NavList:

Andres has solved this problem in the only way possible with the given information.  He has plotted the entire LOP on a world map and made a deduction as to which intersection is the proper one. 

 

In this case, the proper intersection can be easily determined, despite the relative closeness of the two fixes and that they are both over the water.  At first glance, you would assume that the ship wouldn't be near the antarctic coast which is probably a good first assumption.  However, a better way to determine which is the correct fix is to look at the twilight.  Since this is a star fix, there is a finite time in which to shoot it at sea.  Therefore, the difference between GMT of the sight and the LMT of twilight can significantly narrow the range of longitude that the observer must be in, in order to even observe the stars (barring exotic things like night vision or moonlight sights which would not be commonplace).  In fact, the easiest way to figure out an approximate longitude is look at the time of civil twilight in the nautical almanac for that date and compare it to GMT.  That will give you a close enough approximation of Longitude to solve the fix ambiguity in most cases (provided you don't shoot stars due east and west giving fixes on nearly the same meridian).  In this case it was 1500Z and evening civil twilight is around 1800 LMT, you have about 3 hours difference to the East.  Therefore I would assume first that my DR is around 45 degrees East.  Even though i am off by 5 degrees due to my massive rounding, i can safely eliminate the other position as impossible

 

In addition to twilight, another piece of data might be provided by the navigator if there was sufficient manpower to do this in the case where no DR can be provided.  In this particular case, only ONE star is needed to get an approximate position.  This of course is an altitude AND azimuth.  This is the same as using radar off of a single object for a position in pilotage waters and offers the same limitation.  That is to say that in both the case of Radar and Celestial, the observation of azimuth/bearing is not nearly as accurate as altitude/range.  however if no DR is available, you can get a rough position (dare I call it an EP?) with just this data.  If we have two stars and even if a single azimuth determination is made, we can easily determine which of the two fixes is the correct one to choose.

 

The trouble with all of this, and the reason it was not popular, is that you can't easily do this with the mechanical tools available at sea.  Sure a computer can do it all, as I said before, and Andres has demonstrated; but it's not very easy to do with plotting tools and the charts that we have.   I don't have a compass on my ship that can stretch that far on any chart so doing circles of equal altitude isn't an option except in very rare cases in the tropics (less than 2.5 degrees of Zenith distance).

 

Since JK finally explained what he meant my calculating LOPs "directly" I agree with him to a point.  When using a calculator for my celestial triangle reductions (which I do for a vast majority of my work at sea if it is not done on the computer), I certainly could use the DR/GPS fix as the AP for each star and then calculate and plot each star line from the azimuth and intercept from this position.  The trouble with this in a practical sense is the mess it makes when plotting manually.  I prefer to advance my lines in a way we do with radar LOP's.  I advance or retard each AP and plot from there.  This gives me a clean fix near my track.  If I were to try to advance the DR I would have a mess near my track and might be confused with the previously plotted LOP's.  It is even worse when you plot the original lines then try to advance them as you cross your track.  When my computer plots this, it is much easier, and I use his "Combined" method for nearly all of my star fixes.

 

Let me give you an example of this combined method as done on computer and how difficult it would be to plot these.

 

For 23 August 2009, to continue the example as posted previously.  The "DR" was the 1500Z GPS fix at Lat 31 deg 13.3 S; long 040 deg 43.6' E

 

Jupiter @ 17m 7s  Intercept -0.6

Moon (LL) @ 03m 19s Intercept -0.4

Rasalhague @ 19m 20s Inter 0.8

Altair @ 11m 31s Inter 0.3

Peacock @ 27m 28s Inter -0.2

Rig. Kent @ 7m 8s Inter 0.1

Spica @ 21m 43s Inter -0.4

Arcturus @ 14m Inter -0.3

 

As you can see, there are a lot of lines with intercepts so small that my dividers cannot accurately be used (they get rather dicey under 1.0 nm on the scale of plotting sheets.)  This is not to mention that all of these lines must be retarded to get the 1500 fix.  The computer does a good job of drawing the lines and determining the fix to be Lat 31 deg 13.1' S;  Long 040 deg 43.8' E; but plotting this manually would be more difficult than if I used AP's based on whole degrees for entry into tables

 

To be honest, most of my non-star fixes were done semi-automatically using the NA and my Tamaya NC-2 which merely provides Az and Intercept for inputted LHA, Dec, and Lat.  I would routinely use AP's chosen to give me whole degrees of latitude and LHA to make a cleaner plot.  I would sometimes enter everything using the DR as the AP when I was shooting a lone sunline, but that was about it.

 

In conclusion, there are a lot of practical considerations as to why a particular method is used to determine position at sea that goes far beyond the reduction of the celestial triangles, especially when you are using more primitive tools.

 

Jeremy

 

 

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

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

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