Species-Area Relationship Calculator

Calculate species richness from area using MacArthur-Wilson species-area relationship.

Area A (km² or ha) (km²)

Reference area A₀ (km²) (km²)

Species at reference area S₀ (species)

z exponent (0.2-0.35 typical)

Islands: ~0.3, Mainland: ~0.15

Area reduced to fraction (habitat loss)

e.g. 0.5 = 50% habitat remains

Please check your inputs and try again.

Why Habitat Loss Costs Less Than Expected — Initially

The power law means species loss is strongly non-linear. Removing half a habitat does not remove half its species, because the remaining area still contains most common species.

S = cA^z S_remaining/S_original = (fraction remaining)^z

Habitat remaining	Species retained (z = 0.25)	Species lost
90%	97%	3%
50%	84%	16%
10%	56%	44%
1%	32%	68%

The reassuring figures at the top are misleading, because the relationship steepens as area shrinks. Going from 50% to 10% habitat loses far more species than going from 100% to 50% did. The last fragments matter disproportionately.

There is also an extinction debt: species persist in fragments for generations before declining populations finally disappear, so observed richness lags the equilibrium the equation predicts.

What z Tells You

z value	Setting
0.10–0.20	Nested samples within a continuous habitat
0.20–0.35	Typical islands and fragments
> 0.35	Highly isolated islands, poor dispersers
Higher z	Greater sensitivity to area loss

Isolation raises z, because immigration cannot rescue small islands from local extinction. This is why fragmentation is more damaging than simple area loss — the same total area divided into isolated pieces behaves like a set of small islands rather than one large one.

The SLOSS Debate

Whether a Single Large Or Several Small reserve conserves more species has no universal answer. Several small reserves capture more habitat heterogeneity and spread risk; one large reserve supports viable populations of area-demanding species and has less edge effect.

The species-area relationship alone favours a single large reserve, since one big patch holds more species than the same area fragmented. In practice the answer depends on the taxa, the landscape and what is already protected — and connectivity between fragments often matters more than the choice itself.

Worked Examples

Example 1: 100km² vs 1km² reference, S0=10, z=0.25

S=10×(100/1)^0.25=10×3.16

Result: S=31.6 species predicted

10× area ~ 2× species (for z=0.3)

Example 2: 50% habitat loss: S0=100, z=0.25

S_remaining=100×0.5^0.25

Result: S=84 species remain (16% lost)

Much more than expected! Area loss is not linear

Example 3: Why the last fragments matter most

50% to 10% habitat remaining

Result: 84% to 56% of species

The second reduction costs nearly twice as many species as the first, despite removing less absolute area. The curve steepens as area shrinks.

Example 4: Extinction debt

Fragment reduced decades ago

Result: Richness still above equilibrium

Populations decline slowly toward the predicted level. Current richness overstates what the fragment will ultimately support.

Common Mistakes

⚠️

Assuming species loss is proportional to area loss

The relationship is a power law. Losing half the habitat costs about 16% of species at z = 0.25, but the curve steepens sharply thereafter.

⚠️

Ignoring extinction debt

Species persist in fragments for generations before disappearing. Observed richness lags the equilibrium prediction, understating eventual loss.

⚠️

Using one z value everywhere

It varies from about 0.1 for nested samples to above 0.35 for isolated islands. Isolation raises z and sensitivity to area loss.

⚠️

Treating fragmentation as equivalent to area loss

The same area divided into isolated pieces behaves like small islands. Connectivity between fragments substantially affects the outcome.

Frequently Asked Questions

The SLOSS debate?

Single Large Or Several Small reserves? SAR predicts large reserve has more species. But small reserves may protect different species (beta diversity) and allow subpopulations. Modern consensus: large connected reserves best + corridors.

z-value interpretation?

z≈0.25 for oceanic islands. z≈0.12-0.15 for mainland areas (sampling effect). z≈0.35 for true islands with high isolation. Higher z: more rapid species gain with area. Used to estimate extinction debt from habitat fragmentation.

Why isn't species loss proportional to habitat loss?

Because the relationship is a power law. At z = 0.25, half the habitat retains about 84% of species — but the curve steepens as area shrinks further.

What does the z exponent represent?

How steeply species number responds to area. Values near 0.1 apply within continuous habitat and above 0.35 to isolated islands.

What is extinction debt?

Species persisting in a reduced habitat that will eventually be lost as populations decline. Observed richness lags the equilibrium prediction.

Does the species-area relationship settle the SLOSS debate?

It favours a single large reserve, but heterogeneity, risk spreading and connectivity all matter. The answer depends on taxa and landscape context.