engineering surveying (221 be) levelling-technical

ENGINEERING SURVEYING (221 BE)

Levelling-Technical

Sr Tan Liat Choon

Email: [email protected]

Mobile: 016-4975551

LEVELLING

TWO PEG TEST Set 2 marks at 30-40 metre apart, also mark centre point in a relatively flat area

Set level at midpoint and take readings at each end

Determine difference in readings (difference in elevation)

Move level to one end and setup so that level is just in front of rod on point

Read rod by looking backward through scope (x-hair not visible), hold pencil on

rod to determine reading

Read rod at other end in normal manner

Difference in readings should equal of 3

If values are not equal, there is error • Most instruments have adjustment screws • Adjust and repeat test as a check

2

Two Peg Test

3

L / 2 L / 2

x

S1

S1’

Line of Collimation

Horizontal Line

L

A

B

S2

S2’

x

L / 2 L / 2

x

S1

S1’

Line of Collimation

Horizontal Line

L

A

B

= S1’ - S2’

The APPARENT height difference hA

The TRUE height difference hT

= S1 - S2

S1 = S1’ + x and S2

= S2’ + x 4

Two Peg Test

L / 2 L / 2

L

x

S1

S1’

Line of Collimation

Horizontal Line

A

B

S2

S2’

x

The TRUE height difference hT = S1’ - S2’

= S1 - S2 The APPARENT height difference hA

S1 = S1’ + x and S2

= S2’ + x hA = (S1’ + x) - (S2’ + x ) 5

Two Peg Test

L / 2 L / 2

L

x

S1

S1’

Line of Collimation

Horizontal Line

A

B

S2

S2’

x

The TRUE height difference hT = S1’ - S2’


S1 = S1’ + x and S2

= S2’ + x hA = S1’ - S2’ = hT 6

Two Peg Test

hA = hT

Therefore :

This is true since the instrument is the same distance from both staff positions and

the errors x are equal and cancel out

7

Two Peg Test

S3’

S3

A

B

L / 10

Now move the instrument outside the “odd numbered” peg

8

Two Peg Test

S3

S3’

A

B

L / 10

S4

S4’


But the TRUE height difference hT We already know

9

Two Peg Test

S3

S3’

A

B

L / 10

S4

S4’

= S3 - S4

= S1 - S2

then the instrument is OK

If NOT then the error is e =

The APPARENT height difference hA

But the TRUE height difference hT

Therefore if hA = hT

(S1 - S2) - (S3 - S4) / L mm / m 10

Two Peg Test

Summary

Place two pegs about L = 30m (to 40m) apart

Set up level midway between the two pegs

Read staff on each peg, and calculate true height difference

Move level about L / 10 = 3m (or 4m) beyond one of the pegs

Read staff on each peg again, and calculate height difference

Collimation Error e = difference in the differences

and is expressed as a number of mm per L m

Acceptable errors

Uren and Price 1mm per 20m

Wimpey 4mm per 50m

Test should be carried out regularly say once per week or two 11

L / 2 L / 2

L

S1

A B

S2

S3

A B

L / 10

S4

Collimation error,

e = (S1 - S2) - (S3 - S4) mm / Lm

12

DATUM

Could be our own Datum - Assumed Datum

- Arbitrary Datum

- Site Datum Or

A National Datum

In the UK we have a national organisation known as

The Ordnance Survey (O.S.)

The O.S. has established a ZERO Datum at Newlyn in

Cornwall.

- Ordnance Datum

Based on the Ordnance Datum - points of known height above

or below Zero height have been established around the U.K.

These points around the country are known as Bench Marks

Above Assumed Datum

Above Ordnance Datum

13

Levelling

A B

Measured and Calculated

Level of A

Reduced

Level of A

RL A (known) Reduced

Level of B RL B

(unknown)

the Plane of Collimation Height of

DATUM

(HPC)

HPC = RL A + S1

S1

Levelling Staff

HPC = RL A + S1

S2

RL B = HPC - S2

14

A B

C

Some Terminology

RL A RL B RL C

S1

Level staff on A Back Sight (BS) reading is first reading

BS

Levelling

15

RL A RL B

A B

C

RL C

Level staff on A Back Sight (BS) reading is first reading

S2

Level staff on B Fore Sight (FS) reading is last reading

FS

Move instrument to new position

Levelling

16

Move instrument to new position

RL A RL B RL C

A B

C

Level staff stays on B

The instrument has changed its position about point B

Point B is known as a Change Point (CP)

CP

S3 BS

2nd instrument position starts with BS to B

Levelling

17

and finishes with

FS

FS to C

S4 S3 BS

RL A RL B RL C

A B

C

Levelling

18

RL A RL B

A B

C

RL C

BS FS

BS FS

RL A is known

HPC =

HPC

RL A + BS RL B = HPC - FS

(CP)

Now the RL B is known So we can repeat the process

HPC =

HPC

RL B + BS RL C = HPC - FS

Generally : HPC = Known RL + Back Sight

Unknown RL = HPC - Fore Sight

Levelling

19

RL A RL B

A B

C

RL C

BS FS

BS FS

RL A is known

HPC =

HPC

RL A + BS RL B = HPC - FS

RL B + BS RL C = HPC - FS HPC =

HPC

Generally : HPC = Known RL + Back Sight

Unknown RL = HPC - Fore Sight

(CP)

Now the RL B is known So we can repeat the process

Levelling

20

PLANE AND COLLIMATION METHOD

This method is simple and easy

Reduction of levels is easy

Visualization is not necessary regarding the nature of the ground

There is no check for intermediate sight readings

This method is generally used where more number of readings can be taken with less number of change points for constructional work and profile levelling

To check:

∑ BS - ∑ FS = Last RL – First RL

21

PLANE AND COLLIMATION METHOD Determine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is 136.440 gives the Height of Plane and Collimation method and applies the check

22

Station BS IS FS HLC RL Remarks

1

2

3

4

5

0.585

0.350

1.010

1.735

3.295

3.775

137.025

133.600

136.440

136.015

135.290

133.730

133.250

BM A RL=136.440

CPI

6

7

8

9

10

1.735

1.300

1.795

2.575

3.375

3.895

131.440

132.300

131.805

131.025

130.225

129.705

CP 2

11

12

0.635

1.605

130.805

129.835

BM B RL=129.835

Sum of BS=2.670 Sum of FS = 9.275

2.690-9.275 = -6.605 129.835-136.440= -6.605

HLC = RL + BS

= 136.440 + 0.585 = 137.025

RL = HL – BS

Check

(Summation of BS)-(Summation of FS) = Last RL – First RL

2.670 – 9.275 = 129.835 – 136.440

-6.605 = -6.605

How Levelling is Conduct

23


24

Calculation checks

FS - BS = 1st RL - Last RL

IS + FS + (RLs except first)

= (each HPC x number of applications)

Check Misclosure

Allowable Misclosure = 5 N mm. ("Rule of Thumb")

When calculations are checked and

if the misclosure is allowable

Distribute the misclosure

Simple check

Full check

25

RISE AND FALL METHOD This method is complicated and is not easy to carry out

Reduction of levels takes more time

Visualization is necessary regarding the nature of the ground

Complete check is there for all readings

This method is preferable for check levelling where number of change points is more

To check:

∑ BS - ∑ FS = ∑ Rise - ∑ Fall = Last RL – First RL

26

RISE AND FALL METHOD Determine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is 122.156 gives the Height of Rise and Fall method and applies the check

27

Station BS IS FS Rise Fall RL Remarks

1

2

3

4

1.536

0.974

1.124

1.768

2.072

2.700

0.536

0.794

0.932

122.156

121.620

120.826

119.894

BM A RL=122.156m

CP1

CP2

5

6

7

8

9

2.236

1.413

1.994

1.639

1.256

2.362

1.302

0.874

0.825

1.120

0.934

0.539

1.169

0.519

1.238 118.656

119.590

120.129

121.298

121.817

CP 3

CP 4

CP 5

CP 6

CP 7

10 1.468 0.212 121.605 BM B RL=121.605

Sum of BS=12.172 Sum of FS =

12.723

3.161 3.712

12.723-12.172=0.551 3.712-3.161=0.551 122.156-

121.605=0.551

R/F = BS - FS

= 1.536 – 2.072 = 0.536

RL = HL – IS

Check

(Summation of BS)-(Summation of FS) = Last RL – First RL

12.172 – 12.725 = 121.605 – 122.156

-0.551 = -0.551


28

COMPARISON

Plan and Collimation Method • Quicker

• Good for a lot of IFSs

Rise and Hall Method • More accurate

• More calculation

• Intermediate RLs are known

29

ACCURACY IN LEVELLING

For normal engineering works and site surveys

Allowance misclosure = ± 5 √ N mm

Where N = Number of instrument positions

OR

Allowance misclosure = ± 12 √ K mm

Where K = length of levelling circuit in KM

If actual misclosure > allowance misclosure, levelling should be repeated

If actual misclosure < allowance misclosure, misclosure should be equally distributed between the instrument positions

30

CORRECTION IN LEVELLING

Correction = (Misclosure / No. of Station) x n, n+1, n+2 and ………

For example:

Loop 1 = (0.117 / 3) x 1 = 0.039m

Loop 2 = (0.117 / 3) x 2 = 0.078m

Loop 3 = (0.117 / 3) x 3 = 0.117m

31

Example

32

Example

33

Example

34

Example

35

Example

36

Example

37

Example

38

Summary of work:

Check tripod is on stable ground or dig feet well in

Use pond bubble to set approximately vertical standing axis

Eliminate PARALLAX every time we sight the staff

check that the compensators are functioning every

time we sight the staff.

and

LEVELLING WORK

39

At every instrument set up - always start with a BS to a

point of known RL.

At every instrument set up - always finish with a FS.

Either the instrument moves or the staff moves

NEVER BOTH

ALWAYS CLOSE levelling to a point of KNOWN RL

LEVELLING WORK

40

TBM

9.09 m A.A.D.

TBM

10.00 m A.A.D.

Main Gate

Burnaby Building

Approximate North

Start at a TBM outside the main entrance of Burnaby Building and obtain the

RL values of three points before closing onto another TBM near the main gate.

Point 1

Ground level at entrance to structures laboratory

Top of door level at entrance to structures laboratory

41

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD

It is important to complete details at the top of booking forms or

on every page of field books.

42

TBM Level Posn. BS

Key

43

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group

Good Group

TBM 10.00m AAD 10.000 1.546

HPC = RL + BS HPC = 10.000 + 1.546 = 11.546

11.546

We now signal to the staff person to move to the next point.

As the next required point is too far away (it is also round a corner)

we will eventually need to move the instrument.

So, we must move the staff to a change point (CP), to allow us to move the

instrument to a better position later on.

44

TBM CP Level Posn. BS FS

Key

45

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD 1.546 10.000 11.546

C.P.

New staff

position

therefore

a new row.

Each row

represents

a staff

position.

1.562

RL = HPC - FS RL = 11.546 - 1.562 = 9.984

9.984

After we make a FS and we have calculated the new RL we are finished

with that instrument position.

Move the Instrument (about the CP) to a new position where we can see the CP

and also the next point we want the RL value of. 46

TBM CP Level Posn. BS FS IS Key

47

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418

Same staff

position as

last reading

therefore

the same row

HPC = RL + BS HPC = 9.984 + 1.418 = 11.402

11.402

48


This reading is not the first so it is not a BS It is not the last from this position (we can see the next points) so it is not a FS

So it is known as an INTERMEDIATE SIGHT (IS) 49

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390

RL = HPC - IS RL = 11.402 - 1.390 = 10.012

10.012

New staff

position

therefore

a new row

50


51

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012

GL Struct. Lab Door

New staff

position

therefore

a new row

1.281

RL = HPC - IS RL = 11.402 - 1.281 = 10.121

10.121

52


53

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….


Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012

GL Struct. Lab Door 1.281 10.121

Top Struct. Lab Door

New staff

position

therefore

a new row

Requires an inverted staff i.e turn the staff upside down

Read and then book the staff with a sign

-2.420

The negative sign will keep all the calculations correct

RL = HPC - IS RL = 11.402 - (-2.420) = 11.402 + 2.420 = 13.822

13.822

54


The last point required is the TBM. However it is too long a sight

So we need a CP. This will be the last sighting from this position

Therefore we need a FS

55

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1


Top Struct. Lab Door -2.420 13.822

TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP New staff

position

therefore

a new

row

1.321

RL = HPC - FS RL = 11.402 - 1.321 = 10.081

10.081

Last Reading -- FS -- Move the instrument 56


57

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 Same staff

position as

last reading

therefore

the same row

HPC = RL + BS HPC = 10.081 + 1.011 = 11.092

11.092

58


59

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD

New staff

position

therefore

a new row

2.009

RL = HPC - FS RL = 11.092 - 2.009 = 9.083

9.083

60


Before we look more fully at the results we will complete the

second half of the levelling exercise

61


Top of door level at entrance to structures laboratory

Ground level at entrance to structures laboratory

Point 2

62


63


64


65


66


67


68


69


70


71

Summary of Levelling Fieldwork

For levelling fieldwork, the following practice should be adhered to in order to improve the accuracy of the levelling works

Levelling should always start and finish at points of known RL so that misclosure can be detected

Where possible, all sight lengths should be below 60 metres

The staff must be held vertical by suitable use of a bracket bubble

BS lengths =/~ FS lengths

Reading should be booked immediately after they are observed. Important readings, particularly readings at change points, should be checked

The rise and fall method of reduction should be used if possible, especially for control works. This HPC is only a sample

72



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083

The RL value of 9.083m is our measured and calculated value.

It should be 9.09m.

This gives an actual misclosure of 9.083 - 9.09 = -0.007m

This actual misclosure may be because of calculation errors or field errors

Levelling Booking & Calculation

73



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083


3.975 4.892

LHS = 4.892 - 3.975 = 0.917

RHS = 10.000 - 9.083 = 0.917

Therefore LHS = RHS Therefore Calculations are OK

If it is due to calculation errors we MUST NOT continue.

Therefore the first thing we always do after reducing our field booking is:

Carry out Calculation Checks

FS - BS = 1st RL - Last RL Simple Calculation Check:

74



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083

This Simple Check does not check the calculations for RL values calculated from IS

NOT

CHECKED

NOT

CHECKED

NOT

CHECKED


3.975 4.892

Full Calculation Check:

IS + FS + (RLs except first)

= (each HPC x number of applications)

75

1.418

1.546



TBM 10.00m AAD 10.000 11.546

C.P. 1.562 9.984 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083

LHS = IS + FS + (RLs except first)

0.251 4.892 63.103

= 0.251 + 4.892 + 63.103 = 68.246

RHS = (each HPC x number of applications)

= (11.546x1 + 11.402x4 + 11.092x1) = (11.546 + 45.608 + 11.092) = 68.246

Therefore LHS = RHS

Therefore the calculations for all the RL values

are correct. 76



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083

Now we can look at the magnitude of the misclosure

We have already seen that the

Actual misclosure = 9.083 - 9.09 = -0.007m Is this acceptable ?

Rule of Thumb:

Allowable misclosure = ± 5 N mm Where N is the Number of Instrument Positions

which is the same as Number of BS readings

Therefore our Allowable misclosure = ± 5 3 mm = ± 8.66 say ± 9mm

Therefore Actual < Allowable Therefore our Fieldwork is OK 77

We have carried out the calculation checks and have an acceptable misclosure

The final stage is to apply a correction procedure to distribute the actual misclosure

We assume that we made a similar error every time we set up the instrument

There are 3 backsights, so we set up the instrument 3 times

We could divide 7 between 3 like this: 3 2 2

Or like this: 2 3 2

Let use choose the middle method. We will give 2mm to the 1st instrument position,

an extra 3mm to the 2nd position, and an extra 2mm to the 3rd position

The actual misclosure was -7mm, so we need to add 7mm in order to correct it

We can add these 7mm to our Reduced Levels in any way, but it is best to assume

that the 7mm error occurred gradually as a set of small errors,

rather than all in one go.

We cannot divide our 7mm misclosure evenly between 3 positions,

but we can do our best (we do not use fractions of a millimetre)

Or like this: 2 2 3

We must not correct the initial Reduced Level

We apply the same correction to all readings up to and including each foresight 78



TBM 10.00m AAD 1.546 10.000 11.546

C.P. 1.562 9.984 1.418 11.402

Point 1 1.390 10.012


CP 1.321 10.081 1.011 11.092

TBM 9.09m AAD 2.009 9.083

x

We cannot correct the given TBM value

10.000

2

5

5

5

5

7

Corrections are applied with a +ve or -ve sign depending on the sign of the misclosure

9.986

10.017

10.126

13.827

10.086

9.090

We MUST end up with the correct

final reduced level.

79

QUESTION 1) Figure below shows the levelling data obtained from a fieldwork. The benchmark value is 10.00m while the staff reading at BM, CP 1 and CP 2 are 1.546m, 1.418m and 1.011 respectively.

a) Book the Backsight (BS), Intermediate Sight (IS) and Foresight (FS) reading and calculate the Reduced Level at all points.

b) Calculate the Corrected Reduced Level.

BM=10.000m

Fall=0.016

BM

CP 1 CP 2 P 1 P 2

P 3

Rise=0.109 Rise=0.028

Fall=1.139

Rise=1.099 Fall=0.087

S 1

S 2

S 3

80

QUESTION

2) Prepare a set of three-wire levelling notes for the data given and make the page check as well as calculate the height of plane collimation (HPC) by using Height of Collimation Method. The elevation on Benchmark BM ‘X’=185.101m. Rod reading (in metres) are (U=upper wire reading; M=middle wire reading; and L=lower wire reading). The given reading are as follow:

Backsight (BS) on BM ‘X’: U=0.843, M=0.621, L=0.397

Foresight (FS) on Turning Point (TP 1): U=1.604, M=1.332, L=1.062

Backsight (BS) on Turning Point (TP 1): U=1.459, M=1.136, L=0.813

Foresight (FS) on BM ‘Y’: U=0.976, M=0.646, L=0.320

81

T H A N K YO U Q u e s t i o n & A n s w e r

82

engineering surveying (221 be) levelling-technical

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