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Traction Mechanics

By on May 6, 2013

Traction Mechanics is the study of train motion.

The mechanics involves conversion of overhead electrical power/diesel/steam power in rail horse power to haul a train encountering acceleration, train, grade, curve resistance and similarly braking.

Traction Mechanics involves concept of

for detailing the motion of a train.

  • WAG9 hauling a freight train
  • Loco pilot with one hand on master controller, another on Train Brake handle with Loco Inspector
  • Accelerating the Vehicle by Master Controller
  • Passenger Train hauled by Diesel Locomotive
  • Freight Train hauled by Electric Locomotive
  • Less gap in between coaches means lessf air  resistance
  • Air resistance on the front surface
  • Curve Resistance
  • Curve-Grade-Acceleration
  • Assessment of Load to be hauled by Electric Traction in 1920 report
WAG9 hauling a freight train1 Loco pilot with one hand on master controller, another on Train Brake handle with Loco Inspector2 Accelerating the Vehicle by Master Controller3 Passenger Train hauled by Diesel Locomotive4 Freight Train hauled by Electric Locomotive5 Less gap in between coaches means lessf air  resistance6 Air resistance on the front surface7 Curve Resistance8 Curve-Grade-Acceleration9 Assessment of Load to be hauled by Electric Traction in 1920 report10
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Units

The units followed in day to day talk of Raiways are different as compared to  conventional MKS system. Conversion table is given below for better understanding

Item Symbol Railway convention MKS
Time t Hours 3600 sec
Distance D Kilometer 1000 meters
Speed V Kmph 1000/3600=0.277m/sec
Acceleration fa Kmph/sec 0.277m/sec/sec
Weight W Tonne 1000Kg
Force F Kgf 9.8Newton
Energy/Work kWh kWh W-sec or N-m or Joules

Formulas

V=U+at                          D=\frac{V+U}{2}t                                 D=Ut+\frac{1}{2}at^{2}             V^{2}+U^{2}=2aD

where D-distance;V-final speed; U-initial speed; a-acceleration

Force(N-m) = Mass (kg)\times Acceleration (m/sec^{2})

Work done (W)=Force(N-m)\times Distance(m)   where W in N-m or Joules or Watt-sec)

KE = \frac{1}{2}m\left ( V_{1}^{2}-V_{2}^{2} \right )=T\times D

PE=mgh

Rail Horse Power = \frac{TE\times V}{270}    where TE is in Kgf, V in Kmph and HP=75*Kgf*m/sec

OHE Current = \frac{HP\times 735.5}{OHE Voltage\times pF\times \eta }

where  η is efficiency  of Loco

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