EACA2010

Problem
Distance of closest approach of two ellipses
Distance of closest approach of two ellipsoids
Conclusions

Computing the distance of closest approach
between ellipses and ellipsoids

F. Etayo, L. González-Vega, G. R. Quintana

Departamento de MATemáticas, EStadística y COmputación
University of Cantabria, Spain

XII Encuentro de Álgebra Computacional y Aplicaciones,
Santiago de Compostela, 19-21 de julio de 2010
F. Etayo, L. González-Vega, G. R. Quintana EACA2010

Problem
Conclusions

Contents

1 Problem

2 Distance of closest approach of two ellipses

3 Distance of closest approach of two ellipsoids

4 Conclusions


Problem
Conclusions

Introduction

The distance of closest approach of two arbitrary separated
ellipses (resp. ellipsoids) is the distance among their centers
when they are externally tangent, after moving them through
the line joining their centers.

That distance appears when we study the problem of
determining the distance of closest approach of hard particles
which is a key topic in some physical questions like modeling
and simulating systems of anisometric particles, such as liquid
crystals, or in the case of interference analysis of molecules.


Problem
Conclusions

Introduction

Distance of closest approach of two ellipses.


Problem
Conclusions

Previous work

A description of a method for solving the problem in the case of
two arbitrary hard ellipses (resp. ellipsoids) can be found in

X. Z HENG , P. PALFFY-M UHORAY, Distance of closest
approach of two arbitrary hard ellipses in two dimensions,
Phys. Rev., E 75, 061709, 2007.
X. Z HENG , W. I GLESIAS , P. PALFFY-M UHORAY, Distance of
closest approach of two arbitrary hard ellipsoids, Phys.
Rev. E, 79, 057702, 2009.

An analytic expression for that distance is given as a function of
their orientation relative to the line joining their centers.


Problem
Conclusions

Previous work (Zheng,Palffy-Muhoray)
Ellipses case:
1 Two ellipses initially distant are given.
2 One ellipse is translated toward the other along the line
joining their centers until they are externally tangent.
3 PROBLEM: to ﬁnd the distance d between the centers at
that time.
4 Transformation of the two tangent ellipses into a circle and
an ellipse.
5 Determination of the distance d of closest approach of the
circle and the ellipse.
initial ellipses by inverse transformation.


Problem
Conclusions

Previous work (Zheng,Palffy-Muhoray)
Ellipses case:
1 Two ellipses initially distant are given.
2 One ellipse is translated toward the other along the line
joining their centers until they are externally tangent.
3 PROBLEM: to ﬁnd the distance d between the centers at
that time.
4 Transformation of the two tangent ellipses into a circle and
an ellipse. ⇒ Anisotropic scaling
circle and the ellipse.
initial ellipses by inverse transformation.


Problem
Conclusions

Previous work (Zheng,Iglesias,Palffy-Muhoray)

Ellipsoids case:
1 Two ellipsoids initially distant are given.
2 Plane containing the line joining the centers of the two
ellipsoids.
3 Equations of the ellipses formed by the intersection of this
plane and the ellipsoids.
4 Determining the distance of closest approach of the
ellipses
5 Rotating the plane until the distance of closest approach of
the ellipses is a maximum
6 The distance of closest approach of the ellipsoids is this
maximum distance


Problem
Conclusions

Previous work

To deal with anisotropic scaling and the inverse transformation
involves the calculus of the eigenvectors and eigenvalues of the
matrix of the transformation.

Our goal is to ﬁnd when that computation is not required and if
it is, to simplify it. The way in which we do that extends in a
natural way the ellipsoids case.


Problem
Conclusions

Our approach

We use the results shown in:
F. E TAYO, L. G ONZ·LEZ -V EGA , N. DEL RÌ O, A new approach to
characterizing the relative position of two ellipses depending on
one parameter, Computed Aided Geometric Desing 23,
324-350, 2006.
W. WANG , R. K RASAUSKAS, Interference analysis of conics and
quadrics, Contemporary Math. 334, 25-36,2003.
W. WANG , J. WANG , M. S. K IM, An algebraic condition for the
separation of two ellipsoids, Computer Aided Geometric Desing
18, 531-539, 2001.


Problem
Conclusions

Our approach
Following their notation we deﬁne

Deﬁnition
Let A and B be two ellipses (resp. ellipsoids) given by the equations
X T AX = 0 and X T BX = 0 respectively, the degree three (resp.
four) polynomial
f (λ) = det(λA + B)
is called the characteristic polynomial of the pencil λA + B

Two ellipses (or ellipsoids) are separated if and only if their
characteristic polynomial has two distinct positive roots.
The ellipses (or ellipsoids) touch each other externally if and
only if the characteristic equation has a positive double root.


Problem
Conclusions

Our approach

We use the previous characterization in order to obtain the
solution of the problem.

We give a closed formula for the polynomial S(t) (depending
polynomially on the ellipse parameters) whose biggest real root
provides the distance of closest approach:
Ellipses case: d = t0 x 2 + y0
0
2

Ellipsoids case: d = t0 x 2 + y0 + z0
0
2 2


Problem
Conclusions

We consider the two coplanar ellipses given by the equations:

E1 = (x, y) ∈ R2 : a22 x2 + a33 y 2 + 2a23 xy + 2a31 x + 2a32 y + a11 = 0

E2 = (x, y) ∈ R2 : b22 x2 + b33 y 2 + 2b23 xy + 2b31 x + 2b32 y + b11 = 0

We change the reference frame in order to have E1 centered at the
origin and E2 centered at (x0 , y0 ) with axis parallel to the coordinate
ones:
(x cos (α) + y sin (α))2 (x sin (α) − y cos (α))2
E1 = (x, y) ∈ R2 : + =1
a b

(x − x0 )2 (y − y0 )2
E2 = (x, y) ∈ R2 : + =1
c d


Problem
Conclusions

Let A1 and A2 be the matrices associated to E1 and E2 .
Characteristic polynomial of the pencil λA2 + A1 :

H(λ) = det(λA2 + A1 ) = h3 λ3 + h2 λ2 + h1 λ + h0

Compute the discriminant of H(λ), and introduce the change of
variable (x0 , y0 ) = (x0 t, y0 t). The equation which gives us the
searched value of t, t0 , is S(t) = 0 where:

S(t) = discλ H(λ) |(x0 ,y0 )=(x0 t,y0 t) = s4 t8 + s3 t6 + s2 t4 + s1 t2 + s0

Making T = t2 :

S(T ) = s4 T 4 + s3 T 3 + s2 T 2 + s1 T + s0

Searched value of t: square root of the biggest real root of S(T )


Problem
Conclusions

Distance of closest approach of two separated ellipses

Theorem
Given two separated ellipses E1 and E2 the distance of their
closest approach is given as

d = t0 x 2 + y0
0
2

where t0 is the square root of the biggest positive real root of
S(T ) = S(t) |T =t2 = (discλ H(λ) |(x0 ,y0 )=(x0 t,y0 t) ) |T =t2 , where
H(λ) is the characteristic polynomial of the pencil determined
by them and (x0 , y0 ) is the center of E2 .


Problem
Conclusions

Example

Let A and B be the ellipses:
√
2 7 2 3 5
A := (x, y) ∈ R : x + xy + y 2 = 10
8 4 8
1 2 3 1 8 109
B := (x, y) ∈ R2 : x − x + y2 − y = −
4 2 9 9 36


Problem
Conclusions

Example

Polynomial whose biggest real root gives the square of the
instant T = T0 when the ellipses are tangent:

466271425
√ √
B
SA(T ) (T ) = + 9019725 3 T 4 + − 627564237 − 16904535 3 T 3
16 √ 32 √ 2
+ 39363189 3 + 690647377 T 2 + − 1186083
16 256 16 3 − 58434963 T
128
+ 4499761
256

B
The two real roots of SA(T ) (T ) are:

T0 = 0.5058481537; T1 = 0.07113873679


Problem
Conclusions

Positions of A and B(t)

t0 = T0 t1 = T1


Problem
Conclusions

Let A1 and A2 be the matrices deﬁning the separated ellipsoids E1
and E2 as X T A1 X = 0 and X T A2 X = 0 where X T = (x, y, z, 1), and
A = (aij ), B = (bij ), i, j = 1..4
Change the reference frame to have E1 centered at the origin and E2 ,
at (x0 , y0 , z0 ) with axis parallel to the coordinate ones:
P2 Q2 R2
E1 = (x, y, z) ∈ R3 : + 2 + 2 =1
a2 b c

(x − x0 )2 (y − y0 )2 (z − z0 )2
E2 = (x, y, z) ∈ R3 : 2
+ 2
+ =1
d f g2

where
P = x ux2 + 1 − ux 2 cos (α) + (ux uy (1 − cos (α)) − uz sin (α)) y + . . .
Q = (ux uy (1 − cos (α)) + uz sin (α)) x + y uy 2 + 1 − uy 2 cos (α) + . . .
R = (ux uz (1 − cos (α)) − uy sin (α)) x + (uyuz (1 − cos (α)) + ux sin (α)) y + . . .


Problem
Conclusions

Characteristic polynomial of the pencil λA2 + A1 :

H(λ) = det(λA2 + A1 ) = h4 λ4 + h3 λ3 + h2 λ2 + h1 λ + h0

Compute the discriminant of H(λ), and introduce the change of
variable (x0 , y0 , z0 ) = (x0 t, y0 t, z0 t). The equation which gives us the
searched value of t, t0 , is S(t) = 0 where:

S(t) = discλ H(λ) |(x0 t,y0 t,z0 t) = s6 t12 +s5 t10 +s4 t8 +s3 t6 +s2 t4 +s1 t2 +s0

Making T = t2 :

S(T ) = s6 t6 + s5 t5 + s4 T 4 + s3 T 3 + s2 T 2 + s1 T + s0

Searched value of t: square root of the biggest real root of S(T )


Problem
Conclusions


Theorem
Given two separated ellipsoids E1 and E2 the distance of their
closest approach is given as

d = t0 x 2 + y0 + z0
0
2 2

where t0 is the square root of the biggest positive real root of
S(T ) = S(t) |T =t2 = (discλ H(λ) |(x0 t,y0 t,z0 t) ) |T =t2 , where H(λ)
is the characteristic polynomial of the pencil determined by
them and (x0 , y0 , z0 ) is the center of E2 .


Problem
Conclusions

Example
Let A (blue) and B (green) be the ellipsoids:

1 2 1 2
A := (x, y, z) ∈ R3 : x + y + z2 = 1
4 2

1 2 1 1 51
B := (x, y, z) ∈ R3 : x − 2 x + y2 − 3 y + z2 − 5 z = −
5 4 2 2


Problem
Conclusions

Example

Polynomial whose biggest real root gives the square of the
instant T = T0 when the ellipsoids are tangent:

SA(T ) (T ) = 16641 T 2 2725362025 T 4 − 339879840 T 3 + 3362446 T 2 − 11232 T + 9
B

B
The two real roots of SA(T ) (T ) are:

T0 = 0.1142222397; T1 = 0.001153709353


Problem
Conclusions

Positions of A (blue) and B(t) (green)

t0 = T0 t1 = T1


Problem
Conclusions

Ellipses case:
Basic configuration:
Compute the eigenvectors of a 2x2 matrix
Compute the real roots of a 4-degree polynomial
Other configurations: roots of a 8-degree polynomial
Ellipsoids case:
Basic configuration:
Compute the eigenvectors of a 3x3 matrix
Compute the real roots roots of a 6-degree polynomial
Other configurations: roots of a 12-degree polynomial


Problem
Conclusions

Thank you!


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