SME
Ecology · Problem type
Species and area: S = cAᶻ
Use this for “If a reserve keeps half its area and z = 0.25, what share of species remains?”
Chart
S(10,000) = 200
This is an example. Type your own values to start a new problem.
q
=
p
c
Species in 1 km²
example
A
Area
example · km²
S
Species
calculated
S₂
Species left
calculated
Formulas
S
=
c
×
A
z
200
=
20
×
10,
000
0.25
q
=
p
z
0.841
=
0.5
0.25
S₂
=
q
×
S
168 = 0.841 × 200
Assumptions
•
The species–area rule S = cAᶻ holds over this range of areas, with the same c and z.
•
z is about 0.15 to 0.35: higher on islands and isolated reserves than within a mainland.
•
Keeping a share p of the area keeps a share pᶻ of the species, once the losses settle.
•
c is the species in 1 km², so A is in km².
Step-by-step
WE KNOW
A = 10,000 km², c = 20, z = 0.25, p = 0.5
FIND
species (S), share of species kept (q), species left (S₂)
Step 1 · Find species (S)
Use S = c × Aᶻ
Raise
the
area
to
z
and
multiply
by
c
.
S
=
20
×
10,
000
0.25
S
=
20
×
10
S
=
200
Step 2 · Find share of species kept (q)
Use q = pᶻ
S
=
cAᶻ,
so
a
share
p
of
the
area
keeps
pᶻ
of
S
.
q
=
0.5
0.25
q
=
0.841
Step 3 · Find species left (S₂)
Use S₂ = q × S
The kept share of the species there now.
S₂
=
0.841
×
200
S₂
=
168
Check
Put the numbers back in. Both sides should match.
200
=
20
×
10,
000
0.25
✓
0.841
=
0.5
0.25
✓
168 = 0.841 × 200 ✓
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