Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
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.
2 Answers
- CharlesLv 61 decade agoFavorite Answer
Real field will be about 2 degrees.
http://www.davidpaulgreen.com/tec.html
HTH
Charles
[Late edit]: Well, not to be too pedantic, but an eyepiece's apparent field of view is the angular diameter, expressed in degrees (°), of the circle of light that the eye sees. It is analogous to the screen of a television (not the picture seen through it). Eyepiece apparent fields range from narrow (25° - 30°) to extra-wide angle (80° or more).
The true field (or real field) of view is the angle of sky seen through the eyepiece when it's attached to the telescope. The true field can be approximated using the formula:
True field = Apparent field / Magnification
For example, suppose you have an 8" Schmidt-Cassegrain telescope with a 2000mm focal length, and a 20mm eyepiece with a 50° apparent field. The magnification would be 100X (2000mm / 20mm). The true field would be 50 / 100, or 0.5° - about the same apparent diameter as the full Moon or the Sun.
HTH
Charles
- GeoffGLv 71 decade ago
A good approximation is to divide the apparent field of view by the magnification. In this case, the magnification will be 1200/32 = 37.5x, so the field of view will be 70/37.5 = 1.87° = 1° 52'. This should work quite well, though you may notice some coma towards the edge of the field.
This is only an approximation of the actual field of view, which can be calculated by measuring the field stop of the eyepiece, dividing this by the focal length of the telescope, and multiplying by 57.3 (answer in degrees). If you really want to be accurate, you can time the drift of a star across the field of view. This will be most accurate if you use a star right on the celestial equator, such as the easternmost star in Orion's Belt. A star on the celestial equator moves 360° in 24 hours, or 360/24/60 = 0.25 degrees in one minute.
