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Sumner Line of position
From: Andrés Ruiz
Date: 2009 Nov 17, 09:31 +0100

[NavList 10640] Re: AP terminology, WAS: 2-Body Fix -- take three

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So we need to get at least 2 Sumner points, and preferably three to expose plotting and/or math errors. Sounds like at least as much work, if not more, than St. Hilaire.  I don't know, since I've never plotted Sumner lines.  As an aside, we use the same equations with different names in our Great Circle sailings.

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Jeremy, an example of a FIX by Sumner line.

TWO Circles of Position

19/02/2004   20:00:00 Fix

 UT1 date body GHA Dec Ho 19/02/2004 20:00:00 Procyon 334.23 5.22 28.5 19/02/2004 20:00:00 Sirius 347.78 -16.72 19.55

DR at tFix: 42N 30.5W

Choose a latitude near your DR position

B1 = Be - e

B2 = Be + e

Obtain the LHA for these two latitudes:

LHA = ACOS( ( SIN( Ho )-SIN( B )*SIN( Dec ) )/ COS( B )/COS( Dec ) )

LHAprima = -LHA

Do: 0 <= LHA <= 360º

Obtain the longitude for this two points of the LoP:

L = LHA-GHA

Do: -180ºW <= L <= +180ºE

Choose L near your DR longitude

Plot: on plotting sheet or white chart for celnav. No protractor is needed.

For two simultaneous LoP fix: Chauvenet  Vol I, pg 428

Restrictions: the intrinsic for a time sight

• Body near the transit. Line N/S
• Line W/E
• High altitudes and declination 1/COS( B )/COS( Dec ) term

I have the felling, subjectively, that when I use Sumner it is simpler than the Saint-Hilaire. Doing the calculations with a standard calculator and obtaining the fix with a RADAR plotting sheet.

The results are, (see Sumner double altitudes.pdf):

L = 81.583253

Lprima = -30.043253

L = 81.499574

Lprima = -29.959574

SumnerLoP 1:   41.916667         -30.043253        42.083333         -29.959574

L = 54.553448

Lprima = -30.113448

L = 54.313510

Lprima = -29.873510

SumnerLoP 2:   41.916667         -30.113448        42.083333         -29.873510

Sumner2CoPFix:           41.991536         -30.005998

For comparison:

.

TWO Circles of Position

CoP1 = 334.230000   5.220000     28.500000

CoP2 = 347.780000   -16.720000   19.550000

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      22.047719943373227  41.991775    -30.005655   -11.987157   85.157807    41.991775    -30.005655       334.230000   5.220000     347.780000   -16.720000

I1: -11.987157      85.157807

I2: 41.991775 -30.005655

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      22.047719943364704  41.991775    -30.005655   41.991775    -30.005655   -11.987157   85.157807       334.230000   5.220000     347.780000   -16.720000

I1: 41.991775 -30.005655

I2: -11.987157      85.157807

Marq Saint Hilaire: ( 41º 59.5' -30º  0.3')

Andrés Ruiz

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