The Compass

Closing error (misclosure) is the gap by which a closed traverse fails to close when it is plotted from the observed bearings and lengths. Starting at A, the plotted end falls at A′ instead of A; the distance A′A is the closing error. It arises from errors in the measured angles/bearings, errors in the measured lengths, local attraction, and plotting errors.

The accuracy is judged by the relative error of closure: $\dfrac{\text{closing error}}{\text{perimeter}}$ (acceptable for compass work if about 1/300 or smaller).

Graphical adjustment by Bowditch's rule

Bowditch's rule says that the correction to each station is proportional to the length of traverse up to that station, so that the whole closing error is removed at the last station. It can be applied without calculation:

1. Plot the traverse A-B-C-D-E-A′; the closing error is A′A.
2. Draw a straight base line AA′ equal to the perimeter and mark b, c, d, e at distances AB, AB+BC, AB+BC+CD, ... (in any scale).
3. At the end A′ erect a perpendicular A′X equal to the closing error (to the plan scale) and join A to X.
4. The vertical ordinates bb′, cc′, dd′, ee′ drawn from the base to the line AX are the corrections for stations B, C, D, E. They increase uniformly, and the last ordinate equals the full error.
5. On the plan, from each of B, C, D, E draw a line parallel to the closing line A′A and mark off the corresponding correction towards A. Join the new points in order to get the adjusted closed traverse.

                                      X
                                    . |
                                .     |
                          . e'        |
                    . d'  |           |
              . c'  |     |           |
        . b'  |     |     |           |
 A -----b-----c-----d-----e-----------A'
 (base = perimeter, marks at cumulative distances;
  A'X = closing error, drawn at right angles)

The method is satisfactory only when the error is small (about 1/300 of the perimeter); a larger error means a mistake in the field, and the survey should be repeated.

In a link traverse the bearing of the last line, computed from the first bearing and the observed angles, must agree with its observed bearing. The difference is the angular misclosure.

Rule (angles to the right): $\text{FB of next line} = \text{FB of previous line} + 180^\circ + \text{angle to the right}$ (add or subtract $360^\circ$ to bring the result between $0^\circ$ and $360^\circ$).

Step 1: Bearings computed from the observed angles

Computed FB of EF = 288°06′20″; given (observed) FB of EF = 288°05′40″.

Step 2: Angular misclosure

The observed back bearing of EF is 108°05′40″, so its observed FB is 108°05′40″ + 180° = 288°05′40″. Computed FB = 288°06′20″, so the angles are not exactly correct: there is a misclosure of 40″. For a theodolite of least count $C = 20''$ (assumed, as in the companion problem) the permissible misclosure is $C\sqrt{n} = 20''\sqrt{4} = 40''$, so the error is just within the limit and the angles are adjusted equally.

Step 3: Adjustment

Total correction = −0°00′40″ (added to the computed bearing of the last line). It is distributed equally over the 4 observed angles: −0°00′40″ / 4 = −0°00′10″ per angle. The correction to a line is −0°00′10″ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 288°05′40″.

Answer: misclosure = −40″ (observations not exactly correct, but within 40″ permissible). Corrected FBs: AB 61°06′00″, BC 334°12′40″, CD 309°58′00″, DE 17°07′30″, EF 288°05′40″ (BBs are 180° different, see table).

The negative angles are right-turn angles of the same traverse written as $-204^\circ14'30'' \equiv 155^\circ45'30''$ and $-112^\circ50'20'' \equiv 247^\circ09'40''$ (add $360^\circ$), so the angles are B = 93°06′50″, C = 155°45′30″, D = 247°09′40″, E = 90°58′20″.

Rule (angles to the right): $\text{FB of next line} = \text{FB of previous line} + 180^\circ + \text{angle to the right}$ (add or subtract $360^\circ$ to bring the result between $0^\circ$ and $360^\circ$).

Step 1: Bearings computed from the observed angles

Computed FB of EF = 288°06′20″; given (observed) FB of EF = 202°17′20″.

Step 2: Angular misclosure

Permissible misclosure $= C\sqrt{n} = 20''\sqrt{4} = 40''$.

The observed BB of EF is 22°17′20″, so the observed FB of EF is 202°17′20″. Computed FB of EF is 288°06′20″, i.e. computed BB = 108°06′20″. The difference is 85°49′00″, far above 40″. The observations of the angles (or of the last bearing) are not correct. A misclosure of this size cannot be distributed among the angles; there is a blunder (a wrong reading or booking error, or a wrong bearing of the last line), and the angles must be re-observed.

If the last BB had been read as 108°05′40″ (the value of the companion problem on the same angles), the misclosure would be −40″, which equals the permissible 40″; the correction is then −10″ per angle, giving corrected FBs: AB 61°06′00″, BC 334°12′40″, CD 309°58′00″, DE 17°07′30″, EF 288°05′40″.

Answer: misclosure with the given data = 85°49′00″ ≫ 40″ (not acceptable); the computed bearings from the angles are BC 334°12′50″, CD 309°58′20″, DE 17°08′00″, EF 288°06′20″, valid only after the field error is found.

Both compasses measure the magnetic bearing of a line with a magnetic needle, but they differ in the reading system, graduation and sighting arrangement.

Main advantage of the prismatic compass: sighting and reading are done together, so there is less chance of disturbing the instrument and the survey is faster.

Whole circle bearing (WCB)

The bearing of a line measured clockwise from the north end of the meridian, $0^\circ$ to $360^\circ$. Example: a line pointing south-west may have WCB $225^\circ$. This is the system of the prismatic compass.

Quadrantal bearing (QB), or reduced bearing

The acute angle ($0^\circ$ to $90^\circ$) between the line and the nearer end (north or south) of the meridian, measured towards east or west. It is written with the quadrant letters, e.g. N 40° E, S 30° W. This is the system of the surveyor's compass.

Fore bearing (FB) and back bearing (BB)

The bearing of a line in the direction of progress of the survey is its fore bearing; the bearing of the same line in the opposite direction is its back bearing. They differ by exactly $180^\circ$:

Use $+$ if FB $< 180^\circ$ and $-$ if FB $> 180^\circ$. Example: FB of AB = $65^\circ 30'$, BB = $245^\circ 30'$.

Local attraction

The deviation of the magnetic needle from the magnetic meridian caused by magnetic substances near the station, such as iron objects, steel structures, power lines, or rocks containing iron ore. Bearings observed at an affected station are in error by the same amount.

• Detection: the FB and BB of a line differ by $180^\circ$ if neither end is affected; otherwise one or both stations are affected.
• Correction: (i) the line with FB − BB = $180^\circ$ is taken as free; bearings of the other lines are corrected station by station from it, or (ii) the included-angle method is used, because the angle at an affected station is not affected.

Because the FB and BB of every line differ from 180° by 1°30′ to 3°00′, all the stations (A, B, C, D, E) are affected by local attraction (no line is free of it). The included angle method is therefore used, since an angle at one station is not affected by local attraction at that station.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 542°30′; error = +2°30′; correction per angle = −0°30′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line AB has the smallest difference (−1°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 192°15′.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Answer: all stations A to E are affected. Corrected included angles: A = 138°00′, B = 56°00′, C = 145°15′, D = 51°45′, E = 149°00′ (sum 540°). The corrected bearings are as in the last table (FB: AB 192°15′, BC 68°15′, CD 33°30′, DE 265°15′, EA 234°15′).

Assumption: the traverse has no line with FB − BB = 180°, so the line with the smallest discrepancy (AB) is taken as the reference and its mean bearing is adopted.

For an open traverse whose last line has been observed independently, the bearing of the last line computed through the deflection angles must agree with the observed bearing. Right deflection is taken as $+$ and left as $-$.

Rule (deflection angles): $\text{FB of next line} = \text{FB of previous line} \pm \text{deflection}$, with $+$ for a right deflection and $-$ for a left deflection.

Step 1: Bearings computed from the observed angles

Computed FB of FG = 28°33′; given (observed) FB of FG = 29°33′.

Step 2: Angular misclosure

The observed back bearing of FG is 209°33′, so its observed FB is 209°33′ − 180° = 29°33′. The computed FB is 28°33′, so the observations of the deflection angles are not correct: there is a misclosure of +1°00′, which is distributed equally among the 5 angles (12′ each).

Step 3: Adjustment

Total correction = +1°00′ (added to the computed bearing of the last line). It is distributed equally over the 5 observed angles: +1°00′ / 5 = +0°12′ per angle. The correction to a line is +0°12′ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 29°33′.

Corrected deflection angles

The correction is applied to each signed deflection angle (right $+$, left $-$), so a right angle becomes larger and a left angle smaller when the correction is positive.

Answer: misclosure = +1°00′ (observations not correct). Corrected bearings: AB 40°45′, BC 67°34′, CD 1°01′, DE 22°09′, EF 56°15′, FG 29°33′.

The bearing of the last line, computed from the first bearing and the observed deflection angles, is compared with its observed bearing. Right deflection is taken as $+$ and left as $-$.

Rule (deflection angles): $\text{FB of next line} = \text{FB of previous line} \pm \text{deflection}$, with $+$ for a right deflection and $-$ for a left deflection.

Step 1: Bearings computed from the observed angles

Computed FB of GH = 64°00′; given (observed) FB of GH = 63°00′.

Step 2: Angular misclosure

The observed FB of GH is 63°00′ (given directly). The computed FB is 64°00′, so the deflection angles are not correct: misclosure = −1°00′. It is distributed equally among the 6 angles (−10′ each).

Step 3: Adjustment

Total correction = −1°00′ (added to the computed bearing of the last line). It is distributed equally over the 6 observed angles: −1°00′ / 6 = −0°10′ per angle. The correction to a line is −0°10′ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 63°00′.

Corrected deflection angles

The correction is applied to each signed deflection angle (right $+$, left $-$), so a right angle becomes larger and a left angle smaller when the correction is positive.

Answer: misclosure = −1°00′ (observations not correct). Corrected bearings: AB 81°45′, BC 107°05′, CD 69°10′, DE 114°15′, EF 58°35′, FG 343°10′, GH 63°00′.

Differences FB − BB: AB 179°30′, BC 181°00′, CD 178°00′, DE 181°30′, EA 180°00′. Line EA is free of attraction, so stations E and A are free; the other stations B, C and D are suspected.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Line EA has a difference of exactly 180°, so stations E and A are free from local attraction. The other stations are corrected one after another from these.

Step 2: Correct the stations in turn

• Line AB: FB at A = 305°00′ is correct. Correct BB should be 125°00′, observed 125°30′, so the correction at B is −0°30′.
• Line BC: FB at B = 75°30′ with correction −0°30′ gives 75°00′. Correct BB should be 255°00′, observed 254°30′, so the correction at C is +0°30′.
• Line CD: FB at C = 115°30′ with correction +0°30′ gives 116°00′. Correct BB should be 296°00′, observed 297°30′, so the correction at D is −1°30′.

Stations affected by local attraction: B, C, D.

Step 3: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Answer: Corrected bearings (FB): AB = 305°00′, BC = 75°00′, CD = 116°00′, DE = 165°00′, EA = 225°00′.

Line AB has FB − BB = 180°00′, so stations A and B are free. The differences for the other lines are BC +1°00′, CD −2°00′, DE +1°30′ and EA +1°00′, so C, D and E are suspected (and A, B are not). Correcting the stations one by one from AB and from AB backwards does not give the same values (the closing difference is 1°30′), so the included-angle method is used.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 538°30′; error = −1°30′; correction per angle = +0°18′ (equal distribution, since each angle is measured with the same care).

Line AB has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): AB = 305°30′, BC = 75°12′, CD = 115°54′, DE = 164°36′, EA = 224°18′.

A bearing is the horizontal angle between a reference direction (meridian) and a survey line, measured clockwise. Bearings are classified in three ways.

1. According to meridian

• True bearing: measured from the true (geographic) meridian, the line through the geographic North and South poles. It is fixed for a place.
• Magnetic bearing: measured from the magnetic meridian, the direction of a freely suspended magnetic needle. It is read with a compass. True bearing = magnetic bearing ± declination.
• Arbitrary bearing: measured from an arbitrary meridian, usually the direction of the first line of a small survey.
• Grid bearing: measured from the central meridian of a map grid (used in large surveys).

2. According to designation

• Whole circle bearing (WCB): angle from the north end of the meridian clockwise, $0^\circ$ to $360^\circ$ (prismatic compass).
• Quadrantal bearing (QB): acute angle ($0^\circ$–$90^\circ$) from the north or south end towards east or west, e.g. N 35° E, S 50° W (surveyor's compass).

3. According to path followed by the line

• Fore bearing (FB): bearing of a line in the direction of progress of the survey.
• Back bearing (BB): bearing of the line in the opposite direction.

If the FB is less than $180^\circ$, add $180^\circ$; if more, subtract. Example: FB of AB = $120^\circ$, BB = $300^\circ$.

A meridian is the fixed reference direction (a line through a point towards north) from which the bearings of survey lines are measured. It may be the true meridian (towards geographic north), the magnetic meridian (direction of the freely suspended magnetic needle), or an arbitrary meridian.

Magnetic declination is the horizontal angle between the true (geographic) meridian and the magnetic meridian at a place. It is east (positive) when the magnetic north lies to the east of true north and west (negative) when it lies to the west. True bearing = magnetic bearing + declination (east $+$, west $-$).

The relative misclosure (precision) ratio is the linear closing error of a traverse divided by the total length (perimeter) of the traverse, written as 1 in N:

where $\Sigma L$ and $\Sigma D$ are the sums of latitudes and departures. It measures the accuracy of the traverse; a larger N means a better survey.

Local attraction is the disturbance of the magnetic needle from its normal position (the magnetic meridian) caused by magnetic materials near the station, such as iron or steel objects, electric cables, or iron-ore rocks. It makes the bearings observed at that station wrong by a constant amount.

Whole circle bearing (WCB)

The bearing of a line measured clockwise from the north end of the meridian, from $0^\circ$ to $360^\circ$. It is read directly on the prismatic compass. The bearing is stated as a single angle, e.g. $135^\circ 30'$.

Reduced bearing (quadrantal bearing, R.B.)

The acute angle which the line makes with the nearer end (north or south) of the meridian, measured towards the east or west. It is written with the quadrant letters, e.g. N $40^\circ$ E.

            N (0°)
      NW    |    NE
  (IV)      |      (I)
 W (270°)---+---E (90°)
  (III)     |      (II)
      SW    |    SE
            S (180°)

Conversion

Example: WCB $215^\circ 30'$ lies in the third quadrant, so the reduced bearing is S $(215^\circ 30' - 180^\circ)$ W = S $35^\circ 30'$ W. Conversely, S $62^\circ$ E has WCB $= 180^\circ - 62^\circ = 118^\circ$.

In the quadrantal bearing (Q.B.) system, the interior angle at a station is the angle between the back line (the back bearing of the previous line) and the forward line (the fore bearing of the next line). Both are written as reduced bearings, e.g. N $\theta_1$ E and S $\theta_2$ W.

Rules (angle $=$ smaller angle between the two lines)

If the interior angle is the reflex angle, take $360^\circ$ minus the value found. The correct one is identified from a small sketch of the traverse; the interior angles of a closed traverse must total $(2n - 4) \times 90^\circ$.

Example: at B, back line BA = N $30^\circ$ E and forward line BC = S $50^\circ$ E. They lie on the east side of the meridian, one in the NE and the other in the SE quadrant (adjacent across the E–W line), so

(Check with WCB: BA = $30^\circ$, BC = $130^\circ$, difference $= 100^\circ$.)

Quadrantal and whole circle bearings

• Whole circle bearing (WCB): angle of a line from the north end of the meridian, clockwise, $0^\circ$–$360^\circ$ (prismatic compass).
• Quadrantal bearing (QB): acute angle from the nearest N or S end of the meridian towards E or W, written like N $35^\circ$ E (surveyor's compass).

Magnetic declination

The angle between the true meridian and the magnetic meridian at a place. East if the magnetic north is east of true north (positive), west if it is west (negative).

Isogonic line

An imaginary line on a map (or the earth) joining places having the same magnetic declination.

Agonic line

The isogonic line along which the declination is zero, that is, where the magnetic and true meridians coincide. The magnetic bearing equals the true bearing on this line. Declination is east on one side of it and west on the other.

The magnetic declination at a place is not constant; it varies with time. The variations are:

Effect in surveying: because of these variations, bearings taken at different times are not comparable unless the declination at that date is known. For re-survey of an old line, the old declination is compared with the present one and the bearing corrected. Diurnal and irregular changes are small and are neglected in ordinary compass survey, but secular change must be accounted for in work separated by many years.

Local attraction is the deflection of the magnetic needle from the magnetic meridian caused by magnetic material near the compass. It makes the observed bearings at that station wrong by a constant amount.

Sources

• Natural: iron-ore deposits, magnetite, rocks and soil rich in iron.
• Artificial: steel rails, iron pipes, steel girders and bridges, reinforcing bars, electric poles and cables, iron fences, vehicles.
• Small objects carried by the surveyor: steel chain or tape, keys, knife, iron-shod tools, mobile phone, spectacles with metal frame.

Detection

The FB and BB of each line are compared. If they differ by exactly $180^\circ$, both stations are free. If not, one or both stations are affected.

Elimination

1. Before the survey: keep metal objects away; select stations away from known sources of attraction; take bearings of the line from both ends.
2. First method (correction of stations): find a line with FB − BB = $180^\circ$; its stations are free. Starting from it, compare the BB of the next line with the correct value; the difference at the next station is the correction applied to all bearings observed at that station. Continue to the other stations.
3. Second method (included angles): the angle at a station is not affected by local attraction there, because both bearings are in error by the same amount. Compute all included angles, correct their sum to $(2n - 4) \times 90^\circ$, fix the bearing of one reliable line, and calculate the others from the corrected angles.
4. In an open traverse, use the angle method or take back-sights and resights to check.

A compass traverse is a series of connected lines whose lengths are measured with a chain or tape and whose bearings are measured with a compass. It may be closed (a loop) or open.

Reconnaissance and marking

1. Walk over the area and choose stations so that adjacent stations are intervisible, the lines are free of obstructions, and the stations are away from local attraction (power lines, iron objects).
2. Mark the stations with pegs and fix a ranging rod over each. Draw a reference sketch for each station (index sketch).

Observations (at each station)

1. Set up the compass over the station with a plumb bob (or drop a pebble) and level it.
2. Sight the ranging rod at the preceding station (back station) and read the back bearing of the previous line; note it.
3. Sight the next (forward) station and read the fore bearing.
4. Measure the length of the line by chain or tape, with offsets and details if required.
5. Move to the next station and repeat. At the last station, close back to the starting station.

Booking

Enter the station names, the lengths, and the FB and BB in a field book. Check that $\text{FB} - \text{BB} = 180^\circ$ for each line.

Checks and closure

• If FB and BB differ by $180^\circ$, no local attraction; otherwise correct as per local attraction method.
• For a closed traverse, check that interior angles sum to $(2n - 4) \times 90^\circ$ and compute the closing error.
• Plot the traverse by the included-angle or coordinate method, and adjust the closing error by Bowditch's rule.

Factors causing closing error in a compass traverse

1. Errors in measured lengths (incorrect tape length, slope not reduced, sag, wrong tension, miscounted chain lengths).
2. Errors in observed bearings (compass not centred or levelled, sticky needle, reading mistakes, parallax, wrong back-sight).
3. Local attraction at a station that has not been detected or corrected.
4. Errors in plotting (wrong scale, protractor error, thick pencil lines, shrinkage of paper).
5. Booking and calculation mistakes (wrong FB/BB recorded, wrong conversion).

Graphical adjustment of the closing error (Bowditch's rule)

The closing error is distributed so that the correction at a station is proportional to its cumulative distance from the starting point.

1. Plot the traverse A-B-C-D-E-A′; the closing error is the line A′A.
2. Draw a base line equal to the perimeter and mark the points b, c, d, e at the distances AB, AB+BC, AB+BC+CD, ... from A; the last point is A′.
3. At A′ draw a perpendicular A′X equal to the closing error (plan scale) and join AX.
4. Vertical ordinates from b, c, d, e to AX are the corrections for B, C, D, E.
5. On the plan, from each of B, C, D, E, draw a line parallel to A′A and set off the correction towards A. Join the new points to give the adjusted traverse.

                                      X
                                    . |
                                .     |
                          . e'        |
                    . d'  |           |
              . c'  |     |           |
        . b'  |     |     |           |
 A -----b-----c-----d-----e-----------A'
 (base = perimeter, marks at cumulative distances;
  A'X = closing error, drawn at right angles)

Checking an open traverse in compass survey

An open traverse has no closing check, so extra observations are taken:

1. Fore and back bearings of every line are observed; FB − BB should be $180^\circ$ (detects local attraction).
2. Cross-checks to a distant landmark: a prominent object is sighted from several stations; plotted rays must meet at one point.
3. Check lines (tie lines) are measured between non-adjacent stations and compared with the plotted distance.
4. Linking to known points: the traverse is closed on a station of known position or bearing (link traverse) and the closure is checked.
5. Repeat measurement of lengths (forward and back).

Graphical Bowditch adjustment of a closed traverse

Bowditch's rule distributes the closing error in proportion to the lengths of sides, so the correction at a station is proportional to the distance travelled from the start.

1. Plot the traverse A-B-C-D-E-A′. The closing error is A′A.
2. Draw a base line AA′ equal to the perimeter and mark b, c, d, e at the cumulative distances AB, AB+BC, ...
3. Erect A′X perpendicular at A′ equal to the closing error and join AX.
4. Ordinates from b, c, d, e to AX give the corrections of B, C, D, E.
5. On the plan, draw through each station a line parallel to A′A and mark off these corrections; join the shifted points.

                                      X
                                    . |
                                .     |
                          . e'        |
                    . d'  |           |
              . c'  |     |           |
        . b'  |     |     |           |
 A -----b-----c-----d-----e-----------A'
 (base = perimeter, marks at cumulative distances;
  A'X = closing error, drawn at right angles)

The corrected traverse closes at A.

Closing error: when a closed traverse is plotted, the last point falls at A′ instead of the starting point A; the distance A′A is the closing error. The ratio (closing error / perimeter) shows the accuracy of the survey.

Methods of plotting a compass traverse

1. Included (interior) angle method

• Calculate the interior angles from FB and BB, plot the first line to scale, and at each station set out the angle with a protractor to plot the next line. Angular errors accumulate, but the sum of angles is checked first. Suitable for a closed traverse.

2. Bearing (protractor) method

• Draw a meridian line through each station and plot each line directly at its observed bearing with a protractor or a paper protractor, laying off lengths to scale. It is simple and quick; errors do not carry from one angle to the next as much as in the angle method, but accuracy is limited by the protractor.

3. Tangent method

• The offset is calculated using the tangent of the angle (or of the bearing) and set off to scale on a line perpendicular to the meridian. More accurate than the protractor.

4. Co-ordinate (latitude and departure) method

• Compute the latitude $L\cos\theta$ and departure $L\sin\theta$ of every line, find the co-ordinates of each station, and plot the stations on a grid. It is the most accurate method, and the closing error can be adjusted by Bowditch's rule before plotting.

After plotting, the closing error A′A is adjusted by Bowditch's rule.

In a link traverse starting and ending on lines of known bearing, the bearing of the last line computed from the observed deflection angles is compared with its known bearing. Right deflections are taken as $+$ and left deflections as $-$ (as given).

Rule (deflection angles): $\text{FB of next line} = \text{FB of previous line} \pm \text{deflection}$, with $+$ for a right deflection and $-$ for a left deflection.

Step 1: Bearings computed from the observed angles

Computed FB of FG = 336°30′; given (observed) FB of FG = 335°30′.

Step 2: Angular misclosure

Permissible misclosure $= L.C.\times\sqrt{n} = 30'\sqrt{5} = 67'$ (about 1°07′), where $n = 5$ angles. The misclosure of −60′ (1°00′) is less than 67′, so the observations are accepted and the error is distributed equally.

Step 3: Adjustment

Total correction = −1°00′ (added to the computed bearing of the last line). It is distributed equally over the 5 observed angles: −1°00′ / 5 = −0°12′ per angle. The correction to a line is −0°12′ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 335°30′.

Answer: angular misclosure = −1°00′ (observed 335°30′, computed 336°30′), within permissible 1°07′. Adjusted bearings: AB 154°50′, BC 349°58′, CD 81°56′, DE 15°54′, EF 45°02′, FG 335°30′.

The bearings of the first line (PA) and the last line (QC) are known. Starting from PA, the bearing of CQ is computed through the deflection angles (right $+$). The bearing of line QC = 225°45′ is its back-direction, so the FB of CQ = 225°45′ − 180° = 45°45′.

Rule (deflection angles): $\text{FB of next line} = \text{FB of previous line} \pm \text{deflection}$, with $+$ for a right deflection and $-$ for a left deflection.

Step 1: Bearings computed from the observed angles

Computed FB of CQ = 46°30′; given (observed) FB of CQ = 45°45′.

Step 2: Angular misclosure

Permissible misclosure $= L.C. \times \sqrt{n} = 30' \times \sqrt{3} = 52'$ (n = 3 angles). The misclosure of 45′ is less than 52′, so the work is acceptable and the error is distributed equally (−15′ per angle).

Step 3: Adjustment

Total correction = −0°45′ (added to the computed bearing of the last line). It is distributed equally over the 3 observed angles: −0°45′ / 3 = −0°15′ per angle. The correction to a line is −0°15′ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 45°45′.

Answer: permissible angular misclosure = 52′ (actual −45′, acceptable). Adjusted bearings: PA 30°15′, AB 130°30′, BC 266°00′, CQ 45°45′ (QC 225°45′).

Deflection angles: $+$ = right, $-$ = left. A leg's bearing is the previous leg's bearing plus the deflection angle at the station between them.

Step 1: Bearings from the observed deflection angles

Computed bearing of M8-M9 = 35°, observed = 36°.

Step 2: Check of accuracy

Permissible misclosure $= L.C. \times \sqrt{n} = 1^\circ \times \sqrt{4} = 2^\circ$ (n = 4 deflection angles). Since $1^\circ < 2^\circ$, the work is acceptable, but the bearings of the link legs are adjusted.

Step 3: Correction

Correction per angle $= +1^\circ/4 = +0.25^\circ = +0°15′$. The correction to a leg increases by 15′ at each station.

Corrected deflection angles: M2: −69°45′ (left), A: +20°15′ (right), B: −89°45′ (left), M8: +70°15′ (right).

Answer: bearings of the link legs M2-A = 35°, A-B = 55°, B-M8 = 325°; misclosure = +1° (within permissible 2°); corrected bearings M2-A = 35°15′, A-B = 55°30′, B-M8 = 325°45′.

The bearings of PQ (180°00′) and QR (90°30′) are correct, so the angle at Q is known exactly and the other included angles are corrected to make the total $(2n-4)\times 90^\circ$.

Step 1: Angle at Q and angle sum

At Q, the back bearing of PQ is QP $= 180^\circ + 180^\circ = 0^\circ$ (i.e. 360°00′) and the bearing of QR is 90°30′, so the included angle at Q $= 90^\circ 30'$ (the traverse runs anticlockwise, with the interior on the left).

Observed sum $= 59°30′ + 90°30′ + 30°00′ + 320°30′ + 40°30′ = 541°00′$.

Required sum $= (2 \times 5 - 4) \times 90^\circ = 540°00′$.

Error $= +1°00′$ (excess). The angle at Q is already fixed by correct bearings, so the error is distributed equally over the other four angles: $-60'/4 = -15'$ each. (Least count 30′: the permissible error is $30'\sqrt{4} = 60'$, so the error is just permissible.)

New sum $= 59°15′ + 90°30′ + 29°45′ + 320°15′ + 40°15′ = 540°00′$ (check).

Step 2: Corrected bearings

For this anticlockwise traverse, the FB of the next line = BB of the previous line + included angle at the station between them.

Answer: RS = 300°15′, ST = 80°30′, TP = 300°45′ (their back bearings are 120°15′, 260°30′ and 120°45′). The check on PQ closes exactly at 180°00′.

Assumption: the included angles are interior angles of an anticlockwise traverse, the angle at Q (90°30′) being taken from the correct bearings.

Here no line has FB − BB = 180°, so every station (P, Q, R, S, T) is suspected of local attraction. The included-angle method is used because an angle at a station is not affected by local attraction there.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 539°00′; error = −1°00′; correction per angle = +0°12′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line TP has the smallest difference (+0°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 293°15′.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Answer: Corrected bearings (FB): PQ = 189°12′, QR = 120°54′, RS = 349°21′, ST = 338°03′, TP = 293°15′.

Reading of the data: the back bearing of CD is printed as 284°45′; with FB = 4°00′ this would give FB − BB = 79°15′, which is not a possible local-attraction error, so it is read as 184°45′ (the interior angles then add up to 540° within 0°30′). All other values are used as given.

Line EA has FB − BB = 180°00′, so stations E and A are free from local attraction; the included-angle method is used for the remaining lines, because the correction of stations one by one does not close (the closing error of the station corrections is 0°30′).

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 539°30′; error = −0°30′; correction per angle = +0°06′ (equal distribution, since each angle is measured with the same care).

Line EA has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): AB = 51°24′, BC = 184°03′, CD = 5°27′, DE = 165°51′, EA = 251°30′. True bearings (FB): AB = 52°54′, BC = 185°33′, CD = 6°57′, DE = 167°21′, EA = 253°00′.

Note: with the scanned values the angle at C comes out as a reflex angle of 358°30′ (lines CB and CD almost coincide). The sum of the angles is exactly 540°, so the method is applied as printed; if any value in the table was misread, the working above shows how to repeat it.

Every line differs from 180° (by 0°20′, 0°30′, 1°00′, 0°10′, 0°30′, 0°20′), so no station can be called free with certainty. The line DE has the smallest difference, so D and E are the least affected stations; the other stations are corrected using the included angles.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 720^\circ$.

Sum observed = 720°30′; error = +0°30′; correction per angle = −0°05′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line DE has the smallest difference (−0°10′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 321°15′.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Answer: Corrected bearings (FB): AB = 216°00′, BC = 135°40′, CD = 80°40′, DE = 321°15′, EF = 246°10′, FA = 298°35′.

The differences FB − BB are 183°00′, 178°30′, 177°15′, 182°00′, 181°45′, so none of the lines is free from local attraction and all stations A, B, C, D, E are suspected. The included-angle method is used.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 537°30′; error = −2°30′; correction per angle = +0°30′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line BC has the smallest difference (−1°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 192°15′.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): AB = 67°45′, BC = 192°15′, CD = 229°00′, DE = 258°15′, EA = 29°15′.

The bearings are first converted to whole circle bearings (FB of AB = S11°30′W = 191°30′, BC = 67°30′, CD = 32°15′, DE = S82°45′W = 262°45′, EA = S50°15′W = 230°15′). FB − BB = 178°30′, 181°00′, 181°45′, 182°00′ and 177°15′, so no line is free and all stations are suspected. The included-angle method is used; results are shown in quadrantal form.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 540°30′; error = +0°30′; correction per angle = −0°06′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line BC has the smallest difference (+1°00′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = N 67°00′ E.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Corrections are in the whole-circle sense (+ is clockwise).

Answer: Corrected bearings (FB): AB = S 12°36′ W, BC = N 67°00′ E, CD = N 32°39′ E, DE = S 84°48′ W, EA = S 54°12′ W.

No line is free (the differences are 178°30′, 183°00′, 181°45′, 180°30′ and 177°15′), so all stations are suspected of local attraction. Line DE has the smallest difference (0°30′), so D and E are the least affected; the included-angle method is used.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 541°00′; error = +1°00′; correction per angle = −0°12′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line DE has the smallest difference (+0°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 262°30′.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): AB = 188°36′, BC = 74°54′, CD = 30°27′, DE = 262°30′, EA = 230°18′.

Line DE has FB − BB = 180°00′, so stations D and E are free. The other lines show errors (AB −1°30′, BC +3°00′, CD +1°45′, EA −2°45′), so A, B and C are suspected. The line EA differs by −2°45′ although E is free (by DE), which shows that the bearings also contain observational errors and not only local attraction; the corrections of the stations do not agree in the two directions around the traverse, so the included-angle method is used.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 540°30′; error = +0°30′; correction per angle = −0°06′ (equal distribution, since each angle is measured with the same care).

Line DE has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): AB = 188°33′, BC = 64°57′, CD = 30°36′, DE = 262°45′, EA = 230°09′.

Reading of the table: 148°15′ is the adjusted bearing of the known line XP and 36°45′ is the adjusted bearing of the known line TY. Angles are measured to the right at P, Q, R, S, T, so the bearing of the next line = bearing of the previous line + 180° + angle.

Rule (angles to the right): $\text{FB of next line} = \text{FB of previous line} + 180^\circ + \text{angle to the right}$ (add or subtract $360^\circ$ to bring the result between $0^\circ$ and $360^\circ$).

Step 1: Bearings computed from the observed angles

Computed FB of TY = 36°00′; given (observed) FB of TY = 36°45′.

Step 2: Angular misclosure

The computed bearing of TY is 36°00′ against the known 36°45′: angular misclosure = +0°45′ (the computed bearing is 45′ less than the known one). It is distributed equally among the 5 angles (+9′ each).

Step 3: Adjustment

Total correction = +0°45′ (added to the computed bearing of the last line). It is distributed equally over the 5 observed angles: +0°45′ / 5 = +0°09′ per angle. The correction to a line is +0°09′ multiplied by the number of angles used to reach it.

Check: corrected FB of the last line equals the observed value 36°45′.

Answer: angular misclosure = +0°45′. Adjusted bearings: XP 148°15′, PQ 60°54′, QR 109°33′, RS 44°57′, ST 277°51′, TY 36°45′.

The bearings are converted to whole circle bearings (PQ: FB = S37°30′E = 142°30′, BB = N37°30′W = 322°30′; QR: 223°15′ and 44°15′; RS: 287°00′ and 107°45′; ST: 12°45′ and 193°15′; TP: 60°00′ and 239°15′). Line PQ has FB − BB = 180°00′, so P and Q are free. The other lines have differences of −1°00′ (QR), −0°45′ (RS), −0°30′ (ST) and +0°45′ (TP), so stations R, S and T are suspected (Q is free). Station corrections from the free line do not close (difference 1°30′), so the included-angle method is used.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Step 2: Included (interior) angles

The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle.

Sum of interior angles should be $(2n-4)\times 90^\circ = 540^\circ$.

Sum observed = 541°30′; error = +1°30′; correction per angle = −0°18′ (equal distribution, since each angle is measured with the same care).

Line PQ has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line.

Step 3: Corrected bearings from the corrected angles

Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise).

Step 4: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Corrections are in the whole-circle sense (+ is clockwise).

Interior angles from the corrected bearings

Sum = 540°00′ (should be 540°)

Answer: Corrected bearings (FB): PQ = S 37°30′ E, QR = S 43°33′ W, RS = N 73°24′ W, ST = N 11°54′ E, TP = N 58°57′ E.

Let T be the tree on the opposite bank and PQ the base line of 400 m on the near bank. The width of the river is the perpendicular distance of T from the line PQ.

Step 1: Bearings in whole circle form

         T (tree, far bank)
        / \
       /   \
  ----P-----Q---- near bank
        400 m

Step 2: Angles of the triangle PQT

Step 3: Sine rule

Step 4: Width of river

(Check with the single formula: $\dfrac{400 \sin 54^\circ \sin 47^\circ 45'}{\sin 78^\circ 15'} = 244.67$ m.)

Answer: width of the river = 244.67 m ≈ 244.7 m.

Line AB has FB − BB = 180°, so A and B are free. Stations C, D, E are suspected. Declination is 7° West, so true bearing = magnetic bearing − 7°.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Line AB has a difference of exactly 180°, so stations A and B are free from local attraction. The other stations are corrected one after another from these.

Step 2: Correct the stations in turn

• Line BC: FB at B = 139°30′ is correct. Correct BB should be 319°30′, observed 317°00′, so the correction at C is +2°30′.
• Line CD: FB at C = 215°15′ with correction +2°30′ gives 217°45′. Correct BB should be 37°45′, observed 36°30′, so the correction at D is +1°15′.
• Line DE: FB at D = 208°00′ with correction +1°15′ gives 209°15′. Correct BB should be 29°15′, observed 29°00′, so the correction at E is +0°15′.

Stations affected by local attraction: C, D, E.

Step 3: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Answer: Corrected bearings (FB): AB = 59°00′, BC = 139°30′, CD = 217°45′, DE = 209°15′, EA = 318°45′. True bearings (FB): AB = 52°00′, BC = 132°30′, CD = 210°45′, DE = 202°15′, EA = 311°45′.

Line AB has FB − BB = 180°00′, so stations A and B are free; C, D (and the closing line DA) are corrected from it.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Line AB has a difference of exactly 180°, so stations A and B are free from local attraction. The other stations are corrected one after another from these.

Step 2: Correct the stations in turn

• Line BC: FB at B = 48°00′ is correct. Correct BB should be 228°00′, observed 226°00′, so the correction at C is +2°00′.
• Line CD: FB at C = 290°30′ with correction +2°00′ gives 292°30′. Correct BB should be 112°30′, observed 115°15′, so the correction at D is −2°45′.

Stations affected by local attraction: C, D.

Step 3: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

Sum = 360°00′ (should be 360°)

Answer: Corrected bearings (FB): AB = 104°30′, BC = 48°00′, CD = 292°30′, DA = 177°30′.

The included angles at the stations, computed from the observed bearings (FB of the leaving line and BB of the arriving line at the same station), are A = 107°00′, B = 123°30′, C = 64°30′, D = 65°00′. Their sum is 360°00′ = $(2\times4-4)\times 90^\circ$, so the angles need no correction; they agree with the angles obtained from the corrected bearings above.

Bearings are converted to whole circle bearings: AB = S40°30′W = 220°30′ (BB 41°15′), BC = 260°45′ (BB 79°30′), CD = 19°30′ (BB 200°00′), DA = S80°00′E = 100°00′ (BB 280°00′). Line DA has FB − BB = 180°00′, so D and A are free; B and C are suspected.

Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly.

Step 1: Check each line

Line DA has a difference of exactly 180°, so stations D and A are free from local attraction. The other stations are corrected one after another from these.

Step 2: Correct the stations in turn

• Line AB: FB at A = S 40°30′ W is correct. Correct BB should be N 40°30′ E, observed N 41°15′ E, so the correction at B is −0°45′.
• Line BC: FB at B = S 80°45′ W with correction −0°45′ gives S 80°00′ W. Correct BB should be N 80°00′ E, observed N 79°30′ E, so the correction at C is +0°30′.

Stations affected by local attraction: B, C.

Step 3: Corrected bearings

Check: corrected FB − corrected BB = 180° for every line.

Corrections are in the whole-circle sense (+ is clockwise).

Interior angles from the corrected bearings