The diagram above is a section of my LMS layout. When I began planning this layout in 1999, the first consideration was the radius of the curves. These are 1200mm in diameter and therefore 600mm radius or 2 feet. This is probably a minimum if you want to run trains with long locomotives and long carriages, such as the "4-6-2 Coronation class. Anything less is too tight a curve. The next consideration was vertical height. On this two-tier layout, the differential between the upper level station and the lower is 100 mm or 4 inches. Between the two is a gradient that is effectively 19 feet or 5800mm long.

The first question that comes to mind is: why is it so long - couldn't it have been shorter? The answer is "no it couldn't! Well, it could have been, but the price to pay would have been a big problem for operation unless I just ran light engines. The reason is that for every linear unit forward on an inclined plane such as this, the train has to work that much harder than if it was on the flat. A train would probably not get up an incline of more than 1:30 without stalling, or at least having problems with traction. The proportion 1:30 means that for every unit forward i.e (in this case) 30, the track (or road) is inclined by one unit. These units may be millimetres, inches, metres, kilometres or whatever. So if we used inches, 1:30 would be a rise of 1 inch in 30 inches or 2 feet 6 inches. It would be 2 inches of vertical height in 5 feet and so on. It may be seen from this, that the greater the second number (e.g. 1:80) the more gentle is the incline, because the track has more length to achieve its projected height. So 1:80 is less steep than 1:30.

On my layout, I needed to raise the track 100mm or 4 inches. As you may see in the above diagram, the incline begins in the middle of the curve at the bottom and extends to beyond the curve at the top. This was to provide maximum length in which to reach the top level. Simple arithmetic will show that a rise of 100 mm in 5800 is 1:58, i.e 5,800 divided by 100 is 58. However, had the length available to me only been (say) 3000, then 3000 divided by 100 is 30, so the gradient would have been 1:30. This is too steep a gradient for effective train running. Had I had 10 metres or 32 feet, then the gradient would have been 30,000 divided by 100 = 1:300. Some American publications talk of gradients as a percentage, e.g 2 per cent. In my case a 1:58 gradient can be recalculated as 1divided by 58 (0.017) X 100 = 1.72%. A gradient of 1:30 is 3.3%. It has been said that the "ideal' gradient should not be more than 2.5 %.

Arithmetic aside, how is a gradient achieved. Model shops now sell foam track bed that is wedge shaped and thus provides a ready-made incline. However, a cheaper way is to use plywood - this is how I did it. Cut the plywood just wider than the track, say, 50mm to allow for ballasting. (00 track is 16mm) Sand the bottom end down to a wedge shape so that it provides a smooth entry point to the incline. Fasten it down at this end. Assume that you want an incline of 1:48 or a quarter of an inch to the foot. (This is 2%). In metric terms, this is 6mm in 305m, which is actually 1:50. One foot from the bottom of the ramp put a quarter inch block beneath the plywood and glue it down to your baseboard. Every foot from that first block, put another block in, but make it a quarter of an inch higher, i.e 0.25, 0.5, 0.75, 1 inch and so on. It may be seen that to achieve a height of 4 inches, this will take up 16 linear feet. However, if you curve the track at the top and bottom as I did, you can reduce the gradient by increasing the track length. If this process was done in metric (as mine was) the riser blocks were 5mm and the spacing was every 300. This gave me a gradient of 1:60 or 1.6% - in theory.

In reality, on using a method of grade measuring that I discovered after the layout was completed, I discovered that it was close enough. Firstly, the method. Take a length of bar (a long spirit level is ideal) whose length is known. My spirit level is 890mm. Place the spirit level on the track and then raise the lower end of it until the bubble shows that the level is level. Then measure the vertical distance from the bottom of the level to the top of the rail. Let us say that it measured 25mm. Eight hundred and ninety (890) divided by 25 = 35.6. The gradient would therefore be 1:35.6 (1:37 rounded off). So it would be 1:37 only for 890mm. If the spirit level was two metres (2000 mm) long, then the gradient would be 2000/25 = 1:80. The first gradient would give your train a short, fairly steep climb. The second would be a long, lesser, but sustained climb. In terms of prototype gradients, the Lickey Bank between Bromsgrove and Blackwell on the Bristol to Birmingham line is 1:38 (2.6%) for just over 2 miles (3.2km) Shap is about the same and in the South Island of New Zealand, the Trans-Alpine from Christchurch to Greymouth is 1:33 for a lengthy stretch.

Just for fun, imagine that you were going to model Lickey in 00 scale. Bromsgrove at the bottom, Blackwell at the top. Since 00 is 1:76, 2.3km - it would be 30 metres long. Thirty metres is 30,000 mm. Two point six percent of 30,000 is 780 mm. If Bromsgrove station was on the floor, Blackwell station would be 3/4 of a metre high 30 metres away. A 'big lizzie' 4-6-2 with six coaches behind would take a long time to cover the distance, but it would be great to see it coming back, even at scale speed!

Copyright © Peter Baddlely 2004