Introduction

Surveying is based on two fundamental principles. Every survey, large or small, follows them to control error and to fix points correctly.

1. Working from whole to part.
2. Fixing the position of a new point by at least two independent measurements from points already fixed.

(A third working rule, always check the measurements, supports both.)

Principle 1: Working from whole to part

First a framework of control stations is fixed over the whole area with high accuracy (a few widely spaced stations, using precise instruments). Then the area is divided into smaller parts and the details are surveyed inside this framework with less precise methods.

Reason: errors do not accumulate. If we worked from part to whole, small errors of each part would add up and grow. If a mistake occurs in a small part, it stays inside that part and does not spoil the whole survey.

Example: For a large town map, main triangulation stations are fixed first with a theodolite. Inside each big triangle, smaller chain-survey triangles are laid and the buildings, roads and boundaries are located from them by offsets.

   Whole:  big triangle ABC (precise)
           A /\ 
            /  \ 
           / .. \     Part: small triangles
          /______\    and details inside
         B        C

Principle 2: Location of a point by two independent measurements

A point can be located from two known points A and B by any one of these pairs of measurements:

• two distances (A-P and B-P, intersection of two arcs),
• one distance and one angle (e.g. angle at A and distance AP),
• two angles (angles at A and B, intersection of two lines),
• a perpendicular offset from a known line and the distance along that line.

Example: In chain surveying a tree P near line AB is fixed by measuring the distance along AB to the foot of the perpendicular (chainage) and the length of the perpendicular offset. A third measurement, the tie distance to a point on the line, is taken as a check.

Plane surveying

Plane surveying is the type of surveying in which the earth's surface is treated as a plane, so the curvature of the earth is neglected. The level line is taken as a straight line, the angles of a triangle sum to $180^\circ$ and plane trigonometry is used. It is suitable for areas up to about 250 km$^2$ (about 15.5 km $\times$ 15.5 km), where the difference between the arc and its chord is negligible. Examples: road, canal, building-site and small town surveys.

Geodetic surveying

Geodetic surveying is the type of surveying in which the curvature of the earth (its spheroidal shape) is taken into account. Precise instruments and methods are used, the lines are treated as arcs and spherical trigonometry is used. It is used for large areas above about 250 km$^2$, for the fixing of widely spaced control points (latitude, longitude, elevation) and for national mapping. The angles of a triangle sum to more than $180^\circ$ (spherical excess).

Principles of surveying

The two basic principles are:

1. Working from whole to part. A strong main framework (control network) of few stations is first established over the whole area with high accuracy. The smaller parts and details are then surveyed inside it with less accurate methods. Error is therefore controlled, and a mistake in one part does not spread to the others. Working from part to whole would let errors accumulate and enlarge.

   Example: main triangulation or a main traverse is made first, then chain lines and offsets locate details inside each triangle.

2. Location of a point by at least two measurements from fixed reference points. A new point is fixed relative to two known points by two distances (linear intersection), two angles (angular intersection), a distance and an angle, or a perpendicular offset and a chainage. A third measurement is taken as a check.

   Example: Point C is fixed from A and B by measuring distances AC and BC; the check is the distance from C to a point on AB.

Other working rules that follow from these: take every measurement twice or by a different method (check), and keep the accuracy of the work consistent with the purpose of the map.

 Point C fixed from A and B
        C
       /|\ 
    b / | \ a      by distances a and b
     /  |  \ 
    A---+---B      check: tie distance

Surveying is the art and science of determining the relative positions of points on, above or below the earth's surface by measuring distances, directions (angles) and elevations, and of representing them on a map or plan to a suitable scale. It is also used to set out (locate) planned works on the ground.

Difference between plane and geodetic surveying

In short, the two methods differ in whether the earth's curvature is considered or neglected; this depends on the size of the area and the accuracy required.

Plane and geodetic surveying

Primary divisions

1. Plane surveying (small areas, curvature ignored).
2. Geodetic surveying (large areas, curvature considered).

Secondary classifications

1. According to the nature of the field

• Land surveying: topographic, cadastral (property boundaries), city surveying.
• Marine or hydrographic surveying: coastlines, rivers, lakes, depths and tides.
• Astronomical surveying: positions of points from the sun and stars.

2. According to the object (purpose)

• Engineering surveying: for roads, railways, canals, dams, bridges, buildings.
• Military surveying: for defence and strategic planning.
• Mine surveying: underground and surface mine work.
• Geological surveying: rock structures and mineral deposits.
• Archaeological surveying: ancient remains.

3. According to the instruments used

• Chain (tape) survey, compass survey, plane table survey, theodolite (transit) survey, tacheometric survey, levelling, total station/GNSS survey, photogrammetric survey (from aerial photographs), remote-sensing survey.

4. According to the method

• Triangulation (network of triangles), traversing (series of connected lines), trilateration.

5. According to the purpose of the result

• Control survey (fixing control points) and detail survey; route survey; construction (setting-out) survey; reconnaissance, preliminary and location surveys.

6. According to the position of the instrument: ground, aerial and satellite surveying.

RF from the area

Area on map $= 4.56 \times 4.48 = 20.4288$ cm$^2$. Ground area $= 51072$ m$^2$ $= 51072 \times 10^{4}$ cm$^2 = 5.1072 \times 10^{8}$ cm$^2$.

$$\text{RF}^2 = \frac{20.4288}{5.1072 \times 10^{8}} = 4.0 \times 10^{-8}$$ $$\text{RF} = 2 \times 10^{-4} = \frac{1}{5000}$$

Scale

Length for 600 m $= \dfrac{600 \times 100}{5000} = 12$ cm. Here 100 m $= 2$ cm, 10 m $= 0.2$ cm and 1 m $= 0.02$ cm. Since 1 m is only 0.2 mm, a diagonal scale is needed.

Construction

1. Draw a 12 cm line and divide it into 6 main divisions of 2 cm (100 m). Number 0, 100, ... 500, and extend one division to the left of 0.
2. Divide the left division into 10 parts of 0.2 cm (10 m).
3. Draw 10 equal horizontal lines (height, say, 2 cm), erect perpendiculars at each division and draw diagonals from the top of each sub-division to the next lower sub-division at the base. The diagonal steps read 1 m.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+---+---+---+---+---+---+---+-+
   90..0  100  200  300  400  500 (m)

Answer: RF = 1/5000; diagonal scale length = 12 cm, reading 1 m up to 600 m.

RF

Ground area $= 6.304$ ha $= 6.304 \times 10^{4}$ m$^2 = 6.304 \times 10^{8}$ cm$^2$. Map area $= 4.7 \times 4.6 = 21.62$ cm$^2$.

$$\text{RF}^2 = \frac{21.62}{6.304 \times 10^{8}} = 3.4296 \times 10^{-8}$$ $$\text{RF} = 1.852 \times 10^{-4} \approx \frac{1}{5400}$$

Scale length (to read 1 m, up to 500 m)

$$\frac{500 \times 100}{5400} = 9.26\ \text{cm}$$

100 m $= 1.852$ cm, 10 m $= 0.1852$ cm, 1 m $= 0.01852$ cm.

Construction (diagonal scale)

1. Draw a line 9.26 cm long; divide into 5 main divisions of 1.852 cm (100 m each) and extend one more division to the left of 0.
2. Divide the left division into 10 parts (10 m each, 0.1852 cm).
3. Draw 10 horizontal lines at equal spacing and the diagonals from the top of each sub-division to the next lower sub-division. A step on a diagonal reads 1 m.

Indicating 123.400 m

$123.4 = 100 + 20 + 3 + 0.4$. Take the vertical at 100, the diagonal from the 20 sub-division, and the horizontal line level 3, then mark a little (0.4 of the next 1 m division, about 0.007 cm, estimated by eye) above the third line, between the 3rd and 4th lines.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  3 |  .. 123.4 m                     |
  0 +---+---+---+---+---+---+---------+
   90..0  100  200  300  400  500 (m)

Answer: RF = 1/5400; scale length = 9.26 cm.

RF

Map area $= 4.6 \times 4.5 = 20.7$ cm$^2$. Ground area $= 51750 \times 10^{4} = 5.175 \times 10^{8}$ cm$^2$.

$$\text{RF}^2 = \frac{20.7}{5.175 \times 10^{8}} = 4.0 \times 10^{-8}, \qquad \text{RF} = 2 \times 10^{-4} = \frac{1}{5000}$$

Scale (to read 1 m, up to 600 m)

$$\text{Length} = \frac{600 \times 100}{5000} = 12\ \text{cm}$$

100 m $= 2$ cm, 10 m $= 0.2$ cm, 1 m $= 0.02$ cm: a diagonal scale is needed.

Construction

1. Draw a 12 cm line and divide it into 6 main divisions of 2 cm (100 m). Add one more division to the left of 0 and mark 0, 100 ... 600.
2. Divide the left division into 10 parts (10 m each).
3. Draw 10 equal horizontal lines on a rectangle (height about 2 cm).
4. Draw diagonals joining the top of each sub-division with the next lower sub-division on the base. A diagonal step is 1 m.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+----+----+----+----+----+----+
   90..0  100  200  300  400  500  600

Answer: RF = 1/5000; scale length = 12 cm.

A scale is a ruler-like graduated line used to measure ground distances on a map. Plain, diagonal and vernier scales are the three common types, each reading to a smaller unit.

Plain scale

A plain scale has main divisions and one set of sub-divisions. It reads two units, e.g. metres and decimetres, but cannot read a third unit exactly. Example: RF = 1/200, so 1 cm = 2 m; scale length 10 cm = 20 m; the left division is divided into 10 parts, each of 0.2 m.

 m 2    1   0      2    4    6    8   (m)
   |....|...|------|----|----|----|----|
   <-sub-> <-- main divisions (2 m) -->

Reading: 7.4 m = 3 main divisions (6 m) + 7 sub-divisions on the left (1.4 m).

Diagonal scale

A diagonal scale reads three units (e.g. 100 m, 10 m, 1 m). The sub-division is divided further by a set of parallel horizontal lines and diagonal lines drawn from the end of each sub-division; the principle is similar triangles.

Example: RF = 1/4000 (1 cm = 40 m), to read 1 m up to 400 m. Length of scale = 400/4000 m = 10 cm. Divide into 4 main divisions of 2.5 cm (100 m). Divide the left main division into 10 sub-divisions of 0.25 cm (10 m). Draw 10 horizontal lines at equal spacing, and join the top of the left sub-division to the first line of the bottom, and so on. Each horizontal step on a diagonal gives $\frac{1}{10}$ of 10 m = 1 m.

   10m
 10|  /./././././././/|
  8| / /  .. 
  6|/ / 
  4|  
  2|__|_____|_____|___|
   100 0   100  200 300

Measurement of 234 m: 200 m (2 main divisions) + 30 m (3 sub-divisions) + 4 m (4th diagonal line level).

Vernier scale

A vernier scale is a small sliding scale used with a main scale. It reads a fraction of the smallest main division. If $n$ divisions of the vernier equal $(n-1)$ divisions of the main scale, then

$$\text{least count} = \frac{\text{value of 1 main division}}{n}$$

Example: 1 main division = 1 m, $n$ = 10, so least count = 0.1 m. If the 0 of the vernier is just past 4 m and the 3rd vernier line coincides with a main line, the reading is $4 + 3 \times 0.1 = 4.3$ m.

The vernier is mostly used in theodolites, levelling staves and compasses, while the plain and diagonal scales are drawn on maps.

The scale of a map should be chosen after considering the following points:

1. Purpose of the map. Detailed engineering plans (building, road design) need a large scale such as 1:100 to 1:1000; reconnaissance or index maps use a small scale such as 1:50,000.
2. Extent of the area. A large area at a large scale needs huge drawing sheets; the scale must let the area fit on a convenient sheet.
3. Accuracy and least count. The smallest length that can be plotted is about 0.25 mm. The scale must be large enough that the required ground accuracy is at least 0.25 mm $\times$ scale denominator. For 0.1 m accuracy the scale cannot be smaller than 1:400.
4. Amount of detail to be shown. More details need a larger scale.
5. Size of the drawing sheet and ease of handling and storage.
6. Cost and time. A larger scale means more field work, cost and time.
7. Convenience of reading. Round RFs (1:500, 1:1000, 1:2500) make plotting and reading easy.
8. Standard of the department, e.g. the Survey Department of Nepal uses 1:500 to 1:2500 for cadastral and 1:25,000 for topographic maps.

RF

Ground distance = 200 km $= 200 \times 10^5$ cm $= 2 \times 10^7$ cm; map distance = 5 cm.

$$\text{RF} = \frac{5}{2 \times 10^7} = \frac{1}{4{,}000{,}000}$$

Length of the diagonal scale

$$\text{Length} = \frac{600\ \text{km} \times 10^5\ \text{cm}}{4 \times 10^6} = 15\ \text{cm}$$

1 km on the scale $= 10^5/(4\times10^6) = 0.025$ cm; 10 km $= 0.25$ cm; 100 km $= 2.5$ cm.

Construction

1. Draw a line 15 cm long and divide it into 6 equal parts of 2.5 cm; each represents 100 km. Number them 0, 100, ... 500 from the second division (the first, left division is the sub-divided one).
2. Divide the left-most main division into 10 equal parts of 0.25 cm; each is 10 km. Number them 10, 20, ... 90 to the left of 0.
3. At the ends erect perpendiculars of height 2.5 cm (any convenient height) and divide each into 10 equal parts; draw horizontal lines through the points. Complete the rectangle.
4. Join the top of the first sub-division (10 km) to the 0 at the bottom of the main scale line, and then draw parallel diagonals through every other sub-division. Each horizontal step along a diagonal is 1 km (0.025 cm).

 10 |    /  /  /  /  /  /  /  /  /  /|
  8 |   /  /  /                        |
  6 |  /  /                            |
  4 | /  /                             |
  2 |/                                 |
  0 +---+---+---+---+---+---+---+---+---+
   90..10 0   100   200   300   400   500  (km)

Indicating the distances

• (i) 222 km $= 200 + 20 + 2$: take the vertical line at 200, the diagonal that starts at the 20 km sub-division, and the 2nd horizontal line. The point where this diagonal meets the 2nd horizontal line is on the 222 km mark; the length from it back to the vertical at 200 is 222 km.
• (ii) 336 km $= 300 + 30 + 6$: use the vertical at 300, the diagonal from the 30 km sub-division and the 6th horizontal line.

Answer: RF = 1/4,000,000; scale length = 15 cm.

RF

200 km $= 2 \times 10^7$ cm is shown by 5 cm.

$$\text{RF} = \frac{5}{2 \times 10^{7}} = \frac{1}{4{,}000{,}000}$$

Scale length

$$\frac{600 \times 10^{5}}{4 \times 10^{6}} = 15\ \text{cm}$$

So 100 km $= 2.5$ cm, 10 km $= 0.25$ cm and 1 km $= 0.025$ cm.

Construction

1. Draw a 15 cm line and divide it into 6 main divisions of 2.5 cm (100 km). Mark 0 at the end of the first division and 100, 200 ... 500 for the others.
2. Divide the first main division into 10 parts of 0.25 cm (10 km each).
3. On the end verticals mark 10 equal divisions (height 2.5 cm) and draw 10 horizontal lines.
4. Draw the diagonals from the top of each sub-division to the next lower sub-division on the base line. A step on the diagonal between two horizontal lines is 1 km.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+---+---+---+---+---+---+---+-+
   90..0   100  200  300  400  500 (km)

Distances

The required length is measured between the vertical line and the point where the diagonal meets the horizontal line.

Answer: RF = 1/4,000,000; scale length = 15 cm.

Objectives of surveying

Surveying is the measurement of distances, angles and elevations to fix the relative positions of points. Its main objectives are:

1. To prepare plans, maps and sections (topographic, cadastral, engineering) of the ground.
2. To collect data needed for the planning and design of engineering projects such as roads, railways, canals, bridges, dams and buildings.
3. To set out (mark on the ground) the designed work in the correct position and level.
4. To determine areas, volumes and the quantities of earthwork.
5. To fix property boundaries and to settle land disputes.
6. To establish control points for further surveys.
7. To check the work during and after construction (deformation and progress monitoring).

Plane and geodetic surveying

Surveying is the art and science of determining the relative positions of points on, above or below the earth's surface by measuring horizontal distances, angles and vertical distances, and plotting them to scale on a plan or map.

Importance to a civil engineer

Surveying is the first step of almost every civil engineering project.

1. Planning and feasibility. Topographic maps and contour plans help to choose the best site, route and alignment of roads, canals, railways, pipelines and transmission lines.
2. Design. Levels, sections and contours give the data to design earthwork, drainage, slopes and structures.
3. Estimating. Areas and volumes of cutting and filling, and quantities of materials, are calculated from survey data, so cost estimates depend on it.
4. Setting out. The designed centre lines, building corners, foundations, columns and levels are marked on the ground by surveying.
5. Construction control. Alignment, levels and verticality are checked during construction.
6. Land matters. Property boundaries, land acquisition and land records need cadastral surveys.
7. Monitoring. Settlement, deflection and movement of dams, bridges and tall buildings are measured by precise surveys.
8. Water resources and hydropower projects in Nepal depend on river, catchment and contour surveys.

Hence a civil engineer who knows surveying can plan economically, set out accurately and control quality.

Engineering surveying is the branch of surveying done for engineering works. It covers all measurements and setting out needed for the planning, design, construction and monitoring of projects such as roads, bridges, dams and buildings. Accuracy is chosen to suit the work.

The survey principle used here is "make a check on every measurement" together with "fix a quantity by at least two independent measurements".

Two-way linear distance measurement

A line AB is measured twice, once from A to B (forward) and again from B to A (backward), with the same tape and tension.

1. Forward measurement: $L_f$; backward measurement: $L_b$.
2. Difference $d = |L_f - L_b|$ is compared with the allowable limit (for example, 1 in 5000 for ordinary chaining).
3. If within the limit, the mean is adopted:

$$L = \frac{L_f + L_b}{2}$$

4. If the difference is large, a mistake (miscounting a tape length, wrong reading) exists, so the line is measured again.

The second measurement is independent of the first, so a blunder is caught and random errors are reduced by averaging.

Single angle measurement

An angle $\angle AOB$ is measured with a theodolite by:

1. Observing on face left and then on face right, and taking the mean (removes instrumental errors like collimation and trunnion errors).
2. Repeating the measurement (method of repetition) on different zero settings and taking the mean, giving a check on the reading and a smaller error.
3. As a further check, the angle can be measured as $360^\circ$ minus the explement (the other side), or the angle can be compared with a value calculated from measured sides.

Thus every distance and angle is obtained from two independent observations, and the survey is checked.

General principles of surveying

1. Working from whole to part.
2. Locating a new point by at least two independent measurements (distances or angles) from points already fixed.
3. Check every measurement (take it twice or by another method) and keep the accuracy consistent with the purpose.

Why work from whole to part?

In this method a precise control framework is first fixed over the entire area, and the smaller details are then surveyed inside it by less precise methods.

• Errors do not accumulate: each small part is tied to the already correct frame.
• An error made in one part stays inside that part and does not affect the others.
• The big triangles are measured accurately, so the plan fits the ground as a whole.
• If we worked from part to whole, the small errors of each part would add up as we proceed, and the final survey would be distorted.

Example: For a farm, big triangles ABC are measured accurately by chain; then the fields and buildings inside are fixed by offsets from the sides of the triangle.

Why accuracy must be followed

• A survey is used to design and set out works; wrong measurements cause wrong layout, wasted cost and failure of structures.
• The final plan can be only as good as its measurements; there is no use of high accuracy where the purpose does not need it, so the required accuracy is first decided (e.g. 1 in 1000 for rough work, 1 in 5000 or more for good work).
• Appropriate instruments and methods are then selected to reach this accuracy economically.
• Accuracy checks (closing error, repeated measurement) detect blunders before they spread.

Hence, whole to part controls the error and accuracy-based working ensures the result is reliable and economical.

Surveying has the principle that a quantity should be fixed by at least two independent measurements, and each measurement must be checked. In linear measurement this is done by measuring the line twice, in forward and backward directions (two-way measurement).

Procedure

1. Measure line AB from A to B with the tape, keeping it straight, level and under the same tension. Record $L_1$.
2. Measure back from B to A in the same way. Record $L_2$.
3. Find the difference $d = L_1 - L_2$ and compute the relative error $= d / L_{mean}$.
4. Compare with the permissible error (e.g. 1/5000 for chaining on fairly level ground, 1/1000 for rough ground).
5. If the error is within the limit, adopt the mean:

$$L = \frac{L_1 + L_2}{2}$$

6. If it is greater, find the cause (miscounted tape lengths, wrong reading, slack tape) and measure again.

Example

$L_1 = 150.42$ m and $L_2 = 150.38$ m. Difference $= 0.04$ m, relative error $= 0.04/150.4 = 1/3760$. For ordinary work this is acceptable, so $L = 150.40$ m.

The backward measurement is independent of the forward one, so mistakes are detected and accidental errors are reduced by averaging.

Principles of surveying

1. Working from whole to part. A precise control framework is made over the whole area first; the details are then surveyed inside it. Errors are thus localised and do not accumulate.
2. Fixing a point by at least two independent measurements from known points: two distances, two angles, or one angle and one distance (or an offset and a chainage). A third measurement is taken as a check.

A survey is also carried out with checks on all measurements and with accuracy suited to the purpose.

Scales used in surveying

A scale is the ratio of a length on the map to the corresponding length on the ground. It can be expressed as a representative fraction (RF), e.g. 1/50,000, a statement (1 cm = 500 m) or a graphical scale.

1. Plain scale. A line divided into main divisions and sub-divisions; it reads two units (e.g. metres and decimetres).
2. Diagonal scale. Reads three units (e.g. 100 m, 10 m and 1 m) using diagonal lines and the principle of similar triangles.
3. Vernier scale. A sliding scale that reads a fraction of the smallest main division. Least count = value of 1 main division / number of vernier divisions.
4. Comparative scale. Two scales with the same RF but different units, such as a metric and a British scale on one map.
5. Scale of chords. Used for plotting and measuring angles when no protractor is available.
6. Scale of time or speed. Used when time is to be measured from distance travelled at a constant speed.
7. Isometric scale. Used for isometric drawings.

By size, maps are large scale (1:500 to 1:5000, used for detailed plans), medium scale and small scale (1:50,000 and smaller, used for atlases).

A graphical scale (bar scale) is a line drawn on the map and divided into units that represent real ground distances, e.g. a bar marked 0, 100, 200, 300 m. Plain, diagonal and vernier scales are graphical scales.

Importance on the map

1. Distances can be read directly without calculation; the user need not know the RF.
2. It is unaffected by shrinkage or enlargement. Paper expands or shrinks with weather, and a map may be photocopied, enlarged or reduced. A numerical scale (1:5000) would then be wrong, but the graphical scale changes together with the map, so it stays correct.
3. Clear even to non-technical users, who can compare the bar with the map features by eye or with a divider.
4. Easy to use with a divider or strip of paper for measuring curved lengths.
5. Gives an idea of the area size at a glance.
6. Accuracy. A diagonal or vernier graphical scale can read up to a very small unit.

For these reasons, a graphical scale is always shown on a map along with the numerical RF.

The scale of a map is the ratio between a distance measured on the map and the corresponding distance on the ground:

$$\text{Scale (RF)} = \frac{\text{map distance}}{\text{ground distance}} \quad (\text{same units})$$

For example, 1:50,000 means 1 cm on the map = 50,000 cm = 500 m on the ground.

Classification

By the way it is expressed

1. Numerical scale or RF (e.g. 1:25,000).
2. Verbal (statement) scale (1 cm = 250 m).
3. Graphical scale (bar, plain, diagonal, vernier).

By size (range of values)

A larger denominator gives a smaller scale. A large-scale map shows a small area in great detail; a small-scale map shows a large area with few details. In Nepal, cadastral maps are 1:500 to 1:2500 and topographic maps are 1:25,000 and 1:50,000.

Given and basic values

• Distance = 210 km, time = 7 h, so speed $= 210/7 = 30$ km/h.
• RF $= 1/200{,}000$, so 1 km on the ground $= 100{,}000$ cm $/ 200{,}000 = 0.5$ cm on the scale.
• Maximum reading $=$ 30 km $=$ 1 hour $=$ 60 minutes.

Scale length

$$\text{Length} = 30\ \text{km} \times 0.5\ \text{cm/km} = 15\ \text{cm}$$

Distance per minute $= 30/60 = 0.5$ km, so 1 minute $= 0.5 \times 0.5 = 0.25$ cm.

Construction

1. Draw a line 15 cm long and divide it into 6 equal main divisions of 2.5 cm each. Each main division is 10 minutes (5 km).
2. Mark 0 at the end of the first division from the left, and number the others 10, 20, 30, 40, 50 min.
3. Divide the first (left) main division into 10 equal sub-divisions of 0.25 cm, each 1 minute (0.5 km).
4. Draw a parallel line below, and draw the scale as a rectangle (about 0.5 cm high). Write the distance in km on the same scale: 0, 5, 10, ... 25 km.

 min 10   5    0    10   20   30   40   50
     |....|....|----|----|----|----|----|
 km  5    2.5  0     5    10   15   20   25
          <-----  RF 1/200,000  ----->

Distance covered in 29 minutes

29 min = 2 main divisions (20 min) + 9 sub-divisions (9 min).

$$\text{Distance} = 29 \times 0.5 = 14.5\ \text{km}$$

On the scale the length is $29 \times 0.25 = 7.25$ cm. Mark the point 2 main divisions to the right of 0 and 9 sub-divisions to the left of 0. Place the divider on these points.

Answer: distance travelled in 29 minutes = 14.5 km (7.25 cm on the scale); scale length = 15 cm.

RF

150 km on the ground $= 150 \times 10^5$ cm $= 1.5 \times 10^7$ cm is shown by 5 cm.

$$\text{RF} = \frac{5}{1.5 \times 10^{7}} = \frac{1}{3{,}000{,}000}$$

Length of the scale

$$\text{Length for 600 km} = \frac{600 \times 10^5}{3 \times 10^6} = 20\ \text{cm}$$

100 km $= 3.333$ cm; 10 km $= 0.3333$ cm; 1 km $= 0.0333$ cm.

Construction

1. Draw a line 20 cm long and divide it into 6 equal main divisions of 3.333 cm (100 km each). Mark 0 at the end of the first division and 100, 200, ... 500 on the others.
2. Divide the first division into 10 equal parts of 0.333 cm (10 km each); number 10 to 90 to the left of 0.
3. Draw a rectangle of height about 3 cm on the scale, divide the height into 10 equal parts, and draw horizontal lines.
4. Draw the diagonals: join the top of the sub-division marked 10 to the bottom 0, and the others parallel to it. Each horizontal line cuts a diagonal at $\tfrac{1}{10}$ of 10 km $=$ 1 km steps.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +----+----+----+----+----+----+---+
   90..0  100  200  300  400  500 (km)

Distances

• 336 km $= 300 + 30 + 6$: vertical at 300, diagonal from sub-division 30, 6th horizontal line. The intercept from this point to the 300 vertical is 336 km.
• 457 km $= 400 + 50 + 7$: vertical at 400, diagonal from sub-division 50, 7th horizontal line.

Answer: RF = 1/3,000,000; scale length = 20 cm.

RF

1 km $= 1000$ m $= 10^5$ cm on the ground is represented by 25 cm.

$$\text{RF} = \frac{25}{10^{5}} = \frac{1}{4000}$$

Scale length

$$\text{Length for 500 m} = \frac{500 \times 100}{4000} = 12.5\ \text{cm}$$

Therefore 100 m $= 2.5$ cm, 10 m $= 0.25$ cm, 1 m $= 0.025$ cm.

Construction

1. Draw a line 12.5 cm long and divide it into 5 equal main divisions of 2.5 cm (100 m each). Add one more division of 2.5 cm to the left of the zero, so the total drawn is 15 cm. Number 0, 100, ... 500.
2. Divide the left extra division into 10 equal parts of 0.25 cm (10 m each).
3. Draw a rectangle of any convenient height (say 2.5 cm), divide the height into 10 parts and draw 10 horizontal lines.
4. Join the top of the first sub-division (10 m) with the bottom 0 of the main scale; draw parallel diagonals through the other sub-divisions. Each step along a diagonal equals 1 m.

 10|   /  /  /  /  /  /  /  /  /  /|
  5| /  /                          |
  0+---+----+----+----+----+----+--+
  90..0  100  200  300  400  500 (m)

Distances

• 123 m $= 100 + 20 + 3$: vertical at 100, the diagonal from the 20 m sub-division, 3rd horizontal line.
• 256 m $= 200 + 50 + 6$: vertical at 200, diagonal from 50 m, 6th horizontal line.

Answer: RF = 1/4000; length of scale = 12.5 cm (plus the left sub-divided part).

Scale (RF)

1.5 km $= 1500$ m $= 1.5 \times 10^{5}$ cm is shown by 37.5 cm.

$$\text{RF} = \frac{37.5}{1.5 \times 10^{5}} = \frac{1}{4000}$$

so 1 cm on the map $= 40$ m on the ground.

Scale to read 1 m

A diagonal scale is suitable. Take a maximum reading of 500 m.

$$\text{Length} = \frac{500 \times 100}{4000} = 12.5\ \text{cm}$$

100 m $= 2.5$ cm, 10 m $= 0.25$ cm, 1 m $= 0.025$ cm.

Sketch and construction

1. Draw a line 12.5 cm long; divide it into 5 equal parts of 2.5 cm (100 m). Extend by one more division to the left of 0.
2. Divide the left division into 10 equal parts (10 m each).
3. Draw 10 horizontal lines through equal divisions of a vertical 2.5 cm high.
4. Join the top of each sub-division with the next lower sub-division at the base (diagonals). Each horizontal line cuts a diagonal at 1 m steps.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+----+----+----+----+----+----+
   90..0  100  200  300  400  500 (m)

Example: 234 m is 200 m + 30 m + 4 m.

Answer: RF = 1/4000 (1 cm = 40 m); diagonal scale length 12.5 cm.

RF

Map area $= 4.75 \times 5.22 = 24.795$ cm$^2$. Ground area $= 39672 \times 10^{4} = 3.9672 \times 10^{8}$ cm$^2$.

$$\text{RF}^2 = \frac{24.795}{3.9672 \times 10^{8}} = 6.25 \times 10^{-8}, \qquad \text{RF} = 2.5 \times 10^{-4} = \frac{1}{4000}$$

Scale (to read 1 m, up to 400 m)

$$\text{Length} = \frac{400 \times 100}{4000} = 10\ \text{cm}$$

100 m $= 2.5$ cm, 10 m $= 0.25$ cm, 1 m $= 0.025$ cm. This is a diagonal scale.

Sketch and construction

1. Draw a 10 cm line divided into 4 main divisions of 2.5 cm (100 m each); add one division to the left of 0.
2. Divide the left division into 10 parts of 0.25 cm (10 m).
3. Draw 10 horizontal lines (height 2.5 cm) and the diagonals from the top of each sub-division to the next lower sub-division at the base.
4. Each step along a diagonal = 1 m.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+----+----+----+----+--------+
   90..0  100  200  300  400 (m)

Answer: RF = 1/4000; scale length = 10 cm.

RF

Ground area $= 0.55$ ha $= 5500$ m$^2 = 5500 \times 10^{4}$ cm$^2 = 5.5 \times 10^{7}$ cm$^2$. Map area $= 6.11$ cm$^2$.

$$\text{RF}^2 = \frac{6.11}{5.5 \times 10^{7}} = 1.1109 \times 10^{-7}$$ $$\text{RF} = 3.3331 \times 10^{-4} \approx \frac{1}{3000}$$

Scale (to read 1 m, up to 400 m)

$$\text{Length} = \frac{400 \times 100}{3000} = 13.33\ \text{cm}$$

100 m $= 3.333$ cm, 10 m $= 0.3333$ cm, 1 m $= 0.0333$ cm: a diagonal scale.

Construction

1. Draw a 13.33 cm line with 4 main divisions of 3.333 cm (100 m). Add one division to the left of 0.
2. Divide the left division into 10 parts (10 m each).
3. Draw 10 equally spaced horizontal lines on a rectangle 3 cm high.
4. Draw the diagonals from the top of each sub-division to the next lower sub-division on the base. Each diagonal step gives 1 m.

 10 |  /  /  /  /  /  /  /  /  /  /  |
  5 | /  /                            |
  0 +---+----+----+----+--------------+
   90..0  100  200  300  400 (m)

Answer: RF = 1/3000; scale length = 13.33 cm.

Scale of the plan

Map area $= 8 \times 9 = 72$ cm$^2$. Ground area $= 18000 \times 10^{4} = 1.8 \times 10^{8}$ cm$^2$.

$$\text{RF}^2 = \frac{72}{1.8 \times 10^{8}} = 4.0 \times 10^{-7}$$ $$\text{RF} = 6.3246 \times 10^{-4} = \frac{1}{1581.1}$$

So 1 cm on the plan $= 15.81$ m on the ground (approx. 1:1581).

Sketch of a scale reading 1 m

Take a maximum reading of 150 m: length $= 150 \times 100/1581.1 = 9.49$ cm (100 m $= 6.32$ cm, 10 m $= 0.632$ cm, 1 m $= 0.0632$ cm).

1. Draw a line 9.49 cm long and divide it into main divisions of 50 m ($3.16$ cm each): three divisions for 150 m. Add one more division to the left of 0.
2. Divide the left main division into 5 sub-divisions of 10 m each ($0.632$ cm).
3. Draw 10 equally spaced horizontal lines on a rectangle about 2 cm high, and draw diagonals from the top of each sub-division to the next lower sub-division on the base. Each step on a diagonal reads 1 m (a diagonal scale is needed because 1 m is only 0.63 mm).

 10 |  /  /  /  /  /  /  /  /  /  |
  5 | /  /                        |
  0 +---+-----+-----+-------------+
   50..0   50   100   150 (m)

Area on the 1:12500 sheet

Linear ratio $= 1/12500$, so the same ground area shown on the topo sheet is

$$\text{Area on sheet} = \frac{18000 \times 10^{4}\ \text{cm}^2}{12500^{2}} = 1.152\ \text{cm}^2$$

The plan would measure about $1.01\ \text{cm} \times 1.14\ \text{cm}$ on the sheet (each side of 8 cm and 9 cm is reduced by the factor $1581.1/12500$).

Answer: RF $\approx$ 1/1581 (1 cm = 15.81 m); area on the 1:12500 sheet = 1.152 cm$^2$.