From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.num-analysis
Subject: Re: How many Surface Normals ?
Date: 5 Sep 1996 22:17:47 GMT

In article <322AACA9.41C6@icrf.icnet.uk>, islam  <islam@icrf.icnet.uk> wrote:
>Given an implicit Surface F and its normal N
>
>	F(x,y,z) = 0
>	N = i(dF/dx) + j(dF/dy) + k(dF/dz)
>
>By examining F and/or N is there a straight forward/standard
>method to find the maximum number of Normals that may pass
>through a general point ? For examples for a plane the answer
>is 1, for a circle 2, for a conic 4 and for a Quadric up to
>6.

If I understand you correctly, you are given the above surface and a
point P0=(x0,y0,z0), and you want to know how many  points there are
P1=(x1, y1, z1) which lie on the surface, such that the normal to the
surface at  P1  passes through  P0.

This normal is the set of points
	P1 + t*(F_x, F_y, F_z)
where  t  ranges over the real numbers and the partials are evaluated
at  P1. (I'm assuming the surface is nonsingular so that this
gradient is a nonzero vector and the normal line is well-defined.)
Thus your question asks for the number of simultaneous solutions to the
equations
	x0 = x1 + t F_x( x1, y1, z1)
	y0 = y1 + t F_y( x1, y1, z1)
	z0 = z1 + t F_z( x1, y1, z1)
	F(x1, y1, z1) = 0
where the variables are x1, y1, z1, and t. You should expect in general
that four equations in four unknowns will have a finite number of
solutions, although of course there will be degenerate cases.

If F is a polynomial, you have your choice of several approaches.
First of all, the equations may be thought of as describing the
intersection of four (3-dimensional) hypersurfaces in 4-space, which
makes this question the province of algebraic geometry. Indeed, one
can easily eliminate t from these equations and get 3 polynomial
equations to solve, i.e., 3 surfaces in R^3 to intersect (the last of
them being the original surface, the other two dependent on P0). You
might enjoy using computer graphics to look for the solutions in
specific cases.  There is a very general theorem (Bezout's theorem)
which counts the number of points in the intersection of the surfaces.
Subject to certain restrictions, it suggests the number of points of
intersection should never be more than  (deg F)^3  (I think).

You can also compute the points of intersection explicitly. Just as
t  is easily eliminated, one may eliminate  x1  from the three
equations, leaving two equations in  y1  and  z1 (that is, you need
to intersect two curves in the plane). You can use the theory of
resultants for this, as well as to eliminate  y1  from the remaining two
equations to get a polynomial (of high degree!) in  z1  to be
solved. Alternatively, one may use Groebner basis techniques to
find three equations equivalent to the given three but using
respectively 1, 2, and all 3 variables; solving these in turn allows
one to compute the unknowns  x1, y1, and z1. Most computer algebra
packages will allow you to perform these computations symbolically,
so you can investigate the dependence on  P0. 

(If you have a specific  P0  in mind, it may well be quicker to
calculate numerically the points  P1  by using Newton's method, say,
since elimination techniques are very time-consuming.)

I will confess that none of the comments in the last three paragraphs
makes use of the specific nature of the equations to be solved. It
may well be that there is a built-in degeneracy or something coming
from  the construction of the equations from the obviously dependent
polynomials  F, F_x, F_y, and F_z. 

dave