Proper "rule" for pinpoint stars with M43

Here's another calculation and some practical results.

I computed the angular field of view (afv) for one pixel.

afv = 2 arctan (pp / 2f) ; pp = pixel pitch

(see: http://www.edmundoptics.com/resources/application-notes/imaging/understanding-focal-length-and-field-of-view/ )

This is a small number so we can use Taylor approximation, arctan x = x. So

afv = pp / f.

Now, setting this to the rotation of the earth (equatorial), w T (where w is angular speed per second, T = exposure duration), you get

w T = pp / f

or T = (pp/w) / f

w is 7.3 x 10^-5 ; pp is 2.226 x 10^-3 for 16 mpx m4/3 sensor

So pp/w is about 30. For one pixel, T = 30 / f

Test: I took a 15 sec exposure, f = 20 mm. So, this should be 10 pixel "blur." Here's the actual, compared to the calculation.

Star trail, 100%,

Ten pixels

That's pretty close since this is not an equatorial shot...not in the picture. Looks like maybe 2/3 of equatorial speed, which sounds totally reasonable.

I'll post the original just below. On my computer screen, star trails are hardly noticeable. But, on a 24 inch wide print, I find them objectionable, by about a factor of two.

SO, bottom line you get start trails of 30 / f  pixels in length, worst case (equator). PRACTICALLY, how many pixels can you stand? For 13 inch computer display, 10 is fine (equivalent to 300 / f (600 / f if you use FF equivalent). For my 24 inch print, I'd be happier with half that, equivalent to 300 / f (using FF equivalent). JUDGE ON THE BASIS OF YOUR viewing USE. As I sometimes print large, I'm going to use 300 / f (FF equivalent), if I can.

These conclusions are largely consistent with your conclusions. The difference is probably in our standards (large print vs. monitor display).

BTW, your CoC is 6.7 pixels, according to my calculations. That seems a little generous. I think I can plainly see differences of about a half of that.

Here's that image: 20 mm, 15 sec exposure:

15 secs, 20 mm; looks like star trails are about 7 pixels