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Re: Derivation of the time-azimuth formula
From: Lars Bergman
Date: 2024 Oct 25, 13:17 -0700
From: Lars Bergman
Date: 2024 Oct 25, 13:17 -0700
Patrick,
In the PZX-triangle you have four known parts, i.e. the polar distance (=90°-d), LHA, the co-latitude (=90°-L) and Z. Now use the "four parts formula", giving
cos(90°-L)·cos(LHA)=sin(90°-L)·cot(90°-d)-sin(LHA)·cot(Z) ; solve for Z:
cot(Z)=[sin(90°-L)·cot(90°-d)-cos(90°-L)·cos(LHA)]/sin(LHA)=[cos(L)·tan(d)-sin(L)·cos(LHA)]/sin(LHA) ; then invert:
tan(Z)=sin(LHA)/[cos(L)·tan(d)-sin(L)·cos(LHA)]
Lars
