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+----- Thread: 2011 Headway calculation (/showthread.php?tid=978)
RE: 2011 Headway calculation - kball - 11-03-2012
Thanks PJW,
So let me see if I have understood this in my own words. I will write it up later to see if I really have got it.
The headway time for a non stopping train following a stopping train can be calculated by adding the difference from the headway time for a stopping train following a stopping train, minus the time a non stopping train would take to cover the same distance.
Ht (ns f s) = Ht (ns) + [Ht(s f s) - time taken for Ht(ns) to cover the same distance]
It looks more understandable on my sketch pad!
And it only appears to add an extra 110 seconds to the headway time, which would fit in with the question requirements. Making 3mins 10secs headway time.
Regards
KJB
RE: 2011 Headway calculation - PJW - 11-03-2012
I haven't done the maths, but yes I THINK you have understood although I'd have expressed a bit differently (I think you didn't write quite what you meant to)
The headway time for a non stopping train following a stopping train can be calculated by adding:
1. the headway time for a non-stopping train following a non-stopping train,
2. the difference between the time taken by a stopping train and a non-stopping train (i.e. the time taken for a stopping train to decelerate, dwell and re-accelerate to the headway speed, minus the time a non-stopping train would take to cover the same distance).
Ht (f-s) = Ht (f-f) + [t(s) - time taken for t(f) to cover the same distance]
where f=fast (non stop) and s=slow (stopping)
Actually this is a bit of a "cheat" method since it purely relies on consideration of Newton's Laws and does not consider the signalling for the situation prior to the fast train catching up the slow to be at minimum no-stop headway distance. It's ok for this question, but not for a calculation of "slow following slow", when the aspects seen by the 2nd train are relevant.
(11-03-2012, 08:39 PM)kball Wrote: Thanks PJW,
So let me see if I have understood this in my own words. I will write it up later to see if I really have got it.
The headway time for a non stopping train following a stopping train can be calculated by adding the difference from the headway time for a stopping train following a stopping train, minus the time a non stopping train would take to cover the same distance.
Ht (ns f s) = Ht (ns) + [Ht(s f s) - time taken for Ht(ns) to cover the same distance]
It looks more understandable on my sketch pad!
And it only appears to add an extra 110 seconds to the headway time, which would fit in with the question requirements. Making 3mins 10secs headway time.
Regards
KJB
RE: 2011 Headway calculation - kball - 14-03-2012
Thank you again for your comments.
I have updated my answer to version 4.
Tha main change being the revised calcs for a non stopping train following a stopping.
If this looks alright, albeit too much for the exam time allowed, I will have a go at using this information to mark up the layout.
Many thanks
KJB
RE: 2011 Headway calculation - PJW - 17-03-2012
In principle OK, but don't quite agree.
The headway speed is 140km/h which, to your level of approximation, is 40m/s and the braking distance therefore under 1600m rather than the 2000m shown which reflects braking from maximum permissible speed.
Since you were only interested in the differential time between the 2 trains then the consideration of the Sighting Distance SD and the Clearance Distance CD are irrelevant since they are the same in each case. Indeed introducing these just complicates things, particularly as the reference to train length and overlap in an area unrelated to the platform starting signal displays a certain level of confusion.
When on page 7 you state "a non-stopping train would have travelled in the same time 4600m" you do not mean that! Rather you ought to say that "a non stopping train would cover the same 4600m in a shorter time", as indeed you go on to calculate to be 115 seconds compared to the 205 seconds.
Obviously since you have used 2000m rather than 1600m then your relative times of the two trains are out; you have actually made the non-stop one go a distance of 4000m in the time that you claim the stopping train travels 4000m yet the time calculated is consistent with real 3200m.
For the stopping, I agree BDt and ADt are 80 sec; since constant acceleration / deceleration, then the average speed is 20m/s. The non-stop train is travelling at 40m/s so it will catch up by a total of 40+30+40 = 110 seconds. You made it 90 seconds (which is explained by the 800m error at 40m/s = 20 seconds).
As ever don't mix up Max permissible Speed and headway Speed
So Ht4(ns-s) = 90 + 110 = 200s
Ht3(ns-s) = 115 + 110 = 225s
and thus the stopping headway requirement is met by either comfortably, but the non-stop requirement is too onerous for 3 aspect signalling to achieve in practice given that there are bound to be other constraints on placing signals and there is not much tolerance.
(14-03-2012, 02:56 PM)kball Wrote: Thank you again for your comments.
I have updated my answer to version 4.
Tha main change being the revised calcs for a non stopping train following a stopping.
If this looks alright, albeit too much for the exam time allowed, I will have a go at using this information to mark up the layout.
Many thanks
KJB
2011 Calculations - arpan_singhania - 01-05-2012
Hello,
I have done calcutions for 2011 Mod 2 Mainline Paper. Would be grateful if somebody can check and give some feedback based on their experience.
Thanks & Regards,
Arpan Singhania
RE: 2011 Calculations - PJW - 01-05-2012
Arpan-
Happy to, but I won't do until Wed or thurs evening
PJW
(01-05-2012, 11:41 AM)arpan_singhania Wrote: Hello,
I have done calcutions for 2011 Mod 2 Mainline Paper. Would be grateful if somebody can check and give some feedback based on their experience.
Thanks & Regards,
Arpan Singhania
RE: 2011 Calculations - PJW - 02-05-2012
(01-05-2012, 12:51 PM)PJW Wrote: Please see comments in the attached.
Generally a reasonable attempt particularly for the non-stop calculations, but you did fall into some of the common traps:
a) confusion between MAXIMUM PERMISSIBLE speed and HEADWAY speed,
b) Not defining all abbreviations
c) Quoting specific formulae without explaining their derivation and demonstrating a "signalling understanding".
I have attached another attachment that I hope you will find useful. It is intended as a template to guide how calculations for a typical exam paper should be presented / explained.
Regarding the "fast following stopping train" you didn't seem to be calculating what you needed to; certainly you did not give an adequate EXPLANATION of how these worked out the headway. Whereas the calculations you did were ok, they weren't the full story. In particular you did not give the impression that you understood the signalling rationale- you did not mention the 2nd train at all.
Have a look at some of the other attempted calculations in this section.
(01-05-2012, 11:41 AM)arpan_singhania Wrote: Hello,
I have done calcutions for 2011 Mod 2 Mainline Paper. Would be grateful if somebody can check and give some feedback based on their experience.
Thanks & Regards,
Arpan Singhania
RE: 2011 Calculations - arpan_singhania - 02-05-2012
Thank you very much Peter for your comments.
(02-05-2012, 07:44 PM)PJW Wrote:(01-05-2012, 12:51 PM)PJW Wrote: Please see comments in the attached.
Generally a reasonable attempt particularly for the non-stop calculations, but you did fall into some of the common traps.......
(01-05-2012, 11:41 AM)arpan_singhania Wrote: Hello,
I have done calcutions for 2011 Mod 2 Mainline Paper. Would be grateful if somebody can check and give some feedback based on their experience.
Thanks & Regards,
Arpan Singhania
2011 - Mod 2 - Question doubt - NJK - 03-07-2012
Dear Members,
I have one doubt in 2011 Module main line practice layout question paper.
In question 1, part b given as
Determine graphically or by calculation the theoretical best headway (without any allowances) at minimum signal spacing and the given speed for application on layout 1.
Kindly clarify, meaning for "without any allowances" in this question
RE: 2011 - Mod 2 - Question doubt - PJW - 03-07-2012
It means what is the best headway which can be achieved given the precise values given. In reality one would generally make some "less than ideal" asssuumptions to closer reflect reality and these would be bound to detract from the figure that would be achieved in practice.
Hence in a case having a braking distance of for example 750m, 3 aspects could not ever be placed closer than that (excluding modified sequence that ruins the headway anyway) but in the real world often there is some conflict that prevents the precise position so therefore the real spacing will be larger than 750m, perhaps 760m, perhaps 779m, perhaps 805m etc. The headway for the line is governed by the longest spacing, so when calculating the headway for the line one might normally make a blanket estimate and hope that all signals would be placed in a defined range say "750m to 800m" and hence you have made an allowance of 50m and use 800m for the headway calculation. Should onyou end up having to space a particular signal at 805m on this line, then it would actually be outside the allowance made, but at least your initial estimate is far more accurate than if you had chosen the minimum value of 750m.
Similarly if the speed limit on the line is 60mph, then a driver must not exceed that value- the speed of the train is unlikely to be a cpeerfectly constant 60mph and thus will average slightly less, say 55mph. Indeed if there is a form of Automatic Train Protection that intervenes at 60mph, then the train can never ever xceed that value and possibly will be pveted from reaching it die to its safety tolerances, the driver will certainly always aim to keep below the speed of intervention- thus the average speed simply must be less than 60mph. This would be another allowance one might make if wanting a more realistic estimate of achievable headway.
A further thing is if the tmetable requires following trains at exactly 3 minutes, then the signalling must be designed to be better than that; otherwise when train 1 is 15 seconds late, train 2 will also get delayed by at least that amount; in fact if it gets cautionary aspect and has to brake it may actually lose significantly more time by the time it has been braking for most of a signal section and can see that the next signal is actually showing green. Therefore for a reliable 3 minute service, one might design the signalling for a 2.5 minute headway requirement, in order to have a "buffer" for perturbation recovery.
The question in this particular year was clear that it wanted pure unadultorated best theoretically calculated headway value, without any compensatory factor[b] appropriate for a more realistic estimate.
We know that this will mean that the number calculated will not be achievable in practice for reasons as described above, but the examiners were clarifying the basis on which they wanted the student to calculate
(03-07-2012, 12:39 PM)NJK Wrote: Dear Members,at minimum signal spacing and the given speed for application on layout 1.
I have one doubt in 2011 Module main line practice layout question paper.
In question 1, part b given as
Determine graphically or by calculation the theoretical best headway [b](without any allowances)
Kindly clarify, meaning for "without any allowances" in this question