PROFESSOR W. H. ROEVER has described in detail* the application to navigational problems of several graphical methods of solving spherical triangles by the methods of descriptive geometry. He may be quoted as of the opinion that the presentation of the subject of navigation to students (especially to those students whose mathematical preparation leaves something to be desired) gains considerably in concreteness by explaining some of these graphical methods side by side with the solutions by tables or by logarithmic computation which are actually employed in practice.

Table III ........................................ 18

G.S.T.................................... 17 34 02

Long, (west) ....................... 6 01 20

L.S.T..................................... 11 33

Hour angle by the construction

(27°) ................................... -1 48

R.A. of the star ................... 13 21

From the construction, p= 100°, i.e., declination=26°. The unknown “star” is therefore Venus.

As is well known, the principal disadvantage of graphical solutions lies in their comparative inaccuracy. In the case, however, of identifying a star whose altitude and azimuth have been observed, only an approximate solution of the astronomical triangle is necessary. The data are: (i) true altitude, (2) true azimuth reckoned from the elevated pole and expressed as an angle between 0° and 180°, (3) the dead-reckoning latitude. The parts of the astronomical triangle to be found are the hour angle and the declination. The former, keeping in mind its algebraic sign, is subtracted from the local sidereal time to obtain the right ascension of the star.

If the right ascension of the star is determined to within four minutes of time and its declination to within a degree of arc, it may be easily identified. These limits of uncertainty or error will not ordinarily be exceeded by the method outlined below, when the construction is carried out on an ordinary letter-size sheet of paper.

The method, which of course, differs from the usual methods only in the manner of solving the astronomical triangle, is best elucidated by examples.

*American Mathematical Monthly, Vol. XXV, November, 1918, pp. 415-428.

Example 1

P. M. Wed. 18 May, 1927, Lat. 38° 40' N., Long. 90° 20' W., observed (with engineer’s transit) very bright object for Sumner Line. Corrected altitude, 28° 47'-4 True azimuth 281° 15'.

G.C.T. of Observation (19 May).. i^{h} 42^{m} 19" Sid. Time of O^{h} G.C.T. (19 May) .15 42 40

Table III ........................................... 17

G.S.T...................................... 17 25 16

Long, (west) ........................... 6 01 20

L.S.T.................................... ...11 24

Hour angle by the construction

(72°) ...................... 4 48

R.A. of the object..................... 6 36

From the construction p=64°; i.e., declination = 26°. The unknown “star” is therefore Venus.

Example 2

P. M. Wed. 18 May, 1927, Lat. 38° 40' N., Long. 90° 20' W., observed (with engineer’s transit) bright star for Sumner Line. Corrected altitude 34^{0} 26'. 1 True azimuth 147^{0} 20'.

G.C.T. of Observation (19 May).. l^{h} 5l^{m} 04* Sid. Time of O G.C.T. (19 May). .15 42 40

nation = —io°. The unknown star is therefore identified as Spica.

The two cases of the construction illustrated in the two examples given are the only ones that will arise in the star-identification problem. The great merit of the method is that it obviates the awkward interpolations incidental to the use of the star identification tables. For proof of the construction see Loria: Vorlcsungcn iiber Dar- stellcndc Geomtrie, Vol. II, p. 5.

Construction for Example 1

- Draw a circle of any convenient size with center at 0.
- Construct angle AOB = Z, the true Zenith Distance, in this example 61° 12'.6
- Construct the adjacent angle BOC — colatitude, in this example 51
^{0}20'. - Drop a perpendicular AA' from A to BO, D being the point of intersection of this perpendicular with BO.
- With D as center and AA as diameter, draw a semicircle exterior to the circle constructed in step a.
- Construct the angle A'DE equal to the observed azimuth, in this example 78° 45', E being a point on the semicircle constructed in step e.
- From E drop a perpendicular to diameter AA', F being the foot of this perpendicular.
- From F drop a perpendicular to CO, meeting CO in G, and the circle constructed in step a in H and H'.
- With G as center and HH' as diameter, draw a semicircle exterior to the circle constructed in step a.
- From F erect a perpendicular to HH', meeting the semicircle constructed in step i at the point J.
- Measure the angles H'GJ; it is the hour angle, t.
- Measure the angle GOH; it is the polar distance, p.

Construction for Example 2

- Draw a circle of any convenient size with center at 0.
- Construct angle AOB^Z, the true Zenith distance, in this example 55° 33'-9-
- Construct the adjacent angle BOC -= colatitude, in this example 51° 20'.
- Drop a perpendicular AA' from A to BO, D being the point of intersection of this perpendicular with BO.
- With D as center and AA' as diameter, draw a semicircle exterior to the circle constructed in step 0.
- Construct the angle A'DE equal to the observed azimuth, in this example 147° 20', E being a point on the semicircle constructed in step e.
- From E drop a perpendicular to diameter AA', F being the foot of this perpendicular.
- From F drop a perpendicular to CO produced, meeting CO produced in G, and the circle constructed in step a in H and H'.
- With G as center and HH' as diameter, draw a semicircle exterior to the circle constructed in step o.
- From F erect a perpendicular to IIH', meeting the semicircle constructed in step i at the point J.
- Measure the angle H'GJ; it is the hour angle, t.
- Measure the supplement of the angle GOH; it is the polar-distance, p.