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Re: Azimuth Formula Questions
From: Bill B
Date: 2005 Oct 28, 17:37 -0500
From: Bill B
Date: 2005 Oct 28, 17:37 -0500
George wrote: > Here's the expression to use to obtain azimuth from its Tan. It presumes > that southerly latitudes and declinations are negative, that local hour > angle is measured positively Westwards 0 to 360 degrees, and that azimuths > are measured clockwise from North, 0 to 360 degrees. > > az = arctan (sin (hour angle) / (cos (hour angle) sin lat - cos lat tan > dec)) > > Then to put az into the right quadrant, apply the following rules- > > if tan az was negative, add 180 degrees to az. > if hour angle was less than 180 degrees (i.e., to the West of the observer), > add another 180 degrees to az I notice another (longer) formula in the HO229 background section that also does not require Hc: tanZ = (cosd sinLHA)/(cosL sind - sinL cosd cosLHA) L is latitude d is declination Zn is determined as per 229 usual rules: North lat Z > 180, Zn = Z Z < 180, Zn = 360-Z South lat Z > 180, Zn = 180-Z Z < 180, Zn = 180-Z For the mathematically inclined, are there pros and cons for either formula? Bill
