ENGINEERING SURVEYING (221 BE)

Levelling-TechnicalSr Tan Liat Choon

Email: tanliatchoon@gmail.comMobile: 016-4975551

LevellingTwo 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

L / 2 L / 2

xS1

S1’

Line of Collimation

Horizontal Line

L

AB

3

S2

S2’

x

L / 2 L / 2

xS1

S1’

Line of Collimation

Horizontal Line

L

AB

= S1’ - S2’

The APPARENT height difference δhA

The TRUE height difference hTδ

= S1 - S2

S1 = S1’ + x and S2 = S2’ + x 4

Two Peg Test

L / 2 L / 2

L

xS1

S1’

Line of Collimation

Horizontal Line

AB

S2

S2’

x

δThe TRUE height difference hT = S1’ - S2’

= S1 - S2The APPARENT height difference δhA

S1 = S1’ + x and S2 = S2’ + x hA = (S1’ + x) - (S2’ + x )δ5

Two Peg Test

L / 2 L / 2

L

xS1

S1’

Line of Collimation

Horizontal Line

AB

S2

S2’

x

δThe TRUE height difference hT = S1’ - S2’

= S1 - S2The APPARENT height difference δhA

S1 = S1’ + x and S2 = S2’ + x hA = S1’ - S2’δ = δhT6

Two Peg Test

hA δ = δhT

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

AB

L / 10

Now move the instrument outside the “odd numbered” peg

8

Two Peg Test

S3

S3’

AB

L / 10

S4

S4’

= S3 - S4The APPARENT height difference δ hA

δBut the TRUE height difference hT We already know

9

Two Peg Test

S3

S3’

AB

L / 10

S4

S4’

= S3 - S4

= S1 - S2

then the instrument is OKIf NOT then the error is e =

The APPARENT height difference δ hA

δBut the TRUE height difference hT

δ δTherefore if hA = hT

(S1 - S2) - (S3 - S4) / L mm / m10

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 differencesand 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 two11

L / 2 L / 2

L

S1

AB

S2

S3

AB

L / 10

S4

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

12

DatumCould be our own Datum - Assumed Datum

- Arbitrary Datum

- Site DatumOr

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 aboveor 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

AB

Measured and CalculatedLevel of A ReducedLevel of A RL A (known)

ReducedLevel of BRL B(unknown)

the Plane of CollimationHeight of

DATUM

(HPC)

HPC = RL A + S1

S1

Levelling Staff

HPC = RL A + S1

S2

RL B = HPC - S2

14

A BC

Some Terminology

RL A RL BRL C

S1

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

BSLevelling

15

RL A RL B

A BC

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 BRL C

A BC

Level staff stays on B

The instrument has changed its position about point B

Point B is known as a Change Point (CP)

CP

S3BS

2nd instrument position starts with BS to B

Levelling

17

and finishes with

FS

FS to C

S4S3BS

RL A RL BRL C

A BC

Levelling

18

RL A RL B

A BC

RL CBS 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 SightUnknown RL = HPC - Fore Sight

Levelling

19

RL A RL B

A BC

RL CBS FS

BS FS

RL A is knownHPC =

HPC

RL A + BS RL B = HPC - FS

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

HPC

Generally : HPC = Known RL + Back SightUnknown 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

Plane and Collimation MethodDetermine 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 Remarks12345

0.585

0.350

1.0101.7353.295

3.775

137.025

133.600

136.440136.015135.290133.730133.250

BM A RL=136.440

CPI678910 1.735

1.3001.7952.5753.375

3.895 131.440

132.300131.805131.025130.225129.705 CP 2

1112

0.6351.605

130.805129.835 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 = 137.025RL = HL – BS

Check(Summation of BS)-(Summation of FS) = Last RL – First RL2.670 – 9.275 = 129.835 – 136.440-6.605 = -6.605

How Levelling is Conduct

How Levelling is Conduct

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

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

Rise and Fall MethodDetermine 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 Remarks1234

1.5360.974

1.1241.768

2.072

2.700

0.5360.7940.932

122.156121.620120.826119.894

BM A RL=122.156mCP1

CP2

56789

2.2361.4131.9941.6391.256

2.3621.3020.8740.8251.120

0.9340.5391.1690.519

1.238 118.656119.590120.129121.298121.817

CP 3CP 4CP 5CP 6CP 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 = 0.536RL = HL – IS

Check(Summation of BS)-(Summation of FS) = Last RL – First RL12.172 – 12.725 = 121.605 – 122.156-0.551 = -0.551

How Levelling is Conduct

Comparison

Plan and Collimation Method• Quicker• Good for a lot of IFSs

Rise and Hall Method• More accurate• More calculation• Intermediate RLs are known

Accuracy in Levelling

For normal engineering works and site surveysAllowance misclosure = ± 5 √ N mmWhere N = Number of instrument positions

OR

Allowance misclosure = ± 12 √ K mmWhere 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

Correction in Levelling

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

For example:Loop 1 = (0.117 / 3) x 1 = 0.039mLoop 2 = (0.117 / 3) x 2 = 0.078mLoop 3 = (0.117 / 3) x 3 = 0.117m

Example

Example

Example

Example

Example

Example

Example

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 everytime we sight the staff.

and

Levelling Work

At every instrument set up - always start with a BS to apoint of known RL.

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

Either the instrument moves or the staff movesNEVER BOTH

ALWAYS CLOSE levelling to a point of KNOWN RL

Levelling Work

TBM9.09 m A.A.D.

TBM10.00 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 laboratoryTop of door level at entrance to structures laboratory

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good 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.

TBM Level Posn. BSKey

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 GroupGood Group

TBM 10.00m AAD10.0001.546

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 theinstrument to a better position later on.

TBM CP Level Posn. BS FSKey

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good Group 1

TBM 10.00m AAD1.546 10.00011.546C.P.

New staff position therefore a new row.

Each rowrepresentsa staffposition.

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 CPand also the next point we want the RL value of.

TBM CP Level Posn. BS FS ISKey

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good Group 1

TBM 10.00m AAD1.546 10.00011.546C.P.1.562 9.9841.418

Same staff position as last reading therefore the same row

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

11.402

TBM CP Level Posn. BS FS ISKey

This reading is not the first so it is not a BSIt is not the last from this position (we can see the next points) so it is not a FSSo it is known as an INTERMEDIATE SIGHT (IS)

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good Group 1

TBM 10.00m AAD1.546 10.00011.546C.P.1.562 9.9841.418 11.402

Point 11.390

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

10.012

New staff position therefore a new row

TBM CP Level Posn. BS FS ISKey

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good Group 1

TBM 10.00m AAD1.546 10.00011.546C.P.1.562 9.9841.418 11.402Point 11.390 10.012

GL Struct. Lab DoorNew staff position therefore a new row

1.281

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

10.121

TBM CP Level Posn. BS FS ISKey

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

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

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

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1Good Group 1

TBM 10.00m AAD1.546 10.00011.546C.P.1.562 9.9841.418 11.402Point 11.390 10.012

GL Struct. Lab Door1.281 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

TBM CP Level Posn. BS FS ISKey

The last point required is the TBM. However it is too long a sightSo we need a CP. This will be the last sighting from this position

Therefore we need a FS

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

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

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

Building A 2

27/09/11 Group 1Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

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

TBM CP Level Posn. BS FS ISKey

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

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

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

Building A 2

27/09/11 Group 1Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 Same staff position as last reading therefore the same row

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

11.092

TBM CP Level Posn. BS FS ISKey

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

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

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

Building A 2

27/09/11 Group 1Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AADNew staff position therefore a new row

2.009

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

9.083

TBM CP Level Posn. BS FS ISKey

Before we look more fully at the results we will complete the second half of the levelling exercise

TBM CP Level Posn. BS FS ISKey

Top of door level at entrance to structures laboratory

Ground level at entrance to structures laboratory

Point 2

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

TBM CP Level Posn. BS FS ISKey

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

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 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

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

∑ FS - ∑ BS = 1st RL - Last RL

∑ ∑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 RLSimple Calculation Check:

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

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

NOTCHECKED

NOTCHECKED

NOTCHECKED

∑ FS - ∑ BS = 1st RL - Last RL

∑ ∑3.975 4.892

Full Calculation Check:

∑ IS + ∑ FS + ∑ (RLs except first) = ∑ (each HPC x number of applications)

1.418

1.546

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD10.00011.546

C.P.1.562 9.98411.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 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) = 68.246Therefore LHS = RHS

Therefore the calculations for all the RL values are correct.

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 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 readingsTherefore our Allowable misclosure = ± 5 √3 mm = ± 8.66 say ± 9mm

Therefore Actual < Allowable Therefore our Fieldwork is OK

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 assumethat 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

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

xWe cannot correct the given TBM value

10.000

2

5

55

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 correctfinal reduced level.

QUESTIONFigure 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 1CP 2 P 1 P 2P 3

Rise=0.109Rise=0.028

Fall=1.139

Rise=1.099 Fall=0.087

S 1

S 2

S 3

THANK YOUQuest ion & Answer