ENGINEERING SURVEYING (221 BE) Levelling-Technical Sr Tan Liat Choon Email: [email protected] Mobile: 016-4975551
ENGINEERING SURVEYING (221 BE)
Levelling-TechnicalSr Tan Liat Choon
Email: [email protected]: 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