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 Posted: Mon May 08, 2006 7:15 pm
(I need to stop reusing old problems, since this one is also from the aforementioned dead board, except that I posted this one.)
Assuming that the Earth is spherical and rigid, tie a nonelastic cable around a circumference of length one meter more than the circumference. Raise the cable up at a given point as far as it will go. Can a cat pass under the cable? An elephant? Estimate the height and prove your claim.
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Posted: Tue May 09, 2006 2:16 am
Hey, another problem I've heard of razz
As I'm familiar with the problem, does the exclude me from answering?
And I assume you wanted an estimate (read guess, off the top of someone's head), rather than someone working out the actual values?
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Posted: Tue May 09, 2006 5:31 pm
According to Wkipedia, the circumference of the Earth is 40,075.004 km, which is 40,075,004 m.... so the length of the cable would be 40,075,005 m long. When you raise the cable, since it's non-elastic, you wouldn't be able to raise it that much... I'd say half a metre tops, since the cable is only one metre longer than the Earth's circumference. So the cat could go under it, but the elephant would be out of luck.
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Posted: Tue May 09, 2006 7:28 pm
Does a slip of paper with a cat on it count? [Though, my logic may have failed me]
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Posted: Tue May 09, 2006 8:25 pm
We can prove the cat fits without almost any math:
Cinch the cord tight and pinch the ends together. You have a spare meter of cord. Stretch out the spare as far as you can... it'll go to a height of 1/2 meter. Now release your pinch point-- your hand will go up and a tentish shape will form... the cat will surely fit under 1/2 a meter.
Finding the actual answer is pretty (if I've figured this correc tly). I think I know how to do it but unless there's something obvious I'm overlooking it's going to take a few calculations that would take a bit longer then I have to spend... The process spoiler is below in white... there's probably a much better way to think about this problem but... you have to go with whatever comes to mind.
Put a nice big pink flag on the ground where we're pinching the rope together and then insert a VERY long pole (just in case we need the distance) vertically into the ground.
Now if we cut the rope at the pinch point and imagine would would happen as we try to raise one of the two end, we see that the line formed by our hand and where the rope touches the ground should be tangent to the circle formed by the rope's original position on the ground. If we turn ourselves upside down and imagine how the receding earth would look from the end of the rope we see that the pink flag is describing a cycloid. As we continue to watch, the pole will appear to rotate away from us and recede. When the distance along the current tangent line is 1/2 a meter we stop, extend the tangent line, and mark the pole. That height is what we want.
So the modified version of the question is:
Let O be the origin, let p be a point on the cycloid. Let Np be the normal line to the curve at p, and Qp the point where Np crosses the y-axis. For which p does | O Qp | = 1/2?
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Posted: Wed May 10, 2006 5:37 am
so here's my try at it. Take the radius of the earth (which is readily available in any textbook), using the circumference formula figure out the circumference.
C = 2pie*r = 2pie *(6.27*10^6)
C = 40,023,890.14m (approx what was stated before)
increase circumference by one meter
C = 40,023,891.14 and derive radius from this
r2 = C/2pie
r2 = 6,370,000.16
r2 - r1 = 16cm
Thus the height is an average of 16cm above the ground. A cat can easily pass under this, an elephant cant. As to how to prove that this is indeed the height of the cable? I don't know.
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Posted: Wed May 10, 2006 7:45 am
Dave the lost And I assume you wanted an estimate (read guess, off the top of someone's head), rather than someone working out the actual values? Er, it's more the fact that this problem doesn't have an exact solution. Sorry--I guess a physicist would say find the height. Cynthia_Rosenweiss When you raise the cable, since it's non-elastic, you wouldn't be able to raise it that much... I'd say half a metre tops, since the cable is only one metre longer than the Earth's circumference. So the cat could go under it, but the elephant would be out of luck. Are you sure? The half-meter is a lower bound--the height reached when the cable in 'pinched' above the surface, as grey wanderer demonstrated. Surely it is possible to do better. The problem requres just a little bit of calculus, although not necessarily as complicated as grey might believe. poweroutage Thus the height is an average of 16cm above the ground. A cat can easily pass under this, an elephant cant. As to how to prove that this is indeed the height of the cable? I don't know. That's true--increasing circumference of a circle by 1 unit increases the radius by 1/2π units, which in this case is about 16cm. But that's the average, or what would happen if the cable is raised everywhere uniformly to make a new circle. The problem, however, is to raise it at a single point.
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Posted: Wed May 10, 2006 8:00 am
grey wanderer We can prove the cat fits without almost any math: Cinch the cord tight and pinch the ends together. You have a spare meter of cord. Stretch out the spare as far as you can... it'll go to a height of 1/2 meter. Now release your pinch point-- your hand will go up and a tentish shape will form... the cat will surely fit under 1/2 a meter. NON ELASTIC. lol
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Posted: Wed May 10, 2006 11:39 am
rugged grey wanderer We can prove the cat fits without almost any math: Cinch the cord tight and pinch the ends together. You have a spare meter of cord. Stretch out the spare as far as you can... it'll go to a height of 1/2 meter. Now release your pinch point-- your hand will go up and a tentish shape will form... the cat will surely fit under 1/2 a meter. NON ELASTIC. lol ????? I'm not sure I understand your complaint. Could you explain where my post assumes that the cord is elastic? On a seperate note... How much force would be require to hold the cord in place at the point?
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Posted: Wed May 10, 2006 12:58 pm
VorpalNeko poweroutage Thus the height is an average of 16cm above the ground. A cat can easily pass under this, an elephant cant. As to how to prove that this is indeed the height of the cable? I don't know. That's true--increasing circumference of a circle by 1 unit increases the radius by 1/2π units, which in this case is about 16cm. But that's the average, or what would happen if the cable is raised everywhere uniformly to make a new circle. The problem, however, is to raise it at a single point. oooooohhhhhh, ok I didn't understand that that is what the question was asking.
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Posted: Wed May 10, 2006 2:41 pm
VorpalNeko Dave the lost And I assume you wanted an estimate (read guess, off the top of someone's head), rather than someone working out the actual values? Er, it's more the fact that this problem doesn't have an exact solution. Sorry--I guess a physicist would say find the height. Sorry, misread the question. I was thinking of a different, but quite similar one I got. *slaps self*circumference, idiot Rugged: Inelastic means it doesn't stretch, not that it doesn't bend.
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Posted: Mon Jun 05, 2006 6:40 pm
The highest the cable would ever get above the ground would be about 110m because of the curvature of the Earth.The distance from the top of teh wire to the point where it touches ground diagonally would be around 22500.25m.
x=height of top isosceles including curvature of Earth
To do it mathematically(by guess and check razz ) Imagine an isosceles with two sides the length of the Earth's radius (6378100m) and the other side being 2(x^2-1/4)m. With another isosceles triangle on top of it with the two equal side lengths being x^2+1/4m and the other being 2(x^2-1/4)m (overall shape should be a kite). To find out the height of the curvature of the earth you would divide the lower isosceles triangle into two right triangle and find out the length of that line and subtract it from 6378100m or (6378100-((6378100)^2- (x^2-1/4)^2)^1/2)m
If something doesn't add up please tell me
Note: there is a way to find both points of tangency but I do not know enough math to do that without guess and check
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Posted: Tue Jun 06, 2006 6:14 am
[ Message temporarily off-line ]
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Posted: Tue Jun 06, 2006 12:14 pm
[ Message temporarily off-line ]
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