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» Online Classroom   » Celestial Navigation   » Public Discussion of Cel Nav   » Formulas for Calculating Accuracy

   
Author Topic: Formulas for Calculating Accuracy
Aly


 - posted August 31, 2013 05:02 AM      Profile for Aly           Edit/Delete Post 
Hi All,

I calculated a fix from LAN estimating DIP short from a landmark and applying the correction.

The difference between the fix and actual position was:
0°04.1' Lon (@ 32° Lat)
0°15.6' Lat

I calculated the accuracy of the fix from actual via a right triangle:
a = Cos(32°) * 0°04.1' Lon = 3.5 nmi
b = 0°15.6' Lat = 15.6 nmi
c = Sqrt(a^2 + b^2) = 16.0 nmi

Question 1: Is this formula correct for calculating the distance between two points on a sphere or did I oversimplify?

Question 2: Is there a formula or method for calculating the intersections of 2 LOPs without graphing? (ex. via non-graphing calculator, spreadsheet, slide rule...)

Thanks!,
Aly

From: Lyra
David Burch


 - posted September 01, 2013 09:34 AM      Profile for David Burch           Edit/Delete Post 
I do not follow the notation used, but the distance between two points can be computed by a sight reduction with lat-lon-1 = AP and lat-lon-2 = GP. This gives you Hc then distance = z = 90 - Hc.

or better still just use a great circle sailing calculator.

Put your actual position in and the fix position in and calculate distance and bearing between them.

From: Starpath, Seattle, WA
David Burch


 - posted September 01, 2013 09:42 AM      Profile for David Burch           Edit/Delete Post 
There are ways to compute intersection of two straight lines, and various ways to input the data, but not convenient enough to be used often. Also it does not work for high sights. The formulas for general cases are complicated.

An alternative is to plot the lines with an ECS program such as OpenCPN. Noaa has online a list of free echart viewers that can be used for this. Check out the "Free software" link at www.starpath.com/getcharts.

From: Starpath, Seattle, WA
Aly


 - posted September 03, 2013 03:31 AM      Profile for Aly           Edit/Delete Post 
Thank you, that method was more accurate. I think the standard trig formula I modified didn't adjust for the curvature of the sphere in the final result.

I did find the answer to question #2 -- I'll post it here for others' reference.

Nautical Almanac 1984; Sight Reduction Procedures; "Position from intercept and azimuth using a calculator." (p. 282) ["The position of the fix may be calculated from two or more sextant observations as follows."]

Thanks,
Aly

From: Lyra
Aly


 - posted September 03, 2013 03:36 AM      Profile for Aly           Edit/Delete Post 
Correction: *Nautical Almanac 1994 (edits are disallowed)
From: Lyra


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