Two Ellipses That Determine The Size of One

Programming Questions

flugelhorn

Two Ellipses That Determine The Size of One

in Programming Questions • 1 year ago

This sketch has two ellipses; one (myBlue) that moves around the screen & another (myBlue2) that is controlled by the mouse. When they are touching, a new green circle will appear over the moving ellipse (myBlue). I would like the green ellipse's width/height to be determined by the ellipse controlled by the mouse. So, depending on the position of the mouse, the green ellipse will be wider or taller. I would like to have the two Blue ellipses inside the green ellipse at all times.

However, asking the green ellipse to get data from the x & y postion has been proving difficult since it will only grab a number & the size will not be relative to the actual positon of the two ellipses. For example, if the mouse controlled ellipse (myBlue2) is at x position 20, the green ellipse will be 20 pixels wide.

Maybe I've been staring at this too long but how do I determine my desired effect?

Should I be using a different function (beginShape?) or create another class?

Ellipse myBlue;
Ellipse myBlue2;
void setup() {
size(500, 500);
smooth();
myBlue = new Ellipse(25);
myBlue2 = new Ellipse(25);
}
void draw() {
background(255);
myBlue.display();
myBlue2.display();
myBlue.move();
myBlue2.control();
if(myBlue.intersect(myBlue2)) {
myBlue.highlight();
fill(0, 200, 67, 150);
//this is where the problem(???) is
ellipse(myBlue.xpos, myBlue.ypos, myBlue2.xpos+myBlue2.r, myBlue2.ypos);
}
else {
myBlue.display();
}
}
class Ellipse {
float r;
float xpos;
float ypos;
float speedX;
float speedY;
color c;
Ellipse(float tempR) {
xpos = width/2;
ypos = height/2;
speedX = random(-2, 2);
speedY = random(-2, 2);
r = tempR;
c = color(0, 45, 200);
}
void display() {
noStroke();
fill(c);
ellipseMode(CENTER);
ellipse(xpos, ypos, r*2, r*2);
stroke(0, 223, 0);
//line(xpos, ypos, xpos+25, ypos);
c = color(0, 45, 200);
}
void move() {
xpos += speedX;
ypos += speedY;
if(ypos > height || ypos < 0) {
speedY = speedY * -0.95;
}
if(xpos > width || xpos < 0) {
speedX = speedX * -0.95;
}
xpos = constrain(xpos,0,width);
ypos = constrain(ypos,0,height);
}
void control() {
xpos = mouseX;
ypos = mouseY;
xpos = constrain(xpos,0,width);
ypos = constrain(ypos,0,height);
}
void highlight() {
c = color(234, 0, 44);
}
boolean intersect(Ellipse myBlue2) {
float distance = dist(myBlue.xpos, myBlue.ypos, myBlue2.xpos, myBlue2.ypos);
if(distance < myBlue.r + myBlue2.r) {
return true;
}
else {
return false;
}
}
}

Replies(2)

Ale_

Re: Two Ellipses That Determine The Size of One

1 year ago

I think I don´t understand the problem... mmm...

Btw, I think I wouldn´t create two different methods, move() and control()... you can overload move this way:

void move() {
    xpos += speedX; ypos += speedY;
    if(ypos > height || ypos < 0) speedY *= -0.95;
    if(xpos > width || xpos < 0) speedX *= -0.95;
}

void move(int x,int y){   //btw, you don´t need to constrain, mouse is naturally constrained by screen
xpos=x; ypos=y;
}

So it´d be:
myBlue.move();
myBlue2.move(mouseX,mouseY);

I´m not answering the question, but maybe it´s useful... who knows... :-)

PhiLho

Re: Two Ellipses That Determine The Size of One

1 year ago

Like 60rpm.tv, I don't understand where your problem is, your sketch is roughly doing what you describe. Except, perhaps, near the top-left corner, where it doesn't cover perfectly both shapes. I think you must take in account their distance as well, in the dimension of the green ellipse.

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