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    Re: Latitude by Talcott-Horrebow Method
    From: Frank Reed
    Date: 2018 Nov 3, 09:22 -0700

    Ed Popko quoted the equation for this method:
    "Latitude = 1/2(d+d') + 1/2(z-z ')"

    Recognize that this is nothing but the ordinary method for getting latitude by meridian sight. You take one sight facing south (so your "shadow" from the object points north). That implies Lat = z.d. + dec. Then another facing north (so the shadow from the object points south. Now we have Lat = -z.d. + dec. But maybe there's some uncertainty in the refraction tables so you average and get Lat = (dec+dec')/2 + (z.d.-z.d.')/2. That's all this is. Naturally, merdian sights are always as simple as the standard noon case. There's no "new math" here. It's just averaging two meridian sights. And note, too, that you don't have to set it up as a distinct equation. You can just work out the latitude in both (or several) cases directly by "plain vanilla" Lat = z.d. + dec and then average.

    Do we really remove uncertainty in refraction by averaging two sights? Well, yes, to some extent. The difference in tabulated refraction at two altitudes is clearly smaller than the tabulated refraction itself, but note that this works best when the two bodies are relatively close together in altitude, e.g. one 60° high to the north and the other 61° high to the south. The difference, while small, can still be significant. There's no miracle here. Of course, we're talking about astronomical instruments that have been carefully leveled so these observations require no horizon and can be performed deep into the middle of the night when third and fourth magnitude stars are available in large numbers, and finding a pair with nearly the same altitude becomess practical, even easy.

    Yet note that this all depends on the leveling process. A telescope leveled to the nearest tenth of a second of arc?? Good luck with that! This is by far the biggest limiting condition. No matter how acccurately you measure the zenith distances of those stars, if the instrument is some arcseconds out of level, then your latitude will necessarily have the same error. This is not so different from marine celestial navigation. The accuracy of standard sextant sights at sea is primarily limited by the uncertainty of the horizon, which is a giant "inside-out" bubble level for the observer on a ship at sea. Celestial navigation is fundamentally dependent on finding the vertical (or equivalently the level), and its accuracy is fundamentally limited by that process.

    Finally, since the limiting resolution of the human eye is about 1', in order to "see" and reliably record observations to a tenth of second of arc, one would need a scope magnification of 600x (since a tenth of an arcsecond is 600 times smaller than a single minute of arc). With averaging, you could get by with 100x, but clearly the claim of tenth of a second accuracy is a bit crazy. It's a purely theoretical limit when an observer has access to perfect instruments.

    In the early 19th century, the "formal" science of metrology didn't yet exist. The concepts of observational statistics and precision in calculation were only vaguely understood. It's not at all uncommon to see, for example, amplitudes of the Sun (for correcting a magnetic compass) worked out to the nearest minute of arc or sometimes the nearest second of arc. These are meaningless numbers and a reflection of the ignorance of the navigators performing the calculations. They operated under the naive principle that "more is better". This ignorance wasn't universal, and you do sometimes see navigators rounding their results. For example, a navigator might record latitude to the nearest five minutes of arc. But this was "common sense" practice. Even today, many navigation enthusiasts just "don't get it" and invent their own rules of thumb for accuracy.

    Frank Reed

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