Widening of Gauge on Curves – Reasons, Rules and Example

Widening of Gauge on Curves

There are several reasons for widening the gauge length in the case of a sharp curve.

How to use Fillet Gauges? | Radius ...
How to use Fillet Gauges? | Radius Gauges |

➡ Due to the loss of contact between wheel and rail in trailing position

➡ Due to the tendency of the inner wheels to slip in the backward direction. At the same time, the tendency of the outer wheels to skid in a forward direction.

➡ Due to the rigidity of the wheelbase, it is sometimes found on the curve that the rails are tilted outwards so that the actual gauge is more than the theoretical value.

➡ The centrifugal force which is acting in the outward direction tries to take the vehicle outward direction.

To overcome all the above reasons, some widening is required on the gauge on the curve.

How to find out the extra width of gauge required on curve?

There are several formulas that may be used to find out the extra width of the gauge on curve.

Rule: 1

The extra width of gauge (d) =  \frac{(B+L)^{2} \times 125}{R}


d = extra-wide of gauge on curve (in ‘mm’)
B = the rigid wheel base (in ‘m’)
L = Lap of flange (in ‘m’) =  2\times \sqrt{(D+h)h}
R = Radius of curve (in ‘m’)
D = Diameter of wheel (in ‘m’)
h = Depth of wheel flange below rail (in ‘m’)

Rule: 2

First, half the wheelbase and then multiply with lap of flange. After that, this result is to be divided by the radius of the curve plus half the gauge. And finally, multiply the result by 3000 and then the outcome will be the extra wide required in mm. All the above units will be in m.

Mathematical Example

The wheelbase is 4.724 m of a vehicle moving on a broad gauge(B.G). The diameter of the wheels is 1524 mm and the flanges project 32 mm below the top of rail. Calculate the extra wide of gauge required by using rule 1 and 2, where the radius of the curve is 165 m.

Solution: by using rule 1

We know, d =  \frac{(B+L)^{2} \times 125}{R}


B = 4.724 m.
R = 165 m.
D = 1524 mm = 1.524 m.
h = 32 mm = 0.032 m.

L =  2\times \sqrt{(1.524+0.032)0.032} = 0.446 m.

Now, put all the values in the above equation

d =  \frac{(4.724+0.446)^{2} \times 125}{165} = 20.24 mm or, 20 mm.

Solution: by using rule 2

↣ First half the wheel base: B/2 = 4.724/2 = 2.362

↣ Then multiply with lap of flange: 2.362 × L = 2.362 × 0.446 = 1.053

↣ Then divide this result by the radius of the curve plus half the gauge:   \frac{1.053}{R+ \frac{1.676}{2}}

=   \frac{1.053}{165+ \frac{1.676}{2}}

= 0.00635

[Note: The length of the broad gauge is 1.676 m.]

↣ Then multiply the above result by 3000: 0.00635 × 3000 = 19.05 mm Or, 19 mm.

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