>Renato Langersek wrote:

>If 9 microns pixel size are to small for a 10" f10 (which is not my opinion)
>than the use of a focal reducer will make it actually worse!
>Please think before you send a reply!
>Renato.

My reply:

You wrote: "Please think before you send a reply".

I SUGGEST YOU DO A LITTLE RESEARCH BEFORE FLAMING OTHERS.

Anybody on this list will probably confirm that 2 arc-secs per pixel is
about optimum for ccd imagong in average amature seeing conditions
(below 5,000 feet elevation).

Here is the equation:

Arc-secs/pixel = 206,265 secs-per-radian x ccc-pixel-size (in mm)/focal-length.

Your scope has a focal length of 10" x f10 = 100" = 2540mm.
So: Arc-secs/pixel = 206,265 secs-per-radian x .009mm/2540mm = .7308602

This is to small for average seeing conditions at f10.

Using an f6.3 focal reducer will reduce your focal length to 2540 x .63 = 1600.2mm.
So: Arc-secs/pixel = 206,265 secs-per-radian x .009mm / 1600.2mm = 1.1600956.

Q.E.D. The focal reducer gets you closer to the ideal goal of 2 arc-secs/pixel.

Actually you would be better off with an "Optec f3.3" focal reducer; that will
get you: Arc-secs/pixel = 206,265 secs-per-radian x .009mm / 838.2mm = 2.2147279.

This is almost ideal, as it is near the target 2 arc-secs/pixel!!!

You can take ccd images at f10 with 9 micron pixels, and they will probably
look ok, but yu will be over-sampeling and therefore reducing your sensitivity
and introducing unnecessary noise into the image.




Chris Frye
Silver Spring, MD.

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