Field of View Calculation for CCD Imaging

John A McCubbin

"Will it fit?" is the classic question asked by almost all beginning CCD'ers. How do you know? Well first of all you need to know the handiest formula of all, the field of view formula. It can actually be rather confusing in its original form, so I've boiled it down (actually reduced it to its simplest form for you mathemeticians out there) to an easy to understand version.

A field of view is directly related to the size of the chip and the focal length of the telescope you use. The longer the focal length, the less sky is covered by the chip. Changing telescopes is like changing lenses on an SLR camera. The longer the lens, the more magnified the object. You might notice that my primary imaging setup is two piggybacked refractors. Coupled with telecompressers and barlows a wide variety of "lenses" of different focal lengths can be obtained. To know thier coverage, you have to go through theses calculations for each setup.

You have to know these things to find your field of view and if the object will fit:

1. The dimensions of your chip in millimeters
2. The focal length of your telescpe in millimeters
3. The size of the object you want to photograph in acrmin or degrees

Since CCD chips are small and CCD imaging excells in imaging small dim objects, you will probably work most often in arcminutes of sky coverage. These formulas work equally well with film photography also (35mm film is 24mm x 36mm).

These constants are expressed as follows

• S = dimension of one side of the chip (or film negative) in millimeters
• f = focal length of the telescope (this will change with telecompressers or barlows)
• ArcMin = arcminutes of sky covered
• Deg = degrees of sky coverd (if you want to know the coverage in this unit)

The formulas are:

• ArcMin = (S x 3438) / f
• Deg = (S x 57.3) / f

Lets try an example with my ST-8 and Astro-Physics 180EDT working at f/9. The ST-8 has a chip that is 13.8mm on one side and 9.2mm on the other. To calculate the sky covered we have to calculate coverage for each side of the chip. I'm concerned about expressing the value in arcminutes, so we'll use formula one. I want to see how the galaxy M96 will appear on the CCD chip. This helps me frame the image, and understand how it will look. The Sky v.5 lists its size as 7.5 x 5.2 arcmin.

The focal length of the AP180EDT is 180mm x 9 = 1620mm. If your telescope's aperture is expressed in inches (a C-11, for instance, has an aperture of 11 inches), you may convert it to millimeters by multiplying the aperture by 25.4 millimeters per inch.

So the values for the AP180EDT and the ST-8's wide dimension are:

• S = 13.8mm
f = 1620mm

Plugging in the values into the ArcMin formula you get:

• ArcMin = (13.8 x 3438) / 1620
• ArcMin = 29.3 arcminutes of sky covered in that dimension

The other side of the chip would be:

• S = 9.2
• ArcMin = (9.2 x 3438) / 1620
• ArcMin = 19.5 arcminutes of sky covered in that dimension

So the area of the sky covered would be 29.3 x 19.5 arcminutes of sky. Were we correct? I've reduced the picture from its original size for covenience, but the dimesions are accurate. Here's the picture:

Our calculations were accurate, the galaxy fits nicely on the chip and is well framed.

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I consider your heavens, the work of your fingers, the moon and the stars, which you have set in place ... Psalms 8:3