Audited 29 Jul 2026·Last updated 28 Oct 2026·4 citations·Tier 2·0 uses

Telescope Field of View Calculator

Find how much sky an eyepiece shows: true field in degrees and arcminutes from the field stop or apparent field, the barrel's ceiling and star drift time.

Telescope Field of View Calculator

Method
Objective or primary-mirror focal length. A longer focal length gives a smaller true field with the same eyepiece.
mm
Engraved on the eyepiece barrel. Used for the magnification, and for the shortcut method.
mm
Published in the eyepiece's specifications. Typical for a 25 mm Plössl is around 21.5 mm. Used by the accurate method.
mm
The eyepiece's marketed field: about 50° for a Plössl, 68–82° for a wide-field. Used by the shortcut method.
°
About 27 mm for a 1.25-inch focuser and about 46 mm for a 2-inch one. Above this the barrel itself becomes the aperture you are looking through.
mm
Used only for the drift time. Stars near the celestial pole drift more slowly because they trace a smaller circle; at exactly ±90° they never drift out at all.
°
True field of view
1.0265
The angular diameter of the sky visible through the eyepiece, in degrees; the full Moon is about 0.5° for comparison.
True field (arcminutes)
61.593′
Magnification
48×
Apparent field of view
49.2744°
Field stop diameter
21.5 mm
Widest field this barrel can pass
1.2892°
Star drift time across the field
245.6991 s
Reading
1.027° (61.6′) of sky. Computed from the 21.5 mm field stop, which is the accurate method. Your focuser barrel can pass up to 1.289°, so nothing is being clipped here. A star at this declination takes 245.7 s to drift across the field with the drive off.

Background.

Before you swing a telescope toward the full Moon, a sprawling open cluster or the Andromeda Galaxy, the useful question is whether the eyepiece will frame the whole thing. The full Moon is about half a degree across; Andromeda, in long-exposure photographs, stretches to roughly six Moon-widths. A 25 mm eyepiece in a 1200 mm telescope shows about one degree, so the Moon fits twice over and the galaxy does not fit at all.

There are two ways to work that degree out, and they rarely agree exactly. The reliable one starts from the eyepiece's field stop, the metal ring inside the barrel that draws the sharp edge of the view. Because that ring sits at the telescope's focus, its diameter divided by the telescope's focal length is the slice of sky in radians. Tele Vue publishes this relation for its own eyepieces using 57.3 degrees per radian; the calculator uses the unrounded 180/π.

The quick way divides the apparent field printed in the eyepiece's advertising (50° or so for a Plössl, 68–82° for wide-field designs) by the magnification. Real eyepieces have some distortion, so this usually overstates the sky by a few percent; the calculator turns its answer back into the field stop it implies, a number you can hold against the maker's specification.

Alongside the field you get the widest view your focuser barrel can ever pass and the number of seconds a star takes to drift across with the drive switched off.

What is telescope field of view calculator?

True field of view is the angular diameter of the circle of sky visible through an eyepiece, in degrees or arcminutes. It belongs to the telescope and eyepiece together. Apparent field is different: it is how wide the bright circle looks to your eye, a property of the eyepiece alone, from about 50° in a Plössl to around 100° in the widest modern designs. The field stop is the physical aperture that sets the edge of the view, quoted in millimetres. Drift time is the number of seconds a star needs to cross the full field when nothing is tracking it.

How to use this calculator.

  1. Pick the method. Use the field-stop method whenever the eyepiece's specification sheet lists a field stop diameter; fall back to the apparent-field method when the box only gives a degree figure.
  2. Enter the telescope's focal length in millimetres, not its aperture. It is usually printed on the tube or in the manual as, for example, f = 1200 mm. If you use a Barlow lens, multiply by its factor first.
  3. Enter the eyepiece focal length engraved on the barrel, then the field stop or the apparent field depending on the method you chose.
  4. Set the barrel limit to 27 mm for a 1.25-inch focuser or 46 mm for a 2-inch one, and enter the declination of your target if you want its drift time (0° puts it on the celestial equator).
  5. Read the true field, then glance at the implied field stop or implied apparent field. A large gap between that and the maker's figure tells you the eyepiece has noticeable distortion.

The formula.

TFOV = (d_stop ⁄ f_tel) × 180⁄π TFOV ≈ AFOV ⁄ M, M = f_tel ⁄ f_eye t_drift = TFOV × T_sid ⁄ (360 × cos δ)

The field-stop route is pure geometry. A ring of diameter d_stop sitting in the focal plane of a telescope with focal length f_tel subtends an angle d_stop ⁄ f_tel in radians, and multiplying by 180⁄π (57.2958) converts it to degrees: TFOV = (d_stop ⁄ f_tel) × 180⁄π. Nothing about the eyepiece's lens design enters, which is why it is trusted. Strictly the angle is 2·arctan(d_stop ⁄ 2f_tel), but for real eyepiece fields the difference is under 0.01 % at a 1.3° field and under 0.1 % even at 5°, far smaller than the uncertainty in a published stop.

The shortcut is TFOV ≈ AFOV ⁄ M, where the magnification M = f_tel ⁄ f_eye. It holds only if the eyepiece maps angles onto your eye in strict proportion, so its error equals the eyepiece's angular distortion. Each method reports the other's quantity too: the apparent field a stop implies (TFOV × M), or the stop an apparent field implies.

The barrel ceiling is the same geometry applied to the largest stop the barrel can hold. Drift time is t = TFOV × T_sid ⁄ (360 × cos δ). The sky turns 360° in one mean sidereal day, T_sid = 86400 ⁄ 1.00273790935 = 86,164.09 s (23 h 56 min 4 s), from the solar-to-sidereal ratio published by the U.S. Naval Observatory: 15.041 arcseconds of sky per second at the equator. A star at declination δ follows a smaller circle, so its speed falls by cos δ; +45° and −45° drift alike, and declinations beyond ±89.9° are refused because at the pole the time becomes infinite.

A worked example.

Example

Take a 1200 mm telescope with a 25 mm eyepiece sold as having a 52° apparent field, aimed at a target at declination +23.5°, with a 1.25-inch focuser (27 mm limit), using the apparent-field method. The magnification is 1200 ⁄ 25 = 48×. Dividing the apparent field by it gives 52 ⁄ 48 = 1.0833°, which is 65′ of sky. Turned back into a field stop, that field would need 1.0833 ⁄ 57.2958 × 1200 = 22.69 mm. The 27 mm barrel allows up to 27 ⁄ 1200 × 57.2958 = 1.2892°, so nothing is clipped. A typical 25 mm Plössl's published stop is 21.5 mm, not 22.69 mm. Run through the field-stop method, 21.5 ⁄ 1200 × 57.2958 gives 1.0265°, or 61.6′. The shortcut has overstated the field by 5.5 %, the size of the eyepiece's distortion rather than a slip in either calculation, and the smaller figure is the sky you will see. For drift time, a star on the celestial equator would cross the 1.0833° field in 1.0833 × 86,164.09 ⁄ 360 = 259.29 s. At +23.5° the cosine is 0.9171, so the crossing takes 259.29 ⁄ 0.9171 = 282.74 s, about 4 minutes 43 seconds before an untracked star leaves the view.

max Field Stop Mm27
methodapparentField
eyepiece Focal Length Mm25
apparent Field Degrees52
declination Deg23.5
telescope Focal Length Mm1,200
eyepiece Field Stop Mm21.5

Frequently asked questions.

Why does the apparent-field method give a wider field than the field stop for the same eyepiece?
The apparent field quoted by the maker is measured as your eye sees it, and most eyepieces stretch angles slightly toward the edge of the view. Dividing that inflated figure by the magnification overstates the sky. The field stop is a physical ring you could measure with calipers, so distortion cannot affect it.
How can I measure a field stop the manufacturer never published?
Time a drift. Aim at a star near the celestial equator, switch off the drive, and time its crossing through the centre of the field. Multiply the seconds by 360 ⁄ 86,164.09 to get the true field in degrees, then by your focal length and divide by 57.2958. A 245.7 s crossing in a 1200 mm telescope works out to a 21.5 mm stop.
Why can no 1.25-inch eyepiece show more than about 1.29° in my 1200 mm telescope?
The barrel caps the field stop at roughly 27 mm, and 27 ⁄ 1200 × 57.2958 is 1.2892°. A wider apparent field in the same barrel only means a shorter focal length and more magnification, not more sky. A 2-inch focuser lifts the ceiling to about 46 mm, or 2.1963° on the same telescope.
Does a Barlow lens or focal reducer change the true field?
Yes, because it changes the effective focal length. A 2× Barlow turns 1200 mm into 2400 mm and halves the true field of every eyepiece; a reducer shortens the focal length and widens it. Enter the effective focal length, and expect a reducer to darken the edges if its own clear opening is smaller than the field stop.
Why do stars near Polaris barely move across the eyepiece?
Every star circles the celestial pole once per sidereal day, and the circle shrinks with the cosine of declination. The 1.0833° field that takes 259 s to cross on the equator takes about 519 s at 60° and about 1,493 s, close to 25 minutes, at 80°. At exactly 90° the star would never leave.
Can I use these results for a camera at prime focus?
No. With a camera there is no eyepiece, so neither field stop nor apparent field exists; the sensor's width and the telescope's focal length set the framing through an arctangent. The camera field of view calculator handles that case, treating the telescope as a long lens.

How this page was produced

Published by
Quanta Calculator
Primary sources
4 cited below
Method
TFOV = (d_stop ⁄ f_tel) × 180⁄π TFOV ≈ AFOV ⁄ M, M = f_tel ⁄ f_eye t_drift = TFOV × T_sid ⁄ (360 × cos δ)
Published
Last verified

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