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Re: AP terminology, WAS: 2-Body Fix -- take three
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 Atlantic is clear, but this is not the case)

For n=3 observations the problem is determined without any other DR, EP, AS, Chosen position.

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

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

De: navlist@fer3.com [mailto:navlist@fer3.com] En nombre de Anabasis75---.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.

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.

Jeremy

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