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The angular diameter, angular size, apparent diameter, or apparent size is an angular distance describing how large a sphere or circle appears from a given point of view. In the vision sciences, it is called the visual angle, and in optics, it is the angular aperture (of a lens). The angular diameter can alternatively be thought of as the angular displacement through which an eye or camera must rotate to look from one side of an apparent circle to the opposite side. Humans can resolve with their naked eyes diameters down to about 1 arcminute (approximately 0.017° or 0.0003 radians).^[1] This corresponds to 0.3 m at a 1 km distance, or to perceiving Venus as a disk under optimal conditions.

Formula

The angular diameter of a circle whose plane is perpendicular to the displacement vector between the point of view and the center of said circle can be calculated using the formula^[2][3]

${\displaystyle \delta =2\arctan \left({\frac {d}{2D}}\right),}$

in which ${\displaystyle \delta }$ is the angular diameter in degrees, and ${\displaystyle d}$ is the actual diameter of the object, and ${\displaystyle D}$ is the distance to the object. When ${\displaystyle D\gg d}$, we have ${\displaystyle \delta \approx d/D}$,^[4] and the result obtained is in radians.

For a spherical object whose actual diameter equals ${\displaystyle d_{\mathrm {act} },}$ and where ${\displaystyle D}$ is the distance to the center of the sphere, the angular diameter can be found by the following modified formula^{[citation needed]}

${\displaystyle \delta =2\arcsin \left({\frac {d_{\mathrm {act} }}{2D}}\right)}$

The difference is due to the fact that the apparent edges of a sphere are its tangent points, which are closer to the observer than the center of the sphere, and have a distance between them which is smaller than the actual diameter. The above formula can be found by understanding that in the case of a spherical object, a right triangle can be constructed such that its three vertices are the observer, the center of the sphere, and one of the sphere's tangent points, with ${\displaystyle D}$ as the hypotenuse and ${\displaystyle {\frac {d_{\mathrm {act} }}{2D}}}$ as the sine.^{[citation needed]}

The difference is significant only for spherical objects of large angular diameter, since the following small-angle approximations hold for small values of ${\displaystyle x}$:^[5]

${\displaystyle \arcsin x\approx \arctan x\approx x.}$

Estimating angular diameter using the hand

Estimates of angular diameter may be obtained by holding the hand at right angles to a fully extended arm, as shown in the figure.^[6][7][8]

Use in astronomy

In astronomy, the sizes of celestial objects are often given in terms of their angular diameter as seen from Earth, rather than their actual sizes. Since these angular diameters are typically small, it is common to present them in arcseconds (″). An arcsecond is 1/3600th of one degree (1°) and a radian is 180/π degrees. So one radian equals 3,600 × 180/${\displaystyle \pi }$ arcseconds, which is about 206,265 arcseconds (1 rad ≈ 206,264.806247"). Therefore, the angular diameter of an object with physical diameter d at a distance D, expressed in arcseconds, is given by:^[9]

${\displaystyle \delta =206,265~(d/D)~\mathrm {arcseconds} }$.

These objects have an angular diameter of 1″:

• an object of diameter 1 cm at a distance of 2.06 km
• an object of diameter 725.27 km at a distance of 1 astronomical unit (AU)
• an object of diameter 45 866 916 km at 1 light-year
• an object of diameter 1 AU (149 597 871 km) at a distance of 1 parsec (pc)

Thus, the angular diameter of Earth's orbit around the Sun as viewed from a distance of 1 pc is 2″, as 1 AU is the mean radius of Earth's orbit.

The angular diameter of the Sun, from a distance of one light-year, is 0.03″, and that of Earth 0.0003″. The angular diameter 0.03″ of the Sun given above is approximately the same as that of a human body at a distance of the diameter of Earth.

This table shows the angular sizes of noteworthy celestial bodies as seen from Earth: