How do you compute field of view for an eyepiece/scope?
I've got the Orion xt10i. It has a 254 mm aperture, and 1200 mm of focal length. It comes with a 25 mm Plossl - which has a 52 degree apparent filed of view. The field of view of the scope with this eyepiece is around 35 minutes (0.55 degrees). If i get a 32 mm eyepiece with a 70 degree AFOV, what real field of view should i expect?
Note:
I've picked 32 mm, as i'm trying to keep the exit pupil under 7 mm, to maximize brightness (for kids, anyway). This would give me 6.7 mm exit pupil.
What i really want to know is exactly how to compute the real field of view, if it can be done with math. I have 6 eyepieces (so far) that i want to compare. Then, since the difference between theory and practice is that, well, in theory they're the same - i want to actually borrow some of these and see how they work with my scope and my eyes, if at all possible.
Oh. Cool javascript app. I'll have to see how it works.
I've not had much luck timing a star crossing the field of view. It seems easier to pick a pair of fairly bright stars that happen to just fit in a field of view, then look up the separation in a planetarium program. My planetarium programs go down to 12th magnitude, anyway.
OK, so it turns out that vendors list the Field Stop for these eyepieces (and my existing EPs). If it's FS / FL * 57.3, then since FL is constant (1200 mm) - the bigger the FS the bigger the FOV. Shouldn't it depend on magnification or FL of the eyepiece?
Two formulas. While they get similar answers for my existing EPs, they don't for those under consideration.
I knew what time it was when i had just one clock.
It appears that the math is telling me that a $640 31mm Nagler has nearly twice the FOV of my 25mm Sirius plossl. A $100 32mm Q70 beats my 25mm by 1.65x. I'll certainly have to put the 32mm Q70 at the top of my wish list to borrow. So at least the field is narrowed from 5 to 2.