Glitch in Collision Detection

Programming Questions

devstation

Glitch in Collision Detection

in Programming Questions • 1 year ago

New Year Wishes to all.

Please i have this little Challenge, i got this Code with the help of this forum that drops Objects( actually images) at random. I also have a Stationary Object(an Image also). I want to detect when one of these droping objects Touches my Other Stationary Object.

This is what i have so far, it works(so poorly), and does not detect most of these falling objects when they touch my stationary object

Thanks for the help

void setup(){
...
for (int i=0; i < nbrParticles; ++i) {
myFlakes[i] = new Particle();
}
}
void draw(){
background(109);
drop();
}
class Particle
{
float vx, vy, vspin; // velocities - added
float x, y, spin;
boolean show ;
Particle() {
x = random(width);
y = random(height);
show = true;
vspin = random(-maxSpin, maxSpin);
spin = random(TWO_PI);
vx = 0;
vy = 0;
}
void render() {
pushMatrix();
translate(x,y);
rotate(spin);
if(millis()/2 ==2){
image(deathBall, -deathBall.width, -deathBall.height);
}
else{
image(deathBall, -deathBall.width, -deathBall.height);
}
popMatrix();
}
}
Particle[] myFlakes = new Particle[nbrParticles];
public void drop(){
for (int i=0; i < nbrParticles; ++i) {
Particle f = myFlakes[i];
if(f.show){
f.render();
}
// application of forces to velocities
f.vy += random(-drift,+drift); // brownian motion
f.vx += random(-drift,+drift);
f.vy += gravity; // gravity
// velocity affects position
f.x += f.vx; // add vx to x
f.y += f.vy; // add vy to y
// constant spin rate
f.spin += f.vspin;
// apply damping (air friction) to movement velocities
f.vx *= damping; // shortcut for vx = vx * damping
f.vy *= damping;
// wrap around
if (f.y > height+50) {
f.y = -50;
}
if (f.x > width+50) {
f.x = -50;
}
if (f.x < -50) {
f.x = width+50;
}
if(isColiding(floor(marioX), floor(marioY), floor(f.x), floor(f.y))){ //marioX and marioY is the X and Y Location of the Image(the Stationary Object)
gameOver();
}
}
}
public boolean isColiding(float ix,float iy,float ix1,float iy2){
if( floor(iy2) > floor(iy) && floor(ix1)== floor(ix)){
return true;
}
return false;
}

Credit is given to the Original author of the Particle Class, i only tweaked it a little to serve my Purpose

Replies(4)

jbum

Re: Glitch in Collision Detection

1 year ago

I know you're trying to be helpful by omitting code, but it's needed if I'm going to test your code... Could you edit your post and include the whole sketch?

jbum

Re: Glitch in Collision Detection

1 year ago

Think I got the basic idea.

Change this

if (millis()/2 ==2) {
image(deathBall, -deathBall.width, -deathBall.height);
}
else {
image(deathBall, -deathBall.width, -deathBall.height);
}

To this single line,

image(deathBall, -deathBall.width/2, -deathBall.height/2);

and include the /2s - you want to offset by the radii, not the diameters. This will cause the x,y to correspond to the center of the object, instead of it's corner, and will fix your objects to spin about their centers. In general, you want to position objects by their centers, not their corners.

Also insure that marioX, marioY corresponds to the center of your mario object. Once you've done that, your collision detect.

public boolean isColiding(float ix, float iy, float ix1, float iy2) {

return dist(ix,iy,ix1,iy1) < SUM_OF_RADIUS_OF_EACH_OBJECT;

}

This only works if your particles and target are roughly round. If they are square, you can check if they overlap (look up rectangle intersection). If they are non square, things get a bit more complicated.

Btw, this looks like my code (or a bastardization thereof) but I'm not sure where the original came from.

devstation

Re: Glitch in Collision Detection

1 year ago

Thanks for your Prompt response, Made the corrections your proposed, but these colliding objects are neither Round nor Square, they are actually Images, thus the collision is not really what i expect.This is my full sketch and link to my data folder.

Thank you for the time

jbum

Re: Glitch in Collision Detection

1 year ago

It will be a little closer if you use

int d = (mario.width + deathBall.width)/2;

I added /2 so that it adds the radius instead of the width.

Two circles collide when the distance between them is less than the sum of their radii (you were summing their widths, which makes the collision test overlay sensitive by a factor of 2).

There are various techniques you can use to determine if two images intersect, but it's a little complicated, and perhaps more advanced than you are ready for, judging from your indenting habits

(clean up that code, dude! Edit/Auto-format is your friend).

One method, that is suitable for something like this, is to treat the mario and bomb objects as polygons (you would need to work out the coordinates for a simplified polygon that borders each figure), and then do a polygon intersection test. Your bomb is quite simple - basically a circle plus a rect. Mario is a bit more complicated. This method is complex, mathematically, but can be computationally cheaper than looking at the pixels. If you have math phobia, read on.

The second method is to check if any pixels in the images overlap, but this can be quite time consuming, unless you add a lot of optimizations. You should use simple rectangle tests first to skip the obvious cases where intersection can't occur. Similarly, if there is a huge rectangular overlap (say < 50% of the rectangle), that is a simple case where the objects clearly collide. Each object might have a minimum rectangle which encomposes pixels which are all opaque. If those minim rectangles touch, the objects clearly have collided.

For all the remaining cases (in which there is a small overlap that contains both transparent and opaque pixels), you compute the rectangle of the overlap, and then check only those coordinates to see if any non-transparent pixels from both images occupy the same coordinates. This involves looking at the equivalent pixels from the pixel[] array of each image, and doing some basic arithmetic to convert from global coordinates to local coordinates to pixel image index.

Top Reply