Model railroad siding capacity calculator.

How the capacity is calculated

This is a scale conversion and a division. The NMRA's S-3.2 gives the scale ratios (HO 1:87.1, N 1:160, O 1:48 and so on). A car of c prototype feet is c × 12 × 25.4 / ratio millimetres long on the model. Each vehicle also takes up the gap between it and the next, which depends on your couplers, so you enter it. Each car then uses its body length plus the gap, the locomotives are subtracted first, and the cars that fit are the whole number of those lengths in what is left.

The page gives no default for coupler gap as if it were a fact, because couplers vary. The 0.5 inch in the example is an assumption stated in the example, not a measurement of any product; measure two of your coupled cars and use that figure. The answer is a floor: 18.5 cars is 18.

The converter also tells you how long the track would be in real feet. A forum answer of "60 feet to the mile in HO" is the rounded form of 5,280 / 87.1 = 60.62.

Worked examples with real numbers

• 10-foot HO siding, 40-foot cars, one 60-foot locomotive, 0.5 inch gap. A 40-foot car is 5.51 inches at 1:87.1, so each car with its gap takes 6.01 inches. The locomotive takes 8.27 inches plus 0.5 inch. That leaves 111.2 inches of 120, which holds 18 cars with 3.04 inches spare. The siding is 871 feet of real track.
• Same siding in N. A 40-foot car is 3.00 inches at 1:160. With a 0.25 inch gap each car takes 3.25 inches; with no locomotive, 120 inches holds 36 cars.
• Metric. A 3-metre HO yard track is 3,000 mm; a 40-foot car is 140.0 mm, so with a 10 mm gap each takes 150.0 mm and 20 cars fit.

What this calculator leaves out

It treats every car as the same length. A mixed train needs a sum of your own cars. It does not deduct turnout lengths or the fouling point unless you do, so enter only the usable length.