Chapter 28 : Motion Along a Circular Path „„„„„ 579

In such a case, the superelevation is given by the relation,

Gv^2

S

gr

=

where G is the gauge of the track.

Example 28.9. The distance between the rails of the track is 1·67 m. How much the outer
rail be elevated for a curve of 500 m radius, in order that the resultant force may be normal at a
speed of 45 km. p.h.

Solution. Given : Gauge of the track (G) = 1·67 m ; Radius of the curve (r) = 500 m and
speed (v) = 45 km.p.h. = 12·5 m/s

We know that the superelevation,

(^22) 1·67 (12·5)
0·0533 m 53·3 mm
9·8 500
Gv
S
gr
×
== = =
× Ans.
28.9. EQUILIBRIUM SPEED FOR SUPERELEVATION
In last article, we have obtained a relation for the superelevation in railways. The superelevation
obtained, by this relation, is popularly known as equilibrium superelevation. It has been experienced,
all over the world, that different trains pass over the curve with different speeds. It is therefore
obvious that superelevation provided for a particular speed would not suit any other speed. A little
consideration will show, that as the superelevation increases with the square of the speed, therefore
at higher speeds there is a tendency for the train to overturn. Similarly, at lower speeds, there is a
tendency for the train to derail.
It will be interesting to know, that there are so many theories prevalent for the calculation of
superelevation all over the world. In most of the countries (including India) superelevation is provided
in such a way that the faster trains may run safely without the danger of overturning or discomfort to
the passengers due to insufficient superelevation. And at the same time, the slower trains may also
travel safely without the danger of derailment due to excessive superelevation. These days, the
superelevation is, usually, provided for equilibrium speed or weighted average speed under average
conditions. This point is illustrated in the followng example.
Example 28.10. Find the superelevation to be provided on a 1·67 m gauge curved track of
1000 m radius, if the speeds of the trains are as follows :
(a) 15 trains at a speed of 50 km.p.h.
(b) 10 trains at a speed of 60 km.p.h.
(c) 5 trains at a speed of 70 km.p.h.
(d) 2 trains at a speed of 80 km.p.h.
Solution. Given : Gauge of the track (G) = 1·67 m and radius of curved track (r) = 1000 m
We know that the equilibrium speed,
(15 50) (10 60) (5 70) (2 80) 1860
km.p.h.
15 10 5 2 32
v
×+×+×+×

+++
= 58·125 km.p.h. = 16·15 m/s
and superelevation,
(^22) 1·67 (16·15)
0·044 m = 44 mm
9·8 1000
Gv
S
gr
== =
×
Ans.