[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [NSW] 'Errors' in the Central Coast Timetable (longish reply)
wow, that creates more questions than it answers! :-)
--
--
Rgds
J. Boles
Sydney, Australia
jaboles AT bigpond DOT com
decoy.address@spam.trasher.com
"Al" <alpout@optusnet.com.au> wrote in message
3aded555$0$25464$7f31c96c@news01.syd.optusnet.com.au">news:3aded555$0$25464$7f31c96c@news01.syd.optusnet.com.au...
> "Jonathan Boles" <jaboles@bigpond.com> wrote in message
> sTuD6.6649$482.28857@newsfeeds.bigpond.com">news:sTuD6.6649$482.28857@newsfeeds.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.
>
> Um, that is only the average acceleration from rest to that speed. For a
> higher speed it will show a lower acceleration. While this is true (i.e.
the
> faster something gets the slower it accelerates, such as a car in a
changing
> gears), actual pickup rates may be different, depending such things as
> adhesion between rail and wheel.
>
> Also, your results have a V-set taking 43 seconds, or thereabouts, to get
to
> 60km/h, while the K-set takes 45.25 seconds. From experience, I'd be
> surprised if either of them accelerated that slowly (trucks can go faster
than
> that). Any drivers want to time it next time?? Thanks...
>
> Also, using power to calculate acceleration can be misleading. One
advantage
> of electrical motors (and steam engines) is that they produce infinite
torque
> at rest (reciprocating internal combustion engines can't, which is why
cars
> and diesel locos need to idle). Consequently, the torque that is produced
by
> the motor which will get transmitted into tractive effort needs to be
> considered, as it is this torque (exerted by the wheels turning on the
rail)
> which will accelerate the train. You've even supplied the equation for
it,
> Newton's 2nd Law, F=m.a.
>
> Ahah, rememberance from 1st and 2nd year. P=V.I, i.e. power equals
voltage
> times current, but P= T.w (should be omega), i.e. power equals torque
times
> rotational speed.
>
> Therefore V.I = T.w, or T = V.I/w. So when w = 0, T is infinite,
decreasing
> as w increases, thus leading to an acceleration curve which will asymptote
> towards 0. The electrical system will show voltage decreasing as current
> increases (BTW current Australian standards are for household powerpoints
to
> be 240V +/- 10%). Again, any drivers of electrics out there may correct
me,
> but I'll bet that the volt meter drops the most when starting off, then
> settles back towards the supply voltage.
>
> To work out anything more substantial, I would need more info, such as
wheel
> diameters, motor ratings (voltage, current, power etc), and the mass of
the
> train.
>
> Ah, to be an engineer.
>
>
> --
> Al Pout
>
> Men are from Earth.
>
> Women are from Earth.
>
> Deal with it.
>
> PS to Adam Dunning, this is what you'll find engineering is all about,
tying
> in 2 or 3 fields together, such as mechanical and electrical (like I have
> above). Good luck.
>
>