CODEWITHSUNDEEP
math3dgeometryintersection
Pawel Os.
Pawel Os.

Reputation: 633

Circle-circle intersection in 3D

I have two circles, each defined by a centre point c = (x,y,z) a radius r and a normal vector n = (x,y,z) (perpendicular to the circle's plane).

How to calculate the intersection points of two such circles (not necessarily being in the same plane)?

If there is no intersections, but those circles are not parallel, how to get the point, where those circles would intersect if you increased or decreased the radius until they touch?

Upvotes: 4

Views: 4269

Answers (2)

phil1008
phil1008

Reputation: 179

I tried to implement MBo's solution, but it didn't quite work for me. But, I got it working (in Javascript with Three.js) by changing the 'c1' in the last line to 'c2'. My implementation below:

    let c1, r1, c2, r2
    if (circleRadius>=radiusOfCircleOfIntercetion) {
      c1 = projectedSphereCenterOnCirclePlane  // Vector3
      r1 = radiusOfCircleOfIntercetion
      c2 = circleCenter                        // Vector3
      r2 = circleRadius
    }
    else {
      c1 = circleCenter                        // Vector3
      r1 = circleRadius
      c2 = projectedSphereCenterOnCirclePlane  // Vector3
      r2 = radiusOfCircleOfIntercetion
    }
    const cdiff = c1.clone().sub(c2)
    const cdifflen = cdiff.length()
    const cdiffnorm = cdiff.clone().normalize() //unit vector
    const cdiffperp = cdiffnorm.clone().cross(circleNormal)
    const q = cdifflen**2 + r2**2 - r1**2
    const dx = 1/2 * q / cdifflen
    const dy = 1/2 * Math.sqrt(4 * cdifflen**2 * r2**2 - q**2) / cdifflen
    const dxvect = cdiffnorm.clone().multiplyScalar(dx)
    const dyvect = cdiffperp.clone().multiplyScalar(dy)
    const p1 = c2.clone().add(dxvect).add(dyvect)
    const p2 = c2.clone().add(dxvect).sub(dyvect)

Upvotes: 0

MBo
MBo

Reputation: 80327

If n1 x n2 = 0 then normal vectors are (anti)collinear, and planes are parallel.

They are the same if Dot(c1-c2, n1) = 0, otherwise circle intersection is impossible.

Case of two circles in the same plane. I assume that r2>=r1

cdiff = (c1-c2)
cdifflen = cdiff.Length 

if cdifflen > r1 + r2 then no intersection
if cdifflen = r1 + r2 then intersection exists in one point
p = (c1 * r2  + c2 * r1) / (r1 + r2)

if cdifflen < r2 - r1 then no intersection
if cdifflen = r2 - r1 then intersection exists in one point
p = (c1 - c2) * r2 /(r2 - r1)

otherwise there are two intersection points
cdiffnorm = cdiff.Normalized //unit vector
cdiffperp = cdiffnorm * n1.Normalized
q = cdifflen^2 + r2^2 - r1^2
dx = 1/2 * q / cdifflen
dy = 1/2 * Sqrt(4 * cdifflen^2 * r2^2 - q^2) / cdifflen 
p1,2 = c1 + cdiffnorm * dx +/- cdiffperp * dy

Some formulas are explained here

If planes are not parallel, their intersection is a line with direction vector

dl = n1 x n2

Base point of this line might be found using Geometric Tools Engine (file GteIntrPlane3Plane3.h). Then check for intersection of both circles with this line and compare intersection points if exist.

Upvotes: 4

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