The following solution of the astronomical triangle is unique in that no artificial perpendicular is employed to divide the triangle into two similar right spherical triangles. Since Lord Kelvin first compiled his set of tables for use in solving the elements of two similar right spherical triangles many solutions of the astronomical triangle have appeared, each of which makes use of a perpendicular dropped from one apex to the opposite side. Lieutenant Commander Knight perceived that the vertical circle through a heavenly body serves as a natural perpendicular which may be employed in solving the astronomical triangle through the medium of two similar right spherical triangles.

Figure 1 represents the celestial sphere with the elevated pole at P, the zenith at Z, a celestial body at M and the plane of the local horizon defined by NESW Great circles joining P,Z, and M form the astronomical triangle. If the hour circle from P and the vertical circle from Z be continued beyond M to the horizon, intersecting the latter at A and B, respectively, two similar right spherical triangles PAN and UAB are formed which have the angle at A in common. It will be observed that MB defines the altitude of the body and NB its azimuth.

In the larger triangle the angle at P is the supplement of the local hour angle and the side PN is equal to the latitude of an observer at O. Applying Napier’s rules, Knight obtained the remaining elements as follows:

cot PA = cot Z cos (180° — t) (a)

cos PAN = cos L sin (180° — t) (b)

tan NA = sin L tan (180° — t) (c)

If the intersection of the equinoctial with the hour circle be designated D the arc DA equals PA minus PD (= 90°). A combination of DA with the declination (DM) of the body yields side MA of the smaller triangle MAB in which the angle MAB has already been determined. Applying Napier’s rules again h and BA were computed as follows:

sin h = sin MA sin PAN (d)

cos BA=cos MA/cos h (e)

and; z = NA — BA (f)

Knight contemplated a set of tables composed of two parts, the first a table of precomputed values of PA, log sin PAN and NA for integral values of latitude and local hour angle and the second a table of logarithmic sines and cosines ranging from 0° to 180°. Appropriate precepts were devised to handle the various cases which arise.

Inasmuch as the number of steps employed in the solution of the astronomical triangle by Knight’s method are practically the same as those involved in the method of Dreisonstok, no advantage was seen in publishing another book. The solution is interesting, however, and the following modification is submitted as illustrating the possibility of reducing the second table to values of from 0° to 90° and eliminating the use of decimals by substituting log cosecants and secants for sines and cosines.

Throughout this discussion the quantity t is considered as that angle measured from the upper branch of the observer’s meridian, to the eastward or westward, to the body, through 180°. If the first table, Table I, be compiled for values of t from 0° to 90° certain precepts will be required to handle those cases where t exceeds 90°. These precepts will be discussed later.

Equations (a), (b), and (c) above maybe stated in terms of t as follows:

(—) cot PA = cot L-cos t (a’)

cos PA N = cos L sin t (b’)

(—) tan NA = sin L tan t (c’)

The value of arc DA lies between 0° and 90° and since the tangent of DA (PA —90°) is the same as (—) the cotangent of the side PA, equation (a') may again be rewritten:

tan DA = cot L cos t. (a")

It is this value, DA, which is combined with the declination to obtain MA for use in equations (d) and (e). Since the value of t employed in equations (a"), (b') and (c') is less than 90° the value of the side PA lies in the second quadrant, that of the angle PAN in the first quadrant and of the side NA in the second quadrant.

A table of three columns may now be computed for integral values of L and t in which column 1 contains the values of arc DA, column 2 the values of log cosecant of angle PAN raised to the 105 power, and column 3 the values of side NA. Let this table be designated Table I.

By employing log cosecants and log secants raised to the 105 power for sines and cosines, equations (d) and (e) above may be written thus:

log esc h = log esc MA + log esc PAN (d')

log sec BA = log sec MA — log sec h. (e')

A table, Table II, composed of log cosecants and secants raised to the 105 power to eliminate decimals and ranging in value from 0° to 90° is all that is required to complete the solution.

The precepts required to handle the 0° various combinations of latitude, declination, and hour angle may be divided into be oases and explained by reference to Fig. 2.

#### CASE I

t_{1} less than 90°; latitude and declination (D_{1}M_{1}) same name; A_{1}M_{1} less than 90°.

Body at M_{1}. D_{1}A_{1} is positive (equation a"). The declination (D_{1} M_{1}is positive since it is of the same name as the latitude, therefore A_{1} M_{1} is equal to D_{1} M_{1}+D_{1} A_{1 } and, as shown, is less than 90°. Since angle PA_{1}N and M_{1}A_{1} are in the first quadrant h_{1} and B_{1} A_{1} also lie in the first quadrant, Equations d' and e').

Then NB_{1} (or Z_{1})= NA_{1}—B_{1}A_{1}

#### CASE II

t_{1} less than 90°; latitude and declination (D_{1}M_{2}) same name; A_{1}M_{2} greater than 90°.

Body at M_{2}. D_{1}A_{1} is positive and D_{1}M_{2} is positive. Therefore A_{1}M_{2} is equal to D_{1}M_{2 }+D_{1}A_{1} and, as shown, is greater than 90°. Table II is entered with the supplement of A_{1}M_{2} to solve for h_{2} and B_{2}A_{1}. h_{2} by equation (d') lies in the first quadrant while B_{2}A_{1} by equation (e') lies in the second quadrant. The equation for azimuth becomes:

NB_{2} (= Z_{2}) = NA_{1} - (180° - B_{2} A_{1})

#### CASE III

t_{1} less than 90°; latitude and declination opposite names.

Body at M_{3}. D_{1}A_{1} is positive while is negative (declination of opposite name of latitude). Therefore A_{1}M_{3} is equal to D_{1}A_{1}—D_{1}M_{3}. In this case the declination must be less than M_{3}A_{1} for if it were heater the body would be below the horizon. Since the value of D_{1}A_{1} lies between 0° and 90° the value of M_{3}A_{1} will be less than 90°. The angle PA_{1}N and the arc D_{1}A_{1} lie in the first quadrant; therefore h_{3} and B_{3}A_{1} are in the first quadrant and:

NB_{3} (= Z_{3}) = NA_{1} - B_{3}A_{1}

#### CASE IV

t_{2} greater than 90°. Table I is entered with the supplement of t. In this case the declination D_{2}M_{4} must be of the same name as the latitude for the body to be visible above the horizon.

Body at M_{4}. D_{2}A_{2} is negative, for, since t_{2} is greater than 90° equation (a") becomes:

(—) tan DA =cot L cos (180° — t_{2})

Therefore A_{2}D_{2} = M_{4}D_{2}—M_{4}A_{2} and lies in the first quadrant. Angle PA_{2}N and h_{3} lie in the first quadrant, hence B_{4}A_{2} lies between 0° and 90° and equation (c) becomes:

tan (180° — NA_{2}) = sin L-tan t

Therefore,

NB_{4}(= Z_{4}) = (180° - NA_{2}) – B_{4}A_{2}.

For simplicity the arc DA may be termed K, the side AM termed k, the side NA termed C and the side BA termed c. The precepts for a line of position may be grouped thus:

t less than 90°

Lat. and Dec. same name

k = K + d

if k<90°, Z = C—c

if k >90°, use 180° — k; Z = C —(180°—c)

Lat. and Dec. opposite name

k = K — d

Z = C - c

t greater than 90°

Enter Table I with

180° - t

k = d - K

Z = (180° - C) - c

The formulas and tables may readily be adapted to the solution of problems in star identification by substituting the known quantities latitude and azimuth for latitude and hour angle and solving for the unknown quantities declination and hour angle instead of altitude and azimuth. By combining the resultant hour angle with the L.S.T. to determine the right ascension the unknown body may be identified in the *Nautical Almanac.*

Refer to Fig. 1. Triangles ZB'Q and MB'D are two similar right spherical triangles having an included angle at B'. The known parts of the triangle ZB'Q are ZQ, the latitude, and angle QZB', the supplement of the azimuth, Z, of the body M. The azimuth, Z, is that angle measured from the vertical circle passing through the elevated pole of the observer, to the eastward or westward, to the vertical circle of the body, through 180°.

Table I is entered with the known values of L and Z and the resultant values of BB', log esc ZB'Q and QB' are obtained. BB', designated K, is combined with the altitude h0 to obtain side B'M of the triangle MB'D, termed k. Equation (d') becomes:

log esc d = log esc k + log esc ZB'Q.

If side B'D be designated c, equation (c') becomes:

log sec c = log sec k — log sec d

If arc B'Q be termed C, equation (/) may be written:

t(= DQ) = C - c

The various cases under star identification and their corresponding precepts may be grouped as follows:

z equal to or less than 90°

Dec. takes name of Lat.

These formulas for both line of position and star identification will be found to be very rugged under all conditions and the solutions are more readily visualized than in many methods employing a perpendicular which does not exist naturally. The precepts permit of no ambiguity. In solving a meridian altitude for instance, K equals 90° minus the D.R. latitude and he equals k (= d ~K) or the supplement of k.

If k is less than 90°, z = 180°.

If k is greater than 90°, z = 0°.

With the present trend toward a solution from a D.R. position rather than an assumed position the method outlined herein, requiring for its solution the use of two tables, perhaps has few advantages- Its greatest advantage would seem to lie in the fact that by means of slide films and motion pictures the beginner striving to master the art of navigation could visualize more readily the relations of the various elements of the astronomical triangle during the motion of a heavenly body throughout its orbit.

The most important *of strategic lines are those which concern the communications. Communications dominate war.*—Mahan.