如何在给定长度的曲线的每个点绘制法向量？

Question

import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]

x = np.linspace(-1, 1, 100)
y = x**2
ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)

plt.show()

Answer 1

import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]

x = np.linspace(-1, 1, 100)
y = x**2

# Calculating the gradient
L=.1 # gradient length
grad = np.ones(shape = (2, x.shape[0]))
grad[0, :] = -2*x
grad /= np.linalg.norm(grad, axis=0)  # normalizing to unit vector

nx = np.vstack((x - L/2 * grad[0], x + L/2 * grad[0]))
ny = np.vstack((y - L/2 * grad[1], y + L/2 * grad[1]))

# ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)
ax.plot(nx, ny, 'r')
ax.axis('equal')

plt.show()

Answer 2

要绘制 normals，您需要计算每个点的斜率；从那里，你得到可以旋转 pi/2.

的切线向量

这是一种使用 python i/o np 的方法，这使得一开始可能更容易理解。

更改长度将调整法线的大小，以根据您的绘图适当缩放。

import matplotlib.pyplot as plt
import numpy as np
import math

def get_normals(length=.1):
    
    for idx in range(len(x)-1):
        x0, y0, xa, ya = x[idx], y[idx], x[idx+1], y[idx+1]
        dx, dy = xa-x0, ya-y0
        norm = math.hypot(dx, dy) * 1/length
        dx /= norm
        dy /= norm
        
        ax.plot((x0, x0-dy), (y0, y0+dx))    # plot the normals


fig, ax = plt.subplots()
plt.rcParams["figure.figsize"] = [8, 8]

x = np.linspace(-1, 1, 100)
y = x**2
ax.set_ylim(-0.3, 1.06)
ax.plot(x, y)
get_normals()


plt.show()

或更长的法线，向下：get_normals(length=-.3) （使用 ax.set_aspect('equal') 保持角度）

如何在给定长度的曲线的每个点绘制法向量？

How to plot normal vectors in each point of the curve with a given length?

python

geometry

numpy

matplotlib

python-3.x