2011 headway calculations

[prpaoorna]

Hi all,

Please look at the attached calculations for 2011 layout and give your suggestions.

while doing stopping headway calculations do we need to add non stopping headway time to the stopping headway time?
please clarify.


Thanks & Regards,
Prapoorna N

[PJW]

1. Very neatly and clearly presented.

2. In the exam you will not have the time to repeat all the info given on the layout- you will need to be more economical- only show what is really needed and be more brief.

3. You correctly gave the braking distances from 160km/h and 100 km/h.

4. However when progressing to the headway determination, you seem to have failed to have read the question on the paper; it said:
Determine graphically or by calculation the theoretical best headway (without any allowances) at minimum signal spacing.

Hence in this case you were WRONG to design for a technical headway with contingency as the question explicitly told you to calculate the theoretical best and reinforced this by saying without any allowances.

You also then launched into a determination of the best form of signalling that could deliver the headway specified for signalling the layout.
Again in some years this may indeed be what you are asked to do, but you weren't on the 2011 question paper. It wanted you to calculate the effect of placing the signals as close as possible and on that basis determine what headway would result.
You should have performed the calculation using 3-aspect signals spaced at 1976m on the mainline and at 772m on the branch.

Then to do the last part of the question you should compare this figure with the headway requirement specified and see whether the closest 3-aspect signal spacing can achieve the necessary headway and with a sufficient margin that:
a) there can be some flexibility to position those signals at rather more than the minimum in order to get good signal sighting, arrange to protect junctions etc well rather than always having to slavishly position them at 1976m spacing- to stand any chance of achieving all the constraints then you's what the flexibility to permit a signal spacing to be at least 2400m without conflicting with the headway specification
b) allows a little contingency for timetable perturbations as you have explained (but too early) in your answer, but given that there are not really many trains then I think that 5 percent would be more appropriate than 20 percent in this instance.

When you do the sums it may of course be immediately evident that you need 4 aspect signalling even without worrying about any contingency!

So what you did was done well BUT you MUST READ THE QUESTION SET to make sure that you are answering what you were asked, whereas your answer gives the impression that you know how to calculate but are following a pre-programmed pattern based on a previous year's example but launched into doing this without checking what you were actually asked. hence a lesson for you (and many others) in the Exam this year!


STOPPING HEADWAY

1. I liked your diagram; however would have been good to show the station platform as well.

2. The graph was also pretty good- it showed that the driver would start to brake from the sighting point of the double yellow and therefore the average braking rate would actually be rather less than the nominal best brake rate and this is correctly shown by shallower gradient.

3. You went a bit wrong by attempting to tie the distance on the diagram with the horizontal axis of the graph as that is labelled as time; hence there should have been 30sec worth of horizontal line when the train was dwelling at the station separating the falling line and the rising line. You may have been better off to have drawn a speed against distance graph instead.

4. You are correct regarding the deceleration distance (apart from failing to deduct that 20m), but need to explain how this gives the time to decelerate when divided by the constant speed of the constant speed for headway.
Think about it- do a sense check. Assuming a start speed of 38.9m/s if then braked at the maximum amount of 0.5 m/s/s then it would take 77.8 sec to stop; however your example has the train braking at a lesser rate and so it must take longer, so 62sec must simply be incorrect. Hence go back and work out where you went wrong! What you worked out was how long a train that WAS NOT SLOWING would have taken to travel that distance; a train that is slowing down at constant rate will on average be moving at half that speed and so will take double the time......

5.You seem to have decided for some reason that the train will have got up to its headway speed by the time it travels its own length, plus the overlap length plus the defensive driving distance; what makes you think that? Is that compatible with the maximum acceleration rate given of 0.5m/s/s?
Of course not; you know this because you have already all but worked this out for the braking; to get up to 140km/h will definitely be more than 772m (100km/h) but less than 1976m (160km/h), so pretty obviously it won't be going particularly fast after just 384m! Hence you seem to be rather confused.

6. You were however partly right; you do need the time it take to cover that 384m but using given the acceleration rate starting from rest (and this is not the time it takes to get up to full headway speed again).


Therefore the numbers you calculated were erroneous and are not actually telling you what you needed to know. It is evident that you are a little confused- on the other hand there is some good stuff here and you are not that far away from the answer, so you would have been picking up some marks so not totally disastrous if this had been the exam. However I think if you have another critical look at what you produced in the light of my comments (and indeed look at other examples on this website) that you'll be able to see where the woolly thinking is and get things sorted out for the exam itself.

Hi all,

Please look at the attached calculations for 2011 layout and give your suggestions.

while doing stopping headway calculations do we need to add non stopping headway time to the stopping headway time?
please clarify.


Thanks & Regards,
Prapoorna N

[prpaoorna]

Thanks for your valuable comments PJW, I am looking at some other examples as you said.
It would be helpful for us if you answer my earlier question in this thread that you haven't noticed.

"While doing stopping headway calculations, do we need to add non stopping headway time to the stopping headway time?"


Thanks & Regards,
Prapoorna N

1. Very neatly and clearly presented.

2. In the exam you will not have the time to repeat all the info given on the layout- you will need to be more economical- only show what is really needed and be more brief.

3. You correctly gave the braking distances from 160km/h and 100 km/h.

4. However when progressing to the headway determination, you seem to have failed to have read the question on the paper; it said:
Determine graphically or by calculation the theoretical best headway (without any allowances) at minimum signal spacing.

Hence in this case you were WRONG to design for a technical headway with contingency as the question explicitly told you to calculate the theoretical best and reinforced this by saying without any allowances.

You also then launched into a determination of the best form of signalling that could deliver the headway specified for signalling the layout.
Again in some years this may indeed be what you are asked to do, but you weren't on the 2011 question paper. It wanted you to calculate the effect of placing the signals as close as possible and on that basis determine what headway would result.
You should have performed the calculation using 3-aspect signals spaced at 1976m on the mainline and at 772m on the branch.

Then to do the last part of the question you should compare this figure with the headway requirement specified and see whether the closest 3-aspect signal spacing can achieve the necessary headway and with a sufficient margin that:
a) there can be some flexibility to position those signals at rather more than the minimum in order to get good signal sighting, arrange to protect junctions etc well rather than always having to slavishly position them at 1976m spacing- to stand any chance of achieving all the constraints then you's what the flexibility to permit a signal spacing to be at least 2400m without conflicting with the headway specification
b) allows a little contingency for timetable perturbations as you have explained (but too early) in your answer, but given that there are not really many trains then I think that 5 percent would be more appropriate than 20 percent in this instance.

When you do the sums it may of course be immediately evident that you need 4 aspect signalling even without worrying about any contingency!

So what you did was done well BUT you MUST READ THE QUESTION SET to make sure that you are answering what you were asked, whereas your answer gives the impression that you know how to calculate but are following a pre-programmed pattern based on a previous year's example but launched into doing this without checking what you were actually asked. hence a lesson for you (and many others) in the Exam this year!


STOPPING HEADWAY

1. I liked your diagram; however would have been good to show the station platform as well.

2. The graph was also pretty good- it showed that the driver would start to brake from the sighting point of the double yellow and therefore the average braking rate would actually be rather less than the nominal best brake rate and this is correctly shown by shallower gradient.

3. You went a bit wrong by attempting to tie the distance on the diagram with the horizontal axis of the graph as that is labelled as time; hence there should have been 30sec worth of horizontal line when the train was dwelling at the station separating the falling line and the rising line. You may have been better off to have drawn a speed against distance graph instead.

4. You are correct regarding the deceleration distance (apart from failing to deduct that 20m), but need to explain how this gives the time to decelerate when divided by the constant speed of the constant speed for headway.
Think about it- do a sense check. Assuming a start speed of 38.9m/s if then braked at the maximum amount of 0.5 m/s/s then it would take 77.8 sec to stop; however your example has the train braking at a lesser rate and so it must take longer, so 62sec must simply be incorrect. Hence go back and work out where you went wrong! What you worked out was how long a train that WAS NOT SLOWING would have taken to travel that distance; a train that is slowing down at constant rate will on average be moving at half that speed and so will take double the time......

5.You seem to have decided for some reason that the train will have got up to its headway speed by the time it travels its own length, plus the overlap length plus the defensive driving distance; what makes you think that? Is that compatible with the maximum acceleration rate given of 0.5m/s/s?
Of course not; you know this because you have already all but worked this out for the braking; to get up to 140km/h will definitely be more than 772m (100km/h) but less than 1976m (160km/h), so pretty obviously it won't be going particularly fast after just 384m! Hence you seem to be rather confused.

6. You were however partly right; you do need the time it take to cover that 384m but using given the acceleration rate starting from rest (and this is not the time it takes to get up to full headway speed again).


Therefore the numbers you calculated were erroneous and are not actually telling you what you needed to know. It is evident that you are a little confused- on the other hand there is some good stuff here and you are not that far away from the answer, so you would have been picking up some marks so not totally disastrous if this had been the exam. However I think if you have another critical look at what you produced in the light of my comments (and indeed look at other examples on this website) that you'll be able to see where the woolly thinking is and get things sorted out for the exam itself.

Hi all,

Please look at the attached calculations for 2011 layout and give your suggestions.

while doing stopping headway calculations do we need to add non stopping headway time to the stopping headway time?
please clarify.


Thanks & Regards,
Prapoorna N

[PJW]

Prapoorna,

Actually I did notice your question and thought I had answered. However I will now try to do so more clearly.

NO and NOT EXACTLY.

First you need to recognise that there are two different situations:

a) "stopping headway for following stopping trains"- all trains do the same thing, they all stop at the station as a regular service pattern;

b "fast train following stopping train"- there is a stopping train within the stream of non-stopping train and hence, because of its lower speed through the relevant section of line, uses up more than a single train path as given by the non-stop headway calculations.

For situation a) SERVICE OF STOPPING TRAINS

1, Consider train1 about to leave the station; work out how long it will take to clear the overlap of the station staring signal (almost certainly it will continue to be accelerating here as not yet moving at the headway speed, but this is not important)

2. Work out how close to train1 the driver of train2 can be without having to change their driving as a result of seeing a restrictive aspect caused by train1. This will depend on whether there is 3-aspect or 4-aspect signalling, also assumptions about how drivers are trained to drive. A driver due to stop at the station would have to brake for it regardless of any aspect displayed at a signal, but is likely to leave to almost the last moment and therefore utilise their max braking rate. However if they think the platform starter is at red they would be more cautious and since in general signals are spaced at more than braking and also the headway speed may be lower than the max permissible speed the train would not need all of a 3-aspect signal section to stop if the only concern was the station stop, yet if the outer signal were yellow then the driver would brake at it (or even the sighting distance prior to it) and use the whole available length to slow down; thus the actual brake rate would be less than the maximum. Hence I suggest using the distance from the sighting point of the first caution to the station and assume uniform braking over that distance from the initial headway speed to a stand. Work out the time taken to stop over that length.

3. Add the station dwell time, after which train2 is ready to start accelerating. that is a complete cycle as train2 now needs to clear the overlap before the driving of train3 is too adversely affected.

This gives the "succession of stopping trains" headway.


For situation b) STOPPING TRAIN AMONGST NON-STOP SERVICE"

1. Work out how long it takes a train to slow down from its maximum headway speed at its declared maximum brake rate, then dwell in the station and then re-accelerate up to the original headway speed again. [Note this has nothing to do with clearing the overlap]

2. Work out how long a train that WAS NOT STOPPING would have taken to travel the same distance. Clearly to work this out you need to know from your calculations in 1 how far the train travelled during its acceleration and re-acceleration time.

3. Now calculate the difference between time 1 and time 2; this tell you how much ADDITIONAL time is needed because of the train needing to call at the station.

4. it is the time in 3 which you add to the non-stop headway time. This tels you how much "head-start" time train3 (a non stop) needs to allow train2 (a stopping service) in order that it does not catch it up too much. What we want is that once train 2 has lost the time because of its stop, it is then running just sufficiently ahead of train3 that train3 does not encounter restrictive aspects- we have two non-stop trains again but we needed to allow a bigger gap between them initially so that we still had enough gap left after the stopping train had experienced the associated delay.


Hence I think your problem is that you have taken bits of a) and bits of b), haven't realised they are different an muddled them. Hence you "have your wires crossed" and need to untangle. Separate out the two different things.

When you are doing the exam be very careful to read the question to find out what they are asking. The 2012 paper was very unusual and the examiners may try a new twist for 2013.

It will always be testing
i) Your knowledge of aspect sequence and headway
ii) Your knowledge of Newton's equations of motion
but the precise question may vary. If you actually understand the concepts, you'll be able to cope, but do not try to learn standard answers as they may well be inappropriate to the question encountered


Hope this helps a bit

It would be helpful for us if you answer my earlier question in this thread that you haven't noticed.

"While doing stopping headway calculations, do we need to add non stopping headway time to the stopping headway time?"


Thanks & Regards,
Prapoorna N