Skip to main content

Curve resistance and grade compensation

Short answer

Curve resistance is additional tractive resistance on horizontal curves due to turned wheels, tyre slip, lateral friction, and off-tracking. Grade compensation reduces design gradient on curves so total resistance (grade + curve) equals straight resistance. IRC formula: compensation = (30+R)/R % (max 75/R%). Not needed for gradients <4%. Example: R=60m gives 1.25% compensation, reducing 6% gradient to 4.75%.

What are curve resistance and grade compensation?

Curve resistance

Curve resistance is the additional tractive resistance a vehicle has to overcome when it moves along a horizontal curve. When the front wheels are turned through an angle α\alpha, only the component Tcos⁡αT\cos\alpha of the tractive force TT acts along the direction of motion, so the loss is T(1−cos⁡α)T(1-\cos\alpha). The sideways slip of the tyres, the lateral friction needed to resist the centrifugal force and the off-tracking of the rear wheels add to it. It increases as the radius becomes smaller, so a gradient and a sharp curve at the same place can overload the vehicle.

Grade compensation

Grade compensation (grade reduction) is the reduction of the design gradient on a horizontal curve, so that the total resistance (grade + curve) on the curve is not more than the resistance on the straight section.

IRC formula (R in metres):

Grade compensation=30+RR %  (max 75R%)\text{Grade compensation} = \frac{30 + R}{R}\ \%\ \ \text{(max } \frac{75}{R}\%)
  • It is not needed when the gradient is flatter than 4%, and in practice it matters mainly for sharp curves (small RR).
  • Example: R=60R = 60 m: (30+60)/60=1.5%(30+60)/60 = 1.5\%, maximum 75/60=1.25%75/60 = 1.25\%, so adopt 1.25%. If the ruling gradient is 6%, use 6−1.25=4.75%6 - 1.25 = 4.75\% on the curve.
  • Used chiefly on hill roads.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transportation Engineering I chapter 3 questions →