CURVES

COMPOUND CURVES AND SUPER

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 Curves
Compound Curves
A compound curve consist of 2 arcs of different radii bending in
the same direction and lying on the same side of their common
tangent. Then the center being on the same side of the curve.
RS = Smaller radius
RL = Larger radius
TS = smaller tangent length = BT1
TL = larger tangent length = BT2
= deflection angle b/w common
tangent and rear tangent
= angle of deflection b/w common
tangent and forward tangent
N = point of compound curvature
KM = common tangent

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 Curves
Compound Curves
Elements of Compound Curve 180o - (+) = I
180o – = I

KT1 = KN =RS tan()

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 Curves
Compound Curves
Elements of Compound Curve
TS = BT1 = BK + KT1
TS = BT1 = + RS tan ()
TL = BT2 = + RL tan ()

Of the seven quantities RS, RL, TS, TL, , ,

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 Curves
Compound Curves
Problem: Two tangents AB & BC are intersected by a line KM. the
angles AKM and KMC are 140o & 145o respectively. The radius of
1st arc is 600m and of 2nd arc is 400m. Find the chainage of tangent
points and the point of compound curvature given that the
chainage of intersection point is 3415 m.
Solution:
= 180o – 140o = 40o
= 180o – 145o = 35o
= + = 75o
I = 180o – 75o = 105o
KT1 = KN = RL tan () = 600 tan(40o/2)
KT1 = KN = 218.38 m
MN = MT2 = RS tan() = 400 tan(35o/2)
MN = MT2 = 126.12 m
KM = MT2 + MN = 218.38 + 126.12
KM = 344.50 m

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 Curves
Compound Curves
Solution:
Fin BKM, by sin rule
=
BK = = = 204.57 m
BM = = = 229.25 m
TL = KT1 + BK = 218.38 + 204.57 = 422.95 m
TS = MT2 + BM = 126.12 + 229.25 = 355.37 m

LL = = = 418.88 m
LS = = = 244.35 m

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 Curves
Solution: Compound Curves
Chainage of intersection point = 3415 m
Minus TL = - 422.95 m
Chainage of T1 = 2992.05 m
Plus LL = + 418.88 m
Chainage of compound curvature (N) = 3410.93 m
Plus LS = + 244.35 m
Chainage of T2 = 3655.25 m

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SETTING OUT OF COMPOUND CURVE
• Done by deflection angle method from T1 and N
• Three unknown are calculated from four known
• B,T1 and T2 are located as already explained
• Chainage of T1 is calculated from known chainage of B
• Length of first arc is calculated and added to chainage of T1 to
obtain chainage of N
• Similarly length of 2nd arc is calculated and added to chainage of
N to get chainage of T2
• Both deflection angles are calculated
• Theodolite is set at T1 on first branch as explained

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 CONTINUED
• Instrument is shifted at N with vernier set to α/2 behind zero i.e.
(360- α/2 )
• Take a back sight on T1 and plunge telescope now it is directed to T1N
extended
• Telescope is swung by α/2 ,line of sight directed to common tangent
NM and vernier reading 360˚
• set the vernier to ∆1 as calculated for 2nd arc
• Process is continued until end of 2nd arc
• Check: Measure of angle of T1NT2 equals
180˚-(α+β)/2 OR (180˚-ɸ/2)

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 REVERSE CURVES
• Composed of two circular curves ,curving in opposite directions
• Having common tangent at their junction
• Joining point of two arcs is called point of reverse curvature OR
contrary flexure.
• Used when straights are parallel or intersect at small angle
• Used for railways , road with low design speed
• Must be avoided on roads and main railway lines with high design speed
for reasons:
 Involve sudden change in cant
 Curves cant be super elevated properly
 Sudden change in direction is objectionable
 Steering is very dangerous in highways

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 CASES OF REVERSE CURVES
• When the two straights are parallel
• When the two straights are not parallel, central angle α1 ,
α2 and length of common tangent are given
• Straights are non parallel (Tangent points,their distance
apart, angles α1 and α2 are given)
• Straights non parallel, radii unequal(same as case III one
radius is given)
• Straights non parallel, radii unequal(angle of intersection ɸ,
two radii, and tangent length are given

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 TRANSITION CURVES
• Common practice to introduce a curve of varying radius
• Between tangent and circular curve
• Also called spiral or easement curve.
• can also be inserted between two branches of reverse or
compound curves
• When the transition curves are inserted at each end of main
circular curve, the resulting curve is called composite or
combined curve.
• Types of transition curves common in use are :
1. Cubic Parabola (used in railways)
2. Clothoid or Spiral (used in railways)
3. Lemaiscate (used in highways)

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 OBJECTS OF PROVIDING TRANSITION
CURVES
• To accomplish gradually the transition from tangent
to circular curve and from circular curve to tangent.
• To obtain the gradual increase of curvature from
zero at the tangent point to the specified quantity at
the junction .
• To provide a satisfactory means of obtaining a
gradual increase of super elevation from zero on the
tangent the specified amount on the main circular
curve

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 CONDITIONS TO BE FULFILLED
• Should meet the original straight tangentially
• Should meet the circular curve tangentially
• Radius at the junction with the circular curve should
be same as that of circular curve
• The rate of increase of curvature along the
transition should be same as that of increase in
superelevation
• Its length should be such that the full superelevation
is achieved at the junction with circular curve.

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 SUPERELEVATION
• When a vehicle moves from a straight path to
curved forces:
 Centrifugal
 Weight
• Act on COG of vehicle
• Centrifugal force always act in direction
perpendicular to axis of rotation (vertical)so its
direction is horizontal
• Effect of this force is to push the vehicle off the track
• In order to cancel this , outer edge of road or rail is
raised or superelevated

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 CONTINUED
• The amount of superelevation depends upon the weight of vehicle and
radius of the curve.
• It should be applied gradually along the transition curve
• Full superelevation is achieved at the junction of the transition curve
with circular curve
• In applying superelevation, it is common practice in America to raise
the outer edge of the road or outer rail and to depress the inner edge
by half the amount of superelevation
• In England ,outer edge or outer rail is raised by full amount of
superelevation

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LENGTH OF TRANSITION CURVE

It may be determined by:

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 RATE OF CHANGE OF RADIAL ACCELERATION
• Rate should be such that passenger should not experience
any discomfort
• Taken as 30 cm/sec3
• Radial acc. In circular curve =v2/R m/sec3
• Time taken by vehicle to pass over transition curve=L/v sec
• Radial acc. In L/v sec at the rate of 0.3 m/sec3 = L/v * 0.3
m/sec3
• v2/R = L/v * 0.3 OR L= v3 /0.3R in m/sec
• L= v3 /14R if v is speed in km/ hr
• This method is commonly used
• Ratio of centrifugal force and weight is called centrifugal ratio

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 CENTRIFUGAL RATIO
The ratio of the centrifugal force and weight is called the centrifugal
ratio
= P/W=)/(gRW) = /
• The maximum value of the centrifugal ratio for roads is 1/4
¼=/

We know that /(0.3R) by putting value of v from above equation

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 CONTINUED
• The maximum value of the centrifugal ratio for railways is 1/8
1/8 = /

We know that /(0.3R) by putting value of v from above equation

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