## blog

4/9

2012

### Circle and Rotated Rectangle Collision Detection

##### Circle and Rotated Rectangle

I’m going to explain how to implement collision detection for circle and rotated rectangle. Collision detection is determining if object A is hitting object B. A circle has center x y position with a radius. A rectangle contains left top x y position, width, height, and the angle to be rotated with. We assume a rectangle rotates around its center point.

I will use applet, pictures, and code to show this. I read this article as a reference to understanding collision detection between a circle and a rectangle not rotated.

##### Sample Applet

Here is a sample Applet to demonstrate collision detection. If collision is detected, the shapes turn blue. The green outlined shapes represent the state used to calculate. The orange line connects the center of unrotated circle to the closest point on rectangle. You can type in an angle and press the button to rotate. Click and/or drag anywhere to move the circle.

##### Implementing collision detection with a circle and rectangle with angle

The idea is simple. To perform the calculation, We need to have both shapes at the position when rectangle has 0 degree. It gets a little complicated if we rotate the rectangle with an angle. It’s easier to keep the rectangle straight with 0 degree to locate other points with simple addition. Top left is (x, y), top right is (x + width, y), and bottom left is (x, y + height).

Instead of rotating rectangle, rotate the circle back with the angle we were going to rotate rectangle. Put the circle’s center point to where it would be when rectangle’s angle is 0. Single point is easy to rotate. Looking at the rough picture, Blue shapes represent when rectangle is rotated and actually what we will see. Black shapes are used for calculation. We adjust the shapes with 0 degree or in other words, leaving the rectangle with 0 degree and rotating the circle back with rectangle’s angle instead.

The formula I used is below. Variable cx/cy stand for circle’s center point, originX/originY are the point we rotate the circle around, and x’/y’ are the rotated point.

x’ = cos(theta) * (cx – originX) – sin(theta) * (cy – originY) + originX

y’ = sin(theta) * (cx – originX) + cos(theta) * (cy – originY) + originY

Find the closest point from the unrotated circle and the rectangle. All we need to do is if-else comparisons for x and y separately.

For x, we use rectangle’s left x (rx) and right x (rx + width) to compare with circle’s center x (cx). If cx is on the left of rx, then rx is the closest x. Otherwise if cx is on the right of rx + width, then closest x is at rx + width. Finally, if none of 2 cases match, closest x point is cx itself. In the picture below, the red line is where the x will lie on.

The same rule applies for y. If circle’s center y (cy) is above top y (ry), ry is the closest y point. If cy is below the bottom y (ry + height) point, ry + height is the closest y point. If these cases don’t match, the closest y point is cy itself.

Lastly, we need to compute the distance from the unrotated circle’s center to the closest point we found earlier. I used the Pythagorean theorem to find that distance (a^2 + b^2 = c^2). Then you compare it with the circle’s radius. If distance is smaller than radius, the 2 shapes are colliding.

##### Sample Code

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 | // Rotate circle's center point back double unrotatedCircleX = Math.cos(rect.angle) * (circle.x - rect.centerX) - Math.sin(rect.angle) * (circle.y - rect.centerY) + rect.centerX; double unrotatedCircleY = Math.sin(rect.angle) * (circle.x - rect.centerX) + Math.cos(rect.angle) * (circle.y - rect.centerY) + rect.centerY; // Closest point in the rectangle to the center of circle rotated backwards(unrotated) double closestX, closestY; // Find the unrotated closest x point from center of unrotated circle if (unrotatedCircleX < rect.x) closestX = rect.x; else if (unrotatedCircleX > rect.x + rect.width) closestX = rect.x + rect.width; else closestX = unrotatedCircleX ; // Find the unrotated closest y point from center of unrotated circle if (unrotatedCircleY < rect.y) closestY = rect.y; else if (unrotatedCircleY > rect.y + rect.height) closestY = rect.y + rect.height; else closestY = unrotatedCircleY; // Determine collision boolean collision = false; double distance = findDistance(unrotatedCircleX , unrotatedCircleY, closestX, closestY); if (distance < circle.radius) collision = true; // Collision else collision = false; |

1 2 3 4 5 6 7 8 9 10 11 12 13 | /** * Pythagorean theorem * @param fromX * @param fromY * @param toX * @param toY */ public double findDistance(double fromX, double fromY, double toX, double toY){ double a = Math.abs(fromX - toX); double b = Math.abs(fromY - toY); return Math.sqrt((a * a) + (b * b)); } |

circle.x and circle.y, are those the center points of the circle or the top-left corner?

Hi,

circle.x and circle.y on lines #2-5 are the center coordinates of the circle before being rotated.

I cant get it to work in my game.

Can you please have a quick check on my codes?

http://pastebin.com/FGpCtRNq

The comment on your code says your angles are in degrees. Those methods take angles in radians. Another thing it might be is the center of the graphic rotation. My sample’s graphic rotates with respect to the center of the rectangle.

Even though i don’t use Java, this tutorial was really helpful for me. The explanation helped me to solve an issue i was having while programming an collision system between circles and rectangles.

Thank you.

I removed some words java and it worked perfectly in javascript. Thank you from Brazil.

Hi and thanks for a great bit of code.

I’ve ported this to Objective-C and changed the code to assume that it is the rectangles centre that is passed in. This allows it to be used more easily with the Cocos2D framework on iOS.

Here is a link: https://gist.github.com/pkclsoft/8293496

Awesome post – thank you!

I was able to translate your examples to a Javascript Canvas application with success. One thing I had to do differently was support negative and positive radians… I achieved that by converting any negative radians to their positive counterparts.

However, after making that adjustment I then needed to invert the radian value if the rectangle was in fact to be rotated with an originally positive value. Not sure if that makes any sense, it resulted in a few extra lines of code but it’s working great.