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d1, d2, and d are projections of the poles of three faces of the rhombic dodecahedron.

Join Cd, Cd, and C,d, meeting in o1; 0, is the projection of the pole of a face of the octahedron.

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(Fig. 43*, Plate IV.*)-B1, E1, F1, G1, H1, K1, L1, M1, N1, P1, Q1 represent the poles of nearly all the known four-faced cubes lying in the arc of the zone C; B2, &c., in Cd2; B, &c., in Cd; B, &c., in Cad; B, &c., in C2d; and B, &c., in C1d ̧. Six poles of each four-faced cube in the octant at equal distances from C1, C, and C3.

Rules for finding the position of B1, E1, &c., will be given hereafter.

b1, e, f, g, h, k1, and 11; b2, e2, &c., 12; and b,, e3, &c., 13, three poles of each three-faced octahedron, lying at equal distances from o,, in the arcs of zones represented respectively by od, od, and o1d1.

b1, e, f, g, h, k, l, m, n, o1, P1, and 91; by, e2, &c., 92; bg, eg, &c., 43, three poles of each twenty-four-faced trapezohedron, lying at equal distances from 0, in arcs of zones represented by 0,C3, 01C1, and o1C2 respectively.

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Lastly A, B, E, F, G1, H2, K1, L1, M1, N1, P1, Q1, R1, S1, T1, U1; A2, B2, &c., U2; A, B, &c., Us; A4, B4, &c., U4; A, B, &c., U; and A, B, &c., U, six poles of the six-faced octahedron; the poles of each particular six-faced octahedron being similarly situated in each of the six triangles do C3, d201C3, d201C1, do1C2, d11C2, and do,C1 respectively.

2,

130. To find geometrically the position of any pole on the gnomic projection (fig. 43*, Plate IV.*).

In (fig. 44*, Plate IV.*).-Let AC3, AC2, and AC, be three adjacent cubical axes, rectangular at A.

Let AC-1. Take AN in AC, produced equal to n.

AM in AC2 produced equal to m.

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Join C,N, NM, MC3, CC2, C2C1, and CC.

3 2

Then C2MN is the plane 1 mn, and C1Č,С2 is the plane of the gnomic projection.

Through A draw AG perpendicular, CM meeting C2C, in g, AH perpendicular C,N, cutting CC, in h, and AK perpendicular to CC, cutting C12 in k.

h, g, and k are the projections on CCC, of H, G, and K. Join NG, MH, and C,K in the plane NMC,, meeting in F; also join AF. Then, as in § 117, F' is the pole of 1 m n, G of 1 m ∞, H of 1 ∞on, and K of ∞ 1

n

m

n

Therefore on the plane of projection, CCC,, g is the projection of the pole of 1 mo, h of loon, and k of ∞ 1 ; hC2 of the line HM, kC, of the line KC,, gC, of the line GN.

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m

f, where hC2, kC, and gC, meet, will be the pole of 1 m n. Through h, in the plane NAC,, draw hE perpendicular to AC3. Let angle hACA. Then since angle AHN-90°, angle ANH=A.

In triangle NAC, tan ANC1=40

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Hence h, the pole of 1 ∞on, is found by taking the point h

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Again since tan λ, and angle hAC,=,, if the angular

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n.

elements be given, CC, is the chord of 90° and h is the point where the angle λ, protracted from A meets C1C3, considering C, as zero.

The chord of 90° marked as a protractor is obtainable from any mathematical instrument maker, or may be readily marked on the chord of 90° by using any form of protractor.

Similarly it may be shown that gC=C203, and that g is

m+1'

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the point where the angle A, is marked on C2C, as the chord of 90°, C, being zero; and tan λ=1. Also C2 = CC2, k being

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the point where the angle A, is marked on the chord of 90°, C2 being zero, and tan X,

m

n

Join Cig, Ch, and Ck. f, the point where these three lines meet, is the pole of the face of the six-faced octahedron

whose angular elements are ' and A3, or whose indices are 1mn.

131. To construct a map of all the forms of the octahedral system on a face of an octahedron comprised in an octant of the sphere of projection.

(Fig. 43*, Plate IV.*)

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Describe any equilateral triangle

Bisect CC, in d1, C1C, in d2, and CC, in d..

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Then C, is the pole of 1∞∞, C, of 1, and C, of ∞ ∞ 1, three poles of the cube.

d is the pole of ∞ 11, d, of 11, and d. of 11 ∞, poles of the rhombic dodecahedron.

Join Cd, Cad, and Cd, meeting in o.

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Then o is the pole

of the face of the octahedron whose symbol is 111.

To place on this octant six poles of the six-faced octahedron whose indices are 1,,

4.2.

In this case λ=36° 52′, λ=26° 34′, and λ=33° 41′. Graduate each of the lines Cd, C, C1d, Cd, Cod, and Cd, from 0 to 45; considering CC, CC, and CC, as chords of 90°, and making the three points C1, C2, C3 cach zero, as described in § 132.

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Let CF 36° 52'=C,F2=C1F=C2F=C2F=CF6
CH=26° 34'-CH=CH-CH=CH=C1H&
CG1=33° 41'=CgG2 = C1G3=C1 G1 = C'2 G 2 = C1 G &
Then E is the intersection of CF, CH,, CG,

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of C,H,, Cal
C2F2, CgGe
of CG, CF3, CgHc
of CF, CG, C'H.
of C1H1, ('G, ('F;
of CG, CH, ('F

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35

1 3

E1, E2, E3, E4, E., and E will be six poles of the six-faced octahedron whose indices are 1, 4, 2, and angular elements λ=36° 52′, P3-68° 12'. The lines of intersection are not shown in the plate.

(Fig. 43*, Plate IV.*) has marked on it the poles on the octant of a sphere of nearly all the forms of the cubical system which have been observed; all the faces whose poles lie in the same line having their poles on the sphere of projection on the same zone circle.

The angular and linear indices of every form are given in the following table.

Where P1, P2, and P, are the polar distances of each form from the three poles of the poles of the cube, C1, C2, and (',, 0, p, and the supplements of the angles of inclination over the edges of adjacent faces determined as in § 123, 124, 125,

and 126.

§ 124, 125, and 126 show how when these angles or any two

of them are determined from observation, the angular or linear elements can be determined from them.

The linear elements have hitherto been almost universally used as a concise means of expressing any form. Their disadvantages will be explained hereafter.

The angular elements are in reality more concise, because they can express the forms they represent to any degree of accuracy which can be derived from observation.

They have also this great advantage, that by the use of angles alone they can express the relations of any form to another without determining the linear elements at all.

P1

Thus in the following table p, for any form gives the inclination of the face for which it stands to that of the adjacent face of the cube in any combination of these two forms.

Faces of all the twenty-four faced trapezohedrons lie in the same zone Cod. Hence the value of P1 for any of these faces gives the inclination of that face to that of the cube in that

zone.

For instance (fig. 43*, Plate IV.*), m2 is the pole of a face of the twenty-four-faced trapezohedron, for which the value of P3-78° 54', λ=11° 19', linear elements 1, 5, 5; l, is the pole of another twenty-four-faced trapezohedron, where s3= 76° 22′, =14° 2′, linear elements 1, 4, 4.

For ma; P=15° 48'. And for l; P1=19° 28'.

Hence 54° 44'-15° 48′-Om2; 54° 44′—19° 28′=Ol¿; and 19° 28′-15° 48′ =m2l2.

Results procured by simple subtraction when the angular elements are used; but only found by retranslating the linear indices obtained from angular observations of the goniometer back again into angles, by trigonometrical formulæ.

Again, referring to (fig. 43*, Plate IV.*), we see that C1, Uз, Q3, H3, h2, E2, f1, N1, P1, H1 all lie in the same meridional

zone.

The values of p1 for each of these forms enable us to determine the distances of these poles from each other in the zone by simple subtraction of angles.

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Table of all the principal forms of the Cubical System.

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