How can I generate a set of points evenly distributed along the perimeter of an ellipse?

user_102441 0
Viewed 13031

If I want to generate a bunch of points distributed uniformly around a circle, I can do this (python):

r = 5  #radius
n = 20 #points to generate
circlePoints = [
    (r * math.cos(theta), r * math.sin(theta))
    for theta in (math.pi*2 * i/n for i in range(n))
]

However, the same logic doesn't generate uniform points on an ellipse: points on the "ends" are more closely spaced than points on the "sides".

r1 = 5
r2 = 10
n = 20 #points to generate
ellipsePoints = [
    (r1 * math.cos(theta), r2 * math.sin(theta))
    for theta in (math.pi*2 * i/n for i in range(n))
]

Is there an easy way to generate equally spaced points around an ellipse?

math language-agnostic geometry
8 Answers

(UPDATED: to reflect new packaging).

An efficient solution of this problem for Python can be found in the numeric branch FlyingCircus-Numeric of FlyingCircus Python package.

Disclaimer: I am the main author of them.

Briefly, the (simplified) code looks (where a is the minor axis, and b is the major axis):

import numpy as np
import scipy as sp
import scipy.optimize

def angles_in_ellipse(
        num,
        a,
        b):
    assert(num > 0)
    assert(a < b)
    angles = 2 * np.pi * np.arange(num) / num
    if a != b:
        e2 = (1.0 - a ** 2.0 / b ** 2.0)
        tot_size = sp.special.ellipeinc(2.0 * np.pi, e2)
        arc_size = tot_size / num
        arcs = np.arange(num) * arc_size
        res = sp.optimize.root(
            lambda x: (sp.special.ellipeinc(x, e2) - arcs), angles)
        angles = res.x 
    return angles

It makes use of scipy.special.ellipeinc() which provides the numerical integral along the perimeter of the ellipse, and scipy.optimize.root() for solving the equal-arcs length equation for the angles.

To test that it is actually working:

a = 10
b = 20
n = 16

phi = angles_in_ellipse(n, a, b)
print(np.round(np.rad2deg(phi), 2))
# [  0.    17.55  36.47  59.13  90.   120.87 143.53 162.45 180.   197.55
#  216.47 239.13 270.   300.87 323.53 342.45]

e = (1.0 - a ** 2.0 / b ** 2.0) ** 0.5
arcs = sp.special.ellipeinc(phi, e)
print(np.round(np.diff(arcs), 4))
# [0.3022 0.2982 0.2855 0.2455 0.2455 0.2855 0.2982 0.3022 0.3022 0.2982
#  0.2855 0.2455 0.2455 0.2855 0.2982]

# plotting
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca()
ax.axes.set_aspect('equal')
ax.scatter(b * np.sin(phi), a * np.cos(phi))
plt.show()

plot

user_52183540

Do take into consideration the formula for ellipse perimeter as under if the ellipse is squashed. (If the minor axis is three times as small as the major axis)

tot_size = np.pi*(3*(a+b) -np.sqrt((3*a+b)*a+3*b))

Ellipse Perimeter

Admin1
Related