Crest Vertical Curves and Sight Distance: 16-Civ-B10 Traffic Engineering, December 2015, Q4

Question 4: Crest Vertical Curves and 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	Quantity	Value
(a)	Curve length \(L\), grades \(g_1,\ g_2\)	200 m; +4.5 % and −2 %
(a)	Sight-distance heights (stopping)	eye \(h_1 = 1.08\) m, object \(h_2 = 0.60\) m
(b)	Curve length \(L\), grades \(g_1,\ g_2\)	2000 m; +3 % and −2 %
(b)	Sight-distance heights (passing)	eye \(h_1 = 1.08\) m, object \(h_2 = 1.30\) m
(b)	Design speed	100 km/h

Find. (a) The highest design speed for which the 200 m curve delivers adequate stopping sight distance; (b) whether the available passing sight distance on the 2 km curve reaches the value required at 100 km/h, and hence whether a no-passing zone must be signed.

Crest vertical curve of Question 4(a): A = 6.5 %, L = 200 m. The sight line from a 1.08 m driver eye to a 0.60 m object is tangent to the curve and spans S = 142.3 m.

Approach. Both parts invert the same crest-curve sight-distance equation. For a sight line that begins and ends on the curve (\(S < L\)), \(L = A S^{2} / \left[ 200\left(\sqrt{h_1}+\sqrt{h_2}\right)^{2}\right]\); solve it for \(S\), then compare \(S\) with the tabulated requirement — stopping sight distance in (a), passing sight distance in (b).

Part (a) — algebraic difference in grades and the sight-distance constant. The algebraic difference is \[ A = \left| g_1 - g_2 \right| = \left| 4.5 - (-2.0) \right| = 6.5\ \% \] With the AASHTO/TAC metric stopping-sight-distance heights (\(h_1 = 1.08\) m driver eye, \(h_2 = 0.60\) m object), the denominator constant is \[ 200\left(\sqrt{1.08}+\sqrt{0.60}\right)^{2} = 200\,(1.0392 + 0.7746)^{2} = 658 \] which is the familiar \(L = A S^2/658\) form for crest curves.
Solve for the sight distance the curve actually provides. Rearranging and substituting, \[ S = \sqrt{\frac{658\,L}{A}} = \sqrt{\frac{658 \times 200}{6.5}} = \sqrt{20\,246} = \boxed{142.3\ \text{m}} \] The assumption \(S < L\) is satisfied (142.3 m < 200 m), so the formula used is the correct branch and no re-solution with the \(S > L\) expression is needed.
Convert the available sight distance into a design speed. Stopping sight distance on the level is the brake-reaction distance plus the braking distance, \[ \text{SSD} = 0.278\,V t + \frac{V^{2}}{254\,(a/g)} \] with \(t = 2.5\) s and \(a = 3.4\ \text{m/s}^2\), so \(a/g = 0.347\). Setting SSD \(= 142.3\) m gives the quadratic \(V^{2} + 61.2\,V - 12\,526 = 0\), whose positive root is \(V = 85.4\) km/h. Design speeds are selected from the standard increments, so the curve supports the next lower standard value.
Confirm against the design tables and state the answer. The tabulated SSD is 130 m at 80 km/h and 160 m at 90 km/h; the available 142.3 m covers the first but not the second. The \(K\)-value route gives the same verdict independently: \[ K = \frac{L}{A} = \frac{200}{6.5} = 30.8 \quad\text{against}\quad K_{80} = 26,\ K_{90} = 39 \] Both routes agree, so the design speed that the curve can properly serve is \[ \boxed{V = 80\ \text{km/h}} \] with 12 m of sight distance in hand. Posting the curve for 90 km/h would leave it 18 m short of the required 160 m.
Part (b) — available passing sight distance on the 2 km curve. Here \(A = \left|3.0-(-2.0)\right| = 5.0\ \%\), and the object is another vehicle at \(h_2 = 1.30\) m, so the constant changes: \[ 200\left(\sqrt{1.08}+\sqrt{1.30}\right)^{2} = 200\,(1.0392+1.1402)^{2} = 950 \] Hence the sight distance available over the crest is \[ S = \sqrt{\frac{950 \times 2000}{5.0}} = \sqrt{380\,000} = \boxed{616\ \text{m}} \] again with \(S < L\) satisfied comfortably (616 m < 2000 m).
Compare with the passing sight distance required at 100 km/h. The AASHTO/TAC design passing sight distance for a 100 km/h design speed is 670 m. The curve provides 616 m, a shortfall of \[ 670 - 616 = 54\ \text{m} \] so a driver at 100 km/h cannot see far enough over the crest to complete an overtaking manoeuvre safely. Yes — a no-passing zone must be established and signed, with the barrier line and the “Do Not Pass” sign beginning where the sight distance first falls below 670 m and ending where it is restored. For reference, 616 m is the passing sight distance appropriate to a 90 km/h design speed (613 m), so the curve is one design-speed increment short.

Check — which passing-sight-distance criterion applies. Two different criteria exist and they give different answers. The design criterion used above (AASHTO/TAC, eye 1.08 m to a 1.30 m oncoming vehicle, 670 m at 100 km/h) is the one the question intends, because it is the only criterion that uses the 1.08 m and 1.30 m heights the question supplies. The marking criterion in the MUTCD for striping no-passing zones is far shorter — of the order of 210 m at 100 km/h, measured eye-to-eye at 1.08 m — and by that test the 616 m available would be ample. The answer is given on the design criterion; on a real project the governing agency’s marking policy would decide where the barrier line is actually painted.

Part	Quantity	Result
(a)	Algebraic grade difference \(A\), rate of curvature \(K\)	6.5 %, \(K = 30.8\)
(a)	Available stopping sight distance \(S\)	142.3 m
(a)	Design speed supported	80 km/h (needs 130 m; 90 km/h would need 160 m)
(b)	Available passing sight distance \(S\)	616 m
(b)	Required passing sight distance at 100 km/h	670 m
(b)	Is a no-passing zone needed?	Yes — 54 m short

Topic. Sight distance over crest vertical curves; stopping versus passing criteria.

Key relations. \(A = |g_1-g_2|\); \(K = L/A\); for \(S < L\), \(L = A S^{2} / \left[200(\sqrt{h_1}+\sqrt{h_2})^{2}\right]\); for \(S > L\), \(L = 2S - 200(\sqrt{h_1}+\sqrt{h_2})^{2}/A\); SSD \(= 0.278Vt + V^{2}/[254(a/g \pm G)]\). Metric constants: 658 for stopping (1.08 m / 0.60 m) and 950 for passing (1.08 m / 1.30 m).

Why this works. A vertical curve is a parabola, so the vertical offset from the tangent grows with the square of the distance from the point of tangency. Setting the offset at the driver’s position equal to the eye height and the offset at the object’s position equal to the object height, and requiring the sight line to be tangent to the parabola between them, produces the two square-root terms; their sum squared is the only place the heights enter. That is why raising the object height from 0.60 m to 1.30 m increases the constant from 658 to 950 and thereby lengthens the sight distance available on a given curve by about 20 % — an oncoming car is simply easier to see over a crest than a small object lying on the pavement. The \(K\)-value is the same equation repackaged: since \(L/A\) is constant for a given required \(S\), tabulating \(K\) lets a designer size a curve in one multiplication.

Common pitfall. Using the wrong pair of heights — the 0.60 m object belongs to stopping sight distance and the 1.30 m object to passing sight distance, and mixing them changes the answer by 20 %. Equally common is forgetting to verify the \(S < L\) assumption after solving; if the computed \(S\) exceeds \(L\), the other branch of the formula applies and the arithmetic must be repeated.

Source. AASHTO, A Policy on Geometric Design of Highways and Streets, Ch. 3 (sight distance) and Ch. 3 (crest vertical curves, \(K\)-value tables); TAC, Geometric Design Guide for Canadian Roads, Ch. 1.2 (design controls) and Ch. 2.1 (vertical alignment); Garber & Hoel, Ch. 3 and Ch. 15.

Question 4 of 7: Crest Vertical Curves and Sight Distance

Question 4: Crest Vertical Curves and Sight Distance (20 marks — 10 each)