如何平均划分椭圆的周长？

Question

类似的问题asked and answered 以前在这里，但 none 很简单，我可以理解。下面的代码以相等的角度间隔计算椭圆的点，并对相邻点之间的距离求和以获得近似圆周。然后它将圆周分成 10 个假定相等的弧并输出由分割点形成的角度。

from math import sqrt,cos,sin,radians

def distance(x1,y1,x2,y2):
    return sqrt((x2-x1)**2 + (y2-y1)**2)

a = 5
b = 3
x0 = a
y0 = 0
angle = 0
d = 0
while(angle<=360):
    x = a * cos(radians(angle))
    y = b * sin(radians(angle))
    d += distance(x0,y0,x,y)
    x0 = x
    y0 = y
    angle += 0.25
print "Circumference of ellipse = %f" %d
onetenth = d/10
angle = 0
x0 = a
y0 = 0
angle0 = 0.25
for i in range(10):
    dist = 0
    while(dist<onetenth):
        x = a * cos(radians(angle))
        y = b * sin(radians(angle))
        dist += distance(x0,y0,x,y)
        x0 = x
        y0 = y
        angle += 0.25
    print "%d : angle = %.2f\tdifference = %.2f" %(i+1,angle-0.25, angle-angle0)
    angle0 = angle

它给出了输出：

Circumference of ellipse = 25.526979
1 : angle = 43.00       difference = 43.00
2 : angle = 75.50       difference = 32.50
3 : angle = 105.00      difference = 29.50
4 : angle = 137.50      difference = 32.50
5 : angle = 180.75      difference = 43.25
6 : angle = 223.75      difference = 43.00
7 : angle = 256.00      difference = 32.25
8 : angle = 285.50      difference = 29.50
9 : angle = 318.00      difference = 32.50
10 : angle = 361.50     difference = 43.50

但是这些角并没有平分圆周(picture)。我的 logic/code 有什么问题，我该如何改进？

Answer 1

您的程序将椭圆分成等弧长度，而不是等弧。在椭圆上，这是不一样的。在下面的代码中，我将每个线段的距离添加到输出中以验证这一点。

from math import sqrt,cos,sin,radians

def distance(x1,y1,x2,y2):
    return sqrt((x2-x1)**2 + (y2-y1)**2)

a = 5
b = 3
x0 = a
y0 = 0
angle = 0
d = 0
while(angle<=360):
    x = a * cos(radians(angle))
    y = b * sin(radians(angle))
    d += distance(x0,y0,x,y)
    x0 = x
    y0 = y
    angle += 0.25
print("Circumference of ellipse = {:f}".format(d))
onetenth = d/10
angle = 0
x0 = a
y0 = 0
angle0 = 0
for i in range(10):
    dist = 0
    while(dist<onetenth):
        angle += 0.025
        x = a * cos(radians(angle))
        y = b * sin(radians(angle))
        dist += distance(x0,y0,x,y)
        x0 = x
        y0 = y

    print(
        "{} : angle = {:.2f}\tdifference = {:.2f}\tDistance {:.2f}"
        .format(i+1,angle, angle-angle0,dist))
    angle0 = angle

示例输出：

Circumference of ellipse = 25.526979
1 : angle = 42.99   difference = 42.99  Distance 2.55
2 : angle = 75.27   difference = 32.28  Distance 2.55
3 : angle = 104.73  difference = 29.46  Distance 2.55
4 : angle = 137.01  difference = 32.28  Distance 2.55
5 : angle = 180.01  difference = 42.99  Distance 2.55
6 : angle = 223.00  difference = 42.99  Distance 2.55
7 : angle = 255.28  difference = 32.28  Distance 2.55
8 : angle = 284.74  difference = 29.46  Distance 2.55
9 : angle = 317.02  difference = 32.28  Distance 2.55
10 : angle = 360.02 difference = 43.00  Distance 2.55

注意，如果把椭圆改成圆（即a = b = 5），角度和距离就统一了：

Circumference of ellipse = 31.415902
1 : angle = 36.00   difference = 36.00  Distance 3.14
2 : angle = 72.00   difference = 36.00  Distance 3.14
3 : angle = 108.00  difference = 36.00  Distance 3.14
4 : angle = 144.00  difference = 36.00  Distance 3.14
5 : angle = 180.00  difference = 36.00  Distance 3.14
6 : angle = 216.00  difference = 36.00  Distance 3.14
7 : angle = 252.00  difference = 36.00  Distance 3.14
8 : angle = 288.00  difference = 36.00  Distance 3.14
9 : angle = 324.00  difference = 36.00  Distance 3.14
10 : angle = 360.00 difference = 36.00  Distance 3.14

我还对代码做了一些小的调整。首先，我将循环中 angle 变量的增量移动到计算之前。在最后一遍中，angle 在所有计算完成后获得了额外的增量。我还使增量值变小以减少最终结果的误差。

Answer 2

我是问题的 OP，我从评论中意识到我做出了错误的假设。该代码利用椭圆的参数方程来计算 x 和 y 坐标。参数方程中的角度不是与 x 轴形成的角度。我假设它们在代码中是相同的。更正后的代码是

from math import sqrt,cos,sin,atan2,radians,degrees

def distance(x1,y1,x2,y2):
    return sqrt((x2-x1)**2 + (y2-y1)**2)

a = 5
b = 3
x0 = a
y0 = 0
angle = 0
d = 0
while(angle<=360):
    x = a * cos(radians(angle))
    y = b * sin(radians(angle))
    d += distance(x0,y0,x,y)
    x0 = x
    y0 = y
    angle += 0.25
print "Circumference of ellipse = %f" %d
onetenth = d/10
angle = 0
x0 = a
y0 = 0
angle0 = 0
for i in range(10):
    dist = 0
    while(dist<onetenth):
        x = a * cos(radians(angle))
        y = b * sin(radians(angle))
        dist += distance(x0,y0,x,y)
        x0 = x
        y0 = y
        angle += 0.25
    xangle = degrees(atan2(y,x))
    print "%d : angle = %.2f\tdifference = %.2f" %(i+1, xangle, xangle-angle0)
    angle0 = xangle

它通过对点的x和y坐标取反正切来计算与x轴的角度。它给出输出：

Circumference of ellipse = 25.526979
1 : angle = 29.23       difference = 29.23
2 : angle = 66.68       difference = 37.46
3 : angle = 114.06      difference = 47.38
4 : angle = 151.20      difference = 37.13
5 : angle = -179.55     difference = -330.75
6 : angle = -150.13     difference = 29.42
7 : angle = -112.57     difference = 37.56
8 : angle = -65.19      difference = 47.37
9 : angle = -28.38      difference = 36.81
10 : angle = 0.90       difference = 29.28

这些角平分椭圆的圆周almost equally。

如何平均划分椭圆的周长？

How to divide circumference of an ellipse equally?

python

math

geometry

ellipse