Coned wheels: how a train steers itself round a curve

A train has no steering. Nothing in the cab turns the wheels, yet a train follows a curve smoothly for mile after mile. The reason is the shape of the wheel, and the geometry fits in a few lines of algebra.

The problem: one axle, two paths

On a railway wheelset the two wheels are pressed onto a single solid axle, so they turn together at the same rate. On straight track that is fine. On a curve it is a problem, because the outer rail is longer than the inner one. If both wheels were the same diameter, one would have to slip.

Road vehicles use a differential. A railway wheelset has none, and gets the same effect from the tread itself.

The taper

A rail wheel tread is not a flat cylinder. It is a shallow cone, wider in diameter on the side nearest the flange. A taper of 1 in 20 is a well-known figure, though real profiles vary and wear into other shapes. A slope of 1 in 20 is 0.05, so the wheel's radius changes by 5 mm for every 100 mm you move across the tread.

Now picture the wheelset sliding sideways along its axle line towards the outer rail. The outer wheel meets the rail on a part of the cone with a larger radius, and the inner wheel meets it on a smaller one. The outer wheel covers more ground per turn, and the wheelset has steered itself.

The derivation

Use these symbols:

• R is the radius of the curve, measured to the centre of the track.
• G is the distance between the two contact points on the rails. For standard gauge this is about 1.5 m, a little more than the 1,435 mm gauge, since the wheels touch near the middle of the rail head. (See why standard gauge is 1,435 mm.)
• r₀ is the wheel's rolling radius when it is centred on the track.
• The taper is 0.05, written as t.
• y is the sideways shift of the wheelset towards the outer rail.

Step 1: what the track demands. The outer rail follows a circle of radius R + G/2 and the inner one R − G/2. In one turn through a small angle θ, the outer wheel must travel (R + G/2)θ and the inner (R − G/2)θ. The difference is Gθ. As a fraction of the distance Rθ the wheelset travels, that is G/R.

Step 2: what the wheels give. Both wheels turn through the same angle, so the distance each covers is proportional to its rolling radius. After a shift y the outer wheel rolls on r₀ + ty and the inner on r₀ − ty. The difference in radius is 2ty, and the fractional difference is 2ty / r₀.

Step 3: set them equal. The wheels give exactly what the track demands when

2ty / r₀ = G / R, so y = r₀G / (2tR)

This is the same as r₀ × (G/2) / (R × t), if you prefer half the track width. The approximation is that the exact ratio of rail radii, (R + G/2)/(R − G/2), is taken as 1 + G/R, which is accurate at these radii.

A worked example

Take R = 500 m, r₀ = 0.46 m (a wheel about 920 mm across), G = 1.5 m and t = 0.05:

y = (0.46 × 1.5) / (2 × 0.05 × 500) = 0.69 / 50 = 0.0138 m, which is 13.8 mm

Check it. The radii become 0.46 + 0.05 × 0.0138 = 0.46069 m outside and 0.45931 m inside. Their difference is 1.38 mm, which divided by 0.46 gives 0.0030. The track asks for G/R = 1.5 / 500 = 0.0030. Over 100 m of curve, the outer wheel travels 0.3 m further than the inner one, and the tapered tread supplies exactly that.

A smaller taper needs a larger shift, because less radius difference is gained per millimetre.

When the taper runs out: flanges

The wheelset cannot slide far. There is only a little clearance between the flange and the rail, and the free sideways play is often quoted as of the order of a few millimetres up to about a centimetre each way. It varies by railway and wear, so treat 7 mm as an assumption. Then y reaches 7 mm when R = 0.69 / (0.1 × 0.007), about 986 m. Any curve tighter than that pushes the flange against the rail.

There the flange is the backup. It carries the extra sideways force, and the wheel and rail wear and squeal. The 13.8 mm needed at 500 m exceeds the assumed play, so that curve would involve flange contact. The taper steers on gentle curves; the flange takes the sharp ones.

Hunting

The taper has a side effect. Take a wheelset on straight track that is displaced sideways. The cones push it back to the centre, it overshoots, and so it weaves from side to side. This is called hunting, and Klingel's formula gives the wavelength of the weave as 2π × √(r₀e / t), where e is half the contact distance, 0.75 m. With our numbers that is 2π × √(0.46 × 0.75 / 0.05) = 2π × √6.9, about 16.5 m.

The wavelength does not depend on speed, so at 100 km/h (27.8 m/s) the wheelset weaves 27.8 / 16.5 ≈ 1.7 times a second. Real vehicles have suspension and damping to stop the weave growing. A steeper taper steers better but hunts more, so the taper is a compromise.

What to take from it

• The required shift is y = r₀G / (2tR). It halves each time the radius doubles.
• At R = 500 m with a 1 in 20 taper, y = 13.8 mm.
• Flanges take over when the shift exceeds the play. Hunting is the taper's price.