Crest Vertical Curve — Minimum Length and Stationing: 16-Civ-B7 Transportation Planning and Engineering, May 2016, Q1

Question 1: Crest Vertical Curve — Minimum Length and Stationing (20 marks)

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.

Quantity	Symbol	Value
Approach grade	g₁	+1.5 %
Departure grade	g₂	−2.0 %
Road class / design speed	RCU80	rural collector, undivided, 80 km/h
Height of driver's eye	h₁	1.05 m
Height of object	h₂	0.380 m
Station of the PVI (grade intersection)	—	200 + 15
Station interval	—	30 m

Find. (a) the shortest crest curve that still provides the design stopping sight distance for 80 km/h with the stated eye and object heights, and (b) the stations of the BVC and EVC for that curve.

Crest vertical curve joining +1.5 % and −2.0 % grades. The sight line runs from a driver's eye 1.05 m above the pavement to a 0.380 m object, and it is the pavement surface itself that limits it. Vertical scale exaggerated for clarity.

Approach. Establish the algebraic grade difference and the design SSD for 80 km/h, then apply the two standard crest-curve sight-distance expressions (sight distance shorter than the curve, and sight distance longer than the curve), retain the one whose result is self-consistent, and lay the resulting length symmetrically about the PVI to get the BVC and EVC stations.

Part (a) — algebraic difference in grades. The controlling parameter of any vertical curve is the change in grade, taken as a signed algebraic difference so that a crest is positive: $$A = |g_1 - g_2| = |(+1.5) - (-2.0)| = 3.5$$ The curve is therefore a crest of magnitude A = 3.5 (grades in per cent).
Design stopping sight distance for 80 km/h. SSD is the sum of the distance covered during perception–reaction and the braking distance, $$S = 0.278\,V\,t + \frac{V^2}{254\,(a/9.81)}$$ with V = 80 km/h, t = 2.5 s and a deceleration a = 3.4 m/s². Substituting, $0.278(80)(2.5) = 55.6$ m of reaction distance and $80^2/[254(0.3466)] = 72.7$ m of braking distance, giving 128.3 m. This is rounded up in the TAC and AASHTO design-domain tables to the tabulated design value $$\boxed{S = 130\ \text{m}}$$ which is the figure used from here on.
Sight-distance constant for the stated eye and object heights. Both crest formulae share the grouping $$C = 100\left(\sqrt{2h_1}+\sqrt{2h_2}\right)^2 = 100\left(\sqrt{2.10}+\sqrt{0.76}\right)^2$$ so that $C = 100(1.4491+0.8718)^2 = 100(2.3209)^2 = 538.67$. (Had the AASHTO heights 1.08 m and 0.60 m been used, C would be 658 and the required curve would be shorter — 72.00 m against the 106.10 m found below — because a taller target is visible over a sharper crest.)
Trial 1 — assume the sight distance is contained within the curve. For $S \le L$, $$L = \frac{A S^2}{C} = \frac{3.5(130)^2}{538.67} = 109.81\ \text{m}$$ This result is rejected: it returns L = 109.81 m against S = 130 m, so the sight distance is not in fact contained within the curve and the assumption underlying the formula fails.
Trial 2 — sight distance longer than the curve. For $S \ge L$, $$L = 2S - \frac{C}{A} = 2(130) - \frac{538.67}{3.5} = 260 - 153.91$$ $$\boxed{L_{\min} = 106.10\ \text{m}}$$ The check is now consistent, since 106.10 m is less than the 130 m sight distance. The corresponding rate of vertical curvature is $K = L/A = 106.10/3.5 = 30.31$ m per per-cent of grade change.
Confirm the comfort and appearance minima. TAC also requires a curve at least $0.6V = 0.6(80) = 48$ m long for appearance, and a crest curve is not controlled by rider comfort. Since 106.10 m comfortably exceeds 48 m, sight distance governs and the design length stands. For construction, 106.10 m would normally be rounded up to 110 m; the stationing below uses the computed minimum so that part (b) answers exactly what was asked.
Part (b) — chainage of the PVI. With 30 m between full stations, station 200 + 15 corresponds to a running chainage of $$\text{Ch}_{PVI} = 200(30) + 15 = 6000 + 15 = 6015.00\ \text{m}$$
Locate the BVC and EVC. An equal-tangent parabola is symmetric about the PVI, so each tangent length is $L/2 = 106.10/2 = 53.05$ m. Subtracting and adding, $$\begin{aligned} \text{Ch}_{BVC} &= 6015.00 - 53.05 = 5961.95\ \text{m} \\ \text{Ch}_{EVC} &= 6015.00 + 53.05 = 6068.05\ \text{m} \end{aligned}$$ Converting back to 30 m stations, $5961.95/30 = 198.73$, so the BVC lies 21.95 m beyond station 198, and $6068.05/30 = 202.27$, so the EVC lies 8.05 m beyond station 202: $$\boxed{\text{BVC} = 198+21.95, \quad \text{EVC} = 202+8.05}$$ Had the length been rounded up to 110 m, the BVC and EVC would move outward by 1.95 m each, to 198 + 20.00 and 202 + 10.00.

Quantity	Result
Algebraic grade difference, A	3.5 %
Design stopping sight distance, S (80 km/h)	130 m (computed 128.3 m)
Governing case	S > L
Minimum curve length, L	106.10 m (round up to 110 m for construction)
Rate of vertical curvature, K	30.31 m / %
Chainage of PVI	6015.00 m
Station of BVC	198 + 21.95 (chainage 5961.95 m)
Station of EVC	202 + 8.05 (chainage 6068.05 m)

Topic. Sight-distance control of crest vertical curves and equal-tangent parabolic stationing.

Key relations. $A=|g_1-g_2|$; $S = 0.278Vt + V^2/[254(a/9.81)]$; for $S\le L$, $L = AS^2/[100(\sqrt{2h_1}+\sqrt{2h_2})^2]$; for $S\ge L$, $L = 2S - 100(\sqrt{2h_1}+\sqrt{2h_2})^2/A$; $K = L/A$; BVC and EVC lie at $L/2$ either side of the PVI.

Why this works. On a crest the sight line is limited by the pavement itself: the driver's eye and the target object are both above the road surface, and the parabola between them cuts the chord. Setting the chord tangent to the parabola and expressing both offsets through the parabola's constant rate of change of grade produces the two closed-form expressions above, one for each geometric configuration. The two formulae cross exactly where L = S, so evaluating both and keeping the self-consistent root is not a shortcut — it is the correct procedure. Lowering the assumed object height makes the target harder to see over the crest and therefore lengthens the curve required for a given sight distance, which is why the paper's 0.380 m object gives 106.10 m where a 0.60 m object behind the same 1.05 m eye would give only 75.00 m.

Common pitfall. Applying only the $S \le L$ formula and reporting 109.81 m without the consistency check — that result contradicts its own assumption. A second frequent error is treating the station interval as 100 m out of habit; here the paper states 30 m, so station 200 + 15 is chainage 6015 m, not 20 015 m, and every subsequent station is wrong if that is missed.

Source. TAC, Geometric Design Guide for Canadian Roads, Ch. 1.2 (design-domain stopping sight distance) and Ch. 2.1.7 (vertical alignment); Garber & Hoel, Traffic and Highway Engineering, 5th ed., §3.4 (vertical curves).

Question 1 of 7: Crest Vertical Curve — Minimum Length and Stationing

Question 1: Crest Vertical Curve — Minimum Length and Stationing (20 marks)