查找圆弧的算法，其圆心，半径和角度给定3个点

问题描述

给出3点A，B和C

我如何找到以A开始，以C结束并经过B的弧形；它的中心的坐标，半径和r和r'的角度?

How can I find and arc that begins at A, ends at C and pass through B; its center's coordinates, radius and angles for r and r' ?

推荐答案

有几种方法可以做到这一点.这是一种算法:

There are a few ways to do this. Here is one algorithm:

1. 获取您的订单

1. Get your COORDS

A = {xA，yA}

A = {xA,yA}

B = {xB，yB}

B = {xB,yB}

C = {xC，yC}

C = {xC,yC}

d = {xd，yd}

d = {xd,yd}

计算线AB和BC的中点

Calculate Mid-points of lines AB and BC

mid_AB = {(xA + xB)/2，(yA + yB)/2}

mid_AB = { (xA+xB)/2, (yA+yB)/2 }

mid_BC = {(xB + xC)/2，(yB + yC)/2}

mid_BC = { (xB+xC)/2, (yB+yC)/2 }

找到AB和BC线的斜率

Find slopes of lines AB and BC

slope_AB =(yB-yA)/(xB-xA)

slope_AB = (yB-yA)/(xB-xA)

slope_BC =(yC-yB)/(xC-xB)

slope_BC = (yC-yB)/(xC-xB)

构建从中点到AB和BC的直线(感谢您 Yves 抓住负面因素！)

Construct lines running through mid-points PERPENDICULAR to AB and BC (thank you to Yves for catching the negative!)

Slope_perp_AB =-(slope_AB)^(-1)

Slope_perp_AB = -(slope_AB)^(-1)

Slope_perp_BC =-(slope_BC)^(-1)

Slope_perp_BC = -(slope_BC)^(-1)

***带有Slope_perp_AB的线穿过mid_AB

*** Line with Slope_perp_AB runs through mid_AB

***带有Slope_perp_BC的线穿过mid_BC

*** Line with Slope_perp_BC runs through mid_BC

将两个方程式设置为相等，并求解以找到交点！这给你点d = {xd，yd} !!!

Set the two equations equal to each other and solve to find intersection!This gives you point d={xd,yd} !!!

现在计算半径和角度与中心点d无关紧要！

Calculating the radius and angles is now trivial with the center-point d!

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