theodolite notes

Surveying II 10CV44

Problems

1) Following readings were taken from a tachometer felted with an analytic lens find the
horizontal distance AB and RL of pt.B

Station Staff At Vertical angle Hair readings Remarks
A BM -6000‟‟ 1100, 1.153, 2063 RLof BM = 9760

B +5000‟‟ 0.980, 1085, 1190

We know that
RLB = HIA +V – r
HIA = BM + rBM + VBM
We have
S = 2.063 – 1.100 = 0.963
θ = 6000‟
K = 100.00

VBM = 10.01m
HIA = 987.163m

ii) Calculation of V at ‘B’
we have

S = 1.190-0.980=0.21
θ = 8000‟
K = 100.00
V = KS sin2θ / 2
V = 2.89m
RLB = HIA + V – r

= 988.972m

Dept of Civil Engineering, SJBIT Page 51

Surveying II 10CV44

To calculate horizontal distance AB i.e D S = 1.190 – 0.980
D = KScos2 θ S = 0.21
D = 100.00(0.21) cos2(80.00) θ = 800‟‟
D = 20.593 = 20.6

2) The following observations were taken with the tacheometer at station „P‟ to staff at Q
held normal to line of sights

Staff readings are 1.450, 1.920, 2.380 and vertical angle is – 10030‟ RL of P is 201.170 and
height of collominaion axis at „P‟ is 1.315m determine the horizontal distance between P
& Q and RL of Q if k = 100.00 and C is o

Data : S =2.380-1.456=0.93 Page 52
r = 1.920
θ = -10030‟
K = 100.00
C= 0.0
RL of P = 201.170
H.Ip = 1.315m
Staff held normal
WKT:
D = KS cos θ - rsin θ
V = KS sin θ
Solution:
We know that
D = Horizontal distance for staff held normal condition
= (KS+C) cos θ -r sin θ

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

Here K = 100.00
S = 0.93
R = 1.920
θ = 10030‟

D = KScos θ - rsin θ
D = 91.093m
D = 91.1m
HIP = RLP + height at P

= 202.485m

3) A Tachometer was fitted with an analytic lens and was step at a station „D‟ with the
following observations.

Station Bearing Staff ready Vertical angle
A 340‟30‟ 0.800,1.855,2.910 +60 30‟
B 70‟30‟ 0.660,2.200,3.740 -40 20‟

Calculate the distance AB and gradient from pt. A to B

Solution: The horizontal distance A +B. Horizontal distance is obtained by applying cosine rule
is the

DA = D = KS cos2 θ Page 53
Here K = 100 S = 2.910-0.800 = 2.11
θ = 60 30‟

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

DA = 208.296m
DB = 306.241m
Gradient between A & B = RLA = HIA + V – r
V = 23.73m
r = 1.855,
RLA = 121.875m

4) A theodolite was set up in between 2 towers „x‟ & „y‟ the distance of the theodolite station
from x is 10m & from y is 120m observations were taken from theodolite to the top of
lowers X & Y as 33026‟ & 30050‟ resp. the RL of turnout axis of theodolite was
139.675m. Determine the RL of top of towers X & Y.

Solution:

RL of top of tower „X‟ = HIP +hx

= 139.675+hx
hx, D tan θx = 60tan 33026‟

= 39.612

RL of top of tower X = 139.675+39.613
RL of top of tower X = 179.288m

RL of top of tower „X‟ = HIP +hy
Hy, D tan θy = 120tan 30050‟

RL top tower „Y‟ = 211.304m

5) Determine the gradient from pt p to another pt. Q from the following observations with a
tachometer fitted with anal lactic lens the constant of inst was 100.00

Dept of Civil Engineering, SJBIT Page 54

Surveying II 10CV44

Inst At R R
Staff At held normal
WCB P Q
Vertical angle 1300 2200
Staff Read -10032‟ +506‟

1.255 1.300

1.810 2.120

2.365 2.940

6) In a 2 peg test on a dumpy level following readings were taken determine the correct staff
reading on B when the LOS is horizontal at A Also find the collimation error.

Inst at Reading on Remarks

A B P is exactly midway b/w A+B
P 1.555 1.250 Distance b/w A+B = 80000mt
A 1.325 1.010

Solution:
To find
Readies At B,
Collimation = ?

Now, the exact difference in elevation b/w pt.A & pt.B = 1.555-1.250
= 0.305m

But the apparent (actual) difference in elevation between A & B

Here
0.315-0.305=0.010

As the apparent difference in RL is more than the true difference in RL we are taking lesser
reading at B (LOS is inclined downwards) true staff

The true staff reading at B

Dept of Civil Engineering, SJBIT Page 55

Surveying II 10CV44

1.010+0.010 = 1.020

(Reading on B + diff of RL)
Reading at „B‟ is 1.020 (is the correct value)

7) To determine the distance b/w the two stations A & B a tachometer was set up at a point
on line AB and the following observations were made.
i. Staff head at A
Staff readings: 2.225, 2.605, 2.985
Vertical angle : +8024‟
ii. Staff head at B
Staff readings: 1.640, 1.920, 2.200
Vertical angle : -10 10‟
Also find RL of B if RL of A is 315.675m taking k = 100, C = 0

Subtense bar method

Sub tense bar is a rectangular horizontal bar fixed to a tripod stand. The horizontal bar in
attached through a leveling head with foot screws. This helps in keeping the bar truly horizontal
and parallel to the ground.

The horizontal has is attached with 2 targets at the either edge, the distance between the
targets in called „S‟ S = 3m in a standard Bubaline bar.

Sub- tense bar also has a device at the centre which helps in keeping the bar perpendicular to
the line of sight.

Dept of Civil Engineering, SJBIT Page 56

Surveying II 10CV44

Principle of sub-tense bar:

1. While finding the distance the horizontal angle between the two ends in determined as the
distance between targets in taken as „S‟ hence the angle between the midpoint & one end
is and the corresponding distance is S/2

In AB‟C‟

AB‟-distance between A & B‟ = D

B‟C‟ = s/2

B‟AC‟ = β / 2

tan β/2 = s/2

Hence, D = (s/2) / tan (β/2) Page 57
β in radians
D=s/β

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

β in seconds, D = 206265 x s /β
Application of sub - tense bar:

1. Used to measure the base line distance
2. Used to locate control points in traverse survey.
3. To measure horizontal distances in river valley, marshy area rough tannin

Beamen stadia arc / stadia circle:

Beamen stadia are is a device which is fixed concentrically over the vertical uncle of a
theodolite. The stadia are has 2 scales H-scale and V-scale. The stadia are directly gives us the
vertical intercept „V‟ and the horizontal distance „D‟ for a particular vertical angle observe thus
this device reduces the work of calculating „D‟ and „V‟ through the formulae.

When the elevation and distance of an object is required vertical angle to the staff at the
pt. is observed simultaneously, the stadia are also rotates the reading against the version „O‟ in
the H-scale and V-scale will give the horizontal distance and the vertical intercept „V‟.

Dept of Civil Engineering, SJBIT Page 58

Surveying II 10CV44

Unit 5 - CURVE SETTING (Simple curves)

Contents
1. Introduction
2. Classification of Curves

Simple curves
Compound curves
Reverse curve

3. Parts of a Circle
4. Designation of a Curve
5. Setting Out Curves

Linear method of setting out curves
Angular / Instrumental method

Dept of Civil Engineering, SJBIT Page 59

Surveying II 10CV44

Unit 5 – Curves

Introduction

Curves are a geometrical device and a part of a circle. These are used while changing the
direction of movement.

These curves help in providing smooth path in road ways & railway line construction.

Depending on the place in which the curves are provided they are classified as horizontal
curves( Curves provided in horizontal plane) and vertical curves (curves provided in vertical
plane)s

Classification of Curves:

Depending on the technical aspects curves have been classified into)

Basically depending on geometrical properties curves are classified as circular curves &
parabolic curves,

Circular curves are used in highway or road construction and parabolic curves are used in
railway line construction.

Circular curves are classified as

A) Simple curves: A type of circular curve which has only one are of a circle between the
tangents.

B) Compound curves: A kind of circular curves which has more than one are of a circle
having different radio with centers of the arc on one side of common tangent.

C) Reverse curve: It‟s a kind of circular curve which has more than one arc of a circle with
different on equal radii having the centers of are on either side of the common tangent.

Parts of a curve Page 60
R = Radius of simple circular curve
O = Centre of the simple circular curve
T1 V = Back tangent / rare tangent
T2V = Forward tangent
Dept of Civil Engineering, SJBIT

Surveying II 10CV44

T1 = Point of curve
T2 = Point of tangency
V = Point of intersection of tangents
I / φ = It is the intersection angle at v
EF = mid ordinate
T1 T2 = long chord / Major chord
VF= Vertex distance
F = Mid pt. of curve

V

Dept of Civil Engineering, SJBIT Page 61

Surveying II 10CV44

Definitions & Notations

1. Backward tangent: T1 V = (t or T) This is the first tangent for a curve and this is before
the starting of the curve.

2. Forward tangent: T2 V = (t or T) This is the last tangent for a curve which starts after the
curve
T = R tan Δ/2

3. Long/ major chard: T1 T2 = (L)
This is the line joining point of curve & pt. of tangency. This is the longest chard in a

curve.
L = 2R sin Δ/2
4. Mid ordinate – Mp (O0) This is the line joining centre of curve to mid point to the long
chord.
O0 = R (1 - cos Δ/2)
5. Length of curve : T1PT2 = L

This is the circular distance of the curve along the are from : T1 &T2,
We have
L = πRΔ / 180

It is the length of curve and is the subtended angle, if complete circle is considered
then, (circumfered) and the included angle at the centre is 360‟

6. Vertex distance/external distance : PV= (V)

This is the distance between centre of the curve to intersection angle
VP = R (sec Δ/2 – 1)

7. Point of curve T1 (P.C):
This is the point where the curve starts from the straight stretch.

8. Point of tangency: T2 (P.T):
This is the point where the curve ends and changes to straight stretch

Dept of Civil Engineering, SJBIT Page 62

Surveying II 10CV44

Designation of a Curve

The designation depends on the sharpness of the curve. This is called the degree of curve.
Degree of curve is expressed in terms of either

a. Radius : (Adopted is Britain)
b. Degree of curvature : (Adopted in India, USA, Germany)
To express the degree of curvature there are 2 definitions
a. Arc definition: According to this degree of curve is the central angle subtended at the
centre by an arc of 30m length.

D0 = 1718.87 / R (degree)
b. Chord definition: According to this degree of curve is the angle subtended at the centre

by a chord of 30M or 10M length.
R = 1719 / D

Setting Out Curves:
Simple circular curves can be set in two ways.

a. Linear method: Here, only chain and tape is used as a major device for setting out
curves. The method is adopted for small curves and in roadway construction.

b. Angular / Instrumental method: Here theodolite is used as a major device along
with tape and chain the method is adopted in setting out large curves, in railway
line construction and where accuracy is preferred.

Linear method of setting out curves:
Following are the different methods through which simple circular curve is set in the field.

a. Offset from major chord:

Calculation:

O0 = R - √(R2 – (L/2)2)
O1 = √(R2 – x12) - √(R2 – (L/2)2)

Dept of Civil Engineering, SJBIT Page 63

Surveying II 10CV44

In this method offsets are erected or constructed at regular internals from major chard on either
side of the midpoint

These offsets are called ordinate heights and the distance at which they are constructed i.e x1,
x2…………. are called ordinate intervals

Procedure:

1. Tangent points T1, T2 are identified and joining which, major chord is obtained

2. The point of the chord „H‟ is identified and the chord length on either side is divided
into no. of equal parts like MM1, H1H2……..

3. At each point the corresponding ordinate heights like 01, 02………. are constructed
on perpendicular offset.

4. Joining the tip of each ordinate we get the curvature of the curve

By successive bisection of chord:

Dept of Civil Engineering, SJBIT Page 64

Surveying II 10CV44

In this method the major chord is bisected successively to get the midpoints of small curves.
These midpoints when jointed will give the curvature of simple circular curve.

Procedure:

1. After getting T1, T2 they are connected to get the major chord.

2. The midpoint is identified and midordinate is erected to get P.

3. T1 P is taken as an independent curve for which T1 P is the major chord and the midpoint
M1 is identified.

4. At M1 the midordinate O1 further curve is constructed

5. With this T1 P1 and P1 P will now be 2 independent curves for which we set out
midordinates to get curve points P2 P3 .

6. This process can be continued to get number of midodirate points joining which, we get
smooth curve.

Calculations
Midordinate MP = R (1 - cos Δ/2)
= R - √(R2 – (L/2)2)

Ordinate M1P1 = R(1 – cos Δ/4) = R - √(R2 – (L/4)2)

Offset from tangents:

Dept of Civil Engineering, SJBIT Page 65

Surveying II 10CV44

Radial offset method

i. Radial offset method

The method is adopted when the curve is small in radius and for less important works.

Procedure:
I. The pt of intersection „V‟ point of curve PC ie (T1) The point tangency P.T i.e (T2) are
identified in the field.
II. The tangent distance T1V is divided into no. of equal parts At PC, r to T1V is
established to a length equal to radius of the curve
III. At each of the offset interval points like A, B, C……. ordinates, 01, 02, 03… are
established so that 01 i.e AA1 is in line with the radius of the curve.
IV. The line AO, BO are ranged ant and the corresponding ordinate heights like 01, 02…..
are measured from A,B respectively.

As a check, the last ordinate height should be equal to the apex distance or external distance

O1 = √(R2 + x12) – R

On = √(R2 + xn2) – R
Calculation: Let A, B…… be the offset interval points at a distance of x1 x2…. From T1 01, 02
are the ordinate heights at these points.

Perpendicular offset method:

Dept of Civil Engineering, SJBIT Page 66

Surveying II 10CV44

This method is adopted for small circles. This is also adopted when accuracy is preferred.

Procedure:
1. Point of intersection „V‟ and „PC‟ and „PT‟ are identified in the field.

2. The tangent length T1V is divided into no. of equal parts
3. At each of these points like A1, B1,……… perpendicular offsets are constructed of length

01, 02 etc respectively.

4. The end of these offset points like A,B e.t.c. are the points on the curve. Joining which,
we get smooth curve.

Calculation:
Let A1, B1 be the pts along T1V and regular internal. They are at x1, x2……. Distances
respectively from T1.
01, 02……. Are perpendicular offsets at A1, B1. Etc respectively.
On = R - √(R2 – xn2)

Offsets from chord produced (Deflection distance method)

Dept of Civil Engineering, SJBIT Page 67

Surveying II 10CV44

This method is an important one which is used to set out large curves and cures on high
ways or road constructions.

Procedure:

1. The total curve length is divided into no. of equal parts such that joining these parts we
get sub chords and the sub chord length will be equal to the corresponding curve length
ie., T1A = T1A

2. The length of 1st chord C1 is measured along tangent to get A1 At A1 an arc of length o1
is drawn to cut an arc of radius C1 drawn from T1 this will give point A on the curve

3. The chord T1A is produced to B1 so that A B1 = C2 At B1 and ordinate of length O2 is cut
through arc to cut an arc of radius C2 from pt A this gives pt B on the curve.

4. Step 3 is repeated till we reach T2 by getting points on the curve

Data and Calculation: Page 68
T1v = Rare tangent/ Backward tangent
T1A = T1A = C1 = First sub chord
A B1= AB = C2 = Normal chord
O1 = A1a = First ordinate = T1A* 8
O2 = B1b = Second ordinate
= On-1
VT1A = δ = AA1 T1= B1A B2
T1OA = 2 δ
Let the first sub chord T1A Make an angle with the back tangent (race tangent)
Angle V T1A== T1A A1
Now the arc length A1A = 01 = T1A x δ
Here
T1A = T1O x 2 δ

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

T1OA=2A T1 A1
δ = C1 / 2R

O1 = C12 / 2R
δ' = C2 (C1 + C2) / 2R
On = Cn (Cn + Cn-1) / 2R

Problems:

1) Calculate the necessary data required to set out a simple circular curve when two tangents
intersect at an angle of 1320 for a radius of curve – 48m by the method of (a)
Perpendicular offset from tangent (b) Deflection distance method (c) Offset from the long
chord.

Solution:
Data : Intersection angle = 1320

Radius of curve = 48m
Deflection angle = 180 – 132 = 480

(i) Setting out curve by
a. Perpendicular offset from tangent length of first offset interval = 21.3 – 20 = 1.3
Length of subsequent offset interval = 2.5m
Ordinate calculation:

Offset distance Ordinate

Height

X1 O1 0.0176
X2 O2 0.1506
X3 O3 0.4156

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Surveying II 10CV44

b. Deflection distance method Page 70
Curve length = 40.21 mts
Length of first subchord = 40.21-39
= 1.21mts
No of normal chords of 3mts length
= 39/3 = 13
We have 01 = first ordinate
O1 = C12/2R = 0.015 m
O2 = C2 (C1 + C2) / 2R = 0.13m
O3 = O4 = 0.187m

a) Perpendicular offsets from long chord

Length of long chord = 39.04(L)
C/2= 19.52mts
Offset internal = 19.520 – 18

X1 = 1.52mts
Thus interval of subsequent offset points = 3mts

= 18/3 = 6points
Mid ordinate
O0 =

R - √(R2 – (L/2)2)

O0 = 4.14m
Offset distance

X1 = 1.52 Ordinate distance
X2 = 4.52 O1 = 4.12
O2 = 3.935

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

X3 = 7.52 O3 = 3.555
X4 = 10.52 O4 = 2.9813
X5 = 13.52 O5 = 2.2049
X6 = 16.52 O6 = 1.2159
X7 = 19.52 O7 = 0

2) Two straight lines intersecting at an angle of 1450 are to be connected by a circular curve
of radius 250m. If the chain age of intersection is 1000m calculate the necessary data to
set out a curve by the method of radial offsets from tangents. Take peg internal as 10m.

Data
: Xn angle = 1450
R – Radius = 250.00m
Ch. Of Xn = 1000.m

Solution:

Here we have
Δ = 180- 1450 = 350
Length of tangent = T . R tan Δ/2
T = 250tan 35/2 = 78.82m
Length of chord = Perpendicular = 2Rsin Δ/2

Perpendicular = 150.35m

Length of curve Perpendicular = 15.72m

Calculation of ordinates: Page 71
Chain age of T1
= 1000-length of tangent T1
=1000-78.82
=921.18m
Interval of 1st offset = 78.82 – 70.00

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

= 8.82
Interval of subsequent

We have Offset ht/length (m)

Offset 0.150
distance(m) 0.6969
X1 = 8.68 1.639
X2 =18.68 2.97
X3 = 28.68 4.69
X4 = 38.68 6.79
X5 = 48.68 9.26
X6 = 58.68 12.08 = 1213
X7 = 68.68
X8 = 78.68

Check

=R(sec Δ/2 -1)

=250(sec17.5 – 1)

= 12.13 m

3) A circular curve of a radius of 90 + 20 chain is to be set between 2 tangents with the
deflection angle of 480 the peg internal is taken as 1 chain length of 20m calculate the

required data to set out the curve by the method of offsets from chords produced chain
age of pt of Xn 1600m.

Data:
90+20 chain, = 480

Or Radius of curve = 90+20lenks

P.I = chain of 20m

Chain age = 1600m

Here, Page 72
Dept of Civil Engineering, SJBIT

Surveying II 10CV44

90+20 chains means Page 73
90×20 chains + 20 × 0.20

R = 1800+4
R = 1804 mt
Tangent length T = R tan Δ/2 = 803.19m
Length of curve = perpendicular = 1511.31m
Length of chord = perpendicular = 2Rsin Δ/2 = 1467.51m
Calculation of ordinates:
Chainage of T2
= Chainage of pt of Xn – Length of tangent
= 1600 -803.19
=796.81m
Chainage of T2 = 796.81 + Perpendicular

= 796.81+ 1151.31
= 2308.12m
Chainage of 1st pt on the curve
Or length of 1st sub chord
C1 = 800.00 – 796.81
C1 = 3.19m
lllly length of last sub – chord
= 2308.12 – 2300
= 1.12m
Hence normal chords of 20m

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

= 75

4) Two straight lines are intersecting at a chinage of 40+20 links. Calculate the data required
to set out simple circular curve , having degree of curve as D = 20 by the method of
perpendicular offsets from tangents and deflection angle 620.

Data: D = 620

Chain age 40+30 links

We have

Degree of curve = 1718.8 / R = 85.9
Tangent length = R tan Δ/2

= 51.61m

Curve length Perpendicular = 92.95m

Calculation of ordinates

We know that perpendicular offset is given by
On = R - √(R2 – xn2)
Here,
Length of 1st offset interval = 51.61m – 50.00

= 1.61m

Assume peg interval of 5.0m

No. of offsets intervals of 5.0m

Offset distance Offset ht/length (On)
X1 = 1.61 O1 = 0.015
X2 = 6.61 O2 =0.254
X3 = 11.61 O3 =0.788
X4 = 16.61 O4 =1.6212
X5 = 21.61 O5 =2.7626
X6 = 26.61 O6 =4.225
X7 = 31.61 O7 =6.027
X8 = 36.61 O8 =8.192
X9 = 41.61 O9 =10.75
X10 = 46.61 O10 =13.745

Dept of Civil Engineering, SJBIT Page 74

Surveying II 10CV44

Unit 6 – CURVE SETTING (Compound and Reverse curves)

Contents
1. Compound curves

Elements

2. Design of compound curves
3. Setting out of compound curves

4. Reverse curve between two parallel straights
Equal radius
Unequal radius

Dept of Civil Engineering, SJBIT Page 75

Surveying II 10CV44

Unit 6 – CURVE SETTING (Compound and Reverse curves)

A) Compound curves: A kind of circular curves which has more than one are of a circle
having different radio with centers of the arc on one side of common tangent.

Elements of compound curve: Page 76
R 1= Radius of first curve
R2= Radius of second curve
O = Centre of the simple circular curve
T1 V = Back tangent / rare tangent
T2V = Forward tangent
T1 = Point of curve
T2 = Point of tangency
V = Point of intersection of tangents
I / φ = It is the intersection angle at v

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

EF = mid ordinate

T1 T2 = long chord / Major chord
VF= Vertex distance

F = Mid pt. of curve

Definitions & Notations
1.Backward tangent: T1 V = (t or T) This is the first tangent for a curve and this is before the
starting of the curve.

2.Forward tangent: T2 V = (t or T) This is the last tangent for a curve which starts after the
curve

T = R tan Δ/2
3.Long/ major chard: T1 T2 = (L)

This is the line joining point of curve & pt. of tangency. This is the longest chard in a
curve.
L = 2R sin Δ/2
4Co ordinate – Mp (O0) This is the line joining centre of curve to mid point to the long chord.
O0 = R (1 - cos Δ/2)
5 Length of curve : T1PT2 = L

This is the circular distance of the curve along the are from : T1 &T2,
We have
L = πRΔ / 180

It is the length of curve and is the subtended angle, if complete circle is considered
then, (circumfered) and the included angle at the centre is 360‟
6.Vertex distance/external distance : PV= (V)

This is the distance between centre of the curve to intersection angle
VP = R (sec Δ/2 – 1)

7. Point of curve T1 (P.C):

Dept of Civil Engineering, SJBIT Page 77

Surveying II 10CV44

This is the point where the curve starts from the straight stretch.
8.Point of tangency: T2 (P.T):

This is the point where the curve ends and changes to straight stretch

Designation of a Curve

The designation depends on the sharpness of the curve. This is called the degree of curve.
Degree of curve is expressed in terms of either

c. Radius : (Adopted is Britain)
d. Degree of curvature : (Adopted in India, USA, Germany)
To express the degree of curvature there are 2 definitions
c. Arc definition: According to this degree of curve is the central angle subtended at the
centre by an arc of 30m length.

D0 = 1718.87 / R (degree)
d. Chord definition: According to this degree of curve is the angle subtended at the centre

by a chord of 30M or 10M length.
R = 1719 / D
Setting out of compound curves

The compound curve can be set by method of deflection angles. The first branch is set out by
setting the theodolite at PC and the second branch is set out by setting the the theodolite at the
point PCC.

After having known any four parts calculate the rest of three parts by the formulae.

Knowing Ts and Tl Locate points T1 and T2 by linear measurements from the of intersection.

Calculate the length of curves ls and ll .

Calculate the chainage of T1, D and T2 as usual.
For the first curve, calculate the tangential angles etc for setting out the curves by Rankine‟s
method.

Set the theodolite at T1 and set out the first branch of the curve as already explained.

Dept of Civil Engineering, SJBIT Page 78

Surveying II 10CV44

Ater having located the last point D (P C C) shift the theodolite to D and set it there. With the
vernier set to 360-D/2 reading, take a backsight on T1 and plunge the telescope. The line of sight
is thus oriented along T1D produced and if the theodolite is now swing through D!/2, the line of
sight will be directed along the common tangent d1d2. Thus the theodolite is correctly oriented at
d
Calculate the tangential angles for the second branch and set out the curve by observations from
d till T2 is reached

Reverse curve between two parallel straights (Equal radius and unequal
radius).

Reverse curve: It‟s a kind of circular curve which has more than one arc of a
circle with different on equal radii having the centers of are on either side of the
common tangent.

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`

They are used when either the two straights are parallel or their angle of

intersection is too small. They are used in hilly terrains, in railway sidings,

on highways when low speed is required. When r1 and r2 are the radius of
two circular arms total angle of deflection is Δ the central angle is Δ1 and Δ2.
δ1 and δ2 are the angle between the straights.

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R=d/(tanΔ1/2 + tanΔ2/2)

It is definetly an advantage to separate to curves by either a shorter length

Of straight or reversed spiral. The elements of a reverse are not directly
determinate unless some condition or dimension is specified as for example
equal radii R1 = R2 or equal central angle . Frequently a common or equal
radius is required for both parts of the curve in order to use largest possible.

Problem: Two tangents AB and BC intersect at B. Another line DE intersects
AB and BC at D and E such that angle ADE = 1500 and angle DEC = 1400.
The radius of the first curve is 200m and that of the second is 300 m.
calculate all the data necessary for setting out a compound curve if the
chainage of B is 1050 m.

θ1= 1800- 1500 = 300

θ2 = 1800- 1400 = 400

θ = 300+400=700

Length of the tangent for first arc = 200tan150=53.58m

Length of the tangent for second arc = 300tan200=109.19m

Common tangent length = 162.77m

Chainage of T1= 950-164.92=785.08m

Curve length = 104.72m

Chainage of T2= 785.08+104.72= 889.80m
Long curve length = 209.44m

Chainage of T3=1099.24m Page 81
Deflection angle of full chord = 1718.9*20/200=2051‟53”
Deflection angle of sub-chord= 1718.9*4.12/200 = 0040‟34”
Deflection angle of full chord for 30m= 1718.9*30/300=2051‟53”
Deflection angle of final sub-chord = 1718.9/300*29.44=2048‟41”

Dept of Civil Engineering, SJBIT

Surveying II 10CV44

Unit 7 - CURVE SETTING (Transition and Vertical curves)

Contents

1. Introduction
2. Setting out Cubic Parabola and Bernoulli‟s Lemniscates
3. Length of Transition Curve
4. Characteristics of Transition Curve
5. Vertical curves

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Unit 7 – Curve Setting

Introduction

Transition curves: A non - circular curve of varying radius introduced between a straight and a
circular curve for the purpose of giving easy changes of direction of route is called a Transition
Curve.

Characteristics of Transition Curve

1. In order to fit in the transition curve at the ends a circular imaginary curve of slightly
greater radius has to be shifted towards the center. The distance through which the
curve is shifted is known as shift (S) of the curve equal to L2/24R, where L is the
length of each transition curve and R is the desired circular curve

2. Length of the combined curve is equal to

= (R + S) tan Δ/2 + L/2

3. Spiral angle φ1 = L /2R radians
4. The central angle for the curve = Δ - 2φ1
5. Length of circular curve is equal to πR(Δ - 2φ1) / 1800
6. Length of combined curve = πR(Δ - 2φ1) / 1800 + 2L
7. Chainage of beginning of combined curve = chainage of intersection point – total

length for combined curve

8. Chainage of junction point of transition curve and circular curve = chainage of
tangent point + length of transition curve

9. Chainage of other junction point of the circular curve and the other transition curve is
equal to chainage of E + length of circular curve.

10. Chainage of the end point of the combined curve = chainage of T + length of
combined curve

11. The deflection angle for any point on the transition curve distant l from the beginning
of combined curve = L/6R radians

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Length of Transition Curve

1. Length may be assumed on the basis of experience and judgment.

2. Length may be such that super elevation is applied at a uniform rate 1 in 300 to 1
in 1200.

3. The length of the transition curve may be such that super elevation is applied at an
arbitrary time rate of “a” cm/sec. The super elevation attained = (L/v) x a = h.

4. The radial acceleration on the circular curve is L = v3/ CR, where v is the speed in
m/sec , C is the rate of change of radial acceleration in m/sec2 , R is the radius of
curve in meters.

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Setting out Cubic Parabola and Bernoulli’s Lemniscates

Bernoulli‟s Lemniscates is commonly used in road work where it is required to have the curve
transitional throughout. The super elevation continuously increases till the apex is reached. It is
used in preference to following reasons:

1. Radius of curvature decreases more gradually

2. Rate of increase of curvature diminishes towards the transition curve.

3. It corresponds to an autogenous curve of an auto mobile.

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Vertical curves

Vertical curves are introduced at changes of gradient to avoid impact and to maintain good
visibility. They are set out in a vertical plane to round off the angle and to obtain gradual change
of gradient. They are also called as summit curves if they have convexity upwards and valley
curves if they have concavity upwards.

Problems

1. Find the length of the vertical curve connecting two uniform gradients of +0.8% and –0.6%.
the rate of change of grad. being 0.1% per 30m.

Solution: Length of vertical curve = Algebraic difference of two grads / rate of change of grad
= [0.8 – (– 0.6) / 0.1] x 30

= 420m

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Unit 8 - AREAS AND VOLUMES

Contents

1. Measurement of Area
2. Computation of Area by Geometrical figures
3. Area from Offsets
4. Simpson's One-Third Rule
5. Area from Coordinates
6. By Latitude and Meridian Distance (M.D. Or Longitude)
7. By Latitudes and Double Meridian Distances (D.M.D or Double

Longitude)
8. By departure and Total Latitude
9. Area By Coordinate Squares
10.Planimeter
11.Subdivision of an Area
12.Measurements of Volumes
13.Prismoidal Correction (CP)
14.Curvature Correction (CC)
15.Volume from spot levels
16.Mass Diagram
17.Sources of Error

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Unit 8 – Areas and Volumes

Measurement of Area

Foremost among the reasons for making land surveys is for the determination of area.
Land is frequently bought and sold by the area measure. Building costs are often computed on
the basis of per square metre of are. The field work consists of a sries of linear and angular
measurements defining the outline of a piece of land, forming a closed polygon. The following
are some of the prevalent methods of area determination.

By field measurements These may be made by dividing the area into geometrical figures offsets
from base line, double meridian distances and coordinates.

By plan measurements These may be made by computations based on measurements scaled
from plan or by use of a planimeter.

Computation of Area by Geometrical figures

Some of the formulae used for determining the area (A) of various geometric figures are
as follows.

Geometric formulae for area

1. Triangle: A = base half of the perpendicular height

2. Parallelogram: A = base perpendicular height

3. Trapezoid: A = half of the sides perpendicular height

4. Trapezium: A = area as found by dividing the figure into two triangles

5. Regular polygon: A =
length of perimeter half of the perpendicular distancecentre of sides

Also, A = nL2 / 4 cot 2 180 / n

where n is the number of sides and L is the length of one side.

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Area from Offsets
Offsets are made from a base line to an irregular boundary. The irregular field is thus reduced to
a series of trapezoids by the right angled offsets drawn from points at regular intervals along the
base line.there are several formulae to calculate the area of the figures so formed.
Mid-ordinate Rule
The offsets h1, h2, ..., hn are measured at the mid-point of each division.

Area = average ordinate length of thebase
= (h1+h2+ ... +hn)L/n = (h1+h2+ ... +hn)nd/n
= (h1+h2+ ... +hn)d

where n = number of divisions
Average Ordinate Rule
The offsets O1,O2, ... ,On are measured at the end of each division and are spaced apart at equal
distances d.

Area = average ordinate length of base

= (O1+O2+ ... +On)L/(n+1)
= nd ΣO/(n+1)
Trapezoidal Rule
Area of first trapezoid = (O1+O2)d/2
Area of second trapezoid = (O2+O3)d/2
Area of last trapezoid = (On-1+On)d/2
Summing up, A = d [(O1+O2)/2 + (O2+O3)/2 + ... + (On-1+On)/2]

= d [(O1+On)/2 + O2 + O3 + ... + On-1 ]
Simpson's One-Third Rule
Assumption: The boundary between the extremities of three consecutive offsets is a parabolic
arc whose axis is parallel to the offsets. Refer to fig

Area of trapezoid ABCD = (O1+O2)2d/2

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The area of the parabolic segment formed by the chord and the curve is known to be two-thirds
of the area of the parallelogram bounded by the chord, a parallel tangent, and the extended end
offsets.

Area enclosed by segment CIDHC = 2 / 3 area of parallelogram CDEF

= 2/ 3 HI AB

= 2/3[ (O2-O1+O3)2d/2]
Area of first two divisions = (d/3)(O1+4O2+O3)
Similarly, area of next two divisions = (d/3)(O3+4O4+O5)
and area of last two divisions = (d/3)(On-2+4On-1+On)
Summing up,
Total area: A = (d/3) [(O1+On) + 4(O2+O4+ ... + On-1) + 2(O3 + O5 + ... + On-2 )]
The essential condition for the application of simpson‟s rule is the division of the required area
into even number of segments, i.e. The number of offsets should be odd.

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Area from Coordinates

A figure represented by a closed traverse may be so divided that its area can be computed from
the latitudes and departures of the lines,

Rule: Area is equal to one-half of the sum of products obtained by multiplying each Y-
coordinate by the difference between the adjacent X-coordinates. The X-coordinate must always
be taken in the same order around the traverse.

Area = ½ [ YA(XB-XD) + YB(XC-XA) +YC(XA-XC) ]
also, Area = ½ ∑ Nr (Er+1 – Er-1)

This formula is based on the summation of the areas of a series of trapezoids. Some of the other
prevalent methods for computing areas by coordinates are as follows.

By Latitude and Meridian Distance (M.D. Or Longitude)

The meridian passing through the most westerly station is termed as the 'reference meridian'. The
perpendicular distance of the mid-point of a line passing through this from the reference
meridian is reckoned as meridian distance or longitude of the line.

In the given fig. Longitude of BC = longitude of AB + ½ departure of AB + ½ departure of BC.

Rule The area within the lines of a closed traverse is the algebraic sum of the products of the
latitude of each line with its longitude.

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By Latitudes and Double Meridian Distances (D.M.D or Double Longitude)

The double longitude of either of the lines meeting at the most westerly station is the departure
of the line. Double longitude of any other line is the algebraic sum of the double longitude of the
preceding line, the departure of that line and its own departure.

Rule The algebraic sum of the products of the latitude of each line by its double longitude is
twice the required area.

By departure and Total Latitude

The area within the lines of a closed traverse survey is the algebraic sum of the products of the
total latitude of each station and half the algebraic sum of departures of the two lines meeting at a
station.

Area = [area of triangles] + [area of trapezoids]
= [d1l1/2 + d5l4/2 +d3l2/2 – d3l3/2] +[d2(l1+l2)/2 + d4(l3+l4)/2]

= l1(d1+d2)/2 + l2(d2+d3)/2 + l3(-d3+d4)/2 +l4(d4+d5)/2

Area By Coordinate Squares

The plan is marked off in squares of unit area. The complete unit squares are counted and the
partial units are estimated. To simplify the procedure, a tracing paper with squares marked to the
desired scale is placed over the plan and the number of squares and partial units are counted.
This method is very rapid and is advantageous if approximate results are required at the
preliminary stage.

Planimeter

It is a mechanical integrator used for the measurement of areas of figures, plotted to a scale.
There are tw types of planimeters: the Amsler polar planimeter and the rolling planimeter. If
areas that do not have straight line boundaries are drawn to some scale on a map or plan, their
values can be very easily found by the planimeter. Some of the other uses of planimeter are
measuring areas of cross-sections for highways and railways, and checking computed areas in
property surveys.

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The most common type of a planimeter is the polar planimeter. The area is computed by utilizing
the relationship between the tracing arm point, moved over the outline of the figure, and the
connected recording wheel which records the displacement.

A planimeter essentially consists of two bars hinged together. At the extreme end on one of the
bars, a weight is suspended over a 'needle point' or 'anchor point' which is used to anchor the bar
outside the area to be measured. The other bar known as the tracing arm, has a 'tracing point' at
its extreme end. The tracing point is moved as desired, about the needle point. At the other end of
the tracing arm there is a roller which rolls on the surface of the plan as the pointer is moved.
Thus, when in use the planimeter has three contact points on the surface of the plan, the anchor
point, the tracing point, and the roller circumference. A fixed vernier is attached to the roller
drum. A disk is also connected to the roller, by a work drive such that one revolution of the roller
turns the disk trough one part.

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In large planimeters the tracing arm is made adjustable. This arrangement has two distinct
advantages. Firstly, the arm can be adjusted to the unit area of the plan; and secondly, the
planimeter may be tested for any unit area, and if found incorrect , the error can be eliminated by
adjusting the arm. Consider a moving line AB, of length L as shown in the fig. the ends of which
move in a given loci. The ends A and B, of the line AB, are displaced to A1 and B2 respectively.
Let the normal displacement of end A be dz and the rotation of B1 as dӨ.

The area swept ABB2A1, (1)
dA' = Ldz + ½ L× L dӨ

or dA' = Ldz + ½ L2 dӨ

Now suppose a wheel F fixed upon AB, its plane being perpendicular to that line, so that in the
displacement of AB, the wheel rolls when the point F moves perpendicularly to AB. Let dα be

the angle through which the wheel turns upon its axis in passing from F to F'. If r is the radius of
the wheel, rdα is the length of the arc applied to the paper. This length is equal to dz + the arc
L'dӨ

rdα = dz + L'dӨ (2)

Eliminating dz from Eq.(1) (3)
dA' = rLdα + (L2/2 – LL')dӨ (4)
∫dA' = ∫rLdα + ∫(L2/2 – LL')dӨ

When the directing curve AA1 is exterior to the area X,
∫rLdα = rLα = Lz (where z = rα)

∫ dӨ = 0, since AB returns to its original position without having made a circuit about O.

Therefore integrating Eq.(4)

A' = A = Lz (5)

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When the directing curve AA1 is within the area X,
∫ dӨ = 2π, since AB returns to its original position after making a circuit about O.

A = ∫dA' + area of the circle described by OA
By integrating Eq. (4),

A' = Lz + 2π(L2/2 – LL')
A = A' + πR2
= Lz + π(L2 - 2LL' + R2)

= Lz + area of a circle of radius L2 2LL'+R2 (6)

where R = the length of the anchor arm from the hinge to the point.

The sign of 2LL' is negative if the wheel is between the tracing point and the pivot point,
otherwise it is positive.

If r is the radius and n the number of revolutions of the wheel, then
Lz = L 2π r n

= Mn
where M is known as the multiplier constant (= 2πrL)

If B is to travel along a circle of radius R, no rotation of the wheel will be caused and such a
circle is called zero circle. Its area forms a constant to be measured. The area of this zero circle
may be determined by tracing a closed figure, first with the anchor point outside the boundary of
the figure, and then with the anchor point inside. The difference between the two areas thus
obtained will be the area of the zero circle.
Area of zero circle, A = π( L2± 2LL'+R2 )

where L = length of the tracing arm,

L' = distance between the pivot and the wheel, and

R = length of the anchor arm.

Working

To measure an area with a planimeter, the following procedure may be adopted.

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1. The tracing arm is so adjusted as to give the results in the desired units.

2. The anchor point is suitably fixed outside the area to be measured. It should be in such a
position that the tracing point may reach all parts of the boundary.

3. A point is marked on the boundary and the tracing point is placed over it.

4. Observe the wheel, the counting disc and the vernier. The reading is recorded and is
known as initial reading.

5. The tracing point is guided clockwise along the boundary till it returns to the starting
point.

6. The disc and drum readings are noted. This is known as the final reading. The difference
of the final and initial reading gives the required plan area A. Hence,

A= M F.R. I.R.±10N+C (8)

where M = multiplying constant

N = Number of times the zero mark of the dial passes the fixed mark. It is positive if zero
of the dial passes the fixed index in the clockwise direction.

C = a constant marked on tracing arm. It is additive when the anchor point is inside the
figure, and is zero when anchor point is outside.

Planimeter Constant (M)

It is also known as multiplier constant. The value is generally given by the instrument maker.
When not given it can be calculated by,

M = 2πrL

where M is the planimeter constant, L the length of the tracing arm and r the radius of the roller
or wheel.

Adjustments of Planimeter

The planimeter should be examined before application with respect to the following
requirements:

1. The planimeter registering roller must rotate about the axis freely and without vibrations.
This can be done by adjusting the screws near the bearings.

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2. The plane of the registering roller rim must be perpendicular to the axis of the tracing
arm.

Accuracy

The area determined with a planimeter is more accurate when the area is greater and particulary
when the plotting scale is large. It can safely be considered that for an area of about 15cm2 on
paper, the planimeter provides an average precision of 1:400 when both are in good condition.

Subdivision of an Area

Subdividing an Area from a point on its Boundary
Problem The problem essentially consists of finding the coordinates of another point on the
boundary, so that the line joining the two points subdivides the given area as desired.

Let there be an area ABCDEF from which a definite area A1 is to be cut off from a point G on the
boundary. The following procedure may be followed.

1. Calculate the area of the figure ABCDEF from known coordinates.

2. By judgement find a station F so that the area bounded on one side of the line GF is
approximately close to the desired value.

3. Compute the length and bearing of GF and the area of the figure GFABG. Let it be A'

4. Length and bearing of GF is known. The bearing of FH is also known. Hence the angle
GFH can be computed. The actual line of subdivision GH can be found from the relation:
A1 – A' = ½ FH FGsin GFH

or FH = 2 A1 A' / FG sin GFH

once length of FH is known, the coordinates of H can be found.

Subdividing an area by a Line of Given Bearing

Problem The problem consists of finding the coordinates of two points on the boundary of a
given piece of land so that the line joining these points has the given bearing and subdivides the
area as desired.

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Sources of Error
Some souces of errors in computation of area are as follows.

1. Poor selection of offsets and intervals which do not fit properly within an irregular
boundary.

2. Using too large coordinate squares, making it difficult to estimate partial squares.
3. Adjustments of latitudes and departures not made in accordance with true conditions.
4. Poor selection of the origin, resulting in minus values of coordinates and double meridian

distances.
5. Poor setting of the planimeter and failure to check planimeter constant.
6. Inaccurate determination of the planimeter constants.

Measurements of Volumes

Methods of measurements of volume

The most common methods used are cross section method, spot level method, and contour line
method.

Measurements from cross-sections

This is the most commonly used method for computing volumes. The total volume is divided
into a series of solids by the planes of cross sections. The spacing of the cross sections depends
upon the character of the ground and the accuracy desired. When the ground is irregular, the
cross sections must be taken close together. The area of these cross sections can be found by
calculating from the field data or by a planimeter. The volume is computed by the average end
area formula or by the prismoidal formula.

Types of cross sections and areas
The various types of cross sections which may be encountered in a route are :
Level section: This is suitable for a flat terrain
Two-level or Three-level section: This is employed for ordinary ground conditions.
Five-level or multi-level section: This is employed for rough topography.
Transition and side-hill section: These occur in passing from cut to fill and on side-hill locations.

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Volume Formulae

The volume of earthwork may be calculated either by the end area formula or the prismoidal
formula. The prismoidal formula usually yields a volume less than that obtained by the end area
formula. An exception occurs when at one of the sections the centre height is large, with a
narrow base, and at the other consecutive section the centre height is small and with a large base.
The difference between the volume obtained by the two formulae is known as prismoidal
correction.

End Area Formula
This is also known as trapezoidal formula. The volume is assumed to be equal to the average of
the areas of the two ends multiplied by the distance between them.

Volume between the first two sections,
V1 = (A1 + A2)L/2

Volume between next two sections,
V2 = (A2 + A3)L/2
. . . and so on

Total volume, V = V1 + V2 + ... + Vn
or V = [(A1 + An)/2 + A2 + A3 + ... + An-1]L
where A1, A2, A3, ..., An are the end consecutive areas and L is the distance between them.
Prismoidal Formula
This is also known as simpson's rule for volumes. The meanings of the various symbols used in
this formula and its derivation are as given
L = Length of prismoid (the distance between two end areas)
A1, A2 = The two end areas A1B1C1D1A1 and A2B2C2D2A2 as shown in fig.
An = Mid-area EFGHE (the area of cross section midway between the end areas)
h = Perpendicular distance of EF from O.

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