Inflation

[Fijoy George]

Hi,

I am interested in solving the following problem:

I have a closed 3D surface represented by a set of points (nodes) and
connections (edges) between them. I want to find a transformation of this
surface which physically corresponds to the inflation of an inelastic closed
surface. That is, the transformation should maximize the volume enclosed by
the surface, but simultaneously should retain the geodesic distances between
the nodes.

Can this be solved (analytically/using a computer)? Can there be an
alternative mathematical formulation of the problem which may be easier to
solve?

Thanks
Fijoy

[Zdenek Sperling]

"Fijoy George" <tof...@yahoo.co.in> wrote in message
news:crp6ce$ro7$1...@mailhub227.itcs.purdue.edu...

I think that your problem is not solvable.

Just intuitively, because all points of the grid should have stiff
connectors, the grid has to be composed of triangles. Then the surface has
to be constant. The largest volume for a given surface would have a sphere.
However, a network of triangles on a spherical surface has no degree of
freedom, that is the principle of those Fuller domes, therefore a change of
surface shape is impossible. And this must be valid in reverse, too.

I am not sure, if my argument is correct, I am no topologist, not even a
mathematician.

[fijoy]

Perhaps I should illustrate my question with a simple example.

Suppose you take one of those polyethylene covers that you get when you buy
stuff from grocery store. Then (using your mouth) you blow air into it
through its mouth just like you inflate a balloon. The cover will eventually
assume a shape, and it is not spherical (it may have more length than
breadth for example). The volume enclosed will be maximum since you have
blown as much air into it as possible. If you assume that the cover is thick
enough, it will not stretch either, and the geodesic distances will not
alter. The final shape of the cover will be unique no matter how crumbled
the cover was initially.

The probem is to do this inflation mathematically, or using a computer,
given an initial description of the surface.

-Fijoy

"Zdenek Sperling" <sper...@shaw.ca> wrote in message
news:jcXDd.18628$8l.18068@pd7tw1no...

[Hiu Chung Law]

> Thanks
> Fijoy

You may find the following paper related.

Kilian Q. Weinberger. Lawrence K. Saul.
Unsupervised learning of image manifolds by semidefinite programming.
Proceedings of the 2004 IEEE Computer Society Conference on Computer Vision and Pattern Recognition (CVPR 2004),
27 June - 2 July 2004, Volume 2, pages 988-995.

Instead of inflating, however, what the authors did is to ``unfold''
a manifold to be as flat as possible. During unfolding, distances between
points that are "neighbors" are preserved.

This is of course very different from the problem of inflating, but
this "unfolding" operation should keep the "empirical" geodesic distances
(defined by the neighborhood graph) constant.

[Stephen M. Fortescue]

The easiest case to solve is a dihedron made from two circular disks.
When inflated, the circular edge shrinks by becoming crumpled, since
what were diameters of each disk must curve when inflated to contain
the volume of air while retaining their geodesic length. This makes
the idealized shape of a typical mylar balloon.

Let h(s) and r(s) be coordinates along a curved diameter of a disk
as a funtion af arc length. The problem is to maximize volume:

V = Integral Pi r^2 h_s ds, where (r_s)^2 + (h_s)^2 = 1.
V = Pi Integral F(s, r, r') ds,
where F(s, r, r') = r^2 sqrt(1 - (r')^2).

F(s, r, r') must satisfy the Euler differential equation:

F_r - F_sr' - F_rr' r' - F_r'r' r'' = 0.

F_r = 2r (1 - (r')^2)^(1/2)
F_rr' = -2r r' (1 - (r')^2)^(-1/2)
F_r'r' = -r^2 (1/(1 - (r')^2)) (1 - (r')^2)^(-1/2)

2r (1 - (r')^2) + 2r (r')^2 + (r^2 r'')/(1 - (r')^2) = 0
2(1 - (r')^2) + r r'' = 0

g(r) = r' and g dg/dr = r''

2(1 - g^2) + r g dg/dr = 0
4 dr/r = 2g dg / (g^2 - 1)
4 ln r = ln(g^2 - 1) + C
1 - (r/k)^4 = g^2

r' = k x' = sqrt(1 - x^4) where x = r/k
h' = x^2

ds/k = dx / sqrt(1 - x^4)
(ds + dh)/k = dx (1 + x^2) / sqrt(1 - x^4)

Integrate using Abramowitz & Stegun <17.2.7> and <17.2.8> :

sn( s/k | -1) = x = r/k from <16.10.2> :
sqrt(2) r/k = sd( u | .5) where u = sqrt(2) s/k

E( s/k | -1) = (s + h)/k from <17.4.18> :
sqrt(2) h/k = 2 E( u | .5) - u - sqrt(2) sn( u | .5) cd( u | .5)

u=K(.5) makes sd(1.854074677301372 | .5) = sqrt(2), which is maximal.

sqrt(2) h/k = 2 (1.350643881047676) - 1.854074677301372

h/k = 0.59907011736780 = h/r since r/k = 1.

The number agrees with this article which uses a different method:

http://www.sm.luth.se/~norbert/home_journal/electronic/11s_art7.pdf

The next case to try to solve is a square dihedron so as to find the
idealized shape of ravioli.