found for the same moment of absolute time. Their

difference in right ascen

sion gives the included angle, P, at the celestial pole. The declination subtracted from 90°, if it be north, and added to 90°, if it be south, will give the sides, PZ and PS.

In the following examples, we give the right ascension and declination of the bodies, and from

S/C

these the student is required to compute the distance between them.

The right ascensions are given in time. Their difference must be changed to degrees for the included angle.

SECTION VI.

REGULAR POLYEDRONS

A Regular Polyedron is a polyedron having all its faces equa! and regular polygons, and all its polyedral angles equal.

The sum of all the plane angles bounding any polyedral angle is less than four right angles; and as the angle of the equilateral triangle is of a right angle, we have ×3<4,×4<4, and 3×5<4; but×6=4, x74, and so on. Hence, it follows that three, and only three, polyedral angles may be formed, having the equilateral triangle for faces; namely, a triedral angle and polyedra! angles of four and of five faces.

There are, therefore, three distinct regular polyedrons bounded by the equilateral triangle.

1. The Tetraedron, having four faces and four solid angles.

2. The Octaedron, having eight faces and six solid angles. 3. The Icosaedron, having twenty faces and twenty solid angles. With right plane angles we can form only a triedral angle; hence, with equal squares we may bound a solid having six faces and eight equal triedral angles. This solid is called the Hexaedron.

៖

The angle of the regular pentagon being § of a right angle, we have x3<4; but g×4>4; hence, with plane angles equal to those of the regular pentagon, we can form only a triedral angle. The solid bounded by twelve regular pentagons, and having twenty solid angles, is called the Dodecaedron.

There are, then, but five regular polyedrous, viz.: The tetraedron, the octaedron, and the icosaedron, each of which has the equilateral triangle for faces; the hexaedron, whose faces are equal squares, and the dodecaedron, whose faces are equal regular pentagons.

It is obvious that a sphere may be circumscribed about, or inscribed within, any of these regular solids, and conversely: and

that these spheres will have a common center, which may taken as the center of the polyedron.

also be

Any regular polyedron may be regarded as made up of a number of regular pyramids, whose bases are severally the faces of the polyedron, and whose common vertex is its center. Each of these pyramids will have, for its altitude, the radius of the inscribed sphere; and since the volume of the pyramid is measured by one third of the product of its base and altitude, it follows that the volume of any regular polyedron is measured by its surface multiplied by one third of the radius of the inscribed sphere.

PROBLEM.

Given, the name of a regular polyedron, and the side of the bounding polygon, to find the inclination of its faces; the radii of the inscribed and circumscribed spheres; the area of its surface; and its volume.

Let AB be the intersection of two adjacent faces of the polyedron, and C and D the centers of these faces, O being the center of the polyedron. Draw the radii,

OC and OD, of the inscribed, and the radii OA and OB,of the circumscribed sphere; also from C and D let fall the perpendiculars CE and DE, on the edge AB, and draw OE; then will the angle DEC measure the inclination of the faces of the polyedron, and the angle DEO is one half of this inclination.

Let I denote the inclination of the faces, m the number of faces which meet to form a polyedral angle, n the number of sides in each face, and suppose the edge of the polyedron to be unity.

B

The surface of the sphere of which O is the center, and radius unity, will form, by its intersections with the planes, AOE, AOD, DOE, the right-angled spherical triangle dae, right-angled at e In the right-angled triangle DEO, the angle DOE is equal to

and is measured by the arc de. The angle dae, of the spherical

Now, by Napier's Rules we have

cos.dae: = sin.ade cos.de.

or,

and,

we find

Substituting in eq. (1), for the angles dae and ade, their values,

Equation (3) gives the value of the sine of one half of the incli nation of the planes; and by means of this equation we may readily find the radii of the inscribed and circumscribed spheres.

From the triangle AOD, we find

cos. DOA : 1 :: OD : OA

But the angle DOA is measured by the are ad; hence, substa tuting in this last equation the values of cos.DOA and OD, taken from eqs. (2) and (4), we have

Equation (4) gives the value of OD, the radius of the inscribed sphere, and equation (5) gives that of OA, the radius of the circumscribed sphere. The area of one of the faces of the polyedron is equal to one half of the apothegm multiplied by the perimeter. 360° The apothegm, as found above, is equal to cot. ; hence, we 360° 2n

2n

bave in cot. plying this by the number of faces of the polyedron, we shall have the expression for its entire area. The expression for the surface multiplied by one third of the radius of the inscribed sphere, gives the measure of the volume of the polyedron.

for the area of one of the faces; and multi

In what precedes, we have supposed the edge of the polyedron to be unity. Having found the radii of the inscribed and circumscribed spheres, the surfaces, and the volumes of such polyedrons, to determine the radii, surfaces, and volumes of regular polyedrons having any edge whatever, we have merely to remember that the homologous dimensions of similar bodies are proportional; their surfaces are as the squares of these dimensions; and their volumes as the cubes of the same.

Formula (3) gives, for the inclination of the adjacent faces of

The subjoined table gives the surfaces and volumes of the regular oolyedrons, when the edge is unity.