GENERALISATION

of the coordinates of Brueckner 22-11 (6-14)

Applicable to compounds of 6{3,3} and 12{3,3} [ S4A4 and S4*I] with one degree of freedom

R. Webb has pointed out that Brueckner 22-11 (6-14) is a compound of 12 tetrahedra having following coordinates:
X    =    (    3    +    sqrt ( 3 )    )    /    6
Y    =    (    3    -    sqrt ( 3 )    )    /    6
Z    =    sqrt ( 3)    /    3

R calculated from the coordinates    X,    Y    , Z    :
( (3 + sqrt (3) ) / 6 ) ^ 2    +    ( ( 3 - sqrt ( 3 ) ) / 6 ) ^ 2    +    ( ( sqrt ( 3 ) / 3 ) ^ 2    =    1    ( R = 1 )

New value for R :
( 3 + sqrt ( 3 ) ) ^ 2    +    ( 3 - sqrt ( 3 ) ) ^ 2    +    ( 2 * ( sqrt (3 ) ) ^ 2    =    36    ( R = 6 )    [1]

Arc corresponding to the degree of freedom :
Atan (alpha)    =    Y / X

Z = 2 * sqrt(3)

= Section A : First solution =

Putting    3=m    and    sqrt(3)=n    in [1]    we have :
( m + n ) ^ 2    +    ( m - n ) ^ 2    +    12    =    36
2 * (m ^ 2    +    n ^ 2 )    =    24
m ^ 2    +    n ^ 2    =    12

12    can be split up in this way:

 m ^ 2 n ^ 2 X = m + n Y = m -n Alpha Name 6 6 2 * sqrt(6) 0 A =   0.000 000° Compound    A 7 5 sqrt(7) + sqrt(5) sqrt(7) - sqrt(5) 4.797 034° Compound 8 4 2 * sqrt(2) + 2 2 * sqrt(2) - 2 B =   9.736 610° Compound    B 9 3 3 + sqrt(3) 3 - sqrt(3) C = 15.000 000° Compound    C 10 2 sqrt(10) + sqrt(2) sqrt(10) - sqrt(2) D = 20.905 157° Compound    D 11 1 sqrt(11) + 1 sqrt(11) - 1 28.221 345° Compound 12 0 2 * sqrt(3) 2 * sqrt(3) I =   45.000 000° Compound    I

= Section B : Second Solution =

Putting    3 + sqrt ( 3 ) = m    and    3 - sqrt ( 3 ) = n    in [1]    we have :
m ^ 2    +    n ^ 2    +    12    =    36
m ^ 2    +    n ^ 2    =    24

24    can be split up in this way:

 m ^ 2 n ^ 2 X = m Y = n Alpha Name 24 0 2 * sqrt(6) 0 A =   0.000 000° Compound    A 22 2 sqrt(22) sqrt(2) 16.778 655° Compound 20 4 2 * sqrt(5) 2 F =  24.094 843° Compound    F 18 6 3 * sqrt(2) sqrt(6) G = 30.000 000° Compound    G 16 8 4 2 * sqrt(2) H = 35.264 390° Compound    H 14 10 sqrt(14) sqrt(10) 40.202 966° Compound 12 12 2 * sqrt(3) 2 * sqrt(3) I =   45.000 000° Compound    I

= Section C : Compound E =

Calculation of the arc corresponding to the degree of freedom :

Atan ( alpha ) = Y / X

If    alpha = 22.5°    then Y / X    =    sqrt ( 2 ) - 1    [1]
We have    :    X ^ 2    +    Y ^ 2    =    24               [2]

From [1] and [2] we deduce :
X    =    sqrt ( 6 * ( 2 + sqrt ( 2 ) )    =    4. 526 066 877
Y    =    sqrt ( 6 * ( 2 - sqrt ( 2 ) )    =    1. 874 758 285

From Section A, B and C we deduce the following table:

Table of Coordinates :

R  =  6
Z  =  2 * sqrt(3)  =  3.464

Alpha  :
A + I  =  B + H  =  C + G  =  D + F  =  2 * E  =  45.000 000°

 Name X X Y Y Y / X Alpha Compound    A 2 * sqrt(6) 4.899 0 0.000 0 A =   0.000° Compound    B 2 * sqrt(2) + 2 4.828 2 * sqrt(2) - 2 0.828 3 - 2 * sqrt(2) B =   9.736° Compound    C 3 + sqrt(3) 4.732 3 - sqrt(3) 1.268 2 - sqrt(3) C = 15.000° Compound    D sqrt(10) + sqrt(2) 4.576 sqrt(10) - sqrt(2) 1.748 ( 3 - sqrt(5) ) / 2 D = 20.905° Compound    E sqrt (6 * (2 + sqrt(2) ) 4.526 sqrt (6 * (2 - sqrt(2) ) 1.875 sqrt(2) - 1 E = 22.500° Compound    F 2 * sqrt(5) 4.472 2 2.000 1 / sqrt(5) F = 24.095° Compound    G 3 * sqrt(2) 4.243 sqrt(6) 2.449 1 / sqrt(3) G = 30.000° Compound    H 4 4.000 2 * sqrt(2) 2.828 1 / sqrt(2) H = 35.264° Compound    I 2 * sqrt(3) 3.464 2 * sqrt(3) 3.464 1 I  = 45.000°

Stellation Applet (V. Bulatov)

I m a g e s

S t e l l a t e d    d i a g r a m s

D e t a i l s

Stella - files

Compound of 6 tetrahedra - Rotational Freedom

Compound of 12 tetrahedra - Rotational Freedom

Stellation Applet (V. Bulatov) : Cell-configuration

Compound of 6{3,3} in S4A4 (Td) - One degree of freedom

Compound A - Rigid : 6|S4xI/D4D2
Stellation Applet :
{0,1,2,3,4,5,6(2,5,6,7)7(0,1,6,7)8(8)}

Compound B
Stellation Applet :
{0,1,2,3,4,5,6(3,5,6,7,8)7(1,7,8)8(7)}

Compound C
Qbasic : lijnen 14, 15, 16
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(7)}

Compound D
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(7)}

Compound E
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(6)}
Qbasic : lines 14, 15, 16

Compound F
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(6)}

Compound G
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(6)}

Compound H
Qbasic : lijnen 14, 15, 16
Stellation Applet :
{0,1,2,3,4,5,6(2,4,5,6,7)7(2,7,8)8(6)}

Compound I - 1 tetrahedron
{0}

Compound of 12{3,3} in S4*I (Oh) - One degree of freedom

Compound A - Rigid : 6|S4xI/D4D2
{0,1,2,3,4,5,6(1,2,4)7(1,4)8(0)}

Compound B
{0,1,2,3,4,5,6,7,8,9,10,11,12(5,6,8,9,10,11,12,13,14,15,16)13(1,5,7,8,9,10,12,13,14,16)14(3,12,13,16,18)15(3,5,17)}

Compound C - Brueckner 22-11 (6-14)
{0,1,2,3,4,5,6,7,8,9,10,11,12(4,5,7,8,9,10,11,12,13,14,15)13(2,4,6,8,10,12,13,14)14(8,15,16)15(13)}

Compound D
{0,1,2,3,4,5,6,7,8,9,10,11(0,1,2,4,5,6,7,8,9,10,11,12,13,14)12(4,5,7,8,10,11,12,13,14,15,16)13(1,3,5,8,12,13,16)14(11,15)}

Compound E
{0,1,2,3,4,5,6,7,8,9,10(0,1,2,3,4,5,6,7,9,10,11,12,13,14)11,12(4,5,7,8,9,10,11,12,13,15,16)13(1,3,10,12,14,16)14(10,13)}

Compound F
{0,1,2,3,4,5,6,7,8,9,10,11,12(4,5,7,8,9,10,11,12,13,15,16)13(1,3,10,12,14,16)14(13)}

Compound G
{0,1,2,3,4,5,6,7,8,9,10,11,12(4,5,9,10,11,12,13,15,16)13(1,3,6,9,13,15)14(13)}

Compound H
{0,1,2,3,4,5,6,7,8,9,10,11(0,3,4,5,7,8,10,11,12,13,14)12(4,5,9,10,11,12,15,16)13(2,3,6,9,13,15)14(13)}

Compound I - 2 tetrahedra - Stella Octangula
{0,1}

Belgium - Aarschot - 10 May 2005 - G.M. Fleurent