Distance Measurements

EDM (Electronic Distance Measurement) is the measurement of a distance using the velocity and travel time (or phase change) of a modulated electromagnetic wave (light, infrared or microwave) sent from an instrument to a reflector and back.

Principle

The instrument at A transmits a modulated wave to a reflector (prism or transponder) at B. The reflected wave returns to A. The distance is found from the time taken or the phase shift of the returned wave. The wave travels with a known velocity $v = c/n$, where $c = 299{,}792{,}458$ m/s is the velocity in vacuum and $n$ the refractive index of the air.

  EDM at A ------ outgoing wave ------> Prism at B
        <------ reflected wave ------
           Double path = 2D

1. Pulse (time-of-flight) method

A short pulse of energy is sent and the time $t$ between sending and receiving is measured.

$$D = \frac{v\,t}{2}$$

It needs a very accurate clock: a time error of 1 ns gives about 15 cm error in distance. It is used for long ranges and in laser total stations (reflectorless).

2. Phase-comparison method

A continuous carrier wave is modulated with a fixed frequency $f$ (wavelength $\lambda = v/f$). The phase difference $\phi$ between the transmitted and the received wave is measured. The double path $2D$ contains $N$ complete wavelengths plus a fraction:

$$2D = N\lambda + \frac{\phi}{2\pi}\lambda \quad\Rightarrow\quad D = \frac{1}{2}\left(N + \frac{\phi}{2\pi}\right)\lambda$$

The fraction is measured by the instrument, but the whole number $N$ is not known. It is solved by using several modulation frequencies (a coarse and a fine wavelength), so $N$ is resolved automatically. This is the common method in total stations.

Steps in operation

1. Set up the EDM/total station over the point and a reflector over the other point.
2. Enter temperature, pressure and the prism constant.
3. Sight the reflector and press measure.
4. The instrument corrects for atmosphere and gives the slope distance, horizontal distance and height difference.

EDMs give high accuracy (a few mm $\pm$ 1 to 5 ppm), speed and long range.

When a tape or chain is used in the field, the measured length differs from the true horizontal length because of the tape's condition and the field conditions. The following corrections are applied. Let $L$ be the measured length.

1. Correction for incorrect length of tape (standardisation)

$$C_a = \frac{L\,(l' - l)}{l}$$

where $l'$ is the actual tape length and $l$ the nominal length. It is positive if the tape is too long, negative if too short.

2. Correction for temperature

$$C_t = \alpha\,(T_m - T_0)\,L$$

where $\alpha$ is the coefficient of linear expansion, $T_m$ the mean field temperature and $T_0$ the standard temperature. Positive if $T_m > T_0$.

3. Correction for pull (tension)

$$C_p = \frac{(P - P_0)\,L}{A\,E}$$

$P$ = applied pull, $P_0$ = standard pull, $A$ = cross-sectional area, $E$ = Young's modulus. Positive if $P > P_0$.

4. Correction for sag

$$C_s = -\frac{L\,w^{2}\,l_s^{2}}{24\,P^{2}} = -\frac{n\,W^{2}\,l_s}{24\,P^{2}}$$

where $w$ = weight per metre, $l_s$ = length of each span between supports, $W = w\,l_s$. It is always negative when the tape is standardised on the flat and used suspended.

5. Correction for slope

$$C_h = -\frac{h^{2}}{2L} \quad\text{(approx.)} \qquad\text{or}\qquad C_h = -L(1-\cos\theta)$$

where $h$ is the height difference. Always negative.

6. Correction for alignment (tape not in line): always negative.

7. Reduction to mean sea level

$$C_{msl} = -\frac{L\,h}{R}$$

where $h$ is the elevation above MSL and $R$ is the radius of the earth (6370 km). Negative.

True length $= L + C_a + C_t + C_p + C_s + C_h + C_{msl}$.

Errors are classed as cumulative (systematic), which keep the same sign and add up with the number of tape lengths, and compensating (accidental), which are equally likely to be positive or negative and partly cancel out.

Cumulative errors are corrected by calculation; compensating errors are reduced by repeated measurements.

An error is the difference between a measured value and its true value. No measurement is free from error.

Mistakes (blunders) and errors

Types of errors

1. Cumulative (systematic) errors. These have the same sign in all measurements and increase with the amount of work. Their size and sign can be found and a correction applied.

• Examples: tape too long or short, a wrong zero in a levelling staff, sag, temperature change, collimation error in a theodolite.

2. Accidental (random or compensating) errors. These occur due to causes beyond the surveyor's control and are equally likely to be positive or negative. They follow the laws of probability and are reduced by repeating the measurements and taking the mean.

• Examples: estimating the last fraction of a reading, slight movement of the plumb bob, personal limitations of eyesight.

By source

• Personal errors: due to the observer (limited sight, bias, careless reading).
• Instrumental errors: due to imperfect adjustment or construction of the instrument.
• Natural errors: due to wind, temperature, refraction, curvature of the earth.

Systematic errors are removed by calibration and corrections, and accidental errors are minimised by taking many observations.

Method. A tape standardised in catenary has the sag of the standardisation span built into its stated length. We first add this sag back to get the tape's true length, then apply the field sag for supports every 10 m.

Data

• $w = 30$ g/m $= 0.03$ kg/m; pull $P = 15$ kg (same as standard); $L = 86$ m
• Standard: 20 m tape $= 20.006$ m in catenary (one 20 m span)
• Field spans: 8 spans of 10 m + 1 span of 6 m

1. Tape length

Sag in standardisation (20 m span):

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 20^{3}}{24\times 15^{2}} = 0.00133\ \text{m}$$

True (flat) tape length $= 20.006 + 0.00133 = 20.00733$ m.

$$C_a = \frac{86\,(20.00733 - 20)}{20} = +0.0315\ \text{m}$$

2. Sag correction in the field

Each 10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 15^{2}} = 0.000167$ m; for 8 spans $= 0.001333$ m. The 6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 15^{2}} = 0.000036$ m.

$$C_s = -(0.001333 + 0.000036) = -0.0014\ \text{m}$$

Temperature and pull corrections are zero (same as standard, no data).

3. Correct length

$$L_{true} = 86 + 0.0315 - 0.0014 = 86.0302\ \text{m}$$

Answer: corrected length of the line $= 86.0302$ m.

Method. The tape was standardised in catenary, so first the sag included in its standard length is added back to find its true length. Then the sag for the supports at every 15 m is applied.

Data

• $w = 30$ g/m $= 0.03$ kg/m; $P = 15$ kg (same as standard); $L = 86$ m
• Standard: 30 m tape $= 30.006$ m in catenary
• Field spans: 5 spans of 15 m + 1 span of 11 m

1. Tape length

Sag in standardisation (30 m span):

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 30^{3}}{24\times 15^{2}} = 0.0045\ \text{m}$$

True tape length $= 30.006 + 0.0045 = 30.0105$ m.

$$C_a = \frac{86\,(30.0105 - 30)}{30} = +0.0301\ \text{m}$$

2. Sag correction

15 m span: $\dfrac{0.03^{2}\times 15^{3}}{24\times 15^{2}} = 0.000563$ m; for 5 spans $= 0.002813$ m. 11 m span: $\dfrac{0.03^{2}\times 11^{3}}{24\times 15^{2}} = 0.000222$ m.

$$C_s = -(0.002813 + 0.000222) = -0.0030\ \text{m}$$

3. Correct length

$$L_{true} = 86 + 0.0301 - 0.0030 = 86.0271\ \text{m}$$

Answer: corrected length of the line $= 86.0271$ m.

Method. The tape was standardised in catenary, so the sag in the standard length is added back to find its true length. Then the sag for the supports at every 10 m is applied.

Data

• $w = 30$ g/m $= 0.03$ kg/m; $P = 10$ kg (same as standard); $L = 86$ m
• Standard: 20 m tape $= 20.006$ m in catenary (20 m span)
• Field spans: 8 spans of 10 m + 1 span of 6 m

1. Tape length

Sag at standardisation: $\dfrac{0.03^{2}\times 20^{3}}{24\times 10^{2}} = 0.0030$ m. True tape length $= 20.006 + 0.0030 = 20.009$ m.

$$C_a = \frac{86\,(20.009 - 20)}{20} = +0.0387\ \text{m}$$

2. Sag correction

10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 10^{2}} = 0.000375$ m; for 8 spans $= 0.003000$ m. 6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 10^{2}} = 0.000081$ m.

$$C_s = -(0.003000 + 0.000081) = -0.0031\ \text{m}$$

3. Correct length

$$L_{true} = 86 + 0.0387 - 0.0031 = 86.0356\ \text{m}$$

Answer: corrected length of the line $= 86.0356$ m.

Assumption. Standardised fully supported at $10^\circ$C and 5 kg, so there is no standard sag to add back.

Data

• $L = 680$ m; supports every 10 m (68 spans)
• $P_0 = 5$ kg, $P = 16$ kg; $T_0 = 10^\circ$C, $T = 22^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C
• Weight per metre $= A\times\rho = 0.03\ \text{cm}^2\times 10\ \text{g/cm}^3 = 0.3$ g/cm $= 0.03$ kg/m

1. Temperature

$$C_t = 11\times 10^{-6}\times(22 - 10)\times 680 = +0.0898\ \text{m}$$

2. Pull

$$C_p = \frac{(16 - 5)\times 680}{0.03\times 2.1\times 10^{6}} = +0.1187\ \text{m}$$

3. Sag

Per 10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 16^{2}} = 0.000146$ m.

$$C_s = -68\times 0.000146 = -0.0100\ \text{m}$$

4. True length

$$L = 680 + 0.0898 + 0.1187 - 0.0100 = 680.1985\ \text{m}$$

Answer: true length of the line $= 680.1985$ m.

Accuracy is the degree of closeness of a measured value to the true value. It shows how correct the result is.

Precision is the degree of closeness of repeated measurements of the same quantity to one another. It shows how consistent the measurements are, whether or not they are close to the true value.

Example

The true length of a line is 100.000 m.

A target-shooting example: shots grouped tightly but away from the bull's eye are precise but inaccurate; shots grouped around the bull's eye are both.

High precision does not guarantee accuracy, because a systematic error (like a wrong tape length) affects all readings equally. Accuracy needs both low systematic errors and high precision.

Principles of operation of EDM

An EDM instrument sends a modulated electromagnetic wave (infrared, laser or microwave) to a reflector at the other end of the line. The reflected wave returns to the instrument, and the distance is obtained by measuring either the time taken or the phase difference of the wave.

1. Pulse method: $D = \dfrac{v\,t}{2}$, with $v = c/n$ the velocity in air and $t$ the round-trip time.
2. Phase-comparison method: the carrier wave is modulated with frequency $f$ ($\lambda = v/f$). The phase shift $\phi$ is measured and

$$D = \frac{1}{2}\left(N + \frac{\phi}{2\pi}\right)\lambda$$

The integer $N$ is solved by using two or more modulation frequencies. A microprocessor applies the atmospheric corrections and displays the distance.

Sources of errors in EDM

The final error is usually expressed as $\pm(a\ \text{mm} + b\ \text{ppm})$.

Distance-independent errors (constant)

• Zero error (instrument/reflector constant), due to the difference between the electrical centre and the mechanical centre.
• Centring errors of the instrument and the reflector over the stations.
• Phase measurement (resolution) error.
• Error in measuring the height of the instrument and the target.

Distance-dependent errors (proportional)

• Error in the atmospheric refractive index from wrong temperature, pressure and humidity readings. This is the largest source.
• Error in the modulation frequency (instrument oscillator drift).
• Error in the velocity of light used.

Other errors

• Cyclic error (phase non-linearity) inside the instrument.
• Reflection from other objects (ground swing) and multipath.
• Poor line of sight, beam passing close to the ground or heat shimmer.
• Operator mistakes, e.g. wrong prism constant.

Errors are reduced by regular calibration, using accurate temperature and pressure, careful centring, and taking several readings.

Reading of the data. A tape standardised in catenary reads 29.895 m between its end graduations when hanging in catenary at 50 N, so it is too short by 0.105 m as a hanging tape. The sag effect included in this standard must be accounted for to get the true length of the tape, then the 20 m span is corrected for its own sag. The field tension is taken equal to the standard tension, 50 N (not stated).

Data

• Mass per metre $= 0.015$ kg/m, so weight per metre $w = 0.015 \times 9.81 = 0.14715$ N/m
• Pull $P = 50$ N, standard catenary length of 30 m tape $= 29.895$ m
• Recorded span $L = 20$ m

Step 1: Sag of the 30 m tape at standardisation

$$C_{s,30} = \frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}$$

Length of tape fully supported (flat):

$$l_0 = 29.895 + 0.009744 = 29.9047\ \text{m}$$

Step 2: Correction for tape length (for 20 m)

$$C_a = 20\times\frac{29.9047 - 30}{30} = -0.0635\ \text{m}$$

Step 3: Sag correction for the 20 m span

$$C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}} = -\frac{0.14715^{2}\times 20^{3}}{24\times 50^{2}} = -0.0029\ \text{m}$$

Step 4: Horizontal length

$$D = 20 - 0.0635 - 0.0029 = 19.9336\ \text{m}$$

Answer: horizontal length of the span $= 19.9336$ m.

Method. A tape standardised in catenary reads 29.8940 m for its nominal 30 m when hanging at 50 N. To find the true tape length, the sag included in this standard value is added back. Then the 24 m span is corrected for tape length and for its own sag. The field pull is taken as 50 N (the standard pull).

Data

$w = 0.015 \times 9.81 = 0.14715$ N/m, $P = 50$ N, $L = 24$ m.

1. Sag at standardisation (30 m)

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}$$

True tape length $l_0 = 29.8940 + 0.009744 = 29.9037$ m.

2. Tape length correction

$$C_a = 24\times\frac{29.9037 - 30}{30} = -0.0770\ \text{m}$$

3. Sag correction for the 24 m span

$$C_s = -\frac{0.14715^{2}\times 24^{3}}{24\times 50^{2}} = -0.0050\ \text{m}$$

4. Horizontal length

$$D = 24 - 0.0770 - 0.0050 = 23.9180\ \text{m}$$

Answer: horizontal length of the span $= 23.9180$ m.

Method. The tape was standardised in catenary, so the sag in its standard length is added back to find the true tape length. Then the sag for the three 22 m spans is applied.

Data

• $w = 0.03$ kg/m; $P = 5$ kg (same as standard); $L = 66$ m in three equal spans of 22 m
• Standard length of the 30 m tape in catenary $= 30.008$ m

1. Tape length

Sag at standardisation: $\dfrac{0.03^{2}\times 30^{3}}{24\times 5^{2}} = 0.0405$ m. True tape length $= 30.008 + 0.0405 = 30.0485$ m.

$$C_a = 66\times\frac{30.0485 - 30}{30} = +0.1067\ \text{m}$$

2. Sag correction

Each span: $\dfrac{0.03^{2}\times 22^{3}}{24\times 5^{2}} = 0.015972$ m.

$$C_s = -3\times 0.015972 = -0.0479\ \text{m}$$

3. Corrected length

$$L_{true} = 66 + 0.1067 - 0.0479 = 66.0588\ \text{m}$$

Answer: corrected length of the line $= 66.0588$ m.

Assumptions. Standardised fully supported at $20^\circ$C and 100 N, so no standard sag is added back. $g = 9.81$ m/s$^2$.

Data

• $L = 1020$ m, supports every 10 m (102 spans of 10 m)
• $P_0 = 100$ N, $P = 160$ N; $T_0 = 20^\circ$C, $T = 30^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, so $AE = 0.03\times 2.1\times 10^{6}\times 9.81 = 618{,}030$ N
• $\alpha = 11\times 10^{-6}$/$^\circ$C; $w = 24$ g/m $= 0.024\times 9.81 = 0.23544$ N/m

1. Temperature correction

$$C_t = 11\times 10^{-6}\times(30 - 20)\times 1020 = +0.1122\ \text{m}$$

2. Pull correction

$$C_p = \frac{(160 - 100)\times 1020}{618{,}030} = +0.0990\ \text{m}$$

3. Sag correction

Per 10 m span: $\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = \dfrac{0.23544^{2}\times 10^{3}}{24\times 160^{2}} = 0.0000902$ m.

$$C_s = -102\times 0.0000902 = -0.0092\ \text{m}$$

4. True length

$$L_{true} = 1020 + 0.1122 + 0.0990 - 0.0092 = 1020.2020\ \text{m}$$

Answer: true length of the line $= 1020.202$ m.

Assumption. The tape was standardised fully supported ($20^\circ$C, 10 kg), so there is no standard sag to add back.

Data

• $L = 780$ m; supports every 15 m, so $780/15 = 52$ spans
• $P_0 = 10$ kg, $P = 15$ kg; $T_0 = 20^\circ$C, $T = 32^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C
• $w = 0.693/30 = 0.0231$ kg/m

1. Temperature correction

$$C_t = 11\times 10^{-6}\times(32 - 20)\times 780 = +0.1030\ \text{m}$$

2. Pull correction

$$C_p = \frac{(15 - 10)\times 780}{0.03\times 2.1\times 10^{6}} = +0.0619\ \text{m}$$

3. Sag correction

Per span: $\dfrac{0.0231^{2}\times 15^{3}}{24\times 15^{2}} = 0.000334$ m.

$$C_s = -52\times 0.000334 = -0.0173\ \text{m}$$

4. True length

$$L_{true} = 780 + 0.1030 + 0.0619 - 0.0173 = 780.1475\ \text{m}$$

Answer: true length of the line $= 780.1475$ m.

Method. The tape was standardised in catenary (30 m span), so the sag in its stated length is added back to give the true tape length. The 24.726 m is measured as one catenary span at the same pull.

Data

$w = 0.03$ kg/m, $P = 5$ kg, standard length $= 30.015$ m (catenary), $L = 24.726$ m.

1. Tape length

Sag at standardisation: $\dfrac{0.03^{2}\times 30^{3}}{24\times 5^{2}} = 0.0405$ m. True tape length $= 30.015 + 0.0405 = 30.0555$ m.

$$C_a = 24.726\times\frac{30.0555 - 30}{30} = +0.0457\ \text{m}$$

2. Sag correction

$$C_s = -\frac{0.03^{2}\times 24.726^{3}}{24\times 5^{2}} = -0.0227\ \text{m}$$

3. Corrected length

$$L_{true} = 24.726 + 0.0457 - 0.0227 = 24.7491\ \text{m}$$

Answer: corrected length $= 24.7491$ m.

Reading. "Fully support condition" means the tape rests on the ground throughout the measurement, so there is no sag correction. The correction per tape length is made up of the temperature and pull corrections.

Data

• $l = 30$ m; $T_0 = 15^\circ$C, $T = 20^\circ$C; $P_0 = 10$ kgf, $P = 16$ kgf
• $W = 0.8$ kgf; density of steel $= 0.00786$ kgf/cm$^3$; $E = 2.1\times 10^{5}$ kgf/cm$^2$; $\alpha = 7.1\times 10^{-7}$/$^\circ$C

Cross-sectional area

$$A = \frac{W}{\text{density}\times l} = \frac{0.8}{0.00786\times 3000} = 0.03393\ \text{cm}^2$$

$AE = 0.03393\times 2.1\times 10^{5} = 7124.7$ kgf.

1. Temperature correction

$$C_t = 7.1\times 10^{-7}\times(20 - 15)\times 30 = +0.000107\ \text{m}$$

2. Pull correction

$$C_p = \frac{(16 - 10)\times 30}{7124.7} = +0.02526\ \text{m}$$

3. Correction per tape length

$$C = 0.000107 + 0.02526 = +0.02537\ \text{m}$$

Answer: correction per tape length $= +0.0254$ m (the true length of one tape length is 30.0254 m).

(If the tape were suspended at its ends in the field, a further sag correction of $-\dfrac{W^{2}l}{24P^{2}} = -0.0031$ m would apply.)

Reading. The tape was standardised with the specified pull (fully supported) and is used supported at the two ends only, so the sag correction applies. Corrections are temperature, pull and sag.

Data

• $l = 30$ m; $T_0 = 15^\circ$C, $T = 20^\circ$C; $P_0 = 10$ kg, $P = 16$ kg
• $W = 0.8$ kg; density $= 0.00786$ kg/cm$^3$; $E = 2.1\times 10^{6}$ kg/cm$^2$; $\alpha = 7.1\times 10^{-7}$/$^\circ$C

Area of the tape

$$A = \frac{0.8}{0.00786\times 3000} = 0.03393\ \text{cm}^2,\qquad AE = 0.03393\times 2.1\times 10^{6} = 71{,}247\ \text{kg}$$

1. Temperature

$$C_t = 7.1\times 10^{-7}\times 5\times 30 = +0.000107\ \text{m}$$

2. Pull

$$C_p = \frac{(16 - 10)\times 30}{71{,}247} = +0.002526\ \text{m}$$

3. Sag

$$C_s = -\frac{W^{2}\,l}{24\,P^{2}} = -\frac{0.8^{2}\times 30}{24\times 16^{2}} = -0.003125\ \text{m}$$

4. Correction per tape length

$$C = 0.000107 + 0.002526 - 0.003125 = -0.000492\ \text{m}$$

Answer: correction per tape length $= -0.0005$ m (about $-0.5$ mm), so the true length of one tape length is 29.9995 m.

Methods of linear measurement

1. Pacing: counting the paces and multiplying by the average pace length (rough, accuracy about 1/100).
2. Passometer/pedometer and odometer or measuring wheel: record the steps or the revolutions of a wheel (rough).
3. Chaining or taping: direct measurement by a chain, steel tape or invar tape (ordinary to high accuracy, 1/1000 to 1/10,000 or better).
4. Tacheometry: stadia hairs in a theodolite measure the distance by readings on a staff.
5. Subtense bar method: a bar of known length subtends a small angle at the observer.
6. EDM/total station: distance from the time or phase of electromagnetic waves.
7. GNSS/GPS and photogrammetry: distances from coordinates.

Sag correction

When a tape is suspended between two supports, it hangs in the form of a catenary (approximately a parabola) and the measured length along the curve is longer than the horizontal chord. For a tape of weight $W$ (total) with equal spans $l$ and pull $P$:

$$C_s = -\frac{W^{2}\,l}{24\,P^{2}} \text{ per span}$$

or, in terms of weight per metre $w$ and total length $L$ in $n$ spans:

$$C_s = -\frac{n\,w^{2}\,l^{3}}{24\,P^{2}} = -\frac{L\,w^{2}\,l^{2}}{24\,P^{2}}$$

Sign of sag correction

• Negative when the tape was standardised on the flat (fully supported) and is used suspended (in catenary), because the sagging tape reads more than the horizontal distance. The correction is subtracted.
• Positive when the tape was standardised in catenary and is used on the flat (fully supported), because the sag effect included in its standard length must be removed. In this case it is added.
• The correction is not required if the standardisation and the use are in the same conditions, with the same pull and span.

Sag can be reduced by applying a higher pull or reducing the span length, but the pull must be such that sag exactly balances the pull and the elastic stretch (normal tension).

Eight methods of linear measurement are:

1. Pacing: distance = number of paces $\times$ average length of one pace. Rough, about 1/100 accuracy, used for reconnaissance.
2. Passometer or pedometer: a small instrument carried by the surveyor that records the number of steps.
3. Odometer, measuring wheel (perambulator): a wheel that records its revolutions; distance = revolutions $\times$ circumference. Used for road lengths.
4. Chaining: direct measurement with a metric chain (20 m or 30 m) for chain survey.
5. Taping: direct measurement with steel, invar or cloth tapes. Corrections are applied for high accuracy.
6. Tacheometry (stadia): the distance is calculated from the staff intercept between the stadia hairs and the vertical angle, using a theodolite.
7. Subtense bar method: a bar of fixed length (e.g. 2 m) is placed perpendicular to the line and the small angle subtended at the instrument is measured by a theodolite; $D = \dfrac{b}{2}\cot\dfrac{\theta}{2}$.
8. Electronic distance measurement (EDM) and total station: distance from the time or phase of electromagnetic waves. Very accurate and fast.

Additional methods: GNSS/GPS positioning, photogrammetry (from photographs) and LiDAR/laser scanning.

Horizontal distance is required for plotting, but on sloping ground the tape follows the slope, so a method must reduce the slope to horizontal. There are two ways: direct (by stepping) and indirect (measuring along the slope and applying a correction).

1. Direct method: stepping

The line is measured in short horizontal steps, as shown. The tape is held horizontal, with one end resting on the ground and the other end plumbed from the tape by a plumb bob. The ground is measured downhill when possible, so the plumbed end is held at a convenient height. The steps are added to get the total horizontal distance.

  A o------- step 1 (tape horizontal)
            | plumb bob
            o------- step 2
                    |
                    o------- step 3  -> B

Limits: slow, many plumbing errors, only for steep ground and short steps.

2. Indirect method: measuring along the slope and calculating

The slope length $L$ is measured and the horizontal length $D$ is found by one of the following:

• Height difference $h$ from levelling: $D = \sqrt{L^{2} - h^{2}}$, or the slope correction $C_h = -\dfrac{h^{2}}{2L}$ (approx.).
• Slope angle $\theta$ measured with a clinometer or theodolite: $D = L\cos\theta$, correction $C_h = -L(1-\cos\theta)$.
• Hypotenuse allowance: used in chain survey on slopes. At the starting point, an extra length (allowance) is added so that the slope length measured by the chain equals the horizontal distance. For a slope $\theta$ and a 30 m chain, the allowance is $30\sec\theta - 30 = 30(\sec\theta - 1)$. For example, at $\theta = 10^\circ$ the allowance is 0.46 m.

Example

Slope length $L = 30$ m and $h = 2$ m. Then $D = \sqrt{900 - 4} = 29.933$ m (correction $-0.067$ m). The approximate formula gives $-4/60 = -0.067$ m.

Classification of EDM

EDM instruments are classified according to the type of carrier wave used.

By measuring principle: pulse method and phase-comparison method. By range: short (up to 5 km), medium (5 to 100 km), long (more than 100 km).

Significance of electromagnetic energy

• EM waves travel at a very high, known velocity ($c = 299{,}792{,}458$ m/s in vacuum, $v = c/n$ in air), so time or phase measurement gives distance: $D = vt/2$.
• They travel in a straight line and can be reflected by a prism, which allows round-trip measurements.
• A carrier can be modulated (amplitude or frequency) so that wavelength and phase can be measured precisely.
• Velocity changes with atmosphere (temperature, pressure, humidity), so those readings are used for correction.
• Different parts of the spectrum give different ranges and weather behaviour, e.g. microwaves go through haze and rain while light needs clear visibility.

Electromagnetic (EM) energy is transmitted as waves consisting of electric and magnetic fields perpendicular to each other and to the direction of travel. Light, infrared, microwave and radio waves are parts of the EM spectrum.

Characteristics of propagation

1. Velocity. In vacuum EM waves travel at $c = 299{,}792{,}458$ m/s. In air, the velocity is $v = c/n$, where $n$ is the refractive index, slightly greater than 1 (about 1.0003). It varies with temperature, pressure and humidity.
2. Frequency and wavelength:

$$c = f\,\lambda \quad\text{or}\quad v = f\,\lambda$$

3. Straight-line travel. In a uniform medium the wave travels in a straight line. In real air, refraction bends the path slightly.
4. Reflection, refraction, absorption and scattering. At a prism the wave is reflected back parallel to its path; fog, rain, dust and haze absorb or scatter it and reduce the range.
5. Modulation. In EDM the carrier wave is modulated in amplitude (or frequency) with a known frequency so that the measuring wave has a known wavelength.

  Transmitter ~~~~ wave ~~~~> Reflector
              <~~~ return ~~~~
  wavelength  | lambda |   one cycle = 360 degrees

Use in distance measurement

The wave travels the double distance $2D$, which equals $N\lambda$ plus a fraction. Atmospheric corrections are applied to the velocity because it changes with temperature and pressure. Microwaves are unaffected by poor visibility, whereas light waves and infrared are affected, but they are more accurate.

In the phase-comparison method of EDM, a continuous carrier wave is modulated with a known frequency, sent to a reflector and compared with the returned wave. The phase difference gives the distance.

Principle

Let the modulation frequency be $f$ and velocity $v$, so wavelength $\lambda = v/f$. The wave travels from the instrument A to the reflector B and back, a total of $2D$. Hence

$$2D = N\,\lambda + \Delta\lambda, \qquad \Delta\lambda = \frac{\phi}{2\pi}\,\lambda$$ $$D = \frac{1}{2}\left(N + \frac{\phi}{2\pi}\right)\lambda$$

where $N$ is the integer number of complete wavelengths in the double path and $\phi$ the phase difference (in radians) measured by the instrument.

 Sent wave      /\  /\  /\  /\ ...
 Received wave   /\  /\  /\ ...
                  |<-->| phase shift (phi)

Steps

1. The transmitted wave and the received wave are compared in a phase detector, and $\phi$ (the fractional part $\Delta\lambda$) is obtained.
2. The unknown integer $N$ cannot be found from one frequency. Hence several modulation frequencies are used, e.g. a low frequency (long wavelength, e.g. 1500 m) for the coarse distance and a higher frequency (short wavelength, e.g. 10 m) for fine reading. The readings of different wavelengths remove the ambiguity.
3. The microprocessor combines the results, applies atmospheric corrections and displays the distance.

Example

For $\lambda = 20$ m, $N = 10$ and $\phi = 90^\circ$ ($\tfrac{1}{4}$ cycle):

$$D = \frac{1}{2}(10 + 0.25)\times 20 = 102.5\ \text{m}$$

Reading of the doubtful value. The scanned tape weight "0.05 kg/cm" is taken as 0.05 kg per metre (a 30 m tape then weighs 1.5 kg, a realistic value for steel tape of 0.04 cm$^2$ section). The other values are as printed.

Data

• $l = 30$ m; standard pull $P_0 = 5.5$ kg (tape fully supported); $A = 0.04$ cm$^2$; $E = 2.1\times 10^{6}$ kg/cm$^2$
• Measured distance $L = 358.650$ m; field pull $P = 8$ kg; tape supported at the ends only (catenary)
• $w = 0.05$ kg/m, so $W = 0.05\times 30 = 1.5$ kg
• Number of tapes: 11 full tapes (330 m) + a last part of 28.650 m. No temperature or tape-length error is given, so only pull and sag are applied.

1. Correction for pull

$$C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(8 - 5.5)\times 358.650}{0.04\times 2.1\times 10^{6}} = +0.0107\ \text{m}$$

2. Correction for sag

Full tape: $C_s = -\dfrac{W^{2}\,l}{24\,P^{2}} = -\dfrac{1.5^{2}\times 30}{24\times 8^{2}} = -0.043945$ m per tape.

11 full tapes: $-11\times 0.043945 = -0.48340$ m.

Last length of 28.650 m: $C_s = -\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = -\dfrac{0.05^{2}\times 28.65^{3}}{24\times 8^{2}} = -0.03828$ m.

Total sag correction $= -0.48340 - 0.03828 = -0.5217$ m.

3. True length

$$L_{true} = 358.650 + 0.0107 - 0.5217 = 358.1390\ \text{m}$$

Answer: correct length of the line $= 358.139$ m.

Reading. The tape was exactly 30 m (standard) at the start, so the first 825 m has no tape-length error. It was found 2.5 cm too long (30.025 m) at the end, so the remaining 750 m, measured with the longer tape, needs a tape-length correction. The temperature and pull corrections apply to the whole line. No sag is mentioned, so the tape is taken as fully supported.

Data

• Standard: $T_0 = 20^\circ$C, $P_0 = 10$ kg; field: $T = 30^\circ$C, $P = 12$ kg
• $\alpha = 3.5\times 10^{-6}$/$^\circ$C, $A = 0.025$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$
• Total measured length $= 825 + 750 = 1575$ m

1. Tape length correction (750 m part only)

$$C_a = \frac{0.025}{30}\times 750 = +0.6250\ \text{m}$$

2. Temperature correction (whole line)

$$C_t = \alpha\,(T - T_0)\,L = 3.5\times 10^{-6}\times 10\times 1575 = +0.0551\ \text{m}$$

3. Pull correction (whole line)

$$C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(12 - 10)\times 1575}{0.025\times 2.1\times 10^{6}} = +0.0600\ \text{m}$$

4. True length

$$L_{true} = 1575 + 0.6250 + 0.0551 + 0.0600 = 1575.7401\ \text{m}$$

Answer: true length of the line $= 1575.740$ m.

Assumptions. The tape is nominally 30 m long, standardised on the flat (fully supported), so it has no length error. Each bay is measured with the tape in catenary at 120 N. Slope corrections use $h^2/2l$.

Data

• Area $A = 3.24$ mm$^2 = 3.24\times 10^{-6}$ m$^2$; density $= 7700$ kg/m$^3$
• Mass per metre $= 7700\times 3.24\times 10^{-6} = 0.024948$ kg/m, so $w = 0.24474$ N/m
• $E = 15.3\times 10^{4}$ MN/m$^2 = 1.53\times 10^{11}$ N/m$^2$, so $AE = 495{,}720$ N
• $P = 120$ N, $P_0 = 89$ N, $T = 31^\circ$C, $T_0 = 20^\circ$C, $\alpha = 1\times 10^{-6}$/$^\circ$C

Formulas for each bay of length $l$ with level difference $h$

$$C_h = -\frac{h^{2}}{2l},\quad C_t = \alpha\,(T-T_0)\,l,\quad C_p = \frac{(P-P_0)\,l}{AE},\quad C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}}$$

Calculation

Sum of measured lengths $= 117.998$ m.

$$\text{True length} = 29.4743 + 29.2033 + 29.3793 + 29.9048 = 117.9617\ \text{m}$$

Answer: true horizontal length of the base line AB $\approx 117.962$ m (the sum of the bays computed with unrounded values is $117.9616$ m).

Assumptions. The tape was standardised in catenary at 7 kg with the same support spacing (20 m) as in the field, so its stated length is correct for that condition. Only the changes (pull and sag) are corrected. $g = 9.81$ m/s$^2$.

Data

• $L = 360$ m; field pull $P = 5$ kg; standard pull $P_0 = 7$ kg
• $w = 10$ g/m $= 0.01$ kg/m; $A = 0.03$ cm$^2$
• $E = 210\times 10^{3}$ N/mm$^2 = 2.1\times 10^{7}$ N/cm$^2 = 2.1407\times 10^{6}$ kg/cm$^2$
• Spans: $360/20 = 18$ spans of 20 m

1. Pull correction

$$C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(5 - 7)\times 360}{0.03\times 2.1407\times 10^{6}} = -0.0112\ \text{m}$$

2. Sag correction (difference between the sag at 5 kg and that already in the standard at 7 kg)

Sag at 5 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 5^{2}} = 0.02400$ m. Sag at 7 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 7^{2}} = 0.01224$ m.

$$C_s = -(0.02400 - 0.01224) = -0.0118\ \text{m}$$

3. Correct length

$$L_{true} = 360 - 0.0112 - 0.0118 = 359.9770\ \text{m}$$

(If the sag in standardisation is ignored, so that the full sag of $-0.0240$ m is applied, the length is 359.9648 m.)

Answer: correct length of the line $= 359.977$ m.

Method. A tape standardised in catenary has the sag effect included in its stated length. We add this sag back to find the true tape length, then correct the 23 m span for tape length and for its own sag. The field pull is taken equal to the standard pull, 50 N.

Data

$w = 0.015\times 9.81 = 0.14715$ N/m, $P = 50$ N, $L = 23$ m, standard length $= 29.8950$ m.

1. Sag at standardisation (30 m)

$$\frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}$$

True tape length $= 29.8950 + 0.009744 = 29.9047$ m.

2. Tape length correction

$$C_a = 23\times\frac{29.9047 - 30}{30} = -0.0730\ \text{m}$$

3. Sag correction for the 23 m span

$$C_s = -\frac{0.14715^{2}\times 23^{3}}{24\times 50^{2}} = -0.0044\ \text{m}$$

4. Horizontal length

$$D = 23 - 0.0730 - 0.0044 = 22.9226\ \text{m}$$

Answer: horizontal length of the span $= 22.9226$ m.

Reading. The modulus is used as given, $E = 12\times 10^{11}$ N/m$^2$. Standardised fully supported (flat), so no standard sag is added back. Field supports are at the 0 and 30 m ends only, so each tape length is a 30 m catenary span.

Data

• $W = 8.9$ N for 30 m; $A = 2.58$ mm$^2 = 2.58\times 10^{-6}$ m$^2$; $AE = 2.58\times 10^{-6}\times 12\times 10^{11} = 3.096\times 10^{6}$ N
• Standard: 30.005 m at $20^\circ$C, $P_0 = 52$ N
• Field: $T = 31.2^\circ$C, $P = 110$ N, $L = 630$ m (21 tapes of 30 m)
• $\alpha = 12\times 10^{-6}$/$^\circ$C

1. Tape length

$$C_a = \frac{630\times 0.005}{30} = +0.1050\ \text{m}$$

2. Temperature

$$C_t = 12\times 10^{-6}\times(31.2 - 20)\times 630 = +0.0847\ \text{m}$$

3. Pull

$$C_p = \frac{(110 - 52)\times 630}{3.096\times 10^{6}} = +0.0118\ \text{m}$$

4. Sag

Per tape: $\dfrac{W^{2}\,l}{24\,P^{2}} = \dfrac{8.9^{2}\times 30}{24\times 110^{2}} = 0.008183$ m.

$$C_s = -21\times 0.008183 = -0.1718\ \text{m}$$

5. Correct horizontal distance

$$D = 630 + 0.1050 + 0.0847 + 0.0118 - 0.1718 = 630.0296\ \text{m}$$

Answer: correct horizontal distance $= 630.030$ m.