how to calculate the crosssection area of a bundle of tubes/cables First of all, my mathematics skills don't allow me to understand regular formulas as I didn't get the whole mathetics thing to work for my brain until I discovered excel at the age of 20, and how to everything is written there, not just its fancy functions, but the way you write a formula is a way I understand them. So please, if you answer my question and know how to do formula in excel, I prefer them that way :-)
And of course, if you know any good excel/google sheet functions to calculate this, it always appreciated being pointed in the right direction.
My case is, I do design installations for audio, video, and network. And they tend to grow in size, and the amount of cables grows with them. 
To enter the main rack/board/room, we bundle the cables together in a perfect circular stack to get it look neat.
How to calculate the approximate cross-section area for, in this case, 96 cables bundled tightly (not compressing) together, each cable, in this case, is 7.2mm.
I tried to search for some answers first, but it might be that it is there or not, or just my English skill that doesn't allow me to apply the correct search term :-)
 A: If you look at the cross-section of your bundle, it is effectively just a bunch of circles packed together. Since you don't mention wire size, I am going to assume all the wires are the same size. This makes the problem much more tractable. When there is a lot of variation in the wire size, there is no simple way of estimating this.

When all the circles are the same size, then you can fit exactly 6 circles around a center one. These six circles form a hexagonal shape. Putting on another layer adds another $12 = 2\times 6$ circles, and the next layer adds $18 = 3 \times 6$ circles, and so on. So the total number of circles for $n$ layers (not counting the original circle as a layer) is
$$1 + 6(1 + 2 + 3 + ... + n)$$
The sum of the first $n$ natural numbers is given by $\dfrac{n(n+1)}2$. Plugging that in, $n$ layers provides $$1 + 3n(n+1)$$ wires.
This means that if you have some number $N$ of wires to be in the bundle, you want to find the smallest number of layers $n$ such that $1 + 3n(n+1) \ge N$. Once you have that, then by looking at the picture, you can see that the diameter of the bundle will be $2n + 1$ times the diameter of a wire. (If $N$ is less than $1 + 3n(n+1)$, then the diameter will be a bit smaller, since the outer layer will not be complete).
Solving for $n$, we find that we need the least integer $n$ satisfying $$n \ge \frac12 + \sqrt{\frac{4N-1}{12}}$$
If the wire diameter is $d_W$, then the bundle diameter will be
$$d_B = (2n+1)d_w$$ 
Combining this and putting it in an Excel-like formula:
dB = (2 * CEILING(0.5 + SQRT((4*N - 1)/12)) + 1) * dW

Now hopefully somewhere along this you started thinking "that is a hexagon, not a circle". And you are correct. For a small number of layers, this hexagonal pattern is the best you can do. But at some point the flat sides that form from adding new layers leave enough space that additional wires can be inserted inside your hole without adding another full layer. At that point, the formula will start overestimating the diameter of your bundle. 
But the correction for this is much more complicated. So unless you are using an extremely large number of wires, and getting the correct diameter is critical, your best course is probably to ignore this and just live with the overestimation.
