SME
Remote Sensing · Problem type
Percent and yearly rate of area change
Use this for “A wetland shrank from 1,200 km² to 1,050 km² in 10 years. What are its percent change and its yearly rate?”
Diagram
12.5% of 1,200 km² = 150 km²
1,200 km² − 150 km² = 1,050 km²
This is an example. Type your own values to start a new problem.
r (%/yr)
=
100
×
ln
A₂ (km²)
A₁ (km²)
÷
t (years)
ΔA
Change in area
worked out (not typed) · km²
p
Percent change
calculated · %
Formulas
ΔA
=
A₂
−
A₁
−150 = 1,050 − 1,200
p
=
100
×
ΔA
A₁
−12.5
=
100
×
−150
1,200
r
=
100
×
ln(
A₂
÷
A₁
)
t
−1.3353
=
100
×
ln(1,050 ÷ 1,200)
10
Assumptions
•
Both areas come from images classified the same way, so the change is on the ground, not in the method.
•
The percent change compares the change with the first area; a negative change is a loss.
•
The yearly rate r is steady, continuous change, as deforestation studies use: over t years the area is multiplied by e to the power rt ÷ 100.
Step-by-step
WE KNOW
A₁ = 1,200 km², A₂ = 1,050 km², t = 10 years
FIND
change in area (ΔA), percent change (p), rate of change (r)
Step 1 · Find change in area (ΔA)
Use ΔA = A₂ − A₁
The area after less the area before: negative when land cover was lost.
ΔA = 1,050 − 1,200
ΔA = −150 km²
Step 2 · Find percent change (p)
Use p = 100 × ΔA ÷ A₁
The change as a share of the first area, in percent.
p
=
100
×
−150
1,200
p
=
−
15,000
1,200
p
=
−12.5%
Step 3 · Find rate of change (r)
Use r = 100 × ln(A₂ ÷ A₁) ÷ t
The log of the area ratio is the whole continuous change; share it over the years, as a percent.
r
=
100
×
ln(1,050 ÷ 1,200)
10
1,050
1,200
=
0.875
ln(0.875) = −0.1335
r
=
−1.3353
%/yr
Check
Put the numbers back in. Both sides should match.
−150 = 1,050 − 1,200 ✓
−12.5
=
100
×
−150
1,200
✓
−1.3353
=
100
×
ln(1,050 ÷ 1,200)
10
✓
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