[seqfan] Re: Brick sequences
franktaw at netscape.net
franktaw at netscape.net
Mon Oct 26 10:05:25 CET 2009
I'm not sure how to interpret your diagrams. The bricks on each row
are supposed to be offset by 1/2 brick from those on the next row. You
appear to be placing bricks directly on top of each other.
I'm going to assume that you mean the following:
..|=|
.|=|=|
|=|=|
.|=|
...|=|
..|=|=|
.|=|=|
|=|=|
.|=|
I do not believe that either of these is stable. In the first one, the
center of mass for the right brick on the row next to the bottom is
directly over the edge of the bottom brick. A small perturbation to
the right will send it toppling.
Franklin T. Adams-Watters
-----Original Message-----
From: Dmitry Kamenetsky <dmitry.kamenetsky at rsise.anu.edu.au>
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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