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» Online Classroom   »   » Public Discussion of Cel Nav   » NAO REDUCTION EXAMPLE


 - posted November 10, 2007 01:29 PM      Profile for jehan           Edit/Delete Post 
I need help with the following example of a sight reduction page 11-39 example number 2.

Line number 2 gives correctly the following values: 53h0 49h1 88po and Z2 of 89.3.
the following line list the first correction from the aux. table as minus 11 which is correct and the second correction as 00 which again is correct. If I have understood correctly in order to obtain Hc I must apply both corrections....
But the values are 53 49'
minus 11'
plus 00'

this should sum to 53 38' according to my fingers..and the example says 53 48'
Please advise me if I am wrong... as I have been loosing much sleep over this.

From: north bay on

 - posted November 10, 2007 08:17 PM      Profile for HHEW           Edit/Delete Post 
It's been an eon since I looked at the NAO table. I haven't worked my way through the entire problem, bu thave taken a look at the aux table. Looks to me like the correction for an A' of 49 and a P of 88 is zero (0). The way I read it, P of 88 is the second row from the top in the right-hand column (headed Z2). F' of 49 is in the middle of the 5th column from the left. Dropping down in this 49 column to the point where it crosses the row for 88, I see 0'.

The pesky 11' appears just above the 49, but I take that to be for an F' of 11' and not the correction.

FWIW, working this reduction with a calculator I get an Hc of 53* 48.4' , Zn of 178.196.

As I said, I haven't looked at this SR table in a long time, but as a first shot at an explanation, it looks reasonable.

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