NavList:

If the procedures being considered are

• 1) Take 5 sights of each of 3 objects. Average each of the sets of 5 sights and draw 3 LOPs
• 2) Take 3 sights of each of 5 objects. Average each of the sets of 3 sights and draw 5 LOPs

Which one is better?

If the errors are Gaussian then there are some fairly simple expressions that can be obtained for area of the confidence ellipse and root-mean-square distance of possible positions from the Most Probable Position (MPP). Either of these can serve as a useful figure of merit (smaller is better) to evaluate a round of sights and can be found in Appendix C of my paper Probabilities in a Gaussian Cocked Hat . They depend on the azimuths of the objects being observed ( Z₁, Z₂, Z₃,...) and the standard deviation of the error associated each sight ( σ₁, σ₂, σ₃,...). Any conclusions will therefore depend on the specifics of the round of sights. As discussed in the attached document the two key quantities that go into these figures of merit are Tr A₃₃ and |A₃₃|

Let's assume that σ is the same for each of the sights and the azimuths are equally distributed around the horizon at intervals of 120° for case 1 and 72° for case 2. It turns out that the key quantities are equal in both case 1 and case 2 and hence the two procedures are equally good. In a way we might have expected this. More generally for n sights with equal errors of each of m objects distributed at equal azimuth intervals around horizon the key quantities are Tr A₃₃ = mn/(2σ²) and |A₃₃|=(mn/(4σ²))².

On somewhat related note: Both of the figures of merit mentioned above are inversely proportional to the square root of |A₃₃|. From that it follows that if you want to chose 3 objects with which to form a cocked hat, as is done in some sets of tables, then you should choose the 3 objects for which sin²(Z₁-Z₂)+sin²(Z₂-Z₃)+sin²(Z₃-Z₁) is largest.

Robin Stuart