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Nonrectangular grid between two lines

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Jan Kappen
Jan Kappen on 7 May 2015
Edited: 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
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.
Jan Kappen
Jan Kappen on 15 May 2015
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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Answers (1)

Matt J
Matt J on 7 May 2015
Edited: 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

Jan Kappen
Jan Kappen on 18 May 2015
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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