- 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 t_{1} = V_{m} / α
…. ( 1 )

Retardation time t_{3} = V_{m} / β
……( 2 )

Free running time t_{2} = T – ( t_{1}
+ t_{3} ) ….. ( 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_{m}t_{1} / ( 2 × 3600 )
+ V_{m}t_{2} / ( 3600 ) + V_{m}t_{3} / ( 2 × 3600
) ….. ( 4 )

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

S = V_{m}^{2}
/ 7200 α + V_{m} { T – ( t1 + t3 ) } / 3600 + V_{m}^{2}
/ 7200 β

S = V_{m}^{2}
/ 7200 α + V_{m} / 3600 { T – ( t1 + t3 ) } + V_{m}^{2}
/ 7200 β

S = V_{m}^{2}
/ 7200 α + V_{m} / 3600 { T – ( V_{m} / α + V_{m} / β ) } + V_{m}^{2}
/ 7200 β

S = V_{m}^{2}
/ 7200 α + V_{m}T / 3600 – V_{m}^{2} / 3600α – V_{m}^{2} / 3600β ) } + V_{m}^{2}
/ 7200 β

S = V_{m}^{2}
/ 7200 α – V_{m}^{2} / 3600α + V_{m}T /
3600 + V_{m}^{2}
/ 7200 β – V_{m}^{2} / 3600β

S = V_{m}T
/ 3600 – V_{m}^{2}
/ 7200 α – V_{m}^{2} / 3600β

**OR**

Vm^{2}
/ 3600 ( 1 / 2α + 1 / 2β ) – V_{m}T / 3600 + S = 0

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

kVm^{2}
/ 3600 – V_{m}T
/ 3600 + S = 0

kVm^{2}
– V_{m}T
+ 3600S = 0

Now

a = k

b = T

c = 3600S

Solution of the
equation

V_{m} = b ± √ ( b^{2}
– 4ac ) / 2a

V_{m} = T ± √ ( T^{2 } – 4( k ) ( 3600S ) / { 2k }

V_{m}
= T ± √ ( T^{2 } – 4( k ) ( 3600S
) / { 2k }

V_{m} = T ± √ ( T^{2} /
4k^{2 } – ( 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 – √ ( T^{2} /
4k^{2 } – ( 3600S / k )

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