How Much Field Of View For Different Objects?

[Skiddins]

Is there a way I can work out how small, or how much of the field of view of my eyepiece should be filled with an object?

The sizes seem to be given in minutes? such as 17.4' for M5, but then 178'x40' for M31 etc.

How can I work out how big this should be?

I have a variety of eyepieces but the original Celestron 25mm and 9mm get used the most for DSO's. On Celestron's website it say's my scope has 1.7deg angular field of view, is this even relevent?

Skiddins

[bamus]

try stellarium ocular plugins:

Oculars plugin - Stellarium Wiki

[George]

The New CCD Astronomy Home Page

You could mess around with the exit pupil size instead of chip size, might work.....worth a try.

[acey]

True FOV is the apparent FOV divided by magnification. For example if you have a plossl with 50 degree apparent FOV and your magnification is 25, then your true field of view is 2 degrees. There are 60 arcminutes (denoted ') in a degree so 2 degrees equals 120'. If an object has diameter 2' and your true field is, say, half a degree (30'), then the object should fill a fifteenth of the field diameter.

True field can also be worked out directly for a given eyepiece if you know the focal length of the telescope and the diameter of the eyepiece field stop (i.e. the width of the bit of glass at the bottom, furthest from your eye, in the case of a design such as a plossl). True field in radians is then field stop divided by focal length. To convert to degrees you multiply that by 57.3

Example: eyepiece has a 20mm field stop, telescope has 1000mm focal length. True field = 20/1000 = 0.02 radians = 1.15 degrees = 68.8'.

As a further point, note that quoted sizes of DSOs are often larger than what you'll actually see, because they're measured from CCD images which go fainter then the eye can. M31 is regularly quoted as having a diameter of 3 degrees but no one will ever see that much visually.

A traditional way of visually measuring the size of DSOs was to estimate its observed size as a fraction of the field width, which is exactly what you're asking about. There are other ways of finding the true field width apart from what I've suggested (e.g. timing the transit of a star across the field), but if you know what the manufacturer states as the apparent field then dividing by magnification is simplest. If the Celestron site says your scope has a certain field then this must be when a particular eyepiece (presumably the supplied one) is used.

[Macavity]

See above post.

With the exception of Andromeda and the Pleiades, most DSOs will (initially) fit within a field a degree or two. I usually allow a bit extra for "framing" - So that the object seems vaguely recognisable, in context. With a field of approx. 1.5 degrees, I tend not to change eyepieces overmuch, during a DSO hunting session.