Initial Calibration of MEMS Accelerometers, Used for Measuring Inclination and Toolface

Abstract

In the present work, we consider calibrating MEMS accelerometers for the purpose of determining orientation in space. We propose a new objective function whose minimization gives an estimate of the calibration coefficients. The latter takes into account the specifics of measuring toolface and inclination in seek of better accuracy, when the device is used for this purpose. To the best of our knowledge, such an objective function has not been mentioned in the literature. The calibration algorithm is described in detail because, even though, some of the steps are standard from the point of view of a numerical analyst, this could be helpful for an engineer or an applied scientist, looking to make a concrete implementation for applied purposes. On the basis of numerical experiments with sensor data, we compare the accuracy of the proposed algorithm with a classical method. We show that the proposed one has an advantage when sensors are to be used for orientation purposes.

Appendix

Here, we shall give formulas for computing the Jacobian matrix, associated with the linearization of the minimization problem (6).

We have

$$ M.\hat{\mathbf{a }}_i+\mathbf b =\left[ \begin{array}{c} m_{xx}.\hat{a}_{i,x}+m_{xy}.\hat{a}_{i,y}+m_{xz}.\hat{a}_{i,z}+b_x\\ m_{yx}.\hat{a}_{i,x}+m_{yy}.\hat{a}_{i,y}+m_{yz}.\hat{a}_{i,z}+b_y\\ m_{zx}.\hat{a}_{i,x}+m_{zy}.\hat{a}_{i,y}+m_{zz}.\hat{a}_{i,z}+b_z \end{array} \right] . $$

Let us define for $i=\overline{1,n}$ and $j=\overline{1,12}$

$$\begin{aligned} \begin{aligned} J_{2i-1,j}&=\frac{\partial r_{2i-1}}{\partial p_j}=\frac{360^\circ }{2\pi }\frac{\partial }{\partial p_j}tf(a_i)\\&=\frac{360^\circ }{2\pi }\frac{\partial }{\partial p_j}\arctan \left( \frac{m_{xx}.\hat{a}_{i,x}+m_{xy}.\hat{a}_{i,y}+m_{xz}.\hat{a}_{i,z}+b_x}{m_{yx}.\hat{a}_{i,x}+m_{yy}.\hat{a}_{i,y}+m_{yz}.\hat{a}_{i,z}+b_y}\right) \\&=\frac{360^\circ }{2\pi }\frac{\partial }{\partial p_j}\arctan \left( \frac{p_1.\hat{a}_{i,x}+p_2.\hat{a}_{i,y}+p_3.\hat{a}_{i,z}+p_{10}}{p_4.\hat{a}_{i,x}+p_5.\hat{a}_{i,y}+p_6.\hat{a}_{i,z}+p_{11}}\right) \\&=:\frac{360^\circ }{2\pi }\frac{\partial }{\partial p_j}\arctan \left( \frac{A}{B}\right) . \end{aligned} \end{aligned}$$

(13)

Let us denote for further use:

$$ \begin{aligned} B_1&=p_1 \hat{a}_{i,x}+p_2 \hat{a}_{i,y}+p_3 \hat{a}_{i,z}+p_{10},\\ B_2&=p_4 \hat{a}_{i,x}+p_5 \hat{a}_{i,y}+p_6 \hat{a}_{i,z}+p_{11},\\ \end{aligned} $$

$$ \begin{aligned} B_3&=p_7 \hat{a}_{i,x}+p_8 \hat{a}_{i,y}+p_9 \hat{a}_{i,z}+p_{12},\\ B&=\sqrt{B_1^2+B_2^2+B_3^2},~~A=B_1^2+B_2^2. \end{aligned} $$

After straightforward (but rather lengthy) computations that we omit due to lack of space, one can obtain:

$$ \begin{aligned}&J_{2i-1,1}=\frac{360^\circ \hat{a}_{i,x}}{2\pi B\left( 1+\frac{A^2}{B^2}\right) },~ J_{2i-1,2}=\frac{360^\circ \hat{a}_{i,y}}{2\pi B\left( 1+\frac{A^2}{B^2}\right) },\\&J_{2i-1,3}=\frac{360^\circ \hat{a}_{i,z}}{2\pi B\left( 1+\frac{A^2}{B^2}\right) },~ J_{2i-1,10}=\frac{360^\circ }{2\pi B\left( 1+\frac{A^2}{B^2}\right) },\\&J_{2i-1,4}=-\frac{360^\circ A \hat{a}_{i,x}}{2\pi B^2\left( 1+\frac{A^2}{B^2}\right) },~ J_{2i-1,5}=-\frac{360^\circ A \hat{a}_{i,y}}{2\pi B^2\left( 1+\frac{A^2}{B^2}\right) },\\&J_{2i-1,6}=-\frac{360^\circ A \hat{a}_{i,z}}{2\pi B^2\left( 1+\frac{A^2}{B^2}\right) },~ J_{2i-1,11}=-\frac{360^\circ A }{2\pi B^2\left( 1+\frac{A^2}{B^2}\right) },\\&J_{2i-1,7}=J_{2i-1,8}=J_{2i-1,9}=J_{2i-1,12}=0. \end{aligned} $$

Further, if

$$\begin{aligned} \frac{360^\circ }{2\pi }\left( \arccos \left( \frac{\hat{a}_{i,z}}{\sqrt{\hat{a}_{i,x}^2+\hat{a}_{i,y}^2+\hat{a}_{i,z}^2}}\right) \right) >60^\circ \end{aligned}$$

(14)

holds true, we have:

$$ \begin{aligned}&J_{2i,1}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,x}B_1A}{B^4}+\frac{2\hat{a}_{i,x}B_1}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,2}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,y}B_1A}{B^4}+\frac{2\hat{a}_{i,y}B_1}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},\\&J_{2i,3}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,z}B_1A}{B^4}+\frac{2\hat{a}_{i,z}B_1}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,10}=-\frac{360^\circ }{2\pi }\frac{-\frac{2B_1A}{B^4}+\frac{2B_1}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},\\&J_{2i,4}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,x}B_2A}{B^4}+\frac{2\hat{a}_{i,x}B_2}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,5}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,y}B_2A}{B^4}+\frac{2\hat{a}_{i,y}B_2}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},\\&J_{2i,6}=-\frac{360^\circ }{2\pi }\frac{-\frac{2\hat{a}_{i,z}B_2A}{B^4}+\frac{2\hat{a}_{i,z}B_2}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,11}=-\frac{360^\circ }{2\pi }\frac{-\frac{2B_2A}{B^4}+\frac{2B_2}{B^2}}{2\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},\\&J_{2i,7}=\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,x}AB_3}{B^4\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,8}=\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,y}AB_3}{B^4\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},\\&J_{2i,9}=\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,z}AB_3}{B^4\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}},~ J_{2i,12}=\frac{360^\circ }{2\pi }\frac{AB_3}{B^4\sqrt{\frac{A}{B^2}}\sqrt{1-\frac{A}{B^2}}}. \end{aligned} $$

Otherwise, if condition (14) is not fulfilled, then:

$$ \begin{aligned}&J_{2i,1}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,x}B_1B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,2}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,y}B_1B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},\\&J_{2i,3}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,z}B_1B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,10}=-\frac{360^\circ }{2\pi }\frac{B_1B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},\\&J_{2i,4}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,x}B_2B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,5}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,y}B_2B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},\\&J_{2i,6}=-\frac{360^\circ }{2\pi }\frac{\hat{a}_{i,z}B_2B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,11}=-\frac{360^\circ }{2\pi }\frac{B_2B_3}{B^3\sqrt{1-\frac{B_3^2}{B^2}}},\\&J_{2i,7}=-\frac{360^\circ }{2\pi }\frac{-\frac{\hat{a}_{i,x}B_3^2}{B^3}+\frac{\hat{a}_{i,x}}{B}}{\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,8}=-\frac{360^\circ }{2\pi }\frac{-\frac{\hat{a}_{i,y}B_3^2}{B^3}+\frac{\hat{a}_{i,y}}{B}}{\sqrt{1-\frac{B_3^2}{B^2}}},\\&J_{2i,9}=-\frac{360^\circ }{2\pi }\frac{-\frac{\hat{a}_{i,z}B_3^2}{B^3}+\frac{\hat{a}_{i,z}}{B}}{\sqrt{1-\frac{B_3^2}{B^2}}},~ J_{2i,12}=-\frac{360^\circ }{2\pi }\frac{-\frac{B_3^2}{B^3}+\frac{1}{B}}{\sqrt{1-\frac{B_3^2}{B^2}}}. \end{aligned} $$