The three classes of crystal geometric forms are each based on a basic shape, and a series of derived crystal forms are obtained through geometric transformations. For example, the tetrahedron and cube classes are based on a regular tetrahedron and a cube, respectively, and subclasses of the polygon (polyhedron) class are based on an equilateral triangle, a regular quadrilateral, a regular hexagon, and a parallelogram, for each of which a 3D rectangular coordinate system can be established. The vertex coordinates of the derived form are determined through geometric transformation operations, including plane-center extension, edge center extension, three-segment rotation of the edge line, and geometric rotation.
3.1 Derivation of a Plane-Center-Extended Form
A regular tetrahedron comprises six plane diagonals of a cube. Therefore, a tetrahedron can be considered as a half cube with half its vertices: A1, B2, C1, and D2 (Fig. 7). We take the center of a regular tetrahedron as the origin O to establish a 3D Cartesian coordinate system. Let the side length of the cube be 2r; then, the coordinates of the four vertices of a regular tetrahedron are (r, − r, r), (r, r, − r), (− r, r, r), and (− r, − r, − r). If N points are added on the plane-normal line, where N is the number of planes, then a trigonal tritetrahedron, a deltoid-dodecahedron, and a hexatetrahedron can be derived.
As an example, we derive a trigonal tritetrahedron as follows. A trigonal tritetrahedron has 12 planes, 18 sides, and 8 points. It is first extended outward from the central point of the four planes of a tetrahedron. In other words, the normal vectors OA2, OB1, OD1, and OC2 along the four planes of the tetrahedron extend outward by a distance S (Fig. 7b). The minimum value of s is PO, and the maximum is B1O.
A similar triangular relationship for Fig. 7b, with:
\({{PP}_{1}}^{2} + {P}_{1}{O}^{2} = {PO}^{2}\)
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(1)
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\({PO}^{2} + {O}_{1}{P}^{2} = {O}_{1}{O}^{2}\)
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(2)
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Gives:
\({{PP}_{1}}^{2} +{ P}_{1}{O}^{2} + {O}_{1}{P}^{2} = {O}_{1}{O}^{2}\)
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(3)
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Similar triangulation results in:
\(\frac{{B}_{1}{O}_{1}}{{O}_{1}O} = \frac{{O}_{1}P}{OP} = \frac{{P}_{1}O}{{P}_{1}P} = \sqrt{2}\) | (4) |
For which we set:
And:
\({P}_{1}O = \sqrt{2}\)s,\(OP = \sqrt{3}\)s,\({O}_{1}P = \sqrt{6}\)s | (6) |
Then, point P on the plane-centered outward extension vector OB1 is (s, s, s), with s taking the value [r/3, r].
The vertices of the resulting trigonal tritetrahedron are added to the positive tetrahedron by plane-centered outward extension, with four vertices along the plane-normal direction of (s, s, s), (- s, - s, s), (s, - s, - s), (- s, s, - s).
3.2 Derivation of an Extended Crystal Form with the Edge Center
In the case where the center of the geometric form is connected to the long edge center, N points are added at N long edges for outward extension of the derived form. We take the hexahedron as an example (Fig. 8). The hextetrahedron has 24 planes, 36 sides, and 14 points. It is externally extended by trigonal tritetrahedron long edge centers, extending outward by distance t in six directions along the X, Y, and Z axes (the central O of the trigonal tritetrahedron is connected to the central line of the six long sides), respectively. The smallest value of t is r, which yields a hexahedral form.
The 14 vertices of the hextetrahedron are based on the eight vertices of the trigonal tritetrahedron, with the outward extension adding six vertices: (t, 0, 0), (− t, 0, 0), (0, t, 0), (0, − t, 0), (0, 0, t), and (0, 0, − t).
3.3 Derivation of a Coplanar Form with Plane Center and Edge Center
The tetrahedron has 12 planes, 24 sides, and 14 points. The deltoid-dodecahedron can be derived from the hexatetrahedron. In the case where the points C1, Oy, P, and Oz in the hexatetrahedron (Fig. 8) are coplanar, the two adjacent triangles beyond the tetrahedral vertices are coplanar, so that their smallest constituent polygons are transformed into quadrilaterals, and the resultant monomorphism is a deltoid-dodecahedron.
Given that the three point coordinates are (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), an equation can be set as a(x − x1) + b(y − y1) + c(z − z1) = 0 (dot method). The coefficients a, b, and c are determined as follows:
a = (y2 − y1) (z3 − z1) − (z2 − z1) (y3 − y1), | (7) |
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b = (x3 − x1) (z2 − z1) − (x2 − x1) (z3 − z1), | (8) |
c = (x2 − x1) (y3 − y1) − (x3 − x1) (y2 − y1). | (9) |
That is, the general formula for the plane equation over three coordinate points is ax + by + cz + d = 0, where d = − (ax1 + by1 + cz1). As C1, Oy, Oz, and P are coplanar and have spatial point coordinates of C1 (− r, r, r), Oy (0, t, 0), and Oz (0, 0, t), we substitute the solution formula for the to-be-determined coefficients, which gives a = t2 − 2tr, b = c = − tr, d = t2r.
After substituting the coordinates of point P (s, s, s) again, the coplanar expression for the center of the plane and the center of the edges is as follows:
\(s = \frac{tr}{4r - t}\) | (10) |
at which point the hexatetrahedron is transformed into a deltoid-dodecahedron.
3.4 Derivation of a Form by Three-Segment Rotation of the Edge Line
Crystal form derivation by three-segment rotation of the edge line is a unique conversion method. On the basis of the original geometry, the edge of the geometry of a crystal form is divided into three parts, and will be one of the middle parts of external rotation for two points again, and at the same time a point is added to the center of each crystal plane, as explained in Fig. 9. These points are combined with the original vertices to form the geometry. According to the direction of rotation, a polyhedron can be divided into left and right types.
We take the tetartoid as an example. This crystal form has 12 planes, 24 sides, and 20 points, with each plane being a pentagon. Each facet of a tetrahedron is transformed from a central projection into three identical facets, and no facet is parallel to another ones.
In Fig. 9a, the vertices of the tetrahedron are A1, C1, B2, and D2, and the gravity center of the A1B2C1 plane is P. In Fig. 9b, S1 and S2 divide edge A1B2 into three segments, with Ox being the midpoint of edge A1B2. S1S2 is the result of rotation of the middle part of the tetrahedron with edge A1B2 divided into three segments. In Fig. 9c, T1 and T2 divide edge A1C1 into three segments, with O1 being the center point of edge A1C1. Where S1, S2, T1, P, and A1 are coplanar, they form the base pentagon of a tetartoid (Fig. 9b).
If we set a = OOx, b = OxS1, α = ∠S1OxA1, c = b × cos (45 − α), and d = b × sin (45 − α); then O1T1 = b, ∠T1O1A1 = α. The S1 coordinate is (a, − c, d); The S2 coordinate is (a, c, − d); The T1 coordinate is (c, − d, a); The P coordinate is (t, t, t); and The A1 coordinate is (s, − s, s).
\(t = \frac{a{d}^{2} -{ a}^{2}c}{{d}^{2} - 2ac + dc - ad + {c}^{2}}\)
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(11)
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\(s = \frac{a\times {d}^{2}-{a}^{2}\times c}{{d}^{2} - 2ac - dc + ad + {c}^{2}}\)
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(12)
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Based on the formula for the intersection of a line and surface, we have:
The tetartoid has 20 vertices. The four vertices of the tetrahedron have coordinates of (s, − s, s), (s, s, − s), (− s, s, s), and (− s, − s, − s). The four plane centroid coordinates are (t, t, t), (− t, − t, t), (t, − t, − t), and (− t, t, − t). Six sides are divided by 12 points with coordinates (a, c, − d), (a, − c, d), (− c, d, a), (c, − d, a), (− d, − a, − c), (d, − a, c), (− d, a, c), (d, a, − c), (− c, − d, − a), (c, d, − a), (− a, − c, − d), and (− a, c, d).
3.5 Geometric Transformation of the Regular Polygon Class
3.5.1 Regular Polygon Conversion
In the geometry of a regular polygon, the center of the polygon is the center of a circle and the length from the center to the vertex can be rotated at an angle with the radius of the center to the vertex (Fig. 10). That is, x = r × sin(α) and y = r × cos(α), where r is the radius of rotation and α is the angle of rotation. As an example, a positive triangle (equilateral triangle) can be used to derive a positive trigonal prism, a positive trigonal pyramid, and a positive trigonal dipyramid (Fig. 2), as follows: (1) An equilateral triangle (Fig. 10) is rotated 120° each time, taking the values of 0°, 120°, and 240°, respectively. The coordinates of the three vertices of the right triangle are A (r, 0), B (r, r), C (r, r). (2) Then, the vertex coordinates of the positive trigonal prism are (r, 0, h), (r, r, h), (r,0, − h), (r, r, − h), (r, r, − h); where h is the height. (3) The vertex coordinates of the positive trigonal pyramid are (r, 0, 0), (r, r, 0), (r, r,0), (0, 0, h). (4) The vertex coordinates of the positive trigonal dipyramid are (r, 0, 0), (r, r, 0), (r, r,0), (0, 0, h), (0, 0, − h).
Similarly, based on a square quadrilateral (square) with a rotation angle of 90°, we derive a tetragonal prism, a tetragonal pyramid, and a tetragonal dipyramid. The angle of rotation of a regular hexagon is 60°, and the regular hexagonal prism, regular hexagonal pyramid, and regular hexagonal dipyramid are derived.
3.5.2 Conversion of a Compound Regular Polygon
\(x = rsin\left(\alpha \right)\)
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(13)
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\(y = rcos\left(\alpha \right)\)
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(14)
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\(x = {r}_{1}sin\left(\alpha + 0.5\alpha \right)\)
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(15)
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\(y = {r}_{1}cos\left(\alpha + 0.5\alpha \right)\)
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(16)
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A compound regular polygon is a combination of two regular polygons of different side lengths, and the difference between the rotation angles of the two regular polygons is half the rotation angle. The vertices of a compound regular polygon are solved as follows:
We take a compound triangle as an example. As shown in Fig. 11, the vertices of the triangle are A, B, and C, the center of the circle is O, and the midpoint of AB is O1. If we set r1 = OA1 and r = OA, then the range of values of r1 is (0.5r, r).
The coordinates of the seven vertices of the ditrigonal pyramid (Fig. 2) are (r, 0, 0), (\(- \frac{1}{2}\)r1, \(\frac{\sqrt{3}}{2}\)r1, 0), (\(- \frac{1}{2}\)r, \(\frac{\sqrt{3}}{2}\)r, 0), (− r1, 0, 0), (\(- \frac{1}{2}\)r, \(- \frac{\sqrt{3}}{2}\)r, 0), (\(\frac{1}{2}\)r1, \(- \frac{\sqrt{3}}{2}\)r1, 0), (0, 0, h); where h is the height of the single cone. The coordinates of the vertices of the ditrigonal prism and the ditrigonal dipyramid can also be determined.
Similarly, for two regular quadrilaterals (squares) with different side lengths, the ditetragonal prism, the ditetragonal pyramid, and the ditetragonal dipyramid are derived.
Two regular hexagons with different side lengths are transformed into the dihexagonal prism, the dihexagonal pyramid, and the dihexagonal dipyramid.
3.6 Geometric Rotation for Crystal Form Transformation
Two geometric bodies can be rotated at a particular angle to form a new geometric body; for example, a trigonal trapezohedron, tetragonal trapezohedron, or hexagonal trapezohedron. A trigonal trapezohedron consists of two mirrored structures arranged in a trigonal pyramid rotated relative to each other at a particular angle. The vertexes of a trigonal trapezohedron comprise the intersection of the edges of the trigonal pyramid and the opposite plane and the vertices of the trigonal pyramid (Fig. 12). If the height of the trigonal pyramid is h and the radius of rotation of the bottom crosscut is r, then the length of both OD and OD1 is h, and the length of both OA and OA1 is r. The coordinates of vertex D are (0, 0, h), the coordinates of vertex D1 are (0, 0, −h), point A is (r, 0, 0), point B is (r, r, 0), and point A1 is (r × cos(α), r × sin(α), 0).
Points D, A, B, and A1in the trigonal trapezohedron are coplanar, and according to the above-mentioned plane equation, the plane equation for DAB is
\(-\frac{x}{r} - \sqrt{3}y - \frac{z}{ℎ} + 1 = 0\)
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(17)
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The linear (two-point) equation for D1A1 is
\(\frac{x}{r cos\alpha } = \frac{y}{r sin\alpha } = \frac{z + ℎ}{ℎ}\)
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(18)
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The intersection of the DAB plane and the line of D1A1 is A1, with the solution a s(\(\frac{2r{cos}\alpha }{{cos}\alpha +1+\sqrt{3}rsin\alpha }, \frac{2rsin\alpha }{{cos}\alpha +1+\sqrt{3}rsin\alpha }, \frac{2ℎ}{{cos}\alpha +1+\sqrt{3}rsin\alpha } - ℎ\)). The other vertices are solved in the same way.
3.7 Center-Line Extension
A cube is a solid figure composed of six squares of the same size, so it is also termed a regular hexahedron. The tetrahexahedron, rhomb-dodecahedron, pyritohedron, and diploid can be derived from the cube. The tetrahexahedron and rhomb-dodecahedron can be derived from the cube by the method of plane-center extension.
The transformation process from cube to pyritohedron is unique (Fig. 13). The center line of the cube plane is divided into three segments, with the middle segment being outwards so as to add two points to each surface on the basis of the original 8 vertices of the cube, adding 12 vertices in total. Here, the center line of the cube plane refers to the line connecting two opposite points of the crystal edge, and the adjacent center line cannot be parallel. Simultaneously, the distance of outward extension and the length of center-line segmentation are forced to satisfy a certain relation to ensure that the basic component polygon of the geometric form is a pentagon.
If we set the cube edge length as 2r, OOx = a, QOz = b, and TA1 = c, then we have OOz = a and A1P = r. In a pyritohedron, Q, A1, and Ox are coplanar, and triangles QTA1 and A1POx are similar triangles, according to the relationship of similar triangles:
\(\frac{{PO}_{x}}{{PA}_{1}} = \frac{{TA}_{1}}{TQ}\)
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(19)
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Put the above values into this formula, yields c = (a − r) (a − r)/r. From b = r – c, we obtain b = r − (a − r) (a − r)/r. Accordingly, the coordinates of vertex Q of the pyritohedron are (b, 0, a). In the same way, the coordinates of the other vertices can be obtained.
3.8 Other Geometric Transformations Subsection
In the parallelogram subclass (pedions, dihedrons, and parallelohedrons), the lower crystal family is simple and involves only translational geometric transformations, which are not described here.