EXAMPLE III.

To find the furface of the torrid zone extends to the diftance of AI or Ai 23 each fide from AB the diameter.

Here the LKC 23 degrees.
Then, s. 4L S. LLKC:: CK CL
And the height L2CL

1кki, which

degrees on

1586.572.

3173*144.

Hence 25000 X 317314479328600 fquare miles the furface of the torrid zone.

EXAMPLE IV.

Required the convex furface of each of the temperate zones IDF K, idfk, which are included between the frigid and torrid zones.

Therefore 25000 X 2062 2955

3648.8675

1586.572

2062 2955

the height.

= 51557387

=

16500375

= 79328600

fquare miles each temperate zone. Hence the two frigid zones

the two temperate zones =103114775

the torrid zone

their fum is 198943750

or the furface of the whole fphere.

PROBLEM XI.

To find the Lunar Surface ABDCA, included between two Great Circles A BD, ACD of a Sphere.

!

That is, if d = A D the diameter,

and a

the length of the arc BC.

Then ad is the furface.

RULE

II.

As four right angles, to the furface of the sphere;
Or, As one right angle, to a great circle of the sphere;
So is the angle made by the two great circles,
To the furface included by them.

EXAMPLE.

Required the furface included by two great circles forming an angle of 25 degrees; the diameter of the fphere being 10 feet.

1. By rule 1.

First, 3.1415926 × 10 = 31°415926 is the circumf. And 360 25 (:: 72: 5 :: 144: 10):: 31415926: 2-181661 the greatest breadth of the furface.

Then 2.181661 X 102181661 is the furface req. 2. By

If the great circle BCE, whofe poles are A and D, be conceived to be divided into an indefinite number of equal parts, and great circles be conceived to be drawn through the points of division, and through A and D: it is evident that the whole furface will be divided, by the circles, into the fame number of parts, fimilar and equal to one another; and that, therefore, the furface in cluded between any two of thofe circles, will be as the number of parts, or as the arc of the circle BE included by them; wherefore as the whole circumference BE is to the arc BC, or as four right angles to the angle BAC, fo is the furface of the fphere to the lunar furface ABD CA; or alfo as one right angle is to the angle BAC, fo is the area of a great circle of the sphere (4 of the furface) to the lunar furface. Which is rule 2.

Corollary. If d be the diameter, and e the circumference of the fphere, and a the arc BC; then, from the procefs above, a: cd (the furface of the fphere) : ad the lunar furface. Which is rule 1.

2. By rule 2.

Here 31415926 × 10 = 314′15926 is the furface of the fphere.

And 360 25 : (144: 10 :) 31415926 : 21.81661 the furface required.

PROBLEM XII.

To find the Area of a Spherical Triangle; that is, the
Spherical Surface included by the Arcs of Three
Great Circles of the Sphere Interfecting
one another.

As 8 right angles or 720°,
To the furface of the fphere;

Or, As 2 right angles or 180°,

To a great circle of the sphere;

So is the excess of the 3 angles above 2 right angles,
To the area of the triangle.

*

DEMONSTRATION.

That

For having produced all the fides of the triangle ABC (P), till they interfect again, and form the triangle p, which by the prin ciples of the fphere will be equal to the former triangle P; puts the furface of the fphere P+Q+R+T. Then, by the last problem,

180°: A :: s: P+T,

180 BSP + Q,

2

180: c :: s: P+R =

S

=

R; hence

180: A+B+C :: {s: 3P +T+Q+ R

2 PS,

and as 180: A + B + C 180:: s: 2 P ::

S: P=SX

A+B+C - 180

720

Corollary. When P = 0, then (A + B + c or) S = 180; but when P = pdd, half the furface of the fphere, then s- - 180 =360, or s = 540: confequently A+B+c is always between' 180 and 540, that is, greater than 2, and lefs than 6, right angles.

5- 180

That is, pdd x

the area of the triangle;

720

putting p314159,

d the diameter of the fphere,

s the fum of the 3 angles of the triangle.

EXAMPLE.

If the angles be 55, 60, and 85 degrees; what is the trilineal furface; fuppofing the diameter to be IO?

Here 7853982 X 102 78.53982 of the furface of the fphere. And 55 +60 +85 180

= 20.

Then 180: 20, or 9: 1 :: 78.53982: 8.72664, the area of the triangle required.

PROBLEM XIII.

To find the Area of a Spheric Polygon, or to find the Spherical Surface included by Any Number of Interfeiting Great Circles.

As 8 right angles or 720°,
To the furface of the sphere;
Or, As 2 right angles or 180°,

To a great circle of the sphere;

So is the excefs of all the angles, above the product of 180 and 2 lefs than the number of angles, To the area of the fpherical polygon.

That is, putting

the number of angles,

s = the fum of all the angles, d = the diameter of the sphere, P = 3.14159 &c.

What furface is included by the intercepted arcs of five interfecting great circles of a fphere, of 10 feet diameter; fuppofing the fum of the angles formed by thofe arcs to be 640 degrees.

Here the furface of the fphere is 3.1415926 x 102 = 31415926.

And 180 x (n-2) 180 x 3 = 540; which taken from 640, leaves 100.

Then as 720 100 :: 31415926: 43.63323, the area of the polygon required.

For if the polygon be fuppofed to be divided into as many triangles as it has fides, by great circles drawn to all the angles through any point within it, forming at that point the vertical angles of all the triangles. Then, by the laft problem, it will be, in any one triangle,

As 720: s: fum of its angles-180: its area. Theref. by compof. as 720s: 5+ all the vertical angles - 180

fum of all the triangles or area of the polygon. Therefore

=

But all the vertical angles 360 or 180 x 2.

As 720 SS — 180 X (2) SX

area of the polygon. 2. E. D.

s- 180 × (n − 2)

720

the

Corollary. When the polygon iso, then sis 180 × (n−2); and when the polygon is the femi fpheric furface, then

s = 180 × (n = 2) + 360 = 180 x n; confequently s the fum of all the angles of any polygon, is always between 180 × (n-2) and 180 x n, that is, lefs than n times 2 right angles, but greater than a - 2 times 2 right angles, a being the whole number of angles.