椭圆弧按长度等分求圆周上点的坐标

[zwphot]

已知：
以逆时针为准，起点坐标(x1,y1), 终点坐标(x2,y2), 圆心坐标(x0,y0)
长轴半径a, 短轴半径b
要求：
按每段长度s，等分椭圆弧，求椭圆弧上的所有等分点的坐标
求算法

[wanghui0380]

额，要啥算法，直接百度“椭圆极坐标方程”

[wanghui0380]

哦，错了，原来不是等分角度，是等分长度哦。这个估计得先展开，然后求点，在重新映射回来

[stherix]

初想挺简单，后来发现椭圆弧长很难计算，看别人的算法吧 https://blog.csdn.net/baidu_38621657/article/details/87907829

[wanghui0380]

查了一下，如果是等长分割没有算法，只能采用经验法近似解 https://blog.csdn.net/baidu_38621657/article/details/87907829

[wanghui0380]

或许有另外的办法，可以先FFT变成频谱，然后再频谱周期上想想办法

[bloodish]

此处有正解，python to C#应该很简单 how-to-divide-circumference-of-an-ellipse-equally

[zwphot]

how-to-divide-circumference-of-an-ellipse-equally 这个链接我这里打不开，可以直接贴内容出来吗

[bloodish]

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

[bloodish]

[wanghui0380]

引用 8 楼 bloodish 的回复:

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

这个是定积分采样，每次采样0.025增量角度积分计算长度

[wanghui0380]

先定积分采样出总长，然后再分次采样出总长/10(代码是10等分)

[it_gz_xi]

学习一下，说不定哪天能用上