PSN-L Email List Message

Subject: Re: optical mouse sensor
From: Charles R Patton charles.r.patton@........
Date: Sun, 20 Jan 2008 18:30:29 -0800


Dr. Peters,
I’d like to comment on one point in your pendulum paper where you 
mention, “In general there is a tendency for the mean position of the 
displayed waveform to migrate in spite of the absence of actual physical 
migration.” You surmise a Windows problem. I don’t believe it’s in 
Windows. Windows can’t tell what mouse is on the end of the plug – just 
that it’s a mouse, not if it’s made of encoder disks or optical sensor. 
These sensors would require different algorithms to process raw data.

A while back I started looking at the optical mouse as a optical encoder 
for use as a (ham) antenna position indicator. I abandoned the project 
when it became clear that the mouse itself does not put out a 
consistent, repeatable stream of pulses. I attribute this to the method 
of sensing that essentially does a center-of-light or center-of-darkness 
(the microscopic shadows cast from edge lighting of surface roughness) 
then estimates their change in position on the optical cell fields. The 
problem is that this is fuzzy involving coarse resolution a/d’s (that 
are very likely noisy, also) for each cell. So the simple test I did was 
just to run the mouse back and forth along a straight edge to stops. I 
would see the endpoints drift as this was repeated. I tried to develop a 
grating like pattern to turn the estimation into a fixed grating on the 
cells, figuring maybe I could get around the granularity, but didn’t 
finish the project. Nonetheless, sometimes the sensor will sit there and 
jitter or drift, even when it is standing still, which again I attribute 
to poor sensor algorithms in the mouse itself. Some optical mice use 
early versions of the HP sensor that can be re-jumpered to put out 
quadrature pulses instead of the mouse communication bytes. I didn’t do 
this experiment, but your pendulum would be perfect for it – just 
re-jumper one of those units for quadrature, and verify that the total 
is zero at the end of a swing down run. I’m sure you’ll find that it 
won’t be.

Regards,
Charles R. Patton
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