EXPLANATION OF TABLES 5 
TABLE 5.—MERIDIONAL PARTS. 
This table contains the meridional parts, or increased latitudes, for every degree and minute 
to 80°, calculated by the following formula: 
m= log tan (45° ue 3) Gg Gem b-b yet sind fp Petsin’ Ler. sy 
in which 
: : 10800’ p 
the Equatorial radius a = Ee 3487'.74677 (log 3.5362739) ; 
M, the modulus of common logarithms=0.4342945; 
T= 23025851 (log 0.362157) ; 
c, the compression or meridional ellipticity of the earth 
according to Clarke (1880) 599 195 = 0.003407562 (log 7.5324437) ; 
e=-+/2c—c?=0.0824846 (log 8.9163666) ; 
from which 
M =7915’.7044558 (log 3.8984895) ; 
ae = 23/.38871 (log 1.3690072); 
lace 0.053042 (log 8.7246192) ; 
ge _0’.000216523 (log 6.3355038). 
The results are tabulated to one decimal place, which is sufficient for the ordinary problems of 
navigation. 
The practical application of this table is illustrated in Chapters II and V, in articles treating of 
the Mercator Chart and Mercator Sailing. 
TABLE 6.—LENGTH OF DEGREES OF LATITUDE AND LONGITUDE. 
This table gives the length of a degree in both latitude and longitude at each parallel of lati- 
tude on the earth’s surface, in nautical and statute miles and in meters, based upon Clarke’s value 
(1866) of the earth’s compression, In the case of latitude, the length relates to an arc of 
299.15 
which the given degree is the center. 
TABLE 7.—DISTANCE OF OBJECT BY TWO BEARINGS—DEGREES. 
This table has been computed to facilitate the operation of finding the distance from an object 
by two bearings from a given distance run and course. The arguments are given in degrees; the 
first column contains the multiplier of the distance run to give the distance of observed object at 
second bearing; the second, at time of passing abeam. 
The method is explained in Chapter IV. 
TABLE 8.—DISTANCE OF VISIBILITY OF OBJECTS. 
This table contains the distances, in nautical and statute miles, at which any object is visible 
at sea. It is calculated by the formule: 
d=1.15yz, and d’=1.32y2, 
in which d is the distance in nautical miles, d’ the distance in statute miles, and z the height of the 
eye or the object in feet. 
To find the distance of visibility of an object, the distance given by the table corresponding to 
its height should be added to that corresponding to the height of the observer’s eye. 
ExampLe: Required the distance of visibility of an object 420 feet high, the observer being at 
an elevation of 15 feet. 
Dist. corresponding to 420 feet, 23.5 naut. miles. 
Dist. corresponding to 15 feet, 4.4 naut. miles. 
Dist. of visibility, 27.9 naut. miles. 
TABLE 9.—DISTANCE BY VERTICAL ANGLES (distance less than 5 miles). 
This table gives the distance, up to 5 miles, of an object of known height by the vertical angle 
that it subtends at the position of the observer. It was computed by the formula 
tan an", 
where a=the vertical angle; 
h=the height of the observed object in feet; and 
d=the distance of the object, also converted into feet. 
No correction for Dip is applied. ; ; é ' ; 
The employment of this method of finding distance is explained in Chapter IV. 
