meridian, was, from a mean of five sets of observations, 61° 56' 30"; the observed altitude of the moon's lower limb was 32° 4'; the observed altitude of the star 6° 16′; the barometer being 29.2 inches, the thermometer 42° F., and the height of the eye 20 feet; what was the true longitude? Est. time 13h 32m Long. in time 0 2 E.

moon's equatorial hor. par. 58' 14"

moon's semidiameter

15 52

augmentation to 32°
augm. semidiameter

+9

Est. G. time 13 30

16 1

Now to correct the oblique semidiameter by Dr Young's method from Tables I. and II. we have

the

Now to approximate time 1h 30m, and second difference-44", equation of second difference is 5.5, to which and variation 1° 29′ nearly in 3 hours, the final equation in time is about 11' to be subtracted. Whence from 13h 29m 30s this equation of 11 being subtracted, the true apparent time is 13h 29m 19 at Greenwich.

To compute the apparent time at ship.

Long. in time

13 30

3.91 19.00

48h

44.91-11′ 15′′ E.

If the operations had been performed by the ordinary tables without attention to the smaller equations, the longitude would be about 30" E. and the error on that account 10′ 45′′.

Ex. 5.-Let the example to the general formula, page 120, be solved as an exercise.

In this table h is half the sum of the apparent distance and altitudes of the moon, and the sun or a fixed star s, the altitude of the moon, s' that of the sun, or a star and d the apparent distance.

The correction from this table is always to be subtracted from the semidiameter found from the Nautical Almanac.