Question 7 of 7: Three-Dimensional Conformal Coordinate Transformation
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.
Given. Three common points (1, 2, 3) known in both a measured (arbitrary) frame and the control (ground) frame, all coordinates at $\sigma=\pm0.05$ m, and three further points (4, 5, 6) known only in the measured frame.
Find. The seven transformation parameters and the control-frame coordinates of points 4, 5 and 6.
Figure 7.1 — The seven-parameter 3-D conformal transformation maps the measured frame (blue axes, common points 1–3) onto the control frame (red) by one scale $S$, three rotations $(\omega,\varphi,\kappa)$ and three translations $T$.
Check (source defect). The control $Y$ of point 2 is a duplicate of its $X$ (both print as 371.46) and is clearly corrupt. Point 2 still supplies two valid equations ($X_2,Z_2$), so with the six equations of points 1 and 3 the seven parameters are over-determined (redundancy 1) without $Y_2$. Re-solving with $Y_2$ filled at its fitted value changes the transformed coordinates of points 4–6 by less than 1 mm, so the results below are independent of the corrupt value (the fitted $Y_2\approx628.8$ m is what the entry should read).
Approach. The conformal model $X=S\,R(\omega,\varphi,\kappa)\,x+T$ is non-linear in the rotations, so linearise about approximate parameters and iterate the least-squares correction (Gauss–Newton) until it converges, then apply the parameters to points 4–6.
Transformation model. For each common point,
$$\begin{bmatrix}X\\Y\\Z\end{bmatrix}=S\,R(\omega,\varphi,\kappa)\begin{bmatrix}x\\y\\z\end{bmatrix}+\begin{bmatrix}T_x\\T_y\\T_z\end{bmatrix}$$
with unknowns $S,\omega,\varphi,\kappa,T_x,T_y,T_z$ (seven).
Initial approximations. Scale $S_0=1$, rotations $0$, and translations from the centroid difference $T_0\approx(-221,\,+4,\,-1.4)$ m. The near-unit $z\!\to\!Z$ mapping confirms a small rotation and near-unit scale.
Linearise and iterate. Forming the Jacobian $J=\partial(X,Y,Z)/\partial(\text{params})$ at each iteration and solving $\delta=(J^{\mathsf T}J)^{-1}J^{\mathsf T}r$ (equal weights, since every $\sigma=0.05$ m) converges in a few iterations to
$$\boxed{S=1.003558,\quad \omega=-0.224^\circ,\ \varphi=+0.657^\circ,\ \kappa=+2.488^\circ}$$
$$\boxed{T=(-202.415,\ -31.665,\ +7.540)\ \text{m}}$$
Fit quality. The standard error of fit (one degree of freedom) is $\sqrt{r^{\mathsf T}r/1}\approx0.19$ m. Because general least squares gives both coordinate sets $\sigma=0.05$ m, each misclosure has a-priori $\sigma=0.05\sqrt{1+S^2}\approx0.071$ m, so the reference standard deviation is $s_0\approx0.19/0.071\approx2.7$ — larger than 1, but with only one degree of freedom it is a weak statistic; the common points agree to roughly two decimetres.
Transform points 4, 5, 6. Applying the parameters to the measured coordinates:
$$\boxed{P_4=(409.490,\ 437.734,\ 81.134)}$$
$$\boxed{P_5=(453.664,\ 296.726,\ 97.334)},\quad \boxed{P_6=(401.967,\ 708.813,\ 100.842)\ \text{m}}$$
