Weighted Adjustment of a Plane Triangle: 18-Geom-A2 Adjustment of Observations, May 2014, Q2

Question 2: Weighted Adjustment of a Plane Triangle (20 marks)

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Given. Four independent readings of each interior angle of a plane triangle. The three angles must satisfy the single geometric condition $A+B+C=180^\circ$.

Find. Relative weights for A, B, C from their scatter, and the weighted, condition-adjusted angles to the nearest second.

Approach. Take each angle's mean; assign weights inversely proportional to the sample variance of the readings; distribute the $180^\circ$ misclosure among the angles in proportion to $1/w$ (the variance), i.e. the least-precise angle absorbs the most correction.

Means of the four sets. Averaging the seconds columns: $$\bar A=38^\circ47'59'',\quad \bar B=71^\circ22'18'',\quad \bar C=69^\circ50'15''$$
Sample variances (seconds$^2$). Using $s^2=\tfrac{1}{n-1}\sum(x-\bar x)^2$ on the four readings of each angle, $$s_A^2=134.7,\qquad s_B^2=50.7,\qquad s_C^2=124.0$$
Relative weights. Weight is inversely proportional to variance, $w_i\propto1/s_i^2$; normalising to the smallest, $$w_A:w_B:w_C=1.00:2.66:1.09$$ Angle B, being the most consistent, is the most heavily weighted.
Misclosure. $\;\bar A+\bar B+\bar C=180^\circ00'32''$, so the misclosure is $$w_0=+32''\quad(\text{to be removed}).$$
Weighted corrections. The correction to each angle is proportional to its variance (reciprocal of weight): $$v_i=-w_0\,\frac{s_i^2}{\sum s_j^2}\;\Rightarrow\; \boxed{v_A=-13.9'',\ v_B=-5.2'',\ v_C=-12.8''}$$ which sum to $-32.0''$ as required.
Adjusted angles. Applying the corrections: $$\boxed{A=38^\circ47'45'',\quad B=71^\circ22'13'',\quad C=69^\circ50'02''}$$ Their sum is exactly $180^\circ00'00''$.

Angle	Mean	$s^2$ (″$^2$)	Rel. weight	Corr. $v$	Adjusted
A	38°47'59″	134.7	1.00	−13.9″	38°47'45″
B	71°22'18″	50.7	2.66	−5.2″	71°22'13″
C	69°50'15″	124.0	1.09	−12.8″	69°50'02″
Σ					180°00'00″

Concept & Theory

Topic. Weighted adjustment of a single geometric condition (angle sum of a triangle).

Key relations. Weight $w_i\propto1/\sigma_i^2$; standard deviation of a mean $\sigma_{\bar x}=s/\sqrt{n}$; for one condition, correction $v_i=-w_0\,(1/w_i)\big/\sum(1/w_j)$.

Why this works. Minimising $\sum w_i v_i^2$ subject to $\sum v_i=-w_0$ (a Lagrange problem) yields corrections proportional to $1/w_i=\sigma_i^2$: the more scattered an angle, the more of the misclosure it should soak up. Because all three angles were read the same number of times, using $1/s^2$ is equivalent to using $1/\sigma_{\bar x}^2$.

Common pitfall. Making corrections proportional to the weight instead of its reciprocal (that over-corrects the best angle), or rounding intermediate seconds so the adjusted angles fail to close on $180^\circ$.

Source. Wolf & Ghilani, Elementary Surveying §3 (theory of errors) and §9 (angle-condition adjustment); Ghilani, Adjustment Computations §10 (weights).

Question 2 of 7: Weighted Adjustment of a Plane Triangle

Question 2: Weighted Adjustment of a Plane Triangle (20 marks)