16-Civ-B7 Transportation Planning and Engineering · December 2015

Question 5 of 8: Curve Stationing on a Three-Tangent Route and a Curb Return Radius

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

Notes on this paper

Paper format. 98-Civ-B7 Highway Engineering, National Examinations, December 2015. Three hours, open book, any non-communicating calculator. Eight questions of equal value (20 marks each); five solutions constitute a complete paper and only the first five in the answer book are marked. Note 1 invites the candidate to state any assumption made about an ambiguous input, and Note 2 permits any datum that is required but not given to be assumed. All eight questions are solved here, because the set is a study resource rather than a timed attempt.

Reference texts. Garber & Hoel, Traffic and Highway Engineering, 5th ed. (geometric design, sight distance, pavement design); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (superelevation and spiral tables — the paper's Table 2.1.2.5 is TAC page 2.1.2.12); AASHTO, A Policy on Geometric Design of Highways and Streets (Green Book) for runoff distribution and relative-gradient limits; AASHTO, Guide for Design of Pavement Structures (1993) for the flexible pavement equation and layer/drainage coefficients; Asphalt Institute MS-2, Asphalt Mix Design Methods and Mamlouk & Zaniewski, Materials for Civil and Construction Engineers, for mixture volumetrics and binder grading.

Check: assumptions carried through this paper. Under the paper's own Note 2 the following values are assumed and stated where used: the AASHTO maximum relative gradient (0.50 % at 80 km/h) and the 70 % / 30 % split of superelevation runoff either side of the PC for two lanes rotated (Question 2); a truck factor of 0.52 for all trucks on a rural Interstate and a lane-distribution factor of 0.70 for three lanes in one direction (Question 6); and a downhill 2 % ramp grade in Question 4, since the freeway is elevated above the local street. Each is flagged again at the point of use with the sensitivity of the answer to it.

Question 5: Curve Stationing on a Three-Tangent Route and a Curb Return Radius (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.

Part A — stations of the BCs and ECs

Given. Three tangents whose tangent-line stationing runs 0+000 to 0+525 to 0+770 to 1+350, with bearings S 33°30′ E, S 49°00′ E and S 14°30′ W respectively, to be joined by two simple curves of radius 200 m.

Find. The stations of BC₁, EC₁, BC₂ and EC₂ measured along the final, curved route.

Plan of the three-tangent route. Chainage runs along the tangents until the first BC; thereafter each curve replaces two tangent distances of length T with one arc of length L.

Approach. Convert the quadrant bearings to azimuths so the deflection angle at each PI can be differenced directly, compute $T$ and $L$ for each 200 m curve, then chain forward, remembering that at every PI the route is shortened by $2T - L$.

Azimuths and deflection angles. Converting the quadrant bearings, $$\text{S }33^\circ 30' \text{ E} = 146^\circ 30', \quad \text{S }49^\circ 00' \text{ E} = 131^\circ 00', \quad \text{S }14^\circ 30' \text{ W} = 194^\circ 30'$$ so the deflections are $$\Delta_1 = |146^\circ 30' - 131^\circ 00'| = 15^\circ 30' \ (\text{left}), \qquad \Delta_2 = |194^\circ 30' - 131^\circ 00'| = 63^\circ 30' \ (\text{right})$$
Curve elements. With $R = 200$ m for both curves, $$T_1 = 200\tan\left(\tfrac{15^\circ 30'}{2}\right) = 27.219\ \text{m}, \qquad L_1 = 200\left(\tfrac{15.5\pi}{180}\right) = 54.105\ \text{m}$$ $$T_2 = 200\tan\left(\tfrac{63^\circ 30'}{2}\right) = 123.764\ \text{m}, \qquad L_2 = 200\left(\tfrac{63.5\pi}{180}\right) = 221.657\ \text{m}$$
First curve. Chainage to PI₁ is unaltered because no curve precedes it, so $$\text{BC}_1 = 525.000 - 27.219 = \boxed{0+497.781}$$ $$\text{EC}_1 = 497.781 + 54.105 = \boxed{0+551.886}$$
Carry the chainage to the second PI. From EC₁ the route runs along tangent 2, but only the part of that tangent beyond $T_1$ remains to be travelled: $$\text{station PI}_2 = 551.886 + \bigl[(770 - 525) - 27.219\bigr] = 551.886 + 217.781 = 0+769.668$$ Equivalently, the route is 0.333 m shorter than the tangent chainage because $2T_1 - L_1 = 0.333$ m has been cut off at the first PI.
Second curve. $$\text{BC}_2 = 769.668 - 123.764 = \boxed{0+645.904}$$ $$\text{EC}_2 = 645.904 + 221.657 = \boxed{0+867.561}$$ and the route ends at $867.561 + (1350 - 770 - 123.764) = 1+323.797$, which is 26.203 m shorter than the tangent chainage of 1+350 — the sum of $2T-L$ for the two curves.
Feasibility check. The tangent that survives between the two curves is $645.904 - 551.886 = 94.018$ m, comfortably positive, so the curves do not overlap and a simple-curve solution is admissible. Had $T_1 + T_2$ exceeded the 245 m between the PIs, one radius would have had to be reduced or a compound curve introduced.

Question 5A — stations along the final route
Point	Station	Element
BC₁	0+497.781	$\Delta_1 = 15^\circ 30'$, $T_1 = 27.219$ m
EC₁	0+551.886	$L_1 = 54.105$ m
BC₂	0+645.904	$\Delta_2 = 63^\circ 30'$, $T_2 = 123.764$ m
EC₂	0+867.561	$L_2 = 221.657$ m
End of route	1+323.797	26.203 m shorter than the tangent chainage
Intervening tangent	94.018 m	curves do not overlap

Part B — curb return radius that abuts the catch basin

Given. Two curb lines meeting at vertex V with deflection $\Delta = 71^\circ 36'$; the catch basin lies on the future curb line at a distance $8.713\ \text{m}$ from V, and the angle E–V–CB measured from the departing curb line is $21^\circ 14'$.

Find. The curb return radius $R$ for which the arc passes exactly through the catch basin.

Curb return geometry. O is the centre, B and E the tangent points on the two curb lines, and CB the catch basin, which must lie on the arc. The bisector VO makes the half angle with each curb line.

Approach. Place the unknown circle by its centre O, which lies on the bisector of the two curb lines at a distance $R/\cos(\Delta/2)$ from V. The condition that the catch basin lies on the circle is then a single triangle relation V–O–CB, which reduces to a quadratic in $R$.

Angles at the vertex. The interior angle between the two curb lines is $180^\circ - \Delta = 108^\circ 24'$, so the bisector VO makes $$\theta = \tfrac{1}{2}\left(180^\circ - \Delta\right) = 54^\circ 12'$$ with each curb line. The catch basin lies $21^\circ 14'$ from the departing curb line VE, hence $$\phi = \angle(\text{VO},\text{V-CB}) = 54^\circ 12' - 21^\circ 14' = 32^\circ 58'$$
Distance from the vertex to the centre. The tangent length is $T = R\tan(\Delta/2)$ and O lies $R$ perpendicular to each curb line at its tangent point, so $$VO = \sqrt{T^{2}+R^{2}} = R\sqrt{\tan^{2}\left(\tfrac{\Delta}{2}\right)+1} = \frac{R}{\cos(\Delta/2)} = \frac{R}{\cos 35^\circ 48'} = 1.232949\,R$$
Impose that the catch basin lies on the circle. In triangle V–O–CB the side O–CB must equal $R$, so the cosine rule gives $$R^{2} = VO^{2} + d^{2} - 2\,VO\,d\cos\phi$$ with $d = 8.713$ m. Substituting $VO = 1.232949R$, $$R^{2} = 1.520162\,R^{2} + 75.916369 - 2(1.232949)(8.713)\cos 32^\circ 58'\,R$$
Reduce to a quadratic and solve. Collecting terms, $$0.520162\,R^{2} - 18.025946\,R + 75.916369 = 0$$ $$R = \frac{18.025946 \pm \sqrt{324.9347 - 157.9553}}{2(0.520162)} = \frac{18.025946 \pm 12.922051}{1.040324}$$ giving $R = 29.748$ m or $R = 4.906$ m.
Select the correct root. The two roots are the near and far intersections of the ray V–CB with the circle. For $R = 4.906$ m the tangent length is only 3.538 m and the centre lies 6.049 m from V, so the catch basin at 8.713 m would sit beyond the centre; the drawing shows it well inside the return, between V and O. The near root therefore governs: $$\boxed{R = 29.748\ \text{m}}$$ for which $T = 21.455$ m and $VO = 36.678$ m.
Confirm the catch basin falls on the arc, not on its extension. In triangle V–O–CB the angle at CB follows from the sine rule, $\sin(\angle \text{VCB O}) = VO\sin\phi/R = 0.6709$, and taking the obtuse value $137^\circ 52'$ (the near intersection) leaves an angle at O of $$180^\circ - 32^\circ 58' - 137^\circ 52' = 9^\circ 10'$$ measured from the bisector. The arc extends $\Delta/2 = 35^\circ 48'$ each side of the bisector, so the catch basin lies comfortably within the return, about a quarter of the way from the midpoint towards the tangent point E. A 29.75 m return would be rounded to 30 m in construction, which moves the curb about 74 mm off the basin — within the tolerance of a cast-in-place gutter, but worth stating on the drawing.

Question 5B — final results
Quantity	Value
Deflection between curb lines, $\Delta$	71°36′
Half angle at V, $\theta$	54°12′
Angle between bisector and V–CB, $\phi$	32°58′
Quadratic in $R$	$0.520162R^{2} - 18.025946R + 75.916369 = 0$
Roots	29.748 m and 4.906 m
Radius adopted	29.748 m (say 30 m)
Tangent length $T$, centre distance $VO$	21.455 m, 36.678 m
Position of CB on the arc	9°10′ from the bisector, inside the return

Topic. Simple-curve stationing along a route, and the inverse problem of finding a curve radius from a point the curve must pass through.

Key relations. $T = R\tan(\Delta/2)$; $L = R\Delta$ (radians); BC $=$ PI $- T$; EC $=$ BC $+ L$; route shortening at each PI $= 2T - L$; $VO = R/\cos(\Delta/2)$; cosine rule $R^{2} = VO^{2}+d^{2}-2\,VO\,d\cos\phi$.

Why this works. Stationing follows the pavement, not the tangent skeleton, so every curve replaces two tangent distances $T$ by a single arc $L$; because $L < 2T$ for any $\Delta$, the route is always shorter than the tangent chainage and each curve must be chained through before the next PI can be located. Part B inverts the same geometry: the position of the centre is fixed in direction by the bisector and in distance by $R$ alone, so requiring a known point to lie on the circle leaves one equation in one unknown. That the equation is quadratic is physically meaningful — a ray from the vertex cuts a circle twice — and the drawing, not the algebra, decides which root is the curb.

Common pitfall. Subtracting $T_2$ from the tangent station of PI₂ rather than from its station on the curved route, which double-counts the shortening at the first PI. In Part B, the usual error is to assume the catch basin lies on the bisector, in which case $\phi$ would be zero and the answer would come out as the middle-ordinate relation instead.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., §15.4 (horizontal curves and stationing); Wolf & Ghilani, Elementary Surveying, Chapter 24.

← Question 4 Next: Question 6 →