Hi guys, I'm in need of some figures - I don't have anything set up yet, and would like some measurements of the following train lengths (imperial or metric is fine): 3 SD40-2s +25 50' cars = 3 SD40-2s +30 50' cars = 3 SD40-2s +35 50' cars = 5 SD40-2s +25 50' cars = 5 SD40-2s +30 50' cars = 5 SD40-2s +35 50' cars = I need the above to check my sidings - the second grouping would be with two additional helpers. Either truck-mount or body-mount couplers are OK.

SD40-2 = 70ft long, 50ft car = 50ft long. Real railroads use a 3% stretch to account for coupler slack. These numbers should be close when divided by 160 to convert to N scale. 3 SD40-2s +25 50' cars = 1460ft, 1504ft stretched. Scale 9.125ft, 9.4ft stretched 3 SD40-2s +30 50' cars = 1710ft, 1761ft stretched. Scale 10.7ft, 11ft stretched 3 SD40-2s +35 50' cars = 1960ft, 2019ft stretched. Scale 12.25ft, 12.6ft stretched 5 SD40-2s +25 50' cars = 1600ft, 1648ft stretched. Scale 10ft, 10.3ft stretched 5 SD40-2s +30 50' cars = 1850ft, 1906ft stretched. Scale 11.6ft, 11.9ft stretched 5 SD40-2s +35 50' cars = 2100ft, 2163ft stretched. Scale 13.125ft, 13.5ft stretched Using these simple basics you should be able to figure any length out you need.

Don't forget to include the length of the couplers themselves. Doesn't sound like much but multiplied 25 or 35 times, it adds up and will throw off your expectations! Trevor D.

That's what the 3% stretch is for. Some cars will be accurate over the knuckles, some won't. The 3% stretch should more than cover it. It does in real life.

just out of curiosity... why do you want to run 5 sd40-2 with only 35 cars ? a single sd40-2 of the latest kato release easily pulls 25-30 cars. i do run coal trains with 75 slightly over-weighted aeroflow hoppers (approx 38-40 grams each) with 2 sd40-2 leading and one sd40-2 pushing. but then my layout is flat and uses 20" minimum radius.

It's interesting that the question relates to 50ft cars. I have encountered some modellers who prefer 40ft, because, when you sit at the side of the track and count the cars going past, a train of 40 fifty footers doesn't look as long as a train of 50 forty footers. In theory, they ought to be the same length. 5x40=200, same as 4x50=200. I had thought it was associated with that strange optical illusion where in facing arrow heads on the end of a straight line don't seem to be the same length as out facing arrow heads on the same straight. No! I've measured 5 forty footers on the table against 4 fifty footers, and the 5 forty footer train is longer . . . It's the extra couplings! Regards, Pete Davies

The two extra locos are helpers. I'm concerned about the couplers, which is why I wanted to know the measurement of a complete train.

Rail Op, a computer program for operating your model railroad, adds four scale feet to the length of every car and engine to allow for model couplers. 35 x 54 feet is 1890 scale feet or 11.81 feet (3600 cm).