如何创建椭圆形状几何体

Question

如何根据已知轴坐标和峰半径创建椭圆？

从下图：

A点和B点是知道的

R是菲涅尔区计算的结果（以米为单位）。

点X是LineString AB的质心

我还读到： this 和 this 但我不知道如何实现它。

Answer 1

可以举例如下：

static_cast<RBTreeNode *>(x)

Answer 2

我正在努力解决类似的问题。 我也想制作一个菲涅耳区，但是我想在LOS中绘制它，这是连接A点和B点的线。

使用ewcz提供的代码，我添加了一行并绘制了所有内容。

旋转线的结果与椭圆的轴不对应，因此它不对应于LOS。

#!/usr/bin/env python
import math
from shapely.geometry import Point, LineString
from shapely.affinity import scale, rotate
from matplotlib import pyplot as plt

#input parameters
A = Point(0, 0)
B = Point(400, 10)
R = 5

d = A.distance(B)

#first, rotate B to B' around A so that |AB'| = |AB| and B'.y = A.y
#and then take S as midpoint of AB'
S = Point(A.x + d/2, A.y)
#Make a straight line
LOS = LineString([(A.x, A.y), (B.x, A.y)])
#alpha represents the angle of this rotation
alpha = math.atan2(B.y - A.y, B.x - A.x)

#create a circle with center at S passing through A and B'
C = S.buffer(d/2)

#rescale this circle in y-direction so that the corresponding
#axis is R units long
C = scale(C, 1, R/(d/2))

#rotate the ellipse obtained in previous step around A into the
#original position (positive angles represent counter-clockwise rotation)
C = rotate(C, alpha, origin=A, use_radians=True)
f_x, f_y = C.exterior.xy
#plot the ellipse
plt.plot(f_x, f_y)

#rotate the line in the same way as the ellipse
LOS_R = rotate(LOS, alpha, origin=A, use_radians=True)
f_x, f_y = LOS_R.xy
#plot the line
plt.plot(f_x, f_y)

plt.show()

结果图是： Plotted image with matplot