MATLAB: How do i find the intersection between two torus

intersectiontorus

I have 2 torus in three-dimensional space, and i need to find the intersection between them

R=9; % outer radius of torus
r=3; % inner tube radius
th=linspace(0,2*pi,36); % e.g. 36 partitions along perimeter of the tube 
phi=linspace(0,2*pi,18); % e.g. 18 partitions along azimuth of torus
% we convert our vectors phi and th to [n x n] matrices with meshgrid command:
[Phi,Th]=meshgrid(phi,th); 
% now we generate n x n matrices for x,y,z according to eqn of torus
x1= (R+r.*cos(Th)).*cos(Phi) +9;
y1= r.*sin(Th) ;
z1= (R+r.*cos(Th)).*sin(Phi);
daspect([1 1 1]) % preserves the shape of torus

colormap('jet') % change color appearance 

hold on
x2= ((R+r.*cos(Th)).*cos(Phi))*cos(2*pi/3) - (r.*sin(Th))*sin(2*pi/3) - 4.5;
y2= ((R+r.*cos(Th)).*cos(Phi))*sin(2*pi/3) + (r.*sin(Th))*cos(2*pi/3) + 9*sin(pi/3); 
z2= (R+r.*cos(Th)).*sin(Phi);
 
surf(x1,y1,z1); % plot surface
surf(x2,y2,z2);
daspect([1 1 1]) % preserves the shape of torus
colormap('jet') % change color appearance 
title('Torus')
xlabel('X');ylabel('Y');zlabel('Z');

Best Answer

Refer this:

https://in.mathworks.com/matlabcentral/answers/93623-how-do-i-plot-the-line-of-intersection-between-two-surfaces

Related Solutions

MATLAB: Solve function returns expression in terms of z

Hello Ashmika,

You do have a symbolic solution for h, which goes to show that symbolic solutions are not always the ideal.

There are eight solutions in all. Each one contains, among other things, a roots function, the roots of a 4th degree polynomial in z

roots(z^4 + z^3*A(r,phi) + z^2*B(r,phi) +... D(r,phi), z, k)

where the coefficents A(r,phi) etc are complicated functions of r and phi. Here z is just a dummy variable, and the roots are determined by the coefficients A,B,C,D. Four distinct roots are indicated by cycling k through 1,2,3,4 as you can see at the end of each of the first four very long lines of code.

That's four solutions. The other four solutions have the same values for the roots, the only difference being that the expression at the beginning of each line

2*r^2*cos(phi + pi/3)

has positive sign for the first four solutions and negative sign for the last four solutions.

The four roots are merely indicated by the index k. Since this is a polynomial of degree four, it is possible to write each of the four roots explicitly in terms of A,B,C,D but if you think you have long expressions now, those expressions might be several times longer.

A symblic expression that takes up a page of text begins to lose its effectiveness. You can still do calculaton by plugging values or even more symbolic epressions into them, but at some point one might consider just doing a numerical calculation from the beginning.

MATLAB: Error using plot3 – Vectors must be the same length

See if this does what you want:

theta = [0:0.101:pi/2,3*pi/2:0.01:2*pi];
rho = sqrt(cos(theta));
[x,y]=pol2cart(theta,rho);
x1=real(x);
y1=real(y);
x2=[x1,-x1];
y2=[y1,-y1];
z=zeros(size(x2));
figure
plot3(x2,y2,z)
hold on
for th=(29:29:340)
    p1 = plot3(x2,y2,z,'b');
    rotate(p1,[0 1 0],th)
end
rho = 2*sqrt(cos(theta));
[x,y]= pol2cart(theta, rho);
x1=real(x);
y1=real(y);
x2=[x1,-x1];
y2=[y1,-y1];
for the=(20:20:340)
    p2 = plot3(x2,y2,z,'r');
    rotate(p2,[0 1 0],th)
end
hold off
grid on
legend([p1,p2],'b','a')

A few changes were necessary to make it work correctly.

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