[seqfan] Re: Brick sequences

[Dmitry Kamenetsky]

Hi Franklin,

Would you consider the following configurations stable?

..=
.==
==
.=

...=
..==
.==
==
.=

I think the first one might be stable, but not the second one. This shows
that you cannot define stability of a configuration based on the properties
of its individual bricks - you must consider the whole structure. We
probably need to find the center of gravity of the whole structure and see
whether it is near/away from the center of the structure. Let me think some
more about this.

Cheers,
Dmitry
 
----------------original message-----------------
From: franktaw at netscape.net
To: seqfan at seqfan.eu
Date: Sat, 24 Oct 2009 07:44:25 -0400
-------------------------------------------------
 
 
> I came up with an idea for a sequence. Actually, it's several 
> sequences, some of which I'm not sure how to define mathematically.
> 
> Consider a connected structure of bricks, idealized as congruent 
> rectangles. The bricks on the bottom are at integer offsets; those on 
> the next row up are at half-integer offsets, the next row is again at 
> integer offsets, etc. The structure has to stand up.
> 
> For the simplest case, we require that each brick be supported by two 
> bricks under it. If we regard mirror images as the same, this sequence 
> starts, from n=1 (hand-calculated):
> 
> 1,1,2,2,4,6,10
> 
> (I suppose a(0) = 1, but I'm not including it here.)
> 
> For example, the two structures for n=4 are:
> 
> |=|=|=|=|
> 
> and
> 
> .|=|
> |=|=|=|
> 
> (Imagine the lines on the = are farther apart than most fonts will make 
> them.)
> 
> If mirror images are different, the sequence starts:
> 
> 1,1,2,3,5,9,15
> 
> I'm pretty sure that neither of these is in the OEIS: there are 
> sequences matching either, but none with even a vaguely related 
> description.
> 
> Note that, for these two sequences, we could just as well use hexagons 
> as bricks. This won't work for the next one.
> 
> -----
> But now things get more interesting. The structure:
> 
> .|=|
> |=|
> 
> is not stable; the center of gravity of the top brick is right over the 
> edge of the brick below it. A small perturbation will send it 
> tumbling. However,
> 
> .|=|
> |=|=|
> .|=|
> 
> is stable; the weight of the top brick holds the center of gravity for 
> the weight supported by the bricks under it to be over the base brick.
> 
> On the other hand, consider:
> 
> .|=|=|
> |=|.|=|
> 
> Now the two top bricks are braced against each other. I think that, by 
> the definition I want to use, this is actually unstable; perturb the 
> top two bricks a bit to the right, and the left one will start to tilt, 
> pushing the right one further right until the left brick falls.
> 
> At this point I'm not sure how to proceed. How do you determine 
> whether a given structure is stable? If you measure the center gravity 
> of the weight supported by a brick, it must be over another brick; this 
> is certainly a necessary and sufficient condition. But how do you 
> distribute the weight for a brick that has two bricks under it? My 
> second thought (the first thought was clearly wrong) is that the weight 
> should be distributed according to the position of the center of 
> gravity of the weight supported by the brick; if this is 1/4 of the way 
> across the brick, distribute the left-most 3/4 of the weight to the 
> left and the remaining 1/4 to right. I'm pretty dubious that this is 
> correct in general, however.
> 
> Once this question is settled, there will again be two sequences: one 
> for mirror images considered the same, and the other considering them 
> different.
> 
> Franklin T. Adams-Watters
> 
> 
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