Rotate a std::vector as a rigid transformation?

Question

I have a few 3d points, stored in a std::vector<Eigen::Vector3d>. I need to rigidly rotate and translate these points, without changing their relationship to one another. As if moving the cloud as a whole.

Based on this question:

https://stackoverflow.com/questions/50507665/eigen-rotate-a-vector3d-with-a-quaternion

I have this code:

std::vector<Eigen::Vector3d> pts, ptsMoved;


Eigen::Quaterniond rotateBy = Eigen::Quaterniond(0.1,0.5,0.08,0.02);
Eigen::Vector3d translateBy(1, 2.5, 1.5);


for (int i = 0; i < pts.size(); i++)
{
    //transform point                   
    Vector3d rot = rotateBy * (pts[i] + translateBy);
    ptsMoved.push_back(rot);

}

When i view the points and compare them to the original points, however, I get this: (White are the original, green are the transformed).

What i expect, is the cloud as a whole to look the same, just in a different position and orientation. What i get, is a moved and rotated and scaled cloud, that looks different to the original. What am i doing wrong?

EDIT:

If i apply the inverse transform to the adjusted points, using:

std::vector<Eigen::Vector3d> pntsBack;
for (int i = 0; i < ptsMoved.size(); i++)
{
    //transform point       
    Vector3d rot = rotateBy.inverse() * (ptsMoved[i] - translateBy);
    pntsBack.push_back(rot);
}

It gives me an even worse result. (dark green = original points, white = transformed, light green = transformed inverse)

If you apply the inverse transform to the transformed points, do you get back to where you started? — Caleth, Nov 23 '18 at 12:49
Or even better, apply the inverse transform to the camera, and compare *those* pictures — Caleth, Nov 23 '18 at 12:55
thanks for your reply. using `Vector3d rot = rotateBy.inverse() * (ptsMoved[i] -translateBy);` to apply the inverse, I get a very incorrect result. — anti, Nov 23 '18 at 12:56
that isn't the opposite transformation. Iirc it would be `(rotateBy.inverse() * ptsMoved[i]) - translateBy;` — Caleth, Nov 23 '18 at 12:57
To me, the differences look like [parallax](https://en.wikipedia.org/wiki/Parallax) — Caleth, Nov 23 '18 at 12:58
I just made that change :) Still very wrong result, however. I am attaching the result as image to question.. — anti, Nov 23 '18 at 12:58
Apparently, according to your link, it is possible to get the equivalent 3x3 rotation matrix. You could print it, and/or check it is an unitary matrix — Damien, Nov 23 '18 at 13:01
You can also apply rotation + inverse to vectors (1,0,0), (0,1,0) and (0,0,1) — Damien, Nov 23 '18 at 13:09
You could show us the code you used to calculate the quaternion — Damien, Nov 23 '18 at 13:13

score 2 · Accepted Answer · answered Nov 23 '18 at 13:49

2

Your Quaternion is not a unit-Quaternion, therefore you will get unspecified results. If you are not sure your quaternion is normalized, just write

rotateBy.normalize();

before using it. Additionally, if you want to rotate more than one vector it is more efficient to convert the Quaternion to a rotation matrix:

Eigen::Matrix3d rotMat = rotateBy.toRotationMatrix();
// ...
// inside for loop:
    Vector3d rot = rotMat * (ptsMoved[i] - translateBy);

Also, instead of .inverse() you can use .conjugate() for unit quaternions and .adjoint() or .transpose() for orthogonal Matrices.

answered Nov 23 '18 at 13:49

chtz

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Thank you! This has indeed fixed my issue. However, when I try to get the points back to where they started, using `Vector3d rot = rotateBy.conjugate() * (ptsMoved[i] - translateBy);` The rotation looks correct, but the translation is off in one axis. – anti Nov 23 '18 at 13:55
Adjusting teh order of operations has fixed this last part. `Vector3d rot = rotateBy.conjugate() * (ptsMoved[i]); rot = rot - translateBy;` thank you again for your help! – anti Nov 23 '18 at 14:19

Rotate a std::vector as a rigid transformation?

1 Answers1