the vernal equinox to the great circle passing through the pole of the ecliptic and the body.

EXERCISES

(1.) The right ascension of a given star is 25° 35', and its declination is +(north) 63° 26'. Assuming the angle between the celestial equator and the ecliptic to be 23° 27', find the celestial latitude and celestial longitude.

In this figure AB is the celestial equator, AC the ecliptic, P the pole of the equator, P' the pole of the ecliptic. S is the position of the star, and the lines SB and SC are drawn through P and P' perpendicular to AB and AC. AB is the right ascension and BS the declination of the star, while AC is the longitude and SC the latitude of the star.

In the spherical triangle P'PS, it will be seen that P'S is the complement of the celestial latitude, PS the complement of the declination, and PPS is 90° plus the right ascension. It is to be noted that A is the vernal equinox.

(2.) The declination of the sun on December 21st is - (south) 23° 27'. At what time will the sun rise as seen from a place whose latitude is 41° 18′ north?

The arc ZS which is the distance from the zenith to the centre of the sun when the sun's upper rim is on the horizon is 90° 50'. The 50' is made up of the sun's semi-diameter of 16', plus the correction for refraction of 34'.

(3.) The declination of the sun on December 21st is - (south) 23° 27'. At what time would the sun set as seen from a place in latitude 50° 35' north?

In these figures P is the pole of the equator, Z the zenith, EQ the celestial equator. AS is the declination of the sun, ZS=90° 50', PS=90°+declination, PZ-90°-latitude. The problem is to find the angle SPZ. An angle of 15° at the pole corresponds to 1 hour of time.

GEOGRAPHICAL PROBLEMS

102. The meridian of a place is the great circle passing through the place and the poles of the earth.

The latitude of a place is the arc of the meridian of the place extending from the equator to the place.

Latitude is measured north and south of the equator from 0° to 90°. The longitude of a place is the arc of the equator extending from the zero meridian to the meridian of the place. The meridian of the Greenwich Observatory is usually taken as the zero meridian.

Longitude is measured east or west from 0° to 180°.

The longitude of a place is also the angle between the zero meridian and the meridian of the place.

8

In the following problems one minute is taken equal to one geographical mile.

(1.) Required the distance in geographical miles between two places, D and E, on the earth's surface. The longitude of D is 60° 15' E., and the latitude 20° 10' N. The longitude of E is 115° 20' E., and the latitude 37° 20' N.

PB

In this figure AC represents the equator of the earth, P the north pole, and A the intersection of the meridian of Greenwich with the equator. and PC represent meridians drawn through D and E respectively. AB is the longitude and BD the latitude of D; AC the longitude and CE the latitude of E.

Then

(2.) Required the distance from New York, latitude 40° 43' N., longitude 74° o' W., to San Francisco, latitude 37° 48′ N., longitude 122° 28′ W., on the shortest route.

(3.) Required the distance from Sandy Hook, latitude 40° 28' N., longitude 74° 1' W., to Madeira, in latitude 32° 28' N., longitude 16° 55, W., on the shortest route.

(4.) Required the distance from San Francisco, latitude 37° 48' N., longitude 122° 28′ W., to Batavia in Java, latitude 6° 9' S., longitude 106° 53' E., on the shortest route.

(5.) Required the distance from San Francisco, latitude 37° 48' N., longitude 122° 28' W., to Valparaiso, latitude 33° 2' S., longitude 71° 41′ W., on the shortest route.

CHAPTER XI

GRAPHICAL SOLUTION OF A SPHERICAL TRIANGLE

103. The given parts of a spherical triangle may be laid off, and then the required parts may be measured, by making use of a globe fitted to a hemispherical cup.

The sides of the spherical triangle are arcs of great circles, and may be drawn on the globe with a pencil, using the rim of the cup, which is a great circle, as a ruler. The rim of the cup is graduated from 0° to 180° in both directions.

The angle of a spherical triangle may be measured on a great circle drawn on the sphere at a distance of 90° from the vertex of the angle.*

CASE I. Given the sides a, b, and c of a spherical triangle, to determine the angles A, B, and C.

Place the globe in the cup, and draw upon it a line equal to the number of degrees in the side c, using the rim of the cup as a ruler. Mark the extremities of this line A and B. With A and B as centres, and b and a respectively as radii, draw with the dividers two arcs intersecting at C (Fig. 1). Then, placing the globe in the cup so that the points A and C shall rest on the rim, draw the line AC=b, and in the same way draw BC=a.

To measure the angle A place the arc AB in coincidence

* Slated globes, three inches in diameter, made of papier-maché, and held in metal hemispherical cups, are manufactured for the use of students of spherical trigonometry at a small cost.

with the rim of the cup, and make AE equal to 90°. Also make AF in AC produced equal to 90°. Then place the globe in the cup so that E and F shall be in the rim, and note the measure of the arc EF. This is the measure of the

angle A. In the same way the angles B and C can be determined.

CASE II. Given the angles A, B, and C, to find the sides a, b, and c.

Subtract A, B, and C each from 180°, to obtain the sides a', b', and c' of the polar triangle. Construct this polar triangle according to the method employed in Case I. Mark its vertices A', B', and C'. With each of these vertices as a centre, and a radius equal to 90°, describe arcs with the dividers. The points of intersection of these arcs will be the vertices A, B, and C of the given triangle. The sides of this triangle a, b, and c can then be measured on the rim of the cup.