Question 4 of 7: Circular horizontal curve — stationing and deflection angles

Question 4: Circular horizontal curve — stationing and deflection angles

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 simple circular horizontal curve of radius 500 m turning through 86°, whose tangents intersect at station 12+78.230. Stations are 100 m, so the point of intersection lies at chainage 1 278.230 m.

Find. (a) the stationing of the point of curvature and the point of tangency; (b) the deflection angles, and the chords that go with them, for setting the curve out from the PC at every full station.

Approach. Compute the tangent length and the arc length from R and Δ, chain back from the PI to the PC and forward along the arc to the PT, then use the property that the deflection from the back tangent to any point on the curve is half the central angle subtended by the arc to that point.

1. Part (a) — tangent length. $$T = R \tan\frac{\Delta}{2} = 500 \tan 43^\circ = 500 \times 0.932515 = 466.258\ \text{m}$$
2. Curve (arc) length. $$L = R\,\Delta_{\text{rad}} = \frac{\pi R \Delta}{180} = \frac{\pi (500)(86)}{180} = 750.492\ \text{m}$$ For completeness the remaining curve elements are the long chord $LC = 2R\sin(\Delta/2) = 681.998$ m, the external distance $E = R[\sec(\Delta/2) - 1] = 183.664$ m and the middle ordinate $M = R[1 - \cos(\Delta/2)] = 134.323$ m.
3. Station of the PC. The PC lies one tangent length back along the back tangent from the PI: $$\text{Sta PC} = \text{Sta PI} - T = 1\,278.230 - 466.258 = 811.972\ \text{m}\;\Rightarrow\; \boxed{\text{PC} = 8+11.972}$$
4. Station of the PT. Chainage runs along the alignment, which follows the arc and not the tangents, so the PT is reached by adding the curve length to the PC: $$\text{Sta PT} = \text{Sta PC} + L = 811.972 + 750.492 = 1\,562.464\ \text{m}\;\Rightarrow\; \boxed{\text{PT} = 15+62.464}$$ Adding a second tangent length to the PI instead would put the PT at 1 744.488 m, 182.024 m too far ahead; the two tangents are always longer than the arc they enclose, and that difference is the curve's tangent correction.
5. Part (b) — the deflection-angle rule. By the inscribed-angle theorem, the angle between the back tangent at the PC and the chord to any point on the curve is half the central angle subtended by the intervening arc. For an arc $\ell$ measured from the PC, $$\delta = \frac{1}{2}\cdot\frac{\ell}{R}\ \text{(radians)} = \frac{90\,\ell}{\pi R}\ \text{(degrees)} = 0.0572958\,\ell$$ with $\ell$ in metres for $R = 500$ m. The check on the whole computation is that the deflection to the PT must come out at exactly $\Delta/2 = 43^\circ 00' 00''$.
6. Deflection per full station. Successive full stations are 100 m of arc apart, so after the first sub-arc every deflection advances by the same increment: $$\delta_{100} = 0.0572958 \times 100 = 5.729578^\circ = 5^\circ 43' 46.5''$$ That constant increment is the field check: the instrument person accumulates it, and any station whose computed deflection breaks the pattern has been mis-computed.
7. The first and last sub-arcs. The PC falls at 8+11.972, so the arc from the PC to the first full station, 9+00, is only $900 - 811.972 = 88.028$ m; likewise the arc from the last full station, 15+00, to the PT is $1\,562.464 - 1\,500 = 62.464$ m. These two sub-arcs are what make the first and last deflections irregular.
8. Tabulate the deflections and their chords. Each deflection is turned from the back tangent with the instrument at the PC, and the point is fixed by measuring the corresponding chord $$C = 2R\sin\delta$$ from the PC. The results are given in the table below; every chord is slightly shorter than its arc, as it must be.

The table closes exactly on 43°00'00", which is half the 86° deflection, and the last chord closes on 681.998 m, which is the long chord computed independently in step 2. Those two closures together confirm the stationing, the arc length and every deflection in between. In the field, a curve this long would not in practice be set out entirely from the PC — at 750 m the far end is a long sight and the accumulated chord error grows — so the usual procedure is to set out to about the middle, move the instrument to a station on the curve, backsight the PC with the telescope inverted, and continue with the remaining deflections referred to the new setup.