2008 Module 2 Layout Calculation Attempt

[reuben]

I am inviting guidance and correction on the attached attempt on the calculations part of the 2008 paper; especially with a view to make it achievable within 18 minutes for a novice practising to finish the paper in 90 minutes.

Feedback

General:

remember to declare your variables
well laid out

Part a

A good approach: work out everything at the start, for all given speeds, because you can't yet predict which outcomes you might need later in the process, also it's much quicker to do them all when you've got the formulae in your head / the conversion factors in the calculator.

Laying out your answers in a tabular form as you have done is also helpful. You could probably save a bit more time by just writing down the generic formula you used for each step, instead of writing it out in full with all the actual figures - the examiner can see that you're just applying the same formula to different input figues.

Part b(i)

I always recommend a very quick sketch diagram here, showing the 2 successive trains and the relevant quantities. this shows that you understand what you're doing, instead of just plucking a formula out of the air; it also helps to declare your variables.

b(ii)

Again, good to draw a diagram here.

The graph must show both trains - if you just show one train, it doesn't demonstrate the time relationship between two trains. also must make clear if it's showing the back or front of each train (you're interested in the back of Train 1 and the front of Train 2.)

t3, i think is intended to relate to the dwell time - if this is the case, this phase of the journey should be truly vertical on the graph, not the vaguely angled line shown

error 1:

The question requires Ht for fast following stopping - if this is the case, you need to allow for Train 1 to accelerate back to full speed rather than just clearing the overlap - it continues to delay the train behind it until it reaches full speed

error 2:

Your calculation for t4 is wrong, it should allow for acceleration rather than travelling at full speed.

error 3

you need to allow for the train length as well as the overlap length; you've only allowed for one of these, I can't tell which because they're both 180m

error 4:

if you're using the "extra time" method, which i recommend because it's very quick, you must account for the time Train1 would have taken anyway, if travelling at full speed instead of stopping. Your calculated values for t2 and t4 are the absolute times, not the extra times compared to non-stopping. Your method partly deals in absolute times, and partly deals in relative times, the two are incompatible.

The actual Textra value should be (2 * (38.89/0.5) )/2 + 30 = 107.8 seconds.

Note:

the first factor of 2 accounts for the extra time during both deceleration and acceleration, you can do this because the acceleration values (0.5) are the same magnitude, even though opposite sign.

The second factor of 2 accounts for the time the train would have taken anyway, if travelling at full speed.

these two factors of 2 cancel each other out, so your answer is co--incidentally about right!

because of error 1, you're not actually interested in the train accelerating through the overlap only, so errors 2 & 3 no longer apply. You've already accounted for train and overlap length at full speed in part b(i), so under the "extra time" method, you don't need to count them again.

Overall, a good answer, most candidates would not attempt b(ii) at all.

hope this helps

[dilip421]

Hello sir,

We have solved 2008 paper by the guidence of our seniors. please go through our attachement & give the corrections.
we have some doubt,
1) Why we required 20% of contigency...?
2) How we are calculating this 20% contigency...?
3) After taking ratio how to decide its 2aspect,3aspect or 4aspect is any written document is provided...?
4) For stopping train why we have to take the total time taken by the stopping train....?
5) The way of explaining in the 2008 layout attachment is correct to write in the exam or we have to show the diagram...?
6) After decideing the aspect how to start the signalling the layout...?

please explain us or forward any solved signalling layout plan. we are got stuckup in that.

Regards,
GDK

[PJW]

I'll look at but do have a look at the Study DVD module 2 portion.

1/2. It is unlikely that you'd need 20% contingency, but you should make some allowance. It depends on the service- there often needs to be a margin between what train service is to be advertised to the public and what the signalling must be capable of.
If there is a contractual committment to a train every 180 seconds then would certainly wish to make sure that could signal a train say every 160 seconds; otherwise if train 1 is late then all the subsequent ones will be delayed. Hence there is a vast difference between a requirement for the 2 closest trains in a day to be 3 minutes apart but actually only say 6 trains an hour and where every train is 3 minutes apart and there is no period during which there could be recovery to limit the inmpact of late running.

3. I really don't advise using the DGR method giving a ratio at all. If you do use it as a short cut then you must understand what it really means. It is to do with the minimum spacing determined by the braking consideration and the maximum spacing as dictated by the headway consideration. Look at the Study Pack text and also a Powerpoint presentation included on the DVD re Headway and Braking.

4. Don't understand your question- perhaps it'll become clearer when I look at your attachment.

5. On the Study DVD there are attempts at quite a few layouts and indeed there are even some on this forum.

6. This should be explained in the Study Pack; if there is something you do not understand then try to be moree specific in order that we can help (and indeed amend the Study Pack for future)
Also see this post on signal placement

Hello sir,

We have solved 2008 paper by the guidence of our seniors. please go through our attachement & give the corrections.
we have some doubt,
1) Why we required 20% of contigency...?
2) How we are calculating this 20% contigency...?
3) After taking ratio how to decide its 2aspect,3aspect or 4aspect is any written document is provided...?
4) For stopping train why we have to take the total time taken by the stopping train....?
5) The way of explaining in the 2008 layout attachment is correct to write in the exam or we have to show the diagram...?
6) After decideing the aspect how to start the signalling the layout...?

please explain us or forward any solved signalling layout plan. we are got stuckup in that.

Regards,
GDK

[PJW]

Looking at your calculations:
I) Is fine in approach. You could have made the calculation more brief for the branch line as it essentially repeats the above with a different figure.
Just be careful re the accuracy you quote; braking distances are not precise so I'd tend to round up to the next highest 10m.
Certainly you are quoting more precision than you deliver; for e.g. you used in the second part 27.7 and then squared it to give 768m whereas iif you hadn't thrown away places of decimals at that time you'd have calculated the distance as 771m rather than 768m. Had you just quoted 770m then that would have been fine as not claiming excessive precision.

II) As stated you decide to treat the 120sec minimum headway as a signalling design headway. On this paper given that there are only 2 non-stop and 2 stopping trains using D-F and another 1 stopping using B-E then there would seem no need to add any contingency; however it would have been best you stated that assumption.

You then quote the DGR formula. You do need an annotated diagram to explain that derivation. You also need to explain why S = 38.8 x 10.

Similarly just stating that 1.5<DGR<2 therefore need 4 aspects does not demonstrate understanding. You need to say that it is not possible to get two minimum signal spacings of 1975m within the maximum spacing of 3908m needed to achieve the headway requirement. Therefore 3 aspects is not an option, so have to select 4 aspects.

HOWEVER, do read the question. You were asked to calculate the headway at minimum signal spacing, not at this time asked to determine what signalling type to use on your layout (although indeed you will need to determine that sometime before starting it).
So what you should have done was calculate what headway would be with 3 aspects at 1975m and give the answer, also pointing out that this does not achieve the 2 minute headway requested so therefore for your layout you will need to use 4 aspects. However placing 4 aspects at their minimum signal spacing would be excessive and therefore you will calculate the maximum spacing of the 4 aspects that would just achieve the headway specification.

III) Don't just launch into calculations; think about the layout and operations. The branch is a single track. A freight train needs to go to the end sidings at A and a passenger train needs to go to station B, reverse and return and the freight follow it off the branch prior to the same pattern repeating in the next hour.
It doesn't make much sense to do non-stop calculations as if it is a long length of railway with a succession of following trains. You were not asked in the actual question paper to do any calculations, indeed there is no headway specification for the branch (the only one is at a speed higher than 100kph).

To attempt the layout though, of course you'd need to do and show your calculations re the braking distance on the branch.

You also need to provide the signalling to allow the advertised train service to run, yet not provide too many signals for no value.
Hence you need to decide how far the freight must have proceeded before you allow the passenger train to be routed onto the branch to follow behind it. Then make sure that you can safely authorise the passenger to go as far as station B soon enough after the freight train has cleared that area so that the train can get to the station, dwell there for say 5-10 minutes (the driver needs to change ends and open up the other cab and have a little recovery time) and then start back. The freight loco will have run around its train by then so as soon as the passenger train has cleared the station B area (it is looking as if a platform starter at left hand end of that platform would be essential, particularly if intending to work the line under TCB regulations which require a signal at the place of reversal), the freight needs to shunt its loaded wagons say into siding 1 and then collect its new train of now empty vehicles from siding 2 and then proceed along and off the branch in time that the next freight train can enter and repeat the cycle of operations.
In short, determining the appropriate branch signalling requires thought about its operation and perhaps some specific calculations of sunning time of a tran from one place to another which need sto be completed before another move can be made; it is not about plugging in some new numbers into a standard formula.

Page 7. It is not obvious but you seem to have gone back to the mainline portion here. You have quoted a formula out of nowhere and have not explained.

Stopping Train.
You have calculated the 78s deceleration time.
I don't then see why you then used it to determine the braking distance (rather than the earlier formula utilised); indeed "braking distance for average speed of 70kph" doesn't really make sense to me, similarly "braking distance for acceleration" could have been better expressed!

You then work out "the total time" for the stopping train; begs the question total time for what (I think you mean the time for which it is not moving at the headway speed).

You then look at the time difference for a non-stop train but you don't really explain why the figure of 78 is applicable here and also confuse the issue a bit by calling it a "headway time". Again you introduce a figure 96 without explaining; I think this is the T4=96 from 2 pages back, but you never explained the significance of T4 (which also begs the question "why T4 / where are T1, T2, T3?"
However the diagram on the last page helps a bit. You do not define either the horizontal or vertical axes of the graph; I guess that you are plotting velocity vertically and time horizontally given the depiction of the stopping and non-stopping, but then although you write 30sec dwell time it appears that the train doesn't remain stopped for any time at all but immediately accelerated again having come to rest momentarily. There is also a "Headway Distance" marked from nowhere obvious to nowhere obvious, but plotted horizontally on what would appear to be the time axis. I am forced to conclude that you are muddling up a speed/ time presentation with a distance/time presentation; you are not going to be able to show these 3 related quantities on a 2-dimensional presentation.

So in summary there is much that is ok, but I am not convinced from your answer that you really understand everything thoroughly- looks a bit as if you have copied something from here and something from somewhere else. It has got most of the elements of a good answer within it, but does need more clarity and explanation rather than just quoting formulae.

In the immortal word of Eric Morecombe when accused of playing wrong musical notes in a tune: "On the contrary I played all the right notes, but just not necessarily in the right order!".

=====================================================================================
I think I have probably answered all your doubts now.
The approach you seem to be following to look at the "Fast Following Stopping" is to
a) determine how much time a stopping train loses relative to the Fast,
b) say that after the stop and the stopping train has resumed running at headway speed then there must still be a minimum non-stop headway separation between it and the next train,
c) add the times together to work out how much gap there must originally have been between the train which would be stopping and the following train that would not be.
Hence your "total time" is that period where the stopping train diverges from what it would have done should it itself have been a Fast.

Also note that I have merged this thread with an earlier attempt at this same paper calculations, so do look at that attempt and the comments on it as well. When a thread is long, scrolling only goes so far; be aware that some (like this) extend over multiple pages as well.

Also regarding the layout itself, do look at the 2008 layout thread

Hello sir,

We have solved 2008 paper by the guidence of our seniors. please go through our attachment & give the corrections.....

Regards,
GDK

[s.sathish08]

Hi Everyone...

Herewith i have attached 2008 Module 2 Non stop headway calculation,
Please check and review it ..
Your corrections and suggestion is required please...

Thanks
Sathish

[PJW]

It is an extraordinary neat attempt and a real joy to read; I just don't think that you could do like this in exam conditions when it needs to be fast and therefore more of a scribble.

Similarly it was clearly set out in text book fashion; you do need to show the logic but in the exam you won't have time- hence need to be more brief rather than waste time repeating too much of the question.

The examiners are always wary of the use of DGR, but your diagram is very clear to explain this initially. However it is at stage (v) when you divide DGR by B and come to the answer N=1.9 that they will object- what is the significance of N they ask? Whatdo you think you are doing when you divide a conceptual distance DGR by the braking distance; it is a ratio yes, but what is the significance of the number one gets? You need to relate this back to signalling principles (see later).

You then draw a diagram showing 4 aspect signals spaced at D4 which you then claim must lie within the range 0.5B to 0.75B; also that D4<= GDR/3. There is absolutely nothing wrong with these as conclusions, BUT you really have not explained well eenough WHY!!

I did quite like your diagram of the usable range within which a signal can be positioned and satisfy the constraints- however I wouldn't call it a "safety band". The min signal spacing is a safety constraint arissing from needing the relevant BRAKING DISTANCE, but the maximum signal spacing is an OPERATIONAL CAPACITY constraint arsing from the need to achieve the specified headway.

I did not agree with "enable 4-aspect signalling for a more frequent and economical services".
Basically if you position 3 aspect signals as close as you can safely do, they will be 1975m apart on this layout. However as you have shown (well nearly) if you spaced them like this then the smallest headway distance you could achieve would then be 1975x2+389+180+180 and that does not satisfy the headway requirement.
Hence you have to incur a lot more cost and provide many more signals using 4 aspects that can be as close as 988m. However by spacing these out as far as you can such that the signalling only just achieves the headway requirementt, then the solution is as economical as it can be whilst satisfing the other constraints.


NOW LETS LOOK AT THE ACTUAL QUESTION ASKED
a) Determine minimum braking distances for the permissible speeds and braking
characteristics of the traffic on offer.
b) Determine graphically, or by calculation, the headway at minimum signal spacing and the
given speed for:
i) A fast passenger train following another fast passenger train; AND
ii) A fast passenger train following a stopping passenger train.
All calculations and graphs must be shown [20 Marks]

You did a) in your section (iii)- may have been a good idea to have annotated accordingly to show the examiner that you have done it.

I don't think that you actually did b)i).
What you did do was broadly related to it (and indeed very important for the rest of the question when placing signals on the layout), but you did not explicitly answer the actual question the examiners asked in the 2008 question paper.

Obviously you have not attempted b)ii) as only claiming to be doing the non stop headway at present.

There were 20 marks available and you need to aim to get these in less than 18 minutes.
Let's guess that the marks are actually split between the sections as 6+7+7.
You'd have got the first 6.
You'd have got perhaps 3 of the next 7.
You'd have got none of the next 7.
Hence let's say a total of9; the question is: did it take you more than 8 minutes to achieve them- I bet it did!

The analogy is that it was a very smooth sedate ride in a lovely car and hence pleasurable. However not so good if one particularly anxious to get to the destination by a specific time; the car is using so much petrol that it will be ceertainly out of fuel before getting half way there.
So for the exam, think a little less about quality, more about the need to get a good "marks per minute" rating.

Hi Everyone...

Herewith i have attached 2008 Module 2 Non stop headway calculation,
Please check and review it ..
Your corrections and suggestion is required please...

Thanks
Sathish

[s.sathish08]

Thanks for your valuable comments.
I will review my calculation as per the comments given.

Thanks
Sathish

[MEDEV]

Sir,

I have one doubth, in your review..
Once he calculated the Headway (Distance & Time)using 3 aspect & 4 aspect, he could have come the conclusion which aspect signalling meets the headway requirements , then once again do we need to Appy DGR method and show the significance of Ratio "N" in selecting aspects.

Thanks
Dev

You then quote the DGR formula. You do need an annotated diagram to explain that derivation. You also need to explain why S = 38.8 x 10.

Similarly just stating that 1.5<DGR<2 therefore need 4 aspects does not demonstrate understanding. You need to say that it is not possible to get two minimum signal spacings of 1975m within the maximum spacing of 3908m needed to achieve the headway requirement. Therefore 3 aspects is not an option, so have to select 4 aspects.

HOWEVER, do read the question. You were asked to calculate the headway at minimum signal spacing, not at this time asked to determine what signalling type to use on your layout (although indeed you will need to determine that sometime before starting it).
So what you should have done was calculate what headway would be with 3 aspects at 1975m and give the answer, also pointing out that this does not achieve the 2 minute headway requested so therefore for your layout you will need to use 4 aspects. However placing 4 aspects at their minimum signal spacing would be excessive and therefore you will calculate the maximum spacing of the 4 aspects that would just achieve the headway specification.

[PJW]

Certainly you could
a) work out the headway achieved at minimum spacing of 3 aspects; if good enought then that is your answer, if not
b) work out the headway achieved at minimum spacing of 4 aspects.

No need to mention DGR at all.

Sir,

I have one doubt, in your review..
Once he calculated the Headway (Distance & Time)using 3 aspect & 4 aspect, he could have come the conclusion which aspect signalling meets the headway requirements , then once again do we need to Appy DGR method and show the significance of Ratio "N" in selecting aspects.

Thanks
Dev

You then quote the DGR formula. You do need an annotated diagram to explain that derivation. You also need to explain why S = 38.8 x 10.

Similarly just stating that 1.5<DGR<2 therefore need 4 aspects does not demonstrate understanding. You need to say that it is not possible to get two minimum signal spacings of 1975m within the maximum spacing of 3908m needed to achieve the headway requirement. Therefore 3 aspects is not an option, so have to select 4 aspects.

HOWEVER, do read the question. You were asked to calculate the headway at minimum signal spacing, not at this time asked to determine what signalling type to use on your layout (although indeed you will need to determine that sometime before starting it).
So what you should have done was calculate what headway would be with 3 aspects at 1975m and give the answer, also pointing out that this does not achieve the 2 minute headway requested so therefore for your layout you will need to use 4 aspects. However placing 4 aspects at their minimum signal spacing would be excessive and therefore you will calculate the maximum spacing of the 4 aspects that would just achieve the headway specification.

[prpaoorna]

Hi all,

could you please check my calculations?
Line to be covered in between pages is “Non-stopping train will cover the 1698m in 44 sec.”

Regards,
N Prapoorna.