(I) By the first of these examples, geographers have calculated tables showing how many miles answer to a degree of longitude for each degree of latitude; thus in latitude 45°, a degree of longitude is 42-43 miles; in the latitude of London 51°.32', the length of a degree is 37.32 miles, &c.

3. If a ship sail 2757.7 miles directly eastward or westward, so as to alter her longitude 60°, what latitude is she in? Answer. 40° north or south.

4. Suppose a ship to sail from latitude 40° North to latitude 10° North, till her difference of longitude be 60°, and it be required to find her departure, course, and distance sailed.

Here IL the parallel of 40° is the latitude left, sa the parallel of 10' is the latitude arrived at, and MN the parallel of 25° is the middle latitude between them; hence MP is the complement of the middle latitude, and MN is the meridional distance, or the true departure nearly*; Bw is a quadrant or 90°, and WE is the difference of longitude.

(K) Hence the following CONSTRUCTION (Plate III. Fig. 3.):

1. With the chord of 60°, which is equal to the sine of 90°, and centre P, describe the arc wE, on which set off the difference of longitude (60°=3600 miles); draw pw and PE.

2. With the sine of (65°) the complement of the middle latitude, taken from the line of sines on Gunter's scale, describe the arc MN, and draw the straight line MN.

3. Set off the difference of latitude (30°1800 miles) from P to m; draw mn perpendicular to Pm, and equal to MN; lastly, draw Pn, and the construction is finished. The several lines are named as in the figure.

(L) OR THUS.

Make the angle MPN (Plate III. fig. 4.)the complement of the middle latitude, and PN the difference of longitude, then MN will be the departure or meridional distance; for, in the plane triangle MPN, per Plane Trigonometry, radius PN:: sine / MPN: MN, the same proportion as is used in the first example. The triangle men is the same as before.

(M) OR THUS. (Plate III. Fig. 5.)

Draw the meridian AP, in which assume any point P; make the angle at p equal to the complement of the middle latitude, in the opposite quadrant to which the ship sails; set off PN equal to the difference of longitude, and draw MN perpendicular to

* See the Theory of Navigation, Book IV. Chapter II.

AP; lastly, set off the difference of latitude from м to A, and draw AN; then the names of the several lines are as expressed in the figure.

BY CALCULATION. (Plate III. Fig. 3, 4, or 5.)

Rad diff. long:: sine comp. midd. lat. : merid. dist. or departure=3262.7.

Diff. lat rad::departure: tang course=61°.7.
Sine course dep.::rad: distance=37126.2.

CHAP. XII:

THE APPLICATION OF OBLIQUE ANGLED SPHERICAL TRIANGLES TO ASTRONOMICAL PROBLEMS.

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

(N) Given the sun's declination, and the latitude of the place, to find the apparent time of day-break in the morning, and the end of twilight in the evening.

1. In the oblique-angled spherical triangles zenith N.

1. ON the sun's distance from the north pole. 2. Ozenith the sun's distance from the zenith.

3. Zenith N-the complement of the latitude.

4. ON zenith, measured by the arc aÆ the time from noon.

OR, 2. In the oblique-angled spherical triangle

1. Os the sun's distance from the south pole. 2. O Nadir the sun's distance from the nadir. 3. s Nadir the complement of the latitude.

Nadir s.

4. Os Nadir, measured by the arc ag,the time from midnight.

EXAMPLE.

Given the latitude of the place 51°.32′ N., and the sun's declination 10° N., to find the time of day-break in the morning, and the end of twilight in the evening.

In the triangle ◇ zenith N.

ON 80° the sun's distance from the north pole=90°-10°. zenith 108° the sun's distance from the zenith = 18°

+90°.

Zenith N=38°.28' the complement of the latitude.

To find the angle zenith NO, measured by the arc aÆ, the time from noon.

from noon* when the sun is 18° below the horizon. Consequently the day breaks at 2.53'.40′′ in the morning, and twilight ends at 9h.6′.20" in the evening, supposing the sun's declination to undergo no change between the beginning of twilight in the morning, and the ending thereof at night, being about 18 hours.

The same things might have been found from the triangle Os Nadir, for sŎ=90°+10°=100°, Nadir©=180°-108° = 72°, and Nadir s = comp. lat. 38°.28'. Then by the method above find the angle Os Nadir (measured by the arc aq)=43°.25′.12′′=2.53′.40′′ as before, the time from midnight, when the sun is 18° below the horizon.

(O) When the declination of the sun, the latitude and declination being of the same name, is greater than the difference between the complement of latitude and 18°, the parallel of declination ( sss) will not cut the parallel of 18° (TOW) below the horizon: consequently there will be no real night at these times, but constant day or twilight, as is the case at London from the 22d of May to the 21st of July.

(P) Since the sun sets more obliquely at some times of the year than at others, it necessarily follows that he will be longer in descending 18° below the horizon at one season than at another.

When the sun is on the same side of the equinoctial as the visible pole, the duration of twilight will constantly increase as he approaches that pole, till he enters the tropic, at which time

* The NZ represents the time from apparent noon, that is, from the time the comes to the meridian, and not the time from 12 o'clock, as shown by a well regulated clock or time-piece. If the equation of time, given in the second page of each month in the Nautical Almanac, be applied to this apparent time by addition or subtraction, as there directed, you will obtain the true time, or that shown by a clock.

the duration of twilight will be the longest. It will then decrease till some time after the sun passes the equinox, but will increase again before he arrives at the other tropic; therefore, there must be a point between the tropics, where the duration of twilight is the shortest, which point may be found by the following proportion; viz.

Rad tang 9°:: sine of the latitude sine of the sun's declination.*

The declination must always be of a contrary name with the latitude.

PRACTICAL EXAMPLES.

1. Given the sun's declination 10° south, and the latitude of the place 51°.32' N. to find the time of day-break in the morning, and the end of twilight in the evening? Answer. The Os Nadir=73°.35′.34" 4h.54'.22" the time from midnight. Hence, day breaks at 4h.54.22" in the morning, and twilight ends at 7.5'.38" in the evening; admitting the sun's declination to be constant for one day.

2. Given the sun's declination 23°.28' S. and the latitude of the place 51°.32′ N. to find the time of day-break in the morning, and the end of twilight in the evening?

Answer. The Os Nadir=90°.16′ 6.1'.4" the time from midnight when the is 18° below the horizon. Consequently day breaks at 5h.58′.56′′ and twilight ends at 6.1'.4", on the shortest day at London.

3. In the latitude of London, 51°.32′ N., what time does the day break when the sun has 11°.30' N. declination? Answer. 2.41', rejecting seconds.

4. Required the beginning and ending of twilight at London, lat. 51°.32' N., when the sun's declination is 15°.12' N.? Answer. 2.4.12" and 9h.55'.48".

5. Required the time the twilight begins and ends at London, latitude 51°.32', when the sun has 11.30′ South declination?

Answer. 5.2' and 6.58', rejecting seconds.

6. At London, latitude 51°.32′ N., required the sun's declination, day of the month, and duration of twilight when it is the shortest?

Answer. Sun's declination 70.7.25" south, answering to

* Vide L'Hospit. Infinim. Petit. sect. iii. prop. 13. Dr. Gregory's Astron. page 328. Heath's Royal Astron. page 225. Emerson's Miscellanies, page 492. Vince's Astronomy, vol. i. page 18, &c. Also, Leybourn's edition of the Ladies' Diary, vol. iv. from page 314 to 339, wherein the subject is amply discussed.

March 2d and October 11th; between which days, the twilight increases, and from the latter to the former it decreases. The duration of twilight is 1.56'.32"; this is found by taking the difference between the time of sun-rise and day-break.

PROBLEM VIII. (Plate III. Fig. 6.)

(Q) Given the day of the month, the latitude of the place, the horizontal refraction, and the sun's horizontal parallax, to find the apparent time of his centre appearing in the Eastern and Western part of the horizon.

Example. Given the sun's declination at noon on the 21st of June 1822=23°.28'; the latitude of the place=51°.32' N.; the horizontal refraction=33′; the sun's horizontal parallax =9"; to find the apparent time of his rising and setting.

This example being given for the longest day, the sun's declination may be considered the same at his rising as at noon; for the declination does not vary above 5" in 24 hours at this time, though near the equinoxes it varies about 1' in an hour.

Let s be the true point of the sun's rising, and 6 that of his apparent* rising, then b=33′-9′′-32.51" the distance of the sun's centre below the horizon; hence the true zenith distance zs will be increased, by refraction, to zo, but the declination will remain the same, for NS NO.

In the oblique-angled triangle zNO.

90°.32′.51′′ app. dist. of O's centre from the zenith. NO 66°32′. 0" app. dist. of's centre from the north pole. ZN = 38°.28′. 0′′" the complement of the latitude.

2) 195.32.51

ZN=38°.28' cosec

•20617

NO 66°.32' cosec

⚫03749

Sum 97.16.25|| Half sum=97°.46′.25" sine

9.99599

ZZNO 124.16.30-8.17'.6" the

apparent time from noon when the sun's centre appears in the western part of the horizon. Hence the apparent time of

*The apparent time of rising and setting of the heavenly bodies always differs from the true time, for they are elevated by refraction (S. 87, &c.), and depressed by parallax (S. 95, &c.).

All the heavenly bodies, when in the horizon, appear 33′ above their true places,

desaffearing