and centre P, describe the arc we, on which set off the differ ence 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.

(A) 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 : FN :: sine MPN: MN, the same proportion as is used in the first example. The triangle men is the same as before.

(B) 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 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 depar, ture 3262.7.

Diff. lat: rád :: departure ; tang course=61°.7.
Sine course: dep. :: rad; distance=37126 2.

CHAP. VIII.

THE APPLICATION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES TO ASTRONOMICAL PROBLEMS.

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

(C) 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 O ZN. 1. ON the sun's distance from the north pole.

2. Oz the sun's distance from the zenith.

3. ZN the complement of the latitude.

4. ONZ, measured by the arc a=the time from noon. OR, II. In the oblique-angled spherical triangle ✪ Nadir s. 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.

4. Os Nadir, measured by the arc a q, 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.

ON=80° the sun's distance from the north pole=90° — 10°. Qz=108° the sun's distance from the zenith 18° +90°. ZN=38°.28′ the complement of the latitude.

To find the angle Zenith NO, measured by the arc a,the time from noon.

ON= 80° Co-sec ZN=38°.28' 206177 reject

Oz=180°

Co-sec ON=80°. 0′ - 00665 indices.. 113°.14′ 9.96327

ZN 38°.28 Sine

=

Öz=108. 이 Cosine 68°.17.29"9-56807

Half sum

Remainder

5. 14

2

==

<ONZ=136°.34′.58′′ 91.6'.20", time from noon when the sun is 18° below the horizon. Consequently the day breaks at 2.53'.40′ in the morning, and twi light ends at 9.6'.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 ONZ represents the time from apparent noon, that is from the time thecomes to the meridian, and not the time from 12 o'clock, as shewn 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 shewn by a clock.

The

The same things might have been found from the triangle Os Nadir, for s=90° +10° 100°, Nadir =180° — 108° = 72°, and Nadir s=comp. lat.=38°28'. Then by the method above find the angle OSN (measured by the arcaq)=43°.25′.12′′= 2-53′40′′ as before, the time from midnight, when the sun is 18° below the horizon.

(D) 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.

(E) 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 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 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 decli nationt.

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

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′′=4.54'.22" the time from midnight. Hence, day breaks at 4.54'.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′=6h 1′.4′′ the time from midnight when the is 18° below the horizon. Consequently

+ 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 Vince's Astronomy, vol. i. page 18, &c.

492.

day

day breaks at 5.58′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 9.55′.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=7°.7′.25" south, answering to 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. 5.)

(F) 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 upper limb appearing in the Eastern and Western part of the horizon.

Example. Given the sun's declination at noon on the 21st of June 1813 23°.28; the latitude of the place 51°.39 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 l' in an hour.

Let s be the true point of the sun's rising, and b that of his apparent * rising, then b=33′-9′′=32′.51′′ the distance 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. 81, &c.), and depressed by parallax (S. 89, &c.).

All the heavenly bodies, when in the horizon, appear 38′ above their true places, by the effect of refraction (see Table IV.); but they are depressed by parallax. Hence, by these causes combined, a body becomes visible when its distance from the zenith is equal to 90° +33'-horizontal parallax, and as the sun's horizontal parallax is about 9" (see Table VI.) his upper limb will appear in the horizon when it is 32.51" below it, or 90°.52′51′′ from the zenith.

A star has no sensible parallax, and therefore will appear in the horizon when it is 33′ below it, or 90° 33′ from the zenith.

the sun's upper limb 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.

z=90°.32′.51′′ app. dist. of O's upper limb from the zenith. N©=66°.32′.0′′ app. dist. of O from the north pole. Z N 38°.28.0" the complement of the latitude.

ZO 90.32.51

Remainder =7°13'.34 sine - 9*09961

ZZNO 124.16.30-8.17'6" the apparent time from noon when the sun's upper limb appears in the western part of the horizon. Hence the apparent * time of the sun's rising is 3.42'.54", and the apparent time of setting is 8.17.6". The true time of the sun's rising and setting, on this day, has been determined (Prob. II. I. 250.) to be 3h.47′.32" and 8".12′.28"; hence the apparent day is 9.16′′ longer than the astronomical day.

(G) As the refraction causes an error in the rising and setting of the celestial objects, so it will cause an error in the amplitudes, as may be seen by comparing the triangle ABS with the triangle ADb. If the sun's amplitude be taken when his centre is in the visible horizon, an allowance depending on the horizontal refraction, parallax, height of the eye, and latitude of the place should be applied to the observed amplitude, in order to obtain the true amplitude. This inconvenience may be avoided, by observing the amplitude when the altitude of the sun's lower limb is equal to 16'+the dip of the horizon.

PRACTICAL EXAMPLES.

1. Required the apparent time of the sun's rising and setting at Glasgow, latitude 55°.32′ N., longitude 4o. 15′ W., on the

and

* The equation of time on this day at noon (see page II. of the Nautical Almanac) is 1.12" at Greenwich; this applied as directed in the 2d page, page 149 of the Nautical Almanac, will give the true time of the sun's rising and setting, as shewn by a well regulated clock or time-piece.