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.
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.
(24-09-2013, 08:00 PM)PJW Wrote: 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.
(24-09-2013, 07:57 PM)PJW Wrote: 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.
(24-09-2013, 06:14 AM)prpaoorna Wrote: 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