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Topic: Four-leg table on smooth terrain (Read 2258 times) |
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brac37
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Four-leg table on smooth terrain
« on: Apr 23rd, 2009, 4:48pm » |
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Here is another problem, and it is really hard. Consider a table with four legs standing on a terrain. The feet of the table are perfectly in the same plane, but the terrain is not flat. The purpose is to put the table on the terrain such that is does not waver.
Assume any smoothness conditions for the terrain, except that it is perfectly flat. For which arrangements of the four legs can you always put the table such that it is standing stable on all four legs? Ignore the size of the feet.
If the four feet form a square, then you can put the table in a stable manner. Suppose you put it somewhere on the terrain and assume it is wavering. Now turn the table 90 degrees, such that the stable pair of legs and the wavering pair of legs interchange. At some point in between, the wavering pair of legs becomes stable. At that point, the whole table is stable.
I have thought out a positive answer for a symmetric trapezium as well. Tables for which the feet are not on a circle cannot be put on every smooth terrain: take a part of a sphere as the terrain. For all other arrangements of the feet, I do not know the answer.
EDIT: about the smoothness, let us just say that the slope of the terrain is bounded by an arbitrary small positive number.
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| « Last Edit: Apr 24th, 2009, 6:58am by brac37 » |
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Grimbal
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Re: Four-leg table on smooth terrain
« Reply #1 on: Apr 24th, 2009, 4:25am » |
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Imagine the terrain has constant curvature and the table is a lozenge.
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brac37
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Re: Four-leg table on smooth terrain
« Reply #2 on: Apr 24th, 2009, 6:57am » |
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on Apr 24th, 2009, 4:25am, Grimbal wrote:| Imagine the terrain has constant curvature and the table is a lozenge. |
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I do not get your point. A lozenge does not have its four feet on a circle and therefore cannot stand stable on a sphere.
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Grimbal
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Re: Four-leg table on smooth terrain
« Reply #3 on: Apr 24th, 2009, 7:15am » |
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I didn't read your post carefully to the end.
It was an example of a table that will never be stable.
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brac37
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Re: Four-leg table on smooth terrain
« Reply #4 on: Apr 24th, 2009, 11:35am » |
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Ok, here is the solution for the symmetric trapezium. But try it yourself first. And do not quote it unhidden!
| hidden: | Draw a line from above to below somewhere on the terrain. The intersection point of the diagonals of the symmetric trapezium will be on this line.
Let a0 be the angle of the diagonal of the stable legs and f(a0) be the height on the line of the intersection point of the diagonals of the symmetric trapezium. Now saw in one of the stable legs such that the other legs become stable. Let a1 be the angle of the diagonal of the new stable legs and define f(a1) in a similar manner as f(a0) above. Then f(a1) <= f(a0).
Repair the table and put it in such a way that f(a1) - f(a0) is minimal. If f(a1) = f(a0), then we are done, so assume f(a1) = f(a0). We will derive a contradiction.
Define a2 with respect to a1 in a similar manner as a1 is defined with respect to a0. Define a3 in a similar manner as well, et cetera. Since f(a0) - f(a1) is minimal, we have
f(a(i)) - f(a(i+1)) >= f(a0) - f(a1)
Let a be an accumulation point of the sequence a0, a1, a2, .... By the above inequality, we obtain a contradiction to the continuity of f at a. |
Remark: the terrain must be sufficiently smooth in order to obtain that f is a function.
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