iet-201119175041.pptx

Surveying & Levelling – S1C4
Tacheometric Surveying
Done by: Eng.S.Kartheepan (M.Sc, B.Eng, AMIESL,AMIIESL)
Department of Civil Engineering
IET, Katunayake

Introduction
Branch of surveying in which both horizontal & vertical distances between
stations are determined from instrumental observations (Agor, 1996).
Uniqueness!
No need of slope corrections, tension, etc… since no chaining involved. Rapid
& convenient
Good for rugged terrains with deep valleys, mountains, across water bodies…
But low accuracy…
Purpose;
Preparation of contour plans with details.
Widely used in preliminary site locating for railways, roads, canals,
reservoirs, etc…
Institute of Engineering Technology, Department of Civil Engineering
Special Instruction Programme – Semester - I
Surveying & Levelling [Tacheometry Surveying – Chapter . 02, Eng.S.Kartheepan]

Instruments;
1. Tacheometer :- Similar to a theodolite fitted with stadia hairs
to the eye piece. The stadia diaphragm consists of one stadia
hair above and the other at equal distance below the Horizontal
cross-hair. They are kept in the same vertical plane
Tacheometry cont…
as the H & V cross-hairs. Upper stadia
V
. hair
H. Hair / Middle stadia
Lower stadia
2. Stadia Rods :- A graduated rod. The levelling staves can also
be used.

The Stadia Method

Principle of the Stadia

Fixed hair method of Tacheometry:-
Here, the stadia hairs are kept at fixed interval and the intercept
on the stadia rod (stave) varies depending upon the distance
between the staff and the instrument station.
Tacheometry cont…
i2
i3
i1
d1 d2
d3

This is the most common method of
Tacheometry
Fixed hair method of Tacheometry cont…
The intercept is determined by subtracting the lower stadia
reading from the upper.
When the staff intercept is more than the length of staff, only
half of the intercept is used (between the middle hair/cross-hair
and upper/lower stadia hair, then have to multiply by 2).
Caution: staves must hold vertically to get an accurate
measurement.
 Horizontal and V
ertical angles are read to the middle hair.

Principle of Tacheometry
B
C
C’ B’
O’
O
2
AO  AO' 
1
cot  k
BC B'C ' 2
The value of the constant k entirely depends
on the magnitude of the Apex angle ().
In an isosceles triangles, the ratio of the perpendiculars from
the vertex on their bases and their bases is constant…
Proof:
A


Proof
B
C
C’ B’
O’
O


𝟐 
𝟐
A

And
f/i= d/S
d =(f/i)S
also
Since ab is equal to a’b’, by similar triangles
D = d + (f+ c)
D = (f/i)S+ C
D= K.S + C
Principle of Tacheometry

1.Distance and Heighting for Horizontal sights from the
Tacheometer …
1.1 Horizontal distance from the staff station …
Tacheometric Computations
Where, K & C are Tacheometric constants. K - multiplying constant & C -
additive constant.
From similar Δ ’sABF
& A
’B’F we can get;
CF

AB
OF A'B'
CF  f s
i
D  f  s  ( f  d)
i
D  Ks  C
u
v
i
a
c
b
A
B
C
F
f
O
A
’
B’
d
C s
D

Notations:
1. i - Stadia Interval
2. S – Staff intercept (distance between upper stadia and
lower stadia)
3. f – Focus/ Focal length of the lens
4. d – distance between the lens and the instrument
5. v - Image distance
6. u- Object distance
7. D – Horizontal distance between vertical axis of the
instrument to the staff
8. F – Focus point
9. C- Horizontal distance between instrument and focus point
C = f + d
Tacheometric Computations cont…

1.Distance and Heighting for Horizontal sights from the
Tacheometer cont …
1.2 Elevation of the staff station …
Here, the LoS is kept horizontal and the staff is held vertical.
From the cross-hair (middle stadia), the elevation of the staff
station can be obtained exactly as in the ordinary levelling.
The instrument height also must be measured accurately for
this…
Tacheometric Computations cont …
Elevation of the staff station =
(Elevation of the instrument
axis
- Central hair reading)
Inst. Axis = LoS
I h
110.5m
?m

Example:
The stadia readings with horizontal sides on a vertical
staff held in 50m from the tacheometer were 1.285m
(lower stadia) and 1.780m (upper stadia).
The focus length of the object of lens was 25cm. The distance between object
lens and vertical axis of the tacheometer (d) was 15cm. Calculate the stadia
interval (i)?
v
i
a
c
b
A
B
F
f = 25cm
O
A
’
B’
d=15cm
C
C s
u
D = 50m
Ui= 1.780m
Li = 1.285m

Data’s: f = 25cm, d = 15cm, D = 50m, Ui = 1.780m and Li = 1.285m
From the tacheometric equation, Horizontal distance
D = K(S) + C
𝐟
D = (
𝐢
) S + (f + d)
50m = (
𝟎.𝟐𝟓𝐦
𝐢
) (1.780m – 1.285m) + (0.25m + 0.15m)
i = 2.495mm
Answer :

2.Distance and Heighting for Inclined sights
from the Tacheometer

2.Distance and Heighting for Inclined sights from the
Tacheometer …
2.1 Horizontal distance from the staff station …
Staff is held vertical at E.
AB is not perpendicular to LoS OC.
MakeA’B’perpendicular to AB @C.
Tacheometric Computations cont…
In OCF;
OCF  900

BCB '    ACA '
A 'OC  
2
AA 'C  900
 
2
O
h
A
’
B
B’
C
A
V
L
D F
E



 is the V - Angle

2.Distance and Heighting for Inclined sights from
the Tacheometer …
2.1 Horizontal distance from the staff station cont…
Similarly from OCB '; BB 'C  900
 
2
is very small, by ignoring it angles AA'C & BB 'C can assume 900
2
Since

•From  ' s AA'C & BB 'C;
•A'B '  AB cos  S cos
• Slope length OC  ( f  A' B ')  ( f  d )  ( f  S cos)  ( f  d )
i i
• Horizontal length D  L cos
D  f S cos2
  ( f  d ) cos
i
D  K.S.cos2
  C.cos

2.Distance and Heighting for Inclined sights from the
Tacheometer …
2.2 Elevation of the staff station …
2
i
From OCF;
CF  V  Lsin
 ( f  S cos )  ( f  d )sin
 
 
 (
f
.S cos.sin )  ( f  d ).sin
i
 K.S.cos.sin  C.sin 
sin 2
 K.S.  C sin
O
h
A
’
B
B’
C
A
V
L
D F
E



 is the V angle

Reduced level Calculation for angle of elevation
RL of Q
RL of P
Datum Line or MSL or Zero Level
CQ
Surveying & Levelling [Tacheometry Surveying – Chapter . 02]

Reduced level Calculation for angle of depression
RL of P
CQ
V
h
RL of Q
Datum Line or MSL or Zero Level

Determination of the Tacheometric Constant
Procedure:
Select a line about 100m apart on a fairly level ground and mark
the intermediate points from the starting at constant intervals
(20m/30m). These distances must be measured accurately using a
tape.
s3 s4
s1
20 60 80
s2
40

Determination of the Tacheometric Constant cont…
Procedure cont:
Setup the Tacheometer on the starting point, center and level it.
Observe the staves at each intermediate stations, keeping the telescope
horizontally & obtain the intercepts (S1,S2,…).
Substitude the values to the tacheometric equation (D  Ps  Q) and obtain
the values for the constants.
Or obtain the values graphically as follows,
S
D k
C
D  ks C
Y  mX  C

Tacheometric Surveying - Field Procedure…
Two steps: Running the Traverse, Locating Details and Spot
Heights for the Contours
 Set the instrument over the 1st station, center & level it. Measure the instrument height
(from the top of the picket to the trunnion axis) with a tape.
 Sight the back sight (orient)
 Assume or determine the RL of the starting station by holding the staff over a nearby
BM.
Pick the details and Spot heights. (H angle, V angle to the middle hair, stadia readings).
 Take the foresight on the next traverse station. (H,V angles, stadia readings).
 Transfer the instrument and repeat the same process.

Errors in Tacheometric Surveying …
1. Instrumental
2. Due to manipulation and sighting
3. Natural errors
1. Instrumental Errors: Mainly include the followings
# Imperfect Adjustments
# Irregular graduation of the staves.
# Incorrect values of the Tacheometric constants.(It is
advisable to determine the constants before the observations)

Errors in Tacheometric Surveying cont…
2. Human Errors:
# Inaccurate Centering & Levelling of the
instrument.
# Non-verticality of the staves.(use a sprit
bubble)
# Inaccurate estimation of the stadia intercept.
3. Natural Errors:
# Refraction. (bending of LoS)
# Expansion of the staves.

Example:
1) A levelling staff is held vertically at distances of 100m and
300m from the axis of a tacheometer and the staff intercept for
horizontal sights are 0.99m and 3.00m respectively. Find the
constants of the instrument (K & C).
2) After that the above instrument is setup at station A and the
staff is held vertical at a point B. With the telescope inclined at
an angle of elevation of 10º00’ to the horizontal, the readings
on the staff are 2.670, 1.835, 1.000m. Calculate the R.L of B and
its horizontal distance from A. The height of Instrument (H.I)
is 1.42m and R.L of A is 350.5m

Answer (Part – 01)
Data’s : Horizontal distances – 100m and 300m
Staff intercept : 0.99m and 3.00m
D = K(S) + C
At 100m, 100 = K(0.99) + C
At 300m, 300 = K(3.00) + C
0.99m
100m 300m
3.00m
1
2
Equation 2 – 1,
K = 99.5
C = 1.5

Answer (Part – 02)
Data’s : U = 2.670m, L = 1.000m, M = 1.835m, hi = 1.42m,
  10º00’ & RL of A= 350.5m
Sin(10) = 28.67m
ν  K.S sin 2  C sin
2
 [99.5 × 1.670 × Sin (20)]/2 + 1.5
O
h
L = 1.000m
M= 1.835m
U = 2.670m
V
D F
B

= 10º00’

S = 2.670 – 1.000
 1.670m
D  K.S.cos2
  C.cos
 ×1.67× cos2
10 + 1.5 × cos 10
=
i 162.63m
RL of B = RL of A + hi + V – h
RL of B = 350.5 + 1.62 + 28.67 – 1.835
= 321.415m

Self Reducing Tacheometry (RDS Tacheometer)
 The RDS is one of the rare
theodolites Wild produced. It
was a successful attempt to
create a
tacheometer
self reading
that would not
simply give the slope distance
to the assistant holding the
vertical reference staff, but the
horizontally reduced distance
and the height difference.
 The abbreviation RDS stands
for 'Reduction Distance-meter
for a Vertical Staff

Metric type E- pattern staff
 levelling
GVLV)
staff (model
that has an
extension rod at the bottom
end. The staff is read using
the special curved cross-
hairs of the instrument.
 With the extension rod of
the levelling staff set at the
instrument height minus 1
metre (i.e. at an instrument
height of 1.34m the rod was
set at 0.34m)

 the vertical angle of the
telescope is adjusted until the
zero-line of the instrument (the
bottom one) runs over the 1
metre mark of the staff
 The uppermost line now gives
the horizontal distance to the
height-line (the middle
staff (41.3 metres), while the
one)
shows the factor to multiply its
reading with to get the height
difference (+0.1 x 21.7 = 2.17m)

Distance curve
Factor
Height curve
Zero curve

Principle of RDS Tacheometry surveying
 Initially North direction will be determined by Compass and
which is established in the field and then centering and
levelling should be done in RDS.
 The following readings are taken in this approach.
Height curve
Distance curve
Zero curve
Factor
Horizontal angle

Horizontal
Distance (H) =
Distance
Curve
Reading
Zero
Curve
Reading
X 100
reading
Height curve Zero curve
reading
x 100 x ( + Factor)
Vertical
Component =
(V)
Equations in RDS Tacheometry
 Staff height = Instrument height and zero curve reading at
wedge
 Staff height = Instrument height and zero curve reading
not at wedge

Level = Instrument + Vertical
Difference Height Component
– Staff
Height
+ ZeroCurve
Reading
Reduced Level
at Staff station =
Level
+ Difference
Reduced Level
at Instrument
station
Equations in RDS Tacheometry…….
 Staff height = Instrument height and zero curve reading at
wedge
 Staff height = Instrument height and zero curve reading
not at wedge

Example:
A RDS Tacheometer is setup in a location on the ground. After that compass
was set out on the instrument point and then north direction was established.
Metric type of E-Face staff has been kept at the TBM. RL of TBM is obtained as
100.00m. In this setup instrument height is set out to equal the staff height and
zero curve reading has been taken at wedge. The bearing (Horizontal angle) of
staff is 230° 15’ and finally the following recordings were taken in this setup.
 Instrument height – 1.55m
 Factor - +0.10
 Height Curve - 0.004
 Distance curve – 0.088
 Zero Curve – 0.000
From the above data’s, Calculate the followings
1. horizontal distance
2. Vertical component
3. Level difference
4. Calculation the Reduced Level of Instrument location.

RL of TBM is 100.00 m
Horizontal Distance (H) = ( Distance Curve Reading -
Zero Curve Reading) X 100
= (0.088 – 0.000) x 100
= 8.8 m
Vertical Component = ( Height curve reading - Zero curve
reading) x 100 x (+Factor)
= (0.004- 0) x 100 x ( +0.1)
= 0.004 m
(V)
Level Difference = Instrument Height +V
ertical Component–
( Staff Height + Zero Curve Reading )
= 1.55 +0.004 – (1.55 + 0.00)
= 0.004 m
Answer:

Reduced Level at Staff station = Reduced Level at Instrument
station + Level Difference
100.00 m = RL @ A + 0.004 m
RL @ A = 99.996 m
Answer:

Direct
Pacing
Odometer
T
aping
Indirect
Stadia
Substance bar
Range finder
EDM Tacheometry
Electro-
optical
Microwave
Distance
Measuring
Substance bar -Methods of Measuring Distances
Let’s discuss now????

Substance Bar Tacheometry
is
 Subtense bar
identified as horizontal
stave which is used for
measuring both the
horizontal as well as the
vertical distance in places
where chaining is
impossible in surveying.
 Subtense bar can be
used
locations
reached to measure
where some
can not be
the
horizontal distance.

1 m 1 m
Substance Bar

Substance Bar….

Substance Bar…..
 The width of subtense bar should be 2m or 3m length but
commonly 2m width of subtense bar is practiced in
Tacheometry surveying. In this practice staff intersect is
called as Base.
 When the base is horizontal or subtense bar is horizontal
that is known as horizontal base subtense method & angle
is measured with the horizontal circle of the theodolite.
 If the base is vertical that method is called as vertical base
subtense method.

It is a physical bar with sighting targets at both ends. The
distance, d, between the targets is known,
The bar is set up on a tripod at one end of a line, centered on
and perpendicular to it. At the other end is a transit or
theodolite,
Substance Bar…..

Substance Bar Tacheometry

01
02
α
α/2
α/2
o
hi
A (Right Hand Target)
C (Middle Target)
hs
B (Left Hand Target)
Horizontal Distance (D)
Substance Bar Tacheometry…

The schematic diagram of a subtense bar having centre at C fitted with
targets at A and B. Let the separation between targets be s. A theodolite is
set up at O. The bar is kept perpendicular to the line of sight OC by means
of a sighting device at the centre of the bar. The horizontal angle between
the two targets at the ends of the bar is measured, let it be q. The horizontal
distance.
Principle of Horizontal base Substance bar

Triangle AOC & triangle BOC are similar. So, 𝐭𝐚𝐧
𝑎
𝟐
=
𝐀𝐂
𝐎𝐂
𝐀𝐂
𝐭𝐚𝐧 𝑎
𝟐
𝐎𝐂 = =
𝐬
𝟐
𝐭𝐚𝐧 𝑎
𝟐
,
𝟏 𝐒
𝐃 =
𝟐 𝐭𝐚𝐧 𝑎
𝟐
𝟏
𝟐
𝐃 = 𝐒 𝐜𝐨𝐭
𝑎
𝟐
0 S (Staff Intercept)
A (RMT)
α/2
α/2
B (LMT)
H (Horizontal Distance)
C (MT)
Proof – Principle of Substance bar

Example
The subtense bar method is a method that surveyors use to determine a
small distance d between two points P and Q. The subtense bar with length
b is centered at Q and situated perpendicular to the line of sight between P
and Q. Angle θ is measured, then the distance d can be determined. Find d
with θ = 1°23′12″ and b = 2.0000 cm
From the figure, we have
Let b = 2. Convert θ to decimal degrees:
Answer

O
01
02
α/2
α/2
θ
A
C
B
hs
v
E
h
RLof 02
RLof
01
hi
D (Horizontal Distance)
Mean Sea Level (MSL = RL = 0.0 m)
α = Parallactic angle
α/2 = Mean subtense angle
θ = vertical angle or angle of
elevation
Measurement of Horizontal Distance & Elevation on
Sloping Ground by a Subtense Bar

In triangle AOC & BOC are right angle triangle, therefore horizontal
distance
𝟏
𝐎𝐂 = 𝐬 𝐜𝐨𝐭
Consider triangle OCE
𝐬𝐢𝐧 𝜃 = =
𝑎
𝟐 𝟐
𝐂𝐄 𝐯
𝟎𝐂 𝟎𝐂
The vertical distance (V) can be written as follow,
𝐯 = 𝐨𝐜 𝐬𝐢𝐧 𝜃
Horizontal distance (D) can be written as follow,
𝐜𝐨𝐬𝜃 = =
𝟎𝐄 𝐃
𝟎𝐂 𝟎𝐂
𝐃 = 𝐎𝐂 𝐜𝐨𝐬 𝜃
𝑆 𝛼
2 2
𝑣 = cot ⋅ sin 𝜃
𝑠 𝛼
2 2
𝐷 = cot ⋅ cos 𝜃
Sloping Ground by a Subtense Bar……
1
2

Difference in Level from point 01 to point 02 (h)
h= Height of instrument + Vertical Distance – Height of subtense bar
Where,
hi = Height of instrument
v = Vertical Distance
hs = Height of substance bar
• For a valley area, levels ‘h’
h = hi + v – hs
RL of point 02 = RL of point 01 + h
= RL of point 01 + hi + v – hs
h = hi - v – hs
RL of point 02 = RL of point 01 - h
= RL of point 01 + hi - v - hs
Sloping Ground by a Subtense Bar……
3
4

Example
In Substance bar practical, the bar is 2m long and the following
reading were obtained
Right Hand Target (RHT) – 104°28’00”
Middle Target (MT) - 104°49’22”
Left Hand Target (LHT) - 105°10’50”
Vertical Angle - 05°20’
The substance bar is setup 1.37m above the ground and find the
followings,
1. The horizontal distance between the theodolite and the
substance bar in meters?
2. The RL of the substance bar station if the theodolite was set
1.52m above the ground station and whose RL was 56.5m
above the Mean Sea Level

Answer
Mean Angle = (105°10’50”-104°28’00”)/2 = 00°21’25”
𝟐
𝐬
𝑎
𝟐
Horizontal Distance = 𝐃 = 𝐜𝐨𝐭 ⋅ 𝐜𝐨𝐬 𝜃
𝟐 ′
= 𝐜𝐨𝐭 𝟎𝟎°𝟐𝟏 𝟐𝟓" ⋅ 𝐜𝐨𝐬𝟎𝟓 °20’00”
𝟐
= 159.820m
𝐬 𝑎
𝟐 𝟐
V
ertical Distance = 𝐃 = 𝐜𝐨𝐭 ⋅ 𝐒𝐢𝐧 𝜃
𝟐
′
= 𝐜𝐨𝐭 𝟎𝟎°𝟐𝟏 𝟐𝟓" ⋅ 𝐒𝐢𝐧 𝟎𝟓 °20’00”
𝟐
= 14.920m
Difference in Level = Hi + V – Hs
= 1.52 + 14.920 + 1.37
= 15.07m
RL at Substance bar = RL at Theodalite + Difference in Level
= 56.50 +15.07
= 71.57m

Thank you ..

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