How to find verticies after rotation?

dsa157

Hi all!

I have a little square I want to rotate and position it inside a bigger square. I then want to connect verticies from the bigger square to the smaller one. I don't know how to get the position of the rotated vertices.

I found this article: http://math.stackexchange.com/questions/270194/how-to-find-the-vertices-angle-after-rotation

I will try this, but wondered if processing had any built in geometry methods to make it easier.

Thanks in advance.

Dave

Lord_of_the_Galaxy

Processing already has all the trigonometry methods built in. Check the trigonometry section in the reference.
However, there are ways to do this more easily by using the transformation matrix.

dsa157

Thanks for the reply - do you have any examples of using the transformation matrix?

Lord_of_the_Galaxy

Use screenX() and screenY().

dsa157

I'm getting close, but not quite there. Here is my example. I want the small circle centered on the rotated upper left vertex of the small square.

int w=80;
int w2=20;
int counter=1;
float angle =0.0;
float x0 = 0.0;
float y0 = 0.0;
float x1 = 0.0;
float y1 = 0.0;
float p = 0.0;
float q = 0.0;
float p1 = 0.0;
float q1 = 0.0;

void setup() {
  size(400, 400);
}

void draw() {
  background(255);
  // rect1
  counter++;
  noFill();
  p = w/2;
  q = w/2;
  p1 = p - w2/2;
  q1 = q - w2/2;
  frameRate(1);

  pushMatrix();
    rectMode(CENTER);
    translate(p, q);
    rect(x0, y0, w, w);
    translate(-p, -q);
    //line(x0, y0, w, w);
    //line(w, y0, x0, w);
    pushMatrix();
      translate(p, q);
      angle = 15;
      rotate(radians(angle));
      rect(0, 0, w2, w2);
    popMatrix();
    println(p1, q1);
    x1 = ((x0-p1) * cos(angle)) - ((y0-q1) * sin(angle)) + p1;
    y1 = ((x0-p1) * sin(angle)) + ((y0-q1) * cos(angle)) + q1;
    ellipse(x1, y1, 10, 10);
  popMatrix();

}

Lord_of_the_Galaxy

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Lord_of_the_Galaxy

I told you to check out screenX and screenY.
But I don't understand, if that's all you need draw the circle before popMatrix after rotate and the circle should be cantered at the coordinates as though rotate never occurred.

dsa157

I am just using the circle as an indicator in this test code of where i want to calculate. Ultimately, I will be animating the rotation of the small square and I want to connect its vertices to the larger square with lines as it rotates around. I know the point from the large square where the line will originate, but I need the point on the small square where the line will end.

I tried to use ScreenX and ScreenY, but I guess I am not clear on the inputs. In my nested pushMatrix, if I called ScreenX(0,0,0), it still returned 0,0 in world space

changing line 45 to line(0, 0, x1, y1) is what I am looking for (with x1,y1 at the correct vertex);

TfGuy44

No really, use screenX() and screenY().

void setup(){
  size(400,400);
  noFill();
  stroke(255);
  rectMode(CENTER);
}

void draw(){
  background(0);
  pushMatrix();
  translate(width-mouseY, height-mouseX);
  rotate(map(millis()%5000,0,5000,0,TWO_PI));
  rect(0,0,200,200);
  float x = screenX(100,100);
  float y = screenY(100,100);
  popMatrix();
  line(mouseX, mouseY, x, y);
}

dsa157

Thank you!

Lord_of_the_Galaxy

That's what I was talking about.
Read the screenX and screenY reference.