NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2013 Feb 16, 04:40 -0500
I get a different answer.
Because this is sunset, we can use an amplitude of the sun calculation, as if we were calibrating our compass.
Known on this day is the declination = S14 deg 3.7 minutes
Latitude =N41
Amplitude = A
A= arcsin(sin(dec)/cos(lat))
A= arcsin(sin(-14.061666)/cos(41))
A= arcsin(-.3219332015)
A= -18.7798 degrees
Azimuth Z is the complement of Amplitude A so
Z =90-A
Z = 108.7798 Degrees
In SW Quadrant
Sun setting at 360 - Z
Sun setting at 251.2202 degrees
The problem statement is that "you are facing the sun dog", but not the sun. The sun dog must be north of the sun, since we are facing generally west and the sun dog is to the right of the setting sun.
A sun dog forms at 22 degrees to the side of the sun. Therefore, we are facing (251.2202 + 22) degrees
= 273.2202 degrees
I am prepared to be completely corrected!
Regards
Brad
There was a very bright sundog yesterday at sunset here. Photo linked.
Here's a little navigation puzzle: you're walking in a hilly area near sunset on February 10 in latitude 41 N. You see a sundog about 5 degrees high directly ahead of you but you can't see the Sun (behind a hill to your left). In what direction are you walking?
-FER
----------------------------------------------------------------
NavList message boards and member settings: www.fer3.com/NavList
Members may optionally receive posts by email.
To cancel email delivery, send a message to NoMail[at]fer3.com
----------------------------------------------------------------
Linked File: https://www.NavList.net/imgx/sun-and-sundog.jpg