四面体分析

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2019/08/16 12:20
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Tetrahedron TetrahedronFrame TetrahedronNet
TetrahedronProj1 TetrahedronProj2 TetrahedronProj3

polyhdron netThe regular tetrahedron, often simply called "the" tetrahedron, is the Platonic solid P_5 with four polyhedron vertices, six polyhedron edges, and four equivalent equilateral triangular faces, 4{3}. It is also uniform polyhedron U_1 and Wenninger model W_1. It is described by the Schläfli symbol {3,3} and the Wythoff symbol is 3|23. It is an isohedron, and a special case of the general tetrahedron and the isosceles tetrahedron.

The regular tetrahedron is implemented in the Wolfram Language as Tetrahedron[], and precomputed properties are available as PolyhedronData["Tetrahedron"].

The tetrahedron has 7 axes of symmetry: 4C_3 (axes connecting vertices with the centers of the opposite faces) and 3C_2 (the axes connecting the midpoints of opposite sides).

There are no other convex polyhedra other than the tetrahedron having four faces.

TetrahedronNets

The tetrahedron has two distinct nets (Buekenhout and Parker 1998). Questions of polyhedron coloring of the tetrahedron can be addressed using the Pólya enumeration theorem.

The surface area of the tetrahedron is simply four times the area of a single equilateral triangle face

 A=1/4sqrt(3)a^2,

(1)

so

 S=4A=sqrt(3)a^2.

(2)

The height of the regular tetrahedron is

 h=1/3sqrt(6)a

(3)

and the inradius and circumradius are

r = 1/(12)sqrt(6)a

(4)

R = 1/4sqrt(6)a,

(5)

where h=r+R as it must.

Since a tetrahedron is a pyramid with a triangular base, V=1/3A_bh, giving

 V=1/(12)sqrt(2)a^3.

(6)

The dihedral angle is

 alpha=tan^(-1)(2sqrt(2))=2sin^(-1)(1/3sqrt(3))=cos^(-1)(1/3) approx 70.53 degrees.

(7)

The solid angle Omega subtended from a vertex by the opposite face of a regular tetrahedron is given by

Omega = (Delta_i)/(R^2)=3cos^(-1)(1/3)-pi

(8)

= cos^(-1)((23)/(27)),

(9)

or approximately 0.55129 steradians.

The midradius of the tetrahedron is

rho = sqrt(r^2+d^2)=sqrt(1/8)a=1/4sqrt(2)a

(10)

 approx 0.35355a.

(11)

Plugging in for the polyhedron vertices gives

 (1/3sqrt(3)a,0,0),(-1/6sqrt(3)a,+/-1/2a,0), and (0,0,1/3sqrt(6)a).

(12)

TetrahedronAndDual

The dual polyhedron of an tetrahedron with unit edge lengths is another oppositely oriented tetrahedron with unit edge lengths.

Origami tetrahedron

The figure above shows an origami tetrahedron constructed from a single sheet of paper (Kasahara and Takahama 1987, pp. 56-57).

TetrahedralGraph TetrahedralCirculantGraph

It is the prototype of the tetrahedral group T_d. The connectivity of the vertices is given by the tetrahedral graph, equivalent to the circulant graph Ci_(1,2,3)(4) and the complete graph K_4.

tettet

The tetrahedron is its own dual polyhedron, and therefore the centers of the faces of a tetrahedron form another tetrahedron (Steinhaus 1999, p. 201). The tetrahedron is the only simple polyhedron with no polyhedron diagonals, and it cannot be stellated. If a regular tetrahedron is cut by six planes, each passing through an edge and bisecting the opposite edge, it is sliced into 24 pieces (Gardner 1984, pp. 190 and 192; Langman 1951).

Alexander Graham Bell was a proponent of use of the tetrahedron in framework structures, including kites (Bell 1903; Lesage 1956, Gardner 1984, pp. 184-185). The opposite edges of a regular tetrahedron are perpendicular, and so can form a universal coupling if hinged appropriately. Eight regular tetrahedra can be placed in a ring which rotates freely, and the number can be reduced to six for squashed irregular tetrahedra (Wells 1975, 1991).

TetrahedronLengths

Let a tetrahedron be length a on a side, and let its base lie in the plane z=0 with one vertex lying along the positive x-axis. The polyhedron vertices of this tetrahedron are then located at (x, 0, 0), (-d+/-a/2, 0), and (0, 0, h), where

 x=(a/2)/(cos(pi/6))=1/3sqrt(3)a.

(13)

d is then

 d=sqrt(x^2-(1/2a)^2)=1/6sqrt(3)a.

(14)

This gives the area of the base as

 A=1/2a(R+x)=1/4sqrt(3)a^2.

(15)

The height is

 h=sqrt(a^2-x^2)=1/3sqrt(6)a.

(16)

The circumradius R is found from

 x^2+(h-R)^2=R^2

(17)

 x^2+h^2-2hR+R^2=R^2.

(18)

Solving gives

R = (x^2+h^2)/(2h)

(19)

= 1/4sqrt(6)a

(20)

 approx 0.61237a.

(21)

The inradius r is

r=h-R = 1/(12)sqrt(6)a

(22)

 approx 0.20412a,

(23)

which is also

r = 1/4h

(24)

= 1/3R.

(25)

The angle between the bottom plane and center is then given by

phi = tan^(-1)(r/x)

(26)

= tan^(-1)(1/4sqrt(2))

(27)

= cot^(-1)(2sqrt(2))

(28)

 approx 19.47 degrees.

(29)

Given a tetrahedron of edge length a situated with vertical apex and with the origin of coordinate system at the geometric centroid of the vertices, the four polyhedron vertices are located at (x,0,-r)(-d,+/-a/2,-r)(0,0,R), with, as shown above

x = 1/3sqrt(3)a

(30)

r = 1/(12)sqrt(6)a

(31)

R = 1/4sqrt(6)a

(32)

d = 1/6sqrt(3)a.

(33)

TetrahedronCube TetrahedronInequality

The vertices of a tetrahedron of side length sqrt(2) can also be given by a particularly simple form when the vertices are taken as corners of a cube (Gardner 1984, pp. 192-194). One such tetrahedron for a cube of side length 1 gives the tetrahedron of side length sqrt(2)having vertices (0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0), and satisfies the inequalities

x+y+z <= 2

(34)

x-y-z <= 0

(35)

-x+y-z <= 0

(36)

-x-y+z <= 0.

(37)

The following table gives polyhedra which can be constructed by augmentation of a tetrahedron by pyramids of given heights h.

h (r+h)/h result
1/(15)sqrt(6) 7/5 triakis tetrahedron
1/6sqrt(6) 2 cube
1/3sqrt(6) 3 9-faced star deltahedron

ThreadedTetrahedron

Connecting opposite pairs of edges with equally spaced lines gives a configuration like that shown above which divides the tetrahedron into eight regions: four open and four closed (Steinhaus 1999, p. 246).

Michigan artist David Barr designed his "Four Corners Project" in 1976. It is an Earth-sized regular tetrahedron that spans the planet, with just the tips of its four corners protruding. These visible portions are four-inch tetrahedra, which protrude from the globe at Easter Island, Greenland, New Guinea, and the Kalahari Desert. Barr traveled to these locations and was able to permanently install the four aligned marble tetrahedra between 1981 and 1985 (G. Hart, pers. comm.; Arlinghaus and Nystuen 1986).

TetrahedronSquare1 TetrahedronSquare2

By slicing a tetrahedron as shown above, a square can be obtained. This cut divides the tetrahedron into two congruent solids rotated by 90 degrees. The projection of a tetrahedron can be an equilateral triangle or a square (Steinhaus 1999, pp. 191-192).

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