I have recently tried to change the current way I calculate diffuse lighting in my RayTracer. It used to be calculated like this:
float lambert = (light_ray_dir * normal) * coef;
red += lambert * current.color.red * mat.kd.red;
green += lambert * current.color.green * mat.kd.green;
blue += lambert * current.color.blue * mat.kd.blue;
where coef is an attenuation coefficient that starts at 1 for each pixel and is then attenuated for each reflected ray that is generated by this line:
coef *= mat.reflection;
This worked well.
But I decided to try something more realistic and implemented this:
float squared_attenuation = LIGHT_FALLOFF * lenght;
light_intensity.setX ((current.color.red /*INTENSITY*/)/ squared_attenuation);
light_intensity.setY ((current.color.green /*INTENSITY*/)/ squared_attenuation);
light_intensity.setZ ((current.color.blue /*INTENSITY*/)/ squared_attenuation);
red += ALBEDO * light_intensity.getX() * lambert * mat.kd.red;
green += ALBEDO * light_intensity.getY() * lambert * mat.kd.green;
blue += ALBEDO * light_intensity.getZ() * lambert * mat.kd.blue;
where LIGHT_FALLOFF it is a constant value:
#define LIGHT_FALLOFF M_PI * 4
and length it is the length of the vector that goes from the point light center to the intersect point:
inline float normalize_return_lenght () {
float lenght = sqrtf (x*x + y*y + z*z);
float inv_length = 1/lenght;
x = x * inv_length, y = y * inv_length, z = z * inv_length;
return lenght;
}
float lenght = light_ray_dir.normalize_return_lenght ();
The problem is that all this is generating nothing more than a black screen! The main culprit is the length that goes as the divisor in the light_intensity.set. It makes the final color values being some value ^ -5. However, even if I replace it by one (ruining my goal of a realistic light attenuation), I still get color values to close to zero still, hence a black image.
I tried to add another light_sources closer to the objects, however the fact that the models that shall be shaded are made of multiple polygons with different coordinates, make hard to determine a goodl ocation for them.
So I ask. It seems normal to you that this is happening or it seems a bug? For me theory, does not seems strange for me, since the attenuation is quadratic.
Is does not seem, there is a some hint as to where to place the light sources, or to anything as can get an image that is not all black?
Thanks for reading all this!
P.S: Intensity is commented out because on the example that I used to do my code, it was one
So you are assuming that the light has a luminosity of "1" ?
How far away is your light?
If your light is - say - 10 units away, then they contribution from the light will be 1/10, or a very small number. This is probably why your image is dark.
You need to have quite large numbers for your light intensity if you are going to do this. In one of my scenes, I have a light that is about 1000 units away (pretending to be the Sun) and the intensity is 380000!!
Another thing ... to simulate reality, you should be using 1 / length^2. The light intensity falls off with the square of distance, not just with distance.
Good luck!
I am currently working on a project called "Raytracer" in c.
I encounter a problem, the spheres are oval when they are not centered.
Here is an excerpt of my code:
int i;
int j;
t_ray vect;
i = -1;
vect.x = 100. - cam.x;
while (++i < screenx)
{
j = -1;
vect.y = ((screenx / 2.) - i - cam.y) * -1.;
while (++j < screeny)
{
vect.z = (screeny / 2.) - j - cam.z;
}
}
This is likely not a bug, but simply a reality of how perspective projections work. When the camera is directly looking at a sphere, the projection is circular, but as it moves away from the center, it distorts. For more info read this link in the POV-Ray wiki: http://wiki.povray.org/content/Knowledgebase:Misconceptions#Topic_3
In that way the vector has different length on different pixels. You should normalize the vector at the end (dividing the components by the vector length)
It's probably late now, but to give you an answer, your "problem" is in reality called "fish-eye". I encounted this problem too. there're many ways to avoid this problem. The easiest is to increase the distance between the camera and your scene. It's not the cleaner way.
You also can normalize your rays, here are some reasons :
.keep the same distance ratio for every rays
.keep the same angle difference between every ray and its neighbors
.it makes many intersection computations ways easier
For a project we are trying to make a circle into a line (and back again) while it is rotating along a linear path, much like a tire rotates and translates when rolling on a road, or a curled fore finger is extended and recurled into the palm.
In this Fiddle, I have a static SVG (the top circle) that rotates along the linear black path (which is above the circle, to mimic a finger extending) that is defined in the HTML.
I also use d3 to generate a "circle" that is made up of connected points (and can unfurl if you click on/in the circle thanks to #ChrisJamesC here ), and is translated and rotated
in the function moveAlongLine when you click on the purple Line:
function moveAlongLine() {
circle.data([lineData])
.attr("transform", "translate(78.5,0) rotate(-90, 257.08 70) ")
.duration(1000)
circle.on("click", transitionToCircle)
}
The first problem is that the .duration(1000) is not recognized and throws a Uncaught TypeError: Object [object Array] has no method 'duration' in the console, so there is a difference between the static definition of dur in SVG and dynamically setting it in JS/D3, but this is minor.
The other is should the transform attributes be abstracted from one another like in the static circle? in the static circle, the translate is one animation, and the rotation is another, they just have the same star and duration, so they animate together. How would you apply both in d3?
The challenge that I can not get, is how to let it unroll upwards(and also re-roll back), with the static point being the top center of the circle also being the same as the leftmost point on the line.
these seem better:
I should try to get the unfurl animation to occur while also rotating? This seems like it would need to be stepwise/sequential based...
Or Consider an octogon (defined as a path), and if it were to rotate 7 of the sides, then 6, then 5.... Do this for a rather large number of points on a polyhedron? (the circle only needs to be around 50 or so pixels, so 100 points would be more than enough) This is the middle example in the fiddle. Maybe doing this programmatically?
Or This makes me think of a different way: (in the case of the octogon), I could have 8 line paths (with no Z, just an additional closing point), and transition between them? Like this fiddle
Or anything todo with keyframes? I have made an animation in Synfig, but am unsure ho get it to SVG. The synfig file is at http://specialorange.org/filedrop/unroll.sifz if you can convert to SVG, but the xsl file here doesn't correctly convert it for me using xsltproc.
this seems really complicated but potential:
Define a path (likely a bézier curve with the same number of reference points) that the points follow, and have the reference points dynamically translate as well... see this for an concept example
this seems complicated and clunky:
Make a real circle roll by, with a growing mask in front of it, all while a line grows in length
A couple of notes:
The number of points in the d3 circle can be adjusted in the JS, it is currently set low so that you can see a bit of a point in the rendering to verify the rotation has occurred (much like the gradient is in the top circle).
this is to help students learn what is conserved between a number line and a circle, specifically to help learn fractions. For concept application, take a look at compthink.cs.vt.edu:3000 to see our prototype, and this will help with switching representations, to help you get a better idea...
I ended up using the same function that generates the circle as in the question, and did a bit of thinking, and it seemed like I wanted an animation that looked like a finger unrolling like this fiddle. This lead me to the math and idea needed to make it happen in this fiddle.
The answer is an array of arrays, with each nested array being a line in the different state, and then animate by interpolating between the points.
var circleStates = [];
for (i=0; i<totalPoints; i++){
//circle portion
var circleState = $.map(Array(numberOfPoints), function (d, j) {
var x = marginleft + radius + lineDivision*i + radius * Math.sin(2 * j * Math.PI / (numberOfPoints - 1));
var y = margintop + radius - radius * Math.cos(2 * j * Math.PI / (numberOfPoints - 1));
return { x: x, y: y};
})
circleState.splice(numberOfPoints-i);
//line portion
var lineState = $.map(Array(numberOfPoints), function (d, j) {
var x = marginleft + radius + lineDivision*j;
var y = margintop;
return { x: x, y: y};
})
lineState.splice(i);
//together
var individualState = lineState.concat(circleState);
circleStates.push(individualState);
}
and the animation(s)
function all() {
for(i=0; i<numberOfPoints; i++){
circle.data([circleStates[i]])
.transition()
.delay(dur*i)
.duration(dur)
.ease("linear")
.attr('d', pathFunction)
}
}
function reverse() {
for(i=0; i<numberOfPoints; i++){
circle.data([circleStates[numberOfPoints-1-i]])
.transition()
.delay(dur*i)
.duration(dur)
.ease("linear")
.attr('d', pathFunction)
}
}
(Note: This should be in comments but not enough spacing)
Circle Animation
Try the radial wipe from SO. Need to tweak it so angle starts at 180 and ends back at same place (line#4-6,19) and move along the X-axis (line#11) on each interation. Change the <path... attribute to suit your taste.
Line Animation Grow a line from single point to the length (perimeter) of the circle.
Sync both animation so that it appears good on all browsers (major headache!).
we are programming a 2D game in XNA. Now we have polygons which define our level elements. They are triangulated such that we can easily render them. Now I would like to write a shader which renders the polygons as outlined textures. So in the middle of the polygon one would see the texture and on the border it should somehow glow.
My first idea was to walk along the polygon and draw a quad on each line segment with a specific texture. This works but looks strange for small corners where the textures are forced to overlap.
My second approach was to mark all border vertices with some kind of normal pointing out of the polygon. Passing this to the shader would interpolate the normals across edges of the triangulation and I could use the interpolated "normal" as a value for shading. I could not test it yet but would that work? A special property of the triangulation is that all vertices are on the border so there are no vertices inside the polygon.
Do you guys have a better idea for what I want to achieve?
Here A picture of what it looks right now with the quad solution:
You could render your object twice. A bigger stretched version behind the first one. Not that ideal since a complex object cannot be streched uniformly to create a border.
If you have access to your screen buffer you can render your glow components into a rendertarget and align a full-screen quad to your viewport and add a fullscreen 2D silhouette filter to it.
This way you gain perfect control over the edge by defining its radius, colour, blur. With additional output values such as the RGB values from the object render pass you can even have different advanced glows.
I think rendermonkey had some examples in their shader editor. Its definetly a good starting point to work with and try out things.
Propaply you want calclulate new border vertex list (easy fill example with triangle strip with originals). If you use constant border width and convex polygon its just:
B_new = B - (BtoA.normalised() + BtoC.normalised()).normalised() * width;
If not then it can go more complicated, there is my old but pretty universal solution:
//Helper function. To working right, need that v1 is before v2 in vetex list and vertexes are going to (anti???) cloclwise!
float vectorAngle(Vector2 v1, Vector2 v2){
float alfa;
if (!v1.isNormalised())
v1.normalise();
if (!v2.isNormalised())
v2.normalise();
alfa = v1.dotProduct(v2);
float help = v1.x;
v1.x = v1.y;
v1.y = -help;
float angle = Math::ACos(alfa);
if (v1.dotProduct(v2) < 0){
angle = -angle;
}
return angle;
}
//Normally dont use directly this!
Vector2 calculateBorderPoint(Vector2 vec1, Vector2 vec2, float width1, float width2){
vec1.normalise();
vec2.normalise();
float cos = vec1.dotProduct(vec2); //Calculates actually cosini of two (normalised) vectors (remember math lessons)
float csc = 1.0f / Math::sqrt(1.0f-cos*cos); //Calculates cosecant of angle, This return NaN if angle is 180!!!
//And rest of the magic
Vector2 difrence = (vec1 * csc * width2) + (vec2 * csc * width1);
//If you use just convex polygons (all angles < 180, = 180 not allowed in this case) just return value, and if not you need some more magic.
//Both of next things need ordered vertex lists!
//Output vector is always to in side of angle, so if this angle is.
if (Math::vectorAngle(vec1, vec2) > 180.0f) //Note that this kind of function can know is your function can know that angle is over 180 ONLY if you use ordered vertexes (all vertexes goes always (anti???) cloclwise!)
difrence = -difrence;
//Ok and if angle was 180...
//Note that this can fix your situation ONLY if you use ordered vertexes (all vertexes goes always (anti???) cloclwise!)
if (difrence.isNaN()){
float width = (width1 + width2) / 2.0; //If angle is 180 and border widths are difrent, you cannot get perfect answer ;)
difrence = vec1 * width;
//Just turn vector -90 degrees
float swapHelp = difrence.y
difrence.y = -difrence.x;
difrence.x = swapHelp;
}
//If you don't want output to be inside of old polygon but outside, just: "return -difrence;"
return difrence;
}
//Use this =)
Vector2 calculateBorderPoint(Vector2 A, Vector2 B, Vector2 C, float widthA, float widthB){
return B + calculateBorderPoint(A-B, C-B, widthA, widthB);
}
Your second approach can be possible...
mark the outer vertex (in border) with 1 and the inner vertex (inside) with 0.
in the pixel shader you can choose to highlight, those that its value is greater than 0.9f or 0.8f.
it should work.
With the help of the Stack Overflow community I've written a pretty basic-but fun physics simulator.
You click and drag the mouse to launch a ball. It will bounce around and eventually stop on the "floor".
My next big feature I want to add in is ball to ball collision. The ball's movement is broken up into a x and y speed vector. I have gravity (small reduction of the y vector each step), I have friction (small reduction of both vectors each collision with a wall). The balls honestly move around in a surprisingly realistic way.
I guess my question has two parts:
What is the best method to detect ball to ball collision?
Do I just have an O(n^2) loop that iterates over each ball and checks every other ball to see if it's radius overlaps?
What equations do I use to handle the ball to ball collisions? Physics 101
How does it effect the two balls speed x/y vectors? What is the resulting direction the two balls head off in? How do I apply this to each ball?
Handling the collision detection of the "walls" and the resulting vector changes were easy but I see more complications with ball-ball collisions. With walls I simply had to take the negative of the appropriate x or y vector and off it would go in the correct direction. With balls I don't think it is that way.
Some quick clarifications: for simplicity I'm ok with a perfectly elastic collision for now, also all my balls have the same mass right now, but I might change that in the future.
Edit: Resources I have found useful
2d Ball physics with vectors: 2-Dimensional Collisions Without Trigonometry.pdf
2d Ball collision detection example: Adding Collision Detection
Success!
I have the ball collision detection and response working great!
Relevant code:
Collision Detection:
for (int i = 0; i < ballCount; i++)
{
for (int j = i + 1; j < ballCount; j++)
{
if (balls[i].colliding(balls[j]))
{
balls[i].resolveCollision(balls[j]);
}
}
}
This will check for collisions between every ball but skip redundant checks (if you have to check if ball 1 collides with ball 2 then you don't need to check if ball 2 collides with ball 1. Also, it skips checking for collisions with itself).
Then, in my ball class I have my colliding() and resolveCollision() methods:
public boolean colliding(Ball ball)
{
float xd = position.getX() - ball.position.getX();
float yd = position.getY() - ball.position.getY();
float sumRadius = getRadius() + ball.getRadius();
float sqrRadius = sumRadius * sumRadius;
float distSqr = (xd * xd) + (yd * yd);
if (distSqr <= sqrRadius)
{
return true;
}
return false;
}
public void resolveCollision(Ball ball)
{
// get the mtd
Vector2d delta = (position.subtract(ball.position));
float d = delta.getLength();
// minimum translation distance to push balls apart after intersecting
Vector2d mtd = delta.multiply(((getRadius() + ball.getRadius())-d)/d);
// resolve intersection --
// inverse mass quantities
float im1 = 1 / getMass();
float im2 = 1 / ball.getMass();
// push-pull them apart based off their mass
position = position.add(mtd.multiply(im1 / (im1 + im2)));
ball.position = ball.position.subtract(mtd.multiply(im2 / (im1 + im2)));
// impact speed
Vector2d v = (this.velocity.subtract(ball.velocity));
float vn = v.dot(mtd.normalize());
// sphere intersecting but moving away from each other already
if (vn > 0.0f) return;
// collision impulse
float i = (-(1.0f + Constants.restitution) * vn) / (im1 + im2);
Vector2d impulse = mtd.normalize().multiply(i);
// change in momentum
this.velocity = this.velocity.add(impulse.multiply(im1));
ball.velocity = ball.velocity.subtract(impulse.multiply(im2));
}
Source Code: Complete source for ball to ball collider.
If anyone has some suggestions for how to improve this basic physics simulator let me know! One thing I have yet to add is angular momentum so the balls will roll more realistically. Any other suggestions? Leave a comment!
To detect whether two balls collide, just check whether the distance between their centers is less than two times the radius. To do a perfectly elastic collision between the balls, you only need to worry about the component of the velocity that is in the direction of the collision. The other component (tangent to the collision) will stay the same for both balls. You can get the collision components by creating a unit vector pointing in the direction from one ball to the other, then taking the dot product with the velocity vectors of the balls. You can then plug these components into a 1D perfectly elastic collision equation.
Wikipedia has a pretty good summary of the whole process. For balls of any mass, the new velocities can be calculated using the equations (where v1 and v2 are the velocities after the collision, and u1, u2 are from before):
If the balls have the same mass then the velocities are simply switched. Here's some code I wrote which does something similar:
void Simulation::collide(Storage::Iterator a, Storage::Iterator b)
{
// Check whether there actually was a collision
if (a == b)
return;
Vector collision = a.position() - b.position();
double distance = collision.length();
if (distance == 0.0) { // hack to avoid div by zero
collision = Vector(1.0, 0.0);
distance = 1.0;
}
if (distance > 1.0)
return;
// Get the components of the velocity vectors which are parallel to the collision.
// The perpendicular component remains the same for both fish
collision = collision / distance;
double aci = a.velocity().dot(collision);
double bci = b.velocity().dot(collision);
// Solve for the new velocities using the 1-dimensional elastic collision equations.
// Turns out it's really simple when the masses are the same.
double acf = bci;
double bcf = aci;
// Replace the collision velocity components with the new ones
a.velocity() += (acf - aci) * collision;
b.velocity() += (bcf - bci) * collision;
}
As for efficiency, Ryan Fox is right, you should consider dividing up the region into sections, then doing collision detection within each section. Keep in mind that balls can collide with other balls on the boundaries of a section, so this may make your code much more complicated. Efficiency probably won't matter until you have several hundred balls though. For bonus points, you can run each section on a different core, or split up the processing of collisions within each section.
Well, years ago I made the program like you presented here.
There is one hidden problem (or many, depends on point of view):
If the speed of the ball is too
high, you can miss the collision.
And also, almost in 100% cases your new speeds will be wrong. Well, not speeds, but positions. You have to calculate new speeds precisely in the correct place. Otherwise you just shift balls on some small "error" amount, which is available from the previous discrete step.
The solution is obvious: you have to split the timestep so, that first you shift to correct place, then collide, then shift for the rest of the time you have.
You should use space partitioning to solve this problem.
Read up on
Binary Space Partitioning
and
Quadtrees
As a clarification to the suggestion by Ryan Fox to split the screen into regions, and only checking for collisions within regions...
e.g. split the play area up into a grid of squares (which will will arbitrarily say are of 1 unit length per side), and check for collisions within each grid square.
That's absolutely the correct solution. The only problem with it (as another poster pointed out) is that collisions across boundaries are a problem.
The solution to this is to overlay a second grid at a 0.5 unit vertical and horizontal offset to the first one.
Then, any collisions that would be across boundaries in the first grid (and hence not detected) will be within grid squares in the second grid. As long as you keep track of the collisions you've already handled (as there is likely to be some overlap) you don't have to worry about handling edge cases. All collisions will be within a grid square on one of the grids.
A good way of reducing the number of collision checks is to split the screen into different sections. You then only compare each ball to the balls in the same section.
One thing I see here to optimize.
While I do agree that the balls hit when the distance is the sum of their radii one should never actually calculate this distance! Rather, calculate it's square and work with it that way. There's no reason for that expensive square root operation.
Also, once you have found a collision you have to continue to evaluate collisions until no more remain. The problem is that the first one might cause others that have to be resolved before you get an accurate picture. Consider what happens if the ball hits a ball at the edge? The second ball hits the edge and immediately rebounds into the first ball. If you bang into a pile of balls in the corner you could have quite a few collisions that have to be resolved before you can iterate the next cycle.
As for the O(n^2), all you can do is minimize the cost of rejecting ones that miss:
1) A ball that is not moving can't hit anything. If there are a reasonable number of balls lying around on the floor this could save a lot of tests. (Note that you must still check if something hit the stationary ball.)
2) Something that might be worth doing: Divide the screen into a number of zones but the lines should be fuzzy--balls at the edge of a zone are listed as being in all the relevant (could be 4) zones. I would use a 4x4 grid, store the zones as bits. If an AND of the zones of two balls zones returns zero, end of test.
3) As I mentioned, don't do the square root.
I found an excellent page with information on collision detection and response in 2D.
http://www.metanetsoftware.com/technique.html (web.archive.org)
They try to explain how it's done from an academic point of view. They start with the simple object-to-object collision detection, and move on to collision response and how to scale it up.
Edit: Updated link
You have two easy ways to do this. Jay has covered the accurate way of checking from the center of the ball.
The easier way is to use a rectangle bounding box, set the size of your box to be 80% the size of the ball, and you'll simulate collision pretty well.
Add a method to your ball class:
public Rectangle getBoundingRect()
{
int ballHeight = (int)Ball.Height * 0.80f;
int ballWidth = (int)Ball.Width * 0.80f;
int x = Ball.X - ballWidth / 2;
int y = Ball.Y - ballHeight / 2;
return new Rectangle(x,y,ballHeight,ballWidth);
}
Then, in your loop:
// Checks every ball against every other ball.
// For best results, split it into quadrants like Ryan suggested.
// I didn't do that for simplicity here.
for (int i = 0; i < balls.count; i++)
{
Rectangle r1 = balls[i].getBoundingRect();
for (int k = 0; k < balls.count; k++)
{
if (balls[i] != balls[k])
{
Rectangle r2 = balls[k].getBoundingRect();
if (r1.Intersects(r2))
{
// balls[i] collided with balls[k]
}
}
}
}
I see it hinted here and there, but you could also do a faster calculation first, like, compare the bounding boxes for overlap, and THEN do a radius-based overlap if that first test passes.
The addition/difference math is much faster for a bounding box than all the trig for the radius, and most times, the bounding box test will dismiss the possibility of a collision. But if you then re-test with trig, you're getting the accurate results that you're seeking.
Yes, it's two tests, but it will be faster overall.
This KineticModel is an implementation of the cited approach in Java.
I implemented this code in JavaScript using the HTML Canvas element, and it produced wonderful simulations at 60 frames per second. I started the simulation off with a collection of a dozen balls at random positions and velocities. I found that at higher velocities, a glancing collision between a small ball and a much larger one caused the small ball to appear to STICK to the edge of the larger ball, and moved up to around 90 degrees around the larger ball before separating. (I wonder if anyone else observed this behavior.)
Some logging of the calculations showed that the Minimum Translation Distance in these cases was not large enough to prevent the same balls from colliding in the very next time step. I did some experimenting and found that I could solve this problem by scaling up the MTD based on the relative velocities:
dot_velocity = ball_1.velocity.dot(ball_2.velocity);
mtd_factor = 1. + 0.5 * Math.abs(dot_velocity * Math.sin(collision_angle));
mtd.multplyScalar(mtd_factor);
I verified that before and after this fix, the total kinetic energy was conserved for every collision. The 0.5 value in the mtd_factor was the approximately the minumum value found to always cause the balls to separate after a collision.
Although this fix introduces a small amount of error in the exact physics of the system, the tradeoff is that now very fast balls can be simulated in a browser without decreasing the time step size.
Improving the solution to detect circle with circle collision detection given within the question:
float dx = circle1.x - circle2.x,
dy = circle1.y - circle2.y,
r = circle1.r + circle2.r;
return (dx * dx + dy * dy <= r * r);
It avoids the unnecessary "if with two returns" and the use of more variables than necessary.
After some trial and error, I used this document's method for 2D collisions : https://www.vobarian.com/collisions/2dcollisions2.pdf
(that OP linked to)
I applied this within a JavaScript program using p5js, and it works perfectly. I had previously attempted to use trigonometrical equations and while they do work for specific collisions, I could not find one that worked for every collision no matter the angle at the which it happened.
The method explained in this document uses no trigonometrical functions whatsoever, it's just plain vector operations, I recommend this to anyone trying to implement ball to ball collision, trigonometrical functions in my experience are hard to generalize. I asked a Physicist at my university to show me how to do it and he told me not to bother with trigonometrical functions and showed me a method that is analogous to the one linked in the document.
NB : My masses are all equal, but this can be generalised to different masses using the equations presented in the document.
Here's my code for calculating the resulting speed vectors after collision :
//you just need a ball object with a speed and position vector.
class TBall {
constructor(x, y, vx, vy) {
this.r = [x, y];
this.v = [0, 0];
}
}
//throw two balls into this function and it'll update their speed vectors
//if they collide, you need to call this in your main loop for every pair of
//balls.
function collision(ball1, ball2) {
n = [ (ball1.r)[0] - (ball2.r)[0], (ball1.r)[1] - (ball2.r)[1] ];
un = [n[0] / vecNorm(n), n[1] / vecNorm(n) ] ;
ut = [ -un[1], un[0] ];
v1n = dotProd(un, (ball1.v));
v1t = dotProd(ut, (ball1.v) );
v2n = dotProd(un, (ball2.v) );
v2t = dotProd(ut, (ball2.v) );
v1t_p = v1t; v2t_p = v2t;
v1n_p = v2n; v2n_p = v1n;
v1n_pvec = [v1n_p * un[0], v1n_p * un[1] ];
v1t_pvec = [v1t_p * ut[0], v1t_p * ut[1] ];
v2n_pvec = [v2n_p * un[0], v2n_p * un[1] ];
v2t_pvec = [v2t_p * ut[0], v2t_p * ut[1] ];
ball1.v = vecSum(v1n_pvec, v1t_pvec); ball2.v = vecSum(v2n_pvec, v2t_pvec);
}
I would consider using a quadtree if you have a large number of balls. For deciding the direction of bounce, just use simple conservation of energy formulas based on the collision normal. Elasticity, weight, and velocity would make it a bit more realistic.
Here is a simple example that supports mass.
private void CollideBalls(Transform ball1, Transform ball2, ref Vector3 vel1, ref Vector3 vel2, float radius1, float radius2)
{
var vec = ball1.position - ball2.position;
float dis = vec.magnitude;
if (dis < radius1 + radius2)
{
var n = vec.normalized;
ReflectVelocity(ref vel1, ref vel2, ballMass1, ballMass2, n);
var c = Vector3.Lerp(ball1.position, ball2.position, radius1 / (radius1 + radius2));
ball1.position = c + (n * radius1);
ball2.position = c - (n * radius2);
}
}
public static void ReflectVelocity(ref Vector3 vel1, ref Vector3 vel2, float mass1, float mass2, Vector3 intersectionNormal)
{
float velImpact1 = Vector3.Dot(vel1, intersectionNormal);
float velImpact2 = Vector3.Dot(vel2, intersectionNormal);
float totalMass = mass1 + mass2;
float massTransfure1 = mass1 / totalMass;
float massTransfure2 = mass2 / totalMass;
vel1 += ((velImpact2 * massTransfure2) - (velImpact1 * massTransfure2)) * intersectionNormal;
vel2 += ((velImpact1 * massTransfure1) - (velImpact2 * massTransfure1)) * intersectionNormal;
}