Stationing, Grades, and Horizontal and Vertical Curves: 16-Civ-B7 Transportation Planning and Engineering, May 2014, Q3

Question 3: Stationing, Grades, and Horizontal and Vertical Curves (3 + 3 + 3 + 5 + 6 = 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.

Sub-part	Data
(a)	Start station 23+045; measured length 412 m along the line
(b)	Rise 18.50 m over a horizontal run of 435 m
(c)	$R = 750.000$ m; deflection (intersection) angle $I = 11^\circ\,14'\,21''$
(d)	$L = 1200$ m; BVC at station 88+75; $Elev_{BVC} = 789.94$ m; $Elev_{PVI} = 801.64$ m; $Elev_{EVC} = 788.74$ m
(e)	As for (d); locate the crest

Find. The ending station; the grade in per cent to three decimals; the curve length, tangent length and long chord of the circular curve; both tangent grades of the equal-tangent vertical curve; and the station and elevation of its highest point.

Approach. Parts (a) and (b) are direct chainage and grade arithmetic. Part (c) uses the three standard simple-curve formulae, taking care to convert the deflection angle from degrees-minutes-seconds to radians for the arc length. Parts (d) and (e) use the fact that on an equal-tangent curve the PVI is exactly half the curve length from each end, which turns the three given elevations into the two grades directly; the crest then follows from setting the derivative of the parabola to zero.

Part (a) — convert the starting station to a chainage. In metric stationing the figure before the plus sign counts full kilometres and the figure after it counts metres within that kilometre, so station 23+045 is at chainage $$23(1000)+045=23\,045\ \text{m}$$ The two-digit-plus-three-digit form 23+045 identifies the 1 km station convention used across most of Canada.
Add the measured length and re-express as a station. Adding the 412 m line length, $$23\,045+412=23\,457\ \text{m}\quad\Rightarrow\quad\boxed{\text{Station } 23+457}$$ No carry into the next kilometre occurs here; had the total reached 24 000 m the station would have advanced to 24+000.
Part (b) — express the rise as a percentage of the run. Grade is the vertical rise divided by the horizontal distance, expressed in per cent: $$g=\frac{\text{rise}}{\text{run}}\times100=\frac{18.50}{435}\times100=4.252873\ldots$$ Rounded as the question demands, $$\boxed{g=+4.253\ \text{per cent}}$$ The sign is positive because the road rises. A grade of 4.25 per cent is steep but acceptable for a rural arterial in rolling terrain; TAC limits main highways to about 6 per cent in such terrain.
Part (c) — convert the deflection angle to decimal degrees and radians. The intersection angle is $$I=11^\circ+\frac{14'}{60}+\frac{21''}{3600}=11.239167^\circ$$ which in radians is $I=11.239167\times\pi/180=0.1961607\ \text{rad}$. Carrying six decimal places here matters: an error of one second of arc changes the curve length by about 3.6 mm, which is within tolerance, but truncating to $11.2^\circ$ changes it by half a metre.
Curve (arc) length. The arc subtended by $I$ on a circle of radius $R$ is $$L=R\,I_{\text{rad}}=750.000\times0.1961607=\boxed{L=147.120\ \text{m}}$$ Equivalently $L = \pi R I^{\circ}/180$, and as a check the degree of curve (arc definition, 100 m) is $D_a=5729.58/R=7.6395^\circ$, so $L=100I/D_a=147.12$ m.
Tangent length. From the right triangle formed by the radius at the PC, the tangent and the line from the centre to the PI, $$T=R\tan\frac{I}{2}=750.000\times\tan(5.619583^\circ)=750.000\times0.0983959=\boxed{T=73.797\ \text{m}}$$
Long chord. The chord joining the PC to the PT subtends the same angle at the centre, so $$LC=2R\sin\frac{I}{2}=2(750.000)\sin(5.619583^\circ)=1500\times0.0979231=\boxed{LC=146.885\ \text{m}}$$ The chord is 0.235 m shorter than the arc, as it must be, and the ordering $LC < L < 2T$ holds. For completeness, the external distance is $E=R(\sec\tfrac{I}{2}-1)=3.622$ m and the middle ordinate is $M=R(1-\cos\tfrac{I}{2})=3.605$ m.

Figure 3.1 — Simple circular curve, R = 750.000 m, I = 11 deg 14 min 21 s. L = 147.120 m along the arc, T = 73.797 m from PC to PI, LC = 146.885 m. Angles exaggerated for clarity.

Part (d) — locate the PVI on an equal-tangent curve. "Equal tangent" means the two tangent lengths are the same, so the PVI is exactly $L/2 = 600$ m horizontally from the BVC and 600 m from the EVC. That single fact converts the three given elevations into two grades without any further geometry.
Backward (approach) tangent grade. The back tangent rises from the BVC to the PVI over 600 m: $$g_1=\frac{Elev_{PVI}-Elev_{BVC}}{L/2}=\frac{801.64-789.94}{600}=\frac{11.70}{600}=0.019500$$ that is $\boxed{g_1=+1.95\ \text{per cent}}$.
Forward (departure) tangent grade. The forward tangent falls from the PVI to the EVC over the same 600 m: $$g_2=\frac{Elev_{EVC}-Elev_{PVI}}{L/2}=\frac{788.74-801.64}{600}=\frac{-12.90}{600}=-0.021500$$ that is $\boxed{g_2=-2.15\ \text{per cent}}$. Because $g_1$ is positive and $g_2$ negative, this is a crest curve. The algebraic difference is $A=g_2-g_1=-4.10$ per cent, the rate of vertical curvature is $K=L/|A|=1200/4.10=292.7$ m per per cent, and the rate of change of grade is $r=A/L=-0.003417$ per cent per metre, or $-0.342$ per cent per 100 m station.
Part (e) — write the parabola and differentiate. Measuring $x$ from the BVC, the equal-tangent parabola is $$y(x)=Elev_{BVC}+g_1x+\frac{g_2-g_1}{2L}x^{2}$$ Its slope is $dy/dx=g_1+\dfrac{g_2-g_1}{L}x$, and the highest point is where that slope vanishes.
Offset of the high point from the BVC. Setting the slope to zero and solving for $x$, $$x_{\text{high}}=\frac{-g_1L}{g_2-g_1}=\frac{-(0.019500)(1200)}{-0.041000}=\frac{-23.40}{-0.041000}=\boxed{570.73\ \text{m}}$$ Equivalently $x_{\text{high}} = K\,|g_1| \times 100 = 292.68 \times 1.95 = 570.7$ m. The crest falls short of the PVI (at 600 m) because the descending grade is the steeper of the two.
Station of the high point. The BVC is at station 88+75. On the 100 m station convention used in this part of the question, that is chainage $88(100)+75=8875$ m, so $$\text{chainage}_{\text{high}}=8875+570.73=9445.73\ \text{m}\quad\Rightarrow\quad\boxed{\text{Station } 94+45.73}$$ For reference the PVI is at station 94+75 and the EVC at 100+75.
Elevation of the high point. Substituting $x_{\text{high}}$ into the parabola, $$y=789.94+0.019500(570.73)+\frac{-0.041000}{2(1200)}(570.73)^{2}$$ $$y=789.94+11.129-5.565=\boxed{795.505\ \text{m}}$$ As a check, the tangent offset at the crest is 5.565 m below the back tangent, and evaluating the parabola 25 m either side of the crest gives 795.494 m in both directions, confirming a maximum.

Figure 3.2 — Equal-tangent crest vertical curve of Q3(d) and (e). The dashed lines are the two tangents; the crest (red) lies 570.73 m past the BVC, 29.27 m short of the PVI, because the departure grade is the steeper of the two.

Check: the paper uses two different station conventions. Part (a) writes "station 23+045" with a three-digit offset, which is the 1 km station form ($23\,045$ m). Part (d) writes "Station 88+75 m" with a two-digit offset, which is the 100 m station form ($8875$ m). Each part has been solved in its own convention, which is the reading that makes both sets of numbers sensible. If part (d) were instead read as a 1 km station, the BVC chainage would be 88 075 m and the high point would be at station 88+645.73 — the elevation, the grades and the 570.73 m offset are all unaffected, since only the label changes. Stating the convention adopted is worth doing on the answer paper.

Sub-part	Quantity	Result
(a)	Station of the end of the line	23+457 (chainage 23 457 m)
(b)	Grade of the road section	+4.253 per cent
(c)	Curve length, $L$	147.120 m
(c)	Tangent length, $T$	73.797 m
(c)	Long chord, $LC$	146.885 m
(c)	External $E$ / middle ordinate $M$ (bonus)	3.622 m / 3.605 m
(d)	Backward tangent grade, $g_1$	+1.95 per cent
(d)	Forward tangent grade, $g_2$	−2.15 per cent
(d)	$A$, $K$, $r$	−4.10 per cent; $K$ = 292.7 m/per cent; $r$ = −0.342 per cent per 100 m
(e)	Station of the high point	94+45.73 (chainage 9445.73 m)
(e)	Elevation of the high point	795.505 m

Topic. Route-survey arithmetic: stationing, grade, the simple circular curve, and the equal-tangent parabolic vertical curve.

Key relations. Circular curve: $L=RI_{\text{rad}}$, $T=R\tan(I/2)$, $LC=2R\sin(I/2)$, $E=R(\sec\tfrac{I}{2}-1)$, $M=R(1-\cos\tfrac{I}{2})$, $D_a=5729.58/R$. Vertical curve: $g_1=(Elev_{PVI}-Elev_{BVC})/(L/2)$, $g_2=(Elev_{EVC}-Elev_{PVI})/(L/2)$, $A=g_2-g_1$, $K=L/|A|$, $y=Elev_{BVC}+g_1x+\frac{A}{2L}x^{2}$, turning point at $x=-g_1L/A$.

Why this works. A circular curve is completely determined by two parameters, and the exam supplies $R$ and $I$; every other element follows from elementary trigonometry on the isosceles figure formed by the two tangents and the two radii, which is why the three requested lengths need no additional information. The vertical curve is a second-degree parabola rather than an arc because a parabola gives a constant rate of change of grade, which produces a constant vertical acceleration for a vehicle at constant speed and hence a comfortable, uniformly changing sight-distance profile. The equal-tangent property is what makes part (d) trivial: it fixes the PVI at the mid-length, so each tangent grade is simply an elevation difference divided by $L/2$. Because the grade varies linearly along a parabola, the crest is found by linear interpolation of grade to zero, which is exactly the meaning of $x=-g_1L/A$.

Common pitfall. The most damaging error in part (c) is to substitute the deflection angle in degrees into $L=RI$ without converting to radians, which inflates the curve length by a factor of 57.3. In part (e) the recurring error is to assume the high point is at the PVI or at mid-curve; it is only at mid-curve when $|g_1|=|g_2|$, and here it is 29.3 m short of the PVI. A third is to forget that a metric station "88+75" carries a plus-sign offset that must be added to the kilometre or hectometre count before any arithmetic is done, and to add 412 m to "23" rather than to 23 045.

Source. Garber & Hoel, Traffic and Highway Engineering, 5th ed., Ch. 15 (geometric design: horizontal alignment, vertical alignment, stationing); TAC, Geometric Design Guide for Canadian Roads, Ch. 2 (design controls) and Ch. 3 (alignment, $K$-value tables); Wolf & Ghilani, Elementary Surveying, Ch. 24–25 (circular and parabolic curves).

Question 3 of 6: Stationing, Grades, and Horizontal and Vertical Curves

Question 3: Stationing, Grades, and Horizontal and Vertical Curves (3 + 3 + 3 + 5 + 6 = 20 marks)