Maximum number of cubes within a radius Last night I was mining obsidian in Minecraft, which takes a long time (15 seconds for each block). As a result, I would hold down the left mouse button with my left hand while I did something else. In order to maximize the usefulness of this tactic, I would orient myself in such a way as to mine as many blocks as possible without moving. This lead me to wonder...
Excluding the cube the rod starts in, how can I determine the number of
unit cubes that a rod of length $k$ can go through? 

For example, the following picture shows a rod of length $4$ going through $4$ cubes (although a fifth could easily be added).

 A: Let us first consider the two dimensional case to explain the concepts.
A line segment of the length $L$ (I will call it line $L$) is to be laid down 
on a unit square grid such that the number of the intersected grid cells 
is maximized. The cell coordinates are given by their left bottom corners.

The figure shows a line of length 4.8 starting in the cell $(0, 0)$ and
ending in the cell $(4, 3)$. The angle $\alpha$ between this line and the x-axis is $45^{\circ}$.
We can assume without loss of generality that the line always starts in the
cell $(0, 0)$ and goes upwards and right such that $0 \le \alpha \le 45^{\circ}$ and it ends
in a cell $(X, Y)$. When we move a point $P$ along the line we encounter the crossing
points with the cells. A cell can be entered only at its left side (such as $C_1$)
or at its bottom side (such as $C_2$). If a cell is entered at its left side,
the $x$ grid coordinate of $P$ is increased by $1$ and the $y$ grid coordinate is unchanged.
If a cell is entered at its bottom side,
the $y$ grid coordinate of $P$ is increased by $1$ and the $x$ grid coordinate is unchanged.
So, at each entering point into a cell exactly one grid coordinate of $P$ is increased
by $1$. Since the point $P$ starts in $(0, 0)$ and ends in $(X, Y)$, there are exactly
$X + Y$ different entering points,
so there are exactly $X + Y$  crossed cells (not counting the starting cell).
In the figure we have $4 + 3 = 7$ crossed cells.
Since $X + Y \approx L \cos \alpha + L \sin\alpha$, the maximal number of crossed cell will
be achieved with $\alpha = 45^{\circ}$.
There is one degenerate case when the line enters a cell through its bottom
left corner. Then both grid coordinates are increased but only one new cell
is crossed. Since we would get less crossed cells in this case we avoid this case
by displacing the line slightly. This
is always possible because the start and end points are real numbers and there
are at most finite number of considered cells.
It remains to find $(X, Y)$. First we place the starting point at the coordinate origin and let the line pass through the opposite cell corners (parallel to cell diagonals):

If $L = n \sqrt 2 $ ($n$ is an integer) then the line ends at a cell corner $(n, n)$.
Now, it can be displaced such that it does not pass through the corners and ends in the cell
$(n, n)$. The maximal number of intersected cells (not counting the starting cell) is 
$n + n = 2n$, where $n = \lfloor L / \sqrt 2 \rfloor$. 
If $L = n \sqrt 2 + r, 0 < r < \sqrt 2$, then the line ends on the
diagonal of the cell $(n, n)$. Moving the line along $45^{\circ}$ until the next
cell is reached and displacing it, we obtain the end cell $(n+1, n+1)$.
The maximal number of intersected cells is 
$n+1 + n+1 = 2n + 2$.
The three dimensional case is analogous. The cells are now unit cubes, the line segment is the rod.
When moving along the rod we enter the cubes through one of three possible
faces and the corresponding coordinate is increased. E.g. entering a 
cube through its bottom face increases the z-coordinate by $1$. Maximal
number of intersected cubes is achieved when the rod is put through the opposite corners
of the cubes (along longest diagonals) and then slightly displaced to pass through the faces.
The maximum is X + Y + Z, where (X, Y, Z) is the last 
cube (not counting the first cube).
If $L = n \sqrt 3$ then the rod ends at the cube corner $(n, n, n)$.
It can be displaced such that it ends in the cube
$(n, n, n)$. The maximal number of intersected cubes is 
$n + n + n = 3n$.
If $L = n \sqrt 3 + r, 0 < r < \sqrt 3$, then the rod ends at the diagonal of the cube
$(n, n, n)$.
It can be displaced such that it ends in the cube
$(n+1, n+1, n+1)$. The maximal number of intersected cubes is 
now $n+1 + n+1 + n+1 = 3n + 3$.
Let us summarize:
$$L = \mbox{length of the rod}$$
$$n = \lfloor L / \sqrt 3 \rfloor$$
$$r = L - n \sqrt 3$$
$$
max = \begin{cases} 
3n & \mbox{if} \;\; r = 0  \\
3n + 3 & \mbox{if} \;\; r > 0 \;\;   
\end{cases}
$$
