Calculating Angles Between Line Segments (Python) With Math.atan2


Answer :

The easiest and most logically way to get at this problem is using the dot-product.

Try this code (I've commented practically everything):

import math def dot(vA, vB):     return vA[0]*vB[0]+vA[1]*vB[1] def ang(lineA, lineB):     # Get nicer vector form     vA = [(lineA[0][0]-lineA[1][0]), (lineA[0][1]-lineA[1][1])]     vB = [(lineB[0][0]-lineB[1][0]), (lineB[0][1]-lineB[1][1])]     # Get dot prod     dot_prod = dot(vA, vB)     # Get magnitudes     magA = dot(vA, vA)**0.5     magB = dot(vB, vB)**0.5     # Get cosine value     cos_ = dot_prod/magA/magB     # Get angle in radians and then convert to degrees     angle = math.acos(dot_prod/magB/magA)     # Basically doing angle <- angle mod 360     ang_deg = math.degrees(angle)%360      if ang_deg-180>=0:         # As in if statement         return 360 - ang_deg     else:           return ang_deg

Now try your variations of lineA and lineB and all should give the same answer.


An alternative solution using the formula:

enter image description here

where 'm1' is the slope of line 1 and 'm2' the slope of line 2. If line 1 is defined by the points P1 = [x1, y1] and P2 = [x2, y2], then slope 'm' is:

enter image description here

By using the formulas above you can find the angle in degrees between two lines as follows:

def slope(x1, y1, x2, y2): # Line slope given two points:     return (y2-y1)/(x2-x1)  def angle(s1, s2):      return math.degrees(math.atan((s2-s1)/(1+(s2*s1))))  lineA = ((0.6, 3.6), (1.6, 3)) lineB = ((1.6, 3), (2, 3.6))  slope1 = slope(lineA[0][0], lineA[0][1], lineA[1][0], lineA[1][1]) slope2 = slope(lineB[0][0], lineB[0][1], lineB[1][0], lineB[1][1])  ang = angle(slope1, slope2) print('Angle in degrees = ', ang)

Too much work. Take the absolute value of the arccosine of the dot product of the two vectors divided by each of the lengths of the lines.


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