The Altitude and Azimuth Table, Table VI of H. O. 200, is a spherical traverse table, or table of right spheric triangles, and since right spheric triangles can be solved by Napier’s rules, which are not subject to the same limitations as those for plane right triangles, this table may be used even more generally than the familiar plane traverse table.

The trigonometrical problem involved in determining the line of position or Sumner line from assumed hour angle and latitude and an observed altitude, is that of the solution of the astronomical triangle PZM to find the angle at Z and the side ZM, given the angle at P and the sides PM and PZ. The problem of the great circle course and distance is the same regarding M as the position left and Z the destination, or vice versa. And the problem of determining the hour angle and thence the right ascension of the star and declination is, given the angle at Z and the sides ZM and PZ, to find the side PM and the angle at P.

Line of Position

In the method of Lord Kelvin and Commander de Aquino the triangle PZM is divided into right triangles by dropping the perpendicular from M on the side PZ. It will be found more convenient to consider the perpendicular as dropped from the angle Z on the side PM. Then regarding the d of the table as the latitude, Z of the table as the position angle, and a and /? as the

two parts of the azimuth, the rule for the use of the table becomes, using the notation of Fig. i:

Where

f = Hour angle. d = Declination.

L — Latitude. h—Altitude.

Z = Azimuth = a + ?.

M = Position angle of observed body. a = Length of perpendicular. b = Declination of foot of perpendicular.

C= Distance of foot of perpendicular from observed body.

Rule.—C=b — d, ? has the sign of C and Z=a + ? and is always measured from the elevated pole regardless of sign, and to the east or west as the hour angle is east or west. When the hour angle is greater than 90° (six hours) use 180^{0} — t, and in that case use also 180 ° —d.

Having obtained the hour angle from the observed time and the dead reckoning longitude enter the table and take out as plotting or assumed latitude and hour angle the nearest quantities to these, the latitude from the L \ d column and hour angle from t column, opposite these take out b from left-hand column and a from right- hand column on same line. Obtain the plotting longitude from plotting hour angle and Greenwich hour angle and plot on chart. Compute C by subtracting the declination from b and run down the column C and find the degrees of C, on same line with this take

out ?, h_{1} and M. Compute Z from Z=a + /3 and plot the line of bearing on chart and correct h_{i} for the minutes of C by

hc = h1 AC cos M.

This correction may be taken from any table of plane right triangles, as Table 2, Bowditch, entering Table 2 with M as course and AC as distance and taking the correction from the latitude column, or may be obtained by dividing the minutes of C, AC, by the factor found between the values of h and M in Table VI.

This correction is always subtractive.

The value of Z obtained from Z = a+f3 is exact when the quantities used are found in the middle column headed a, and is sufficiently close for practical purposes in all cases, and the value of M is sufficiently exact for correcting the altitude.

Lay off the difference between the observed altitude properly corrected and the computed altitude just obtained on the line of bearing, toward the observed body if the observed altitude is the greater and away if the less and draw line of position at right- angles to line of bearing.

It will be noticed that three or four pairs of angles may be found in the tables, all near the dead reckoning values, any of these may be used, and these will give the same line of position, but determined from different plotting positions and different altitude differences. Where the dead reckoning position is found much in error where intersecting lines of position are used, the necessity of correcting the position by rectifying the lines of position may be avoided by choosing other angles giving plotting position nearer the intersection first found, thus getting greater accuracy as well as saving computation and the use of additional tables. Something of a check on the dead reckoning position may be obtained by noting the value of a, the perpendicular, which must be less than go° —h.

Example 1.—January n, 1918, a. m., sextant altitude of sun’s lower limb, 12^{0} 10' 40" bearing S. and E.; W. T., 8^{h} 53™ 06^{s}; C—W, 7^{h} 55^{m} 50^{s}; C. C.,-i^{m} 55^{s}; height of eye, 18 feet; I. C., + 1' 30"; approximate position, latitude 38° N., longitude 122^{0 }16' W.

With dead reckoning latitude = 38° and t~ 50° 30' look in table and on page 253 take out 38° 05' and 52^{0} 16', the latitude and hour angle of the plotting position, and opposite them note <1 = 51.8° and b = 52°, which, combined with declination 21° 51.5', gives C = 73° 51.5'. Opposite C 73° 51.5' by interpolation take out h_{c}= 12° 34.4' and 13 = 79-9°, then the difference between the true altitude and h_{c }is the altitude difference 14.3' to be laid off from plotting position away from sun on line of bearing, Z= 131.7° E.

It will be seen that all this requires no computation, but will be done by the navigator mentally at one opening of the tables when he is ready to plot his position on the chart.

Example 2.—May 23, 1918, during p. m. twilight; sextant altitude of star Leonis (Regulus) is 66° 18.2' bearing S. and W.; W. T. obs., 6^{h} 45^{m} i8^{a}; C—W, 4" i8^{m} 02^{s}; C. C., -3” 45-5^{8}; I- C., 0.0; height of eye, 39 feet; dead reckoning position, latitude 34° 38' N., longitude 65° 12' W.

With dead reckoning latitude 34^{0} 30' and t g° 30' on page 212 take out 34^{0} 37' latitude in h/d column and 9^{0} 44' hour angle in Z/t column the latitude and hour angle of plotting position, opposite them note a=844° and £ = 35°; combine b with d and get C, C—b—d=22^{0} 38.1' and corresponding to this value of C find P—71-5° and h_{c} — 66° 03.4', page 213. The difference between this h_{c}—66° 03.4' and the observed h_{0} = 66° 11.6' is 8.2' which lay off toward the star on bearing Z=a+/? = N. 155.9^{0} W., and draw line of position at right-angles to that bearing.

Identification of Unknown Stars

I he problem of identification of an unknown star is almost the same as the determination of the line of position.

In the astronomical triangle PMZ if we have observed the altitude and azimuth of the unknown star M we have the angle Z and the two sides ZM and ZP. If we let fall a perpendicular from M on the side PZ we divide the astronomical triangle into two right spheric triangles and may solve for declination and hour angle using the tables as in observations of known bodies for altitude and azimuth.

These modifications will suggest themselves as the circumstances arise and need not be committed to memory.

The method of using the tables is as follows: Having taken the altitude and compass bearing of the star, observe the time, correct the altitude, compute the azimuth, measured from the elevated pole and either east or west, and the Greenwich sidereal time, and applying the longitude by dead reckoning obtain the approximate local sidereal time.

Enter table and find nearest values to h in the h/d column and Z in h/t column. If Z is greater than 90° use 180° —Z.

Opposite the nearest values to h and Z, take out C in column marked C at top. Mark C negative if Z is greater than 90°. To this value of C add the latitude to get b. If this b is greater than 90°, take it from 180. Opposite b find d and t in columns marked h/d and Z/t at top. If b is negative mark d opposite name to the latitude. If b is greater than 90° take t greater than 90° also, that is, take 180^{0} — t for t. To get the right ascension of the star reduce this t to time and take this hour angle in time from the local sidereal time. Then from the table of apparent places of stars for the upper transit at Greenwich in the Nautical Almanac, identify the star, and get its exact right ascension and declination.

Example J.—November 5, 1918, in dead reckoning latitude 36° N., longitude 122^{0} 15' W., about 1 a. 111., observed altitudes of two unknown stars as follows: At watch time 13^{h} 03^{m} 35^{s}, Star No. 1, altitude 24° 51' bearing N. 30° E. At watch time 13^{1}' 05^{m }35^{s}, Star No. 2, altitude 13° 24' 20" bearing N. 142^{0} E. C—W, 8^{h} 07™ 45^{s}; C. C., + 1^{m} 30^{s}; I. C., o; Height of eye, 20 feet. Identify stars and plot lines of position.

Great Circle Course and Distance To use the table to get great circle course and distance regard M of the triangle PZM as the point of departure and Z as the point to be arrived at, then entering the table with the latitude of the point of destination and the difference of longitude take out b opposite the nearest values of the angles, obtain C by subtracting the latitude of the point of departure from b and take out M the course and compute 90°—h, the distance from h in the columns in which the difference of longitude and the latitude of destination were found. Since the actual latitudes and differences of longitude will not be found in the table, the values of the course and distance will be approximate, but the course will be close enough for all navigational purposes.

Commander de Aquino is preparing a new table with the perpendicular to every 15 minutes of arc, and omitting interpolation factors, and I understand others, in Europe, are preparing still more voluminous tables, but the present tables are within the limit of accuracy of observation in ordinary navigation.