Engineering survey leveling report

1. Levelling
Prepared by: Razhan Sherwan
Lecture 4
Erbil Polytechnic University
Engineering Technical College
Highway Engineering Dept.
Engineering Survey -1

2. Type of Benchmarks
Benchmarks: A benchmark is a relatively permanent reference point,
the elevation of which is known (assumed or known). It is used as a
starting point for levelling or as a point upon which to close for a check.
The following are the different types of benchmarks used in surveying:
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(a) Permanent Benchmark: These are the benchmarks established by state
government agencies. They are established with reference to GTS benchmarks.
They are usually on the corner of plinth of public buildings.
(b) Arbitrary Benchmark: These are reference points whose R.L.s are arbitrarily
assumed. They are used in small works such bench mark may be assumed as 100
or 50 m
(c) Temporary Benchmark: They are the reference points established during the
levelling operations when there is a break in work, or at the end of day’s work the
value of reduced levels are marked on some permanent objects such as stones,
trees etc.

3. BM construction
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4. Benchmark construction
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5. Permanent adjustment of level
The following relationship between the lines are desirable
• The line of collimation should be parallel to the axis of the bubble
• The line of collimation should coincide with the axis of the
telescope
• The axis of the bubble should be perpendicular to the vertical
axis. That is, the bubble should remain in the central position for
all the directions of the telescope.
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6. Following procedure is adopted to make the line of
collimation parallel to the axis of the bubble tube.
Two peg test
Stage 1
On fairly level ground, two points A and B are marked a distance of Lm
apart. In soft ground, two pegs are used, on hard surfaces nails or paint
may be used. The level is set up midway between the points at C and
carefully levelled. A levelling staff is placed at A and B and staff
readings S1 (at B) and S2 (at A) are taken.
Test 1
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Permanent adjustment of level

7. The two readings are:
S1 = (S1‘ + x) and S2 = (S2‘ + x)
S1‘ and S2‘ are the staff readings that would have been obtained if the
line of collimation was horizontal, x is the error in each reading due to the
collimation error, the effect of which is to tilt the line of sight by angle α.
Since AC = CB, the error x in the readings S1 and S2 will be the same.
The difference
between readings S1 and S2 gives:
S1 - S2 = (S1‘ + x) – (S2‘ + x) = S1‘ - S2‘
This gives the true difference in height between A and B. This
demonstrates that if a collimation error is present in a level, the effect of
this cancels out when height differences are computed provided
readings are taken over equal sighting distances.
:. True Difference (ΔH1) = S1-S2
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8. Stage 2
The level is then moved so that it is L/10m from point B at D and readings
S3 and S4 are taken.
The difference between readings S3 and S4 gives the apparent difference
in height between A and B. If the level is in perfect adjustment then:
S1 – S2 = S3 – S4
However this is not always the case and that an error term (e) needs to
be estimates ΔH2 = S3-S4
Collimation Error (e) per Lm= (S3 – S4) – (S1 – S2)/L
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9. • If the results of these tests show that the collimation error is less
than 1mm per 20m (or some specified value). If the collimation error
is greater than this specified value then the
level has to be adjusted. This is normally done by the manufacturer or a
trained technician.
• (1) (ΔH FALSE− ΔH TRUE) = the amount of collimation error.
• (2) If ΔH FALSE > ΔH TRUE then the line of sight is inclined up
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10. Example
When checking a dumpy level, the following readings
were obtained in the two-peg test:
Level set up midway between two staff stations A and B 400ft apart; staff
readings on A 5.75ft and on B 4.31ft.
Level set up 40ft behind B and in line AB; staff reading on B 3.41ft and on
A 4.95ft.
Complete the calculation and state the amount of instrumental error.
Describe the necessary adjustments to the following types of level, making
use of the above readings in each case:
Find the Collimation error and check the staff readings.
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11. Solution:
Find Collimation Error:
Collimation error (e)= (4.95 – 3.41) - (5.75 – 4.31)=+0.1ft
+ sign means the line of sight is inclined up
Error per foot length = 0.1/400 = 2.5x10^-4 foot per foot length
Check Staff readings:
Staff reading at A = 4.95 -(2.5x10^-4 x440)=4.84ft
Staff reading at B = 3.41 – (2.5x10^-4 x40) = 3.40ft
:.Difference in level = 4.84-3.40=1.44ft = ΔH1
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12. Procedure of leveling
1. The instrument must be check before use!
2. The instrument and level must be stable settled-up
3. The bubble tube must be leveled before the reading
• Beware of sun exposure
4. The instrument must be set up approximately in the middle
between two staffs
• Prevents curvature effects
• Reduce collimation error
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13. 5- Readings must be taken 30-50 cm above the ground
• Surface refractions
6. Staff should be set up vertically
7. Be careful when crossing rivers (large water surfaces)
• Use “same-time” (mutual) observations
• Repeat it during different times of the day
Procedure of Leveling
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14. Levelling signs
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Plumb the staff Wave the Staff High Staff
Raise the staff All right Pick up the
instrument
Raise the target
Lower the target Clamp the target

15. Carrying and Setting up Level

16. 1) The height of collimation system (H.I. method)
In this system, the R.L. of height of collimation (H.I) is found out for
every set-up of the level and then the reduced levels of the points are
worked out with the respective plane of collimation as described below.
1) Determine the height of collimation for the first set up of the level by
adding B.S. to the R.L. of B.M.
2) Obtained the R.L. of the intermediate points and first change point by
subtracting the staff readings (I.S. and F.S. from the R.L. of height of
collimation (H.I).
R.L. of a point= height of collimation H.I .- I.S or F.S
Height of collimation(HI)= R.L. of B.M.+B.S.)
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Method of Booking Level Reading:

17. Note
• When the instrument is shifted and set up at new position a new
plane of collimation is determined by addition of B.S. to the R.L of
change point. Thus the levels from two set-ups of the instruments
can be correlated by means of B.S. and F.S. taken on C.P. or T.P
• Find out the R.L.s of the successive points and the second C.P. by
subtracting their staff readings from this height of collimation.
• Repeat the procedure until all the R.Ls are worked out.
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18. The height of collimation method
Station Reading Height of collimation
(H.I)
Reduced
Level (RL)
Remarks
B.S I.S F.S
1
2
3
4
Observation table:-
BM ID Weather
Observer Booker
Date Sheet No.
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19. Arithmetical check: The difference between the sum of the back
sights and the sum of the fore sights should be equal to the difference
between the last and first reduced levels.
ΣB.S - Σ F.S.= LAST R.L –FIRST R.L
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20. 2) The Rise and fall system
In this system, there is no need to determine height of collimation .The
difference of level between consecutive points are obtained as
described below.
1) Determine the difference in staff readings between the consecutive
point comparing each point after the first with that immediately
proceeding it.
2) Obtained the rise or fall from the difference of their staff reading
accordingly to the staff reading at the point is smaller or greater than
that of proceeding point. (+ is rise and – is fall)
3) Find out the reduced level of each point by adding the rise to or
subtracting fall from the R.L. of a proceeding point.
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Method of Booking Level Reading:

21. Rise and fall method
Observation table:-
Station Readings Rise Fall Reduce Level (R.L) Remark
B.S I.S F.S
1 + -
2 + -
3 + -
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22. Arithmetic check:- The difference between the sum of back sight
and the sum of fore sight= difference between the sum of rise and the
sum of fall = the difference between
the last R.L. and the first R.L.
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ΣB.S-ΣF.S =ΣRISE -ΣFALL =LAST RL- FIRST RL

23. Inverted sights:
Figure at the bottom shows inverted sights to the underside of a
structure.
When the B.M of staff station is above the line of collimation (or line of
sight) the staff is held inverted on the point and reading is taken this
reading being negative is entered in observation table with minus
sign,
• To avoid confusion, ‘Staff inverted’ should be
written in the remarks column against the entry of the reading.
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24. Example:
Staff readings were taken as below:
Staff held at (A) and reading (2.13) m, the noted staff moved
at (B) and reading (0.39) m noted level moved forward while
staff remained at (B) and reading (1.99) m noted staff then
taken to (C, D, E and F) reading (0.70, 2.12, 3.40, and 2.74)
m noted respectively. Point F was being last station of
levelling. Calculate the reduced level (RL) of the points B,
C, D, and F, known that point A is B.M. with R.L= 50.00 m.
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25. 1) Height of collimation
Staff
station
Staff reading
H.I R.L
Notes
(Remarks)
B.S I.S F.S
A A is B.M
B 1.99 0.39 C.P
C 0.70
D 2.12
E 3.40
F 2.74
Sum
Ckeck
2.13 50.00
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26. H.I (A) = RL (A) + B.S (A) = 50.00+ 2.13 = 52.13 m
RL (B) = H.I (A) – F.S (B) = 52.13 – 0.39 = 51.74 m
H.I (B) = RL (B) + B.S (B) = 51.74 + 1.99 = 53.73 m
RL (C) = 53.03 m, RL (D) = 51.61 m, RL (E) = 50.33 m, RL
(F) = 50.99 m
Check the table:
Sum B.S – Sum F.S = Last RL – First RL
In order to check the wholly table including the I.S and mid
RL, the following cumbersome check must be carried out:
Sum of all the RL – first RL= (sum of each HPC
multiplied by the number of IS or FS Taken from it) −
(sum of IS and FS)
257.7 = 52.13 x 1 + 53.73 x 4 – (6.22+3.13)
= 257.7 m
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27. 1) Height of collimation
Staff
station
Staff reading
H.I R.L
Notes
(Remarks)
B.S I.S F.S
A A is B.M
B 0.39
C 0.70
D 2.12
E 3.40
F
Sum
Ckeck
2.13 50.00
52.13
+
- 51.74 C.P
+ 53.73
-
-
-
-
53.03
51.61
50.33
50.99
Check? ? !
+
4.12
1.99
+ 2.74
3.13
-
0.99
-
0.99 Correct
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28. Station Staff reading Rise Fall RL Remark
B.S I.S F.S
A 2.13 50 B.M
B 1.99 0.39
C 0.7
D 2.12
E 3.4
F 2.74
SUM
Check
2) Rise and fall method:
51.74
53.03
51.61
50.33
50.99
0.99
0.99
0.99
-2.7
3.69
3.13
4.12
1.74
1.29
0.66
-1.42
-1.28
RL B=?? C.P
-
-
- -
-
-
-
+
+
+
-
Correct
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29. Rise or fall = B.S – F.S or I.S
= 2.13 – 0.39 = + 1.74 (Rise)
R.L (B) = R.L (A) – Rise (+) or fall (-)
= 50 + 1.74 = 51.74
Check the table:
Sum B.S – Sum F.S = Sum Rise – Sum Fall = Last RL – First RL
4.12 – 3.13 = 3.69 – 2.7 = 50.99 – 50
0.99 = 0.99 = 0.99
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30. Example:
The Figure below shows inverted sights at B, C and D to the underside
of a structure. It is obvious from the drawing that the levels of these
points are obtained by simply adding the staff readings to the HPC to
give B = 65.0, C = 63.0 and D = 65.0; E
is obtained in the usual
way and equals 59.5.
However, the problem of
inverted sights is completely
eliminated if one simply
treats them as negative
quantities.
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31. Station Staff reading HI RL Remark
B.S I.S F.S
A 2.0 BM
B -3.0 Inverted staff
C -1.0 Inverted staff
D -3.0 Inverted staff
E 2.5
SUM
Check
62 60
65
63
65
59.5
2 2.5
0.5 0.5 Correct
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32. You are never a loser till
you quit trying.
The End of The Lecture
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