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Fw: Re: The Great Circle Challenge
From: Gary LaPook
Date: 2014 Dec 31, 00:50 -0800
From: Gary LaPook <NoReply_LaPook@fer3.com>
To: garylapook@pacbell.net
Sent: Monday, December 29, 2014 10:41 PM
Subject: [NavList] Re: The Great Circle Challenge
From: Gary LaPook
Date: 2014 Dec 31, 00:50 -0800
I am attaching the pages from HO 214 and from HO 229 that are used to compute the GC route. The initial course is 180 - 153.9 = 26.1 degrees
Doing it with inspection tables.
HO 214
Starting position (AP.) Lat. 41 30.0' S Lon. 174 30' E
Destination lat (decl) 21 57.0' N Lon 1 59 40' W Dlo 25 50' (LHA)
Entering the table with the closest tabulated values Lat 41 Dec 22 contrary LHA 26
Azimuth angle of 153.9 and tab Hc 22 31.8 delta
dec + .93 , delta t +.34, signs determined by
inspection, comparing the tabulated value with the next value for declination and for the next
value of LHA. Since we are using the declination of 22 since it is closest, the difference of
declination is only -03', and since the declination is between 22 and 21 we compare with decl 21
to get the sign of the delta dec correction factor. Since we are using an LHA of 26 the difference
of LHA (t) is -10' and the LHA is between 26 and 25 we compare LHA .25 to get the sign of the
delta t correction factor. These delta values are the equivalent to the delta dec value in HO 229
except HO 214 uses decimals of a
degree and HO 229 uses minutes.
So, looking at the multiplication table in HO 214 we see that the delta dec +.93 times the 3
minutes of declination difference produces a correction of +2.8' and the difference of 10' of LHA
produces a correction of +3.4.
SO:
Tab Hc 22 31.8
Delta d + 2.8
Delta t + 3.4
--------------
22 38.0'
Now we look at the Delta
latitude correction table, entering with the azimuth angle and the
difference in latitude. And extract the correction for the DR latitude of -27.0'.
22.38.0'
- 27.0'
-----------
22 11.0 for the Hc which we subtract from 90 degrees (89 60.0')
22
11.0'
---------------
67 49.0
67 X 60 = 4020.0
+ 49.0
4069.0 NM
You can do the same calculation with HO 229 using the printed delta dec correction and figuring
our your own delta t correction by subtracting the Hc for dec 22, LHA 26 Lat 41, 22 31.8, delta
dec +.56.
From that for the next LHA value of 25, 22 51.4 - 22 31.8 producing the
correction factor of +19.6 which is
the equivalent of the + .34 delta t number in HO 214
Using the multiplication tables in HO 229
Tab Hc 22 31.8
Declination corr +2.8
LHA corr + 3.3
22 37.9'
Applying the same DR latitude corr -27.0
-----------
22 10.9' almost identical as the HO 214 Hc.
gl
From: Gary LaPook <NoReply_LaPook@fer3.com>
To: garylapook@pacbell.net
Sent: Monday, December 29, 2014 10:41 PM
Subject: [NavList] Re: The Great Circle Challenge
Here is a link to those tables:
gl
File:
File:
File:
File:
