NavList:

Paul,

1) I have not been able to understand these arcs. Cannot find any connection to the hour angle determination.

2) As the result is (or at least seems to be) the log hav of the hour angle, the only explanation that I have arrived at is that some kind of non-standard log sec table have been used, namely one giving y = log (c · sec x) with c = 1.3418. Then you get y = 0.1277 + log sec x. Many different formulas could be derived for solving the hour angle, but with

hav(HA) = sec(LAT) · sec(DEC) / [sec(s) · sec(90° - (s - ALT))],
where s = (COLAT + DEC + ALT) / 2,

you get the log of the numerator = 0.1277 + log sec(10°8') + 0.1277 + log sec(7°23') = 0.2658,
s = 52°36', and log denominator = 0.1277 + log sec(52°36') + 0.1277 + log sec(55°21') = 0.7172,
the same numbers as Gerbault's, resulting in 9.5486.

Above solution seems rather far-fetched, but is at least one possibility.

3) Yes, I think it is Johnson of cloudy-weather fame. Gerbault uses some of Johnson's nomenclature, e.g. LT = longitude in time, and SMT = ship mean time. But I haven't found any table giving the shown values in the books by Johnson that I have been able to access on the internet.

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The fact that Gerbault calculate both COLAT+DEC and COLAT-DEC indicates that he is using some other formula, but I have not been able to find out which one. I have re-read two books by Gerbault, one on his Atlantic crossing and one of his circumnavigation, translated to Swedish, and I don't know the original titels. At the end of the first book, in an appendix for "those who know the sea", there is one paragraph about his navigation. He says he use a drum sextant of the same type as used in the Royal Navy torpedo boats. Also that he uses the tables of sub-lieutenant Johnson, which facilitates very quick reductions to sufficient accuracy. He has no real chronometer, but two clocks of "torpedo boat type". He also applies the new navigation methods of Sumner's lines.

At some other place, in one of the books, he writes that as an engineer he has no problem with math, and that he tried star-moon observations (lunars?) to be able to recover time, to be prepared if the clocks started to run apart.

Lars, 59°N 18°E