Demoland Population

Continuing from here.

I want to base my system on Alexis's tech/dev system, which in turn is inspired by the Civilization games. I'll tweak as necessary.

I wanted an equation (surprise!) to determine the population and tech level of a hex. Plugging Alexis's thresholds into Excel yields the following, where $\rho$ is the population density of a hex, and $t$ is the tech level:
\[\rho = 3.03\exp\left(0.69 t\right)\]
I have to give some consideration to the difference between a hex population and a hex density. Density, of course, is dependent on how large the area in question is. Since the hex itself, which is the discrete unit of area, is very large (about 346 sq.mi), a low density can still yield a large population only if the entire hex is assumed to be occupied. This is usually not the case, depending on the topography.

However, for now I will be treating them as the same. The local perceived density, then, is how many people you can travel to see in one day (1 hex).

To keep things more simple, I'll pick a tech level of 8, which has a hex density of 765. That gives a total population of 52,020 for the 68 hexes (at around 2 people per sq.mi, that's practically deserted depending on how spread out they are).

So how many of those live in cities? I really like Lyman Stone's work on this, in lieu of the oft-quoted Medieval Demographics Made Easy. He suggests an urbanization rate of about 5.1% in Europe. I've tried to find some estimates of that number for earlier eras (specifically the Roman era, my go-to), but apparently Roman demographics is a battleground of academic contention, with some estimates as high as 30%. To that end, I think it's most useful to randomly select a rate between 5-15%.

Rolling for a rate: 6.8%. That puts 3537 people in the 10 cities. Seems a little low. Perhaps some of the cities (well, towns, really) will disappear, or be little more than tiny hamlets.

Now, obviously not every city has 353.7 people. So where are they all distributed? Zipf''s Law! Generally speaking, this states that the largest city is twice as large as the second largest city, and so on all the way down. Formally stated, the equation is as follows, where $P_i$ is the $i$-th largest city, $K$ is a constant, and $R_i$ is the rank of the $i$-th largest city:
\[P_i = K R_i^{-1}, \log(P_i) = \log(K) - log(R_i)\]
What we need to find is $K$, once we decide which cities are the largest, in order.

1. Andox
2. Malis
3. Gerlin
4. Derl
5. Ffith
6. Cadewin
7. Betryn
8. Llen
9. Norys
10. Kenor

So we know that the following equation will hold true:

\[\sum_{i=1}^{10} P_i = \sum_{i=1}^{10} {K \over 11-i} = 3537\]

Huh. That was easier than I thought. $K \approx 1207.6$, so the population of each city is $P_i = {1207.6 \over i}$:

Andox	1208
Malis	604
Gerlin	403
Derl	302
Ffith	242
Cadewin	201
Betryn	173
Llen	151
Norys	134
Kenor	121

That adds up to 3539, which is close enough. Interesting. So most of these are barely hamlets. This is in keeping with the low tech of the area though - remember that most people will be scattered across the countryside.

Another problem arises: Andox has more people than the average density (765), and Malis isn't far behind. Does this mean that everyone in the Andox hex lives in the city proper? That doesn't make any sense, given the earlier discussion of urbanization rates. The easiest thing to do is work backwards from the rate to find the number of people who have moved in to crowd around the city. That maintains the overall total. However, I foresee several problems with this.

1. Does the local tech level increase/decrease?
2. Not all hexes have settlements. There are 58 "empty" hexes now, each of which should have about 765 people.
3. With larger numbers at a higher tech level, you could easily have millions of people crammed into a single hex. That works out to thousands per sq.mi - almost a city in its own right!

What is needed, then, is a more robust system for determining the remaining urban population. I suspect my answer lies in the infrastructure system. And hence, a problem for a new post.