Python 画欧拉螺线
欧拉螺线（Euler spiral）/ 羊角螺线（clothoid）之前有写过，它的特性是 An Euler spiral is a curve whose curvature changes linearly with its curve length。 它的曲率随着它的曲线长度线性地改变。
An Euler spiral is a curve whose curvature changes linearly with its curve length (the curvature of a circular curve is equal to the reciprocal of the radius). Euler spirals are also commonly referred to as spiros, clothoids, or Cornu spirals.
线性改变总是一种有意思的特质，clothoid 也因为它的这个性质，存在于很多地方，比如过山车 ，公路的连接：
Clothoid 是使人们上下颠倒的环行游乐设施的理想曲线。在此之前，人们始终认为这种游乐设施的逻辑形状是圆形。实际上，如果过山车绕一圈成环，则该环的入口具有相当大的曲率，从而向乘客施加强大的离心力。另一方面，当汽车到达圆上的最高点时，其加速度会降低，以致乘客有可能跌倒。
但 clothoid 并不好画，看它的中文 Wikipedia 页面：
x = C(t)\\ y = S(t)
其中 C(t), S(t) 是 菲涅耳积分 （Fresnel integral）
S(x)=\int _{0}^{x}\sin(t^{2})\,dt=\sum _{{n=0}}^{{\infty }}(-1)^{n}{\frac {x^{{4n+3}}}{(4n+3)(2n+1)!}} \\ C(x)=\int _{0}^{x}\cos(t^{2})\,dt=\sum _{{n=0}}^{{\infty }}(-1)^{n}{\frac {x^{{4n+1}}}{(4n+1)(2n)!}}.
所以想要画它，思路之一就是使用 Fresnel 积分，比如这个 StackOverflow 提供的答案：
第二个思路是直接 sin, cos 积分：
Computing Clothoid Curves in five lines of Python
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
def clothoid_ode_rhs(state, s, kappa0, kappa1):
x, y, theta = state[0], state[1], state[2]
return np.array([np.cos(theta), np.sin(theta), kappa0 + kappa1*s])
def eval_clothoid(x0,y0,theta0, kappa0, kappa1, s):
return odeint(clothoid_ode_rhs, np.array([x0,y0,theta0]), s, (kappa0, kappa1))
x0,y0,theta0 = 0,0,0
L = 10
kappa0, kappa1 = 0, 0.4
s = np.linspace(0, L, 1000)
sol = eval_clothoid(x0, y0, theta0, kappa0, kappa1, s)
xs, ys, thetas = sol[:,0], sol[:,1], sol[:,2]
plt.scatter(xs, ys, s = 0.5, c = 'k')
plt.xlabel('x (m)');
plt.ylabel('y (m)');
plt.title('An awesome Clothoid');
import numpy as np
from math import cos, sin, pi, radians, sqrt
from scipy.special import fresnel
import matplotlib.pyplot as plt
def spiral_interp_centre(distance, x, y, hdg, length, curvEnd):
'''Interpolate for a spiral centred on the origin'''
# s doesn't seem to be needed...
theta = hdg # Angle of the start of the curve
Ltot = length # Length of curve
Rend = 1 / curvEnd # Radius of curvature at end of spiral
# Rescale, compute and unscale
a = 1 / sqrt(2 * Ltot * Rend) # Scale factor
distance_scaled = distance * a # Distance along normalised spiral
deltay_scaled, deltax_scaled = fresnel(distance_scaled)
deltax = deltax_scaled / a
deltay = deltay_scaled / a
# deltax and deltay give coordinates for theta=0
deltax_rot = deltax * cos(theta) - deltay * sin(theta)
deltay_rot = deltax * sin(theta) + deltay * cos(theta)
# Spiral is relative to the starting coordinates
xcoord = x + deltax_rot
ycoord = y + deltay_rot
return xcoord, ycoord
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
# This version
xs = []
ys = []
for n in range(-100, 100+1):
x, y = spiral_interp_centre(n, 50, 100, radians(56), 40, 1/20.)