UNIVERSITY OF DAR ES SALAAM

COLLEGE OF ENGINEERING AND TECHNOLOGY

DEPARTMENT OF TRANSPORTATION AND GEOTECHNICAL ENGINEERING
TR 325: Railway Engineering

Track Layout / Geometry Tutorial

Example 1:
Calculate the superevelation and maximum permissible speed for a two degree (2°) broad
gauge (BG) transitioned curve on a high-speed route with a maximum sanctioned speed
of 110 km/h. The speed for calculating the equilibrium superelevation as decided by the
chief engineer is 80km/h and the booked speed of goods trains is 50km/h.

i) The degree of a curve (D) is the angle subtended at its centre by a 30.5 m or 100
ft arc.
180𝐿 1750
𝑆𝑜 𝑓𝑜𝑟 𝐿 = 30.5 𝑚 → 𝐷 = =( )
𝜋𝑅 𝑅
1750 1750
𝑅= = = 875𝑚
𝐷 2
𝐺𝑉 2
ii) Superelevation for equillibrium speed, 𝑒 = 127𝑅
G = 1750 mm (c/c distance of 52-kg rail), V = 80kph and R = 875m
1750 ∗ 802
𝐶𝑎 = 𝑒 = = 100.8𝑚𝑚
127 ∗ 875

iii) Superelevation for maximum sanctioned speed (110kph);

𝐺𝑉 2 1750 ∗ 1102
= = 190.6𝑚𝑚
127𝑅 127 ∗ 875

Cant deficiency (Cd) = 190.6 -100.8 = 89.8 mm (which is less than 100mm hence
permissible)

iv) Superelevation for goods trains with a booked speed of 50kph

𝐺𝑉 2 1750 ∗ 502
= = 39.4𝑚𝑚
127𝑅 127 ∗ 875

Cant excess = 100.8 – 39.4 = 61.4 mm (which is less than 75 mm and hence permissible)

v) Maximum speed potential or safe speed of the curve as per theoretical

considerations being a high-speed route:
(𝐶𝑎 + 𝐶𝑑 ) ∗ 𝑅 (100.8 + 89.8) ∗ 875
𝑉=√ =√ = 110.1𝑘𝑝ℎ
13.76 13.76

vi) The maximum permissible speed on the curve is the least of;
a. Maximum sanctioned speed i.e., 110kph
 b. Maximum or safe speed over the curve based on theoretical
considerations, i.e., 110.1kph
c. There is no speed constraint due to the transition length of the curve

Therefore, the maximum permissible speed over the curve is 110kph and the
superelevation to be provided is 100.8 mm or approx. 100mm.

Table for limiting values of various parameters concerning curves

Table of various lengths of transition curves to be considered when calculating

speed
Example 2:
Calculate the maximum permissible speed on a curve of a high-speed BG Group A having
the following particulars: degree of the curve = 1°, superelevation (Ca) = 80 mm, length
of transition curve = 120m, maximum speed likely to be sanctioned for the section = 160
kph
1750 1750
i) Radius of the curve, 𝑅 = 𝐷 = 1 = 1750𝑚

ii) Safe speed over the curve as per theoretical considerations this being a high-
speed route, assume Cd = 100mm.
(𝐶𝑎 + 𝐶𝑑 ) ∗ 𝑅 (80 + 100) ∗ 1750
𝑉=√ = √ = 151.3𝑘𝑝ℎ
13.76 13.76

iii) Speed based on transition length;

a. Rate of change of cant (not to exceed 55 mm /sec)
198 ∗ 𝐿 198 ∗ 120
𝑉𝑚 = = = 297.0𝑘𝑝ℎ
𝐶𝑎 80

b. Rate of gain of cant deficiency (not to exceed 55 mm/sec)

198 ∗ 𝐿 198 ∗ 120
𝑉𝑚 = = = 237.6𝑘𝑝ℎ
𝐶𝑑 100
Cant gradient;
120 ∗ 1000
= 1 𝑖𝑛 = 1 𝑖𝑛 1500
80
(which is not steeper than 1 in 720)

iv) Maximum permissible speed is the least of the following;

a. Maximum sanctioned speed of the section, i.e., 160kph
b. Safe speed based on theoretical considerations, i.e., 151.3kph
c. Speed based on the transition length, i.e., 237.6kph

Therefore, the maximum permissible speed over the curve is 151.3 kph or about 150 kph.
As the controlling factor in this case is the safe speed based on the theoretical
considerations (and not the rate of chage of Cd) no further analysis is necessary.

Example 3:

Calculate the maximum permissible speed on a 1° curve on a route with a maximum

sanctioned speed of 130kph. The superelevation provided is 50 mm and the transition
length 60m. The transition length of the curve cannot be increased due to the proximity
of a yard.
1750 1750
i) Radius of the curve, 𝑅 = 𝐷 = 1 = 1750𝑚

ii) Safe speed on the curve as per theoretical considerations,

 (𝐶𝑎 + 𝐶𝑑 ) ∗ 𝑅 (50 + 100) ∗ 1750
𝑉=√ =√ = 138.3𝑘𝑝ℎ
13.76 13.76

iii) Speed based on transition length:

(a) Rate of change of cant (not to exceed 55 mm/sec)
198 ∗ 60
𝑉𝑚 = = 237.6𝑘𝑝ℎ
50
(b) Rate of change of cant deficiency (not to exceed 55mm/sec)
198 ∗ 𝐿 198 ∗ 60
𝑉𝑚 = = = 118.8𝑘𝑝ℎ
𝐶𝑑 100
(c) Cant gradient
60 ∗ 1000
= 1 𝑖𝑛 = 1 𝑖𝑛 1200
50
(which is not steeper than 1 in 720)

iv) Maximum sanctioned speed on the curve is the least of the following;
(a) Maximum speed sanctioned for the section, i.e., 130kph
(b) Safe speed based on theoretical considerations, i.e., 138.3kph
(c) Speed based on transition length, i.e., 118.3 kph

In this case, the speed has to be restiricted to 118.3kph because of the constraint of
transition length. A cant deficiency of 100 mm has been assumed, which is its maximum
possible value. On the field, the cant deficiency may be somewhat lower giving a lower
rate of change of Cd for the given transition length and a higher permissible speed. The
optimum value of this maximum permissible speed can be found from the following
equation;

Equilibrium superelevation = actual cant + cant deficiency for maximum permissible

speed for a given transition length

𝐺𝑉 2 198𝐿
= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑐𝑎𝑛𝑡 +
127𝑅 𝑉

Where G = 1750 mm, R = 1750 mm, L = 60 m and V is the maximum permissible speed.
Therefore,

1750𝑉 2 198 ∗ 60
= 50 +
127 ∗ 1750 𝑉

Solving this equation, V = 133 kph. This value however cannot be more than the MSS of
the section, i.e., 130kph. Therefore, the maximum permissible speed over the curve is
130kph.
198𝐿 198∗60
With V = 130kph, 𝐶𝑑 = = = 91.4𝑚𝑚
𝑉 130
This is less than 100mm. Therefore the maximum permissible speed over the circular
curve is 130kph and that over the transition curve is 118kph.

Example 4:
A BG branch line track takes off as a contrary flexure through a 1-in-12 turnout from a
main line track of a 3° curvature. Due to the turnout, the maximum permissible speed on
the branch line is 30kph. Calculate the negative superelevation to be provided on the
branch line track and the maximum permissible spped on the main line track (when it
takes off from a straight track).
(i) For a branch line track, the degree of the curve is 4 − 3 = 1°
1750 1750
𝑅𝑎𝑑𝑖𝑢𝑠 = = = 1750𝑚
𝐷 1
𝐺𝑉 2 1676𝑥302
𝑒= = = 6.8𝑚𝑚
127𝑅 1270𝑥1750

After rounding it off to a higher multiple of 5, it is taken as 10 mm.

(ii) The value of negative superelevation for a branch line track,

𝑥 = 𝑒 − 𝐶𝑑 = 10𝑚𝑚 − 75𝑚𝑚 = 65𝑚𝑚 (𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒)

(iii) The superelevation to be provided on the main line track is 65 mm, which is the
same as the superelevation of the branch line track but in the opposite
direction.
(iv) The maximum permissible speed is calculated by taking the actual
superelevation of the main line track (65 mm) and adding it to the cant
deficiency (75mm), and then using this value of superelevation, i.e., 140mm
(65 + 75) in the formula for equillibrium speed. The main line track has a in the
formula for equillibrium speed. The main line track has a 3° curve, i.e., 1750/3
= 583.3 m radius.
Therefore, the maximum permissible speed on the main line track,
𝐺𝑉 2 1676𝑥𝑉 2
𝑒= = = 140𝑚𝑚
127𝑅 127𝑥583.3
or
127𝑥583.3𝑥140
𝑉=√ = 78.7𝑘𝑝ℎ
1676
Alternatively, the maximum permissible speed can also be calculated as
(𝐶𝑎 + 𝐶𝑑 ) ∗ 𝑅
𝑉=√ = 0.27√(65 + 75)𝑥583.3 = 77.16𝑘𝑝ℎ
13.76
Therefore the maximum permissible speed on the main line track is 77.16kph.
After rounding it off to a lower multiple of 5, it becomes 75 kph.
Example 5:
A curve of 600 m radius on a BG section has a limited transition of 40 m length. Calculate
the maximum permissible speed and superelevation for the same. The maximum
sectional speed (MSS) is 100kph.
In normal situation, a curve of 600 m radius will have quite a long transition curve for an
MSS of 100kph. However as the transition curve has been restricted to 40 m, the cant
should be so selected that the speed on the main circular curve is equal to the speed on
the transition curve as a whole.
For the circular curve, the maximum speed is calculated from;

𝑉 = 0.27√𝑅(𝐶𝑎 + 𝐶𝑑 )

The most favourable value of speed is obtained when Ca = Cd

For the transition curve, the maximum change of cant is taken as 55 mm/sec and the
maximum speed is then calculated;
𝐶𝑎 𝑥𝑉𝑚 198𝐿
𝐿= 𝑜𝑟 𝑉 =
198 𝐶𝑎
Therefore,
198𝐿
0.27√𝑅(𝐶𝑎 + 𝐶𝑑 ) =
𝐶𝑎
Or
198𝑥40
0.27√600𝑥2𝐶𝑎 =
𝐶𝑎

On solving this equation 𝐶𝑎 = 89.50𝑚𝑚 ≅ 90𝑚𝑚

On limiting the value of Cd to 75mm,

Maximum speed = 0.27√𝑅(𝐶𝑎 + 𝐶𝑑 ) = 0.27√600(90 + 75) = 84.95 𝑜𝑟 𝑎𝑝𝑝𝑟𝑜𝑥. 85𝑘𝑝ℎ

90
Cant gradient = 40𝑥1000
= 1 𝑖𝑛 444

This is within the permissible limits of 1:360

Therefore, the maximum permissible speed is 85kph and the superelevation to be

provided is 90mm.