(G) This solution is analogous to Problem 8th in the Requisite Tables published by order of the Commissioners of longitude; or to Problem VII. Chap. II. Book III. of Dr. Mackay's Navigation. If you have the Requisite Tables* by you, the sum of the three logarithms above may be looked for in the column of rising, and the hours and minutes answering thereto will be the time from noon. Or, increase the index by 5, and the sum is the log. versed sine of the same angle. Or, the log. versed sine of an arc increased by the logarithm of 5 (with 9 for its index), will give double the logarithmical sine of half the required angle. (P. 131.)

EXAMPLE II.

In latitude 39°.54' North, longitude 80°.39.45" West of Greenwich, suppose the altitude of the sun's lower limb to be 15°.40.57" on the 9th of May 1822, at 5h.34.21" P.M. per watch. Required the error of the watch.

1. To find the time at Greenwich, nearly.

80°.39.45" x 45h.29′.39′′ the difference of time betw ten the clocks at Greenwich and the place of observation. Hence, 5h.34.21"+5h.22′.39"=10h.57' P.M. the time at Gree: wich.

The XVIth of the Requisite Tables, or the XXXV, XXXVI, and XXXVII Tables in Dr. Mackay's Navigation.

The XVIth of the Requisite Tables is divided into three columns. The FIRST column, called half elapsed time, is only a table of secants with 10 rejected from the index; the time being reduced to degrees.

The SECOND, called middle time, is the logarithmical sine of half the elapsed time, increased by the logarithm of 2, and the index of the sum diminished by 5. 'The third, called column of rising, is a table of logarithmical versed sines with their indices diminished by 5.

Thus for 1 hour of time, or 15 degrees.

co-sec of 15°10'58700,

10

Log. elap. time 0.58700

sine of 15°-9.41300, ver. sine of 15°-8.53243 log. of 2 =0.30103,

-5

Proceed in the same manner for any other time.

+ Table IV.

Nautical Almanac.

Table VI.

4. To find the true time. In the triangle szN,

reject indices.

Complement of latitude Nz=50°. 6'. 0" cosecant - 11511

Complement of altitude
Polar distance

(H) In the practical application of this problem, it will be proper to take several altitudes of the sun within a minute or two of each other, and to note the corresponding times per watch. The mean of these altitudes, and of the times, will be more correct than a single altitude and time, and will in a great measure counteract the errors arising from the imperfection of the instrument.

EXAMPLE III.

On the 4th of March 1822, in latitude 45°.37′ North, and longitude 169.59'.30" West, suppose the following altitudes of the sun's lower limb to be observed:

Times per watch P.M. Altitudes of the sun.

Required the apparent time of observation, and the error of the watch.

18

The sun's declination at Greenwich at noon being 6°.31.59" S. (Naut. Alm.) and on the 5th 6°.8'.53" S. also the O's semidiameter 16.9".

Answer. The time at Greenwich is 14h.15.4", the O's declination at that time = 6°.18'.16" S. true altitude of the 's centre 24°.57.45"; hence the apparent time of observation =2.55'.50", and the watch is 44" too slow.

す

EXAMPLE IV.

On the 15th of September 1822, in latitude 33°.56' South, and longitude 84°.6.15" West, suppose the mean time per watch of several altitudes of the sun's lower limb to be 8h.11.5" A.M. and the mean of the altitudes themselves 24°.43′; required the error of the watch?

The O's declination at noon (Naut. Alm.) on the 16th being 2°.46′.24" N. and on the 17th 20.23.11", also the O's semidiameter 15'.56".

Answer. The time at Greenwich on the 16th is 1h.47′.30" P.M. the O's declination at that time=2°.44'.40" N. true alt. of the O's centre 24°.57′.2"; hence the apparent time is 3h.48'. 54" from noon, or 8b.11'.6" A.M. and the watch is right.

PROBLEM XV. (Plate III. Fig. 1.)

(I) Given the latitude of the place, the declination and altitude of a known fixed star, to find the hour of the night when the observation was made.

EXAMPLE I.

In latitude 7°.45' South, and longitude 30°.18' East of Greenwich, on the 7th of September 1822, suppose the altitude of the star Procyon, when east of the meridian, to be 28°.12'; required the true time of the day? The declination of Procyon being 5°.40'.29" N.

Procyon's alt (28°.12-refraction 1.47")28°.10.13′′

(K) The hour / found by calculation, as above, is the distance of the star from the meridian of the place of observation; consequently when the star is eastward of the meridian at the time of observation, this hour angle must be subtracted from the time of the star's culminating, and when westward it must be added.

*'s right asc. (1822, Tab. VIII.)+24h.
O's right ascen. Sept. 7th, 1822, (Naut Alm.)
Time of 's culminating nearly

O's right ascension, Sept. 7th, 1822, at noon
O's right ascension, Sept. 8th, 1817, at noon

31.23.59" 11. 2. 0

20. 27. 59

11h 2′ 0′′ 11. 5. 36

Increase of the 's right ascension in 24 hours ≈ 0. 3. 36

Then 24h,3'.36": 24h::20h.27.′59′′: 20h.24.55" time

of Procyon's culminating at Greenwich. Then,

15° 10": 30°.18' 20", hence 20h.24.55"+20" 20.25′.15" : the time of the 's culminating in longitude 30°.18′. E.

Lastly. 20°.25′.15′′-4.2′.1"-16h.237.14" the correct time.

(L) In order to attain the greatest accuracy several altitudes of a star should be observed in succession, and if two or more separate stars be observed, and the error of the watch be deduced from each star, the medium result will be more correct than that derived from the observation of a single star.

If an equal number of stars can be observed on each side of the meridian, and nearly equidistant therefrom, the errors arising from the imperfection of the instrument, &c. will be rendered almost insensible.

EXAMPLE II.

On the 29th of January 1822, in latitude 53°.24' North, and longitude 25°.18' West; suppose the following altitudes of Procyon to the west, and Alphacca to the east of the meridian, were observed by two separate persons at the same instant of time; required the error of the watch.

The right ascension of Alphacca being 15b.27'.9" declination 27°.19.12′′ N., and of Procyon 7h.29.59" and 5°.40.29" N. The sun's right ascension at noon on the 29th-20h.46′.11′′ (Naut. Alm.) and on the 30th=20.50.17".

Mean 42. 8.55

Mean 14. 58.38 Mean 19. 53. 55 Answer. The hour by Alphacca's corrected altitude is 3". 40'.47", and the culminates at 18h.37.31", hence the correct time 14.56'.54".

The hour by Procyon's corrected altitude is 4h.15′.55", and the culminates at 10h.41'.42", hence the correct time=14". 57.37". The mean time between the two is 14h.56′.55", consequently the watch is 1'.43" too fast.

EXAMPLE III.

In latitude 48°.56' North, longitude 66° West, suppose the altitude of Aldebaran, when west of the meridian, to be 22°.20'. 15" on the 15th of April 1822; required the apparent time of observation?

The right ascension of Aldebaran being 4.25'.43′′ and declination 16°.8'.36" N. Also, the O's right ascension at noon on the 15th (Naut. Alm.) = 1h.32′.23", and on the 16th= 1h.36'.5".

Answer. The hour by the 's corrected altitude=41.57'. 33", and the culminates at 2.52'.9", hence the correct apparent time=7h,49′.42".

(M) SCHOLIUM.

The three preceding problems (XIII, XIV, and XV.) are all included in one triangle (Zenith NS Plate III. Fig. 1.), but they have been considered separately, on account of their importance, and to prevent confusion.

Any three of the five quantities, viz. the sun's or star's declination, the latitude of the place, the altitude of the object, the azimuth, and the hour of the day being given, the rest may be found.

A sixth quantity might be taken into consideration, were it materially useful, viz. the opposite to the complement of the latitude, being formed by the intersection of the vertical circle Zsc, and the azimuth circle NSD.

PRACTICAL EXAMPLES.

1. Given the sun's declination 19°.39′ North, in North lati

Y