The Maths

F.A.R

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.

Outer Districts

2.5 Storey Row Housing

Ground area	= (8 + 84) × (8 + 10 + 10 + 2 +10 + 10)
	= 4600m²

F.A.R	= total floor area / ground area
	= 4200 / 4600
	= 0.913

4 Storey Block

Ground area	= (10 + 48 + 2 + 10) × (10 + 48 + 2 + 10)
	= 4900m²

F.A.R	= total floor area / ground area
	= 7680 / 4900
	= 1.567

Average F.A.R

Assuming that each resident requires 40m² of living space within 2.5 storey row houses, then the ground space required per person for housing is given by:

And assuming each resident requires 30m² for non-residential uses within a 4 storey block, then the ground space required per person for non-residential uses is given by:

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

Central District

6 Storey Block

Ground area	= (12 + 48 + 2 + 10) × (12 + 48 + 2 + 10)
	= 5184m²

F.A.R	= total floor area / ground area
	=  11520 / 5184
	= 2.222

Population

Outer Districts

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

From this we must remove the extra space required for the trolleybus lines.

Building area per outer district	= area - transport
	= 282,743 - 9,600
	= 273,143m²

Removing the central car parks, there are effectively 30 outer districts.  This gives us a total building area of:

Now with an F.A.R of 1.11, and allowing 70m² per capita (40m² for housing, 30m² for work, services etc.) this gives a population of:

Central District

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

From this we must remove the extra space required for the trolleybus lines.

Building area of central district	= area - transport
	= 708,821 - 60,800
	= 648,021m²

Now with an F.A.R of 2.22, and allowing 70m² per capita for housing, work, services etc., this gives a population of:

Total Population

Adding the populations of the outer and central districts we get:

or approximately 150,000.

Construction Phase

The population of 1 whole outer district is:

With parking provided inside districts, this population shrinks to:

This is a difference of:

Allowing 25m² per car, this provides sufficient parking space for:

or 37 cars per 100 residents.  This should be more than enough in a city designed around public transport and carsharing, and more parking can be provided in the peripheral car parks if needed.

Public Transport

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.

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.

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

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.

or, rounding up, 5 minutes.

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

This gives us a longest journey of:

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 minute, and 90 seats per double articulated bus, then the total capacity for the morning rush-hour is:

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

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 60 second intervals, 11 trolleybuses are needed per line.  There are 8 lines and so 11 × 8 = 88 trolleybuses are needed, plus spares.

These 88 trolleybuses moving along 8 trolley lines replace the 50,000 private cars that would be needed in a conventional city and the 25+ road lanes these cars would require.

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