Safe speed from stopping criterion on curve

Step 1: radius of the curve from grade compensation

IRC grade compensation is $\dfrac{30+R}{R}\%$, subject to a maximum of $\dfrac{75}{R}\%$ ($R$ in m). Here the compensation is $6-4.5=1.5\%$.

• From $\dfrac{30+R}{R}=1.5$: $R=60$ m. But then the maximum allowed is $75/60=1.25\% < 1.5\%$, so a compensation of 1.5% is not possible with $R=60$ m.
• From the limit $\dfrac{75}{R}=1.5$: $R=50$ m. Then $\dfrac{30+50}{50}=1.6\% > 1.5\%$, so the limit governs and the compensation is exactly 1.5%. Consistent.

So $R=50$ m. The curve length is $L=60$ m.

Step 2: available sight distance

Assumptions: two-lane carriageway 7.0 m (lane 3.5 m); the driver keeps to the centre of the inner lane; the obstruction is 2 m from the pavement edge. The clearance from the driver's path is $m=2+1.75=3.75$ m, and the radius of the path is $R_i=50-1.75=48.25$ m.

For $S<L$: $m=R_i(1-\cos\theta)$, where $\theta=\dfrac{S}{2R_i}$.

$$\cos\theta=1-\frac{3.75}{48.25}=0.9223\Rightarrow\theta=22.74^\circ=0.3969\ \text{rad}$$ $$S=2R_i\theta=2(48.25)(0.3969)=38.3\ \text{m}$$

$S<L=60$ m, so the assumption is valid. This is the stopping sight distance available.

Step 3: speed from stopping criterion

Take the compensated gradient $n=4.5\%$ as a downgrade (critical case), $f=0.35$, $t=2.5$ s:

$$S=0.278Vt+\frac{V^2}{254(f-n)}\Rightarrow 38.3=0.695V+\frac{V^2}{254(0.305)}$$

Solving the quadratic gives $V=33.8$ km/h.

Answer: the safe speed on the curve is about 34 km/h (about 36 km/h if the 4.5% grade is an upgrade). If the limit of $75/R$ is ignored and $R=60$ m is used, the available sight distance is 42.0 m and the speed is about 36 km/h.