Equal-Tangent Vertical Parabolic Curve (Sag): 18-Geom-A1 Surveying, December 2015, Q4

Question 4: Equal-Tangent Vertical Parabolic Curve (Sag) (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. Equal-tangent vertical (sag) curve data:

Parameter	Value
Back grade $g_1$	$-3.00\%$
Forward grade $g_2$	$+5.00\%$
PVI station	$62{+}00$
PVI elevation	$600.60$ ft
Curve length $L$	$800$ ft

Find. Station and elevation of the BVC and the EVC, and the elevation of the first full station lying on the curve.

Figure 4 — Equal-tangent vertical curve; dashed lines are the grade tangents meeting at the PVI, the solid curve is the parabola. A falling grade meeting a rising grade forms a sag curve (low point on the curve).

Approach. The BVC and EVC lie $L/2$ each side of the PVI along the tangents; elevations there come from the straight grades. The first full (even-hundred) station on the curve is then evaluated from the parabola tangent-offset equation.

Beginning of vertical curve (BVC). The curve is symmetric about the PVI, extending $L/2 = 400$ ft (i.e. $4{+}00$) each way. Moving back along the $-3.00\%$ grade raises the elevation: $$\text{BVC} = 62{+}00 - 4{+}00 = \boxed{58{+}00},\quad \text{Elev} = 600.60 - (-0.03)(400) = \boxed{612.60\ \text{ft}}$$
End of vertical curve (EVC). Moving forward along the $+5.00\%$ grade: $$\text{EVC} = 62{+}00 + 4{+}00 = \boxed{66{+}00},\quad \text{Elev} = 600.60 + (0.05)(400) = \boxed{620.60\ \text{ft}}$$
First full station on the curve. The BVC falls exactly on station $58{+}00$, so the first full station lying within the curve is $59{+}00$, at $x = 100$ ft from the BVC. The rate of grade change is $$r = \frac{g_2 - g_1}{L} = \frac{0.05-(-0.03)}{800} = +0.0001\ \text{ft}^{-1}$$ On the parabola $y = \text{Elev}_{\text{BVC}} + g_1 x + \tfrac{r}{2}x^2$ (grades as decimals): $$y_{59+00} = 612.60 + (-0.03)(100) + \tfrac{0.0001}{2}(100)^2 = 612.60 - 3.00 + 0.50 = \boxed{610.10\ \text{ft}}$$ For reference, the low point (grade $=0$) sits at $x = -g_1/r = 300$ ft, i.e. station $61{+}00$, safely inside the curve as expected for a sag.

Point	Station	Elevation (ft)
BVC (beginning)	$58{+}00$	$612.60$
EVC (end)	$66{+}00$	$620.60$
First full station on curve	$59{+}00$	$610.10$

Concept & Theory

Topic. Equal-tangent parabolic vertical curve — endpoints and a computed curve elevation (sag case).

Key relations. $y=\text{Elev}_{\text{BVC}}+g_1x+\tfrac{r}{2}x^2$ with $r=(g_2-g_1)/L$; grade $g(x)=g_1+rx$; turning point $x=-g_1/r$.

Why this works. A vertical curve is a second-degree parabola, so the grade changes at a constant rate $r$, giving constant vertical acceleration and a comfortable ride. Endpoint elevations lie on the straight tangents; any interior elevation adds the tangent-offset term $\tfrac{r}{2}x^2$. Here $g_1<0

Common pitfall. Getting the sign of the tangent rise wrong when going back from the PVI (a $-3\%$ grade means the BVC is higher, not lower), or treating $58{+}00$ (the BVC itself) as the "first full station on the curve." Keep grades as decimals in the parabola.

Source. Wolf & Ghilani, Elementary Surveying §25 (vertical curves); AASHTO Green Book (sag-curve sight-distance/drainage design). Units are US feet/stations as printed.

Question 4 of 7: Equal-Tangent Vertical Parabolic Curve (Sag)

Question 4: Equal-Tangent Vertical Parabolic Curve (Sag) (20 marks)