Angular Size Calculator
Find the angle an object spans from its size and distance, or size or distance from the angle, with sphere and flat geometry and the small-angle error.
Angular Size Calculator
Background.
Held 70 cm from the eye, roughly at arm's length, an object only 6.33 mm wide is enough to blot out the full Moon. Most people guess something far bigger, because what the eye actually judges is angular size, the angle an object spans as seen from where you stand, and the Moon's is only about half a degree. That angle depends on true size and distance together, which is how the Sun, about 400 times wider than the Moon but also about 390 times farther away, ends up looking almost the same.
Give the calculator any two of size, distance and angle and it returns the third. Size and distance can be in any unit, kilometres for a planet or millimetres for a coin, as long as both use the same one: the angle depends only on their ratio, and a size you solve for comes back in the distance's unit.
You also choose the geometry, because a flat object and a sphere are not seen the same way. Across a disc, a crater or the face of a building, the edges are the ends of a chord, and the angle is twice the arctangent of half the size over the distance. For a planet, moon or star your sight lines graze the limb of a sphere, the distance runs to its centre, and the angle becomes twice the arcsine. For the Moon the two differ by 0.019 arcseconds out of about 1,865, but standing on a sphere's surface the arcsine correctly gives 180 degrees while the arctangent gives 90.
Alongside the exact answer you get the familiar small-angle shortcut, angle in arcseconds ≈ 206,264.806 × size ÷ distance, and how far it is off for your numbers.
What is angular size calculator?
Angular size, or angular diameter, is the angle an object subtends at the observer, measured in degrees, arcminutes or arcseconds: one degree is 60 arcminutes (′) and one arcminute is 60 arcseconds (″), so a degree holds 3,600″. A radian, the natural unit of the trigonometry, is 648,000 ÷ π = 206,264.806″. For scale, the full Moon and the Sun each span a little over half a degree, about 31 to 32′, while a fist at arm's length covers roughly 10°. The Moon's own figure changes through each month, from about 29.4′ when it is farthest to 33.5′ when it is nearest.
How to use this calculator.
- Choose what to solve for: the angle from size and distance, the physical size from angle and distance, or the distance from angle and size.
- Choose the geometry. Sphere suits a planet, moon or star, with the distance measured to its centre; flat suits a disc, crater, building or any width measured across the line of sight.
- Enter size and distance in the same unit. An angle goes in as arcseconds: multiply degrees by 3,600 or arcminutes by 60.
- Read the result in arcseconds, arcminutes and degrees, and check the shortcut error before quoting the small-angle figure.
- For the Moon or a planet, use the distance on the date you care about; a mean distance gives only a mean answer.
The formula.
Draw a line from the eye to the centre of the object. For a flat object of width d at distance D, each half of the object makes a right triangle with that line, so tan(θ⁄2) = (d⁄2) ÷ D and θ = 2·arctan(d ⁄ 2D). For a sphere of diameter d with its centre at distance D, the sight line to the edge touches the surface at right angles to a radius, so sin(θ⁄2) = (d⁄2) ÷ D and θ = 2·arcsin(d ⁄ 2D). Solving for size or distance simply rearranges these: d = 2D·tan(θ⁄2) or D = d ⁄ (2·tan(θ⁄2)), with sine in place of tangent for a sphere. Multiplying radians by 206,264.806 gives arcseconds.
The small-angle shortcut drops the trigonometry: θ ≈ d ⁄ D radians. Against the flat relation it always reads slightly high, by about (d⁄2D)² ÷ 3; against the sphere it reads slightly low, by about half that. The flat error reaches 0.1% at a size-to-distance ratio of 0.1096, an angle of 6.2727°, and the sphere error at a ratio of 0.1548, or 8.8776°. Below those angles the shortcut is fine for most purposes.
Some inputs have no answer and are refused: a size or distance of zero or less; an angle of 648,000″ (180°) or more, since nothing outside you subtends half a turn; and, for a sphere, a size more than twice the distance, which would place you inside it. A size exactly twice the distance is allowed and returns 180°, the view from the surface.
A worked example.
How far away must a 25 mm coin be to cover the full Moon exactly? The Moon's mean angular diameter, using flat geometry for a 3,474.8 km Moon at 384,400 km, is 1864.5267 arcseconds, or 31.0754 arcminutes. Solving for distance: half the angle is 932.2634″, which divided by 206,264.806 is 0.0045197 radians, and its tangent is 0.0045197711. Then D = 25 ÷ (2 × 0.0045197711) = 2765.6268 mm, about 2.77 m. The same figure follows with no trigonometry at all, because a fixed angle means a fixed ratio of size to distance: 25 × 384,400 ÷ 3,474.8 = 2765.6268 mm. That is roughly four times arm's length. Held at 70 cm, the same coin spans 2.0461°, almost four Moons across, which is why the Moon in a photograph so often looks smaller than it felt at the time.
Frequently asked questions.
Should I choose the sphere or the flat setting?
Why does the Moon look bigger near the horizon?
Why do the Sun and Moon look almost the same size?
How much does the Moon's apparent size change over a month?
When is the small-angle approximation good enough?
Can I work out how big a distant object is from its angle?
References& sources.
- [1]NASA JPL Solar System Dynamics, 'Planetary Satellite Physical Parameters'. The Moon (body 301): mean radius 1737.4 ± 0.1 km and GM 4902.800 ± 0.001 km³ s⁻², referenced to ephemeris DE440. Source of the default physical size. Open access. Retrieved 2026-07-29.
- [2]NASA JPL Solar System Dynamics, 'Astrodynamic Constants'. Astronomical unit = 149 597 870 700 m, attributed to IAU 2012 Resolution B1. Used with the nominal solar radius as the independent cross-check on this page's geometry. Open access. Retrieved 2026-07-29.
- [3]Mamajek, E.E. et al. (2016). 'IAU 2015 Resolution B3 on Recommended Nominal Conversion Constants for Selected Solar and Planetary Properties.' The Astronomical Journal 152:41. Nominal solar radius = 6.957 × 10⁸ m, exact by definition; nominal terrestrial equatorial radius = 6.3781 × 10⁶ m. With the exact astronomical unit this yields a solar semidiameter of 959.23″, against the traditional almanac value of about 959.6″ — a 0.33″ disagreement recorded rather than resolved. Open access via IOPscience. Retrieved 2026-07-29.
- [4]Urban, S.E. & Seidelmann, P.K. (eds.) (2013). Explanatory Supplement to the Astronomical Almanac, 3rd edition, University Science Books, §7. Semidiameters and apparent diameters, and the distinction between the geometric and the apparent (irradiation-biased) semidiameter that produces the 959.6″ versus 959.3″ disagreement above. Print / bibliographic reference.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 4 cited below
- Method
- flat: θ = 2·arctan(d ⁄ 2D) sphere: θ = 2·arcsin(d ⁄ 2D) shortcut: θ ≈ 206264.806 · d ⁄ D
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
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