Cross Section and Sketch Maps

Lessons

Senior 6 · Geography

Cross Section and Sketch Maps

4 lessons

Lesson 2 of 4

Vertical Exaggeration, Gradient, Amplitude and Intervisibility

25 min

Four measurements from one section

A cross section is more than a picture. Once it is drawn you can measure from it: how much the vertical scale has stretched the land, how steep each slope is, how far the land rises and falls, and whether one place can be seen from another. These four measurements (vertical exaggeration, gradient, amplitude of relief and intervisibility) are the calculations most often set on a map extract. This lesson works all four on Map 1 and the section A–B from the previous lesson, so keep Figure 3 in mind; Map 1 is repeated here.

1. Vertical exaggeration

Along A–B the land rises and falls by 163 m over 5.6 km. Drawn at the same scale in both directions, that relief would be a line barely thicker than the pencil. So the vertical scale of a section is made larger than the horizontal scale, which stretches the land upwards. Figure 4 shows the same section three times.

Figure 4: the section A–B at three vertical exaggerations.

Definition

Vertical exaggeration (VE): the number of times the vertical scale of a section is larger than its horizontal scale. Both scales must first be written as representative fractions (RF):

\text{VE} = \frac{\text{vertical RF}}{\text{horizontal RF}} = \frac{\text{denominator of the horizontal RF}}{\text{denominator of the vertical RF}}

Example: horizontal 1:50 000 and vertical 1:5 000 give VE = 10, often written "×10".

Worked example 1: the VE of Figure 3, and what it does to slopes

Problem. Figure 3 was drawn with a vertical scale of 1 cm to 50 m on a 1:50 000 map. (a) Find the VE. (b) A real slope of 5° is drawn on this section. What angle does it appear to have?

Solution. (a) Convert the vertical scale to an RF: 50 m = 5 000 cm, so 1 cm to 50 m is 1:5 000.

\text{VE} = \frac{50\,000}{5\,000} = 10

(b) Exaggeration multiplies every height by 10 but leaves distances unchanged, so it multiplies the tangent of each slope angle by 10:

\tan\theta' = \text{VE}\times\tan\theta = 10\times\tan 5^\circ = 0.875, \qquad \theta' = \tan^{-1}(0.875) = 41.2^\circ

A gentle 5° slope looks like a 41° slope. So never measure a real slope angle with a protractor on an exaggerated section.

Warning

Convert before you divide. "1 cm to 50 m" is not 1:50. Change the ground distance to centimetres first: 50 m × 100 = 5 000 cm, so the RF is 1:5 000. A VE of 1 000 or 0.1 is a sign that the units were mixed.

2. Gradient

The gradient of a slope compares how far the land rises with how far it runs. On a map you take both values from the contours and the scale, so you can find a gradient without drawing a section at all.

Definition

Vertical interval (VI): the difference in height between two points. Horizontal equivalent (HE): the horizontal distance between them on the ground, measured on the map and converted with the scale.

Gradient: the ratio of VI to HE, with both in the same units:

\text{gradient} = \frac{\text{VI}}{\text{HE}}

It is written in three ways: as "1 in $n$ " where $n = \frac{\text{HE}}{\text{VI}}$ ; as a percentage, $\frac{\text{VI}}{\text{HE}}\times 100$ ; or as an angle, $\theta = \tan^{-1}\left(\frac{\text{VI}}{\text{HE}}\right)$ . Example: a rise of 10 m over 200 m is 1 in 20, 5%, or 2.9°.

Worked example 2: from the trig station to the bridge

Problem. Find the average gradient from the trig station on Hill X (1,671 m) to the bridge over River Z (1,508 m), in all three forms. Then find the gradient of the steepest part of the east face, between the 1,560 m and 1,540 m contours.

Solution.

VI = 1 671 − 1 508 = 163 m.
HE: the trig station is 1.90 km from A and the bridge 3.70 km, so HE = 1.80 km = 1 800 m.

n = \frac{1\,800}{163} = 11.0 \quad\Rightarrow\quad \text{gradient} = 1 \text{ in } 11

\frac{163}{1\,800}\times 100 = 9.1\%, \qquad \theta = \tan^{-1}\left(\frac{163}{1\,800}\right) = 5.2^\circ

This is an average. The contours show that the slope is not uniform. Between 1,560 m (2.86 km) and 1,540 m (2.95 km), VI = 20 m and HE = 0.09 km = 90 m:

n = \frac{90}{20} = 4.5 \quad\Rightarrow\quad 1 \text{ in } 4.5,\ \ 22\%,\ \ \tan^{-1}\left(\frac{20}{90}\right) = 12.5^\circ

Important

A gradient given as "1 in $n$ " gets steeper as $n$ gets smaller: 1 in 4.5 is much steeper than 1 in 11. When a question asks for the steepest part of a slope, look for where the contours are closest together.

3. Amplitude of relief

Definition

Amplitude of relief (also called relative relief or local relief): the difference in height between the highest and the lowest points of a section or of an area.

\text{amplitude} = \text{highest point} - \text{lowest point}

Example: along A–B the highest point is 1,671 m and the lowest 1,508 m, so the amplitude is 163 m.

Worked example 3: amplitude of the whole map

Problem. What can you say about the amplitude of relief of the whole area of Map 1?

Solution. The highest point on the map is the trig station, 1,671 m. The lowest ground lies along River Z. The river flows south, so its bed falls below 1,508 m downstream of the bridge; but no 1,500 m contour appears anywhere, so it stays above 1,500 m. The amplitude is therefore more than $1\,671 - 1\,508 = 163$ m and less than $1\,671 - 1\,500 = 171$ m. A precise value would need a spot height at the southern edge of the map. Giving a range with its reason is better than inventing one number.

4. Intervisibility

A mobile-network company placing a relay mast, an army choosing a lookout, and a surveyor setting up a line of beacons all ask the same question: can this point be seen from that one?

Definition

Intervisibility: two points are intervisible if the straight line joining them, the line of sight, passes above all the ground between them. Ground that an observer cannot see is called dead ground.

Example: in Figure 5 the trig station on Hill X and the spot height on Hill Y are intervisible, but the bridge is in dead ground.

Graphical method. Draw the section through both points. Rule a straight line from one to the other. If the ground rises above the line anywhere, the points are not intervisible.

Calculation method. Find the likely obstacles between the points: a crest, or the shoulder at the top of a convex slope. At each one, work out the height of the line of sight by proportion. If an observer at height $h_{1}$ looks at a target at height $h_{2}$ a distance $D$ away, the line of sight at a distance $d$ from the observer is at height

h = h_1 + (h_2 - h_1)\times\frac{d}{D}

If $h$ is lower than the ground at the obstacle, the view is blocked.

Figure 5: lines of sight on the section A–B, for an observer whose eye is 1.5 m above the trig station. The thick red line marks dead ground: the east face of Hill X below its shoulder, the whole valley of River Z and the lower west face of Hill Y (from about 2.0 km to 4.4 km from A), and everything beyond the summit of Hill Y (after about 5.0 km).

Worked example 4: testing three views by calculation

(a) Trig station on Hill X (1,671 m, 1.90 km) to the spot height on Hill Y (1,607 m, 5.00 km). $D = 3.10$ km. The obstacles are the shoulder of Hill X at the 1,660 m contour (2.10 km, so $d = 0.20$ km) and the 1,600 m contour on Hill Y (4.86 km, $d = 2.96$ km).

h_{2.10} = 1\,671 + (1\,607 - 1\,671)\times\frac{0.20}{3.10} = 1\,666.9\ \text{m} > 1\,660\ \text{m}

h_{4.86} = 1\,671 - 64\times\frac{2.96}{3.10} = 1\,609.9\ \text{m} > 1\,600\ \text{m}

The line clears both, so the two summits are intervisible.

(b) Trig station to the bridge (1,508 m, 3.70 km). $D = 1.80$ km; test the same shoulder, $d = 0.20$ km:

h_{2.10} = 1\,671 - 163\times\frac{0.20}{1.80} = 1\,652.9\ \text{m} < 1\,660\ \text{m}

The ground is 7 m above the line: the bridge is not intervisible with the trig station. The convex east face hides its own lower slopes, which is why a convex slope always creates dead ground below its shoulder.

(c) A to B. Both ends are at 1,540 m and Hill X rises to 1,671 m between them, so they are clearly not intervisible. No calculation is needed when the obstacle is higher than both points.

Tip

Exam technique. For intervisibility, state your method, show the line of sight on your section, and name the obstacle ("the view is blocked by the convex shoulder of Hill X at about 1,660 m"). A bare "yes" or "no" earns little. For a gradient, always show VI and HE with their units before you divide.

Everything Geography by Thapelo: "Geography mapwork: Cross section/mapping and intervisibility". A worked mapwork example that ends by drawing a line of sight on the finished section. Compare the method with the graphical method above.

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Practice

Practice (use Map 1 and Figures 3 and 5; ignore eye height unless told otherwise)

A section is drawn from a 1:25 000 map with a vertical scale of 1 cm to 20 m. Find its vertical exaggeration.
You want a VE of 8 on a section from a 1:50 000 map. What vertical scale must you use? Give it as an RF and as "1 cm to … m".
Find the average gradient from A (1,540 m) to the trig station on Hill X, as "1 in $n$ " and as an angle.
Find the gradient from the bridge to the summit of Hill Y as "1 in $n$ ", as a percentage and as an angle.
What is the amplitude of relief along the part of the section from the bridge to B?
Is the summit of Hill Y visible from A? Show your working.
A surveyor standing at the bridge says she can see the spot height on the summit of Hill Y. Test her claim by calculation, using the 1,600 m contour on Hill Y as the obstacle.
A student measures the east face of Hill X on Figure 3 with a protractor and reads 50°. Explain why this is not the real angle, and find the real angle that a 50° slope on this section represents.

Show answersHide answers

20 m = 2 000 cm, so the vertical RF is 1:2 000. $\text{VE} = \frac{25\,000}{2\,000} = 12.5$ .
Vertical RF denominator = $\frac{50\,000}{8} = 6\,250$ , so 1:6 250. Then 6 250 cm = 62.5 m: 1 cm to 62.5 m.
VI = 1 671 − 1 540 = 131 m; HE = 1.90 km = 1 900 m. $n = \frac{1\,900}{131} = 14.5$ : 1 in 14.5. Angle $\tan^{-1}\left(\frac{131}{1\,900}\right) = 3.9^\circ$ .
VI = 1 607 − 1 508 = 99 m; HE = 5.00 − 3.70 = 1.30 km = 1 300 m. 1 in 13.1; $\frac{99}{1\,300}\times 100 = 7.6\%$ ; $\tan^{-1}\left(\frac{99}{1\,300}\right) = 4.4^\circ$ .
Highest 1,607 m (Hill Y), lowest 1,508 m (bridge): 99 m.
No. $D = 5.00$ km; at the trig station, $d = 1.90$ km: $h = 1\,540 + 67\times\frac{1.90}{5.00} = 1\,565.5$ m, far below the ground at 1,671 m. Hill X blocks the view.
$D = 1.30$ km and $d = 4.86 - 3.70 = 1.16$ km: $h = 1\,508 + 99\times\frac{1.16}{1.30} = 1\,596.3$ m. The ground there is 1,600 m, about 4 m above the line, so the claim is wrong: the rounded top of Hill Y hides its own spot height. (With her eye 1.5 m above the ground the line is still about 3.5 m too low.)
Figure 3 has a VE of 10, so every slope is drawn far steeper than it is. $\tan\theta = \frac{\tan 50^\circ}{10} = 0.119$ , so $\theta = 6.8^\circ$ . Real angles must be found from VI and HE, not from the drawing.

Summary

VE = horizontal RF denominator ÷ vertical RF denominator; it steepens every slope on the drawing.
Gradient = VI ÷ HE in the same units: "1 in $n$ ", a percentage or an angle.
Amplitude of relief = highest point − lowest point.
Points are intervisible if the line of sight clears all the ground between them; convex slopes create dead ground.

Sources and credits

Written by Ganzaa. Map 1, Figures 4 and 5, the worked examples and the exercises were drawn and written by Ganzaa; the dead ground in Figure 5 was computed from the drawn section.

Video: "Geography mapwork: Cross section/mapping and intervisibility", Everything Geography by Thapelo, embedded from YouTube.

1Four measurements from one section

21. Vertical exaggeration

32. Gradient

43. Amplitude of relief

54. Intervisibility

Four measurements from one section

1. Vertical exaggeration

2. Gradient

3. Amplitude of relief

4. Intervisibility