A contour cross-section is a side view of the ground along a chosen line. It is built from distances and heights, not from an imagined picture of a hill. Mark every contour crossing in order, transfer those distances to a baseline, and plot the corresponding elevations. A reliable profile preserves both the position and the height of each supplied point.
The maps here are original fictional teaching maps. Every necessary distance and elevation is also written in a table, so no external map sheet is required. The diagrams can resize on a screen: use the supplied distances or scale bars, not a ruler held against your phone. Stated centimetre lengths refer to the paper construction described in the exercise.
Read the map before plotting
A contour joins points at the same elevation. The contour interval is the vertical difference between successive contours, not the horizontal space separating them. On the first map, labelled contours are 40, 60, 80 and 100 metres above the stated datum. Their interval is therefore 20 m.
The straight line AB is the transect. It crosses the nested contours on the western side, passes through a supplied spot height of 110 m, and crosses them again on the eastern side. Both crossings of a given contour must appear in the profile. A height encountered twice is not a duplicate to delete.
Complete map description: A is the western endpoint and B is the eastern endpoint of a 2,000 m transect. Nested closed contours rise inwards. Along AB, the 40 m contour meets A and B; 60 m is crossed at 200 and 1,700 m from A; 80 m at 450 and 1,450 m; and 100 m at 700 and 1,200 m. The supplied hilltop spot height is 110 m at 950 m from A. These distances define the exercise precisely.
| Distance from A, metres | Position on paper, centimetres | Elevation, metres | Source of height |
|---|---|---|---|
| 0 | 0.0 | 40 | A on a contour |
| 200 | 0.8 | 60 | Contour crossing |
| 450 | 1.8 | 80 | Contour crossing |
| 700 | 2.8 | 100 | Contour crossing |
| 950 | 3.8 | 110 | Supplied spot height |
| 1,200 | 4.8 | 100 | Contour crossing |
| 1,450 | 5.8 | 80 | Contour crossing |
| 1,700 | 6.8 | 60 | Contour crossing |
| 2,000 | 8.0 | 40 | B on a contour |
Worked case 1: transfer the points and construct the profile
Task: Draw AB at horizontal scale 1:25,000 and vertical scale 1 cm representing 50 m. Since 25,000 cm = 250 m, each horizontal centimetre represents 250 m. AB therefore occupies 2,000/250 = 8 cm on paper. A point 450 m from A is placed 450/250 = 1.8 cm from the left endpoint.
Place a strip of paper along the original map transect and mark the endpoints and crossings. Label the height of every mark. On a digital or supplied-data question, the table serves the same purpose. Do not space the marks equally: the first gap is 200 m, the next is 250 m, and the final gap is 300 m.
Draw a horizontal distance axis and a vertical elevation axis. For this construction, let elevation zero be the baseline. With 1 cm representing 50 m, the 40 m points are 0.8 cm above it, the 60 m points 1.2 cm, the 80 m points 1.6 cm, the 100 m points 2.0 cm and the 110 m spot height 2.2 cm.
Transfer each horizontal mark vertically to its stated elevation. Join the points to show the inferred ground surface. The completed profile rises from 40 m at A to the known 110 m point, then descends to 40 m at B. The illustrated straight segments are a transparent classroom interpolation between known points; they are not measurements of every small surface irregularity.
A gently joined hand-drawn curve is also suitable if it passes through the known points and does not invent extra hills or depressions. The 110 m top is justified by the supplied spot height. Without that spot height, the 100 m contour alone would not justify labelling the summit exactly 110 m.
Worked case 2: calculate vertical exaggeration in matching units
Task: Find the vertical exaggeration of the profile just constructed. The horizontal scale is 1:25,000. The vertical scale is 1 cm to 50 m = 1 cm to 5,000 cm, or 1:5,000.
When scales are written 1:H horizontally and 1:V vertically, vertical exaggeration = H/V. Here VE = 25,000/5,000 = 5, often written 5×. Equivalently, compare the ground distances represented by one centimetre: 250 m horizontally divided by 50 m vertically is five.
The height dimension is drawn five times as large, relative to horizontal distance, as it would be with equal scales. The hill’s real height does not increase. Reversing the denominators gives 0.2 and describes vertical compression rather than the supplied exaggerated profile. Using 25,000/50 mixes centimetres with metres and creates a meaningless result.
A baseline at 20 m instead of zero could make the drawing fit a smaller area, provided the axis labels made that clear. Moving the baseline alone does not change the vertical scale or VE. Changing the distance represented by each vertical centimetre does.
Worked case 3: compare slopes without trusting apparent angles
Task: Compare the slope between 40 and 60 m near A with the corresponding slope near B. On the western side, the rise is 20 m over a horizontal ground distance of 200 m. Its gradient magnitude is 20/200 = 1/10, or 1 in 10.
On the eastern side, the elevation changes by 20 m over 2,000 − 1,700 = 300 m. The gradient magnitude is 20/300 = 1/15, or 1 in 15. The western segment is steeper because the same vertical change occurs over a shorter horizontal distance. Walking from A towards B, the western change is uphill and the eastern change is downhill.
Do not measure an angle on the exaggerated profile and call it the real slope angle. Its vertical dimensions have been enlarged relative to its horizontal ones. Calculate real gradients from ground distances and elevations. Close contour spacing indicates steeper ground when contour interval and map scale are the same.
Three independent practice tasks with solutions
Practice 1: construct a new hill profile
Map B data: A new transect CD is 1,200 m long at horizontal scale 1:20,000. Closed contours rise inwards at intervals of 20 m. Their crossings and the central spot height are listed below. Use a vertical scale of 1 cm representing 40 m, and take elevation zero as the baseline.
| Distance from C, metres | Elevation, metres | Paper position, centimetres | Height above baseline, centimetres |
|---|---|---|---|
| 0 | 40 | 0 | 1.00 |
| 200 | 60 | 1 | 1.50 |
| 400 | 80 | 2 | 2.00 |
| 600 | 90, spot height | 3 | 2.25 |
| 800 | 80 | 4 | 2.00 |
| 1,000 | 60 | 5 | 1.50 |
| 1,200 | 40 | 6 | 1.00 |
Solution: One horizontal centimetre represents 200 m, so the profile is 6 cm long. Plot the paper coordinates in the last two columns. The profile is symmetric about the 90 m point at 3 cm. The vertical scale is 1:4,000, so VE = 20,000/4,000 = 5×. The change from 80 to 90 m is only 10 m because 90 is a spot height, not the next contour.
Practice 2: change only the vertical scale
Task: Redraw the original AB profile at the same horizontal scale, but use 1 cm representing 25 m vertically. What changes? Solution: Vertical scale becomes 1:2,500, so VE = 25,000/2,500 = 10×. Heights measured from the zero baseline double on paper: the 110 m point is now 4.4 cm above it. Horizontal positions and real elevations remain unchanged.
Practice 3: find a missing vertical scale
Task: A different profile has horizontal scale 1:50,000 and must have VE = 10. Find a suitable vertical scale in words. Solution: 10 = 50,000/V, so V = 5,000. A vertical scale of 1:5,000 means 1 cm representing 50 m. Check: each horizontal centimetre represents 500 m; 500/50 = 10.
Frequently asked questions
What is the difference between contour interval and spacing?
Interval is the vertical elevation difference between successive contours. Spacing is their horizontal separation on the map and can vary along a slope.
Should I mark a contour that the transect crosses twice?
Yes. Each crossing has its own horizontal position, even though both have the same height.
Can I assume the summit equals the highest contour?
No. A summit inside a closed contour generally rises above it. Use supplied spot heights or clearly state what the contours alone allow you to infer.
Does vertical exaggeration change the elevation labels?
No. It changes the drawing scale. A 110 m point remains 110 m whether VE is one, five or ten.
Must a profile start at sea level?
No. A labelled truncated elevation axis is possible. Keep its interval uniform and distinguish its starting value from the vertical scale.
Can I measure centimetres from the diagram on my screen?
No. Screen resizing changes displayed dimensions. Use the supplied distance table or a correctly interpreted scale bar, and construct the specified paper dimensions yourself.
Why is my profile steep-looking even where contours are far apart?
A large vertical exaggeration can make gentle ground look steep. Compare actual rise and horizontal ground distance before interpreting the real slope.
Further map-work practice
The Geography topical revision collection is existing navigation; its resource contents may require membership. For background on contour mapping, see the US Geological Survey’s topographic mapping explanation. All map geometry, distances, heights, profiles and exercises in this lesson were created for this lesson.
WhatsApp Us Now