How to rotate a pointcloud data and align parallel with y-z axis?

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Mathanraj
Mathanraj 2023 年 9 月 11 日
コメント済み: William Rose 2023 年 9 月 12 日
I have a point cloud data with x,y and z values. It is little inclined in the y-axis little bit.I want to fit and well align with the axis. I have tried PCA based rotation but it causes over rotation. I have attached the image shows inclined data and excel file of xyz values. Please help me to rotate the data and align perfect with the axis. Thanks for your support.

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Matt J
Matt J 2023 年 9 月 11 日
編集済み: Matt J 2023 年 9 月 11 日
yz=readmatrix('CS_DATA.xls','Range',[1,2]);
[~,v1]=min(yz*[+1;+1]);
[~,v2]=min(yz*[-1;+1]);
[x1,y1,x2,y2]=deal(yz(v1,1), yz(v1,2), yz(v2,1), yz(v2,2));
t=atan2d(y2-y1,x2-x1); %rotation angle
YZ=(yz-mean(yz))*[cosd(t), -sind(t);sind(t), cosd(t)] + mean(yz);
scatter(yz(:,1), yz(:,2)); hold on; scatter(YZ(:,1), YZ(:,2)); hold off
legend('Initial','Corrected', Location='east')
ylabel Z; xlabel Y
  7 件のコメント
Mathanraj
Mathanraj 2023 年 9 月 12 日
@William Rose Thanks for your detailed explanation and it really helps me to understand the concept of the Matt J's code. Yes, The data was collected by the scanner.
About the sheared look of the data, I used "axis equal" to rectify this issue. when I am using this command I can get the actual fitted look of the data without any shear. Is it okay to use this command?
And, sorry for the late reply.
William Rose
William Rose 2023 年 9 月 12 日
Yes it is fine and good to use axis equal.

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その他の回答 (2 件)

Constantino Carlos Reyes-Aldasoro
Constantino Carlos Reyes-Aldasoro 2023 年 9 月 11 日
This answer may solve your problem
https://uk.mathworks.com/matlabcentral/answers/123763-how-to-rotate-entire-3d-data-with-x-y-z-values-along-a-particular-axis-say-x-axis

William Rose
William Rose 2023 年 9 月 11 日
xyz=xlsread('CS_DATA.xls');
C=cov(xyz);
[V,D]=eig(C); % eigenvectors and eigenvalues of the covariance matrix
diag(C);
The eigenvectors are the directions of the axes of the best fit ellipse. The square roots of the eigenvalues are the lengths of the axes. Multiply by 2 so they'll be easier to see on the plot.
ellAxLen=2*diag(D).^0.5;
Display the axis lengths and the standard deviations along x,y,z. The similarity suggests that we are on the right track.
disp([std(xyz);ellAxLen'])
0.0005 0.0499 0.0865 0.0007 0.0997 0.1731
Plot the data
figure(1)
plot3(xyz(:,1),xyz(:,2),xyz(:,3),'k.');
xlabel('X'); ylabel('Y'); zlabel('Z'); grid on; hold on
Make matrices to represent the endpoints of the best fit axes:
ax1=[mean(xyz);mean(xyz)+ellAxLen(1)*V(:,1)'];
ax2=[mean(xyz);mean(xyz)+ellAxLen(2)*V(:,2)'];
ax3=[mean(xyz);mean(xyz)+ellAxLen(3)*V(:,3)'];
Add the axes to the plot
plot3(ax1(:,1),ax1(:,2),ax1(:,3),'-r','LineWidth',2);
plot3(ax2(:,1),ax2(:,2),ax2(:,3),'-g','LineWidth',2);
plot3(ax3(:,1),ax3(:,2),ax3(:,3),'-b','LineWidth',2);
When you do it, you will be able to use the 3d rotate tool to spin the plot around and see the axes better.
Use the eigenvector directions to rotate the data. Scale the eigenvectors to make a rotation matrix.
R1=zeros(3,3);
for i=1:3
R1(:,i)=V(:,i)/norm(V(:,i));
end
Rotate the data
xyzRot=(inv(R1)*(xyz-mean(xyz))')'+mean(xyz);
Plot the rotated points:
figure(2); plot3(xyzRot(:,1),xyzRot(:,2),xyzRot(:,3),'k.');
title('Rotated Point Cloud');
xlabel('X'); ylabel('Y'); zlabel('Z'); grid on;
After viewing the point cloud above from different angles, I think it is over-rotated about the X axis (middle panel below). See screenshot below of 3 views. The rotations about Y and Z look reasonable.
  1 件のコメント
William Rose
William Rose 2023 年 9 月 11 日
編集済み: William Rose 2023 年 9 月 11 日
[edit: correct typo and add description of alternate model]
A more complicated way would be to make a model I-beam with 6 adjustable parameters: L,W,D and orientation, which could be represented as 3 Euler angles, etc. (Assume the beam is made of infinitely thin plates.) Compute the sum squared distances from each point to the nearest part of the model I-beam. This would be tricky since you'd have to check distance from each point to the top plate, the bottom plate, and the middle plate, and take whichever distance is smallest. You may need 3 additional paramers for the center locaiton, if the points are not randomly sampled relative to the center.
Or the model beam is made of 3 infinite plates: 2 parallel, and one perpendicular to the other 2. The beam model has 4 parameters, not including orientation: its center location (x0,y0,z0) and the paprallel plate separation L. Infinite plates will make it easier to compute the distance to each of the three plates.
Then you use a fitting function like fmincon() to find the unknown parameters, including the 3 angles that minimize the sum squared distance. Then you would rotate by the inverse of the rotation specified by the 3 Euler angles.
Better yet, just eyeball it.

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