Here I have given one method of finding the longitude, illuftrated by a fufficient number of examples, all of which are reduced to the year 1809, in order that the reader, or teacher, may have fufficient time to furnish himself with a N. A. for that year, which is now printed. But as many would wish to have some other method of reducing the diftance, that, by comparing them together, they may not only have the advantage of proving their calculations, but also of making choice of which they prefer to work by; the fecond method I fhall prefent the Reader with, is chiefly deduced from that invented by Mr. Witchell, late Mafter of the Royal Academy at Portsmouth, as it is fhort, and requires but four places of figures in the logarithms, befides the index; the preparations in both methods being exactly the fame.

RULE.

Firft. Add the fun, or ftar's and moon's apparent altitudes together, half the fum; fubtract the lefs from the greater, and half the difference; then add together, the co tang. of half the fum, the tang. of half the difference, and the co-tang, of half the apparent distance; their fum (rejecting 20 in the index) will be the log. tang. of an angle, which call A.

Secondly. When the fun or ftar's altitude is greater than the moon's, take the difference between angle A, and half the apparent diftance; but if lefs, take their fum. Then add together the co-tang, of this fum or difference, the co-tang. of fun or ftar's apparent altitude, and the prop. log. of the correction of the fun or ftar's altitude; their fum (rejecting 20 in the index) will be the prop. log. of the first correction.

Thirdly. If the fum of ang'e A and half the distance was taken in the laft article, take now their difference, but if their difference, now take their fum; then add together the co-tang, of the fum, or difference, the co-tang. of the moon's apparent altitude, and the prop. log. of the correction of the moon's apparent altitude; their fum (rejecting 20 in the index) will be the proportional logarithm of the fecond correction.

Fourthly. When the angle A is less than half the apparent diftance, the first correction must be added to, and the second subtracted from, the apparent diftance; but when the angle A is greatest, their fum must be added to the apparent distance, when the fun or ftar's altitude is lefs than the moon's; but when the moon's altitude is leaft, their fum must be subtracted to give the corrected diftance. Fifthly. In Table XXVI. look for the corrected dift. in the top column, and the correction of moon's alt. in the left-hand fide column; take out the number of feconds that and under the former and oppofite to the latter. Look again in the fame Table for the corrected diftance in top column, and the second correction in the left-hand fide column; take out the number of feconds that stand under the former and oppofite the latter, the difference be

2 G 2

tween

tween these two numbers will be the third correction, which muft be added to the corrected diftance, if lefs than 90°, but fubtracted from it, if more than 90°; the fum, or difference, will be the true distance.

To illuftrate this laft method of reducing the apparent diftance to the true distance, I fhall take the apparent altitudes and diftances as they stand in the firft examples, worked by the former method.

EXAMPLE I. See Example I. p. 231.

Given, the apparent diftance of the fun and moon's centres. 105° 10' 0", the fun's apparent altitude 12° 51', that of the moon 43° 32', and horizontal parallax at reduced time 57′ 49′′. Required the true diftance of their centres by Mr. Witchell's method?

M. S.

O's refrac. 4 6 D's hor. par. at red. ti. 57 49 P.L.

0 4932

O's parallax

O 1397

EXAMPLE II. See Example p. 232.

Given, the apparent diftance of the fun and moon's centres 68°. 42' 11", the fun's apparent altitude 32° 0' 12", apparent altitude of the moon 24° 0' 10", the fun's correction 1' 23", the moon's correction 530". What is the true diftance of their centres by Mr. Witchell's method?

EXAMPLE III. See Example p. 233

Given, the apparent diftance of the moon's centre from the star Regulus 31° 18' 30", the apparent altitude of the star 20° 10′ 55′′, that of the moon 31° 18′ 30′′, the ftar's correction 2′ 34′′, that of the moon's correction 48′ 33′′. What is the true distance of their centres by Mr. Witchell's method?

Another Method.

First. From half the sum of the apparent altitudes of the fun and moon, or moon and ftar, and the apparent distance, fubtract the fun or ftar's apparent altitude; the difference call the firft remainder, the moon's apparent altitude taken from the half fum leaves the fecond remainder.

Secondly. To the log. fine of thirty degrees add the log. fine of the apparent diftance, the log. co-fine of the moon's apparent altitude, the log. fecant of the half fum, the log. co-fecant of the first remainder, and the prop. log. of the moon's correction; reject the tens in the index, the remainder will be the prop. log. of the firft correction.

Thirdly. To the log. fine of thirty degrees add the log. fine of the apparent diftance, the log. co-fine of the fun or ftar's apparent altitude, the log. fecant of the half fum, the log. co-fecant of the fecond remainder, and the prop. log. of the fun or star's correction; reject the tens in the index, the remainder will be the prop. log. of the fecond correction.

The difference between the correction of the moon's altitude, and the firft correction, call the difference of corrections.

Enter Table XXVI. with the apparent distance at the top, and the moon's correction in the left-hand fide column, the corresponding number will be the third correction; in the fame column, and corresponding to the difference of corrections, you may find the fourth correction.

Fifthly. Subtract the moon's, the fecond, and fourth corrections from the apparent distance, to the remainder add the fun or ftar's, the first and third correction; the fum will te the true distance. EXAMPLE I. See Example p. 231.

Given, the apparent diftance of the fun and moon's centres 105° 10, the fun's apparent altitude 12° 51', that of the moon 43° 32', the fun's correction 3' 57", and the moon's correction 40′ 55": Required the true distance?

30° o' Sine

9 6990

9 6990 D's cor. 40′ 55′′

Ap. dift. 105 10 Sine

9 9846

9 9846 2d cor.

49

D's ap alt. 43 32 Co-fine 9 8603

C's apalt, 1251 Co-fine

9 9890

Dif. cor. 23 32

EXAMPLE II. See Example p. 2.

Given, the apparent distance of the fun and moon's centres 68° 42' 11", the fun's apparent altitude 32° 0' 12", apparent altitude of the moon 24° 0' 10", the fun's correction 1' 23", the moon's 51′ 30. Required the true diftance?

30° 0' 0" Sine 9 6999 Ap. dift. 68 42 11 Sine 9 9693 0 10 Co-fi. 9 9607 O 12 Co-fi.

9 6990 D's cor. 57' 30" 99693 2d cor.

4th cor. O I

D's ap.alt. 24
O's ap.alt. 32

9.9284

Sum

52 31

68 42 II

O's cor.

D's cor.

16 Secanto 3335
4 Co-fe.o 2964
6 Co-fec.
I 23 P.L.
51 30 P.L.

5435 2d

o 2073 O's cor. + 1 23 21:43 ft. cor. + 28 22 3d cor. + 9

cor. 2 2518

ift. cor.

28 22 P.L.

8024 1'0"P.L. True dift. 63 19 34

Diff. of cor. 23 8

EXAMPLE III. See Example p. 233.

Given, the apparent distance of the moon's centre from the ftar Regulus 31° 18' 30", the apparent altitude of the moon 19° 44′ 50′′, the apparent altitude of the ftar 20° 10′ 55′′, the ftar's correction 2' 34", the moon's correction 48′ 33′′. What is the true distance of their centres ?

The difference in this last method is that there is no variety of cafes.

Questions for Exercife.

Suppofe, on the 23d of May 1805, in longitude 9° weft of Greenwich, by account at 3 h. 41 m. 15 f. P.M. by a watch well regulated, the distance of the fun and moon's neareft limbs fhould

be