Finding distance between two curves at diffrent points

7 次查看（过去 30 天）

显示更早的评论

Praveen Patnaik 2020-3-12

2
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/510506-finding-distance-between-two-curves-at-diffrent-points

评论： darova 2021-2-14

采纳的回答： darova

1.Suppose the curves are defined, how do i find the distance between two curves at diffrent points.

Curve 1= f(x)

curve 2 = g(x)

Do i need to define points on one of the curve to find out the required distance???

For exmple I am posting an image.

Important note-

I need distance at diffrent points between two curves and not single point nor am I looking for minimum distance between curves.

Perpendicular distance from tangent as shown in image.

2 个评论
显示无隐藏无

darova 2020-3-12

Do you have data of curves? Or only image?

Praveen Patnaik 2020-3-12

It is just a general concept.... These are just images. Suppose the function of these curves are defined . Then how will we do it???

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

darova 2020-3-12

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/510506-finding-distance-between-two-curves-at-diffrent-points#answer_419859

在 MATLAB Online 中打开

With use of polyxpoly

clc,clear
x = 0:0.1:10;
y1 = sin(x);                    % data 1
y2 = sqrt(x);                   % data 2
% index of point
ix = 70;
x0 = x(ix);
y0 = y2(ix);
% tangent vector
dy = diff(y2(ix:ix+1));
dx = diff(x(ix:ix+1));
% normal vector
xv = [0 dy]*50+x0;
yv = [0 -dx]*50+y0;
% intersection point
[xc,yc] = polyxpoly(xv,yv,x,y1);
plot(x,y1,x,y2)
hold on
plot(xv,yv)
plot(xc,yc,'or')
hold off
axis equal

28 个评论
显示 26更早的评论隐藏 26更早的评论

Ron Herman 2020-11-17

编辑：Ron Herman 2020-11-17

在 MATLAB Online 中打开

I need the following output

I am unable to get the following output for some reason I an unable to draw horizontal line in all 3 points

(horizontal line from middle point is not getting constructed)

For some reason index =50 or at ind=50 no line is being constructed. I donot the reason for this yet.

clc,clear
r = 10;                     % arc length
R = 55;                     % radius of a circle
aa = 60*pi/180;              % arc angle
ap = 0*pi/180;             % arc position angle
k=0;
%xa=60;
% a=10;
% b=50;
ind=[25 50 75];
t = linspace(0,pi);
[x,y] = pol2cart(t,R);      % circle data
t1 = linspace(0,aa)-aa/2+ap;
[x1,y1] = pol2cart(t1,r); % arc data
xc=65;
yc=15;
%shifting the arc mid point to (65,15)
mp=length(x1)/2;
delx=xc-x1(mp);
dely=yc-y1(mp);
x1=x1+delx;
y1=y1+dely;
XC = ind*nan;
YC = ind*nan;
for i = 1:length(ind)
    ix = ind(i);
    x0 = x1(ix);
    y0 = y1(ix);
    
    % minor difference
    dy = diff(y1(ix:ix+1));
    dx = diff(x1(ix:ix+1));
    
    % horizontal
    xv = [dx -dx]*1500+x0;
    yv = [0 0]*1500+y0;
    
    %line(xv,yv)
    % intersection point
    [xc,yc] = polyxpoly(xv,yv,x,y);
    if ~isempty(xc)
        XC(i) = xc
        YC(i) = yc
    end
end
plot(x,y,x1,y1)
hold on
plot([XC;x1(ind)],[YC;y1(ind)],'.-r') % x1(ind)& y1(ind)are points on the curve
hold off                               % XC and YC are intersection points  
axis equal

darova 2021-2-14

DId you get the answer?

请先登录，再进行评论。

类别

AI and Statistics Curve Fitting Toolbox Smoothing

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

Community Treasure Hunt

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

Start Hunting!

Translated by