Crest vertical curve — high point and 50-m station elevations: 16-Civ-B7 Transportation Planning and Engineering, May 2015, Q7

Question 7: Crest vertical curve — high point and 50-m station elevations

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 crest curve specified by its rate of vertical curvature rather than its length.

Curve data
Parameter	Symbol	Value
Back grade	$g_1$	+4.0 percent
Forward grade	$g_2$	−2.0 percent
Algebraic grade change	$A = \|g_2 - g_1\|$	6.0 percent
Rate of vertical curvature	$K = L/A$	90 m per percent
Design speed	$V$	110 km/h
Chainage of PVI	—	0+400.000 (400 m; 1 000 m stations)
Elevation of PVI	—	150.000 m

Find. The elevation of the high point, and the elevation of the curve at every even 50-m station between the PVC and the PVT.

Approach. Recover the curve length from $K$ and $A$, work back from the PVI to the PVC along the back tangent, write the parabola from the PVC, and evaluate it at the high point and at each 50-m station.

Curve length from the rate of vertical curvature. $K$ is the length of curve per percent of grade change, so $$L = K A = 90 \times 6.0 = 540\ \text{m}$$ with $A = |{-2.0} - ({+4.0})| = 6.0$ percent.
Stationing of the PVC and PVT. The PVI is at mid-length, so each tangent point is $L/2 = 270$ m from it: $$\text{Sta PVC} = 400 - 270 = 130\ \text{m} \;\Rightarrow\; \boxed{\text{PVC at } 0+130.000}$$ $$\text{Sta PVT} = 400 + 270 = 670\ \text{m} \;\Rightarrow\; \boxed{\text{PVT at } 0+670.000}$$
Elevations of the tangent points. Working back down the +4 percent grade from the PVI and forward down the −2 percent grade, $$Y_{PVC} = 150.000 - 0.04(270) = 150.000 - 10.800 = 139.200\ \text{m}$$ $$Y_{PVT} = 150.000 - 0.02(270) = 150.000 - 5.400 = 144.600\ \text{m}$$
Write the parabola from the PVC. With $x$ in metres from the PVC, $$Y(x) = 139.200 + 0.04\,x - \frac{0.06}{2(540)}x^2 = 139.200 + 0.04\,x - 5.5556 \times 10^{-5}x^2$$ and evaluating at $x = 540$ returns $139.200 + 21.600 - 16.200 = 144.600$ m, the PVT elevation already found — the equation is correct.
Locate the high point. The grade falls linearly from +4 percent at the PVC, and the summit is where it reaches zero: $$x_{\max} = \frac{g_1}{A}\,L = \frac{4.0}{6.0}(540) = 360\ \text{m} \;\Rightarrow\; \text{Sta } 130 + 360 = 490\ \text{m}$$ $$Y = 139.200 + 0.04(360) - 5.5556\times10^{-5}(360)^2 = 139.200 + 14.400 - 7.200 = \boxed{146.400\ \text{m at } 0+490.000}$$
Evaluate the even 50-m stations. The even 50-m stations lying between the PVC at 130 m and the PVT at 670 m are 150, 200, …, 650. Substituting $x = \text{station} - 130$ into the parabola gives the elevations tabulated below; the PVC, the PVT and the high point are carried in the same table so the profile can be plotted directly from it.
Check against the tangent offset. At mid-length the curve lies below the PVI by $AL/800 = (6.0)(540)/800 = 4.05$ m, so the elevation at station 0+400 should be $150.000 - 4.050 = 145.950$ m — which is exactly the tabulated value. That single check validates the whole table, because any error in $L$, in the PVC elevation or in the curvature term would show up in it.

Elevations on the crest curve
Station	Chainage (m)	$x$ from PVC (m)	Elevation (m)
PVC 0+130.000	130	0	139.200
0+150.000	150	20	139.978
0+200.000	200	70	141.728
0+250.000	250	120	143.200
0+300.000	300	170	144.394
0+350.000	350	220	145.311
0+400.000 (below the PVI)	400	270	145.950
0+450.000	450	320	146.311
0+490.000 (high point)	490	360	146.400
0+500.000	500	370	146.394
0+550.000	550	420	146.200
0+600.000	600	470	145.728
0+650.000	650	520	144.978
PVT 0+670.000	670	540	144.600

The crest curve with its tangents dashed. The summit is at station 0+490, not beneath the PVI at 0+400, because the back grade is two thirds of the total grade change.

The design speed is given but never used in the computation above, which is a hint that it is there to be checked against. For a crest curve with the sight distance shorter than the curve, the AASHTO metric relation with an eye height of 1.08 m and an object height of 0.60 m is $L = AS^2/658$, so the curve as specified serves $$S = \sqrt{\frac{658\,L}{A}} = \sqrt{\frac{658 \times 540}{6.0}} = 243.4\ \text{m}$$ against a design stopping sight distance of 220 m at 110 km/h on a level grade. The curve therefore satisfies the control with 23.4 m of sight distance to spare. Put in terms of the design parameter itself, the required rate of vertical curvature is $K = S^2/658 = 220^2/658 = 73.6$, so the specified $K = 90$ carries a 22 percent surplus and the minimum acceptable curve would have been $L = 73.6 \times 6.0 = 441$ m — there is about 99 m of curve in hand. The curve would not, however, meet the 120 km/h control, which needs $K = 250^2/658 = 95$; if the alignment were ever to be posted or designed for 120 km/h, this curve would have to be lengthened.

Check — assumptions stated under the paper's Note 1. (i) Stations are 1 000 m, as the question states in parentheses, so 0+400.000 is chainage 400 m and the PVC at 0+130.000 is chainage 130 m. (ii) The sight-distance check uses the AASHTO/TAC design stopping sight distance on a level grade (220 m at 110 km/h) with the standard 1.08 m eye and 0.60 m object heights; a grade-corrected value would be slightly longer on the descending side. (iii) The elevations are reported to the millimetre because the question gives the PVI elevation to the millimetre; that is a stationing convention, not a claim of construction accuracy.

Final results
Quantity	Value
Algebraic grade change, $A$	6.0 percent
Curve length, $L = KA$	540 m
PVC	station 0+130.000, elevation 139.200 m
PVI	station 0+400.000, elevation 150.000 m (off the curve)
PVT	station 0+670.000, elevation 144.600 m
High point	station 0+490.000, elevation 146.400 m
Elevation on the curve beneath the PVI	145.950 m (offset $AL/800 = 4.05$ m)
Even 50-m station elevations	tabulated above, 0+150 through 0+650
Stopping sight distance provided	243.4 m ≥ 220 m required at 110 km/h — adequate
Rate of vertical curvature	$K = 90$ against $K = 73.6$ required (22 % surplus)

Topic. Crest vertical curves, the rate of vertical curvature $K$, and the sight-distance control.

Key relations. $L = KA$; PVC and PVT at $L/2$ either side of the PVI; $Y(x) = Y_{PVC} + \tfrac{g_1}{100}x + \tfrac{g_2-g_1}{100}\tfrac{x^2}{2L}$; summit at $x = g_1 L / A$; PVI offset $= AL/800$; for a crest with $S < L$, $L = AS^2/658$ in metric units with eye height 1.08 m and object height 0.60 m, so $K = S^2/658$.

Why this works. $K$ is the whole design control in one number: because $L = KA$ and the crest sight-distance relation is also linear in $A$ for a given $S$, the sight distance a crest curve provides depends only on $K$ and not on the individual grades. That is why design tables list a minimum $K$ against design speed rather than a minimum length, and why the first thing to do with a crest curve is to compare its $K$ with the tabulated value.

Common pitfall. Putting the summit beneath the PVI. It is beneath the PVI only when the grades are equal and opposite; here the back grade is two thirds of the grade change, so the summit is two thirds of the way along. A second is using the sag headlight formula, with its 120 and 3.5 constants, on a crest curve.

Source. AASHTO, A Policy on Geometric Design of Highways and Streets, chapter 3 — crest vertical curves, tables of design $K$ values and stopping sight distance; TAC, Geometric Design Guide for Canadian Roads, Part 2 (vertical alignment and sight distance); Garber & Hoel, Traffic and Highway Engineering, geometric design chapter.

Question 7 of 7: Crest vertical curve — high point and 50-m station elevations

Question 7: Crest vertical curve — high point and 50-m station elevations