3D view of a hexakis octahedron (
animation )
The disdyakis dodecahedron (from Greek ἑξάκις hexakis "six" and octahedron Disdyakisdodekaeder ( Greek δίς dis "twice", δυάκις dyakis "twice" and dodecahedron convex polyhedron , which consists of 48 irregular triangles composed and belongs to the Catalan bodies . It is dual to the truncated cuboctahedron and has 26 vertices and 72 edges.
Emergence Rhombic dodecahedron as a base If pyramids with the flank lengths and are placed on the 12 boundary surfaces of a rhombic dodecahedron (edge length ) , a hexakis octahedron is created, provided the following condition is met:
a
{\ displaystyle a}
b
{\ displaystyle b}
c
(
<
b
)
{\ displaystyle c \, (<b)}
a
3
6th
<
b
<
2
9
a
15th
{\ displaystyle {\ tfrac {a} {3}} {\ sqrt {6}} <b <{\ tfrac {2} {9}} a {\ sqrt {15}}}
For the above-mentioned minimum value of the pyramids on top have the height 0, so that only the rhombic dodecahedron with the edge length remains.
b
{\ displaystyle b}
a
{\ displaystyle a}
The special hexakis octahedron with equal surface angles at the edges and arises when is.
a
{\ displaystyle a}
b
{\ displaystyle b}
b
=
2
a
(
2
-
1
)
{\ displaystyle b = 2a \, ({\ sqrt {2}} - 1)}
Assumes the aforementioned maximum value, the hexakis octahedron degenerates into a deltoidal icosity tetrahedron with the edge lengths and .
b
{\ displaystyle b}
a
{\ displaystyle a}
b
{\ displaystyle b}
If the maximum value is exceeded , the polyhedron is no longer convex.
b
{\ displaystyle b}
Truncated cuboctahedron as a base
Construction of the triangle on the truncated cuboctahedron
By connecting the centers of three edges that meet in every corner of the truncated cuboctahedron, a triangle is created whose circumference is at the same time the incircle of the triangle, the boundary surface of the hexakisoctahedron. In this special type, all face angles are the same (≈ 155 °) and there is a uniform edge sphere radius .
Let d be the edge length of the truncated cuboctahedron, then the resulting side lengths of the triangle are given by
a
=
2
7th
d
60
+
6th
2
{\ displaystyle a = \, {\ frac {2} {7}} \, d \, {\ sqrt {60 + 6 {\ sqrt {2}}}}}
b
=
3
7th
d
12
+
6th
2
{\ displaystyle b = \, {\ frac {3} {7}} \, d \, {\ sqrt {12 + 6 {\ sqrt {2}}}}}
c
=
2
7th
d
30th
-
3
2
{\ displaystyle c = \, {\ frac {2} {7}} \, d \, {\ sqrt {30-3 {\ sqrt {2}}}}}
Formulas In the following denote the longest edge of the hexakis octahedron ( ).
a
{\ displaystyle a}
a
>
b
>
c
{\ displaystyle a> b> c}
Regular
The basis is the truncated cuboctahedron (dual Archimedean solid).
Sizes of a hexakis octahedron with edge length a
volume
V
=
a
3
28
6th
(
986
+
607
2
)
{\ displaystyle V = {\ frac {a ^ {3}} {28}} \, {\ sqrt {6 \, (986 + 607 {\ sqrt {2}})}}}
Surface area
A.
O
=
3
7th
a
2
543
+
176
2
{\ displaystyle A_ {O} = {\ frac {3} {7}} \, a ^ {2} \, {\ sqrt {543 + 176 {\ sqrt {2}}}}}
Inc sphere radius
ρ
=
a
2
402
+
195
2
194
{\ displaystyle \ rho = {\ frac {a} {2}} \, {\ sqrt {\ frac {402 + 195 {\ sqrt {2}}} {194}}}}
Edge ball radius
r
=
a
4th
(
1
+
2
2
)
{\ displaystyle r = {\ frac {a} {4}} \, (1 + 2 {\ sqrt {2}})}
Face angle
cos
α
=
-
1
97
(
71
+
12
2
)
{\ displaystyle \ cos \, \ alpha = - {\ frac {1} {97}} \, (71 + 12 {\ sqrt {2}})}
Sizes of the triangle
Area
A.
=
a
2
112
543
+
176
2
{\ displaystyle A = {\ frac {a ^ {2}} {112}} \, {\ sqrt {543 + 176 {\ sqrt {2}}}}}
2. Side length
b
=
3
14th
a
(
1
+
2
2
)
{\ displaystyle b = \, {\ frac {3} {14}} \, a \, (1 + 2 {\ sqrt {2}})}
3. Side length
c
=
a
14th
(
10
-
2
)
{\ displaystyle c = \, {\ frac {a} {14}} \, (10 - {\ sqrt {2}})}
1. Angle
cos
α
=
1
12
(
2
-
2
)
{\ displaystyle \ cos \, \ alpha = \, {\ frac {1} {12}} \, (2 - {\ sqrt {2}})}
2. Angle
cos
β
=
1
8th
(
6th
-
2
)
{\ displaystyle \ cos \, \ beta = \, {\ frac {1} {8}} \, (6 - {\ sqrt {2}})}
3. Angle
cos
γ
=
1
12
(
1
+
6th
2
)
{\ displaystyle \ cos \, \ gamma = \, {\ frac {1} {12}} \, (1 + 6 {\ sqrt {2}})}
Rhombic The basis is the rhombic dodecahedron (edge length ).
a
{\ displaystyle a}
General
Sizes of a hexakis octahedron with edge lengths a , b
volume
V
=
8th
9
a
2
3
(
2
a
+
6th
b
2
-
4th
a
2
)
{\ displaystyle V = {\ frac {8} {9}} a ^ {2} {\ sqrt {3}} \ left (2a + {\ sqrt {6b ^ {2} -4a ^ {2}}} \ right )}
Surface area
A.
O
=
8th
a
9
b
2
-
4th
a
2
{\ displaystyle A_ {O} = 8a \, {\ sqrt {9b ^ {2} -4a ^ {2}}}}
Pyramid height
k
=
1
3
9
b
2
-
6th
a
2
{\ displaystyle k = {\ frac {1} {3}} {\ sqrt {9b ^ {2} -6a ^ {2}}}}
Inc sphere radius
ρ
=
a
(
2
a
+
6th
b
2
-
4th
a
2
)
27
b
2
-
12
a
2
{\ displaystyle \ rho \, = {\ frac {a \, (2a + {\ sqrt {6b ^ {2} -4a ^ {2}}})} {\ sqrt {27b ^ {2} -12a ^ {2 }}}}}
Dihedral angle a )
cos
α
1
=
9
b
2
-
2
a
(
4th
a
+
3
6th
b
2
-
a
2
)
18th
b
2
-
8th
a
2
{\ displaystyle \ cos \, \ alpha _ {1} = {\ frac {9b ^ {2} -2a \, (4a + 3 {\ sqrt {6b ^ {2} -a ^ {2}}})} {18b ^ {2} -8a ^ {2}}}}
Dihedral angle b )
cos
α
2
=
3
b
2
-
4th
a
2
9
b
2
-
4th
a
2
{\ displaystyle \ cos \, \ alpha _ {2} = {\ frac {3b ^ {2} -4a ^ {2}} {9b ^ {2} -4a ^ {2}}}}
Dihedral angle c )
cos
α
3
=
3
b
2
4th
a
2
-
9
b
2
{\ displaystyle \ cos \, \ alpha _ {3} = {\ frac {3b ^ {2}} {4a ^ {2} -9b ^ {2}}}}
Sizes of the triangle
Area
A.
=
a
6th
9
b
2
-
4th
a
2
{\ displaystyle A = {\ frac {a} {6}} \, {\ sqrt {9b ^ {2} -4a ^ {2}}}}
3. Side length
c
=
1
3
9
b
2
-
3
a
2
{\ displaystyle c = {\ frac {1} {3}} {\ sqrt {9b ^ {2} -3a ^ {2}}}}
1. Angle
sin
α
=
a
b
9
b
2
-
4th
a
2
9
b
2
-
3
a
2
{\ displaystyle \ sin \, \ alpha = {\ frac {a} {b}} \, {\ sqrt {\ frac {9b ^ {2} -4a ^ {2}} {9b ^ {2} -3a ^ {2}}}}}
2. angle
sin
β
=
9
b
2
-
4th
a
2
9
b
2
-
3
a
2
{\ displaystyle \ sin \, \ beta = {\ sqrt {\ frac {9b ^ {2} -4a ^ {2}} {9b ^ {2} -3a ^ {2}}}}}
3. Angle
sin
γ
=
9
b
2
-
4th
a
2
3
b
{\ displaystyle \ sin \, \ gamma = {\ frac {\ sqrt {9b ^ {2} -4a ^ {2}}} {3b}}}
Special
Sizes of a hexakis octahedron with edge length a
volume
V
=
32
9
a
3
3
(
2
-
2
)
{\ displaystyle V = {\ frac {32} {9}} a ^ {3} {\ sqrt {3}} \, (2 - {\ sqrt {2}})}
Surface area
A.
O
=
16
a
2
26th
-
18th
2
{\ displaystyle A_ {O} = 16a ^ {2} {\ sqrt {26-18 {\ sqrt {2}}}}}
Inc sphere radius
ρ
=
2
a
3
+
2
21st
{\ displaystyle \ rho = 2a \, {\ sqrt {\ frac {3 + {\ sqrt {2}}} {21}}}}
Face angle a, b )
cos
α
1
,
2
=
-
1
7th
(
2
+
3
2
)
{\ displaystyle \ cos \, \ alpha _ {1, \, 2} = - {\ frac {1} {7}} \, (2 + 3 {\ sqrt {2}})}
Face angle c )
cos
α
3
=
-
3
14th
(
3
+
2
)
{\ displaystyle \ cos \, \ alpha _ {3} = - {\ frac {3} {14}} \, (3 + {\ sqrt {2}})}
Sizes of the triangle
Area
A.
=
a
2
3
26th
-
18th
2
{\ displaystyle A = {\ frac {a ^ {2}} {3}} {\ sqrt {26-18 {\ sqrt {2}}}}}
2. Side length
b
=
2
a
(
2
-
1
)
{\ displaystyle b = 2a \, ({\ sqrt {2}} - 1)}
3. Side length
c
=
a
35
-
24
2
3
{\ displaystyle c = a \, {\ sqrt {\ frac {35-24 {\ sqrt {2}}} {3}}}}
1. Angle
sin
α
=
2
57
+
37
2
438
{\ displaystyle \ sin \, \ alpha = \, 2 \, {\ sqrt {\ frac {57 + 37 {\ sqrt {2}}} {438}}}}
2. Angle
sin
β
=
4th
23
-
3
2
438
{\ displaystyle \ sin \, \ beta = \, 4 \, {\ sqrt {\ frac {23-3 {\ sqrt {2}}} {438}}}}
3. Angle
sin
γ
=
1
3
6th
-
2
2
{\ displaystyle \ sin \, \ gamma = \, {\ frac {1} {3}} {\ sqrt {6-2 {\ sqrt {2}}}}}
Occurrence
The hexakisoctahedron occurs naturally as a crystal form . It is the general planar shape of the hexakisoctahedral crystal class m 3 m.
The hexakis octahedron is also used as a dice (D48).
Web links
<img src="https://de.wikipedia.org//de.wikipedia.org/wiki/Special:CentralAutoLogin/start?type=1x1" alt="" title="" width="1" height="1" style="border: none; position: absolute;">
