eclipse of the sun, made at two places, the longitude of one of which is known.

1. At one of the places let, at the time of observation of the beginning of the eclipse, the sun's semi-diameter + the moon's semi-diam. S. The moon's semi-diameter is to be increased according to its altitude (Props. 21 and 22).

Let the true latitude and longitude of the moon at the time of observation be found from the Nautical Almanac. These should be corrected for the errors of the tables if known by observation.

The apparent latitude (L) of the moon and the parallax in longitude are to be found by Prop. 22. At the same time the apparent diameter of the moon may be found.

By the right angled triangle formed by S, L and the apparent difference of longitude of the moon and sun, we can find this apparent diff. its log. being equal to the log.

√ (S2 — L2) = & log (S + L) + log (S—L).

The apparent difference of longitude between the sun and moon being known, the true difference is known, by help of the moon's parallax in longitude.

Hence the interval between the time of observation and the time of conjunction as seen from the centre of the earth, is known, by the help of the sun's and moon's motions in longitude given in the Nautical Almanac. Therefore the time of the true conjunction is known at one of the places.

2. By a similar proceeding the time of the true conjunction will be known from the observation of the beginning at the other place.

The difference of these times is the difference of longitudes. The mode of proceeding is the same for determining the difference of longitudes by observations of the end of the eclipse.

From

a It is convenient to carry on the computation so that the effect of any errors in the data on the result may be known, and thus the degree of the accuracy of the conclusion estimated. This may be done as follows:

Let ds the error in the sum of the semi-diameters,

the above also may be understood the mode of proceeding in determining the longitude by an occultation of a star.

Prop. XXV. Considering the earth as an oblate spheroid, O the equatoreal diameter exceeding the polar diameter by only a small quantity, to find the arc of the meridian intercepted between the equator and a given place.

APN (fig. 66) is the elliptic quadrant, N the pole, AOM the circular quadrant, and OT, TPR are tangents.

Let AC=m, CN=n, the lat. of the place P(= OPR) = 7, TOB=c, AB=x, and AP=%,

Also x=m-m.cos TOB=m-m.cos c. Hence dx = m. dc.

To develope the right hand side of this equation, regarding only the first power of the small quantity s; let cl=k. Then sin

Then if the computation be carried through with these quantities annexed to the sum of the semi-diameters, latitude of the moon, and horizontal parallax; we shall have, calling T the time of conjunctiou computed as in the above solution from the eastern observation, and T' that from the western.

The time of conjunction at eastern place....T+ads+bdl+cdp,
The time of conjunction at western place....T'+a'ds+b'd l+c'd p.
Where a, b, c, a', b', c', are coefficients resulting from the computation.
Hence the diff. of longitude =

T—T'+(a—a').ds+(b—b').dl+(c—c').dp.

From the magnitude of the coeff. (a—a'), &c., the accuracy of the result may be estimated. Thus if (a-a'), &c., be very small quantities considerably less than unity, the observations are adapted to give the difference of longitude to considerable accuracy. Vid. Conn. des Tem. 1811. p. 458.

Thus sink is of the same order of magnitude as s, and therefore (since cos (l-k) = cos l. cos k + sin l. sink = cos l+k. sin l), regarding only the first powers of 8, we have ks. sin l. cosl.

=

Now sinc=sin(l— k) — sin (l — s. sin l. cos l) = sin l—s. sin 7. cos. Hence we have

Hence integrating

=m.dl.(1-8-4.8.cos 27).

z=m, (l′′. sin 1′′-sl.sin 1"-.e. sin 27),

"denoting the latitude in seconds.

Prop. XXVI. Given (a) the length of the arc of the meridian between L' and L'+D, and (b) the length of the arc between L' and L'+D, to find the equatoreal and polar diameters (m) and (n).

We first find ms or m -n, and then m as follows; s is called the compression. Let D" denote the seconds in D; then by the preceding prop.

a=m. {D". sin l'-s.D". sin 1"-.8. sin (2L'+2D) +3.

8. sin 2 L/}

(1) b=m. {D". sin l′′-s. D′′. sin 1"-.s. sin (2 L" + 2D) + 3.

8. sin 2 L"}

b-a.sm. {sin (2L'+2D)—sin 2L-sin(2L"+2D)+sin2L"} . Buta sin(2L'+2D)-sin 2L-2 sin D. cos (2L'+D) sin(2L"+2D'—sin 2L"-2 sin D. cos (2L"+D) cos(2L+D)-cos (2L"+D) = 2 sin (L'+L"+D). sin (L"-L')

a See Trig. p. 21. formulæ (13), (14).

Hence

b-a=3m. s. sin D. sin (L"-L'). sin (L"+L'+D)

and m=

α

sin D

+ms+ 3/2 .ms.cos (2L/+D). Also n=m-ms.

Example. Colonel Mudge found the degree commencing lat. 51° 32′ north 121640 yards. Major Lambton found the degree commencing lat. 12° 33′ north = 120975 yards.

By the above formulæ, m-n=22167 yards, m=6972238 yards 3961,5 miles, n = 6950071 yards = 3948,9 miles, and