The volume of 1000 coins and their container
Volume of coins
Let's start with the basics of naming things: Coins are cylinders of diameter \$D\$ and height \$h\$. A stack of \$n\$ coins has a height of \$H=n\times h\$. Each coin has a volume of $$V_1=\pi \frac {D^2}{4}\times h$$
We can separate that into stacks, but the total volume will be \$V_{1000}=1000\times V_1\$
Defining the container
That's good, right? Well, containers are square-ish, so the volume of our container is different, and there needs to be air in between. Let's assume the box can be described in terms of \$V_c=a\times b\times c\$.
Now, getting the coins into the box means we are stacking them. This can get rather complicated, like in the coins in a fountain case, and there are even more irregular cylinder stacking problems around. To make it simpler, we can just use a form factor \$ff\$ and use a simple trick of math to get the container size from the volume of the coins alone: We define the formfactor as the fraction the coins take in a unit-cell and then stack unit cells till we have the container filled. $$ff=\frac {V_{1000}}{V_c}$$ $$V_c=\frac {V_{1000}}{ff}$$
Form factors
Let's try a simple stacking. Each stack of coins of diameter D occupies a square column of D side length. So each cylinder occupies 78.6 % of each unit cell. $$\frac{h*\pi*\frac {D^2}4}{h*D^2}=0.786$$
But a tighter packing is available, and the best available is present in the bulk, as you can see in the red cell in this picture. However, such stacking also includes less good packing like the yellow cell, and empty areas of \$\frac D 2 \times 0.732 D \$ at the rim in each even row. Math goes complicated. The yellow cell (present 4 times) offers a form factor if the order of \$0.7854 + 0.03134=0.8167\$ and the red one is, \$0.7854 + 4 \times 0.03134=0.9108\$. The lower edge has a form factor of \$ 0.7854 + 2 \times 0.03134 =0.848\$ - for a given volume, how much of each type is there will shift, and that doesn't account for the empty volume yet... so we can assume something in the order of \$84.81\pm 6.27 \%\$ form factor, depending on how many coins we actually have.
Unit-cells of type 1 (ff=0.8188) 2 (ff=0.9108) and 3 (ff=0.848)
$$ff_\text{sorted}\simeq\{ 0.786 ... 0.91 \}$$
If we account for really bad stacking, we can assume that we got an unordered pile with about 20% more hollow space than the lowest stacking. $$ff_\text{unordered}\simeq\{ 0.586 \}$$
Volume of the cavity to house 1000 coins
Ok, we have our form factors. 58.6 % as the lower bound for a loose heap or unsorted poured into a box, as good as 80 % for coins that still are mostly squished in by the edges of the container to lay almost like bar stock in the simple "one bar per unit cell" configuration. 84% seems more likely though. It could be as good as 91.1 % for an infinite number of coins in an infinitely large container in perfect stacking, but we are FAR away from this number.
How much volume took 1000 coins again? Let's chuck in numbers. Let's take those of Carcer, and we get the following boxes:
- Pt: 26 in³ in coins \$\to\$ The Box is 32.50 in³ for somewhat well-sorted and 44.37 in³ for totally unsorted
- Au: 28 in³ in coins \$\to\$ The Box is 35.00 in³ for somewhat well-sorted and 47.78 in³ for totally unsorted
- Ag: 53 in³ in coins \$\to\$ The Box is 66.25 in³ for somewhat well-sorted and 90.44 in³ for totally unsorted
- Cu: 62 in³ in coins \$\to\$ The Box is 77.50 in³ for somewhat well-sorted and 105.8 in³ for totally unsorted
total Box sizes
How much is that? Let's assume the box base has a floor space of about a 2-by-4 - 8 in² - and you get simple heights for your stacks:
The Platinum-filled box is about 4.06 to 5.54 inches deep, Gold 4.38 to 5.97, Silver 8.28 to 11.31, and Copper 9.69 to 13.23.
If we assume the 84.81 % average for the sorted form factors, then our boxes are 3.83, 4.12, 7.81, and 9.14 inches deep respectively.
Accounting about an inch for wall thickness and lid to each side, that makes roughly cubic items of four-by-six base with six inches height in the flattest and 15 inches in the biggest configuration, though that box would be better described with a 15-inch by eight-inch base and 6-inch height. Possibly the lid might account for another 1-2 inches to add a nice bent top, possibly some more.
