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Is there efficient way to find intersection between two vectors?
Ray is infinite so one way is to turn one vector to ray, find intersection between that ray and other vector. Then if there is intersection, find distance between intersection and first vector.
If it is 0 then there is intersection between two vectors, if i don't miss something.
But is there better way in three.js?
That is more a math question than a three.js one, as you mentioned you could use existing methods on the Ray object and work your way out from there, but I doubt this would be optimal as you are concerned about performance.
The right approach is probably to determine the shortest distance between two segments and if this distance is zero then they intersect. You can find a similar discussion on that thread: The Algorithm to Find the Point of Intersection of Two 3D Line Segment
And here I found a C algorithm that does exactly that, it should translate to Javascript without much efforts:
typedef struct {
double x,y,z;
} XYZ;
/*
Calculate the line segment PaPb that is the shortest route between
two lines P1P2 and P3P4. Calculate also the values of mua and mub where
Pa = P1 + mua (P2 - P1)
Pb = P3 + mub (P4 - P3)
Return FALSE if no solution exists.
*/
int LineLineIntersect(
XYZ p1,XYZ p2,XYZ p3,XYZ p4,XYZ *pa,XYZ *pb,
double *mua, double *mub)
{
XYZ p13,p43,p21;
double d1343,d4321,d1321,d4343,d2121;
double numer,denom;
p13.x = p1.x - p3.x;
p13.y = p1.y - p3.y;
p13.z = p1.z - p3.z;
p43.x = p4.x - p3.x;
p43.y = p4.y - p3.y;
p43.z = p4.z - p3.z;
if (ABS(p43.x) < EPS && ABS(p43.y) < EPS && ABS(p43.z) < EPS)
return(FALSE);
p21.x = p2.x - p1.x;
p21.y = p2.y - p1.y;
p21.z = p2.z - p1.z;
if (ABS(p21.x) < EPS && ABS(p21.y) < EPS && ABS(p21.z) < EPS)
return(FALSE);
d1343 = p13.x * p43.x + p13.y * p43.y + p13.z * p43.z;
d4321 = p43.x * p21.x + p43.y * p21.y + p43.z * p21.z;
d1321 = p13.x * p21.x + p13.y * p21.y + p13.z * p21.z;
d4343 = p43.x * p43.x + p43.y * p43.y + p43.z * p43.z;
d2121 = p21.x * p21.x + p21.y * p21.y + p21.z * p21.z;
denom = d2121 * d4343 - d4321 * d4321;
if (ABS(denom) < EPS)
return(FALSE);
numer = d1343 * d4321 - d1321 * d4343;
*mua = numer / denom;
*mub = (d1343 + d4321 * (*mua)) / d4343;
pa->x = p1.x + *mua * p21.x;
pa->y = p1.y + *mua * p21.y;
pa->z = p1.z + *mua * p21.z;
pb->x = p3.x + *mub * p43.x;
pb->y = p3.y + *mub * p43.y;
pb->z = p3.z + *mub * p43.z;
return(TRUE);
}
I'm doing Quaternion mixing via SLERP, but my implementation does something very buggy. You can see it on this video: video
Here is the code:
cosTheta = quatDot(a, b);
if (cosTheta > 0.9){
var quat = new Vec4(
a.x + (b.x - a.x)*t,
a.y + (b.y - a.y)*t,
a.z + (b.z - a.z)*t,
a.w + (b.w - a.w)*t
);
normalizeQuat(quat);
return quat;
}
cosTheta = Math.min(Math.max(cosTheta, -1), 1); //clamp
var theta = Math.acos(cosTheta) * t;
var v2 = quatMinus(b, quatExtend(a, cosTheta));
normalizeQuat(v2);
return quatSum(quatExtend(a, cosTheta), quatExtend(v2, Math.sin(theta)));
Here I have my help functions:
function quatDot(a,b){
return a.x*b.x + a.y*b.y + a.z*b.z + a.w*b.w;
return a.x*b.x + a.y*b.y + a.z*b.z + a.w*b.w;
}
function quatMultiply(q,p){
return new Vec4(
q.w * p.x + q.x * p.w + q.y * p.z - q.z * p.y,
q.w * p.y + q.y * p.w + q.z * p.x - q.x * p.z,
q.w * p.z + q.z * p.w + q.x * p.y - q.y * p.x,
q.w * p.w - q.x * p.x - q.y * p.y - q.z * p.z);
}
function quatExtend(q, t){
return new Vec4(
q.x * t, q.y * t, q.z * t, q.w * t);
}
function quatMinus(q, p){
return new Vec4(
q.x - p.x,
q.y - p.y,
q.z - p.z,
q.w - p.w
);
}
function quatSum(q, p){
return new Vec4(
q.x + p.x,
q.y + p.y,
q.z + p.z,
q.w + p.w
);
}
I tried many implementations from websites but there is always buggy movement. When I tried simple linear interpolation, then animation is smooth, but weird accelerating.
Here's a easy bug: if cosTheta < 0.9 and t= 0, then theta = 0, which means Math.sin(theta) = 0, which means quatExtend(v2, Math.sin(theta)) is the zero quaternion, and you end up returning quatExtend(a, cosTheta), which is not a unit quaternion.
I would expect a correct implementation to take the inverse (or conjugate) of one of the quaternions at least once, and never to call normalizeQuat. But that's just intuition. To get past intuition, try writing a unit test!
By the way, if we're doing a code review, it's confusing that cosTheta is not the cosine of theta. Also, the if (cosTheta > 0.9) block smells like premature optimization; you can probably get rid of it. :)
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 };
I'm working on an Android game and I need to bounce 2 circles of each other (like 2 pool balls bouncing off each other). The collision is an elastic collision, and I need to calculate only 1 circles (called a Particle in my code) new velocity after the collision (the other circle, called a Barrier in my code will remain stationary and will not move because of a collision).
I am using a formula that I found on Wikipedia (http://en.wikipedia.org/wiki/Elastic_collision), but my end result for the new velocity of the particle after the collision is exactly the same as the velocity before the collision?
This is def wrong but I cant see where I am going wrong. Can anyone spot where I am going wrong?
I have just used a Java program to simulate my velocities and locations for the 2 circles as I dont wanna try it in my main Android game at the moment for fear of "breaking something"
Here is what I have so far (like I mentioned, this is just a simulation in NetBeans for the moment and I will use a menthod in my Android game to keep things a bit tidier):
double randomNextDouble = (new Random()).nextDouble();
System.out.println("Random.nextDouble: " + randomNextDouble);
double mathPI = Math.PI * 2;
System.out.println("Math PI: " + mathPI);
// get a random direction for the Particle to move towards
double direction = (new Random()).nextDouble() * Math.PI * 2;
System.out.println("Direction: " + direction);
// Then set the Particle's velocity - Increase to make Particles move faster
int velocity = 10;
System.out.println("Velocity: " + velocity);
// Then calculate the xv and the yv
// Velocity value on the x and y axis
double xv = (velocity * Math.cos(direction));
double yv = (velocity * Math.sin(direction));
System.out.println("\nXV: " + xv + "\nYV: " + yv);
// Genareting a random number for the Particle and Barrier's positions on screen
double Xmin = 0;
double Xmax = 300;
double Ymin = 0;
double Ymax = 300;
double randomNumber1 = Xmin + (int)(Math.random() * ((Xmax - Xmin) + 1));
double randomNumber2 = Ymin + (int)(Math.random() * ((Ymax - Ymin) + 1));
double randomNumber3 = Xmin + (int)(Math.random() * ((Xmax - Xmin) + 1));
double randomNumber4 = Ymin + (int)(Math.random() * ((Ymax - Ymin) + 1));
// Setting the Particle and Barrier's radius
double particleRadius = 8;
double barrierRadius = 16;
// Setting up the Particle and Barrier's mass
double particleMass = 100;
double barrierMass = 200;
// Assigning a random number to the Particle to simulate its position on screen
double particleX = randomNumber1;
double particleY = randomNumber2;
System.out.println("\nParticle X: " + particleX + " Particle Y: " + particleY);
// Assigning a random number to the Barrier to simulate its position on screen
double barrierX = randomNumber3;
double barrierY = randomNumber4;
System.out.println("Barrier X: " + barrierX + " Barrier Y: " + barrierY);
double distanceXToBarrier = barrierX - particleX;
System.out.println("\nBarrier X - Particle X: " + distanceXToBarrier);
double distanceYToBarrier = barrierY - particleY;
System.out.println("Barrier Y - Particle Y: " + distanceYToBarrier);
// Get the distance between the Particle and the Barrier
// Used for collision detection
double distance = Math.sqrt((distanceXToBarrier * distanceXToBarrier) + (distanceYToBarrier * distanceYToBarrier));
System.out.println("\nDistance: " + distance);
// Check to see if the Particle and Barrier has collided
if (distance <= particleRadius + barrierRadius)
{
System.out.println("Distance is less than 2 Radii");
}
else
System.out.println("Distance is NOT less than 2 Radii");
// Velx = (v1.u) * u + (v1 - (v1.u) * u)
// Vely = (v1.u) * u + (v1 - (v1.u) * u)
// Where v1 = xv and yv respectively
// Break it into 2 equations
// (v1.u) * u AND
// (v1 - (v1.u) * u)
//
// u = normalised Vector
// To normalize you just devide the x, y, z coords by the length of the vector.
// This then gives you the Unit Vector.
//
//Normalize the vector
double particleXNormalized = particleX * (1.0 / distance);
double particleYNormalized = particleY * (1.0 / distance);
System.out.println("\nParticle X Normalised: " + particleXNormalized +
"\nParticle Y Normalised: " + particleYNormalized);
// Calculating the first part of the eqaution
// (v1.u)
double v1DotUForX = xv * particleXNormalized;
double v1DotUForY = yv * particleYNormalized;
System.out.println("\nv1.u for X: " + v1DotUForX +
"\nv1.u for Y: " + v1DotUForY);
// The first part of the equation
// (v1.u) * u
double part1X = v1DotUForX * particleXNormalized;
double part1Y = v1DotUForY * particleYNormalized;
System.out.println("\nPart 1 for X: " + part1X +
"\nPart 1 for Y: " + part1Y);
// The second part of the equation
// (v1 - (v1.u) * u)
double part2X = (xv - (v1DotUForX) * particleXNormalized);
double part2Y = (yv - (v1DotUForY) * particleYNormalized);
System.out.println("\nPart 2 for X: " + part2X +
"\nPart 2 for Y: " + part2Y);
// Solving for:
// (((mass 1 - mass2) / (mass1 + mass2) * (v1.u) * u + ((2mass2) / (mass1 + mass2) * ((v1.u) * u))) +
// (v1 - (v1.u) * u))
double newXV = ((((particleMass - barrierMass) / (particleMass + barrierMass)) * part1X) + (((2 * barrierMass) / (particleMass + barrierMass)) * part1X) + part2X);
double newYV = ((((particleMass - barrierMass) / (particleMass + barrierMass)) * part1Y) + (((2 * barrierMass) / (particleMass + barrierMass)) * part1Y) + part2Y);
System.out.println("\nNew XV: " + newXV + "\nNew YV: " + newYV);
Looking at your algorithm, you appear to have made errors in the implementation. Why are you normalizing the coordinates of the particle? Shouldn't you be doing that to the velocity? In the usual equations, u is velocity, not position.
And why do you give the particle a random velocity (xv, yv) that has nothing to do with the two random coordinates you set up for the particle and barrier? (Surely the velocity should be some multiple of (barrier - particle) vector?)
There is a special way of mapping a cube to a sphere described here:
http://mathproofs.blogspot.com/2005/07/mapping-cube-to-sphere.html
It is not your basic "normalize the point and you're done" approach and gives a much more evenly spaced mapping.
I've tried to do the inverse of the mapping going from sphere coords to cube coords and have been unable to come up the working equations. It's a rather complex system of equations with lots of square roots.
Any math geniuses want to take a crack at it?
Here's the equations in c++ code:
sx = x * sqrtf(1.0f - y * y * 0.5f - z * z * 0.5f + y * y * z * z / 3.0f);
sy = y * sqrtf(1.0f - z * z * 0.5f - x * x * 0.5f + z * z * x * x / 3.0f);
sz = z * sqrtf(1.0f - x * x * 0.5f - y * y * 0.5f + x * x * y * y / 3.0f);
sx,sy,sz are the sphere coords and x,y,z are the cube coords.
I want to give gmatt credit for this because he's done a lot of the work. The only difference in our answers is the equation for x.
To do the inverse mapping from sphere to cube first determine the cube face the sphere point projects to. This step is simple - just find the component of the sphere vector with the greatest length like so:
// map the given unit sphere position to a unit cube position
void cubizePoint(Vector3& position) {
double x,y,z;
x = position.x;
y = position.y;
z = position.z;
double fx, fy, fz;
fx = fabsf(x);
fy = fabsf(y);
fz = fabsf(z);
if (fy >= fx && fy >= fz) {
if (y > 0) {
// top face
position.y = 1.0;
}
else {
// bottom face
position.y = -1.0;
}
}
else if (fx >= fy && fx >= fz) {
if (x > 0) {
// right face
position.x = 1.0;
}
else {
// left face
position.x = -1.0;
}
}
else {
if (z > 0) {
// front face
position.z = 1.0;
}
else {
// back face
position.z = -1.0;
}
}
}
For each face - take the remaining cube vector components denoted as s and t and solve for them using these equations, which are based on the remaining sphere vector components denoted as a and b:
s = sqrt(-sqrt((2 a^2-2 b^2-3)^2-24 a^2)+2 a^2-2 b^2+3)/sqrt(2)
t = sqrt(-sqrt((2 a^2-2 b^2-3)^2-24 a^2)-2 a^2+2 b^2+3)/sqrt(2)
You should see that the inner square root is used in both equations so only do that part once.
Here's the final function with the equations thrown in and checks for 0.0 and -0.0 and the code to properly set the sign of the cube component - it should be equal to the sign of the sphere component.
void cubizePoint2(Vector3& position)
{
double x,y,z;
x = position.x;
y = position.y;
z = position.z;
double fx, fy, fz;
fx = fabsf(x);
fy = fabsf(y);
fz = fabsf(z);
const double inverseSqrt2 = 0.70710676908493042;
if (fy >= fx && fy >= fz) {
double a2 = x * x * 2.0;
double b2 = z * z * 2.0;
double inner = -a2 + b2 -3;
double innersqrt = -sqrtf((inner * inner) - 12.0 * a2);
if(x == 0.0 || x == -0.0) {
position.x = 0.0;
}
else {
position.x = sqrtf(innersqrt + a2 - b2 + 3.0) * inverseSqrt2;
}
if(z == 0.0 || z == -0.0) {
position.z = 0.0;
}
else {
position.z = sqrtf(innersqrt - a2 + b2 + 3.0) * inverseSqrt2;
}
if(position.x > 1.0) position.x = 1.0;
if(position.z > 1.0) position.z = 1.0;
if(x < 0) position.x = -position.x;
if(z < 0) position.z = -position.z;
if (y > 0) {
// top face
position.y = 1.0;
}
else {
// bottom face
position.y = -1.0;
}
}
else if (fx >= fy && fx >= fz) {
double a2 = y * y * 2.0;
double b2 = z * z * 2.0;
double inner = -a2 + b2 -3;
double innersqrt = -sqrtf((inner * inner) - 12.0 * a2);
if(y == 0.0 || y == -0.0) {
position.y = 0.0;
}
else {
position.y = sqrtf(innersqrt + a2 - b2 + 3.0) * inverseSqrt2;
}
if(z == 0.0 || z == -0.0) {
position.z = 0.0;
}
else {
position.z = sqrtf(innersqrt - a2 + b2 + 3.0) * inverseSqrt2;
}
if(position.y > 1.0) position.y = 1.0;
if(position.z > 1.0) position.z = 1.0;
if(y < 0) position.y = -position.y;
if(z < 0) position.z = -position.z;
if (x > 0) {
// right face
position.x = 1.0;
}
else {
// left face
position.x = -1.0;
}
}
else {
double a2 = x * x * 2.0;
double b2 = y * y * 2.0;
double inner = -a2 + b2 -3;
double innersqrt = -sqrtf((inner * inner) - 12.0 * a2);
if(x == 0.0 || x == -0.0) {
position.x = 0.0;
}
else {
position.x = sqrtf(innersqrt + a2 - b2 + 3.0) * inverseSqrt2;
}
if(y == 0.0 || y == -0.0) {
position.y = 0.0;
}
else {
position.y = sqrtf(innersqrt - a2 + b2 + 3.0) * inverseSqrt2;
}
if(position.x > 1.0) position.x = 1.0;
if(position.y > 1.0) position.y = 1.0;
if(x < 0) position.x = -position.x;
if(y < 0) position.y = -position.y;
if (z > 0) {
// front face
position.z = 1.0;
}
else {
// back face
position.z = -1.0;
}
}
So, this solution isn't nearly as pretty as the cube to sphere mapping, but it gets the job done!
Any suggestions to improve the efficiency or read ability of the code above are appreciated!
--- edit ---
I should mention that I have tested this and so far in my tests the code appears correct with the results being accurate to at least the 7th decimal place. And that was from when I was using floats, it's probably more accurate now with doubles.
--- edit ---
Here's an optimized glsl fragment shader version by Daniel to show that it doesn't have to be such a big scary function. Daniel uses this to filter sampling on cube maps! Great idea!
const float isqrt2 = 0.70710676908493042;
vec3 cubify(const in vec3 s)
{
float xx2 = s.x * s.x * 2.0;
float yy2 = s.y * s.y * 2.0;
vec2 v = vec2(xx2 – yy2, yy2 – xx2);
float ii = v.y – 3.0;
ii *= ii;
float isqrt = -sqrt(ii – 12.0 * xx2) + 3.0;
v = sqrt(v + isqrt);
v *= isqrt2;
return sign(s) * vec3(v, 1.0);
}
vec3 sphere2cube(const in vec3 sphere)
{
vec3 f = abs(sphere);
bool a = f.y >= f.x && f.y >= f.z;
bool b = f.x >= f.z;
return a ? cubify(sphere.xzy).xzy : b ? cubify(sphere.yzx).zxy : cubify(sphere);
}
After some rearranging you can get the "nice" forms
(1) 1/2 z^2 = (alpha) / ( y^2 - x^2) + 1
(2) 1/2 y^2 = (beta) / ( z^2 - x^2) + 1
(3) 1/2 x^2 = (gamma) / ( y^2 - z^2) + 1
where alpha = sx^2-sy^2 , beta = sx^2 - sz^2 and gamma = sz^2 - sy^2. Verify this yourself.
Now I neither have the motivation nor the time but from this point on its pretty straightforward to solve:
Substitute (1) into (2). Rearrange (2) until you get a polynomial (root) equation of the form
(4) a(x) * y^4 + b(x) * y^2 + c(x) = 0
this can be solved using the quadratic formula for y^2. Note that a(x),b(x),c(x) are some functions of x. The quadratic formula yields 2 roots for (4) which you will have to keep in mind.
Using (1),(2),(4) figure out an expression for z^2 in terms of only x^2.
Using (3) write out a polynomial root equation of the form:
(5) a * x^4 + b * x^2 + c = 0
where a,b,c are not functions but constants. Solve this using the quadratic formula. In total you will have 2*2=4 possible solutions for x^2,y^2,z^2 pair meaning you will
have 4*2=8 total solutions for possible x,y,z pairs satisfying these equations. Check conditions on each x,y,z pair and (hopefully) eliminate all but one (otherwise an inverse mapping does not exist.)
Good luck.
PS. It very well may be that the inverse mapping does not exist, think about the geometry: the sphere has surface area 4*pi*r^2 while the cube has surface area 6*d^2=6*(2r)^2=24r^2 so intuitively you have many more points on the cube that get mapped to the sphere. This means a many to one mapping, and any such mapping is not injective and hence is not bijective (i.e. the mapping has no inverse.) Sorry but I think you are out of luck.
----- edit --------------
if you follow the advice from MO, setting z=1 means you are looking at the solid square in the plane z=1.
Use your first two equations to solve for x,y, wolfram alpha gives the result:
x = (sqrt(6) s^2 sqrt(1/2 (sqrt((2 s^2-2 t^2-3)^2-24 t^2)+2 s^2-2 t^2-3)+3)-sqrt(6) t^2 sqrt(1/2 (sqrt((2 s^2-2 t^2-3)^2-24 t^2)+2 s^2-2 t^2-3)+3)-sqrt(3/2) sqrt((2 s^2-2 t^2-3)^2-24 t^2) sqrt(1/2 (sqrt((2 s^2-2 t^2-3)^2-24 t^2)+2 s^2-2 t^2-3)+3)+3 sqrt(3/2) sqrt(1/2 (sqrt((2 s^2-2 t^2-3)^2-24 t^2)+2 s^2-2 t^2-3)+3))/(6 s)
and
y = sqrt(-sqrt((2 s^2-2 t^2-3)^2-24 t^2)-2 s^2+2 t^2+3)/sqrt(2)
where above I use s=sx and t=sy, and I will use u=sz. Then you can use the third equation you have for u=sz. That is lets say that you want to map the top part of the sphere to the cube. Then for any 0 <= s,t <= 1 (where s,t are in the sphere's coordinate frame ) then the tuple (s,t,u) maps to (x,y,1) (here x,y are in the cubes coordinate frame.) The only thing left is for you to figure out what u is. You can figure this out by using s,t to solve for x,y then using x,y to solve for u.
Note that this will only map the top part of the cube to only the top plane of the cube z=1. You will have to do this for all 6 sides (x=1, y=1, z=0 ... etc ). I suggest using wolfram alpha to solve the resulting equations you get for each sub-case, because they will be as ugly or uglier as those above.
This answer contains the cube2sphere and sphere2cube without the restriction of a = 1. So the cube has side 2a from -a to a and the radius of the sphere is a.
I know it's been 10 years since this question was asked. Nevertheless, I am giving the answer in case someone needs it. The implementation is in Python,
I am using (x, y, z) for the cube coordinates, (p, q, r) for the sphere coordinates and the relevant underscore variables (x_, y_, z_) meaning they have been produced by using the inverse function.
import math
from random import randint # for testing
def sign_aux(x):
return lambda y: math.copysign(x, y)
sign = sign_aux(1) # no built-in sign function in python, I know...
def cube2sphere(x, y, z):
if (all([x == 0, y == 0, z == 0])):
return 0, 0, 0
def aux(x, y_2, z_2, a, a_2):
return x * math.sqrt(a_2 - y_2/2 - z_2/2 + y_2*z_2/(3*a_2))/a
x_2 = x*x
y_2 = y*y
z_2 = z*z
a = max(abs(x), abs(y), abs(z))
a_2 = a*a
return aux(x, y_2, z_2, a, a_2), aux(y, x_2, z_2, a, a_2), aux(z, x_2, y_2, a, a_2)
def sphere2cube(p, q, r):
if (all([p == 0, q == 0, r == 0])):
return 0, 0, 0
def aux(s, t, radius):
A = 3*radius*radius
R = 2*(s*s - t*t)
S = math.sqrt( max(0, (A+R)*(A+R) - 8*A*s*s) ) # use max 0 for accuraccy error
iot = math.sqrt(2)/2
s_ = sign(s) * iot * math.sqrt(max(0, A + R - S)) # use max 0 for accuraccy error
t_ = sign(t) * iot * math.sqrt(max(0, A - R - S)) # use max 0 for accuraccy error
return s_, t_
norm_p, norm_q, norm_r = abs(p), abs(q), abs(r)
norm_max = max(norm_p, norm_q, norm_r)
radius = math.sqrt(p*p + q*q + r*r)
if (norm_max == norm_p):
y, z = aux(q, r, radius)
x = sign(p) * radius
return x, y, z
if (norm_max == norm_q):
z, x = aux(r, p, radius)
y = sign(q) * radius
return x, y, z
x, y = aux(p, q, radius)
z = sign(r) * radius
return x, y, z
# measuring accuracy
max_mse = 0
for i in range(100000):
x = randint(-20, 20)
y = randint(-20, 20)
z = randint(-20, 20)
p, q, r = cube2sphere(x, y, z)
x_, y_, z_ = sphere2cube(p, q, r)
max_mse = max(max_mse, math.sqrt(((x-x_)**2 + (y-y_)**2 + (z-z_)**2))/3)
print(max_mse)
# 1.1239159602905078e-07
max_mse = 0
for i in range(100000):
p = randint(-20, 20)
q = randint(-20, 20)
r = randint(-20, 20)
x, y, z = sphere2cube(p, q, r)
p_, q_, r_ = cube2sphere(x, y, z)
max_mse = max(max_mse, math.sqrt(((p-p_)**2 + (q-q_)**2 + (r-r_)**2))/3)
print(max_mse)
# 9.832883321715792e-08
Also, I mapped some points to check the function visually and these are the results.
Here's one way you can think about it: for a given point P in the sphere, take the segment that starts at the origin, passes through P, and ends at the surface of the cube. Let L be the length of this segment. Now all you need to do is multiply P by L; this is equivalent to mapping ||P|| from the interval [0, 1] to the interval [0, L]. This mapping should be one-to-one - every point in the sphere goes to a unique point in the cube (and points on the surface stay on the surface). Note that this is assuming a unit sphere and cube; the idea should hold elsewhere, you'll just have a few scale factors involved.
I've glossed over the hard part (finding the segment), but this is a standard raycasting problem. There are some links here that explain how to compute this for an arbitrary ray versus axis-aligned bounding box; you can probably simplify things since your ray starts at the origin and goes to the unit cube. If you need help simplify the equations, let me know and I'll take a stab at it.
It looks like there is a much cleaner solution if you're not afraid of trig and pi, not sure if it's faster/comparable though.
Just take the remaining components after determining the face and do:
u = asin ( x ) / half_pi
v = asin ( y ) / half_pi
This is an intuitive leap, but this seems to back it up ( though not exactly the same topic ), so please correct me if I'm wrong.
I'm too lazy to post an illustration that explains why. :D