RULE. From the observed altitude, subtract the refraction (Table IV.) If the star be observed at sea, subtract the dip of the horizon (Table V.) See R. 94.

EXAMPLE I.

The observed altitude of Spica Virginis was 20°.39'.40", the error* of the quadrant 19" subtractive, and the height of the eye 18 feet above the level of the sea; required the correct altitude.

Observed altitude of Spica Virginis 20°.39'.40"
Refraction, Table IV.

-2'.32"

Or, 20°.39.40"-(2′.32"+4.3" +19")=20°.32′.46". Answer. 2. Suppose the observed altitude of Regulus to be 45°.13.15", the height of the eye 14 feet above the level of the sea, and the error of the quadrant 5'.6′′ additive, required the true altitude.

Answer. 45°.13′.50′′.

PROBLEM X.

(G) Given the observed altitude of the sun's lower or upper limb, to find the true altitude of its centre.

RULE. To the observed altitude apply the semidiameter (taken from page III. of the month in the Nautical Almanac) by addition or subtraction, according as the lower or upper limb has been observed; from this result subtract the refraction (Table IV.), and then add the parallax in altitude § (Table VI.). If the altitude be taken at sea, the dip of the

Observations taken with a quadrant are liable to errors, arising from the bending and elasticity of the index, and the resistance it meets with in turning round its centre. These errors, though they cannot in all cases be avoided, may be pretty accurately allowed for by a correct observer.

The sun's upper limb is the upper edge of its face, or the uppermost extremity of the vertical diameter; and the lower limb is the lower edge, or the lower extremity of the vertical diameter. § S. 95.

S. 87. et seq.

horizon (Table V.) must be deducted, the last result will be the true altitude of the sun's centre.

EXAMPLE 1.

On the 13th of March 1822, if the altitude of the sun's lower limb, observed at sea, be 18°.40'; required the true central altitude, the height of the eye being 22 feet.

Observed altitude of the O's lower limb-18°.40'
O's semidiameter, Naut. Alm.

= + 16′. 6"

18°.56'. 6"

Refraction for 19° of altitude (Table IV.) = -2.44"

Or, (16'.6"+8")-(2'.44"+4'.28")=9′.2′′ and 18°.40′+9′.2′′ =18°.49.2". Answer.

NOTE. The index error, if any, of the quadrant must be applied to the observed altitude, previous to the other corrections.

2. On the 3d of September 1822, suppose at sea the altitude of the sun's upper limb to be 28°.31′.30", the index error 40′′ additive, height of the eye 12 feet, and the sun's semidiameter 15.54"; what is the true altitude of the sun's centre?

Answer. 28°.11′.19′′.

PROBLEM XI.

(H) Given the observed altitude of the moon's lower or upper limb, to find the true altitude of its centre.

RULE. Find the moon's semidiameter and horizontal parallax for the time and place of observation (C. 271.), and increase the semidiameter by the augmentation answering to the moon's altitude (Table VII.).

*

To the observed altitude apply the augmented semidiameter by addition or subtraction, according as the lower or upper limb has been observed, and, if the observation has been made at sea, subtract the dip of the horizon (Table V.); the result will be the apparent altitude of the moon's centre.

*Here, as in the preceding problem, the index error of the quadrant, if any, must be applied to the observed altitude, previous to the other corrections.

To the cosine of the moon's apparent altitude, add the logarithm of the horizontal parallax in seconds (found above), the sum, rejecting 10 from the index, will be the logarithm of the moon's parallax in altitude in seconds (T. 96.), from which take the refraction of the moon in altitude (Table IV.), and the remainder will be the moon's correction, which added to the apparent altitude will give the true altitude.

EXAMPLE I.

On the 18th of July 1822, in longitude 113°.10′ W. at 20b.45′.30′′, if the observed altitude of the moon's lower limb be 45°.22.3", error of the quadrant 58" subtractive, and the eye 21 feet above the level of the sea; what is the true central altitude?

Time at the given place
Longitude 113°.10 W. in time

201.45'.30" 7h.32'.40" W.

28h.18'.10"

24

Time at Greenwich on the 19th of July 4h.18'.10"

D's semidiameter at noon, 19th=16′.9", Horizontal parallax
D's semidiameter at midnight, 19th=16′.3", Horizontal parallax

D's semidiam. at 4h. 18'.10"

=

16'. 9"

Difference 0.21"

:

12h: 21":4.18.10" 7" Horizontal parallax at noon 59'.15"

= 16. 7" Horizontal paral. at 4". 18′.10"-59′ 8′′ D's augmentation (Table VII.) == 11"

60

Lastly, 45°.33.1" +40′.51′′ = 46°.13'.52", true altitude of the moon's centre.

2. On the 22d of April 1822, in longitude 105° E. of Greenwich, at 13h.30', if the observed altitude of the moon's upper

limb be 23°.48', and the eye 18 feet above the level of the sea; required the moon's true central altitude.

The moon's semidiameter and horizontal parallax at noon (Naut. Alm.), being 16′.39′′ and 61'.8′′; and at midnight, 16.38" and 61'.4".

Answer. The time at Greenwich is 6.30'; D's true semidiameter 16.44", horizontal parallax 61'.6′′; apparent altitude of the D's centre 23°.27′.13′′; the D's correction 53′.52′′, and the true altitude of her centre 24°.21.5".

PROBLEM XII.

(I) Given the sun's meridian altitude to find the latitude of the place of observation.

RULE. Reduce the sun's declination to the meridian of the place of observation* (B. 270.)

Subtract the sun's corrected altitude (G. 277.) from 90°, the remainder is the zenith distance.

If the sun be south of the observer when the altitude is taken, call the zenith distance north; but, if north, call it south.

Then, if the zenith distance and declination have the same name, their sum is the latitude; but, if they have contrary names, their difference is the latitude, and it is always of the same name with the greater of the two quantities.

EXAMPLE I.

On the 17th of October 1822, in longitude 52°W., suppose the meridian altitude of the sun's lower limb to be 28°.40', the observer at sea, the sun to the south of him, and the eye 14 feet above the surface of the water. Required the latitude of the place of observation?

Obser. alt. O's lower limb 28°.40'
Semidiameter

=

+16'. 6"

Time at given place
Longitude 52° w.

Ob. O' =3h.28'

*The same rule, with a little variation, will answer for a planet; and for the moon, correcting for the declination as at C. 271. See Dr. Mackay's Navigation, 1st Edition, page 151 to 155.

2. On the 18th of March 1822, in longitude 106° E, suppose the meridian altitude of the sun's lower limb, when north of the observer, to be 74°.56', and the height of the eye 16 feet above the level of the sea, required the latitude.

The sun's semidiameter (Ñaut. Alm.) being 16′.5′′; the declination at noon on the 17th 1o.26′.59′′ S., and on the 18th 1°.3'.17" S.

Answer. The sun's true altitude 75°.8'.3"; time at Greenwich 16h.56' on the 17th, sun's corrected declination 1°.10.19′′S. latitude 16°.2.16" S.

3. On the 28th of September 1822, suppose the meridian altitude of Arcturus* to be 36°.18' north of the observer, its declination 20°.6'.50" N, and the eye 20 feet above the level of the sea, required the latitude.

Answer. 33°.40.44" S.

CHAP. XI.

THE APPLICATION OF RIGHT-ANGled spherICAL TRIANGLES

TO ASTRONOMICAL PROBLEMS.

(K) The celestial sphere is represented by Plate III. fig, 1. 1. Let the circular space, South, Zenith, North, Nadir, represent the brazen meridian + of a celestial, or terrestrial globe, having its north pole elevated above the horizon.

2. Imagine the globe to be cut in halves by the brass meridian, and the semi-globe to be of transparent glass with the circles of the sphere drawn on it. Now if a sheet of paper be put upon the section, and a light be placed in the point aries (the eye being in libra), the shadows of all the most useful circles of the sphere, will form a plane figure similar to Figure I. Plate III.

3. NAS will present the axis of the globe, N the north pole, s the south pole.

4. EAQ the equator.

5. Ho the horizon.

6. Zenith A the quadrant of altitude screwed on the zenith, and passing through aries. Or, Zenith a Nadir, the prime vertical passing through aries a.

7. WOAO the ecliptic, n its north pole, m its south pole.

See the note on D. 273.

+ It is here presumed that the learner has some knowledge of the globes, and of the circles described thereon. It would be a very good exercise to solve the succeeding problems both by the globes and by calculation.