What is the longest (or largest) object that can be moved along a corridor of width W and around a...

In article <3511F3DC.8A2882@adinfo.com>, Paul Bishop <paulb@adinfo.com>
wrote:
>If you dealt with length of an object, the "biggest" object that you
>can manipulate down the corridor and around the corner is the
>following:
>
>A flat piece of material with height H, negligible width W, and a
>maximum length of L = 2*W*(square root of 2).
>
>The volume then, if one dimension is negligible would be approaching
>zero, depending on how thin the width is.
>
>How did I figure this out?
>L is the length of the base of the biggest right angle isoceles
>triangle formed in the corridor with the apex being the right angle
>corner, and the sides the walls.  The other corner happens to be the
>midpoint of the base.
>
>ASCII DIAGRAM:
>_____
>\   |
>_\  |
> |\ |
> | \|

If you allow for your object to be non-convex, you can fit a "longer"
object around the corner.  Imagine an object similar to the one
described above, but with a single bend in it along a vertical axis.
Obviously, in order to fit down the straight sections, the length of
the object is limited based on the angle of the bend, such that its
effective width is no more than W (as shown below):

ASCII diagram of object only:

`.          .' ^
  `.      .'   |W
    `.  .'     |
      `'       v

When this object gets down to the corner, you can start turning it so
that one end starts going around the corner (obviously, you want the
bend on the outside wall).  When you're exactly halfway around, your
object will look something like this in the hallway:

+-------
|   .'
| ,'
| |+----
|| |
|| |
|  |

I think you can maximize the length of this object by choosing the
maximum angle so that the object (with dimension W as in my first
diagram) still fits around the corner this way, and I think that
happens for the object with ends A,C and corner B as shown in the
diagram below:

+--c------C
|    /   /
|   /   /
a  B---/---
| /|  /
|/ | b
x  |/
|  y
| /|
|/ |
A  |

when the perpendicular distance (Bb) between the two diagonal lines is
W.  In this case, by is also W and By = Ax is W*sqrt(2).  aB = ax = W,
so Aa is W*(1+sqrt(2)).  AB is thus W*sqrt(1+1+2*sqrt(2)+2) =
W*sqrt(3+2*sqrt(2)).  Ay = Bx = W*sqrt(2) so AC = 2*Ay + 2*by =
W*(2+2*sqrt(2)), longer than your object by 2W.

[ Editor: 1+1+2*sqrt(2)+2 = 4+2*sqrt(2), not 3+2*sqrt(2). ]

Of course, if we're just going for maximum length, we should also make
the object be of negligible height, i.e., a long skinny rod.  By doing
so, and moving the object along the path a "diagonal" (that is, what
would be a diagonal if it was unfolded) moves, you can move an object
of length sqrt(H^2+AC^2) where AC is the answer at the end of the
previous paragraph.

More variations:
Suppose you have more than one corner to take the object around, and
they don't all turn in the same direction; assume the distance between
corners is much longer than the width, and longer than any object you
want to move, so you will be completely finished turning one corner
before you start turning the next one.  If we are going for maximum
length, my bent rod answer still works, since you can turn the object
around in the straight sections.  What impact does this have on the
maximum volume problem?

-- Joseph W. DeVincentis <devjoe@wilma.che.utexas.edu> (March 1998)