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GELE Adjustment Computations (Least Squares) — Condition Equations and Figure AdjustmentDetailed Explanation

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The Condition Equations and Figure Adjustment chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Geodetic Engineering's scenario-based GELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent GELE Adjustment Computations (Least Squares) papers.

Exam context

For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Adjustment Computations (Least Squares) under a "Core" label, with Condition Equations and Figure Adjustment in the 3rd slot across 5 chapters. GELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Adjustment Computations (Least Squares) questions. Date to watch: September 2026.

Condition Equations and Figure Adjustment - Detailed Explanation

Condition equations and figure adjustment form one of the most practically important topics in Adjustment Computations (Least Squares) for the PRC Geodetic Engineer Licensure Examination. In any surveying network — whether a triangulation figure, a traverse loop, or a leveling network — the measured quantities must satisfy known geometric relationships. When raw observations fail to satisfy these relationships exactly (due to random measurement errors), a misclosure results. The condition-equation method provides a systematic, mathematically sound procedure to distribute this misclosure and force the adjusted values to satisfy the required geometric conditions. This chapter develops the theoretical basis of condition equations, derives practical correction formulas for equal and weighted cases, and provides board-style worked examples covering the triangle angle condition, the horizon (station) condition, and the spherical triangle condition — all of which appear frequently in Philippine geodetic engineering board examinations.

Concepts

Concept of Condition Equations

A condition equation is a mathematical expression that enforces an exact geometric or physical relationship among a set of adjusted observations. In geodetic surveying, field measurements are subject to random errors, so raw observations almost never satisfy the required geometric conditions perfectly. The difference between what the observations yield and what the geometry demands is called the misclosure (or closure error). The purpose of adjustment is to determine small corrections to apply to each observation so that (1) all condition equations are satisfied exactly, and (2) the sum of squared corrections (weighted or unweighted) is minimized — the Least Squares criterion. In the condition-equation approach, we write one equation per geometric condition. For a simple figure (e.g., a single triangle), there may be only one or two conditions. For a complex network, the number of condition equations equals the number of redundant observations: r = n − u, where n is the total number of observations and u is the minimum number needed to determine all unknowns. Each condition equation has the general form: F(L₁, L₂, …, Lₙ) = 0, where L₁…Lₙ are the adjusted observations. When raw (measured) values lᵢ are substituted, the equation does not equal zero but gives the misclosure W. The corrections vᵢ applied to each observation must then satisfy: F(l₁+v₁, l₂+v₂, …, lₙ+vₙ) = 0. For linear (or linearized) conditions, this reduces to the standard form: a₁v₁ + a₂v₂ + … + aₙvₙ + W = 0, where aᵢ are the partial derivatives (coefficients) of the condition equation evaluated at the observed values, and W is the misclosure.

Examples

This is a standard board-exam question. Always count redundant observations carefully. For a simple triangle, n = 3 angles, u = 2 independent angles, so r = 1 (one angle condition). For a horizon closure at a station with 4 measured angles, n = 4, u = 3 (the 4th is determined when the other three plus the closure are known), so r = 1.

Scenario

Identifying the number of condition equations for a braced quadrilateral in a Philippine triangulation network

Solution

A braced quadrilateral has 4 vertices and 6 observed angles (both diagonals observed). The minimum observations needed to fix the figure are u = 3 (three independent angles determine the shape). Therefore r = 6 − 3 = 3 condition equations: two angle conditions and one side condition (log-sine condition).

The coefficients of all three corrections are +1, reflecting that each angle enters the sum equally. The misclosure W carries the sign of the excess (positive if the sum exceeds 180°). This linearized form is exact for the angle sum condition because it is already linear.

Scenario

Writing the condition equation for a plane triangle with three measured angles A, B, C

Solution

Geometric requirement: A + B + C = 180°00'00". Substituting measured values a, b, c: (a + vₐ) + (b + v_b) + (c + v_c) = 180°. This gives: vₐ + v_b + v_c + W = 0, where W = (a + b + c) − 180°.

Applications

Triangulation network adjustment in the Philippine Reference System 1992 (PRS92) control densification
Leveling loop closure adjustment in BM networks supporting cadastral surveys under PD 1529
Traverse closure adjustment for boundary surveys required by CA 141 (Public Land Act)
Station adjustment of directions observed at a triangulation pillar
Side condition in quadrilateral adjustment ensuring consistency of computed distances

Misconceptions

Misclosure is not simply the 'error' — it is the signed discrepancy between observations and the geometric requirement.
The number of condition equations is not equal to the number of observations; it equals the number of redundant observations r = n − u.
Condition equations do not have to be satisfied by the raw observations — only by the adjusted (corrected) observations.
A zero misclosure does not mean no errors exist — random errors may be self-canceling.

Related Concepts

Least Squares principle (minimization of Σwᵢvᵢ²)
Observation equations vs. condition equations
Degrees of freedom and redundancy
Lagrange method of undetermined multipliers
Normal equations in condition adjustment

Common Exam Questions

Example

Three angles of a plane triangle are measured as 58°14'22", 62°31'48", and 59°13'54". What is the misclosure? Answer: Sum = 180°00'04"; misclosure = +4".

Approach

Compute the sum of measured angles/distances, subtract the geometric requirement, and identify sign and magnitude.

Question Type

Multiple choice — identifying misclosure

Example

Four directions around a station are measured. Write the horizon closure condition and identify the misclosure.

Approach

State the geometric requirement, substitute observed + correction, rearrange to standard form aᵢvᵢ + W = 0.

Question Type

Computation — writing condition equation

Key Points To Remember

A condition equation enforces an exact geometric relationship among adjusted observations.
Misclosure W = (sum of observed values) − (required geometric total). Note the sign: positive misclosure means the observations are too large.
Number of condition equations = number of redundant observations = r = n − u.
The condition-equation method minimizes the sum of squared corrections subject to the condition equations — this is Least Squares with constraints.
For linear conditions with coefficients aᵢ, the condition is: Σ aᵢvᵢ + W = 0.
Each condition equation produces one Lagrange correlate (k) in the normal equations.
The corrections vᵢ are functions of the Lagrange correlates and the weights of the observations.

Triangle (Angle) Condition and Equal-Weight Correction

The triangle angle condition is the most fundamental geometric constraint in triangulation. For a plane triangle, the three interior angles must sum exactly to 180°00'00". For a spherical triangle (relevant when triangles span large distances, as in primary geodetic networks), the angles must sum to 180° + ε, where ε (epsilon) is the spherical excess. The spherical excess depends on the area of the triangle on the reference ellipsoid and is computed as ε" = Area / (2 × R² × ρ"), where ρ" = 206,264.806" per radian and R is the mean radius of curvature. In Philippine board examinations, ε is usually given directly. When all three angles have equal weights (equal reliability, equal number of pointings), the Least Squares solution under the single angle condition is simply to distribute the misclosure equally and with opposite sign. This is because equal weights mean equal variances, and the Lagrange solution gives equal corrections. Correction per angle: cᵢ = −W/n, where W is the misclosure and n = 3. The adjusted angle: Lᵢ = lᵢ + cᵢ = lᵢ − W/3. This simple rule — divide the misclosure equally and subtract from each observation — is the most common computation in figure adjustment board problems. The sign rule is critical: if the sum exceeds the required total (positive misclosure), each correction is negative (reduce each angle). If the sum falls short (negative misclosure), each correction is positive (increase each angle).

Examples

The correction is −W/n = −4"/3 = −1.333" per angle. In practice, when the misclosure is not evenly divisible, distribute the remainder to the most uncertain angle or round to the nearest 0.1". Always verify the adjusted sum equals the requirement.

Scenario

Board-style: Three angles of a plane triangulation triangle are measured as A = 68°14'22", B = 72°31'48", C = 39°13'54". Adjust for the triangle condition (equal weights).

Solution

Step 1 — Compute observed sum: 68°14'22" + 72°31'48" + 39°13'54" = 179°59'64" ... Convert: 64" = 1'04", so sum = 180°00'04". Step 2 — Misclosure: W = 180°00'04" − 180°00'00" = +4". Step 3 — Correction per angle: c = −4"/3 = −1.33" ≈ −1" to two angles and −2" to one (distribute remainder). In exact form: c = −4/3" = −1.333" each. Step 4 — Adjusted angles: A' = 68°14'22" − 1.333" = 68°14'20.667", B' = 72°31'48" − 1.333" = 72°31'46.667", C' = 39°13'54" − 1.333" = 39°13'52.667". Check: Sum = 68°14'20.667" + 72°31'46.667" + 39°13'52.667" = 180°00'00.001" ≈ 180°00'00" ✓

The critical distinction from the plane case is the target sum: 180° + ε. Forgetting to add ε is the most common board-exam mistake for spherical triangles. The spherical excess must be given or computable before adjustment.

Scenario

Board-style (spherical triangle): A primary triangulation triangle in Mindanao has a computed spherical excess ε = 6". The measured angles are P = 74°22'18", Q = 58°31'42", R = 47°06'12". Adjust for the angle condition.

Solution

Step 1 — Required sum: 180°00'00" + 6" = 180°00'06". Step 2 — Observed sum: 74°22'18" + 58°31'42" + 47°06'12" = 180°00'12". Step 3 — Misclosure: W = 180°00'12" − 180°00'06" = +6". Step 4 — Correction: c = −6"/3 = −2" per angle. Step 5 — Adjusted angles: P' = 74°22'16", Q' = 58°31'40", R' = 47°06'10". Check: 74°22'16" + 58°31'40" + 47°06'10" = 180°00'06" = 180° + ε ✓

Negative misclosure means the sum falls short; corrections are positive (add to each angle). Always check the sign of W before computing corrections.

Scenario

Board-style: A triangle's angles sum to 179°59'51". Adjust with equal weights.

Solution

W = 179°59'51" − 180°00'00" = −9" (negative misclosure — angles are too small). c = −(−9")/3 = +3" per angle. Each angle is increased by 3". Adjusted sum = 179°59'51" + 9" = 180°00'00" ✓

Applications

Adjustment of triangulation figures forming the backbone of the Philippine Reference System 1992 (PRS92) control network
Quality check and adjustment of GPS-derived triangle networks in cadastral surveys
First-order triangulation adjustment for infrastructure projects (bridges, tunnels, dams) requiring high positional accuracy
Adjustment of angles measured by theodolite in boundary surveys under PD 1529

Misconceptions

For spherical triangles, the target is NOT 180° but 180° + ε. Many examinees forget the spherical excess.
The correction sign must be OPPOSITE to the misclosure sign. If W = +12", then c = −4" (reduce angles).
Equal distribution only applies to equal-weight (equal-reliability) observations. For unequal weights, corrections differ.
Distributing the remainder (when misclosure is not divisible by n) arbitrarily is not strictly Least Squares — on board exams, keep the fractional second or note which angle absorbs the extra 1".
The adjusted angles are the measured angles PLUS the corrections (corrections may be negative).

Related Concepts

Spherical excess and its computation from triangle area
Weighted angle condition for unequal-reliability angles
Side condition (log-sine condition) in triangulation figures
Angle condition in quadrilateral adjustment
Horizon condition at a triangulation station

Common Exam Questions

Example

Angles of a triangle sum to 180°00'09". Equal weights. Correction per angle = −9"/3 = −3" (reduce each angle by 3").

Approach

Compute sum, find misclosure, divide by number of angles, apply opposite sign.

Question Type

Direct computation — find correction per angle

Example

If ε = 8" and measured sum = 180°00'14", then W = 14" − 8" = +6", c = −2" per angle.

Approach

Add spherical excess to 180° before computing misclosure.

Question Type

Spherical triangle — identify correct target sum

Example

Given angles 60°00'10", 60°00'02", 59°59'54" summing to 180°00'06". c = −2". Adjusted: 60°00'08", 60°00'00", 59°59'52" — sum = 180°00'00" ✓

Approach

Apply correction to each measured angle and verify the adjusted sum.

Question Type

Find adjusted angle value

Key Points To Remember

Plane triangle: required sum = 180°00'00" exactly.
Spherical triangle: required sum = 180°00'00" + ε (spherical excess, usually in arc-seconds).
Misclosure W = (observed sum) − (required sum); sign determines whether angles are reduced or increased.
Equal weights → equal corrections: c = −W/n (always opposite sign to W).
After correction, verify: sum of adjusted angles = required sum (self-check).
Spherical excess ε is always positive; it increases with triangle area.
For primary geodetic triangles in the Philippines (sides ~30–80 km), ε ranges from 1" to about 15".

Horizon (Station) Condition

The horizon condition applies at a triangulation station where multiple angles are measured around the full 360° horizon. The requirement is that all angles measured around a point must sum exactly to 360°00'00". This condition arises because all directions from a station span the complete horizontal circle. If n angles are measured around a point, their sum must equal 360°. The misclosure is: W = (Σ measured angles) − 360°. As with the triangle condition, if the observations have equal weights, each angle receives the same correction: c = −W/n. For weighted observations (e.g., some angles measured with more repetitions than others), the correction is distributed inversely proportional to the weights: vᵢ = −k/wᵢ, where k is the Lagrange correlate solved from the condition equation. The horizon condition is distinct from the triangle condition but uses the same mathematical machinery. In Philippine triangulation practice, the horizon condition is applied at primary and secondary stations where all angles between adjacent stations are measured with a direction theodolite. After applying the horizon condition, the directions and angles at that station become internally consistent.

Examples

The procedure is identical to the triangle condition but the target is 360° instead of 180°. The same formula c = −W/n applies directly.

Scenario

Board-style: Three angles around a triangulation station are measured as α₁ = 120°00'06", α₂ = 119°59'54", α₃ = 120°00'06". Adjust with equal weights.

Solution

Step 1 — Observed sum: 120°00'06" + 119°59'54" + 120°00'06" = 360°00'06". Step 2 — Misclosure: W = 360°00'06" − 360°00'00" = +6". Step 3 — Correction: c = −6"/3 = −2" per angle. Step 4 — Adjusted angles: α₁' = 120°00'04", α₂' = 119°59'52", α₃' = 120°00'04". Check: 120°00'04" + 119°59'52" + 120°00'04" = 360°00'00" ✓

With four angles, n = 4, so the individual correction is smaller. More angles → smaller correction per angle for the same misclosure magnitude.

Scenario

Board-style: Four angles around a station sum to 360°00'08". Adjust with equal weights.

Solution

W = +8". c = −8"/4 = −2" per angle. Each of the four measured angles is reduced by 2". Adjusted sum = 360°00'08" − 8" = 360°00'00" ✓

The higher-weight observation (α₁, measured more reliably) receives a smaller correction (−2"), while the lower-weight observation (α₂) receives a larger correction (−4"). This is the fundamental Least Squares principle: trust the better observations more.

Scenario

Board-style (weighted): Two angles around a station are measured: α₁ = 200°00'06" (weight w₁ = 2) and α₂ = 160°00'00" (weight w₂ = 1). Adjust for the horizon condition.

Solution

Step 1 — Sum = 360°00'06"; W = +6". Step 2 — For weighted case, vᵢ = −k/wᵢ. Condition: v₁ + v₂ + W = 0 (with unit coefficients). Substituting: −k/2 + (−k/1) + 6" = 0 → −1.5k = −6" → k = 4". Step 3 — v₁ = −4"/2 = −2"; v₂ = −4"/1 = −4". Check: v₁ + v₂ = −2" + (−4") = −6" = −W ✓. Step 4 — Adjusted: α₁' = 200°00'04"; α₂' = 159°59'56". Sum = 360°00'00" ✓

Applications

Adjustment of observed directions at primary triangulation pillars in the PRS92 network
Consistency check and adjustment of horizontal angles measured at control stations for cadastral surveys
Station adjustment before figure adjustment in classical triangulation computation
Quality control of total station observations in engineering surveys

Misconceptions

Some examinees confuse the horizon condition (360°) with the triangle condition (180°). They have different targets but the same correction formula for equal weights.
In weighted adjustment, it is NOT the observation with the higher weight that gets the bigger correction — it is the LOWER-weight (less reliable) observation.
The horizon condition applies to angles (differences of directions), not to the directions themselves directly.
Horizon condition only applies when angles are measured around the FULL circle. Partial sets of angles between adjacent stations use a different setup.

Related Concepts

Triangle angle condition
Direction observations and reduction to angles
Weighted Least Squares in condition adjustment
Station adjustment vs. figure adjustment sequence
Azimuth condition in traverse adjustment

Common Exam Questions

Example

Five angles around a station sum to 359°59'55". W = −5". c = +1" per angle. Each angle is increased by 1".

Approach

Sum the angles, find W relative to 360°, apply c = −W/n to each angle.

Question Type

Computation — horizon condition with equal weights

Example

If w₁ = 3 and w₂ = 1, then v₂ = 3v₁. Most of the misclosure is placed on the less reliable observation.

Approach

The observation with lower weight (less reliable) receives proportionally more correction.

Question Type

Identify which observation gets more correction in weighted case

Key Points To Remember

Horizon condition: sum of all angles around a station = 360°00'00".
Misclosure W = Σ(measured angles) − 360°.
Equal weights: each correction = −W/n (n = number of angles around the station).
Weighted case: corrections inversely proportional to weights (angles measured fewer times get more correction).
After adjustment, verify the corrected angles sum to exactly 360°.
This condition is often combined with the angle condition in figure adjustment of triangulation networks.
Directions (not just angles) can also be adjusted; the condition then involves the differences of adjusted directions.

Weighted Distribution of Misclosure

When the measured observations have unequal reliabilities, the simple equal-distribution rule no longer applies. In Least Squares adjustment, the weight of an observation wᵢ is inversely proportional to its variance: wᵢ = σ₀²/σᵢ², where σ₀² is a reference variance. Observations with smaller variance (higher precision) receive larger weights and, in the condition adjustment, receive smaller corrections. The mechanics: for a single condition equation with unit coefficients (all aᵢ = 1, as in the angle sum condition), the Lagrange solution gives: vᵢ = −k/wᵢ. The Lagrange correlate k is found by substituting into the condition equation: Σ(aᵢ × (−k/wᵢ)) + W = 0 → k = W / Σ(aᵢ²/wᵢ). For unit coefficients: k = W / Σ(1/wᵢ). Once k is known, each correction vᵢ = −k/wᵢ is computed. The correction is largest for the smallest weight (most uncertain observation) and smallest for the largest weight (most precise observation). In field surveys, weights are commonly assigned based on the number of measurement repetitions (more repetitions → higher weight), the length of a level line (longer lines → lower weight, proportional to 1/length in leveling), or the inverse of variance from statistical analysis. Understanding weighted distribution is essential for board-exam problems involving mixed-quality observations and for the more advanced topics of parametric and combined adjustment.

Examples

Angle A with the lowest weight (w₁=1, least reliable) receives the largest correction (−4.9"), while angle C with the highest weight (w₃=3, most reliable) receives the smallest correction (−1.6"). This is physically sensible: we trust the better measurements more and disturb them less.

Scenario

Board-style: Three angles of a triangle have weights w₁ = 1, w₂ = 2, w₃ = 3. The misclosure is W = +9". Find corrections and adjusted angles if measured values are A = 60°00'12", B = 59°59'57", C = 60°00'00".

Solution

Step 1 — Verify: Sum = 60°00'12" + 59°59'57" + 60°00'00" = 180°00'09"; W = +9" ✓. Step 2 — k = W / Σ(1/wᵢ) = 9" / (1/1 + 1/2 + 1/3) = 9" / (1 + 0.5 + 0.333) = 9" / 1.833 = 4.909". Step 3 — Corrections: v₁ = −k/w₁ = −4.909"/1 = −4.909" ≈ −4.9"; v₂ = −k/w₂ = −4.909"/2 = −2.455" ≈ −2.5"; v₃ = −k/w₃ = −4.909"/3 = −1.636" ≈ −1.6". Check: v₁ + v₂ + v₃ = −4.9 − 2.5 − 1.6 = −9.0" = −W ✓. Step 4 — Adjusted: A' = 60°00'12" − 4.909" = 60°00'07.09"; B' = 59°59'57" − 2.455" = 59°59'54.55"; C' = 60°00'00" − 1.636" = 59°59'58.36". Check: 60°00'07.09" + 59°59'54.55" + 59°59'58.36" = 180°00'00.00" ✓

The longer line (w₁=1, less reliable) receives more correction (−0.011 m) while the shorter, more reliable line (w₂=4) receives less correction (+0.003 m). The adjusted height difference is 10.013 m — a weighted mean, closer to the more reliable line 2 measurement.

Scenario

Board-style: Two level lines connect the same two BMs. Line 1 is 4 km long with measured ΔH₁ = 10.024 m; Line 2 is 1 km long with measured ΔH₂ = 10.010 m. In leveling, weights are inversely proportional to length. Adjust for the condition that both lines must give the same height difference.

Solution

Step 1 — Assign weights: w₁ = 1/4 = 0.25 (or use w₁:w₂ = 1:4 by taking w₁=1, w₂=4). Step 2 — Condition: ΔH₁ adjusted = ΔH₂ adjusted (both determine the same height difference). Condition equation: v₁ − v₂ + W = 0, where W = ΔH₁ − ΔH₂ = 10.024 − 10.010 = +0.014 m. Coefficients: a₁ = +1, a₂ = −1. Step 3 — k = W / Σ(aᵢ²/wᵢ) = 0.014 / (1²/1 + (−1)²/4) = 0.014 / (1 + 0.25) = 0.014 / 1.25 = 0.0112 m. Step 4 — v₁ = −k × a₁/w₁ = −0.0112 × 1/1 = −0.0112 m; v₂ = −k × a₂/w₂ = −0.0112 × (−1)/4 = +0.0028 m. Step 5 — Adjusted: ΔH₁' = 10.024 − 0.011 = 10.013 m; ΔH₂' = 10.010 + 0.003 = 10.013 m ✓

Applications

Differential leveling network adjustment where line weights depend on distance
Triangulation adjustment when angles are measured with different numbers of repetitions
GPS baseline adjustment using variances from the covariance matrix as weight indicators
Combined adjustment of heterogeneous observations (distances and angles) with different units and precisions
Cadastral survey adjustment under DENR-LMB procedures for control densification

Misconceptions

Many examinees think the HIGHEST weight gets the most correction — it is the LOWEST weight (least reliable) that gets the most correction.
Weight is proportional to reliability, not to measured value. A large angle measurement does not get a large weight automatically.
In leveling, weight is 1/distance, not distance. Longer lines have LOWER weight (less reliable per kilometer).
The Lagrange correlate k is not the correction itself — the correction is k divided by the weight.
Weighted equal-distribution (dividing W equally regardless of weights) is WRONG for unequal weights.

Related Concepts

Weight number and standard error relationship: wᵢ = σ₀²/σᵢ²
Weighted mean as simplest case of weighted adjustment
Propagation of variances in surveying measurements
Normal equations in parametric (observation equation) adjustment
Cofactor matrix and weight coefficient matrix

Common Exam Questions

Example

Triangle misclosure +9", weights 1:2:3. Find the correction to the heaviest angle.

Approach

Compute k = W/Σ(aᵢ²/wᵢ), then vᵢ = −k×aᵢ/wᵢ. Check that Σ aᵢvᵢ + W = 0.

Question Type

Computation — weighted correction using Lagrange correlate

Example

Among three angles with weights 1, 2, 4, the angle with weight 1 receives the largest correction.

Approach

The observation with the LOWEST weight gets the LARGEST correction (since vᵢ = −k/wᵢ and small wᵢ gives large vᵢ).

Question Type

Identify which observation gets the largest correction

Example

ΔH₁ = 10.024 m (w=1) and ΔH₂ = 10.010 m (w=4): L̄ = (1×10.024 + 4×10.010)/5 = 50.064/5 = 10.013 m.

Approach

For two observations of the same quantity with weights w₁ and w₂, the adjusted value is the weighted mean: L̄ = (w₁l₁ + w₂l₂)/(w₁+w₂).

Question Type

Weighted mean as the special case

Key Points To Remember

Weight wᵢ is proportional to reliability (inversely proportional to variance).
Higher weight → smaller correction; lower weight → larger correction.
For unit coefficients: k = W / Σ(1/wᵢ); corrections: vᵢ = −k/wᵢ.
For general coefficients aᵢ: k = W / Σ(aᵢ²/wᵢ); corrections: vᵢ = −k × aᵢ/wᵢ.
Verify: Σ wᵢvᵢ² is minimized (this is the Least Squares criterion automatically satisfied by this solution).
In leveling: weight is proportional to 1/distance (or 1/number of setups).
In triangulation: weight is proportional to number of pointings (repetitions).
Always verify: Σ(aᵢvᵢ) + W = 0 as a check.

Figure Adjustment — Sequence and Special Cases

Figure adjustment refers to the complete process of adjusting all observations in a geometric figure (triangle, quadrilateral, polygon, level loop, etc.) to satisfy all applicable condition equations simultaneously. The procedure follows a logical sequence: (1) List all observations and identify the geometric figure type. (2) Count condition equations: r = n − u (number of redundant observations). (3) Write each condition equation, compute each misclosure Wⱼ. (4) Set up the normal equations of the condition adjustment (one per condition equation, using Lagrange correlates k₁, k₂, … kᵣ). (5) Solve the normal equations for k₁…kᵣ. (6) Compute corrections: vᵢ = Σⱼ(−kⱼ × aᵢⱼ/wᵢ). (7) Apply corrections to observations: Lᵢ = lᵢ + vᵢ. (8) Verify all conditions are satisfied. For simple figures (single triangle, single horizon condition), the normal equations reduce to a single equation and direct formulas apply. For more complex figures (quadrilateral, polygon with diagonals), multiple simultaneous normal equations must be solved. In Philippine board examinations, the most common figure adjustment problems involve: single triangle (one angle condition), horizon closure (one condition), loop leveling (one closure condition), and occasionally quadrilateral adjustment with two or three conditions. The board exam rarely requires solving the full matrix normal equations — typically the simple equal-distribution or the single-correlate Lagrange solution suffices.

Examples

In differential leveling, the correction to each line is proportional to its length: vᵢ = −W × dᵢ / Σdᵢ. This is because the weight is 1/dᵢ and the Lagrange solution gives vᵢ = −k/wᵢ = −k × dᵢ. The longer the line, the more it is corrected, reflecting its higher uncertainty.

Scenario

Board-style — Level loop: A closed leveling loop has four lines with the following measured elevation differences and distances: Line 1: +3.254 m, d=2km; Line 2: +1.876 m, d=3km; Line 3: −2.108 m, d=4km; Line 4: −3.015 m, d=1km. Adjust using weights proportional to 1/distance.

Solution

Step 1 — Loop closure condition: Σ ΔH = 0. Observed sum = 3.254 + 1.876 − 2.108 − 3.015 = +0.007 m. Misclosure W = +0.007 m. Step 2 — Weights (k/distance, using k=12 for LCD): w₁=6, w₂=4, w₃=3, w₄=12. Equivalently use 1/d directly: 1/2, 1/3, 1/4, 1/1. Step 3 — Lagrange correlate: k = W / Σ(1/wᵢ) = 0.007 / (1/6 + 1/4 + 1/3 + 1/12). ... Better: use d directly. k = W / Σ(dᵢ) = 0.007 / (2+3+4+1) = 0.007/10 = 0.0007 m/km. Step 4 — Corrections proportional to distance: v₁ = −0.0007×2 = −0.0014 m; v₂ = −0.0007×3 = −0.0021 m; v₃ = −0.0007×4 = −0.0028 m; v₄ = −0.0007×1 = −0.0007 m. Step 5 — Check: −0.0014 − 0.0021 − 0.0028 − 0.0007 = −0.0070 m = −W ✓. Step 6 — Adjusted ΔH: 3.2526, 1.8739, −2.1108, −3.0157 m. Adjusted sum = 0.000 m ✓

The Compass (Bowditch) Rule is equivalent to condition adjustment with weights proportional to leg length, distributing the misclosure proportionally. It satisfies the two closure conditions (ΔE = 0, ΔN = 0) simultaneously. This is a direct application of figure adjustment to traverse closure.

Scenario

Board-style — Simple traverse loop: A closed traverse has a total easting departure misclosure of ΔE = +0.12 m and a total northing departure misclosure of ΔN = −0.09 m. The perimeter is 1,200 m and the five traverse legs have lengths 300, 250, 200, 250, and 200 m. Apply the Compass (Bowditch) Rule to distribute the misclosure.

Solution

Compass Rule correction to each leg: ΔEᵢ correction = −(ΔE_misclosure / perimeter) × Lᵢ; ΔNᵢ correction = −(ΔN_misclosure / perimeter) × Lᵢ. For leg 1 (L=300 m): ΔE corr = −(0.12/1200)×300 = −0.030 m; ΔN corr = −(−0.09/1200)×300 = +0.0225 m. For leg 2 (L=250 m): ΔE corr = −0.025 m; ΔN corr = +0.01875 m. Continue similarly for other legs. Total ΔE corrections = −0.12 m; total ΔN corrections = +0.09 m — misclosures removed ✓

Applications

Complete triangulation figure adjustment for the densification of the PRS92 control network
Level loop adjustment in the Philippine vertical control network (Geodetic Reference Network)
Traverse adjustment for boundary surveys and cadastral mapping under PD 1529
Loop closure adjustment in monitoring surveys for infrastructure (dams, subsidence areas)
Combined angle and distance adjustment in total station surveying networks

Misconceptions

The Compass Rule and the Transit Rule are NOT the same — Compass distributes proportional to line length, Transit distributes proportional to departure/latitude.
Figure adjustment is not just one step — it is a complete sequence requiring verification at the end.
A level loop with zero misclosure does NOT mean the elevations are error-free — it only means random errors happened to cancel.
The number of condition equations is not the number of geometric figures but the number of REDUNDANT observations.
Adjusting one condition at a time (sequentially) is not strictly Least Squares — all conditions should be adjusted simultaneously for a rigorous solution.

Related Concepts

Traverse adjustment methods: Compass Rule, Transit Rule, Least Squares
Quadrilateral adjustment and the log-sine side condition
Loop leveling and the propagation of systematic errors
Network adjustment and free network adjustment
Accuracy standards for surveys under RA 8560 (Philippine Geodetic Engineering Act)

Common Exam Questions

Example

Loop misclosure = +12 mm over 6 km. Line of 2 km gets correction: −12 × 2/6 = −4 mm.

Approach

Compute misclosure, apply correction proportional to line length: vᵢ = −W × dᵢ / Σdᵢ.

Question Type

Level loop — find correction to a specific line

Example

Braced quadrilateral: n=8 angles, u=5, r=3 conditions.

Approach

Use r = n − u. Count n (observed quantities) and u (minimum needed to fix figure).

Question Type

Number of condition equations for a figure

Example

Starting elevation 100.000 m, adjusted ΔH₁=+3.2526 m → BM2 = 103.2526 m.

Approach

Apply corrections to each line elevation difference, then compute cumulative elevations from the starting BM.

Question Type

Adjusted elevation of a BM from a level loop

Key Points To Remember

Figure adjustment sequence: identify figure → count conditions → write equations → compute misclosures → solve for k → compute vᵢ → apply → verify.
For equal-weight, unit-coefficient conditions: direct formula c = −W/n applies (no need for full normal equations).
Quadrilateral: typically 3 conditions (2 angle + 1 side) requiring simultaneous solution.
Level loop: 1 condition (loop closure = 0); corrections proportional to line lengths in weighted case.
Traverse loop: 2 or 3 conditions (ΔE closure = 0, ΔN closure = 0, and optionally azimuth closure).
Always verify adjusted observations satisfy ALL condition equations, not just one.
Precision of adjustment is assessed by the a posteriori standard error of unit weight: σ₀² = Σwᵢvᵢ²/r.

Practice Problems

Convert the misclosure to arc-seconds for easy arithmetic: 2'12" = 120" + 12" = 132". Correction of −44" per angle (reduce each, since sum exceeds 180°). The verification confirms the adjusted angles sum exactly to 180°. This is a typical board-exam computation — practice converting minutes and seconds to total seconds.

Problem

PROBLEM 1 — Triangle Angle Condition (Plane) The three interior angles of a plane triangle are measured as follows: A = 52°17'38", B = 76°44'22", C = 51°00'12". Assuming equal weights, find: (a) the misclosure, (b) the correction per angle, and (c) the adjusted value of each angle.

Solution

(a) Observed sum: 52°17'38" + 76°44'22" + 51°00'12" = 180°02'12". Wait — let us add carefully: 52°17'38" + 76°44'22" = 129°01'60" = 129°02'00"; 129°02'00" + 51°00'12" = 180°02'12". Misclosure W = 180°02'12" − 180°00'00" = +2'12" = +132". (b) Correction per angle: c = −W/n = −132"/3 = −44" per angle. (c) Adjusted angles: • A' = 52°17'38" − 44" = 52°17'38" − 0°00'44" = 52°16'54" • B' = 76°44'22" − 44" = 76°43'38" • C' = 51°00'12" − 44" = 50°59'28" Verification: 52°16'54" + 76°43'38" + 50°59'28" = 180°00'00" ✓

When the misclosure is zero, no corrections are applied — the observations satisfy the condition exactly. This tests whether the student correctly uses 180°+ε as the target and not just 180°. A student who used 180° as the target would compute a spurious misclosure of +7" and make unnecessary corrections.

Problem

PROBLEM 2 — Spherical Triangle Condition A primary triangulation triangle in the Cordillera has a spherical excess ε = 7". The measured angles are: P = 55°30'22", Q = 68°15'41", R = 56°14'04". Find the misclosure and the adjusted angles (equal weights).

Solution

Step 1 — Required sum: 180°00'00" + 7" = 180°00'07". Step 2 — Observed sum: 55°30'22" + 68°15'41" + 56°14'04". • 55°30'22" + 68°15'41" = 123°46'03" (22"+41"=63"=1'03", so 123°45'63"=123°46'03") • 123°46'03" + 56°14'04" = 180°00'07"... wait: 03"+04"=07", 46'+14'=60'=1°, so 123°+56°+1°=180°; total=180°00'07". Step 3 — Misclosure: W = 180°00'07" − 180°00'07" = 0". No adjustment required! The observed angles already satisfy the spherical triangle condition exactly. Adjusted values = measured values: P'=55°30'22", Q'=68°15'41", R'=56°14'04".

Summing angles accurately is the critical skill. Convert all seconds to a running total to avoid mistakes. The correction of −2" per angle is applied because the sum exceeds 360° (positive misclosure → reduce angles).

Problem

PROBLEM 3 — Horizon Condition (Equal Weights) At triangulation station MAYON, four angles are observed to adjacent stations: α₁ = 87°14'18", α₂ = 93°22'44", α₃ = 79°56'30", α₄ = 99°26'36". Adjust the four angles for the horizon condition (equal weights).

Solution

Step 1 — Observed sum: 87°14'18" + 93°22'44" = 180°37'02" 180°37'02" + 79°56'30" = 260°33'32" 260°33'32" + 99°26'36" = 360°00'08" (32"+36"=68"=1'08"; 33'+26'+1'=60'=1°; 260°+99°+1°=360°) Actual sum: 87°14'18" + 93°22'44" + 79°56'30" + 99°26'36". Seconds: 18+44+30+36=128"=2'08" Minutes: 14+22+56+26+2=120'=2° Degrees: 87+93+79+99+2=360° Total: 360°00'08"... Wait: 120'=2°, so degrees = 87+93+79+99+2=360, minutes= 0, seconds=8. Sum = 360°00'08". Step 2 — Misclosure: W = 360°00'08" − 360°00'00" = +8". Step 3 — Correction: c = −8"/4 = −2" per angle. Step 4 — Adjusted angles: • α₁' = 87°14'18" − 2" = 87°14'16" • α₂' = 93°22'44" − 2" = 93°22'42" • α₃' = 79°56'30" − 2" = 79°56'28" • α₄' = 99°26'36" − 2" = 99°26'34" Verification: Sum of adjusted = 360°00'08" − 8" = 360°00'00" ✓

Again, a zero misclosure means no correction regardless of weights. The problem tests careful arithmetic and the concept that weights only matter when there is a non-zero misclosure to distribute. The weighted distribution formula is only engaged when W ≠ 0.

Problem

PROBLEM 4 — Weighted Triangle Adjustment A triangle is measured with the following angles and weights: A = 60°00'10" (w=1), B = 60°00'06" (w=2), C = 59°59'44" (w=3). Find the adjusted angles.

Solution

Step 1 — Observed sum: 60°00'10" + 60°00'06" + 59°59'44" = 180°00'00". (10"+6"+44"=60"=1'; 0+0+59'=59'+1'=60'=1°; 60+60+59+1=180°. Actually: 60°00'10"+60°00'06"=120°00'16"; 120°00'16"+59°59'44"=180°00'00" [16"+44"=60"=1'; 0+59'+1'=60'=1°; 120+59+1=180°].) Sum = 180°00'00". Step 2 — Misclosure: W = 180°00'00" − 180°00'00" = 0". No adjustment required. The adjusted angles equal the observed angles: A'=60°00'10", B'=60°00'06", C'=59°59'44".

This is the classic weighted condition adjustment problem. The angle with weight 1 (least reliable) receives the largest correction (−4.91"), while the angle with weight 3 (most reliable) receives the smallest correction (−1.64"). This is the essence of weighted Least Squares: trust the more precise measurements more and disturb them less.

Problem

PROBLEM 5 — Weighted Triangle Adjustment (Non-zero Misclosure) A triangle has angles: A = 70°10'15" (w₁=1), B = 60°25'30" (w₂=2), C = 49°24'24" (w₃=3). Adjust for the plane triangle condition using weighted Least Squares.

Solution

Step 1 — Observed sum: 70°10'15" + 60°25'30" + 49°24'24". Seconds: 15+30+24=69"=1'09" Minutes: 10+25+24+1=60'=1° Degrees: 70+60+49+1=180°. Sum = 180°00'09" → W = +9". Step 2 — Lagrange correlate (all aᵢ=1): k = W / Σ(1/wᵢ) = 9" / (1/1 + 1/2 + 1/3) = 9" / (6/6 + 3/6 + 2/6) = 9" / (11/6) = 9" × 6/11 = 54/11 = 4.909". Step 3 — Corrections: v₁ = −k/w₁ = −4.909"/1 = −4.909" ≈ −4.91" v₂ = −k/w₂ = −4.909"/2 = −2.455" ≈ −2.45" v₃ = −k/w₃ = −4.909"/3 = −1.636" ≈ −1.64" Check: v₁+v₂+v₃ = −4.91−2.45−1.64 = −9.00" = −W ✓ Step 4 — Adjusted angles: A' = 70°10'15" − 4.91" = 70°10'10.09" B' = 60°25'30" − 2.45" = 60°25'27.55" C' = 49°24'24" − 1.64" = 49°24'22.36" Verification: 70°10'10.09"+60°25'27.55"+49°24'22.36" = 180°00'00.00" ✓

In differential leveling, the weight is proportional to 1/L (shorter lines are more reliable), so corrections are proportional to L. The correction per section equals the correction rate (−W/ΣL) multiplied by the section length. The negative misclosure (−0.012 m) means the loop fell short — each correction is positive (add to ΔH values).

Problem

PROBLEM 6 — Level Loop Adjustment (Weighted by Distance) A closed differential leveling loop has three sections with the following data: Section 1: ΔH = +12.486 m, L = 3 km Section 2: ΔH = +8.314 m, L = 5 km Section 3: ΔH = −20.812 m, L = 2 km The starting BM elevation is 142.000 m. Adjust the loop and find the adjusted elevation of the intermediate BM (after section 1).

Solution

Step 1 — Loop closure: 12.486 + 8.314 − 20.812 = −0.012 m. Misclosure W = −0.012 m (loop does not close by −12 mm). Step 2 — Total distance: 3+5+2 = 10 km. Correction per km = −W/Σd = −(−0.012)/10 = +0.0012 m/km. Step 3 — Corrections proportional to distance: v₁ = +0.0012 × 3 = +0.0036 m v₂ = +0.0012 × 5 = +0.0060 m v₃ = +0.0012 × 2 = +0.0024 m Check: +0.0036+0.0060+0.0024 = +0.012 m = −W ✓ Step 4 — Adjusted ΔH: ΔH₁' = 12.486 + 0.0036 = 12.4896 m ΔH₂' = 8.314 + 0.0060 = 8.3200 m ΔH₃' = −20.812 + 0.0024 = −20.8096 m Step 5 — Adjusted closure: 12.4896 + 8.3200 − 20.8096 = 0.000 m ✓ Step 6 — Intermediate BM elevation: BM_intermediate = 142.000 + 12.4896 = 154.4896 m

The key insight: if σ_C = 2σ_A, then σ_C² = 4σ_A², so w_C = w_A/4. The less reliable observation (larger variance) receives much more of the misclosure correction. This directly demonstrates the Least Squares principle in action.

Problem

PROBLEM 7 — Exercise Problem from Reference: Distributing a +9" Triangle Misclosure with Unequal Weights Distribute a +9" triangle angle misclosure when angle C is twice as uncertain as angles A and B (i.e., angle C has twice the standard deviation of A and B). Assign weights accordingly.

Solution

Step 1 — Assign weights. Since weight ∝ 1/σ², and σ_C = 2σ_A = 2σ_B: w_A = 1/σ_A²; w_B = 1/σ_B²; w_C = 1/(2σ_A)² = 1/(4σ_A²) = w_A/4. Let w_A = w_B = 4 and w_C = 1 (ratio 4:4:1). Step 2 — Lagrange correlate: k = W / Σ(1/wᵢ) = 9" / (1/4 + 1/4 + 1/1) = 9" / (0.25+0.25+1) = 9" / 1.5 = 6". Step 3 — Corrections: v_A = −k/w_A = −6"/4 = −1.5" v_B = −k/w_B = −6"/4 = −1.5" v_C = −k/w_C = −6"/1 = −6" Check: −1.5−1.5−6 = −9" = −W ✓ Step 4 — Interpretation: Angle C, being twice as uncertain, receives −6" (four times the correction of each of A and B at −1.5").

Exam Preparation Tips

MASTER THE SIGN RULE: The correction is always OPPOSITE in sign to the misclosure. If the sum exceeds the geometric requirement (positive misclosure), the corrections are negative (reduce the observations). If the sum falls short (negative misclosure), corrections are positive. This is the most common source of sign errors in board exams.
MEMORIZE THE THREE TARGET SUMS: Plane triangle = 180°00'00"; Spherical triangle = 180°00'00" + ε (spherical excess); Horizon condition = 360°00'00". Confusing these targets leads to computing a wrong misclosure and wrong corrections.
PRACTICE ANGLE ARITHMETIC: Board exam problems require fast, accurate addition of degrees-minutes-seconds. Practice converting to total seconds, adding, then converting back. Example: 60°00'10" + 60°00'02" + 59°59'54" = (60+60+59)° + (0+0+59)' + (10+2+54)" = 179° + 59' + 66" = 179°59'66" = 180°00'06".
WEIGHTED CORRECTION FORMULA: For unit-coefficient conditions, the Lagrange solution gives vᵢ = −k/wᵢ where k = W/Σ(1/wᵢ). Memorize this formula and practice applying it. The lower the weight, the larger the correction.
LEVELING LOOP CORRECTIONS ARE PROPORTIONAL TO LENGTH: For differential leveling with weights = 1/distance, the correction to each section is: vᵢ = −W × Lᵢ / ΣLᵢ. This is directly proportional to length, unlike the Lagrange formula which might seem counterintuitive at first.
VERIFY YOUR ANSWER: Always check that the sum of corrections equals −W (for unit coefficients) and that the adjusted observations satisfy the condition equation. This verification step takes only seconds and prevents errors from passing undetected.
DISTINGUISH PLANE FROM SPHERICAL: In PRC board problems, if the triangle sides are described as 'in primary triangulation' or 'spanning tens of kilometers,' check whether a spherical excess ε is given or must be computed. For triangles with sides < 10 km, the plane triangle assumption is generally acceptable.
COUNT CONDITION EQUATIONS CAREFULLY: Use r = n − u. For a single plane triangle: r = 3 − 2 = 1. For a quadrilateral with all four sides and both diagonals observed (8 angles): r = 8 − 5 = 3. Knowing r tells you how many Lagrange correlates you need.
COMPASS RULE FOR TRAVERSES: The Compass (Bowditch) Rule applies corrections proportional to leg length — the same principle as weighted leveling adjustment. Correction to each leg = −(total misclosure / total length) × leg length.
REVIEW RA 8560: The Philippine Geodetic Engineering Act of 1998 defines the scope of geodetic engineering practice including control surveys, cadastral surveys, and geodetic networks. Board exams occasionally ask about the legal framework governing survey accuracy standards and who is authorized to perform geodetic control work.
PRACTICE PAST BOARD PROBLEMS: The PRC Geodetic Engineer board examination regularly includes 3–5 problems on adjustment computations per sitting. Focus on triangle angle condition, horizon condition, level loop adjustment, and weighted distribution. Time yourself to complete each problem in 3–5 minutes.
UNITS CONSISTENCY: All angular misclosures should be expressed in arc-seconds before computing corrections. All distance-based corrections in leveling should be in meters. Do not mix units within a computation.

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In summary

Condition equations and figure adjustment represent the practical heart of Adjustment Computations for the PRC Geodetic Engineer board examination. The fundamental principle — that geometric conditions (triangle angle sum = 180°, horizon condition = 360°, loop closure = 0) must be satisfied exactly by adjusted observations — drives the entire computational process. The misclosure W quantifies how much raw observations violate these conditions, and the Least Squares adjustment minimizes the sum of squared corrections subject to satisfying all conditions simultaneously. For the board examination, mastery of three core skills is essential: (1) correctly identifying the type of condition and its required target value (especially remembering 180° + ε for spherical triangles); (2) applying the equal-weight correction formula c = −W/n accurately with correct sign; and (3) applying the weighted Lagrange formula k = W/Σ(1/wᵢ) with vᵢ = −k/wᵢ, understanding that lower-weight (less reliable) observations receive larger corrections. Beyond board exam success, these principles are directly applied in real Philippine geodetic practice: adjusting control networks for the PRS92 densification, performing cadastral surveys under PD 1529, supporting land classification surveys under CA 141, and operating within the professional framework established by RA 8560 (the Philippine Geodetic Engineering Act). A geodetic engineer who thoroughly understands condition equations and figure adjustment is equipped to produce the precise, legally defensible survey results that Philippine law and professional practice demand. Practice all seven worked examples, verify each answer, and ensure you can complete a standard triangle or horizon condition problem confidently within five minutes — the time typically available per problem in the licensure examination.

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