Crest, average and schedule speed; speed-time curve

Crest, average and schedule speed

• Crest speed ($V_m$): the maximum speed reached by the train during a run.
• Average speed ($V_a$): distance between two stops divided by the actual running time (stop time not included). $V_a = \dfrac{\text{distance}}{\text{running time}}$
• Schedule speed ($V_s$): distance between two stops divided by running time plus stop time. $V_s = \dfrac{\text{distance}}{\text{running time} + \text{stop time}}$, so $V_s < V_a < V_m$.

Numerical: speed-time curve for the run

Given: $V_a = 42$ km/h, $D = 1400$ m $= 1.4$ km, $\alpha = 1.7$ km/h/s, $\beta = 3.3$ km/h/s. A simplified trapezoidal curve is assumed.

Running time:

$$T = \frac{3600 \times 1.4}{42} = 120\ \text{s}$$

With $K = \frac{1}{2}\left(\frac{1}{1.7}+\frac{1}{3.3}\right) = 0.4456$:

$$\begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600D}{K}} \\ &= \frac{120}{0.8913} - \sqrt{\left(\frac{120}{0.8913}\right)^2 - \frac{3600\times1.4}{0.4456}} \\ &= 134.64 - \sqrt{18128.5 - 11309.7} = 134.64 - 82.58 \\ &= 52.07\ \text{km/h} \end{aligned}$$

Times and distances:

 V (km/h)
 52.07 |      _________________
       |     /                 \
       |    /                   \
       |   /                     \
       |  /                       \
     0 +-/-------------------------\--- t (s)
       0  30.63             104.22  120

Answer: Crest speed $= 52.07$ km/h; acceleration 30.63 s, free run 73.59 s, braking 15.78 s.