SME
Geographic Information Systems (GIS)
Geoprocessing
Use this for “On a 10 m DEM a cell’s neighbours stand 112 m east, 100 m west, 106 m north and 98 m south. Find its slope and aspect.”
Diagram
East:
112 − 100
2 × 10
=
0.6;
north:
106 − 98
2 × 10
=
0.4.
Slope
=
tan⁻¹
√
0.6
2
+
0.4
2
=
35.8°
(72.1%).
Downhill is −east, −north: the slope faces 236° from north (SW).
This is an example. Type your own values to start a new problem.
c
Cell size
example · m
zE
Elevation east
example · m
zW
Elevation west
example · m
zN
Elevation north
example · m
zS
Elevation south
example · m
gE
East gradient
worked out (not typed)
gN
North gradient
worked out (not typed)
θ
Slope
calculated · °
s
Slope (percent)
calculated · %
α
Aspect
calculated · °
Formulas
gE
=
zE − zW
2 ×
c
0.6
=
112 − 100
2 × 10
gN
=
zN − zS
2 ×
c
0.4
=
106 − 98
2 × 10
s
=
100
×
√
gE
2
+
gN
2
72.111
=
100
×
√
0.6
2
+
0.4
2
θ
=
tan⁻¹(
s
100
)
35.7958
=
tan⁻¹(
72.111
100
)
α
=
the
bearing
downhill
for
east
gE
and
north
gN
236.3099 = the bearing downhill for east 0.6 and north 0.4
Assumptions
•
The centre cell’s own height does not enter: its four neighbours, two cells apart, set the slope.
•
The ground is treated as one tilted plane across the 3 × 3 window.
•
Aspect is the way the slope faces (downhill), in degrees clockwise from north; flat ground has none.
Step-by-step
WE KNOW
c = 10 m, zE = 112 m, zW = 100 m, zN = 106 m, zS = 98 m
FIND
east gradient (gE), north gradient (gN), slope (percent) (s), slope (θ), aspect (α)
Step 1 · Find east gradient (gE)
Use gE = (zE − zW) ÷ (2 × c)
The rise from west to east over the two cells between them: metres of climb per metre.
gE
=
112 − 100
2 × 10
gE
=
12
20
gE = 0.6
Step 2 · Find north gradient (gN)
Use gN = (zN − zS) ÷ (2 × c)
The rise from south to north over the two cells between them.
gN
=
106 − 98
2 × 10
gN
=
8
20
gN = 0.4
Step 3 · Find slope (percent): s
Use s = 100 × √(gE² + gN²)
The two gradients add like the sides of a right triangle to the steepest climb, written per 100 m.
s
=
100
×
√
0.6
2
+
0.4
2
s
=
100
×
√
0.36 + 0.16
s
=
100
×
√
0.52
s
=
100
×
0.7211
s
=
72.111%
Step 4 · Find slope (θ)
Use θ = tan⁻¹(s ÷ 100)
The slope angle is the one whose tangent is the rise per metre.
θ
=
tan⁻¹(
72.111
100
)
θ
=
35.7958°
Step 5 · Find aspect (α)
Use α = the bearing downhill for east gE and north gN
Water runs against both gradients (east −gE, north −gN). tan⁻¹ of their sizes gives the angle off north or south; turn it into the right quadrant.
α
=
the
bearing
downhill
for
east
0.6
and
north
0.4
tan⁻¹(
0.6
0.4
)
=
56.3099°
180° + 56.3099° = 236.3099°
α
=
236.3099°
Check
Put the numbers back in. Both sides should match.
0.6
=
112 − 100
2 × 10
✓
0.4
=
106 − 98
2 × 10
✓
72.111
=
100
×
√
0.6
2
+
0.4
2
✓
35.7958
=
tan⁻¹(
72.111
100
)
✓
236.3099 = the bearing downhill for east 0.6 and north 0.4 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.