Leveling

Profile levelling (longitudinal sectioning) is levelling along a fixed line, such as the centre line of a road, canal, railway, or pipeline, to find the reduced levels of points along it at regular intervals (usually 10 m, 20 m, 30 m) and at every change of slope. The levels are plotted against chainage to get the longitudinal section (profile) of the ground.

Cross-section levelling is levelling along lines at right angles to the centre line at each chainage, to find the ground levels on both sides. Cross-sections show the ground profile across the alignment and give the width to be cleared and the volume of earthwork.

Field procedure

1. Peg the centre line at the required intervals and mark the chainages.
2. Set the level at a convenient point; take a BS on the BM and then IS on every peg along the centre line (the staff is held on the ground at each chainage and at breaks of slope); take FS on a change point when the instrument must be shifted.
3. For cross-sections, at each chainage set a line at right angles (with a cross-staff or prismatic square) and take staff readings at fixed offsets (e.g. 5 m, 10 m, 15 m) on the left and right, as IS.
4. Close the survey on another BM (or the first BM) and apply the arithmetic check and misclosure adjustment.

 Longitudinal section          Cross-section at ch. 120
 (RL vs chainage)                 left    CL    right
      ______                       \      |     /
  ___/      \___ ground              \____|____/
 -------------------- datum          15  5  0  5  15 m
 0   30  60  90 120

Uses in civil engineering

• To fix the formation level, gradient and depth of cutting/filling of a road, railway or canal.
• To compute earthwork quantities from the cross-sections (area and volume).
• To design the longitudinal gradient of sewers, water pipes and drains.
• To select the best alignment of roads and canals.
• To find the capacity of reservoirs and to plan bridge sites.

Two peg test is the field test used to find the collimation error of a level: the error caused by the line of collimation not being exactly parallel to the axis of the bubble tube when the bubble is central. The line of sight then is inclined up or down, and the staff readings contain an error proportional to the sight distance.

When permanent adjustment is required: when the test shows a collimation error larger than the permissible limit (about 10 mm in 100 m for ordinary levelling; 5 mm or less for precise work), or when the instrument has been damaged or dropped. If the error is small, it is eliminated in the field by keeping the backsight and foresight distances equal.

Field procedure

1. Choose two points A and B about 60–100 m apart on level ground. Drive pegs or mark them.
2. First setup: set up the level at the mid-point F of AB. Level it and read the staff on A ($a_1$) and B ($b_1$). Equal sights cancel the collimation error, so the true difference of level is $h = a_1 - b_1$.
3. Second setup: move the level to a point O on the line AB produced (or between A and B), about 2–5 m from one peg, say B. Level it and read the staff on B ($b_2$, near) and on A ($a_2$, far).
4. The apparent difference is $a_2 - b_2$. The error in the far staff reading $e = (a_2 - b_2) - h$.

 Setup 1:       A       F       B      equal sights
                |  <----+---->  |
 Setup 2:       A                O B    near B
                |  <------------>|-|

5. If the apparent difference equals the true difference, there is no error. If $e$ is positive, the line of collimation is inclined upward; if negative, downward.

Adjustment

• Let $d_A$ and $d_B$ be the distances of the instrument from A (far) and B (near). The error per metre of sight is $k = e/(d_A - d_B)$. The correct readings for a horizontal line of sight are $a_2' = a_2 - k\,d_A$ on A and $b_2' = b_2 - k\,d_B$ on B (check: $a_2' - b_2' = h$).
• Without moving the instrument, bring the horizontal cross-hair to the correct reading $a_2'$ on the far staff A with the vertical capstan screws of the diaphragm (top and bottom screws, loosening one and tightening the other).
• Check the bubble stays central and repeat the test until the error is within the limit.

Collimation error $= \dfrac{e}{\text{difference of distances}}$ (for example 0.004 m in 60 m).

Entering the readings. The level was set up three times (it was changed after the 4th and 7th readings), so:

• Setup 1: 1st reading 2.191 = BS on TBM1; 2nd (2.505) and 3rd (2.325) = IS; 4th (1.496) = FS on change point CP1.
• Setup 2: 5th (3.019) = BS on CP1; 6th (2.513) = IS; 7th (2.811) = FS on change point CP2.
• Setup 3: 8th (1.752) = BS on CP2; 9th (3.824) = FS on TBM2.

HI $=$ RL of point $+$ BS, and RL of a point $=$ HI $-$ staff reading.

Level field book (height of instrument method)

Arithmetic check

$\Sigma BS - \Sigma FS = 6.962 - (8.131) = -1.169$ ; last RL $-$ first RL $= 1448.701 - 1449.870 = -1.169$. Both are equal, so the reduction is correct.

Adjustment of misclosure

Computed RL of the last point = 1448.701 m; known RL = 1448.710 m.

$$\text{Closing error} = 1448.701 - 1448.710 = -0.009\ \text{m}$$

The correction at a point is $-(-0.009) \times d / 200$, where $d$ is the distance of the point from TBM1. The 7 staff points are taken at equal spacing, so the 200 m is divided into 6 equal parts of 33.33 m.

Answer: the computed RL of TBM2 is 1448.701 m against the known 1448.710 m, a closing error of −0.009 m (9 mm low). Adjusted RLs: TBM1 1449.870, point 2 1449.557, point 3 1449.739, CP1 1450.569, point 6 1451.077, CP2 1450.780, TBM2 1448.710 m.

Assumption: the distances to the intermediate points are not given, so equal spacing between successive staff points is assumed for the proportional adjustment.

Entering the readings. The level was changed after the 4th and 8th readings, so there are three set-ups:

• Set-up 1: BS 2.191 on TBM1; IS 2.505, 2.325; FS 1.496 on CP1 (4th reading).
• Set-up 2: BS 3.019 on CP1; IS 2.513 and $-1.685$; FS 2.811 on CP2 (8th reading).
• Set-up 3: BS 1.752 on CP2; FS 3.824 on TBM2.

The reading $-1.685$ m is taken with the staff held inverted (for example against a ceiling or beam): the point is 1.685 m above the line of sight, so it is entered with a negative sign and the usual rule RL $=$ HI $-$ reading still applies. RL of TBM1 $= 1449.870$ m; RL of TBM2 (known) $= 1448.710$ m.

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 6.962 - (8.131) = -1.169$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 5.713 - 6.882 = -1.169$ ; last RL $-$ first RL $= 1448.701 - 1449.870 = -1.169$

All three are equal, so the reduction is correct.

Adjustment of misclosure

Computed RL of the last point = 1448.701 m; known RL = 1448.710 m.

$$\text{Closing error} = 1448.701 - 1448.710 = -0.009\ \text{m}$$

The correction at a point is $-(-0.009) \times d / 7$, where $d$ is the distance from TBM1. The distances between the points are not given, so the 8 staff points are taken at equal spacing (the 'distance' column then counts the intervals, 0 to 7).

Answer: the computed RL of TBM2 is 1448.701 m against the known 1448.710 m, so the closing error is -0.009 m. Adjusted RLs: TBM1 1449.870, point 2 1449.557, point 3 1449.739, CP1 1450.569, point 6 1451.076, point 7 1455.275, CP2 1450.781, TBM2 1448.710 m.

Reciprocal levelling is the method of finding the difference of level between two points, such as the banks of a wide river or a valley, where it is not possible to set the instrument at the mid-point, by taking two sets of readings, one with the instrument close to each point.

Procedure

1. Set up the level near A (on one bank), and take staff readings on A (near, $a_1$) and B (far, $b_1$).
2. Shift the level close to B (other bank) and take readings on B (near, $b_2$) and A (far, $a_2$).
3. Keep the instrument and staff in the same state and the time between the two sets short, so the atmospheric conditions are the same.

        Set 1: level near A        Set 2: level near B
   A  [L]  ---------> B          A <--------- [L]  B
   a1 (near)   b1 (far)          a2 (far)    b2 (near)

Significance

• A single setting with unequal sight lengths contains the errors due to collimation (line of collimation not parallel to axis of bubble), curvature of the earth, and atmospheric refraction. Each is proportional to the distance.
• In reciprocal levelling the far sight in each set is the same length, so the errors have the same magnitude and opposite effect on the two differences; they cancel in the mean:

$$H = \frac{(a_1 - b_1) + (a_2 - b_2)}{2}$$

• It gives the correct difference of level without balancing sights and without measuring the width of the river.

Check on precision

The two apparent differences should agree. The difference between them, $(a_1 - b_1) - (a_2 - b_2)$, equals twice the combined error; if it is within the permissible error of the work (a few millimetres for a short crossing; the allowable closing error $\pm 24\sqrt{K}$ mm with K in km for ordinary levelling is a guide), the work is acceptable. If it is large, the observations are repeated, preferably at a time of steady atmosphere (early morning or late afternoon).

When and why used: reciprocal levelling is used when the difference of level between two points that are far apart and separated by an obstacle (a wide river, lake, valley) has to be found and the instrument cannot be set midway so that the backsight and foresight are equal. A single setting would include the errors of collimation, curvature and refraction, which are proportional to the sight distance. By taking readings from both banks, these errors cancel.

Derivation: let A and B be the two points, level 1 near A and level 2 near B. Let $e$ be the combined error in the far staff reading (error due to collimation + curvature $-$ refraction), which has the same value and the same sign for both far sights because the distance is the same. The error in the near sights is negligible.

Level near A: staff readings $a_1$ (near, on A) and $b_1$ (far, on B). Level near B: $a_2$ (far, on A) and $b_2$ (near, on B).

First set:

$$H_1 = a_1 - (b_1 - e) = (a_1 - b_1) + e$$

Second set:

$$H_2 = (a_2 - e) - b_2 = (a_2 - b_2) - e$$

The true difference H is the same in both, so $H_1 = H_2$:

$$(a_1 - b_1) + e = (a_2 - b_2) - e \;\Rightarrow\; e = \frac{(a_2 - b_2) - (a_1 - b_1)}{2}$$

Adding $H_1$ and $H_2$ gives:

$$H = \frac{(a_1 - b_1) + (a_2 - b_2)}{2}$$

So the true difference of level is the mean of the two apparent differences. $H$ is the rise of B above A (BS $-$ FS is a rise), so RL of B $=$ RL of A $+ H$; a negative $H$ means B is lower than A.

Best method: the instrument is set near each bank in turn and readings are taken on the two staffs with the same instrument, the same staff, and the same observer, in as short a time as possible, and the mean taken of several sets. The errors removed are:

1. Error due to collimation (line of sight not horizontal),
2. Error due to curvature of the earth, and
3. Error due to refraction of the atmosphere, assuming that it is the same for the two sets.

Reduction of levels is the calculation of the reduced level (RL) of each point from the staff readings entered in the level book. Two methods are used.

1. Height of instrument (HI) or height of collimation method

• RL of the instrument line = RL of the BM + BS.
• RL of any point = HI $-$ staff reading (IS or FS) on it.
• At a change point, a new HI = RL of the CP + new BS.

$$\text{HI} = \text{RL} + \text{BS}, \qquad \text{RL} = \text{HI} - \text{IS or FS}$$

Arithmetic check: $\Sigma BS - \Sigma FS = \text{last RL} - \text{first RL}$.

It is quicker, suitable when many intermediate sights are taken (profile and cross-section levelling), but intermediate sights are not checked.

2. Rise and fall method

• The difference between consecutive staff readings gives a rise or a fall.
• If the second reading is less than the first, it is a rise: rise $=$ previous reading $-$ present reading, and RL $=$ previous RL $+$ rise.
• If it is greater, it is a fall, and RL $=$ previous RL $-$ fall.

Arithmetic check:

$$\Sigma BS - \Sigma FS = \Sigma \text{Rise} - \Sigma \text{Fall} = \text{last RL} - \text{first RL}$$

It checks every reading (including the intermediate sights), but takes more time. It is used for fly levelling and for accuracy.

Example (HI): BM RL 100.000, BS 1.500 gives HI 101.500; FS 0.800 on point P gives RL 100.700.

Arranging the readings. The staff is 4 m long, so the instrument must be shifted when the reading nears 4 m: after 3.825 and after 3.455. The first reading ($-0.855$ m) is negative: the staff was held inverted (for example under a beam), so the point A is above the line of sight. The set-ups are:

• Set-up 1: BS $-0.855$ on A; IS 1.545, 2.353, 3.115; FS 3.825 on CP1.
• Set-up 2: BS 0.455 on CP1; IS 1.380, 2.055, 2.855; FS 3.455 on CP2.
• Set-up 3: BS 0.585 on CP2; IS 1.015, 1.850, 2.755; FS 3.945 on B.

RL of A $= 380.500$ m. The 15 readings are on 13 points (the FS and BS at a change point are on one point), 30 m apart.

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 0.185 - (11.225) = -11.040$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.000 - 11.040 = -11.040$ ; last RL $-$ first RL $= 369.460 - 380.500 = -11.040$

All three are equal, so the reduction is correct.

Gradient of the line AB

There are 13 points, so 12 intervals of 30 m: $AB = 12 \times 30 = 360$ m.

$$\text{Fall from A to B} = 380.500 - 369.460 = 11.040\ \text{m}$$ $$\text{Gradient} = \frac{11.040}{360} = \frac{1}{32.6}\ (\text{falling from A to B})$$

Answer: RL of B = 369.460 m; the line AB falls at a gradient of 1 in 32.6 (about 1 in 33).

Assumptions: 12 readings at 15 m interval. The instrument was moved whenever the reading nearly reached the limit of the 4 m staff and the next reading drops sharply. The shifts are therefore taken after the readings 2.805, 3.255 and 3.895 (the change points), and 1.685 is the last back sight. Set-ups: (0.895, 1.305, 2.805), (0.965, 2.695, 3.255), (1.120, 2.825, 3.450, 3.895), (1.685, 2.050). The last reading 2.050 is on A, whose RL is 1250.250 m, so levels are worked from A backwards (rise/fall is first found with an arbitrary RL and then shifted so that A = 1250.250).

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 4.665 - 12.005 = -7.340 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 7.340 = -7.340 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1250.250 - 1257.590 = -7.340 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Setting out B and C from the last instrument position

The last set-up has HI = RL of A + FS on A = 1250.250 + 2.050 = 1252.300 m (this is also the HI of the last set-up from the row with BS 1.685: 1252.300 m, which agrees).

Staff reading required = HI - required RL.

$$\begin{aligned} \text{Reading at B} &= 1252.300 - 1250.800 = 1.500 \text{ m}\\ \text{Reading at C} &= 1252.300 - 1251.250 = 1.050 \text{ m} \end{aligned}$$

Both readings are within the 4 m staff length, so B and C can be set out without shifting the instrument. A peg is driven at B and C until the staff reads these values.

Answer: staff reading at B = 1.500 m, at C = 1.050 m. RL of the first point = 1257.590 m.

Assumptions: 12 readings at 15 m interval. The instrument was moved whenever the reading nearly reached the limit of the 4 m staff and the next reading drops sharply. The shifts are therefore taken after the readings 2.805, 3.255 and 3.895 (the change points), and 1.685 is the last back sight. Set-ups: (0.895, 1.305, 2.805), (0.965, 2.695, 3.255), (1.120, 2.825, 3.450, 3.895), (1.685, 2.050). The last reading 2.050 is on A, whose RL is 1250.250 m, so levels are worked from A backwards (rise/fall is first found with an arbitrary RL and then shifted so that A = 1250.250).

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 4.665 - 12.005 = -7.340 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 7.340 = -7.340 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1250.250 - 1257.590 = -7.340 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Setting out B and C from the last instrument position

The last set-up has HI = RL of A + FS on A = 1250.250 + 2.050 = 1252.300 m (this is also the HI of the last set-up from the row with BS 1.685: 1252.300 m, which agrees).

Staff reading required = HI - required RL.

$$\begin{aligned} \text{Reading at B} &= 1252.300 - 1250.800 = 1.500 \text{ m}\\ \text{Reading at C} &= 1252.300 - 1251.000 = 1.300 \text{ m} \end{aligned}$$

Both readings are within the 4 m staff length, so B and C can be set out without shifting the instrument. A peg is driven at B and C until the staff reads these values.

Answer: staff reading at B = 1.500 m, at C = 1.300 m. RL of the first point = 1257.590 m.

Assumptions: 12 readings at 15 m interval. The instrument was moved whenever the reading nearly reached the limit of the 4 m staff and the next reading drops sharply. The shifts are therefore taken after the readings 2.800, 3.255 and 3.895 (the change points), and 1.685 is the last back sight. Set-ups: (0.895, 1.305, 2.800), (1.960, 2.690, 3.255), (2.120, 2.825, 3.450, 3.895), (1.685, 2.050). The last reading 2.050 is on A, whose RL is 50.250 m, so levels are worked from A backwards (rise/fall is first found with an arbitrary RL and then shifted so that A = 50.250).

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 6.660 - 12.000 = -5.340 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 5.340 = -5.340 \text{ m}\\ \text{Last RL} - \text{First RL} &= 50.250 - 55.590 = -5.340 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Setting out B and C from the last instrument position

The last set-up has HI = RL of A + FS on A = 50.250 + 2.050 = 52.300 m (this is also the HI of the last set-up from the row with BS 1.685: 52.300 m, which agrees).

Staff reading required = HI - required RL.

$$\begin{aligned} \text{Reading at B} &= 52.300 - 50.800 = 1.500 \text{ m}\\ \text{Reading at C} &= 52.300 - 51.000 = 1.300 \text{ m} \end{aligned}$$

Both readings are within the 4 m staff length, so B and C can be set out without shifting the instrument. A peg is driven at B and C until the staff reads these values.

Answer: staff reading at B = 1.500 m, at C = 1.300 m. RL of the first point = 55.590 m.

Assumptions: the staff positions are 30 m apart; the readings 3.825 and 3.455 are foresights on change points, and the next readings (0.455 and 0.585) are backsights on the same points after the instrument is shifted. This gives 13 points, with A to B = 12 x 30 = 360 m. RL of B is known, so the levels are worked back from B.

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 1.895 - 11.125 = -9.230 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 9.230 = -9.230 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1380.500 - 1389.730 = -9.230 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Gradient of the line AB

RL of A = 1389.730 m, RL of B = 1380.500 m, difference = 9.230 m (A is higher).

$$\text{Gradient} = \frac{9.230}{360} = \frac{1}{39.00}$$

Answer: RL of A = 1389.730 m; the line AB falls from A to B at 1 in 39.0 (2.56 %).

Assumptions: the staff positions are 30 m apart; the readings 3.825 and 3.455 are foresights on change points, and the next readings (0.455 and 0.585) are backsights on the same points after the instrument is shifted. This gives 13 points, with A to B = 12 x 30 = 360 m. RL of A = 380.500 m is known. The third reading is printed as 2.353 and has been used as printed (in a similar paper it is 2.335).

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 1.895 - 11.125 = -9.230 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 9.230 = -9.230 \text{ m}\\ \text{Last RL} - \text{First RL} &= 371.270 - 380.500 = -9.230 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Gradient of the line AB

RL of A = 380.500 m, RL of B = 371.270 m, difference = 9.230 m (falling from A to B).

$$\text{Gradient} = \frac{9.230}{360} = \frac{1}{39.00}$$

Answer: RL of B = 371.270 m; the line AB falls at 1 in 39.0 (2.56 %).

Principle methods of levelling

1. Direct (spirit) levelling: a level and levelling staff are used; the difference in level is found from staff readings with a horizontal line of sight. It is the most accurate and most used. Types: simple, differential, fly, check, profile, cross-section, reciprocal.
2. Trigonometric levelling: the difference of elevation is found from the vertical angle (theodolite) and the distance: $h = D \tan\alpha$. It is used in hilly ground and for inaccessible points.
3. Barometric levelling: based on the variation of atmospheric pressure with altitude. Used for reconnaissance.
4. Hypsometric (boiling point) levelling: based on the variation of the boiling point of water with altitude.

Errors eliminated by balancing of sights

Balancing of sights means keeping the backsight and foresight distances equal from each instrument position. The following errors are eliminated:

• Error due to collimation (line of collimation not parallel to the axis of the bubble tube).
• Error due to curvature of the earth.
• Error due to atmospheric refraction.
• Error due to non-adjustment of the bubble tube (the same effect as collimation error).

These errors are proportional to the sight distance, so equal sights give equal errors that cancel in the difference of readings.

Example: an instrument has a collimation error of 0.002 m per 20 m. If the BS and FS are both 40 m, each reading is wrong by 0.004 m and the difference is correct. If BS = 20 m and FS = 60 m, the difference is wrong by $0.002 \times (60 - 20)/20 = 0.004$ m.

Basic principles of levelling

1. A horizontal line of sight is established with a level (the line of collimation is made horizontal by centring the bubble).
2. The vertical distance from this line of sight to the points whose elevations are required is measured with a graduated levelling staff held vertically on them.
3. The difference of staff readings gives the difference of level between two points: the point with the smaller reading is higher.
4. If the RL of one point is known, the RL of the other is: RL = known RL $\pm$ difference of level, usually found through the height of instrument or the rise and fall method.
5. When the points are far apart or at different heights, the work is carried out in steps using change points; sights are balanced to remove the errors of collimation, curvature and refraction.

Reciprocal levelling

Used when two points A and B are on opposite sides of a river or valley and the level cannot be set at the middle.

   Set 1: level near A            Set 2: level near B
   A [L] ------------> B          A <------------ [L] B
   a1(near)      b1(far)          a2(far)      b2(near)

1. Set the level near A; read the staff at A ($a_1$) and at B ($b_1$).
2. Move the level near B; read the staff at A ($a_2$) and B ($b_2$).
3. The true difference of level is the mean of the two apparent differences:

$$H = \frac{(a_1 - b_1) + (a_2 - b_2)}{2}$$

The error in the far reading (collimation, curvature, refraction) is

$$e = \frac{(a_2 - b_2) - (a_1 - b_1)}{2}$$

The mean eliminates these errors because the far sight is of the same length in each set. The two sets should be taken at about the same time under the same atmospheric conditions, and the mean of several sets is used for precise work.

Reading the data. The staff is 4 m long, so the readings 3.938 and 3.844 (nearly 4 m) show where the instrument was shifted. The sequence is therefore:

• Setup 1: BS 0.585 on A; IS 0.936, 1.953, 2.846, 3.644; FS 3.938 on change point CP1.
• Setup 2: BS 0.962 on CP1; IS 1.035, 1.689, 2.534; FS 3.844 on change point CP2.
• Setup 3: BS 0.956 on CP2; IS 1.579; FS 3.016 on B.

The ground slopes continuously down (readings increase). The RL of B is known, so the RL of A is found first:

$$\text{RL of A} = \text{RL of B} + (\Sigma FS - \Sigma BS) = 1120.450 + (10.798 - 2.503) = 1128.745\ \text{m}$$

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 2.503 - (10.798) = -8.295$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.000 - 8.295 = -8.295$ ; last RL $-$ first RL $= 1120.450 - 1128.745 = -8.295$

All three are equal, so the reduction is correct.

The last RL computed from A equals the given RL of B (1120.450 m), which confirms the work.

Gradient between the first and last point

The FS and BS at a change point are taken on the same point, so the 14 readings fall on 12 points at 30 m spacing: 11 intervals, distance $= 11 \times 30 = 330$ m.

$$\text{Fall} = 1128.745 - 1120.450 = 8.295\ \text{m}, \qquad \text{Gradient} = \frac{8.295}{330} = \frac{1}{39.8}$$

Answer: RL of A = 1128.745 m; falling gradient from A to B of about 1 in 40 (1 in 39.8).

Note: if each of the 14 readings were counted as a separate 30 m station (13 intervals, 390 m), the gradient would be 1 in 47.0.

a) and b)

Notation: station 1 is near A and station 2 is near B. At station 1: staff on A (near) $a_1 = 1.486$, staff on B (far) $b_1 = 1.726$. At station 2: staff on A (far) $a_2 = 1.191$, staff on B (near) $b_2 = 1.416$. The rise of B above A is BS $-$ FS.

Apparent differences of level and the true difference

$$H_1 = a_1 - b_1 = 1.486 - 1.726 = -0.240\ \text{m}, \qquad H_2 = a_2 - b_2 = 1.191 - 1.416 = -0.225\ \text{m}$$ $$H = \frac{H_1 + H_2}{2} = -0.2325\ \text{m}\ (\text{B is lower than A by } 0.2325\ \text{m})$$

Reduced level of A

$$\text{RL of A} = \text{RL of B} - H = 1160.18 + 0.2325 = 1160.4125\ \text{m} \approx 1160.413\ \text{m}$$

Total error in the far reading

$$e = \frac{H_2 - H_1}{2} = \frac{-0.225 - (-0.240)}{2} = +0.0075\ \text{m}$$

Combined curvature and refraction correction (AB = 300 m, R = 6370 km)

$$\frac{d^2}{2R} = \frac{0.3^2}{2 \times 6370} = 0.00706\ \text{m}, \qquad \text{refraction} = \frac{1}{7} \times 0.00706 = 0.00101\ \text{m}$$ $$\text{Combined correction} = 0.00706 - 0.00101 = 0.00606\ \text{m} \approx 0.0061\ \text{m}$$

Collimation error and precision ratio

The error in the far reading is made up of the collimation error and the combined curvature and refraction error, so

$$\begin{aligned} \text{Collimation error} = e - \text{(C \& R)} = 0.0075 - 0.00606 = 0.00144\ \text{m in 300 m} \end{aligned}$$ $$\text{Precision ratio} = \frac{0.00144}{300} = \frac{1}{207637}$$

The error is positive, so the far readings are too large (line of collimation inclined upward). The total error in the far sight, including curvature and refraction, is 0.0075 m in 300 m, or 1 in 40,000.

Answer: RL of A = 1160.413 m; combined curvature and refraction correction = 0.0061 m; collimation error = 1.44 mm in 300 m (about 0.48 mm per 100 m), precision ratio about 1 in 208,000.

Notation: station 1 is near A and station 2 is near B. At station 1: staff on A (near) $a_1 = 1.486$, staff on B (far) $b_1 = 1.726$. At station 2: staff on A (far) $a_2 = 1.191$, staff on B (near) $b_2 = 1.416$. The rise of B above A is BS $-$ FS.

Apparent differences of level and the true difference

$$H_1 = a_1 - b_1 = 1.486 - 1.726 = -0.240\ \text{m}, \qquad H_2 = a_2 - b_2 = 1.191 - 1.416 = -0.225\ \text{m}$$ $$H = \frac{H_1 + H_2}{2} = -0.2325\ \text{m}\ (\text{B is lower than A by } 0.2325\ \text{m})$$

Reduced level of A

$$\text{RL of A} = \text{RL of B} - H = 1260.18 + 0.2325 = 1260.4125\ \text{m} \approx 1260.413\ \text{m}$$

Total error in the far reading

$$e = \frac{H_2 - H_1}{2} = \frac{-0.225 - (-0.240)}{2} = +0.0075\ \text{m}$$

Combined curvature and refraction correction (AB = 300 m, R = 6370 km)

$$\frac{d^2}{2R} = \frac{0.3^2}{2 \times 6370} = 0.00706\ \text{m}, \qquad \text{refraction} = \frac{1}{7} \times 0.00706 = 0.00101\ \text{m}$$ $$\text{Combined correction} = 0.00706 - 0.00101 = 0.00606\ \text{m} \approx 0.0061\ \text{m}$$

Collimation error and precision ratio

The error in the far reading is made up of the collimation error and the combined curvature and refraction error, so

$$\begin{aligned} \text{Collimation error} = e - \text{(C \& R)} = 0.0075 - 0.00606 = 0.00144\ \text{m in 300 m} \end{aligned}$$ $$\text{Precision ratio} = \frac{0.00144}{300} = \frac{1}{207637}$$

The error is positive, so the far readings are too large (line of collimation inclined upward). The total error in the far sight, including curvature and refraction, is 0.0075 m in 300 m, or 1 in 40,000.

Answer: RL of A = 1260.413 m; combined curvature and refraction correction = 0.0061 m; collimation error = 1.44 mm in 300 m (about 0.48 mm per 100 m), precision ratio about 1 in 208,000.

Reading of the data. The first reading is printed as 0.680 [?]; it is read as 0.680 m. The printed "(B)" stands against the change point where the 10th reading (FS 3.540) and the 11th reading (BS 0.835) are taken, so the point B (RL 1180.750 m) is this change point (CP2); the RLs of all the other points are found from it. If the 1180.750 m belonged to the first point instead, every RL below would change by the same constant and the gradient would be unchanged.

The staff is 4 m, so the instrument was shifted after the readings 3.380 and 3.540 (6th and 10th readings). The set-ups are:

• Set-up 1: BS 0.680; IS 1.455, 1.855, 2.880, 2.800; FS 3.380 on CP1.
• Set-up 2: BS 1.055 on CP1; IS 1.860, 2.265; FS 3.540 on CP2 (point B).
• Set-up 3: BS 0.835 on B; IS 0.945, 1.530; FS 2.445 on the last point.

The 14 readings are on 12 points, 30 m apart. Taking B (9th point) as RL 1180.750 m, the HI of set-up 3 is $1180.750 + 0.835 = 1181.585$ m, and the other RLs follow by working back through the rises and falls.

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 2.570 - (9.365) = -6.795$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.080 - 6.875 = -6.795$ ; last RL $-$ first RL $= 1179.140 - 1185.935 = -6.795$

All three are equal, so the reduction is correct.

Gradient of the line joining the first and last points

Distance $= 11 \times 30 = 330$ m.

$$\text{Fall} = 1185.935 - 1179.140 = 6.795\ \text{m}, \qquad \text{Gradient} = \frac{6.795}{330} = \frac{1}{48.6}\ (\text{falling})$$

Answer: RLs of the points 1 to 12 = 1185.935, 1185.160, 1184.760, 1183.735, 1183.815, 1183.235, 1182.430, 1182.025, 1180.750, 1180.640, 1180.055, 1179.140 m. The line from the first to the last point falls at a gradient of 1 in 48.6 (about 1 in 49).

Assumptions: the readings 4.685 (B) and 4.485 (C) are foresights on change points, and the next readings 0.625 and 0.975 are backsights on the same points after the instrument is shifted. Chainage is taken from A at 20 m per staff position, so A to B = 100 m, B to C = 60 m, C to D = 80 m (total 240 m).

B is the 6th reading, so its RL is fixed first (1200.800 m) and the others are found upwards and downwards from it.

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 1.985 - 12.760 = -10.775 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 10.775 = -10.775 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1194.325 - 1205.100 = -10.775 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Gradient of the line AD (first to last point)

RL of A = 1205.100 m, RL of D = 1194.325 m. Difference in level = 10.775 m (falling). Horizontal distance = 240 m.

$$\text{Gradient} = \frac{10.775}{240} = \frac{1}{22.27}$$

Answer: RLs: A 1205.100, B 1200.800, C 1196.940, D 1194.325 m. The line AD falls at about 1 in 22.3 from A to D (4.49 %).

Principle methods of levelling

1. Direct (spirit) levelling: staff readings with a level.
2. Trigonometric levelling: from the vertical angle and the distance.
3. Barometric levelling: from the atmospheric pressure.
4. Hypsometric (boiling point) levelling: from the boiling point of water.
5. Stadia levelling (tacheometric) and GPS/total station levelling are also used in modern work.

Direct (spirit) levelling

The differences in level are found with a level and a vertical staff. A horizontal line of sight is obtained from the level.

1. Set the level at a convenient point, level it with the foot screws and bubble.
2. The BS is read on a BM of known RL; the HI is found: $HI = RL + BS$.
3. The staff is held at the other points (IS), and finally the FS is read at a change point; $RL = HI - \text{staff reading}$.
4. The instrument is moved ahead; the staff stays at the CP and a new BS is taken. The process continues up to the last point.
5. The readings are booked in a level field book and checked: $\Sigma BS - \Sigma FS = \text{last RL} - \text{first RL}$.

It is the most accurate method and is used for fixing benchmarks, roads, canals, and building sites.

Profile levelling (longitudinal sectioning) is levelling along a given line, such as the centre line of a road, railway, canal or pipeline, to determine the elevations of the ground at regular intervals and at changes of slope, in order to draw the longitudinal section of the ground along the line.

Procedure

1. Peg the centre line at equal intervals (10, 20 or 30 m) and note the chainage of each peg and of the breaks of slope.
2. Set up the level on firm ground within about 100 m of a BM and take a BS on the BM.
3. Hold the staff on each peg along the line within reach and take an intermediate sight (IS); take a FS at a change point when the level has to be shifted; take a BS on that point from the new position.
4. Close the levelling on a BM (or the starting BM) to check the work.
5. Reduce the levels (HI or rise and fall method), apply the arithmetic check, and distribute any closing error.
6. Plot the RLs against the chainage (the vertical scale larger than the horizontal) to get the profile.

 RL
  |   ____
  |__/    \__     ___  <- ground profile
  |          \___/
  +------------------- chainage

Balancing of sight

Balancing of sights means keeping the backsight distance equal to the foresight distance from each instrument position.

• The instrument is placed midway between the BS and FS points as far as possible.
• Errors due to the line of collimation not being horizontal, curvature of the earth and atmospheric refraction are proportional to distance. For equal sights these errors are equal in the BS and FS readings and cancel when the difference is taken.
• When equal distances for each setup cannot be kept (for example on steep ground), the sum of the BS distances is kept equal to the sum of the FS distances.

It is also followed in fly levelling and differential levelling generally.

Fly levelling is levelling in which only BS and FS readings are taken, to find the elevation of a distant BM or to check levels previously established. The points to be considered are:

1. Balance the sights: keep the BS and FS distances approximately equal to remove the errors of collimation, curvature and refraction.
2. Sight length: limit the length of sight to 60–100 m (usually not more than 100 m) and avoid very short sights; it should be such that the staff can be read with confidence.
3. Change points: choose firm, well-defined points (such as pegs with a nail or a stone) and keep the staff on the same spot between the FS and the BS; the staff should be held vertical (use a staff bubble).
4. Starting and closing: start from a BM of known RL and close on another BM, or return to the starting BM (a loop), to check the work.
5. Readings should be taken to the nearest millimetre, entered at once in the field book and checked: $\Sigma BS - \Sigma FS =$ last RL $-$ first RL.
6. Instrument: it must be well adjusted (two-peg test); the bubble should be centred before each reading; avoid sighting at the bottom (< 0.5 m) of the staff because of refraction, and avoid reading in strong wind or heat shimmer.
7. Closing error must be within the permissible limit, $\pm C\sqrt{K}$ mm (K in km), for example $\pm 24\sqrt{K}$ for ordinary work and $\pm 12\sqrt{K}$ for more accurate work; if exceeded, repeat the levelling.
8. The work should be done quickly, in one direction, with the same instrument and observer, and in the same weather conditions.

Errors in levelling are classified as instrumental, natural, and personal errors.

1. Instrumental errors

• Collimation error: the line of collimation is not parallel to the axis of the bubble tube; removed by balancing sights or by permanent adjustment (two-peg test).
• Bubble tube not adjusted, sluggish bubble.
• Staff not correctly graduated; wear and tear of the staff foot, bent staff, or loose joint (telescopic staff).
• Parallax due to improper focusing.

2. Natural errors

• Curvature of the earth: makes the staff readings too high, $C_c = 0.0785\,d^2$ (d in km, correction in m).
• Refraction: bends the line of sight downward, so readings are too low, $C_r = \dfrac{1}{7}C_c$.
• Temperature changes causing unequal expansion of the instrument and heat shimmer; wind causing vibration; settlement of the tripod or of the staff in soft ground.

3. Personal errors

• Bubble not centred at the time of reading.
• Staff not held vertical; wrong reading (e.g. reading 1.234 as 1.324) or inverted figures; wrong booking.
• Staff not held on the same point at the change point.
• Mistakes in calculation (wrong addition or subtraction, wrong sign).
• Wrong focusing or sighting in a hurry.

Minimising the errors: balance the sights, test and adjust the level, keep the sights short (about 60–100 m), hold the staff vertical, avoid the lowest part of the staff, and apply the arithmetic checks.

Closing error (misclosure) in a level circuit is the difference between the computed RL of the closing benchmark (or of the starting BM, if the levelling returns to it) and its known RL. It results from the accumulation of instrumental, natural and personal errors.

$$\text{Closing error} = \text{computed RL of last point} - \text{known RL of last point}$$

Permissible closing error: depends on the class of work and the project specification, for example about $\pm 24\sqrt{K}$ mm for ordinary levelling and $\pm 12\sqrt{K}$ mm for more accurate levelling (K = length of the circuit in km). If it is exceeded, the levelling must be repeated.

Adjustment

When the error is within the limit, it is distributed to the levels of the intermediate points. The correction is proportional to the distance (or to the number of instrument setups) from the starting point:

$$\text{Correction at a point} = -\,e \times \frac{\text{distance of the point from the start}}{\text{total length of the circuit}}$$

In terms of setups: correction at the $n$-th setup $= -\,e \times n/N$ where N is the total number of setups. The correction is applied to the CPs and BMs, and to the intermediate points of the same setup as the CP before them.

Example: computed RL of the end BM $= 100.036$ m, known RL $= 100.000$ m, 4 setups. Error $= +0.036$ m. The corrections at the CPs after the 1st, 2nd, 3rd and 4th setups are $-0.009$, $-0.018$, $-0.027$ and $-0.036$ m.

Effect of curvature

The earth is a sphere, so a level surface curves away from the horizontal line of sight. A horizontal line of sight at the instrument departs from the level surface through the instrument, so the staff reading at a distance $d$ is greater than the true reading by the curvature correction.

Let O be the centre of the earth, R its radius, A the instrument position on the level surface, and AC the horizontal line of sight (tangent at A), of length $d$ to the staff. The staff point B lies on the level surface vertically below C, and $BC = h_c$ is the curvature error. In the triangle OAC, right-angled at A, $OA = R$ and $OC = R + h_c$:

$$(R + h_c)^2 = R^2 + d^2 \;\Rightarrow\; 2Rh_c + h_c^2 = d^2$$

$h_c^2$ is negligible, so the curvature correction is

$$h_c = \frac{d^2}{2R}$$

With $R = 6370$ km and d in km, $h_c = 0.0785\,d^2$ metres. This correction is subtracted from the staff reading.

Effect of refraction

Light rays from the staff pass through layers of air of different density and bend towards the earth, so the staff appears higher and the reading is less than the true one. The path is nearly a circular arc with a radius of about 7 R, so

$$h_r = \frac{d^2}{2 \times 7R} = \frac{1}{7}\,h_c = 0.0112\,d^2\ \text{m}$$

This correction is added to the reading.

Combined correction

$$h = h_c - h_r = \frac{d^2}{2R} - \frac{d^2}{14R} = \frac{6}{7}\frac{d^2}{2R} = \frac{3d^2}{7R} = 0.0673\,d^2\ \text{m}$$

(d in km). Thus the staff reading is made too high by $0.0673\,d^2$ metres; the combined correction is subtracted. For $d = 100$ m, $h = 0.0673 \times 0.01 = 0.0007$ m, which is negligible in ordinary levelling, and is eliminated by equal BS and FS distances.

Personal errors are mistakes or errors caused by the surveyor or the staff holder, not by the instrument or the weather.

• Bubble not centred exactly at the time of the reading, or the instrument not levelled properly.
• Staff not held vertical (leaning forward or backward), so the reading is too high; it can be avoided by waving the staff and taking the least reading or by using a staff bubble.
• Parallax from improper focusing of the eyepiece and the objective.
• Wrong reading of the staff (e.g. reading a wrong metre/decimetre figure, or reading an inverted staff), and wrong booking in the field book.
• Staff not held on the same point at the change point when the level is shifted.
• Moving the staff or the tripod during observations.
• Calculation mistakes while reducing the levels.

They are avoided by careful observation, checking the bubble before each reading, and using the arithmetic checks.

Given: staffs at C and D, 100 m apart. Setup 1: instrument at the mid-point F, readings C = 2.000 m, D = 3.000 m. Setup 2: instrument on the line DC produced, near C, with DF = 120 m, so CF = 120 − 100 = 20 m; readings C = 1.500 m, D = 2.750 m.

Step 1: True difference of level (mid-point setup)

Equal sight lengths remove the collimation error, so

$$\text{True difference (D below C)} = 3.000 - 2.000 = 1.000\ \text{m}$$

Step 2: Apparent difference from setup 2

$$\text{Apparent difference} = 2.750 - 1.500 = 1.250\ \text{m}$$

Step 3: Error and direction

$$e = 1.250 - 1.000 = 0.250\ \text{m for a difference of sight lengths } (120 - 20) = 100\ \text{m}$$

The error per metre of sight is $0.250/100 = 0.0025$ m/m (1 in 400; angle about 8′36″). The far staff D reads more than it should, so the line of collimation is inclined upward.

Step 4: Readings for a horizontal line of sight (instrument near C)

Error at C (20 m): $0.0025 \times 20 = 0.050$ m. Error at D (120 m): $0.0025 \times 120 = 0.300$ m.

$$C = 1.500 - 0.050 = 1.450\ \text{m}, \qquad D = 2.750 - 0.300 = 2.450\ \text{m}$$

Check: $2.450 - 1.450 = 1.000$ m, equal to the true difference.

Answer: the line of collimation is inclined upward; the readings for a horizontal line of sight are C = 1.450 m and D = 2.450 m. The cross-hair is then adjusted to read 1.450 on C (or 2.450 on D) with the bubble centred.

Data: AB = 40 m. Setup 1: instrument at the mid-point, A = 1.464, B = 1.154. Setup 2: instrument near B on the inner side (between A and B), A = 1.642, B = 1.321.

Assumption: the distance of the setup from B is not stated, so the instrument is taken as practically at B (error at the near staff B negligible). The whole error then appears on the far staff A (distance 40 m).

Step 1: True difference of level

$$\text{True difference} = 1.464 - 1.154 = 0.310\ \text{m}\ (\text{B higher than A})$$

Step 2: Apparent difference from the second setup

$$1.642 - 1.321 = 0.321\ \text{m}$$

Step 3: Collimation error and precision

$$e = 0.321 - 0.310 = 0.011\ \text{m in 40 m}$$ $$\text{Collimation precision} = \frac{0.011}{40} = 0.000275 = \frac{1}{3636}\ \ (\text{i.e. } 27.5\ \text{mm per 100 m})$$

The usual permissible limit for ordinary levelling is about 10 mm per 100 m (1 in 10,000). The error is larger, so the instrument needs adjustment.

The far staff A reads too high (the apparent difference is greater than the true one), so the line of collimation is inclined upward.

Step 4: Correct reading on A for the second setup

$$A = 1.642 - 0.011 = 1.631\ \text{m}, \qquad B = 1.321\ \text{m}$$

Check: $1.631 - 1.321 = 0.310$ m = true difference.

Answer: collimation precision = 1 in 3636 (error 0.011 m in 40 m, line of collimation inclined upward); the staff reading on A for a horizontal line of sight is 1.631 m (B = 1.321 m).

If the setup is a distance $x$ from B, the correction is $0.011 \times (40 - x)/(40 - 2x)$ at A and $0.011\, x/(40 - 2x)$ at B.

Data: A and B are 50 m apart. Mid-point setup F gives the difference of staff readings $B - A = 0.250$ m (B reads more, so B is lower than A by 0.250 m). Setup 2: 5 m from A and 45 m from B; readings A = 1.500 m, B = 1.752 m.

Step 1: Apparent difference and error

$$\text{Apparent difference} = 1.752 - 1.500 = 0.252\ \text{m}$$ $$e = 0.252 - 0.250 = 0.002\ \text{m for the difference of sight lengths } (45 - 5) = 40\ \text{m}$$

Step 2: Magnitude and direction of the collimation error

$$\text{Error per metre} = \frac{0.002}{40} = 0.00005\ \text{m/m} = \frac{1}{20\,000}$$

(that is 5 mm per 100 m, less than the usual limit of 10 mm per 100 m, so the instrument is within tolerance). The far staff B reads too high, so the line of collimation is inclined upward by about 10″ ($0.00005 \times 206265''$).

Step 3: Correct readings with a horizontal line of sight (setup 2)

Error at A (5 m): $0.00005 \times 5 = 0.00025$ m. Error at B (45 m): $0.00005 \times 45 = 0.00225$ m.

$$A = 1.500 - 0.00025 = 1.49975\ \text{m}, \qquad B = 1.752 - 0.00225 = 1.74975\ \text{m}$$

Check: $1.74975 - 1.49975 = 0.250$ m.

Step 4: Collimation precision ratio

$$\text{Precision ratio} = \frac{e}{\text{difference of distances}} = \frac{0.002}{40} = \frac{1}{20\,000}$$

Answer: error = 0.00005 m per metre (1 in 20,000), line of collimation inclined upward; correct readings A = 1.49975 m (≈ 1.500), B = 1.74975 m (≈ 1.750).

a) Collimation precision

Data: A and B are 100 m apart. Setup 1 at the mid-point P: A = 1.587 m, B = 1.289 m. Setup 2 at Q, 10 m from B on the inner side (so Q is 90 m from A): A = 1.356 m, B = 1.046 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference} = 1.587 - 1.289 = 0.298\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference} = 1.356 - 1.046 = 0.310\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 90 m and B = 10 m, so the difference of sight lengths is 80 m.

$$e = \text{apparent} - \text{true} = 0.310 - 0.298 = +0.012\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.012}{80} = 0.00015\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{6667}$$

This is 15.0 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too large and the error grows with distance (the far staff reads too much), so the line of collimation is inclined upward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= 0.00015 \times 90 = +0.0135$ m; error in reading on B $= 0.00015 \times 10 = +0.0015$ m.

$$A = 1.356 - (+0.0135) = 1.3425\ \text{m}, \qquad B = 1.046 - (+0.0015) = 1.0445\ \text{m}$$

Check: A $-$ B $= 1.3425 - 1.0445 = 0.298$ m, equal to the true difference.

b) Correct readings

From Step 4: the correct staff readings for the second setup are A = 1.3425 m and B = 1.0445 m.

c) Making the line of collimation parallel to the axis of the bubble

1. Keep the level at the second setup (near one peg) and centre the bubble with the foot screws.
2. Calculate the correct staff reading for a horizontal line of sight on the far staff (as found above).
3. Using the capstan-headed screws at the top and bottom of the diaphragm, move the horizontal cross-hair (loosen one screw and tighten the other) until it reads the correct value on the far staff.
4. Check that the bubble is still central. If it has moved, centre it with the foot screws and repeat the adjustment.
5. Repeat the two peg test; the error should now be within the permissible limit.

Answer: the error is 0.012 m in 80 m (1 in 6667, about 15 mm per 100 m), which is not within the permissible limit; the line of collimation is inclined upward. Correct readings: A = 1.3425 m, B = 1.0445 m.

Data: A and B are 60 m apart; with the level at the mid-point the true difference of level is 0.320 m (reading on A is greater than on B by 0.320 m, since the apparent difference is also positive). Setup 2: level at Q on AB, 6 m from B (so 54 m from A): readings A = 1.387 m, B = 1.069 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference of readings (A} - \text{B)} = 0.320\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference} = 1.387 - 1.069 = 0.318\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 54 m and B = 6 m, so the difference of sight lengths is 48 m.

$$e = \text{apparent} - \text{true} = 0.318 - 0.320 = -0.002\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.002}{48} = 0.000042\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{24000}$$

This is 4.2 mm per 100 m of sight, which is within the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs no adjustment.

The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is inclined downward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= -0.000042 \times 54 = -0.00225$ m; error in reading on B $= -0.000042 \times 6 = -0.00025$ m.

$$A = 1.387 - (-0.00225) = 1.38925\ \text{m}, \qquad B = 1.069 - (-0.00025) = 1.06925\ \text{m}$$

Check: A $-$ B $= 1.38925 - 1.06925 = 0.320$ m, equal to the true difference.

Answer: correct readings A = 1.389 m (1.38925), B = 1.069 m (1.06925); collimation precision 1 in 24,000 (the line of collimation is inclined downward, but the error is small, 4.2 mm per 100 m).

Data: C and D are 50 m apart. Instrument at E (mid-point): C = 1.455 m, D = 1.860 m. Instrument at F with CF = 5 m and DF = 45 m: C = 1.500 m, D = 1.925 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference (C} - \text{D)} = 1.455 - 1.860 = -0.405\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (C} - \text{D)} = 1.500 - 1.925 = -0.425\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are C = 5 m and D = 45 m, so the difference of sight lengths is 40 m.

$$e = \text{apparent} - \text{true} = -0.425 - -0.405 = -0.020\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.020}{40} = 0.0005\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{2000}$$

This is 50.0 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too large and the error grows with distance (the far staff reads too much), so the line of collimation is inclined upward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on C $= 0.0005 \times 5 = +0.0025$ m; error in reading on D $= 0.0005 \times 45 = +0.0225$ m.

$$C = 1.500 - (+0.0025) = 1.4975\ \text{m}, \qquad D = 1.925 - (+0.0225) = 1.9025\ \text{m}$$

Check: C $-$ D $= 1.4975 - 1.9025 = -0.405$ m, equal to the true difference.

Answer: the closing (collimation) error is 0.020 m in 40 m, that is 0.0005 m per metre or 1 in 2000 (25 mm over a 50 m sight); the line of collimation is inclined upward. The precision of the instrument is 1 in 2000, which is poor and needs adjustment. With respect to a horizontal line of sight the readings are C = 1.4975 m and D = 1.9025 m (the readings 1.500 and 1.925 less the errors 0.0025 and 0.0225).

Data: A and B are 60 m apart. Setup 1 at the mid-point P: A = 1.585 m, B = 1.287 m. Setup 2 near A, 6 m inside AB (so 54 m from B): A = 1.355 m, B = 1.045 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference (A} - \text{B)} = 1.585 - 1.287 = 0.298\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (A} - \text{B)} = 1.355 - 1.045 = 0.310\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 6 m and B = 54 m, so the difference of sight lengths is 48 m.

$$e = \text{apparent} - \text{true} = 0.310 - 0.298 = +0.012\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.012}{48} = 0.00025\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{4000}$$

This is 25.0 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is inclined downward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= -0.00025 \times 6 = -0.0015$ m; error in reading on B $= -0.00025 \times 54 = -0.0135$ m.

$$A = 1.355 - (-0.0015) = 1.3565\ \text{m}, \qquad B = 1.045 - (-0.0135) = 1.0585\ \text{m}$$

Check: A $-$ B $= 1.3565 - 1.0585 = 0.298$ m, equal to the true difference.

Procedure to make the line of collimation horizontal

1. Keep the level at the second setup (near one peg) and centre the bubble with the foot screws.
2. Calculate the correct staff reading for a horizontal line of sight on the far staff (as found above).
3. Using the capstan-headed screws at the top and bottom of the diaphragm, move the horizontal cross-hair (loosen one screw and tighten the other) until it reads the correct value on the far staff.
4. Check that the bubble is still central. If it has moved, centre it with the foot screws and repeat the adjustment.
5. Repeat the two peg test; the error should now be within the permissible limit.

Answer: collimation precision = 1 in 4000 (0.012 m in 48 m, i.e. 25 mm per 100 m), more than permissible; the line of collimation is inclined downward. Correct readings for the second setup: A = 1.3565 m, B = 1.0585 m.

Data: A and B are 50 m apart. Setup 1 at the mid-point: A = 1.875 m, B = 1.790 m. Setup 2 at a point 5 m beyond B on the line AB produced (so B is 5 m and A is 55 m from the instrument): A = 1.630 m, B = 1.560 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference (A} - \text{B)} = 1.875 - 1.790 = 0.085\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (A} - \text{B)} = 1.630 - 1.560 = 0.070\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 55 m and B = 5 m, so the difference of sight lengths is 50 m.

$$e = \text{apparent} - \text{true} = 0.070 - 0.085 = -0.015\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.015}{50} = 0.0003\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{3333}$$

This is 30.0 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is inclined downward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= -0.0003 \times 55 = -0.0165$ m; error in reading on B $= -0.0003 \times 5 = -0.0015$ m.

$$A = 1.630 - (-0.0165) = 1.6465\ \text{m}, \qquad B = 1.560 - (-0.0015) = 1.5615\ \text{m}$$

Check: A $-$ B $= 1.6465 - 1.5615 = 0.085$ m, equal to the true difference.

Answer: with a horizontal line of sight the readings in the second setup are A = 1.6465 m and B = 1.5615 m; the line of collimation is inclined downward and the precision is 1 in 3333 (30 mm per 100 m), which needs adjustment.

Reading of the data (assumptions).

• "The distance to stations A and B was 60 m" is taken as 60 m from the mid-point level to each station, so AB = 120 m.
• With the level at the mid-point, A = 1.855 m and B = 1.625 m, so A reads 0.230 m more than B. In the second set-up the printed readings are A = 2.385 and B = 2.655, which would give A − B = −0.270 m, of opposite sign to 0.230 m; this is impossible for the same two stations, so the two readings are taken as interchanged: A = 2.655 m and B = 2.385 m (apparent difference +0.270 m).
• "Near to station B" is taken as the level practically at B (distance from B about 0), so A is 120 m away.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference (A} - \text{B)} = 1.855 - 1.625 = 0.230\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (A} - \text{B)} = 2.655 - 2.385 = 0.270\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 120 m and B = 0 m, so the difference of sight lengths is 120 m.

$$e = \text{apparent} - \text{true} = 0.270 - 0.230 = +0.040\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.040}{120} = 0.000333\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{3000}$$

This is 33.3 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too large and the error grows with distance (the far staff reads too much), so the line of collimation is inclined upward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= 0.000333 \times 120 = +0.04$ m; error in reading on B $= 0.000333 \times 0 = +0$ m.

$$A = 2.655 - (+0.04) = 2.615\ \text{m}, \qquad B = 2.385 - (+0) = 2.385\ \text{m}$$

Check: A $-$ B $= 2.615 - 2.385 = 0.230$ m, equal to the true difference.

Answer: collimation error = 0.040 m in 120 m, i.e. 0.000333 m per metre (1 in 3000, 33 mm per 100 m); the sign is upward (the line of collimation is inclined upward, the far staff A reads 0.040 m too much). Correct readings in the second set-up: A = 2.615 m, B = 2.385 m.