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# Crash Course before the Mobius ride

HS Thomas
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Crash course :
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The Tucker boys , Johnny and Jack ( I think I mentioned the scoundrels before) , release identical steel balls down their respective tracks at exctly the same time. Jack's track is straight while Dave's is cycloidal in shape. Will the balls hit their targets at the same time or will one of them hit the target first ? If the latter which ball ?

Mobius ride
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Judy and Dan are the first to try out the latest ride at the amusement park an enormous mobius strip. If the cars are travelling at the same speed , will there ever be a collision .

P.S. The CODE tag is not working as well as it used to.
The second diagram looks much better in edit mode.
[ December 04, 2003: Message edited by: HS Thomas ]

fred rosenberger
lowercase baba
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Posts: 12234
36
if the curved shape is 1/2 a cycloid, and it's positioned correctly (i think the tangent at the high endpoint should be a vertical line, and at the low endpoint should be a horizontal line), that's the fastest way for a marble to get down the ramp. this was solved by Sir Isaac Newton using calculus of variations.

Stan James
(instanceof Sidekick)
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Interesting. My intuition would vote for the straight ramp because any curve burns off some energy changing direction.

HS Thomas
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I created these diagrams under Red Hat Mozilla
Last night I viewed the same under MS IE and they looked really abominable.
Sorry, not sure what to do about that. :0 Glad tidings , I get my old PC back next week.

Jim Yingst
Wanderer
Sheriff
Posts: 18671
[Stan]: Interesting. My intuition would vote for the straight ramp because any curve burns off some energy changing direction.
In fact, changing direction doesn't burn off any energy. The cycloid does have a longer path to travel, true, but it gives you a good jump-start at the beginning by plunging straight down. The increased speed at the start turns out to more than offset the increased path length. Though I'd agree it's far from obvious.
[HS]: The CODE tag is not working as well as it used to.
Mmmm, it's long given slightly differnet behavior under different browsers. If you look carefully at the HTML generated by UBB, some of the font tags end up improperly nested, and different browsers treat this illegal HTML differently. I spent a while trying to pin down just what the behavior was, and which part of the UBB code (in Perl) was responsible for it, but it gets pretty messy to look at, and few people seem to have problems with it most of the time...
Judy and Dan are the first to try out the latest ride at the amusement park an enormous mobius strip. If the cars are travelling at the same speed , will there ever be a collision .
I don't think we have enough information. Will they always be traveling at the same speed? Should we assume that after the initial conditions, gravity is the only force that will cause the cars to slow down or speed up? Most any actual ride would have significant use of brakes and motors to control spped - not just gravity. But if we assume that only gravity is relevant, then yes, there must eventually be a collision. First, I can't tell from the diagram which sides of the Möbius stip the cars are on. Both sides actually link up and are the same side, but the question is, do the cars meet in a head-on collision, or are they going the same direction at least? Obviously if they're going opposide directions on a closed track, they will collide. Even if they aren't though - it's clear that D has higher total energy (kinetic plus potential) because at the start, both have equal kinetic energy (same spped), and D has higher potential energy (because it's higher). From this we can prove that for any given position on the track, the speed of D when it passes that point will be greater than the speed of J at that same point. (Since at that point, both cars have equal height and thus equal potential energy, D's greater total energy must be manifest as greater kinetic energy.) So at every postion, D is faster than J. Thus D will complete a round trip in less time than J will (assuming both have enough energy to make the complete loop, and they don't lose any due to friction). So after some indefinite amount of time, D would have completed one more loop than J - which means it would have had to pass J, which would mean it would collide instead. Unless your Möbius stip has a passing lane.

HS Thomas
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Posts: 3404
[Stan]: Interesting. My intuition would vote for the straight ramp because any curve burns off some energy changing direction.
The ball on the cycloidal path hits the target first as it speeds up more rapidly than the ball on the straight path in the early stages of descent.The cycloidal creates the fastest track according to scientists.
Originally posted by Jim Yingst:

I spent a while trying to pin down just what the behavior was, and which part of the UBB code (in Perl) was responsible for it, but it gets pretty messy to look at, and few people seem to have problems with it most of the time..

You might be glad to know that the diagrams looks a treat under Mandrake Konqueror.

I don't think we have enough information. Will they always be traveling at the same speed? Should we assume that after the initial conditions, gravity is the only force that will cause the cars to slow down or speed up? Most any actual ride would have significant use of brakes and motors to control spped - not just gravity. But if we assume that only gravity is relevant, then yes, there must eventually be a collision. First, I can't tell from the diagram which sides of the Möbius stip the cars are on. Both sides actually link up and are the same side, but the question is, do the cars meet in a head-on collision, or are they going the same direction at least? Obviously if they're going opposide directions on a closed track, they will collide. Even if they aren't though - it's clear that D has higher total energy (kinetic plus potential) because at the start, both have equal kinetic energy (same spped), and D has higher potential energy (because it's higher). From this we can prove that for any given position on the track, the speed of D when it passes that point will be greater than the speed of J at that same point. (Since at that point, both cars have equal height and thus equal potential energy, D's greater total energy must be manifest as greater kinetic energy.) So at every postion, D is faster than J. Thus D will complete a round trip in less time than J will (assuming both have enough energy to make the complete loop, and they don't lose any due to friction). So after some indefinite amount of time, D would have completed one more loop than J - which means it would have had to pass J, which would mean it would collide instead. Unless your Möbius stip has a passing lane.

The cars are travelling at the same speed . Both cars will cover the same ground so where speed is lost both cars will eventually lose the same amount of speed. Where speed is gained , both cars will gain the same amount. Of course, they started from the same point but only one after the other. So no, they won't collide.Everytime on a helter-skelter ride or any other this theory is always being put to the test, IMHO. The answer would have to take the assumption that starting and stopping speeds of both cars are the same.
The diagram didn't help unfortunately. Sorry.
[ December 09, 2003: Message edited by: HS Thomas ]

Jim Yingst
Wanderer
Sheriff
Posts: 18671
Ummm... I'm having trouble making sense of this. When you say "the cars are travelling at the same speed", do you mean that at the instant shown on the picture, Dan and Judy have the same speeds? (That's what I had assumed.) Or do you mean that by the time Dan reaches the spot Judy is at now, he will have the same speed Judy had when at that spot, and when Judy reaches the spot Dan is at, she will have the same speed Dan has now? Or more generally, for any given positiuon on the track, every time a car passes that point, it has the same speed as every other car passing the same point? This is how most amumsement rides work, so if I assume this is what you meant, then most of your other comments make sense. But in this case the two cars would not start out with the same speed shown in the diagram, as Judy at point J is lower in the loop than Dan at D is; Judy must have picked up speed on the way there. This contradicts what I understood by "the cars are travelling at the same speed"; hence my confusion. At this point I really can't tell what problem is actually intended to be under discussion here - but I stand by my answer for the problem as it was originally presented.

HS Thomas
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Posts: 3404
Sorry ,Jim. I could have made it clearer if I marked 2 spots on the level to indicate their relative starting positions and stated that the cars travel at the same speed. Because the riders travel over every spot on the rail before returning to their starting positions(relative) and they are moving in the same direction, I think there is a 100% guarantee they won't collide. That is if they do not start too close.
The diagram is horrible ;
If, as you assumed, they started from the spots marked with J and D, this begs the answer to be examined again.
D has a fast fall followed by a slower rise (somewhat acelerated by the previous fast fall) and then a fast side swish and then a s-l-o-w rise to the top again.
J has a s-l-o-w rise and a similar fast fall followed by a slower rise (somewhat acelerated by the previous fast fall) and a fast side swish to the starting position indicated.

As I see it they still won't collide. Could we both be right ?
Though if they don't start too close to each other then there is no chance of colliding. But if they start close they may collide.
A third opinion is required.
People may never get on an amusement ride again !
[ December 09, 2003: Message edited by: HS Thomas ]