The smallest difference between 2 Angles

问题

Given 2 angles in the range -PI -> PI around a coordinate, what is the value of the smallest of the 2 angles between them?

Taking into account that the difference between PI and -PI is not 2 PI but zero.

Example:

Imagine a circle, with 2 lines coming out from the center, there are 2 angles between those lines, the angle they make on the inside aka the smaller angle, and the angle they make on the outside, aka the bigger angle. Both angles when added up make a full circle. Given that each angle can fit within a certain range, what is the smaller angles value, taking into account the rollover

回答1:

This gives a signed angle for any angles:

a = targetA - sourceA
a = (a + 180) % 360 - 180

Beware in many languages the modulo operation returns a value with the same sign as the dividend (like C, C++, C#, JavaScript, full list here). This requires a custom mod function like so:

mod = (a, n) -> a - floor(a/n) * n

Or so:

mod = (a, n) -> (a % n + n) % n

If angles are within [-180, 180] this also works:

a = targetA - sourceA
a += (a>180) ? -360 : (a<-180) ? 360 : 0

In a more verbose way:

a = targetA - sourceA
a -= 360 if a > 180
a += 360 if a < -180

回答2:

x is the target angle. y is the source or starting angle:

atan2(sin(x-y), cos(x-y))

It returns the signed delta angle. Note that depending on your API the order of the parameters for the atan2() function might be different.

回答3:

If your two angles are x and y, then one of the angles between them is abs(x - y). The other angle is (2 * PI) - abs(x - y). So the value of the smallest of the 2 angles is:

min((2 * PI) - abs(x - y), abs(x - y))

This gives you the absolute value of the angle, and it assumes the inputs are normalized (ie: within the range [0, 2π)).

If you want to preserve the sign (ie: direction) of the angle and also accept angles outside the range [0, 2π) you can generalize the above. Here's Python code for the generalized version:

PI = math.pi
TAU = 2*PI
def smallestSignedAngleBetween(x, y):
    a = (x - y) % TAU
    b = (y - x) % TAU
    return -a if a < b else b

Note that the % operator does not behave the same in all languages, particularly when negative values are involved, so if porting some sign adjustments may be necessary.

回答4:

I rise to the challenge of providing the signed answer:

def f(x,y):
  import math
  return min(y-x, y-x+2*math.pi, y-x-2*math.pi, key=abs)

回答5:

For UnityEngine users, the easy way is just to use Mathf.DeltaAngle.

回答6:

Arithmetical (as opposed to algorithmic) solution:

angle = Pi - abs(abs(a1 - a2) - Pi);

回答7:

There is no need to compute trigonometric functions. The simple code in C language is:

#include <math.h>
#define PIV2 M_PI+M_PI
#define C360 360.0000000000000000000
double difangrad(double x, double y)
{
double arg;

arg = fmod(y-x, PIV2);
if (arg < 0 )  arg  = arg + PIV2;
if (arg > M_PI) arg  = arg - PIV2;

return (-arg);
}
double difangdeg(double x, double y)
{
double arg;
arg = fmod(y-x, C360);
if (arg < 0 )  arg  = arg + C360;
if (arg > 180) arg  = arg - C360;
return (-arg);
}

let dif = a - b , in radians

dif = difangrad(a,b);

let dif = a - b , in degrees

dif = difangdeg(a,b);

difangdeg(180.000000 , -180.000000) = 0.000000
difangdeg(-180.000000 , 180.000000) = -0.000000
difangdeg(359.000000 , 1.000000) = -2.000000
difangdeg(1.000000 , 359.000000) = 2.000000

No sin, no cos, no tan,.... only geometry!!!!

回答8:

An efficient code in C++ that works with both radians and degrees is:

inline double getAbsoluteDiff2Angles(const double x, const double y, const double c)
{
    // c can be PI (for radians) or 180.0 (for degrees);
    return c - fabs(fmod(fabs(x - y), 2*c) - c);
}

来源：https://stackoverflow.com/questions/1878907/the-smallest-difference-between-2-angles

标签

language-agnostic

geometry

angle