Fill cylinder inside a matrix

3 vues (au cours des 30 derniers jours)
DG
DG le 12 Fév 2019
Commenté : Matt J le 3 Jan 2021
Given a 3d matrix:
vx = 1; % mm
vy = 1; % mm
vz = 1; % mm
nx = 500; %Number of elements in x direction
ny = 500; %Number of elements in y direction
nz = 100; %Number of elements in z direction
x = linspace(-vx/2,vx/2,nx);
y = linspace(-vy/2,vy/2,ny);
z = linspace(-vz/2,vz/2,nz);
[X,Y,Z] = ndgrid(x,y,z);
mat= zeros(size(X)); %Matrix to fill with cyliinders
and 2 random points on the edge of the cube,
How can I make a cylinder of radius R between those 2 points?
I want to fill up the binary matrix with 1 where the cylinder is placed and 0 everywhere else.
Thank you for your help.

Réponse acceptée

Matt J
Matt J le 13 Fév 2019
Modifié(e) : Matt J le 13 Fév 2019
Suppose the two end points are r1=[x1,y1,z1], r2=[x2,y2,z2]
[X,Y,Z] = ndgrid(x,y,z);
dXYZ=[X(:),Y(:),Z(:)].'-r1(:);
u=r1-r2;
u=repmat(u(:)./norm(u),1,nx*ny*nz);
mat=vecnorm( cross(dXYZ,u) ,2,1 )<=R;
mat=reshape(mat,nx,ny,nz);
  3 commentaires
Matt J
Matt J le 13 Fév 2019
Modifié(e) : Matt J le 13 Fév 2019
Yes, u is the direction vector of the central line of the cylinder. The left hand side of the inequality is the formula for the distance of points in your ndgrid to that line.
Matt J
Matt J le 3 Jan 2021
It should be giving you filled cylinders.

Connectez-vous pour commenter.

Plus de réponses (1)

Matt J
Matt J le 12 Fév 2019
Modifié(e) : Matt J le 12 Fév 2019
Choose a coordinate (xcenter,ycenter) where the cylinder will be centered and do,
mat = ( (x-xcenter).^2 + (y-ycenter).^2 <= R^2 );
  1 commentaire
DG
DG le 12 Fév 2019
But I want cylinder between two locations. This won't work for a tilted cylinder, I think. Something like the attached picture. In this case, if I take a slice in z axis, it will form some kind of ellipse and not a circle.
Hope this makes it clear.
Thank you!

Connectez-vous pour commenter.

Catégories

En savoir plus sur Computational Geometry dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by