Module 2 - 2009 - layout 1 Calcs

[michaelmcnulty]

Guys,

Any chance of some feedback on my calcs which are attached?

Im not 100% sure if my non stopping calcs are squeaky clean!

Cheers,

Mick

[PJW]

Guys,

Any chance of some feedback on my calcs which are attached?

Im not 100% sure if my non stopping calcs are squeaky clean!

Cheers,

Mick

They are clear and brief, but you ought to put a little more of explanation in them generally. 

a)
You did show how you converted km/h to m/s but you only needed to show that once.  You could have used the time you saved to explain what BD stand for and define what you mean by u and a.
You didn't need to wok out for 40 km/h.
Be careful when you throw away significant figures and then square the result; all your braking distances are slightly short and of course it is important that never underestimated.  Also quoting braking distances to fractions of a metre is ridiculous anyway (let alone when the error you introduced is an order of magnitude more- there is a difference between precision and accuracy).  Be sensible; always round up calculated distances to the next higher ten metre, so give the answers as 1510m and 770m. [Actually if you keep all the figures you'll notice that even 1510m is  bit of an underestimate but at least it is really only claiming a lower level of precision].  Alternatively always ROUND UP; if you had squared 38.9 or even 39 then you'd have definitely been right side.

b)
Whereas you' get away with quoting Newton's equations, don't just put HT3=(S+2B+O+L)/V;  you haven't defined what any of the terms mean nor explain where this equation comes from.  Strongly suggest you draw an annotated  diagram.You did get the "right" answer but you'd be losing marks because of your lack of explanation- for example having introduced an unknown "S" you then substitute it for 8 x 33.3 with no hin where those figures come from.

Similar comments re the stopper.
You claim the deceleration time is (2 x 1505)/ 33.3
I think that you are meaning that the train will braking at the yellow and therefore decelerating over the length of a signal section and on average will be travelling on average at half the initial speed, but you need to state this.

Your entry for TIME = 78.49s is surely the time that a train going at constant speed would take to cover the same distance as the stopper does to brake and reaccelerate; I am sur ethat you meant this but the issue is that you didn't state.

It is also missing a few words (or diagram!) which justifies why adding the "increase in journey time" to the non-stop headway time gives the answer you were asked for.  Also a mistake to call it "stopper headway time" as the headway between two stopping trains is something different.

So in summary your calculations seem to show that you know what you are doing, but it would be worth taking a little longer in order to get a few more marks.  It is a balance of course; don't take too long over the calcs that you  eat too much into the time in which you should be signaling the layout, but I think you needed a bit more justification amongst the numbers.

You should also when explaining your choice of signaling have stated the range of signal spacing to be used on the layout because it may be impossible (or indeed not even desirable) to place 3 aspect signals at the bare minimum braking distance