绘制椭圆轨道

Question

我试图编写一个代码，使用椭圆r = a（1-e ^ 2）/（1 + e * cos（theta））的等式绘制对象的椭圆路径。我也希望将这些数据放入数组中供其他用途。

from numpy import *#Imports Python mathematical functions library
import matplotlib.pyplot as plt #Imports plot library
from pylab import *

a = 5
e = 0.3
theta = 0
while theta <= 2*pi:
    r = (a*(1-e**2))/(1+e*cos(theta))
    print("r = ",r,"theta = ",theta)
    plt.polar(theta, r)
    theta += pi/180

plt.show()

代码为r和theta吐出正确的值，但是图是空白的。出现极坐标图窗口，但没有绘制任何内容。

请帮忙。提前谢谢。

Answer 1

不要为每一点致电plt.polar一次。相反，调用它一次，将所有数据作为输入：

import numpy as np #Imports Python mathematical functions library
import matplotlib.pyplot as plt #Imports plot library
cos = np.cos
pi = np.pi

a = 5
e = 0.3
theta = np.linspace(0,2*pi, 360)
r = (a*(1-e**2))/(1+e*cos(theta))
plt.polar(theta, r)

print(np.c_[r,theta])

plt.show()

---

顺便说一句，numpy可以做一个双线计算，而不是使用while循环：

theta = np.linspace(0,2*pi, 360)   # 360 equally spaced values between 0 and 2*pi
r = (a*(1-e**2))/(1+e*cos(theta))  

这将theta和r定义为numpy数组（而不是单个值）。

Answer 2

我认为你需要做points.append([theta,r])然后在最后plt.polar(points) ...这也是一个有点整洁的设计

from numpy import *#Imports Python mathematical functions library
import matplotlib.pyplot as plt #Imports plot library
from pylab import *

a = 5
e = 0.3
theta = 0

points = []
while theta <= 2*pi:
    r = (a*(1-e**2))/(1+e*cos(theta))
    print("r = ",r,"theta = ",theta)
    points.append((theta, r))
    theta += pi/180
#plt.polar(points) #this is cool but probably not what you want
plt.polar(*zip(*points))
plt.show()