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Extra widening: reasons and derivation

Short answer

Extra widening is additional carriageway width on horizontal curves due to off-tracking (rear wheels track inside), steering difficulty, front overhang swing, and clearance needs. Total widening We = Wm + Wps, where mechanical widening Wm = n*l²/(2R) and psychological widening Wps = V/(9.5√R). Methods: provide on inner edge only, equally on both edges, or gradually along transition curve.

Define extra widening. State the necessity/reasons for extra widening of pavement at horizontal curves and derive an expression for the extra widening required (mechanical and psychological widening); mention the methods of providing it.

Extra widening is the additional width of carriageway provided on a horizontal curve, over and above the normal width of the straight road.

Need (reasons)

  1. Off-tracking: the rear wheels of a long vehicle do not follow the front wheels; they track inside, so the vehicle occupies more width.
  2. Difficulty of steering: on a curve the driver tends to keep a greater clearance from the pavement edge (psychological effect).
  3. Wider swing path of the front overhang, so the vehicle may encroach on the adjacent lane.
  4. Greater clearance is needed between two vehicles that pass or overtake on a curve, because their lateral position varies with speed.
  5. Therefore, extra widening We=Wm+WpsW_e = W_m + W_{ps} (mechanical + psychological).

Mechanical widening (WmW_m)

Let ll = length of wheelbase (rear axle to front axle), R1R_1 = radius of the path of the outer front wheel, R2R_2 = radius of the path of the outer rear wheel, RR = mean radius of the curve, and nn = number of traffic lanes.

            O  (centre of curve)
            |\
            | \
       R2   |  \  R1
            |   \
            |    \
  rear axle +-----+ front wheel
            <- l ->

The rear axle points towards the centre OO, so the centre, the outer rear wheel and the outer front wheel form a right triangle with R1R_1 as the hypotenuse:

R22=R12−l2Wm=R1−R2=R1−R12−l2≈l22R\begin{aligned} R_2^2 &= R_1^2 - l^2 \\ W_m = R_1 - R_2 &= R_1 - \sqrt{R_1^2 - l^2} \approx \frac{l^2}{2R} \end{aligned}

For nn lanes, each vehicle gives the same off-tracking, so

Wm=n l22RW_m = \frac{n\, l^2}{2R}

Psychological widening (WpsW_{ps})

Empirical formula (IRC):

Wps=V9.5RW_{ps} = \frac{V}{9.5\sqrt{R}}

where VV is the design speed in km/h and RR is the radius of the curve in m.

Total extra widening

We=n l22R+V9.5RW_e = \frac{n\, l^2}{2R} + \frac{V}{9.5\sqrt{R}}

IRC takes l=6.1l = 6.1 m for the design vehicle. Widening is generally provided for R<300R < 300 m, since for larger radii it is negligible (NRS 2070/IRC practice).

Methods of providing

  1. On the inner edge only - preferred on hill roads, where the widening is given on the inner (hill) side.
  2. Equally on both edges (inner and outer) - common on plain roads, so the centre line stays central.
  3. Introduce the widening gradually: along the transition curve, so that the full width is reached at the start of the circular curve; on a simple circular curve without a transition, about two-thirds of the widening is attained on the tangent and one-third on the curve (the same as the superelevation run-off), giving a smooth edge.
  4. Add the widening to the pavement and keep the shoulders as usual.

Example: for V=60V = 60 km/h, R=100R = 100 m, 2 lanes, l=6.1l = 6.1 m: We≈2(6.1)2/200+60/(9.5×10)=0.37+0.63≈1.0W_e \approx 2(6.1)^2/200 + 60/(9.5\times10) = 0.37 + 0.63 \approx 1.0 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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