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2005 Module 2 Layout Calculations
#21
Hi PJW,

In the recent papers like 2010 and 2011 solved papers contingency is not taking. In the comming exam 2013 we have to fallow contingency or not.

with regards,
kirankumar

(29-01-2013, 08:42 AM)PJW Wrote: It really doesn't matter. The 10% figure is "plucked out of the air" and so, whereas I agree there is a small amount of ambiguity, any discussion whether the calculation is based on 136s or 135s is irrelevant- personally I'd do the division and therefore utilise 136s, but IT DOESN'T MATTER.
You have the concept- now move on.

(29-01-2013, 04:55 AM)kiran218 Wrote: Dear PJW,

If contingency is provided for headway is 10% and headway time is 150sec. Practical headway time we have to take 150/1.1=136 or 150-10%=135 which is we have to fallow in the exams. what is the difference between two.

with regards,
kirankumar

(28-01-2013, 03:43 PM)PJW Wrote: The whole point is that it is a judgement call, depending on all the circumstances; for IRSE exam I would not be overly concerned whether put a 10 second contingency or any reasonable percentage say from 2 - 20 per cent. Understand WHY we put a margin and the factors that determine what might be appropriate; don't worry re the ABSOLUTE VALUE in any particular circumstance!

(28-01-2013, 11:33 AM)kiran218 Wrote: Hi NJK,

1o percent of contigency is required for headway time. Under what conditions we have to fallow this?

With Regards,
KiranKumar

(27-08-2012, 05:26 AM)NJK Wrote: Dear Members,

I have attempted module 2 2005 layout calculation. Kindly review it and give your comments for my further enhancement.

Thanks in advance.

Regards,
NJK
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#22
You will find that out when you read the question paper in your 10 minute reading time at the beginning of the module paper in the exam itself!

Very much doubt that the examiners have even begun to set the question for 2013 yet, I certainly don't know what they are thinking and wouldn't tell you if I did.

Personally if I were an examiner I would vary the question from year to year. 2012 was quite different to previously; if I were a betting man then I would go for it being different again, but in a different way....

The whole purpose is to check what candidates truly understand and how they cope with someething that they are confronted with, not whether they are capable of mechanically followng a pre-set reheared process.


(29-01-2013, 10:36 AM)kiran218 Wrote: Hi PJW,

In the recent papers like 2010 and 2011 solved papers contingency is not taking. In the comming exam 2013 we have to fallow contingency or not.

with regards,
kirankumar
PJW
Reply
#23
Herewith, my attempt at the 2005 Calculations for Mainline. Comments please?


Attached Files
.pdf   IRSE-MOD2-2005-Calcs-DAP.pdf (Size: 169.63 KB / Downloads: 21)
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#24
Please see attached commented version.
I think these are pretty much what you should do in the exam, the couple of arithmetical slips are of little consequence.
Slightly marred by feeling the need to think about freight and the decision to use the longest acceptable 3 aspect spacing to demonstrate the headway was achievable and then finding that it barely was (or actually not achievable given correction of the maths!).


(05-04-2014, 03:55 PM)dorothy.pipet Wrote: Herewith, my attempt at the 2005 Calculations for Mainline. Comments please?


Attached Files
.pdf   DAP 2005 calcs.pdf (Size: 237.78 KB / Downloads: 48)
PJW
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#25
Hi,

In sheet 4 of DAP calculations, extra time is calculated by half of the accelerating/decelerating time.

can someone please let me know the reason for considering half of the time.

Thanks in advance.
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#26
(25-06-2014, 02:19 PM)jenni.joseph9 Wrote: Hi,

In sheet 4 of DAP calculations, extra time is calculated by half of the accelerating/decelerating time.

can someone please let me know the reason for considering half of the time.

Thanks in advance.

Here we are comparing two times to cover the same distance:
(1) at constant speed
(2a) while accelerating from rest
(2b) while declerating to rest.
Since the accerleration rate and braking rate are identical, 2a=2b.

now if a train is proceding at constant speed the time is t1=distance/speed

But if it is constant acceleration we would use the AVERAGE speed, t2=distance/average speed.
The average speed (v-u)/2 in this case is half the (initial) speed.

It follows that t1=0.5 t2
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#27
Hi,

thanks alot.

Can we take average speed as (v+u)/2 as one value is always zero. ( either initial or final speed is zero).
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#28
Where acceleration is constant the average speed is always (v+u)/2.
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#29
Hi,

That answers my query.

Thanks a bunch.
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