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field stop measurement for eyepiece fov-works?

Started by gnowell, 06/17/2010 04:25AM
Posted 06/17/2010 04:25AM | Edited 06/17/2010 04:38AM Opening Post
Televue gives 46mm for the field stop of the Pan 41 (an eyepiece I really like)
Pentax gives 46.5mm for the field stop of the XW40

For a c14, 3910mm fl

Pan 41--
57.3*46=2635.8
2635.8/3910=.674 degrees afov
or .674 * 60 = 40.4 arc minutes tfov

XW 40
57.3*46.5=2664.5
2664.5/3910=.681 degrees afov
.681 * 60 = 40.86 degrees

Both companies publish the field stop specs

TV claims they are offering the largest possible fov in a two inch eyepiece barrel, but they are edged out. They do however claim 68 degrees for their eyepiece, but it calculates to 67 (rounding the decimal down).

Pentax claims 70 degrees but publishes the field stop which gives 68 degrees.

Now, either both companies are fudging, TV by 1 degree and Pentax by 2, or perhaps there is some additional something to this formula that can be jiggered? The companies both publish their field stops so they're not going to be surprised that some anal retentive person will run the numbers. Using 57.29 for the radians isn't going to help.

Should the eyepieces just say "more or less what we put on the sticker on the side"??? Is the field stop calculation the "absolute last word"? Or does it have an approximative element?

Greg N

p.s. I can't believe my sacred XW40 got shoehorned down to a Pan Optic 68 degrees! I would rather believe the sticker.




"Scope will get you through times of no money better than money will get you through times of no scope." --Freewheelin' Franklin
Posted 06/17/2010 10:54AM #1
These equations relating apparent FOV and the field
stop diameter are only an approximation, and do
not include effects of geometric distortion. Typically
one assumes that apparent FOV and the field stop
diameter are proportional (i.e. linear relationship),
but there can also be terms in the equation proportional
to the square, cubed, etc. of the field stop diameter.
For example, the Panoptics are noted for having
significant geometric distortion, so these higher order terms
become important.
Posted 06/17/2010 03:38PM #2
<>

What numbers are required? Why do any math at all?

The only things you are going to see with either eyepiece are those that are inside the FOV defined by the field stop. The field stop defines the largest possible FOV for the eyepiece. The eyepiece with the largest field stop wins.

Whether or not that FOV is distortion free doesn't really matter. You won't see anything more than the field stop allows. If it is a poorly designed eyepiece, you may see less.

dan
Posted 06/18/2010 07:52PM #3
Hi Greg.

Several years ago, I asked David Nagler about the differences you get between the fs-based calculations for tfov and the afov-based calculations. If I'm quoting him correctly, he said that the formula tfov = fs * 57.3 / focal length is the correct calculation for tfov. It's not an approximation. The division of afov by magnification is the approximation. The reason is that different eyepieces with the same true field can have different apparent fields, depending on how light rays converge on your eye after traveling through the various eyepiece elements.

John