520. A Lune is a portion of the surface of a sphere included between two semicircumferences of great circles; as ACBD. The Angle of a lune is the angle between the semi-circumferences which form its sides; as the angle CAD, or the angle COD. B 521. On the same, or on equal, spheres, lunes of equal angles are equal, as they are evidently superposable. 522. A Spherical Wedge, or Ungula, is the part of a sphere bounded by a lune and the planes of its sides; as AOBCD. The diameter AB is called the Edge of the ungula, and the lune ACBD is called its Base. PROPOSITION VI. THEOREM. 523. The area of a lune is to the surface of the sphere as the angle of the lune is to four right angles. Let ACBD be a lune, and ECDH the great circle whose poles are A and B; let L be the area of the lune, and S the surface of the sphere, and 4 the angle E CAD, or the angle of the lune. H D B since angle A is measured by arc CD, and ECDH is the circumference. Apply a common measure to CD and ECDH, and suppose it is contained n times in CD and m times in ECDH, then CD n m Through these points of division pass great circles; then the lune ACBD will contain n equal lunes, and the entire sphere m equal lunes, The student can supply the proof for the case when CD and ECDH are incommeasurable. 524. COR. Let A denote the numerical measure of the angle of a lune referred to a right angle as the unit, and T the area of the tri-rectangular triangle. Then since the surface of the sphere is expressed by 8 T, we have (by 523) L A or L = 2 A × T. 8 T 41 That is, if the unit of measurement for angles is the right angle, the area of a lune is equal to twice its angle multiplied by the area of the tri-rectangular triangle. = = For example, if A= 60° of a right angle, its area would be of the area of the tri-rectangular triangle. Then if the surface of the sphere were 120 square inches, the area of the tri-rectangular triangle is 15 square inches, of which is 20 square inches or the area of the lune. EXERCISES. 1. Show that in a spherical triangle, each side is greater than the difference between the other two. 2. If the radius of the sphere is 12, what is the linear length of the sides of the triangle whose angular measures are 40°, 60°, and 80° ? 3. Find the area of a lune when the angle is 135°, and the surface of the sphere 300 square inches. PROPOSITION VII. THEOREM. 525. If two arcs of great circles BAB' and CAC" intersect each other on the surface of a hemisphere, the sum of the opposite triangles ABC and AB'C' is equivalent to a lune whose angle is equal to the angle BAC included between the given arcs. Draw the diameters AA', BB', CC". Since A'BA is a semicircle, it is equal to BAB'; subtract the portion BA, and we have arc A'B= AB'; likewise arc A'C = AC', and BC = B'C', both being measures of the 526. The area of a spherical triangle is proportional to its spherical excess. Let A, B, C be the numerical measures of the angles of the spherical triangle ABC; let the right angle be the unit of angular measure, and the tri-rectangular triangle T be the unit of areas. A B To prove that Area ABC (A+B+C-2) × T. = Continue any side, say AB, so as to complete the great circle, and produce the other sides until they meet this circle in B' and A'. likewise and or Area ABC + A'BC = lune ABA'C = 2 A× T; ABC + AB'C = lune ABCB' = 2 B × T, ABC+A'B'C= lune ACBC" = 2 C × T. By addition, 3 ABC+A'BC + AB'C + A'B'C=(2A+2 B+2C) × T. But ABC + A'BC + AB'C + A'B'C' = the hemisphere = 4 T. Therefore 2 ABC +4 T = (2 A + 2 B + 2 C) × T, ABC+2 T (A + B + C) × T, = That is, the greater the excess of A+B+C over 2 right angles, the greater will be the area. 527. COR. spherical excess. Q.E.D. The area of a spherical polygon is equal to its THE SPHERE. 528. A Zone is a portion of the surface of a sphere included between parallel planes. The circumference of the circles which bound the zone are called the Bases, and the distance between their planes the Altitude. One of the bases may be a tangent plane. 529. A Spherical Segment is a portion of the volume of the sphere included between two parallel planes; the planes are the Bases, and their distance apart is the Altitude. D A G H 530. Let the sphere be generated by the revolution of the semicircle ACDEFB about its diameter AB as an axis; and let CG and DH be drawn perpendicular to the axis. The arc CD generates a zone whose altitude is GH, and the figure CDHG generates a spherical segment whose altitude is GH. The circumferences generated by the points C and D are the bases of the zone, and the circles generated by CG and DH are the bases of the segment. E F B PROPOSITION IX. THEOREM. 531. The area generated by the revolution of a straight line about an axis in its plane is equal to the projection of the line on the axis, multiplied by the circumference of a circle whose radius is the length of the perpendicular erected at the middle point of the line and terminating in the axis. Let AB be the straight line revolving about the axis FM, CD its projection on FM, and EF the perpendicular erected at the middle point of AB, terminating in the axis. To prove that area generated by AB= CD x 2π × EF. Draw AG parallel to CD, and EH perpendicular to CD. |