Swallow Posted January 23, 2011 Share Posted January 23, 2011 Hi people, I hope there are a few people who can 100% answer the question above. After being dissapointed at the NMRA for failing to give a standard for weight of locomotives I turn to the wider community. I am seeking a formula or a series of Formula's for working out the correct weight of a locomotive. Specifically that of 00/H0 but im sure the theory can be applied to all scales. I have seen people suggest the old 'finding stall current, finding at which point amps start to rise quickly and then remove the last weight increment added but (and if anyone can confirm this) This method does not allow for the actual weight of a train the locomotive is to pull? If you add weight to a model locomotive to a point below, or just below the stall point (wheels stop spinning) isn't it safe to assume that it won't be able to pull much because 95% of the power generated from the motor, when applied to the wheels is just to overcome the weight of the locomotive? The formula I'm seeking goes something along the lines of... If a motor produces x amount of Torque, or/and a motor produces x amount of Watts, and/or is connected to a reduction/increase gearbox of x to x and is expected to pull X amount of weight (train) up a grade no steeper than x% how much weight is required for OPTIMUM performance? There are alot of words in there and I am no engineer or mathmatic guru. I am hoping someone out there is able to comprehend the above and come up with a formula, maybe there are variables I have missed out that need to be included. Basically, someone must be able to workout what is required based on data, particularly that of a motor which usually comes with a data sheet Things I can think of that need to be considered are... (amps, volts, watts, torque, tractive effort, force applied to trains on gradients aka effect of gravity, effects of reduction gearboxs, weight of train, number of driving wheels etc etc) I do believe that there is something called the Co-efficient of rail/track used in prototypical locomotive design whereby the relationship between the track steel and the wheel is considered but I am assuming that this formula, what ever it is will produce results negligable of considering for the main question in hand. Kind regards everyone Link to post Share on other sites More sharing options...
Coombe Barton Posted January 23, 2011 Share Posted January 23, 2011 I'll start off by saying there isn't an effective formula. If you reduce from the prototype then, in 4mm, you'd be looking at 1/76.2 x 1/76.2 x 1/76.2 the weight of the prototype, that's 2.226 grammes per tonne. For a 43.8 tonne loco (that's the loco of 3205) that turns out somewhere in the region of 98.1 grammes. The straight from the box models turn out at around 200g (just weighed it). That's overscale by a factor of 2. But then we're into motors. And gear trains. And friction. For such a small scale the calculations and errors and uncertainties of the build, gear meshing, lubrication and that small piece of dog hair (in this house an inevitability) caught somewhere in the works.And you can't rely in measuring the input voltage and current either - unless you want to spend lots on accurate meters. And then there's the different losses of motors of the same type. And wagons - oft quoted is 25g per axle, that's the 4mm equivalent of 22 tonnes whatever the loading. I think there's a much simpler way of getting it running right than to spend the time looking for formulae, and that's experimentation - get it running free then add weight and no more until it pulls the stuff you want on your layout without struggling. If it starts emitting blue smoke then you've put in too much. And this is speaking from someone who develops mathematical models to represent the real world every day (and who's working on one right now to teach next week). Don't bother - it ain't worth it. Experiment instead. Link to post Share on other sites More sharing options...
Jellicoe Posted January 23, 2011 Share Posted January 23, 2011 Oops! I think you're off there: it's 76.2 x 76.2 x 76.2 = 2.260 grammes per tonne. Link to post Share on other sites More sharing options...
bertiedog Posted January 23, 2011 Share Posted January 23, 2011 i remember a very famous member of the NMRA and publisher, Linn Westcott, who after a question about this came up at a convention, said ".....we just run model trains,,...don' t we?" The reason the NMRA do not answer it is there is no answer, you cannot scale the effects. Modelling can give results in Lab experiments, where scale items are tested, but only after complex adjustments to allow for density, and other un-scalable factors. Locos can weight as much as the motor will take, and given a reasonable gear ratio choice, apply that that the track. Stephen. Link to post Share on other sites More sharing options...
Coombe Barton Posted January 23, 2011 Share Posted January 23, 2011 Oops! I think you're off there: it's 76.2 x 76.2 x 76.2 = 2.260 grammes per tonne. A linear scale of 1:76.2 is the ratio of 4mm to the foot, 4 in 304.8, with an inch being defined as 25.4mm exactly. Mass is related to volume, and the volume of a 4mm to the foot model is 1/(76.23) of the prototype. This is assuming that the density of the materials used is similar for both prototype and model. 76.2 cubed is 442450.728 One tonne is a thousand kilogrammes or a million grammes 1,000,000 divided by 442450.7 = 2.26 grammes per scale tonne. A long ton of 2240lb (1016.5kg) is only 1.605% heavier than a metric tonne which was used for my calculations, not really an issue here. Link to post Share on other sites More sharing options...
Allegheny1600 Posted January 23, 2011 Share Posted January 23, 2011 Hi people, I have seen people suggest the old 'finding stall current, finding at which point amps start to rise quickly and then remove the last weight increment added but (and if anyone can confirm this) This method does not allow for the actual weight of a train the locomotive is to pull? If you add weight to a model locomotive to a point below, or just below the stall point (wheels stop spinning) isn't it safe to assume that it won't be able to pull much because 95% of the power generated from the motor, when applied to the wheels is just to overcome the weight of the locomotive? Hi Swallow, I think that John (Coombe Barton!) has given the best all round answer that you're going to get here (or anywhere!) however, I thought I'd just point out that in the situation you mention above, the loco is actually 'tethered'! This means that under test, the motor will/should be able to spin the driving wheels at all weights. If you overload the loco body such that the motor will no longer spin the drivers - you'll soon be looking to buy a new motor (& another and another etc!!!). HTH, John E. Link to post Share on other sites More sharing options...
jazz Posted January 24, 2011 Share Posted January 24, 2011 The answer is simple. Too little, it spins, too much you may loose the motor. Link to post Share on other sites More sharing options...
Kempenfelt Posted January 24, 2011 Share Posted January 24, 2011 An interesting question! One of our club members (Graham) has come up with his own formula which has been derived from scaling the weight similar to how John (Combe Barton) describes. Graham also has a standard formula for gearing, the intention being that any combination of his loco's when run together will run at the same speed, even if it is his 1f and Jubilee (This was all calculated way before the invention of DCC by the way). This result's in all of Graham's loco's and stock share a common standard and allows each loco to pull loads relative to it's size. This method clearly has it's advantages (as expressed in relation to the gearing) and is fine when all of Graham's stock is run together, the problems start to arise however when the stock is mixed and matched with others. Graham's stock is all very light because it is based around scaling the weights of the full size item (i'm guessing around 30 grams for a 4 wheel open wagon, therefore very close to John's calc's). Therfore it is farly common for Graham's loco's to struggle on a relatively small load of someone elses stock. I am not at all knocking Graham's approach toward's weighting his loco's as it suits him and gives him a consistent working standard in which to aim for. Therefore I would argue that it's not so much a correct weight for loco's that you are after, but instead a consistant weight per scale ton that you ideally want to achieve. The only other factor that will need to be considered is what you're intentions are for your loco's and stock. I.e. are the loco's to be used exclusively with your own stock or would you like to run it on a group/visiting layout's and mixing with other people stock? Not really answering the original question but I hope this helps. Best Wishes Paul Link to post Share on other sites More sharing options...
34theletterbetweenB&D Posted January 24, 2011 Share Posted January 24, 2011 I weight my steam locos to give some relationship to power class and have a 'formula' for the first cut at weighting. All locos up to power class 3 250g on the driven wheels, or as nearly as can be got to that figure (pure pragmatism for good electrical pick up) Add 50g per power class above class 3. The final weight is a matter of the motor being able to slip the wheels when the model is held back (modern can motors wth 30: 1 or higher ratios typically have little trouble) and adjustment so that the loco 'fits' tractively with the rest of the fleet. Coefficient of friction varying dependent on the tyre material can have a perceptible effect. All as rough and ready as the OO track standard... Link to post Share on other sites More sharing options...
Miss Prism Posted January 25, 2011 Share Posted January 25, 2011 For 4mm scale, try 4g per prototype ton. Link to post Share on other sites More sharing options...
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