Mobile Comms: Number of Cells in Hexagonal Pattern

[Master1022]

Homework Statement

Why is the number of cells in the radius $D$ determined by $N_c = D_R^2$ where $D_R = \sqrt{i^2 + j^2 + ij}$

Relevant Equations

area

Hi,

I was looking at some notes and trying to understand the following statement which refers to the diagram below.
"The number of cells in the radius $D$ determined by $N_c = D_R^2$ where $D_R = \sqrt{i^2 + j^2 + ij}$
where $i$ and $j$ are the number of cells along the $u$ and $v$ axes respectively.

From what I understand, B is at (2, 2) in (u, v) coordinates and radius $D$ is $= \sqrt{2^2 + 2^2 + (2)(2)} \cdot R \sqrt{3}$. Therefore, the number of cells within the radius $D$ should be $12$. I cannot see how this is the case, no matter how I try to encircle cells...

I think I am missing something quite simple. Any help would be greatly appreciated.

 

[lewando]

I think that the formula N_c = i² +j² +ij refers to the number of hexagonal cells in a "cluster", where clusters are distanced by i and j.

[edited for additional clarity]
For i=2, j=2, this results in clusters that share cells, fractionally. Add the fractional cells to the non-shared cells and you will get 12.

 

[Master1022]

Thank you very much for your reply @lewando !

I think that the formula N_c = i² +j² +ij refers to the number of hexagonal cells in a "cluster", where clusters are distanced by i and j.

[edited for additional clarity]
For i=2, j=2, this results in clusters that share cells, fractionally. Add the fractional cells to the non-shared cells and you will get 12.

I am struggling to picture this on the image... Is there any chance you could edit/add something to the image posted to show what you are saying? I will keep trying to think about it in the meantime.

 

[lewando]

Perhaps if you study this image you will see what I mean by fractional cells. The black dots are distanced by i=2, j=2. [edited to match your original orientation]

 

[lewando]

FWIW, with a change of grid origin, you can get a different cluster pattern that eliminates the fractional cells:

 

[Master1022]

Thank you very much @lewando for taking the time to produce those pictures! They are extremely helpful!