Field of View vs. Apparent F.O.V.

[Ed Isenberg]

There's a term everyone uses that has me confused, and that is
"apparent" field of view. The Meade catalog, for example, says (on page
111) that all of its Ultra Wide Angle lenses have an APPARENT field of
view of 84 degrees. In the text on the same page, however, it says the
14mm lens would give an ACTUAL field of view of 0.6 degrees in an 8"
f/10 SCT, but a full 1 degree used with a 10" f/4.5 Newtonian reflector.

So, what is "apparent field of view," how could it be the same for
eyepieces of different sizes, and how does one calculate the ACTUAL
field of view given the focal length and f ratio of a specific
telescope? Also, does it matter (other than FL and F-ratio) what type of
telescope one is using (i.e.., refractor, reflector, SCT, MCT)?

It seems odd to me that I can estimate with a glance and a hand gesture
how far in degrees two objects in the sky are apart from each other, but
can't even give a wild guess what field of view I'm going to see when I
look through a specific eyepiece on my telescope.

Thanks for any help.

Ed

[Ed Isenberg]

Robert Lane wrote:

> In article <3AADEDC3...@earthlink.net>, Ed Isenberg

> <edise...@earthlink.net> wrote:
>
> > So, what is "apparent field of view," how could it be the same for
>

> I don't have the formula handy, but that's not what you need.
> Here goes:
> A 13-inch TV and a 35-inch TV, tuned to the same show, will have
> the same true field of view. If you view them from the same
> distance, they will have different apparent fields of view. TFOV is
> the amount of sky that is visible in the eyepiece. AFOV is the arc
> that the eyepiece _image_ occupies in your personal vision field.

Thanks. It took a moment for me to figure that out, but it helped.

I received the formula (pretty simple really) from someone named Doug
via E-mail. Since it doesn't look as if he posted his reply, he may not
wish his E-mail address splattered on the newsgroup to avoid spam, so I
won't give it here. Anway, here's his excellent explanation:

> True field of view = apparent field of view divided by the
magnification
> provided by that eyepiece. In the case of the 14mm eyepiece with the
84
> degree apparent field and the 8" SCT with the 2000mm fl. 2000 divided
by 14
> = 143 times. So true field = 84 divided by 143 = 0.587 or close to
0.6.

Thanks to both of you,

Ed

[Ed Isenberg]

Robert Lane wrote:

> In article <3AADEDC3...@earthlink.net>, Ed Isenberg
> <edise...@earthlink.net> wrote:
>

> > So, what is "apparent field of view," how could it be the same for
>

[Mike Doolan]

There's a program (free) called Sky Charts - Charts du Ciel that will show
exactly what your eye piece will cover in the sky. You can input upto 10
different eypieces and see how much sky each will cover. It will give you a
good idea of the effect of the eye piece before you buy it as well as being
a pretty good piece of astronomy software. I got it form
www.astrostuff.org/astropc

Mike