I read in this set of slides on p.11 that the IMU measurements, such as acceleration are affected by bias and noise, as expressed in this equation:

$_B \mathbf {\tilde a} _{WB}(t) =\ \mathbf R_{BW}(t)(_W \mathbf a_{WB}(t) -\ _W \mathbf g) + \mathbf b^a(t) + \mathbf n^a(t)$

where $b^a(t)$ is the bias of the accelerometer, $n^a(t)$ is the noise of the accelerometer, ${\tilde a}(t)$ is the measured acceleration, and a(t) is the true acceleration. As for notations:

Left subscript denotes the reference frame in which the quantity is expressed.

Right subscript {Q}{Frame1}{Frame2} denotes Q of Frame2 with respect to Frame1.

Last of all, noises and bias are all in the body frame.

What I would like to ask is what does the $R$ term in the equation mean and do? I'm guessing it is a rotation, but why is it needed? Sorry if this question seems trivial, but I'm new to all this and my physics is terrible... Thank you!


1 Answer 1


Maybe this answer is a bit late, but it could be useful for someone else.

Since the true acceleration is defined in the world frame, but the IMU's measured acceleration is defined in the body frame (the same as the IMU), you need the rotation matrix $_B\mathbf{\tilde{a}}_{WB}(t)$ to convert from one acceleration into the other. If you don't consider this matrix, the calculations will be wrong because you'd be mixing measurements observed from different perspectives.

With a practical example, let's suppose you are holding your phone (which we also suppose that already includes an IMU) with the screen pointing up, and you define a vector normal to the screen as a forward direction (this would be the so-called body frame). Then, if you move the phone in that direction, the IMU could measure an acceleration forward. However you, as an external observer, would observe that the phone moved upward rather than forward, because the notion of forward is different from your perspective. That apparent inconsistency is caused by the different coordinate frames we define and is fixed by knowing the relative transformation between them; in this case, that transformation is the rotation matrix we were talking before.


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