Cities: Skylines II - Practical Engineering: Efficient Grids

The grid tool claims “to create a perfect grid”. It’s certainly got exactly-90° angles, but is it really perfect? Let’s calculate what’s optimal.

Efficient Grids Guide

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Introduction

Which of the two square grids below do you think has more zoning cells?

Each covers the same footprint: 1008 metres for each edge road, about 100 hectares each.

The one on the left sure looks like the obvious choice. It leaves exactly 12 zoning cell widths between the roads, the most you can get without leaving a gap. That’s one of the easiest lengths to draw with the road tool, since you just have to line up the zoning circles. It even happens to be the widest width you can get with the parallel road mode.

But if you count them all up, the one on the right actually has more zoning cells despite the big holes in the middle of each block. The right is cheaper to build and maintain, too, since it has less roadway.

The left has 12²×9² = 11664 cells; the right has (16² – 4²)×7² = 11760 cells.

Can we do better? Let’s calculate it (Spoiler: Yes, we can)!

Optimal Square Grids

First, let’s define some variables:

x is the length of the road on one side of a block, as reported by the road tool.
z is the length of zonable frontage.
h is the length of the side of the hole in the middle of the zoning.
w is the width of the road being used.

Using this, the zonable area in each block is the space inside the block (z²) minus the part too far from the roads (h²). But both of these are directly related to the block length, as the frontage is the block size minus the road width (z = x – W), and relatedly the hole is the frontage minus how far the zoning goes (h = z – 2 × 6 x 8m = x – W – 96m).

The density, which we’re trying to optimize, is thus

(z² – h²) / x²= ((x – W)² – (x – W – 96)²) / x²= 192×(x – W + 48) / x²

How do we maximize it? Calculus! The maximum is where the derivative is zero:

0 = d/dx 192×(x – W + 48m) / x²
0 = (384W – 192x + 18432) / x³
0 = 384W – 192x + 18432
x = 2W + 96m

Plugging that answer into the previous formula for the hole simplifies interestingly:

h = x – W – 96m
h = 2W + 96m – W – 96m
h = (2W – W) + (96m – 96m)
h = W

In other words, to make an optimal square grid, leave a hole in the middle the same size as the road you’re using.

Let’s look at how that works for a small road:

Block Size – Density – Road Needed

112m (14u) – ³⁶⁄₄₉ ≈ 73.5% – ²⁸⁄₉ ≈ 3.11 m/cell
128m (16u) – ¾ = 75% – ⁸⁄₃ ≈ 2.67 m/cell
144m (18u) – ²⁰⁄₂₇ ≈ 74.1% – ¹²⁄₅ = 2.40 m/cell

So with 16m roads, by using the optimal square grid you get about 2% more zoning cells in the same area yet pay about 14% less for roads (both up-front and in upkeep), compared to the square grid without the hole in the middle.

Final answer

Thus for the best square grid, use a block size of