All the important parameters of a truncated octahedron such as normal distances & solid angles subtended by the faces, inner radius, outer radius, mean radius, surface area & volume have been calculated by using HCR's formula for regular polyhedron. This formula is a generalized dimensional formula which is applied on any of the five platonic solids i.e. reguler tetrahedron, regular hexahedron (cube), regular octahedron, regular dodecahedron & regular icosahedron to calculate their important parameters. It can be used in analysis, designing & modelling of regular n-polyhedrons.

1. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
Mr Harish Chandra Rajpoot
M.M.M. University of Technology, Gorakhpur-273010 (UP), India Dec, 2014
Introduction: A truncated octahedron is a solid which has 6 congruent square & 8 congruent regular
hexagonal faces each having equal edge length. It is obtained by truncating a regular octahedron (having 8
congruent faces each as an equilateral triangle) at the vertices to generate 6 square & 8 regular hexagonal
faces of equal edge length. For calculating all the parameters of a truncated octahedron, we would use the
equations of right pyramid & regular octahedron. When a regular octahedron is truncated at the vertex, a right
pyramid, with base as a square & certain normal height, is obtained. Since, a regular octahedron has 6 vertices
hence we obtain 6 truncated off congruent right pyramids each with a square base.
Truncation of a regular octahedron: For ease of calculations, let there be a regular octahedron with edge
length & its centre at the point C. Now it is truncated at all 6 vertices to obtain a truncated octahedron.
Thus each of the congruent equilateral triangular faces with edge length is changed into a regular
hexagonal face with edge length (see figure 2) & we obtain 6 truncated off congruent right pyramids with
base as a square corresponding to 6 vertices of the parent solid. (See figure 1 showing a right pyramid with
square base & normal height being truncated from the regular octahedron).
Figure 2: Each of the congruent equilateral triangular
faces with edge length 𝟑𝒂 of a regular octahedron is
changed into a regular hexagonal face with edge
length 𝒂 by truncation of vertices.
Figure 1: A right pyramid with base as a square with side length
𝒂 & normal height h is truncated off from a regular octahedron
with edge length 𝟑𝒂

2. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
Analysis of Truncated octahedron by using equations of right pyramid & regular octahedron
Now consider any of 6 truncated off congruent right pyramids having base as a square ABDE with side length
, normal height & an angle between any two consecutive lateral edges (see figure 3 below)
Normal height ( ) of truncated off right pyramid: We know that the normal height of any right pyramid
with regular polygonal base is given as
√
√ √(√ ) ( ) ( )
Figure 3: Normal distance ( ) of square faces is always greater than the normal distance ( ) of regular hexagonal
faces measured from the centre C of any truncated octahedron.
⇒ √
√
√
( )
Volume ( ) of truncated off right pyramid: We know that the volume of a right pyramid is given as
( ( )) ( )

3. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
( * ( *
√
√ √
√
( )
Normal distance ( ) of square faces from the centre of truncated octahedron: The normal distance
( ) of each of the square faces from the centre C of truncated octahedron is given as
( )
⇒ ( ( ) )
⇒
( )
√ √
( )
( )
√ √
√
⇒ √ ( )
It’s clear that all 6 congruent square faces are at an equal normal distance from the centre of any
truncated octahedron.
Solid angle ( ) subtended by each of the square faces at the centre of truncated octahedron: we
know that the solid angle ( ) subtended by any regular polygon with each side of length at any point lying
at a distance H on the vertical axis passing through the centre of plane is given by “HCR’s Theory of Polygon”
as follows
(
√
)
Hence, by substituting the corresponding values in the above expression, we get
(
( √ )
√ ( √ )
)
(
√
√
√
) ( *
( * ( )
Normal distance ( ) of regular hexagonal faces from the centre of truncated octahedron: The
normal distance ( ) of each of the regular hexagonal faces from the centre C of truncated octahedron is
given as

4. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
( )
⇒ ( ( ) )
⇒
( )
√
√ ( )
⇒ √ ( )
It’s clear that all 8 congruent regular hexagonal faces are at an equal normal distance from the centre of
any truncated octahedron.
It’s also clear from eq(III) & (V) i.e. the normal distance ( ) of square faces is greater than the
normal distance ( ) of regular hexagonal faces from the centre of truncated octahedron i.e. hexagonal faces
are much closer to the centre as compared to the square faces in any truncated octahedron.
Solid angle ( ) subtended by each of the regular hexagonal faces at the centre of truncated
octahedron: we know that the solid angle ( ) subtended by any regular polygon is given by “HCR’s Theory of
Polygon” as follows
(
√
)
Hence, by substituting the corresponding values in the above expression, we get
(
( √ )
√ ( √ )
)
(
√
√ (√ ) )
( √ ) (
√
*
(
√
* ( )
It’s clear that the solid angle subtended by each of the regular hexagonal faces is greater than the solid angle
subtended by each of the square faces at the centre of any truncated octahedron.
Important parameters of a truncated octahedron:
1. Inner (inscribed) radius( ): It is the radius of the largest sphere inscribed (trapped inside) by a
truncated octahedron. The largest inscribed sphere always touches all 8 congruent regular hexagonal
faces but does not touch any of 6 square faces at all since all 8 hexagonal faces are closer to the
centre as compared to all 6 square faces. Thus, inner radius is always equal to the normal distance
( ) of hexagonal faces from the centre & is given from the eq(V) as follows

5. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
√
Hence, the volume of inscribed sphere is given as
( ) ( √ )
2. Outer (circumscribed) radius( ): It is the radius of the smallest sphere circumscribing a given
truncated octahedron or it’s the radius of a spherical surface passing through all 24 vertices of a given
truncated octahedron. It is calculated as follows (See figure 3 above).
In right
⇒
( )
√
√
( *
In right
⇒ √( ) ( ) √(
√
* ( √ ) ( )
√ √ ( )
Hence, the outer radius of truncated octahedron is given as
√
Hence, the volume of circumscribed sphere is given as
( ) ( √ )
3. Surface area( ): We know that a truncated octahedron has 6 congruent square & 8 regular
hexagonal faces each of edge length . Hence, its surface area is given as follows
( ) ( )
We know that area of any regular n-polygon with each side of length is given as
Hence, by substituting all the corresponding values in the above expression, we get

6. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
( * ( * √ ( √ )
( √ )
4. Volume( ): We know that a truncated octahedron with edge length is obtained by truncating a
regular octahedron with edge length at all its 6 vertices. Thus, 6 congruent right pyramids with
square base are truncated off from the parent regular octahedron. Hence, the volume (V) of the
truncated octahedron is given as follows
( ) ( )
(
√
( ) ) ( ) ( )
(
√
) (
√
) ( ( ))
√ √ √
√
5. Mean radius( ): It is the radius of the sphere having a volume equal to that of a given truncated
octahedron. It is calculated as follows
( ) √ ⇒ ( )
√
(
√
)
(
√
)
It’s clear from above results that
Construction of a solid truncated octahedron: In order to construct a solid truncated octahedron with
edge length there are two methods
1. Construction from elementary right pyramids: In this method, first we construct all elementary right
pyramids as follows
Construct 6 congruent right pyramids with square base of side length & normal height ( )
√
Construct 8 congruent right pyramids with regular hexagonal base of side length & normal height ( )
√
Now, paste/bond by joining all these right pyramids by overlapping their lateral surfaces & keeping their apex
points coincident with each other such that all the edges of each square base (face) coincide with the edges of

7. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
four hexagonal bases (faces). Thus, a solid truncated octahedron, with 6 congruent square & 8 congruent
regular hexagonal faces each of edge length , is obtained.
2. Machining a solid sphere: It is a method of machining, first we select a blank as a solid sphere of certain
material (i.e. metal, alloy, composite material etc.) & with suitable diameter in order to obtain the maximum
desired edge length of truncated octahedron. Then, we perform facing operation on the solid sphere to
generate 6 congruent square & 8 congruent regular hexagonal faces each of equal edge length.
Let there be a blank as a solid sphere with a diameter D. Then the edge length , of a truncated octahedron of
maximum volume to be produced, can be co-related with the diameter D by relation of outer radius ( ) with
edge length ( )of a truncated octahedron as follows
√
Now, substituting ⁄ in the above expression, we have
√ √
√
Above relation is very useful for determining the edge length of a truncated octahedron to be produced from
a solid sphere with known diameter D for manufacturing purposes.
Hence, the maximum volume of truncated octahedron produced from the solid sphere is given as follows
√ √ (
√
* √
√ √
√
Minimum volume of material removed is given as
( ) ( ) ( )
√
(
√
*
( ) (
√
*
Percentage ( ) of minimum volume of material removed
(
√
*
(
√
*

8. Mathematical analysis of truncated octahedron
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
It’s obvious that when a truncated octahedron of maximum volume is produced from a solid sphere then
about of material is removed as scraps. Thus, we can select optimum diameter of blank as a solid
sphere to produce a truncated octahedron of maximum volume (or with maximum desired edge length)
Conclusions: let there be any truncated octahedron having 6 congruent square & 8 congruent regular
hexagonal faces each with edge length then all its important parameters are calculated/determined as
tabulated below
Congruent
polygonal faces
No. of
faces
Normal distance of each face from the
centre of the given truncated octahedron
Solid angle subtended by each face at the
centre of the given truncated octahedron
Square 6 √ ( *
Regular
hexagon
8
√
(
√
*
Inner (inscribed)
radius ( ) √
Outer (circumscribed)
radius ( ) √
Mean radius ( ) (
√
)
Surface area ( ) ( √ )
Volume ( ) √
Note: Above articles had been developed & illustrated by Mr H.C. Rajpoot (B Tech, Mechanical Engineering)
M.M.M. University of Technology, Gorakhpur-273010 (UP) India Dec, 2014
Email: rajpootharishchandra@gmail.com
Author’s Home Page: https://notionpress.com/author/HarishChandraRajpoot
Courtesy: Advanced Geometry by Harish Chandra Rajpoot