Question 6 of 7: Sag vertical curve — PVI, PVT, low point and grade points

Question 6: Sag vertical curve — PVI, PVT, low point and grade points

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. An equal-tangent parabolic sag curve.

Find. The station and elevation of the PVI, of the PVT and of the lowest point, and the station and elevation of the points on the curve where the grade is −3, −2, −1 and 0 percent.

Approach. Write the curve as a parabola in the distance $x$ measured from the PVC, so that the elevation, the grade and the location of the turning point all follow from one expression, then evaluate it at each station the question asks for.

1. Set up the parabola. An equal-tangent vertical curve has a constant rate of change of grade, so with $x$ measured in metres from the PVC and grades in percent, $$Y(x) = Y_{PVC} + \frac{g_1}{100}x + \frac{A}{100}\cdot\frac{x^2}{2L} = 320.000 - 0.04\,x + \frac{0.05}{600}x^2$$ and the grade at any point is $$g(x) = g_1 + \frac{A}{L}x = -4.0 + \frac{5.0}{300}x \ \text{percent}$$
2. The PVI. On an equal-tangent curve the PVI lies at mid-length along the back tangent: $$\text{Sta PVI} = 2\,600 + \tfrac{L}{2} = 2\,600 + 150 = 2\,750 \;\Rightarrow\; \boxed{\text{PVI at } 2+750.000}$$ $$Y_{PVI} = 320.000 + \left(\frac{-4.0}{100}\right)(150) = 320.000 - 6.000 = \boxed{314.000\ \text{m}}$$ Note that the PVI is on the tangents, not on the curve.
3. The PVT. The PVT is one further half-length ahead, along the forward tangent from the PVI: $$\text{Sta PVT} = 2\,750 + 150 = 2\,900 \;\Rightarrow\; \boxed{\text{PVT at } 2+900.000}$$ $$Y_{PVT} = 314.000 + \left(\frac{1.0}{100}\right)(150) = 314.000 + 1.500 = \boxed{315.500\ \text{m}}$$ Evaluating the parabola at $x = L$ gives $320.000 - 12.000 + 7.500 = 315.500$ m, which confirms both the equation and the arithmetic.
4. The lowest point. The low point is where the grade reaches zero, which for a sag curve is the turning point of the parabola: $$x_{\min} = \frac{-g_1}{A}\,L = \frac{4.0}{5.0}(300) = 240\ \text{m}$$ $$\text{Sta} = 2\,600 + 240 = 2\,840 \;\Rightarrow\; \boxed{\text{low point at } 2+840.000}$$ $$Y = 320.000 - 0.04(240) + \frac{0.05}{600}(240)^2 = 320.000 - 9.600 + 4.800 = \boxed{315.200\ \text{m}}$$ The low point lies within the curve because $g_1$ and $g_2$ have opposite signs; it is at 80 percent of the length, reflecting the fact that the back grade is four fifths of the total grade change.
5. Stations at the intermediate grades. Inverting the grade expression, $$x = \frac{g - g_1}{A}\,L = \frac{g + 4.0}{5.0}(300) = 60\,(g + 4.0)$$ so each one percent of grade recovered corresponds to exactly 60 m of curve. That gives $x = 60$, 120, 180 and 240 m for the grades −3, −2, −1 and 0 percent respectively.
6. Elevations at those stations. Substituting each $x$ into the parabola, $$\begin{aligned}Y(60) &= 320.000 - 2.400 + 0.300 = 317.900\ \text{m} \\Y(120) &= 320.000 - 4.800 + 1.200 = 316.400\ \text{m} \\Y(180) &= 320.000 - 7.200 + 2.700 = 315.500\ \text{m} \\Y(240) &= 320.000 - 9.600 + 4.800 = 315.200\ \text{m}\end{aligned}$$ The last of these is the low point already found, and the third happens to equal the PVT elevation exactly — a coincidence of these particular grades, but a useful check that the parabola has been evaluated correctly.

A useful property visible in the figure is the offset from the PVI to the curve, which for an equal-tangent parabola is always $AL/800 = (5.0)(300)/800 = 1.875$ m at mid-length; adding it to the PVI elevation of 314.000 m gives 315.875 m on the curve at station 2+750, and the same offset measured from the two tangent points is a quick way of sketching the whole curve by hand.

The question does not ask for a sight-distance check, but the curve invites one and it takes two lines. The rate of vertical curvature is $K = L/A = 300/5 = 60$ m per percent. Inverting the sag headlight criterion $L = AS^2/(120 + 3.5S)$ for the sight distance actually served gives $S = 240$ m, and the corresponding design controls are $K = S^2/(120+3.5S) = 54.4$ for the 220 m stopping sight distance required at 110 km/h and 62.8 for the 250 m required at 120 km/h. The curve therefore satisfies the headlight control comfortably at 110 km/h, with 20 m of sight distance in hand, and falls just short of the 120 km/h control — which is a sensible place for a 300 m sag curve on a highway of this class to sit.