A cross-section is built from two lists: where the line crosses each contour, and how high that contour is. Plot them as points, join smoothly and check the slopes.
This lesson sits in contours, height and cross-sections. The steepness calculation it uses is taught in inferring slope steepness from contour spacing.
What are the steps?
- Draw the transect line on the map from P to Q.
- Lay a strip of paper along the line. Mark each contour crossing and write its height.
- Draw the graph: horizontal axis for distance, vertical axis for height.
- Place the strip on the horizontal axis and plot each height above its mark.
- Join the points smoothly and label the axes, scale and landforms.
Worked example
This is a worked example. The map scale is 1:50 000 and the contour interval is 20 m. A transect from P to Q is 7 cm long, and the contour crossings are recorded from P.
| Distance from P (cm) | Height (m) |
|---|---|
| 0 | 20 |
| 1.0 | 40 |
| 1.6 | 60 |
| 2.0 | 80 |
| 2.4 | 100 |
| 3.5 | 100 |
| 4.4 | 80 |
| 5.2 | 60 |
| 6.0 | 40 |
| 7.0 | 20 |
Axes. Horizontal: 1 cm on paper for 1 cm on the map, which is 500 m on the ground (1 × 50 000 = 50 000 cm = 500 m). Vertical: 1 cm for 20 m, so the graph is 5 cm tall for 100 m.
Shape. The profile rises from 20 m to 100 m, stays at or above 100 m between 2.4 and 3.5 cm (the summit area), then falls back to 20 m.
Gradients.
- Western side: from 0 to 2.4 cm is 2.4 × 500 = 1 200 m for an 80 m rise, so 80 ÷ 1 200 = 1 in 15.
- Eastern side: from 3.5 to 7.0 cm is 3.5 × 500 = 1 750 m for an 80 m fall, so 80 ÷ 1 750 = about 1 in 22.
Written reading. “The western slope is steeper than the eastern slope, at about 1 in 15 against 1 in 22.”
Vertical exaggeration. Horizontal 1 cm is 500 m and vertical 1 cm is 20 m, so the exaggeration is 500 ÷ 20 = 25 times. State this beside the graph.
The mistake: the flat-topped summit
The two 100 m crossings at 2.4 and 3.5 cm do not mean the top is flat at exactly 100 m. The transect passes over a summit area higher than 100 m but below the next contour, 120 m.
A student who joins the two points with a straight line at 100 m draws a flat top. Draw a gentle rise above 100 m between those marks and note that the summit height is between 100 m and 120 m.
Check yourself
Using the same map, the transect crosses the 40 m and 60 m contours at 1.0 cm and 1.6 cm from P. What is the gradient between these points?
Answer
Distance: 1.6 − 1.0 = 0.6 cm, and 0.6 × 500 = 300 m.
Rise: 60 − 40 = 20 m. Gradient: 20 ÷ 300 = 0.067, about 1 in 15.
This matches the western side overall, so the slope is fairly steady in this stretch.
What to study next
Use a finished profile to give a reasoned answer in explaining a route choice using relief and gradient. Then try the contours practice set.
The contour profile and slope reasoning trainer gives plotting drills. For a teacher to check your graph, see online one-to-one Geography tuition.