Question 2 of 7: Sight Distance on a Horizontal Curve and a Sag Vertical Curve

Question 2: Sight Distance on a Horizontal Curve and a Sag Vertical Curve (8 + 12 = 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.

Find. (a) the highest posted speed the available sight line across the building corner can support; (b) the minimum length of the sag curve that keeps the headlight-illuminated distance at least equal to the design stopping sight distance for 80 km/h.

Approach. Part (a): convert the physical clearance to a middle ordinate measured from the driver's own path, invert the middle-ordinate relation for the sight distance, then invert the AASHTO/TAC stopping-sight model for the speed it supports and round down to a posted value. Part (b): apply both branches of the headlight sight-distance criterion, keep the branch whose own assumption is satisfied, and then compare against the tabulated K-value, comfort and appearance minima before adopting a design length.

1. Part (a) — convert the clearance into a middle ordinate. The sight line is a chord of the path the driver actually follows, which is the centre of the inside lane. The obstruction is 3.00 m outside the pavement edge, and the pavement edge is half a lane from that path: $$M = 3.00 + \frac{3.30}{2} = 4.65\ \text{m}$$ Dropping the half-lane term is the classic error here; it would understate M by 36 percent and overstate the permissible speed.
2. Invert the middle-ordinate relation for the sight distance. For a sight line that lies wholly within the curve, $$M = R_{v}\left[1 - \cos\!\left(\frac{S}{2R_{v}}\right)\right] \quad\Longrightarrow\quad S = 2R_{v}\arccos\!\left(1 - \frac{M}{R_{v}}\right)$$ Substituting R_v = 500 m and M = 4.65 m, $$\boxed{S = 2(500)\arccos(0.99070) = 136.49\ \text{m}}$$ The familiar degree form S = (R/28.65)·arccos(1 − M/R) returns 136.48 m, confirming the arithmetic.
3. Convert the available sight distance into a speed. The design stopping sight distance on a level grade is $$SSD = 0.278\,V\,t + \frac{0.039\,V^{2}}{a}$$ with the standard perception–reaction time t = 2.5 s and deceleration a = 3.4 m/s². Setting SSD = 136.49 m and solving the quadratic 0.011471 V² + 0.695 V − 136.49 = 0 gives $$\boxed{V = 82.9\ \text{km/h}}$$
4. Round down to a design and posted speed. The tabulated design values bracket the answer: 130 m of stopping sight distance is required at 80 km/h and 160 m at 90 km/h, and 136.5 m falls between them. The curve therefore serves 80 km/h with a 6.5 m (5 percent) margin but falls 23 m short of the 90 km/h requirement: $$\boxed{\text{post the section at } 80\ \text{km/h}}$$ If a higher speed were required, the remedy is not signing but geometry: the building corner would have to be set back to M = 5.9 m (a clearance of 4.25 m) to reach the 160 m needed for 90 km/h, or the sight line cleared by removing the obstruction.
5. Part (b) — establish the controls for the sag curve. For RCU80 the design speed is 80 km/h, so the design stopping sight distance is S = 130 m. The algebraic difference in grades is $$A = |g_{2} - g_{1}| = |{+}1.0 - ({-}2.0)| = 3.0\ \text{percent}$$ and the sag is genuine (the grade turns from falling to rising), so the headlight criterion governs rather than a crest sight line. Note the arithmetic identity behind the standard formula: 200h = 200(0.60) = 120 and 200 tan 1° = 3.49 ≈ 3.5, so the values this paper supplies reproduce the tabulated coefficients exactly.
6. Apply the branch that assumes the sight distance is shorter than the curve. When S < L the beam strikes the pavement within the curve and $$L = \frac{A\,S^{2}}{200\,(h + S\tan\beta)} = \frac{3.0\,(130)^{2}}{200\,(0.60 + 130\tan 1^{\circ})} = \frac{50\,700}{573.83} = 88.35\ \text{m}$$ This result is self-contradictory: 88.35 m is less than S = 130 m, so the assumption S < L on which the formula rests does not hold and the value must be discarded as the governing answer.
7. Apply the branch that assumes the sight distance exceeds the curve. When S > L, $$L = 2S - \frac{200\,(h + S\tan\beta)}{A} = 2(130) - \frac{573.83}{3.0} = 260 - 191.28$$ $$\boxed{L_{\min} = 68.7\ \text{m}}$$ Here 68.7 m is indeed less than 130 m, so this branch is internally consistent and it is the true minimum length satisfying the headlight criterion.
8. Check the other length controls before adopting a design value. Three further minima apply at 80 km/h:

9. Adopt the design length. The K-value method is the form TAC and AASHTO tabulate, and it deliberately applies the S < L expression at all values of A so that a designer never has to test which branch applies; it returns L = 29.39 × 3.0 = 88.2 m here. Since the two branches straddle the answer and construction lengths are rounded anyway, $$\boxed{\text{adopt } L = 90\ \text{m}}$$ which satisfies every control listed above with margin. The strict mathematical minimum, quoted where the question asks for the minimum based on SSD, is 68.7 m.

Final Results.