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What is the Correct Weight For HO Wagons?
Club member Albert asks readers:
“Can someone enlighten me as to the right weight for an HO wagon so that it will on the track especially around curves and on turnouts? I guess what I am asking is – 30 foot = total ounces of car – 40 foot = total ounces of car – 50 foot = total ounces of car. I am hoping someone can help. Thanks.”
5 Responses to What is the Correct Weight For HO Wagons?
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The NMRA Website has the chart that recommends for different cars. What I can tell you that in a student engineer Model Railroad Club I know took 2 coal hoppers and loaded them down with metal plates under the load to represent the car’s weight as if it was a full load. Their HO Engine could barely pull those 2 much less any others Yes I believe they accomplished a close weight for a loaded hopper car But the didn’t take into account that the engine is way underpowered to the same in it’s proto type it represented. An example is: a proto type pair of engines use 2 engines to pull a 75 car train but to do that in an HO model can only do 1.4 to 1/2 a train that size. The only way you can increase the train length in HO is to add another engine or lighten the weight of all your cars and then you run into the reason of needing the weight in them in the first place. It would take quite a model train engine to represent the actual proto type. I for one weigh an HO Car by feel and not necessarily by the recommended weight from the NMRA Standards although after years of experience I am not to far off on the proper weight that is recommended. Some clubs are real strict to approve your car to be run on their layout that includes proper weights. Just use common sense when adding weight. Remember the engine in HO no matter how expensive it is will pull an equal amount of cars that the proto type will. Cheap beginners train cars usually are very under weight and that causes derailments. That and the plastic trucks and wheels may not be able to stand much weight added.and will just wear out your trucks and bearing points of the axles as they are not made to carry the weight
from Newman
I believe the NRMA recommendation is not to be taken as “car full” or “car empty” figure.. The figure given is the “best fit” for a car of a given length for model rail operation for optimum performance on our tighter curves and heavier gradients.
Most commercial wagons are too light and if placed in a longish train tend to derail on poor to average track or tight curves. Also, if there is a mix of weights or bogie/4 wheelers there is a greater tendency to derail if not weight balanced.
To try and replicate true “loaded/unloaded” weights will wear out your locos (very expensive because they will need both re-motoring and re wheeling) and the wheel bearings on your wagons (very annoying and time consuming replacing them),
Use the NMRA standards!!!!!!!!!!!!!!!!!!! 1 ounce + 1/2 oz per inch of length. ie a 50′ box =4.375oz.
Irrespective of size, it is the weight of the loaded car that you need to scale. Thus, in HO (1:87 IIRC) a 40 tonne load would weigh (40×10^6)/(87^3) = approx 61 gr (~2.1 oz). a 100 tonne load would be about 15 gr (~5 oz). Currently I am modelling in Z scale (1:220) this my 10 tonne coal truck which weihs 1/2 oz (empty) would weigh about (220^3)/(16×2240) ton (empty), nearly 300 tons! That assumes, of course, that all the thicknesses of material are scaled as well as the overall dimensions.
Turning to locomotives, a DRG BR03.10 weighs about 100 tonnes. Thus the scale weight of an H0 model would be (100×10^6) gr / (87^3) wich comes out at 151 gr (5.3 oz).
However, there is a much more important factor that affects the apparent pulling power of the locomotive — friction. Prototype wagons use roller bearings (or similar) on their axles. As there is little of no surface frictions effect (they are rolling, after all) on a level straingt track the wagon offers very little resistance once it is moving at a steady speed. On gradients and curves there is resistance (gravitational for gradient and frictional on curves) and it is this resistance (and momentum) that require powerful locos. Most models use needle or rod-through-a-hole bearings.. Because these are not friction free this has to be taken into account. Thus model locomotives cannot pull 200 exle trains.
On my layout, my Bavarian pacifics can just manage a nine coach train, i.e. 42 unpowered axles, but to do that the baseboard has to be exactly level. A half a degree slope across the semicircle end and I am down to 7 coaches. You guys with bigger locos, of course, have rubber tyres to hel with traction but my locos are metal on metal.
Hope all that helps. I have no idea how much empty wagons weigh but the same rules apply to them as to the loads they carry.
CM
Listen to Jerry, Albert.