How to map atan2() to degrees 0-360

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.

Solution using Modulo

A simple solution that catches all cases.

degrees = (degrees + 360) % 360;  // +360 for implementations where mod returns negative numbers

Explanation

Positive: 1 to 180

If you mod any positive number between 1 and 180 by 360, you will get the exact same number you put in. Mod here just ensures these positive numbers are returned as the same value.

Negative: -180 to -1

Using mod here will return values in the range of 180 and 359 degrees.

Special cases: 0 and 360

Using mod means that 0 is returned, making this a safe 0-359 degrees solution.

(x > 0 ? x : (2*PI + x)) * 360 / (2*PI)

Just add 360° if the answer from atan2 is less than 0°.

Or if you don't like branching, just negate the two parameters and add 180° to the answer.

(Adding 180° to the return value puts it nicely in the 0-360 range, but flips the angle. Negating both input parameters flips it back.)

@erikkallen is close but not quite right.

theta_rad = atan2(y,x);
theta_deg = (theta_rad/M_PI*180) + (theta_rad > 0 ? 0 : 360);

This should work in C++: (depending on how fmod is implemented, it may be faster or slower than the conditional expression)

theta_deg = fmod(atan2(y,x)/M_PI*180,360);

Alternatively you could do this:

theta_deg = atan2(-y,-x)/M_PI*180 + 180;

since (x,y) and (-x,-y) differ in angles by 180 degrees.

I have 2 solutions that seem to work for all combinations of positive and negative x and y.

1) Abuse atan2()

According to the docs atan2 takes parameters y and x in that order. However if you reverse them you can do the following:

double radians = std::atan2(x, y);
double degrees = radians * 180 / M_PI;
if (radians < 0)
{
    degrees += 360; 
}

2) Use atan2() correctly and convert afterwards

double degrees = std::atan2(y, x) * 180 / M_PI;
if (degrees > 90)
{
    degrees = 450 - degrees;
}
else
{
    degrees = 90 - degrees;
}

@Jason S: your "fmod" variant will not work on a standards-compliant implementation. The C standard is explicit and clear (7.12.10.1, "the fmod functions"):

> if y is nonzero, the result has the same sign as x

thus,

fmod(atan2(y,x)/M_PI*180,360)

is actually just a verbose rewriting of:

atan2(y,x)/M_PI*180

Your third suggestion, however, is spot on.

This is what I normally do:

float rads = atan2(y, x);
if (y < 0) rads = M_PI*2.f + rads;
float degrees = rads*180.f/M_PI;

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

var angle = (Math.atan2(x,y) * (180/Math.PI) + 360) % 360;

An alternative solution is to use the mod () function defined as:

function mod(a, b) {return a - Math.floor (a / b) * b;}

Then, with the following function, the angle between ini(x,y) and end(x,y) points is obtained. The angle is expressed in degrees normalized to [0, 360] deg. and North referencing 360 deg.

    function angleInDegrees(ini, end) {
        var radian = Math.atan2((end.y - ini.y), (end.x - ini.x));//radian [-PI,PI]
        return mod(radian * 180 / Math.PI + 90, 360);
    }

angle = Math.atan2(x,y)*180/Math.PI;

I have made a Formula for orienting angle into 0 to 360

angle + Math.ceil( -angle / 360 ) * 360;

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.

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)))

The R packages geosphere will calculate bearingRhumb, which is a constant bearing line given an origin point and easting/northing. The easting and northing must be in a matrix or vector. The origin point for a wind rose is 0,0. The following code seems to readily resolve the issue:

windE<-wind$uasE
windN<-wind$vasN
wind_matrix<-cbind(windE, windN)
wind$wind_dir<-bearingRhumb(c(0,0), wind_matrix)
wind$wind_dir<-round(wind$wind_dir, 0)

theta_rad = Math.Atan2(y,x);
if(theta_rad < 0)
  theta_rad = theta_rad + 2 * Math.PI;    //if neg., add 2 PI to it
theta_deg = (theta_rad/M_PI*180) ;        //convert from radian to degree

//or
theta_rad = Math.Atan2(y,x);
theta_rad = (theta_rad < 0) ? theta_rad + 2 * Math.PI : theta_rad;
theta_deg = (theta_rad/M_PI*180) ;

-1 deg becomes (-1 + 360) = 359 deg
-179 deg becomes (-179 + 360) = 181 deg

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];
}