180°-λ, of Imn, 180°+λ, of Imn, 270°-λ, of mĨn, 270°+λ, of mĨn, and 360°—λ, of 1 mn. P3 The north polar distances of these eight faces will each be λ, the longitude of 1 m ñ, 90°—À ̧ of î 1 ñ, 90°+λ ̧ ̈ ̃of m1n, 180°—λ, of Ï î ñ, 180°+λ, of Ï î ñ, 270°—À ̧ of mÏñ, 270° +λ of m 1 ñ, and 360°—λ3 of 1 m̃ ñ. The north polar distances of these eight faces will each be 180°-P3 A will be the longitude of 1 nm, 90°—λ2 of n 1 m, 90° +λ2 of ñ 1m, 180°—λ, of Inm, 180°+λ, of Iñm, 270°—λ, of 1 m, 270° +λ, of n I m, 360°-λ, of 1 ñ m. 2 The north polar distances of these eight faces will each be p The eight similar faces in the southern hemisphere will have the same longitudes as those corresponding to them in the northern, the eight north polar distances being each equal 180°-P2. A, will be the longitude of mn 1, 90°-λ, of n m 1, 90° + λ, of n m 1, 180°-λ, of mn 1, 180°+λ, of mñ 1, 270°-λ, of π m 1, 270° +λ, of nm 1, and 360°-λ of mñ 1. P1 will be the north polar distance of each of these eight faces. The corresponding eight faces of the southern hemisphere will have the same longitudes as the corresponding ones in the northern, 180°-p1 being the north polar distance of these eight faces. Hence the 48 faces or poles of the six-faced octahedron can be expressed in terms of P1, A1, P2, A2, and P3, Ag; and, as all other forms of the cubical system can be derived from those of the six-faced octahedron, all faces of those forms can be similarly expressed. lu 118. Given P, and A, to determine p, and A1, and also A in terms of the former. P2 and From the spherical triangle C1ƒC, (fig. 40*, Plate IV.*), we have by the formulæ of spherical trigonometry, cos fC=cos CC, cos C1f+ sin C1C, sin C1f cos ƒC1C3; but the spherical angle ƒCC, is measured by the arc gC, at the equator. Hence, substituting the values of these arcs given in the previous section, we have cos p1=cos 90° cos p+ sin 90° sin p, cos λ, =sin p2 cos g. 3 Again, in the spherical triangle ƒgC3, we have sin fg sin fCg, = sin fC, sin fg C3 ; but spherical triangle fCg is measured by arc kC2, and fyC, is 90°; hence or, From the spherical triangle C12f, we have 2 2 cos C2f=cos C1C2 cos C1f+sin C1C2 sin C1f cos C2C1f, cos P2 From the spherical triangle Cafg, we have sin C2f_sin C2 If cos P3 sin P2 or sin P2 sin 90°-13 = sin 90° sin λ2 119. To find the angle between the poles of two faces in terms of their polar distances and longitudes. Let CF be the polar distance of F (fig. 41, Plate IV.*), CL its longitude, Cif the polar distance, and Cl the longitude of f. Also let CFP, C1f=P3; C2L=L3, C2l=λ ̧, and Fƒ=0. Then in spherical triangle C1Fƒ cos Ff=cos CF cos Cif+ sin C1F sin C1f sin FC1ƒ. Then angle FC1f is measured by arc Ll=LC,-ÏC3. Hence cos 0=cos P, cos P3+ sin P, sin p, cos (L-λg). To adapt this to logarithmic computation cos 0=cos pcos P2+ sin P, tan p ̧ cos (L ̧—λ)}. Then cos 0=cos P3 {cos P3+tan a . 120. To find the distance between any two poles on the sphere of projection in terms of the three polar distances from C1, C2, and C. 3 $119. cos 0=cos P3 cos p3+ sin P, sin p, cos (L-λ3) =cos P3 cos p2+ sin P, sin p, (cos L, cos λ ̧ + sin Lg sin λ) but § 118, 3 =cos P2 cos p2+ sin P, sin p, cos L, cos λ ̧ 3 Hence cos 0=cos P, cos p2+ cos P2 cos p2+ cos P1 cos P1 The same formula which we obtained by geometry of three dimensions, § 109. 121. To find the polar distances and longitudes in terms of the indices. Referring to § 117 and (fig. 40*, Plate IV.*), C1, C2, and C, are poles of the cube, ƒ is a pole of 1 m n, g of 1 m ∞, n h of 1 ∞on, k of ∞ 1 C3 of 1 ∞ ∞, C2 of ∞ 1 ∞, and C1 of ∞ ∞ 1. m' ƒC3=P1, ƒC2=P2, ƒC1=P3, C2k=λ1, Сgh=λ2, C3J=λ3. Then A, is the distance between the poles of ∞ 1 and ∞ 1 ∞, p, that between 1 mn and 1 ∞ ∞. cot Pi sin λ 1 1 Pi 1+ + 1 1 + = 2 m n2 = sec2 λ, m 1 tan Pi sec 1 m 1 n 1+ m m=sec λ1 cot P1 Again λ, is the distance between 1 ∞on and 1 ∞∞, P2 between 1mn and ∞ 1 ∞. n m that Also, is the distance between 1 m∞ and 1∞∞, p, that between 1 mn and ∞ ∞ 1. tan Pan2 sec2 Ag Hence the indices being given, the polar distances and longitudes can be determined, or the polar distances and longitudes being given the indices can be determined. 122. To find the polar distances of any two adjacent poles of faces of the six-faced octahedron, or of the supplement of the angle over the edge of any two adjacent faces, in terms of the indices. Let be the angle between any two poles adjacent to the arc CO (figs. 31* and 32*, Plate IV.*), adjacent to OD, and adjacent to CD. For the faces nm 1, m n 1, 1 1 mn mn √ (d++) (+2+1) Similarly for 1 nm and 1 m n we have 2 1 1 1+ + The same is true over every arc CO in (figs. 31* and 32*, Plate IV.*). 1 1 m2 |