east side.
I think you've even confused the poor wind now.
Top gets 10 from rear, the bottom (greek....) blade, gets only 6m/s from front (belt flow) so motion = 4 CCW. From cart at windspeed, 6ms CW.
I have no idea what you mean by that. Wind is the motion of an air mass relative to something else's motion. None of these motions are absolute. Blah blah blah blah...
The answer, in case you missed it, was 4 m/s CW looking from the east side.
No. Try again. Back to the drawing board.
(1) Real wind = constant difference.
(2) Stationary and windspeed on belt opposite rotations.
(3) Motion up belt can stop paddle at some point (4ms up the belt in this case) Never in wind.