Nonlinear system Circle problem

7 次查看（过去 30 天）
Aldo
Aldo 2016-11-11
评论： Torsten 2016-11-11
I am supposed to determine with 3 points the midpoint and the radius of a circle
clear all; close all; clf;
x=[5 8 4]'; y=[9 8 3]';
X = (x(3)-x(2))/2; Y=(y(1)-y(3))/2; R= ; p=[X Y R]';
for iter=1:5
f=[ (x(1)-X)^2+(y(1)-Y)^2-R^2
(x(2)-X)^2+(y(2)-Y)^2-R^2
(x(3)-X)^2+(y(3)-Y)^2-R^2 ];
disp(norm(f))
J=[ 2*(x(1)-X) 2*(y(1)-Y)
2*(x(2)-X) 2*(y(2)-Y)
2*(x(3)-X) 2*(y(3)-Y) ];
dp=-J\f; p=p+dp;
X=p(1); Y=p(2); R=p(3);
end, p
plot(x,y,'o')
This is my code thus far, is my X=(x(3)-x(2))/2; Y=(y(1)-y(3))/2 correct, what should 'R=' be?
Best Regards Aldo
  5 个评论
显示 3更早的评论
Aldo
Aldo 2016-11-11
编辑：Aldo 2016-11-11
How should the Jacobian be? Shouldn't it be J_11 J_12 J_21 J_ 22. Shouldn't R be considered a constant?
Best Regards
Torsten
Torsten 2016-11-11
You have three equations in the unknowns X,Y and R. Thus the Jacobian is (3x3):
[df1/dX df1/dY df1/dR
df2/dX df2/dY df2/dR
df3/dX df3/dY df3/dR]
Best wishes
Torsten.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

KSSV
KSSV 2016-11-11

类别

在 Help Center 和 File Exchange 中查找有关 Spline Postprocessing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by