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    Re: Longitude from Local Apparent Noon Altitudes
    From: Frank Reed
    Date: 2012 Aug 19, 08:29 -0700

    Bruce, you wrote:
    "I'm curious that there must be a way to do a "best fit" regression analysis"

    Sure. You can do a least squares quadratic fit to the data. Such a regression analysis might sounds like it would be significantly more accurate, but it's amazing how close an "eyeball" fit will usually be to the least squares solution. And of course an eyeball fit requires no computing device. The data can also be analyzed as a set of standard lines of position and again you would use a least squares analysis to find the best fix (once again, that requires a computing device). The difference in the resulting position is generally trivial, but it's an interesting way to understand how it all works. An advantage of the latter solution is that you can also simultaneously generate an error ellipse which will show the relative confidence that you should have in the longitude compared to the latitude. With so many sights here you could expect about 2 nautical miles accuracy in the resulting longitude and about 0.2 in the latitude, assuming no systematic errors (counting systematic errors, I would expect typically +/-3 in longitude, +/-1 in latitude at the one standard deviation level).

    It's important to remember that this is a type of "running fix". You, the observer, weren't moving in this case, but the Sun was. This time of year the Sun is moving south --away from you-- at a rate of nearly one knot. Today it's just about 1 nautical mile in 72 minutes. Your final sight is 42 minutes after the first, so ADD 0.6 minutes of arc to that final sight to "back out" the motion of the Sun (the Sun would have been higher at the end of the run if it were not travelling south away from you). For sights earlier in the set, add 0.1 minutes for every seven minutes after the start of the observations. It's easy to do this graphically: draw a line above the graph of the sights sloping upward at a rate of 0.1 minutes in 7.2 minutes of time. Then take out that offset from the graph of the sights below. Find the axis of symmetry of the sights with this offset applied.


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