68 USE OF THE TABLES 
* Table II Observe, now, equations (8) and (9). Table I gives us secant a 
(column A) and cosecant a (column C). All that is necessary to obtain the values 
of h, and Z’’ is to get the cosecant and cotangent of (d+6). This value of (d+b) 
is the basis of Table II. It is obtained by finding the algebraic sum of d (the 
declination) and b (the value found in the first column of Table I). 
Table II, then, is merely a log cosecant and cotangent table of angles from 
0° to 90° and given for every minute. It contains two columns, B and D. 
Column B is the log cosecant of these angles multiplied by 10° power. 
Column D is the log cotangent of these same angles to three places and multi- 
plied by 10? power. 
Adding the value of B, taken from Table II, to the value of A, taken from 
Table I, gives us the log cosecant h,. (See equation 8). Now, since the first 
column (B) of Table II is already a log cosecant column, the value cf h,° (the 
computed altitude) may be found at the top of this column corresponding to its 
log. The minutes are found to the left of the table. 
Similarly, adding the value of D, Table II, to the value of C, taken from 
Table I, gives us the log tangent of Z’’. (See equation 9.) Now, since the second 
column (D) of Table II is a log cotangent column, and we are dealing with the 
log tangent Z’’, it is but necessary to find this value of the log tangent in column 
D and the complement is the value of Z’’. This value of Z’’ may be found at the 
top of the column containing its corresponding log. The tenths of a degree are 
found to the right of the table. 
For simplicity and space, Table I is carried only to 90°. For values over 90°, 
subtract angle from 180° and enter tables with supplement. 
GRAPHIC ILLUSTRATION OF SOLUTION 
In equation 
(8) Cosec h=sec a cosec (d+b). 
(9) Tan Z’’=cosec a cot (d+b). 
Let A=log sec a. 
C=log cosec a. 
B=log cosec (d+). 
D=log cot (d+b). 
b=natural value of side b in degrees and minutes. 
d=declination of body. 
Then use the following arrangement for quick solutions: 
with ‘Enter Table I. 
d (from Nautical Almanac) 
Equation 8 | Equation 9 
be (Grom ‘Table )) 222s 5_ A (Table I) | C (Table I) Z’ (Table I) 
d+b6 (algebraic sum) ___-- B (Table II) | D (Table IT) 
Ae (table BD 02a 3 3a A+B Ca Disk = 2 Z’’ Table IT) 
Z= Algebraic 
sum ‘ 
EXPLANATION IN DETAIL 
1. G. A. T. is found from midnight in the usual manner. From this the G. H. A. 
is computed as follows: for the sun, 
G. H. A.=G. C. T.—1254+ Eq. of T. 
For star, planet, or moon, 
G. H. A.=G. 8. T.—R. A. 
(Add 24 to the G.'S. T. if necessary to perform this subtraction). 
2. Convert the G: H. A. to degrees (see short method p. IV). 
3. Apply an assumed longitude [minus (—) if west, and plus (+) if east] such 
that the resultant local hour angle will be an integral degree. If west longitude, 
subtract the smaller from the larger. 
