Railway alignments comprise three basic elements; straights, circular curves and spiral transitions.
This article explains the basics of how to apply cant and how to set out the alignment of a transition. Initially I need to explain a few definitions:
Also called “super elevation”. Cant is a measure of the height of one rail vertically above the other. The maximum value of cant in the UK (at present) is 150 mm or 110 mm through platforms.
Cant deficiency is the amount of additional cant in excess of that actually provided, which would need to be added to the track to enable a train to run around a curve at the permitted speed in perfect equilibrium. It is directly proportional to the sensation of a sideways force caused by the centripetal acceleration experienced by a passenger in the train. Curves are not normally canted for trains to run in equilibrium. The recognised practice is for only two thirds of the equilibrium cant for the speed of the line to be applied to the track, leaving one third as cant deficiency. The proportions can be varied to suit difficult design situations. The maximum allowable cant deficiency in the UK (at present) is 110 mm on welded track and 90 mm on jointed (fish-plated) track. Higher values will be permitted on TGV routes including the Channel Tunnel High Speed Rail Link where very high standards of maintenance will apply.
A transition takes the form of a clothoid spiral. As one moves along a transition the radius varies inversely proportional to the distance travelled. This sounds simple but in practice the calculations are so lengthy that it is only since the advent of computers that engineers have been able to design transitions accurately. Previously a cubic parabola was adopted as an acceptable approximation.
The cubic parabola has the simple formula: y = Kx³
The use of a clothoid or a cubic parabola is optional for lower speed lines on Network Rail but for high-speed lines the clothoid is now mandatory.
Figure 1: The Cubic Parabola
The cubic parabola has some very convenient properties, which make it easy to use:
- The shift between a straight and curve at the tangent point is given by the formula: Shift (S) = L² ÷ (24 x radius of curve)
- The tangent point occurs halfway along the baseline of the transition.
- The radius at the tangent point is twice the radius of the curve.
- The offset at the end of the transition is four times the shift at the tangent point.
- At the tangent point the alignment of the transition passes exactly halfway between the straight and the curve, i.e. the offset from the baseline is half the shift.
To be able to build a model railway at all we have to accept a very considerable compression of the prototype. The radii we use are very tight. Generally, six coupled mainline steam locomotives can only just get around a 90m radius curve, dead slow, with someone walking alongside to ensure the wheels don’t start to lift over the rails. In 4mm scale that is 3’-10½” radius. To meet the requirements of Her Majesty’s Inspectors, strictly speaking, any 4mm scale curve of less than 8’-6” radius should be fitted with a continuous check rail. At the other extreme the minimum radius around which our 4mm scale HST can run at 100 mph is 55 feet. With few exceptions model railways should have a scale speed limit of 15 mph with some parts as low as 5 mph.
In theory one could apply the maximum cant of 150mm and assume the maximum cant deficiency (for jointed track) of 90mm to get the maximum possible speed around a sharp curve. In practice this is not possible. Sharp curves are most usually found in busy locations where trains are frequently stopped or slowed below the permitted speed by adverse signals. The engineer must take great care in his design not to create a situation where the cant is greater than is needed for the majority of trains. This causes crushing of the low rail due to transfer of weight of the tilted train and loss of steerage, which causes sidewear to the outside rail. Also relatively small amounts of excessive speed on sharp curves causes very rapid rises in the cant deficiency, which increases in proportion to the square of the speed. Changes to the applied cant must be approved by the Chief Civil Engineer and it is usual for them to be cautious.
In the context of 4mm scale model railways the best that can be achieved with a 3 foot radius curve would be 10 mph. 15 mph can be achieved with a 5 foot radius curve.
Above: Northam Curve at Southampton takes the Waterloo to Bournemouth main line around a curve of a little over 90 deg. The radius is 200 m (8ft-7in in 4mm scale or 15ft-1in in O Gauge) with 90mm of cant. The design is theoretically safe for a speed of 35 mph. However, the Chief Civil Engineer cautiously limited the permanent speed restriction to 25mph.
Let us run through an example of say a 5 foot radius curve to straight transition on a 4mm scale layout running at a scale 15 mph. Now immediately we run into difficulty because engineering design on BR was metricated in the 1960’s, but speeds remain signed in mph due to trade union resistance and distances remain measured in miles and chains because they are written into the original acts of Parliament. So please bear with me.
The formula, which relates speed to cant and radius is: V = 0.29 x sqrt(R x (C + D))
11.82 = R x (C+D) ÷ V²
V is the maximum speed in kph
R is the radius in metres
C is the cant in mm
D is the cant deficiency in mm
(Reference: BR Engineering Handbook No.3, which is now absorbed into Network Rail Standards. Also reference “British Railway Track” published by The Permanent Way Institution)
In our example:
V = 24.14 kph (or 15 mph)
R = 116.129m (scaling up our 5ft radius curve to full size, 5ft x 76.2)
Therefore by substituting in the above formula: (C + D) = 60mm
Applying the two thirds rule: Cant = 0.666 x 60 = 40mm
(Had this not been a nice round number it would be rounded to the nearest 5mm for practical purposes)
The desirable transition length is given by the formula: L = 0.007536 x V x (The larger of C or D)
(This formula is derived from the requirement that the desirable rate of change of cant or rate of change of cant deficiency should not exceed 33mm per sec)
Where L is in metres, V is in kph and C or D is in mm.
Therefore: L = 0.007536 x 24.14 x 40 = 7.277 m
Network Rail design rules require that the minimum length of any element in an alignment must not be less than 20 metres (30m on TGV lines). Although we have calculated our transition needs to be 7.277m in length we will have to increase it to 20m to comply with the rules.
Another factor we have to consider is twist in the track. Twisting the track will make our rolling stock wobble. The limit for design in standard gauge is 1 in 400. That is for every 400mm measured along the track the cant cannot increase or decrease by more than 1mm. Although model railways have larger flanges than prototypical I would not recommend this value be exceeded because usually our models do not have the benefit of a sprung suspension. P4 modellers might have a view on this but the closer one works to true scale standards the more appropriate the limits of the prototype become.
Now as I mentioned earlier it is not possible to install a transition shorter than 20m so this requirement will over ride our calculations. In general all transitions on our model railways are going to require to be a scale 20m in length.
Finally we have to check that the twist is within the 1 in 400 limit:
Twist = transition length ÷ change in cant = 20m ÷ 40mm = 1 in 500
So we are well within the limit. However, the twist would have been too severe if we had used the calculated transition length of 7.277m. In that case the twist would have been 1 in 182, which is not permitted.
So let’s put this theory into practice on our model railway:
The transition length in 4mm scale is 20m ÷ 76.2 = 262.5mm.
This should be measured along the rails but for practical purposes we can skip the lengthy calculation and accept a value of 260mm measured along the x axis of the transition.
The shift (See figure.1) between the straight and the curve is given by the formula: S = L² ÷ (24 x R)
S is the shift in mm
L is the length of transition in mm
R is the radius of the curve in mm
Therefore: Shift = 260² ÷ (24 x 5ft x 304.8)= 1.85 mm
(where 304.8 is the conversion factor from feet to millimetres)
Let’s pause for a moment. In summary I am suggesting that most model railway transitions are going to be 260mm in length measured along the x axis (straight) and that the cant is never more than a prototype 40mm. This translates to an elevation of the track at 1 in 36.
Clearly transitions have to be drawn out on the model railway baseboard very carefully. A point to bear in mind when designing parallel tracks is that both tracks have the same value of shift S but different radii of curves. The outside track will have a slightly longer transition. It is necessary to calculate both transitions independently on real railways, but the error at 4mm scale is so small that this really is nit picking.
For Graham Plowman’s Ashprington Road layout, the transitions were designed by computer using software, which I had developed for prototype railway design. The designs were plotted full size onto paper using Autocad to create templates, which were then laid on the baseboard and the alignment pricked through.
The subject of how to construct canted track has been well covered in the model press so I will not cover that ground again.
A cant measuring device made from Meccano by the author about 35 years ago. It works on the principle of a pendulum hanging on a triangular frame. The pointer scale is calibrated in prototype inches and is suitable for any gauge of model representative of a standard gauge railway.
14 January 2020
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