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CELE Surveying (Geomatics) — LevelingDetailed Explanation

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Want to really understand Leveling before tackling CELE Surveying (Geomatics) questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.

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 Leveling is the 2nd 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).

Leveling - Detailed Explanation

Leveling is the surveying operation that determines differences in elevation between points on the Earth's surface. It is indispensable in civil engineering practice — from establishing benchmark networks mandated by government agencies, to setting grades for roads, drainage systems, and building foundations. In the PRC Civil Engineer Licensure Examination, leveling problems appear consistently under the Surveying cluster, testing your ability to compute heights of instrument (HI), elevations of turning points (TP) and intermediate foresight (IFS) stations, apply the arithmetic check, and account for curvature and refraction over long sights. Mastery of these topics is non-negotiable for board passers.

Concepts

Fundamental Principles of Leveling

Leveling relies on the principle that a level instrument establishes a horizontal line of sight — a plane perpendicular to the direction of gravity at the instrument station. By reading a graduated rod held vertically on points of known and unknown elevation, the surveyor transfers elevations from point to point. Key terminology you must internalize: • Benchmark (BM): A monumented point of known elevation, often established by NAMRIA (National Mapping and Resource Information Authority) or DPWH. • Backsight (BS): A rod reading taken on a point of KNOWN elevation; this establishes the Height of Instrument (HI). A BS always ADDS to an elevation. • Foresight (FS): A rod reading taken on a point of UNKNOWN elevation; this lowers the line from HI to determine the new elevation. An FS always SUBTRACTS from HI. • Height of Instrument (HI): The elevation of the line of sight above the datum. NOT the height of the instrument above the ground. • Turning Point (TP): A stable intermediate point where both an FS (from the old instrument position) and a BS (from the new instrument position) are taken. It transfers the elevation forward. • Intermediate Foresight (IFS): A rod reading taken between TPs; it gives the elevation of a point but does NOT change the HI. It is NOT used in the arithmetic check. The two master equations: HI = Elevation_known + BS Elevation_new = HI − FS These two equations govern ALL differential leveling computations.

Examples

The HI (101.525 m) is above the benchmark because the rod reading places the line of sight that distance above BM-1. Subtracting the foresight (rod reading on the lower TP-1) gives a lower elevation of 99.207 m. A large FS means the ground is lower; a small FS means the ground is higher — remember this intuition for quick field checks.

Scenario

A benchmark BM-1 has an elevation of 100.000 m. A backsight reading of 1.525 m is taken on BM-1. A foresight of 2.318 m is read on TP-1. Find: (a) HI and (b) elevation of TP-1.

Solution

Step 1: Compute HI. HI = Elev_BM1 + BS = 100.000 + 1.525 = 101.525 m Step 2: Compute elevation of TP-1. Elev_TP1 = HI − FS = 101.525 − 2.318 = 99.207 m

Notice that a large BS on TP-1 (3.041 m) significantly raised the HI — the instrument is now much higher than the line of sight in the previous setup. A small FS (0.892 m) on BM-2 gives a relatively high elevation (101.356 m), confirming BM-2 is near the elevation of the HI. This interplay between BS and FS values is what board examiners test.

Scenario

From TP-1 (Elev = 99.207 m), a new BS of 3.041 m is read. A foresight of 0.892 m is read on BM-2. Find HI at the new instrument position and the elevation of BM-2.

Solution

Step 1: New HI. HI = 99.207 + 3.041 = 102.248 m Step 2: Elevation of BM-2. Elev_BM2 = 102.248 − 0.892 = 101.356 m

Applications

Setting out grades for road construction projects (e.g., DPWH projects under RA 8975 or RA 9184 procurement).
Establishing floor-level benchmarks for building construction to comply with National Building Code elevations.
Drainage design — determining flow directions and pipe invert elevations.
Irrigation canal layout — slope and water-surface elevation determination.
Foundation setting-out surveys for bridges and structures.

Misconceptions

Confusing HI with the physical instrument height above the tripod base — HI is an ELEVATION above datum, not a distance above the ground.
Thinking that a large backsight always means the ground is higher — a large BS simply means the line of sight is far above the backsight point.
Applying FS readings from intermediate foresight stations in the arithmetic check — IFS readings are NOT included in ΣBS or ΣFS for the check.
Forgetting that the TP must be a stable, well-defined point — using loose gravel or soft soil as a TP introduces gross errors.

Related Concepts

Benchmark establishment and NAMRIA vertical control network
Differential (spirit) leveling procedure
Profile leveling and intermediate foresights
Arithmetic check for level runs
Curvature and refraction corrections
Two-peg test for instrument collimation error

Common Exam Questions

Example

BM elevation = 215.400 m, BS = 1.850 m, FS = 2.640 m. Find new elevation. HI = 215.400 + 1.850 = 217.250 m; Elev = 217.250 − 2.640 = 214.610 m.

Approach

Apply HI = Elev + BS, then Elev_new = HI − FS. Tabulate each setup in a level notes format.

Question Type

Direct HI and elevation computation

Example

HI = 101.500 m, new point elevation = 103.200 m. Find FS. FS = HI − Elev = 101.500 − 103.200 = −1.700 m — NEGATIVE foresight means the rod top is BELOW the line of sight, indicating an inverted rod reading (rod held upside-down on overhead structure). Recognizing this is a common board trap.

Approach

Rearrange the formulas: BS = HI − Elev_known; FS = HI − Elev_unknown. Identify what is known and solve algebraically.

Question Type

Finding missing BS or FS given elevations

Key Points To Remember

BS (backsight) is read on a KNOWN point — it raises the HI above that known elevation.
FS (foresight) is read on an UNKNOWN point — subtract from HI to get the new elevation.
HI is the elevation of the line of sight, NOT the physical height of the tripod.
A Turning Point (TP) carries BOTH an FS (closing the old setup) and a BS (opening the new setup).
Intermediate Foresights (IFS) give elevations of intermediate stations but do NOT establish a new HI and are NOT included in the arithmetic check.
Elevations are always referenced to the adopted vertical datum — in the Philippines, Philippine Vertical Datum 1963 (PVD63) or Mean Sea Level (MSL).

Differential Leveling and the Arithmetic Check

Differential leveling is the step-by-step process of carrying an elevation from a known benchmark to one or more unknown points through a series of instrument setups, each linked by a turning point. Level notes are organized in a table with columns: Station | BS | HI | IFS | FS | Elevation ARITHMETIC CHECK (also called the 'page check' or 'summation check'): For any level run or closed loop: ΣBS − ΣFS = Elev_last − Elev_first This check verifies the ARITHMETIC of your computations — it catches addition or subtraction errors in the table. It does NOT detect field measurement blunders (e.g., misread rod). Note carefully: • ONLY BS and FS (on TPs and the final BM) enter ΣBS and ΣFS. • Intermediate foresights (IFS) are EXCLUDED from the sums. • For a closed loop returning to the starting BM, Elev_last − Elev_first = 0, so ΣBS = ΣFS. • The MISCLOSURE of a loop is detected by comparing the computed final elevation with the known closing BM elevation, NOT by the arithmetic check.

Examples

The arithmetic check confirms all additions and subtractions in the table are correct (both differences equal +0.555 m). The net elevation ROSE by 0.555 m from BM-A to BM-B, consistent with ΣBS > ΣFS. In the board exam, if this check fails, you have an arithmetic error — go back and recheck each HI and elevation computation systematically.

Scenario

A level run starts at BM-A (Elev = 50.000 m). The following readings are taken: Setup 1: BS on BM-A = 1.420 m, FS on TP-1 = 2.315 m Setup 2: BS on TP-1 = 2.850 m, FS on TP-2 = 1.635 m Setup 3: BS on TP-2 = 1.180 m, FS on BM-B = 0.945 m Find all HIs, elevations, and apply the arithmetic check.

Solution

LEVEL NOTES TABLE: Station | BS | HI | FS | Elevation BM-A | 1.420 | 51.420 | | 50.000 TP-1 | 2.850 | 52.105 | 2.315 | 49.105 TP-2 | 1.180 | 50.340 | 1.635 | 49.155 (wait — let me redo carefully) Setup 1: HI₁ = 50.000 + 1.420 = 51.420 m Elev_TP1 = 51.420 − 2.315 = 49.105 m Setup 2: HI₂ = 49.105 + 2.850 = 51.955 m Elev_TP2 = 51.955 − 1.635 = 50.320 m Setup 3: HI₃ = 50.320 + 1.180 = 51.500 m Elev_BM-B = 51.500 − 0.945 = 50.555 m ARITHMETIC CHECK: ΣBS = 1.420 + 2.850 + 1.180 = 5.450 m ΣFS = 2.315 + 1.635 + 0.945 = 4.895 m ΣBS − ΣFS = 5.450 − 4.895 = +0.555 m Elev_BM-B − Elev_BM-A = 50.555 − 50.000 = +0.555 m ✓ Check passes.

This example is the classic board trap: many examinees include IFS readings in ΣFS, causing the arithmetic check to fail and giving wrong elevations. Remember: intermediate foresights give you station elevations but are excluded from the summation check.

Scenario

A level run with IFS: BM (Elev = 100.000 m), BS = 2.100 m on BM, IFS = 1.450 m on Sta. 1+00, IFS = 0.880 m on Sta. 2+00, FS = 2.750 m on TP-1. Find all elevations and apply the arithmetic check.

Solution

HI = 100.000 + 2.100 = 102.100 m Elev_Sta.1+00 = 102.100 − 1.450 = 100.650 m (IFS — no new HI) Elev_Sta.2+00 = 102.100 − 0.880 = 101.220 m (IFS — no new HI) Elev_TP-1 = 102.100 − 2.750 = 99.350 m (FS — closes this setup) ARITHMETIC CHECK: ΣBS = 2.100 m (only TPs and BM get BS) ΣFS = 2.750 m (only TPs and closing BM get FS) ΣBS − ΣFS = 2.100 − 2.750 = −0.650 m Elev_TP1 − Elev_BM = 99.350 − 100.000 = −0.650 m ✓ The IFS values (1.450 and 0.880) do NOT appear in either sum — they are excluded.

Applications

Establishing elevation control for highway project surveys (DPWH standard practice).
Closing level loops to detect and distribute errors in precise engineering surveys.
Profile leveling for LRT/MRT track alignment and grade verification.
Flood control and drainage studies requiring accurate ground elevation profiles.
Quality control during earthwork operations — comparing design grade to actual ground elevation.

Misconceptions

Including intermediate foresights (IFS) in the arithmetic check — they must be excluded.
Confusing the arithmetic check (tests computation accuracy) with the loop misclosure check (tests field measurement accuracy).
Using ΣBS − ΣFS = 0 for a level line (not a closed loop) — that equation applies only when start and end elevations are identical.
Forgetting to update the HI at each new setup — the HI from Setup 1 cannot be used in Setup 2.

Related Concepts

Profile leveling and IFS stations
Loop misclosure and error distribution (transit rule, equal-weight distribution)
Allowable misclosure standards (NSCP, DPWH, NAMRIA specifications)
Two-peg test — detecting collimation error in the level instrument
Precise leveling (first-order, second-order) vs. ordinary leveling

Common Exam Questions

Example

ΣBS = 9.340 m, ΣFS = 7.120 m, Elev_initial = 200.000 m. Find Elev_final. Elev_final = 200.000 + (9.340 − 7.120) = 200.000 + 2.220 = 202.220 m.

Approach

Set up the standard table (Station, BS, HI, IFS, FS, Elevation). Compute HI for each setup (Elev + BS), then each new elevation (HI − FS or HI − IFS). Sum only BS and FS columns (exclude IFS). Verify ΣBS − ΣFS = Elev_last − Elev_first.

Question Type

Complete the level notes table and apply the arithmetic check

Example

ΣBS = 12.500 m, ΣFS (partial, missing one FS) = 9.800 m, ΔElev = +1.250 m. Find missing FS. ΣBS − ΣFS_total = 1.250; 12.500 − ΣFS_total = 1.250; ΣFS_total = 11.250 m; Missing FS = 11.250 − 9.800 = 1.450 m.

Approach

Use ΣBS − ΣFS = ΔElev. Substitute known sums and solve for the missing BS or FS value. This is algebra applied to the summation equation.

Question Type

Find a missing reading that makes the check pass

Key Points To Remember

ΣBS − ΣFS = Elev_final − Elev_initial — this must hold exactly (purely arithmetic).
IFS readings are EXCLUDED from ΣBS and ΣFS in the arithmetic check.
The arithmetic check catches addition/subtraction errors only — NOT field blunders.
Loop misclosure = Elev_computed_closing_BM − Elev_known_closing_BM; this reflects actual measurement error.
Allowable misclosure for ordinary leveling: ±12√K mm (K in km) per standard practice; for precise leveling: ±4√K mm.
Each HI is valid only until the next foresight (TP or final point) — do not use an old HI for a new instrument position.

Profile and Cross-Section Leveling

PROFILE LEVELING produces a longitudinal section of the ground surface along a route centerline (road, pipeline, canal). The procedure is essentially differential leveling with many intermediate foresight readings taken at regular stations (typically every 20 m or every full station of 20 m in Philippine practice, or at breaks in slope). Station notation: Stations are labeled in multiples of 20 m (Philippine practice) or 100 ft (old US practice). In SI, stations are labeled as 0+000, 0+020, 0+040, etc., or sometimes as km markers. The ground profile is then plotted and compared against the design grade line to compute cut (excavation) and fill depths at each station. CROSS-SECTION LEVELING produces ground profiles perpendicular to the centerline at each station. These transverse profiles, combined with the design template (road cross-section), allow computation of earthwork volumes using the end-area method or prismoidal formula. For cross-sections: • Rod readings are taken at the centerline, and at regular offsets left (L) and right (R) of the centerline. • Elevations are computed using the same HI equation. • The 'five-level' cross-section: CL, L-5, L-10, R-5, R-10 (or at slope-stake positions for irregular terrain). Key distinction for the board exam: • Profile leveling uses IFS for centerline stations → many readings per setup. • Cross-section leveling also uses IFS for offset points. • ONLY TP readings enter the arithmetic check.

Examples

Notice that Sta. 0+040 is the only station in cut — the ground is above the design grade. All other stations require fill. The TP reading (3.125 m) is the largest rod reading, confirming the TP is the lowest point. The arithmetic check would use only ΣBS and this FS (3.125 m), NOT the IFS values.

Scenario

A level is set up with HI = 152.380 m. Rod readings are taken at the following stations along the centerline: Sta. 0+000 (IFS = 2.450 m), Sta. 0+020 (IFS = 1.985 m), Sta. 0+040 (IFS = 1.210 m), Sta. 0+060 (FS on TP = 3.125 m). The design grade at all stations is 150.500 m. Find ground elevations and determine cut or fill.

Solution

Ground elevations (using Elev = HI − rod reading): Sta. 0+000: 152.380 − 2.450 = 149.930 m Sta. 0+020: 152.380 − 1.985 = 150.395 m Sta. 0+040: 152.380 − 1.210 = 151.170 m Sta. 0+060 (TP): 152.380 − 3.125 = 149.255 m Design grade = 150.500 m at all stations. Cut (+) or Fill (−): Sta. 0+000: 149.930 − 150.500 = −0.570 m → FILL 0.570 m Sta. 0+020: 150.395 − 150.500 = −0.105 m → FILL 0.105 m Sta. 0+040: 151.170 − 150.500 = +0.670 m → CUT 0.670 m Sta. 0+060: 149.255 − 150.500 = −1.245 m → FILL 1.245 m

Applications

Design of longitudinal profiles for national highways (DPWH Road Design Guidelines).
Earthwork quantity estimation for bid documents under RA 9184 (Government Procurement Reform Act).
Setting stakes for cut and fill during road construction.
Pipe invert elevation determination for gravity sewer and drainage systems.
Railway and LRT alignment surveys.

Misconceptions

Using IFS readings to compute a new HI — IFS readings ONLY give station elevations, not a new HI.
Confusing cut and fill directions — CUT means digging (ground above grade), FILL means placing material (ground below grade).
Forgetting to reset HI when a new TP is established in a long profile level run.
Applying the arithmetic check including IFS readings — only BS (on TPs/BM) and FS (on TPs/closing BM) are summed.

Related Concepts

Differential leveling fundamentals
Earthwork computation — end-area method, prismoidal formula
Mass diagram (mass-haul curve) for earthwork optimization
Slope staking for irregular terrain
Grade line design and vertical curves

Common Exam Questions

Example

HI = 85.420 m, IFS at Sta. 3+40 = 1.760 m. Find ground elevation. Elev = 85.420 − 1.760 = 83.660 m.

Approach

Identify the current HI. Apply Elev = HI − rod reading for each IFS station. Remember IFS readings do not change HI. Only FS on a TP changes the HI (requiring a new BS from the TP).

Question Type

Compute ground elevation from profile leveling data with IFS readings

Example

Ground elev = 18.350 m, design grade = 19.000 m. Difference = 18.350 − 19.000 = −0.650 m → FILL 0.650 m.

Approach

Compute ground elevation from leveling data. Compare with design (grade) elevation. Cut = ground − design (if positive); Fill = design − ground (if positive).

Question Type

Determine cut or fill depth at a station

Key Points To Remember

Profile leveling = longitudinal ground profile along the route centerline.
Cross-section leveling = transverse ground profiles at each station.
IFS readings at intermediate stations DO NOT change the HI.
Ground elevation at any station = HI − rod reading at that station.
Cut = ground elevation − grade elevation (ground is ABOVE design grade).
Fill = grade elevation − ground elevation (ground is BELOW design grade).
Station numbering in Philippine practice: metric, every 20 m (0+000, 0+020, 0+040...).

Curvature and Refraction Corrections

For short sights (< 300 m), the Earth's surface can be treated as flat and the line of sight as perfectly horizontal. However, for long sights — common in geodetic leveling, trigonometric leveling, and precise surveys — two physical effects introduce significant errors: 1. CURVATURE OF THE EARTH (h_c): The Earth curves away below a truly horizontal line of sight. A distant rod will read HIGHER than the true difference in elevation would indicate (the rod intercepts the line of sight above where a level surface would be). Curvature effect (exact, based on Earth radius R ≈ 6,371 km): h_c = K²/(2R) ≈ 0.0785 K² (m, K in km) 2. ATMOSPHERIC REFRACTION (h_r): The atmosphere bends (refracts) the line of sight downward toward the denser air near the ground. This partially cancels the curvature effect by making distant objects appear HIGHER. Refraction correction: h_r ≈ 0.0112 K² (m, K in km) [approximately 1/7 of curvature] 3. COMBINED (NET) CORRECTION: Since curvature makes readings too high and refraction reduces this error but does not eliminate it, the net correction is: h_cr = h_c − h_r = 0.0785K² − 0.0112K² ≈ 0.0673K² ≈ 0.0675K² (m, K in km) The combined correction h_cr = 0.0675K² is SUBTRACTED from the rod reading to obtain the corrected rod reading, because the apparent rod reading is too large (the line of sight hits the rod above the true horizontal level). Corrected rod reading = Observed rod reading − h_cr OR equivalently, the corrected elevation of the distant point = computed elevation − h_cr Note: The constant 0.0675 is the value universally used in Philippine board exams and textbooks (Fayos, La Putt, Gillesania). Some references give 0.0673 — accept either unless the problem specifies.

Examples

For a 2 km sight, the combined correction is 0.270 m — significant enough to affect precise elevation work. This correction must be SUBTRACTED from the rod reading to get the true elevation difference. In differential leveling with balanced backsight and foresight distances, curvature and refraction cancel and can be ignored.

Scenario

Find the combined curvature-and-refraction correction for a sight distance of 2 km.

Solution

Given: K = 2 km h_cr = 0.0675 × K² = 0.0675 × (2)² = 0.0675 × 4 = 0.270 m

At 3.5 km, the error from ignoring curvature and refraction is 0.827 m — enormous for engineering work. This scenario is typical of geodetic leveling or trigonometric leveling over long distances. The board exam will test whether you (a) use K in km, (b) square it, and (c) subtract (not add) the correction.

Scenario

A surveyor reads a rod at a distance of 3.5 km. The observed rod reading is 2.847 m. The instrument elevation (HI) is 500.000 m. Find the corrected elevation of the rod point.

Solution

Step 1: Compute h_cr. K = 3.5 km h_cr = 0.0675 × (3.5)² = 0.0675 × 12.25 = 0.827 m Step 2: Corrected rod reading. Corrected FS = Observed FS − h_cr = 2.847 − 0.827 = 2.020 m (The observed reading is too large by 0.827 m due to curvature and refraction.) Step 3: Corrected elevation of rod point. Elev = HI − Corrected FS = 500.000 − 2.020 = 497.980 m (Using the uncorrected reading: Elev = 500.000 − 2.847 = 497.153 m — an error of 0.827 m!)

This example shows the individual components. For 800 m, the combined correction is only 43 mm — small but not negligible for precise work. The board exam sometimes asks for curvature only, refraction only, or combined — know all three formulas and their constants.

Scenario

A sight of 800 m is taken. Find the curvature correction alone, the refraction correction alone, and the combined correction.

Solution

K = 800 m = 0.800 km Curvature only: h_c = 0.0785 × (0.800)² = 0.0785 × 0.64 = 0.050 m Refraction only: h_r = 0.0112 × (0.800)² = 0.0112 × 0.64 = 0.007 m Combined correction: h_cr = 0.0675 × (0.800)² = 0.0675 × 0.64 = 0.043 m Verification: h_c − h_r = 0.050 − 0.007 = 0.043 m ✓

Applications

Geodetic leveling for national vertical control networks (NAMRIA).
Trigonometric leveling over long distances (mountainous terrain, bridge surveys).
Reciprocal leveling across rivers or ravines where balanced sights are impossible.
Precise spirit leveling for dam and reservoir projects.
GPS-derived ellipsoidal heights converted to orthometric heights (geoid undulation).

Misconceptions

Using K in meters instead of km in the formula h_cr = 0.0675K² — this gives an answer 1,000,000 times too large.
Adding the correction to the rod reading instead of subtracting — curvature makes readings too LARGE, so subtract.
Confusing the direction of the correction: curvature makes the distant point appear LOWER than it really is (or equivalently, the rod reading appears too HIGH).
Ignoring curvature for sights less than 100 m — while negligible for ordinary leveling, some board problems specifically ask for the value even at short distances.
Using 0.0785 (curvature only) instead of 0.0675 (combined) or vice versa when the problem specifies which correction to apply.

Related Concepts

Reciprocal leveling (eliminates curvature and refraction)
Trigonometric leveling with curvature and refraction correction
Geodetic leveling and vertical datum
Atmospheric refraction in horizontal distance measurement
Geoid and ellipsoid — distinction between ellipsoidal and orthometric heights

Common Exam Questions

Example

K = 1.5 km: h_cr = 0.0675 × 1.5² = 0.0675 × 2.25 = 0.152 m.

Approach

Convert sight distance to km. Apply h_cr = 0.0675K². State the direction of correction (subtract from rod reading).

Question Type

Compute h_cr for a given sight distance

Example

Observed elev of point (from HI − FS) = 215.000 m, K = 4 km. Corrected elevation = 215.000 + 0.0675×16 = 215.000 + 1.080 = 216.080 m. (Adding to elevation is equivalent to subtracting from FS.)

Approach

Compute h_cr. Subtract from observed rod reading (making it smaller) OR add to the computed elevation (making it higher). This is where many examinees make sign errors.

Question Type

Correct an observed elevation or rod reading for curvature and refraction

Example

h_cr = 0.300 m. K² = 0.300/0.0675 = 4.444. K = √4.444 = 2.108 km = 2,108 m.

Approach

Set h_cr = given value. Solve K² = h_cr / 0.0675. Take the square root to get K in km, then convert to meters if required.

Question Type

Find the sight distance at which curvature and refraction equals a given value

Key Points To Remember

Curvature makes distant rod readings TOO HIGH (line of sight is above the true level surface).
Refraction partially corrects this by bending the line of sight downward.
Net combined correction: h_cr = 0.0675K² (m, K in km) — this is SUBTRACTED from the rod reading.
K MUST be in kilometers in the formula — convert meters to km before computing.
The formula is parabolic (K²) — doubling the sight distance quadruples the correction.
For reciprocal leveling (reading from both ends), curvature and refraction effects cancel each other.
Refraction ≈ 1/7 × curvature (h_r ≈ 0.0112K², h_c ≈ 0.0785K²).

Two-Peg Test (Collimation Error Detection)

The two-peg test checks whether the level's line of sight is truly horizontal when the bubble is centered — this error is called the collimation error or C-factor error. If the instrument has a collimation error, the line of sight tilts slightly upward or downward even when the instrument appears level. PROCEDURE: 1. Set two pegs (A and B) approximately 60 m apart. 2. Set up the level at the midpoint M (30 m from each peg). • Read rod on A: r_A1. Read rod on B: r_B1. • True elevation difference: Δh_true = r_A1 − r_B1 (errors cancel at equal distances). 3. Move the level very close (within 3–5 m) to peg B. Read rod on B: r_B2. Read rod on A: r_A2. • Near reading r_B2 has negligible collimation error. • Expected rod reading on A (if no error): r_A_expected = r_B2 − Δh_true • Collimation error per meter: C = (r_A2 − r_A_expected) / d_AB 4. If C ≠ 0, adjust the line of sight using the instrument's capstan screws until the correct reading is obtained on the far rod. The two-peg test is a standard field calibration procedure in Philippine DPWH and NAMRIA surveys, and board exam problems involving it typically ask you to find the true elevation difference and/or identify the collimation error per meter of sight.

Examples

The collimation error of 0.895 mm/m means for every 100 m of sight, the instrument introduces a 89.5 mm error. The line of sight tilts upward (reads too high on the far rod). Adjusting the instrument to read the correct value (1.660 m) on peg A eliminates this systematic error.

Scenario

Two pegs A and B are 80 m apart. From midpoint M (40 m from each): rod on A = 1.628 m, rod on B = 0.914 m. Level moved near peg B (distance from B = 4 m, distance from A = 76 m): rod on B = 0.946 m, rod on A = 1.728 m. Determine: (a) true elevation difference, (b) correct rod reading on A from near-B position, (c) collimation error.

Solution

(a) True elevation difference (from midpoint setup, errors cancel): Δh = r_A − r_B = 1.628 − 0.914 = +0.714 m (A is 0.714 m higher than B) (b) Correct rod reading on A from near-B position: From near B: r_B2 = 0.946 m (nearly error-free, 4 m distance) If A is 0.714 m higher than B: r_A_correct = r_B2 + Δh = 0.946 + 0.714 = 1.660 m (c) Actual rod reading on A from near-B position = 1.728 m Error in far reading = 1.728 − 1.660 = +0.068 m (reading too high → line of sight tilts upward) Sight distance to A = 76 m Collimation error = 0.068/76 = 0.000895 m/m = 0.895 mm/m To correct: adjust the instrument to read 1.660 m on peg A rod (not 1.728 m).

Applications

Pre-survey instrument calibration for DPWH and NAMRIA leveling projects.
Quality assurance in precise (first and second order) leveling operations.
Determining when a level needs factory adjustment or recalibration.
Minimizing systematic errors in long level runs by ensuring balanced sight distances.

Misconceptions

Thinking the midpoint readings give HI or absolute elevations — they only give the TRUE relative elevation difference.
Using the eccentric (near-one-peg) readings for both A and B to compute Δh — only the near reading is error-free; the far reading contains the collimation error you are trying to find.
Confusing the direction of adjustment — if the far reading is too high, the line of sight tilts upward and must be adjusted down.

Related Concepts

Curvature and refraction corrections
Differential leveling procedure
Instrument adjustment and calibration
Systematic vs. random errors in surveying
Balanced backsight and foresight distances in precise leveling

Common Exam Questions

Example

From midpoint: r_A = 2.105 m, r_B = 1.843 m. Δh_true = 2.105 − 1.843 = 0.262 m (A higher than B).

Approach

Use the midpoint readings where collimation errors cancel. Δh_true = r_A − r_B from the midpoint setup.

Question Type

Find the true elevation difference between two pegs from a two-peg test

Example

Near peg B: r_B = 1.200 m; Δh = B is 0.300 m higher than A; r_A_correct = 1.200 + 0.300 = 1.500 m.

Approach

From the near-peg setup, the near reading is error-free. Use the true Δh to compute the expected far reading. Compare with actual far reading to find the error.

Question Type

Find the correct rod reading on the far peg from the eccentric setup

Key Points To Remember

At equal BS and FS distances, collimation errors CANCEL — always balance sight distances in precise leveling.
The two-peg test uses the midpoint setup to determine the TRUE elevation difference (errors cancel at equal distance).
The eccentric setup (near one peg) reveals the collimation error by comparing the actual far-rod reading with the expected value.
Collimation error is expressed in mm per meter or mm per km of sight.
The two-peg test should be performed at the start of each project survey and at regular intervals.

Practice Problems

The net elevation DECREASED by 0.860 m from BM-1 to BM-2 (ΣFS > ΣBS), meaning the terrain generally descended. The arithmetic check confirms the table is arithmetically correct. In an actual field survey, BM-2 would have a known elevation and you would compare 199.140 m to that known value to find the loop misclosure.

Problem

PROBLEM 1 (Board-Style): A level run proceeds from BM-1 to BM-2 with the following data: BM-1 elevation = 200.000 m Setup 1: BS on BM-1 = 1.625 m; FS on TP-1 = 2.840 m Setup 2: BS on TP-1 = 3.215 m; FS on TP-2 = 0.965 m Setup 3: BS on TP-2 = 2.480 m; FS on TP-3 = 3.610 m Setup 4: BS on TP-3 = 1.125 m; FS on BM-2 = 1.890 m Determine: (a) elevation of each TP, (b) elevation of BM-2, (c) verify using the arithmetic check.

Solution

SOLUTION: (a) & (b) Tabular computation: Setup 1: HI₁ = 200.000 + 1.625 = 201.625 m Elev_TP1 = 201.625 − 2.840 = 198.785 m Setup 2: HI₂ = 198.785 + 3.215 = 202.000 m Elev_TP2 = 202.000 − 0.965 = 201.035 m Setup 3: HI₃ = 201.035 + 2.480 = 203.515 m Elev_TP3 = 203.515 − 3.610 = 199.905 m Setup 4: HI₄ = 199.905 + 1.125 = 201.030 m Elev_BM2 = 201.030 − 1.890 = 199.140 m (c) Arithmetic Check: ΣBS = 1.625 + 3.215 + 2.480 + 1.125 = 8.445 m ΣFS = 2.840 + 0.965 + 3.610 + 1.890 = 9.305 m ΣBS − ΣFS = 8.445 − 9.305 = −0.860 m Elev_BM2 − Elev_BM1 = 199.140 − 200.000 = −0.860 m ✓ CHECK PASSES.

This is a typical board exam profile leveling problem combining leveling computation with grade line comparison. Stations 0+060 and 0+080 require significant excavation (cut), while Stations 0+020 and 0+040 need minor fill. The TP at 0+100 requires fill as well. Note that the IFS readings at intermediate stations do NOT affect the HI — the HI remains 85.625 m throughout this setup.

Problem

PROBLEM 2 (Board-Style): A profile level run is established with one instrument setup (HI = 85.625 m). The following rod readings are taken: Sta. 0+000: BS on BM = 1.425 m (BM elev = 84.200 m) — this IS the setup Sta. 0+020: IFS = 2.115 m Sta. 0+040: IFS = 1.875 m Sta. 0+060: IFS = 1.340 m Sta. 0+080: IFS = 0.910 m Sta. 0+100: FS on TP-1 = 2.650 m The design grade drops at 0.5% from Sta. 0+000 (design elev = 84.000 m). Determine: (a) HI, (b) ground elevation at each station, (c) cut (+) or fill (−) at each station.

Solution

(a) HI = 84.200 + 1.425 = 85.625 m ✓ (given, confirming our computation) (b) Ground elevations (Elev = HI − IFS): Sta. 0+000: 85.625 − 1.425 = 84.200 m [this is the BM, correct] Sta. 0+020: 85.625 − 2.115 = 83.510 m Sta. 0+040: 85.625 − 1.875 = 83.750 m Sta. 0+060: 85.625 − 1.340 = 84.285 m Sta. 0+080: 85.625 − 0.910 = 84.715 m Sta. 0+100 (TP-1): 85.625 − 2.650 = 82.975 m (c) Design grade at 0.5% downgrade from Sta. 0+000 (design elev = 84.000 m): Grade drop per meter = 0.005 m/m Sta. 0+000: design = 84.000 m Sta. 0+020: design = 84.000 − 0.005×20 = 83.900 m Sta. 0+040: design = 84.000 − 0.005×40 = 83.800 m Sta. 0+060: design = 84.000 − 0.005×60 = 83.700 m Sta. 0+080: design = 84.000 − 0.005×80 = 83.600 m Sta. 0+100: design = 84.000 − 0.005×100 = 83.500 m Cut (+) / Fill (−) [= Ground − Design]: Sta. 0+000: 84.200 − 84.000 = +0.200 m → CUT Sta. 0+020: 83.510 − 83.900 = −0.390 m → FILL Sta. 0+040: 83.750 − 83.800 = −0.050 m → FILL Sta. 0+060: 84.285 − 83.700 = +0.585 m → CUT Sta. 0+080: 84.715 − 83.600 = +1.115 m → CUT Sta. 0+100: 82.975 − 83.500 = −0.525 m → FILL

At 4.5 km, the combined correction is 1.367 m — a substantial error that would cause serious problems in engineering design. The observed rod reading (1.450 m) is too large because the line of sight is above the true level surface at the rod location. Subtracting h_cr from the rod reading (or adding to the elevation) gives the corrected result. This problem illustrates why curvature and refraction are critical for long-distance leveling.

Problem

PROBLEM 3 (Board-Style): A surveyor takes a rod reading of 1.450 m on a point that is 4.5 km away. The HI at the instrument is 312.500 m. (a) Find the combined curvature-and-refraction correction. (b) Find the corrected elevation of the distant point. (c) What would the elevation be if the correction were ignored?

Solution

(a) Combined curvature-and-refraction correction: K = 4.5 km h_cr = 0.0675 × K² = 0.0675 × (4.5)² = 0.0675 × 20.25 = 1.367 m (b) Corrected elevation: Corrected FS = Observed FS − h_cr = 1.450 − 1.367 = 0.083 m Corrected Elev = HI − Corrected FS = 312.500 − 0.083 = 312.417 m (Alternatively: Uncorrected Elev + h_cr = 311.050 + 1.367 = 312.417 m) (c) If correction ignored: Elev = HI − Observed FS = 312.500 − 1.450 = 311.050 m Error from ignoring correction = 312.417 − 311.050 = 1.367 m

The collimation error of 0.964 mm/m means the line of sight tilts upward. The actual far rod reading (1.588 m) is 53 mm too high compared to what it should be (1.535 m). At 55 m, this is already significant; at longer sight distances, the error would be proportionally larger. The instrument should be adjusted until the rod on B reads 1.535 m from the near-A position. DPWH specification typically allows no more than 1 mm/km for precise leveling.

Problem

PROBLEM 4 (Board-Style): In a two-peg test, pegs A and B are set 60 m apart. The level is set up at the midpoint M (30 m from each peg). Rod readings: rod on A = 2.365 m, rod on B = 1.798 m. The level is then moved and set up 5 m from peg A (55 m from peg B). New readings: rod on A = 2.102 m, rod on B = 1.588 m. (a) Determine the true elevation difference between A and B. (b) Determine the correct rod reading on the far peg (B) from the near-A setup. (c) What is the collimation error in mm per meter of sight?

Solution

(a) True elevation difference (midpoint setup — errors cancel): Δh = r_A − r_B = 2.365 − 1.798 = +0.567 m (Peg A is 0.567 m higher than peg B) (b) Correct rod reading on B from near-A setup: From near A (5 m): r_A = 2.102 m (nearly error-free, short sight) Since A is 0.567 m higher than B, rod on B should read LESS by 0.567 m: r_B_correct = r_A − Δh = 2.102 − 0.567 = 1.535 m (c) Actual rod reading on B from near-A setup = 1.588 m Error in far reading = 1.588 − 1.535 = +0.053 m (reads TOO HIGH) Sight distance to B = 55 m Collimation error = 0.053 m / 55 m = 0.000964 m/m = 0.964 mm/m Per kilometer: 0.964 mm/m × 1000 m/km = 964 mm/km (very large — instrument needs adjustment!)

This problem tests multiple skills: (1) correctly excluding IFS from the arithmetic check and elevation computation, (2) computing misclosure, and (3) applying the allowable misclosure formula. The run's actual misclosure of 70 mm far exceeds the allowable 27.4 mm, indicating possible blunders (misread rod, unstable TP, or instrument malfunction). The surveyor must re-run the level line. If the misclosure had been within tolerance, it would be distributed among the TPs by the transit rule or equal-distribution method.

Problem

PROBLEM 5 (Board-Style, Synthesis): A level run from BM-X (Elev = 500.000 m) to BM-Y (known Elev = 502.850 m) yields the following: ΣBS = 15.320 m, ΣFS = 12.540 m. An intermediate foresight sum (IFS only) = 4.885 m. (a) Compute the elevation of BM-Y as derived from the level run. (b) Determine the loop misclosure. (c) If the run length is 5.2 km, what is the allowable misclosure using ±12√K mm?

Solution

(a) Elevation of BM-Y from level run: Note: IFS readings do NOT enter the elevation computation via ΣBS−ΣFS. ΔElev = ΣBS − ΣFS = 15.320 − 12.540 = +2.780 m (IFS sum of 4.885 m is NOT included here) Computed Elev_BM-Y = 500.000 + 2.780 = 502.780 m (b) Loop misclosure: Misclosure = Computed Elev − Known Elev = 502.780 − 502.850 = −0.070 m = −70 mm (The run falls 70 mm short of the known elevation — negative misclosure) (c) Allowable misclosure: K = 5.2 km Allowable = ±12√K = ±12√5.2 = ±12 × 2.280 = ±27.4 mm Actual misclosure = 70 mm >> Allowable = 27.4 mm → The level run FAILS to meet ordinary leveling standards and must be RE-RUN.

Exam Preparation Tips

MEMORIZE the two master equations first: HI = Elev + BS and Elev = HI − FS. All leveling problems reduce to these two formulas.
ALWAYS tabulate level notes in proper column format (Station | BS | HI | IFS | FS | Elevation) even on scratch paper — it prevents errors and makes the arithmetic check easy.
For the arithmetic check, circle your BS and FS values only (not IFS) and sum them separately. If ΣBS − ΣFS ≠ Elev_last − Elev_first, you have an arithmetic error — do NOT proceed.
Convert K to KILOMETERS before applying h_cr = 0.0675K². If K is given in meters, divide by 1000 first. Forgetting this conversion is the most common curvature-refraction error.
Know the sign rule for curvature and refraction: the correction is SUBTRACTED from the rod reading (or ADDED to the computed elevation). Remember — curvature makes the rod reading appear too large.
Recognize IFS questions: if a problem gives multiple rod readings between TPs, those are IFS. Compute their elevations using the CURRENT HI but do NOT include them in ΣBS or ΣFS.
For two-peg test problems: midpoint setup gives TRUE Δh (errors cancel); near-peg setup reveals the collimation error. Always compute the 'expected' far rod reading and compare with the actual.
Know the allowable misclosure formulas: ±12√K mm (ordinary), ±4√K mm (precise), with K in km. These appear directly in board exam questions about acceptable accuracy.
In profile leveling problems, always check whether a grade is ascending or descending before computing cut/fill. A 0.5% downgrade means elevation drops 0.005 m per meter of horizontal distance.
During the actual board exam, if a leveling table is given with a blank cell, use the arithmetic check relationship first to see if the missing value can be determined from ΣBS − ΣFS = ΔElev before computing cell by cell.
Practice mental arithmetic for HI and elevation chains — board exam time pressure is real. If Elev = round number, a large BS gives a round HI — check your numbers for reasonableness.
Review the distinction between elevation (of a point) and height of instrument (HI) — many students confuse these in complex multi-setup problems with numerous intermediate points.

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In summary

Leveling is one of the most straightforward yet most error-prone topics in the PRC Civil Engineer board examination. The mathematics is simple — two formulas, HI = Elev + BS and Elev = HI − FS — but the complexity lies in correctly organizing multi-setup level notes, rigorously applying the arithmetic check (while excluding IFS readings), and making the curvature-refraction correction with the correct sign and units. For exam success, internalize these critical rules: 1. BS always adds (to establish HI from a known point); FS always subtracts (to find an unknown elevation from HI). 2. IFS readings give station elevations but do NOT change the HI and are NOT included in the arithmetic check. 3. The arithmetic check (ΣBS − ΣFS = ΔElev) catches arithmetic errors; the loop misclosure detects field measurement errors — these are different checks. 4. Curvature and refraction: h_cr = 0.0675K² (K in km) is subtracted from the rod reading (or added to the computed elevation). 5. The two-peg test midpoint setup gives the TRUE elevation difference (errors cancel); the near-peg setup reveals the collimation error. With diligent practice using the worked examples and board-style problems in this chapter, you will be well-prepared to handle any leveling question the PRC examinations present. Consistent tabulation, careful arithmetic, and disciplined unit conversion are your best defenses against errors under exam pressure. Magsipag, magtiyaga — and master this fundamental surveying skill.

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