Elongated triangular orthobicupola: Difference between revisions
Dedhert.Jr (talk | contribs) I don't get it: what's the point of adding the relation with this Archimidean solid? Is it because that their same construction but different polyhedrons, only? |
Dedhert.Jr (talk | contribs) symmetry and dihedral angle |
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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}} |
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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<math display="block"> \left(\frac{5\sqrt{2}}{3} + \frac{3\sqrt{3}}{2}\right)a^3 \approx 4.955a^3. </math> |
<math display="block"> \left(\frac{5\sqrt{2}}{3} + \frac{3\sqrt{3}}{2}\right)a^3 \approx 4.955a^3. </math> |
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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 </math>. 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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<math display="block"> \begin{align} |
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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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==Related polyhedra and honeycombs== |
==Related polyhedra and honeycombs== |
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| volume = 46 | issue = 3 | page = 177 |
| volume = 46 | issue = 3 | page = 177 |
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| url = https://go.gale.com/ps/i.do?id=GALE%7CA340298118 |
| 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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| 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> |
}}</ref> |
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Revision as of 14:43, 11 March 2024
Elongated triangular orthobicupola | |
---|---|
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 or cantellated triangular prism is one of the Johnson solids (J35). As the name suggests, it can be constructed by elongating a triangular orthobicupola by inserting a hexagonal prism between its two halves.
Construction
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 36th Johnson solid [3]
Properties
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 </math>. 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
The elongated triangular orthobicupola forms space-filling honeycombs with tetrahedra and square pyramids.[5]
References
- ^ 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".