Question 3 of 7: Grades and safe speed on a sag vertical curve

Question 3: Grades and safe speed on a sag vertical curve (20 marks — (a) 15, (b) 5)

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 symmetrical parabolic vertical curve whose PVC and PVI are both at elevation 100.00 m, 80 m apart, with the curve itself 100.60 m at the PVI station.

Find. (a) the initial and final grades; (b) the maximum safe speed in each direction under the AASHTO sight-distance standard.

Approach. Recover $L$ from the PVC–PVI spacing, $G_1$ from the two tangent elevations and $A$ from the mid-curve offset $e_m=AL/800$; identify the curve as a sag, solve the headlight sight-distance pair for $S$, and read $S$ back to a speed on each approach grade.

1. Get the curve length from the PVC-to-PVI distance. The PVI of a symmetrical parabola sits at mid-length, so $$ \tfrac{L}{2}=1203.000-1123.000=80\ \text{m}\quad\Longrightarrow\quad L=160\ \text{m}. $$
2. Read the initial grade straight off the two tangent elevations. Both the PVC and the PVI are at 100.00 m, hence $$ G_1=\frac{100.00-100.00}{80}=\boxed{0\ \%} $$ The back tangent is level — not a degenerate case, simply a curve that begins flat.
3. Use the mid-curve offset to recover the second grade. For a parabola the vertical distance between the PVI and the curve at mid-length is $e_m=AL/800$ with $A$ in percent. Here $e_m=100.60-100.00=0.60$ m, so $$ A=\frac{800\,e_m}{L}=\frac{800(0.60)}{160}=3.0\ \% \quad\Longrightarrow\quad G_2=G_1+A=\boxed{+3.0\ \%} $$ Because the curve sits above the PVI, $A$ is positive and this is a sag curve; the tangent elevation at the PVT is $100.00+0.03(80)=102.40$ m.
4. Choose the AASHTO control for a sag curve. Sag curves are governed by headlight sight distance: the beam leaves the headlight $H=0.60$ m above the pavement and diverges upward by $\beta=1^\circ$, so the sheet's pair applies with $200H=120$ and $200\tan\beta\approx3.5$. Trying the $S\lt L$ branch first, $$ L=\frac{A S^{2}}{120+3.5S}\;\Rightarrow\;3S^{2}-560S-19200=0 \;\Rightarrow\;S=216.26\ \text{m}, $$ which contradicts its own premise because $216.26\gt L=160$ m.
5. Switch to the consistent branch. With $S\ge L$ the sheet gives $L=2S-\dfrac{120+3.5S}{A}$, so $$ 160=2S-\frac{120+3.5S}{3}\;\Rightarrow\;2.5S=600 \;\Rightarrow\;\boxed{S=240.0\ \text{m}} $$ and $S=240\ \text{m}\ge L=160\ \text{m}$ is self-consistent, so 240 m is the sight distance the geometry actually delivers.
6. Convert the sight distance into a speed, direction by direction. Travelling on the $-3\ \%$ side the grade term reduces the available deceleration, so with $f=0.29$ at 100 km/h $$ \text{SSD}=\frac{27.778^{2}}{2(9.81)(0.29-0.03)}+2.5(27.778) =151.26+69.44=220.70\ \text{m}\;\lt\;240\ \text{m}. $$ At 110 km/h the same expression gives 266.73 m, which exceeds 240 m, so the descending direction is limited to 100 km/h.
7. Check the ascending direction. Following AASHTO practice the upgrade is credited with no help ($G=0$), which gives 205.06 m at 100 km/h — comfortably inside 240 m — and 246.34 m at 110 km/h, which is not. The ascending direction is therefore also limited to 100 km/h, and the maximum safe speed for the section in both directions is $$ \boxed{V_{max}=100\ \text{km/h}} $$ The descending direction is the critical one (220.70 m of demand against 205.06 m going the other way), so any future regrading that steepens $G_1$ will bite there first.