Crest Speed, Average Speed & Schedule Speed of an Electric Train

Crest Speed of Electric Train

The maximum speed attained by an electric train during running is termed as its "crest speed". It is denoted by V_m.

Average Speed of Electric Train

Average speed is defined as the mean of the speeds attained by the electric train from start to stop. In other words, average speed is defined as the ratio of the distance covered by the train between two stops to the actual time of run. The average speed of train is denoted by V_a and is given by,

$$\mathrm{\mathrm{Average\: speed,}V_{a}\:=\:\frac{\mathrm{Distance\: between \:two\: stops\: (\mathit{D})}}{\mathrm{Actual\: time\: of\: run (\mathit{T_{\mathrm{run}}})}}\:\mathrm{kmph}}$$

Schedule Speed of Electric Train

The schedule speed of an electric train is defined as the ratio of the distance between two stops to the actual time of the run including the time for stop. The schedule speed of a train is denoted by V_s and is given by,

$$\mathrm{\mathrm{Schedule\: speed,}V_{S}\:=\:\frac{\mathrm{Distance\: between \:two\: stops\: (\mathit{D})}}{\mathrm{Actual\: time\: of\: run \:+\:Time\: for\: stop}}}$$

$$\mathrm{\therefore V_{S}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathrm{run}}}\:+\:\mathit{T_{\mathrm{stop}}}}\:=\:\frac{\mathit{D}}{\mathit{T_{S}}}}$$

Where,T_s is the schedule time.

Numerical Example

An electric train has maximum speed of 80 kmph and the distance between the two stops is 6 km. Determine the average speed and schedule speed over the run assuming −

• Free running of 300 seconds.

• Duration of stops as 60 seconds.

Solution

Given,

• Maximum speed,V_m = 80 kmph

• Distance between two stops,D = 6 km

• Actaul run time,T_run = 300 seconds = $\frac{300}{3600}$hours

• Stop time,T_stop = 60 seconds = $\frac{60}{3600}$ hours

Therefore, the average speed of the train is,

$$\mathrm{\mathrm{Average\: speed}\:V_{\mathit{a}}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathit{run}}}}\:=\:\frac{6}{\mathrm{\left ( 360/3600 \right )}}\:=\:72\:\mathrm{kmph}}$$

Now, the schedule time is given by,

$$\mathrm{\mathit{T_{\mathit{S}}}\:=\:\mathit{T_{\mathrm{run}}}\:=\:\mathit{T_{\mathrm{stop}}}\:=\:300\:+60\:=\:360\:\mathrm{sec}\:=\:\frac{360}{3600}\:\mathrm{hours}}$$

Therefore, the schedule speed of the train is given by,

$$\mathrm{\mathrm{Schedule\: speed}\:V_{\mathit{S}}\:=\:\frac{\mathit{D}}{\mathit{T_{\mathit{S}}}}\:=\:\frac{6}{\mathrm{\left ( 360/3600 \right )}}\:=\:60\:\mathrm{kmph}}$$