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In several branches of applied mathematics the problem arises to describe the intersection of two cones in three space.

I have searched and found a few references that discuss the problem for cones with parallel axes. I am interested in the general case.

Assume that one cone has vertex at the origin with a certain "cone angle" (is that the phrase?) at the vertex and an axis some vector through the origin. Situate the second cone at point (b,0,0) with a (perhaps) different cone angle and axis some arbitrary vector through (b,0,0). Describe the locus of points where they intersect.

The answer ought to be a polynomial in the various parameters. I want a fully symbolic answer, no numerical methods. It will probably be a fairly large polynomial. Of course, the intersection might be empty, which simply means that when certain values are plugged in for the parameters, the polynomial has no real solution.

Has this been done?

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I am not sure I understand the question, see the remarks of Qfwfq. Here is something related which might nevertheless be interesting for some branches of applied mathematics: an algorithm for parameterizing the intersection of two arbitrary quadrics, given in implicit form with integer coefficients. There are several research reports and even an implementation. You can even input your coefficients on-line on the web and see the result. Maybe there is something easier for the special case of two circular cones.

ADDITION: Apparently they changed the structure of their web pages; The work of these people is based on solid algebraic work, which might be what you are looking for, in their publications; then (what I find remarkable), they often go through and implement their results (see the web page about the work in their lab; I hope the links work now. in particular the work about Constant-complexity geometric problems and algebraic invariants). Those algorithms, while producing "numerical results", do so in a completely reliable way, in contrast to what you conventionally expect from numerical floating-point computation. O.k. you don't want computations; you also don't want the "sum of squares=0" solution, but what DO you want then? I suspect you are interested in the real solutions, so there will be case distinctions, 0, 1, 2 components etc. How should a potential solution to your problem look like? A single polynomial? In which variables? What should the solutions to the polynomial describe?

When you write: "The solution should describe directly what the intersection "looks like" from the point of view of the origin". Does that mean you want the projection of the intersection curve from the origin? It will be some intervals on the circle which is the projection of the whole primary cone. Do you want the endpoint of these intervals?

Another group that you might check out is the work of the EXACUS project in Saarbrücken.

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  • $\begingroup$ Thanks for the link. However, when I tried it, it failed to work. In any event, it seems to be numerical computing, which I don't want. Also, the condition that all coefficients be integers is too restrictive. $\endgroup$
    – rhlewis
    Commented Jan 29, 2013 at 14:00
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Let me take the opportunity of this year-old question being kicked to the front page to adorn Günter Rote's answer with an image from the web page (broken link) (edit: but see Wayback Machine and the current page) to which he refers:


            
"Parameterization of intersections of quadrics: theory, algorithms and implementation"

(source: Wayback Machine) (source: current)

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If $U = (u_1, u_2, u_3)$ is a unit vector, a (double) cone with vertex $(0,0,0)$, axis in direction $U$ and opening angle $\arccos(\alpha)$ is the zero-set of the polynomial $(u_1 x_1 + u_2 x_2 + u_3 x_3)^2 - \alpha^2 (x_1^2 + x_2^2 + x_3^2)$. The double cone with vertex $(b,0,0)$, axis in direction $V = (v_1, v_2, v_3)$ and opening angle $\arccos(\beta)$ is the zero-set of the polynomial $(v_1 (x_1 - b) + v_2 x_2 + v_3 x_3)^2 - \beta^2 (x_1^2 + x_2^2 + x_3^2)$. The resultant of these polynomials with respect to $x_1$ is a polynomial of degree 4 in $x_2$ and $x_3$, with some 392 terms.

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  • $\begingroup$ Probably I'm not understanding the question, but if we are over the reals why not use the polynomial $[(u_1 x_1 + u_2 x_2 + u_3 x_3)^2 - \alpha^2 (x_1^2 + x_2^2 + x_3^2)]^2+[(v_1 (x_1 - b) + v_2 x_2 + v_3 x_3)^2 - \beta^2 (x_1^2 + x_2^2 + x_3^2)]^2=0$, that describes the same zero set and has few terms? $\endgroup$
    – Qfwfq
    Commented Jan 28, 2013 at 19:21
  • $\begingroup$ Thanks Robert. In the applications I have in mind, there are two cameras or sensors, one at the origin and one at (b,0,0). The one at the origin is primary. The solution should describe directly what the intersection "looks like" from the point of view of the origin. Projecting the intersection to the xy plane or yz plane is not immediately helpful. I set up some polynomial equations in a reasonable coordinate system (though maybe not the best) and got the resultant. It has 4201 terms. $\endgroup$
    – rhlewis
    Commented Jan 29, 2013 at 14:10

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