About the Angular Size Calculator
This angular size calculator works out how big an object appears in the sky — its angular diameter — from its real size and its distance. It also works in reverse: give an angle and a distance to find the true size, or an angle and a size to find how far away something is. Results come in degrees, arcminutes and arcseconds, the units astronomers use for the Moon, planets, galaxies and double stars.
Amateur astronomers use it to check whether a target will fit in a telescope or camera field of view, students use it for small-angle-formula homework, and photographers and surveyors use the same geometry on Earth to estimate the size or distance of distant objects. Distances and sizes can be entered in metres, kilometres, miles, astronomical units, light-years or parsecs.
The calculator uses the exact formula δ = 2·arctan(d / 2D), which is accurate at any angle, and also shows the small-angle approximation that is fine for anything smaller than a few degrees.
How to use the angular size calculator
- 1Choose what to solve for: angular size, actual size or distance.
- 2Enter the two known values and pick their units.
- 3For an angle input, choose degrees, arcminutes or arcseconds.
- 4Read the answer, plus the angle in all three units and a comparison with the full Moon.
Formula and method
An object of diameter d at distance D subtends an angle δ at your eye. Splitting the view into two right triangles gives the exact relationship tan(δ/2) = d/(2D), which the calculator rearranges to solve for the angle, the size or the distance. Size and distance are converted to metres first, so they can use different units.
For small angles tan(x) ≈ x in radians, giving δ ≈ d/D radians, or 206,265 × d/D arcseconds (206,265 is the number of arcseconds in a radian). This small-angle formula is within 0.1% below about 3°, which covers almost every astronomical object. One degree is 60 arcminutes (′) and one arcminute is 60 arcseconds (″).
- δ
- Angular diameter
- d
- Actual (linear) diameter of the object
- D
- Distance to the object
- 206,265
- Arcseconds per radian
Worked examples
Angular size of the Moon at its mean distance
δ = 2 · arctan(3,474.8 ÷ (2 × 384,400)) = 0.5179°, or about 31.1 arcminutes — just over half a degree. A little fingernail held at arm’s length spans roughly 1°, enough to cover the Moon twice over.
Jupiter at opposition (628.7 million km)
Jupiter’s 142,984 km equatorial diameter at 628.7 million km subtends about 46.9″ — too small to see as a disk by eye, but a clear disk at 40× in a telescope.
Distance to a 1.8 m person who appears 1° tall
D = 1.8 ÷ (2 × tan 0.5°) ≈ 103.1 m. Surveyors and hunters use this same relationship with reticle markings to range distances.
Size of an object spanning 0.5° at the Moon’s distance
d = 2 × 384,400 × tan(0.25°) ≈ 3,354.5 km, so an object 0.5° wide at the Moon’s distance would be about 3,355 km across — slightly smaller than the Moon itself.
Frequently asked questions
What is angular size?+
Angular size (angular diameter) is how large an object appears, measured as the angle it covers in your field of view. It depends on both the true size and the distance, which is why the Sun and Moon look almost the same size (about 0.5°) despite the Sun being 400 times wider.
How big is the Moon in the sky?+
About 31 arcminutes on average, or roughly 0.52°. It varies from about 29.4′ at apogee to 33.5′ at perigee because the Moon’s orbit is elliptical; a “supermoon” is a full Moon near the larger end.
What is the small-angle formula?+
For small angles, δ (in radians) ≈ size ÷ distance, or δ (arcseconds) ≈ 206,265 × size ÷ distance. It is accurate to better than 0.1% for angles under about 3°, so it is used for nearly all astronomical objects.
How many arcseconds are in a degree?+
There are 60 arcminutes in a degree and 60 arcseconds in an arcminute, so 3,600 arcseconds per degree. A full circle is 1,296,000 arcseconds and one radian is about 206,265 arcseconds.
What is the smallest angular size the eye can see?+
A person with good vision resolves about 1 arcminute. That is why planets such as Jupiter (30–50″) look like points without optical aid, while a telescope at 40× or more reveals their disks.