Question 6 of 7: Circular curve stationing and deflection-angle layout

Question 6: Circular curve stationing and deflection-angle layout (6 + 14 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. A simple circular curve with the point of intersection of the tangents at station 12+78.230, a radius of 500 m and a central (deflection) angle of 86 degrees 28 minutes. Stationing is metric, so a full station is 100 m and the station 12+78.230 denotes a chainage of 1278.230 m. The paper prints “horizintal” and “Detrmine”; these are typographical errors in the original and are reproduced above as printed.

Find. (a) The chainage of the point of curvature and the point of tangency; (b) the deflection angle from the back tangent at the PC to every full station on the curve, in the form a survey party would set out with a total station or theodolite.

Approach. Convert the central angle to decimal degrees, compute the tangent distance and arc length from the standard circular-curve relations, subtract and add them to the PI chainage to place the PC and PT, then compute the deflection angle to each full station as half the central angle subtended by the arc from the PC.

1. Part (a) — convert the central angle. Working in decimal degrees throughout avoids sexagesimal arithmetic errors: $$\Delta = 86^{\circ}28' = 86 + \frac{28}{60} = 86.466667^{\circ} = 1.509128\ \text{rad}$$
2. Compute the tangent distance. The tangent distance is the distance from the PI back to the PC, and forward from the PI to the PT, along the respective tangents: $$T = R\tan\frac{\Delta}{2} = 500\tan(43.233333^{\circ}) = 500(0.940158) = \boxed{470.079\ \text{m}}$$
3. Compute the length of curve. The curve length is the arc, not the chord: $$L = R\,\Delta_{\text{rad}} = 500 \times 1.509128 = \boxed{754.564\ \text{m}}$$ For completeness the long chord is $LC = 2R\sin(\Delta/2) = 684.971$ m, the external distance is $E = R[\sec(\Delta/2)-1] = 186.276$ m and the middle ordinate is $M = R[1-\cos(\Delta/2)] = 135.715$ m.
4. Place the PC. The PC lies one tangent distance back along the back tangent from the PI, so its chainage is the PI chainage less T: $$\text{PC} = 1278.230 - 470.079 = 808.151\ \text{m} \Rightarrow \boxed{\text{PC} = 8+08.151}$$
5. Place the PT. The PT must be found by adding the arc length to the PC chainage, never by adding a second tangent distance to the PI chainage, because chainage runs along the curve once the curve begins: $$\text{PT} = 808.151 + 754.564 = 1562.715\ \text{m} \Rightarrow \boxed{\text{PT} = 15+62.715}$$ Adding $T$ to the PI chainage instead would give 1748.309 m, which is 185.594 m too far; that difference, $2T - L$, is the tangent-versus-arc discrepancy and is the classic error in this calculation.
6. Part (b) — establish the deflection-angle relation. By the inscribed-angle theorem, the angle between the back tangent at the PC and the chord to any point P on the curve is half the central angle subtended by the arc from the PC to P. Writing $\ell$ for that arc length, $$\delta = \frac{1}{2}\cdot\frac{\ell}{R}\ \text{radians} = \frac{90\,\ell}{\pi R}\ \text{degrees}$$ For $R = 500$ m this gives a convenient working rate of $0.0572958^{\circ}$ per metre of arc, or $5^{\circ}43'46''$ per 100 m station.
7. Tabulate the deflection angles at the full stations. The full stations lying on the curve are 9+00 through 15+00. For each, the arc from the PC is $\ell = \text{station} - 808.151$, and $\delta$ follows from Step 6. The results are set out in the table below. The instrument is set over the PC and oriented on the PI along the back tangent; each deflection is then turned from that reference and the chord distance taped or measured to fix the point.
8. Apply the closing check. The total deflection from the PC to the PT must equal exactly half the central angle: $$\delta_{\text{PT}} = \frac{90 \times 754.564}{\pi \times 500} = 43.23333^{\circ} = 43^{\circ}14'00'' = \frac{\Delta}{2}\quad\checkmark$$ This is the standard field check and it closes exactly, confirming the stationing in part (a) and every deflection in part (b).