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This is a "Venus +1" scenario. I shot the altitude of Venus high in the south at meridian passage and the Sun before and after the Venus sights, low in the southeast. Sights were taken on 20 Nov 2023 before 8:45am local standard time using a Davis Mk 15 sextant with a sight tube (no magnification). Index correction -1' (tested before sights only!). Height of eye 25 feet. Zero observer motion. It probably doesn't matter, but it was cold, 36°F, and air pressure was 30.30 in.Hg.

Shooting Venus is tricky. You have to find it with binoculars first. Then it helps to try to get it aligned with some foreground feature. In this case I kept Venus in view in binoculars and moved (slowly walked) until it was aligned with a chip of paint high on a flagpole about thirty feet south of me. After that I could find it relatively easily naked eye (two weeks ago this was still possible, but Venus is fading fast). Even with this alignment, that doesn't mean it was easy with a sextant. This is where I realized I could do better by avoiding any optical losses. If I could look at Venus directly --no scope, no mirrors, then I could get a good sight. So I removed the scope and held the sextant in my left hand upside-down. I aimed directly at Venus but kept it to the right of the horizon glass of the sextant. Then I slid the index arm until I could see the horizon beside Venus. From there I could adjust the micrometer to align Venus with the horizon. A full-horizon glass might have actually worked better here. Best of all might have been a sextant with no clear glass at all on the horizon (like a Davis Mk. 3). Venus remained outside the field of view of the horizon and visible plainly with no loss of light from any optics. It didn't move while adjusting the sextant, so I never risked losing sight of it. Adjusting the micrometer caused the horizon to move up and down to align with Venus. This is a little awkward at first, but it's a handy trick. It also works well with fainter stars in twilight.

Raw data: three Venus altitudes, separated by roughly one minute each, centered approximately on 13:40 UT:
 45°00'
 45°03'
 44°58'
Even in that small number of sights, it's clear that there is considerable scatter. This was quite close to the expected time of meridian passage so the real variability of altitudes over just a few minutes would have been undetectable with a plastic sextant. But never mind... Average and go...

From that average altitude, I can get my latitude. So you tell me: what do you get for a best estimate of latitude? Are there corrections you can ignore? Does the result depend on the cold weather that day? Should we worry about the parallax of Venus? How sensitive is this latitude to my rough estimate of UT? Note: this isn't a homework puzzle requiring an answer to every question. I'm bringing up these things for conversation. :)

Before and after the Venus sights, I shot the Sun (LL, of course!). The raw altitudes were:
  16°42' at  13:37:05 UT,
  17°30' at 13:44:05 UT.
Averaging the altitudes and the times gives
  17°06' at 13:40:35.
I can look up the Sun's GHA and Declination at that time. From the zenith distance, ZD, of the Sun, the Latitude (by Venus, already determined)), and the Dec of the Sun, I can calculate the Sun's dLon (its difference of longitude from my location):
  cos(dLon) = cos(ZD) / cos(Dec) / cos(Lat) - tan(Dec) × tan(Lat).
Since the Sun is still climbing, I can add that dLon onto the Sun's GHA: Lon = GHA + dLon. What do we get? What is my longitude? Is this one impacted by the cold weather?

Notice once again that you don't need to do any plotting of LOPs, and the "intercept method" would be over-kill. Just calculate the dLon (also known as HA) directly. Etymologically, HA comes from "Hour Angle" but that's only jargon; its meaning is simple: HA is just dLon, the difference in longitude between me, the observer, and the subSun point (and of course the longitude of the subSun point is simply the Sun's GHA). You can calculate the latitude directly from the Venus meridian sight. And since the Sun sight when averaged can be considered simultaneous with the latitude by Venus, we can calculate the longitude directly using that little bit of trigonometry above. It's a fast and easy position fix.

Frank Reed

PS: A reminder: that "bit of trig" is nothing exotic. It's the standard great circle distance equation, also known as the fundamental law of cosines of spherical trigonometry, re-arranged for the case where distance away (identical to ZD in celestial navigation) is already known, and we're solving for the dLon, the difference in longitude. If you've attended any of my Modern Celestial Navigation workshops, you've seen this "bit of trig" split up as A,B,C:
  A = cos(ZD) / cos(Dec) / cos(Lat),
  B = tan(Dec) × tan(Lat),
  C = (A - B),
and finally
  dLon = HA = acos(C)
(on common calculators, that acos or "inverse cosine" is often accessed via shift+cos). We split the calculation up like this --as A,B,C-- for better error management. If something looks funny, you can go back and check each step.