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        riddles >> easy >> Convex Polygon Stability
        
(Message started by: william wu on Oct 17th, 2003, 4:35am)
Title: Convex Polygon Stability
Post by william wu on Oct 17th, 2003, 4:35am
Denote the edge of a polygon as being stable if the polygon does not tip over when placed on that edge. Does every convex polygon have a stable edge? Justify your answer.
Title: Re: Convex Polygon Stability
Post by Sir Col on Oct 17th, 2003, 5:14am
Nice problem.

I've had a quick think about it, and although I've not solved it, this is what I've come up with so far...
::[hide]
The CoG is inside all convex polygons.
If the CoG is above an edge, it will be stable; more formally: if a line that is perpendicular to an edge passes through the CoG and the edge, it will be stable.
An alternative way of looking at this is by joining the CoG to each edge, to form triangles. We can see that the apex of an unstable triangle is not above its base.
[/hide]::
Title: Re: Convex Polygon Stability
Post by James Fingas on Oct 17th, 2003, 7:19am
Building on Sir Col's answer,[hide] for each edge, we can draw a rectangle in which the CoG could be for that edge to be stable.

The convex polygon will be stable if the CoG is inside at least one such rectangle. I think that the entire interior of the polygon is covered by these rectangles.[/hide]

The simplest proof of this, however, uses the laws of thermodynamics ;D
Title: Re: Convex Polygon Stability
Post by towr on Oct 17th, 2003, 8:10am

on 10/17/03 at 07:19:41, James Fingas wrote:
The simplest proof of this, however, uses the laws of thermodynamics ;D
That's about what I would suggest, ::[hide]if there isn't a stable edge it keeps on roling.[/hide]::
Title: Re: Convex Polygon Stability
Post by william wu on Oct 27th, 2003, 8:45am
Yes, arguing the impossibility of [hide]perpetual motion[/hide] is the slickest way to do it.
Title: Re: Convex Polygon Stability
Post by Sir Col on Oct 27th, 2003, 10:35am
What do you mean impossible? Because of this puzzle, I'm so close to designing the perfect polygon that never stops rolling. Dang! I'll have to make my millions as a school teacher after all.  ::)
Title: Re: Convex Polygon Stability
Post by william wu on Jan 25th, 2004, 9:51pm
I was talking about this problem with my friend David, and he didn't like the perpetual motion answer. After some consideration I don't like it either. It seems we are confusing the world of mathematics with the real world. The 2nd law of thermodynamics tells us that perpetual motion is not possible in the real world, but in the shiny world of problem sets and frictionless pendulums and rotating polygons, why not? I think we must use Sir Col's proof instead. What do you guys think?
Title: Re: Convex Polygon Stability
Post by John_Gaughan on Jan 25th, 2004, 10:39pm

on 01/25/04 at 21:51:42, william wu wrote:
The 2nd law of thermodynamics tells us that perpetual motion is not possible in the real world, but in the shiny world of problem sets and frictionless pendulums and rotating polygons, why not? I think we must use Sir Col's proof instead. What do you guys think?

I agree. Sir Col had a good answer, but I don't think it was exactly a proof. I'm interested in seeing how to prove it since I suck at geometric proofs and want to learn more.

The concept is trivial, just by visualizing a convex polygon I know instantly that his reasoning is correct, I just don't know how to prove it.
Title: Re: Convex Polygon Stability
Post by towr on Jan 26th, 2004, 1:36am

on 01/25/04 at 21:51:42, william wu wrote:
The 2nd law of thermodynamics tells us that perpetual motion is not possible in the real world, but in the shiny world of problem sets and frictionless pendulums and rotating polygons, why not?
Because the polygon starts from rest. Any polygon would keep roling if it had sufficient starting energy and there was no friction. Moreso, if it was on any side on which it would start tipping over it would perpetually move in some way (as no energy is drained from the system).
You don't have to get the real world into this, physics is quite sufficient, and I'm sure you know that's hardly ever about the real world :P Aside from that 'tipping over' is a real-world concept..

What is necessary for an object to tip over? (supposing no other effects like wind, brownian motion, molecular attraction to the surface etc)
The net potential energy must decrease. It should be obvious that at some point the potential energy cannot decrease further. For every edge the polygon is resting on there is some finite amount of potential energy in the polygon. One, or multiple, of these must have the minimum amount (which means tipping over can only increase it, which makes it impossible).
The only thing the rolling (with friction) argument does, is weaken the constraints on the process. From testing each edge with no kinetic energy at the start; basicly draining all energy immediately once it has tipped over. It is weakened to testing each edge with some kinetic energy but draining the energy from the system more slowly. The end result is the same, the polygon stops rolling once the potential energy increases more than the current level of kinetic energy.
Title: Re: Convex Polygon Stability
Post by towr on Jan 26th, 2004, 2:15am

on 01/26/04 at 01:36:03, towr wrote:
What is necessary for an object to tip over? (supposing no other effects like wind, brownian motion, molecular attraction to the surface etc)
The net potential energy must decrease.
I'm not forgetting there is another necessary condition for tipping over from rest btw, it's just irrelevant. Either one alone is sufficient grounds for claiming there is at least one edge for which the polygon won't tip over.
The other condition, as Sir Col pointed out, is that the center of gravity mustn't lie over the edge it is resting on. Which brings me to this question.

Can you proof there are at least two edges for which a convex polygon won't tip over (from rest)?
And if so, how about three, or more? (If you can proof any kind of relation between the number of non-tip edges and N-gons that'd be even better)
Title: Re: Convex Polygon Stability
Post by rmsgrey on Jan 26th, 2004, 5:04am
The "must have a stable edge or will keep rolling" argument seems to be assuming that all unstable edges are unstable in the same direction - after all, a pendulum starts at rest, and, in the absence of damping efects, will swing indefinitely...
It's possible for two adajcent edges to both be unstable away from each other, but it turns out to be impossible for them to be unstable towards each other (the center of gravity would have to be to the left of the perpendicular to the right hand edge at the common vertex, and to the right of the left edge's but since the angle between the edges is less than 180o the two perpendiculars are on the worng sides of each other). Obviously, then, if a convex polygon at least one edge which is unstable in each direction (clockwise and anticlockwise), it must also have at least one stable edge somewhere to separate them where they meet towards each other. So an unstable polygon must be unstable in the same direction all the way round. But toppling from an edge in the direction in which it's unstable lowers the center of gravity, so after one full rotation, the center of gravity must be lower than it was originally - contradiction.

The existence of triangles with one unstable edge means that, in the general case, convex polygons can't be required to have more than two stable edges. In fact, it's generally possible to replace an unstable edge of a polygon with two unstable edges by adding a new vertex sufficiently close to the midpoint of the original edge (both the movement of the center of gravity due to the added area and the change in the critical regions due to the change in angle can be made aritrarily small) so it is always possible to construct an N-gon with N-2 unstable edges. The only question remaining is whether you can manage N-1.
Title: Re: Convex Polygon Stability
Post by SWF on Jan 26th, 2004, 7:02pm

on 01/26/04 at 02:15:19, towr wrote:
Can you proof there are at least two edges for which a convex polygon won't tip over (from rest)?

No, but it is not hard to find one that is stable on only one side.
Title: Re: Convex Polygon Stability
Post by towr on Jan 27th, 2004, 12:06am

on 01/26/04 at 19:02:55, SWF wrote:
No, but it is not hard to find one that is stable on only one side.
Could you perhaps draw one, and show us?
Title: Re: Convex Polygon Stability
Post by SWF on Jan 27th, 2004, 6:41pm
This polygon is stable only on the red side.  Weight is not distributed uniformly, but is instead heavily biased toward the red side.  The yellow dot is the center of gravity, and is, of course, inside the polygon. In each of the orientations shown the center of gravity is not directly above the base, so it will tip to an adjacent side.
Title: Re: Convex Polygon Stability
Post by towr on Jan 28th, 2004, 12:47am
Well yes, if you go distribute weight however you want sure.. But I would think it natural to assume it's uniformly distributed. It's not like a polygon has any depth to begin with, so no 'real' mass either..

But ok.. try to disprove any convex polygon with mass distributed uniformly has two stable edges..


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