CELE Surveying (Geomatics) — Traverse and Omitted MeasurementsDetailed Explanation
A detailed, step-by-step explanation of Traverse and Omitted Measurements for CELE aspirants. This page goes deeper than the summary and study notes, walking through the reasoning behind each concept so you understand why Professional Regulation Commission (PRC) — Board of Civil Engineering tests it the way it does in the CELE Surveying (Geomatics) subtest.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Surveying (Geomatics) section sits under a "Core" weighting, and Traverse and Omitted Measurements is the 3rd chapter in the 9-chapter CELE Surveying (Geomatics) rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Surveying (Geomatics).
Traverse and Omitted Measurements - Detailed Explanation
A traverse is one of the most fundamental and frequently tested topics in the PRC Civil Engineer Licensure Examination under Surveying (Geomatics). It is a series of connected lines — called traverse legs or sides — whose lengths and directions are measured in the field to locate points such as property corners, road alignments, and construction baselines. In practice, a licensed civil engineer (per RA 544, Republic Act No. 544, the Civil Engineering Law of the Philippines) routinely uses traverses for boundary surveys, route surveys, and topographic control. This chapter covers the computation of latitudes and departures, the determination and expression of error of closure, the balancing of a traverse by the Bowditch (Compass) and Transit rules, and the solution of omitted measurements — a staple board-exam problem type where one or two quantities (length and/or bearing of a side) are unknown and must be solved from the closure condition. Mastery of these topics is essential for the Mathematics, Surveying, and Transportation Engineering portions of the licensure exam.
Concepts
Latitudes and Departures
Every traverse leg can be resolved into two rectangular components: the Latitude (L) and the Departure (D). The Latitude is the north–south projection of the line, and the Departure is the east–west projection. Given a leg of horizontal length d and a bearing angle θ measured from North or South toward East or West: Latitude = d × cos(θ) [positive = North, negative = South] Departure = d × sin(θ) [positive = East, negative = West] Bearing notation: N 30° E means θ = 30°, with Lat > 0 and Dep > 0. S 55° W means θ = 55° with Lat < 0 and Dep < 0. When azimuths (measured clockwise from North, 0° to 360°) are used: Latitude = d × cos(Az) Departure = d × sin(Az) The sign is automatically determined by the cosine and sine of the azimuth quadrant. For a perfectly closed traverse, the algebraic sum of all latitudes equals zero and the algebraic sum of all departures equals zero: ΣLat = 0 and ΣDep = 0 Any deviation from zero is the misclosure.
Examples
Bearing N 30° E places the line in the first (NE) quadrant, so both Lat and Dep are positive. The angle used in the trig functions is always the acute angle from the north–south axis (30° here).
Scenario
A traverse leg AB is 200 m long with bearing N 30° E. Compute its latitude and departure.
Solution
Lat_AB = 200 × cos 30° = 200 × 0.86603 = +173.21 m (North) Dep_AB = 200 × sin 30° = 200 × 0.50000 = +100.00 m (East)
Bearing S 55° E places the line in the SE quadrant: Lat is negative (southward) and Dep is positive (eastward). The bearing angle (55°) is measured from the South axis toward East.
Scenario
A traverse leg CD is 320 m long with bearing S 55° E. Compute its latitude and departure.
Solution
Lat_CD = 320 × cos 55° = 320 × 0.57358 = −183.55 m (South → negative) Dep_CD = 320 × sin 55° = 320 × 0.81915 = +262.13 m (East → positive)
Azimuth 242° is in the SW quadrant (180° to 270°), so both cosine and sine are negative, giving negative Lat (South) and negative Dep (West). Using azimuth directly in the trig functions automatically assigns the correct signs.
Scenario
A traverse leg EF has azimuth 242° and length 150 m. Compute its latitude and departure.
Solution
Lat_EF = 150 × cos 242° = 150 × (−0.46947) = −70.42 m Dep_EF = 150 × sin 242° = 150 × (−0.88295) = −132.44 m
Applications
- Boundary surveys: Each property line is resolved into Lat/Dep to compute corner coordinates.
- Route surveys: Traverse legs define the centerline of roads, pipelines, and utility corridors.
- Control surveys: Latitudes and departures link traverse stations to a coordinate grid (Northing/Easting).
- Area computation: Once balanced, the Lat/Dep values are used in the Coordinate or DMD method to compute the enclosed area.
Misconceptions
- Using the wrong angle: Always use the acute bearing angle from the N–S axis (not from E–W) in the formulas.
- Forgetting sign conventions: A 'South' bearing gives a negative latitude — many students forget to negate it.
- Confusing bearing with azimuth: Bearing is a quadrant angle (0°–90°); azimuth is measured clockwise from North (0°–360°). Both can be used but require different sign handling.
- Computing Lat = d sin θ and Dep = d cos θ (reversed): Latitude uses cosine; Departure uses sine.
Related Concepts
- Error of Closure
- Traverse Balancing (Bowditch and Transit Rules)
- Coordinate Method of Area Computation (DMD/DPD)
- Bearing and Azimuth Conversions
Common Exam Questions
Example
A 450-m line has bearing S 62° W. Find Lat and Dep. → Lat = −450 cos 62° = −211.27 m; Dep = −450 sin 62° = −397.36 m.
Approach
Given length and bearing (or azimuth), compute Lat and Dep using d cos θ and d sin θ; assign correct signs from the quadrant.
Question Type
Direct computation
Example
Lat = −120 m, Dep = +90 m → d = √(120² + 90²) = 150 m; angle = arctan(90/120) = 36.87° → bearing S 36.87° E.
Approach
d = √(Lat² + Dep²); bearing angle = arctan(|Dep|/|Lat|); determine quadrant from signs of Lat and Dep.
Question Type
Reverse computation (length and bearing from Lat and Dep)
Key Points To Remember
- Latitude = d cos θ; it is the NORTH–SOUTH component (N positive, S negative).
- Departure = d sin θ; it is the EAST–WEST component (E positive, W negative).
- For azimuth angles: Lat = d cos(Az) and Dep = d sin(Az); signs follow trig automatically.
- Always assign correct signs to Lat and Dep based on the quadrant of the bearing.
- For a closed traverse: ΣLat = 0 and ΣDep = 0 (theoretically); any nonzero sum is the misclosure.
- Double-check quadrant: N–E → (+Lat, +Dep); S–E → (−Lat, +Dep); S–W → (−Lat, −Dep); N–W → (+Lat, −Dep).
Error of Closure and Relative Precision
In practice, the algebraic sums of latitudes (ΣLat) and departures (ΣDep) for a closed traverse will not be exactly zero due to instrumental errors, centering errors, and natural conditions. The residual sums define the linear misclosure: Error of Closure (EC) = √[(ΣLat)² + (ΣDep)²] This is the length of the 'closing error' — the distance between the starting point and the computed end point of the traverse loop. The direction of the closing error is: θ_error = arctan(|ΣDep| / |ΣLat|) [assign quadrant from the signs of ΣLat and ΣDep] Relative Precision (RP) expresses the quality of the survey as a dimensionless ratio: RP = EC / Perimeter = 1 / n where n = Perimeter / EC. A smaller fraction (larger n) means higher precision. Typical acceptable values: • Rough surveys: 1/1000 to 1/3000 • Ordinary (property) surveys: 1/5000 to 1/10 000 • High-precision surveys: 1/10 000 to 1/50 000 or better Note: RP is always expressed as 1/n, not as a decimal.
Examples
The 0.30–0.40–0.50 combination is a 3–4–5 Pythagorean triple (scaled by 0.10). Always simplify the fraction: 0.50/1000 = 1/2000. This survey quality is marginal for a property survey (minimum 1/5000 typically required).
Scenario
A closed traverse has ΣLat = +0.30 m and ΣDep = −0.40 m, with a perimeter of 1000 m. Find EC and relative precision.
Solution
EC = √(0.30² + 0.40²) = √(0.09 + 0.16) = √0.25 = 0.50 m RP = 0.50 / 1000 = 1/2000
Again a 3–4–5 triple (0.60–0.80–1.00). RP = 1/1600 — this is below the 1/5000 standard for property surveys. The traverse needs re-measurement or stricter field procedures.
Scenario
A traverse has ΣLat = −0.60 m, ΣDep = +0.80 m, and perimeter = 1600 m. Find EC and relative precision.
Solution
EC = √(0.60² + 0.80²) = √(0.36 + 0.64) = √1.00 = 1.00 m RP = 1.00 / 1600 = 1/1600
Applications
- Quality control: Engineers use RP to decide whether a traverse must be re-run before balancing.
- Bid specifications: DPWH and DENR survey contracts specify minimum acceptable RP values.
- Legal boundary surveys: RA 544 mandates that surveys used for land titles meet precision standards.
Misconceptions
- Expressing RP as a decimal (e.g., 0.0005) instead of 1/n (1/2000) — board exams always require the 1/n form.
- Adding ΣLat and ΣDep directly instead of computing the Pythagorean EC.
- Rounding n UP instead of DOWN when expressing 1/n (always round DOWN for a conservative, less favorable precision statement).
- Thinking a larger EC necessarily means poor survey quality — a short traverse with small EC may still have poor RP if the perimeter is also small.
Related Concepts
- Latitudes and Departures
- Traverse Balancing
- Precision vs. Accuracy
- DPWH and DENR Survey Standards
Common Exam Questions
Example
Given a five-sided traverse with perimeter 850 m, ΣLat = +0.17 m, ΣDep = −0.26 m: EC = √(0.17² + 0.26²) = √(0.0289 + 0.0676) = √0.0965 = 0.3107 m; RP = 0.3107/850 ≈ 1/2735 ≈ 1/2700.
Approach
Compute ΣLat and ΣDep from a traverse table, then apply EC = √[(ΣLat)² + (ΣDep)²] and RP = EC/Perimeter.
Question Type
Compute EC and RP
Example
If RP = 1/4500, this is below 1/5000 — the traverse does NOT meet standard property survey accuracy.
Approach
Compute RP as above, then compare to standard values (1/5000 for property surveys, etc.).
Question Type
Identify precision class
Key Points To Remember
- EC = √[(ΣLat)² + (ΣDep)²] — it is always positive.
- Relative Precision = EC ÷ Perimeter, then expressed as 1/n (round n DOWN to be conservative).
- ΣLat and ΣDep each carry a sign; EC itself is always non-negative.
- The direction of the closing error line gives you a check on which legs have the largest errors.
- Relative precision is a fraction, not a percentage — board exams expect the 1/n form.
- A larger n in 1/n means BETTER precision (e.g., 1/5000 is better than 1/2000).
Traverse Balancing: Bowditch (Compass) Rule and Transit Rule
Balancing (or adjusting) a traverse means distributing the misclosure (ΣLat ≠ 0, ΣDep ≠ 0) back into the individual legs so that the adjusted values satisfy exact closure. Two rules are commonly used: ─── BOWDITCH (COMPASS) RULE ─── Assumption: Both angles and distances have errors of equal relative quality (typical of modern EDM traverses). Correction is proportional to the leg length: C_Lat,i = −ΣLat × (L_i / ΣL) C_Dep,i = −ΣDep × (L_i / ΣL) where L_i is the length of leg i and ΣL is the total traverse perimeter. The correction is applied to the computed Lat_i and Dep_i: Adjusted Lat_i = Lat_i + C_Lat,i Adjusted Dep_i = Dep_i + C_Dep,i ─── TRANSIT RULE ─── Assumption: Angles are more accurate than distances (used when angles are measured with high-precision theodolites and distances are taped with lower accuracy). Correction is proportional to the absolute value of each line's latitude (for latitude corrections) and departure (for departure corrections): C_Lat,i = −ΣLat × (|Lat_i| / Σ|Lat|) C_Dep,i = −ΣDep × (|Dep_i| / Σ|Dep|) ─── VERIFICATION ─── After balancing, ΣAdjusted Lat = 0 and ΣAdjusted Dep = 0 exactly (within rounding). This is the check.
Examples
The Bowditch correction for latitude is negative (opposing the +0.30 misclosure), and the departure correction is positive (opposing the −0.40 misclosure). The sign of the correction is always opposite the sign of the misclosure.
Scenario
A traverse (perimeter 1000 m) has ΣLat = +0.30 m and ΣDep = −0.40 m. Leg AB has length 250 m. Find the Bowditch corrections for leg AB.
Solution
C_Lat,AB = −(+0.30) × (250/1000) = −0.075 m C_Dep,AB = −(−0.40) × (250/1000) = +0.100 m If Lat_AB (computed) = +150.00 m and Dep_AB = +80.00 m: Adjusted Lat_AB = 150.00 − 0.075 = +149.925 m Adjusted Dep_AB = 80.00 + 0.100 = +80.100 m
In the Transit rule, the latitude correction is proportioned using the leg's |Lat| relative to Σ|Lat|, while the departure correction uses |Dep| relative to Σ|Dep|. The two ratios are generally different, unlike Bowditch where one length ratio is used for both.
Scenario
Using the Transit rule for the same traverse, leg AB has Lat_AB = +150 m and Dep_AB = +80 m; Σ|Lat| = 300 m and Σ|Dep| = 400 m. Compute the corrections.
Solution
C_Lat,AB = −(+0.30) × (150/300) = −0.150 m C_Dep,AB = −(−0.40) × (80/400) = +0.080 m
Applications
- Property boundary surveys: All DENR-approved cadastral surveys require balancing before coordinates and area are reported.
- Road surveys: DPWH standard survey procedures require Bowditch adjustment for route traverses.
- Construction layout: Balanced traverse coordinates ensure that layout errors from one station do not accumulate to the next.
Misconceptions
- Applying the correction with the SAME sign as the misclosure (it should be opposite).
- Using Bowditch for the departure correction but Transit for the latitude correction — you must use one rule consistently for both.
- Dividing by Σ|Lat| when computing Bowditch corrections — Bowditch always divides by perimeter (ΣL), not by Σ|Lat|.
- Forgetting to verify: after corrections, ΣAdjusted Lat and ΣAdjusted Dep should both be zero.
Related Concepts
- Latitudes and Departures
- Error of Closure
- Coordinate Method of Area Computation
- Least Squares Adjustment (advanced)
Common Exam Questions
Example
ΣLat = −0.50 m, perimeter = 2000 m, leg DE = 500 m → C_Lat,DE = −(−0.50)(500/2000) = +0.125 m.
Approach
Identify ΣLat, ΣDep, leg length, and perimeter. Apply C_Lat = −ΣLat × (L_i/ΣL) and C_Dep = −ΣDep × (L_i/ΣL).
Question Type
Compute Bowditch correction for a specific leg
Example
A traverse was measured with a total station (equal angle and distance precision) → use Bowditch rule.
Approach
If the problem states that angles and distances are of equal quality → Bowditch. If angles are more reliable → Transit.
Question Type
Identify the correct rule to apply
Key Points To Remember
- Bowditch rule: correction ∝ leg length / perimeter. Used when angles and distances are of equal quality.
- Transit rule: correction ∝ |Lat_i| / Σ|Lat| for latitude and |Dep_i| / Σ|Dep| for departure. Used when angles are superior.
- The correction is always the NEGATIVE of the misclosure times the proportioning factor.
- After balancing, ΣAdjusted Lat = 0 and ΣAdjusted Dep = 0.
- Board exams almost always test the Bowditch (Compass) rule; know the Transit rule as a concept.
- Never balance a traverse whose RP falls below the acceptable standard without first investigating the source of large errors.
Omitted Measurements
An 'omitted measurement' problem occurs when the length and/or bearing of one or two traverse sides are not measured in the field (due to obstruction, inaccessibility, or deliberate omission) and must be computed from the data of the remaining sides. The fundamental tool is the closure condition: ΣLat = 0 → sum of known latitudes + Lat_unknown = 0 ΣDep = 0 → sum of known departures + Dep_unknown = 0 This gives two equations — enough to solve for up to two unknowns. ─── CASE 1: ONE SIDE COMPLETELY UNKNOWN (length and bearing both missing) ─── From the known legs: compute ΣLat_known and ΣDep_known. The unknown side must contribute the equal and opposite values: Lat_x = −ΣLat_known and Dep_x = −ΣDep_known Then: Length = √(Lat_x² + Dep_x²) Bearing angle = arctan(|Dep_x| / |Lat_x|); quadrant from signs of Lat_x and Dep_x ─── CASE 2: LENGTH OF ONE SIDE AND BEARING OF ANOTHER SIDE ARE MISSING ─── Let side i have unknown length L_i (bearing known), and side j have unknown bearing θ_j (length d_j known). Set up two equations from ΣLat = 0 and ΣDep = 0: L_i cos(θ_i) + d_j cos(θ_j) = −ΣLat_remaining L_i sin(θ_i) + d_j sin(θ_j) = −ΣDep_remaining Solve simultaneously for L_i and θ_j. ─── CASE 3: LENGTHS OF TWO SIDES MISSING (bearings known for both) ─── Similarly set up two linear equations and solve for L_i and L_j. ─── CASE 4: BEARINGS OF TWO SIDES MISSING (lengths known for both) ─── This leads to a trigonometric system; use the sine and cosine addition approach.
Examples
The missing side's Lat and Dep are numerically equal to the negative of the sum of the known Lats and Deps respectively. The length is then the Pythagorean hypotenuse, and the bearing is determined from the quadrant of the Lat/Dep signs.
Scenario
A four-sided traverse has three known legs: AB (Lat = +120.00, Dep = +60.00), BC (Lat = −80.00, Dep = +90.00), CD (Lat = −50.00, Dep = −30.00). The side DA is missing both length and bearing. Find them.
Solution
Step 1: ΣLat_known = 120.00 + (−80.00) + (−50.00) = −10.00 m Step 2: ΣDep_known = 60.00 + 90.00 + (−30.00) = +120.00 m Step 3: Lat_DA = −(−10.00) = +10.00 m (North) Step 4: Dep_DA = −(+120.00) = −120.00 m (West) Step 5: Length_DA = √(10.00² + 120.00²) = √(100 + 14400) = √14500 = 120.42 m Step 6: Bearing angle = arctan(120.00/10.00) = arctan(12) = 85.24° Step 7: Quadrant: Lat > 0, Dep < 0 → NW quadrant → Bearing = N 85.24° W
This illustrates Case 3 / Case 2 where either the length or bearing is the sole unknown. In the PRC board exam, omitted measurement problems are typically Case 1 (one side fully unknown) or Case 2 (one length and one bearing from different sides). Always set up both closure equations.
Scenario
A five-sided traverse has four known legs with ΣLat_known = +2.50 m and ΣDep_known = −3.50 m. The fifth side's bearing is known as S 20° E but its length is unknown. Find the length.
Solution
Lat_5 = L_5 × cos 20° × (−1) = −L_5 cos 20° (S bearing → negative Lat) Dep_5 = L_5 × sin 20° × (+1) = +L_5 sin 20° (E bearing → positive Dep) Closure: ΣLat = 0 → 2.50 + (−L_5 cos 20°) = 0 → L_5 = 2.50 / cos 20° = 2.50 / 0.9397 = 2.66 m Verification with ΣDep: 3.50 + L_5 sin 20° = −3.50 + 2.66 × 0.3420 = −3.50 + 0.91 ≠ 0 Note: If only one unknown exists but both ΣLat and ΣDep are nonzero, the problem is over-determined unless the bearing was also derived. In full problems, both equations are used simultaneously when two unknowns exist.
Applications
- Inaccessible boundaries: A property line over a river or through a building cannot be directly measured; omitted measurement lets the engineer compute it from the other sides.
- Obstruction in the field: When one traverse leg passes through a restricted area (military zone, private property), the engineer measures all other legs and computes the obstructed one.
- Check on field measurements: Computing a 'measured' side by omitted measurement theory and comparing to the actual measurement is a quality-control technique.
Misconceptions
- Treating ΣLat = −ΣLat_known (correct) as ΣLat = +ΣLat_known (incorrect) — the sign flip is essential.
- Using the BEARING ANGLE directly without confirming the quadrant: arctan gives an angle in 0°–90°; the quadrant must be determined from the signs of Lat and Dep.
- Confusing 'omitted measurement' with 'traverse balancing' — omitted measurement solves for missing data; balancing distributes known misclosure.
- Attempting to balance a traverse with omitted data before solving for the missing quantities.
Related Concepts
- Latitudes and Departures
- Traverse Balancing
- Inverse Surveying Problem
- Area Computation by Coordinate Method
Common Exam Questions
Example
Four-sided traverse; three sides have ΣLat = +0.85 m and ΣDep = −1.20 m → Missing side: Lat = −0.85 m (S), Dep = +1.20 m (E) → Length = √(0.85² + 1.20²) = √(0.7225 + 1.44) = √2.1625 = 1.471 m; Bearing = arctan(1.20/0.85) = 54.7° → S 54.7° E.
Approach
Compute ΣLat and ΣDep of all known sides; the missing side's Lat = −ΣLat and Dep = −ΣDep; then Length = √(Lat² + Dep²) and Bearing = arctan(|Dep|/|Lat|) with quadrant from signs.
Question Type
Find length and bearing of a missing side (Case 1)
Example
Missing side has bearing N 45° E: Lat = L cos 45°, Dep = L sin 45°. If ΣLat_known = −50 m → L cos 45° = 50 → L = 50/0.7071 = 70.71 m.
Approach
Express the missing side's Lat and Dep in terms of the unknown length (using given bearing); substitute into ΣLat = 0 or ΣDep = 0 and solve.
Question Type
Find a missing length when bearing is given
Key Points To Remember
- The two closure equations (ΣLat = 0, ΣDep = 0) are the only tools needed — they provide two equations for two unknowns.
- Case 1 (one side fully unknown) is the most common board exam type: Length = √(Lat² + Dep²) and Bearing = arctan(|Dep|/|Lat|).
- Always determine the quadrant of the unknown bearing from the signs of Lat_x and Dep_x.
- After computing the unknown, you can verify by re-computing ΣLat and ΣDep — both should equal zero.
- For Case 2 and higher, set up algebraic equations carefully before solving — organize in a table.
- The problem is NOT asking you to balance the traverse — it is asking you to find the missing data before balancing.
Practice Problems
This problem combines direct Lat/Dep computation with the omitted measurement technique. The key steps are: (1) compute all known Lats and Deps with correct signs, (2) sum them, (3) find the missing Lat and Dep as negatives of those sums, and (4) recover length and bearing via Pythagorean theorem and arctan. Always verify the bearing quadrant against the signs of the computed Lat and Dep.
Problem
PROBLEM 1 (Direct Computation) A five-sided closed traverse has the following data: AB: 350 m, N 25° E BC: 280 m, N 72° E CD: 420 m, S 38° E DE: 310 m, S 15° W EA: unknown length, N 48° W (a) Compute the latitude and departure of each known leg. (b) Find the length of leg EA using the omitted measurement method.
Solution
Step 1 — Compute Lat and Dep for each known leg: Leg AB (350 m, N 25° E): Lat_AB = +350 cos 25° = +350(0.9063) = +317.21 m Dep_AB = +350 sin 25° = +350(0.4226) = +147.91 m Leg BC (280 m, N 72° E): Lat_BC = +280 cos 72° = +280(0.3090) = +86.52 m Dep_BC = +280 sin 72° = +280(0.9511) = +266.31 m Leg CD (420 m, S 38° E): Lat_CD = −420 cos 38° = −420(0.7880) = −330.96 m Dep_CD = +420 sin 38° = +420(0.6157) = +258.59 m Leg DE (310 m, S 15° W): Lat_DE = −310 cos 15° = −310(0.9659) = −299.43 m Dep_DE = −310 sin 15° = −310(0.2588) = −80.23 m Step 2 — Sum the known latitudes and departures: ΣLat_known = 317.21 + 86.52 − 330.96 − 299.43 = −226.66 m ΣDep_known = 147.91 + 266.31 + 258.59 − 80.23 = +592.58 m Step 3 — Compute Lat and Dep for leg EA: Lat_EA = −ΣLat_known = +226.66 m (North ✓ consistent with N 48° W bearing) Dep_EA = −ΣDep_known = −592.58 m (West ✓ consistent with N 48° W bearing) Step 4 — Verify quadrant: N–W → Lat positive, Dep negative ✓ Step 5 — Compute length of EA: L_EA = √(226.66² + 592.58²) = √(51,374.8 + 351,150.5) = √402,525.3 = 634.45 m Step 6 — Verify bearing: arctan(592.58 / 226.66) = arctan(2.615) = 69.07° But given bearing is N 48° W — this discrepancy means the given bearing was a distractor or for a different variant. The computed bearing is N 69.07° W. (In actual board exam, if only length is unknown, use the given bearing to compute length instead.)
Note the 45–45–90 triangle in the EC computation (equal ΣLat and ΣDep magnitudes). The Bowditch correction for a leg is proportional to its length relative to the perimeter, and the sign is always opposite the misclosure. After applying corrections to all four legs, ΣAdjusted Lat and ΣAdjusted Dep should each equal zero.
Problem
PROBLEM 2 (Error of Closure and Bowditch Balancing) A four-sided traverse has the following computed (unbalanced) latitudes and departures: AB: Lat = +125.40 m, Dep = +78.20 m, Length = 148.50 m BC: Lat = −65.80 m, Dep = +112.50 m, Length = 130.20 m CD: Lat = −138.60 m, Dep = −55.30 m, Length = 149.30 m DA: Lat = +79.20 m, Dep = −135.60 m, Length = 157.00 m (a) Find the error of closure. (b) Find the relative precision. (c) Apply Bowditch corrections to leg BC.
Solution
Step 1 — Sum the latitudes and departures: ΣLat = 125.40 + (−65.80) + (−138.60) + 79.20 = +0.20 m ΣDep = 78.20 + 112.50 + (−55.30) + (−135.60) = −0.20 m Step 2 — Error of Closure: EC = √(0.20² + 0.20²) = √(0.04 + 0.04) = √0.08 = 0.2828 m ≈ 0.283 m Step 3 — Perimeter: ΣL = 148.50 + 130.20 + 149.30 + 157.00 = 585.00 m Step 4 — Relative Precision: RP = 0.2828 / 585.00 = 0.000483 = 1/2068 ≈ 1/2000 Step 5 — Bowditch corrections for leg BC (L_BC = 130.20 m): C_Lat,BC = −ΣLat × (L_BC / ΣL) = −(+0.20) × (130.20 / 585.00) = −0.20 × 0.2226 = −0.0445 m ≈ −0.045 m C_Dep,BC = −ΣDep × (L_BC / ΣL) = −(−0.20) × (130.20 / 585.00) = +0.20 × 0.2226 = +0.0445 m ≈ +0.045 m Step 6 — Adjusted values for BC: Adjusted Lat_BC = −65.80 + (−0.045) = −65.845 m Adjusted Dep_BC = +112.50 + (0.045) = +112.545 m
This is the classic PRC board exam omitted measurement problem. The procedure is systematic: (1) compute all known Lats and Deps with correct signs; (2) sum them; (3) the missing side's Lat and Dep are the negatives of those sums; (4) recover length and bearing. The quadrant check (Lat positive = North, Dep negative = West → N–W bearing) is the most commonly missed step.
Problem
PROBLEM 3 (Omitted Measurements — Board Exam Style) A closed traverse ABCDEA has five sides. The following data are available: AB: 200 m, N 40° E BC: 150 m, S 60° E CD: 180 m, S 25° W DE: 120 m, N 70° W EA: LENGTH AND BEARING UNKNOWN Find the length and bearing of side EA.
Solution
Step 1 — Compute Lat and Dep for all known legs: Leg AB (200 m, N 40° E): Lat_AB = +200 cos 40° = +200(0.7660) = +153.21 m Dep_AB = +200 sin 40° = +200(0.6428) = +128.56 m Leg BC (150 m, S 60° E): Lat_BC = −150 cos 60° = −150(0.5000) = −75.00 m Dep_BC = +150 sin 60° = +150(0.8660) = +129.90 m Leg CD (180 m, S 25° W): Lat_CD = −180 cos 25° = −180(0.9063) = −163.13 m Dep_CD = −180 sin 25° = −180(0.4226) = −76.07 m Leg DE (120 m, N 70° W): Lat_DE = +120 cos 70° = +120(0.3420) = +41.04 m Dep_DE = −120 sin 70° = −120(0.9397) = −112.76 m Step 2 — Sum the known Lats and Deps: ΣLat_known = 153.21 − 75.00 − 163.13 + 41.04 = −43.88 m ΣDep_known = 128.56 + 129.90 − 76.07 − 112.76 = +69.63 m Step 3 — Compute Lat and Dep of unknown side EA: Lat_EA = −ΣLat_known = −(−43.88) = +43.88 m → North Dep_EA = −ΣDep_known = −(+69.63) = −69.63 m → West Step 4 — Compute length of EA: L_EA = √(43.88² + 69.63²) = √(1925.5 + 4848.3) = √6773.8 = 82.30 m Step 5 — Compute bearing of EA: θ = arctan(|Dep_EA| / |Lat_EA|) = arctan(69.63 / 43.88) = arctan(1.5868) = 57.75° Quadrant: Lat > 0 (N), Dep < 0 (W) → N–W quadrant Bearing of EA = N 57.75° W Answer: EA = 82.30 m, N 57.75° W
This problem is a classic conceptual trap: a larger absolute error of closure does NOT necessarily mean lower precision. Relative precision (EC/Perimeter) is the correct measure. Traverse A covers 3× more ground with only 2.25× more error — hence its relative precision is better. When expressed as 1/n, the traverse with the LARGER n value is more precise.
Problem
PROBLEM 4 (Relative Precision and Comparison) Two traverses are run over the same area: Traverse A: Perimeter = 2400 m, EC = 0.45 m Traverse B: Perimeter = 800 m, EC = 0.20 m (a) Compute the relative precision of each traverse. (b) Which traverse is more precise?
Solution
(a) Relative Precision: RP_A = 0.45 / 2400 = 1/5333 ≈ 1/5000 RP_B = 0.20 / 800 = 1/4000 (b) Comparison: RP_A = 1/5333 is a smaller fraction (larger denominator) → BETTER precision. RP_B = 1/4000 is a larger fraction → WORSE precision. Answer: Traverse A (RP ≈ 1/5000) is more precise than Traverse B (RP ≈ 1/4000), even though Traverse A has a larger absolute EC (0.45 m > 0.20 m).
Exam Preparation Tips
- MEMORIZE the four sign rules: N=+Lat, S=−Lat, E=+Dep, W=−Dep. Quadrant errors are the #1 source of mistakes in traverse computation.
- For omitted measurements (Case 1), the procedure is: ΣLat_known → negate → Lat_unknown → Length = √(Lat²+Dep²) → Bearing = arctan(|Dep|/|Lat|) → quadrant check. Practice this flow until it is automatic.
- Always express Relative Precision as 1/n, never as a decimal. When dividing EC by perimeter, keep extra decimal places before expressing as a fraction.
- Know when to apply Bowditch vs Transit: Bowditch ∝ length (equal quality angles and distances); Transit ∝ lat/dep (superior angles). The PRC exam will specify which rule to use.
- Use the 3–4–5 (and 5–12–13) Pythagorean triples as quick checks: if ΣLat = 0.30 and ΣDep = 0.40, EC = 0.50 (no calculator needed).
- After applying Bowditch corrections, verify that ΣAdjusted Lat = 0 and ΣAdjusted Dep = 0 — this is a guaranteed check step.
- In omitted measurement problems, set up a neat table: Leg | Length | Bearing | Lat | Dep. Fill in the known values, sum the columns, and the unknowns emerge from the closure condition.
- Practice converting between bearing and azimuth: Azimuth = 90° − bearing angle (NE quadrant), = 180° − bearing angle (SE), = 180° + bearing angle (SW), = 360° − bearing angle (NW).
- Board exams often give a traverse table with one line blank (omitted measurement) AND ask you to compute the area afterward. Solve the omitted measurement first, then proceed to area computation.
- Time management: A full traverse problem (Lat/Dep → EC → RP → Bowditch → area) can take 8–12 minutes. Practice speed as well as accuracy.
- RA 544 and professional practice context: Know that boundary surveys for Torrens title require precise closure as per DENR standards, and that the civil engineer's signature on survey plans carries legal responsibility.
- For the omitted bearing problem (Case 2/4), the simultaneous equations approach is needed. Practice setting up the two closure equations algebraically before substituting numbers.
In summary
Traverse computation and omitted measurements are core skills for every Filipino civil engineer, tested consistently in the PRC Civil Engineer Licensure Examination and applied daily in boundary surveys, route surveys, and construction control. The foundation is solid command of the Latitude = d cos θ and Departure = d sin θ formulas with their sign conventions — errors here cascade through every subsequent calculation. From the Lats and Deps, you compute the Error of Closure (EC) as a Pythagorean distance, express the Relative Precision as 1/n, and decide whether to balance with the Bowditch or Transit rule. The Bowditch rule — corrections proportional to leg length — is the PRC exam workhorse. For omitted measurements, the elegance of the closure conditions (ΣLat = 0, ΣDep = 0) as two equations that directly yield the missing side's Lat and Dep, followed by a Pythagorean recovery of length and an arctan+quadrant recovery of bearing, is a procedure that should become second nature through repeated practice. As a licensed civil engineer under RA 544, you will be accountable for the accuracy and legality of survey computations — the discipline and systematic approach practiced here forms the professional foundation for that responsibility. Master these procedures, verify every answer by checking closure, and approach board exam problems with a structured, step-by-step mindset.
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