In what position does a floating wooden cube come to rest: flat on a face, on an edge, or with a corner up?
physics/floating.cube New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
In what position does a floating wooden cube come to rest: flat on a face, on an edge, or with a corner up? [ Editor: problem statement reconstructed from the posting. ]
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In article <Cnuxvn.Dzx@cnsnews.colorado.edu>, I
<thompson@spot.Colorado.EDU> wrote:
>
>Intriguing... All right, a floating object diplaces it's weight in
>water. So if a cube displaces a fixed amount of water, in what
>position does it float so that its center of gravity rides the lowest?
>
>By my calculation it depends on whether the density of the wood is
>greater or less than half that of water. If the density is less than
>one half, the block will float on its side. If it's more, the block
>will float with a corner up. If it's exactly half, the block will not
>show any preference.
This is incorrect. Isaac Kuo pointed out to me that it doesn't work in
practice, and I quickly realised my mistake. David Karr's followup is
correct. The center of gravity should not be as _low_ as possible, but
rather as _close_to_the_surface_as_possible_, loosely speaking. So the
block will float flat in general.
>This might explain why objects that float, then sink, tend to "tip up"
>before going under. (It wouldn't be fair to use sinking ships as an
>example, since that often involves some internal compartments filling
>with water before others).
I didn't know how right I was.
-- Benjamin Rhinelander Thompson <thompson@spot.Colorado.EDU>
[ Editor: David Karr's followup was not among the collected postings. ]
[ Editor: from a 1994 thread, where the puzzle was posted by Mark
Kantrowitz <mkant+@cs.cmu.edu> as "I throw a wooden cube into a tub of
water. Does the cube float corner up or side up, and why?" ]
First off, I'll only talk about what happens to a very long square rod.
That way, only perturbations along one rotational axis need be
considered. I'm not positive how to generalize to the cubical case.
If the rod is exactly 1/2 the density of water, then it will float edge
up. If one perturbs the square clockwise from a corner up position,
then one sees that the center of buoyancy moves right of the cog. It
is stable when it is edge up. If one perturbs the square clockwise
from a side up position, the center of buoyancy moves left. It is
unstable when it is face up.
I suspect that in the cubical case, the cube will still float edge up
(as opposed to corner up).
If the rod is very much less dense than water, then it will float side
up. It is obviously stable when side up (to a naval engineer). When
corner up, it is unstable.
This pretty straight-forwardly generalizes to floating side up in the
cubical case.
If the rod is almost as dense as water, then it will float side up. I
found this one hard to figure out, but the side up configuration is
definitely stable while the corner up configuration isn't.
This too should generalize to floating side up in the cubical case.
-- Isaac Kuo <isaackuo@lhasa.berkeley.edu>
David Karr pointed out to me that surface tension is a factor. But we
can ignore it if the cube is large.
I find that the critical densities are at 1/2 +- 1/2sqrt(3) . My guess
is that between these the cube will always come to equilibrium with one
edge along the surface of the water, but I don't know how to prove it
(yet).
-- Dave Ring <dwr2560@tamuts.tamu.edu>
I haven't solved to floating cube problem, but I think I've solved the
analogous problem in two dimensions. Here we are given a horizontal
line, "the waterline" and must position a unit square ABCD, such that
a. the "submerged part of the square" (i.e., the region inside
the square and below the waterline) has area equal to a given
value d (the "density", 0 < d < 1), and
b. the "energy" of the configuration is minimized, where
the energy is the vertical distance between the square's
"center of gravity" [which is identical to its geometrical
center] and its center of buoyancy [which is the center of
gravity of the submerged portion].
Let us imagine the density of the square gradually increasing from 0.
According to my calculations (which I haven't been able to simplify
enough to make them worth posting):
1. When the density of the square is greater than zero but less
than
1/2 - sqrt(3)/6 [ = 1/2 - 1/2sqrt(3) = 0.2113...]
the minimuum energy is achieved with the square floating so
that one pair of sides is vertical and the other is
horizontal. So far, this agrees with David Ring's result.
B+-----------+A
| |
| |
| |
| |
--------| - - - - - |-------waterline
C+-----------+D
2. As the density exceeds 1/2 - sqrt(3)/6, the orientation above
ceases to be a local energy minimum and becomes a local
maximum. The true energy minimum is now achieved with the cube
tilting slightly to one side or the other, but still with two
corners above the waterline.
3. When the density is exactly 1/4, the minimum energy is achieved
with one corner, say C, on the waterline, and one side, say AD,
half submerged, as shown in the fairly crufty graphic below.
B+-__
/ --__
| --__
/ -+A
| /
/ |
----C+-__- - - - /----------
--__ |
--__ /
-+D
4. As the density increases past 1/4, the square tilts further
and C rises above the waterline, but does not reach the same
height as A until ...
5. When the density reaches 9/32 [= 0.28125], the orientation in
which the sides are all at 45 degree angles to the waterline,
(formerly a metastable local energy maximum) become the energy
minimum.
B+
/ \
/ \
/ \
/ \
C+ +A
--------\ - - - - - - - /--------
\ /
\ /
\ /
+D
This is the minimum energy orientation for all densities from
9/32 through 1 - 9/32 = 23/32, including the density 1/2, for
which minimum energy is achieved with two opposited corners,
say A and C, on the waterline.
6. For densities in the range from 23/32 to 3/4 (= 24/32), the
square begins to tilt away from the 45 degree orientation
Three corners, say A, D, and C, remain below the waterline, but
A and C are no longer at the same height. This case is
symmetrical to case 4.
7. At density approaches 3/4, one of the submerged corners, say A,
rises back to the waterline. At d = 3/4, we have the case
symmetrical to case 3. Two sides are fully submerged, one is
fully exposed, and one, say BC is exactly half submerged.
B+-__
/ --__
| --__
--------/ - - - - - - -+A----------
| /
/ |
C+-_ /
--__ |
--__ /
-+D
8. As the density increases from 3/4 towards 1/2 + sqrt(3)/6, the
exposed side becomes more nearly horizontal (symmetrical to
case 2).
9. For densities at least 1/2 + sqrt(3)/6 but less than 1, the
square floats with its sides vertical and horizontal, as in
case 1.
B+-----------+A
--------| - - - - - |-------waterline
| |
| |
| |
| |
C+-----------+D
The floating cube is more complicated to analyze than the floating
square, since more possible orientations must be considered. For a
unit cube with density 0.5, I get the following values for the
difference, delta, in height between the center of gravity and the
center of buoyancy for the three obvious orientations:
Face-up (edges vertical and horizontal) =>
delta = 1/4 = 0.25
Edge-up (edges horizontal and 45 degrees to horizontal) =>
delta = sqrt(2)/6 = 0.2357...
Corner-up (long diagonal vertical) =>
delta = 13*sqrt(3)/96 = 0.2345...
The corner-up orientation is slightly better than the other two (if I
calculated right). I suspect it is the true minimum energy
orientation, but haven't checked this. I also suspect, but haven't
checked, that face-up is optimal for densities sufficiently near zero
or one. I won't hazard a guess about the sequence of optimal
orientations for intermediate densities.
-- Jim Saxe <saxe@src.dec.com>