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

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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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