4. In the right-angled spherical triangle A B C, right-angled at A, given A B 97° 20′ and B C 94° 13′, to find the other parts.

Ans. . B 55° 2′ 52′′; / C 96° 0′ 20′′ and A C 54° 49′ 36′′.

5. In the right-angled spherical triangle A B C, right-angled at A, given A B 114° 22′ and ▲ C 108° 19′, to find the other parts.

Ans. B 49° 37′ or 130° 23′; A C 46° 57′ 45′′ or 123° 2′ 15′′ and B C 106° 21' 15" or 73° 38′ 45".

6. In the right-angled spherical triangle A B C, right-angled at A, given A B 95° 52′, B 76° 13′, to find the other parts.

Ans. C 95° 41′ 50′′; A C 76° 8′ 49′′ and B C 91° 24′ 9′′.

QUADRANTAL SPHERICAL TRIANGLES

A quadrantal spherical triangle has one of its sides 90°, hence its name. Napier's Rules for circular parts are used in the solution of these triangles and in considering the "Circular Parts" the quadrantal side is omitted in the same manner as the right angle in right-angled spherical trigonometry, and the "Circular Parts" are the two angles adjacent to the quadrantal side, which does not separate them, the complements of the other angle and the two remaining sides. The amplitude is the commonest example in nautical astronomy of the quadrantal triangle.

In the quadrantal triangle P Z X (Fig. 4), the quadrantal side being Z X, given P X 70°, and P Z 40°, to find Z.

P X is the middle part and Z and P Z the opposite parts.

I.

Exercises in Quadrantal Triangles.

S

Fig. 4.

In the quadrantal triangle A B C, given the quadrantal side (a) and ▲ B = 85° 20′, and side c = 81° 36', to find the other parts.

Ans. LA 90° 41′, C 81° 34', and side b 85° 23'.

Ans. a 22° 31′, c 86° 6′, ▲ A 22° 13′.

C = 100° 14′, to find the

3. Given a = 90°, b = 101° 15', C = 79° 36', to find the other parts.

Ans.

LA 87° 57′, ▲ B 101° 26′, C 79° 47′.

F 2

The diagrams used to demonstrate the problems in Nautical Astronomy are generally drawn on the Stereographic Projection of the sphere on the plane of the rational horizon, as it is generally agreed that this projection shows the positions of the celestial objects, as viewed by an observer at sea, in the most natural way.

The first circle drawn is the rational horizon and is called the primitive, and this circle contains the projection.

Right circles are great circles seen edgewise and appear as straight lines, such as the meridian of the observer and the prime vertical. These right circles are always a diameter of the primitive.

Oblique circles are great circles which lie between the right circles and the primitive, such as hour circles, the equinoctial and parallels of declination.

The six o'clock hour circle is that great circle which passes through the pole and cuts the rational horizon in the true east and west points. It is, therefore, obvious that when the latitude is o° the rational horizon is also the six o'clock hour circle. The radius of the six o'clock hour circle increases as the secant of the latitude.

The poles of the meridian of the observer are at the east and west points; and the poles of the prime vertical are at the north and south points.

The poles of all other circles vary in position according to the position of the observer.

Celestial bodies are said to be circumpolar when they neither rise nor set, but circle round the pole of the observer always above his horizon. This is always the case when the latitude (the elevation of the pole) is greater than the polar distance of the object.

We shall now construct a figure for latitude 40°; declination 10° N.; easterly hour-angle 4 hours, using any scale of chords, tangents, semitangents and secants.

Arrange the data thus :

FIG. I. With the chord of 60° in the compasses and from Z as centre, describe the circle N W S E, which represents the rational horizon. Draw the prime vertical W Z E, bisect the prime vertical, and draw NZ S, the meridian of the observer, and extend it both ways.

To locate the Pole (P), take the semi-tangent of co-latitude (50°) in the

dividers and place one foot on Z and the other foot will find P, the pole, on the meridian N Z S.

To draw the Equinoctial W Q E

Take from the scale the secant of the co-latitude (50°) and with one foot of the compasses on W or E, the other foot will find C, the centre of the equinoctial, on the meridian produced, and C W or C E is the radius; draw the circle W Q E, and it will be the equinoctial.

To draw the Parallel of Declination

Take from the scale the semi-tangent of 130°, and with one foot of the compasses on Z the other will find the point d' in the direction of N; also take the semi-tangent of 30° and with one foot on Z the other will find d" in the direction of S on the meridian N Z S. If the declination were greater than the latitude it would be laid off in the direction of N; bisect d' d" in ƒ, and ƒ is the centre of the parallel of declination and ƒ d' its radius; draw the circle d d d and it is the required parallel.

To draw the four-hours Easterly Hour Circle

Take from the scale the secant of the latitude and with one foot of the compasses on P the other foot will find x on the meridian N ZS; through x draw a line parallel to W Z E, which mark M N'; on this line will be found the centres of all the hour circles for that latitude. The point x is the centre of the six o'clock hour circle.

Lay off from P, with a protractor, the complement of the hour angle (30°) and draw P F, then F is the centre of the four o'clock easterly hour circle

and P F is its radius; draw the circle P X, and it will be the required hour circle.

Now draw Z X, the zenith distance, and the figure is complete.

If the 1, 2, 3, 4, 5 and 6 o'clock easterly hour circles be continued through the pole till they meet the primitive they will become the II, 10, 9, 8, 7 and 6 o'clock westerly hour circles respectively.

It, therefore, follows that the centres of the easterly hour circles from I to 6 and the westerly hour circles from 6 p.m. to midnight lie to the west of the meridian on the line M x N'; and the centres of the westerly hour circles from 1 to 6 and the easterly hour circles from midnight to 6 a.m. lie to the east of the meridian on the line M x N'.

It is practically impossible to draw the figures for latitude by ex-meridian altitude and latitude by the pole star to scale, owing to the arcs being so small, and it is usual to exaggerate the figures in these cases.

Construct a figure for latitude o°; declination 40° N.; hour-angle 4 hours

FIG. 2. With the chord of 60° describe the circle N W S E, the rational horizon. Draw W Z E, the prime vertical and equinoctial, and N Z S, the observer's meridian, and produce it to d'.

N.B. The rational horizon is also the six o'clock hour circle in this

case.

To draw the four o'clock Hour Circle

From Z towards R lay off the semi-tangent of 60° 4 hours and with the secant of the complement of the hour-angle 30° = 2 hours, in the compasses, place one foot on N and the other foot will find the centre, C, of the four o'clock hour circle on W Z E, and P c is its radius. Now draw N R, the required hour circle.

To draw the Parallel of Declination

From Z lay off the semi-tangent of 140° towards N on the meridian produced, which mark d', also lay off the semi-tangent of 40° and mark d"; bisect d' d" in ƒ and with the radius ƒ d' draw the parallel of declination d d" d.

Draw Z X, the zenith distance, and the figure is complete.

In practice the arc of the hour circle from X to R would be omitted.

The centre of the hour circle, Fig. 2, can also be found in the same way

as in Fig. 1. Lay off from the meridian at N, by means of a protractor, the angle Z N C, and C is the required centre as before.

FIG. 3 is the figure constructed from the above data in the same manner as Fig. 1. But it is necessary to remark that the arc X' X in the figure is an arc of a small circle, whereas it should be an arc of a great circle, as shown in Fig. 3A, which, however, is not the same triangle as in Fig. 3, and is simply put in to show the method of finding the centre and radius of the great circle X' X.

Proceed as follows: Join the positions X' X, Fig. 3A, by a straight line; bisect it in the point C, and draw C C' at right angles to X' X. With the chord of 60° in the dividers and one leg on X' or X, the other leg will find the centre of the required circle on the line C' C; draw the arc X'N X and it will be an arc of a great circle.