NavList:

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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