calculate the normal for a plane passing through more than three points
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Hi! I have a plane passing through three points (triangular plane) and from which I determine the normal.
P = [ 0.034488 0.036484 0.006912; ...
0.019104 0.055041 0.047894; ...
-0.008596 0.123650 0.033786];
N = cross(P(1,:) - P(2,:), P(3,:) - P(2,:));
N = N/norm(N);
Is it possible to do the same thing with a circular plane ('V1_in')?
- If it is not possible to do this, how can I create the matrix equal to 'P' using the 'V1_in' plane?
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Risposta accettata
Bruno Luong
il 29 Set 2023
load('V1_in.mat')
[~,~,V]=svd( V_1-mean(V_1));
N=V(:,3)
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Bruno Luong
il 29 Set 2023
% Check againts other method for 3 points
P = [ 0.034488 0.036484 0.006912; ...
0.019104 0.055041 0.047894; ...
-0.008596 0.123650 0.033786];
N = cross(P(1,:) - P(2,:), P(3,:) - P(2,:));
N = N/norm(N)
[~,~,V]=svd( P-mean(P));
N=V(:,3)% opposite sign, which is arbitrary choice for a 2D plane
Più risposte (3)
Torsten
il 29 Set 2023
Modificato: Torsten
il 29 Set 2023
If the points in V1_in.mat all lie in a common plane, you can arbitrarily pick three points and define these points as P.
If the points are only "approximately" in a common plane, you have to determine this plane via regression (like @Bruno Luong did).
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Matthew Blomquist
il 29 Set 2023
Hi Alberto,
If you think of the circular plane as a collection of points with coordinates (x, y, z), you can create a bunch of "triangular planes" by selecting any three coordinates to create a triangle. If all of the coordinates lie in the same plane, computing the normal of any triangular plane will be the same as computing the normal of the circular plane.
So, you can use the same code, but substitute three points from the circular plane (in V1_in) to your P matrix. Also, I often use the functions "patch" and "quiver3" to help visualize the normals.
Hope that helps!
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Matt J
il 30 Set 2023
Modificato: Matt J
il 30 Set 2023
Use planarFit() from this FEX package,
fitObj=planarFit(V_1')
fitObj =
planarFit with properties:
normal: [0.2147 -0.1801 0.9599]
distance: 82.5216
[hL,hD]=plot(fitObj); %Visualize the fit
legend([hL,hD],'Plane fit', 'XYZ samples', 'Location','northoutside','FontSize',15);
axis padded;
view(-70,15);drawnow
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