Re: C Class Rumours

[mauried@commslab.gov.au (Maurie Daly)]

In article <mauried.286.37944D90@commslab.gov.au> mauried@commslab.gov.au (Maurie Daly) writes:
>From: mauried@commslab.gov.au (Maurie Daly)
>Subject: Re: C Class Rumours
>Date: Tue, 20 Jul 1999 10:21:04 GMT

>In article <7n0imq$3kg$1@mirv.unsw.edu.au> G.Lambert@unsw.edu.au (Geoff Lambert) writes:
>>From: G.Lambert@unsw.edu.au (Geoff Lambert)
>>Subject: Re: C Class Rumours
>>Date: Tue, 20 Jul 1999 00:28:39 GMT

>>David Johnson <trainman@ozemail.com.au> wrote:

>>>Keith Malcolm wrote:

>>>> The "dead" weight of a steam loco is no water in the boiler and an empty
>>>> tender. The "live" weight is a full tender and 3/4 full boiler.

>>>Not necessarily.  If a steam locomotive was to fail climbing Molonglo Gorge and need to be pushed
>>>out by the Exploder, the tender would be full, as would the boiler.  How heavy would it be?  How
>>>much resistance would the Pistons have?

>>I don't think the practice described here for engines with traction
>>motors was ever adopted for steam.  I have heaps (hundreds) of WTT's
>>in my collection, and I've never noticed anything like this for steam.
>>(But then, I never noticed that a dead diesel was heavier than a live
>>one either and I have all the NSW WTTs 1950-1998)

>>However, the usual formulas for loss of power at any particular speed
>>for steam locomotives assumed that this was about 8% between the
>>piston and the wheel.  This power, consumed in friction, etc., had an
>>equivalent manifestation as a force.  It is at least partially valid
>>to turn this idea around and work out the force offered by the
>>machinery when a loco is moving "dead".  For a 38-class, one would
>>expect a force of about 3,000 lb on this basis.  The next step, to
>>equate this to a notional increased mass, is just about as dodgy as it
>>is in doing so for a diesel electric, because the whole thing is speed
>>and grade dependent.  Indeed, I would say to do so is meaningless, and
>>I really think the practice for diesels is meaningless too (if that's
>>the rationale behind it for diesels, which it mightn't be).

>>But, if we assume that the total resistance (including air) of a
>>38-class, in lb/ton is given by the "Abbot formula"  of R = 5 +
>>0.003*V *V, where V is in mph and we assume the 38-class is about 195
>>ton (I forget exactly what it actually is), then 3000 lb represents
>>the resistance at about 55 mph.  But I don't know how to translate
>>this back into "equivalent dead weight"

>>All of this assumes that the wheels remain connected to the pistons
>>during "dead-haulage" and that the valves are set so that
>>counter-pressure doesn't develop in the cylinders.  Neither is
>>necessarily true.

>>Geoff Lambert


>Your pretty right Geoff in that most of the argument is academic these days.
>The worst case scenerio for DE hauled train with a "dead" loco attached is 
>trying to start with full load on a severe grade , ie the full load is set by 
>the grade.
>Locos of 2000 HP and over with axle loads of less than 23 tonnes have 
>sufficient starting torque to spin the wheels if too much power is applied.
>The dead loco in this case supplies an additional starting load in that we 
>must suuply enuf pull to overcome the inertia and brush pressure of the 
>stationary traction motors.
>If you try to spin any large DC series motor , even a car starter motor will do
>you will find that the initial resistance to movement is higher than when you 
>actually get the motor spinning.
Very few trains these days are anywhere near full loads so the problem rarely 
if ever arises.

cheers
MD