Field of view calculator
Enter your telescope and camera or eyepiece. We draw your frame at true scale over the Moon, a planet or a deep-sky object, and work out the numbers that matter — with every formula shown below.
Your equipment is remembered on this device only.
Results
- Field of view
- 2,24° × 1,50°
- 135′ × 90′
- Image scale
- 1,29″/px
- Image: 6248 × 4176 px
- Diagonal
- 2,70° (162′)
- Effective focal length
- 600 mm · f/7,5
- Dawes limit
- 1,45″
- Rayleigh limit
- 1,73″
Sampling
Under-sampled
1,29″/px is coarser than the ideal 0,83–1,25″/px for 2,5″ of blur, so stars may look square and fine detail is lost. That is normal and fine for wide fields; for detail, use a longer focal length, a Barlow or smaller pixels, or drizzle when stacking.
Rule used: the image scale should be between 1/3 and 1/2 of the blur, so each star spans 2–3 pixels. Blur is your seeing, or the telescope's Dawes limit if that is larger.
True-scale preview
Drawn to scale. Sun, Moon and planet sizes are computed for the chosen date. Deep-sky objects are catalogue sizes drawn as dashed ellipses with the long axis horizontal; faint outer parts often extend beyond them. Rotate the frame to try other framings.
The formulas
Every number above comes from these formulas. F is the effective focal length and D the aperture, both in millimetres.
Effective focal length
F = focal length × reducer or Barlow factor
A 0.8× reducer turns 600 mm into 480 mm; a 2× Barlow turns it into 1200 mm.
Focal ratio
f-ratio = F ÷ D
Lower numbers are 'faster': for a nebula or galaxy, each pixel collects light more quickly.
Image scale
scale (″/px) = 206.265 × pixel size (µm) × binning ÷ F (mm)
206,265 is the number of arcseconds in a radian. This is how much sky each pixel sees.
Camera field of view
field = 2 × atan(sensor size ÷ (2 × F))
Applied to the sensor's width, height and diagonal. For fields of a few degrees it equals sensor size ÷ F × 57.3°, the shortcut most calculators use; the exact form stays right for camera lenses and very wide fields.
Ideal sampling
blur ÷ 3 ≤ scale ≤ blur ÷ 2
Blur is the seeing (star size, FWHM, in arcseconds) or the Dawes limit, whichever is larger. 2–3 pixels across each star keeps them round without wasting light.
Magnification
M = F ÷ eyepiece focal length
A 1200 mm telescope with a 25 mm eyepiece magnifies 48×.
True field, with a field stop
true field = 2 × atan(field stop ÷ (2 × F)) ≈ field stop ÷ F × 57.3°
The field stop is the ring inside the eyepiece that sharply edges the view, so this is the exact method.
True field, without a field stop
true field ≈ apparent field ÷ M
A quick estimate. Distortion in wide-angle eyepieces can make it a few percent off.
Exit pupil
exit pupil = D ÷ M = eyepiece focal length ÷ f-ratio
The width of the beam leaving the eyepiece. About 7 mm is the most a dark-adapted young eye takes in; 0.5–1 mm is the useful minimum for planets.
Useful magnification range
D ÷ 7 ≤ M ≤ 2 × D
The low end gives a 7 mm exit pupil. The high end (50× per inch) is the usual practical ceiling; beyond it the image only gets bigger and dimmer.
Dawes limit
Dawes = 116 ÷ D (arcseconds)
The closest pair of similar stars a telescope can just split, from W. R. Dawes's 19th-century observations.
Rayleigh limit
Rayleigh = 1.22 × λ ÷ D ≈ 138 ÷ D (arcseconds, λ = 550 nm)
The theoretical diffraction limit for green light, slightly looser than Dawes.
Questions
What is the difference between apparent field and true field?
The apparent field is how wide the view looks through the eyepiece on its own (for example 52° or 82°). The true field is the patch of sky you actually see through the telescope, which is the apparent field divided by the magnification. A 68° eyepiece at 50× shows about 1.4° of sky — almost three full Moons.
Why doesn't binning change my field of view?
Binning combines neighbouring pixels (2×2, 3×3) into one bigger pixel. The sensor still covers the same patch of sky, so the field stays the same; what changes is the image scale (arcseconds per pixel) and the number of pixels in the final image.
Does a focal reducer make my images brighter?
For extended objects such as nebulae and galaxies, yes: a reducer shortens the focal length, so the same light lands on fewer pixels and each pixel fills faster. It also widens the field and makes the image scale coarser. The total light collected from the object is set by the aperture and doesn't change.
What image scale should I aim for?
Around 1/3 to 1/2 of your typical seeing. Under average skies with 2–3″ seeing, about 1–1.5″ per pixel is a sweet spot. Wide-field images of large nebulae are often deliberately under-sampled at 2–4″/px, and that's fine.
Why might my real photos frame slightly differently from this preview?
Real focal lengths often differ by a few percent from the label, and a reducer's actual factor depends on its spacing to the sensor. Plate-solve one of your images to get your true focal length, then type it in here for an exact match.