- 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 ]
Two methods to check if a point exists within an ellipsoid
14 views (last 30 days)
Show older comments
I have a need to determine if a point, measured in [x, y, z] from the origin, is within an ellipsoid also centered at the origin with semi-axes [a1, a2, a3] that is rotated around x/y/z by tilt/roll/yaw angles. Most answers I have found suggest it is easier to rotate the point backward into a non-rotated ellipsoid rather than rotating the equation of the ellipsoid. Then after the point is rotated, I can use the new x'/y'/z' in the standard equation of an ellipsoid: (x'/a1)^2+(y'/a2)^2+(z'/a3)^2=ans. If ans is less than or equal to 1, then the point is inside or on the ellipsoid. I am using a right handed system, for reference.
This is where I started and I believe I have the solution. However, I would like to also solve the problem by rotating the ellipoid, for verification and personal curiosity reasons.
What I have now is:
% 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
% rotations are intrinsic and in the order roll, tilt, yaw
Rx=rotx(r_angles(1))
Ry=roty(r_angles(2))
Rz=rotz(r_angles(3))
% total rotation matrix
R=Rz*Rx*Ry
% since rotation matrices inverses are the same as their transpose and
% the inverse of each matrix is the same as rotating in the opposite
% direction, i can apply the transpose rotation matrices in the reverse
% order to rotate the point backward.
% R'=Ry'*Rx'*Rz'
p1=R'*p0
% now I can check if the rotated point is inside the origin centered and
% aligned ellipsoid
% point in ellipsoid <=1; yes
pie=(p1(1)/axes(1))^2+(p1(2)/axes(2))^2+(p1(3)/axes(3))^2
% an answer above 1 suggests this point is not inside the ellipsoid
If I attempt to rotate the ellipsoid instead, I can start with the general equation for an ellipsoid using matrix notation.
(p-v)'*A*(p-v)=1
v is a vector with the center of the ellipsoid. Since my ellipsoid is at the origin, we can elimiante v.
p'*A*p=1
This is where my confusion comes in. I have seen some resources say that the general equation can be rotated in two ways:
1: By applying the rotation matrices as follows to A, which is defined as a square symmetric matrix with the inverse square of each axis on its diagonals.
(p-v)' * R' * A * R * (p-v) = 1
or
p' * R' * A * R * p = 1
2: Another method defines A differently. It has A defined as a positive definite matrix with eigen spaces(vectors?) as the principal axes of the ellipsoid and the eigenvalues are the squared inverses of the semiaxes. Which means that A is defined as: A=Q * D * Q'
D is the diagonal matrix with the inverse square of each axis on the diagonal. Q is the matrix with columns that represent the axes direction of the ellipsoid. I interpret this to mean that Q = R, since the ellipsoid started along the x/y/z axes. Which means:
p' * R * D * R' * p = 1
% continuing from here I can attempt both methods to see which works
% we can use p0 since it is not being rotated
% for method 1 we need to define the diagonal matrix
D=diag(axes.^(-2))
pie2=p0'*R'*D*R*p0
% for method 2 we just switch the locations of R' and R
pie3=p0'*R*D*R'*p0
Since method 2 agrees with the backward rotation method above, I am inclined to say that it is the correct method. However, I am unsure if I simply implemented the first method incorrectly. I would like some confirmation or correction of my methods. Where did I go wrong, or what am I missing to prove I did it right?
0 Comments
Answers (1)
Simar
on 1 Aug 2024
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
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
0 Comments
See Also
Categories
Find more on Creating and Concatenating Matrices in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!