I have a cuboid with a width of 500mm, a height of 500mm and a length of 1000mm.
I need to know how many 38mm balls i will need to fill the cuboid.
Any help would be greatly appreciated.
Thank you.
[Math] how many items can i fit in the box
algebra-precalculus
Related Solutions
In general, a parallelotope (the generalization of parallelograms to many dimensions) can be specified by $n+1$ points: a vertex $u$ and all the vertices $v_1,\ldots,v_n$ adjacent to that vertex - and, as long as long as there is no hyperplane on which all vertices lie, any collection of $n+1$ vertices really does extend to a parallelotope (since, essentially, one can just take the Minkowski sum of the edge vectors from $u$ to each $v_i$ to get a volume).
It seems like a box, for you, is just a parallelotope in which the faces are perpendicular to one another - they can be specified the same way, except they additionally require that the edge vectors from $u$ to each $v_i$ be pairwise perpendicular - which doesn't really bring down the number of vertices required, but it means that, if we specify $u$ then $v_1$, we would know that $v_2$ lies on some specific hyperplane, then that $v_3$ lives on a codimension two subspace, and so on - bringing down the degrees of freedom. In general, you have $n$ dimensions of choice for $u$ and $v_1$, then $n-1$ for $v_2$ and $n-2$ for $v_3$ and so on, until you only have a single dimension of choice for $v_n$.
All told, you get $n(n+1)$ degrees of freedom in choosing a parallelotope in $n$ dimensions, but only $\frac{n(n+3)}2=n+(1+2+3+\ldots+n)$ degrees of freedom in choosing a box - which is probably a more meaningful measurement than "number of points required."
I'm showing below how to arrange 14 spheres into each gap. Computing $x$ and $y$ via Pythagoras' theorem is not difficult: $$ (2+0.4)^2+x^2=(2\sqrt2-0.4)^2,\quad (2x)^2+y^2=0.8^2. $$ You may check that $y\le0.4$.
Best Answer
if i am not wrong the total numbers of balls will be "4394"
{ 1000/38= 26(approx) }
13 balls width wise { 500/38= 13(approx) }
and 13 height wise { 500/38= 13(approx) } which will lead you to 26*13*13 = 4394 balls (approx)