Easy way to keeping angles between -179 and 180 degrees

Question

Is there an easy way to convert an angle (in degrees) to be between -179 and 180? I'm sure I could use mod (%) and some if statements, but it gets ugly:


//Make angle between 0 and 360
angle%=360;

//Make angle between -179 and 180
if (angle>180) angle-=360;

It just seems like there should be a simple math operation that will do both statements at the same time. I may just have to create a static method for the conversion for now.

After consulting the docs I am with Matthew Whited - angle % 360 yields a value between -359 and +359, not 0 and +359. Hence your solution fails to normalize initial values smaller then -179. — Daniel Brückner, Feb 23 '10 at 19:43
@Daniel: Good point. The given example is not good enough. Feel free to suggest something that is better. — User1, Feb 23 '10 at 20:15

score 128 · Answer 1 · answered Feb 24 '10 at 01:19

128

// reduce the angle  
angle =  angle % 360; 

// force it to be the positive remainder, so that 0 <= angle < 360  
angle = (angle + 360) % 360;  

// force into the minimum absolute value residue class, so that -180 < angle <= 180  
if (angle > 180)  
    angle -= 360;

answered Feb 24 '10 at 01:19

President James K. Polk

40,516
21
95
125

3

That's the correct answer with the least calculations necessary. – tzot Aug 27 '10 at 07:50
Seems like the best solution here to me. – Steve Nov 02 '11 at 06:07
6

@PlatinumAzure: No, it is not redundant. Suppose the angle is -5000? – President James K. Polk Feb 16 '12 at 22:57
8

I'm rather shocked more people accepted the old double-while-loop approach over this... – Gurgadurgen Feb 14 '15 at 03:00
You could even remove the if statement by using `result -= 360 * ((result - 1) / 180);` Assuming this is all int math. – Gurgadurgen Feb 14 '15 at 03:37
what if my angle is a float rather than an integer? – Charbel Oct 04 '15 at 19:55
9

Note for anyone coming across this in the future (like me), in Java the `%` operator is remainder rather than modulus, so in languages like Python, the second line `angle = (angle + 360) % 360` isn't needed for it to work for negative numbers (although it doesn't do any harm). See: http://stackoverflow.com/questions/5385024/mod-in-java-produces-negative-numbers – Steve Jul 05 '16 at 15:09
This is a great approach that works with any "de-iterate" situation, not just angles. I'm wondering if there's a simple way to modify it, if you wanted to change the inclusive/exclusive boundaries. In other words `( ( value % maxValue ) + maxValue ) % maxValue` results in 0 <= n < maxValue, but say you wanted 0 <= n <= maxValue (in the case of angles, to include 360). Is there a better way than simply plopping `if( value == maxValue ){ return value; }` in front of the code? – Beejor Mar 11 '19 at 01:55
My hunch was that the first line (reduce) is redundant and is taken care of by the second line. Testing in another language (Rust) seems to indicate that that is true so I think you could shorten this further. – Peter Roehlen Feb 10 '20 at 05:08
@PeterRoehlen: possibly true in Rust, but not in java. The first reduce brings the angle it into a range where the addition can flip the sign. [Try it with -10000034](https://tio.run/##7VLLTsMwELznK/aClAjSpipCSG34g4pDj4iDEzthi2NH9iYSQvn2YOdVRKt@AXvxesbeWY/3xFoWn/hn39dNJjGHXDJr4cBQwXcALiwxcjgqgleDJSomQ79hqpQimg75GABI5@QOtk/JLljo9RoKbXIBSEAaMgH0IaDWFglbAUZUTpML8wBWO4oRJLBPp2J7X@xCKRyTe09GNwQV6UGsQoVVUwHLrJYNCWiZbLy0Re7W4eln@Xjz7DqYG0jBbZfaWMziLx6PFuLcXpyO/cyoEdQYNZIj2v3196A5Fij4dX//Dbxu4DS4k4@tRg7eifBIBlX59g7MlPa3kccvS6Ja6YZWtTtCUoXLZMebxMf2MYp2ty4sX3VxoQu6oO9/AA "Java (JDK) – Try It Online") – idbrii Dec 07 '21 at 20:45
It just works, not sure why this is pancaked under 7 inferior solutions. – Justas Nov 24 '22 at 15:21

Seth · Answer 2 · 2010-02-24T00:00:16.833

Try this instead!

atan2(sin(angle), cos(angle))

atan2 has a range of [-π, π). This takes advantage of the fact that tan θ = sin θ / cos θ, and that atan2 is smart enough to know which quadrant θ is in.

Since you want degrees, you will want to convert your angle to and from radians:

atan2(sin(angle * PI/180.0), cos(angle * PI/180.0)) * 180.0/PI

Update My previous example was perfectly legitimate, but restricted the range to ±90°. atan2's range is the desired value of -179° to 180°. Preserved below.

Try this:

asin(sin(angle)))

The domain of sin is the real line, the range is [-1, 1]. The domain of asin is [-1, 1], and the range is [-PI/2, PI/2]. Since asin is the inverse of sin, your input isn't changed (much, there's some drift because you're using floating point numbers). So you get your input value back, and you get the desired range as a side effect of the restricted range of the arcsine.

Since you want degrees, you will want to convert your angle to and from radians:

asin(sin(angle * PI/180.0)) * 180.0/PI

(Caveat: Trig functions are bazillions of times slower than simple divide and subtract operations, even if they are done in an FPU!)

+1 for taking advantage of limited-range functions for normalizing. A bit of overhead is involved, but this is by far the most concise answer that is actually correct. — Platinum Azure, Feb 23 '10 at 19:43
Even better now, since the value is in the correct range. However, now you're doing three trig functions instead of two! — Seth, Feb 23 '10 at 21:22
if you use http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/complex/Complex.html you can do Complex(0.0, angle).exp().log() which is arguably more concise. see http://stackoverflow.com/a/29237626/415404 for explanation. — Gus, Sep 12 '15 at 22:24

score 12 · Answer 3 · answered Apr 08 '14 at 23:08

12

I know that years have passed, but still.

This solution contains no loops, no subtracting, no modulo (allows to normalize to radians interval). Works for any input, including negative values, big values, edge cases.

double normalizedAngle = angle - (ceil((angle + M_PI)/(2*M_PI))-1)*2*M_PI;  // (-Pi;Pi]:
double normalizedAngle = angle - (ceil((angle + 180)/360)-1)*360;           // (-180;180]:

double normalizedAngle = angle - (floor((angle + M_PI)/(2*M_PI)))*2*M_PI;  // [-Pi;Pi):
double normalizedAngle = angle - (floor((angle + 180)/360))*360;           // [-180;180):

answered Apr 08 '14 at 23:08

alexburtnik

7,661
4
32
70

works good, no loops and portable :) – Richard Jan 05 '22 at 16:48
For some reason, whenever I write this (very elegant) version in Swift, it take the compiler three minutes to grok it – p10ben Apr 17 '23 at 06:46

Astronomino · Answer 4 · 2014-04-08T22:36:43.017

11

This works with both negative and decimal numbers and doesn't require loops, nor trigonometric functions:

angle -= Math.floor(angle / 360 + 0.5) * 360

The result is in the [-180, 180) interval. For (-180, 180] interval, you can use this instead:

angle -= Math.ceil(angle / 360 - 0.5) * 360

edited Apr 08 '14 at 22:36

answered Apr 08 '14 at 22:22

Astronomino

141
1
3

2

Just a note that `angle` must be a `float` for this or integer math will kick in. Use `angle -= Math.ceil((double)angle / 360 - 0.5) * 360` or something for best security. I love the answer, though. Should be #1 – Cory-G Jun 24 '14 at 21:26
I also like how this doesnt cause float-drift if the angle is already within the constraint. – Khlorghaal Dec 24 '18 at 10:56
This answer is the best combination of concise and mathematically simple. – Robert Penner Aug 09 '19 at 19:28

score 10 · Answer 5 · edited Dec 10 '13 at 20:23

10

~~Not that smart, too, but no if.~~

~~angle = (angle + 179) % 360 - 179;~~

~~But I am not sure how Java handles modulo for negative numbers. This works only if -1 modulo 360 equals 359.~~

UPDATE

Just checked the docs and a % b yields a value between -(|b| - 1) and +(|b| - 1) hence the code is broken. To account for negative values returned by the modulo operator one has to use the following.

angle = ((angle + 179) % 360 + 360) % 360 - 179;

But ... no ... never ... Use something similar to your initial solution, but fixed for values smaller then -179.

edited Dec 10 '13 at 20:23

ANeves

6,219
3
39
63

answered Feb 23 '10 at 19:21

Daniel Brückner

59,031
16
99
143

It should be "angle = (angle + 179) % 360 - 179". Otherwise 180 gets converted to -180. – Rômulo Ceccon Feb 23 '10 at 19:25
Too cryptic IMO. ... I absolutely agree. Still thinking about a better solution. – Daniel Brückner Feb 23 '10 at 19:27
1

this fails if the starting angle is less then -179 – Matthew Whited Feb 23 '10 at 19:27
I'm not sure being cryptic can be regarded as a problem - with commonly used math, you just want it to be efficient. Of course, two additions and a mod probably isn't better than a mod, a comparison, and sometimes an addition. – Cascabel Feb 23 '10 at 19:27
@Matthew Whited Because -1 modulo 360 yields -1? – Daniel Brückner Feb 23 '10 at 19:29
@Daniel, The code is fine, there needs to be a check before that line of course for out of range angles (less than -179 , greater than 180), but that needs to be done somewhere anyway, no matter how you code up this piece. – amischiefr Feb 23 '10 at 19:41
-1 is great than -179 and less then 180 so it would be find. But try it with -181 (which is less then -179) – Matthew Whited Feb 23 '10 at 19:47
from math import pi angle = (angle + 180) % (360)-180 – Alexis Paques Mar 22 '19 at 12:33

score 8 · Accepted Answer · answered Feb 23 '10 at 19:42

8

I'm a little late to the party, I know, but...

Most of these answers are no good, because they try to be clever and concise and then don't take care of edge cases.

It's a little more verbose, but if you want to make it work, just put in the logic to make it work. Don't try to be clever.

int normalizeAngle(int angle)
{
    int newAngle = angle;
    while (newAngle <= -180) newAngle += 360;
    while (newAngle > 180) newAngle -= 360;
    return newAngle;
}

This works and is reasonably clean and simple, without trying to be fancy. Note that only zero or one of the while loops can ever be run.

answered Feb 23 '10 at 19:42

Platinum Azure

45,269
12
110
134

1

And it handles angles with non-integer degrees! – Aniko Feb 23 '10 at 20:10
1

could you start with newAngle = angle % 360 at the beginning so you don't have to worry about normalizeAngle(Integer.MAX_VALUE) ? that would fall in the "don't take care of edge cases" category probably. – Jimmy Feb 23 '10 at 21:28
51

-1: This answer is ridiculously slow for angles that are not near 0. This is the solution for people who oppose division on religious grounds. – President James K. Polk Feb 24 '10 at 01:10
10

Its pretty slow for a large number. :( – Peter Lawrey May 14 '10 at 06:21
17

normalizeAngle(360000000000) the "go get a sandwich" angle – Glenn Maynard May 15 '16 at 23:02
3

How is this cleaner than the original? Since after all, that is the question being asked. – MCMastery May 15 '18 at 16:13

score 3 · Answer 7 · answered Feb 23 '10 at 19:31

Maybe not helpful, but I always liked using non-degree angles.

An angle range from 0 to 255 can be kept in bounds using bitwise operations, or for a single byte variable, simple allowed to overflow.

An angle range from -128 to 127 isn't quite so easy with bitwise ops, but again, for a single-byte variable, you can let it overflow.

I thought it was a great idea many years back for games, where you're probably using a lookup table for angles. These days, not so good - the angles are used differently, and are float anyway.

Still - maybe worth a mention.

+1 that is a very interesting idea, and if more granularity is needed than more bits can be used -- two bytes for example give 65536 distinct angles. — President James K. Polk, Feb 25 '10 at 02:03

score 3 · Answer 8 · answered Feb 23 '10 at 23:38

3

A short way which handles negative numbers is

double mod = x - Math.floor((x + 179.0) / 360) * 360;

Cast to taste.

BTW: It appears that angles between (180.0, 181.0) are undefined. Shouldn't the range be (-180, 180] (exclusive, inclusive]

answered Feb 23 '10 at 23:38

Peter Lawrey

525,659
79
751
1,130

not completely correct. If angle is 270 degrees, I get an unexpected result of -90 degrees – letsdev-cwack Aug 02 '15 at 14:38
3

@letsdev-cwack -90 is what I would expect to get. What would you expect and why? – Peter Lawrey Aug 03 '15 at 17:25

score 3 · Answer 9 · answered Jun 24 '14 at 21:23

3

Here is an integer-only solution:

int normalize(int angle)
{
    angle %= 360;
    int fix = angle / 180; // Integer division!!
    return (fix) ? angle - (360 * (fix)) : angle;
}

Sometimes being clever is just more fun, Platinum Azure.

answered Jun 24 '14 at 21:23

Cory-G

1,025
14
26

score 3 · Answer 10 · answered Feb 04 '15 at 08:05

I have made a formula for orientation of circular values

to keep angle between 0 and 359 is:

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

but to keep between -179 to 180 formula can be:

angle + Math.ceil( (-angle-179) / 360 ) * 360

this will give its orientation shift about -179 keeping actual angle intact

generalized formula for shifting angle orientation can be:

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

Vlad · Answer 11 · 2012-09-14T15:52:37.263

Well, one more solution, this one with just one division and no loops.

static double normalizeAngle(double angle)
{
    angle %= 360.0; // [0..360) if angle is positive, (-360..0] if negative
    if (angle > 180.0) // was positive
        return angle - 360.0; // was (180..360) => returning (-180..0)
    if (angle <= -180.0) // was negative
        return angle + 360.0; // was (-360..180] => returning (0..180]
    return angle; // (-180..180]
}

pkaeding · Answer 12 · 2010-02-23T19:36:18.537

1

How about

(angle % 360) - 179

This will actually return different results than the naive approach presented in the question, but it will keep the angle between the bounds specified. (I suppose that might make this the wrong answer, but I will leave it here in case it solves another persons' similar problem).

edited Feb 23 '10 at 19:36

answered Feb 23 '10 at 19:22

pkaeding

36,513
30
103
141

3

I think you mean `( (angle + 180) % 360 ) - 180` – Cascabel Feb 23 '10 at 19:23
And there's the off-by-one issue (fixed in Daniel's answer). – Cascabel Feb 23 '10 at 19:28

Matthew Whited · Answer 13 · 2010-05-14T04:38:21.850

1

int angle = -394;

// shortest
angle %= 360;
angle = angle < -170 ? angle + 360 : (angle > 180 ? angle - 380 : angle);

// cleanest
angle %= 360;
if (angle < -179) angle += 360;
else if (angle > 180) angle -= 360;

edited May 14 '10 at 04:38

answered Feb 23 '10 at 20:12

Matthew Whited

22,160
4
52
69

Your 'shortest' answer should have `angle < -179` instead of `angle < -170`. – Cory-G Jun 24 '14 at 21:37

score 0 · Answer 14 · answered Mar 25 '16 at 15:26

It is better to use library functions. They handle special cases like NaN and infinities.

public static double normalizeAngleDegrees(double angle) {
    return Math.toDegrees(Math.atan2(Math.sin(Math.toRadians(angle)), Math.cos(Math.toRadians(angle))));
}

score 0 · Answer 15 · answered Oct 05 '17 at 02:24

Here is my contribution. It seems to work for all angles with no edge issues. It is fast. It can do n180[360000359] = -1 almost instantaneously. Notice how the Sign function helps select the correct logic path and allows the same code to be used for different angles.

Ratch

n180[a_] := 
 If[Abs[Mod[a, If[Sign[a] == 0, 360, Sign[a] 360]]] <= 180, 
  Mod[a, If[Sign[a] == 0, 360, Sign[a] 360]], 
  Mod[a, If[Sign[a] == 0, 360, -Sign[a] 360]]]

Frank Vel · Answer 16 · 2020-09-17T20:38:48.773

I don't know much Java, but I came across the same problem in Python. Most of the answers here were either for integers so I figured I'd add one that allows for floats.

def half_angle(degree):
    return -((180 - degree) % 360) + 180

Based off the other answers I'm guessing the function would looks something like this in Java (feel free to correct me)

int halfAngle(int degree) {
    return -Math.floorMod(180 - degree, 360) + 180
}

double halfAngle(double degree) {
    // Java doesn't have a built-in modulus operator
    // And Math.floorMod only works on integers and longs
    // But we can use ((x%n) + n)%n to obtain the modulus for float
    return -(((180 - degree) % 360 + 360) % 360) + 180
}

Replace int with float to your liking.

This works for any degree, both positive and negative and floats and integers:

half_angle(180) == 180
half_angle(180.1) == -179.9  // actually -179.89999999999998
half_angle(-179.9) == -179.9
half_angle(-180) = 180
half_angle(1) = 1
half_angle(0) = 0
half_angle(-1) = -1

Math explanation:

Because the modulo operator is open at the upper end, the value x % 360 is in the range [0, 360), so by using the negative angle, the upper end becomes the lower end. So -(-x%360) is in (-360, 0], and -(-x%360)+360 is in (0, 360].

Shifting this by 180 gives us the answer.

Easy way to keeping angles between -179 and 180 degrees

16 Answers16

Linked