# How do I calculate the volume of a given quantity of coins?

## Question

A rectangular chest is filled to the brim with 1,000 unevenly piled gold coins, each weighing one third of an ounce. How big is that chest? Is there a formula I can use for routinely calculating the volume of a given quantity of coins?

## Context

Some of the players in my group are, bless their hearts, quite detail oriented. As their DM, I love this, but it does mean I have to get certain things right. For my own satisfaction, and for the sake of creating a plausible world, I like to know this kind of information. Yes, I could say "the chest looks large enough to fit 1,000 coins" but some of my players are the type of people who will still want to know the size, and I want to honor their commitment to my world with good information.

## Resources

We are playing 5e. Please feel free to fill in any information missing from 5e with info from previous editions, however 5e takes preemption wherever there might be conflict. If no coin dimension is described RAW in any previous edition, you may assume any reasonable dimension of a round coin that makes sense for the given weight. For calculating packing density of coins and other inputs that might not be provided RAW, feel free to use real-world figures.

## Reverse engineer it from known metal densities (and marvel at how tiny and dense the currency is)

Given we know the standard weight of all currency as described in 5e D&D (0.32 oz or 9.07 grams, as there are 50 coins to a pound regardless of denomination), we can reference the metal density to find out what the volume of coinage is, and therefore work out how much space a given value of coins must occupy.

• Copper - 0.324 lb/in3 - ~0.062 in3/coin - 27,993.6 coins per cubic foot
• Silver - 0.379 lb/in3 - ~0.053 in3/coin - 32,745.6 coins per cubic foot
• Gold - 0.698 lb/in3 - ~0.028 in3/coin - 60,307.2 coins per cubic foot
• Platinum - 0.775 lb/in3 - ~0.026 in3/coin - 66,960 coins per cubic foot

Since pure trade bars and the actual currency have the same value per weight, I am assuming that the coins are pure and have the same densities (but, as explored here, this is not necessarily consistent with how currency has been described or depicted in D&D's history). The values given here also don't account for the fact that the coins are probably not shaped so that they perfectly tessellate. As this blog post from dmsworkshop.com helpfully summarises, assuming round coins that are about 1/16th of an inch thick and roughly 1 inch in diameter (varies depending on denomination):

To save you some math, the ideal packing density of coins is 78.6% (if neatly ordered in stacks) or about 60% if loose. This means that an unorganized heap of coins (like those stuffed into a sack) will contain about 40% empty space.

Non-round shapes would improve the packing density to varying degrees depending on the exact shape of the coin, but using round coins seems a sensible default. So taking that into consideration:

• Copper - ~22,000 coins per cubic foot (neatly stacked) - ~16,800 coins per cubic foot (loose jumble)
• Silver - ~25,740 coins per cubic foot (neatly stacked) - ~19,650 coins per cubic foot (loose jumble)
• Gold - ~47,400 coins per cubic foot (neatly stacked) - ~36,180 coins per cubic foot (loose jumble)
• Platinum - ~52,630 coins per cubic foot (neatly stacked) - ~40,180 coins per cubic foot (loose jumble)

Using these values we can calculate that, for instance, a coffer filled to the brim with 1,000 neatly stacked gold coins would be slightly over 36 cubic inches in capacity (for instance, 6 inches × 6 in. × 1 in. internal dimensions), or loosely piled somewhat less densely at about 48 cubic inches (for instance, 6 in. × 6 in. × 1.33 in.).

You're probably noticing that those are very small dimensions. Metals, especially precious metals, are dense, and sensibly shaped coins made of these metals are small - much smaller than we are apt to imagine them (see Brythan's answer for a helpful comparison to US coinage). Even if you use a chest loosely filled with 10,000sp (equivalent value to 1,000gp) that only comes to about 880 cubic inches - roughly 1 ft. × 1 ft. × 6 in. internally, but weighing in at a hefty 200 lbs!

• By the way: one cubic foot of volume is about 28 liters, which is also about the volume of one Roman amphora. In 2016 near Seville in Spain, 19 Roman amphora were discovered containing about 50,000 Roman bronze coins. So .. it seems the size of D&D coins are very much smaller than Roman coins, because 19 amphora of loose D&D copper coins would amount to about 319,000 coins. Apr 8, 2021 at 16:46

For people accustomed to metric, here are the numbers from this answer recalculated:

• Copper - 8.96 g/cm3 - ~0.988 cm3/coin - 1010 coins per liter
• Silver - 10.49 g/cm3 - ~.8648 cm3/coin - 1156 coins per liter
• Gold - 19.32 g/cm3 - ~.4696 cm3/coin - 2130 coins per liter
• Platinum - 21.45 g/cm3 - ~.4229 cm3/coin - 2364 coins per liter

Density numbers from Density of metals, except platinum.

Fifty coins mass one pound. So one coin is .32 oz or 9.071847 grams.

Using the same 78% stacked and 60% jumbled estimates:

• Copper - 790 coins per liter stacked; 607 coins per liter loose.
• Silver - 901.9 coins per liter stacked; 693.8 coins per liter loose.
• Gold - 1661 coins per liter stacked; 1278 coins per liter loose.
• Platinum - 1844 coins per liter stacked; 1418 coins per liter loose.

Just for comparison, here are the sizes of some United States coins.

• Penny: 19.05 mm diameter; 1.52 mm thickness; .433 cm3.
• Nickel: 21.21 mm diameter; 1.95 mm thickness; .689 cm3.
• Dime: 17.91 mm diameter; 1.35 mm thickness; .340 cm3.
• Quarter: 24.26 mm diameter; 1.75 mm thickness; .809 cm3.
• Half dollar: 30.61 mm diameter; 2.15 mm thickness; 1.58 cm3.
• Dollar: 26.49 mm diameter; 2 mm thickness; 1.10 cm3.

Volume calculated from diameter and thickness.

The penny is closest in size to the gold (and platinum) coin. So if you wanted a visual, you could go to a US bank and get twenty rolls of pennies for $10. That would be a little smaller than the gold coins would be. Perhaps if you leave them in the paper, the size might be closer. Then find a box (chest) large enough to hold the pennies. Perhaps a jewelry box with a lock. The quarter is more in line with the size of silver coins. Note that there are only forty quarters in a roll, so you would need twenty-five rolls ($250) to make a thousand coins.

The dollar coin is perhaps closest to the size of copper coins. At twenty-five coins per roll, you would need forty rolls (\$1000) to make a thousand coins.

• The irony of US currency in an answer about converting to metric is great. +1 and thanks for being useful and saving me some work. May 15, 2019 at 7:43
• I was considering including metric values in my answer as well (you can see I started with a mention of coin mass in grams...) but figured as D&D already uses imperial measurements for everything, that should be understandable enough. Plus I'm lazy. The comparison to actual coin sizes is a very useful aid! May 15, 2019 at 10:16

# 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.

I believe gygax based the Coins on those of medieval Britain, the gold guinea being the standard size of 0.29 grams and .96 inches in diameter. Thus 50 coins to a pound

0.29 grams is the mass based on the standard ounce, not the Troy ounce.

• Hi, welcome to RPG.SE. This is an interesting factoid, but it sounds more like a comment or maybe the answer to a different question as it does not answer the question about coin volumes. Stack Exchange is not like a traditional discussion forum; we have direct questions and answers. Be sure to check out the tour and help center for more information. May 15, 2021 at 6:49