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| subject: | Re: ATM Robo Misalignment Calculations, did I do this right? |
From: "James Lerch"
To: "Jeff Anderson-Lee" ,
"ATM List"
Cc:
Reply-To: "James Lerch"
----- Original Message -----
From: "Jeff Anderson-Lee"
> > Couder/Video = Still using the Couder mask, however replaced the eyeball
> > with a video camera and Television monitor. Note, this was just a
single
> > impromptu run.
>
> It would be interesting to formalize this and capture several images at
each
> point where the human decided the grays were equal and see what the
computer
> analysis of those positions was.
>
Hi Jeff,
When the optic returns from testing (and depending on the results!) Yes, I
agree this would be interesting!
> How do these dimensions relate to the apparent pixel width of the
apparatus?
Each pixel represents about 0.04" of an inch of surface area, In each
frame of video data I grab a vertical line of 11 pixels (I've played with
the arrangement of pixels, nothing I did seemed to make a measurable change
in results). Anyways, with my single row of 11 pixels, in effect each zone
radius is the average intensity of a 0.04" wide x 0.44" tall
strip of pixels.
As far as positional accuracy (finding the mirror center in software, and
calculating the correct zone radius) the software is usually within a few
pixels. I should be pretty close as far as measuring the correct zone
radius location.
I did one experiment, where I assumed a perfect surface (parabolic) and
using my knife edge readings, I then 'solved' for the zone radiuses (using
the outer zone as the common point). The result of this little experiment
showed I would have had to mis-located zone 1 more than an inch in radius!
Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)
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