webinar on speed

1. RAISING OF SPEED TO 130/160 KMPH
in existing track

2. Speed and curve
• For particular curve of Radius R
Cant deficiency
Speed Cant + Transition length
Cant excess

5. Suitability of curves for diff speeds.
SN Deg-ree
of cur-
ve
Radius Maximum
permissible
speed with
Ca=165
mm
Cd=75 mm
Maximum
permissi-ble
speed with
Ca=165 mm
Cd=100 mm
Maximum
Speed
potential
Ca=185 MM
CD=75mm
Maximum
Speed potential
Ca=185 MM
CD=100mm
Remark
1 0.5 3500 247 260 257 269
2 1.0 1750 174 183 182 190
3 1.25 1400 156 164 162 170
4 1.5 1166.6 142 150 148 155
5 2.0 875 123 130 128 134
6 2.5 700 110 116 115 120
7 3.0 583.3 101 106 105 110
8 3.5 500 93.5 98 97 101
9 4.0 437.5 87.5 91 91 95

6. Raising of speed to 130 KMPH with goods train running
at 75 kmph
Rad-ius
of
Curve
Degree
of
curve
Requ-ired
Eq cant for
130 kmph
Eq cant
for goods
train
Aver-age
of two
Provide
cant
cd ce TL
Required
1750 1 133.07 44.29 88.68 90 43.1 45.71 93.6
1500 1.166 155.25 51.67 103.46 105 50.7 53.3 109.2
1400 1.25 166.34 55.36 110.85 105 61.3 6.9 109.2
1250 1.4 186.29 62.0 124.15 125 613 63.0 130
1000 1.75 232.87 77.50 155.19 150 82.8 72.5 156
875 2.0 266.14 88.58 177.36 165 101.1 76.41 --
700 2.5 332.67 110.72 221.7 185 147.6 74.27 --

7. Raising of speed to 130 KMPH with goods train running
at 100 kmph
Rad-ius
of Curve
Degree
of curve
Eq cant
for 130
kmph
Eq cant for
goods train
at 100 kmph
Average
of two
Provide
cant
cd ce TL
Required
1750 1 133.07 78.74 105.9 90 43.0 11.25 94
1500 1.166 155.25 91.86 123.5 110.5 45.2 18.1 115
1400 1.25 166.34 98.42 132.04 120 16.3 21.5 125
1250 1.4 186.29 110.23 148.02 130 56.2 19.7 135
1000 1.75 232.87 137.74 185.3 160 72.8 22.2 167
875 2.0 266.14 157.48 211.8 165 101.1 7.5 172
700 2.5 332.67 196.85 264.7 185 147.6 -
(11.85)
193

8. For 160 KMPH with goods train running at 75 kmph
Rad-ius
of curve
(M)
Deg of
curve
Requi-red
Eq cant
for 160
kmph
Eq cant
for goods
train at 75
kmph
AVG of
two
Provid
e cant
mm
Cd Cant
exc-
ess
Transition
length
required
1750 1 201.57 44.29 122.9 120 81.57 75.7 154
1600 1.09375 220.47 48.44 134.4 122 98.47 73.5 156
1400 1.25 251.97 65.36 153.7 130 121.2 74.6 166.4
1250 1.4 282.20 62.01 172.1 165 117.2 103.0 211
1000 1.75 352.75 77.51 215.1 185 167.7 107.5 236.8
875 2 403.15 88.58 245.3 185 218.2 96.4 236.8

9. For 160 KMPH with goods train running
at 100 kmph
Radius
of curve
(M)
Deg of
curve
Required
Eq cant for
160 kmph
Eq cant for
goods train
at 100
kmph
AVG of
two
Provide
cant
mm
Cd Cant
excess
Transition
required
1750 1 201.57 78.7 140.10 140 61.85 61.2 179.2
1600 1.0937
5
235.17 91.9 163.5 160 75.1 68.1 204.8
1400 1.25 251.97 98.4 175. 160 91.9 61.5 204.8
1250 1.4 282.0 110.2 196.2 185 97.2 74.7 236.8
1000 1.75 352.75 137.8 245.2 185 167.7 47.2 236.8
875 2 503.93 196.8 280.3 185 218.1 (-11.8) 236.8

10. Our objective ?
• Since, goods train will run at 75 kmph on the same track
for, may be another 10-15 years.
• We need to assess each curve and redesign,
wherever required, for 130/160 kmph for passenger
trains and 65/75 kmph for goods train.

11. Cant, Cant excess and Cant deficiency
• 1. Cant - Raising of outer rail wrt to inner is called Cant=C
• 2. Equilibrium Cant: when cant provided balances the loads
on both the rail Ceq = GV²/127R
• 3. Cant Excess: when provided cant is more than the
required cant for a speed, Ce= C-Creq
• 4. Cant deficiency : when provided cant is less than the
required cant for a speed, Cd = Creq - C

12. Maximum cant which can be provided ?
• In a highly canted curved track, a vehicle standing or
travelling at a very low speed may overturn about the inner
rail. Overturning occurs on account of following factors.
1. Absence of centrifugal force (due to very slow speed), causing
substantial offloading on outer rail,
2. Wind pressure on train due to wind blowing from outside towards
inside of curve.
3. Vibration and other disturbing/curving forces.

13. In worst case when vehicle is standing,
V=0 what max’m Cant can be provided
• Centrifugal Force = 0 , we have only weight components
• Overturning about inner Rail: take moment about inner rail.
• WH x H = Wv x G/2 ---(1)
• Where WH= Horizontal component of Weight of vehicle
• WV= Vertical component of Weight of vehicle
• H= Height of CG of vehicle above Rail level
• Here, WH=W sinϴ =W.SE/G, WV=W as angle theta is very small
• Hence from equation (1) above W.SE.H/G = W.G/2
• Or SE= G²/2H
• For G=1750 & H=1676 SE works out to be 304mm with factor of safety 3
H

14. Maximum Cant ?
• In this we have not considered effect of winds. Hence,
considering additional factor of safety of min 1.5 to take care of
possible wind load etc. The maximum cant can be about
200mm.
• On maintainability account- higher cant means sleepers inclined
and causing extra lateral forces on ballast on inner rail side
resulting into ballast rolling down and finally disturbance to track
parameters.
• However, in addition to overturning about inner Rail, we need to
check for offloading of outer wheel resulting into derailment.

15. Maximum Cant Contd..
• When the vehicle start moving on canted track after being stationary.
The rail guiding force acting on the outer leading wheel will be
significant. The Y/Q ratio for the outer leading wheel thus become
adverse and wheel may mount the rail.
• Flatter curve Y i.e. Flange force is less higher cant can be
provided compared to sharper curve where flange forces increase.
• In places where high wind are experienced, it is not desirable to
provide cants with maximum value. (check with chartet formula etc.
for Y/Q ratio for each situation)
• Maximum cant in Indian Railway = 165 mm (can go up to 185 mm in
future)

16. Cant excess:
• Can we have cant excess = 165 mm ?
• The permissible limit of cant excess is controlled by -
(i)Permissible axle load thus allowable rail stresses
(ii) and excess wear on inner rail.
• The limits of 75 mm are wrt permissible axle load and allowable
stresses due to extra load on inner rail.
• In general higher values of cant deficiency can be adopted
compared to cant excess considering the axle load of high
speed passengers train vis a vis low speed goods train but with
higher axle load.

17. Cant Deficiency: -
• Cant deficiency occurs for trains moving at higher speed
than equilibrium speed. Limits on Cd depends on three
factors.
• (a) Safety against overturning about outer Rail
• (b) Safety against derailment.
• (c ) Maintainability and comfort.

18. Safety against overturning
• In case of train moving at higher speed than the equilibrium speeds the
excess of unbalanced lateral outward acceleration will tend to overturn the
vehicle about outer rail.
• On this consideration the maximum permissible cant deficiency is given as
( with factor of safety of 4) Calculation
• Cd = G²/8H,
• Where G=1750,
• H is Height of CG of vehicle above rail level
• Taking H= 1676 mm Cd =228mm

19. B. Safety against derailment:-
• It can be seen that before condition of overturning
around outer rai is reached, the wheel may climb on
outer rail and derail.
• As per C & M -1 report of RDSO for this in case of BG,
CD upto 150 mm can be permitted without danger of
derailment.

20. (C ) Maintainability and comfort:-
• Higher CD higher outward lateral forces tendency to distort
the track and alignment and excess forces on rail and fastenings.
• Higher CD Unbalanced lateral acceleration Discomfort to
the pass.
• ( i.e. Δ P may be more than 0.4 to 0.7 m/sec2 , but not more than 1.0
m/sec2).

21. Unbalanced Lateral Acc and CD
• Unbalanced lateral acceleration is related to Cd as under:
• Δ P = g x Cd/G ; where g is acceleration due to gravity, G –
dynamic gauge
• If we workout for G=1750 then range of Cd = Δ P *G/g
comes to 71 mm to 125 mm for
Δ P = 0.4 To 0.7 m/sec2 . In case of high speed this unbalanced
lateral acc will compress outer spring of coach more which
further adds to discomfort to passenger.

22. Key Design Parameters for
Transition length
Rate of Change of
Cant Deficiency (RCd
) mm/s,
Actual Cant (RCa
) mm/s,
Cant Gradient (i) mm/m;

23. Transition Length:
se
SE Diagram
Change of CD
Change of unbalanced
Lat acceleration
Rate of change of
Lateral acceleration
Ula or Δ P = 0.4 to 0.7 m/sec²
cd
Δ P/ Δ t, R/ula =0.3 M/sec³ range 0.2 to 0.4
Cant gradient =se/L
L
L
L

24. •Comfort Criteria: Rate of Change of Cd (RCd
)
•Rate of Change of ULA i.e Δ P/ Δ t less than 0.03g / s
taken as 0.3m/s3 ,range 0.2 to 0.4 m/s³
•Rate of Change of Cd over Transition
where, Δ P/(L/V) is 0.3 m/ s3
and is 35mm/s for 0.2
Length of Transition Curve
RCd
=
1750mm
9.81m
s2
0.3m
s3
é
ë
ê
ù
û
ú @ 53.5mm s

25. •Comfort Criteria:
•Rate of Change of Cd (RCd
) for lateral forces and
and Rate of Change of Ca (RCa
) for vertical forces.
• Rate of Change of Cd or Ca, normally shall not
exceed 35 mm/sec
• Under Exceptional Circumstances it can
be increased to 55 mm/sec
Length of Transition Curve

26. • Safety Criteria (Twist)
• Cant gradient causes twist in track
• Critical - longest rigid wheel base
• Cant gradient (i)
• Limited to 1.4 mm/m or 1 in 720
• In Exceptional Cases 2.8 mm/M or 1 in 360
• Future Layouts with 1 in 1200
Length of Transition Curve

27. •Length of transition will be maximum of
L1 = 0.008 Ca* Vm (m, mm, kmph, RCa
=35 mm/s)
or
L2 = 0.008 Cd*Vm (m, mm, kmph, RCd
=35 mm/s)
or
L3 = 0.72 Ca (m, mm, i = 1 in 720)
Length of Transition Curve

28. •In exceptional circumstances, minimum
length of transition will be maximum of
2/3rd of L1
or
2/3rd of L2
or
½ of L3
Length of Transition Curve

29. Finally transition length is based on
• 1. Rate of change of Cd = 35mm/sec or 55 mm/sec which is based on
rate of change of ULA of 0.3 m/sec3 permitted range is 0.2 to 0.4
m/sec3 and is purely on comfort consideration.
• 2. Rate of change of Ca i.e. unbalanced vertical acceleration and is
kept same as above.
• 3. Cant gradient of 1 in 720 or 1 in 360 and determines the designed
twist value.

30. Further Values of Cd, Ca and Ce
• 1. Cd is taken as 75 to 100 mm whereas we can go upto 125 mm
without any safety issue.
• 2. Ca is kept as 165 mm can be taken easily upto 200 mm without any
safety issue.
• 3. Ce is kept as 75 mm this can also be increased to 100 mm but
stresses will have to be checked for the concerned track structure.

31. Shift
Transition Curves
Inserting Transition Curves
How transitions are introduced.

32. Shift Due to Transition Curve
Circular
Curve Without
Transition
Transition Curve
C D B
E
F G
A L/2 L/2
S/2
S
Extended Circular
Curve
Circular
Curve With
Transition
H Tangent
Shift = S = L2/24R BG=L2/6R
DE=L2/8R

33. RAISING OF SPEED TO 130/160 KMPH
STEPS TO BE FOLLOWED:
1. Collect the curve data along with other features such as obligatory points,
location of bridge, ROB, FOB, PF, any other permanent structure which may
affect the realignment of curve. Curve passing through Yard and Point taking
off from curve etc.
2. Collect the train running data both for goods train and passenger train with
their speeds and the type of load being carried.
3. Any plan for upgrading the speed of goods train.
4. Work out the equilibrium speed by Russian or Shramm’s formulafor mixed
traffic

34. RAISING OF SPEED TO 130/160 KMPH …
contd..
5. Find out the equilibrium cant for this worked out equilibrium speed.
6. Check for cant excess for goods train and cant deficiency for high
speed passenger train.
7. Also check the effect of cant excess and cant deficiency for how
many number of trains and its load. Study the likely adverse affect
of cant excess and cant deficiency on track.
8. Workout the required transition length for the proposed cant and
check whether any modification is required in existing transition.
9. If feasible otherwise, the transition length worked out in step 8 can
be reduced to 2/3rd of its calculated normal transition length.

35. RAISING OF SPEED TO 130/160 KMPH …
contd..
10. If curve is not suitable after exercising all step up to 9 then try to
increase the radius of the curve after taking a decision on proposed
cant and proposed degree of curvature.
11. For general guidance if the speed of goods train proposed to run is
taken as 75 KMPH then for 160 KMPH the curve upto radius 1500 m
only are suitable other sharper curves will have to be redesigned to
increase the radius to approx.1500 m
12. Such curves identified in step 11 will be again checked for any
obstruction or infringement for realigning the curve with increased
radius.

36. • Example:
•
• JHANSI- AGRA SECTION DN ROAD EQUILLIBRIUM SPEED CALCULATION
•
• Nos. of Goods trains Sumo loaded per day n1 = 16.14 nos. speed v1 = 60 Kmph
• Nos. of Goods trains Sumo empty per day n2 = Nil . speed v2 = Nil
• Nos. of Goods trains other loaded per day n3 = 9.71 nos. speed v3 = 75 Kmph
• Nos. of Goods trains other empty per day n4 = 0.14 nos. speed v4 = 75 Kmph
• Nos. of Mail/Exp. trains per day n5 = 25.57 nos. speed v5 = 110 Kmph
• Nos. of Passenger trains per day n6 = 2 nos. speed v6 = 100 Kmph
• Nos. of Shatabdi/ Rajdhani per day n7 = 3.71 nos. speed v7 = 130 Kmph

37. Example contd…
Vav = n1 x w1 x v1² + n2 x w2 x v2²+ n3 x w3 x v3² + n4 x w4 x v4² + n5 x w5 x v5²+ n6 x w6 x v6² + n7 x w7 x v7²
n1 x w1 + n2 x w2 + n3 x w3 + n4 x w4 + n5 x w5 + n6 x w6 + n7 x w7
Vav = 16.14x 5000x60 ² + 0+ 9.71x5000x75²+0.14x1200x75²+25.57x1200x110²+2x1200x100²+3.71x1200x130²
16.14 x 5000 + 0+ 9.71 x 5000 +0.14x1200 + 25.57x1200 + 2x1200+ 3.71x 1200
Vav = 78.73 Kmph

38. Example contd…
• CURVE NO. 79 From Km.1226.500 to Km. 1227.760 = 1.260 Km
•
• Existing Details: -
•
• Degree = 1.3º, Radius ( R ) = 1346.15 m, CANT = 60mm., T.L. = 120 m. (CSTM end) & 120 m
(NDLS end)
•

39. Main Parameters/Constraints in Raising the speed
• 1. Increasing the Cant
• 2. Increasing the transition length
• 3. Increasing the Radius of curve
• 4. Turnout taking off from curved main line; circular or
transition portion.
• 5. Obligatory points

40. 1. Increasing the cant
• Following needs to be observed:
• 1. Not exceeding the maximum allowed
• 2. Maintain the cant gradient in transition length
• 3. Rate of change of Cd and Ca within limits in transition
length.

41. 2. How to increase the transition.
• A) Theoretical method; Steps to be followed:
• 1. Find the original beginning of the curve: from curve register find
out the length of the transitions on either side, length of circular
curve and radius of curve; say they are L1 , L2, Cl and R
• 2. Find φ1t= L1/2R and φ2t = L2/2R in radians and convert
them in degrees; L1 and L2 may be same in most of the cases.
Find deflection angle between tangents,
Δ = 360 x C/(2 ∏R) + φ1t + φ2t
Cross check this calculated delta with existing curve on ground as below
Δ = 4x (∑V )/c , where c is chord length for measurement of versine

42. Sketch

43. Locate correct Beginning and end of curve
Contd..
3. Establish the correct beginning of original circular curve;
the distance from intersection point to the beginning of
circular curve is given as
TL = R tan Δ/2 , and correct beginning of ST
TL1 = (R +S1)tan Δ/2 + L1/2 ;
Similarly correct ending of curve from intersection
point TS will be given as
TL2 = (R+S2) tan Δ/2 + L2/2 ,here a cross check can
be done to assess the actual disturbance of original curve.
Correct intersection of two tangents can be
established with the help of total station and correct
beginning and end of curve are to be located if they are found
to be disturbed.

44. Sketch

45. B. Based on versine measurement; Field
method (approximate method)
• Details given in IRICEN publication on curves;
• 1. Chainage of centre of curve = n – SS of Ve upto last station
• FSVe up to last station
• N = FSVe/V ( versine of curve as per degree of curvature)
• Beginning of curve = CC – (N+L)/2 where L is the length of proposed
transition length.
• End of curve = CC + (N+L)/2

46. Sketch

47. How to increase the transition.. Contd..
4. Work out the shift at crown = (L2² - L1²)/24R
4. Work out the shift due to increase in transition at every station and
check the feasibility and repercussion of such shift on ground for any
infringement. etc. This can be done by use of software

48. 3. How to increase the radius of curve
• Increase in radius involve downward shifting of the entire
curve.
• The shift of curve at apex
= (R2-R1)Sec Δ/2 +(R1-R2)
Altogether a new curve is to
Be laid if shift is large.

49. Sketch

50. How to increase the radius of curve
Steps to be followed.
1. Work out the correct location of start and end of existing
curve as explained earlier
2. Calculate revised transition length with proposed speed and
new proposed radius of curve.
3. Calculate new beginning and end of curve i.e. shift of original
station to new station away from original direction towards
straight portion =((R2-R1)+(L2²/24R-L1²/24R))*tanΔ/2 + (L2-
L1)/2

51. How to increase the radius of curve contd…
• Finding shift from original curve:
• Ox1 = x1³/6R2 L2 similarly we can workout other offsets.
• At point 5 the shift is calculated as
• Ox5’ = x5³/6R2 L2 - x1’³/6R1L1
• Likewise workout the shift at each station
to decide further course of action.
Or go for complete laying of new curve
If shits are large.

52. 4. Turnout taking off from curved main line;
circular or transition portion.
• Two cases :
• 1. Turnout taking off circular portion of curve
• 2. Turnout taking of transition part of the curve

53. Safe Speed on Curve para- 403
• BROAD GAUGE :- V on circular portion= 0.27 √ R (Ca + Cd)
• (This is on the assumption that the centre to centre distance
between railheads is 1750 mm.)
• Speed in transition portion is governed by rate of change of Cd or Ca
which should not exceed 35/55 mm/sec ; Vmax1 = L/Cd/0.008 or
Vmax2 = L/Ca/0.008 whichever is minimum.
• For optimum speed Vmax1 = Vmax2 ie. Ca= Cd

54. Turnout taking off from transition
1. Turnout taking off from beginning of transition
2. Turnout taking off from end of transition
3. Turnout taking off from middle of transition

55. Provision in IRPWM for turnout taking off transition
• Para 410. No Change of Super-elevation over Turnouts:-
There should be no change of cant between points 20 metres
outside the toe of the switch and the nose of the crossing
except in cases where points and crossings have to be taken off
from the transitioned portion of a curve.
• Normally, turnouts should not be taken off the transitioned
portion of a main line curve. However, in exceptional cases,
when such a course is unavoidable a specific relaxation may be
given by the Chief Track Engineer of the Railway.
• In such cases, change of cant and/or curvature may be
permitted at the rates specified in Para 405 or such lesser rates
as may be prescribed.

56. Case 1- Turnout taking off from beginning of transition

57. Steps to be followed

58. Steps contd…

59. Contd….

60. Turnout taking off in between transition

61. Steps to be followed

62. turnout taking off from end of transition

64. Steps to be followed

65. Thanks
For
Your patient listening

66. Cant Deficiency
G
SE
θ
θ
CENT FORCE=MVh²/R
W
W Sinθ
Consider figure OBK
MVh²/R = W Sinθ1 = W* AK/G =W*(AB+BK)/G
= Mg*(CD + SE)/G or
GVh²/gR = (CD + SE) or
GVh²/gR – SE = CD
Now consider figure below, Taking moment about outer rail,
(MVh²/R – W Sinθ)*H = WG/2 Cos θ
H*MVh²/R – mg*H* SE/G = mgG/2 (since θ is small, cos θ =1)
Vh²/gR – SE/G = G/2H
G Vh²/gR – SE = G²/2H = CD
W Cosθ back

67. Cant Deficiency
G
SE
θ
θ
CENT FORCE=MVh²/R
W
W Sinθ
Consider figure OBK
MVh²/R = W Sinθ1 = W* AK/G =W*(AB+BK)
= Mg*(CD + SE)/G or
GVh²/gR = (CD + SE) or
GVh²/gR – SE = CD
Now consider figure below,
(MVh²/R – W Sinθ)= unbalanced force = M * ΔP
Vh²/R – g SE/G)= Δ P
g*(GVh²/gR – SE)/G= Δ P
g *CD /G = Δ P
Or alternately, CD = SEreq-SEeq
= GVh²/gR- GVeq²/gR
=G/g(Vh²/R- Veq²/R )
= G/g(Δ P )
back

68. lateral acceleration is comfortable at 0.10 g. and the rate of attainment
is 0.03 g per .
Considering the average vehicle roll tendency and allowing for variability
in tracks and vehicles, the rate of change of unbalanced lateral
acceleration acting on the passenger should not exceed 0.03 g per
second. In difficult situations, an acceleration of 0.04 g per second may
be acceptable.

69. AREMA 3.11c
• The desirable length of the spiral for main tracks where the
alignment is being entirely reconstructed or where the cost of
the realignment of the existing track will not be excessive
should be such that when passenger cars of average roll
tendency are to be operated the rate of change of the
unbalanced lateral acceleration acting on a passenger will not
exceed 0.03 g per sec .
• In this case longitudinal slope of the outer rail with respect to
the inner rail will not exceed 1/744
• which is based on an 85-foot long car

70. AREMA 3.11d
• It is recognized that in the case of realignment of existing
tracks, calculated length of spiral the construction of which
would result in excessive costs. Therefore, in such cases it is
felt that the length should be such that, with average roll
tendency of passenger cars operated on the track, the rate of
change of the unbalanced lateral acceleration acting on a
passenger will not exceed 0.04 g per sec