A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Brad Morris
Date: 2009 Apr 29, 08:57 -0400
I am a bit disappointed that you would use regression analysis. Typically this is a least squares fit, but I am uncertain as to your exact methodology.
Perhaps I misunderstood. I thought your method was a mark-1 eyeball method.
Dave Walden has shown that a least squares curve fit provides a reasonable determination of latitude and longitude.
If regression analysis is required for each line on either side of LAN, then there is no inherent advantage over a curve fit.
Please Jim, just try another case. This test we have devised is tough, agreed. That was the intent. Imagine the poor navigator who has lost his electronics
and now wants to obtain a noon fix. Will he sit down to manually perform a least squares fit (linear or curve)? I think not!
I'll attach my results from your first data set. To do more, I would have to automate it, since I'm lazy. Rather than prolong this event, I fudged and used regression analysis to fit lines through your data. The scatter was beyond what my eyeballs could handle.
The results are: Lat, 56°12.2'N, Lo, 9°53.6'W.
As I examine the plot, I realize that I have committed the sin of extrapolation,
i. e., the intersection of the lines is outside of the data range. In this case, I should have used data over a longer period. That should have moved the time of LAN a bit earlier. If I use the plot from my Navigation paper, the time difference between maximum altitude and LAN is 4m50s. That moves LAN earlier by 15s, consistent with the above.
I enjoyed the learning, and I got quite a bit of that.
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