Question 5 of 7: Sight distance and crest vertical curves

Question 5 — Sight distance and crest vertical curves (a) 6 marks, (b) and (c) 7 marks each — 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.

(a) Perception-reaction time and braking distance

Perception-reaction time is the interval between the instant an object or condition requiring a response becomes visible to the driver and the instant the driver's foot begins to apply pressure to the brake pedal. It is conventionally decomposed into the four PIEV stages: perception, in which the light from the object reaches the eye and registers; intellection (or identification), in which the driver recognises what the object is; emotion (or judgement), in which the driver decides what to do about it; and volition, the muscular act of initiating the response. Nothing decelerates the vehicle during this interval — it continues at its approach speed — so the distance covered, the brake reaction distance, is simply

with $V$ in km/h, $t$ in seconds and $d_1$ in metres. The value adopted for design is not an average but a high percentile of the driving population under conditions of low expectancy: AASHTO and the TAC Geometric Design Guide both use 2.5 s, against a measured median of roughly 0.7 s to 1.5 s for an alert driver anticipating a signal. The generous margin covers the older driver, the unexpected hazard, and the fact that sight distance deficiencies cannot be corrected once the road is built. The 2.5 s figure is stated explicitly in the note beneath the table printed on page 4 of this paper, and it reproduces the brake-reaction-distance column exactly.

Braking distance is the distance travelled from the first application of the brakes until the vehicle comes to rest, and it is obtained from the work–energy principle: the kinetic energy of the vehicle is dissipated by the retarding force over that distance. In the deceleration form used by AASHTO 2001, and again matching the note on page 4,

with $a=3.4$ m/s², a deceleration comfortably within the capability of virtually all drivers and vehicles on a wet surface without loss of steering control. Where a coefficient of friction and a grade are supplied instead, the equivalent friction form is used,

with $G$ positive upgrade and negative downgrade. The two forms differ by about one per cent at 100 km/h, and the deceleration form is the one to use for anything keyed to the printed table. The sum of the two components is the stopping sight distance, $\text{SSD}=d_1+d_2$. Note that braking distance grows with the square of speed while reaction distance grows linearly, so at low speeds reaction dominates — at 50 km/h the split is 34.8 m against 28.7 m — while at 120 km/h braking is twice the reaction component.

(b) Minimum crest length for a 50 km/h design speed

Given. Entering grade $g_1=+2$ per cent; departing grade $g_2=-4$ per cent; design speed 50 km/h; driver eye height $h_1=1050$ mm; object height $h_2=500$ mm.

Find. The minimum length $L$ of the crest vertical curve that provides the stopping sight distance for 50 km/h.

Approach. Take the SSD from the printed table, then apply the crest-curve sight-distance relation. Because there are two branches depending on whether the sight line lies wholly within the curve, assume $S first and switch branches if the result contradicts the assumption.

1. Establish the algebraic difference in grades. For a crest curve,
   $$A=\left|g_2-g_1\right|=\left|-4-(+2)\right|=6\ \text{per cent}$$
   The sign convention matters: a crest exists because the grade change is negative, and $A$ is the magnitude of that change, not the difference of magnitudes.
2. Read the stopping sight distance from the printed table. At 50 km/h the table gives a Calculated SSD of 63.5 m and a Design SSD of 65 m. Design uses the Design column, so $S=65$ m.
3. Test the $S<L$ branch. When the whole sight line lies within the curve,
   $$L=\frac{A\,S^{2}}{100\left(\sqrt{2h_1}+\sqrt{2h_2}\right)^{2}}$$
   With $\sqrt{2(1.050)}=1.4491$ and $\sqrt{2(0.500)}=1.0000$, the bracket is 2.4491 and its square 5.998, so
   $$L=\frac{6\left(65\right)^{2}}{100\left(5.998\right)}=\frac{25350}{599.8}=42.3\ \text{m}$$
   This contradicts the assumption, since $S=65>L=42.3$ m. The sight line therefore extends beyond the curve onto the tangents and the other branch governs.
4. Apply the $S>L$ branch. With the sight line beginning and ending on the tangents,
   $$L=2S-\frac{200\left(\sqrt{h_1}+\sqrt{h_2}\right)^{2}}{A}$$
   Here $\sqrt{1.050}=1.02470$ and $\sqrt{0.500}=0.70711$, so the bracket is 1.73180, its square 2.9991, and the constant term is $200(2.9991)=599.83$. Hence
   $$L=2\left(65\right)-\frac{599.83}{6}=130.0-99.97$$
   $$L=\boxed{30.0\ \text{m}}$$
   This is self-consistent, since $S=65>L=30.0$ m, so 30.0 m is the answer. The corresponding rate of vertical curvature is $K=L/A=30.0/6=5.0$.
5. Check the answer against the minimum-length criteria that apply independently of sight distance. AASHTO and TAC both require a crest curve to be at least $0.6V$ long for driver comfort and appearance, which at 50 km/h is $0.6(50)=30$ m. The sight-distance requirement of 30.0 m and the comfort requirement of 30 m coincide almost exactly here, so either criterion gives the same design and 30 m is adopted. At a lower grade change the comfort criterion would have governed, which is why it must be checked rather than assumed slack.

(c) Design speed supported by a given 125 m crest

Given. Curve length $L=125$ m; entering grade $g_1=+3$ per cent; departing grade $g_2=-1$ per cent; driver eye height $h_1=1080$ mm; object height $h_2=625$ mm.

Find. The design speed for which this existing curve provides ample stopping sight distance.

Approach. Invert the crest-curve relation to get the sight distance the curve actually delivers, then find the largest tabulated design speed whose required SSD is no greater than that. The problem runs backwards from part (b), so the branch test must be applied in reverse.

1. Establish $A$ and test which branch applies.
   $$A=\left|-1-(+3)\right|=4\ \text{per cent}$$
   Assuming $S and inverting that branch, with $\sqrt{2(1.080)}=1.46969$ and $\sqrt{2(0.625)}=1.11803$ giving a squared bracket of 6.6963:
   $$S=\sqrt{\frac{100\left(\sqrt{2h_1}+\sqrt{2h_2}\right)^{2}L}{A}}=\sqrt{\frac{669.63\times125}{4}}=144.7\ \text{m}$$
   Since 144.7 m exceeds the 125 m curve, the assumption fails and the $S>L$ branch governs.
2. Solve the $S>L$ relation for the available sight distance. Rearranging $L=2S-200(\sqrt{h_1}+\sqrt{h_2})^{2}/A$ for $S$, and with $\sqrt{1.080}=1.03923$, $\sqrt{0.625}=0.79057$ so that $(\,\cdot\,)^{2}=3.34817$:
   $$\frac{200\left(3.34817\right)}{4}=167.41\ \text{m}$$
   $$S=\frac{L+167.41}{2}=\frac{125+167.41}{2}=146.2\ \text{m}$$
   This is consistent, since $S=146.2>L=125$ m.
3. Convert the available sight distance to a standard design speed. Comparing 146.2 m against the Design SSD column of the printed table: 80 km/h requires 130 m, which is available; 90 km/h requires 160 m, which is not. The curve therefore supports
   $$\boxed{V=80\ \text{km/h}}$$
   The Calculated column gives the same verdict (129.0 m at 80 km/h against 155.5 m at 90 km/h), so the answer does not depend on which column is used.
4. Cross-check with the continuous root and with the K-value table. Solving the SSD relation continuously for the speed that exactly consumes 146.2 m,
   $$0.278\,V\left(2.5\right)+0.039\,\frac{V^{2}}{3.4}=146.2$$
   $$0.011471\,V^{2}+0.695\,V-146.2=0\quad\Longrightarrow\quad V=86.6\ \text{km/h}$$
   Design speeds are adopted in standard increments and always rounded down to the next available value, so 86.6 km/h confirms 80 km/h rather than 90. Independently, the rate of vertical curvature is
   $$K=\frac{L}{A}=\frac{125}{4}=31.25$$
   against the AASHTO metric crest K-values of 26 at 80 km/h and 39 at 90 km/h. $K=31.25$ clears 26 but not 39, so the K table, the SSD table and the continuous root all agree on 80 km/h — three independent routes to the same answer, which is as much confirmation as this question admits.

Results.