Question 4 of 7: Stopping Sight Distance and Crest Vertical Curves

Question 4: Stopping Sight Distance and Crest Vertical Curves (20 marks — (a) 4, (b) and (c) 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. The paper's own AASHTO 2001 metric table (reproduced verbatim), plus the two curve geometries.

Find. (a) definitions; (b) the minimum crest length $L$ that delivers the design stopping sight distance for a 100 km/h design speed with a raised object height; (c) the design speed that a given 150 m crest can safely serve.

Approach. Read the required stopping sight distance off the paper's own table, then apply the crest-curve sight-distance relation in the branch that is self-consistent ($S < L$ or $S > L$). Part (c) inverts the same relation for $S$ and then inverts the AASHTO stopping-sight-distance equation for the speed.

1. Part (a) — Definitions. Perception–reaction time (often expanded as the PIEV process: perception, intellection, emotion, volition) is the interval between the moment a hazard first becomes visible and the moment the driver's foot begins to develop braking force. Nothing is decelerating during it, so the vehicle covers the full brake reaction distance $d_{1} = 0.278\,V t$ at the initial speed. AASHTO adopts $t = 2.5$ s for design, a value near the 90th percentile of measured driver response for an expected event; emergency-response studies give a median nearer 1.5 s, and the design value deliberately carries the slower drivers. Braking distance is the distance covered while the vehicle decelerates from the initial speed to a stop, obtained from the work–energy principle: the kinetic energy $\tfrac{1}{2}mv^{2}$ is dissipated by the resisting force over the stopping distance, giving $d_{2} = V^{2}/[254(f \pm G)]$ in metric units, or, in the deceleration form AASHTO now prefers, $d_{2} = 0.039V^{2}/a$. Their sum is the stopping sight distance, and the split matters: at 100 km/h reaction accounts for 69.5 m and braking 114.7 m of the 184.2 m total, so a design that skimps on reaction time loses over a third of the safety margin.
2. Part (b) — Required sight distance and rate of vertical curvature. From the paper's table at $V = 100$ km/h the design stopping sight distance is $S = 185$ m (the calculated value is 184.2 m). The algebraic difference in grades for this crest is $$A = |g_{1} - g_{2}| = |{+2.5} - ({-3.5})| = 6.0\ \%$$ Assuming the sight distance is shorter than the curve, the crest sight-distance relation is $$L = \frac{A\,S^{2}}{100\left(\sqrt{2h_{1}} + \sqrt{2h_{2}}\right)^{2}}$$ Evaluating the height term with $h_{1} = 1.070$ m and $h_{2} = 1.295$ m, $$\sqrt{2(1.070)} = 1.4629, \qquad \sqrt{2(1.295)} = 1.6094$$ $$100\left(1.4629 + 1.6094\right)^{2} = 100(3.0722)^{2} = 943.85$$
3. Part (b), continued — minimum curve length. Substituting, $$L = \frac{6.0 \times 185^{2}}{943.85} = \frac{205\,350}{943.85} = \boxed{217.6\ \text{m}}$$ The assumption must now be checked: $S = 185\ \text{m} \lt L = 217.6\ \text{m}$, so the $S \lt L$ branch was the right one. Testing the alternative branch confirms it — $L = 2S - 200(\sqrt{h_{1}}+\sqrt{h_{2}})^{2}/A = 370 - 157.3 = 212.7$ m, which is greater than $S = 185$ m and therefore violates its own premise. Expressed as a rate of vertical curvature the answer is $K = L/A = 217.6/6.0 = 36.3$ m per per-cent, against the AASHTO $K = 105$ for a standard 600 mm object at 100 km/h; the raised 1295 mm target is what makes this curve so much shorter. Had the calculated 184.2 m been used instead of the 185 m design value, $L$ would be 215.7 m — a 0.9 per cent difference, so rounding up to a constructed $L = 220$ m covers either reading.
4. Part (c) — Sight distance available on the given 150 m curve. Here $L$ is known and $S$ is the unknown, with standard heights $h_{1} = 1.080$ m and $h_{2} = 0.600$ m and $$A = |{+4} - ({-2})| = 6.0\ \%$$ The height term is now $$100\left(\sqrt{2(1.080)} + \sqrt{2(0.600)}\right)^{2} = 100(1.4697 + 1.0954)^{2} = 658.0$$ Rearranging the same relation for $S$, $$S = \sqrt{\frac{100\,L\left(\sqrt{2h_{1}}+\sqrt{2h_{2}}\right)^{2}}{A}} = \sqrt{\frac{150 \times 658.0}{6.0}} = \sqrt{16\,450} = \boxed{128.3\ \text{m}}$$ Again $S = 128.3\ \text{m} \lt L = 150\ \text{m}$, so the branch is consistent.
5. Part (c), continued — design speed. Inverting the AASHTO metric stopping-sight-distance equation with $t = 2.5$ s and $a = 3.4\ \text{m/s}^{2}$, $$S = 0.278\,V t + \frac{0.039\,V^{2}}{a} \;\Longrightarrow\; 0.011471\,V^{2} + 0.695\,V - 128.3 = 0$$ $$V = \frac{-0.695 + \sqrt{0.695^{2} + 4(0.011471)(128.3)}}{2(0.011471)} = \boxed{79.7\ \text{km/h}}$$ so the curve serves a design speed of essentially 80 km/h. Reading the table directly gives the same picture: 128.3 m of available sight distance lies just below the 130 m design value tabulated at 80 km/h and comfortably above the 105 m required at 70 km/h.