Rotating a point about another point 2D

Question

I m trying to make a card game where the cards fan out  Right now to display it Im using the Allegro API which has a function   al draw rotated bitmap OBJECT TO ROTATE CENTER X CENTER Y X          Y DEGREES TO ROTATE IN RADIANS     so with this I can make my fan effect easily  The problem is then knowing which card is under the mouse  To do this I thought of doing a polygon collision test  I m just not sure how to rotate the 4 points on the card to make up the polygon  I basically need to do the same operation as Allegro   for example  the 4 points of the card are   card x  card y  card x   card width  card y   card height   I would need a function like   POINT rotate point float cx float cy float angle POINT p        Thanks

User · Answer

The coordinate system on the screen is left-handed, i.e. the x coordinate increases from left to right and the y coordinate increases from top to bottom. The origin, O(0, 0) is at the upper left corner of the screen.

A clockwise rotation around the origin of a point with coordinates (x, y) is given by the following equations:

where (x', y') are the coordinates of the point after rotation and angle theta, the angle of rotation (needs to be in radians, i.e. multiplied by: PI / 180).

To perform rotation around a point different from the origin O(0,0), let's say point A(a, b) (pivot point). Firstly we translate the point to be rotated, i.e. (x, y) back to the origin, by subtracting the coordinates of the pivot point, (x - a, y - b). Then we perform the rotation and get the new coordinates (x', y') and finally we translate the point back, by adding the coordinates of the pivot point to the new coordinates (x' + a, y' + b).

Following the above description:

a 2D clockwise theta degrees rotation of point (x, y) around point (a, b) is:

Using your function prototype: (x, y) -> (p.x, p.y); (a, b) -> (cx, cy); theta -> angle:

POINT rotate_point(float cx, float cy, float angle, POINT p){

     return POINT(cos(angle) * (p.x - cx) - sin(angle) * (p.y - cy) + cx,
                  sin(angle) * (p.x - cx) + cos(angle) * (p.y - cy) + cy);
}

User · Answer

First subtract the pivot point  cx cy   then rotate it  then add the point again   Untested   POINT rotate point float cx float cy float angle POINT p      float s   sin angle     float c   cos angle         translate point back to origin    p x -  cx    p y -  cy        rotate point   float xnew   p x   c - p y   s    float ynew   p x   s   p y   c        translate point back    p x   xnew   cx    p y   ynew   cy    return p

User · Answer

float s   sin angle      angle is in radians float c   cos angle      angle is in radians   For clockwise rotation    float xnew   p x   c   p y   s  float ynew   -p x   s   p y   c    For counter clockwise rotation    float xnew   p x   c - p y   s  float ynew   p x   s   p y   c

User · Answer

I struggled while working MS OCR Read API which returns back angle of rotation in range  -180  180   So I have to do an extra step of converting negative angles to positive  I hope someone struggling with point rotation with negative or positive angles can use the following  def rotate origin  point  angle        quot  quot  quot      Rotate a point counter-clockwise by a given angle around a given origin       quot  quot  quot        Convert negative angles to positive     angle   normalise angle angle         Convert to radians     angle   math radians angle         Convert to radians     ox  oy   origin     px  py   point            Move point  p  to origin  0 0       px   px - ox      py   py - oy            Rotate the point  p       qx    math cos angle     px  -  math sin angle     py      qy    math sin angle     px     math cos angle     py             Move point  p  back to origin  ox  oy      qx   ox   qx     qy   oy   qy          return  qx  qy    def normalise angle angle        quot  quot  quot  If angle is negative then convert it to positive   quot  quot  quot      if  angle    0   amp   abs angle      angle   -1            angle   360   angle     return angle

User · Answer

This is the answer by Nils Pipenbrinck  but implemented in c  fiddle   https   dotnetfiddle net btmjlG  using System   public class Program       public static void Main                    var angle   180   Math PI 180          Console WriteLine rotate point 0 0 angle new Point X 10  Y 10   Print                static Point rotate point double cx  double cy  double angle  Point p                double s   Math Sin angle           double c   Math Cos angle              translate point back to origin          p X -  cx          p Y -  cy             rotate point         double Xnew   p X   c - p Y   s          double Ynew   p X   s   p Y   c             translate point back          p X   Xnew   cx          p Y   Ynew   cy          return p             class Point               public double X          public double Y           public string Print                return    X   Y                        Ps  Apparently I can   t comment  so I   m obligated to post it as an answer

User · Answer

If you rotate point  px  py  around point  ox  oy  by angle theta you ll get   p x   cos theta     px-ox  - sin theta     py-oy    ox  p y   sin theta     px-ox    cos theta     py-oy    oy  this is an easy way to rotate a point in 2D

[c++] Rotating a point about another point (2D)

a 2D clockwise theta degrees rotation of point (x, y) around point (a, b) is:

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