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Nonrectangular grid between two lines

Asked by Jan Kappen on 7 May 2015
Latest activity Edited by Matt J
on 18 May 2015
Hey guys, this might be a simple problem but I'm going crazy here:
Assume you have two lines or direction vectors r,s from the point p (e.g. the origin):
p = [0 0 0]; r = [2 1 1]; s = [.2 0.1 .5];
Interpreting these vectors span a plane in R^3 (red vector is the normal vector):
This works fine with a normal meshgrid. What I really want is to create a grid "between" the r and s axis, and not a rectangular grid between the x and y values.
I can plot points along the vectors, so I have the coordinate for each star or diamond. Let's call them a and b (which are 3d vectors):
rn = r/norm(r); sn = s/norm(s);
a = linspace(p,p+rn,10);
b = linspace(p,p+sn,10);
plot3(a(1,:),a(2,:),a(3,:),'*')
plot3(b(1,:),b(2,:),b(3,:),'d')
How can I define a grid between a and b?
PS, this is only for visualization, I know that planes have an infinity spread ;)
Thank you very much! Jan

  2 Comments

Matt J
on 7 May 2015
I don't really see how you got this to work
a = linspace(p,p+rn,10);
You have vector input arguments, but All input arguments of linspace are supposed to be scalars according to the documentation.
Hi Matt, you are right. I wrote the script at home with Octave which does support the vector arguments. I didn't know Matlab doesn't.

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1 Answer

Answer by Matt J
on 7 May 2015
Edited by Matt J
on 18 May 2015

[i,j]=ndgrid(linspace(0,1,10));
xyz = bsxfun(@plus, p, i(:)*r + j(:)*s)
scatter3(xyz(:,1),xyz(:,2),xyz(:,3))

  1 Comment

Thank you! But this is "just" a replacement for the vectorized linspace version above, or did I get something wrong?
Anyways, in the end I realized, that a plane (for which the grid was used) between the two vectors does not look good, so I needed a rectangular plane which is parallel to the vectors, like:
Additionally a surf plot didn't look good, too. So I could simply create the meshgrid manually (planeSpread just makes the plane a bit larger):
normal = cross(r,s)/norm(cross(r,s));
d = dot(p,normal);
edge1 = p-planeSpread(1)*r-planeSpread(2)*s;
edge2 = p+planeSpread(1)*r-planeSpread(2)*s;
edge3 = p+planeSpread(1)*r+planeSpread(2)*s;
edge4 = p-planeSpread(1)*r+planeSpread(2)*s;
xx = [edge1(1) edge4(1); edge2(1) edge3(1)];
yy = [edge1(2) edge4(2); edge2(2) edge3(2)];
zz = (d-(xx*normal(1)+yy*normal(2))/normal(3); % x'*n0 = d -> x1*n1+x2*n2+x3*n3=d
mesh(xx,yy,zz)
Yes this is a bit ugly but at least I realized how the grid things work (I mean the structure of a meshgrid) and mesh does not need a rectangular grid (rectangular in the sense of being parallel to the coordinate axis)
Thank you.

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