IRSE Exam Forum

Hello Peter

1 I have completed the G23 Activity Layout Calculation that is based on the 2004 layout. However my train separation for headway, Green to Red and N values are very different from those given in Appendix W, Activity 1B Headways and Braking Calculations. I would appreciate it if you can have a look at it.

2 For the same 2004 paper, I have computed the stopping requirement from C to Station D and from Junction B to Station D although for the latter I did it more out of practice and as an after thought after I did from C to Station D. I am glad to have done so as the Stopping requirement from the Junction B to Station D is the more onerous of the two. I have appended another file to show the graphs, the time taken and the distance covered.

3 Speeds for turnouts. So far in my computations for stopping requirements, I have used the same turnout speed for both normal and reverse movements. Or can I use the maximum permitted speed if the move through the point is facing, normal and trailing normal ie. Without any diverging movement.

Thank you for your comments and do look forward to your reply

(09-03-2009 10:49 AM)alexgoei Wrote: [ -> ]Hello Peter

1 I have completed the G23 Activity Layout Calculation that is based on the 2004 layout. However my train separation for headway, Green to Red and N values are very different from those given in Appendix W, Activity 1B Headways and Braking Calculations. I would appreciate it if you can have a look at it.

2 For the same 2004 paper, I have computed the stopping requirement from C to Station D and from Junction B to Station D although for the latter I did it more out of practice and as an after thought after I did from C to Station D. I am glad to have done so as the Stopping requirement from the Junction B to Station D is the more onerous of the two. I have appended another file to show the graphs, the time taken and the distance covered.

3 Speeds for turnouts. So far in my computations for stopping requirements, I have used the same turnout speed for both normal and reverse movements. Or can I use the maximum permitted speed if the move through the point is facing, normal and trailing normal ie. Without any diverging movement.

Thank you for your comments and do look forward to your reply

Just got home after all day meeting at Derby, tomorrow is my last real day in office, then two days meeting in Madrid and then my last day of NR employment; hence may well not have time to look at attachments until weekend. May be that others may be able to respond sooner.....

However I can tell you that turnout speed applies for the divergence (both in facing and trailing direction) but not for the straight. Occasionally a turnout is actually made as a Y (with neither line straight) and in which case there is a speed restriction on both routes- generally equal speed. Similarly there can be curved turnouts where there is a shallow curve and a sharper curve of the same direction and for which there are separate speed restrictions. However the general answer to your question is YES.

(09-03-2009 09:41 PM)PJW Wrote: [ -> ]

(09-03-2009 10:49 AM)alexgoei Wrote: [ -> ]Hello Peter

1 I have completed the G23 Activity Layout Calculation

2 For the same 2004 paper, I have computed the stopping requirement .

3 Speeds for turnouts.

Thank you for your comments and do look forward to your reply

Just got home after all day meeting at Derby, tomorrow is my last real day in office, then two days meeting in Madrid and then my last day of NR employment; hence may well not have time to look at attachments until weekend. May be that others may be able to respond sooner.....

Hello Peter,

Thanks for the reply.

Take your time for the others. Absolutely no problem.

(11-03-2009 06:31 AM)alexgoei Wrote: [ -> ]1 I have completed the G23 Activity Layout Calculation that is based on the 2004 layout. However my train separation for headway, Green to Red and N values are very different from those given in Appendix W, Activity 1B Headways and Braking Calculations. I would appreciate it if you can have a look at it.

2 For the same 2004 paper, I have computed the stopping requirement from C to Station D and from Junction B to Station D although for the latter I did it more out of practice and as an after thought after I did from C to Station D.

You seem to have put the Excel calculations for stopping and effectively the same sheet but in .pdf; did you intend to post the non-stopping calcs for item 1?

Looking at the stopping calcs, starting with platform starter at 3640. I agree that with acceleration of 0.5m/s/s train will be travelling at 11.11m/s after 22.22s. Distance travelled is 0.5 x 0.5 x 22.22 x 22.22 so I agree your figures for section "g" (given your sensible approximation).

Not sure that I understand your sections "h - j". As per previous reply for train leaving the Down Main platform there is no speed retriction but let us say we are considering train from Down Loop- you are correct that train may not accelerate more until rear of train has passed completely over the points. This means rear of train beyond 3700 and hence front of train beyond 3900. Therefore the distance to be travelled at constant speed is 3900 - (3640 + 125) = 135m taking just over 12s; after that the acceleration which you denote as "k" could take place. I agree your figures for "k" EXCEPT that you have clearly calculated the acceleration to MAXIMUM PERMISSIBLE, whereas you should have calculated only to the TIMETABLED HEADWAY speed of 100km/h = 27.78m/s

Now working backwards from the station stop obviously "e" is equivalent to "g". I note that you are working on the basis of a speed restriction at 3180 (since this is straight turnout then actually the restriction would be the facing points into the loop at 3330); I make these 460m apart so deducting the 125m gives 335m whereas you have put 315m in for "d". I guess this may be intended as a 20m defensive driving gap prior to the signal at 3640, but you should then have added this 20m in for the calculations upon leaving the station (you didn't seem to do this from what I can tell). This apart I agree this section; obviously it is only the front of the train that must be considered for respecting the decreasin speed profile.

I don't understand platform at 2551- to me there is nothing there (other than the box in which you are supposed to write your candidate number!). If it had been a station then for a stopping train it would be starting from rest there! However if we treat "b" and "c" together the figures match "k" so I agree the totals, (apart from the 33.33 rather than 27.78m/s confusion as before).

I agree the 1115m for deceleration from maximum permissible speed (you are correct that you must use this for spacing signals). You have stated that signal protecting platform at 2551 whereas I think 3640 - 1115 = 2525m would be minimum. Having difficulty reconciling this with the diagram showing braking starting at 2190 (actually this would seem reasonable if you are claiming that signals spaced at something like 30% over the minimum permissible spacing. By the way I'd prefer not to place a signal that close to the end of the tunnel and if we were going for a 3 aspect solution it is certainly worth making the spacing for the sections just prior to and just beyond a platform to be shorter than others (since will be the bottleneck for stopping trains) so even if the line were generally given signals spaced at 130% braking, then putting more like 2600m would be better on both headway and signal sighting grounds.

Finally when coming to calculate headway need to be clear that you are calculating the time between train 1 passing a specific place and train 2 also passing that same place. The way of achieving this may be by calculating how long it takes train 1 to go from a specific signal until its rear later clears the overlap of the next signal, recognising that this is the first time that the initial signal can show a proceed aspect and then deciding how close to that signal train 2 could be to it without being unduly restricted by it. If we were to assume that train 2 would not be adversely affected by seeing the signal at yellow when 350m from it, we'd need to add in the travelling time of train 2 from this place until it passes the signal that we are using for the datum for our calculation.

So hope this help a bit to clarify. Also be aware for the exam a few other things:
a) you need to explain your workings. I must admit that I thought the reason why you supplied the spreadsheet was because I'd be able to the logic formulae (but in truth many were just numbers inserted with no explanation where they'd come from)
b) I wouldn't have worried re the loop line at all but just the through platform to prove could deliver headway. having the availability of a parallel line must aid (precisely how much depends on the overlap arrangements)

Please feel free to respond if anything I have written doesn't seem to make sense / you need more explanation etc

Hello Peter,

You are absolutely correct. It was supposed to have been the non stopping calculations for the G23 Activity Calculations. Please have a look as the train separation for headway, Green to Red and N values are very different from those given in Appendix W, Activity 1B Headways and Braking Calculations.

Thank you for your clarifications. 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 Sorry for the typo - there is indeed no platform at chainage 2551. 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.

4 The more onerous Stopping Requirement is from Junction B (Branch) to the platform at Station D.

Finally is there any need to work out if the Branch (which is single line operation) can support a stopping headway of 4 minutes?

Look forward to your comments.

Thank you and Regards

(19-03-2009 04:17 AM)alexgoei Wrote: [ -> ]Hello Peter,

Finally is there any need to work out if the Branch (which is single line operation) can support a stopping headway of 4 minutes?

Thank you and Regards

Answering for now just the quick and easy bit! DEFINITELY NOT.
It would of course be possible to signal the line to allow a "flight" of trains in one direction at 4 minute headway between each one of them for say 20 minutes, then when the last of them has cleared the section then timetable all the return trips at 4 minute intervals. This would be a way of maximising traffic that could be carried on a single line (and more applcable to heavy freight moves than passengers). This would presuppose some form of storage facility- I suppose there could be a large fan of sidings within the power station beyond the track depicted on the plan for example.
However look at the traffic requirement- 6 a day freight between C&B, 2 per hour stopping passenger between A&E; obviously 4 minute stopping headway is not required.
What you should do though is determine how youe would be able to fit on the single line section 4 passenger trains per hour and have enough spare capacity within that hour's operation to get at least one freight path (i.e. during one hour schedule a train from D to B and in the next hour the return working from B to D). Therefore some form of consideration of how long it would take passenger train leaving D to clear the points at 1040 (NB plan typo!) to permit a train from A to be signalled and then travel sufficiently to clear points at 2780 to then allow a freight being held on the Up branch to then travel to clear junction B thus freeing up the line for the next passenger train DOES need to be undertaken. You don't necessarily need an accurate figure to give the maximum line capacity; do just enough to show that it could be made to work. Such a consideration should then influence your decision of how to signal the branch in the area of junction B; for example it could mean that operating the points here from a Ground Frame might not be sensible as it would just add too much delay for the operation to be feasible........

(19-03-2009 04:17 AM)alexgoei Wrote: [ -> ]Hello Peter,

You are absolutely correct. It was supposed to have been the non stopping calculations for the G23 Activity Calculations. Please have a look as the train separation for headway, Green to Red and N values are very different from those given in Appendix W, Activity 1B Headways and Braking Calculations.

Look forward to your comments.
Thank you and Regards

You have only given me the results rather than your workings which led to them so I am having to do some reverse engineering- that is why in the exam you MUST show examiners HOW you get to the result.

However by deducting the "Green to Red" figures from the "120 second headway" ones gives 895m in each case (except 890m in top row). For the stopping headway you have said you are assuming 225m overlap and 200m train length so that makes me think that you must have assumed a Sighting allowance of 895-225-200= 470m. At the headway speed of 100km/h this suggests you are giving the driver 17 seconds observation time. That really is rather "over the top" and would explain the discrepancy between the figures. The Minimum Reading Time for a signal does depennd on a variety of factors including signal complexity and obscurations but tends to be around 8-9 seconds. For calculating headway it is sensible to recognise that drivers nowadays are encouraged to brake upon observing a cautionary aspect and thus I think 12 seconds is sensible to reflect that some signals do have more than the minimum sighting. Where signals have extremely good sighting I think drivers do use some common sense and don't start braking when they see a cautionary aspect in the far distance.

So did I guess right; how did you calculate the penultimate column?

(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