(Optimization) How to quickly get coordinates (coordinates are integer) of all possible points inside a specific cylinder?

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Let's say I know coordinates of the center points of bottom surface and top surface; radius (R) and height (H) of cylinder. I need to write down all points which are inside the cylinder and also have integer coordinates.
The calculation speed could not be too slow since I have lots of calculation to run and need my simulation work down in rational time!
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Roger Stafford has given a right method. Is any one can optimize the method for improving calculation speed? Since I need to do lots of that calculation. _____________________________________________________________________________

Accepted Answer

Roger Stafford
Roger Stafford on 4 Oct 2014
Edited: Roger Stafford on 4 Oct 2014
The hard part of your question is formulating the equations of the cylinder's bounding surfaces.
If P1 = [x1,y1,z1] and P2 = [x2,y2,z2] are the center points you referred to, and R is the cylinder's radius, then the equation of any point P = [x,y,z] on the infinite cylinder's surface is:
dot(P-P1,P-P1)-dot(P-P1,P2-P1)^2/dot(P2-P1,P2-P1) = R^2
The equation of the infinite plane of the end containing P1 is:
dot(P-P1,P2-P1) = 0
and for the other end
dot(P-P2,P2-P1) = 0.
Nested for-loops can find all the points within the cylinder with integer-valued coordinates:
in = zeros(ceil(pi*(R+1)^2*(norm(P2-P1)+2)),3);
k = 0;
H2 = dot(P2-P1,P2-P1);
for x = floor(-R+min(x1,x2)):ceil(R+max(x1,x2))
for y = floor(-R+min(y1,y2)):ceil(R+max(y1,y2))
for z = floor(-R+min(z1,z2)):ceil(R+max(z1,z2))
P = [x,y,z];
if dot(P-P1,P-P1)-dot(P-P1,P2-P1)^2/H2<=R^2 & ...
dot(P-P1,P2-P1)>=0 & dot(P-P2,P2-P1)<=0
k = k + 1;
in(k,:) = P;
end
end
end
end
in(k+1:end,:) = [];
[Corrected]
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More Answers (1)

Image Analyst
Image Analyst on 4 Oct 2014
Why are the (x,y) coordinates at some z level any different than those at the end caps? Why can't you just use repmat() to copy them?
  5 Comments
Image Analyst
Image Analyst on 5 Oct 2014
Possibly the easiest way is to just rotate the cap so that it is parallel to the x-y plane (it's all in 1 z level), then use repmat(). Then rotate back.
Yi
Yi on 5 Oct 2014
Yeah, that would definitely be much easier for calculation. But how could we know which point is gonna have integer coordinates after rotating back? since in the rotation, a point with integer coordinates might no longer have integer coordinates. My point is that if we are not calculating points with integer coordinates any more, how do we know all possible points are calculated and which point should be calculate?

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