# Train, grade, curve and Acceleration Resistance

Trains resistance is defined in terms of force required to encounter resistance arising due to vehicle, track, grade, curve, acceleration, wind at different time and place etc. Study of these resistances and its impact in train motion is important to develop strategies for reducing it. It is measured in terms of Kg/tonne.

Primarily, train Resistance is bifurcated into internal and external resistance. The internal resistance is internal to the train and prevailing track geometry over the entire train run. External resistance is situational in nature.

### Internal Resistance

Internal resistance plays different role during start and running and subdivided so.

#### Starting Resistance

Starting resistance is to overcome the inertia and low temperature of the bearing, tightening of couplers. Resistance drops rapidly as the train speed increases. It may be noted that the entire train does not start at a time. There is always some slackness in the coupling and by pushing the train backward, all the coupler are brought in slack position. While pulling in such condition, the entire train starts one by one with tightening of the couplers. There is no derived formula for starting resistance. Based on the different measurements, empirically Train and Loco Resistance are taken as 4 and 6 kg/tonne.

#### Running resistance

Running resistance is given by an empirical formula **F _{rr} = a+bV+cV2 **where the dependency of constants a, b & c are constants, V is speed in Kmph and F

_{rr }is train resistance in Kg/Tonne. Constant factors a, b and c accounts for the entire possible variable which has its role in the running resistance having bearing on weight alone or weight & speed both. The factor a, b and c is worked out based on the actual fit-in equation for the behavior of acceleration or retardation curve i.e. speed versus time. These are explained as follows;

#### Constant Factor a

It is independent of speed and multiplied solely with load to get the resistance. It is primarily consists of

Rolling: It results due to rolling resistance between the wheel tread and head of the rail. Behavior of rolling resistance is different to friction resistance. Coefficient of rolling resistance on-line like coefficient of friction is given by µ_{r} and is very small of the order of .0001—0002 as compared to coefficient of friction between steel to steel body which is around 0.73-0.75 in laboratory conditions. A certain amount of flattening of the surfaces in contact also contributes. Rolling resistance increases with an increase in weight. Empirically, resistance on this account is taken as 0.0016W where W is the weight of the vehicle.

Track: Due to loading, track is depressed in proportion to the loading and stiffness of the track structure. At any moment, the track is having dips and ups on which the wheel is moving. This absorbs some energy and adds to the resistance. The amplitude of dips is small in case of well cushioned ballast track and so the track resistance as compared to ballast deficient track, loose joints etc.

Journal Bearing: The term appears from the time when journal bearing were commonly used. Now, roller bearings are only used which has helped in reducing the bearing resistance considerably. Bearing resistance is higher at the start and tapers down considerably up to a speed of 20 kmph and then slow drop. Beside reduces resistance, other advantages are smooth start, reduces cases of hot boxes, less maintenance cost. During winter, the starting resistance goes up due to low temperature and higher bearing friction.

#### Constant Factor b

The resistance due to this cause is directly proportional to the speed of the vehicle. Even on tangent track, the flange of the wheel touches the inner face of the rail in a sinusoidal motion thus creating a retarding force due to sliding friction. Track irregularities and load distribution influence the hunting causing frequency of sinusoidal motion and affecting resistance to the motion. This frictional resistance goes up on curved track and with increasing speed. So it is the lateral displacement of the wheel during the run and energy loss due to sliding friction results into this cause. It is easy to evaluate factor a and c through instrumentation and tests, therefore contribution of this cause is worked out by subtracting the contribution of these two factors from the total resistance. Empirically, resistance on this account is taken as 0.00008WV.

#### Constant Factor c

This resistance factor c due to this cause is directly proportional to the square of the speed and directly with the cross-sectional area of the vehicle therefore; its contribution is more visible with speed going more than 80 kmph. This resistance can be written as cAV^{2}. The factor accounts for the resistance of quite air which the vehicle envelope has to displace continuously during the run. The envelope, thus, is not only the frontal cross-section but also the space in-between vehicle, fittings under the vehicle, skin resistance of the sides, turbulence and draft created at the rear end. The effect of these additional cross-sections is generally not mentioned separately but included in c when worked out for a specific design of locomotive and trailing stock. Empirically, Train resistance on this account is taken as 0.0000006WV^{2}

#### Managing the effect of air resistance

- It has been estimated that there is little gain in design train envelope for air resistance up-to a speed of 80 kmph.
- Egg or naturally suspended shape is naturally occurring most streamlined shapes and more nearer the actual shape is towards that, the better it is
- Rounded or triangular shape in the front, reduced inter-vehicle space, shrouding of under-floor mounting fittings are some of the practice tried by IR
- A smooth exterior surface with minimum protrusions helps in reducing skin resistance.

#### Empirical values for different types of rolling stock

Different values of a, b and c forming the formula a+bv+cv2 are as given below:

- Locomotive: 0.647+ (13.17/W) +0.00933V+ (0.057/Wn)
^{2} - Locomotive (WAP5) 1.34819+.02153V +0.00008358V
^{2 }(Source Feasibility Study by METI, Japan) - Passenger: 1.43+0.0054V+0.000253V
^{2}(Source Feasibility Study by METI, Japan) - Passenger (LHB) 0.69873+.021533V+0.0000835V
^{2} - BOXN (E): 1.333973+0.021983V+0.000242V
^{2} - BOXN (L): 0.64388+0.010472V+0.00007323V
^{2} - Conraj: 0.8+0.011V+0.00035V
^{2} - Motor Coach: 2.35 + (0.02933-.00049w)V+(0.03722/w)V
^{2} - Trailer Coach: 1.347 + 0.00385V + 0.000165V
^{2} - MEMU Trailer: 0.6855 + 0.02112V + 0.000082V
^{2}

Running train resistance as given by SNCF

1.5/10/axle load + V^{2}/120*axle load

Graphical representation

### Air resistance of Container trains

Container trains popularly called in Indian Railways as CONRAJ works at a maximum permissible speed of 100 kmph and factor c plays an important role in train resistance as given below:

Passenger: 0.000253

BOXN (E): 0.000242

BOXN (L): 0.00007323

Conraj: 0.00035

Trailer Coach: 0.000165

Train resistance of conraj is similar to BOX(N) empty train but with load between 2000-3000T. The train resistance at 100kmph is 5.4 Kg/Tonne which further gets aggravated due to container of different height and empty wagons in between which increases the frontal cross-section. Rail Horse Power required to haul a 2000T train at 100kmph on flat territory will be

Train Resistance: 2000*5.4=10800Kg; RHP= 10800*100/270=4000HP. A train resistance of 5.4 kg/tonne is on the high side of a loaded train. It is important to place all empty wagons in one hook which will only add to the train resistance per tonne and may be difficult to haul at maximum permissible speed with an existing fleet of locomotive.

### External Resistance

External resistances are those which are not fixed and depend on varying terrain (gradient and curve), prevailing conditions of air (speed and humidity), self generation for lighting and air conditioning, brake binding etc.

#### Grade Resistance

Grade resistance arises due to effort made work against gravitational force. While climbing on a grade, resistive force of WxSinθ exists. Sinθ can be written as G is defined as Vertical height/longitudinal distance (1/100 is the vertical height of 1m for every 100m longitudinal travel). Grade resistance is then written as WG

Contribution of grade resistance plays an important role in limiting the hauling capacity in a section. In view of this ruling gradient is defined as the maximum gradient in a section based on which the hauling capacity is decided. Length and location of ruling gradient can help in increasing hauling capacity wherein the attaching speed can help in overcoming the grade resistance more than hauling capacity of the train.

#### Curve Resistance

Curve resistance is caused due to

1. The flange of the outer wheel of the leading axle rubs against the inner face of the outer rail causing resistance due to sliding friction.

2. The outer wheel rotates faster than the inner wheel causing transverse slip thus adding to sliding friction

3. Less super elevation causes increased pressure on outer rail whereas higher super-elevation results in increased pressure on the inner rail thus increasing curve resistance.

Curve resistance is given as = Constant*Force of Sliding friction*Gauze/Radius of curvature

And empirical formula is given as 0.4C where C is the degree of the curve; 2^{0} curve will offer a curve resistance of 0.4*2 *W kg

Curve resistance is quite often compensated or offset by a reduction in the gradient to be within the limits of ruling gradient and is called compensated gradient. A section is having ruling gradient 1/200. For a curve to exist in the section of say 3^{0} (583meters), the grade shall be limited by flattening to (5/1000)-(0.4×3/1000)= 3.8/1000=1/263 so that it is within the limits of ruling gradient.

#### Resistance due to acceleration

This is the force exerted by the locomotive to accelerate the rolling stock and calculated as per Newton Second Law of Motion.

F_{a}=Accelerating Force=ma; or = W*a Newton;

Where W is weight of train and locomotive in tonne and a is acceleration in m/sec^{2}

And also expressed in Tonnef as

** **F_{a} = W*fa/g kgf

*This is ok when there are no rotating parts but in actual practice the wheels, motor armature, gears have to be accelerated in an angular direction therefore weight is multiplied by a factor K which is taken as a 1.2, 1.1-1.15 and 1.03 for locomotive, EMU and freight train respectively so *

* F _{a = }(W/g)*K*fa*

#### Resistance due to self generation

Self and End-on-generation system is provided on coaching trains for lighting and air conditioning on trains. The train lighting coach is provided with 1 no. 4.5kW(6HP) and AC coach with 2 nos. of 25kW(33.5HP) alternator. The train resistance offered by these machines can be found out as follows:

Train Resistance (in Kgf)*Speed (kmph)/270=HP; let the redundancy factor is e and efficiency of generation is n, then we get

Train resistance (in kgf)/coach= HP*270*e /v*n

Proceedings of the AREA 1942 Vol 43 has given empirical derivation for the resistance of self generation as

R_{4kW} = 87.5/(0.94V) and R_{15kW }= 59.2/(0.925*V) where R is pound/ton and V in mph

#### Wind Resistance

Wind resistance is different to air resistance which is for quite air. Wind speed of 5 Km/h is very common, and therefore, to be added to speed of the vehicle if moving in the opposite direction. Lateral flow adds to the flange resistance.

Humidity also affects air resistance. The density of humid air is less as compared to dry air therefore will offer less resistance as compared to dry air. There is a myth that density of humid air is more which is not so, but this myth is due to poor visibility due to water molecules construed as higher density of air. With humidity going up means nitrogen and oxygen molecules are replaced by hydrogen and oxygen molecules which is having less molecular weight therefore weight is less for the same volume in case of humid air. This is an opinion only as no validation records of measurement of wind resistance with varied humidity is available.

Dear Mr. Jain,

Thank you for detailed insights. Could you share any insight into the difference in air resistance between an aircon LHB coach and non-Aircon open window trains run by IR ?

The coefficient “c”for non-aircon i.e. open window will be more due to air getting more lateral surface to react. I don’t have the actual figure.

For wap 7 what is the Different values of a, b and c the formula a+bv+cv2 ?

Starting resistance which u have shown pertain to journal bearings. I have seen somewhere that stating resistance for roller bearings is 5 lbs/ton.please confirm if I am correct

There is huge difference of resistance per ton when calculated by the formula 0.0016 W+o.00008 VW +0.0000006 VWW and the resistances given for each type of stock. Could u kindly explain

The value of the resistance mentioned are as adopted by RDSO for all its calculation. I am doubtful whether RDSO has ever tried to validate the formula by its own studies with the introduction of new technology. So you may be correct.

Can you please give an example of the deceleration (m/s/s) of a train, based on the formulae above, given that the train is just rolling without applying any traction?

I have tried to do this my self but I am not sure that I have calculated correctly.

Assume a=1.3, b=0.02 and c= 0.0002 and v = 160 km/h . This gives Frr = 1.3 + 3.2 + 5.12 = 9.62 kg/tonne. Assume that the train weighs 280 ton => 9.62*280 = 2694 kg. Now, I need a force (N) and convert kg to Newton by multiplying by 10 (earth gravitation) => 26940 N. This can now be converted to a deceleration (m/s/s) with the formula a = F/m = 26940/280000 =0.1 m/s/s.

The Loco has a pulling force of 235000 N. This gives (using a = F/m), an acceleration of 235000/280000 = 0.84 m/s/s. When the train runs at full speed (160 km/h) it would still have an acceleration power of 0.84 – 0.1 = 0.74 m/s/s. Have I used your formulae correctly?

r u sure the loco having pulling force of 235000 N at 160 km/ph. You will get the answer, if you work out the HP of the locomotive. As regards, calculation, it is ok.

Dear Sir,

How to use these formulas for a formation that has 1Driving Power Car + 4 Trailer Coach + 1Driving Trailer Car do we have to add below 2

Motor Coach: 2.35 + (0.02933-.00049w)V+(0.03722/w)V2

Trailer Coach: 1.347 + 0.00385V + 0.000165V2

were the driving trailer car will fit in this equation?