Question 1 of 7: Sag vertical curve and night sight distance on a wildlife-collision section

Question 1: Sag vertical curve and night sight distance on a wildlife-collision section (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. A vertical curve joining a descending grade to an ascending grade on a rural two-lane highway, with the headlight geometry the question supplies and the friction values printed on appendix page 15.

Find. The curve type and its governing design controls; the sight distance the curve actually provides at night; whether that satisfies the stopping requirement at the design and posted speeds; and the engineering explanation for the wildlife collisions together with practical countermeasures.

Approach. Classify the curve from the algebraic sign of the grade change, invert the headlight sight-distance relation to obtain the sight distance the 200 m curve actually delivers, compare it with the stopping sight distance demanded at 100 and 90 km/h, and interpret the shortfall in terms of night-time animal detection.

1. Classify the curve from the algebraic grade change. The controlling parameter is the algebraic difference in grades, $$A = G_2 - G_1 = (+2.5) - (-3.5) = +6.0\ \%$$ A positive difference means the grade increases through the curve, so the roadway falls, flattens and rises again. The curve is therefore a sag (valley) vertical curve of rate $$K = \frac{L}{A} = \frac{200}{6.0} = \boxed{33.3\ \text{m per }1\ \%\text{ of grade change}}$$
2. State the design controls that apply to a sag curve. A crest curve is controlled by the driver eye height and an object on the pavement, because the pavement itself cuts the line of sight. On a sag the line of sight is never cut in daylight — the whole curve is visible — so four other controls govern, in the order in which they usually bite: (i) headlight sight distance, the length of pavement illuminated by a beam leaving a headlight 0.60 m above the surface and diverging 1° upward, which is the criterion used below; (ii) rider comfort, limiting the radial acceleration to about 0.3 m/s², which typically requires only about half the headlight length; (iii) drainage, which caps K near 51 on a curbed section so that water does not pond in the flat portion; and (iv) general appearance, a minimum length so that the sag does not read as a kink. Headlight control governs here.
3. Write the headlight relation and solve for the sight distance the curve provides. With the sight distance S contained within the curve, appendix page 15 gives $$L = \frac{A\,S^{2}}{200H + 200 S \tan\beta} = \frac{A\,S^{2}}{120 + 3.491\,S}$$ Substituting L = 200 m and A = 6.0 turns this into a quadratic in S: $$6.0\,S^{2} - 698.15\,S - 24\,000 = 0$$ Solving for the positive root, $$S = \frac{698.15 + \sqrt{698.15^{2} + 4(6.0)(24\,000)}}{2(6.0)} = \boxed{144.1\ \text{m}}$$
4. Confirm that the correct branch was used. The relation above is valid only while the sight distance lies inside the curve. Since S = 144.1 m is less than L = 200 m, the assumption holds. For completeness, the alternative branch (sight distance longer than the curve) is $$L = 2S - \frac{120 + 3.491\,S}{A} \quad\Longrightarrow\quad S = \frac{AL + 120}{2A - 3.491} = 155.1\ \text{m}$$ which would require S to exceed 200 m and does not, so that branch is self-inconsistent and is discarded. Always test both branches and keep the one that satisfies its own inequality.
5. Compute the stopping sight distance the road must provide. The braking distance is taken on the descending approach, where the −3.5 % grade lengthens the stop: $$SSD = 0.278\,V\,t + \frac{V^{2}}{254\,(f + G)}$$ At the design speed of 100 km/h with f = 0.30 and G = −0.035, $$SSD_{100} = 0.278(100)(2.5) + \frac{100^{2}}{254(0.30 - 0.035)} = 69.5 + 148.6 = \boxed{218.1\ \text{m}}$$ At the posted speed of 90 km/h with f = 0.31, $$SSD_{90} = 0.278(90)(2.5) + \frac{90^{2}}{254(0.31 - 0.035)} = 62.6 + 116.0 = \boxed{178.5\ \text{m}}$$
6. Compare provision with demand and quantify the deficiency. The curve delivers 144.1 m against 218.1 m required at the design speed and 178.5 m required even at the posted speed, so it is deficient at both. The same conclusion follows from the rate of curvature: the appendix table of K factors for sag curves under headlight control asks for K between 37 and 50 at 100 km/h and between 30 and 40 at 90 km/h, whereas the curve supplies K = 33.3. Rebuilding to the required standard means lengthening the curve to $$L_{req} = \frac{A\,S^{2}}{120 + 3.491\,S}\Big|_{S = 218.1} = 323.8\ \text{m}\ \ (100\ \text{km/h}), \qquad L_{req}\big|_{S = 178.5} = 257.3\ \text{m}\ \ (90\ \text{km/h})$$ so the existing 200 m curve is 38 % short of the length needed at the design speed and 22 % short of the length needed even at the posted speed.
7. Explain the collision pattern and set out countermeasures. Deer and moose move at dusk and after dark, exactly when the sag curve stops being generous and becomes the most restrictive geometry on the road. On a sag the beam is thrown into the rising pavement ahead, so the illuminated length collapses to 144 m; a driver travelling at the posted 90 km/h needs 178.5 m to stop and at the design 100 km/h needs 218.1 m. An animal standing in the lane is therefore inside the stopping distance before it is inside the light. A large animal is also dark, low in contrast and often below the 0.60 m beam axis at first, which erodes the effective detection distance further. Countermeasures, in increasing order of cost: post an advisory or reduced night speed of 70 km/h (at which the required stopping distance falls to about 120 m, inside the illuminated length) together with wildlife warning signs and, if warranted, an animal-detection activated flasher; install continuous fixed lighting through the sag, which removes the headlight control entirely; clear and maintain the roadside so that animals are visible before they reach the pavement and provide the appropriate clear zone; and, as the permanent fix, reconstruct the profile to a 325 m curve (K = 54) or install wildlife exclusion fencing funnelling animals to an underpass or overpass at the crossing point identified by the collision data.