17-07-2012, 01:32 PM
Broadly this looks ok and certainly you have got the answers to be different from each other this time which is good!
Hopefully drawing the graphs have strengthened your understanding.
The manner in which you have calculated t2 however does basically say:
a) the distance over which the train is given warning to brake is 2400m [TRUE]
b) the driver knows they only need 500m to come to a stand at their maximum service braking rate [TRUE]
c) therefore the driver will not act upon the cautionary aspects and will carry on at the norma speed regardless until the very lst instant when they will apply the brakes and come to a stand at the signal [NOT TRUE].
In reality drivers will always be more cautious than that and nowadays they are tauht that they must brake at cautionary aspects, significantly reduce speed, then coast buntil they get closer to their stopping place and then brake again but not at their maximum rate so that they have "something in hand" so as to be certain that even in poor conditions that they will indeed be able to stop by the required place.
The easiest way to make some approximation to this braking profile is to say instead:
a) the distance over which the train is given warning to brake is 2400m [TRUE]
b) the driver knows they only need 500m to come to a stand at their maximum service braking rate [TRUE],
c) the driver will utilise all the 2400m to stop by selecting a lower average braking rate than their train would be capable and hence have a more gentle reduction in speed (and is defensive in that this technique places lower reliance on friction between wheel and rail so less liable to slide in poor adhesion conditions, also if the driver has misjudged his position slightly or weight of the train and runs the risk of not being able to stop in time, can always select a higher brake rate and thus recover the situation- if they left braking to what they judged the last possibloe instant but had misjudged then they would have no escape and they would SPAD!)
Therefore you can calculate the time taken (t1+t2) by saying that d =2400, a= unknown, v=22.35m/s and u=0.
Another small thing is that t6 is actually the ADDITIONAL time after having cleared the overlap in order to regain full speed. Actually you didn't need to calculate t5 and t6 separately (as you'll see you actually deducted the value for t5 when calculating t6 but then add t5 to t6 in your end equation!). Also what you calculated isn't quite the stopping headway- you have calculated the time taken for a train to travel the distance commencing at the sighting point of the relevant signal approaching the station stop until regaining normal speed afterwards. What you need to do is compare this to the time the train would have taken to trravel the same distance if it had not been stopping to calculate how much EXTRA time is needed.
This is the time that a stopping train loses. To calculate the initial separation between trains needed prior to the stop in order that they are still separated by the minimum non-stop hyeadway after the station then you must add your calculated figure to the non-stop headway figure.
This is the "fast train following stopping train" headway.
If however you are wanting to calculate the "stopping train following another stopping train" headway, then you do not need to worry about the place at which the train has reaccelerated to its normal speed but just look at the separation in time between a train being at the sighting point for the restrictive signal, slowing down, dwelling, reaccelerating again to clear the overlap which then permits the following train to be driven in the same manner.
You actually got very close to obtaining the answe and certainly in the exam would have got a lot of credit for your explanation and working out, but you didn't quite calculate what you were intending to.
Hopefully drawing the graphs have strengthened your understanding.
The manner in which you have calculated t2 however does basically say:
a) the distance over which the train is given warning to brake is 2400m [TRUE]
b) the driver knows they only need 500m to come to a stand at their maximum service braking rate [TRUE]
c) therefore the driver will not act upon the cautionary aspects and will carry on at the norma speed regardless until the very lst instant when they will apply the brakes and come to a stand at the signal [NOT TRUE].
In reality drivers will always be more cautious than that and nowadays they are tauht that they must brake at cautionary aspects, significantly reduce speed, then coast buntil they get closer to their stopping place and then brake again but not at their maximum rate so that they have "something in hand" so as to be certain that even in poor conditions that they will indeed be able to stop by the required place.
The easiest way to make some approximation to this braking profile is to say instead:
a) the distance over which the train is given warning to brake is 2400m [TRUE]
b) the driver knows they only need 500m to come to a stand at their maximum service braking rate [TRUE],
c) the driver will utilise all the 2400m to stop by selecting a lower average braking rate than their train would be capable and hence have a more gentle reduction in speed (and is defensive in that this technique places lower reliance on friction between wheel and rail so less liable to slide in poor adhesion conditions, also if the driver has misjudged his position slightly or weight of the train and runs the risk of not being able to stop in time, can always select a higher brake rate and thus recover the situation- if they left braking to what they judged the last possibloe instant but had misjudged then they would have no escape and they would SPAD!)
Therefore you can calculate the time taken (t1+t2) by saying that d =2400, a= unknown, v=22.35m/s and u=0.
Another small thing is that t6 is actually the ADDITIONAL time after having cleared the overlap in order to regain full speed. Actually you didn't need to calculate t5 and t6 separately (as you'll see you actually deducted the value for t5 when calculating t6 but then add t5 to t6 in your end equation!). Also what you calculated isn't quite the stopping headway- you have calculated the time taken for a train to travel the distance commencing at the sighting point of the relevant signal approaching the station stop until regaining normal speed afterwards. What you need to do is compare this to the time the train would have taken to trravel the same distance if it had not been stopping to calculate how much EXTRA time is needed.
This is the time that a stopping train loses. To calculate the initial separation between trains needed prior to the stop in order that they are still separated by the minimum non-stop hyeadway after the station then you must add your calculated figure to the non-stop headway figure.
This is the "fast train following stopping train" headway.
If however you are wanting to calculate the "stopping train following another stopping train" headway, then you do not need to worry about the place at which the train has reaccelerated to its normal speed but just look at the separation in time between a train being at the sighting point for the restrictive signal, slowing down, dwelling, reaccelerating again to clear the overlap which then permits the following train to be driven in the same manner.
You actually got very close to obtaining the answe and certainly in the exam would have got a lot of credit for your explanation and working out, but you didn't quite calculate what you were intending to.
(17-07-2012, 05:23 AM)savitha kandasamy Wrote: hi,
I made modifications to the answer & tried to understand the concept by plotting in graph. Kindly review it.
savitha
PJW