•MARCH 2015 Research, Planning & Coordination Department 2

For giving motion to a train, following resistances must

be exceeded:

1. Locomotive resistance (Rl)

2. Wagon resistance (Rv)

3. Curve resistance (Rk)

4. Slope(ramp) resistance (Rr)

Total resistance R=Rl+Rv+Rr+Rk

TRACTION POWER

MOTION OF A TRAIN

•MARCH 2015 Research, Planning & Coordination Department 3

We symbolize traction force with F:

R>F The train can not move

R=F The train moves with a constant speed and

without any acceleration.

R<F The move with acceleration and speed

increases by the time

TRACTION POWER

MOTION OF A TRAIN

•MARCH 2015 Research, Planning & Coordination Department 4

Rl= (rl).(Gl)

rl== 0,65 + (13,15/P) + (0,00932 . V) + (0,004526 . A . V²)/(P.N)

• P = average axle load (tonnes)

• N = number of axle

• V = speed (km/h)

• A = locomotive area of view(m²)

• G L = mass of locomotive (tonnes) ifade eder.

Rl unit is Kp and rl unit is kp/tonnes

TRACTION POWER

LOCOMOTIVE RESISTANCE (Rl)

•MARCH 2015 Research, Planning & Coordination Department 5

Rv= (rv).(Gv)

Freight Wagon: r v = 2 + 0,057 . (V²/100)

Passenger Wagon: r v = 1,968 + 0,00932 . V +

0,000161. V²

• G v = mass of wagon

• V = train speed (km/h)

Rv unit is Kp and rv unit is kp/tonnes

TRACTION POWER

LOCOMOTIVE RESISTANCE (Rv)

•MARCH 2015 Research, Planning & Coordination Department 6

R k = r k . (G l + G v)

rk= 650/(R-55)

• R is the curve with minimum radius among the

all curves.

Rk unit is Kp and rk unit is kp/tonnes

TRACTION POWER

CURVE RESISTANCE (Rk)

•MARCH 2015 Research, Planning & Coordination Department 7

TRACTION POWER

SLOPE RESISTANCE (Rk)

R r = rr . (G l + G v)

rr = i

Rk unit is Kp and rk unit is kp/tonnes

•MARCH 2015 Research, Planning & Coordination Department 8

100 tonnes and 6 axle-locomotive will carry 700 tonnes burden with 40

km/h speed. What is the power of this locomotive on alignment and on

% 05, % 010, % 015, % 020, %025 slopes?

(Locomotive view area is 10 m2)

13,15 0,004526 . A . V2

r l = 0,65 + ---------- + 0,00932 . V + ------------------------------ Kp/ton

P P . N

13,15 0,004526 . 10 . 402

r l = 0,65 + ---------- + 0,00932 . V + ------------------------------ = 2,535 Kp/ton

16,66 16,66 . 6

TRACTION POWER

EXAMPLE

•MARCH 2015 Research, Planning & Coordination Department 9

R l = r l . G l

= 2,535 . 100

= 253 Kp

V2

r v = 2 + 0,057 -------- Kp/ton

100

402

r v = 2 + 0,057 -------- = 2,912 Kp/ton

100

R v = r v . G v

= 2,912 . 700

= 2038 Kp

TRACTION POWER

EXAMPLE

•MARCH 2015 Research, Planning & Coordination Department 10

•R r = rr . (G l + G v) and on alignment R r = 0

•On % 05 slope Rr = 5 . (100 + 700) = 4000 Kp

•On % 010 slope Rr = 10 . (100 + 700) = 8000 Kp

•On % 015 slope Rr = 15 . (100 + 700) = 12000 Kp

•On % 020 slope Rr = 20 . (100 + 700) = 16000 Kp

•On % 025 slope Rr = 25 . (100 + 700) = 20000 Kp

TRACTION POWER

EXAMPLE

•MARCH 2015 Research, Planning & Coordination Department 11

R = R L + R v + R r = T

On alignment R = R L + R v olur ve R = 253 + 2038 = 2291 Kp

On % 05 slope R = RL + Rv + Rr and R = 253 + 2038 + 4000 =6291 Kp

On % 010 slope R = RL + Rv + Rr and R = 253 + 2038 + 8000 = 10291 Kp

On % 015 slope R = RL + Rv + Rr and R = 253 + 2038 + 12000 = 14291 Kp

On % 020 slope R = RL + Rv + Rr and R = 253 + 2038 + 16000 = 18291 Kp

On % 025 slope R = RL + Rv + Rr and R = 253 + 2038 + 20000 = 22291 Kp

TRACTION POWER

EXAMPLE

•MARCH 2015 Research, Planning & Coordination Department 12

P=T.V/367 (kW)

On alignment P=250 kW

On % 05 slope P = 686 kW

On % 010 slope P = 1122 kW

On % 015 slope P = 1558 kW

On % 020 slope P = 1994 kW

On % 025 slope P = 2430 kW

TRACTION POWER

EXAMPLE

•MARCH 2015 Research, Planning & Coordination Department 13

%00 %05 %010 %015 %020 %025

RL (Kp) 253 253 253 253 253 253

Rv (Kp) 2038 2038 2038 2038 2038 2038

Rr (Kp) 0 4000 8000 12000 16000 20000

R = T(Kp) 2291 6291 10291 14291 18291 22291

P (KW) 250 686 1122 1558 1994 2430

TRACTION POWER

EXAMPLE