Counts distribution scheme

There are 9 pixels, organized in a symmetric distribution of counts:

10 20 10

20 30 20

10 20 10

Each parallelepiped is sized as 2 by 2 by 2. There are 9 of them, 3 by 3 configuration. The total size of the domain is 2 by 6 by 6.

2 is the short dimension (going orthogonal to the imaginery piece of paper).

The bins are cut right down the center lines on the paper. Due to symmetry, each bin gets exactly the same distribution (the same math too).

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For one of the four bins:

The counts: 10; 20; 30; 20

Paralleliped volume: 8; 8; 8; 8

C0 values are: 10/8; 20/8; 30/8; 20/8

C1 values are: 1.25; 2.5; 3.75; 2.5

The fractions : 100%; 50%; 25%; 50%

C2 is: (1.25 * 100% + 2.5 * 50% + 3.75 * 25% + 2.5 * 50%) / (100% + 50% + 25% + 50%)
= 4.6875 / 2.25 = 2.08333 ...

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Total counts = 2.08333 * 4

Bin volume: 2 * 3 * 3 = 18

Total counts with normalization = total counts * bin volume = 150