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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

2037.0. "The Iron Shroud" by TPOVC::BUCHANAN (A small bird called Offlogic) Wed Apr 10 1996 03:01

    	A horror story written in the Eighteenth Century describes the fate
    of a prisoner stuck in a shrinking cell. For simplicity, let's say the
    cell is cuboidal. Each night, the height and breadth of the cell become
    smaller (although the length remains the same).
    
    	It's easy enough to imagine a mechanism for that shrinking in 2
    dimensions. The question is: can one construct a shrinking cell, in
    which the length, breadth and height can all reduce to zero, by
    suitable shrinking of the walls. Or must the walls block one another?
    
    An De Ru
T.RTitleUserPersonal
Name
DateLines
2037.1Works for any prism, I thinkWIBBIN::NOYCEEV5 issues 4 instructions per meterWed Apr 10 1996 09:4239
Here's the mechanism for shrinking in 2 dimensions:
                        _
                        |
                        |
       <----------------|
             |     |    |
             |     V    |
             |->      <-|
             |     _    |
             |     |    |
             |---------------->
             |
             |
             V

Add the third dimension, so each line becomes a wall.  We want to add front and
back walls that can shrink too.  Nestle the back wall into the corner formed
by the top and right-hand walls above, so those walls can extend behind the
back wall.  The bottom and left-hand walls will move across the face of the
back wall.  Similarly, nestle the front wall into the opposite corner, formed
by the bottom and left-hand walls above, so that those walls (which touch
the face of the back wall) can extend beyond the front wall.


                        _
                        |
              FFFFFFFFFF|FFF
              F         |  F
       <----------------|  F
          BBB|FBBBB|BBBB|  F
          B  |F    V   B|  F
          B  |->      <-|  F
          B  |F    _   B|  F
          B  |FFFFF|FFFF|FFF
          B  |---------------->
          B  |         B
          BBB|BBBBBBBBBB
             |
             V
2037.2exitTPOVC::BUCHANANA small bird called OfflogicWed Apr 10 1996 10:3011
    	Ah, but do your front and back walls approach one another?
    
    	I should mention that when the hero of the tale begins to get
    squashed, he finds a message written low down on one wall by the
    designer of the cell, who was incarcerated in the cell himself as soon
    as he had completed the construction. The designer was not able to find
    a way out, nor does he reveal in his anguished little paragraph how
    he solved the design problem.
    
Cheers,
    Andrew
2037.3HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Wed Apr 10 1996 11:2412
I seem to recall a movie hero (batman ?  James Bond ?) saving himself and
the beautiful girl, when they were about to be crushed by some sort of
scrap metal compactor.

I won't tell you how they did it, in case you see the movie.

Only kidding, I'll tell you right now.  The hero pulls out his combination
pen-radio and merely holds it horizontally, and it stops the walls !

/Eric

2037.4Did you have a different solution in mind?WIBBIN::NOYCEEV5 issues 4 instructions per meterWed Apr 10 1996 13:029
 >   	Ah, but do your front and back walls approach one another?

Yes, why can't they?  The front wall pushes only on two other walls that can
slide past the back wall, and vice versa.

Or were you asking about the mechanism outside the walls to make them move
in the proper paths?  When the size is zero, each wall has one corner touching
the center.  The wall needs to move in a straight line toward that position,
with speeds such that they all reach it at the same time.
2037.5AUSS::GARSONachtentachtig kacheltjesWed Apr 10 1996 20:039
    re .0
    
    Cheat solution: How about pairs of parallel planks where adjacent pairs are
    perpendicular as the 2D solution - this is like the construction that kids
    make from icecream sticks - and then the 3D solution is obtained merely by
    removing pairs of planks. Obviously this allows only discrete rather than
    continuous shrinkage in the third dimension but the steps can be made
    arbitrarily small. In any case it is certainly adequate to extinguish the
    prisoner.
2037.6exitTPOVC::BUCHANANA small bird called OfflogicThu Apr 11 1996 06:1222
    	Let the cell be cubic, and have vertices at the eight points given
    by ( {0,1}, {0,1}, {0,1} ). The cell floor is bounded by ( {0,1}, {0,1}
    ). We will shrink the cell to the origin:
    
    floor:	stationary
    N wall:	(-1,-1,0)	diagonal slide SW
    E wall:	(-1,0,0)	slide W
    S wall:	(0,0,-1)	sinks downwards into floor
    W wall:	(0,-1,-1)	sinks downwards into floor, while slide S
    ceiling:	(-1,-1,-1)	sinks downwards to floor, while slide SW
    
    One problem with this is that the roof may fall off part way through.
    One way to fix this is to have another wall like the S wall, but
    located (0,-1,0) (ie. to the W). This takes the weight of the roof as
    it slides SW. The W wall slides between the two S facing walls.
    
    Thus you end up with a cross-shaped trench, which is angled at the N
    end.
    
    Cheers,
    Andrew