The Maths
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
nonresidential uses at a F.A.R of 1.8, then the ground space
required per person for nonresidential uses is given by:
Ground per person (nonres) 
= 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 
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.
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 rushhour, 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 rushhour 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
rushhour 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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