I have just realised that in b)ii) I divided 40 by 0.5 and got 20!
looing at it again now, I would need to add 160 seconds to the headway time for both 4 and 3 aspect which would exceed the 4 min required headway on both.
In conclusion I susupect that I do not need to calculate the time taken to reaccelerate back to line speed, but insteat the time taken for a departing train to clear the overlap.
I will take another look this eveneing and recalculate.
Your comments on the rest will still be appreciated.
Thanks
KJB
looing at it again now, I would need to add 160 seconds to the headway time for both 4 and 3 aspect which would exceed the 4 min required headway on both.
In conclusion I susupect that I do not need to calculate the time taken to reaccelerate back to line speed, but insteat the time taken for a departing train to clear the overlap.
I will take another look this eveneing and recalculate.
Your comments on the rest will still be appreciated.
Thanks
KJB