的等式为:绘制椭圆给出的椭圆的focii
sqrt((x-a1)**2 + (y-b1)**2) + np.sqrt((x-a2)**2 + (y-b2)**2) = c
的focii是(a1, b1)
和(a2, b2)
。 c
也是已知的。如何使用matplotlib在python中绘制这个?
感谢您的帮助。
的等式为:绘制椭圆给出的椭圆的focii
sqrt((x-a1)**2 + (y-b1)**2) + np.sqrt((x-a2)**2 + (y-b2)**2) = c
的focii是(a1, b1)
和(a2, b2)
。 c
也是已知的。如何使用matplotlib在python中绘制这个?
感谢您的帮助。
可以参数化在一些变量t
表示椭圆。例如,您可以查看Wikipedia以了解如何完成此操作。
在下面的代码中,我根据您提供的参数导出了参数形式所需的参数。
# Example focii and sum-distance
a1 = 1
b1 = 2
a2 = 5
b2 = 7
c = 9
# Compute ellipse parameters
a = c/2 # Semimajor axis
x0 = (a1 + a2)/2 # Center x-value
y0 = (b1 + b2)/2 # Center y-value
f = np.sqrt((a1 - x0)**2 + (b1 - y0)**2) # Distance from center to focus
b = np.sqrt(a**2 - f**2) # Semiminor axis
phi = np.arctan2((b2 - b1), (a2 - a1)) # Angle betw major axis and x-axis
# Parametric plot in t
resolution = 1000
t = np.linspace(0, 2*np.pi, resolution)
x = x0 + a * np.cos(t) * np.cos(phi) - b * np.sin(t) * np.sin(phi)
y = y0 + a * np.cos(t) * np.sin(phi) + b * np.sin(t) * np.cos(phi)
# Plot ellipse
plt.plot(x, y)
# Show focii
plt.plot(a1, b1, 'bo')
plt.plot(a2, b2, 'bo')
plt.axis('equal')
plt.show()
这给了你所需要的:
需要2所列出或X阵列,Y,使得元件满足椭圆方程
通常的椭圆绘制方案参数由中央(或焦点)角的椭圆方程以使X,Y的功能单值在0角度2PI
我显示在Drawing elliptical orbit in Python (using numpy, matplotlib) Y中一个黑客“感觉出”的x范围,则片的溶液作为X的函数一起双重ÿ解决方案对于每个X
只是将代码放入等式中的最小修改数将会失败为A1 = A2
符号溶液花费一分钟左右运行时
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
# sqrt((x-a1)**2 + (y-b1)**2) + np.sqrt((x-a2)**2 + (y-b2)**2) = c
coeffs = [1, 0, -1, 0, 4]
xs = [coeffs[0], coeffs[2]]
def ysolv(coeffs):
x,y,a1,b1,a2,b2,c = symbols('x y a1 b1 a2 b2 c', real = True)
ellipse = sqrt((x-a1)**2 + (y-b1)**2) + sqrt((x-a2)**2 + (y-b2)**2) - c
y_sols = solve(ellipse, y)
print(*y_sols, sep='\n')
num_coefs = [(a, f) for a, f in (zip([a1,b1,a2,b2,c], coeffs))]
y_solsf0 = y_sols[0].subs(num_coefs)
y_solsf1 = y_sols[1].subs(num_coefs)
print(y_solsf0, '\n', y_solsf1)
f0 = lambdify([x], y_solsf0)
f1 = lambdify([x], y_solsf1)
return f0, f1
f0, f1 = ysolv(coeffs)
y0 = [f0(x) for x in xs]
y1 = [f1(x) for x in xs]
def feeloutXrange(f, midx, endx):
fxs = []
x = midx
while True:
try: f(x)
except:
break
fxs.append(x)
x += (endx - midx)/200
return fxs
midx = (min(xs) + max(xs))/2
xpos = feeloutXrange(f0, midx, max(xs))
xnegs = feeloutXrange(f0, midx, min(xs))
xs_ellipse = xnegs[::-1] + xpos[1:]
y0s = [f0(x) for x in xs_ellipse]
y1s = [f1(x) for x in xs_ellipse]
ys_ellipse = y0s + y1s[::-1] + [y0s[0]] # add y start point to end to close drawing
xs_ellipse = xs_ellipse + xs_ellipse[::-1] + [xs_ellipse[0]] # added x start point
plt.plot(xs_ellipse, ys_ellipse)
plt.show()
(-c*sqrt((a1**2 - 2*a1*a2 + a2**2 + b1**2 - 2*b1*b2 + b2**2 - c**2)*(a1**2 + 2*a1*a2 - 4*a1*x + a2**2 - 4*a2*x + b1**2 - 2*b1*b2 + b2**2 - c**2 + 4*x**2))*(-b1 + b2 + c)*(b1 - b2 + c) + (b1**2 - 2*b1*b2 + b2**2 - c**2)*(-a1**2*b1 + a1**2*b2 + 2*a1*b1*x - 2*a1*b2*x + a2**2*b1 - a2**2*b2 - 2*a2*b1*x + 2*a2*b2*x - b1**3 + b1**2*b2 + b1*b2**2 + b1*c**2 - b2**3 + b2*c**2))/(2*(-b1 + b2 + c)*(b1 - b2 + c)*(b1**2 - 2*b1*b2 + b2**2 - c**2))
(c*sqrt((a1**2 - 2*a1*a2 + a2**2 + b1**2 - 2*b1*b2 + b2**2 - c**2)*(a1**2 + 2*a1*a2 - 4*a1*x + a2**2 - 4*a2*x + b1**2 - 2*b1*b2 + b2**2 - c**2 + 4*x**2))*(-b1 + b2 + c)*(b1 - b2 + c) + (b1**2 - 2*b1*b2 + b2**2 - c**2)*(-a1**2*b1 + a1**2*b2 + 2*a1*b1*x - 2*a1*b2*x + a2**2*b1 - a2**2*b2 - 2*a2*b1*x + 2*a2*b2*x - b1**3 + b1**2*b2 + b1*b2**2 + b1*c**2 - b2**3 + b2*c**2))/(2*(-b1 + b2 + c)*(b1 - b2 + c)*(b1**2 - 2*b1*b2 + b2**2 - c**2))
sqrt(-48*x**2 + 192)/8
-sqrt(-48*x**2 + 192)/8
其他的答案所使用的参量的变换方法
我特别喜欢表示sympy求解所述方程你而不是有人解决它
符号表达式只需要找到一次f或特定的椭圆参数化,则该符号表达式可以简单地硬编码:
"""
for Ellipse equation:
sqrt((x-a1)**2 + (y-b1)**2) + sqrt((x-a2)**2 + (y-b2)**2) = c
sympy solution to Ellipse equation, only have to run once to get y_sols
symbolic expression to paste into ysolv below
#def symEllipse():
# x,y,a1,b1,a2,b2,c = symbols('x y a1 b1 a2 b2 c', real = True)
# ellipse = sqrt((x-a1)**2 + (y-b1)**2) + sqrt((x-a2)**2 + (y-b2)**2) - c
# y_sols = solve(ellipse, y)
# print(*y_sols, sep='\n')
"""
coeffs = [1, 1, -1, -1, 3]
xs = [coeffs[0], coeffs[2]]
def ysolv(coeffs):
x,y,a1,b1,a2,b2,c = symbols('x y a1 b1 a2 b2 c', real = True)
y_sols = [
(-c*sqrt((a1**2 - 2*a1*a2 + a2**2 + b1**2 - 2*b1*b2 + b2**2 - c**2)*
(a1**2 + 2*a1*a2 - 4*a1*x + a2**2 - 4*a2*x + b1**2 - 2*b1*b2 + b2**2
- c**2 + 4*x**2))*(-b1 + b2 + c)*(b1 - b2 + c) + (b1**2 - 2*b1*b2 +
b2**2 - c**2)*(-a1**2*b1 + a1**2*b2 + 2*a1*b1*x - 2*a1*b2*x +
a2**2*b1 - a2**2*b2 - 2*a2*b1*x + 2*a2*b2*x - b1**3 + b1**2*b2 +
b1*b2**2 + b1*c**2 - b2**3 + b2*c**2))/(2*(-b1 + b2 + c)*
(b1 - b2 + c)*(b1**2 - 2*b1*b2 + b2**2 - c**2)),
(c*sqrt((a1**2 - 2*a1*a2 + a2**2 + b1**2 - 2*b1*b2 + b2**2 - c**2)*
(a1**2 + 2*a1*a2 - 4*a1*x + a2**2 - 4*a2*x + b1**2 - 2*b1*b2 + b2**2
- c**2 + 4*x**2))*(-b1 + b2 + c)*(b1 - b2 + c) + (b1**2 - 2*b1*b2 +
b2**2 - c**2)*(-a1**2*b1 + a1**2*b2 + 2*a1*b1*x - 2*a1*b2*x +
a2**2*b1 - a2**2*b2 - 2*a2*b1*x + 2*a2*b2*x - b1**3 + b1**2*b2 +
b1*b2**2 + b1*c**2 - b2**3 + b2*c**2))/(2*(-b1 + b2 + c)*
(b1 - b2 + c)*(b1**2 - 2*b1*b2 + b2**2 - c**2))
]
num_coefs = [(a, f) for a, f in (zip([a1,b1,a2,b2,c], coeffs))]
y_solsf0 = y_sols[0].subs(num_coefs)
y_solsf1 = y_sols[1].subs(num_coefs)
print(y_solsf0, '\n', y_solsf1)
f0 = lambdify([x], y_solsf0)
f1 = lambdify([x], y_solsf1)
return f0, f1
作为一名物理学家,如果对这个问题有任何理论上的简化,我就这样做。这就是为什么我喜欢这个答案。我想知道你给我看的这段代码是否已经在某个模块中实现了。它似乎不是。谢谢你的时间。 –