Elongated triangular orthobicupola: Difference between revisions
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{{Short description|Johnson solid with 20 faces}} |
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{{Infobox polyhedron |
{{Infobox polyhedron |
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|image= |
| image = Elongated triangular orthobicupola.png |
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|type=[[Johnson solid|Johnson]]<br>[[pentagonal orthobirotunda|''J'' |
| type = [[Johnson solid|Johnson]]<br>{{math|[[pentagonal orthobirotunda|''J''{{sub|34}}]] – '''''J''{{sub|35}}''' – [[elongated triangular gyrobicupola|''J''{{sub|36}}]]}} |
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|faces= |
| faces = 8 [[triangle]]s<br>12 [[Square (geometry)|square]]s |
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|edges=36 |
| edges = 36 |
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|vertices=18 |
| vertices = 18 |
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|symmetry= |
| symmetry = <math> D_{3h} </math> |
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|vertex_config= |
| vertex_config = <math> \begin{align} |
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&6 \times (3 \times 4 \times 3 \times 4) + \\ |
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|dual=- |
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&12 \times (3 \times 4^3) |
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\end{align} </math> |
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}} |
}} |
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In [[geometry]], the '''elongated triangular orthobicupola''' or '''cantellated triangular prism''' is one of the [[Johnson solid]]s (''J''<sub>35</sub>). As the name suggests, it can be constructed by elongating a [[triangular orthobicupola]] (''J''<sub>27</sub>) by inserting a [[hexagon]]al [[prism (geometry)|prism]] between its two halves. The resulting solid is superficially similar to the [[rhombicuboctahedron]] (one of the [[Archimedean solid]]s), with the difference that it has threefold [[rotational symmetry]] about its axis instead of fourfold symmetry. |
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In [[geometry]], the '''elongated triangular orthobicupola''' is a polyhedron constructed by attaching two regular [[triangular cupola]] into the base of a regular [[hexagonal prism]]. It is an example of [[Johnson solid]]. |
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{{Johnson solid}} |
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== |
== Construction == |
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The elongated triangular orthobicupola can be constructed from a [[hexagonal prism]] by attaching two regular [[triangular cupola]]e onto its base, covering its hexagonal faces.{{r|rajwade}} This construction process known as [[Elongation (geometry)|elongation]], giving the resulting polyhedron has 8 [[equilateral triangle]]s and 12 squares.{{r|berman}} A [[Convex set|convex]] polyhedron in which all faces are [[Regular polygon|regular]] is [[Johnson solid]], and the elongated triangular orthobicupola is one among them, enumerated as 35th Johnson solid <math> J_{35} </math>.{{r|francis}} |
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The volume of ''J''<sub>35</sub> can be calculated as follows: |
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== Properties == |
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''J''<sub>35</sub> consists of 2 cupolae and hexagonal prism. |
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An elongated triangular orthobicupola with a given edge length <math> a </math> has a surface area, by adding the area of all regular faces:{{r|berman}} |
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<math display="block"> \left(12 + 2\sqrt{3}\right)a^2 \approx 15.464a^2. </math> |
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Its volume can be calculated by cutting it off into two triangular cupolae and a hexagonal prism with regular faces, and then adding their volumes up:{{r|berman}} |
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It has the same [[Point groups in three dimensions|three-dimensional symmetry groups]] as the [[triangular orthobicupola]], the dihedral group <math> D_{3h} </math> of order 12. Its [[dihedral angle]] can be calculated by adding the angle of the triangular cupola and hexagonal prism. The dihedral angle of a hexagonal prism between two adjacent squares is the [[internal angle]] of a regular hexagon <math> 120^\circ = 2\pi/3</math>, and that between its base and square face is <math> \pi/2 = 90^\circ </math>. The dihedral angle of a regular triangular cupola between each triangle and the hexagon is approximately <math> 70.5^\circ </math>, that between each square and the hexagon is <math> 54.7^\circ </math>, and that between square and triangle is <math> 125.3^\circ </math>. The dihedral angle of an elongated triangular orthobicupola between the triangle-to-square and square-to-square, on the edge where the triangular cupola and the prism is attached, is respectively:{{r|johnson}} |
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The two cupolae makes 1 cuboctahedron = 8 tetrahedra + 6 half-octahedra. |
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<math display="block"> \begin{align} |
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1 octahedron = 4 tetrahedra, so total we have 20 tetrahedra. |
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\frac{\pi}{2} + 70.5^\circ &\approx 160.5^\circ, \\ |
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\frac{\pi}{2} + 54.7^\circ &\approx 144.7^\circ. |
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\end{align} </math> |
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What is the volume of a tetrahedron? |
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Construct a tetrahedron having vertices in common with alternate vertices of a |
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cube (of side <math>\frac{1}{\sqrt{2}}</math>, if tetrahedron has unit edges). The 4 triangular |
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pyramids left if the tetrahedron is removed from the cube form half an |
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octahedron = 2 tetrahedra. So |
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<math>V_{tetrahedron} = \frac{1}{3} V_{cube} = \frac{1}{3} \frac{1}{{\sqrt{2}}^3} = \frac{\sqrt{2}}{12}</math> |
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The hexagonal prism is more straightforward. The hexagon has area <math>6 \frac{\sqrt{3}}{4}</math>, so |
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{{reflist|refs= |
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<ref name="berman">{{cite journal |
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<math>V_{prism} = \frac{3 \sqrt{3}}{2}</math> |
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| last = Berman | first = Martin |
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| year = 1971 |
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| title = Regular-faced convex polyhedra |
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| journal = Journal of the Franklin Institute |
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| volume = 291 |
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| issue = 5 |
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| pages = 329–352 |
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| doi = 10.1016/0016-0032(71)90071-8 |
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| mr = 290245 |
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}}</ref> |
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<ref name="francis">{{cite journal |
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Finally |
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| last = Francis | first = Darryl |
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| title = Johnson solids & their acronyms |
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| journal = Word Ways |
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| date = August 2013 |
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| volume = 46 | issue = 3 | page = 177 |
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| url = https://go.gale.com/ps/i.do?id=GALE%7CA340298118 |
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}}</ref> |
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<ref name="johnson">{{cite journal |
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<math>V_{J_{35}} = 20 V_{tetrahedron} + V_{prism} = |
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| last = Johnson | first = Norman W. | authorlink = Norman W. Johnson |
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| year = 1966 |
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| title = Convex polyhedra with regular faces |
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| journal = [[Canadian Journal of Mathematics]] |
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| volume = 18 |
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| pages = 169–200 |
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| doi = 10.4153/cjm-1966-021-8 |
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| mr = 0185507 |
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| s2cid = 122006114 |
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| zbl = 0132.14603| doi-access = free |
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}}</ref> |
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<ref name="rajwade">{{cite book |
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numerical value: |
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| last = Rajwade | first = A. R. |
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| title = Convex Polyhedra with Regularity Conditions and Hilbert's Third Problem |
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| series = Texts and Readings in Mathematics |
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| year = 2001 |
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| url = https://books.google.com/books?id=afJdDwAAQBAJ&pg=PA84 |
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| page = 84–89 |
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| publisher = Hindustan Book Agency |
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| isbn = 978-93-86279-06-4 |
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| doi = 10.1007/978-93-86279-06-4 |
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}}</ref> |
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}} |
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<math>V_{J_{35}} = 4.9550988153084743549606507192748</math> |
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{{Reflist}} |
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==External links== |
==External links== |
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* {{MathWorld2|title2=Johnson solid|urlname2=JohnsonSolid| urlname=ElongatedTriangularOrthobicupola | title=Elongated triangular orthobicupola}} |
* {{MathWorld2|title2=Johnson solid|urlname2=JohnsonSolid| urlname=ElongatedTriangularOrthobicupola | title=Elongated triangular orthobicupola}} |
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{{Johnson solids navigator}} |
{{Johnson solids navigator}} |
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{{Polyhedron-stub}} |
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[[Category:Johnson solids]] |
[[Category:Johnson solids]] |
Latest revision as of 02:15, 1 April 2024
Elongated triangular orthobicupola | |
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Type | Johnson J34 – J35 – J36 |
Faces | 8 triangles 12 squares |
Edges | 36 |
Vertices | 18 |
Vertex configuration | |
Symmetry group | |
Properties | convex |
Net | |
In geometry, the elongated triangular orthobicupola is a polyhedron constructed by attaching two regular triangular cupola into the base of a regular hexagonal prism. It is an example of Johnson solid.
Construction
[edit]The elongated triangular orthobicupola can be constructed from a hexagonal prism by attaching two regular triangular cupolae onto its base, covering its hexagonal faces.[1] This construction process known as elongation, giving the resulting polyhedron has 8 equilateral triangles and 12 squares.[2] A convex polyhedron in which all faces are regular is Johnson solid, and the elongated triangular orthobicupola is one among them, enumerated as 35th Johnson solid .[3]
Properties
[edit]An elongated triangular orthobicupola with a given edge length has a surface area, by adding the area of all regular faces:[2] Its volume can be calculated by cutting it off into two triangular cupolae and a hexagonal prism with regular faces, and then adding their volumes up:[2]
It has the same three-dimensional symmetry groups as the triangular orthobicupola, the dihedral group of order 12. Its dihedral angle can be calculated by adding the angle of the triangular cupola and hexagonal prism. The dihedral angle of a hexagonal prism between two adjacent squares is the internal angle of a regular hexagon , and that between its base and square face is . The dihedral angle of a regular triangular cupola between each triangle and the hexagon is approximately , that between each square and the hexagon is , and that between square and triangle is . The dihedral angle of an elongated triangular orthobicupola between the triangle-to-square and square-to-square, on the edge where the triangular cupola and the prism is attached, is respectively:[4]
Related polyhedra and honeycombs
[edit]The elongated triangular orthobicupola forms space-filling honeycombs with tetrahedra and square pyramids.[5]
References
[edit]- ^ Rajwade, A. R. (2001). Convex Polyhedra with Regularity Conditions and Hilbert's Third Problem. Texts and Readings in Mathematics. Hindustan Book Agency. p. 84–89. doi:10.1007/978-93-86279-06-4. ISBN 978-93-86279-06-4.
- ^ a b c Berman, Martin (1971). "Regular-faced convex polyhedra". Journal of the Franklin Institute. 291 (5): 329–352. doi:10.1016/0016-0032(71)90071-8. MR 0290245.
- ^ Francis, Darryl (August 2013). "Johnson solids & their acronyms". Word Ways. 46 (3): 177.
- ^ Johnson, Norman W. (1966). "Convex polyhedra with regular faces". Canadian Journal of Mathematics. 18: 169–200. doi:10.4153/cjm-1966-021-8. MR 0185507. S2CID 122006114. Zbl 0132.14603.
- ^ "J35 honeycomb".