16-Civ-A6 Highway Design, Construction, and Maintenance · May 2017

Question 7 of 7: Vertical-curve adequacy and advance placement of a ramp sign

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2017 — 16-Civ-A6 Highway Design, Construction and Maintenance. Three-hour closed-book paper (Casio or Sharp approved calculator only). Seven questions, all of equal value at 20 marks; the candidate submits five, and only the first five in the answer book are marked. NOTE 1 invites a written statement of any assumption, and NOTE 2 permits any datum that is required but not given to be assumed. All seven questions are worked below, because the set is a study resource rather than an examination script.

Reference texts.

Garber, N. J. and Hoel, L. A., Traffic and Highway Engineering, 5th ed., Cengage — the exam's equation sheet, the friction/side-friction table, Table 4.2 and Table 4.4 axle-load equivalency factors, Table 4.5 reliability Z_R values and the structural-layer-coefficient table are all reproduced from this text.
AASHTO, Guide for Design of Pavement Structures, Washington DC, 1993 — the flexible-pavement design equation and nomograph, and the source of the LEF tables.
AASHTO, A Policy on Geometric Design of Highways and Streets (the “Green Book”) — stopping sight distance, horizontal sightline offset and vertical-curve K values.
Transportation Association of Canada, Geometric Design Guide for Canadian Roads (TAC GDG) — the governing Canadian geometric-design reference; its design-domain values parallel the AASHTO relations used here.
Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) — advance placement of guide signs.

Canadian context. These are Engineers Canada national examinations, so the Canadian frame governs practice: geometric design in British Columbia follows the TAC Geometric Design Guide and signing follows the MUTCDC. The paper, however, supplies an AASHTO equation sheet and AASHTO tables and names the AASHTO method explicitly in Questions 4 and 5, so every calculation below is carried out with the data the paper gives; where a Canadian practice differs in emphasis it is noted in the concept block rather than substituted for the examiner's method.

Assumptions used throughout (declared under NOTE 1). Gravitational acceleration $g = 9.81\ \text{m/s}^{2}$; perception–reaction time $t_{pr} = 2.5\ \text{s}$ (the AASHTO design value — the paper does not state one); coefficients $f$ and $f_{s}$ read from the paper's own Table 1 at the initial vehicle speed; drainage coefficients $m_{2} = m_{3} = 1.0$ where the question is silent.

Question 7: Vertical-curve adequacy and advance placement of a ramp sign (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. Part (a) supplies a vertical curve by its PVC, PVI and PVT rather than by its grades and length, so both must be recovered first. Part (b) supplies a chain of driver tasks that must be completed before the exit is reached.

Given data
Item	Symbol	Value
(a) PVC station / elevation	—	$1+234.000$ / $1\,100.00\ \text{m}$
(a) PVI station / elevation	—	$1+324.000$ / $1\,100.00\ \text{m}$
(a) PVT elevation	—	$1\,102.00\ \text{m}$
(a) Design speed	$V$	$100\ \text{km/h} = 27.778\ \text{m/s}$
(a) Friction at 100 km/h (Table 1)	$f$	$0.29$
(b) Freeway / ramp speed	—	$120$ / $30\ \text{km/h}$
(b) Legibility distance	—	$50\ \text{m}$
(b) Reading / decision time	—	$1\ \text{s}$ / $1\ \text{s}$
(b) Grade	$G$	$-2\ \%$
(b) Coefficient of friction	$f$	$0.28$
(b) Coefficient of road adhesion	—	$0.60$

Find. (a) Whether the curve as staked provides stopping sight distance at 100 km/h; (b) how far in advance of the ramp exit the directional sign must stand.

Part (a). Because the PVI has the same elevation as the PVC, the approach grade is zero and the curve is a sag, controlled by the headlight criterion.

Approach (a). Recover the curve length from the PVC–PVI distance, recover the two grades from the three elevations, identify the curve type from their signs, compute the stopping sight distance the design speed requires, and compare the length provided with the length the governing sight-distance criterion demands.

Recover the curve length. On a symmetrical parabolic curve the PVI sits midway between the PVC and the PVT, so half the length is the PVC-to-PVI distance: $$\frac{L}{2}=1\,324.000-1\,234.000=90\ \text{m}\quad\Rightarrow\quad L=180\ \text{m}$$ and the PVT therefore lies at station $1+414.000$.
Recover the two grades. The PVI has exactly the same elevation as the PVC, so the approach tangent is level: $$G_{1}=\frac{1\,100.00-1\,100.00}{90}=0\ \%$$ $$G_{2}=\frac{1\,102.00-1\,100.00}{90}\times 100=+2.2222\ \%$$ $$A=|G_{2}-G_{1}|=2.2222\ \%$$ Because the grade increases through the curve, this is a $\boxed{\text{sag curve}}$, and the governing control is the headlight criterion, not the crest sight line.
Compute the required stopping sight distance. At the design speed, taking the level approach grade, $$\text{SSD}=\frac{27.778^{2}}{2(9.81)(0.29)}+27.778(2.5)=135.61+69.44=\boxed{205.06\ \text{m}}$$
Apply the sag (headlight) criterion. The equation sheet gives $L_{m}=KA$ with $$K=\frac{\text{SSD}^{2}}{120+3.5\,\text{SSD}}=\frac{205.06^{2}}{120+3.5(205.06)}=\frac{42\,050}{837.7}=50.20$$ $$L_{m}=50.20(2.2222)=\boxed{111.54\ \text{m}}$$ which is comfortably less than the $180\ \text{m}$ provided.
Check the other branch, since $S \gt L$. The $L_{m}=KA$ form assumes the whole sight line lies on the curve, but here $S=205.06\ \text{m}$ exceeds $L=180\ \text{m}$, so the $S \ge L$ headlight expression applies: $$L=2S-\frac{200(H+S\tan\beta)}{A}=2(205.06)-\frac{200\left[0.6+205.06\tan 1^\circ\right]}{2.2222}$$ $$L=410.11-376.13=33.98\ \text{m}$$ Both branches demand far less than the length provided, so the conclusion does not depend on which is used.
State the verdict, and cross-check against the AASHTO rate. The curve provides a rate of vertical curvature $$K_{\text{provided}}=\frac{L}{A}=\frac{180}{2.2222}=81.0$$ against the AASHTO design value of $K=45$ for a 100 km/h sag curve. $$\boxed{\text{Yes — the curve is adequate, by a wide margin.}}$$ It is in fact considerably longer than the sight-distance criterion requires; on a sag that is not waste, because rider comfort, drainage and appearance commonly govern the length rather than sight distance.

Part (b). The driver must finish reading, decide, and then decelerate from 120 km/h to the 30 km/h ramp speed, all before reaching the exit.

Approach (b). Follow the driver forward through the task chain — the sign becomes legible, is read, a decision is made, and only then does braking begin — and require that the vehicle has slowed to the ramp speed exactly at the exit. The sign then sits one legibility distance downstream of where the chain starts.

Distance covered while reading. The driver is still at the freeway speed $120\ \text{km/h}=33.333\ \text{m/s}$ throughout the reading and decision phases, so $$d_{\text{read}}=33.333(1.0)=33.33\ \text{m}$$
Distance covered while deciding and reacting. The second second is spent deciding and moving the foot to the brake, still at full speed: $$d_{\text{decide}}=33.333(1.0)=33.33\ \text{m}$$ Together these replace the usual $2.5\ \text{s}$ perception–reaction allowance, because the question specifies the driver's task times explicitly.
Deceleration distance from freeway speed to ramp speed. The driver does not stop, only slows to the ramp speed, and the $2\ \%$ downgrade works against the brakes: $$d_{\text{brake}}=\frac{V_{0}^{2}-V_{t}^{2}}{2g\,(f+G)}=\frac{33.333^{2}-8.333^{2}}{2(9.81)(0.28-0.02)}=\frac{1\,111.11-69.44}{5.1012}$$ $$d_{\text{brake}}=\boxed{204.20\ \text{m}}$$
Place the sign. Measuring back from the exit, the whole chain occupies $$33.33+33.33+204.20=270.87\ \text{m}$$ but that chain begins where the sign first becomes legible, which is $50\ \text{m}$ upstream of the sign itself. The sign therefore stands $$D=270.87-50=\boxed{220.87\ \text{m}}\ \text{ahead of the exit}$$ say $221\ \text{m}$, or $225\ \text{m}$ rounded up to a practical staking increment.
Confirm the manoeuvre is within the available friction. The deceleration implied is $$a=g\,(f+G)=9.81(0.26)=2.551\ \text{m/s}^{2}$$ while the maximum the surface can deliver is $$a_{\max}=g\,(0.60-0.02)=5.690\ \text{m/s}^{2}$$ so the driver uses only $44.8\ \%$ of the available adhesion. The manoeuvre is comfortable and no skidding is implied — which is exactly why the problem supplies both coefficients: $0.28$ is the comfortable design deceleration and $0.60$ is the physical limit against which it is checked.

Final results
Quantity	Symbol	Value
(a) Curve length	$L$	$180\ \text{m}$
(a) Grades	$G_{1}$ / $G_{2}$	$0\ \%$ / $+2.2222\ \%$
(a) Grade difference / curve type	$A$	$2.2222\ \%$, sag
(a) Required stopping sight distance	$\text{SSD}$	$205.06\ \text{m}$
(a) Rate of vertical curvature required	$K$	$50.20$
(a) Minimum length required	$L_{m}$	$111.54\ \text{m}$
(a) Rate provided	$K_{\text{prov}}$	$81.0$ (AASHTO needs 45)
(a) Verdict	—	Adequate
(b) Reading + decision distance	—	$66.67\ \text{m}$
(b) Deceleration distance, 120 → 30 km/h	$d_{\text{brake}}$	$204.20\ \text{m}$
(b) Advance placement of the sign	$D$	$220.87\ \text{m}$
(b) Fraction of available adhesion used	—	$44.8\ \%$

Check: in part (a) the curve is a sag, not a crest — the PVI elevation equals the PVC elevation, so $G_{1}=0$ and the grade rises through the curve. Applying the crest constant $K=\text{SSD}^{2}/658$ instead would give $K=63.9$ and $L_{m}=142.0\ \text{m}$, still adequate here, but it is the wrong criterion and would give a wrong answer on a curve nearer the limit. In part (b) the $50\ \text{m}$ legibility distance is subtracted, not added: the driver begins reading before reaching the sign, so that distance is already part of the 270.87 m chain.

Topic. Sag-curve adequacy under the headlight criterion, and advance placement of a guide sign from the driver task chain.

Key relations. $L/2=\text{Sta}_{PVI}-\text{Sta}_{PVC}$; $G=\Delta\text{elev}/(L/2)$; $A=|G_{2}-G_{1}|$; $L_{m}=KA$ with $K=\text{SSD}^{2}/(120+3.5\,\text{SSD})$ for sags and $K=\text{SSD}^{2}/658$ for crests; the $S\ge L$ sag branch $L=2S-200(H+S\tan\beta)/A$ with $H=0.6\ \text{m}$ and $\beta=1^\circ$; and $d=(V_{0}^{2}-V_{t}^{2})/[2g(f+G)]$ for a deceleration between two speeds.

Why this works. A sag curve is limited at night, not by day: in daylight the sight line over a sag is unobstructed, but after dark the driver sees only as far as the headlight beam reaches, and the beam — mounted $0.6\ \text{m}$ above the pavement and aimed $1^\circ$ upward — strikes the rising pavement ahead. That geometry, not the eye and object heights of the crest case, produces the $120+3.5\,\text{SSD}$ denominator. Sign placement follows a different but equally mechanical logic: every task the driver must finish before the decision point consumes distance at the speed prevailing while it is performed, so the placement is simply the sum of those distances, with the legibility distance subtracted because reading starts before the sign is reached.

Common pitfall. Classifying the curve from the sign of $G_{2}$ alone, rather than from the change $G_{2}-G_{1}$, and so applying the crest constant to a sag. In part (b), adding the perception–reaction time and the stated reading and decision times, which double-counts about $2.5\ \text{s}$ (some $83\ \text{m}$ at freeway speed); and using the stopping-distance formula instead of the speed-change formula, which would size the sign for a full stop the driver never makes.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §15.6 (vertical curves, crest and sag criteria) and §3.6 (deceleration and sign legibility); AASHTO Green Book, Ch. 3; TAC, Manual of Uniform Traffic Control Devices for Canada, Part B (advance placement of guide signs).

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