Talk:Truncated octahedron

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Polyhedra

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I'm too primitive to get it[edit]

The truncated octahedron is the only tridimensional primitive parallelohedra.

(I'll correct that to –hedron.) Does tridimensional here mean 'in 3space' or something else? What does primitive mean? —Tamfang (talk) 05:56, 21 January 2011 (UTC)[reply]

Definitely means 3-space. I found primary parallelohedron [1]. Tom Ruen (talk) 18:19, 21 January 2011 (UTC)[reply]

suggestion: dotted lines for invisible edges?![edit]

may I suggest to use dotted lines for the invisible edges in the drawings of the projections? not using them leads to puzzling effects: so e.g. in the column "edge 4-6" ot the table of "orthogonal projections" you think to see pentagons as surface-polygons (in the first row...) thanx! --HilmarHansWerner (talk) 06:18, 2 January 2017 (UTC)[reply]

Additional information[edit]

compared to similar articles (eg Octahedron) there is some missing basic information about dimensions...

where side length = a

Circumsphere (vertex) radius = 0.5 * sqrt(10) * a ~= 1.5811a

Midsphere (mid-edge) radius = 1.5 * a == 1.5a

Insphere(1) (square face) radius = sqrt(2) * a ~= 1.4142a

Insphere(2) (hexagonal face) radius = 0.5 * sqrt(6) * a ~= 1.2247a

Volume = (1/3) * sqrt(2) * (3a)^3 - sqrt(2) * a^3

also it only requires 6 (not 12) rectangles, of sides [a, 3a] with the short edges along where octagons meet, passing through the center, each corner mapping a vertex.

I'd edit them in, but templates and formatting aren't my forte, so any help is appreciated
74.214.226.120 (talk) 17:39, 19 January 2017 (UTC)[reply]