This page is my contribution to the discussion at The Geometry Junkyard: http://www.ics.uci.edu/~eppstein/junkyard/teabag.html on the teabag problem. This is "What is the maximum volume that a teabag can hold?" where of course, the teabag is a unit square, sealed on all four edges, and made of paper that does not stretch or shear.
So, here is where I got to:
|Start 28th of November, 1997|
The discussion at The Geometry Junkyard: http://www.ics.uci.edu/~eppstein/junkyard/teabag.html on the teabag problem caught my interest, so after a bit of scribbling, origami and number-crunching, I managed to improve the solution provided by Dan Hirschberg by 8% or so. For the configuration below, the maximum comes out as 0.190205 ish. It is clear that this can be improved using more pleats, and there is probably a curved surface that can be approximated arbitrarily closely by such a polyhedron with pleats.
The calculations and plots were done with Maple on a Macintosh.
For the impatient
The solution is an improvement on Dan Hirschberg's solution by making some of the vertical sides bow outwards, and by adding small pleats to accommodate the wrinkling that you see in the wine box liners. (BTW: I have been told that these were an Australian innovation. They are locally known as "casks" -- how embarrassing!)
I only considered an octant's-worth of the teabag and doubled its size. Here is the relevant picture, consisting of bits of the above two pictures superimposed:
With this configuration, there are two parameters, for example, the size and height of the top face. The parameters I used were the x-coordinate of one of the top corners and the x-coordinate of the extreme corner, to which I applied standard undergraduate multivariate calculus. The zeros of the partial derivatives were located graphically, as Maple was not up to doing it analytically or numerically. If anyone wants a look at the Maple worksheet I will mail it to them.
top: [[0 , 0 ], [.4672670844, 0 ], [0 , .4672670844]] ramp: [[.4672670844, 0 ], [0 , .4672670844], [1 , 1 ]] sides: [[.4672670844, 0 ], [1 , 1 ], [1 , .0538357131]] [[0 , .4672670844], [1 , 1 ], [.0538357131, 1]] pleat: [[.4672670844, 0 ], [.8032538609, 0 ], [1 , .0538357131]] [[1 , 0 ], [.8032538609, 0 ], [1 , .0538357131]] pleat: [[0 , .4672670844], [0 , .8032538609], [.0538357131, 1 ]] [[0 , 1 ], [0 , .8032538609], [.0538357131, 1 ]]
top: [[0 , 0 , .4859245692], [.4672670844, 0 , .4859245692], [0 , .4672670844, .4859245692]] ramp: [[.4672670844, 0 , .4859245692], [0 , .4672670844, .4859245692], [.9186563185, .9186563185, 0 ]] sides: [[.4672670844, 0 , .4859245692], [.9186563185, .9186563185, 0 ], [.6921669652, 0 , 0 ]] [[0 , .4672670844, .4859245692], [.9186563185, .9186563185, 0 ], [0 , .6921669652, 0 ]] pleat: [[.4672670844, 0 , .4859245692], [.6383312521, 0 , .1967461391], [.6921669652, 0 , 0 ]] [[.6383312521, 0 , 0 ], [.6383312521, 0 , .1967461391], [.6921669652, 0 , 0 ]] pleat: [[0 , .4672670844, .4859245692], [0 , .6383312521, .1967461391], [0 , .6921669652, 0 ]] [[0 , .6383312521, 0 ], [0 , .6383312521, .1967461391], [0 , .6921669652, 0 ]]
top: .05304802763 ramp: .1036926086 sides: .01673246402 .01673246402 Total: .1902055642
Throwing in a few more parameters, a configuration with pleats slightly opened or with couple of extra pleats on each side could be accommodated. Given the shape of an inflated cask liner, each of these will probably give a bit more volume.
In the limit, we can probably get a surface like that sketched on the right. To get this we could specify a space curve and use some geometry to find how the rest of the surface falls, assuming the limiting case of infinitely many infinitesimal pleats. The maths behind this is very similar to what I did above, except my curve wasn't curved. Then using some variational techniques, it should be possible to find equations satisfied by this curve to give maximal volume. [Note: the sketched is probably not foldable, as I just threw in some parabolas as the curves.]
It is too long since I did anything of this type to try to attempt it now. Anyone else is welcome to try!
This will probably not be optimal, though. It should be possible to make the corners of the teabag more conical, giving a completely smooth surface. I think I have a way of increasing the number of sides above 6, by putting pleats in different places, but I have more pressing things to do.
I have been trying to think of constraints on the smooth parameterisation that will give us some PDEs that some software can be thrown at. The non-stretch & non-shear properties of paper, with the possibility of infinitesimal pleats translate (at a guess) into the Jacobean's eigenvectors being orthogonal with eigenvalues 1 & a, where a is less than or equal to 1. [Yes, I don't mean the Jacobean of the parameterisation, it won't have eigenvectors. There needs to be a transformation into the surface's local 2d coordinates thrown in.]
This (once formulated) will give a necessary condition on the parameterisation. My conjecture is that it is sufficient.
|End 28th of November, 1997|
|Start 27th of December, 1997|
I have thought of a way to use curved surfaces without having to use infinitesimal pleats, so the surface can be actually made from paper, not just approximated arbitrarily closely. A teabag of volume 0.2055 or so can be obtained. This will not be the optimum - it is just the optimum for the configuration I considered. This stuff is described in less detail than that above, as I don't have a huge amount of time to polish it.
This consists of two quarter cones in place of the two sloped sides in the configuration considered in the 28th of November stuff above. This image (double size, again) shows what happens to the square as it is draped in this shape. Two small curvilinear triangle bits stick out of the octant. To fit into the bit like this in the next octant, one of these triangles needs to be reflected in the vertical coordinate plane.Even in the degenerate case where there are no triangular facets, the volume comes out as pi * 21/2 * (31/2 - 1)/16, or about 0.203. The configuration has only one parameter. I took this to be the angle at the tip of the cone of the curved surface, before rolling. Thus, the degenerate case has this angle pi/4 radians. The maximum volume is attained with this angle about 0.72400695527 radians. Here are some data of this configuration:
vertex of top triangle: [0.115830154289240,0,0.615237562203822] cone meets x-axis: [0.677545175856641,0,0] tip of cone [0.893537822705192,0.893537822705192,0] tip of "out of quadrant" triangle: [0.658577511221745, -.078467142316903, 0] Volume of triangular prism: .00412720571907975 Volume under ramp: .0396993568802905 Volume under each cone: .0808426326513601 Total volume: .205511827902090
This can be improved with a different (i.e. not circular) profile to the cone, and might get around to squeezing the extra few millilitres into the teabag this way. This makes the maths more difficult -- for circular cones it is easy to relate the "unrolled angle" to the angle with the axis.
I suspect that adding additional curved pleats will also help. Again, I suspect that modelling infinitesimal curved pleats would soak up vast amounts of my time for not much gain.
I have an upper bound of (-4 + 5 * 21/2 + 61/2) * 23/4 / (24*pi1/2) or approximately 0.2182575424. I know how to improve this marginally, but have not yet done the calculation. Details may appear next year. However, I really have better things to do!
|End 27th of December, 1997|
|Start 22nd of April, 1998|
Of course, the simplest upper bound on the volume is that of a sphere of surface area 2. This has volume 21/2 3-1 pi-1/2 or approximately .265961520
A new upper bound of
or approximately 0.217584104961869 can be obtained. This is the volume of the surface formed by joining four conical surfaces of slant height 1/2 and slope 30 degrees onto a sphere of such a radius that the total area is 2. (The area of the spherical surface showing will be 2-Pi/2.) It is not too hard to see why this is an upper bound, if you assume:
The upper bound of 0.21825... above (now redundant) was a similar configuration, but with more of the surface devoted to the sphere.
So the state of the art on the volume of a square teabag is: 0.2115 +/- 3% .
|End 22nd of April, 1998|
|Start 18-20 November, 1998|
I did a quick calculation for a configuration along the lines of Tom Longtin's suggestion based on a McDonald's fries cup. See the Junkyard's Teabag page for his original correspondence.
This is along the lines of some musings in my correspondence with Tom. The configuration is as pictured, where the curved surface sweeps through an angle of 2*theta of a circular arc. Then it can be shown (via a horrible integral which Maple did for me) that the volume is given by
Which achieves a maximum value of approximately 0.1703844172 when theta is approximately 1.047 radians. This is the value used in the image. If the chip cup is to succeed, a non-circular profile is required.
|End 18-20 November, 1998|