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Re: time of meridian passage accuracy
From: Andrés Ruiz
Date: 2009 Sep 29, 09:41 +0200

Gary LaPook in [NavList 9968] is right, but calculating the position at LAN has the taste of old times, when navigation was more art and less science. Do you remember the film “master & commander” taking shoots at noon aboard the HMS Surprise?.

George Huxtable, [NavList 9958], has put his finger on the pulse, the problem is longitude, because the rate of change of GHA.

My opinion is the same as Joseph Schultz, [NavList 9964]. This fact is problem for the beginners when take a prestigious book as a reference.

Recapitulating:

Avoid the EoT, BOWDITCH says: [NavList 9914]

“To calculate latitude and longitude at LAN, the navigator seldom requires the time of meridian passage to accuracies greater than one minute”

At time of LAN:

LHA = 0 for upper culmination and 180º for lower.

Then |L| = |GHA|

Sin H = sin Dec sin B +/- cos Dec cos B

Where B is the latitude

B = f( Dec, H )

L = f( GHA )

And with some simplifications the error e is function of the rates of change of the variables:

eB = f( dDec/dt ) since at culmination dH/dt = 0 and is near transit.

eL = f( dGHA/dt )

This is the academic point. Douglas Denny, [NavList 9939], compare practical navigation and theoretical astronomy.

For practical navigation, the question is: are you near the coast or shallow waters? Is the error of 1 min acceptable? Are you in blue water with no dangers around? Of course, the position obtained in such a way is better than nothing.

A numerical example: (Position from observation of a single body. Jim N. Wilson)

30/12/1982  19:55:45

Sun

Number of shoots = 26

a0 = -664.781491

a1 = 69.933785

a2 = -1.752351

Mean Square Error

mse = 0.000257

Maximun Altitude:

t (Hmax) = 19.954279 h = 19:57:15

Hmax = 32.957642 º

LAN (James Wilson): (full equation)

t LAN = 19.929029 h = 19:55:45

Dec = -23.149089 º

GHA = 118.285394 º

Checks:

LHA = 0.000000 º

Hc = 32.956525 º

Z = 180.000004 º

heye = 1.86 m

IC = 1.5

Latitude

Dec = -23.149088 = -23º  8.9'

Hs = 32.956525 =  32º 57.4'

Ho = 33.189697 =  33º 11.4'

Culmination = S

B LAN = 33.661215 =  N33º 39.7'

Longitude

30/12/1982  19:55:45

GHA = 118.287461

L = W118.287461 = W118º 17.25’

For other UT  - Peter Hakel has  provided his solution: [NavList 9961]

30/12/1982  19:55:42

Dec = -23.149091 = -23º  8.9'

B = 33.661212 =  33º 39.7'

GHA = 118.274965

L = W118.274965 = W118º 16.50’

30/12/1982  19:55:44

Dec = -23.149089 = -23º  8.9'

B = 33.661214 =  33º 39.7'

GHA = 118.283296

L = W118.283296 = W118º 17.00’

Introducing an error of 1 min:

30/12/1982  19:56:45

Dec = -23.149041 = -23º  8.9'

B = 33.661262 =  33º 39.7'

GHA = 118.537378

L = W118.537378 = W118º 32.24’

In my humble opinion, 1 min of error in time of LAN is unacceptable for longitude calculations, but is not a problem for latitude.

Andrés Ruiz

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