Hi Steve,
According to my understanding you want to understand how to determine if a point is within a rotated ellipsoid by rotating the ellipsoid itself and are seeking confirmation on the methods. The approach and calculations seem correct, and inclination toward method 2 is indeed correct. Let's go through the methods to confirm findings and clarify any confusion-
Method 1: Rotating the Ellipsoid Matrix
- The general equation for an ellipsoid centred at origin is: [ p' A p = 1 ] where ( p ) is the point vector and ( A ) is the ellipsoid matrix.
- For an ellipsoid aligned with the coordinate axes, ( A ) is a diagonal matrix with the inverse squares of the semi-axes: [ A = \text{diag}(1/a1^2, 1/a2^2, 1/a3^2) ]
- When the ellipsoid is rotated by a rotation matrix ( R ), the new ellipsoid matrix ( A' ) is given by: [ A' = R' A R ]
- The equation of the rotated ellipsoid becomes: [ p' A' p = 1 ]
Let's verify implementation of Method 1:
1. Define the diagonal matrix ( D ):
D = diag(axes.^(-2));
2. Compute the rotated ellipsoid matrix ( A' ):
D = diag(axes.^(-2));
3. Check if the point ( p0 ) satisfies the ellipsoid equation:
pie2 = p0' * A_rotated * p0 ;
Code for Method 1 appears to be correct. The result (pie2 = 0.1871) indicates that the point is inside the ellipsoid, which contradicts the backward rotation method.
Method 2: Using Eigen Decomposition
In Method 2, you correctly interpret that ( Q ) should be the rotation matrix ( R ), and ( D ) is the diagonal matrix with the inverse squares of the semi-axes. The equation of the rotated ellipsoid is: [ p' R D R' p = 1 ]
Let's verify implementation of Method 2:
1.Define the diagonal matrix ( D ):
D = diag(axes.^(-2));
2. Compute the ellipsoid equation with rotation:
pie3 = p0' * R * D * R' * p0;
Code for Method 2 is also correct. The result (pie3 = 1.8992) indicates that the point is outside the ellipsoid, which agrees with the backward rotation method.
Implementation of Method 2 is correct and aligns with the backward rotation method. Method 1, as implemented , is also correct but seems to give a different result. This discrepancy might be due to numerical precision or differences in interpretation. To further verify, cross-check with additional points and see if the results consistently match between the backward rotation method and Method 2.
Here's the MATLAB code for both methods for verification:
% point vector measured at x/y/z
p0 = [-70, -100, 300]';
% ellipsoid axes a1/a2/a3
axes = [1300, 100, 700];
% rotation angles tilt, roll, yaw degrees
r_angles = [4, -1, 120];
% rotation matrices for the ellipsoid
Rx = rotx(r_angles(1));
Ry = roty(r_angles(2));
Rz = rotz(r_angles(3));
% total rotation matrix
R = Rz * Rx * Ry;
% Backward rotation method
p1 = R' * p0;
pie = (p1(1)/axes(1))^2 + (p1(2)/axes(2))^2 + (p1(3)/axes(3))^2;
% Method 1
D = diag(axes.^(-2));
A_rotated = R' * D * R;
pie2 = p0' * A_rotated * p0;
% Method 2
pie3 = p0' * R * D * R' * p0;
% Display results
disp(['Backward Rotation Method: ', num2str(pie)]);
disp(['Method 1: ', num2str(pie2)]);
disp(['Method 2: ', num2str(pie3)]);
Both the backward rotation method and Method 2 should consistently give the same result, confirming the correctness of your approach.
Hope it helps!
Best Regards,
Simar
