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Topic: Stabilise the square! (Read 3349 times) |
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towr
wu::riddles Moderator Uberpuzzler
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Re: Stabilise the square!
« Reply #25 on: Dec 19th, 2005, 2:40pm » |
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I hope you don't mind I trimmed some of the white off that image, it was a bit large.
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Joe Fendel
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Re: Stabilise the square!
« Reply #26 on: Dec 19th, 2005, 2:46pm » |
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Not at all. I don't know how to make a jpeg except by making a powerpoint and saving it as a jpeg. Like the farmer said to the raincloud, thanks for the crop assistance.
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Joe Fendel
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Re: Stabilise the square!
« Reply #27 on: Dec 19th, 2005, 2:56pm » |
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And I think I can see how to do it with 35 non-crossing, but I'm scared to make another jpeg...
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towr
wu::riddles Moderator Uberpuzzler
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Re: Stabilise the square!
« Reply #28 on: Dec 19th, 2005, 3:37pm » |
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You could get irfanview, it's a free image viewer with some basic manipulation. You can just open the file, select a rectangular area, and then choose 'crop' from the edit menu, and save it. Or otherwise, just post it, and I'll fix it in the morning
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Icarus
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Re: Stabilise the square!
« Reply #29 on: Dec 19th, 2005, 7:24pm » |
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If you have Windows, you can also use Paint (under the accessories menu). It allows you to crop images and save them in a number of formats.
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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towr
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Re: Stabilise the square!
« Reply #30 on: Dec 20th, 2005, 12:28am » |
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Actually Paint doesn't necessarily allow you to safe in a number of formats. There's some chance it'll just be a bmp, but with a different extension if you're not carefull.
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Barukh
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Re: Stabilise the square!
« Reply #31 on: Dec 20th, 2005, 4:00am » |
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on Dec 19th, 2005, 2:56pm, Joe Fendel wrote:And I think I can see how to do it with 35 non-crossing, but I'm scared to make another jpeg... |
| It can be done with further sticks.
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Joe Fendel
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Re: Stabilise the square!
« Reply #32 on: Dec 20th, 2005, 5:32am » |
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on Dec 20th, 2005, 4:00am, Barukh wrote: "Further"? As in fewer?
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Barukh
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Re: Stabilise the square!
« Reply #33 on: Dec 20th, 2005, 7:58am » |
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on Dec 20th, 2005, 5:32am, Joe Fendel wrote: Yes, of course!
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Grimbal
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29 and no crossings
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Barukh
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Re: Stabilise the square!
« Reply #36 on: Dec 22nd, 2005, 4:04am » |
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on Dec 21st, 2005, 3:43pm, Grimbal wrote:29 and no crossings |
| Indeed, very nice, but can be improved further. By the way, how do you show that a configuration is stable?
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towr
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Re: Stabilise the square!
« Reply #37 on: Dec 22nd, 2005, 4:59am » |
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on Dec 22nd, 2005, 4:04am, Barukh wrote:By the way, how do you show that a configuration is stable? |
| That can be difficult. But as long as you're just working with triangles, each triangle is stable if the sides are stable. And a square is stable if two opposite points are stable.
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Grimbal
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Re: Stabilise the square!
« Reply #38 on: Dec 22nd, 2005, 5:19am » |
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on Dec 22nd, 2005, 4:04am, Barukh wrote: Indeed, very nice, but can be improved further. By the way, how do you show that a configuration is stable? |
| It can be improved without crossing? With crossings I can bring it down to 21 in an obvious way. Stabillity. There are 29 sticks, each stick has 3 degrees of freedom. Each connection removes 2 degrees of freedom. There are 42 connections (k sticks joining counts as k-1 connections). That makes 29*3-42*2 = 87 - 84 = 3 degrees of freedom. That is just enough to move the whole figure. This assumes there is no redundant stick. That is the difficult point. A better way to show it is: - one stick is a rigid construct. - 3 rigid constructs that are connected 2 by 2 make a rigid construct if the connection points are not aligned (and therefore distinct). - my figure can be built that way.
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« Last Edit: Dec 23rd, 2005, 12:51am by Grimbal » |
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Barukh
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Re: Stabilise the square!
« Reply #39 on: Dec 22nd, 2005, 5:55am » |
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on Dec 22nd, 2005, 5:19am, Grimbal wrote:It can be improved without crossing? |
| Yes.
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JocK
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Re: Stabilise the square!
« Reply #40 on: Dec 22nd, 2005, 10:36am » |
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on Dec 22nd, 2005, 5:19am, Grimbal wrote: It can be improved without crossing? With crossings I can bring it down to 21 in an obvious way. |
| With crossing you should be able to bring it down below 21.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Wonderer
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Re: Stabilise the square!
« Reply #41 on: Dec 22nd, 2005, 9:55pm » |
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on Dec 21st, 2005, 3:43pm, Grimbal wrote:29 and no crossings |
| Very nice!
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Wonderer
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Re: Stabilise the square!
« Reply #42 on: Dec 22nd, 2005, 9:55pm » |
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on Dec 22nd, 2005, 5:55am, Barukh wrote: Really?! how?
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Barukh
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Re: Stabilise the square!
« Reply #43 on: Dec 23rd, 2005, 12:36am » |
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on Dec 22nd, 2005, 9:55pm, Wonderer wrote:Really?! how? |
| See the following page and a link given in the references which has interesting results on higher polygons. The solution shown there passes "Grimbal's test" of rigidity, although frankly, I don't understand it fully.
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Wonderer
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Re: Stabilise the square!
« Reply #44 on: Dec 23rd, 2005, 3:07am » |
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on Dec 23rd, 2005, 12:36am, Barukh wrote: See the following page and a link given in the references which has interesting results on higher polygons. The solution shown there passes "Grimbal's test" of rigidity, although frankly, I don't understand it fully. |
| unbelievable...
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SMQ
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Re: Stabilise the square!
« Reply #45 on: Dec 23rd, 2005, 6:07am » |
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on Dec 23rd, 2005, 12:36am, Barukh wrote:although frankly, I don't understand it fully. |
| The solution, or Grimbal's test? Grimbal's test basically says a triangle formed of rigid elements is itself a rigid -- no mystery there, I hope. For the non-crossing solution, the bottom section extends two sides of the square to form two horizontal phantom lines (dashed) of length sqrt(2), then the top section used two sqrt(1),sqrt(2),sqrt(3) right triangles to stabilize the figure. Because of the symmetry the vertical center link must be vertical, and because it's of length 1 the square must be a perfect square. The crossing solution uses the fact that a right triangle with legs 1 + sqrt(2) and 1 - sqrt(2) has a hypotenuse of sqrt(3), and uses two crossed "diamonds" to stabilize the figure. Hope that helps. --SMQ
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--SMQ
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Sjoerd Job Postmus
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I found yet another solution, one that assumes that sticks MAY NOT cross, ever... Not even while de-stabalizing. The bottom line is stable, duh. Look at the segment left to the square, it can only fall over to the right. But, because the link, the other segment must too, but it can only fall over to the left. So, you can't move either of them in a direction. So, it's stable.
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