Question 5 of 7: Best-Fit Parabola for a Vertical Curve
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Geom-A2, Adjustment of Observations and Data Analysis. Three-hour, CLOSED-BOOK exam; any non-communicating calculator permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Least-squares adjustments are carried out in the Canadian survey frame (heights on CGVD2013, horizontal on NAD83(CSRS)); US-foot units are retained in Q4 wherever the printed question uses them.
Reference texts: Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley, 2017); Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981).
Method note. Every weighted least-squares result below was formed from the normal equations $N\hat{x}=A^{\mathsf T}W\,\ell$; the reference (unit-weight) standard deviation is $s_0=\sqrt{v^{\mathsf T}Wv/(n-u)}$ with $n$ observations and $u$ unknowns.
Question 5: Best-Fit Parabola for a Vertical Curve (20 marks)
Given. Nine measured elevations at full stations 10+00 through 18+00 along an existing sag vertical curve.
Find. The least-squares parabola $y=a_2x^2+a_1x+a_0$, the adjusted (on-curve) elevations, and the residuals.
Figure 5.1 — Measured station elevations (dots) and the least-squares best-fit parabola (red). The curve is a sag with its low point near station 14+46.
Approach. Let $x$ be the station count from the start ($x=0$ at 10+00, $x=8$ at 18+00) and fit a second-degree polynomial by least squares (nine observations, three unknowns), then evaluate the curve and residuals at each station.
Set up the observation equations. Each station gives $a_2x_i^2+a_1x_i+a_0=y_i+v_i$. Stacking all nine into $A\hat a=\ell+v$ with $\hat a=[a_2,a_1,a_0]^{\mathsf T}$ (equal weights, since all elevations share the same precision).
Solve the normal equations. $\hat a=(A^{\mathsf T}A)^{-1}A^{\mathsf T}\ell$ gives
$$\boxed{y=0.222511\,x^2-1.983420\,x+51.234545}\qquad(x=\tfrac{\text{sta}-(10{+}00)}{100})$$
Low point. The vertex is at $x=-a_1/(2a_2)=4.457$, i.e. station $14{+}45.7$, elevation $46.81$ m — a genuine sag curve since $a_2>0$.
Adjusted elevations and residuals. Evaluating $y(x_i)$ and $v_i=y(x_i)-y_i^{\text{obs}}$ station by station gives residuals no larger than $61$ mm, listed below. The residuals change sign repeatedly along the curve with no run of one sign at either end, confirming a good parabolic fit with no systematic trend.
