How to map atan2() to degrees 0-360

willc2_23973 0
Viewed 164755

atan2(y, x) has that discontinuity at 180° where it switches to -180°..0° going clockwise.

How do I map the range of values to 0°..360°?

here is my code:

CGSize deltaPoint = CGSizeMake(endPoint.x - startPoint.x, endPoint.y - startPoint.y);
float swipeBearing = atan2f(deltaPoint.height, deltaPoint.width);

I'm calculating the direction of a swiping touch event given the startPoint and endPoint, both XY point structs. The code is for the iPhone but any language that supports atan2f() will do.

atan2 quartz-2d math
16 Answers

Here's some javascript. Just input x and y values.

var angle = (Math.atan2(x,y) * (180/Math.PI) + 360) % 360;
user_21745590 
double degree = fmodf((atan2(x, y) * (180.0 / M_PI)) + 360, 360);

This will return degree from 0°-360° counter-clockwise, 0° is at 3 o'clock.

user_20197140

A formula to have the range of values from 0 to 360 degrees.

f(x,y)=180-90*(1+sign(x))* (1-sign(y^2))-45*(2+sign(x))*sign(y)

     -(180/pi())*sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))
user_105728670

For your application I suspect you don't need exact degrees and would prefer a more approximate compass angle, eg 1 of 16 directions? If so then this code avoids atan issues and indeed avoids floating point altogether. It was written for a video game so uses 8 bit and 16 bit integers:

/*

                                           349.75d         11.25d, tan=0.2034523
                                              \             /
                                               \   Sector  /      
                                                \    0    /  22.5d tan = ?2 - 1
                                             15      |      1   33.75
                                                     |         /   45d, tan = 1
                                        14           |            2 _56.25
                                                     |             /  67.5d, tan = 1 + ?2
                                     13              |               3
                                                     |                __ 78.75
                                                     |                
                                    12---------------+----------------4 90d tan = infty
                                                     |                __ 101.25
                                                     |                
                                     11              |               5
                                                     |               
                                        10           |            6
                                                     |          
                                             9       |      7
                                                     8



*/

// use signs to map sectors:
static const int8_t map[4][5] = {  /* +n means n >= 0, -n means n < 0 */
  /* 0: +x +y */ {0, 1, 2, 3, 4},
  /* 1: +x -y */ {8, 7, 6, 5, 4},
  /* 2: -x +y */ {0, 15, 14, 13, 12},
  /* 3: -x -y */ {8, 9, 10, 11, 12}
};

int8_t sector(int8_t x, int8_t y) { // x,y signed in range -128:127, result 0:15 from north, clockwise.
  int16_t tangent; // 16 bits
  int8_t quadrant = 0;
  if (x > 0) x = -x; else quadrant |= 2; // make both negative avoids issue with negating -128 
  if (y > 0) y = -y; else quadrant |= 1;
  if (y != 0) {
    // The primary cost of this algorithm is five 16-bit multiplies.
    tangent = (int16_t)x*32;   // worst case y = 1, tangent = 255*32 so fits in 2 bytes.
    /*
       determine base sector using abs(x)/abs(y).
       in segment:
           0 if         0 <= x/y < tan 11.25   -- centered around 0     N
           1 if tan 11.25 <= x/y < tan 33.75   --                 22.5  NxNE
           2 if tan 33.75 <= x/y < tan 56.25   --                 45    NE
           3 if tan 56.25 <= x/y < tan 78.75   --                 67.5  ExNE
           4 if tan 78.75 <= x/y < tan 90      --                 90    E
    */
    if (tangent > y*6  ) return map[quadrant][0]; // tan(11.25)*32
    if (tangent > y*21 ) return map[quadrant][1]; // tan(33.75)*32
    if (tangent > y*47 ) return map[quadrant][2]; // tan(56.25)*32
    if (tangent > y*160) return map[quadrant][3]; // tan(78.75)*32
    // last case is the potentially infinite tan(90) but we don't need to check that limit.
  }
  return map[quadrant][4];
}
user_2108300
Related