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Old 07-14-18, 03:50 AM   #77
Sean C
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Join Date: Jun 2017
Location: Norfolk, VA
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Well, the good news is that your math is correct. The bad news is that it seems your noon altitude is wrong.


The equation for figuring out the predicted altitude of a celestial body is:


asin(sin(Dec)·sin(Lat)+cos(Dec)·cos(LHA)·cos(Lat))


"Dec" is the declination of the body, "Lat" is your latitude and "LHA" is the local hour angle of the body. By definition, at local noon, the LHA is 0°. So, plugging in the other figures from your example gives us:


asin(sin(-23°04')·sin(46°41')+cos(-23°04')·cos(0°)·cos(46°41') = 20°15'


In other words, the altitude you should have measured given these parameters is 20°15'. Subtracting this from 90° (or 89°60') gives 69°45'. Subtracting the declination, 23°04', from 69°45' gives 46°41' ... your DR latitude.


Now, why your altitude is so wrong is another question altogether.
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