P " 18 5 7
PV " 26 2 32

Calling P the polar angle and PV the obliquity of vortex.

[Illustration: Fig. 13]

To find the arc AR.

By combining the two proportions already given, we have by logarithms:

M.R.V. minor = 3256 Log. 3.512683
M.S.D. of moon = 940" " 2.973128
P.S.D. of earth = 3950 A. C. 6.403403
Radius 10.000000
T.S.D. of moon 885".5 A. C. 7.052811
Log. Cosine arc AR = 28d 57' 3" 9.942025
---------

As the only variable quantity in the above formula is the "True"
semi-diameter of the moon at the time, we may add the Constant logarithm
2.889214 to the arithmetical complement of the logarithm of the true
semi-diameter, and we have in two lines the log. cosine of the arc AR.

We must now find the arc RK equal at a maximum to 2d 45'. The true
longitude of the moon's node being 79d 32', and the moon's longitude,
per Nautical Almanac, being 58d 30', the distance from the node is 21d
2', therefore, the correction is

-2d 45' x sin 21d 2'
-arc RK = --------------------- = -59' 13"
R

To find the correction for displacement.

True longitude of sun at date 100d 30'
" of moon " 58 30
Moon's distance from quadrature 48 0

As the moon is less than 90d from the sun this correction is also
negative, or

-90' x sin 48d
Arc Kq = --------------- = -1d 6' 46".
R

Arc AR = 28d 57' 3"
RK = - 0d 39' 13"
Kq = - 1d 6' 46"
Sum = 26d 51' 4" = corrected arc AQ.

We have now the necessary elements in the Nautical Almanac, which we
must reduce for the instant of the vortex passing the meridian in
Greenwich time.

July 2d.
Meridian passage, local time, at 9h. 5m. A.M.
" in Greenwich time 2d. 3h. 1m.
Right ascension same time 56d 42' 45"
Declination north " 18 00 1
Obliquity of the vortex " 26 2 32
Polar angle " 18 5 7
Arc AQ " 26 51 4

[Illustration: Fig. 14]

PA = 17d 59' 59" } P = 128d 37' 38"
PV = 26 2 32 }
VA = 89 3 0 V = 47 59 44
VQ = 62 11 56 A = 20 3 42
PQ = 47 14 22 Q = 26 2