Slope and Aspect from a Gridded DEM: 18-Geom-B1 Digital Terrain Modelling, December 2017, Q4

Question 4: Slope and Aspect from a Gridded DEM (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 5×5 gridded DEM, cell size $\Delta = 20\text{ m}$ (both directions), with north up. The centre cell (highlighted) has elevation $z_0 = 20\text{ m}$. Its four cardinal neighbours are:

Neighbour	North (N)	South (S)	East (E)	West (W)
Elevation (m)	20	40	40	20

Find. The slope angle and the aspect (azimuth of steepest descent) at the centre point, using the 4-neighbour (rook) finite differences only.

The 5×5 DEM (20 m cells). The centre cell ($z_0=20$ m) is highlighted; its four cardinal neighbours used for the 4-neighbour gradient are outlined N/S/E/W.

Approach. Estimate the two partial derivatives of the surface at the centre from the opposite cardinal neighbours (central differences), combine them into a gradient magnitude for the slope angle, and take the azimuth of the downslope direction for the aspect.

East–west gradient (central difference). With $x$ positive to the east, $$\frac{\partial z}{\partial x} = \frac{z_E - z_W}{2\,\Delta} = \frac{40 - 20}{2(20)} = 0.500$$ The surface rises 0.5 m per metre toward the east.
North–south gradient (central difference). With $y$ positive to the north, $$\frac{\partial z}{\partial y} = \frac{z_N - z_S}{2\,\Delta} = \frac{20 - 40}{2(20)} = -0.500$$ The negative sign shows the surface falls toward the north (it rises to the south).
Slope angle from the gradient magnitude. The steepest slope is the magnitude of the gradient vector: $$\beta = \arctan\!\sqrt{\left(\tfrac{\partial z}{\partial x}\right)^2 + \left(\tfrac{\partial z}{\partial y}\right)^2} = \arctan\!\sqrt{0.5^2 + 0.5^2} = \arctan(0.7071)$$ $$\boxed{\beta = 35.26^\circ}$$
Aspect as the azimuth of steepest descent. The gradient vector $(\partial z/\partial x,\ \partial z/\partial y) = (+0.5,\ -0.5)$ points uphill — toward the east and south, i.e. the south-east. Aspect is the direction the slope faces (downhill), the opposite azimuth. Measuring the downslope azimuth clockwise from north, $$\alpha = \operatorname{atan2}\!\big(-\tfrac{\partial z}{\partial x},\ -\tfrac{\partial z}{\partial y}\big) = \operatorname{atan2}(-0.5,\ +0.5) = -45^\circ \equiv \boxed{315^\circ \ (\text{NW})}$$ Consistent with the data: the ground climbs to the SE (E and S neighbours are higher), so it faces the NW.

Quantity	Value
$\partial z/\partial x$ (east +)	$+0.500$
$\partial z/\partial y$ (north +)	$-0.500$
Gradient magnitude	$0.7071$
Slope angle $\beta$	$35.26^\circ$
Aspect $\alpha$ (descent azimuth)	$315^\circ$ — north-west

Concept & Theory

Topic. First-derivative terrain parameters (slope and aspect) by finite differences on a grid.

Key relations. $\dfrac{\partial z}{\partial x}=\dfrac{z_E-z_W}{2\Delta}$, $\dfrac{\partial z}{\partial y}=\dfrac{z_N-z_S}{2\Delta}$; slope $\beta=\arctan\sqrt{z_x^2+z_y^2}$; aspect $\alpha=$ azimuth of $-\nabla z$ (steepest descent), clockwise from north.

Why this works. A central difference between opposite neighbours is a second-order-accurate estimate of the local partial derivative; the gradient vector points up the steepest ascent, so its magnitude is the tangent of the slope angle and its reverse azimuth is the aspect. The 4-neighbour scheme uses only the cardinal cells (the 8-neighbour Horn method also weights the diagonals).

Common pitfall. Sign/orientation errors — forgetting that grid rows increase southward so $z_N$ is the row above, and confusing aspect (downhill, faces the valley) with the uphill gradient azimuth. Aspect is undefined on a flat cell ($z_x=z_y=0$).

Source. Li, Zhu & Gold, ch. 6 (terrain derivatives); Burrough & McDonnell, Principles of GIS, ch. 8.

Question 4 of 12: Slope and Aspect from a Gridded DEM

Question 4: Slope and Aspect from a Gridded DEM (12 marks)