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

["Chris Downs" <cvdowns@ozemail.com.au>]

Arse about - I read this message after I'd posted another on this topic.

Are K-set (& S, R, C & L) all only 135kW (180hp) per motor?

The single deckers (2 motors/motor car) had 360hp (270kW) motors.  I thought
the single deck cars re-motored with 4 motors had 200hp (150kW) motors and
that the same went for the Sputnik cars?  I also always thought the DD cars
were the same at 200hp (150kW) motors.

At 180hp per motor for DD stock unmodified single deck sets (85t for 2 cars)
should have had performance only very slightly below double deck replacement
stock (81t for 2 cars, 5% less weight) [starting adhesion aside], something I
never observed in either acceleration (at low or high speed) or in top speed.

Or is there more to this power issue?

Chris

Jonathan Boles <jaboles@bigpond.com> wrote in message
sTuD6.6649$482.28857@newsfeeds.bigpond.com">news:sTuD6.6649$482.28857@newsfeeds.bigpond.com...
>
> 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