SPHERICAL MEASUREMENTS

719. The portion of the surface of a sphere included between two parallel planes is called a Zone.

The perpendicular distance between the planes is the altitude of the zone, and the circumferences of the sections made by the planes are called the bases of the zone. If one of the parallel planes is tangent to the sphere, the zone is called a zone of one base.

ABCD is a zone of the sphere.

B

720. The portion of the surface of a sphere bounded by two semicircumferences of great circles is called a Lune.

The angle between the semicircumferences which form its boundaries is called the angle of the lune.

ABCD is a lune of which BAD is the angle.

721. Lunes on the same sphere, or on equal spheres, having equal angles may be made to coincide, and are equal.

722. A convenient unit of measure for the surfaces of spherical figures is the spherical degree, which is equal to of the surface of a hemisphere.

Like the unit of arcs, it is not a unit of fixed magnitude, but depends upon the size of the sphere upon which the figure is drawn.

It may be conceived of as a birectangular spherical triangle whose third angle is an angle of one degree.

The distinction between the three different uses of the term degree should be kept clearly in mind; an angular degree is a difference of direction between two lines, and it is the 360th part of the total angular magnitude about a point in a plane (§ 35); an arc degree is a line, which is the 360th part of the circumference of a circle (§ 224); a spherical degree is a surface, which is the 360th part of the surface of a hemisphere, or the 720th part of the surface of a sphere.

Proposition XXI

723. Represent an axis and a line oblique to it, but not meeting it; draw lines from the extremities and middle point of this line perpendicular to the axis; from the nearer extremity draw a line parallel to the axis; also a line perpendicular to the given line at its middle point and terminating in the axis. If the given line revolves about the axis, what kind of a surface will it generate? To what is this surface equivalent? (§ 628) By means of the proportion of lines from similar right triangles, express the surface in terms of the projection of the given line on the axis and the circumference of a circle whose radius is the perpendicular from the middle point of the given line. Would this result hold true, if the line should meet the axis or be parallel to it?

Theorem. The surface generated by a straight line revolving about an axis in its plane is equivalent to the rectangle formed by the projection of the line on the axis and the circumference whose radius is a perpendicular erected at the middle point of the line and terminated by the axis.

Data: Any line, as AB, revolving about an axis, as MN; its projection upon MN, as CD; and EO perpendicular to AB at its middle point and terminating in the axis.

To prove surface AB rect. CD. 2 TEO.
Proof. Draw EF MN and AK || MN.

C

E

B

K

If AB neither meets nor is parallel to MN it generates the lateral surface of a frustum of a cone of revolution whose slant height is AB and axis CD; surface AB rect. AB · 2πEF.

.. § 628,

If AB meets axis MN, or is parallel to it, a conical or a cylindri cal surface is generated, and the truth of the theorem follows. Therefore, etc.

Q.E.D.

MILNE'S GEOM. -23

Proposition XXII

724. Draw a semicircumference and inscribe in it a regular semipolygon. How does the sum of the projections of the sides of the polygon on the diameter of the semicircle compare with the diameter? How do the perpendiculars to the sides of the polygon at their middle points compare in length, if they terminate in the diameter? If the figure is revolved about the diameter as an axis, to what is the surface generated by the perimeter of the semipolygon equivalent? How does the perimeter of the semipolygon at its limit compare with the semicircumference, if the number of its sides is indefinitely increased? What is the limit of the perpendicular to the middle point of a side of the semipolygon? How, then, does the surface of a sphere compare with the rectangle formed by its diameter and the circumference of a great circle?

Theorem. The surface of a sphere is equivalent to the rectangle formed by its diameter and the circumference of a great circle.

Data: A sphere, whose center is 0, generated by the revolution of the semicircle ABCD about the B diameter AD.

Denote the surface of the sphere by S, and its radius by R.

A

H

E

D

Proof. Inscribe in the semicircle half of a regular polygon of an even number of sides, as ABCD, and let s' denote the surface generated by its sides.

Draw BE and CF 1 AD, and the perpendiculars from 0 to the chords AB, BC, and CD.

§§ 202, 200, these perpendiculars are equal, and bisect the chords. Then, § 723,

and

surface AB

rect. AE. 2 πOH,

But the sum of the projections AE, EF, and FD equals the diameter AD;

Now, if the number of sides of the inscribed semipolygon is indefinitely increased,

§ 392, the semiperimeter will approach the semicircumference as its limit;

and

OH will approach R as its limit,

s' will approach s as its limit.

But, however great the number of sides of the semipolygon,

725. Cor. I. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.

726. Cor. II. § 725, area = AD × 2 πR = 2 R × 2 πR = 4 πR2; that is, the area of the surface of a sphere is equal to the area of four great circles.

727. Cor. III. Let R and R' denote the radii, D and D' the diameters, and 4 and 4' the areas of the surfaces of two spheres. Then, § 726, A=4πR2, and 4' = 4πR";

that is, the areas of the surfaces of two spheres are to each other as the squares of their radii, or as the squares of their diameters.

728. Cor. IV. Area of a zone, as BC = EF × 2 πR;

that is, the area of a zone is equal to the product of its altitude by the circumference of a great circle.

729. Cor. V. Zones on the same sphere, or on equal spheres, are to each other as their altitudes.

730. Cor. VI. § 728, area of a zone of one base, as

that is, the area of a zone of one base is equal to the area of a

circle whose radius is the chord of the generating arc.

Proposition XXIII

731. Divide the surface of a sphere into hemispheres by a great circle; on one of the hemispheres form two opposite triangles by drawing two great circle arcs to intersect; complete the circumferences of which these arcs are parts. By comparing one of these opposite triangles with a triangle on the other hemisphere that completes a lune of which the other of the given triangles is a part, discover how the sum of the given triangles compares with a lune whose angle is the angle between the given arcs.

Theorem. If two arcs of great circles intersect on the surface of a hemisphere, the sum of the two opposite triangles thus formed is equivalent to a lune whose angle is the angle between the given arcs.

Data: Opposite A, as AEB and DEC, formed by two great circle arcs, as AED and BEC, on the hemisphere E-ABDC.

To prove ▲ AEB + ▲ DEC

lune AEBF.

Proof. Produce arcs AED and BEC around

the sphere intersecting as at F.

E

§ 656, arc DE = arc AF (each being the supplement of arc AE),

arc CE

and, § 693,

... § 710,

arc BF (each being the supplement of arc BE),

Adding A AEB to each side of this expression of equivalence,

Δ ΑΕΒ + Δ DEC = Δ ΑΕΒ + Δ AFB.

Hence,

▲ AEB +▲ DEC≈ lune AEBF.

Proposition XXIV

Q.E.D.

732. On the surface of a sphere draw a lune whose angle is to four right angles as 3:12; from the vertex of its angle as a pole describe the circumference of a great circle. What is the ratio of the arc included between the sides of the lune to the whole circumference? Divide the circumference into 12 equal parts and through the points of division and the poles pass great circle arcs. Into how many equal lunes do these arcs divide the surface of the sphere? The given lune? How, then, does the ratio of the given lune to the surface of the sphere compare with the ratio of the angle of the lune to four right angles?