Calculate angle (clockwise) between two points
Question:
I have been not using math for a long time and this should be a simple problem to solve.
Suppose I have two points A: (1, 0) and B: (1, -1).
I want to use a program (Python or whatever programming language) to calculate the clockwise angle between A, origin (0, 0) and B. It will be something like this:
angle_clockwise(point1, point2)
Note that the order of the parameters matters. Since the angle calculation will be clockwise:
- If I call angle_clockwise(A, B), it returns 45.
- If I call angle_clockwise(B, A), it returns 315.
In other words, the algorithm is like this:
- Draw a line (line 1) between the first point param with (0, 0).
- Draw a line (line 2) between the second point param with (0, 0).
- Revolve line 1 around (0, 0) clockwise until it overlaps line 2.
- The angular distance line 1 traveled will be the returned angle.
Is there any way to code this problem?
Answers:
Check out the cmath python library.
>>> import cmath
>>> a_phase = cmath.phase(complex(1,0))
>>> b_phase = cmath.phase(complex(1,-1))
>>> (a_phase - b_phase) * 180 / cmath.pi
45.0
>>> (b_phase - a_phase) * 180 / cmath.pi
-45.0
You can check if a number is less than 0 and add 360 to it if you want all positive angles, too.
Numpy’s arctan2(y, x) will compute the counterclockwise angle (a value in radians between -π and π) between the origin and the point (x, y).
You could do this for your points A and B, then subtract the second angle from the first to get the signed clockwise angular difference. This difference will be between -2π and 2π, so in order to get a positive angle between 0 and 2π you could then take the modulo against 2π. Finally you can convert radians to degrees using np.rad2deg.
import numpy as np
def angle_between(p1, p2):
ang1 = np.arctan2(*p1[::-1])
ang2 = np.arctan2(*p2[::-1])
return np.rad2deg((ang1 - ang2) % (2 * np.pi))
For example:
A = (1, 0)
B = (1, -1)
print(angle_between(A, B))
# 45.
print(angle_between(B, A))
# 315.
If you don’t want to use numpy, you could use math.atan2 in place of np.arctan2, and use math.degrees (or just multiply by 180 / math.pi) in order to convert from radians to degrees. One advantage of the numpy version is that you can also pass two (2, ...) arrays for p1 and p2 in order to compute the angles between multiple pairs of points in a vectorized way.
Use the inner product and the determinant of the two vectors. This is really what you should understand if you want to understand how this works. You’ll need to know/read about vector math to understand.
See: https://en.wikipedia.org/wiki/Dot_product and https://en.wikipedia.org/wiki/Determinant
from math import acos
from math import sqrt
from math import pi
def length(v):
return sqrt(v[0]**2+v[1]**2)
def dot_product(v,w):
return v[0]*w[0]+v[1]*w[1]
def determinant(v,w):
return v[0]*w[1]-v[1]*w[0]
def inner_angle(v,w):
cosx=dot_product(v,w)/(length(v)*length(w))
rad=acos(cosx) # in radians
return rad*180/pi # returns degrees
def angle_clockwise(A, B):
inner=inner_angle(A,B)
det = determinant(A,B)
if det<0: #this is a property of the det. If the det < 0 then B is clockwise of A
return inner
else: # if the det > 0 then A is immediately clockwise of B
return 360-inner
In the determinant computation, you’re concatenating the two vectors to form a 2 x 2 matrix, for which you’re computing the determinant.
Here’s a solution that doesn’t require cmath.
import math
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
v1 = Vector(0, 1)
v2 = Vector(0, -1)
v1_theta = math.atan2(v1.y, v1.x)
v2_theta = math.atan2(v2.y, v2.x)
r = (v2_theta - v1_theta) * (180.0 / math.pi)
if r < 0:
r += 360.0
print r
Chris St Pierre: when using your function with:
A = (x=1, y=0)
B = (x=0, y=1)
This is supposed to be a 90 degree angle from A to B. Your function will return 270.
Is there an error in how you process the sign of the det or am I missing something?
A formula that calculates an angle clockwise, and is used in surveying:
f(E,N)=pi()-pi()/2*(1+sign(N))* (1-sign(E^2))-pi()/4*(2+sign(N))*sign(E)
-sign(N*E)*atan((abs(N)-abs(E))/(abs(N)+abs(E)))
The formula gives angles from 0 to 2pi,start from the North and
is working for any value of N and E. (N=N2-N1 and E=E2-E1)
For N=E=0 the result is undefined.
A verified 0° to 360° solution
It is an old thread, but for me the other solutions didn’t work well, so I implemented my own version.
My function will return a number between 0 and 360 (excluding 360) for two points on the screen (i.e. ‘y’ starts at the top and increasing towards the bottom), where results are as in a compass, 0° at the top, increasing clockwise:
def angle_between_points(p1, p2):
d1 = p2[0] - p1[0]
d2 = p2[1] - p1[1]
if d1 == 0:
if d2 == 0: # same points?
deg = 0
else:
deg = 0 if p1[1] > p2[1] else 180
elif d2 == 0:
deg = 90 if p1[0] < p2[0] else 270
else:
deg = math.atan(d2 / d1) / pi * 180
lowering = p1[1] < p2[1]
if (lowering and deg < 0) or (not lowering and deg > 0):
deg += 270
else:
deg += 90
return deg
in radians, clockwise, from 0 to PI * 2
static angle(center:Coord, p1:Coord, p2:Coord) {
var a1 = Math.atan2(p1.y - center.y, p1.x - center.x);
var a2 = Math.atan2(p2.y - center.y, p2.x -center.x);
a1 = a1 > 0 ? a1 : Math.PI * 2 + a1;//make angle from 0 to PI * 2
a2 = a2 > 0 ? a2 : Math.PI * 2 + a2;
if(a1 > a2) {
return a1 - a2;
} else {
return Math.PI * 2 - (a2 - a1)
}
}
I have been not using math for a long time and this should be a simple problem to solve.
Suppose I have two points A: (1, 0) and B: (1, -1).
I want to use a program (Python or whatever programming language) to calculate the clockwise angle between A, origin (0, 0) and B. It will be something like this:
angle_clockwise(point1, point2)
Note that the order of the parameters matters. Since the angle calculation will be clockwise:
- If I call angle_clockwise(A, B), it returns 45.
- If I call angle_clockwise(B, A), it returns 315.
In other words, the algorithm is like this:
- Draw a line (line 1) between the first point param with (0, 0).
- Draw a line (line 2) between the second point param with (0, 0).
- Revolve line 1 around (0, 0) clockwise until it overlaps line 2.
- The angular distance line 1 traveled will be the returned angle.
Is there any way to code this problem?
Check out the cmath python library.
>>> import cmath
>>> a_phase = cmath.phase(complex(1,0))
>>> b_phase = cmath.phase(complex(1,-1))
>>> (a_phase - b_phase) * 180 / cmath.pi
45.0
>>> (b_phase - a_phase) * 180 / cmath.pi
-45.0
You can check if a number is less than 0 and add 360 to it if you want all positive angles, too.
Numpy’s arctan2(y, x) will compute the counterclockwise angle (a value in radians between -π and π) between the origin and the point (x, y).
You could do this for your points A and B, then subtract the second angle from the first to get the signed clockwise angular difference. This difference will be between -2π and 2π, so in order to get a positive angle between 0 and 2π you could then take the modulo against 2π. Finally you can convert radians to degrees using np.rad2deg.
import numpy as np
def angle_between(p1, p2):
ang1 = np.arctan2(*p1[::-1])
ang2 = np.arctan2(*p2[::-1])
return np.rad2deg((ang1 - ang2) % (2 * np.pi))
For example:
A = (1, 0)
B = (1, -1)
print(angle_between(A, B))
# 45.
print(angle_between(B, A))
# 315.
If you don’t want to use numpy, you could use math.atan2 in place of np.arctan2, and use math.degrees (or just multiply by 180 / math.pi) in order to convert from radians to degrees. One advantage of the numpy version is that you can also pass two (2, ...) arrays for p1 and p2 in order to compute the angles between multiple pairs of points in a vectorized way.
Use the inner product and the determinant of the two vectors. This is really what you should understand if you want to understand how this works. You’ll need to know/read about vector math to understand.
See: https://en.wikipedia.org/wiki/Dot_product and https://en.wikipedia.org/wiki/Determinant
from math import acos
from math import sqrt
from math import pi
def length(v):
return sqrt(v[0]**2+v[1]**2)
def dot_product(v,w):
return v[0]*w[0]+v[1]*w[1]
def determinant(v,w):
return v[0]*w[1]-v[1]*w[0]
def inner_angle(v,w):
cosx=dot_product(v,w)/(length(v)*length(w))
rad=acos(cosx) # in radians
return rad*180/pi # returns degrees
def angle_clockwise(A, B):
inner=inner_angle(A,B)
det = determinant(A,B)
if det<0: #this is a property of the det. If the det < 0 then B is clockwise of A
return inner
else: # if the det > 0 then A is immediately clockwise of B
return 360-inner
In the determinant computation, you’re concatenating the two vectors to form a 2 x 2 matrix, for which you’re computing the determinant.
Here’s a solution that doesn’t require cmath.
import math
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
v1 = Vector(0, 1)
v2 = Vector(0, -1)
v1_theta = math.atan2(v1.y, v1.x)
v2_theta = math.atan2(v2.y, v2.x)
r = (v2_theta - v1_theta) * (180.0 / math.pi)
if r < 0:
r += 360.0
print r
Chris St Pierre: when using your function with:
A = (x=1, y=0)
B = (x=0, y=1)
This is supposed to be a 90 degree angle from A to B. Your function will return 270.
Is there an error in how you process the sign of the det or am I missing something?
A formula that calculates an angle clockwise, and is used in surveying:
f(E,N)=pi()-pi()/2*(1+sign(N))* (1-sign(E^2))-pi()/4*(2+sign(N))*sign(E)
-sign(N*E)*atan((abs(N)-abs(E))/(abs(N)+abs(E)))
The formula gives angles from 0 to 2pi,start from the North and
is working for any value of N and E. (N=N2-N1 and E=E2-E1)
For N=E=0 the result is undefined.
A verified 0° to 360° solution
It is an old thread, but for me the other solutions didn’t work well, so I implemented my own version.
My function will return a number between 0 and 360 (excluding 360) for two points on the screen (i.e. ‘y’ starts at the top and increasing towards the bottom), where results are as in a compass, 0° at the top, increasing clockwise:
def angle_between_points(p1, p2):
d1 = p2[0] - p1[0]
d2 = p2[1] - p1[1]
if d1 == 0:
if d2 == 0: # same points?
deg = 0
else:
deg = 0 if p1[1] > p2[1] else 180
elif d2 == 0:
deg = 90 if p1[0] < p2[0] else 270
else:
deg = math.atan(d2 / d1) / pi * 180
lowering = p1[1] < p2[1]
if (lowering and deg < 0) or (not lowering and deg > 0):
deg += 270
else:
deg += 90
return deg
in radians, clockwise, from 0 to PI * 2
static angle(center:Coord, p1:Coord, p2:Coord) {
var a1 = Math.atan2(p1.y - center.y, p1.x - center.x);
var a2 = Math.atan2(p2.y - center.y, p2.x -center.x);
a1 = a1 > 0 ? a1 : Math.PI * 2 + a1;//make angle from 0 to PI * 2
a2 = a2 > 0 ? a2 : Math.PI * 2 + a2;
if(a1 > a2) {
return a1 - a2;
} else {
return Math.PI * 2 - (a2 - a1)
}
}