Curves
Curves
Curves
VALLEY CURVE
COMPOUND CURVE
REVERSE CURVE
TRANSITION 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
………………….(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:-
………………..( 17 )
Hence,
…………..( 20 )
…………………..( 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.
2. Two theodolite method :-
.
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.
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.
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.
.
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.
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:-
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
…………….( 2 )
seconds.
Radial acceleration attained in L/v seconds at the rate of
0.3m/sec3
……………( 3 )
If v = speed in km/hr.
………..( 5 )
On railways
……( 6 )
…………( 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.
…………….( 9)
………….(10)
Rejecting all the terms of the expressions 9 and 10 except the first
we have, …………….( 11 )
………….(13)
…………..( 15)
Characteristics of a transition curve:-
……....(19 )
Also, ………..( 20 )
……( 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 )
…………………………(24)
……….. ( 25 )
….(26) ……….(27)
…………( 28 )
Elements of True spiral:-
Using the same notation the elements are:-
The coordinates of any point:
…………. ( 29 )
………( 30)
…..(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:-
The length of the curve may also be found out by another way
The central angle subtended by the circular arc
E1E’1 = ∆ degrees
in case of
of true spiral.
12. Finally compute the offsets from chords produced from
for circular curve.
Tabulate the result as under
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.
Spiralling compound curve:-
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
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,