NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Andrés Ruiz
Date: 2009 Nov 17, 09:30 +0100
The problem is undetermined.
Peter in [NavList 10563] explains this question.
Is it possible to obtain the two mathematical solutions
but not the fix.
For obtaining the Fix with n=2 observations, additional
information is needed:
- Another shoot
- DR position
- Or rough information
about where we are. (if one solution is in the Sahara an the other in the
For n=3 observations the problem is determined without
any other DR, EP, AS,
For n>=4 observations the problem is overdetermined
For anybody who want to play with CelestialFix.exe, I
have attached the input file, (is a txt file editable with notepad)
TWO Circles of Position
CoP1 = 262.103255 8.896379 22.151667
CoP2 = 338.760532 -60.880271 57.716667
TWO Circles of Position
CoP1 = 262.103255 8.896379 22.151667
CoP2 = 338.760532 -60.880271 57.716667
Vector Solution for the
Intersection of two Circles of Equal Altitude
THE JOURNAL OF NAVIGATION
(2008), 61, 355-365. The Royal Institute of Navigation
Andrés Ruiz González - Navigational Algorithms -
http://sites.google.com/site/navigationalalgorithms/
iter Err Be Le B1 L1 B2 L2 GHA1f dec1f GHA2f dec2f
0 3078.946281246031504 -31.230905 40.717797 -58.464734 87.910376 -31.230905 40.717797 262.047917 8.893050 338.738269 -60.866380
1 1.083375217608083 -31.229957 40.738884 -58.447603 87.912810 -31.229957 40.738884 262.039722 8.909401 338.714244 -60.866215
2 0.000264296945961 -31.229958 40.738889 -58.447596 87.912812 -31.229958 40.738889 262.039724 8.909408 338.714235 -60.866216
3 0.000000092865235 -31.229958 40.738889 -58.447596 87.912812 -31.229958 40.738889 262.039724 8.909408 338.714235 -60.866216
I1: -58.447596 87.912812
I2: -31.229958 40.738889
An analytical solution of the
two star sight problem of celestial navigation
James A. Van Allen.
NAVIGATION Vol. 28, No. 1, 1981
iter Err Be Le B1 L1 B2 L2 GHA1f dec1f GHA2f dec2f
0 3078.946281246026047 -31.230905 40.717797 -31.230905 40.717797 -58.464734 87.910376 262.047917 8.893050 338.738269 -60.866380
1 1.083375217609392 -31.229957 40.738884 -31.229957 40.738884 -58.447603 87.912810 262.039722 8.909401 338.714244 -60.866215
2 0.000264296951883 -31.229958 40.738889 -31.229958 40.738889 -58.447596 87.912812 262.039724 8.909408 338.714235 -60.866216
3 0.000000092857165 -31.229958 40.738889 -31.229958 40.738889 -58.447596 87.912812 262.039724 8.909408 338.714235 -60.866216
I1: -31.229958 40.738889
I2: -58.447596 87.912812
This solution is obtained starting with George’s
favourite point: (B,L)=(0,0)
Fix(-31.229956267426363, 40.738880679580390) (-31º
13.8' 40º 44.3') LopLSSR
Fix(-31.229957799037262, 40.738889251954326) (-31º
13.8' 40º 44.3') KaplanSR_Bm = 0
Well, I am agree with JK about the terminology used
in navigation is not so fortunate in some cases. For example “circle of
position”, why circle?, if it is a circumference.
JK says in [NavList 10635] “iterations are
required to get exact points on the LOP”, yes, but iterations are required
to improve the solution if the starting point for calculation gives an unacceptable
error, in Sumner, Marq, matrix SR, and others.
Andrés Ruiz
Navigational Algorithms
http://sites.google.com/site/navigationalalgorithms/
De:
navlist@fer3.com [mailto:navlist@fer3.com] En nombre de Anabasis75@aol.com
Enviado el: viernes, 13 de
noviembre de 2009 22:31
Para: navlist@fer3.com
Asunto: [NavList 10640] Re: AP
terminology, WAS: 2-Body Fix -- take three
So here is my
challenge. I need a demonstration
from you to try to understand how you are solving these triangles without the
third point of the triangle. I can even supply you with any amount of
real world information you may require to demonstrate 1) how the Sumner line is
easier/better than the St. Hilaire line; and 2) how you can easily draw an LOP
on a mercator projection without the use of a prime point position
(DR/AP whatever you want to call it). I only ask that you don't use a computer
to do this because we all know that computers can do this easily. I just
want to know how we do this as simple sailors with tables a
universal plotting sheet, and some plotting tools.
Here's your chance:
August 23, 2009
Altair at Ho 22 deg 09.1' shot at
UTC 15h 11m 31 seconds
Rigel Kentaurus Ho 57 deg 43.0'
shot at UTC 15h 07m 08 seconds
Ship's course 245 deg true, speed
17.8 knots.
Find position at 1500 UTC
All altitudes are ready to go.
I look forward to your reply.
Jeremy
--
NavList message boards: www.fer3.com/arc
Or post by email to: NavList@fer3.com
To , email NavList+@fer3.com
