# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: 3 sights SR example
From: UNK
Date: 2017 Jun 30, 10:43 -0700

Yves,

You wrote

>>  But I have observed that some deviation from analytics may occur when the position triangle -or more generally the polygon- is going to limit conditions. For instance, when the triangle has 2 angles growing significantly, tending to Pi/2 (or when a high DOP is anticipated). <<

No discrepancy occurs between statistical MPP and Symmedian Point (SP).

If two angles in the Cocked Hat are approaching Pi/2, the third one becomes 0. In other words, you end up with two parallel LOPs cutting the third one perpendicularly. Statistically, this situation can be interpreted as two observations with the same (or opposite) azimuth, which can be averaged into a new LOP running in the middle of the old ones. Intersecting it with the remaining LOP obtains the MPP where common sense would have expected it: At the bisection point of the base line of this "Cocked Hat".

To find the solution via the SP, all you need to remember is that the SP in a rectangular triangle bisects the height drawn from the point at the right angle to its base line. In the degenerate triangle of your example you have in fact two right angles. The respective heights coincide with the base line. Done.

You have a more interesting example?

By the way, it is not essential that the 3rd LOP runs perpendicular to the other two. Step 1 of the proof remains the same and step 2 will be based on the general method of constructing the SP. The details are left to the reader.

Best regards,

Herbert Prinz

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