0, and

like frustrums of the two solids estimated from O where x = this whether the radii y, y2, remain constant for a given value of x or vary in any way whatever, that is, whether the perimeters ABC abc continue of the same magnitude or be variable, only that they preserve their similarity. Therefore we have

V' : V2 = U : U2 = y2 : y2 = x2 : x2;

2

.. (507), V : V2 = frustrum[U] : fr.[U2] = ‡x3 : ‡Ã3=OP3 : Op3, and the proposition is proved for this more general case.

29

Lastly, let the similar solids be any whatever, and assume any two like diameters for the axes of x, x2, the origins dividing them proportionally. It follows from the definition of similar solids given above, that the generating planes U, U2, perpendicular tox, T2, will have corresponding positions in their respective solids V, V1, when any diameter in U is to a like diameter in U, as the abscissa z terminating in U is to the abscissa T, terminating in U2, and that U and U2 will be similar figures;

29

U : U2= x2 : x2.

But from (300) it is manifest that (507) is applicable in this case also;

3

and V: V2 = fr.[U]: fr.[U2]=fr3 : ‡r;=r3 : x2, Q. E. D.

2

EXERCISES.

1o. A cylindrical cistern, capped with a hemispherical dome, is 15 feet deep and 10 in diameter. Required its capacity.

2o. A hollow cylinder, the side of which is one inch thick, is set into a cubical box, touching its sides and equalling it in height, and within the cylinder is placed a hollow sphere also an inch in thickness and tangent to the cylinder. What is the capacity of the sphere, it requiring just 10 gallons of water to fill the space between the box and the cylinder?

3o. What is the capacity of a cask, regarded as a frustrum of an ellipsoid, the bung diameter being 30 inches, the head diameters 25 each, and the length 40 inches; and how much will it differ from its inscribed double conical frustrum?

4°. What is the capacity of a paraboloidal cistern, having the dimensions in 1°?

5o. Wishing to ascertain the weight of a marble column 30 feet high, I take the semidiameters at the elevations 0, 10, 20, 30 feet, and find them to be 3, 4, 3, 2, feet; the specific gravity of marble is 27, water being 1, and a cubic foot of water weighs 1000 oz. avoirdupois. What is the weight?

6o. The earth may be regarded as an oblate spheroid, generated by the revolution of an ellipse about its shorter diameter of 7899 170 miles, while the equatorial diameter is 7925 648, according to Sir J. F. W. Herschel. Required the excess of volume over the inscribed sphere, and the quantity of water on the surface; allowing the sea to be to the land as 4 to 3, and its mean depth to be 24 miles.

7°. What must be the dimensions of a tub to hold 10 cubic feet, the depth and two diameters being as the numbers 4, 5, 6?

8°. What must be the dimensions of a paraboloidal kettle to contain 12 gallons, the diameter across the top being to the depth as 7 to 8?

BOOK SECOND.

SPHERICAL GEOMETRY.

SECTION FIRST.

Spherical Trigonometry.

PROPOSITION I.

A SPHERE, being a solid bounded by a curve surface (518) everywhere equally distant from a point within, called the centre, may be generated by the revolution of a semicircle about its diameter.

For, let the semicircumference, PeEP',

revolve about its diameter, PoOP'; it follows that every point, e, in PeEP' will, while describing the circumference, ee'e" maintain a constant distance,

E

from the centre, O; whence the surface T described will be that of a sphere.

Fig. 118.

T

Cor. 1. The radius, OE, perpendicular to the axis, (519) POP', will describe a Great Circle, the circumference of which, EE'E" denominated the Equator, will be everywhere 90° distant from its poles, P, P'.

Cor. 2. Any other perpendicular, oe, will describe a (520) Small Circle, also perpendicular to the same axis, and having the same poles, P, P'.

Cor. 3. Every section of a sphere by a plane is a circle. (521) Cor. 4. The perpendicular, P'T, through the extremity (522)

of the diameter, POP', will generate a plane (P'‚TT'T” ... ) tangent to the sphere. Hence, a plane passing through the extremity of any radius, and perpendicular to it, is tangent to the sphere; and, conversely, a tangent plane is perpendicular to the radius, drawn to the point of contact.

Cor. 5. All great circles mutually bisect each other; as (523) the Meridians, PEP', PE'P', since they have a common diameter, PP'.

Cor. 6. Every great circle bisects the sphere.

(524) Cor. 7. A small circle divides the sphere into unequal (525) parts, and is less the more distant it is from the centre.

Cor. 8. A Lune, or the spherical surface embraced by (526) two meridians, is to the whole surface of the sphere as its equatorial arc to the total equator; thus,

or

Lune PEP'E'P: Sph. Surface = arcEE': 360°,
Lune 4πr2 = e° : 360°;

Cor. 9. The Spherical Wedge or Ungula, PEP'E'O, is (527) to the whole sphere as its equatorial arc is to the total equator; Ungula,, : • Tr3 = e° : 360°.

Cor. 10. Every meridian is perpendicular to its equator; (528) as PE'P' to EE'E".

....

Cor. 11. A Spherical Angle is identical with that em- (529) braced by the tangents to its sides, and is measured by the arc of its equator intercepted by these sides; as the angle EP'E' TP'T' measured by EE'.

=

Definition. A Spherical Triangle is the surface em- (530) braced by three arcs of great circles.

PROPOSITION II.

A spherical triangle is equal to the sum of its three (531) angles diminished by a semicircumference, multiplied into the square of the radius of the sphere.

Produce one side, AB, of the spherical triangle, T, so as to form the circumference, ABPQ, intersected in P and Q by the productions of the other sides, AC, BC; then, by the addition of the triangles, X, Y, Z, to T, there will be formed the lunes

A

T B

Y

X

Z

Fig. 119

T + X, T+Y, having the angles A, B, and to the surface T+ Z, which may be readily shown by (523) to be equal to a lune with the angle, C; and, from (526), there results,

.. T

=

A° + B° + C° — 180°

180°

•πr2 = (A+B+C-T)r2. (531)

Cor. 1. The sum of the three angles of a spherical tri- (532) angle is always greater than two right angles and less than six; since for the existence of T we must obviously have

Cor. 2. Similar spherical triangles are to each other as (533) the squares of the radii of their respective spheres, or as the squares of their homologous sides.

Cor. 3. A spherical polygon is measured by the sum of (534) its angles diminished by as many semicircumferences as it has sides save two, multiplied into the square of the radius of the sphere.

For the number of sides being n, the polygon may be divided into (n-2) triangles, T, T2, T3, whose angles A, B, C; A2, B2, C2; A, B, C3;..., make up the angles of the polygon, and there results

2

(T+ T2+ T3 + ••• )n~2 = [(A+B+C−π)+(A2+B2+C2—π)+.....]r2, or P2 = [S-(n − 2)π] • r2, S = sum of angles.

* For, producing the arcs CP, CQ, till they meet in R, and drawing the diameters (?) AOP, BOQ, COR, we have only to show that the triangle PRQ, which completes the lune CPRQ, is equal to ▲ ACB. In order to this, take the point S, equally distant from A, B, C, or the pole of the small circle passing through these points, and draw the diameter SOT; then the arcs SA, SB, SC, TP, TQ, TR, will be all equal; also (?) 4 ASB = PTQ, BSC QTR, ASC= PTR, and it may therefore be shown by superposition that

▲ QTP = ASB, RTQ = BSC, RTP = ASC;

Fig. 1192.