Question 2 of 7: Exit-ramp curve adequacy and sight clearance

Question 2: Exit-ramp curve adequacy and sight clearance (20 marks — (a) 10, (b) 10)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A one-lane exit ramp curves through $90^\circ$ on an 80 m inner-edge radius with $e=0.06$, falling 3 m from PC to PT, designed for 40 km/h with a 3.5 s perception–reaction time.

Find. (a) whether the curve would serve a 60 km/h design speed; (b) the width that must be cleared inside the inner edge at 40 km/h.

Approach. Move the radius from the inner edge to the travelled path, compare the radius 60 km/h needs with what exists, then build the 40 km/h sight distance on the actual ramp grade and convert it into a middle ordinate re-datumed to the inner edge.

1. Place the vehicle on its travelled path. The 80 m is quoted to the inner edge, and a vehicle in a single 4.0 m lane runs down the middle of it, so the design path radius is $$ R_v=R_i+\tfrac{w}{2}=80+2.0=82.0\ \text{m}. $$
2. Ask what radius 60 km/h would need. From the friction table $f_s=0.15$ at 60 km/h, and $V_0=16.667\ \text{m/s}$: $$ R_{req}=\frac{V_0^{2}}{g\,(f_s+e)}=\frac{16.667^{2}}{9.81(0.15+0.06)} =\boxed{134.84\ \text{m}} $$ which is far more than the 82.0 m the ramp provides, so the curve is not adequate for a 60 km/h design speed.
3. Show the shortfall as a friction demand, which is how a reviewer will judge it. Rearranging the same relation for the side friction the curve would actually call on, $$ f_{s,demand}=\frac{V_0^{2}}{g R_v}-e=\frac{16.667^{2}}{9.81(82.0)}-0.06 =0.2853, $$ nearly twice the 0.15 that a 60 km/h pavement can be relied on to supply. Checking the tabulated speeds downwards, 50 km/h needs 89.38 m (still too much) while 40 km/h needs only 54.72 m, so the ramp is correctly signed for its stated 40 km/h design speed and nothing higher.
4. Recover the ramp grade from the two elevations. The curve length along the travelled path is $$ L=\frac{\pi\Delta R_v}{180}=\frac{\pi(90)(82.0)}{180}=128.81\ \text{m}, $$ and the 3 m fall from PC to PT spreads over that length, so $$ G=\frac{100-103}{128.81}=-0.02329\quad(-2.329\ \%). $$
5. Compute the stopping sight distance the ramp must provide. With $f=0.38$ at 40 km/h, $V_0=11.111\ \text{m/s}$ and the stipulated $t_{pr}=3.5$ s on the downgrade, $$ \text{SSD}=\frac{V_0^{2}}{2g(f+G)}+V_0t_{pr} =\frac{11.111^{2}}{2(9.81)(0.38-0.02329)}+11.111(3.5) $$ $$ =17.64+38.89=\boxed{56.53\ \text{m}} $$ Note how the long 3.5 s reaction dominates: it alone accounts for 38.89 m of the 56.53 m.
6. Turn the sight distance into a middle ordinate. Everything inside that offset must be cleared of obstructions: $$ M_s=R_v-R_v\cos\!\left[\frac{90\,\text{SSD}}{\pi R_v}\right] =82.0\left[1-\cos(19.749^{\circ})\right]=4.82\ \text{m}. $$
7. Re-datum the offset to the inside edge, which is what the question asks for. $M_s$ is measured from the travelled path, and the inner edge lies half a lane inboard of it, so $$ \text{clearance from the inside edge}=M_s-\tfrac{w}{2}=4.82-2.00 =\boxed{2.82\ \text{m}} $$ A strip about 2.8 m wide inside the inner edge, held over the whole $90^\circ$ arc, must be kept free of barriers, sign structures, noise walls and vegetation taller than about 0.6 m.