##### Things To Be Known:

THINGS TO BE KNOWN BY GUARDS

Ruling Gradient is the term always the train guards come across. This is an attempt to make it clear for better understanding. Ruling gradient means the steepest gradient in the railroad which should be adequate enough to haul the maximum load at maximum permissible speed for the given hauling power of the engine.

The curves on the railroad provide extra resistance to the movement of trains and reduce the allowable gradient. If a curve happens to fall on a ruling gradient, and the train goes uphill, it will not be possible for a train to overcome the resistance caused by the gradient and the curve. In such cases either the train load is decreased or the gradient is made flattened.

This flattening of the grade is called grade compensation on curves. For Indian Railways, on BG the permitted grade compensation is 4% per degree of the curve. (An indication board will be provided on every curve with information like degree of curve, length of the curve, direction either left or righted).

Guards must know about the grade compensation also. For example, the ruling gradient is 1 in 100(1%) of the section, the curve resistance is 4% or 0.04 per degree curve. Then the curve resistance for 2-degree curve 0.08. For the given 2 degrees curve the compensation is

1– 0.08= 0.92.

Therefore the gradient permitted on 2-degree curve is 100/0.92=108.7. It means you can be provided a grade of 1 in 108.7 instead of 1 in 100 which comparatively flatter. Indian Railways have On BG; the permitted yard gradient is 1 in 400. Even if it has a 2-degree curve, it can be reduced to in 1 in 368 which is more than the allowed level of 1 in 270. So the grade compensation is not necessary for the yards. This is the reason why gradients are negotiated by curves.

Next thing is the coefficient of friction. It depends upon the speed of the locomotive and the condition of the rail surface. The higher the speed of the locomotive, the lower the coefficient of friction, which is about 0.1 for high-speed trains and 0.2 for low-speed trains. The condition of the rail surface, whether wet or dry, smooth or rough, etc., also plays an important role in deciding the value of the coefficient of friction. If the surface is very smooth, the coefficient of friction will be very low.

The important factors affecting the hauling power of the locomotive are the resistance due to wave action and track irregularities, resistance due to the wind. These two are proportional to the velocity and total load of the train. They are formulated as

Rwt= 0.00008WV and Rwind=0.0000006WV2

Where W is the total weight of the train and V is the velocity of the train.To sum up, the hauling power of locomotive depends upon the rolling resistance, the resistance due to speed, resistance due to wind and resistance due to the ruling gradient and curves.Tractive effort of the locomotive = Total train resistances

Hauling capacity of the locomotive is proportional to the number of the driving wheels, axle load and the coefficient of friction. The number of axles and the axle load for various locomotives can be found in the previous chapters.

Hauling capacity = no. of axles X Axle load X coefficient of friction. And it is expressed in terms of tonnes and given in the Working Time Table on Locomotive parameters.

Calculate the maximum permissible load that a BG locomotive with six pairs of driving wheels bearing an axle load of 22 t each can pull on a straight level track at a speed of 80km/h. Also calculate the reduction in speed if the train has to run on a rising gradient of 1 in 200. What would be the further reduction in speed if the train has to negotiate a 40 curve on the rising gradient? Assume the coefficient of friction to be 0.2 Solution

Hauling power of the locomotive = number of pairs of driving wheels X wt exerted on each pair X coefficient of friction =6 X 22 X 0.2= 26.4t. The total resistance negotiated by the train on a straight level rack at a speed of 80 km/h. R = Resistance due to friction + resistance due to wave action and track irregularities + resistance due to wind =0.0016w+0.00008wv +0.0000006wv2

Substituting the value of V = 80 km/h,R = 0.01184 W

Assuming total resistance = hauling power,W X 0.01184 = 26.4 t

Or

W= 26.4/0.01184 = 2229.72t Approx. 2230t

On a gradient of 1 in 200, there will be an additional resistance due to gradient equal

to W*%of slope. Since hauling power = total resistance,

26.4 =0.0016W + 0.00008WV + 0.0000006WV2 +W0.5/100

= W(0.0016+0.00008V+0.0000006V2+0.005)

Since W=2230t,

26.4 = 2230(0.0016+0.00008V+ 0.0000006V2)

On solving the equation further,

V=48.13km/h

Reduction in speed = 80 – 48.13 = 31.87km/h

On a curve of 40 on a rising gradient of 1 in 200, curve resistance will be equal to
R = 0.0004 X degree of curve X wt

= 0.0004 X 4 X W = 0.0016W

Hauling power of locomotive = total resistance, therefore,

26.4 = 0.0016W+0.00008WV+0.0000006WV2+ 0.005W+ 0.0016W

By substituting the value of W = 2230 t in the equation and solving further,V= 43.68km/h

Further reduction in speed = 48.13-43.68=4.45 km/h. Therefore,Maximum permissible train load = 1115t

Reduction in speed due to rising gradient = 31.87t, Further reduction in speed due to curvature= 4.45km/h

Compute the steepest gradient that a train of 20 wagons and a locomotive can negotiate given the following data: weight of each wagon =20t, weight of locomotive =150t , tractive effort of locomotive = 15t, rolling resistance of locomotive= 3 kg/t, rolling resistance of wagon= 2.5 kg/t, speed of the train= 60 km/h
Solution:

Rolling resistance due to wagons = rolling resistance of wagon X weight of wagon X number of wagons
= 2.5 X 20 X 20= 1000 kg=1t

Rolling resistance due to locomotive
=rolling resistance of locomotive X wt of locomotive

= 3 X 150=450kg=0.45t

Total rolling resistance = rolling resistance due to wagons + rolling resistance due to locomotive =1.00+0.45t=1.45t
Total weight of train = weight of all wagons +wt of locomotive
=20 X 20+150 =550t

Total train resistance= rolling resistance +resistance dependent on speed + resistance due to wind + resistance due to gradient
= 1.45+0.00008WV + 0.0000006WV2+W/g

=1.45+ 0.00008 X 550 X 60+0.0000006 X 550 X 602+(550/g)

=1.45 +2.64+1.19 +(550g)=5.28+(550/g)

Where g is the gradient. Tractive effort of locomotive = Total train resistance 15=5.28 +(550/g)

Or

g=56.5 =1/56= 1 in 56

Therefore, the steepest gradient that the train will be able to negotiate is I in 56.

Calculate the maximum permissible trainload that can be pulled by a locomotive with four pairs of driving wheels with an axle load of 28.42t each on a BG track with a ruling gradient of 1 in 200 and a maximum curvature of 3, traveling at a speed of 48.3 km/h. Take the coefficient of friction to be 0.2

##### Solution

Hauling capacity of locomotive =no. of pairs of driving wheels X axle load X coefficient of friction -=4 X 28.42 X 0.2=22.736t

(b) Total resistance of train = resistance due to friction + resistance due to speed +resistance due to wind +resistance due to gradient +resistance due to curve

= 0.0016W + 0.00008WV +0.0000006WV2+W(l/g) +0.0004WD

= 0.0016W + 0.00008W x 48.3+0.0000006W x (48.3)2 +W x (1/200) x 0.0004 x W X 3

( c ) Hauling capacity = total resistance

22.73 = 0.01306 W

Or

W = 1740 t

Therefore the maximum weight of the train is 1740 t.