Electrical Revolution: Trapezoidal Speed

The speed – time curve of the main line service is replaced by trapezoidal curve.
The running and coasting period in the trapezoidal curve is simply replaced by constant speed as shown in the figure.
The area of the speed – time curve represents total distance travelled by the train.

Let

α = Acceleration in km / second

β = Retardation in km / second

V_m = Crest speed in km / hour

T = Total time in second

Acceleration time t₁ = V_m / α …. ( 1 )

Retardation time t₃ = V_m / β ……( 2 )

Free running time t₂ = T – ( t₁ + t₃ ) ….. ( 3 )

= T – ( V_m / α + V_m / β )

Total distance of run in km ( S )

= Distance travelled during acceleration +

Distance travelled during free run +

Distance travelled during braking

= Triangle Area OAB + Rectangle Area ABCD + Triangle Area CDE

= V_mt₁ / ( 2 × 3600 ) + V_mt₂ / ( 3600 ) + V_mt₃ / ( 2 × 3600 ) ….. ( 4 )

Substitute value of ( 1 ), ( 2 ) and ( 3 ) in the equation ( 4 )

S = V_m² / 7200 α + V_m { T – ( t1 + t3 ) } / 3600 + V_m² / 7200 β

S = V_m² / 7200 α + V_m / 3600 { T – ( t1 + t3 ) } + V_m² / 7200 β

S = V_m² / 7200 α + V_m / 3600 { T – ( V_m / α + V_m / β ) } + V_m² / 7200 β

S = V_m² / 7200 α + V_mT / 3600 – V_m² / 3600α – V_m² / 3600β ) } + V_m² / 7200 β

S = V_m² / 7200 α – V_m² / 3600α + V_mT / 3600 + V_m² / 7200 β – V_m² / 3600β

S = V_mT / 3600 – V_m² / 7200 α – V_m² / 3600β

Vm² / 3600 ( 1 / 2α + 1 / 2β ) – V_mT / 3600 + S = 0

Let us consider that k = ( 1 / 2α + 1 / 2β )

kVm² / 3600 – V_mT / 3600 + S = 0

kVm² – V_mT + 3600S = 0

Now

a = k

b = T

c = 3600S

Solution of the equation

V_m = b ± √ ( b² – 4ac ) / 2a

V_m = T ± √ ( T² – 4( k ) ( 3600S ) / { 2k }

V_m = T ± √ ( T² / 4k² – ( 3600S / k )

If we use positive sign for the solution of the equation, the value of V_m should be much higher than the expected.
The negative sign is used in practice.

V_m = T – √ ( T² / 4k² – ( 3600S / k )

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Electrical Revolution