Caption: A diagram giving---for the cognoscenti---a proof by inspection that the altitude above due north of the north celestial pole (NCP) A_N_NCP equals the observer's latitude on Earth L: i.e.,
Note that a right angle can be rotated into a right angle.
Features:
L = δ ± [A(±) - 90°]: For finding latitude with south latitude counted as a negative number. This formula is used in celestial navigation.
λ = (360°/P)(t_UT - t_UT_G)
where:
P is the period:
sidereal day
for fixed stars,
apparent solar day
for the Sun.
t_UT is the
marine chronometer
time of the transit
of meridian.
t_UT_G is the
prime meridian (Greenwich meridian)
time of the transit
of meridian
which you read from some source
(e.g., a nautical almanac)
but is about your local time as appropriate for the
astronomical objects???.