Scholium. Conceive a polyedron, all of whose faces touch the sphere ; this polyedron may be considered as formed of pyramids, each having for its vertex the centre of the sphere, and for its base one of the polyedron's faces. Now it is evident that all these pyramids will have the radius of the sphere for their common altitude: so that each pyramid will be equal to one face of the polyedron multiplied by a third of the radius: hence the whole polyedron will be equal to its surface multiplied by a third of the radius of the inscribed sphere. It is therefore manifest, that the solidities of polyedrons circumscribed about the sphere are to each other as the surfaces of those polyedrons. Thus the.property, which we have shown to be true with regard to the circumscribed cylinder, is also true with regard to an infinite number of other bodies. We might likewise have observed that the surfaces of polygons, circumscribed about the circle, are to each other as their perimeters. PROPOSITION XVI. PROBLEM. a A If a circular segment be supposed to make a revolution about a diameter exterior to it, required the value of the solid which it describes. В. E Let the segment BMD revolve about AC. On the axis, let fall the perpendiculars M BE, DF; from the centre C, draw CI D perpendicular to the chord BD; also draw the radii CB, CD. The solid described by the sector BCD с is measured by a CB”.EF (Prop. XIV. Sch. 1). But the solid described by the isosceles triangle DCB has for its measuren.CI'EF (Prop. XII. Cor.); hence the solid described by the segment BMD=37.EF.(CBP-CI). Now, in the rightangled triangle CBI, we have CB__CI?=BI’=BD? ; hence the solid described by the segment BMD will have for its measure ..EF. BD’, or 17.BDP.EF: that is one sixth of n into π the square of the chord, into the distance between the two perpendiculars let fall from the extremities of the arc on the axis. Scholium. The solid described by the segment BMD is to the sphere which has BD for its diameter, as f..BD'.EF is to foBD3, or as EF to BD. PROPOSITION XVII. THEOREM. Every segment of a sphere is measured by the half sum of its bases multiplied by its altitude, plus the solidity of a sphere whose diameter is this same altitude. B Let BE, DF, be the radii of the two bases of the segment, EF its altitude, the В. segment being described by the revolution of the circular space BMDFE about M the axis FE. The solid described by the segment BMD is equal to 1.BDP.EF D E (Prop. XVI.); and the truncated cone described by the trapezoid BDFE is equal to 47.EF.(BE?+DF+BE.DF) (Prop. VI.); hence the segment of the sphere, which is the sum of those two solids, must be equal to 1.7.EF.(2BE+2DF2+2BE.DF+BD?) But, drawing BO parallel to EF, we shall have DO=DF-BE, hence DO DF-2DF.BE+BE? (Book IV. Prop. IX.); and consequently BD?=B02+DO?=EF?+ DF?_2DF.BE+BE?. Put this value in place of BD’ in the expression for the value of the segment, omitting the parts which destroy each other; we shall obtain for the solidity of the segment, AEF.(3BE+3DF2+ EF), an expression which may be decomposed into two parts; the r.BE?+0.DF one 47.EF.(3BEP+3DF), or EF.(*-! , 2 being the half sum of the bases multiplied by the altitude ; while the other 1EF3 represents the sphere of which EF is the diameter (Prop. XIV. Sch.): hence every segment of a sphere, &c. Cor. If either of the bases is nothing, the segment in question becomes a spherical segment with a single base ; hence any spherical segment, with a single base, is equivalent to half the cylinder having the same base and the same altitude, plus the sphere of wkich this altitude is the diameter, ܪ General Scholium. Let R be the radius of a cylinder's base, H its altitude : the solidity of the cylinder will be aR$ x H, or rR’H. Let R be the radius of a cone's base, H its altitude: the solidity of the cone will be aRx }H, or },R’H. Let A and B be the radii of the bases of a truncated cone, Q H its altitude : the solidity of the truncated cone will be 1A.H. (A + B + AB). Let R be the radius of a sphere ; its solidity will be SAR". Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base: the solidity of the sector will be RRH. Let P and Q be the two bases of a spherical segment, H its P+Q altitude: the solidity of the segment will be .H+.H? 2 If the spherical segment has but one base, the other being nothing, its solidity will be {PH+H3. a 1. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles. These arcs are named the sides of the triangle, and are always supposed to be each less than a semi-circumference. The angles, which their planes form with each other, are the angles of the triangle. 2. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 3. A spherical polygon is a portion of the surface of a sphere terminated by several arcs of great circles. 4. A lune is that portion of the surface of a sphere, which is included between two great semi-circles meeting in a common diameter. 5. A spherical wedge or ungula is that portion of the solid sphere, which is included between the same great semi-circles, and has the lune for its base. 6. A spherical pyramid is a portion of the solid sphere, included between the planes of a solid angle whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes. 7. The pole of a circle of a sphere is a point in the surface equally distant from all the points in the circumference of this circle. It will be shown (Prop. V.) that every circle, great or small, has always two poles. a PROPOSITION I. THEOREM. other two. B COB, to be drawn; these planes will form С a solid angle at the centre 0; and the angles AOB, AOC, COB, will be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid angle is less than the sum of the other two (Book VI. Prop. XIX.); hence any side of the triangle ABC is less than the sum of the other two, PROPOSITION II. THEOREM, NP The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points. А Let ANB be the arc of a great circle which joins the points A and B ; then will it be the shortest path between them. 1st. If two points N and B, be taken on D+ the arc of a great circle, at unequal distances from the point A, the shortest distance from B to A will be greater than the shortest distance from N to A. B For, about A as a pole describe a circumference CNP. Now, the line of shortest distance from B to A must cross this circumference at some point as P. But the shortest distance from P to A whether it be the arc of a great circle or any other line, is equal to the shortest distance from N to A; for, by passing the arc of a great circle through P and A, and revolving it about the diameter passing through A, the point P may be made to coincide with N, when the shortest distance from P to A will coincide with the shortest distance from N to A : hence, the shortest distance from B to A, will be greater than the shortest distance from N to A, by the shortest distance from B to P. If the point B be taken without the arc AN, still making AB greater than AN, it may be proved in a manner entirely similar to the above, that the shortest distance from B to A will be greater than the shortest distance from N to A. If now, there be a shorter path between the points B and A, than the arc BDA of a great circle, let M be a point of the shortest distance possible: then through M draw MA, MB, arcs of great circles, and take BD equal to BM. By the last theorem, BDA<BM+MA; take BD=BM from each, and there will remain ADAM. Now, since BM=BD, the shortest path from B to M is equal to the shortest path from B to D: hence if we suppose two paths from B to A, one passing through M and the other through D, they will have an equal part in each ; viz. the part from B to M equal to the part from B to D. But by hypothesis, the path through M is the shortest path from B to A: hence the shortest path from M to A must be less than the shortest path from D to A, whereas it is greater since the arc MA is greater than DA: hence, no point of the shortest distance between B and A can lie out of the arc of the great circle BDA. PROPOSITION III. THEOREM. The sum of the three sides of a spherical triangle is less than the circumference of a great circle. Let ABC be any spherical trian C gle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be semicircumferences, since two great circles always bisect B D each other (Book VIII. Prop. VII. A Cor. 2.). But in the triangle BCD, we have the side BC<BD+CD (Prop I.); add AB + AC to both; we shall have AB+AC+BC<ABD+ACD, that isto say, lessthan a circumference. E PROPOSITION IV. THEOREM The sum of all the sides of any spherical polygon is less than the circumference of a great circle. Take the pentagon ABCDE, for F example. Produce the sides AB, DC, D till they meet in F; then since BC is B less than BF+CF, the perimeter of the pentagon ABCDE will be less G than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till E A they meet in G; we shall have ED<EG+DG; hence the perimeter of the quadrilateral AEDF is less than that of the triangle AFG; which last is itself less than the circumference of a great circle; hence, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference. a |