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LEVELLING | Civil Engineers, Study notes of Civil Engineering

For booking and reducing the levels of points, there are two systems, namely the height of instrument or height of collimation method and rise and fall method.

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LEVELLING
3.1 LEVELLING
Levelling is an operation in surveying performed to determine the difference in levels of two points.
By this operation the height of a point from a datum, known as elevation, is determined.
3.2 LEVEL SURFACE
A level surface is the equipotential surface of the earth’s gravity field. It is a curved surface and
every element of which is normal to the plumb line.
3.3 DATUM
A datum is a reference surface of constant potential, called as a level surface of the earth’s gravity
field, for measuring the elevations of the points. One of such surfaces is the mean sea level surface
and is considered as a standard datum. Also an arbitrary surface may be adopted as a datum.
3.4 LEVEL LINE
A line lying in a level surface is a level line. It is thus a curved line.
Line of sight
Horizontal line
Staff held ver tical
Mean sea-level
Fig. 3.1
A level in proper adjustment, and correctly set up, produces a horizontal line of sight which
is at right angles to the direction of gravity and tangential to the level line at the instrument height.
It follows a constant height above mean sea level and hence is a curved line, as shown in Fig. 3.1.
Over short distances, such as those met in civil engineering works, the two lines can be taken
to coincide. Over long distances a correction is required to reduce the staff readings given by the
horizontal line of sight to the level line equivalent. Refraction of the line of sight is also to be taken
into account. The corrections for the curvature of the level line Cc and refraction Cr are shown in
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LEVELLING

3.1 LEVELLING

Levelling is an operation in surveying performed to determine the difference in levels of two points. By this operation the height of a point from a datum , known as elevation , is determined.

3.2 LEVEL SURFACE

A level surface is the equipotential surface of the earth’s gravity field. It is a curved surface and every element of which is normal to the plumb line.

3.3 DATUM

A datum is a reference surface of constant potential, called as a level surface of the earth’s gravity field, for measuring the elevations of the points. One of such surfaces is the mean sea level surface and is considered as a standard datum. Also an arbitrary surface may be adopted as a datum.

3.4 LEVEL LINE

A line lying in a level surface is a level line. It is thus a curved line.

Line of sight

Horizontal line

Staff held vertical

Mean sea-level

Fig. 3. A level in proper adjustment, and correctly set up, produces a horizontal line of sight which is at right angles to the direction of gravity and tangential to the level line at the instrument height. It follows a constant height above mean sea level and hence is a curved line, as shown in Fig. 3.1.

Over short distances, such as those met in civil engineering works, the two lines can be taken to coincide. Over long distances a correction is required to reduce the staff readings given by the horizontal line of sight to the level line equivalent. Refraction of the line of sight is also to be taken into account. The corrections for the curvature of the level line Cc and refraction C (^) r are shown in

59

60 SURVEYING

Fig. 3.2. The combined correction is given by

C d R

c r = −^

2 ...(3.1)

where

C (^) cr = the correction for the curvature and refraction, d = the distance of the staff from the point of tangency, and R = the mean earth’s radius. For the value of R = 6370 km and d in kilometre, the value of Ccr in metre is given as

C^0.^067 d^2 cr =^ − ...(3.2)

Fig. 3.

3.5 DIRECT DIFFERENTIAL OR SPIRIT LEVELLING

Differential levelling or spirit levelling is the most accurate simple direct method of determining the difference of level between two points using an instrument known as level with a levelling staff. A level establishes a horizontal line of sight and the difference in the level of the line of sight and the point over which the levelling staff is held, is measured through the levelling staff.

Fig. 3.3 shows the principle of determining the difference in level ∆ h between two points A and B, and thus the elevation of one of them can be determined if the elevation of the other one is known. S (^) A and S (^) B are the staff readings at A and B , respectively, and h (^) A and h (^) B are their respective elevations.

Fig. 3.

L e v e l li n e

L i n e o f si g h t

H o r iz o n ta l lin e

M e a n s e a - le v e l

C r C c r C c

62 SURVEYING

Section-

Section-

Reduced level (R.L.): Reduced level of a point is its height or depth above or below the assumed datum. It is the elevation of the point.

Rise and fall: The difference of level between two consecutive points indicates a rise or a fall between the two points. In Fig. 3.3, if ( S (^) A – S (^) B ) is positive, it is a rise and if negative, it is a fall. Rise and fall are determined for the points lying within a section.

Section: A section comprises of one back sight, one fore sight and all the intermediate sights taken from one instrument set up within that section. Thus the number of sections is equal to the number of set ups of the instrument. (From A to B for instrument position 1 is section-1 and from B to C for instrument position 2 is section-2 in Fig. 3.4). For booking and reducing the levels of points, there are two systems, namely the height of instrument or height of collimation method and rise and fall method. The columns for booking the readings in a level book are same for both the methods but for reducing the levels, the number of additional columns depends upon the method of reducing the levels. Note that except for the change point, each staff reading is written on a separate line so that each staff position has its unique reduced level. This remains true at the change point since the staff does not move and the back sight from a forward instrument station is taken at the same staff position where the fore sight has been taken from the backward instrument station. To explain the booking and reducing levels, the levelling operation from stations A to C shown in Fig. 3.4, has been presented in Tables 3.1 and 3.2 for both the methods. These tables may have additional columns for showing chainage, embankment, cutting, etc., if required.

Table 3.1 Height of instrument method

Station B.S. I.S. F.S. H.I. R.L. Remarks A S (^) 1 H.I. A = h (^) A B.M. = ha h (^) A + S 1 a S 2 h (^) a = H.I. AS 2 b S 3 h (^) b = H.I. AS 3 B S (^) 5 S 4 H.I. B = h (^) B = H.I. AS 4 C.P. – h (^) B + S 5 c S 6 h (^) c = H.I. (^) B – S 6 C S 7 H (^) C = H.I. (^) B – S 7 Σ B.S. Σ F.S.

Check : Σ B.S. – Σ F.S. = Last R.L. – First R.L.

In reducing the levels for various points by the height of instrument method, the height of instrument (H.I.) for the each section highlighted by different shades, is determined by adding the elevation of the point to the back sight reading taken at that point. The H.I. remains unchanged for all the staff readings taken within that section and therefore, the levels of all the points lying in that section are reduced by subtracting the corresponding staff readings, i.e., I.S. or F.S., from the H.I. of that section. In the rise and fall method, the rises and the falls are found out for the points lying within each section. Adding or subtracting the rise or fall to or from the reduced level of the backward

LEVELLING 63

Section-

Section-

station obtains the level for a forward station. In Table 3.2, r and f indicate the rise and the fall, respectively, assumed between the consecutive points.

Table 3.2 Rise and fall method

Station B.S. I.S. F.S. Rise Fall R.L. Remarks A S 1 h (^) A B.M. = h (^) a a S 2 r 1 = h (^) a = h (^) A + r 1 S 1 – S 2 b S 3 f 1 = S 2 – S 3 h (^) b = h (^) af 1 B S 5 S 4 f 2 = S 3 – S 4 h (^) B = h (^) bf 2 C.P. c S 6 f 3 = S 5 – S 6 h (^) c = h (^) Bf 3 C S 7 r 2 = H (^) C = h (^) c + r 2 S 6 – S 7 Σ B.S. Σ F.S. Σ Rise Σ Fall Check : Σ B.S. − Σ F.S. = Σ Rise − Σ Fall = Last R.L. − First R.L.

The arithmetic involved in reduction of the levels is used as check on the computations. The following rules are used in the two methods of reduction of levels.

( a ) For the height of instrument method ( i ) Σ B.S. – Σ F.S. = Last R.L. – First R.L. ( ii ) Σ [H.I. × (No. of I.S.’s + 1)] – Σ I.S. – Σ F.S. = Σ R.L. – First R.L. ( b ) For the rise and fall method Σ B.S. – Σ F.S. = Σ Rise – Σ Fall = Last R.L. – First R.L.

3.6 COMPARISON OF METHODS AND THEIR USES

Less arithmetic is involved in the reduction of levels with the height of instrument method than with the rise and fall method, in particular when large numbers of intermediate sights is involved. Moreover, the rise and fall method gives an arithmetic check on all the levels reduced, i.e., including the points where the intermediate sights have been taken, whereas in the height of instrument method, the check is on the levels reduced at the change points only. In the height of instrument method the check on all the sights is available only using the second formula that is not as simple as the first one. The height of instrument method involves less computation in reducing the levels when there are large numbers of intermediate sights and thus it is faster than the rise and fall method. The rise and fall method, therefore, should be employed only when a very few or no intermediate sights are taken in the whole levelling operation. In such case, frequent change of instrument position requires determination of the height of instrument for the each setting of the instrument and, therefore, computations involved in the height of instrument method may be more or less equal to that required in the rise and fall method. On the other hand, it has a disadvantage of not having check on the intermediate sights, if any, unless the second check is applied.

LEVELLING 65

and the combined error is given by

( ) ( ) 2

e = b^1 − a^1 − b^2 − a^2 ...(3.7)

where e = e (^) l + e (^) c – er ...(3.8) e (^) l = the collimation error assumed positive for the line of sight inclined upward, ec = the error due to the earth’s curvature, and er = the error due to the atmospheric refraction. We have ec – er = the combined curvature and refraction error = 0.067 d^2. The collimation error is thus given by e (^) l = e – 0.067 d^2 in metre …(3.9)

where d is the distance between A and B in kilometre.

3.9 TRIGONOMETRIC LEVELLING

Trigonometric levelling involves measurement of vertical angle and either the horizontal or slope distance between the two points between which the difference of level is to be determined. Fig. 3. shows station A and station B whose height is to be established by reciprocal observations from A on to signal at B and from B on to signal at A. Vertical angles α (angle of elevation) and β (angle of depression), are measured at A and B , respectively. The refracted line of sight will be inclined to the direct line AB and therefore, the tangent to the refracted line of sight makes an angle υ with AB. The vertical angle α is measured with respect to this tangent and to the horizontal at A.

Similarly, from B the angle of depression β is measured from the horizontal to the tangent to the line of sight. The point C lies on the arc through A , which is parallel to the mean sea level surface. d is the geodetic or spheroidal distance between A and B , and could be deduced from the geodetic coordinates of the two points. Angle BAC between AB and chord AC is related to angle θ between the two verticals at A and B which meet at the earth’s centre, since AC makes an angle of θ/2 with the horizontal at A.

If the elevations of A and B are h (^) A and h (^) B , respectively, then the difference in elevation ∆ h between A and B , is found out by solving the triangle ABC for BC. Thus in triangle ABC

υ

θ α

θ

υ

θ α

sin 180 90

sin

o o

BC AC

cos( )

) 2

sin(

α θ υ

υ

θ α

  • AC ...(3.10)

66 SURVEYING

The correction for curvature and refraction at A and B is (θ / 2 υ ) and this refers angles a and β to chords AC and BD , respectively.

Thus angle of elevation BAC = α + θ / 2 υ and angle of depression DBA = β θ / 2 + υ. Since AC is parallel to BD,BAC = ∠ DBA, Therefore, ∠ BAC + ∠ DBA = ( α + θ / 2 υ ) + ( β θ / 2 + υ )

or 2 ∠ BAC = α + β

or ∠ BAC =

α + β = ∠ DBA.

This is the correct angle of elevation at A or angle of depression at B and is mean of the two angles α and β. In addition, we can write Eq. (3.10) as

è

υ^ α

υ D^ â^ è/2 B

A è/2 C

Mean sea-level

R

h (^) A

h (^) B

Vertical at A

Vertical at (^) B

Horizontal at A

Horizontal at B

Line of sight

d

90 °+è/

Fig. 3.

BC = AC tan ( α + θ /2 – υ)

or  

  

∆ = ^ + 2

h AC tan α^ β.

Therefore, h (^) B = h (^) A + D h

+ ^ +

tan

α β hA AC … (3.11)

Note that in Eq. (3.11) it is assumed that α is the angle of elevation and β is angle of depression. In practice, only magnitudes need to be considered, not signs provided one angle is elevation and the other is depression.

The coefficient of refraction K in terms of the angle of refraction υ and the angle θ subtended at the centre of the spheroid by the arc joining the stations, is given by

θ

υ K = …(3.12)

68 SURVEYING

C A B A B D d d d d

θ θ

Fig. 3.

3.12 EYE AND OBJECT CORRECTION

The heights of the instrument and signal at which the observation is made, are generally not same and thus the observed angle of elevation θ′ as shown in Fig. 3.9, does not refer to the ground levels at A and B. This difference in height causes the observed vertical angle θ′ to be larger than that θ which would have been observed directly from those points. A correction (ε) termed as eye and object correction , is applied to the observed vertical angle to reduce it to the required value.

The value of the eye and object correction is given by

d

h (^) shi ε = radians …(3.15)

d

h (^) shi = (^206265) seconds …(3.16)

Example 3.1. The following readings were taken with a level and 4 m staff. Draw up a level book page and reduce the levels by the height of instrument method.

0.578 B.M.(= 58.250 m), 0.933, 1.768, 2.450, (2.005 and 0.567) C.P., 1.888, 1.181, (3. and 0.612) C.P., 0.705, 1.810.

Solution: The first reading being on a B.M., is a back sight. As the fifth station is a change point, 2. is fore sight reading and 0.567 is back sight reading. All the readings between the first and fifth readings are intermediate sight-readings. Similarly, the eighth station being a change point, 3.679 is fore sight reading, 0.612 is back sight reading, and 1.888, 1.181 are intermediate sight readings. The last reading 1.810 is fore sight and 0.705 is intermediate sight-readings. All the readings have been entered in their respective columns in the following table and the levels have been reduced by height of instrument method. In the following computations, the values of B.S., I.S., H.I., etc., for a particular station have been indicated by its number or name.

Section-1: H.I. 1 = h 1 + B.S. 1 = 58.250 + 0.578 = 58.828 m h 2 = H.I. 1 – I.S. 2 = 58.828 – 0.933 = 57.895 m h 3 = H.I. 1 – I.S. 3 = 58.828 – 1.768 = 57.060 m h 4 = H.I. 1 – I.S. 4 = 58.828 – 2.450 = 56.378 m h 5 = H.I. 1 – F.S. 5 = 58.828 – 2.005 = 56.823 m

hs

A

B

h (^) i

B

B ′′

hi d

ε θ θ′

Fig. 3.

LEVELLING 69

Section-2: H.I. 5 = h 5 + B.S. 5 = 56.823 + 0.567 = 57.390 m h 6 = H.I. 2 – I.S. 6 = 57.390 – 1.888 = 55.502 m h 7 = H.I. 2 – I.S. 7 = 57.390 – 1.181 = 56.209 m h 8 = H.I. 2 – F.S. 8 = 57.390 – 3.679 = 53.711 m Section-3: H.I. 8 = h 8 + B.S. 8 = 53.711 + 0.612 = 54.323 m h 9 = H.I. 8 – I.S. 9 = 54.323 – 0.705 = 53.618 m h 10 = H.I. 8 – F.S. 10 = 54.323 – 1.810 = 52.513 m Additional check for H.I. method: Σ [H.I. × (No. of I.S.s + 1)] – Σ I.S. – Σ F.S. = Σ R.L.

  • First R.L. [58.828 × 4 + 57.390 ´ 3 + 54.323 × 2] – 8.925 – 7.494 = 557.959 – 58.250 = 499.709 ( O. K .)

Table 3.

Station B.S. I.S. F.S. H.I. R.L. Remarks 1 0.578 58.828 58.250 B.M.=58.250 m 2 0.933 57. 3 1.768 57. 4 2.450 56. 5 0.567 2.005 57.390 56.823 C.P. 6 1.888 55. 7 1.181 56. 8 0.612 3.679 54.323 53.711 C.P. 9 0.705 53. 10 1.810 52. Σ 1.757 8.925 7.494 557. Check : 1.757 – 7.494 = 52.513 – 58.250 = – 5.737 ( O.K. ) Example 3.2. Reduce the levels of the stations from the readings given in the Example 3.1 by the rise and fall method.

Solution: Booking of the readings for reducing the levels by rise and fall method is same as explained in Example 3.1. The computations of the reduced levels by rise and fall method is given below and the results are tabulated in the table. In the following computations, the values of B.S., I.S., Rise ( r ), Fall ( f ), etc., for a particular station have been indicated by its number or name.

( i ) Calculation of rise and fall Section-1: f 2 = B.S. 1 – I.S. 2 = 0.578 – 0.933 = 0. f 3 = I.S. 2 – I.S. 3 = 0.933 – 1.768 = 0. f 4 = I.S. 3 – I.S. 4 = 1.768 – 2.450 = 0. r 5 = I.S. 4 – F.S. 5 = 2.450 – 2.005 = 0.

LEVELLING 71

( ii ) the gradient of the line joining the first and last points. 0.420, 1.115, 2.265, 2.900, 3.615, 0.535, 1.470, 2.815, 3.505, 4.445, 0.605, 1.925, 2.885. Solution: Since the readings have been taken along a line on a continuously sloping ground, any sudden large change in the reading such as in the sixth reading compared to the fifth reading and in the eleventh reading compared to the tenth reading, indicates the change in the instrument position. Therefore, the sixth and eleventh readings are the back sights and fifth and tenth readings are the fore sights. The first and the last readings are the back sight and fore sight, respectively, and all remaining readings are intermediate sights.

The last point being of known elevation, the computation of the levels is to be done from last point to the first point. The falls are added to and the rises are subtracted from the known elevations. The computation of levels is explained below and the results have been presented in the following table.

(i) Calculation of rise and fall Section-1: f 2 = B.S. 1 – I.S. 2 = 0.420 – 1.115= 0. f 3 = I.S. 2 – I.S. 3 = 1.115 – 2.265= 1. f 4 = I.S. 3 – I.S. 4 = 2.265 – 2.900= 0. f 5 = I.S. 4 – F.S. 5 = 2.900 – 3.615= 0. Section-2: f 6 = B.S. 5 – I.S. 6 = 0.535 – 1.470 = 0. f 7 = I.S. 6 – I.S. 7 = 1.470 – 2.815 = 1. f 8 = I.S. 7 – I.S. 8 = 2.815 – 3.505 = 0. f 9 = I.S. 8 – F.S. 9 = 3.505 – 4.445 = 0. Section-3: f 10 = B.S. 9 – I.S. 10 = 0.605 – 1.925 = 1. f 11 = I.S. 10 – F.S. 11 = 1.925 – 2.885 = 0. (ii) Calculation of reduced levels h 10 = h 11 + f 11 = 155.272 + 0.960 = 156.232 m h 9 = h 10 + f 10 = 156.232 + 1.320 = 157.552 m h 8 = h 9 + f 9 = 157.552 + 0.940 = 158.492 m h 7 = h 8 + f 8 = 158.492 + 0.690 = 159.182 m h 6 = h 7 + f 7 = 159.182 + 1.345 = 160.527 m h 5 = h 6 + f 6 = 160.527 + 0.935 = 161.462 m h 4 = h 5 + f 5 = 161.462 + 0.715 = 162.177 m h 3 = h 4 + f 4 = 162.177 + 0.635 = 162.812 m h 2 = h 3 + f 3 = 162.812 + 1.150 = 163.962 m h 1 = h 2 + f 2 = 163.962 + 0.695 = 164.657 m

72 SURVEYING

Table 3.

Station Chainage (^) B.S. I.S. F.S. Rise Fall R.L. Remarks (m) 1 0 0.420 164. 2 20 1.115 0.695 163. 3 40 2.265 1.150 162. 4 60 2.900 0.635 162. 5 80 0.535 3.615 0.715 161.462 C.P. 6 100 1.470 0.935 160. 7 120 2.815 1.345 159. 8 140 3.505 0.690 158. 9 160 0.605 4.445 0.940 157.552 C.P. 10 180 1.925 1.320 156. 11 200 2.885 0.960 155.272 Elevation = 155.272 m Σ 1.560 10.945 0.000 9. Check : 1.560 – 10.945 = 0.000 – 9.385 = 155.272 – 164.657 = – 9.385 ( O.K. )

( iii ) Calculation of gradient The gradient of the line 1-11 is

= (^) distancebetweenpoints 1 - 11

differenceoflevelbetweenpoints 1 - 11

= 1 in 21.3 ( falling ) Example 3.4. A page of level book is reproduced below in which some readings marked as (×), are missing. Complete the page with all arithmetic checks.

Solution: The computations of the missing values are explained below. B.S. 4 – I.S. 5 = f 5 , B.S. 4 = f 5 + I.S. 5 = – 0.010 + 2.440 = 2. B.S. 9 – F.S. 10 = f 10 , B.S. 9 = f 10 + F.S. 10 = – 0.805 + 1.525 = 0. B.S. 1 + B.S. 2 + B.S. 4 + B.S. 6 + B.S. 7 + B.S. 9 = ΣB.S. 3.150 + 1.770 + 2.430 + B.S. 6 + 1.185 + 0.720 = 12. B.S. 6 = 12.055 – 9.255 = 2.

74 SURVEYING

Table 3.7.

Station B.S. I.S. F.S. Rise Fall R.L. Remarks 1 3.150 221. 2 1.770 3.850 0.700 220.400 C.P. 3 2.200 0.430 219. 4 2.430 1.850 0.350 220.320 C.P. 5 2.440 0.010 220. 6 2.800 1.340 1.100 221.410 C.P. 7 1.185 2.010 0.790 222.200 C.P. 8 -2.735 3.920 226.120 Staff held inverted 9 0.720 1.685 4.420 221.700 C.P. 10 1.525 0.805 220. Σ 12.055 12.266 6.610 6. Check : 12.055 – 12.266 = 6.610 – 6.365 = 220.895 – 221.100 = – 0.205 ( O.K. )

Example 3.5. Given the following data in Table 3.8, determine the R.L.s of the points 1 to

  1. If an uniform upward gradient of 1 in 20 starts at point 1, having elevation of 150 m, calculate the height of embankment and depth of cutting at all the points from 1 to 6.

Table 3.

Station Chainage (m) B.S. I.S. F.S. Remarks B.M. – 10.11 153.46 m 1 0 3. 2 100 1. 3 200 6.89 0. 4 300 3. 5 400 11.87 3. 6 500 5. Solution: Reduced levels of the points by height of instrument method H.I. (^) B.M. = R.L. (^) B.M. + B.S. (^) B.M. = 153.46 + 10.11 = 163.57 m h 1 = H.I. (^) B.M. – I.S. 1 = 163.57 – 3.25 = 160.32 m h 2 = H.I. (^) B.M. – I.S. 2 = 163.57 – 1.10 = 162.47 m h 3 = H.I. (^) B.M. – F.S. 3 = 163.57 – 0.35 = 163.22 m H.I. 3 = h 3 + B.S. 3 = 163.22 + 6.89 = 170.11 m

LEVELLING 75

h 4 = H.I. 3 – I.S. 4 = 170.11 – 3.14 = 166.97 m h 5 = H.I. 3 – F.S. 5 = 170.11 – 3.65 = 166.46 m H.I. 5 = h 5 + B.S. 5 = 166.46 + 11.87 = 178.33 m h 6 = H.I. 5 – F.S. 6 = 178.33 – 5.98 = 172.35 m Levels of the points from gradient Since the gradient is 1 in 20, for every 100 m the rise is 5 m. Level of point 1, h ′ 1 = 150 m (given) Level of point 2, h ′ 2 = 150 + 5 = 155 m Level of point 3, h ′ 3 = 155 + 5 = 160 m Level of point 4, h ′ 4 = 160 + 5 = 165 m Level of point 5, h ′ 5 = 165 + 5 = 170 m Level of point 6, h ′ 6 = 170 + 5 = 175 m Height of embankment and depth of cutting At point 1 h 1 – h ′ 1 = 160.32 – 150.00 = + 10.32 m (embankment) At point 2 h 2 – h ′ 2 = 162.47 – 155.00 = + 7.47 m (embankment) At point 3 h 3 – h ′ 3 = 163.22 – 160.00 = + 3.22 m (embankment) At point 4 h 4 – h ′ 4 = 166.97 – 165.00 = + 1.97 m (embankment) At point 5 h 5 – h ′ 5 = 166.46 – 170.00 = – 3.54 m (cutting) At point 6 h 6 – h ′ 6 = 172.35 – 175.00 = – 2.65 m (cutting) The computed values of the height of embankment and depth of cutting are tabulated below.

Table 3.

Station Chainage B.S. I.S. F.S. H.I. R.L.

Gradient Embankment/cutting (m) level Height (m) Depth (m) B.M. – 10.11 163.57 153.46 – 1 0 3.25 160.32 150 10. 2 100 1.10 162.47 155 7. 3 200 6.89 0.35 170.11 163.22 160 3. 4 300 3.14 166.97 165 1. 5 400 11.87 3.65 178.33 166.46 170 3. 6 500 5.98 172.35 175 2.

Example 3.6. The readings given in Table 3.10, were recorded in a levelling operation from points 1 to 10. Reduce the levels by the height of instrument method and apply appropriate checks. The point 10 is a bench mark having elevation of 66.374 m. Determine the loop closure and adjust the calculated values of the levels by applying necessary corrections. Also determine the mean gradient between the points 1 to 10.

LEVELLING 77

i.e., the correction at station 1 0.0 m the correction at station 2 + 0.005 m the correction at station 6 + 0.010 m the correction at station 8 + 0.015 m the correction at station 10 + 0.020 m The corrections for the intermediate sights will be same as the corrections for that instrument stations to which they are related. Therefore,

correction for I.S. 3 , I.S. 4 , and I.S. 5 = + 0.010 m correction for I.S. 7 = + 0.015 m correction for I.S. 9 = + 0.020 m Applying the above corrections to the respective reduced levels, the corrected reduced levels are obtained. The results have been presented in Table 3.11.

Table 3.

Station Chainage (^) B.S. I.S. F.S. H.I. R.L. Correction Corrected (m) R.L. 1 0 0.597 68.830 68.233 – 68. 2 20 2.587 3.132 68.285 65.698 + 0.005 65. 3 40 1.565 66.720 + 0.010 66. 4 60 1.911 66.374 + 0.010 66. 5 80 0.376 67.909 + 0.010 67. 6 100 2.244 1.522 69.007 66.763 + 0.010 66. 7 120 3.771 65.236 + 0.015 65. 8 140 1.334 1.985 68.356 67.022 + 0.015 67. 9 160 0.601 67.755 + 0.020 67. 10 180 2.002 66.354 + 0.020 66. Σ 6.762 8. Check : 6.762 – 8.641 = 66.354 – 68.233 = – 1.879 ( O.K. )

Gradient of the line 1- The difference in the level between points 1 and 10, ∆ h = 66.324 – 68.233 = –1.909 m The distance between points 1-10, D = 180 m

Gradient = – 180

= 1 in 94.3 ( falling ) Example 3.7. Determine the corrected reduced levels of the points given in Example 3.6 by two alternative methods.

78 SURVEYING

Solution: Method-

From Eq. (3.3), the correction L

c =− el

The total correction at point 10 (from Example 3.6) = + 0.020 m The distance between the points 1 and 10 = 180 m

Correction at point 2 = 20 180

  • × = + 0.002 m

Correction at point 6 = 100 180

  • × = + 0.011 m

Correction at point 8 = 140 180

  • × = + 0.016 m

Correction at point 10 = 180 180

  • × = + 0.020 m

Corrections at points 3, 4, and 5 = + 0.011 m Correction at point 7 = + 0.016 m Correction at point 9 = + 0.020 m The corrections and the corrected reduced levels of the points are given in Table 3.12.

Table 3.

Station R.L. Correction Corrected R.L. 1 68.233 – 68. 2 65.698 + 0.002 65. 3 66.720 + 0.011 66. 4 66.374 + 0.011 66. 5 67.909 + 0.011 67. 6 66.763 + 0.011 66. 7 65.236 + 0.016 65. 8 67.022 + 0.016 67. 9 67.755 + 0.020 67. 10 66.354 + 0.020 66.

Method- In this method half of the total correction is applied negatively to all the back sights and half of the total correction is applied positively to all the fore sights.

Total number of back sights = 4 Total number of fore sights = 4

Correction to each back sight =  

×

= + 0.0025 m