TL;DR: Does anybody know the width of a single crosshair as a fraction of the full field of view on a Skywatcher 1.25 inch 12.5mm illuminated reticle?

Explanation (in case there's a different question I should have asked instead).

I'm doing drift alignment. Apparently, my particular camera + telescope combination means that each pixel is approximately 1.06 arcseconds wide/long. Therefore, if I want to take an exposure of length T minutes, I need to achieve a drift of less than 1.06/T arcseconds in declination per minute.

I have an eyepiece with illuminated crosshairs. Apparently, the field of view is approximately 2400 arcseconds. If I can keep a star sitting within the crosshair for t minutes, and the crosshair is taking up a fraction c of the total field of view, then I've got the drift down to at most 2400c/t arcseconds per minute.

This is why I want to know the value of c. I estimated it by eye to be around 1/500 (i.e. about 1/10 of the inter-crosshair distance, which is about 1/50 of the full field of view - it's a double cross-hairs), but I wondered if this was a known quantity.

Clearly, the smaller c is, the less time I need to spend checking for drift, so extra points for anyone who can tell me that c < 1/500.

  • 2
    $\begingroup$ You can get a pretty good estimate if you are able to take the eyepiece out of the telescope an look at it and estimate the physical width of the crosshair. The width at the focal plane divided by the focal length of the telescope will be pretty close to the angular width in radians. If it's 0.5 mm wide and the focal length is 1200 mm, then it's 0.00042 radians or 0.024 degrees or 86 arcseconds wide. Of course if you are looking for a few arcseconds only, then the crosshair will have to be extremely thin and so this won't work. Can't you just time an "occultation" of a star drifting behind it? $\endgroup$
    – uhoh
    Apr 5 at 23:25
  • $\begingroup$ Drift is 15 arcseconds per one second of time times the cosine of declination. $\endgroup$
    – uhoh
    Apr 5 at 23:26
  • 2
    $\begingroup$ @uhoh: "Can't you just time an 'occultation' of a star drifting behind it?" Good idea... I'll have to wait for a clear night then! $\endgroup$ Apr 6 at 6:31

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