# How to determine reasonable population density of a three-dimensional volume

I'm attempting to rough-estimate potential population density of three-dimensional megastructures in space. Assume a filled spherical volume (i.e. death star, not Dyson sphere), where all internal space is available. What would be a generally reasonable population-per-cubic-mile, assuming conventional transportation of resources (by which I mean no replicating food or teleporting out waste)? At what point would waste heat become a major concern for this structure in deep space? I'm currently just looking to land my estimate in the right order of magnitude, so perfect accuracy is less important.

Edit: It occurs to me that the estimate could also apply to large-scale arcologies on a planet, but I can't find that answer either, so consider that a potential alternate estimate method should you be more familiar with it.

• "At what point would waste heat become a major concern for this structure in deep space?" Waste heat is a major problem from the very beginning. Look a photo of the International Space Station, and notice the large radiators. When one of the radiators failed all other activity was dropped and the crew went to perform three emergency EVAs to repair it. – AlexP Oct 15 '18 at 16:28
• Look at existing high-density 3D "structures" like Manhattan and Hong Kong. – RonJohn Oct 15 '18 at 16:37
• @AlexP: The ISS is far, far smaller than a small moon. Given the amount of care the Empire had for OH&S, they might simply consider a lack of insulation a good way to get rid of waste heat, even from the reactor core. It avoids unshielded vents. – nzaman Oct 15 '18 at 16:57
• @nzaman: You physics is backwards. The ratio between volume (where heat is produced) and area (where heat is dissipated) goes up when the radius increases. It is harder to get rid of waste heat on a large ship than on a small ship. And your idea about insulation is strange -- the ship is in a vacuum, and the vacuum is very good insulator by itself. That's why spacecraft (such as the ISS) need active radiators in order to dump waste heat. – AlexP Oct 15 '18 at 17:24
• Is your structure purely residential, or it has some industry and agriculture as well? – Alexander Oct 15 '18 at 17:47

Let's run the calculations for heat dissipation as the only limiting factor. We assume that the sphere is a maximum density human habitat. All life support (oxygen, food, water) is coming from outside of this sphere. There is a practical limit, though, how quickly the heat could be pumped from out of this sphere that it stays cool.

Sphere volume: $$V = \frac43\pi r^3$$

Sphere surface: $$S = 4\pi r^2$$

$$n$$ - number of humans

$$V_H$$ - volume per human

$$P_H$$ - power per human (own metabolic plus lighting, devices and appliances)

$$F$$ - heat dissipation factor, watts per square meter of surface

*

With number of humans and other variables: $$nV_H = \frac43\pi r^3$$ and $$nP_H = 4 \pi r^2 F$$

$$nP_H = 4\pi\sqrt[\frac32]{\frac{3nV_H}{4\pi}}F$$

$$\sqrt[3]{\frac{n}{V_H^2}} = \frac{4\pi}{P_H}\sqrt[\frac32]{\frac 3{4\pi}}F$$

$$\frac n{V_H^2} = \frac{4 \times 3^2 \pi F^3}{P_H^3}$$

• this is the formula that ties up all the factors together. Let's try to get some practical numbers using it.

Let's assume $$P_H = 200W$$ and $$F = 1000 \frac{W}{m^2}$$

For 1,000,000 people: $$V_H = 8.4 m^3$$ (similar to first class sleeping railroad car) and $$r = 126m$$ (tiny, actually!)

For 1,000,000,000 people, $$V_H = 266 m^3$$ (more than double typical cruise ship's space) and $$r = 3990m$$

Practically, 1000 watt per square meter estimate is rather low. With advanced techniques we can transfer heat more efficiently. 200 watt per person means a very basic accommodations - lighting and small electronics, but no hot meals and no heating devices.

Overall, it appears like (without any energy intensive processes), heat dissipation is not going to be a major limiting factor for a space station size.

• So it sounds like heat dissipation will only become a concern in the event of proximity to an outside heat source like a star (in which case it's a matter or surface area rather than scaling with volume), and relatively trivial to deal with in deep space? – Knight Porter Oct 15 '18 at 20:14