## 12/03/2021

### Trapezoidal Speed - time Curve

• 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

Vm = Crest speed in km / hour

T = Total time in second

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

Retardation time t3 = Vm / β ……( 2 )

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

= T – ( Vm / α + Vm / 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

= Vmt1 / ( 2 × 3600 ) + Vmt2 / ( 3600 ) + Vmt3 / ( 2 × 3600 )  ….. ( 4 )

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

S = Vm2 / 7200 α + Vm { T – ( t1 + t3 ) } / 3600 + Vm2 / 7200 β

S = Vm2 / 7200 α + Vm / 3600 { T – ( t1 + t3 ) } + Vm2 / 7200 β

S = Vm2 / 7200 α + Vm / 3600 { T – ( Vm / α + Vm / β ) } + Vm2 / 7200 β

S = Vm2 / 7200 α + VmT / 3600 – Vm2 / 3600α – Vm2 / 3600β ) } + Vm2 / 7200 β

S = Vm2 / 7200 α – Vm2 / 3600α + VmT / 3600 + Vm2 / 7200 β – Vm2 / 3600β

S = VmT / 3600Vm2 / 7200 α  – Vm2 / 3600β

OR

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

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

kVm2 / 3600 VmT / 3600 + S = 0

kVm2 VmT + 3600S = 0

Now

a = k

b = T

c = 3600S

Solution of the equation

Vm = b ± √ ( b2 – 4ac ) / 2a

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

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

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

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

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

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