3d line-intersection code not working properly - math

I created this piece of code to get the intersection of two 3d line-segments.
Unfortunately the result of this code is inaccurate, the intersection-point is not always on both lines.
I am confused and unsure what I'm doing wrong.
Here is my code:
--dir = direction
--p1,p2 = represents the line
function GetIntersection(dirStart, dirEnd, p1, p2)
local s1_x, s1_y, s2_x, s2_y = dirEnd.x - dirStart.x, dirEnd.z - dirStart.z, p2.x - p1.x, p2.z - p1.z
local div = (-s2_x * s1_y) + (s1_x * s2_y)
if div == 0 then return nil end
local s = (-s1_y * (dirStart.x - p1.x) + s1_x * (dirStart.z - p1.z)) / div
local t = ( s2_x * (dirStart.z - p1.z) - s2_y * (dirStart.x - p1.x)) / div
if (s >= 0 and s <= 1 and t >= 0 and t <= 1) and (Vector(dirStart.x + (t * s1_x), 0, dirStart.z + (t * s1_y)) or nil) then
local v = Vector(dirStart.x + (t * s1_x),0,dirStart.z + (t * s1_y))
return v
end
end

This is example of Delphi code to find a distance between two skew lines in 3D. For your purposes it is necessary to check that result if small enough value (intersection does exist), check that s and t parameters are in range 0..1, then
calculate point using parameter s
Math of this approach is described in 'the shortest line...' section of Paul Bourke page
VecDiff if vector difference function, Dot id scalar product function
function LineLineDistance(const L0, L1: TLine3D; var s, t: Double): Double;
var
u: TPoint3D;
a, b, c, d, e, det, invdet:Double;
begin
u := VecDiff(L1.Base, L0.Base);
a := Dot(L0.Direction, L0.Direction);
b := Dot(L0.Direction, L1.Direction);
c := Dot(L1.Direction, L1.Direction);
d := Dot(L0.Direction, u);
e := Dot(L1.Direction, u);
det := a * c - b * b;
if det < eps then
Result := -1
else begin
invdet := 1 / det;
s := invdet * (b * e - c * d);
t := invdet * (a * e - b * d);
Result := Distance(PointAtParam(L0, s), PointAtParam(L1, t));
end;
end;

As far as I can tell your code is good. I've implemented this in javascript at https://jsfiddle.net/SalixAlba/kkrc9kcf/
and it seems to work for all the cases I can think of.
The only changes I've done is to change things to work in javascript rather than lua. The final condition was commented out
function GetIntersection(dirStart, dirEnd, p1, p2) {
var s1_x = dirEnd.x - dirStart.x;
var s1_y = dirEnd.z - dirStart.z;
var s2_x = p2.x - p1.x;
var s2_y = p2.z - p1.z;
var div = (-s2_x * s1_y) + (s1_x * s2_y);
if (div == 0)
return new Vector(0,0);
var s = (-s1_y * (dirStart.x - p1.x) + s1_x * (dirStart.z - p1.z)) / div;
var t = ( s2_x * (dirStart.z - p1.z) - s2_y * (dirStart.x - p1.x)) / div;
if (s >= 0 && s <= 1 && t >= 0 && t <= 1) {
//and (Vector(dirStart.x + (t * s1_x), 0, dirStart.z + (t * s1_y)) or nil) then
var v = new Vector(
dirStart.x + (t * s1_x),
dirStart.z + (t * s1_y));
return v;
}
return new Vector(0,0);
}
Mathmatically it makes sense. If A,B and C,D are your two lines. Let s1 = B-A, s2 = C-D. A point of the line AB is given by A + t s1 and a point on the line CD is given by C + s s2. For an intersection we require
A + t s1 = C + s s2
or
(A-C) + t s1 = s s2
You two formula for s, t are found by taking the 2D cross product with each of the vectors s1 and s2
(A-C)^s1 + t s1^s1 = s s2^s1
(A-C)^s2 + t s1^s2 = s s2^s2
recalling s1^s1=s2^s2=0 and s2^s1= - s1^s2 we get
(A-C)^s1 = s s2^s1
(A-C)^s2 + t s1^s2 = 0
which can be solved to get s and t. This matches your equations.

Related

Plotting a graph for 2 elements in scilab within a For loop

I am currently writing a program to show the relation between the Relaxation Factor and number of iterations it takes to achieve a solution using the Successive OverRelaxation Method
This is the concerned For loop:
for (w = 0:0.05:2)
T = -inv(D + w*L)*(w*U + (w -1) * D);
C = inv(D + w*L) * w *B ;
X = zeros(B);
for(i = 1:1:MaxIter)
X = T * X + C;
err = A * X - B;
if (abs(err) < abs(tol))
break
end
end
disp("Relaxation Factor = " + string(w) +" No. of iterations = " + string(i));
I wish to draw a plot showing the relation between Relaxation factor and no. of iterations(between w and i). How should I proceed?
Just proceed as follow
I=zeros(W)
for k = 1:size(W,"*")
w=W(k)
T = -inv(D + w*L)*(w*U + (w -1) * D);
C = inv(D + w*L) * w *B ;
X = zeros(B);
for(i = 1:1:MaxIter)
X = T * X + C;
err = A * X - B;
if (abs(err) < abs(tol))
I(k)=i
break
end
end
end
plot(W,I)

how to detect point on/around a line (with some offset)

Drawn a line from a point A to point B. Let d be offset. Let C be point to be tested.
I am going to do a kind of hit testing around the line with offset.
How can i do the hit testing around the line with the given offset.
Ex: A = (10,10), B (30,30), offset = 2. choose C as any point. Please Refer the image in the link please.
http://s10.postimg.org/6by2dzvax/reference.png
Please help me.
Thanks in advance.
Find offset for C.
e.g. dx1 and dy1. If dy1/dx1=dy/dx then your C hits the line.
For segment you should also check if whether dx1 < dx or dy1 < dy.
In other words, you want to check if that point C is inside a certain rectangle, with dimensions 2*d and |A-B|+2*d. You need to represent the line as u*x+v*y+w=0, this can be accomplished by
u = A.y-B.y
v = B.x-A.x
w = A.x*B.y - A.y * B.x
Then the (signed) distance of C from that line would be
d = (u*C.x + v*C.y +w) / sqrt( u*u+v*v)
You compare abs(d) to your offset.
The next step would be to check the position of C in the direction of the line. To that end you consider the orthogonal line u2*x+v2*y+w2=0 with
u2 = v
v2 = -u
w2 = -u2*(A.x+B.x)/2 - v2*(A.y+B.y)/2
and the distance
d2 = (u2 * C.x + v2 * C.y + w2 ) / sqrt( u2*u2+v2*v2 )
This distance must be compared to something like the length of the line+offset:
abs(d2) < |A-B| / 2 + offset
A convenient trick is to rotate and translate the plane in such a way that the segment AB maps to the segment (0, 0)-(0, L) (just like on the image), L being the segment length.
If you apply the same transform to C, then it a very simple matter to test inclusion in the rectangle.
That useful transform is given by:
x = ((X - XA).(XB - XA) + (Y - YA).(YB - YA)) / L
y = ((X - XA).(YB - YA) - (Y - YA).(XB - XA)) / L
maybe you can use this function to count the shortest distance of the point to the line. If the distance is <= offset, then that point is hitting the line.
private double pointDistanceToLine(PointF line1, PointF line2, PointF pt)
{
var isValid = false;
PointF r = new PointF();
if (line1.Y == line2.Y && line1.X == line2.X)
line1.Y -= 0.00001f;
double U = ((pt.Y - line1.Y ) * (line2.Y - line1.Y )) + ((pt.X - line1.X) * (line2.X - line1.X));
double Udenom = Math.Pow(line2.Y - line1.Y , 2) + Math.Pow(line2.X - line1.X, 2);
U /= Udenom;
r.Y = (float)(line1.Y + (U * (line2.Y - line1.Y ))); r.X = (float)(line1.X + (U * (line2.X - line1.X)));
double minX, maxX, minY , maxY ;
minX = Math.Min(line1.Y , line2.Y );
maxX = Math.Max(line1.Y , line2.Y );
minY = Math.Min(line1.X, line2.X);
maxY = Math.Max(line1.X, line2.X);
isValid = (r.Y >= minX && r.Y <= maxX) && (r.X >= minY && r.X <= maxY );
//return isValid ? r : null;
if (isValid)
{
double result = Math.Pow((pt.X - r.X), 2) + Math.Pow((pt.Y - r.Y), 2);
result = Math.Sqrt(result);
return result;
}
else {
double result1 = Math.Pow((pt.X - line1.X), 2) + Math.Pow((pt.Y - line1.Y), 2);
result1 = Math.Sqrt(result1);
double result2 = Math.Pow((pt.X - line2.X), 2) + Math.Pow((pt.Y - line2.Y), 2);
result2 = Math.Sqrt(result2);
return Math.Min(result1, result2);
}
}

Intersection between a line and a sphere

I'm trying to find the point of intersection between a sphere and a line but honestly, I don't have any idea of how to do so.
Could anyone help me on this one ?
Express the line as an function of t:
{ x(t) = x0*(1-t) + t*x1
{ y(t) = y0*(1-t) + t*y1
{ z(t) = z0*(1-t) + t*z1
When t = 0, it will be at one end-point (x0,y0,z0). When t = 1, it will be at the other end-point (x1,y1,z1).
Write a formula for the distance to the center of the sphere (squared) in t (where (xc,yc,zc) is the center of the sphere):
f(t) = (x(t) - xc)^2 + (y(t) - yc)^2 + (z(t) - zc)^2
Solve for t when f(t) equals R^2 (R being the radius of the sphere):
(x(t) - xc)^2 + (y(t) - yc)^2 + (z(t) - zc)^2 = R^2
A = (x0-xc)^2 + (y0-yc)^2 + (z0-zc)^2 - R^2
B = (x1-xc)^2 + (y1-yc)^2 + (z1-zc)^2 - A - C - R^2
C = (x0-x1)^2 + (y0-y1)^2 + (z0-z1)^2
Solve A + B*t + C*t^2 = 0 for t. This is a normal quadratic equation.
You can get up to two solutions. Any solution where t lies between 0 and 1 are valid.
If you got a valid solution for t, plug it in the first equations to get the point of intersection.
I assumed you meant a line segment (two end-points). If you instead want a full line (infinite length), then you could pick two points along the line (not too close), and use them. Also let t be any real value, not just between 0 and 1.
Edit: I fixed the formula for B. I was mixing up the signs. Thanks M Katz, for mentioning that it didn't work.
I believe there is an inaccuracy in the solution by Markus Jarderot. Not sure what the problem is, but I'm pretty sure I translated it faithfully to code, and when I tried to find the intersection of a line segment known to cross into a sphere, I got a negative discriminant (no solutions).
I found this: http://www.codeproject.com/Articles/19799/Simple-Ray-Tracing-in-C-Part-II-Triangles-Intersec, which gives a similar but slightly different derivation.
I turned that into the following C# code and it works for me:
public static Point3D[] FindLineSphereIntersections( Point3D linePoint0, Point3D linePoint1, Point3D circleCenter, double circleRadius )
{
// http://www.codeproject.com/Articles/19799/Simple-Ray-Tracing-in-C-Part-II-Triangles-Intersec
double cx = circleCenter.X;
double cy = circleCenter.Y;
double cz = circleCenter.Z;
double px = linePoint0.X;
double py = linePoint0.Y;
double pz = linePoint0.Z;
double vx = linePoint1.X - px;
double vy = linePoint1.Y - py;
double vz = linePoint1.Z - pz;
double A = vx * vx + vy * vy + vz * vz;
double B = 2.0 * (px * vx + py * vy + pz * vz - vx * cx - vy * cy - vz * cz);
double C = px * px - 2 * px * cx + cx * cx + py * py - 2 * py * cy + cy * cy +
pz * pz - 2 * pz * cz + cz * cz - circleRadius * circleRadius;
// discriminant
double D = B * B - 4 * A * C;
if ( D < 0 )
{
return new Point3D[ 0 ];
}
double t1 = ( -B - Math.Sqrt ( D ) ) / ( 2.0 * A );
Point3D solution1 = new Point3D( linePoint0.X * ( 1 - t1 ) + t1 * linePoint1.X,
linePoint0.Y * ( 1 - t1 ) + t1 * linePoint1.Y,
linePoint0.Z * ( 1 - t1 ) + t1 * linePoint1.Z );
if ( D == 0 )
{
return new Point3D[] { solution1 };
}
double t2 = ( -B + Math.Sqrt( D ) ) / ( 2.0 * A );
Point3D solution2 = new Point3D( linePoint0.X * ( 1 - t2 ) + t2 * linePoint1.X,
linePoint0.Y * ( 1 - t2 ) + t2 * linePoint1.Y,
linePoint0.Z * ( 1 - t2 ) + t2 * linePoint1.Z );
// prefer a solution that's on the line segment itself
if ( Math.Abs( t1 - 0.5 ) < Math.Abs( t2 - 0.5 ) )
{
return new Point3D[] { solution1, solution2 };
}
return new Point3D[] { solution2, solution1 };
}
Don't have enough reputation to comment on M Katz answer, but his answer assumes that the line can go on infinitely in each direction. If you need only the line SEGMENT's intersection points, you need t1 and t2 to be less than one (based on the definition of a parameterized equation). Please see my answer in C# below:
public static Point3D[] FindLineSphereIntersections(Point3D linePoint0, Point3D linePoint1, Point3D circleCenter, double circleRadius)
{
double cx = circleCenter.X;
double cy = circleCenter.Y;
double cz = circleCenter.Z;
double px = linePoint0.X;
double py = linePoint0.Y;
double pz = linePoint0.Z;
double vx = linePoint1.X - px;
double vy = linePoint1.Y - py;
double vz = linePoint1.Z - pz;
double A = vx * vx + vy * vy + vz * vz;
double B = 2.0 * (px * vx + py * vy + pz * vz - vx * cx - vy * cy - vz * cz);
double C = px * px - 2 * px * cx + cx * cx + py * py - 2 * py * cy + cy * cy +
pz * pz - 2 * pz * cz + cz * cz - circleRadius * circleRadius;
// discriminant
double D = B * B - 4 * A * C;
double t1 = (-B - Math.Sqrt(D)) / (2.0 * A);
Point3D solution1 = new Point3D(linePoint0.X * (1 - t1) + t1 * linePoint1.X,
linePoint0.Y * (1 - t1) + t1 * linePoint1.Y,
linePoint0.Z * (1 - t1) + t1 * linePoint1.Z);
double t2 = (-B + Math.Sqrt(D)) / (2.0 * A);
Point3D solution2 = new Point3D(linePoint0.X * (1 - t2) + t2 * linePoint1.X,
linePoint0.Y * (1 - t2) + t2 * linePoint1.Y,
linePoint0.Z * (1 - t2) + t2 * linePoint1.Z);
if (D < 0 || t1 > 1 || t2 >1)
{
return new Point3D[0];
}
else if (D == 0)
{
return new [] { solution1 };
}
else
{
return new [] { solution1, solution2 };
}
}
You may use Wolfram Alpha to solve it in the coordinate system where the sphere is centered.
In this system, the equations are:
Sphere:
x^2 + y^2 + z^2 = r^2
Straight line:
x = x0 + Cos[x1] t
y = y0 + Cos[y1] t
z = z0 + Cos[z1] t
Then we ask Wolfram Alpha to solve for t: (Try it!)
and after that you may change again to your original coordinate system (a simple translation)
Find the solution of the two equations in (x,y,z) describing the line and the sphere.
There may be 0, 1 or 2 solutions.
0 implies they don't intersect
1 implies the line is a tangent to the sphere
2 implies the line passes through the sphere.
Here's a more concise formulation using inner products, less than 100 LOCs, and no external links. Also, the question was asked for a line, not a line segment.
Assume that the sphere is centered at C with radius r. The line is described by P+l*D where D*D=1. P and C are points, D is a vector, l is a number.
We set PC = P-C, pd = PC*D and s = pd*pd - PC*PC + r*r. If s < 0 there are no solutions, if s == 0 there is just one, otherwise there are two. For the solutions we set l = -pd +- sqrt(s), then plug into P+l*D.
Or you can just find the formula of both:
line: (x-x0)/a=(y-y0)/b=(z-z0)/c, which are symmetric equations of the line segment between the points you can find.
sphere: (x-xc)^2+(y-yc)^2+(z-zc)^2 = R^2.
Use the symmetric equation to find relationship between x and y, and x and z.
Then plug in y and z in terms of x into the equation of the sphere.
Then find x, and then you can find y and z.
If x gives you an imaginary result, that means the line and the sphere doesn't intersect.
I don't have the reputation to comment on Ashavsky's solution, but the check at the end needed a bit more tweaking.
if (D < 0)
return new Point3D[0];
else if ((t1 > 1 || t1 < 0) && (t2 > 1 || t2 < 0))
return new Point3D[0];
else if (!(t1 > 1 || t1 < 0) && (t2 > 1 || t2 < 0))
return new [] { solution1 };
else if ((t1 > 1 || t1 < 0) && !(t2 > 1 || t2 < 0))
return new [] { solution2 };
else if (D == 0)
return new [] { solution1 };
else
return new [] { solution1, solution2 };

Quadratic Bézier Curve: Calculate Points

I'd like to calculate a point on a quadratic curve. To use it with the canvas element of HTML5.
When I use the quadraticCurveTo() function in JavaScript, I have a source point, a target point and a control point.
How can I calculate a point on the created quadratic curve at let's say t=0.5 with "only" knowing this three points?
Use the quadratic Bézier formula, found, for instance, on the Wikipedia page for Bézier Curves:
In pseudo-code, that's
t = 0.5; // given example value
x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;
p[0] is the start point, p[1] is the control point, and p[2] is the end point. t is the parameter, which goes from 0 to 1.
In case somebody needs the cubic form:
//B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3
x = (1-t)*(1-t)*(1-t)*p0x + 3*(1-t)*(1-t)*t*p1x + 3*(1-t)*t*t*p2x + t*t*t*p3x;
y = (1-t)*(1-t)*(1-t)*p0y + 3*(1-t)*(1-t)*t*p1y + 3*(1-t)*t*t*p2y + t*t*t*p3y;
I created this demo :
// x = a * (1-t)³ + b * 3 * (1-t)²t + c * 3 * (1-t)t² + d * t³
//------------------------------------------------------------
// x = a - 3at + 3at² - at³
// + 3bt - 6bt² + 3bt³
// + 3ct² - 3ct³
// + dt³
//--------------------------------
// x = - at³ + 3bt³ - 3ct³ + dt³
// + 3at² - 6bt² + 3ct²
// - 3at + 3bt
// + a
//--------------------------------
// 0 = t³ (-a+3b-3c+d) + => A
// t² (3a-6b+3c) + => B
// t (-3a+3b) + => c
// a - x => D
//--------------------------------
var A = d - 3*c + 3*b - a,
B = 3*c - 6*b + 3*a,
C = 3*b - 3*a,
D = a-x;
// So we need to solve At³ + Bt² + Ct + D = 0
Full example here
may help someone.
I edited talkhabis answer (cubic curve) so the curve is displayed with the right coordinates. (Couldn't comment)
The Y-coordinates needed to be changed (-p[].y+150). (A new variable for that might be a nicer and more efficient solution, but you get the idea)
// Apply points to SVG and create the curve and controllers :
var path = document.getElementById('path'),
ctrl1 = document.getElementById('ctrl1'),
ctrl2 = document.getElementById('ctrl2'),
D = 'M ' + p0.x + ' ' + (-p0.y+150) +
'C ' + c0.x + ' ' + (-c0.y+150) +', ' + c1.x + ' ' + (-c1.y+150) + ', ' + p1.x + ' ' + (-p1.y+150);
path.setAttribute('d',D);
ctrl1.setAttribute('d','M'+p0.x+','+(-p0.y+150)+'L'+c0.x+','+(-c0.y+150));
ctrl2.setAttribute('d','M'+p1.x+','+(-p1.y+150)+'L'+c1.x+','+(-c1.y+150));
// Lets test the "Bezier Function"
var t = 0, point = document.getElementById('point');
setInterval(function(){
var p = Bezier(p0,c0,c1,p1,t);
point.setAttribute('cx',p.x);
point.setAttribute('cy',-p.y+150);
t += 0.01;
if(t>=1) t=0;
},50);
// OK ... Now tring to get "y" on cruve based on mouse "x" :
var svg = document.getElementById('svg'),
point2 = document.getElementById('point2');
svg.onmousemove = function(e){
var x = (e.pageX - 50)/2,
y = (e.pageY - 50)/2;
// "-50" because of "50px margin" on the left side
// and "/2" because the svg width is 300 units and 600 px => 300 = 600/2
// Get the x,y by mouse x
var p = YBX(p0,c0,c1,p1,x);
point2.setAttribute('cx',p.x);
point2.setAttribute('cy',-p.y+150);
}
http://jsfiddle.net/u214gco8/1/
I also created some C-Code to test the results for the cubic curve. Just enter the X and Y coordinates in the main function.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void bezierCurve(int x[] , int y[])
{
double xu = 0.0 , yu = 0.0 , u = 0.0 ;
int i = 0 ;
for(u = 0.0 ; u <= 1.0 ; u += 0.05)
{
xu = pow(1-u,3)*x[0]+3*u*pow(1-u,2)*x[1]+3*pow(u,2)*(1-u)*x[2]
+pow(u,3)*x[3];
yu = pow(1-u,3)*y[0]+3*u*pow(1-u,2)*y[1]+3*pow(u,2)*(1-u)*y[2]
+pow(u,3)*y[3];
printf("X: %i Y: %i \n" , (int)xu , (int)yu) ;
}
}
int main(void) {
int x[] = {0,75,50,300};
int y[] = {0,2,140,100};
bezierCurve(x,y);
return 0;
}
https://ideone.com/glLXcB
Just a note: If you are using the usual formulas presented here then don't expect t = 0.5 to return the point at half of the curve's length.. In most cases it won't.
More on this here under "§23 — Tracing a curve at fixed distance intervals" and here.

Area of Intersection between Two Circles

Given two circles:
C1 at (x1, y1) with radius1
C2 at (x2, y2) with radius2
How do you calculate the area of their intersection? All standard math functions (sin, cos, etc.) are available, of course.
Okay, using the Wolfram link and Misnomer's cue to look at equation 14, I have derived the following Java solution using the variables I listed and the distance between the centers (which can trivially be derived from them):
double r = radius1;
double R = radius2;
double d = distance;
if(R < r){
// swap
r = radius2;
R = radius1;
}
double part1 = r*r*Math.acos((d*d + r*r - R*R)/(2*d*r));
double part2 = R*R*Math.acos((d*d + R*R - r*r)/(2*d*R));
double part3 = 0.5*Math.sqrt((-d+r+R)*(d+r-R)*(d-r+R)*(d+r+R));
double intersectionArea = part1 + part2 - part3;
Here is a JavaScript function that does exactly what Chris was after:
function areaOfIntersection(x0, y0, r0, x1, y1, r1)
{
var rr0 = r0 * r0;
var rr1 = r1 * r1;
var d = Math.sqrt((x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0));
var phi = (Math.acos((rr0 + (d * d) - rr1) / (2 * r0 * d))) * 2;
var theta = (Math.acos((rr1 + (d * d) - rr0) / (2 * r1 * d))) * 2;
var area1 = 0.5 * theta * rr1 - 0.5 * rr1 * Math.sin(theta);
var area2 = 0.5 * phi * rr0 - 0.5 * rr0 * Math.sin(phi);
return area1 + area2;
}
However, this method will return NaN if one circle is completely inside the other, or they are not touching at all. A slightly different version that doesn't fail in these conditions is as follows:
function areaOfIntersection(x0, y0, r0, x1, y1, r1)
{
var rr0 = r0 * r0;
var rr1 = r1 * r1;
var d = Math.sqrt((x1 - x0) * (x1 - x0) + (y1 - y0) * (y1 - y0));
// Circles do not overlap
if (d > r1 + r0)
{
return 0;
}
// Circle1 is completely inside circle0
else if (d <= Math.abs(r0 - r1) && r0 >= r1)
{
// Return area of circle1
return Math.PI * rr1;
}
// Circle0 is completely inside circle1
else if (d <= Math.abs(r0 - r1) && r0 < r1)
{
// Return area of circle0
return Math.PI * rr0;
}
// Circles partially overlap
else
{
var phi = (Math.acos((rr0 + (d * d) - rr1) / (2 * r0 * d))) * 2;
var theta = (Math.acos((rr1 + (d * d) - rr0) / (2 * r1 * d))) * 2;
var area1 = 0.5 * theta * rr1 - 0.5 * rr1 * Math.sin(theta);
var area2 = 0.5 * phi * rr0 - 0.5 * rr0 * Math.sin(phi);
// Return area of intersection
return area1 + area2;
}
}
I wrote this function by reading the information found at the Math Forum. I found this clearer than the Wolfram MathWorld explanation.
You might want to check out this analytical solution and apply the formula with your input values.
Another Formula is given here for when the radii are equal:
Area = r^2*(q - sin(q)) where q = 2*acos(c/2r),
where c = distance between centers and r is the common radius.
Here is an example in Python.
"""Intersection area of two circles"""
import math
from dataclasses import dataclass
from typing import Tuple
#dataclass
class Circle:
x: float
y: float
r: float
#property
def coord(self):
return self.x, self.y
def find_intersection(c1: Circle, c2: Circle) -> float:
"""Finds intersection area of two circles.
Returns intersection area of two circles otherwise 0
"""
d = math.dist(c1.coord, c2.coord)
rad1sqr = c1.r ** 2
rad2sqr = c2.r ** 2
if d == 0:
# the circle centers are the same
return math.pi * min(c1.r, c2.r) ** 2
angle1 = (rad1sqr + d ** 2 - rad2sqr) / (2 * c1.r * d)
angle2 = (rad2sqr + d ** 2 - rad1sqr) / (2 * c2.r * d)
# check if the circles are overlapping
if (-1 <= angle1 < 1) or (-1 <= angle2 < 1):
theta1 = math.acos(angle1) * 2
theta2 = math.acos(angle2) * 2
area1 = (0.5 * theta2 * rad2sqr) - (0.5 * rad2sqr * math.sin(theta2))
area2 = (0.5 * theta1 * rad1sqr) - (0.5 * rad1sqr * math.sin(theta1))
return area1 + area2
elif angle1 < -1 or angle2 < -1:
# Smaller circle is completely inside the largest circle.
# Intersection area will be area of smaller circle
# return area(c1_r), area(c2_r)
return math.pi * min(c1.r, c2.r) ** 2
return 0
if __name__ == "__main__":
#dataclass
class Test:
data: Tuple[Circle, Circle]
expected: float
tests = [
Test((Circle(2, 4, 2), Circle(3, 9, 3)), 0),
Test((Circle(0, 0, 2), Circle(-1, 1, 2)), 7.0297),
Test((Circle(1, 3, 2), Circle(1, 3, 2.19)), 12.5664),
Test((Circle(0, 0, 2), Circle(-1, 0, 2)), 8.6084),
Test((Circle(4, 3, 2), Circle(2.5, 3.5, 1.4)), 3.7536),
Test((Circle(3, 3, 3), Circle(2, 2, 1)), 3.1416)
]
for test in tests:
result = find_intersection(*test.data)
assert math.isclose(result, test.expected, rel_tol=1e-4), f"{test=}, {result=}"
print("PASSED!!!")
Here here i was making character generation tool, based on circle intersections... you may find it useful.
with dynamically provided circles:
C: {
C1: {id: 'C1',x:105,y:357,r:100,color:'red'},
C2: {id: 'C2',x:137,y:281,r:50, color:'lime'},
C3: {id: 'C3',x:212,y:270,r:75, color:'#00BCD4'}
},
Check FULL fiddle...
FIDDLE

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