Curves.pdf

CURVES
Curves are usually employed in lines of
communication in order that the change of direction
at the intersection of the straight line shall be
gradual.
The lines connected by the curves are
tangential to it and are called as tangents or
straights.

NECESSITY OF CURVES:-
Straight route of road or track is always desirable, since it
provides economy in construction, transportation and
maintenance.
But when there is change in alignment or gradient of road or
track, then it becomes a need to provide curves under following
circumstances
1. Excessive cutting and filling can be prevented by providing
the change in alignment by curves.
2. The obstruction which came in the way of straight alignment
can be made easier by providing by pass with the help of
curves.
3. In the straight route gradient are made more comfortable
and easy providing diversions with help of curves.
4. In the straight route costly land comes in the way then it can
avoided by providing diversions with the help of curves.

TYPES OF CURVES:-
VERTIAL CURVES
HORIZONTAL CURVES
SIMPLE CURVE
COMPOUND CURVE
REVERSE CURVE
TRANSITION CURVE
SUMMIT CURVE
VALLEY CURVE
COMBINED CURVE

1. SIMPLE CURVE;-
A simple curve consist of a single arc
connecting two straights or tangents.
simple curve is normally represented by the length of its
radius or by the degree of curve.

2. COMPOUND CURVE;-
A compound curve consist of two
arcs of different radii curving in the same direction and
lying on the same side of their common tangent , their
centers being on the same side of the curve

3. REVERSE CURVE;-
A reverse curve is composed of
two arcs of equal or different radii bending or curving in
opposite direction with common tangent at their junction,
their centers being in opposite sides of the curve.

NOMENCLATURE OF CURVE
A curve is designated by the angle subtended by a
chord of a specified length or by the radius.
In England the usual method of designating the curve
is by its radius e.g. 400 m radius curve. Sometimes it is
designated by a number of degrees subtended at the
centre by a chord of 100 links e.g. 2º curve.
In India so far the standard chord was 100ft. With metric
conversion this may be changed to 30 m. However the
length of 20 m is more convenient.

Let ,
R = the radius of a curve in m
D = the degrees of the curves
MN = the chord 30 m long
P = its mid point

IN the ∆ OMP, OM=R; MP=1/2MN= 15 m. MOP= D/2
Then, ………..(1)
When D is small, sin (D/2) may be taken approximately equal
to D/2 in radians.
If MN =100 links (20m) and R is expressed in links degree of
the curve can be shown as
With R= 600m = 3000 links
A curve of 600 m radius is equivalent to 1.910 curve.

Notation for circular curve:-
1. The straight lines AB and
BC, which are connected by
the curve are called the
tangents or straights to the
curve.
2. The point B at which the
two tangent lines AB and BC
intersect is known as the
point of intersection (P.I.) or
the vertex (V).

3. If the curve deflects to the right of direction of progress of
survey (AB), it is called as right hand curve, if to the left , it is
called as left hand curve.
4. The tangent line AB is called the first tangent or rear tangent
(also called the back tangent)
5. The tangent line BC is called as the second tangent or
forward tangent.
6. The points (T1 and T2) at which the curve touches the
straights are called tangent point(T.P.). The beginning of the
curve (T1) Is called the point of curve.(P.C.) or the tangent
curve (T.C.). The end of the curve (T2) is known as the point
of tangency(P.T.) or the curve tangent(C.T.).
7. The ‫ے‬ABC between the tangent lines AB and BC is called
the angle of intersection (I). The ‫ے‬B'BC (i.e. the angle by
which the forward tangent deflects from the rear tangent) is
known as the deflection angle (ø) of the curve.

8. The distance from the point of intersection to the tangent point
is called the tangent distance or tangent length.
(BT1 and BT2).
9. The line T1T2 joining the two point (T1 and T2) is known as the
long chord.(L).
10.The arc T1FT2 is called the length of the curve.(l).
11.The mid point F of the arc T1FT2 is known as the apex or the
summit of the curve and lies on the bisector of the angle of
the intersection.
12. The distance BF from the point of the intersection to the apex
of the curve is called the apex distance of external distance.
13.The angle T1OT2 subtended at the centre of curve by the arc
T1FT2 is known as the central angle, and is equal to the
deflection angle.(ø)
14.The intercept EF on the line OB between the apex (F) of the
curve and the midpoint (E) of the long chord is called the
versed sine of the curve.

ELEMENTS OF THE CURVE;-
T1BT2 + B’BT2 = 1800 or I + Ø =1800 ……….…(2)
The angle T1OT2 = 1800 - I = Ø ……………(3)
Tangent length = BT1 = BT2 = OT1 tan(Ø/2) = R tan(Ø/2) … ..(4)
Length of the chord (L) = 2T1E = 2OT1 sin (Ø/2)=2R sin (Ø/2)…(5)
Length of the curve(l) = length of the arc T1FT2
= R X Ø (in radians) = ( RØ) / 1800 …………….(6)
If the curve be designated by the degrees of the curve(D),
Length of the curve =(30 Ø)/D (30 m chord) ………….(6a)
= (20 Ø)/D (20 m chord) ………....(6b)
Apex distance = BF = BO – OF = OT1 sec (Ø/2) – OF
….……(7)

Versed sine of the curve = EF = OF – OE=OF – OT1 COS(Ø/2)
……(8)
Methods of curve ranging:-
The methods of setting out curves may be divided in two classes
according to the instrument employed.
1. Linear or chain and tape method:-
Linear methods are those in
which the curve is set out with chain and tape only.
2. Angular or instrumental method:-
instrumental methods are those in
which theodolite with or without a chain is employed to set out
curve.

Peg interval :-
It is the usual practice to fix pegs at equal interval
on the curve as along the straight. The interval between the
peg is usually 20 to 30 m. strictly speaking this interval must be
measured as the arch intercept between them. However as it is
necessarily measured along the chord, the curve consist of
series of a chord rather than of arcs. In other words, the length
of the chord is assume to be equal to be that of the arc. In
order that the difference in length between the arc and chord
may be negligible, the length of the chord should not be more
than 1/20th of the radius of the curve.
The length of unit chord (peg interval) is,
therefore, 30m for flat curve, 20m for sharp curve, and 10 m or
less for very sharp curve. When the curve is of a small radius,
the peg interval are considered to be along the arc and the
length of the corresponding chords are calculated to locate the
pegs.

Location of tangent points :-
To locate the tangent point T1 and T2 proceed as fallows:-
1. Having fixed the direction of the tangents, produce them so as
to meet at the point B.
2. Set up the theodolite at the intersection point B and measure
the angle of intersection T1BT2. Find the deflection angle Ø
from the relation I + Ø =1800

3. Calculate the tangent length from the formula (4)
i.e. Tangent length = BT1 = BT2 = OT1 tan(Ø/2) = R tan(Ø/2)
4. Locate T1 by measuring the tangent length backward along the
rear tangent AB from the intersection point B.
5. Similarly, locate the T2 by measuring the same distance along
the forward tangent BC from B.

Chain and tape (or linear) Methods of setting out curves
1. By offsets or ordinates from the long chord

Let AB and BC = the tangents to the curve T1DT2
T1 and T2 = tangent points
T1T2 = the long chord of length L.
ED = O0 = the offsets at the midpoint of T1T2 (the versed sine)
PQ = Ox = the offsets at a distance x from E so that EP = x
OT1 = OT2 = OD = R = The radius of the curve.
The exact formula for the offset at any point on the long chord
may be deduced as follows:-
Draw QQ1 parallel to T1T2, meeting ED at Q1 join OQ cutting
TT1 in P1.
Now in the ∆OT1E, OT1 = R; T1E = (L/2) ;
OE = OD – ED = R – O0,
OT12 = T1E2 + OE2 or R2 = (L/2)2 + (R – O0)2
………………….(9)

Of the three quantities O0, L and R, two quantities L and R, or L
and O0 are usually known. The remaining unknown may be
calculated from the above formula.
From the ∆OQQ1, OQ2 = OQ12 + OQ12.
But QQ1= OE + EQ1 =OE + Ox = (R-O0) + Ox
QQ1 = x; OQ = R.
R2 = x2 + {( R - O0) +Ox } 2
Or Ox + (R – O0) = √ R2 – Ox2
Hence Ox = √ R2 – x2 − (R – O0) (exact) ………..(10 )
When the radius of curve is large as compared with the length of
the long chord, the offsets may be calculated from the
approximate formula, which may be deduced as fallows,
In this case PQ is very nearly equal to the radial ordinate QP1
Then QP1 X 2R = T1P X PT2
Now, T1P = x; T1T2 = L hence PT2 = L – x; QP1 = Ox.
(Approximate) ………..(11)

Procedure :-
To set out the curve ,
i. Divide the long chord into an even number of equal parts.
ii. Set out the offsets as calculated from formula no.10 at each
of the points of division, thus obtaining the required points on
the curve. Since the curve is symmetrical along ED, the
offsets for the right half of the curve will be same as those for
the left half.
If the offsets are calculated from formula (11), the long chord
should be divided into a convenient number of equal parts
and the calculated offsets set out at each points of division.
This method is usually adopted for setting out short curves.
e.g. curves for street kerbs.

2. By successive bisections of arcs:-
Let,T1andT2 be the tangent points. Join T1and T2 and bisect at
E.
Set out the offset ED ,

Thus determining the point D on the curve. Join T1D and
DT2 and bisect them at F and G respectively.
At F and G set out the offsets FH and GK each equal to
R [1 – cos(ø/4)], thus obtaining two more points H and K on
the curve, By repeating the process, set out as many points
as required.
3. By offsets from the tangents:-
In this method the offsets are set
out either radially or perpendicular to the tangents BA and BC
according as the centre (O) of the curve is accessible or
inaccessible.
a) By Radial offsets:-
Let T1 be the first tangent point
EE1 = Ox= the radial offset at E at a distance of x from T1
along the tangent AB.

Let, T1 be the first tangent point
EE1 = Ox= the radial offset at E at a distance of x from T1
along the tangent AB.
Now in the ∆OT1E, OT1 = R; T1E = x; OE = OE1+EE1 =R + Ox
Now, OE2 = OT12 + T1E2
(R + Ox)2 = R2 + x2. i.e. Ox = √R2 – x2 – R ……………..(12)
a) By Radial offsets:-

When the radius is large, the offsets may be calculated by the
approximate formula, which may be deduced as follows :
ET12 = EE1 × (2R + EE1) i.e. x2 = Ox × (2R – Ox)
Since Ox2 is very small as compared with 2R, it may be neglected.
Ox = ( x2 / 2R ) ……………….( 13 )
The formula may be obtained from the exact ones thus :
Expanding the factor = √R2 – x2 , we have
Neglecting the other terms except the first two, we get,

b) By offsets perpendicular to tangents:-
Let, EE1 be the perpendicular offset at a distance x measured
along the tangent AB from the tangent point T1 so that T1E =
x.

Through E1 draw E1E2 parallel to BT1, meeting OT at E2.
Then E1E2 = T1E = x; T1E2 = EE1 = Ox;
OE2 = OT1 – T1E2 = (R - Ox)
Now, from ∆OE1E2, OE12 = E1E22 + OE22
i.e. R2 = x2 ( R – Ox) or Ox = R - √ R2 – x2 exact ……(14)
From which, the corresponding approximate formula may be
obtained by expanding the factor √ R2 – x2 . Thus, we have
Neglecting higher powers of R2 we get
(app)……..( 14 )

Procedure:-
To set out the curve
i. Locate the tangent points T1 and T2, by measuring a distance
equal to the tangent length R tan( ø/2), backward along the
tangent BA from the point of intersection(B) and the same
distance forward along the tangent BC.
ii. Measure equal distances, say, 20 or 30m along the tangent
T1B from T1.
iii. Set out the offsets calculated from formula(14) or (13)
perpendicular to T1B at each distance, thus obtaining the
required points in the curve.
iv. Continue the process until the apex of the curve is reached.
v. Set out the remaining half of the curve from the second
tangent.

4) By offsets from chord produced :-

Let,
AB = the rear tangent ; T1 = the first tangent point.
E,F,G etc. = the successive points on the curve.
T1E1 = T1E = the first chord of length b1.
EF,FG, etc = the successive chords of length b2 ,b3 etc each
being equal to the length of unit chord.
BT1E = α in radian = the angle between the tangent BT1 and the
first chord TE1.
E1E = O1 = the offset from the tangent BT1.
E2F = O2 = the offset from the preceding chord T1E produced.
Draw the tangent DEF1 at E1 meeting the rear tangent at D.
Produce T1E to E2 so that EE2 = b2.
Let, F1 be the point of intersection of DEF1 and E2F. The formula
for the offsets may be deduced as follows:
The angle subtended at the centre (O) of the curve by the arc
T1E is obviously equal to 2α.

But the chord T1E = arc T1E very nearly
= R × 2 α or α = (T1E / 2R)
Similarly, the chord E1E = arc E1E nearly.
The first offset (O1) = E1E = T1E × α = ……( 15 )
Now ‫ے‬E2EF1 = ‫ے‬DET1 (vertically opposite);
‫ے‬DET1 = ‫ے‬DT1E, since DT1 = DE, both being tangent to the circle
‫ے‬E2EF1 = ‫ے‬DT1E = ‫ے‬E1T1E
The ∆s E1T1E and E2EF1 being nearly isosceles, may be
considered approximately similar.
F1F being the offset from the tangent at E, is equal to

Now the second offset (O2) E2F = E2F1 + F1F
………………..( 17 )
Similarly, the third offset (O3) …………( 18 )
Since b2 = b3 = b4 =…………
Each of the successive offsets O4, O5, etc. except the last one
(On) is equal to O3. Since the length of the last chord is usually
less than the length of unit chord (b2),
The last offset ………...( 19 )

Mode of Procedure:-
i. Having the fixed directions of the tangents AB and BC, locate
the first tangent point T1 by measuring backward distance equal
to the length ( R tan(Ø/2) ), along the tangent from the point of
intersection (B).
Similarly, mark the other tangent point T2 by measuring forward
the same distance along the tangent BC.
ii. Measure the distance equal to the length of the first chord (T1E)
along T1B. Thus marking the point E1.
iii. With the zero end of the chain (or tape) pinned down at T1,
swing the portion of the chain (=T1E1) around the point T1
through the calculated offset O1, thus fixing the first point E on
the curve.
iv. Pull the chain forward in the direction of T1E produced, until
EE2 equals the length (1 chain or ½ chain) of the second chord
(b2).
v. Hold fast the zero end of the chain at E and swing the chain
around E through the second calculated offset O2, thus locating
the second point F on the curve.

vi. Repeat the process until the end of the curve is reached. The
last point thus fixed should coincide with the previously
located point T2. If not, find the closing error. If it is large, the
whole curve must set out again . But if it is small, all the points
are moved sideways by an amount proportional to the square
of their distances from the beginning of the curve (T1) thus
distributing the closing error among all the points.
This method is largely used for road curves. It gives better
result than those obtained by the preceding method. It can be
used in confined situations, since all the work is done in the
immediate proximity of the curve. The most serious objection
to this method is that if any point is inaccurately fixed, its error
is carried forward through all the subsequent points.

INSTRUMENTAL OR ANGULAR METHODS
1. Rankine’s method of tangential angle:-
.

In this method the curve is set out by the tangential angles(often
called the deflection angles) with a theodolite and a chain or
tape.
The derivation of the formula for calculating the deflection angle
it as fallows :-
Let,
AB = the rear tangent to the curve.
T1 and T2 = the tangent points
D,E,F, etc. = the successive points on the curve.
δ1, δ2,δ3, etc = the tangential angles which each of the
successive chord T1D, DE. EF etc makes with
the respective tangents at T1, D, E, etc.
∆1,∆2,∆3, etc = the total tangential or deflection angles
(between the rear tangent AB and each of the
lines T1D, DE, EF, etc.
c1,c2,c3 etc = the lengths of the chord T1D, DE, EF, etc.
R = radius of the curve
Chord T1D = arc T1D (very small ) = c1.

BT1D = δ1 =½ T1OD i.e. T1D = 2δ1
NOW,
Hence,
…………..( 20 )
Similarly, and so on.
…………………..( 20 a)
Since the chord lengths c2, c3, …….cn-1 is equal to the length of
the unit chord (peg interval), δ2=δ3= δ4= δn-1.
Now, the total tangential (deflection) angle (∆1) for the first chord
(T1D) = BT1D. ∆1 = δ1.
The total tangential angle (∆2) for the second chord (DE) =BT1E.
But BT1E = BT1D + DT1E.

Now the angle DT1E is the angle subtended by the chord DE in
the opposite segment and therefore, equals the tangential angle
(δ2) between the tangent at D and the chord DE.
Therefore, ∆2 = δ1 + δ2 = ∆1 + δ2
Similarly, ∆3 = δ1 + δ2 + δ3 = ∆2 + δ3
∆n = δ1 + δ2 + δ3+………... + δn
∆n = ∆n-1 + δn ……………………………..( 21 )
Check;- The total deflection angle (BT1T2) = ∆n = ( ø/2 )
where ø is the deflection angle of the curve.
From the above, it will be seen that the deflection angle (∆) for any
chord is equal to the deflection angle for preceding chord plus
the tangential angle for the chord/
If the degree of the curve (D) be given, the deflection angle for
30m chord is equal to ½ D degrees, and that for the sub chord
is equal to (c1×D)/60 degrees, where c1 is the length of the first
chord ,

If the degree of the curve (D) be given, the deflection angle for
30m chord is equal to ½ D degrees, and that for the sub
chord is equal to (c1×D)/60 degrees,
where c1 is the length of the first chord,
Hence,
………….. ( 22 )
In the case of left hand curve each of the values ∆1,∆2,∆3, etc
should be subtracted from 3600 to obtain required value to
which the vernier of the instrument is to be set.

Procedure:-
To set out a curve
i. Set up the theodolite over first tangent point (T1) and level it.
ii. With both plates clamped at zero, direct the telescope to the
ranging rod at the point of intersection B and bisect it.
iii. Release the vernier plate and set the vernier A to first
deflection angle (∆1), the telescope being thus directed along
T1D.
iv. Pin down the zero end of the chain or tape at T1, and holding
the arrow at the distance on the chain equal to the length of
the first chord, swing the chain around T1 until the arrow is
bisected by the cross-hairs, thus fixing the first point D on the
curve.
v. Unclamp the upper plate and set the vernier to the second
deflection angle ∆2, the line of sight bring now directed along
T1E.

vi. Hold the zero end of the chain at D and swing the other end
around D until the arrow held at other end is bisected by the
line of sight , thus locating the second point on the curve.
vii. Repeat the process until the end of the curve is reached.
Check:-
The last point thus located must coincide with the
previously located tangent point T2. If not, find the distance
between them which is actual error. If it is within the
permissible limit, the last few pegs may be adjusted, if it is
exceeds the limit, the entire work must be checked.

This method is used when the ground is not
favourable for accurate chaining e.g. rough ground. It is based on
the fact that angle between the tangent and the chord is equal to
the angle which that chord subtends in the opposite segment.
Let, D, E, etc. be the points on the curve . The angle (∆1)
between the tangent T1B and the chord T1D = BT1D = T1T2D.
Similarly, the angle BT1E = ∆2 = T1T2E, the total tangential or
deflection angles ∆1, ∆2, ∆3 being calculated as in the first method.
Procedure:-
To set out the curve
i. Set up theodolite over T1 and another over T2.
ii. Set the vernier of each instrument to zero.
iii. Direct the instrument at T1 to the ranging rod at the point of
intersection B and bisect it.
iv. Direct the instrument at T2 to the first tangent point T1 and

v. Set the vernier of each of the instrument to read the deflection
angle ∆1. Thus the line of sight of the instrument at T1 is
directed along the T1D and that of the other instrument at T2
along T2D. Their point of intersection gives the required point
on the curve.
vi. Move the ranging rod until it is bisected by the crosshairs of
both instruments, thus locating the point D on the curve.
vii. To obtain the second point on the curve, set the vernier of
each of the instruments to the second deflection angle ∆2 and
proceed as before.
If the first tangent point T1 cannot be sighted from the
instrument at T2 the ranging rod at the point of intersection B
may be sighted. The procedure will be then be as follows:-
i. With both plates of the second instrument clamped at zero,
bisect the signal at B
ii. Release the vernier plate and swing the telescope (clockwise)
through (3600 – ø/2), thus directing the line of sight along T2T1.

iii. To obtain the first point on the curve, set the vernier to the first
deflection angle ∆1. the vernier reading will then be
(3600 – ø/2) + ∆1 instead of ∆1 as in the first case.
iv. The rest of the procedure is exactly the same as before.
Instead of sighting the intersection point B, any point K in the
forward direction of the tangent line T2C may be used.
In this case, however, the angle through which the telescope has
to be turned, after having the signal at K with both plates
clamped at zero, is equal to (1800 – ø/2). The line of
collimation is thus directed along the line T2T1. To obtain the
first point on the curve, the vernier reading must be
(1800 – ø/2) + ∆1.
It will seen that in this method no chain or tape is used to fix the
points on the curve, but they are located by the intersection of
the lines of the sight of the two instruments. The method is
simple and accurate, but it is expensive, since two surveyors
and two instruments are required to use this method.

Obstacles in setting out curves:-
The following obstacles occurring in common practice will now be
considered:-
1) When the point of the intersection of the tangent lines is
inaccessible:-
e.g. when the intersection falls in the lake, river. or wood.

Let, AB and BC be the two tangent lines intersecting at the point
B, and T1 and T2 the tangent points. It is required to determine
the value of the deflection angle (ø) between the tangents and
to locate the tangent point T1 and T2.
Procedure:-
a) Fix points M and N suitably on the tangents AB and BC resp.
so that M and N are intervisible, and the line MN runs on
moderately level ground in order that accurate chaining may
be possible. If the ground beyond the curve is not suitable, the
points may be fixed inside the curve as at M and N. Measure
MN accurately.
b) Set up the instrument at M and measure the angle AMN(ϴ1)
between AB and MN.
Transfer the instrument to N and measure the angle CNM (ϴ2)
between BC and MN.
Now in the ∆ ∠BMN, ∠BMN = α =1800 − ∠AMN = 1800 − ϴ1,
∠BNM = β = 1800 − ∠CNM = 1800 − ϴ2.

The deflection angle (ø) = ∠BMN + ∠BNM = α + β
or = 3600 − sum of measured angles.
= 3600 − (ϴ1 + ϴ2).
c) Solve the triangle BMN to obtain the distance BM and BN.
d) Calculate the tangent length BT1 and BT2from the formula
T= R ( tan ø/2).
e) Obtain the distances MT1 and NT2.
MT1 = BT1 – BM and NT2 = BT2 – BN.
f) Measure the distance MT1 from M along the tangent line BA.
Thus locating the first tangent point T1.
similarly, locate the second tangent point T2 by measuring the
distance NT2 from N along the tangent BC.
If the point fixed inside the curve, the procedure is same as
above, except for the distances to be measured from the

If the point fixed inside the curve, the procedure is same as
above, except for the distances to be measured from the
points M1 and N1 to locate the tangent point T1 and T2.
MT1 and NT2 being respectively equal to (BM1 – BT1)
and ( BN1 = BT2).
When it is found impossible to obtain a clear line MN, a traverse is
run between M and N to find the length and bearing of the line
MN. From the known bearing of the tangent lines and the
calculated bearing of the line MN, the angles α and β may
easily be obtained. The distances BM and BN are then
calculated as before.

2. When the curve can not be set out from the tangent points,
vision being obstructed:-
.

2. When the curve can not be set out from the tangent points,
vision being obstructed:-
As a rule, the whole curve is set out
from the first tangent point T1. But this is possible only when
the curve is short and ground is moderately level and free
from all the obstructions. However, it is often found that this
cannot be done on account of great length of the curve, or
obstructions intervening the line of sight such as buildings,
cluster of trees, plantations, etc. In such a case, the
instrument requires to be set up at one or more points along
the curve.
Procedure ( first method ):-
Suppose the first four points have been located
by the deflection angles from the instrument at T1, when it is
found necessary to shift the instrument.

Let, d be the last point located from T1 and its deflection angle ∆.
i. Shift the instrument and set it up at d.
ii. With both plates clamped at zero, take a backsight on T1 and
transit the telescope.
iii. To locate the next point e, set the vernier to read the deflection
angle ∆5, thus directing the line of sight along de.
iv. Using the same tabulated deflection angles, continue the
setting out of the curve from d as already explained.
Proof:-
Draw the tangent d1dd2 at d, meeting the first tangent BT1 at
d1. Produce T1d to K.
Kdd2 = d1dT1 (vertically opposite); d1dT1 = d1T1d, since
d1T1 = d1d. But d1T1d = ∆4, so, Kdd2 = d1T1d = ∆4.
The tangential angle for the chord de = d2de = δ5 = dT1e.
Kde = ∆4 + δ5 = ∆5.
The total tangential (or deflection) angle for the chord
de =d1T1e = d1T1d + dT1e = ∆4 + δ5 = ∆

Thus is well be seen that when the telescope is transited after a
backsight was taken with the vernier reading zero, the line of
sight directed along dK and when the telescope is turned
through the angle ∆4, it is in the direction of tangent at d.
To use this method instrument must be in perfect adjustment
Second method:-
Suppose d is the last point located from the
instrument at T1.
i. Suppose the instrument is at T1, fix any point K in the line T1d
produced.
ii. Move the instrument to d and with the vernier set to zero
bisect K.
iii. Release the vernier plate and set the vernier to ∆5, to locate
the next point e.

When the
obstacle to chaining
occurs:-

Third method:-
Suppose that the first four points have been
located from T1, d being the last point located and that the
portion of the curve between d and f cannot be chained
across.
In such a case, the following procedure may be adopted:-
Let g be the next station on the curve which can be seen from T1.
i. Calculate the length of long chord T1g from the formula,
T1g = 2R sin ∆4.
ii. Set the vernier A to the deflection angle ∆4, thus directing the
line of sight along T1g. Measure the distance T1g along this
direction, and locate the point g on the curve. Locate the
remaining points in the usual way. Alternatively, set out the
curve T2g in the reverse direction from the second tangent
point T2.
iii. The point e and f, which were left out, will be located sfter the
obstruction is removed.

In the figure it is shown that a compound curve which is
tangential in three straights AB, BC and KM at T1,T2 and N
respectively. The two circular arcs T1N and NT2 having centres
at the point N, OS and OL being in straight line. The arc having a
smaller radius may be first or second.
Let, the tangents AB and BC intersect at the point B, AB and KM
and BC at M.
Notation:-
Let , Rs = the smaller radius OsT1.
RL = the larger radius OLT2.
Ts = the smaller tangent length (BT1)
TL = the larger tangent length (BT2)
ø = the deflection angle between the rear tangent (AB) and
forward tangent (BC)
α = the deflection angle between the rear tangent (AB) and
common tangent (KM) = ‫ے‬BKM.
β = the deflection angle between the forward tangent (AB)
and common tangent (KM) = ‫ے‬BMK.

ts = the length of the tangent to the arc (NT1) having a
smaller radius
tL = the length of the tangent to the arc (NT2) having a
larger radius.
Elements of the compound curve:-
Ø = α + β
KN = KT1 = ts = Rs tan ( α/2 )
MN = MT2 = tL = RL tan ( β/2 )
Length of common tangent (KM) =
KM = KN + MN = Rs tan (α/2)+ RL tan ( β/2 )
In the ∆BKM,
Now ………….(A)
…………..(B)
***

Substituting the values of ts and tL in the equation A and B we
get,
Of the seven quantities Rs, RL, Ts, TL, Ø, α and β four must be
known. Then remaining three may be calculated from the above
equations,
The following equation gives the relationship between the seven
elements involved in compact form
Ø = α + β

Procedure :-
The curve may be setout by the method of deflection angles from
the two points T1 and N, the first branch from T1 and second
from N.
1. The four quantities of the curve being known, calculate the
other three.
2. Locate B, T1 and T2 as already explained, obtain the
chainage of T1 from the known chainage of B.
3. Calculate the length of the first arc and add it to the chainage
of T1 to obtain the chainage of N. Similarly, compute the
length of the second arc which added to the chainage of
chainage of N gives the chainage of T2.
4. Calculate the deflection angles for both the arcs.
5. With the theodolite set up over T1 set out the first branch
already explain in Rankine’s method.
6. Shift the instrument and set up at N, with the vernier set to
(α/2) behind zero i.e. (360 - α/2 ), take a backsight on T1 and
plunge the telescope which is thus directed along T1N

produced. ( if the telescope is now swing through an angle α/2
the line of sight will be directed along the common tangent NM
and the vernier will read 360 )
6. Set the vernier to the first deflection angle ∆1 as calculated
for the second arc.
7. Repeat the process until the end of the second arc is
reached i.e. T2.
Check :-

TRANSITION CURVE
A curve of varying radius is known as
‘transition curve’. The radius of such curve varies
from infinity to certain fixed value. A transition curve
is provided on both ends of the circular curve. The
transition curve is also called as spiral or easement
curve.

OBJECTS OF PROVIDING TRANSITION CURVES:-
1. To accomplish gradually the transition from the tangent to
the circular curve, and from the circular curve to the
tangent.
2. To obtain a gradual increase of curvature from zero at the
tangent point to the specified quantity at the junction of the
transition curve with the circular curve.
3. To provide the superelevation gradually from zero at the
tangent point to the specified amount on the circular curve.
4. To avoid the overturning of the vehicle.

REQUIREMENT OF IDEALTRANSITION CURVES:-
1. It should meet the original straight tangentially.
2. It should meet the circular curve tangentially.
3. Its radius at the junction with the circular curve should be
the same as that of the circular curve.
4. The rate of increase of curvature along the transition curve
should be same as that of in increase of superelevation.
5. The length should be such that the full superelevation is
attained at the junction with the circular curve.
The types of transition curve which are in common use are
1. A Cubic parabola
2. A clothoid or spiral
3. A lemniscate
the first being used on railways and third one on highways.

SUPERELEVATION:-
When a vehicle passes from a straight path
to curved one, the forces acting on it are
i. The weight of the vehicle.
ii. The centrifugal force both acting through the CG of vehicle.
Since the centrifugal force is always acts in the direction
perpendicular to the axis of rotation which is vertical, its
direction is always horizontal. The effect of the centrifugal force
is to push the vehicle off the track or rail. In order to
counteract the action, the plane of rails or the road surface
is made perpendicular to the resultant of centrifugal force
and weight of the vehicle.
In other word, the outer rail is superelevated or raised above
the inner one. Similarly the road should be “banked”, i.e.
the outer edge of the road should be raised above the inner
one, the raising of the outer rail or outer edge above the
inner one, being called as superelevation or cant.
The amount of cant is depend on vehicle and radius of curve.

Let,
W = weight of the vehicle
P = the centrifugal force.
v = the speed of the vehicle in m/s.
g =the acceleration due to gravity = 9.81 m/s2
R = the radius of the curve in m
h = the superelevation in m

b = the width of the road in m
G = the distance between centers of the rails in m.
Then the equilibrium, the resultant of the weight and the
centrifugal force must be equal and opposite to the reaction
perpendicular to the road or rail surface.
If ϴ be the inclination of the road or rail surface, the inclination of
the resultant to the vertical is also ϴ, therefore we have
Hence the amount of superelevation h
on roads
on railways

The amount of superelevation is limited about 1/12th of the
gauge, 1/10th being permitted under special circumstances.
The maximum superelevation recommended for broad gauge
(1676 mm) , meter gauge (1000 mm) and narrow gauge
( 762 mm) are 140mm (165 mm) , 90 mm ( 102 mm ) and 65
mm ( 75 mm ) respectively

LENGTH OF TRANSITION CURVE:-
The length of the transition curve may be
determined in the following ways
1. By an arbitrary gradient :-
The length may be such that the superelevation is applied
at a uniform rate of 1in n, the value of n varying from 300 to
1200.
Therefore, L = nh …………..( 1 )
where, L = the length of transition curve in m
h = the superelevation in m
1 in n = the rate of canting
2. By the time rate:- The transition curve may be such a
length that the cant is applied at an arbitrary time rate of a
cm per second, a varying from 2.5 to 5 cm.
Let, L = the length of transition curve in m
h’ = the amount of superelevation in cm.
v = the speed in m/sec
a = the time rate ( cm/sec )

Time taken by vehicle in passing over the transition curve
Superelevation attained in this time
…………….( 2 )
3. By the rate of change of radial acceleration:-
This rate should be such that the passenger should
not experience any sensation of discomfort when the train
travelling over the curve. It is taken as 30 cm per sec3 .
Now the radial acceleration on the circular curve (m/sec2)
Time taken by the vehicle to pass over the transition curve
seconds.

Radial acceleration attained in L/v seconds at the rate of
0.3m/sec3
……………( 3 )
If v = speed in km/hr.
The ratio of centrifugal force and the weight is called the
centrifugal ratio.
Centrifugal ratio ……….( 4 )
The maximum value of the centrifugal ratio on roads is taken as
1/4 and for railways as 1/8.
On roads

Now from the formula
………..( 5 )
On railways
……( 6 )
The equation 5 or 6 is to be used only when the full centrifugal
force is developed and when the rate of gain of radial
acceleration is 0.3 m /sec.

Ideal transition curve:-
The equation of the ideal transition curve (clothoid spiral) may
be as follows:-
Let, TB = the initial tangent
T = the beginning of the transition curve
E = the point of junction of the transition curve with the
circular curve.

M = any point on the transition curve, l m along it from T.
r = the radius of the transition curve at M.
R = the radius of the circular curve.
Ø = the inclination of the tangent to the transition curve
at M to the initial tangent TB.
Ø1 = the angle between the tangent TB and the tangent
to the transition curve at the junction point E.
( this angle is known as spiral angle)
L = the length of the transition curve.
α is the deflection angle i.e. angle MTB between the tangent
at T and line from T to any point (M) on the curve
The fundamental requirement of the spiral curve is that its
radius vary inversely as the distance( l ) from the beginning of
the curve.
Therefore,
Now for all curve

Integrating we get, …………( 7 )
The constant of integration being zero, since ø = 0, when l=0.
At the junction point E, l = L, r = R, and ø = ø1.
Substituting the value of m in equation 7, we get
…………( 8 )
where

If the curve is to set out by offsets from the tangent at the
commencement of the curve (T), it is necessary to calculate the
rectangular (cartesian) coordinates, the ‘the axes of the
coordinates’ being the tangent at T as the x axis and a line
perpendicular to it as the y axis.
Let, M and N be the two points at a distance δl apart on the
curve. Let the coordinates of M and N be (x,y)and (x+ δx, y+ δy)
And the respective inclinations of the tangents at M and N to the

initial tangent (TB) at T Ø and Ø + δØ.
…………….( 9)
………….(10)
Rejecting all the terms of the expressions 9 and 10 except the first
we have, …………….( 11 )
Which is the equation of curve parabola, the length of the curve
being measured along x axis the cubic parabola is known as
Froude’s transition or easement curve.

Taking the first two terms of the expressions 9 and 10
We get,
…………….( 12)
………….(13)
From which the coordinates of any point on the true or clothoid
spiral can be obtained.
If we take the first term only of expansion 10, we get the equation
for cubic parabola.
…………………..( 14 )
…………..( 15)

Characteristics of a transition curve:-
Let, TB = the original tangent
T = the commencement of the transition curve.

E = the end of the transition curve
EE2 = the tangent to both the transition and the circular curve
at E.
Y = EE1 = the offset to the junction point (E) of both the curves.
X = TE1 = the x coordinate of E.
EE’ = the redundant circular curve.
T1 = the point of intersection of the line (OE’)
perpendicular to the tangent to the circular curve at
E’ and the origin at tangent TB.
S= E’T1= the shift of the circular curve.
N = the point in which OE’ cuts the transition curve.
Ø1= the spiral angle (EE2B) between the common tangent
EE2 and the original tangent TB.
R = the radius of the circular curve (OE)
a) Now EE’ =R Ø1; but Ø1= L/2R EE’ = L/2
EE’ nearly equal to EN, EN = L/2 ………… ( 16 )

i.e. The shift (E’T1) bisect the transition curve at N.
Hence TN = L/2 ……………( 17 )
b) Draw ED perpendicular to OE’
Now S = E’T1 = Y – R (1 – cosØ1) …...……….(18)
or S = Y – 2R sin2 (Ø1/2)
……....(19 )
Also, ………..( 20 )
i.e. the transition curve bisect the shift.
c) The total tangent length BT :-
a) True spiral (clothoid) BT = BT1 + T1T
……( 21 )

b) cubic parabola:- in this case of cubic parabola the length
of the curve is measured along the x axis (TB)
Therefore, TE = L = TE1 = X
…………( 22 )
The amount S tan (∆/2) is called the shift increment , and
( x - R sinØ1 ) is the spiral extension.
Thus it will be noticed that when the transition curve is inserted
in between the tangent and the circular curve, the length of the
combined curve is greater than that of simple curve (= R tan
∆/2) by an amount depending upon the transition curve used.
In the case of cubic parabola this increase is equal to
while in the case of true spiral or clothoid, it equals

Elements of cubic parabola:-
Let,x = TM1 = the distance of any point m on the curve
measured along the tangent TB from the
commencement T on the curve.
y = M1M = the perpendicular offset to point M
l = the lengthy of transition curve TM.
X = TE1 = the distance of the junction point E of transition
curve with the circular curve from T measured
along TB
Y = E1E = the perpendicular offset to the junction point E.
Ø = ‫ے‬MM2B= the angle between tangent line AB and tangent to
the transition curve at any point M.
Ø1 = ‫ے‬EE2B= the angle between TB and tangent at E.
R = the radius of the circular curve.
α = ‫ے‬MTB = the deflection angle to any point M.
α1 = ‫ے‬ETB = the deflection angle to the junction E.

…………………..(23)
…………………………(24)
……….. ( 25 )
If the degree (D) of the curve given instead of radius, the
corresponding values α, αn, Ø1, may be found by substituting the
value of R in terms of D, i.e. R=1719/D in equation 24 and 25
….(26) ……….(27)
…………( 28 )

Elements of True spiral:-
Using the same notation the elements are:-
The coordinates of any point:
…………. ( 29 )
………( 30)
The coordinate of the end (E) of the curve:
…..(31)
……….( 32)
The expression for deflection are same as cubic parabola.
The total tangent length

Elements of cubic spiral:-
Using the same notation the elements are:-
Total tangent length

Combined Curve:- when the transition curves are inserted at
each end of the main circular curve, the resulting curve is
known as combined curve.

Length of combined curve:-
The angle subtended at the centre by the circular arc
(EE’) = (∆ - Ø1) degrees.
length of circular arc = EE’ ………..(1A)
the length of the combined curve …(1B)
The length of the curve may also be found out by another way
The central angle subtended by the circular arc
E1E’1 = ∆ degrees
Length of circular curve arc E’E’1 …………(2A)
Length of combined curve ………….(2B)

The calculation required for setting out the combined curve may
be made in the following steps:-
Data:- a) the deflection angle ( ∆) between the straights,
b) the radius (R) of the circular curve
c) the length of the transition curve (L)
d) the chainage of the point of intersection of the two
straights.
1. Find the shift (S) of the circular curve from
2. Compute the total tangent length from the formula according
as a cubic parabola or spiral is used.
3. Calculate the spiral angle Ø1 from radians
4. Calculate the length of the circular curve from formula 1A and
length of the combined curve from formula 1B also check
result from formula 2B.

5. Find the chainage of the beginning (T) of the combined curve
by subtracting the total tangent length from the given chainage
of point of intersection B.
6. Obtain the chainage of the junction point E of the transition
curve with the circular curve by adding the length of the curve
to the chainage of T.
7. Determine the chainage of the other junction point (E’) of the
circular arc with the other transition curve by adding the length
of the circular arc to the chainage of E.
8. Obtain the chainage of the end point (T’) of the combined
curve by adding the length of the combined curve to the
chainage of T.
9. Calculate the deflection angle for the transition curve from
minutes or
And also for circular curve from bearing in

the mind that the points stacked on the compound curve with
through chainage so that there will be sub-chords at each end
of the transition curve and of the circular curve.
10. Find the total tangential angles for the circular curve from
and check the result s by observing if the ∆n
equals 0.5(∆ - 2Ø1 ).
11. Calculate the offsets for the transition curve from in
case of cubic parabola and from
in case of
of true spiral.
12.Finally compute the offsets from chords produced from
for circular curve.

Tabulate the result as under
Station
no.
Chainage Length
of chord
Deflection
angle α or δ in
minutes
Total
tangential
angle ∆
∆ in
º ̍ ̋
Actual
Instrumen
t
Reading
º ̍ ̋
Remar
ks

Setting out the combined curve by deflection angle:-
The first transition curve may be set out from T (i) by the
deflection angles or (ii) by the tangent offsets; and the circular
curve from the junction point E. the second transition curve
may be setout from T’, checking on the junction point E’
previously located.

1. Having the fixed the tangent AB and BC, locate the tangent
point T by measuring the backward the total tangent length
from B and other tangent point T’ by measuring the forward
the same distance from B along the forward tangent BC.
2. From T measure along TB, the distances equal to the ½ L ,
2/3L, and L, the peg these points, which are lettered T1,
E2, and E3 resp.
3. Set up theodolite over T and with both plates clamped at
zero, bisect B.
4. Release the vernier plate and set the vernier to the first
deflection angle as obtained , thus directing the line of sight
to the first point on the transition curve.
5. Pin down the zero end of the tape at T, and holding the
arrow at the distance on the chain equal to the length of the
first chord, swing the chain around T until the arrow is
bisected by the cross-hairs, thus fixing the first point on the
transition curve.

6. Repeat the procedure until the end of the curve (E) is
reached. Check the location of E by measuring the
distance EE2 which should be 4S.
7. To set out circular curve, shift the instrument and set it up to
E.
8. With the vernier set to 2/3 Ø1, behind zero, for right hand
curve, take backsight on T and plunge the telescope which
is thus directed along forward direction ( if the telescope is
now swing through an angle 2/3 Ø1 the line of sight will be
directed along the common tangent and the vernier will
read 360 )
9. Transit the telescope and set the vernier to first deflection
angle for the circular curve and hence the first point on the
curve is obtained.
10.Continue the setting out the circular curve upto E in the
usual way.
11.Set out the other transition curve from T’ as before.

Setting out the Transition curve by tangent offset:-
a) Cubic parabola:-
i) From T measure the x coordinate of points along TB.
ii) Locate the points on the curve by setting out the respective
offsets perpendicular to TB at each distance.
b) Cubic spiral:-
i) Each point is located by swinging the chord length from the
preceding point through the calculated offset.
Setting out the transition curve by offsets from tangent
(TT1) and from the circular arc (E1E):-
This method is based upon the fact that the
often from the circular arc (E1E) to the transition curve at a
distance x from E is equal to the offset from the tangent
(TT1) to the transition curve at a distance x from T, the
transition curve at a distance x from T, the tangent offsets
being calculated from

In this method, therefore, half the transition curve is set out by
means of offsets from the tangent (TT1) and the remaining half
by means of offsets from the circular arc E1E.

When it is required to insert transition curve in between two
arcs of a compound curve, the following procedure may be
adopted:-
1. With the given radii of two circular arcs, the maximum
speed and distance between the rail heads, calculate the
amount of superelevation for each arc by the relation

VERTICAL CURVES
When two different or contrary gradients meet,
they are connected by a curve in vertical plane is called
a vertical curve.
It is advisable to introduce a vertical curve in road
and in railway work in order to round off the angle and to
obtain a gradual change in grade so that abrupt change
in grade is avoided at the apex.

REQUIREMENTS
1. It gives adequate visibility and safety to the traffic.
2. It gives gradual change in grade or slope.
3. It gives adequate comfort to the passengers.

Types of Vertical Curves
1. An up grade followed by a down grade.
2. A down grade followed by an up grade.
3. An up grade followed by another up grade.
4. A down grade followed by another down grade.

1. An up grade followed by a down
grade.

2. A down grade followed by an up
grade

3.An up grade followed by another up
grade.

4. A down grade followed by another
down grade.

PROPERTIES OF PARABOLA:-
In fig,
OX and OY = the axes of the rectangular co-ordinates passes
through the point (O) of the vertical curve.
O = origin of the co-ordinates.

OA and OB = the tangent to the vertical curve
+g1% = the grade of the tangent OB
-g2% = the grade of the tangent AB
M = Any point on the curve whose co-ordinates are x
and y
Now, it may be shown that the equation of the parabola with
respect to OX and OY is y = cx2 + g1x
Now, OK = x; KM = y; KN = g1x; and NM = KN – KM
NM = g1x – y = -cx2
From which it fallows that vertical distance from tangent to any
point on the curve varies as the square of its horizontal distance
from the point of commencement of the curve (the point of the
tangency). This vertical distance is called as tangent correction.
y = -cx2
When x=2l=L

When x=l,
By knowing c, the tangent corrections may be computed in the
following way:-
By tangent correction:-
Let the chainage and elevation of the apex is given
1) The length of the curve on either side of the A being l m,
determine the chainages of points of tangency (O and B)
chainage of O = chainage of A – l.
chainage of B = chainage of A + l.
2) by knowing the grades of the tangents OA and AB, and the
elevation of A, compute the elevation of the tangent points O and
B.
elevation of O = elevation of A – ( lg1/100 )
elevation of B = elevation of A – ( lg2/100 )

3) Compute the tangent correction from y=cx2 for the stations
on the curve yx’ = cx’ 2
4) Determine the elevations of the corresponding stations on the
tangent OAG
Elevation of the tangent at any station = elevation of the point
of tangency (O) + x’g1.
5) Find the elevations of the stations on the curve by adding
algebraically the tangent corrections to the elevations of the
corresponding stations on the tangent OA .
Elevation of the station at a distance x’ on the curve =
elevation of the station on the tangent ± tangent correction yx’
The result may be tabulated as under.
Station Chainage Tangent or
grade
elevation
Tangent
correction
Elevation
of the
curve
Remarks

Location of Highest or Lowest Point
The position and elevation of the highest point
(Summit) or the lowest point at the sag may be calculated as
follows :
In fig. 1 & 2, let P be the required point at a distance x
from the beginning A of the curve. The tangent to the curve at
this point P being a horizontal line, its slope is zero.
The general equation of the parabola is y= cx2 + g1x.
The slope of the tangent at any point on the parabola
= dy/dx.
Now, dy/dx = 2cx + g1

At summits where speeds of 100 kmph are contemplated
the requirement of visibility i.e. the sight line will lead to
longer curves than one obtained by the above formula.
Sight distances:-
Let two points on the curve at a height ‘h’ metres from
the ground be intervisible and let the distance between be
S. A value of 1.1 m is usually taken as the eye level height
above the road surface for an observer sitting in a car. Sight
distances are laid down in the interest of road safety and
the choice for any distance depends on the nature of the
road and the speed of the traffic using it.

There are three cases to consider,
(a) sight distance equal to the length of the curve.
S = 2L and EF = h = (g1-g2 /400)L
Given h, g1, g2 , L may be determined and the
offsets computed from
y = (g1-g2 /400L) x
2

(b) Sight distance longer than the curve.
h = GK + KJ = GK + EF = GK + AF
Height of A above O = g1 /100L
Height of B above O = g1 /100L+ g2 /100L
= L/100 (g1+g2 ).
.

the angle between OB and horizontal
= L/100 x (g1-g2 )/2L
= (g1+g2 )/ 200 radians.
Angle KOG = g1/ 100 - (g1+g2 )/ 200
= (g1 -g2 )/ 200 radians.
But OK = S/2 –L.
Therefore
KG = (S/2 –L) ((g1 -g2 )/ 200
h = KG + AF
= (S/2 –L) (g1 -g2 / 200) + = L/100 x (g1-g2 )
= ((S-L)/400) (g1-g2 )

(C) Sight distance less than length of the curve.
The offsets from the tangent MFG are given by
the equation
y= cx2
Therefore ,
h = c (S/2) 2
Therefore ,
c = 4h/S2

At point J, x =L .
JO = 4h /S2 x L2
But JO = EF = AF = L/400 x (g1-g2 )
Therefore,
4h /S2 x L2 = L/400 x (g1-g2 ).
Therefore,
L = (g1-g2 ) x S2 /1600 x h ...................
(3)

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