# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Ex-Meridian Exercise
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 Ruiz

2012/1/23 Greg Rudzinski

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.

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