how to get intersection point of two curves?
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I want to get intersect point of two curve. I am use InterX for intersection of curves but I get error which is attacted below.
%program
syms Rb
R1=200;
R3=100;
R2=(R1+R3)/2;
theta=135;
Rb=solve((tan(acos(Rb/R1))-tan(acos(Rb/R2))-(acos(Rb/R1))+(Rb/R2))*(180/pi)==180-theta,Rb);
disp(Rb)
alpha=(tan(acos(Rb/R2))-acos(Rb/R2))*(180/pi);
disp(alpha)
t=linspace(0,2*pi);
x1=Rb*(cos(t-alpha)+t.*sin(t-alpha));
y1=Rb*(sin(t-alpha)-t.*cos(t-alpha));
x2=R1*cos(t);
y2=R1*sin(t);
plot(x1,y1,x2,y2)
hold on
q1=InterX([x1;y1],[x2;y2])
plot(x1,y1,x2,y2,P(1,:),P(2,:),'ro')
%error
140.5685985955217381715139492076
0.9117041268384028699662544052952
Error using symengine (line 59)
Array sizes must match.
Error in sym/privBinaryOp (line 903)
Csym = mupadmex(op,args{1}.s, args{2}.s, varargin{:});
Error in .* (line 238)
X = privBinaryOp(A, B, 'symobj::zip', '_mult');
Error in InterX (line 60)
S1 = dx1.*y1(1:end-1) - dy1.*x1(1:end-1);
Error in solve_equation (line 40)
q1=InterX([x1;y1],[x2;y2])
0 个评论
采纳的回答
Star Strider
2018-11-10
Try this:
syms Rb
R1=200;
R3=100;
R2=(R1+R3)/2;
theta=135;
Rb=vpasolve((tan(acos(Rb/R1))-tan(acos(Rb/R2))-(acos(Rb/R1))+(Rb/R2))*(180/pi)==180-theta,Rb);
disp(Rb)
alpha=(tan(acos(Rb/R2))-acos(Rb/R2))*(180/pi);
disp(alpha)
t=linspace(0,2*pi);
x1=Rb*(cos(t-alpha)+t.*sin(t-alpha));
y1=Rb*(sin(t-alpha)-t.*cos(t-alpha));
x2=R1*cos(t);
y2=R1*sin(t);
x1 = double(x1);
y1 = double(y1);
x2 = double(x2);
y2 = double(y2);
plot(x1,y1,x2,y2)
hold on
% q1=InterX([x1;y1],[x2;y2])
[D,I] = pdist2([x1;y1]',[x2;y2]','euclidean','Smallest',1);
[minD,idx] = min(D);
P = [x1(I(idx)); y1(I(idx))];
plot(x1,y1,x2,y2,P(1,:),P(2,:),'ro')
It uses the pdist2 function to return the smallest distance between the points in the two sets, approximating the intersection.
3 个评论
Star Strider
2018-11-10
I wanted to see if I could code an acceptable alternative using built-in functions.
Bruno Luong
2018-11-10
polyxpoly but it requires the Mapping toolbox. Without the toolbox you might end up doing what this FEX does.
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