I want to process accelerometer and gyroscope data. To correctly track my position I have to account for the Earth's rotation and gravity. I want to know the position in my rotating Earth frame but the data is relative to an inertial frame i.

Since the Earth is rotating, I have this equation for the velocity:

$$\dot{x}_i = \dot{x}_r + \omega \times x_r$$

I derive this equation to get the acceleration:

$$ \begin{aligned} \ddot{x}_i &= \ddot{x}_r + \omega \times \dot{x}_r + \dot{\omega} \times x_r + \omega \times (\dot{x}_r + \omega \times x_r)\\ \ddot{x}_i &= \ddot{x}_r + 2\omega \times \dot{x}_r + \dot{\omega} \times x_r + \omega \times (\omega \times x_r) \end{aligned} $$

For a constant rotation I get the following result as my relative acceleration:

$$\ddot{x}_r = \ddot{x}_i - 2\omega \times \dot{x}_r - \omega \times (\omega \times x_r)$$

The $\ddot{x}_i$ represents the measured acceleration + gravity. I think.

However I have looked up a few papers on the navigation equation. What I pieced together was this derivation, starting from the same velocity equation:

$$\dot{x}_i = \dot{x}_r + \omega \times x_r$$

they take this derivative:

$$\frac{d\dot{x}_i}{dt} = \frac{d\dot{x}_r}{dt} + \frac{d(\omega \times x_r)}{dt}$$

or with a constant acceleration:

$$\frac{d\dot{x}_i}{dt} = \frac{d\dot{x}_r}{dt} + \omega \times \frac{dx_r}{dt}$$

which turns into:

$$\frac{d\dot{x}_i}{dt} = \frac{d\dot{x}_r}{dt} + \omega \times (\dot{x}_r + \omega \times x_r)$$

This expression is rewritten to:

$$ \begin{aligned} \ddot{x}_i &= \ddot{x}_r + \omega \times \dot{x}_r + \omega \times ( \omega \times x_r)\\ \ddot{x}_r &= \ddot{x}_i - \omega \times \dot{x}_r - \omega \times ( \omega \times x_r) \end{aligned} $$

With $\ddot{x}_i$ equal to the measured specific force + gravity.

Between what I derive and what I find in the papers there is a factor 2 difference in the term $\omega \times \dot{x}_r$.

Yes, there is a difference in the total and partial derivative of the $\dot{x}_r$ vector. But the force as measured by a MEMS-accelerometer and gyroscope can only correspond to one of either solutions.

Which subtlety am I missing/forgetting here?


1 Answer 1


Your expression ("I get the following result...") is correct - you correctly obtained the terms for the Coriolis and centrifugal accelerations, respectively (both "fictitious" accelerations that appear in a rotating frame of reference).

I can't comment on your "papers on navigation equation" - but if they give a different answer, either your interpretation of these papers is wrong, or the papers themselves are. Without appropriate detail and references I can't tell you which of these it is.

UPDATE I could not see all the pages in the linked book, but what I saw looked wrong. By contrast, this paper has the same equations that you derived (including the factor 2x for the Coriolis term). It seems to me that your book is wrong. You might want to contact the authors...

  • $\begingroup$ Yes, the equation derived is what I was taught and is what I am finding in the textbook I have. Yet the papers on this topic give a different result and unfortunately my mechanics knowledge is covered under a lot of dust. Most papers just give the set of navigation equations to solve without deriving them. This book does go into detail and is more less what I wrote in the question. Starting at page 23. books.google.be/… $\endgroup$
    – user965972
    Dec 21, 2016 at 20:34

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