Mathemnati- 
cal Geogra- 
phy- 
Develope- 
ment of dif- 
ferent bo- 
dies. 
General 
principle. 
PLATE 
CCLXVI. 
Fig. 9. 
“ment. 
‘this purpose, 
166 
representation of different countries, and oats: A in 
finding the longitude and latitude of any given place, 
are the characteristic properties of this projection; and, 
to the majority of these who have most occasion to con- 
sult. such maps, these properties are by far the most 
important. 
“Secr. III. Construction of Maps by Developement. 
In the various methods of projecting the sphere, as 
explained in the preceding Section, the reader cannot 
fail to observe, that, besides the inaccurate representa- 
tion of different portions of the earth’s surface, common 
in some measure to them all, they are also attended in 
practice with considerable difficulty and inconvenience. 
This difficulty increases with the scale of the projec- 
tion ; and, in the delineation of small portions of the 
-earth’s surface, it becomes so great as almost to prevent 
the application of any of them, to the construction of 
such.maps. To remedy-this defect, geographers have 
had recourse to the method of developement, or that 
which supposes.the earth’s surface to be spread out on 
a plane. But asa sphere or spheroid is a body that 
does not adit of its surface being so extended, it must 
be supposed to be converted into some other body, as 
much as possible resembling the sphere, and whose sur- 
face is at the same time susceptible of such a develope- 
The only bodies of this kind with which the 
sphere can be at all compared, are the cone and cylin« 
er; and accordingly, both have been employed for 
I. Of the Cone. 
_ The principle of this developement, or projection as 
it is sometimes called, may be shortly explained thus, 
Let WNES (Fig. 9.) be the sphere of which it is proposed 
to develope any portion, as the fourth part WNE, WE 
the equator, SN the meridian at right angles to WNES, 
and ML the radius of the middle parallel, or that which 
divides the part to be developed into two equal parts in 
the direction of latitude, in this case 45°. Draw MA 
and M’A at right angles to the radii CM and CM’, and 
meeting SN produced in A; that is, make MA and 
M’A the cotangents of the latitude of the middle pa- 
rallel ; then AM and AM’ will be the sides of a cone, tati« 
gent to the sphere at M and M’, and of which any zone, 
extending to a moderate distance on each side of MM’, 
may, without any sensible error, be considered as equal 
to the corresponding zone of the sphere. From A with 
the radius AM describe the arch M m M’; then, if the 
radius CN and the arch Mm M’ be both divided into 
equal parts, arches described from A through the for- 
mer will be the projected parallels of latitude, and 
straight lines drawn from the same point through the 
latter will represent meridians, both at greater or less 
distances from one another, according to the number of 
divisions. The arch M m M’ will be the parallel of 45°. 
By this projection may be obtained a tolerably accu- 
rate representation of a small portion of the globe; but 
when it is extended to a donidobable space, as the fourth 
part of the whole sphere, the countries towards the 
pole and the equator are extended a great deal beyond 
their true limits, in the direction of their latitude.  Va- 
rious methods of remedying this defect have been a- 
dopted or recommended by different geographers ; but 
the simplest, as well as the most successful, is that 
known by the name of Flamstead’s pro jeclion, It was 
so denominated at first from its aavitinee) and it still re. 
tains the name, though since his time it has undergone 
GEOGRAPHY. 
various alterations. In its most improved form, the Mat 
construction is as follows. 
Draw an indefinite straight line NS (Fig. 10.) to res _ 
present the middle meridian of the map, and from the 
point M, near the middle of the line, set off on both 
sides towards N and S equal distances of any conve- 
nient length, to represent degrees of latitude. Suppose, 
for example, that the map is to contain 60 of 
latitude, viz. from the 20th to the 80th parallel, and 
that this extent is to be equal to three inches ; then each 
degree will be equal to 34, or .05 in. 5 the point M will 
be in the parallel of 50°, and the distance of 3 or .5 in. 
set off towards N and S§ will give the points through 
which the parallels of 40°, 30° and 20° must pass on 
the one side, and those of 60°, '70° and 80° on the other, 
The centre C of these parallels) will be in the line NS 
towards N, and may be found thus. tn 
Let d be the length of the assumed degree of lati 
Z the latitude of the middle el, or M, and 
d’ the length of an arch of 1° to radius 1; then ~~ 
co.lxd - 
% asin“ 
Now, in this example, d = .05 in., / = 50°, and d zz 
ote = 01745329 ; therefore, 
_. 05 X cotan, 50. 
MC = —Oi7agga9 inches. ms 
The computation is performed most. conveniently by 
logarithms, thus: sth 
P nt Ri) 
d': cotan. 1::d: MC = & 
td OB AE Te +e oe » 2:698970 
Log. cot. 50°to Rad. 1... + 1923818 
Bererss 
Log. .01745829 ..... 4. 2.241876) © 
Logi MC. 02 Pt 2). 20880907 
and MC = 2.4038 inches. ’ 
From M, therefore, set off towards N, MC = 2.4 in 
and from C as a centre through each of the divisions in 
NS, describe arches for the parallels of latitude. ~~ 
To find the meridians, take any parallel, as the mid 
dle.one passing through M, and from the table of dea 
grees of longitude, p. 151, take the length co - 
correspond- 
ing to the latitude of the parallel, multiply it by the © 
length of the assumed degree, and that product again’ 
by the number of degrees to which the map is to ex- 
tend on each side of the middle meridian ; the last pro- 
duct will express the distance from 'M at which the ex 
treme meridian will intersect the middle parallel. ‘Thus, 
in the present example, let the map extend 40° on each’ 
side of NS; that is, let it include 80° of longitude, 
and the cal ion becomes— rion 
Degree of longitude in lat. 50....'.. 5 64279 — 
Length of the assumed degree in inches . | .05 
0821395. 
Half longitude of the map... . 40 
12855800. 
From M, therefore, set off on the middle both 
ways, 1.285 in. and it will give the limits of the map 
on that parallel. Corresponding points being deter. 
mined, in the same way, on each 
the other peaiiese 
the curves passing through these points will be the 
meridians bounding the map on the east and west.. To 
find the other meridians, divide the arch of each paral« 
lel between the middle and extreme meridians, into as _ 
P 
FE 
Fig. 
