eco-city design walkable centres density efficient buildings road layout car-lite districts freight variations sustainable farming resiliency quality of life

# Density

F.A.R is a measure of density, and is found by dividing the total areas of all the floors by the area of the ground occupied (including the building plots, gardens and streets.)  The higher the F.A.R, the higher the density.

## 2.5 Storey Row Housing

 Total floor area = floor area × number of storeys = (2 × 84 × 9) × 2.5 = 3780m²

 Ground area = (6 + 84) × (12 + 9 + 10 + 1.5 + 10 + 9) = 4635m²

 F.A.R = total floor area / ground area = 3780 / 4635 = 0.8155

## 4.5 Storey Block

 Total floor area = floor area × number of storeys = ((50 x 50) - (30 x 30)) × 4.5 = 7200m²

 Ground area = (5 + 50) × (5 + 50) = 3025m²

 F.A.R = total floor area / ground area = 7200 / 3025 = 2.38

## 6 Storey Block

 Total floor area = floor area × number of storeys = ((65 x 65) - (45 x 45)) × 6 = 13200m²

 Ground area = (9.5 + 65) × (9.5 + 65) = 5550m²

 F.A.R = total floor area / ground area = 13200 / 5550 = 2.378

## Outer Districts

Assuming that each resident requires 40m² of living space at a F.A.R of 0.9, then the ground space required per person for housing is given by:

 Ground per person (residential) = floor area / F.A.R = 40 / 0.8155 = 49.05m²

And assuming each resident requires 30m² for non-residential uses at a F.A.R of 1.8, then the ground space required per person for non-residential uses is given by:

 Ground per person (non-res) = floor area / F.A.R = 30 / 2.38 = 12.61m²

Now we can calculate the average F.A.R:

 Average F.A.R = floor area per person/ total ground area per person = (40 + 30) / (49.05 + 12.61) = 1.135

The area of an outer district is given by r² where r, the radius, is 300m.

 Area of one district = ∏ × 300² = 282,743m²

From this we must remove the extra space required for the bus lanes.

 Area of bus lanes = 10 × 600 = 6000m²

And the extra space required for the circular road.

 Area of circular road = Outer area - inner area = (∏ × 155²) - (∏ × 145²) = 9424m²

 Building area per outer district = area - bus lanes - circular road = 282,743 - 6000 - 9424 = 267,319m²

With an average F.A.R of 1.145, and allowing 70m² per capita (40m² for housing, 30m² for work, services, etc.) this gives each outer district a population of:

 Population = building area × F.A.R / floor space per capita = 267,319 × 1.135 / 70 = 4,334

## Central District

The area of the central district is given by r² where r, the radius, is 475m.

 Area of central district = ∏ × 475² = 708,821m²

From this we must remove the extra space required for the bus lanes.

 Area of bus lanes = 8 × 10 × 475 = 38,000m²

 Building area of central district = area - bus lanes = 708,821 - 38,000 = 670,821m²

If the central district has a F.A.R of 2.1, and allowing 70m² per capita for housing, work, services, etc., this gives a population of:

 Population (central) = building area (central) × F.A.R / floor space per capita = 670,821 × 2.1 / 70 = 20,124

Note that the 4.5 storey block above achieves a F.A.R 13% above 2.1.  This extra space could be used for larger interior courtyards, or wider streets, but I've chosen to use it for linear parks threading through the central district.

## Total Population

Adding the populations of the 24 full outer districts, 8 three quarter outer districts and the central district we get:

 Total population = 30 × 4,334 + 20,124 = 150,144

## Longest Journey

The longest journey occurs when travelling from one of the outermost districts to another of the outermost districts which lies on a separate loop.

Longest journey = wait + ride + change platform + wait + ride

With buses running every 3 minutes, the average wait is 1.5 minutes or 90 seconds.

Now we need to calculate how long it takes to ride to the central station.  We will assume a top speed of 30mph, or 13.33m/s, an acceleration of 1.33m/s², and a dwell time of 15 seconds.

 Time of acceleration = v / a = 13.33 / 1.33 = 10s

 Distance of acceleration = ½ × a × t² = ½ × 1.33 × 10² = 66.66m

 Distance covered at top speed = 600 - 2 × 66.6 = 466.66m

 Time spent at top speed = 466.66 / 13.33 = 35s

Using this data, together with the dwell time, we can calculate how long it takes for the trolley bus to travel between stations.

 Total time = 15 + 10 + 35 + 10 = 70s

The central district, being wider than the 600m of other districts, takes longer to cross.

 Extra time = extra distance / v = (475 + 300 - 600) / 13.33 = 13.125s

The ride to the centre consists of 4 station to station rides, with 1 requiring extra time due to the greater diameter of the central district.

 Ride = 4 ×70 + 13.125 = 293s = 4.88mins

or, rounding up, 5 minutes.

Changing platforms at the central station would take about 1 minute.

This gives us a longest journey of:

 Longest journey = wait + ride + change platform + wait + ride = 1.5 + 5 + 1 + 1.5 + 5 = 14 minutes

In the middle of the night (where the buses run every 12 minutes, including waiting an extra minute at the central station for you to change platforms) the longest journey would be 11 to 23 minutes, averaging 17 minutes.

## Capacity

In the morning rush-hour, how many people can the trolleybus network transport to the city centre?  With buses running every 90 seconds, and assuming 135 passengers per double articulated bus, then the total capacity for the morning rush-hour is:

 Capacity = number of passengers per bus × buses per hour × trolley lines = 135 × 40 × 8 = 43,200

or just under a third of the 130,000 residents of the outer districts.  The 20,000 residents of the central district are, of course, already there.

The double articulated buses in Curitiba, Brazil, are actually capable of transporting 270 passengers, but I halved this number for a more comfortable passenger experience.

Also, since work places are spread throughout the city, rather than being concentrated at the centre, not everyone will have to travel through the city centre to get to their place of work, reducing the necessary rush-hour capacity.  E.g. children would attend the school in their own district, walking to school, or schools further out, riding away from the centre instead of towards it.

## Size of the Fleet

How many double articulated trolleybuses are needed for the whole city?

Each trolleybus does a complete circuit in 9 × 70 + 2 × 13.125 = 656.25s = 10.94mins or, rounding up, 11 minutes.  Thus to provide service at 90 second intervals, 8 trolleybuses are needed per line.  There are 8 lines and so 8 × 8 = 64 trolleybuses are needed, plus spares.

## Maximum Capacity

But what if these 64 trolleybuses aren't enough at rush hour?

The trolleybuses can run with headways as low as 60 seconds, increasing the fleet size to 88 trolleybuses plus spares.

 Capacity = number of passengers per bus × buses per hour × trolley lines = 135 × 60 × 8 = 64,800

or just under half of the 130,000 residents of the outer districts.

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