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How can I find volume of intersection of two conical frustums? Both objects are on the same X axis. I'm not sure yet how exactly they will intersect because each frustum is one user on the map and I need to find probability that two users saw each other.

I would appreciate some example because I think I will not get to final solution myself.


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The intersection of two parallel conical frustums on the same axis is either (a) another conical frustum or (b) two conical frustums glued together at their bases. What variables should we assume are a 'given' in the problem - e.g. the radii of the tops and bottoms and their centers' positions on the $x$-axis? – anon Aug 2 '11 at 11:39
Anon, yes I have radii for tops and bottoms, and centers (In my case that's latitude and longitude). The height of conical frustums can by any number, I think it doesn't metter for me. – Dimitry Aug 2 '11 at 22:49
Is my question so complex that no-one can give me an answer? Or I just need to provide some more details to describe it better? I've spend few days searching in google and on couple forums and can't find any example. Any help would be appreciated. – Dimitry Aug 4 '11 at 8:36
I'm sorry, I figured either someone else would answer or you'd figure it out - assuming I understand your question correctly. I'll open my MSPaint and work on an explanation. – anon Aug 4 '11 at 12:08
up vote 1 down vote accepted

A cone is simply a triangle rotated about an axis to make a solid. We'll place it on the $y$-axis instead of the $x$-axis because it looks better standing upright. Now if you slice a smaller cone off the top, then you're left with a conical frustum. This new solid is now the same as an isosceles trapezoid rotated about an axis down the middle. Therefore, if we take a look at an intermediate slice of the intersection of two frustums, we'll get essentially one of four possible situations (up to otherwise superfluous orientation and proportion):

conical frustum intersections

You can see how in the right two cases, the intersection is two conical frustums glued together, base to base, while in the left two cases the intersection is just another frustum. Do keep in mind, however, that the two main frustums don't necessarily look "upside-down" relative to each other - but they will make one of these situations, regardless. (Actually, I'm lying: there's a fifth possible case where one frustum completely contains the other, but the problem is trivial then... and makes my nice picture less symmetrical.) Instead of going over all four possible situations, I'll only lay out the two tools you need in order to do the problem yourself, hopefully. The first is this:

$\hskip 1.1 in $ frustum height

If you know the top has length (diameter for the frustum) $A$, and you know the bottom has length $B$, and you know the purple line is $u$ distance down from the top and $v$ distance up from the bottom, then the purple line has length equal to the weighted average $(Au+Bv)/(u+v)$. You can use this to find the base diameters of the intersection frustum based on the top and bottom diameters of the main two frustums.

Lastly, a frustum with top diameter $A$, bottom diameter $B$, and height $h$, has volume that can be calculated with the formula $\frac{\pi h}{12}(A^2+AB+B^2)$. If you don't understand how to resolve your intersecting frustums problem now, tell me where you're stuck and I'll try to elaborate.

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Anon, thank you for such detailed explanation. I think I've solved my problem, at latest I'm getting some result. – Dimitry Aug 11 '11 at 16:25

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