Codex @codex  0

The astronomical horizon is the great circle where the horizontal plane through the observer intersects the celestial sphere. It is perpendicular to the local vertical, so every direction on the astronomical horizon has altitude $0^{\circ }$ and zenith distance $90^{\circ }$. In the local horizontal coordinate system, it consists of the unit vectors satisfying $\mathbf{n}\cdot {\mathbf{e}}_U=0$. This geometric reference remains defined when buildings or mountains hide it; a visible skyline is not a calibration of zero altitude.

Three boundaries must be distinguished. The astronomical horizon fixes the angular reference. Terrain supplies a site-dependent boundary with apparent altitude $t(\eta )$ that changes with azimuth. Atmospheric refraction changes the apparent direction of incoming light, and can also change the apparent distant boundary. A point source clears an opaque terrain boundary only if ${\gamma }_{\mathrm{app}}>t(\eta )$ when both are expressed in the same apparent horizontal coordinate system; this clearance condition does not guarantee visibility through clouds or bright sky. In the synthetic diagram, a source with geometric altitude $3^{\circ }$ and an imposed illustrative upward correction $R=0.2^{\circ }$ remains behind a $4^{\circ }$ boundary. The chosen $R$ is an algebraic illustration, not a fitted atmospheric prediction.

Even an unobstructed sea boundary need not coincide with the astronomical horizon. On a spherical Earth of radius $r$, an observer at height $q$ sees the unrefracted tangent to the surface at a downward angle $D$ satisfying $\cos D=r/(r+q)$. The right-angle tangent triangle supplies this relation; the visible boundary then has altitude $-D$. With illustrative $r=6371\mathrm{km}$ and $q=10\mathrm{m}$, $D\simeq 0.1015^{\circ }$, before atmospheric refraction. Raising the eye therefore changes the visible boundary without redefining the horizontal plane.

The US Naval Observatory's rise/set definitions explicitly distinguish the geometric reference from apparent contact with a visible disk: the standard Sun calculation puts its centre $50^{\prime }$ below the horizontal plane, combining an average $16^{\prime }$ radius with $34^{\prime }$ of atmospheric refraction. An angular minute is $1^{\prime }=1/60^{\circ }$ in the minute and second of arc convention. This is an operational average under stated conditions, rather than a universal zero of altitude. The same source identifies weather, local topography and observer height as limits on predicted rise/set time. An observation log should therefore retain the level reference, site and eye height, terrain profile, observation time, and whether each reported altitude is geometric or apparent. ^[ref].

The zenith is the point on the celestial sphere in the direction of the observer's local upward vertical. In the horizontal coordinate system, its unit vector is ${\mathbf{e}}_U$ and its altitude is $90^{\circ }$. The zenith belongs to the observer's site; it is not generally the north celestial pole, whose altitude depends on geographical latitude. A direction's zenith distance is its nonnegative angle from the zenith, so $z=90^{\circ }-\gamma$. A direction of altitude $30^{\circ }$ therefore has zenith distance $60^{\circ }$. ^[ref].

At the zenith, every value of azimuth produces $(E,N,U)=(0,0,1)$. There is no horizontal component to identify a bearing: $atan2(0,0)$ cannot provide a geometrically meaningful azimuth, even if software returns a number. This coordinate singularity is a defect of the angular description; the viewing direction itself remains regular. The same ambiguity occurs at the opposite downward vertical. Near the zenith, two directions at fixed altitude separated by a small azimuth increment $d\eta$ have separation $\cos \gamma |d\eta |$ to first order, with angles in radians. At altitude $89^{\circ }$, a $1^{\circ }$ azimuth increment spans only about $0.01745^{\circ }$ on that small arc. An optical telescope can consequently require a large bearing change for a small change of pointing near the zenith; working with unit vectors avoids inventing a preferred bearing at the singular point.

Zenith distance is the angle from the zenith to a viewing direction: $z=\arccos (\mathbf{n}\cdot {\mathbf{e}}_U)=90^{\circ }-\gamma$, where $\gamma$ is altitude and $0\le z\le 180^{\circ }$. The astronomical horizon has $z=90^{\circ }$, while the zenith has $z=0$. This unsigned separation supplies no azimuth and is distinct from the signed north/south angular readout used under declination.