NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Ex-Meridian Exercise
From: Andrés Ruiz
Date: 2012 Jan 25, 17:44 +0100
Navigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
From: Andrés Ruiz
Date: 2012 Jan 25, 17:44 +0100
Hello,
I have two books for the licence of yacht captain (1968 and 1975), that includes the ex-meridian sights.
Spanish books differentiate between circunmeridian (t < 16 min) and ex-meridian (t >= 16 min)sights, depending on the time from meridian passage.
Here is the example by three methods.
15/01/2012
10:21:00 UT1
Geocentric equatorial coordinates
Moon
Dec = -9.009300 = -9º 0.6'
GHA = 80.077112 = 80º 4.6'
DR
Be = 28.666667 = 28º 40.0'
Le = -87.166667 = -87º 10.0'
Altitude
Hs = 51.183333 = 51º 11.0'
Ho = 51.788832 = 51º 47.3'
Hs = 51.1833 º = 51º 11.0'
ie = 0.0000 '
heas = 0.0000 m
T = 10.0000 ºC
P = 1010.0000 hPa
Hoi = 51.1833 º
dip = 0.0000 º
Ha = 51.1833 º = 51º 11.0'
R = 0.0130 º
OB = 0.00168790 º
HP = 0.9841 º
PA = 0.61853196 º
SD = 0.0000 º
Aug = 0.00000000 º
SDag = 0.000000 '
Ho = 51.7888 = 51º 47.3'
Method 1 - Latitude by Ex-meridian sight at2/60
B = 28.579961 = 28º 34.8'
Upper transit
Culmination = S
LHA = 352.910446 = 352º 54.6'
a = 2.784005
ma = 7.089554
t = 28.358218
C = 37.314413
Hlan = 52.410739
Method 2 - BOWDITCH 1906 (attached)
B = 28.583413 = 28º 35.0'
Method 3 - Newton-Raphson iterative method
0 0.001452 0.498876 0.498876 28.583492
1 0.000001 0.498875 0.498875 28.583413
2 0.000000 0.498875 0.498875 28.583413
B = 28.583413 = 28º 35.0'
--
Andrés RuizNavigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
2012/1/23 Greg Rudzinski <gregrudzinski@yahoo.com>
Linked image shows an ex-meridian plot crossed with a corresponding LOP using the formula (t^2)(a)/60 with a=2.8
The Moon example is from my most recent flight from Los Angeles to Orlando. See index card working notes.
Linked File: https://www.NavList.net/imgx/f1-IMG_1040.JPG
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