Curves

Data: $g_1 = -2.5\% = -0.025$, $g_2 = +3.5\% = +0.035$ (a valley or sag curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1268.50 m at chainage 780 m (above the IP, as it must for a sag curve).

Length of the curve

$e = 1268.50 - 1267.00 = 1.50$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_2 - g_1)L}{8}$:

$$L = \frac{8e}{g_2 - g_1} = \frac{8\times 1.50}{0.035 + 0.025} = 200.00\ \text{m}$$

Elements

• PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
• RL of PVC $= 1267.00 + 0.025\times 100 = 1269.500$ m; RL of PVT $= 1267.00 + 0.035\times 100 = 1270.500$ m
• $r = \dfrac{g_2 - g_1}{L} = 0.000300$ per m

Levels at full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2$$

Lowest point

$$x = \frac{-g_1}{r} = \frac{0.025}{0.000300} = 83.33\ \text{m}$$

Chainage $= 680.00 + 83.33 = 763.33$ m and RL $= 1268.458$ m.

Check: level at chainage 780 m (x = 100 m) $= 1268.500$ m = 1268.500 m as required.

Answer: L = 200 m; lowest point at chainage 763.33 m, RL 1268.458 m.

Data: $g_1 = +0.5\% = +0.005$, $g_2 = -3.5\% = -0.035$ (a summit curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1266.00 m at chainage 780 m (below the IP, as it must for a summit curve).

Length of the curve

$e = 1267.00 - 1266.00 = 1.00$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_1 - g_2)L}{8}$:

$$L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.00}{0.005 + 0.035} = 200.00\ \text{m}$$

Elements

• PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
• RL of PVC $= 1267.00 - 0.005\times 100 = 1266.500$ m; RL of PVT $= 1267.00 - 0.035\times 100 = 1263.500$ m
• $r = \dfrac{g_2 - g_1}{L} = -0.000200$ per m

Levels at full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2$$

Highest point

$$x = \frac{-g_1}{r} = \frac{0.005}{0.000200} = 25.00\ \text{m}$$

Chainage $= 680.00 + 25.00 = 705.00$ m and RL $= 1266.562$ m.

Check: level at chainage 780 m (x = 100 m) $= 1266.000$ m = 1266.000 m as required.

Answer: L = 200 m; highest point at chainage 705.00 m, RL 1266.562 m.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a summit curve, so the curve lies below the IP:

$$e = 1266 - 1267 = -1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|-0.5-(3.5)|} = \frac{800}{4} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a summit (crest) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 780 - 100 = 680\ \text{m}\\ \text{Ch. of EVC} &= 780 + 100 = 880\ \text{m}\\ \text{RL of BVC} &= 1267 - \frac{(3.5)}{100}\times 100 = 1263.5\ \text{m}\\ \text{RL of EVC} &= 1267 + \frac{(-0.5)}{100}\times 100 = 1266.5\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = 3.5\%$, $g_2 = -0.5\%$, $L = 200$ m, so the offset is $y = \frac{(-4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1266.5 m) equals the RL of EVC (1266.5 m), and the curve at the IP chainage lies 1 m below the IP, at RL 1266 m.

Step 4: Highest point

At the highest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{-3.5\times 200}{-4} = 175$ m from BVC.

• Chainage = 680 + 175 = 855 m
• RL = 1263.5 + (3.5/100)(175) + (-4)(175)²/(200 × 200) = 1266.562 m

Answer: $L = 200$ m, BVC at 680 m (RL 1263.5 m), EVC at 880 m (RL 1266.5 m); curve levels as tabulated.

Data: $g_1 = +3.5\% = +0.035$, $g_2 = -2.75\% = -0.0275$ (summit curve), IP at chainage 2800.00 m, RL 1470.00 m. The curve must pass through RL 1468.75 m at chainage 2800.00 m, i.e. at the IP vertical.

Length of the curve

The offset from the IP to the curve at mid-point is $e = 1470.00 - 1468.75 = 1.25$ m. For an equal-tangent parabolic curve

$$e = \frac{(g_1 - g_2)L}{8} \quad\Rightarrow\quad L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.25}{0.035 + 0.0275} = 160.00\ \text{m}$$

Each tangent length $= L/2 = 80.00$ m.

Elements

• PVC chainage $= 2800 - 80 = 2720.00$ m; PVT chainage $= 2880.00$ m
• RL of PVC $= 1470.00 - 0.035\times 80 = 1467.200$ m; RL of PVT $= 1470.00 - 0.0275\times 80 = 1467.800$ m
• $r = \dfrac{g_2 - g_1}{L} = -0.0003906$ per m

Levels at the full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1 x + \frac{r}{2}x^2$$

Highest point

$$x = \frac{-g_1}{r} = \frac{0.035}{0.0003906} = 89.60\ \text{m}$$

Chainage $= 2720.00 + 89.60 = 2809.60$ m, and RL $= 1467.200 + 0.035(89.60) + \frac{-0.0003906}{2}(89.60)^2 = 1468.768$ m.

Check: level at the IP chainage (x = 80 m) $= 1468.750$ m = 1468.750 m as required.

Answer: L = 160 m; highest point at chainage 2809.60 m, RL 1468.768 m.

Chainage of IP $= 2+030$ km $= 2030$ m.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 1518.3 - 1517.6 = 0.7\ \text{m}\ \Rightarrow\ |e| = 0.7\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.7}{|4.75-(1.25)|} = \frac{560}{3.5} = 160\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 80$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 2030 - 80 = 1950\ \text{m}\\ \text{Ch. of EVC} &= 2030 + 80 = 2110\ \text{m}\\ \text{RL of BVC} &= 1517.6 - \frac{(1.25)}{100}\times 80 = 1516.6\ \text{m}\\ \text{RL of EVC} &= 1517.6 + \frac{(4.75)}{100}\times 80 = 1521.4\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = 1.25\%$, $g_2 = 4.75\%$, $L = 160$ m, so the offset is $y = \frac{(3.5)x^2}{200\times 160}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1521.4 m) equals the RL of EVC (1521.4 m), and the curve at the IP chainage lies 0.7 m above the IP, at RL 1518.3 m.

Step 4: Highest and lowest points

Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1516.6 m) and the highest level is at the EVC (RL 1521.4 m).

Answer: $L = 160$ m, BVC at 1950 m (RL 1516.6 m), EVC at 2110 m (RL 1521.4 m); curve levels as tabulated.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a summit curve, so the curve lies below the IP:

$$e = 1474.5 - 1475 = -0.5\ \text{m}\ \Rightarrow\ |e| = 0.5\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.5}{|3-(5)|} = \frac{400}{2} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a summit (crest) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 3500 - 100 = 3400\ \text{m}\\ \text{Ch. of EVC} &= 3500 + 100 = 3600\ \text{m}\\ \text{RL of BVC} &= 1475 - \frac{(5)}{100}\times 100 = 1470\ \text{m}\\ \text{RL of EVC} &= 1475 + \frac{(3)}{100}\times 100 = 1478\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = 5\%$, $g_2 = 3\%$, $L = 200$ m, so the offset is $y = \frac{(-2)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1478 m) equals the RL of EVC (1478 m), and the curve at the IP chainage lies 0.5 m below the IP, at RL 1474.5 m.

Step 4: Highest and lowest points

Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1470 m) and the highest level is at the EVC (RL 1478 m).

Answer: $L = 200$ m, BVC at 3400 m (RL 1470 m), EVC at 3600 m (RL 1478 m); curve levels as tabulated.

The second grade is taken as written, $-35\%$. (If it was meant to be $-3.5\%$, the same method gives $L = 800\times 2.5/0.5 = 4000$ m.)

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a summit curve, so the curve lies below the IP:

$$e = 2474.5 - 2477 = -2.5\ \text{m}\ \Rightarrow\ |e| = 2.5\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 2.5}{|-35-(-3)|} = \frac{2000}{32} = 62.5\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a summit (crest) curve with equal tangent lengths $L/2 = 31.25$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 4500 - 31.25 = 4468.75\ \text{m}\\ \text{Ch. of EVC} &= 4500 + 31.25 = 4531.25\ \text{m}\\ \text{RL of BVC} &= 2477 - \frac{(-3)}{100}\times 31.25 = 2477.938\ \text{m}\\ \text{RL of EVC} &= 2477 + \frac{(-35)}{100}\times 31.25 = 2466.062\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -3\%$, $g_2 = -35\%$, $L = 62.5$ m, so the offset is $y = \frac{(-32)x^2}{200\times 62.5}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (2466.062 m) equals the RL of EVC (2466.062 m), and the curve at the IP chainage lies 2.5 m below the IP, at RL 2474.5 m.

Step 4: Highest and lowest points

Both grades are falling, so the grade never becomes zero within the curve and there is no turning point; the highest level is at the BVC (RL 2477.938 m) and the lowest level is at the EVC (RL 2466.062 m).

Answer: $L = 62.5$ m, BVC at 4468.75 m (RL 2477.938 m), EVC at 4531.25 m (RL 2466.062 m); curve levels as tabulated.

A composite curve is a circular curve with a transition (spiral) curve at each end. The deflection angle method sets out the transition curves from T1 and T2 and the central circular curve from the junction points.

                 IP
                /  \
               /    \
      T1 ~~~S1 ______ S2~~~ T2
          (spiral) (arc) (spiral)
         |<-L->|<-Lc->|<-L->|

Elements to compute

$R$ = radius of the circular arc, $L$ = length of transition (from $L = V^3/(\alpha R)$), $\Delta$ = deflection angle.

• Shift: $s = \dfrac{L^2}{24R}$
• Spiral angle: $\phi_s = \dfrac{L}{2R}$ rad
• Tangent length: $T_s = (R + s)\tan\dfrac{\Delta}{2} + \dfrac{L}{2}$
• Central angle of the circular arc: $\Delta_c = \Delta - 2\phi_s$; length $L_c = R\Delta_c$ (in radians)
• Chainages: T1 = IP − $T_s$; S1 = T1 + L; S2 = S1 + $L_c$; T2 = S2 + L

Setting out

1. First transition (from T1). Set the theodolite at T1 with zero reading on the IP. For a peg at distance $l$ along the curve (cubic spiral $y = x^3/6RL$):

$$\delta = \frac{l^2}{6RL}\ \text{rad} = \frac{573\,l^2}{RL}\ \text{minutes}$$

The pegs are set out by turning angle $\delta$ and measuring the chord (taken equal to the arc). The last peg at $l = L$ is S1 with $\delta_L = \phi_s/3$. Check the position of S1 from the data.

2. Circular curve (from S1). Shift the instrument to S1. Take a back-sight on T1. The angle between the chord S1T1 and the common tangent at S1 is $\phi_s - \phi_s/3 = \dfrac{2\phi_s}{3}$. So, after the back-sight, plunge the telescope and turn it through $\dfrac{2\phi_s}{3}$ to get the direction of the tangent at S1, and set the horizontal circle to zero there. Then set out the arc by Rankine's deflection angles from this tangent:

$$\delta = \frac{1718.9\,c}{R}\ \text{minutes}$$

for a chord $c$ (first sub-chord, full chords, last sub-chord). The cumulative angle at S2 must equal $\Delta_c/2$, which checks the work.

3. Second transition (from T2). Set up at T2, back-sight the IP (the tangent), and set out pegs backwards from T2 towards S2 using the same deflection angles $\delta = 573\,l^2/(RL)$ minutes, where $l$ is the distance of the peg from T2. The pegs should meet the circular curve at S2.

Check: The pegs of the circular curve from S1 and the pegs of the second transition from T2 must meet at S2.

The method needs only a theodolite and a tape, but it is slow because of the shifting of the instrument at the junction points.

The question lists the data with elevation and chainage interchanged. Taken as: IP at chainage 2600 m, RL 1300 m, and the curve passes through RL 1301 m at the IP chainage.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 1301 - 1300 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 2600 - 100 = 2500\ \text{m}\\ \text{Ch. of EVC} &= 2600 + 100 = 2700\ \text{m}\\ \text{RL of BVC} &= 1300 - \frac{(-3.5)}{100}\times 100 = 1303.5\ \text{m}\\ \text{RL of EVC} &= 1300 + \frac{(0.5)}{100}\times 100 = 1300.5\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -3.5\%$, $g_2 = 0.5\%$, $L = 200$ m, so the offset is $y = \frac{(4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1300.5 m) equals the RL of EVC (1300.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 1301 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175$ m from BVC.

• Chainage = 2500 + 175 = 2675 m
• RL = 1303.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1300.438 m

Answer: $L = 200$ m, BVC at 2500 m (RL 1303.5 m), EVC at 2700 m (RL 1300.5 m); curve levels as tabulated.

Transition curve (cubic spiral)

A transition curve connects a straight with a circular curve so that the radius changes gradually from $\infty$ to $R$. The radius at a distance $l$ from the start is inversely proportional to $l$:

$$\rho \propto \frac{1}{l} \quad\Rightarrow\quad \rho\, l = RL$$

where $R$ is the radius of the circular curve and $L$ is the length of the transition curve.

        y
        |        . P(x, y)
        |     .  
        |  .    deflection angle = delta
   T1 --+-----------------------> x   (tangent)

Derivation

Take T1 (start of the curve) as the origin and the tangent as the x-axis. For a point P at a distance $l$ along the curve, the tangent angle (spiral angle) $\phi$ is given by $d\phi = \dfrac{dl}{\rho} = \dfrac{l\,dl}{RL}$:

$$\phi = \int_0^l \frac{l\,dl}{RL} = \frac{l^2}{2RL}$$

The coordinates of P are, for small $\phi$ ($\cos\phi \approx 1$, $\sin\phi \approx \phi$):

$$dx = dl,\quad dy = \phi\,dl = \frac{l^2}{2RL}\,dl$$ $$x \approx l, \qquad y = \int_0^l \frac{l^2}{2RL}\,dl = \frac{l^3}{6RL}$$

This is the equation of a cubic parabola.

The deflection angle $\delta$ at T1 between the tangent and the line T1P:

$$\tan\delta = \frac{y}{x} = \frac{l^3}{6RL\cdot l} = \frac{l^2}{6RL}$$

As $\delta$ is small, $\tan\delta \approx \delta$ (in radians):

$$\delta = \frac{l^2}{6RL}\ \text{radians}$$

Converting radians to minutes: $1\ \text{rad} = \dfrac{180}{\pi}\times 60 = 3437.75'$:

$$\delta = \frac{3437.75}{6}\cdot\frac{l^2}{RL} = 572.96\,\frac{l^2}{RL} \approx \boxed{\frac{573\,l^2}{RL}\ \text{minutes}}$$

Hence proved.

At the end of the transition ($l = L$): $\delta_L = \dfrac{L}{6R}$ rad $= \dfrac{\phi_s}{3}$, where $\phi_s = \dfrac{L}{2R}$ is the spiral angle. This is the usual check on the last peg.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 1268 - 1267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 780 - 100 = 680\ \text{m}\\ \text{Ch. of EVC} &= 780 + 100 = 880\ \text{m}\\ \text{RL of BVC} &= 1267 - \frac{(-3.5)}{100}\times 100 = 1270.5\ \text{m}\\ \text{RL of EVC} &= 1267 + \frac{(0.5)}{100}\times 100 = 1267.5\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -3.5\%$, $g_2 = 0.5\%$, $L = 200$ m, so the offset is $y = \frac{(4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1267.5 m) equals the RL of EVC (1267.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 1268 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175$ m from BVC.

• Chainage = 680 + 175 = 855 m
• RL = 1270.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1267.438 m

Answer: $L = 200$ m, BVC at 680 m (RL 1270.5 m), EVC at 880 m (RL 1267.5 m); curve levels as tabulated.

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 50$ km/h = 13.889 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

i) Length of transition curve

$$L_s = \frac{v^3}{C R} = \frac{13.889^3}{0.5\times 300} = 17.86\ \text{m}$$

Adopt $L_s = 20$ m (rounded up to a multiple of 5 m).

ii) Super-elevation

Equilibrium super-elevation for the full speed, taking the road width $B$ as the distance between the outer and inner edges:

$$\frac{h}{B} = \frac{v^2}{gR} = \frac{13.889^2}{9.81\times 300} = 0.0655 \quad\Rightarrow\quad h = 8\times 0.0655 = 0.524\ \text{m}$$

(Slope $1$ in $15.3$. In practice the value is limited to the maximum permitted by the road code, e.g. 7 %, and the balance is taken by side friction.) The super-elevation is attained gradually over $L_s$ so that the outer edge rises uniformly from TS to SC.

iii) Chainages of tangent and junction points

$$\begin{aligned} s &= \frac{L_s^2}{24R} = 0.056\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 1.910^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 183.24\ \text{m}\\ L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 294.16\ \text{m} \end{aligned}$$

iv) Deflection angles for the first two points

Transition curve from TS, $\alpha = 573\,l^2/(R L_s)$ minutes (points at multiples of 10 m):

Circular curve from SC, $\delta = 1718.87\,c/R$ minutes (points at multiples of 20 m):

Answer: $L_s = 20$ m; $h = 0.524$ m; TS = 3421.76, SC = 3441.76, CS = 3735.92, ST = 3755.92 m; deflection angles as tabulated.

Why a parabola?

A parabola is used for vertical curves because it gives a constant rate of change of grade along the curve, so the change in slope is uniform and riding is smooth. Its equation is simple ($y = ax^2 + bx$), offsets from the tangents vary as the square of the distance and are very easy to compute, the curve is symmetrical about the IP for equal tangents, and a true circle and a parabola are almost identical for the very flat curves used on roads.

Computation

Chainage of IP $= 1+780$ km $= 1780$ m (the kilometre part has been taken as 1 km and the curve is worked in the local chainage 1780 m).

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 3268 - 3267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 1780 - 100 = 1680\ \text{m}\\ \text{Ch. of EVC} &= 1780 + 100 = 1880\ \text{m}\\ \text{RL of BVC} &= 3267 - \frac{(-3.5)}{100}\times 100 = 3270.5\ \text{m}\\ \text{RL of EVC} &= 3267 + \frac{(0.5)}{100}\times 100 = 3267.5\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -3.5\%$, $g_2 = 0.5\%$, $L = 200$ m, so the offset is $y = \frac{(4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (3267.5 m) equals the RL of EVC (3267.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 3268 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175$ m from BVC.

• Chainage = 1680 + 175 = 1855 m
• RL = 3270.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 3267.438 m

Answer: $L = 200$ m, BVC at 1680 m (RL 3270.5 m), EVC at 1880 m (RL 3267.5 m); curve levels as tabulated.

Chainage of IP $= 1+780$ km $= 1780$ m. Peg interval is not stated; the usual 30 m is used.

An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve.

Step 1: Length of the curve

The curve is a sag curve, so the curve lies above the IP:

$$e = 1267.7 - 1267 = 0.7\ \text{m}\ \Rightarrow\ |e| = 0.7\ \text{m}$$

For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence

$$L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.7}{|4.75-(1.25)|} = \frac{560}{3.5} = 160\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 80$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 1780 - 80 = 1700\ \text{m}\\ \text{Ch. of EVC} &= 1780 + 80 = 1860\ \text{m}\\ \text{RL of BVC} &= 1267 - \frac{(1.25)}{100}\times 80 = 1266\ \text{m}\\ \text{RL of EVC} &= 1267 + \frac{(4.75)}{100}\times 80 = 1270.8\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = 1.25\%$, $g_2 = 4.75\%$, $L = 160$ m, so the offset is $y = \frac{(3.5)x^2}{200\times 160}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1270.8 m) equals the RL of EVC (1270.8 m), and the curve at the IP chainage lies 0.7 m above the IP, at RL 1267.7 m.

Step 4: Highest and lowest points

Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1266 m) and the highest level is at the EVC (RL 1270.8 m).

Answer: $L = 160$ m, BVC at 1700 m (RL 1266 m), EVC at 1860 m (RL 1270.8 m); curve levels as tabulated.

Data: deflection angle $\Delta = 60^\circ$, design speed $V = 40$ km/h $= 11.111$ m/s, centrifugal ratio $P/W = 1/8$, rate of change of radial acceleration $\alpha = 0.30$ m/s$^3$, chainage of IP = 1+030 m (1030.00 m), peg interval 20 m.

1. Radius, length of transition and shift

Centrifugal ratio $= \dfrac{V^2}{gR}$:

$$R = \frac{V^2}{g\times(1/8)} = \frac{11.111^2}{9.81/8} = 100.68\ \text{m}$$ $$L = \frac{V^3}{\alpha R} = \frac{11.111^3}{0.30\times 100.68} = 45.42\ \text{m}$$ $$s = \frac{L^2}{24R} = \frac{45.42^2}{24\times 100.68} = 0.854\ \text{m}$$

2. Tangent length and spiral angle

$$T_s = (R + s)\tan\frac{\Delta}{2} + \frac{L}{2} = (100.68 + 0.854)\tan 30^\circ + 22.71 = 81.33\ \text{m}$$ $$\phi_s = \frac{L}{2R}\ \text{rad} = 12°55'24", \qquad \Delta_c = \Delta - 2\phi_s = 34°09'13"$$ $$L_c = R\Delta_c = 100.68\times 0.59609 = 60.01\ \text{m}$$

3. Chainages

4. Setting out the circular curve by Rankine's method (peg interval 20 m)

Set the theodolite at S1, backsight T1, and set the tangent to the circular curve (turn the telescope to the common tangent, which makes $2\phi_s/3$ with the chord S1T1 for a cubic spiral). Then set out the circular curve from S1 by deflection angles from this tangent:

$$\delta = \frac{1718.9\,C}{R}\ \text{minutes}$$

The first chord is a sub-chord up to the next 20 m peg.

Check: the final cumulative deflection $= \Delta_c/2 = 17°04'36"$, and it equals 17°04'37".

Answer: R = 100.68 m, L = 45.42 m, shift = 0.854 m, $T_s$ = 81.33 m, $L_c$ = 60.01 m; the Rankine deflection angles are tabulated above.

Data: width of road $B = 8$ m, $\Delta = 60^\circ$, $R = 330$ m, chainage of IP = 3605.0 m, $\alpha = 0.5$ m/s$^3$, $V = 50$ km/h $= 13.889$ m/s, peg interval 5 m.

(a) Length of the transition curve

$$L = \frac{V^3}{\alpha R} = \frac{13.889^3}{0.5\times 330} = 16.24\ \text{m}$$

(b) Superelevation

For full balancing of the centrifugal force, $\tan\theta = e = \dfrac{V^2}{gR}$:

$$e = \frac{13.889^2}{9.81\times 330} = 0.0596 \approx 1\text{ in }16.8$$

Rise of the outer edge over the width of the road:

$$h = e\times B = 0.0596\times 8 = 0.477\ \text{m}$$

(c) Chainages of the junction points

$$s = \frac{L^2}{24R} = 0.0333\ \text{m},\qquad T_s = (R+s)\tan 30^\circ + \frac{L}{2} = (330 + 0.0333)(0.57735) + 8.119 = 198.664\ \text{m}$$ $$\phi_s = \frac{L}{2R} = 1°24'35",\qquad \Delta_c = 60^\circ - 2\phi_s = 57°10'51",\qquad L_c = R\Delta_c = 329.338\ \text{m}$$

(d) Layout of the transition curve by deflection angles (5 m pegs, from T1)

$$\delta = \frac{573\,l^2}{RL}\ \text{minutes}$$

At the end of the curve $\delta = \phi_s/3 = 0°28'12"$. The second transition is set out in the same way from T2 towards S2 (measured backwards along the tangent), and the circular curve from S1 by Rankine's deflection angles $\delta = 1718.9\,c/R$ minutes.

Answer: L = 16.24 m; superelevation = 0.477 m (1 in 16.8); T1 = 3406.34 m, S1 = 3422.57 m, S2 = 3751.91 m, T2 = 3768.15 m.

A parabolic vertical curve with a constant rate of change of grade is set out using the tangent-offset method.

Step 1: Length of the curve

Rate of change of grade $r = 0.1\%$ per 20 m, algebraic change in grade $= 0.5 - (-1) = 1.5\%$.

$$L = \frac{|g_2-g_1|}{r}\times 20 = \frac{1.5}{0.1}\times 20 = 300\ \text{m}$$

Step 2: BVC, EVC and their levels

The curve is a sag (valley) curve with equal tangent lengths $L/2 = 150$ m.

$$\begin{aligned} \text{Ch. of BVC} &= 1500 - 150 = 1350\ \text{m}\\ \text{Ch. of EVC} &= 1500 + 150 = 1650\ \text{m}\\ \text{RL of BVC} &= 1250 - \frac{(-1)}{100}\times 150 = 1251.5\ \text{m}\\ \text{RL of EVC} &= 1250 + \frac{(0.5)}{100}\times 150 = 1250.75\ \text{m} \end{aligned}$$

Step 3: RL of curve points

Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %):

$$y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L}$$

with $g_1 = -1\%$, $g_2 = 0.5\%$, $L = 300$ m, so the offset is $y = \frac{(1.5)x^2}{200\times 300}$ and pegs are taken at multiples of 30 m.

Check: the last curve RL (1250.75 m) equals the RL of EVC (1250.75 m), and the curve at the IP chainage lies 0.562 m above the IP, at RL 1250.562 m.

Step 4: Lowest point

At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{1\times 300}{1.5} = 200$ m from BVC.

• Chainage = 1350 + 200 = 1550 m
• RL = 1251.5 + (-1/100)(200) + (1.5)(200)²/(200 × 300) = 1250.5 m

Answer: $L = 300$ m, BVC at 1350 m (RL 1251.5 m), EVC at 1650 m (RL 1250.75 m); curve levels as tabulated.

Data: carriageway width $B = 8$ m, $\Delta = 60^\circ$, $R = 300$ m, chainage of IP = 3605 m, $v = 70$ km/h = 19.444 m/s, rate of change of radial acceleration $C = 0.5$ m/s³.

i) Length of transition curve

$$L_s = \frac{v^3}{C R} = \frac{19.444^3}{0.5\times 300} = 49.01\ \text{m}$$

Adopt $L_s = 50$ m (rounded up to a multiple of 5 m).

ii) Super-elevation

Equilibrium super-elevation for the full speed, taking the road width $B$ as the distance between the outer and inner edges:

$$\frac{h}{B} = \frac{v^2}{gR} = \frac{19.444^2}{9.81\times 300} = 0.1285 \quad\Rightarrow\quad h = 8\times 0.1285 = 1.028\ \text{m}$$

(Slope $1$ in $7.8$. In practice the value is limited to the maximum permitted by the road code, e.g. 7 %, and the balance is taken by side friction.) The super-elevation is attained gradually over $L_s$ so that the outer edge rises uniformly from TS to SC.

iii) Chainages of tangent and junction points

$$\begin{aligned} s &= \frac{L_s^2}{24R} = 0.347\ \text{m},\quad \theta_s = \frac{L_s}{2R} = 4.775^\circ\\ T_s &= (R+s)\tan\frac{\Delta}{2} + \frac{L_s}{2} = 198.41\ \text{m}\\ L_c &= \frac{\pi R}{180}(\Delta - 2\theta_s) = 264.16\ \text{m} \end{aligned}$$

iv) Deflection angles for the first two points

Transition curve from TS, $\alpha = 573\,l^2/(R L_s)$ minutes (points at multiples of 10 m):

Circular curve from SC, $\delta = 1718.87\,c/R$ minutes (points at multiples of 20 m):

Answer: $L_s = 50$ m; $h = 1.028$ m; TS = 3406.59, SC = 3456.59, CS = 3720.75, ST = 3770.75 m; deflection angles as tabulated.