Hi
I did the same calculation as OzEclipse but did not get the same result. Joe, could you explain how you made your calculation ? I explain the maths behind mine
here (that's in french but Google can translate it).
No matter the crop factor, if you take into account the physics under the optical light path (Airy disk) and the CFA pattern of the DSLRs, you will arrive to the following rule that is a good compromise between star trails close to the equator and allowable exposure time :
Simplified NPF rule:
t = [35*N + 30*p(µm) ] / f (mm)
Where :
- N is the aperture
(and not the number of pixel drift as Joe's formula)
- f is the focal length in millimeters
- p is the pixel size, in microns
For example :
- 5D mk II (pixels=6.4µm) with a 16 mm lens open at 2.8 : t=[35x2.8+30x6.4]/16=18 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 2.8 : t=[35x2.8+30x4.7]/16=15 s
- Nokia Lumia 1020 (pixel=1.1µm), focal length=5.9 mm, aperture=2.2 : t=[35x2.2+30x1.1]/5.9=19 s (but this phone is not able to shoot such a long exposure !)
This simplified formula is based on declination of 60°. It is very close to the formula of OzEclipse except it adds N and P instead of multiplying them.
Should you need a more precise formula to take into account the declination of the stars, and really experience NO trail anywhere on your shot, then it becomes :
Full NPF rule : t=[17*N + 14*p(µm) + f(mm)/10 ] / [f(mm)*cos(declination)]
Therefore, using the same equipment and at the equatorial plan (cos(dec)=1) which is the most defavourable place :
- 5D mk II (pixels=6.4µm) with a 16 mm lens open at 2.8 : t=[17x2.8+14x6.4+16/10]/16=9 s
- 50D (pixels=4.7 µm) with a 16 mm lens open at 2.8 : t=[17x2.8+14x4.7+16/10]/16=7 s
- Nokia Lumia 1020 (pixel=1.1µm), focal length=5.9 mm, aperture=2.2 : t=[17x2.2+14x1.1+5.9/10]/5.9=9 s (but again, this phone is not able to shoot such a long exposure !)
Fred