_. vide by 360. ‘Thus, to find the area of a se; 
GEOGRAPHY. 
, 35 ; 
a- required nearly = 5 X ,00633=.0086925. The true 
area, as found from the Table of sines, is .008705. 
To find the area of a segment of a zone, terminated at 
both extremities by meridians ; multiply the area of 
the whole zone, by the! of the segment, and di- 
ent of 
the zone between 43° and 43° 85’, terminated by two 
meridians 6° 20’ distant from one another, multiply 
.0036925 by 6° 20’, and divide by 360, that is the area 
~ 6°20’ 380 
of the segment = 360° 21600 % 0086925= .000065. 
To find the area of any particular country or district, 
en coun- divide the country into segments of zones, by parallels 
of latitude, and find the area of each segment separate 
ly; the sum of these areas will be the area required. 
In some cases, this operation may be considerably 
abbreviated without affecting, in any great degree, the 
accuracy of the result. Let it be required, for exam-. 
ple, to find the area of Portugal. Instead of dividing 
the whole surface into ents of zones of different 
lengths, according to the difference in the extent of the 
country from west to east, we may suppose the whole 
to consist of one segment, of a uniform length and 
breadth, viz. between 37° and 42° north latitude, and 
between 7° and 9° west longitude. By this arrange- 
ment, indeed, the eastern boundary cuts off a part of 
Tralos Montes and Beira, and the western a part of 
Estremadura ; but, in liew of these, the former includes 
a portion of Andalusia in Spain, and the latter a part 
the Atlantic ocean. Supposing, therefore, these ex- 
changes to be nearly equivalent, the area may be found 
thus :— 
' Take the sum of the areas of the zones between 37° 
and 429, which is .03366, multiply by 2 the length of 
the segment, and divide by 360, that is, 
et ge oe oe (08366 
pine belay bopenati To 
Taking the circumference of the globe at 24856.148 
English miles, the radius is 7911.964, or, making the 
circumference 21600 hical miles, the radius be- 
comes 6875.499. By the former the area of the globe 
is 196660948 English square miles, and by the latter 
148510778.4 geographical square miles. The area of a 
zone is found by multiplying these numbers by the frac- 
tional valueof the zone, Thus the zone included between 
the parallelsof 56° and 57°is equal tol 96660948 x .004815 
=946922.46 English square miles, or 148510778.4x 
-004815=715079.39 geographical square miles. Thus 
also the area of Portugal is equal to 196660948 x 
-000187=36775.6 English square miles, or 148510778.4 
X .000187=27771.5 geographical square miles. 
_ There is another i gaa connected with the figure 
and dimensions of the earth, which, though properly 
belonging to trigonometry, may, from its application to 
the present subject, and the facility with which it can 
be solved, be properly introduced here, viz. to find the 
most distant point of the globe visible to the eye at any 
elevation ; or to determine the extent of the visible ho- 
rizon from any ley point. 
Let ABG (Fig. 3.) represent the circumference of 
the globe, an GB a diameter produced to E a given 
elevation above B, it is required to find the most dis- 
tant point visible to the eye at E, supposing the emi- 
nence BE to be in a level country, or on the sea coast. 
_ Through E draw EF a tangent to ABG in D, then 
Ely a gg seen from E ; the arch 
BD is the measure of the distance required in degrees, 
VOL. X. PART I. 
153 
minutes, or seconds, and DE the tangent of that arch Mathemati- 
to the radius CB or CD. But in very small arches, as “ 
BD must always be, even though E were the summit of 
the highest mountain on the lobe, the tangent hardly 
differs from the arch itself; therefore ED may, with- 
out any sensible error, be considered as the distance re- 
quired, Now ED’? = GE.BE (see Grometry) = (since 
BE is very small compared to GB) GB.BE nearly, and 
therefore ED = ,/GB.BE. Hence, if d represent the 
diameter of the globe in English miles, f the given 
height of the eye in fect, that is the height in 
5280 
miles, and d’ the distance required in miles, we have 
wa Jay FJ 4 
3280 580 xf. Ifthe diameter of 
the globe be assumed = 7912 English miles, the for- 
mula becomes 
= wl a9 X SHV149,84, Kf = 1.224126 x of. 
In every case, therefore, the square root of the height 
in feet multiplied by 1.224126 will give the radius of 
the visible horizon in English miles, supposing the ray 
of light to come from the verge of the He to the 
eye In a straight line, viz. in the direction of the tangent 
E. This, however is not exactly the case, the ray 
by the refractive power of the atmosphere, being bent 
downwards, so as to meet BE at a point below E; that 
is the point D is visible to the eye at an elevation less 
than BE. From E therefore, the horizon extends to a 
greater distance than D, and consequently the value of 
d', as found by the preceding formula, is too little by a 
quantity hay aa to the refraction, and which is 
ound to vary from 4 to , of the whole distance, ac« 
cording to the state of the atmosphere with regard to 
weight and humidity. In a medium state, the refrac- 
tion is about ;,, or .0714, which may therefore be con« 
sidered as a near approximation to the truth in all or- 
dinary cases, With this correction the preceding formula 
becomes d’ = 1,224126 x 1.0714 x / f= 1.3115 / f: 
and reduced to the form of a rule, it may*be stated 
thus: Multiply the square root of the height of the eye, 
in feet, by 1.3115, and the quotient will be the radius of 
the viasble horizon in English miles. 
_, Example 1. Required the distance or radius of the 
visible horizon to the eye, situated six feet above the 
surface of the sea. 
Here d’ = 1.3115 4/6 = 1.3115 % 2.449 = 3.2118 
English miles, the distance required, 
rg as 2. At what distance may a mountain 21440 
feet in height be seen, the eye being on the surface of 
the sea. 
_ In this case d! = 1.31154/21440 = 1.9115 x 146.42 
= 192 English miles. Hence the summit of Chimbo- 
razo, the highest of the Andes, ought to be seen at a 
distance of 192 miles, if its height be as it is stated, 
21,440 feet. 
7. (OR 
CHAP. IT. 
Or THE GLozEs, 
Sect. I, Construction of the Globes, 
As soon as geographers had discovered the spherical 
figure and diurnal revolution of the earth, they would 
naturally be led toa very simple method, of representing 
U 
Geogra- 
- ra 
—_—— 
