Angular Size Converter

Small-angle astronomy: how big objects look on the sky from Earth.

Worked example — Angular Size Converter

Angular size (radians)	0.0090395 rad
Angular size (degrees)	0.51792°
Angular size (arcmin)	31.075′
Angular size (arcsec)	1864.5′′
Small-angle approximation error	0.00068094%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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why D/d is never exactly right 🖖

θ = 2 arctan(D/2d) is exact for a flat object of width D held square-on at distance d, and that is what this tool computes. A sphere is a different problem: the sight lines graze it short of its widest circle, so a ball of diameter D subtends 2 arcsin(D/2d) instead. Both collapse to D/d once D is small against d, which is why the distinction never surfaces in astronomy — for the Moon, D/d ≈ 0.0090 and the plain ratio is already within 0.0007% of the truth. It surfaces exactly where the small-angle rule fails anyway: 0.25% out at 10°, 10% at 60°. And at 60° the choice between disc and ball is worth 17.5%, which is the larger of the two mistakes. Close range is where you have to know which shape you are looking at.

It's about how big, not how far 🖖

Angular size measures how large something looks, not its true size — an angle, not a length. A far-off skyscraper and a coin held at arm's length can appear identical because size and distance trade off in the ratio D/d. The Sun is about 400 times wider than the Moon but also about 400 times farther away, so both span roughly 0.5° in our sky — the coincidence that makes total solar eclipses possible.

In an expanding universe, farther looks bigger 🖖

The small-angle idea assumes flat, static space, but cosmology breaks it. Because the universe expanded while ancient light traveled to us, the angular diameter distance is not monotonic: it peaks around redshift z ≈ 1.6. A galaxy of fixed size therefore shrinks on the sky only out to that point — beyond it, more distant (younger) galaxies actually appear larger, not smaller.

Where your answer lands

The space station is a bigger target than Jupiter

An angle is a ratio, and the ratio does not care which term is large. A 109-metre spacecraft 420 km overhead beats a 140,000-kilometre planet four astronomical units away, and it is not close. Every row below is a real diameter and a real distance run through the same 2·arctan(D/2d) the tool runs, so you can retype any line and get the same figure back.

What you are looking at	Diameter	Distance	Angular size
A golf ball held at arm’s length	42.67 mm	0.70 m	3.492°
The Sun	1,392,700 km	1.00 AU	32.00′
The full Moon	3,475 km	384,400 km	31.08′
The space station passing overhead	109 m	420 km	53.53″
Jupiter at opposition	139,822 km	4.20 AU	45.90″
One astronomical unit, seen from one parsec	149,597,871 km	3.26 ly	1.0000″
The Sun, seen from Proxima Centauri	1,392,700 km	4.25 ly	7.150 mas
That same golf ball, lying on the Moon	42.67 mm	384,400 km	22.90 µas

The parsec row is the one that certifies the other seven. A parsec is defined as the distance at which one astronomical unit subtends one arcsecond, so that line has to come out at 1.0000″ — and it does, to twelve digits, from the same formula as everything above and below it. Nothing inside the table fixes that value; the definition of the unit does. Two rows cannot be typed straight in: the distance menu has no parsec, so use 3.26 ly, and the diameter menu has no millimetres, so the golf ball is 0.04267 m. And the bottom line is the one worth keeping. A golf ball on the Moon spans 22.9 microarcseconds, and the ring the Event Horizon Telescope resolved around M87’s black hole is 42 microarcseconds across — less than twice as wide.

References (6)

The parsec, and the self-check in the middle of the table: IAU, "Measuring the Universe" — a parsec is the distance at which one astronomical unit subtends one arcsecond
Moon diameter and mean distance: NASA Science, Earth’s Moon: Facts
Jupiter radius 69,911 km and 5.2 au from the Sun: NASA Science, Jupiter: Facts
ISS 109 m across the solar arrays: NASA, International Space Station Facts and Figures
Distance to Proxima Centauri: Gaia Collaboration 2021, "Gaia Early Data Release 3: The Gaia Catalogue of Nearby Stars", A&A 649, A6
The 42 µas ring, for the last line of the note: Event Horizon Telescope Collaboration 2019, "First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole", ApJL 875, L1

Problem solved in full

The Moon 3474.8 km across and 384 400 km away 5 steps
The Moon is 3474.8 km across and 384 400 km away. Work out how big it looks in all four units — and then settle the eclipse coincidence with the same formula.
1. The geometry is a right triangle drawn on half the Moon: the half-diameter opposite, the distance adjacent. That gives half the angle, so double it at the end.
  
  $tan \frac{θ}{2} = \frac{D /2}{d} = \frac{1737.4}{384 400} = 0.0045198$
2. The arctangent returns radians, which is the honest unit here and the one every later formula wants. Everything after this is unit conversion.
  
  $θ = 2 arctan (0.0045198) = 0.0090395 rad$
3. Degrees, arcminutes and arcseconds are factors of 180/π, 60 and 60. Half a degree is the figure worth remembering — it is why a little fingernail held at arm's length covers the Moon.
  
  $0.0090395 \times \frac{180}{π} = 0.5179 2^{\circ} = 31.07 5^{'} = 1864. 5^{''}$
4. The small-angle approximation drops the arctangent and just divides. Here it is wrong by 7 parts in a million, which is why astronomers use it without apology; it stays inside a per cent out to about 10°.
  
  $θ \approx \frac{D}{d} = 0.0090395 ⟹ error = 6.8 \times 1 0^{- 6} = 0.00068%$
5. Now run the same two lines on the Sun — 1.39 million km at 1 AU. It comes to 0.5334°.
  
  $θ_{⊙} = 2 arctan \frac{696 350}{1.496 \times 1 0 ^{8}} = 0.533 4^{\circ}, \frac{0.5334}{0.51792} = 1.030$
Answer

The tool prints 0.0090395 rad, 0.51792°, 31.075′ and 1864.5″ for the Moon. The Sun works out at 0.5334°, just 3% wider — from a body 400 times larger and 390 times further away. That near-cancellation is why total solar eclipses are possible at all, and why they are so marginal: the Moon's distance varies by about a tenth over its orbit, so sometimes its disc covers the Sun completely and sometimes it falls short and leaves a ring. Nothing in the physics requires the two to match. Click the Sun preset and read the two figures side by side — the whole phenomenon lives in a 3% margin.

$θ = 2 arctan \frac{D}{2 d} : 0.5179 2^{\circ} (Moon) vs 0.533 4^{\circ} (Sun)$

By the 42Math team · Updated 2026-07-14

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Astronomy & Space lessons →

Example problems

Moon - Moon angular diameter is about half a degree from Earth.
Sun - Sun appears roughly half a degree wide at 1 AU.
ISS pass - ISS has small apparent size and requires high magnification for detail.