2008 Module 2 Layout Calculation Attempt

[PJW]

I wrote:

It starts well up including calculation of the braking distances.

However should not just QUOTE a formula for non stop headway; unlike the formula for braking (which is well known being a Newton Law of Motion), the headway formula is industry, and indeed signalling system, specific so you need to explain its derivation in your answer.

This is best done by a diagram showing how close train 2 can run behind train 1 without being affected by restrictive aspects as a consequence.
At present you have just apparently pulled the formulae for 3 aspect and 4 aspect headway "from nowhere" so you won’t get so many marks for this portion.

You have come up with 2 figures of 121sec and 96sec; you were asked for “the headway at minimum signal spacing”, so you do need to explain why you have given these two alternatives and state which of these would be the appropriate one for this layout.

You should have discussed whether 3 aspect siganlling, being the considerably cheaper of the two, would actually have been adequate for this layout- bearing in mind the various geographical constraints of the layout as well as the speeds and the level of service;
a) if it is, then this is the figure to give and the work to calculate the number for 4 aspects was wasted.

b) if however it is not suitable then you do need to calculate for 4 aspects, but actually I think you should then say that it is unlikely that you’d actually need to place at their minimum spacing- you could need better headway than 3 aspects can give, yet not need the absolute best that could be achieved by 4 aspects spaced at half braking distance.


I am just beginning to look at your non stopping calcs- you were asked for a fast train following a slow one. Your drawing is not quite right as it suggests that the stopping train never goes as fast as a non-stopping train. If this really was the case then the concept of a headway is meaningless- a train running at 160km/h will always catch up a train running at 150 km/h and so there can be no constant time delay between them and therefore no headway can possibly be defined.

When a train in the question is called “fast” what it actually means is that it does not stop en route, whereas one that is called “slow” does stop at aone or more intermediate stations and therefore will take longer to reach its ultimate destination.

Hence for a passenger waiting at station A (where all trains stop) wanting to go to station E (where all trains stop) might see on a train departure board a train with destination E (FAST) and a train with destination E (SLOW) because that is the overall affect on their journey (also the words are short and do not take up too much space on a train indicator display!).

However the “slow” train will still go as fast as the “fast” train as it accelerates away from A and when running at constant speed, UNTIL it needs to start slowing down for station B, Similarly after it has stopped it will eventually pick up speed again and then be just as fast as the “fast” train until there is a need for it to start slowing down for station C etc.

Also your diagram shows correctly the deceleration distance and acceleration distance, but what is that middle horizontal portion?
think your wording is "train length", but that isn’t correct on the diagram.
The diagram is always showing the same place on the train- let us assume the front. Once train is stopped in the station it does not move as time passes and therefore the speed is zero- neither the speed nor the distance is changing and so there is nothing to depict on the drawing as that does not have an axis showing time. You seem to have confused
1. a speed v distance and
2. a speed v time depiction.

Hence although I have not yet had time to look into your workings, this introduction is basically telling me that you don’t quite understand and are confused.

Even if all the numbers turn out to be actually correct, this will be because you have learnt to follow a process, you can turn the handle and the mechanism spews out the desired number as a calculating machine, but from a ‘does this person really “get it” perspective’ there is evidence here that you don’t and thus you would not be getting all the marks you’d be hoping for.

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.

[PJW]

You responded:

I have two questions on this reply.

1.
Required non stopping headway is 2min. I have gone for 4-aspect signalling as 3-aspect signalling exceeded 120 sec. so 4-aspect calculation is not wasted here.

2.
In stopping calculations diagram was wrong w.r.t speed as you said.
But I have shown train length on x-axis because train stops on platform with in dwell time(as the time passes for 30 sec. it’s speed is zero as I took distance instead of time it should be train length). In fact I have taken train length also for calculating extra distance travelled by a stopping train w.r.t stopping train. Am I wrong here?

Even though it is process oriented I analysed so many times for understanding the actual meaning of stopping calculations then I sent finalised calculations to you.

I stopped at the juncture of considering train length as an extra distance. No one has taken that.

Clearly I have taken acceleration, deceleration along with train length for calculating T(extra).

Why don’t you people consider train length as an extra distance?

Please answer this question.

I wrote:
It starts well up including calculation of the braking distances.
.............................................................
Hence although I have not yet had time to look into your workings, this introduction is basically telling me that you don’t quite understand and are confused.

Even if all the numbers turn out to be actually correct, this will be because you have learnt to follow a process, you can turn the handle and the mechanism spews out the desired number as a calculating machine, but from a ‘does this person really “get it” perspective’ there is evidence here that you don’t and thus you would not be getting all the marks you’d be hoping for.

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.

[PJW]

Prapoorna,

1. I agree that in this case that you did need to select 4 aspect signalling.
What I was meaning is that you should have EXPLAINED more. It did not seem that you were answering the question as asked. The problem for you is that if the examiners think similarly when marking your exam paper you won't gain all the marks that otherwise you might. Hence it is in your interests to make things very clear to them.


2. I am having difficulty understanding your issue. I am sure that we can agree that all places along the length of the train are always moving at the same speed as each other. Your diagram is showing speed against distance; this can only have meaning if the line plotted always considers the same place on the train.

Perhaps it is clearer if you actually plot both the front and the back on the same diagram.

a) For the non-stopping train this isn't interesting as the front and back are always lieing on the same horizontal line.

b) However you are right to a certain extent regarding the visibility of the front and the back for the stopping train. Consider the place at which the falling line reaches the horizontal axis and let us agree that this point reflects the FRONT of the train. Yes the train has a certain length and therefore the BACK of the train also reaches speed=0 simultaneously and thus could be drawn a certain distance to its left, intersecting the x axis.
Indeed if you consider the train earlier in its approach it will always be a horizontal line of the same length and therefore various positions of the BACK of the train will form a diagonal line parallel to the one which you have drawn and it is only when the train was originally running at its constant speed that that line will effectively disappear (as for the "fast" train, the points on the graph covered by the FRONT and BACK are perfectly the same).

So we agree that the train length can be depicted horizontally.

Where we disagree is that I say that the acceleration will start at the SAME PLACE as the deceleration ended.

Effectively what you have drawn is
1. the BACK of the train whilst decelerating and then "jumped" to depict 2. the FRONT of the train accelerating.

If you had completed the drawing to show also
1. the FRONT when decelerating then it would have intersected the x axis at the place you show the acceleration from, and also to show
2. the BACK when accelerating it would start from the x axis at the place where the deceleration ended.

In such a diagram then I agree that you would have been correct to show the train length as you did; however as your original drawing then it is plain confusing and causes you to mislead yourself.

I agree that the average speed of the slowing train with constant brake rate is half that of the intial headway speed.

I also agree the time taken, but note that it is somewhat quicker to calculate this by a slightly different method. The deceleration is 0.5m/s/s which means that it take 1 second for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/3 then it takes 39s to stop!
This approach might save you valuable time in the exam.

Anyway your calculations are ok and you have calculated the distance correctly (note that despite your diagram, this is the braking distance plus the acceleration distance with no train length addition).
This is fine because you are looking for the DIFFERENCE between what a stopping train does compared with a Non stopping. The dwell time does influence this; the train length does not.

Of course in the Non-Stopping calculations, one of the factors IS train length and when you are adding the "stopping alllowance" (your T_extra) to the "non-stop headway" then that train length is indeed still incorporated within the combined figure.

So whereas I did find your diagram exhibited confusion, actually your calculations seem good, apart from this train length misunderstanding. The distance over which the travel times of the two trains should be compared is simply 759 x2 = 1518m ; there is no train length to add so I disagree your 1698m.
Hence instead of the non-stop train taking 44 secs as you have calculated ,then it should be 39 secs.
Actually this is pretty self-evident if you think on it; you stated that the stopping train would be going on average at half the headway speed when slowing down and again when accelerating- both these take 39 secs; twice half of 39 = 39s.

However you then give your answer primarily relating to 3-aspect signalling, but your earlier response to me stated that you believed that you needed to provide 4-aspects because of the non-stop headway calculations. Obviously if you have decided that you need to provide 4-aspects then you should be answering all the question from that perspective.

Other than that though, it does seem as if you have grasped the concept. Yes I have criticised your answer, but overall it was quite good and my motivation is that by pointing out defects now then you'll learn from them and hence not lose marks at exam time.

I hope therefore that you are now happier.
Still not convinced that I have addressed your plea;

[i] I stopped at the juncture of considering train length as an extra distance. No one has taken that.

Clearly I have taken acceleration, deceleration along with train length for calculating T(extra).

Why don’t you people consider train length as an extra distance?

Please answer this question.[i]

but if I haven't please ask again in different words to give me a better way of appreciating what you are meaning.

You responded:

I have two questions on this reply.

1.
Required non stopping headway is 2min. I have gone for 4-aspect signalling as 3-aspect signalling exceeded 120 sec. so 4-aspect calculation is not wasted here.

2.
In stopping calculations diagram was wrong w.r.t speed as you said.
But I have shown train length on x-axis because train stops on platform with in dwell time(as the time passes for 30 sec. it’s speed is zero as I took distance instead of time it should be train length). In fact I have taken train length also for calculating extra distance travelled by a stopping train w.r.t stopping train. Am I wrong here?

Even though it is process oriented I analysed so many times for understanding the actual meaning of stopping calculations then I sent finalised calculations to you.

I stopped at the juncture of considering train length as an extra distance. No one has taken that.

Clearly I have taken acceleration, deceleration along with train length for calculating T(extra).

Why don’t you people consider train length as an extra distance?

Please answer this question.

I wrote:
It starts well up including calculation of the braking distances.
.............................................................
Hence although I have not yet had time to look into your workings, this introduction is basically telling me that you don’t quite understand and are confused.

Even if all the numbers turn out to be actually correct, this will be because you have learnt to follow a process, you can turn the handle and the mechanism spews out the desired number as a calculating machine, but from a ‘does this person really “get it” perspective’ there is evidence here that you don’t and thus you would not be getting all the marks you’d be hoping for.

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.

[Peter]

The way that I would rationalise when train length is important or not is to consider what circumstances the values that you calculate would be different if you did the same calculation for a very long or very short train.

Clearly, acceleration and speed are going to be independent of train length because otherwise the trian would be stretching or shrinking. If you look at this thread, it has the derivation of the basic formulae for laws of motion which apply to all sizes of object, and hence all lengths of train!

On the other hand, with things to do with distance or time, the length of the train will be important as the whole train has to pass (hence adding train length into the calculation for headway time and headway distance).

Not sure if this approach helps or hinders. Please ask further if it is the latter.

Peter

[prpaoorna]

thanks and apologies, doubt is clarified.

[prpaoorna]

Hi PJW/Peter,

I found a mistake in stopping calculations.
Deceleration time is supposed to be 38.89/a=78sec. hence
deceleration distance=78*19.45=1517m (because avg.speed is 19.45m/s), but calculation shows deceleration time as 19.45/a=39sec. so deceleration distance was the 759m instead of 1517m
what is the correct approach here?

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.

[mahathiammu]

Hi all,
Could u plz take a look of the calculations and comment.
Also can someone clarify whether contingency in headway time be considered or not while calculating the headway.

Thanks

[Peter]

Hi all,
Could u plz take a look of the calculations and comment.
Also can someone clarify whether contingency in headway time be considered or not while calculating the headway.

Thanks

Comments as they come up in no particular order of importance.

You have calculated the sighting using the headway speed. You should use the linespeed for this as the reduced headway speed is recognising that the trains may not be running at the max. However, sighting, like braking calcs, must use the worst case.

Albeit arithmetically correct, your conversion factor for kilometres per hour to ms-1 uses the scale of 5/18, rather than 1000/3600. You should probably explain (at least once), your shorthand.

You have attempted an explanation of the N=DGR/BD formula but I think this is sightly incomplete. Personally, I would not rely on using this and trying to explain it. Far easier to write a simple explanation of why the numbers you have calculated mean that 2, 3 or 4 are appropriate or not appropriate from first principles. Eg HD = x and spacing for y aspect signalling are compatible or incompatible. You have effectively shown that anyway by the calculation of the headway time for 4 aspect. If you repeat this for three aspect, you will see that it does not quite fit, therefore you have the justification for what you have selected.

You can use the N formula for your own satisfaction to see where you start justifying your decision, but I'd avoid it.

One minor point in the DGR formula - You have stated that if N < 1.5 then the calculation is wrong. This is not necessarily so, it normally means that you cannot achieve the headway with the braking characteristics even with 4 aspect signalling.

On the question of contingency, you can either add it into the require Ht ( eg say achieve better than 120 s so that if things go wrong you have some to spare) or work out using it with no contingency and then say that you will adjust the distance by a factor to allow for contingency. There is no right or wrong way to do it, you just need to make sure that you note that it should be included and make reasonable judgements.

Will have a look at th stopping element a bit later for you.

Peter

[PJW]

You are absolutely right. I see that I wrote:

The deceleration is 0.5m/s/s which means that it take 1 second for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 39s to stop!

Let's change that to read:
The deceleration is 0.5m/s/s which means that it take 2 seconds for the speed to be reduced by 1 m/s. Therefore with initial speed of basically 39m/s then it takes 78s to stop. Somehow I followed your incorrect lead- to be honest whilst replying to your other post yesterday the thought did go through my mind that this one was wrong but I then failed to go back and check so you beat me to it.

Hi PJW/Peter,

I found a mistake in stopping calculations.
Deceleration time is supposed to be 38.89/a=78sec. hence
deceleration distance=78*19.45=1517m (because avg.speed is 19.45m/s), but calculation shows deceleration time as 19.45/a=39sec. so deceleration distance was the 759m instead of 1517m
what is the correct approach here?

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.

[PJW]

Another attempt received for this year's calculation; comments attached