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

Question 4: Crest Vertical Curves and Stopping Sight Distance (20 marks — 10 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. Part (a): a crest curve of length L = 185 m joining g₁ = +3.5 % to g₂ = −1.25 %, with standard sight heights (driver eye 1.08 m, object 0.60 m). Part (b): a crest curve joining +4.5 % to −3.5 % at a design speed of 90 km/h, with a driver eye height of 1.20 m and an object height of 0.20 m. The AASHTO 2001 table above is supplied by the paper.

Find. (a) the standard design speed the existing 185 m curve supports; (b) the minimum curve length that delivers the 90 km/h stopping sight distance for the stated non-standard sight heights.

Approach. Verify the printed table against the published AASHTO relations, then apply the crest sight-distance relation $L = AS^{2}/[100(\sqrt{2h_1}+\sqrt{2h_2})^{2}]$ — inverted for S in part (a), solved for L in part (b) — and cross-check part (a) against the AASHTO rate-of-vertical-curvature (K-value) table.

1. Check the printed table before using it. The published metric relations are $d_1 = 0.278Vt$ with $t = 2.5$ s and $d_2 = 0.039V^{2}/a$ with $a = 3.4$ m/s². At 100 km/h these give $0.278(100)(2.5) = 69.5$ m and $0.039(100)^{2}/3.4 = 114.7$ m, summing to 184.2 m — the printed row exactly; at 20 km/h they give 13.9, 4.6 and 18.5 m, again exact. Every one of the twelve metric rows reproduces, so the table is sound and can be used as the authority for both parts. This costs one line and it is the only free cross-check the page offers.
2. Part (a) — algebraic difference and the sight-distance relation. The algebraic difference in grade is $$A = |g_1 - g_2| = |3.5 - (-1.25)| = 4.75\,\%$$ With the standard sight heights the crest relation for the usual case $S \lt L$ reduces to the familiar constant, $100(\sqrt{2 \times 1.08} + \sqrt{2 \times 0.60})^{2} = 658$, so $$L = \frac{A S^{2}}{658} \qquad\Longrightarrow\qquad S = \sqrt{\frac{658\,L}{A}}$$
3. Part (a) — available sight distance. Substituting the given curve, $$S = \sqrt{\frac{658(185)}{4.75}} = \sqrt{25\,627.4} = \boxed{160.1 \text{ m}}$$ The assumption $S \lt L$ holds (160.1 m against 185 m), so this branch of the relation was the right one and no re-solution is needed.
4. Part (a) — convert to a standard design speed. Entering the design-SSD column of the printed table, 90 km/h requires 160 m and 100 km/h requires 185 m. The available 160.1 m clears the first and falls well short of the second, so the curve supports a design speed of $\boxed{90 \text{ km/h}}$. The independent route agrees: the rate of vertical curvature is $$K = \frac{L}{A} = \frac{185}{4.75} = 38.95 \text{ m per }\%$$ against the AASHTO metric crest requirement of $K = 39$ for 90 km/h and $K = 52$ for 100 km/h. Two independent tables landing on the same standard increment is the strongest confirmation available here.
5. Part (b) — grade difference and the modified constant. Reading the grades as a crest, entering at +4.5 % and departing at −3.5 %, $$A = 4.5 + 3.5 = 8.0\,\%$$ The sight heights are not the standard pair, so the constant must be rebuilt from first principles: $$100\left(\sqrt{2h_1} + \sqrt{2h_2}\right)^{2} = 100\left(\sqrt{2.40} + \sqrt{0.40}\right)^{2} = 100(1.5492 + 0.6325)^{2} = 476.0$$ A lower eye height and a much lower object both shorten the sight line over a given crest, so the constant falls from 658 to 476 and the required curve grows in proportion.
6. Part (b) — minimum length. The design stopping sight distance for 90 km/h is 160 m from the printed table. Hence $$L = \frac{A S^{2}}{100(\sqrt{2h_1}+\sqrt{2h_2})^{2}} = \frac{8.0(160)^{2}}{476.0} = \boxed{430.3 \text{ m}}$$ Again $S \lt L$ (160 m against 430.3 m), so the branch is right. The alternative $S \gt L$ formula returns $L = 260.5$ m, which contradicts its own premise (160 m is not greater than 260.5 m) and is therefore inadmissible — a check worth making explicitly, because on flat curves it is the governing branch.
7. Part (b) — adopt a constructible length. The rate of vertical curvature implied is $K = 430.3/8.0 = 53.8$ m per %, far above the standard 39 for 90 km/h precisely because of the lowered sight heights. Rounding up to a 20 m multiple, adopt $L = 440$ m, which gives $K = 55$ and a small reserve of sight distance.