The following examples illustrate how the graphs in Table 6 are used for each of the three types of problem (see Figure A1). It is assumed in these examples that an allowance will need to be made for the provision of LOS within the DSA but outside the NSA (see section 6.5).

Problem type 1 : How much population can a given site area accommodate at a given plot ratio (PR)?

Example:

Given :

Site Area

14ha

Ppf

3.5

GFApf

50m^{2}

PR

6.5

therefore

m^{2} GFApp

= 50÷3.5

=

14.3

and

NSApp

= 14.3÷6.5

=

2.2

Strike a vertical line on the graph from 14ha on the X-axis. This meets the 2m^{2} NSApp curve at population 32 000 and the 3m^{2} NSApp curve at 25 000. The required population is therefore about 30 600.

Problem type 2 : What site area is required for a fixed population at a given plot ratio?

Example:

Given :

Population

20 000

Ppf

2.8

GFApf

60m^{2}

PR

5

therefore

GFApp

= 60÷2.8

=

21.4m^{2}

and

NSApp

= 4.3

Strike a horizontal line on the graph from 20 000 population on the Y-axis. This meets the 4m^{2} NSApp curve at 13.4ha, and the 5.0m^{2} NSApp curve at 15.7ha. The required area is therefore around 14.1ha.

Problem type 3 : What Plot Ratio is required to accommodate a fixed population on a fixed site area?

Example:

Given :

Population

10 000

Site Area

7ha

Ppf

3.0^{}

GFApf

60m^{2}

therefore

GFApp

20m^{2}

Strike a horizontal line on the graph, from 10 000 population on the Y axis to meet a vertical line from 7ha on the X axis. These meet on the between the 4 and 5m^{2} NSApp curves, say NSApp is 4.3m^{2}.

NSApp

=

GFApp÷PR

i.e.

4.3

=

20÷PR

therefore

PR

=

4.65

It should be noted when interpolating between the curves for NSApp that some allowance may need to be made for the fact that the intervals between curves are not regular. Otherwise minor discrepancies may result, compared with the correct values.