Re: [NSW] 'Errors' in the Central Coast Timetable

["Jonathan Boles" <jaboles@bigpond.com>]

> 2-car K sets are referred to by some in the rail system as "wheelbarrows",
> ie. you really have to push them to get them moving. Bear in mind the V
set
> that you are referring to has 4 cars (2 motor cars) that are relatively
more
> powerful than any old suburban rollingstock, whilst the K set only has one
> motor car.

Yes but remember that the motor car to trailer car ratio is still 1:2 in
both situations.

Here are some figures:
Mass of K set driver: 47t    From memory, power = approx 135kWx4 = 540kW
Mass of K set trailer: 41t
Mass of V set driver: 59t    Power: 4x160kW = 640kW
Mass of V set trailer: 40t

So 540kW has to accelerate 88t on a K set and 640kW has to accelerate 99t on
a V set.

Now for some equations. P=power (Wats), W=work, F=force, r=distance, t=time,
v=speed (ms^-1 [=m / s]), m=mass (kg), a=acceleration(ms^-2 [=m / s / s]).

Of course kg are used because these values are in the MKS system.

v=r/t
F=ma
W=Fr
P=W/t
>From these eq'ns, P=mav, or Power = Mass x Acceleration x Speed

Lets say a K set and a V set both have to accerate to 60kmh^-1, which is
equal to approximately 16.67ms^-1.

Now
P(K set)=540,000W,  m(K set)=88,000kg,  a(K set)=?,  v(K set)=16.67ms^-1
P(V set)=640,000W,  m(V set)=99,000kg,  a(V set)=?,  v(V set)=16.67ms^-1

By making a (Accelertion) the subject of the equation, a= P / mv
so a(K set) = 540,000 / (88,000 x 16.67) = Approximately 0.3681 ms^-2
and a(V set) = 640,000 / (99,000 x 16.67) = Approximately 0.3878 ms^-2

Obviously the V set accelerates at only 0.0197ms^-2 faster than the K set,
and the K set isn't as much of a 'wheelbarrow' as you might think.

--
--
Rgds
J. Boles
Sydney, Australia

jaboles AT bigpond DOT com
decoy.address@spam.trasher.com