MATLAB: Cross Section of PointCloud

cross-sectionpoint cloud

Hello everyone,

I have a 3D point cloud, and I want to find the cross section of it by slicing it with a plane that is defined by a point and a perpendicular vector. Then I wish to detect and store the points close to the plane on either side. Any ideas on how to do this? Thanks in advance.

Best Answer

Ok. Given a tolerance in the form of a distance, then just compute the distance to the plane!

Retain any points that lie within the tolerance.

Personally, I'd probably color code the points as plotted, based on how far they lie off the plane, but within the tolerance.

If you know how to define a plane, based on a point and the normal to the plane, then the distance is given by a simple dot product. So if pointInPlane is the point, and normalToPlane is the normal vector (assume it is of unit length)

D = dot(P-pointInPlane,normalToPlane);

If you do this carefully, the above computation is vectorized. Even simpler, you can do it as a simple matrix multiply, without even needing the function dot. After all, a matrix*vector multiply is just a dot product anyway. Use find to choose the points that are closer than the indicated tolerance. Then I'd just use a tool like scatter to plot the points with my chosen color mapping.

Related Solutions

MATLAB: How to create a point cloud for a sphere

See the documentation for the scatter3 function. It has examples that illustrate how to do what you want.

MATLAB: Find the distance from a point to a plane (known two vectors)

No, we do not know that 2 vectors can create a plane. If these two vectors are used to define the normal vector of the plane, you need an additional point, which is element of the plane. If v1 is the normal and v2 is a point of the plane (or the other way around), the plane is well defined also. But "v1" and "v2" sounds more like two directional vectors, such that the normal is:

v1x2 = cross(v1, v2);
N    = v1x2 / norm(v1x2)

However, please clarify this at first. If you have the plane defined by a point P and a normal vector N, the distance of the point Q is very easy to obtain:

% Vector from P to Q:
PQ = Q - P;
% Dot product between line from Q to P1 and normal of the plane:
Dist = dot(PQ, N);

This can be found directly using the Hesse normal form. The same formula allows to determine the distance between two skew lines in 3D: The 2 lines are defined by two points P1 and P2 on them and the directional vectors N1 and N2. Then create a plane E through e.g. P1 and the normal cross(N1, N2) / norm(cross(N1, N2)). Now the distance between the two lines is the distance between P2 and E.

Best Answer

Related Solutions

MATLAB: How to create a point cloud for a sphere

MATLAB: Find the distance from a point to a plane (known two vectors)

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