double Angle; // radians
double Apex[3];
double Axis[3];
double Point[3];
// send to origin
Point[0] -= Apex[0];
Point[1] -= Apex[1];
Point[2] -= Apex[2];
Apex[0] = 0.0;
Apex[1] = 0.0;
Apex[2] = 0.0;
// get another point on the axis of the cone
double Point2[3] = { Apex[0] + Axis[0], Apex[1] + Axis[1], Apex[2] + Axis[2] };
// get the vector from the second point to the given point
double V[3] = { Point[0] - Point2[0], Point[1] - Point2[1], Point[2] - Point2[2] };
// find the distance from the cone's axis to the point
double Crossed[3] = CrossProduct(Axis, V);
double CrossedMagnitude = Magnitude(Crossed);
double AxisMagnitude = Magnitude(Axis);
double d = CrossedMagnitude / AxisMagnitude;
// find orthogonal distance of point down axis via pythagoras
double V2[3] = { Apex[0] - Point[0], Apex[1] - Point[1], Apex[2] - Point[2] };
double p = sqrt((V2[0] * V2[0] + V2[1] * V2[1] + V2[2] * V2[2]) - (d * d));
// we've reduced the problem to 2D trig
Distance = d * cos(Angle) - p * sin(Angle);Distance from point to cone
Author
I'm having a bit of trouble finding the shortest distance from a 3D point to the surface of an infinite cone defined by an apex, interior semi-angle, and opening axis. I have what I believe to be the solution for a double cone (one opening from either side of the apex), but can't quite wrap my head around how to restrict it to a single cone. Here's pseudocode for what I have thus far:
I didn't check your existing code for correctness (since you stated it seems to be working ok), but as for the double-cone problem, if the point lies behind the plane formed by the apex and the cone axis, then the closest point is the apex.
Does that help at all?
Does that help at all?
Author
Quote:Original post by jyk
I didn't check your existing code for correctness (since you stated it seems to be working ok), but as for the double-cone problem, if the point lies behind the plane formed by the apex and the cone axis, then the closest point is the apex.
Does that help at all?
I thought about this, but couldn't convince myself that it was true. For very very short and fat cone, a point out towards the base of the cone but still above the plane formed by the apex/axis would have a closest point somewhere on the cone closer to the edge, wouldn't it? I'll see if I can whip up some mspaint to illustrate...
Quote:Original post by jouleyYeah, you're right - I wasn't thinking clearly :)Quote:Original post by jyk
I didn't check your existing code for correctness (since you stated it seems to be working ok), but as for the double-cone problem, if the point lies behind the plane formed by the apex and the cone axis, then the closest point is the apex.
Does that help at all?
I thought about this, but couldn't convince myself that it was true. For very very short and fat cone, a point out towards the base of the cone but still above the plane formed by the apex/axis would have a closest point somewhere on the cone closer to the edge, wouldn't it? I'll see if I can whip up some mspaint to illustrate...
Author
Well, if it helps anyone to visualize:
(It can be simplified by a triangle and a point on a plane parallel to the axis.)
(It can be simplified by a triangle and a point on a plane parallel to the axis.)
This paper shows how to discount the fact that the point is part of the double cone -
http://www.geometrictools.com/Documentation/IntersectionLineCone.pdf
http://www.geometrictools.com/Documentation/IntersectionLineCone.pdf
Author
Quote:Original post by ckin2001
This paper shows how to discount the fact that the point is part of the double cone -
http://www.geometrictools.com/Documentation/IntersectionLineCone.pdf
Hmm... It does deal with the double cone, but it also makes the assumption that the angle of the cone is acute. It then rids itself of the second cone with a simple dot product to check for being on the same side of the plane defined by the apex and the axis, as jyk mentioned above. This works for acute angles, but, unless I've missed something (if I have, please explain!), it's the obtuse ones I'm worried about.
The closest distance can be found by doing:
c = (bSingleCone) ? acos(((Point-Apex).Axis)\(|Point-Apex|*|Axis|)) : acos(abs(((Point-Apex).Axis))\(|Point-Apex|*|Axis|)); // from dot product definitionif(abs(c - a) >= Pi/2) return |Point-Apex|;else return |Point-Apex| * sin(abs(c - a));
Author
Well, that certainly is a whole lot more simple! I see the trick, too, I think, about the important angle being that between the Axis and the Apex-Point. I also understand the abs(c-a) makes the result always positive, so removing the abs( ) will give a negative result for points inside the cone, which is the behavior I'm after.
Works like a charm, thanks!
Works like a charm, thanks!
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