Linearized Observation Equation for a Distance: 18-Geom-A2 Adjustment of Observations, December 2019, Q4

Question 4: Linearized Observation Equation for a Distance (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. A measured distance $\ell_{AB}=132.823$ m and approximate plane coordinates $A_0=(X_A,Y_A)=(1023.151,\,873.018)$, $B_0=(X_B,Y_B)=(1094.310,\,985.163)$. The four coordinates are treated as unknowns to be improved by small corrections $dX_A,dY_A,dX_B,dY_B$.

Find. The distance observation equation linearized about $A_0,B_0$ — its coefficients and constant term.

Figure 4.1 — Distance AB in the plane; the observation equation relates a change in $\ell_{AB}$ to the coordinate corrections at each endpoint.

Approach. The distance is a non-linear function of the coordinates, $\ell=\sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}$; a first-order Taylor expansion about the approximate coordinates gives a linear relation between the coordinate corrections and the observation residual.

Compute the approximate distance and its components. $\Delta X=X_B-X_A=71.159$ m, $\Delta Y=Y_B-Y_A=112.145$ m, hence $$\ell_0=\sqrt{\Delta X^2+\Delta Y^2}=\sqrt{71.159^2+112.145^2}=132.816\ \text{m}.$$
Form the partial derivatives (direction cosines). With $\partial\ell/\partial X_B=\Delta X/\ell_0$ and $\partial\ell/\partial Y_B=\Delta Y/\ell_0$ (and equal-and-opposite at A), $$\frac{\Delta X}{\ell_0}=\frac{71.159}{132.816}=0.5358,\qquad \frac{\Delta Y}{\ell_0}=\frac{112.145}{132.816}=0.8444.$$
Assemble the linearized observation equation. Writing $v$ for the distance residual and $k=\ell_{AB}-\ell_0$ for the constant term, $$-0.5358\,dX_A-0.8444\,dY_A+0.5358\,dX_B+0.8444\,dY_B=k+v,$$ $$k=\ell_{AB}-\ell_0=132.823-132.816=\boxed{+0.007\ \text{m}\ (+7\ \text{mm})}.$$
State the prototype form. Equivalently, in Ghilani’s notation with the coefficients as direction cosines $\alpha=\Delta X/\ell_0$, $\beta=\Delta Y/\ell_0$, $$\boxed{\,-\alpha\,dX_A-\beta\,dY_A+\alpha\,dX_B+\beta\,dY_B=(\ell_{AB}-\ell_0)+v\,}$$ which is one row of the design matrix $A$ for a distance observation.

Term	$dX_A$	$dY_A$	$dX_B$	$dY_B$	constant $k$
Coefficient	−0.5358	−0.8444	+0.5358	+0.8444	+0.007 m

Concept & Theory

Topic. Linearization of a distance observation for coordinate (parametric) adjustment.

Key relations. $\ell=\sqrt{(X_B-X_A)^2+(Y_B-Y_A)^2}$; partials $\partial\ell/\partial X_B=(X_B-X_A)/\ell_0=\cos(\text{azimuth complement})$, etc.; observation equation $\sum(\partial\ell/\partial p)\,dp=(\ell_{\text{obs}}-\ell_0)+v$.

Why this works. Coordinates enter distance non-linearly, so the normal equations cannot be formed directly; expanding to first order about a good approximation replaces the curve by its tangent plane, and the coefficients are simply the direction cosines of the line — the rate at which the length grows as each endpoint slides along X or Y. The constant term is the current misclosure $\ell_{\text{obs}}-\ell_0$, which iteration drives toward zero.

Common pitfall. Dropping the sign reversal between the two endpoints (moving A along $+X$ shortens the line, moving B along $+X$ lengthens it), or forgetting that the coefficients are dimensionless direction cosines, not the raw coordinate differences.

Source. Ghilani, Adjustment Computations §14 (coordinate adjustment of distances) — Eq. for the distance observation equation.

Question 4 of 7: Linearized Observation Equation for a Distance

Question 4: Linearized Observation Equation for a Distance (20 marks)