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I'm struggling with a seemingly simple problem in 2D motion. Basically, the question is, given accelerations in $x$ and $y$ ($a_x$ and $a_y$) as well as the angular velocity ($\omega$), how can we find the trajectory of the motion? Also, how can we report the motion like a computer mouse, i.e. in the reference frame of the sensor?

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closed as unclear what you're asking by ja72, Nathaniel, Chris White, Alan Rominger, Manishearth Jul 5 '13 at 14:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

You do know that, for a mouse, $\omega r_{ball}=\sqrt{v%x^t+v_y^2}$? Theres a similar relation for acceleration. My apologies if you did infact mention this, mathjax isn't working for me atn. – Manishearth Mar 15 '12 at 17:44
Fwiw, a mouse has no accelerometer/gyroscopes. It jas two wheels in contact with the roller. These wheels are attached a slotted wheel each. Lihht is passed through the wheel and a detector measures it. The frequency of oscillation of the light signal is proportional to the speed. Mind you, this is for a mechanical mouse. An optical mouse uses some nifty technique (akin to barcode scanners) that I forgot. – Manishearth Mar 15 '12 at 17:50
@Manishhearh thanks for your comments. My question does not really concern existing computer mice. I am just thinking about building one only with accelerometers and gyros. – Shapul Mar 15 '12 at 18:07
Why the gyroscope? Just double-integrate x,y. Also, truncate small velocities to zero, to prevent drift. – Mike Dunlavey Mar 16 '12 at 12:39
@Shapul, if you would like to determine the position from this, you would have to do some numerical integration. By the way Dunlavey has a good point, since with this method you will be bound to have (small) errors and therefore drift. – fibonatic Jul 4 '13 at 21:09

1 Answer 1

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In this hypothetical situation, you can transform the unit vectors in the global reference frame $\hat{x}$ and $\hat{y}$ using the same rotation matrix, to the unit vectors in the transformed coordinate system:

$$ \hat{x}' = R \hat{x}\\ \hat{y}' = R \hat{y} $$

This is what you did implicitly by transforming $x\hat{x}+y\hat{y}$

In the same way, you can do the back transformation

$$ \hat{x}=R^{-1}\hat{x}'\\ \hat{y}=R^{-1}\hat{y}'$$

Which can be used for the back-transformation of the dislpacement in the local frame to displacement in the global frame.

A practical issue will be that errors in you integrated angle $\alpha$ will accumulate, making the mouse annoying to use.

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Ah, yes, I think you are right. It is not really complicated but in my confusion after returning back to basic kinematics after many years, I totally missed it. You are right accumulation of errors in integration. Real sensors will also have drift making integration (and double integration) really difficult. – Shapul Mar 15 '12 at 19:48
@Shapul I'm glad I could help! – Bernhard Mar 15 '12 at 20:32

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