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it lies between 0.2 and 0.4. At the midpoint, 0.3, an increase or decrease of area by

a factor of 10 results in a doubling or halving respectively of the number of species

(by virtue of 100.3=2). Thus, whenA=1, S=Cirrespective of the value ofz; and

whenA=10 andz=0.3, S= 2 C.

The relationship is the same if we count the number of species on nested areas of

progressively larger size on a continent. Here the value ofztends to be lower, usu-

ally around 0.15. It implies that a reduction of area by a factor of 10 reduces the

number of species by a factor of only 1.4 (100.15=1.41). The difference between that

328 Chapter 18

A species–area curve takes the form S=CAz, where:

S=number of species

A=area, in this case always expressed as km^2

C=expected number of species on an area of 1 km^2

z=slope of the curve relating species number to area

Taking the data of Table 18.1 as our example, first convert area and species number to logarithms.

Any base will do but we will use logs to the base e. Label log area as xand log species number as y:

xy

11.126 2.3026

7.193 1.9459

7.003 1.7918

6.098 1.7918

4.745 1.3863

2.996 1.6094

2.303 0.6931

2.197 0.6931

2.079 1.0986

2.079 0.6931

1.792 0.6931

We now calculate these:

n= 11

mean x=4.510 mean y=1.336

∑x=49.61 ∑y=14.70

∑x^2 =315.2 ∑xy=82.58

(∑x)^2 /n=223.7_____ (∑x)(∑y)/n=66.30_____

SSx=∑x^2 −(∑x)^2 /n=91.5 SSxy=∑xy−(∑x)(∑x)/n=16.28

The constants of the species–area curve are now solved:

z=SSxy/SSx

=16.28/91.5

=0.18

C=antilog(∑y/n−z∑x/n)

=exp(1.336 −0.18 ×4.51)

=1.70

Thus S=1.7A0.18

Box 18.3Estimating the
constants of a
species–area curve.