Electronic – calculating inclination using accelerometer

Let's put together a simple mathematical model of an accelerometer - from this, we can work out some calibration options.

Ignoring non-linearity and other nasty effects, the output measurement of an accelerometer is given by:

$$\hat{\mathbf{f}} = \mathbf{M} \mathbf{f} + \mathbf{b}_a + \mathbf{n}_a$$

where \$\hat{\mathbf{f}}\$ is the actual measurement, \$\mathbf{b}_a\$ is the accelerometer bias, \$\mathbf{n}_a\$ is a random noise vector, \$\mathbf{f}\$ is the true specific force (i.e. acceleration) and \$\mathbf{M}\$ is the Scale Factor/Misalignment Matrix.

The individual elements of the SFA matrix are: $$ \mathbf{M} = \begin{bmatrix} S_x && \gamma_{xy} && \gamma_{xz} \\ \gamma_{yx} && S_{yy} && \gamma_{yz} \\ S_x && \gamma_{zy} && S_{zz} \\ \end{bmatrix} $$

So, each scale factor is represented by an \$S\$ and each cross-axis sensitivity is represented by a \$\gamma\$.

Ideally, if the scale factor is 1 and there is no cross-axis sensitivity, then then the resulting matrix is \$\mathbf{M} = \mathbf{I}\$.

Representing it like this allows us to develop a compensation model. If we happen to know \$\mathbf{M}\$ and \$\mathbf{b}_a\$ and assume \$\mathbf{n}_a\$ to be small (i.e. close to zero), we can make a good estimate of the "true" acceleration from the measurements: $$ \mathbf{f} = \mathbf{M}^{-1}\left(\hat{\mathbf{f}} - \mathbf{b}_a\right) $$

The trick is, of course, working out \$\mathbf{M}\$ and \$\mathbf{b}_a\$.

I'll describe a procedure called the six position test, which is an easy and cheap way to calibrate an accelerometer. Step 1 is to mount the accelerometer in a rectangular box with perfectly \$90^\circ\$ sides (or as close as you can get). Place this on a perfectly level surface (or, again, as close as you can get) - you'd be surprised how good you can do this.

At this point, we know what the value should be: gravity on the z-accelerometer: $$ \mathbf{f}_1 = \begin{bmatrix} 0 \\ 0 \\ g\end{bmatrix} $$

So, this becomes: $$ \hat{\mathbf{f}}_1 = \mathbf{M} \mathbf{f}_1 + \mathbf{b}_a + \mathbf{n}_a $$ Noting that \$\hat{\mathbf{f}}_1\$ will close, but not the same as \$\mathbf{f}_1\$

If we put the box on it's head, the force acting is \$-g\$: $$ \mathbf{f}_2 = \begin{bmatrix} 0 \\ 0 \\ -g\end{bmatrix} $$

And when placed on one side: $$ \mathbf{f}_3 = \begin{bmatrix} 0 \\ -g \\ 0\end{bmatrix} $$

And so on for the remaining three sides.

Now, let's write out one of the equations longhand: $$ \hat{\mathbf{f}}_1 = \mathbf{M} \mathbf{f}_1 + \mathbf{b}_a + \mathbf{n}_a = \begin{bmatrix} S_{xx} f_x + \gamma_{xy} f_y + \gamma_{xz} f_z + b_x \\ \gamma_{yx} f_x + S_{yy} f_y + \gamma_{yz} f_z + b_y \\ \gamma_{xz} f_x + \gamma_{yz} f_y + S_{zz} f_z + b_z \end{bmatrix} $$

And even longer hand (for the first one): $$ \hat{\mathbf{f}}_1 = \begin{bmatrix} f_x S_{xx} + f_y \gamma_{xy} + f_z \gamma_{xz} + 0 \gamma_{yx} + 0 S_{yy} + 0 \gamma_{yz} + 0 \gamma_{xz} + 0 \gamma_{yz} + 0 S_{zz} + 1 b_x + 0 b_y + 0 b_z \\ 0 S_{xx} + 0 \gamma_{xy} + 0 \gamma_{xz} + f_x \gamma_{yx} + f_y S_{yy} + f_z \gamma_{yz} + 0 \gamma_{xz} + 0 \gamma_{yz} + 0 S_{zz} + 0 b_x + 1 b_y + 0 b_z \\ 0 S_{xx} + 0 \gamma_{xy} + 0 \gamma_{xz} + 0 \gamma_{yx} + 0 S_{yy} + 0 \gamma_{yz} + f_x \gamma_{xz} + f_y \gamma_{yz} + f_z S_{zz} + 0 b_x + 0 b_y + 1 b_z \end{bmatrix} $$

So we can create a stacked vector of the unknowns $$ \mathbf{z} = \mathbf{A} \mathbf{\beta} $$ Where $$ \mathbf{z} = \begin{bmatrix} \hat{\mathbf{f}}_1 \\ \hat{\mathbf{f}}_2 \\ \vdots \\ \hat{\mathbf{f}}_6 \end{bmatrix} $$

And $$ \mathbf{\beta} = \begin{bmatrix} S_{xx} \\ \gamma_{xy} \\ \gamma_{xz} \\ \gamma_{yx} \\ S_{yy} \\ \gamma_{yz} \\ \gamma_{xz} \\ \gamma_{yz} \\ S_{zz} \\ b_x \\ b_y \\ b_z \\ \end{bmatrix} $$

The design matrix is (for one set of measurements): $$ \hat{A}_1 = \begin{bmatrix} f_x && f_y && f_z && 0 && 0 && 0 && 0 && 0 && 0 && 1 && 0 && 0 \\ 0 && 0 && 0 && f_x && f_y && f_z && 0 && 0 && 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 && f_x && f_y && f_z && 0 && 0 && 1 \end{bmatrix} $$

Now, once this is setup, one may solve for \$\mathbf{\beta}\$ (and hence sensitivity and bias) via least squares.

A similar procedure may be performed with a robotic arm if you can precisely control the angles - it simply replies knowing the precise gravity at that angle which, if you know the angle, is easy to calculate.

I've worked on a project to implement the Kalman filter on an embedded system that was similar in hardware to the iNemo unit from STMicroelectronics.

Even if you can find these IMU (Inertial Measuring Unit) with 90% chance you will have to implement your algorithm by yourself; or if you're lucky, you can find someone that has the code. The problem is that this filter requires a lot of computation, and in our best experiment (using fixed point variables and trying to optimize the code) we were able to run it 45 times per second, in a STM32 at 72 MHz.

So maybe there is one, but as far as I know requires a good microcontroller or maybe a FPGA\ASIC.