This, and related questions, have come up repeatedly ever since D&D first came out. Recently, I saw a post on a forum stating that a character who has doubled in size would be twice as tall and weigh twice as much. Sadly, there were a number of responses before someone finally pointed out the fallacy in that assertion.
Welcome To 3D
The physical world has three dimensions. Doubling one of those dimensions will double the object’s mass (and therefore its weight) but its shape would then be distorted. To have a creature double in height, while retaining the same proportions, it is necessary to double all three dimensions. The creature’s mass will double each time you double a dimension. In order to double all three dimensions, the creature’s mass will also double three times. This will result in the creature’s mass becoming eight times as large (2 x 2 x 2 = 8).
In order to visualize this, stack one six-sided die (d6) on top of another. If the two dice are the same size and weight then the two together will weigh twice as much as either by itself. Clearly the two, stacked in this manner, no longer form a cube. Stack two more dice, one atop the other, and place them next to the first stack. The combined shape is now two dice high, two dice wide, one die thick and weighs four times as much as a single die. Clearly this set of dice does not form a cube either. Take four more dice, arrange them like the first, and place them behind the first four. It becomes obvious from any angle that these eight dice, stacked together in this manner, form a cube that is twice as high, twice as wide, and twice as deep as a single die. It took eight dice to accomplish this so the weight of the combined object must be eight times as much as a single die as well.
Applying This Method to Any Height Variation
It’s easy to calculate the weight difference if an object is doubled or halved in size. But how do you handle other size changes? For example, what if a spell increased a creature from 4′ tall to 5′ tall. You simply divide the new height by the old height and multiply the result by itself for each of the three dimensions. That sounds complicated but all it really means is (5’/4′) x (5’/4′) x (5’/4′), where 5′ is the new height and 4′ is the old height. Another way to write this is (5/4)^3. If the creature weighed 100 pounds when it was 4′ tall, it would weigh 195.3125 pounds (195 pounds 5 ounces) if increased to 5′ tall.
Applying This Method to Statues
In order to determine the weight of a statue, you must:
- Determine the height and weight of the original creature.
- Determine the height of the statue you want to create.
- Calculate the new weight of the creature based on the new height (using the method described above).
- Adjust the weight based on the density of the material used.
Note: If science makes your eyes glaze over, skip the next two paragraphs. :)
Density is measured in grams per cubic centimeter (g/cm^3) and is based on the density of water (1.00 g/cm^3) at 25 degrees Celsius (if memory serves). The human body is mostly water and therefore has effectively the same density as water. I treat all other organic creatures as having the same density as well.
Copper, silver, gold, and platinum have densities of 8.92, 10.50, 19.30, and 21.45 g/cm3 respectively. However, these metals, in pure form, don’t lend themselves well to sculpting or crafting. They will have various impurities added in order to make them easier to work with. Furthermore, D&D assumes that coins made from each of these metals all weigh the same. In my world, I prefer to also have all coins be of the same size. That means that they each must have the same density. For a round coin, 1″ in diameter and 1/10″ thick, to weigh 1/50 of a pound, the metal must have a density of 10.44 g/cm^3. Solid granite and marble, two other materials commonly used to create statues and other sculptures, have densities of 2.691 and 2.56 g/cm^3, respectively. For our purposes, brass and bronze (both composed of copper with various impurities) have the same density as copper.
Since the original creature has a density of 1.00, all you need to do is multiply the weight of the creature by the density of the material used to create the statue.
If you want to create a 1′ tall, solid gold statue of a female elf, and you assume that a 5′ tall female elf would weigh 100 pounds, you would first adjust for the difference in height: 100 x (1’/5′)^3 = 0.8 pounds. Then you would multiply that by the density of gold (10.44) to get 8.352 pounds. Note that statues usually have a base. Therefore, I would round the figure up to 10 pounds to include a small base at the bottom.
If you instead want to create a life-size sold gold statue of the same subject, you would forego the adjustment for size (since the size remains unchanged) and simply multiply by the density of gold: 100 pounds x 10.44 = 1,044 pounds, plus whatever you feel is necessary for the base.