dlaidig
commented
4 months ago I can assure that the operation is correct. Consider the following example:
- Assume that the global magnetic field is [0.71 0 -0.71] (to keep it simple, I'm using a norm of 1 and a dip angle of 45° in this example).
- Starting from initial alignment of the local and global coordinate system, we rotate the sensor 45° around the y-axis.
- The x-axis of this sensor is now pointing 45° down, which means that this sensors now measures a magnetic field of [1 0 0].
- How do you need to rotate the local measurement vector [1 0 0] in order to obtain the global vector [0.71 0 -0.71]?
I've also formulated a very similar example in https://depositonce.tu-berlin.de/items/a1990767-cacd-4749-bb74-5ade7ff16d12:
vincenttate13
Hi @dlaidig, I have a question with respect to the rotation applied to the mag vector in order to yield magEarth.
From the updateMag function in vqf.cpp, magEarth is obtained by rotating the raw magnetometer measurement vector by the 6D orientation quat: getQuat6D(accGyrQuat); quatRotate(accGyrQuat, mag, magEarth);
If I understand the intention of this operation, why do you not need to invert accGyrQuat prior to applying it to mag? As an example, if you began with the IMU's local coordinate system in line with the global inertial coordinate system, then rolled the IMU +45 degrees, accGyrQuat should reflect a +45 degree rolled rotation. If the intention of this operation is to transform the mag vector into the inertial frame, would you not need to apply a -45 degree rolled rotation such that the magnetometer's local z-axis would realign with the global z-axis? Thanks!