Question 1 of 7: Spiralled Horizontal Curve and Superelevation Development

Question 1: Spiralled Horizontal Curve and Superelevation Development (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.

Find. Whether R = 800 m is an acceptable radius at 80 km/h; the spiral parameter A, the transition (spiral) length and the design superelevation rate; the superelevation development diagram drawn at 1:500 horizontal and 1:5 vertical with the pavement rotated about the centreline; and the chainages of the TS, SC and CS (and, for completeness, the ST).

Approach. Test the chosen radius against the tabulated minimum for 80 km/h at e_max = 0.06, read the design superelevation and the minimum spiral parameter from the same row of TAC Table B.3.1.4b, convert the spiral parameter to a spiral length, build the spiral geometry (p, k, T_s), and then chain forward from the PI; finally distribute the superelevation over the spiral with the tangent runout placed ahead of the TS.

1. Part (a) — test the chosen radius against the minimum for 80 km/h. The point-mass equation that underlies the TAC table balances the centripetal demand between superelevation and side friction, $$R_{\min} = \frac{V^{2}}{127\,(e_{\max} + f)}$$ where V is the design speed in km/h and f is the TAC maximum side-friction factor, 0.14 at 80 km/h. Substituting, $$R_{\min} = \frac{80^{2}}{127\,(0.06 + 0.14)} = \frac{6400}{25.4} = 252.0\ \text{m}$$ which is exactly the value Table B.3.1.4b prints at the foot of the 80 km/h column, minimum R = 250 m. The chosen radius is more than three times that: $$\boxed{R = 800\ \text{m} \ge R_{\min} = 250\ \text{m} \quad\text{(ratio 3.2)} }$$ so the radius is appropriate and no substitute is required. It is worth noting the other end of the check as well: the table still assigns R = 800 m a real superelevation rate rather than "NC" or "RC", so the curve is flat enough to be safe but sharp enough to need a designed cross-fall.
2. Part (b) — read the design superelevation and spiral parameter from the table. Entering Table B.3.1.4b at the 80 km/h column and the R = 800 m row gives $$e = 0.036\ \text{m/m}, \qquad A = 175\ \text{m}$$ The two-lane and the three/four-lane columns both print 175 m at this radius, so the two-lane value applies without interpolation. The rate is comfortably below the 0.06 maximum, as expected for a radius 3.2 times the minimum.
3. Convert the spiral parameter into a spiral length. The note under the table defines the parameter through $$L_{s} = \frac{A^{2}}{R} = \frac{175^{2}}{800} = \frac{30\,625}{800}$$ $$\boxed{L_{s} = 38.281\ \text{m} \approx 38.3\ \text{m}}$$ This is the minimum spiral; the table's note permits longer transitions, and a designer would normally round up to 40 m for staking. The calculations below carry the tabulated minimum so that the stationing is reproducible.
4. Build the spiral geometry. The spiral angle, the shift of the circular arc away from the tangent and the abscissa of the shifted arc's origin are $$\theta_{s} = \frac{L_{s}}{2R}, \qquad p = \frac{L_{s}^{2}}{24R}, \qquad k = \frac{L_{s}}{2} - \frac{L_{s}^{3}}{240R^{2}}$$ Substituting L_s = 38.281 m and R = 800 m, $$\theta_{s} = 0.023926\ \text{rad} = 1.3708^{\circ}, \qquad p = 0.0763\ \text{m}, \qquad k = 19.140\ \text{m}$$ The shift is only 76 mm, which is why the spiral barely changes the tangent length — but it must still be carried, because the stationing is quoted to the millimetre.
5. Compute the tangent distance from the PI. For a spiral–circular–spiral curve, $$T_{s} = (R + p)\tan\frac{\Delta}{2} + k = (800.076)\tan 10^{\circ} + 19.140$$ $$\boxed{T_{s} = 141.075 + 19.140 = 160.215\ \text{m}}$$
6. Compute the circular arc between the two spirals. Each spiral consumes θ_s of the total deflection, so the arc subtends Δ − 2θ_s: $$L_{c} = R\,(\Delta - 2\theta_{s}) = 800\,(0.349066 - 0.047851) = 240.971\ \text{m}$$ and the whole curve is $$L_{\text{total}} = 2L_{s} + L_{c} = 2(38.281) + 240.971 = 317.534\ \text{m}$$
7. Part (d) — chain the stations forward. Working back from the PI to the TS and then forward along the curve, $$\text{TS} = \text{PI} - T_{s} = 50\,000.000 - 160.215 = 49\,839.785\ \text{m}$$ $$\text{SC} = \text{TS} + L_{s}, \qquad \text{CS} = \text{SC} + L_{c}, \qquad \text{ST} = \text{CS} + L_{s}$$ With 1000 m stations these become $$\boxed{\text{TS } 49{+}839.785, \quad \text{SC } 49{+}878.066, \quad \text{CS } 50{+}119.037, \quad \text{ST } 50{+}157.319}$$ Check the closure: ST − TS = 317.534 m, which equals 2L_s + L_c exactly. Note that the ST is not at PI + T_s = 50+160.215; the curve is 2.9 m shorter than twice the tangent distance, and confusing the two is the commonest stationing error on spiralled curves.
8. Part (c) — set out the superelevation runoff and runout. With a spiral present, the superelevation runoff is developed over the spiral itself, so $$L_{r} = L_{s} = 38.281\ \text{m}$$ and full superelevation is reached exactly at the SC. The tangent runout, over which the outside lane is lifted from the normal crown to level, is scaled from the same relative gradient: $$L_{t} = L_{r}\,\frac{e_{NC}}{e_{d}} = 38.281 \times \frac{0.020}{0.036} = 21.267\ \text{m}$$ Because the outside edge rises at one constant rate through both the runout and the runoff, the section becomes a single plane (the reverse-crown point) at the fraction e_NC/e_d = 0.556 of the runoff, that is 21.267 m past the TS.
9. Locate the eight control chainages of the diagram. Working outwards from the TS and inwards from the ST, $$\boxed{\begin{aligned} \text{NC (start of runout)} &= 49{+}818.517 \\ \text{TS (outside lane level)} &= 49{+}839.785 \\ \text{RC (single plane at 2 percent)} &= 49{+}861.052 \\ \text{SC (full } e = 0.036) &= 49{+}878.066 \end{aligned}}$$ and, by symmetry on the departure side, CS 50+119.037, RC 50+136.051, ST 50+157.319 and the end of the runout at 50+178.586.
10. Compute the edge offsets and check the relative gradient. With rotation about the centreline each edge moves by half the width times the cross-fall: $$\Delta h = e\,\frac{w}{2} = 0.036 \times \frac{7.5}{2} = 0.135\ \text{m at full superelevation}$$ and 0.020 × 3.75 = 0.075 m at the normal crown. The rate at which the outside edge climbs relative to the centreline profile is $$\frac{\Delta h}{L_{r}} = \frac{0.135}{38.281} = 0.00353 = 1 \text{ in } 284$$ which is flatter than the TAC limit of 1 in 200 at 80 km/h, so the minimum spiral from the table is long enough to develop the superelevation comfortably and no lengthening is needed.
11. Fix the drawing scales. At 1:500 horizontally the runoff plots as 38.281/500 = 0.0766 m = 76.6 mm, and at 1:5 vertically the 0.135 m edge offset plots as 0.135/5 = 0.0270 m = 27 mm. The vertical exaggeration is therefore 100:1, which is what makes a 0.135 m rotation legible beside a 360 m length of road. Figure 1.2 is drawn to those proportions.

Final Results.