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New Dice [message #19911] Sun, 31 October 2010 15:26 UTC Go to previous message
avatar Magic  is currently offline Magic
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Registered: August 2008
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Hi all,

As discuss in the topic "Designing dice for dice snobs", I designed today a new Average D6 where you have to make the average of 4 faces of an octahedron, to find the result of a regular D6.
index.php?t=getfile&id=5704&private=0
The faces of the octahedron are numbered 1, 1, 1, 1, 2, 4, 8, and 10.
Here are the operations :
(1 + 1 + 1 + 1)/4 = 4/4 = 1
(1 + 1 + 2 + 4)/4 = 8/4 = 2
(1 + 1 + 2 + 8)/4 =12/4 = 3
(1 + 1 + 4 +10)/4 =16/4 = 4
(1 + 1 + 8 +10)/4 =20/4 = 5
(2 + 4 + 8 +10)/4 =24/4 = 6


The positions of the virtual 1, 2, 3, 4, 5 and 6 are the regular positions of a D6 (opposite sides sum to 7).
This die is 21.5 x 21.5 x 21.5 mm. The diameter of the wire of the frame is 3 mm so you can print it in Stainless Steel. The price will be less than $25.

It has not been prototyped yet (as soon as I will do, it will be in the "It arrived" section).

More informations soon about other variations of this design.

[EDIT]
if you choose 1, 1, 1, 1 for the four upper faces of the octahedon, you still have several choices for the lower faces (all numbers being larger or equal to 1):
- 1, 5, 9, 9 (all numbers are in the form 4k+1, but I did not choose this one because 1 is repeated again and 9 appears twice)
- 2, 4, 10, 8 (opposite numbers sum to 12, the one I chosed)
- 3, 3, 11, 7 (all numbers are in the form 4k+3 unlike the four 1 of the opposite faces, I did not chose this one also because 3 appears twice)
- 4, 2, 12, 6 (I prefered the other one because the maximum number is smaller)
- 5, 1, 13, 5 (all numbers are in the form 4k+1, I did not choose this one because 1 is repeated again and 5 appears twice)

  • Attachment: Avg_both.jpg
    (Size: 78.48KB, Downloaded 693 time(s))

[Updated on: Sat, 27 November 2010 07:36 UTC]

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