When IMU is gravity aligned, yaw is not observable from linear acceleration data.

However, when IMU has non null pitch, the sensor is not gravity aligned anymore and the gravity acceleration gets remapped on other axes.

Question: how do you compute the yaw angle from linear acceleration data when the pitch is not null (and therefore yaw is observable)?

Thank you in advance


2 Answers 2


If you assume the IMU lies on a fixed incline plane with pitch $\theta$, and you define yaw $\psi$ as the rotation between the IMU and the downhill direction, then the answer is

$$\psi = \text{atan2}(-a_y, a_x)$$

where $a_x$ and $a_y$ are the X and Y accelerometer readings respectively. Note that this is independent of the inclined plane pitch $\theta$ (which is not equal to the pitch as measured by the IMU; the IMU may be facing perpendicularly to the descent direction).

There are many definitions of yaw but the one we are using here is with respect to the downhill direction of the inclined plane. If that inclined plane were to rotate around the $z$ axis, the yaw as defined using e.g. Euler ZYX angles would change, but our yaw would stay the same. I suspect this is the disconnect between the question and the other answers. To be clear, this computation for $\psi$ is only the "local" yaw with respect to the incline. Getting the angle with respect to a fixed north or world $x$-axis given just an accelerometer reading and no other information is not possible. This is because gravity points down, and rotations around that axis (the down axis) will not cause a change in the accelerometer reading.

To derive this expression for $\psi$, note that the orientation of the IMU, $R$, in this inclined plane scenario only really has one degree of freedom: the yaw. The orientation can be described by first pitching downwards by an angle of $\theta$, then rotating around the new body $z$-axis by $\psi$. We can compute an expression for this orientation given $\theta$ and $\psi$ by multiplying an elementary rotation around the $y$ axis and an elementary rotation about the $z$ axis. See https://en.wikipedia.org/wiki/Rotation_matrix for their definitions. Since the rotation around $z$ is intrinsic (w.r.t. the new $z$-axis), we post multiply $R_y(\theta)$ with $R_z(\psi)$.

$$ R = R_y(\theta)R_z(\psi) = \begin{bmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\psi & -\sin\psi & 0 \\ \sin\psi & \cos\psi & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} \cos\theta \cos\psi & - \cos\theta \sin\psi & \sin\theta \\ \sin\psi & \cos\psi & 0 \\ -\sin\theta \cos\psi & \sin\theta \sin\psi & \cos\theta \end{bmatrix} $$

Now what does the accelerometer measure? It measures the gravity vector in the body frame*. The gravity vector in the world frame is $\begin{pmatrix} 0 & 0 & g \end{pmatrix}^\top$, so in the body frame, it is $$ R^\top \begin{pmatrix} 0 \\ 0 \\ g \end{pmatrix} = \begin{pmatrix} -g\sin\theta \cos\psi \\ g\sin\theta \sin\psi \\ g\cos\theta \end{pmatrix} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} $$

Now note that $$ -\frac{a_y}{a_x} = \tan \psi $$ and we have a simple way to compute $\psi$.

Note that when both $a_x$ and $a_y$ are zero, i.e. $\theta = 0$, we are not able to get any information about $\psi$. This corresponds to the intuition that on flat ground, the accelerometer cannot tell you about yaw.

*I am assuming there is no linear acceleration, which is a very common assumption. If the linear acceleration is known, it can be subtracted. For example, if this is a mobile robot, linear acceleration is typically close to zero or it can be estimated using the control inputs.


Yaw is not observable from gravity regardless of pitch.

  • $\begingroup$ How can you be so sure? The gravity acceleration if pitch is not null, gets remapped on other axes. So the yaw information should be there. $\endgroup$
    – Employee
    Commented Jan 7, 2020 at 1:19
  • $\begingroup$ This paper elaborates on yaw extraction from linear acceleration data - ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=893168 (page of interest is page 4) $\endgroup$
    – Employee
    Commented Jan 7, 2020 at 1:45
  • $\begingroup$ Close your eyes and spin around. You can not tell which way you were originally facing by tilting your head. Yaw is not observable from gravity. $\endgroup$
    – holmeski
    Commented Jan 7, 2020 at 11:39
  • 1
    $\begingroup$ Yaw rotations are defined in an intrinsic rotation system, so the example you gave with gravity is not relevant since that assumes an extrinsic rotation (or intrinsic with zero pitch. Stand on an inclined plane with your eyes closed and you can tell which way the inclination is would be a better analogy. $\endgroup$
    – 50k4
    Commented Jan 8, 2020 at 7:43
  • 1
    $\begingroup$ I've tested the approach, and it actually works fantastic. If pitch is non null, gravity acceleration gets remapped on other axis and the relative yaw of the sensor in respect to the inclined plane where it is lying on is beautifully observable. $\endgroup$
    – Employee
    Commented Feb 10, 2020 at 2:12

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