Aljosha said:
Hey guys, this is getting better and better! ;-)
Lots of good thoughts here. Looks like you guys have found the way how to do it. I assume that this EX-cludes fitting ball-bearing wheels thoughout, but IN-cludes additional weight in ALL wagons?!
Well I think I've got the cheapest solution with the reverse super. Sounds like you have a similar problem so maybe give it a go.
We?ve certainly got away from the original reverse superelevation topic, but I like discussing this aspect too.
The idea of scale mass being a cube of the linear scale can?t be taken too exactly. Changing scale from 1:20 to the 1:24 that I model in, means the ?mass scale? goes from 1:8000 to 1:14000. So should a 50t loco weigh 6 kg or 3.5 kg ?
The prototype 3?-6? gauge Garratts I ?modelled? at 1:24, were around 70-90 tonnes so the model should weigh 5 to 6.5 kg. Mine are about half that.
(BTW, why does anyone bother with calling it 1:20.3 scale? are you accurate to 1.5% ? I?m not.)
I?ve done a lot of testing to try to measure the tractive effort of my locos and the ?drag? of my rollingstock.
I?ve even come up with a rough equation to let
me approximate how many wagons any loco can haul.
If you?re not interested in the maths and like pictures, go read another thread now.
Locos
[H1][/H1] Most of my locos have an adhesion factor of around 20-30%. ie if you measure the mass of the loco and take say 25% of that figure, that is how much drag you can overcome and keep moving. (This assumes all the weight is on the drivers). Real steam locos had an adhesion of around 15%, but diesels can manage up to around 30%, so the model figure is close to the prototype.
Wagons [H1][/H1] My model measurements give drag values around 4% of the mass of the wagon (on level track). This is for plastic wheels, mainly with steel axles running in styrene or aluminium ?bearings?. Those of you with steel wheels and better bearings should have a lower figure.
(Interestingly, when I worked in the real railways, we used a first approximation that the drag was equal to a 0.5% grade. ie only a tenth of my models. )
Grades [H1][/H1] The drag due to the grade is the grade in %/100 times the mass of the train (incl locos).
ie on a 3% grade a 5kg train has a drag of 3/100*5 = 0.15 kg
Curves [H1][/H1] A VERY ROUGH approximation is: the curve drag = 0.04/curve radius in m times mass of train (incl locos)
Example [H1][/H1] So taking the derailment example I started with, of my Garratt weighing 2.5 kg and a train weighing 7 kg. The TE will be 0.25x 2.5 kg = say 620g
The wagon drag is 7kg * 0.04 = 280g.
A 4% grade drag is 0.04*(7+2.5) =380g
So on straight track we have a TE of 620g and a drag of 660g, so it shouldn?t work, but it does JUST. So my approximations are a bit on the conservative side.
Now if you want to rearrange all these equations you can determine what mass of train you can pull on what grade or curve:
Mass of wagons that can be hauled (kg) =
(0.25 * mass on loco drivers (kg)) divided by (0.04 + 0.04/rad(m) + %grade/100)
(This ignores the mass of loco on the grade to simplify it.)
Trying the example again:
The mass I should be able to haul up a straight, 4% grade = .25*2.5 / (.04+4%/100)=7.8 kg.
Those of you with steel wheels may be able to reduce the two 0.04 factors. To determine what it is for you, just take a loco and see how much mass it can haul on the level straight. You may have to put some bricks in the wagons! Then your coefficient to replace MY 0.04 will be 0.25 times loco mass divided by wagon mass.
End of lesson.