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    Re: Burton's "Nautical Tables, High Latitude Edition"
    From: Brad Morris
    Date: 2013 Feb 19, 15:40 -0500

    Hi John

    Indeed, that is the answer I got when following the instructions and performing the tabular lookup. 

    I decided to take it one step more.  Since the notes provide the equations that the tables represent, we can attempt to solve this directly.  Here is the first surprise.  There isn't a hidden log function in the tabular representation of the equations, the tabular result MATCHES the equation result without transformation!  There is very regularly a hidden log function in nautical tables, just not these!

    I re-wrote the equations so as to make them easier to perform by calculator.

    A=cot(HA)*tan(Lat)
    A=tan(Lat)/tan(HA)

    B=Cosecant(HA)*tan(dec)
    B=tan(dec)/sin(HA)

    Lat = 77d 40m = 77.6667d
    HA = 2h 24m = 2.4h *15d/h= 36d
    Dec = 14d 4m = 14.0667d

    Therefore
    A= tan(77.6667)/tan(36)
    A=6.29505

    B=tan(14.0667)/sin(36)
    B=.42628

    Since these are natural numbers and not logs, the subtraction is really that, a subtraction and not a division!

    C = A-B = 5.86877

    Since the error term was given as 11m, we can just find the longitude directly

    Corr= C*11m = 64.55651m

    Add to the uncorrected longitude to find the new longitude

    New=Old+Corr
    New=166d 40m + 64.55651m
    New=167.742608
    167d 44.5m

    This is to be compared to the tabular result we both obtained of
    167d 45.7m.   In other words, off by 1.2m.

    I suspect that the deviation between calculator result and table lookup result would grow if I didn't so carefully select values that required little or no interpolation when performing the table lookup. 

    I also find it interesting that Burton indicates only 2 decimal places are required for this equation!

    Cheers
    Brad


    On Feb 18, 2013 7:10 PM, "John Brown" <jdb0302@yahoo.co.uk> wrote:

    Hi Brad

    You wrote:

    "Are the ABC Tables still part of Burton's?"

    ...So far as I know, Burton's are out of print, but they were included in the 8th edition. They are also in Burton's 4-figure tables, Norie's, and others.

    and again:

    "Here's an example of a problem I had in mind.

    On day 0, you measure your longitude using the sun and determine it to be E166d40m. The latitude was S77d51m. The sun's Hour Angle was 2h24m when you determined your longitude, and the declination was S14d4m. Using your sledge-meter (a device similar to your car's odometer), you travel south 69.1 miles. You held a straight line the entire time. This takes several days and it was cloudy the whole time. You expect your position to now be S78d51m E166d40m, that is, 1 degree further south.

    You observe the sun crossing your meridian, the latitude is revealed to be S78d40m, that is, 11 arc minutes in error. What is your longitude?

    According to Burton's "High Latitude Edition", there is a solution, using the ABC tables. The very tables just posted to the list."

    ...You travel south 69.1 miles and expect to be 1 degree further south, so the ice must be moving north?

    Anyway, here goes...

    From initial obs:

    HA 2h 24m, or 36d
    Lat 77d 51'S
    A = +6.40
    Dec S 14d 04'
    B = -.426
    A+B = +5.97 = C

    Z = 39d in NW quadrant, so Zn = 321d and p/l runs NE/SW

    After several days:

    DR posn at time of mer pass is 78d 51' S, 166d 40' E
    Obs Lat by mer alt is 78d 40' S, that is, 11' N of DR lat
    To find the correction to the DR long, multiply the lat difference by 'C'
    11' x 5.97 = 65'.7

    The transferred position line runs NE/SW. If the obs lat is N, the obs long is E

    The longitude is 166d 40' E + 1d 05.7' = 167d 45'.7

    Not having done this for many years, I am standing by, with trepidation, to be shown the error of my ways.

    Best Regards

    John

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