NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Azimuth Equation
From: Bill B
Date: 2005 Oct 19, 15:04 -0500
From: Bill B
Date: 2005 Oct 19, 15:04 -0500
> > I'm using the azimuth equation out of the Nautical Almanac: > cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc] > > To find Hc: sin-1[(sinDec sinLat)+(cosDec cosLHA cos Lat)] > In entering the equation, if the Declination or latitude is South, or the > Longitude is West, the value is negative. Todd After a nights sleep, and looking at the almanac, have rethought the problem. Your signing of Lat, West Lon, and declination is correct. Remember LHA must always be a positive number from 0 to 359.xxx. If GHA+W Long is negative, add 360. If GHA + E Long is equal to or greater than 360, subtract 360. Some will tell you when using a modern calculator, a negative LHA or LHA greater than 360 will work fine, the calculator knows the appropriate trig value of those angles. The formula for Hc looks correct: sinHC=(sinDec sinLat)+(cosDec cosLHA cos Lat) cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc] seems strange to me. Perhaps it folds the formula for Hc in the formula below. For Z and Zn (azimuth) I use: cosZ = (sinDec - (sinLat sinHc))/(cosLat cosHc) If sinLHA is negative, Z=Zn If sinLHA is positive, Zn = 360-Z Purely by inspection if, LHA is >0 to <180, the body will be in the SW or NW quadrant. If LHA is >180 to <360, the body will be in the NE or SE quadrant, which tells you whether to use Z as Zn, of subtract it from 360. Hope that helps, Bill