Traversing

When two quantities of a closed traverse are missing, they can be found from $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. If the lines with missing data are not adjacent, the known lines lying between them are shifted so that the two unknown lines come together. This is allowed because the closure of a traverse depends only on the lines taken, not on their order.

Procedure

1. Compute latitude $L\cos\theta$ and departure $L\sin\theta$ of every line with complete data.
2. Sum them: $\Sigma\text{Lat}$ and $\Sigma\text{Dep}$ of all the known lines.
3. Draw (or imagine) the known lines placed one after the other, and the two unknown lines at the ends so that the figure closes. The closing line then has

$$\text{Lat}_c=-\Sigma\text{Lat},\qquad \text{Dep}_c=-\Sigma\text{Dep},\qquad l=\sqrt{\text{Lat}_c^2+\text{Dep}_c^2},\qquad \tan\phi=\frac{\text{Dep}_c}{\text{Lat}_c}$$

4. The two unknown lines and this closing line form a triangle (the unknown lines are made adjacent, see the sketch). Solve the triangle:
   • Both lengths missing: use the sine rule, the angles being known from the known bearings.
   • Both bearings missing: use the cosine rule, all three sides being known.
   • One length and one bearing missing: use the sine/cosine rule or the quadratic in the unknown length (two possible solutions; select the one that suits the sketch).
5. Find the missing bearings from the triangle angles, then restore the lines to their original positions. Check by computing $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$.

 Order in field:  AB(?) BC CD DE(?) EA
 Reordered:       AB(?) DE(?) | BC CD EA (known)

            P
           /\
   AB(?)  /  \  DE(?)
         /    \
        /______\
       Q    l    R
 l = resultant of the known lines BC, CD, EA

If more than two quantities are missing, they cannot be found by calculation and the lines must be re-measured in the field.

Plotting a coordinated traverse means locating every station on a gridded sheet from its independent northing and easting, using a convenient scale.

1. Choose a scale so that the whole traverse fits on the sheet (for example 1 cm = 10 m). Find the range of N and E, subtract the minimum values and divide by the scale to get the size needed.
2. Draw the grid of squares (the grid sheet) and label the grid lines with the independent northing (vertical axis) and easting (horizontal axis) values, choosing an origin near the south-west corner so that all coordinates are positive.
3. Convert each station's independent coordinates to plotting units. For example, with scale 1 cm = 10 m, a station at $(N, E)$ is at $(E-E_0)/10$ cm along the east axis and $(N-N_0)/10$ cm along the north axis from the origin.
4. Mark each station with a fine pencil dot using a scale and protractor-free method (read the coordinate directly on the grid), and circle it with the station name.
5. Join the stations in order with thin lines and check that each plotted line length, measured with the scale, agrees with the field length. Use this as a check for blunders.
6. Plot details (offsets and radiations) from the stations, ink the map and add title, scale, north arrow, legend and border.

Advantage: the stations are located independently by coordinates, so no error accumulates, and the closing check is automatic.

Balancing angular misclosure in a link traverse

A link traverse starts and ends on stations of known coordinates, and the bearings of the first and last reference lines are known. The angular misclosure is therefore found by carrying the bearing through the traverse.

1. Starting from the known bearing of the first reference line, compute the bearing of every line using the observed angles (bearing of a line $=$ bearing of previous line $\pm180^\circ+$ angle, as the angles are measured clockwise or anticlockwise).
2. The computed bearing of the last reference line is compared with its known bearing:

$$\text{Angular misclosure}=\text{computed bearing}-\text{known bearing}$$

3. If it is within the allowable limit (for example $\pm k\sqrt{n}$, $n$ = number of angles), distribute it equally, the correction per angle being $-\text{misclosure}/n$. The first bearing receives one correction, the second two, and so on, so the last bearing receives $n$ times the correction and then equals the known value.

After the angles are balanced, the lengths are balanced through the linear misclosure of coordinates.

Transit rule (balancing latitudes and departures)

Transit rule: the total error in latitude (or departure) is distributed in proportion to the latitude (or departure) of each line, taken without sign.

$$C_L=-\Sigma L_{at}\times\frac{|\text{lat of line}|}{\Sigma|\text{lat}|},\qquad C_D=-\Sigma D_{ep}\times\frac{|\text{dep of line}|}{\Sigma|\text{dep}|}$$

It is used when angles are measured more precisely than lengths (for example with a precise theodolite and ordinary taping). The assumption is that the error in a line's latitude depends on the latitude itself, so a line running mainly north-south gets a large latitude correction and a small departure correction. It is less used than Bowditch's rule because it can give large corrections to short lines that happen to run along an axis.

Why angular and linear accuracy must be matched

A traverse is only as accurate as its weaker measurement. If the angles are measured to $1''$ but the lengths are taped to 1 in 1000, the extra effort on angles is wasted, and the reverse is equally true. An angular error $\alpha$ moves a station sideways by $L\alpha$ at a distance $L$, which acts like a linear error. So both are made to give the same position error, giving a balanced and economical survey (same accuracy for both lets Bowditch's rule be used).

Expressions

Linear precision is the closing error divided by the perimeter, $1/N$. An angular error $\alpha$ (in radians) gives a sideways error $L\alpha$ over a length $L$, so the equivalent linear precision is $\alpha$ itself:

$$\frac{1}{N}=\alpha\ (\text{rad})=\frac{\alpha''}{206265}$$

Hence

$$\text{angular to linear: } N=\frac{206265}{\alpha''},\qquad \text{linear to angular: } \alpha''=\frac{206265}{N}$$

Examples

• Angular to linear: angles measured to $20''$ need distances to $1/N=20/206265$, i.e. $N=206265/20\approx10313$, about 1 in 10,000.
• Linear to angular: lengths taped to 1 in 5000 need angles to $\alpha''=206265/5000=41.3''$, about $40''$ (a $1'$ least-count theodolite is enough for 1 in 3438, and a $20''$ one for 1 in 10,000).
• A quick rule: $1'\leftrightarrow1/3438$, $1''\leftrightarrow1/206265$.

Method: compute latitudes and departures from the given WCBs, find the closing error against the known coordinates of Y, check it against the permissible limit, then distribute it by Bowditch's rule and compute the adjusted coordinates.

Step 1: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Step 2: Closing error

Required change in coordinates from X to Y:

$$\begin{aligned} \Delta N_{true}&=443.610-562.510=-118.900\ \text{m}\\ \Delta E_{true}&=443.610-175.250=268.360\ \text{m} \end{aligned}$$

Computed: $\Sigma \text{Lat}=-250.530$ m, $\Sigma \text{Dep}=288.723$ m.

$$\begin{aligned} e_N&=-250.530-(-118.900)=-131.630\ \text{m}\\ e_E&=288.723-(268.360)=20.363\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=133.196\ \text{m} \end{aligned}$$

Perimeter (total length) $\Sigma L=461.50$ m, so the relative error is $e/\Sigma L = 1:3.5$.

The permissible closing error is 1:500, i.e. at most 0.923 m, but the error here is 133.196 m (about 1:3.5). The given coordinates of Y and the field data do not agree, so in a real survey this traverse would be rejected and re-measured. The Bowditch adjustment is shown below only to complete the procedure.

Step 3: Bowditch correction

$$C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}$$

Step 4: Adjusted coordinates

Start from X and add the adjusted latitude and departure of each line:

Check: the computed coordinates of Y agree with the given Y (443.610 N, 443.610 E).

Answer: adjusted coordinates A (511.884 N, 254.808 E), B (488.781 N, 346.933 E), C (394.648 N, 352.630 E); closing error 133.196 m (1:3.5).

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of DE and EA are unknown, but all lengths are known. The two unknown lines together must cancel the resultant of the other lines, so DE, EA and that resultant (the closing line) form a triangle whose three sides are known. The angles of this triangle are found by the cosine rule.

Step 1: Resultant of the known lines

$$\begin{aligned} \Sigma\text{Lat}&=-54.105\ \text{m},\qquad \Sigma\text{Dep}=+423.651\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=+54.105,\ \text{Dep}=-423.651\\ r&=\sqrt{54.105^2+423.651^2}=427.092\ \text{m}\\ \phi&=\tan^{-1}\frac{-423.651}{+54.105}=277^\circ 17' \end{aligned}$$

Step 2: Triangle of DE, EA and the closing line

Sides: $r=427.092$, $l_i=240.30$ (DE), $l_j=265.40$ (EA).

$$\cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{427.092^2+240.30^2-265.40^2}{2\times427.092\times240.30}=0.82683\ \Rightarrow\ \alpha=34^\circ 14'$$

$\alpha$ is the angle between the closing line and DE, so the bearing of DE is $\phi\pm\alpha$:

• $\phi+\alpha=311^\circ 30'$ and $\phi-\alpha=243^\circ 03'$, giving two geometrically possible triangles (the figure can lie on either side of the closing line).
• The bearing of EA follows from the remaining vector: for DE = 311°30', the vector left for EA is (-105.135, -243.688), so EA = 246°40'; for DE = 243°03', EA = 307°54'.

Both pairs satisfy closure; they are mirror images about the closing line, and the field sketch of the traverse decides which one applies. The first pair is adopted here, as it follows the general direction of the neighbouring lines.

Step 3: Check with the chosen bearings

$$\Sigma\text{Lat}=-54.105+(+159.240)+(-105.135)=-0.000,\qquad \Sigma\text{Dep}=+423.651+(-179.963)+(-243.688)=-0.000$$

Both are zero, so the traverse closes.

Answer: bearing of DE = 311°30' (N 48°30' W); bearing of EA = 246°40' (S 66°40' W).

Closed and open traverses

• Closed traverse: a traverse that ends where it began (a closed loop) or that starts and ends on stations of known position (a closed link/connecting traverse). It gives a mathematical check on the field work.
• Open traverse: a traverse that starts at a known station and ends at a point of unknown position, so there is no closing check. It is used for roads, canals, pipelines and similar route surveys, and the angles and lengths must be checked by repeat measurement.

Transit rule

Transit rule: the total error in latitude (or departure) is distributed in proportion to the latitude (or departure) of each line, taken without sign.

$$C_L=-\Sigma L_{at}\times\frac{|\text{lat of line}|}{\Sigma|\text{lat}|},\qquad C_D=-\Sigma D_{ep}\times\frac{|\text{dep of line}|}{\Sigma|\text{dep}|}$$

It is used when angles are measured more precisely than lengths (for example with a precise theodolite and ordinary taping). The assumption is that the error in a line's latitude depends on the latitude itself, so a line running mainly north-south gets a large latitude correction and a small departure correction. It is less used than Bowditch's rule because it can give large corrections to short lines that happen to run along an axis.

Example

For the traverse with latitudes and departures below, $e_N=+0.26$ m, $\Sigma|\text{Lat}|=230.34$ and $e_E=-0.13$ m, $\Sigma|\text{Dep}|=398.33$. The correction to AB latitude is $-(+0.26)\times85.20/230.34=-0.096$ m.

The corrected latitudes and departures sum to zero. Note that the lines with larger latitudes/departures get the larger corrections.

Method: compute latitudes and departures from the given WCBs, find the closing error against the known coordinates of Y, check it against the permissible limit, then distribute it by Bowditch's rule and compute the adjusted coordinates.

Step 1: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Step 2: Closing error

Required change in coordinates from X to Y:

$$\begin{aligned} \Delta N_{true}&=1043.610-1162.510=-118.900\ \text{m}\\ \Delta E_{true}&=1043.610-775.250=268.360\ \text{m} \end{aligned}$$

Computed: $\Sigma \text{Lat}=-250.530$ m, $\Sigma \text{Dep}=288.723$ m.

$$\begin{aligned} e_N&=-250.530-(-118.900)=-131.630\ \text{m}\\ e_E&=288.723-(268.360)=20.363\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=133.196\ \text{m} \end{aligned}$$

Perimeter (total length) $\Sigma L=461.50$ m, so the relative error is $e/\Sigma L = 1:3.5$.

The permissible closing error is 1:500, i.e. at most 0.923 m, but the error here is 133.196 m (about 1:3.5). The given coordinates of Y and the field data do not agree, so in a real survey this traverse would be rejected and re-measured. The Bowditch adjustment is shown below only to complete the procedure.

Step 3: Bowditch correction

$$C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}$$

Step 4: Adjusted coordinates

Start from X and add the adjusted latitude and departure of each line:

Check: the computed coordinates of Y agree with the given Y (1043.610 N, 1043.610 E).

Answer: adjusted coordinates A (1111.884 N, 854.808 E), B (1088.781 N, 946.933 E), C (994.648 N, 952.630 E); closing error 133.196 m (1:3.5).

Closed and open traverses

• Closed traverse: a traverse that ends where it began (a closed loop) or that starts and ends on stations of known position (a closed link/connecting traverse). It gives a mathematical check on the field work.
• Open traverse: a traverse that starts at a known station and ends at a point of unknown position, so there is no closing check. It is used for roads, canals, pipelines and similar route surveys, and the angles and lengths must be checked by repeat measurement.

Bowditch's rule

Bowditch (compass) rule: the closing error in latitude and departure is distributed among the lines in proportion to their lengths.

$$C_L=-\Sigma L_{at}\times\frac{l}{\Sigma l},\qquad C_D=-\Sigma D_{ep}\times\frac{l}{\Sigma l}$$

where $l$ is the length of the line and $\Sigma l$ the perimeter (or total length of the traverse).

Assumptions:

• Angular and linear measurements are of equal accuracy, so errors are due to chance only.
• The error in a line is proportional to $\sqrt{l}$ (the correction is taken proportional to $l$, which is a simplification of this).
• Angles have been balanced first, so the remaining error is in lengths.

It is simple, and is the most commonly used rule in plane surveying.

Example

A four-sided closed traverse has ΣL = 494.58 m with computed latitudes and departures below. The errors are $e_N=\Sigma\text{Lat}=+0.26$ m and $e_E=\Sigma\text{Dep}=-0.13$ m, so $e=0.291$ m and the precision is $1:1701$.

Check: the sum of corrections is $-e_N$ and $-e_E$, so the corrected latitudes and departures sum to zero. For line AB: $C_{lat}=-(+0.26)\times99.97/494.58=-0.053$.

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Here the bearings of BC and CD are unknown, but all lengths are known. The two unknown lines together must cancel the resultant of the other lines, so BC, CD and that resultant (the closing line) form a triangle whose three sides are known. The angles of this triangle are found by the cosine rule.

Step 1: Resultant of the known lines

$$\begin{aligned} \Sigma\text{Lat}&=-47.583\ \text{m},\qquad \Sigma\text{Dep}=-705.255\ \text{m}\\ \text{Closing vector}&:\ \text{Lat}=+47.583,\ \text{Dep}=+705.255\\ r&=\sqrt{47.583^2+705.255^2}=706.858\ \text{m}\\ \phi&=\tan^{-1}\frac{+705.255}{+47.583}=86^\circ 08' \end{aligned}$$

Step 2: Triangle of BC, CD and the closing line

Sides: $r=706.858$, $l_i=432.00$ (BC), $l_j=324.00$ (CD).

$$\cos\alpha=\frac{r^2+l_i^2-l_j^2}{2\,r\,l_i}=\frac{706.858^2+432.00^2-324.00^2}{2\times706.858\times432.00}=0.95181\ \Rightarrow\ \alpha=17^\circ 52'$$

$\alpha$ is the angle between the closing line and BC, so the bearing of BC is $\phi\pm\alpha$:

• $\phi+\alpha=104^\circ 00'$ and $\phi-\alpha=68^\circ 17'$, giving two geometrically possible triangles (the figure can lie on either side of the closing line).
• The bearing of CD follows from the remaining vector: for BC = 104°00', the vector left for CD is (+152.088, +286.086), so CD = 62°00'; for BC = 68°17', CD = 110°17'.

Both pairs satisfy closure; they are mirror images about the closing line, and the field sketch of the traverse decides which one applies. The first pair is adopted here, as it follows the general direction of the neighbouring lines.

Step 3: Check with the chosen bearings

$$\Sigma\text{Lat}=-47.583+(-104.505)+(+152.088)=-0.000,\qquad \Sigma\text{Dep}=-705.255+(+419.169)+(+286.086)=0.000$$

Both are zero, so the traverse closes.

Answer: bearing of BC = 104°00' (S 76°00' E); bearing of CD = 62°00' (N 62°00' E).

Computation of a closed theodolite traverse (closed loop) proceeds as follows.

1. Angular check and adjustment: the sum of interior angles must equal $(2n-4)\times90^\circ$. The angular error is spread equally over the angles (or according to the weights of the stations) if it is within the permissible limit.
2. Bearings: starting from the observed or assumed bearing of one line, find the bearings of the others from the adjusted angles. Check that the first line's bearing returns.
3. Latitudes and departures: $\text{Lat}=L\cos\theta$, $\text{Dep}=L\sin\theta$ for each line.
4. Linear misclosure and precision:

$$e_N=\Sigma\text{Lat},\quad e_E=\Sigma\text{Dep},\quad e=\sqrt{e_N^2+e_E^2},\quad \frac{e}{\Sigma L}=\frac1N$$

Accept if the precision is within the required limit (e.g. 1 in 5000), else repeat field work. 5. Balancing: distribute $e_N$ and $e_E$ among the lines by Bowditch's rule (or the transit rule) to get corrected latitudes and departures, whose sums are zero. 6. Final lengths and bearings: from the corrected values, $L=\sqrt{\text{Lat}^2+\text{Dep}^2}$ and $\theta=\tan^{-1}(\text{Dep}/\text{Lat})$. 7. Independent coordinates: starting from the known coordinates of one station (or an assumed origin), add the corrected latitudes and departures in sequence, checking that the starting station is reproduced. 8. Tabulate all the above in a Gale's traverse table, then plot by coordinates.

Closed loop and closed link traverse

• Closed loop traverse: the traverse starts and ends at the same station and forms a polygon. Checks: sum of interior angles $=(2n-4)\times 90^\circ$ and $\Sigma L=0,\ \Sigma D=0$ (latitudes and departures).
• Closed link (connecting) traverse: the traverse starts at one known station and ends at another known station, both with known coordinates, and with known reference bearings at the two ends. Checks: the computed bearing of the last line must equal its known value, and the computed coordinates of the last station must equal the known ones.

Angular misclosure

The difference between the sum of the observed interior angles and the theoretical value $(2n-4)\times90^\circ$ in a closed loop (or between the computed and known bearing of the last line in a link traverse).

Relative closing error

The linear error of closure divided by the total length of the traverse:

$$\frac{e}{\Sigma L}=\frac{\sqrt{(\Sigma\text{Lat})^2+(\Sigma\text{Dep})^2}}{\Sigma L}=\frac{1}{N}$$

It is expressed as 1 in $N$ and compared with the specified precision.

Balancing angular misclosure in a link traverse

A link traverse starts and ends on stations of known coordinates, and the bearings of the first and last reference lines are known. The angular misclosure is therefore found by carrying the bearing through the traverse.

1. Starting from the known bearing of the first reference line, compute the bearing of every line using the observed angles (bearing of a line $=$ bearing of previous line $\pm180^\circ+$ angle, as the angles are measured clockwise or anticlockwise).
2. The computed bearing of the last reference line is compared with its known bearing:

$$\text{Angular misclosure}=\text{computed bearing}-\text{known bearing}$$

3. If it is within the allowable limit (for example $\pm k\sqrt{n}$, $n$ = number of angles), distribute it equally, the correction per angle being $-\text{misclosure}/n$. The first bearing receives one correction, the second two, and so on, so the last bearing receives $n$ times the correction and then equals the known value.

After the angles are balanced, the lengths are balanced through the linear misclosure of coordinates.

The traverse P-Q-R-S-P is anticlockwise, so each deflection is to the left and bearing of a line = bearing of previous line − left deflection. Unknowns: length QR, bearing SP (length SP is given), deflections at P and S, and the coordinates.

Step 1: Bearings that can be fixed

• PQ = S45°W = 180° + 45° = 225°00'
• QR = 225°00' − 85°00' = 140°00'
• RS = 140°00' − 100°00' = 40°00'

Latitude and departure of PQ and RS:

$$\begin{aligned} \text{PQ}:&\ 436.80\cos225^\circ=-308.86,\quad 436.80\sin225^\circ=-308.86\\ \text{RS}:&\ 499.20\cos40^\circ=+382.41,\quad 499.20\sin40^\circ=+320.88 \end{aligned}$$

Their sums are $a=73.545$ m and $b=12.015$ m.

Step 2: Length of QR and bearing of SP

Closure requires $\text{PQ}+\text{QR}+\text{RS}+\text{SP}=0$. With QR of unknown length $x$ at $140^\circ$, the vector SP is $-(a+x\cos140^\circ,\ b+x\sin140^\circ)$ and its length is 516.084 m:

$$(a+x\cos140^\circ)^2+(b+x\sin140^\circ)^2=516.084^2$$ $$x^2+2(a\cos140^\circ+b\sin140^\circ)x+(a^2+b^2-516.084^2)=0$$ $$x^2+(-97.231)x+(-260789.4)=0\ \Rightarrow\ x=561.600\ \text{m}\ \ (\text{the negative root is rejected})$$

So QR = 561.600 m.

Then SP: latitude $=-(-356.665)=+356.665$, departure $=-(+373.005)=-373.005$.

$$L_{SP}=\sqrt{+356.665^2+(-373.005)^2}=516.084\ \text{m (agrees with the given 516.084 m)},\qquad \theta_{SP}=\tan^{-1}\frac{-373.005}{+356.665}=313^\circ 43'$$

(North-west quadrant, N 46°17' W.)

Step 3: Missing deflection angles

• At P: bearing SP − bearing PQ = 313°43' − 225°00' = 88°43' left
• At S: bearing RS − bearing SP = 40°00' − 313°43' + 360° = 86°17' left

Check: 88°43' + 85°00' + 100°00' + 86°17' = 360°00', as required for a closed anticlockwise traverse.

Step 4: Coordinates with respect to R (R = 2345 mE, 6789 mN)

Going round R to S to P to Q and back to R:

The traverse closes on R, so the work is checked.

Answer: QR = 561.600 m; SP bearing = 313°43'; deflection at P = 88°43' L, at S = 86°17' L; bearings PQ 225°00', QR 140°00', RS 40°00'. Coordinates: S (7171.409 N, 2665.880 E), P (7528.075 N, 2292.875 E), Q (7219.210 N, 1984.011 E).

General specification of horizontal control for a major traverse

Typical specifications adopted for a major (primary) traverse are:

The exact limits depend on the specification of the controlling agency (such as the Survey Department of Nepal) for the order of the survey.

Why angular accuracy is better on longer sides

An angular error comes mainly from centering and targeting errors, which are fixed linear amounts $e$ at the station or the target. The angle error is $e/L$ in radians, so it is smaller when $L$ is large. A short side therefore gives larger angular errors, and for the same angular error the position error at the far end is $L\alpha$, which is large on a long side, so angular accuracy on long sides must be high and is easy to achieve.

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Two quantities are unknown, the length of CD and the bearing of EA, so the two closure equations are enough to find them.

Step 1: Latitudes and departures of the known lines

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

The given bearings in whole circle form are AB = S52°30'E = 127°30', BC = N48°45'E = 48°45', CD = N18°15'W = 341°45', DE = S78°30'W = 258°30'. Let the unknown length of CD be $x$ (bearing 341°45'). Its latitude is $x\cos\theta$ and departure $x\sin\theta$. The line EA has length 238.80 m and must close the traverse, so

$$(16.610+x\cos\theta)^2+(189.257+x\sin\theta)^2=238.80^2$$

with $\cos\theta=0.94970$ and $\sin\theta=(-0.31316)$ for 341°45'.

Step 2: Solve for the length

$$x^2+2(16.610\times0.94970+189.257\times(-0.31316))x+(16.610^2+189.257^2-238.80^2)=0$$ $$x^2+(-86.988)x+(-20931.47)=0$$ $$x=\frac{-(-86.988)\pm\sqrt{91292.8}}{2}=194.567\ \text{or}\ -107.580$$

The negative root has no meaning, so length of CD = 194.567 m.

Step 3: Bearing of EA

Latitude of EA $=-(16.610+194.567\times0.94970)=(-201.390)$ m and departure $=-(189.257+194.567\times(-0.31316))=(-128.325)$ m.

$$\tan\theta=\frac{(-128.325)}{(-201.390)}\ \Rightarrow\ \theta=212^\circ 30'$$

The signs of latitude and departure show the line lies in the south-west quadrant, so the whole circle bearing is 212°30' (S 32°30' W).

Step 4: Check

Sums are nil, so the traverse closes.

Answer: length of CD = 194.567 m; bearing of EA = 212°30'.

Method: check and adjust the angles, find the bearings from the known bearing, compute latitudes and departures, balance them by Bowditch's rule and compute the coordinates from the given station.

Step 1: Angular check and adjustment

For a closed traverse of $n=5$ sides, the sum of interior angles must be $(2n-4)\times 90^\circ=540^\circ$.

Observed sum = 539°58'50", so the angular error is 0°01'10" (deficiency). The correction is distributed equally: 14.00" (added) per angle.

Step 2: Bearings

The traverse is taken as clockwise (interior angles on the right), so

$$\text{Bearing of next line}=\text{Bearing of previous line}+180^\circ-\text{interior angle at the station}$$

Starting from AB = 41°00'39":

Check: carrying the bearing round the traverse brings back AB = 41°00'39", as given.

Step 3: Latitudes, departures and closing error

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Closing error: $e_N=-0.071$ m, $e_E=-0.035$ m,

$$e=\sqrt{e_N^2+e_E^2}=0.0789\ \text{m},\qquad \text{relative error}=\frac{e}{\Sigma L}=\frac{0.0789}{250.37}=\frac{1}{3175}$$

Step 4: Bowditch correction and adjusted lines

Correction to a line $=-\text{(total error)}\times L/\Sigma L$. The corrected latitudes and departures give the final length $=\sqrt{\text{Lat}^2+\text{Dep}^2}$ and bearing $=\tan^{-1}(\text{Dep}/\text{Lat})$:

Step 5: Coordinates (D = 3000.000 N, 5000.000 E)

Answer: the final adjusted lengths and bearings are as tabulated above, and the traverse closes with a relative error of 1 in 3175.

Principles of theodolite traversing

• Working from the whole to the part: a framework of control stations of high accuracy is fixed first, and detail or minor traverses are fixed from it.
• Closure and checks: the traverse should start and end on known points or return to the start so that angular and linear errors can be checked.
• Balanced accuracy: angular and linear measurements are made to equal accuracy, and errors are distributed over the whole figure (Bowditch or transit rule).
• Independent coordinates: stations are fixed from coordinates referred to a common origin, so errors do not accumulate in plotting.
• Adequate redundancy: angles are measured on both faces and repeated, and lengths measured twice, to detect blunders.

Planning before traversing

1. Study of the purpose and accuracy required (type of map, scale, specifications) and the available maps and existing control.
2. Reconnaissance: walk over the area and decide the route and number of stations.
3. Selection of stations: firm ground, clear intervisibility of the adjacent stations, sides as long and as equal as possible, few stations, easy access to details, free from disturbance and traffic, not on slopes where the instrument cannot be levelled.
4. Marking and referencing with pegs, nails, concrete pillars and witness marks, with a sketch of each location.
5. Selection of instruments and methods: theodolite or total station, tape or EDM, and the method of angle measurement (included angle, deflection angle, direct bearing).
6. Organisation: party size, schedule, safety, field books, and the method of computation and checking.

A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station.

Known data: M2 = (1000 N, 1000 E), M8 = (1095 N, 1042 E); bearing M1-M2 = 105°, bearing M8-M9 = 36°.

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line M2-M1 (back bearing 285°00', since M1-M2 = 105°) and using the observed angles, the bearing of the last line comes out as 35°00', while its known value is 36°00' (M8-M9).

Angular misclosure = computed − known = 35°00' − 36°00' = -1°00'.

The error is spread equally over the 4 angles: correction per angle = +0°15'. The k-th bearing therefore receives k times this correction.

Step 2: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Starting from M2 (1000.000 N, 1000.000 E), the unadjusted coordinates of M8 are (1095.149 N, 1041.981 E), whereas the known coordinates are (1095.000 N, 1042.000 E).

Step 3: Linear misclosure

$$\begin{aligned} e_N&=1095.149-1095.000=+0.149\ \text{m}\\ e_E&=1041.981-1042.000=-0.019\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.150\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.150}{132.850}=\frac{1}{883} \end{aligned}$$

Step 4: Adjustment by Bowditch's rule

Bowditch's rule: the correction to each line is proportional to its length:

$$C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}$$

Step 5: Final coordinates

The computed coordinates of M8 now equal its known coordinates, so the traverse is balanced.

Answer: m1 (1028.951 N, 1020.494 E); m2 (1060.041 N, 1065.829 E) (linear misclosure 0.150 m, 1 in 883).

Significance of traversing

Traversing is a method of control survey in which a series of connected lines are measured in length and direction. It is used because:

• It gives horizontal control for topographic, engineering and cadastral surveys when the area is long and narrow, wooded or built-up, where triangulation is difficult.
• Only a few intervisible stations are needed, so the field work is quick and cheap.
• Coordinates of the stations are obtained and the work can be checked (angle sum, closing error) and balanced.
• It is the basis for plotting details and for setting out works such as roads, canals and buildings.

Accuracy parameters

Horizontal control

• Angular misclosure, limited to $k\sqrt{n}$, where $n$ is the number of angles and $k$ is the least count related constant (about $10''$ to $1'$, depending on the order of the survey).
• Linear misclosure, given as the relative closing error $e/\Sigma L=1/N$, for example 1 in 10,000 for a major traverse and 1 in 3000 to 1 in 5000 for minor traverses.

Vertical control

• Heights of stations are fixed by levelling (or trigonometric levelling) along the traverse, with the misclosure limited to $C\sqrt{K}$ mm, where $K$ is the length in km and $C$ is about 12 for third order and 25 for fourth order or ordinary work, as per the specification adopted.
• Double-run levelling and closure back to the starting benchmark are used as checks.

The bearings are given directly, so the Gale's table starts from the latitudes and departures. Bowditch's rule is used to balance the traverse, and the coordinates are then run from C (1000 N, 1500 E).

Step 1: Latitude and departure

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Step 2: Closing error

$$\begin{aligned} e_N&=\Sigma\text{Lat}=-0.117\ \text{m},\qquad e_E=\Sigma\text{Dep}=+0.346\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.3657\ \text{m}\\ \text{Relative error}&=\frac{e}{\Sigma L}=\frac{0.3657}{559.50}=\frac{1}{1530} \end{aligned}$$

Step 3: Bowditch correction

$$C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}$$

The corrections are in the last two columns above (their sums cancel the errors). Adjusted values:

Step 4: Coordinates of all points

Starting from C = (1000.000 N, 1500.000 E), add the adjusted latitude and departure of CD, DE, EA, AB in turn:

Check: C plus adjusted BC returns to C.

Answer: A (972.637 N, 1333.995 E), B (1030.036 N, 1367.756 E), C (1000.000 N, 1500.000 E), D (993.467 N, 1565.935 E), E (920.688 N, 1542.047 E); misclosure 0.3657 m, 1 in 1530.

For a closed traverse $\Sigma\text{Lat}=0$ and $\Sigma\text{Dep}=0$. Two quantities are unknown, the length of BC and the bearing of EA, so the two closure equations are enough to find them.

Step 1: Latitudes and departures of the known lines

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Let the unknown length of BC be $x$ (bearing 151°24'). Its latitude is $x\cos\theta$ and departure $x\sin\theta$. The line EA has length 359.60 m and must close the traverse, so

$$((-100.618)+x\cos\theta)^2+((-244.066)+x\sin\theta)^2=359.60^2$$

with $\cos\theta=(-0.87798)$ and $\sin\theta=0.47869$ for 151°24'.

Step 2: Solve for the length

$$x^2+2((-100.618)\times(-0.87798)+(-244.066)\times0.47869)x+((-100.618)^2+(-244.066)^2-359.60^2)=0$$ $$x^2+(-56.984)x+(-59620.10)=0$$ $$x=\frac{-(-56.984)\pm\sqrt{241727.5}}{2}=274.321\ \text{or}\ -217.337$$

The negative root has no meaning, so length of BC = 274.321 m.

Step 3: Bearing of EA

Latitude of EA $=-((-100.618)+274.321\times(-0.87798))=341.467$ m and departure $=-((-244.066)+274.321\times0.47869)=112.751$ m.

$$\tan\theta=\frac{112.751}{341.467}\ \Rightarrow\ \theta=18^\circ 16'$$

The signs of latitude and departure show the line lies in the north-east quadrant, so the whole circle bearing is 18°16' (N 18°16' E).

Step 4: Check

Sums are nil, so the traverse closes.

Answer: length of BC = 274.321 m; bearing of EA = 18°16'.

Cases of omitted measurement

The sum of latitudes and the sum of departures of a closed traverse are zero. Two equations are available, so two unknowns (length or bearing) can be found. Let the closing line be the vector that closes the known part: $\Sigma L_{at}$ and $\Sigma D_{ep}$ of the known lines, with length $\sqrt{\Sigma L_{at}^2+\Sigma D_{ep}^2}$ and bearing $\tan^{-1}(\Sigma D_{ep}/\Sigma L_{at})$ (taken in the opposite sense for the closure).

The affected lines may be adjacent or non-adjacent. If they are not adjacent, the known lines between them are shifted (a closed traverse keeps its closure if the order of its lines is changed), which brings the unknown lines together so that the same triangle method applies.

Plotting by the coordinates method (outline)

1. Compute the independent coordinates of all stations (from the adjusted latitudes and departures).
2. Choose a scale, draw a grid sheet and label the grid lines with northing and easting.
3. Mark each station at its $(E,N)$ position from the grid lines, join the stations in order and check that the plotted lengths agree with the field lengths.
4. Plot details from the stations and complete the map with title, scale and north arrow.

Plotting by coordinates is preferred to protractor plotting because errors do not accumulate.

Coordinates are listed as E, N in the question; in the working below they are written as (N, E). The reference bearings come from the coordinates of the control stations:

$$\begin{aligned} \theta_{AB}&=\tan^{-1}\frac{E_B-E_A}{N_B-N_A}=\tan^{-1}\frac{-711.87}{+397.67}\ \Rightarrow\ 299^\circ 11'20''\\ \theta_{XY}&=\tan^{-1}\frac{E_Y-E_X}{N_Y-N_X}=\tan^{-1}\frac{+661.00}{-131.14}\ \Rightarrow\ 101^\circ 13'18'' \end{aligned}$$

A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station.

Known data: A (2094.72 N, 1769.15 E), B (2492.39 N, 1057.28 E), X (1747.32 N, 2334.85 E), Y (1616.18 N, 2995.85 E).

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line AB (299°11'20" from the coordinates) and using the observed angles, the bearing of the last line comes out as 101°11'10", while its known value is 101°13'18" (X-Y from the coordinates).

Angular misclosure = computed − known = 101°11'10" − 101°13'18" = -0°02'08".

The error is spread equally over the 6 angles: correction per angle = +0°00'21". The k-th bearing therefore receives k times this correction.

Step 2: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Starting from A (2094.720 N, 1769.150 E), the unadjusted coordinates of X are (1747.249 N, 2334.614 E), whereas the known coordinates are (1747.320 N, 2334.850 E).

Step 3: Linear misclosure

$$\begin{aligned} e_N&=1747.249-1747.320=-0.071\ \text{m}\\ e_E&=2334.614-2334.850=-0.236\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.247\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.247}{1262.380}=\frac{1}{5121} \end{aligned}$$

Step 4: Adjustment by the transit rule

The transit rule: the correction to latitude (or departure) of each line is proportional to the size of that latitude (or departure):

$$C_{lat}=-e_N\frac{|\text{lat}|}{\Sigma|\text{lat}|},\qquad C_{dep}=-e_E\frac{|\text{dep}|}{\Sigma|\text{dep}|}$$

Step 5: Final coordinates

The computed coordinates of X now equal its known coordinates, so the traverse is balanced.

Answer: 1 (2214.743 N, 1939.400 E); 2 (2355.934 N, 2071.721 E); 3 (2090.461 N, 2262.173 E); 4 (1815.654 N, 2120.640 E) (linear misclosure 0.247 m, 1 in 5121).

Reference bearings from the coordinates of the control stations:

$$\theta_{AB}=\tan^{-1}\frac{30.000-18.820}{20.000-42.360}=153^\circ 26'06'',\qquad \theta_{MN}=\tan^{-1}\frac{65.667-100.000}{70.600-50.000}=300^\circ 57'50''$$

(The signs of the differences place AB in the south-east and MN in the north-west quadrant.) At B the reference line is BA, with bearing 333°26'06".

A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station.

Known data: A (42.360 N, 18.820 E), B (20.000 N, 30.000 E), M (50.000 N, 100.000 E), N (70.600 N, 65.667 E).

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line BA and using the observed angles, the bearing of the last line comes out as 300°59'06", while its known value is 300°57'50" (M-N from the coordinates).

Angular misclosure = computed − known = 300°59'06" − 300°57'50" = +0°01'15".

The error is spread equally over the 4 angles: correction per angle = -0°00'19". The k-th bearing therefore receives k times this correction.

Step 2: Latitudes and departures

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Starting from B (20.000 N, 30.000 E), the unadjusted coordinates of M are (49.966 N, 100.025 E), whereas the known coordinates are (50.000 N, 100.000 E).

Step 3: Linear misclosure

$$\begin{aligned} e_N&=49.966-50.000=-0.034\ \text{m}\\ e_E&=100.025-100.000=+0.025\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.043\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.043}{90.025}=\frac{1}{2117} \end{aligned}$$

Step 4: Adjustment by Bowditch's rule

Bowditch's rule: the correction to each line is proportional to its length:

$$C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}$$

Step 5: Final coordinates

The computed coordinates of M now equal its known coordinates, so the traverse is balanced.

Answer: X (10.014 N, 59.980 E); Y (29.932 N, 69.929 E) (linear misclosure 0.043 m, 1 in 2117).

Method: check and adjust the angles, find the bearings from the known bearing, compute latitudes and departures, balance them by Bowditch's rule and compute the coordinates from the given station.

Step 1: Angular check and adjustment

For a closed traverse of $n=4$ sides, the sum of interior angles must be $(2n-4)\times 90^\circ=360^\circ$.

Observed sum = 360°00'05", so the angular error is 0°00'05" (excess). The correction is distributed equally: 1.25" (subtracted) per angle.

Step 2: Bearings

The traverse is taken as clockwise (interior angles on the right), so

$$\text{Bearing of next line}=\text{Bearing of previous line}+180^\circ-\text{interior angle at the station}$$

Starting from CD = 314°58'04":

Check: carrying the bearing round the traverse brings back CD = 314°58'04", as given.

Step 3: Latitudes, departures and closing error

$$\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta$$

Closing error: $e_N=-0.090$ m, $e_E=+0.068$ m,

$$e=\sqrt{e_N^2+e_E^2}=0.1127\ \text{m},\qquad \text{relative error}=\frac{e}{\Sigma L}=\frac{0.1127}{239.50}=\frac{1}{2125}$$

Step 4: Bowditch correction and adjusted lines

Correction to a line $=-\text{(total error)}\times L/\Sigma L$. The corrected latitudes and departures give the final length $=\sqrt{\text{Lat}^2+\text{Dep}^2}$ and bearing $=\tan^{-1}(\text{Dep}/\text{Lat})$:

Step 5: Coordinates (C = 500.000 N, 500.000 E)

Answer: the final adjusted lengths and bearings are as tabulated above, and the traverse closes with a relative error of 1 in 2125.

Consecutive and independent coordinates

Consecutive (dependent) coordinates: the latitude and departure of a line, i.e. the coordinates of its end point measured from the preceding station taken as origin. Example: line AB with $L=100$ m and bearing $60^\circ$ gives $\Delta N=100\cos 60^\circ=+50$ m and $\Delta E=100\sin 60^\circ=+86.6$ m from A.

Independent (total) coordinates: the northing and easting of a station measured from one common origin (the reference station or a national grid). They are the algebraic sum of consecutive coordinates from the origin. Example: if A is $(1000,\,1000)$, then B is $(1050,\,1086.6)$, and C is B plus the consecutive coordinates of BC.

Independent coordinates are used for plotting because all stations refer to one origin, so plotting errors do not accumulate.

Plotting a traverse by grid coordinates

1. Choose a scale so that the whole traverse fits on the sheet (for example 1 cm = 10 m). Find the range of N and E, subtract the minimum values and divide by the scale to get the size needed.
2. Draw the grid of squares (the grid sheet) and label the grid lines with the independent northing (vertical axis) and easting (horizontal axis) values, choosing an origin near the south-west corner so that all coordinates are positive.
3. Convert each station's independent coordinates to plotting units. For example, with scale 1 cm = 10 m, a station at $(N, E)$ is at $(E-E_0)/10$ cm along the east axis and $(N-N_0)/10$ cm along the north axis from the origin.
4. Mark each station with a fine pencil dot using a scale and protractor-free method (read the coordinate directly on the grid), and circle it with the station name.
5. Join the stations in order with thin lines and check that each plotted line length, measured with the scale, agrees with the field length. Use this as a check for blunders.
6. Plot details (offsets and radiations) from the stations, ink the map and add title, scale, north arrow, legend and border.

Advantage: the stations are located independently by coordinates, so no error accumulates, and the closing check is automatic.