Question 5 of 7: Stopping Sight Distance on a Grade

Question 5: Stopping Sight Distance on a Grade (20 marks — (a) and (c) 2 each, (b) and (d) 8 each)

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.

Find. The brake reaction distance, the braking distance and total stopping sight distance on the 4.5 per cent upgrade, the direction in which $f$ moves on ice, and the stopping sight distance for the degraded-friction downgrade case.

Approach. Stopping sight distance is the sum of a constant-speed reaction distance and a work–energy braking distance in which the grade adds to or subtracts from the available friction.

1. Part (a) — Brake reaction distance. Nothing decelerates during perception and reaction, so the car covers $$d_{1} = 0.278\,V t = 0.278(110)(2.5) = \boxed{76.45\ \text{m}}$$ The constant 0.278 is simply $1/3.6$, converting km/h to m/s; carrying the conversion exactly gives $30.556 \times 2.5 = 76.39$ m, a 0.08 per cent difference that is immaterial here. Almost 80 m elapses before the brakes even begin to act — at highway speed that is longer than the entire braking distance at 60 km/h.
2. Part (b) — Braking distance and total stopping sight distance. Equating the initial kinetic energy to the work done by the resisting forces, and noting that on an upgrade the component of gravity along the slope assists the brakes, $$d_{2} = \frac{V^{2}}{254\,(f + G)}$$ Substituting $f = 0.25$ and $G = +0.045$, $$254(0.25 + 0.045) = 254(0.295) = 74.93$$ $$d_{2} = \frac{110^{2}}{74.93} = \frac{12\,100}{74.93} = \boxed{161.48\ \text{m}}$$ Adding the reaction distance gives the total stopping sight distance, $$\mathrm{SSD} = d_{1} + d_{2} = 76.45 + 161.48 = \boxed{237.93\ \text{m}}$$ For comparison, the AASHTO design value tabulated in Question 4 for 110 km/h on the level is 220 m; the higher figure here reflects the lower assumed friction ($f = 0.25$ against the equivalent $a = 3.4\ \text{m/s}^{2}$, i.e. $f \approx 0.35$), partly offset by the assisting upgrade.
3. Part (c) — Effect of ice on the coefficient of friction. The coefficient of friction decreases, and decreases sharply. A thin water or ice film separates the tyre rubber from the pavement aggregate, destroying the adhesion and hysteresis mechanisms that generate longitudinal grip. Typical wet-pavement values fall to 0.30–0.40, packed snow to about 0.20, and glare ice to 0.10 or below — which is precisely the value the paper hands over in part (d). Because $d_{2}$ varies as $1/(f + G)$, halving $f$ more than doubles the braking distance, and this non-linearity is why winter speed reductions are set aggressively rather than proportionally.
4. Part (d) — Degraded friction on a downgrade. Two things change together: friction falls to $f = 0.10$ and the grade reverses to $G = -0.030$, so gravity now works against the brakes. The effective resistance collapses to $$f + G = 0.10 - 0.030 = 0.070$$ so the braking distance becomes $$d_{2} = \frac{110^{2}}{254(0.070)} = \frac{12\,100}{17.78} = \boxed{680.54\ \text{m}}$$ and, with the unchanged reaction distance, $$\mathrm{SSD} = 76.45 + 680.54 = \boxed{756.99\ \text{m}}$$ That is 3.2 times the SSD of part (b), and the braking component alone has grown by a factor of 4.2. No practical rural highway alignment provides 757 m of sight distance, so the engineering conclusion is unavoidable: on an icy 3 per cent downgrade a 110 km/h operating speed cannot be made safe by geometry, and the response must be operational — variable speed limits, winter maintenance, and advance warning — rather than geometric.