minimum distance between two arbitrary circles in space - MathOverflow

minimum distance between two arbitrary circles in space

2012-04-12T18:55:55Z

as per the title, i am working out the problem with maxima, but i am surprised by how complicated this rapidly turns out to unfold for such a "simple" question.

monstrous equations, maybe someone has a bright idea? thanks.

Answer by Robert Israel for minimum distance between two arbitrary circles in space

2012-04-12T19:32:34Z

Use rational parametrizations of the circles (e.g. $\cos(\theta) = \frac{1-s^2}{1+s^2}$, $\sin(\theta) = \frac{2s}{1+s^2}$) with parameters $s$ and $t$ (hoping that the minimum is not when the parameter is $\infty$), let $F(s,t)$ be the square of the distance between the points on the circles for parameter values $s$ and $t$, and take the resultant of $\partial F/\partial s$ and $\partial F/\partial t$ with respect to one of $s$ and $t$. You should get a polynomial (I think of degree $20$) whose real roots will correspond to critical points.

Answer by Joseph O'Rourke for minimum distance between two arbitrary circles in space

2012-04-12T20:58:59Z

See "Finding the distance between two circles in three-dimensional space," by C.A. Neff, in the IBM Journal of Research and Development, Volume 34, Number 5, Page 770 (1990). IBM link. From the Abstract:

We show, by combining a theorem about solvable permutation groups and some explicit calculations with a computer algebra system, that, in general, the distance between two circles is an algebraic function of the parameters defining them, but that this function is not solvable in terms of radicals.

Another source is David Eberly's "Distance Between Two Circles in 3D," PDF link. His unpublished note gives explicit instructions for the computation, without analyzing its algebraic complexity too carefully. He does say that a critical equation

can be reduced to a polynomial of degree 8 whose roots ... are the candidates to provide the global minimum...

Answer by Beni Bogosel for minimum distance between two arbitrary circles in space

2012-04-12T21:14:07Z

I didn't manage to solve the problem (edit: in the meantime an answer was posted which says a precise formula using radicals cannot be found), but I can post a proof that the line joining the points where the minimal/maximal distance is achieved is perpendicular to the tangent line at the circles in those contact points. (inspired by the comment of Gerhard Paseman)

To do this, choose $\vec{a}$ and $\vec{d}$ the position vectors of the centers and $\vec{b},\vec{c}$, respectively $\vec{e},\vec{f}$ be two pairs of orthogonal unit vectors which span the planes of the first and respectively the second circle. Denote by $r,s$ the radii of the two circles. Consider the circles parametrized as (in fact, the argument works for any parametrization) $$ p(\theta)=\vec{a}+r\cos\theta\ \vec{b}+r\sin\theta\ \vec{c}, \ \theta \in [0,2\pi] $$ $$ q(\tau)=\vec{d}+s\cos\tau\ \vec{e}+s\sin\tau\ \vec{f}, \tau \in [0,2\pi]$$ and denote $F(\theta,\tau)=|p(\theta)-q(\tau)|^2$. Then the pair of points which realize the minimal/maximal distance must satisfy $$ \frac{\partial F}{\partial \theta}=\frac{\partial F}{\partial \tau}=0. $$

We have $$ \frac{\partial F}{\partial \theta}=2\sum_{i=1}^3 [p_i(\theta)-q_i(\tau)]p_i'(\theta)=2 (p(\theta)-q(\tau))\cdot p'(\theta) $$ $$ \frac{\partial F}{\partial \tau}=-2\sum_{i=1}^3 [ p_i(\theta)-q_i(\tau) ] q_i'(\tau)=-2 (p(\theta)-q(\tau))\cdot q'(\tau) $$ where "$\cdot$" is the usual dot product. Therefore when $\theta,\tau$ correspond to the minimum/maximum value, the partial derivatives vanish and $p'(\theta)\perp (p(\theta)-q(\tau))$ and $q(\tau)'\perp (p(\theta)-q(\tau))$ where $p'(\theta),q'(\tau)$ are the tangent vectors in the contact points and $p(\theta)-q(\tau)$ is the vector connecting the points where minimal/maximal distance is achieved.