Equal-Tangent Vertical Parabolic Curve: 18-Geom-A1 Surveying, December 2013, Q7

Question 7: Equal-Tangent Vertical Parabolic Curve (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 (crest) curve data:

Parameter	Value
Back grade $g_1$	$+2.50\%$
Forward grade $g_2$	$-1.75\%$
PVI station	$44{+}25$
PVI elevation	$368.96$ ft
Curve length $L$	$800$ ft

Find. Station and elevation of the BVC, the EVC, and the high point.

Figure 7 — Equal-tangent crest curve; dashed lines are the grade tangents meeting at the PVI, the solid curve is the parabola, and the high point is where the grade is zero.

Approach. The BVC and EVC lie $L/2$ each side of the PVI along the tangents; the high point is where the (linearly varying) grade reaches zero, and its elevation follows from the parabola equation.

BVC and EVC. The curve is symmetric about the PVI, extending $L/2=400$ ft each way: $$\text{BVC} = 44{+}25 - 4{+}00 = \boxed{40{+}25},\quad \text{Elev}=368.96-0.025(400)=\boxed{358.96\ \text{ft}}$$ $$\text{EVC} = 44{+}25 + 4{+}00 = \boxed{48{+}25},\quad \text{Elev}=368.96+(-0.0175)(400)=\boxed{361.96\ \text{ft}}$$
Locate the high point. The rate of grade change is $r=(g_2-g_1)/L=(-1.75-2.50)/800=-0.0053125\ \%/\text{ft}$. The high point (grade $=0$) lies at $x$ from the BVC: $$x = \frac{g_1 L}{g_1-g_2} = \frac{2.50(800)}{2.50+1.75} = 470.588\ \text{ft}$$ $$\text{Station} = 40{+}25 + 4{+}70.588 = \boxed{44{+}95.588}$$
High-point elevation. On the parabola $y=\text{Elev}_{\text{BVC}}+g_1 x+\tfrac{r}{2}x^2$ (grades as decimals): $$\text{Elev} = 358.96 + 0.025(470.588) + \tfrac{-0.000053125}{2}(470.588)^2 = \boxed{364.842\ \text{ft}}$$ Because $x=470.6$ ft $<L$, the high point genuinely lies on the curve, as expected where a rising grade meets a falling one (a crest).

Point	Station	Elevation (ft)
BVC	$40{+}25$	$358.96$
EVC	$48{+}25$	$361.96$
High point	$44{+}95.588$	$364.842$

Concept & Theory

Topic. Equal-tangent parabolic vertical curve — endpoints and turning point.

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 $x=-g_1/r=g_1L/(g_1-g_2)$.

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, well-drained ride. The extreme (high or low) point is where $g(x)=0$; it lies on the curve only when $0\le x\le L$. A crest curve ($g_1>0>g_2$) has a high point; a sag curve has a low point that governs drainage-inlet placement.

Common pitfall. Using the forward grade with the wrong sign for the EVC elevation, or forgetting that the high point is offset from mid-curve unless $g_1=-g_2$. Keep grades as decimals in the parabola equation.

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

Question 7 of 7: Equal-Tangent Vertical Parabolic Curve

Question 7: Equal-Tangent Vertical Parabolic Curve (20 marks)