(19-03-2009, 04:17 AM)alexgoei Wrote: In the interest of facilitating the review, I have decided to resubmit the graphs, one from C to Station D and the other from Junction B to Station D. Salient points to highlight in these graphs are:
1 Takes on board your earlier comments about the Speed Limit for turnouts as applying to the divergence or convergence only. So you will find that in the graph from C to Station D, it is a simple straight line deceleration although the impact on the signal at braking distance to the platfrom at Station D is minimal.
2 When the rear of the train clears chainage 4090 leaving Station D platform, the total distance is 450 metres. This includes 25 metres from the platform stopping position to the signal at the end of the platform (assumed), the length of the train, 200 metres, and the overlap of 225 metres.
3 Based on the braking distance of 1115 metres, this would be the protecting signal (if using 3 aspect signalling) for entering into the platform at Station D.
Think about your revised graph:
A) the train should be shown to decelerate from its HEADWAY speed of 27.78m/s. Calculating suggests that this would take 56sec and cover 772m at deceleration rate of 0.5m/s/s. The signal however would as you correctly say have been placed at a distance of 1115m (as there must be braking from the MAXIMUM speed of 33.33m/s). Hence for our train there is more distance available than actually needed for its braking. You need to decide whether your calculations are going to assume that:
i) the train continues past the signal at Y but initially maintains speed at 27.78m/s before then braking at the last possible time in the last 772m at 0.5m/s/s prior to the stopping place (assuming from what you have written before this would be at 3640-25m representing defensive driving)
ii) the train brakes on passing the signal, not at 0.5m/s/s but at a lower uniform rate that just allows it to come to a stand within the 1115m it has available to it (well 1090m if you are modelling a 25m defensive driving approach to the signal)
iii) the train brakes initially at 0.5m/s/s, then runs at some reduced constant speed, say 15m/s, before then braking again at 0.5m/s/s from that speed to come to a stand 25m prior to signal.
Option iii) is the most realistic but the most complicated to calculate;
I think that option ii) may be the best for the exam,
older texts suggest i) which perhaps was realistic before the days of defensive driving techniques.
B) I agree that the signal in rear can clear when the rear of the first train has reached 4090. However what your graph does not depict that the first train continues to accelerate up to 27.78 m/s- it is not constrained to run for a time at 11.11m/s (however if you are considering a train leaving the loop platform then it is indeed constrained to run at 11.11m/s for an initial period, but only until rear of train has passed 3700).
Hence diagrams do need further amendment and this affects the calcs
PJW