CELE Surveying (Geomatics) — Route, Topographic and Modern SurveyingExam Answer Templates

Marks

1

Topic

Topographic Surveying and Contours

Difficulty

easy

Template Id

T1

Examiner Tip

The word 'imaginary' is not required but 'equal elevation' and 'map' must appear. One tight sentence earns the full mark.

Model Answer

A contour line is an imaginary line on a map that connects all points having the same elevation above a reference datum.

Question Type

very_short_answer

Answer Structure

• One sentence: define contour line using 'same elevation' and 'reference datum' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'same height above sea level' only — misses the generality of 'reference datum'
• Defining it as a physical line on the ground rather than on a map
• Omitting that points are connected/joined

Key Phrases To Include

• same elevation
• imaginary line
• reference datum
• equal elevation

Marks

1

Topic

Topographic Surveying and Contours

Difficulty

easy

Template Id

T2

Examiner Tip

The keyword 'vertical' is the discriminator. Board questions use this term to test whether students confuse vertical interval with horizontal map spacing.

Model Answer

The contour interval is the constant vertical distance (difference in elevation) between two successive contour lines on a map.

Question Type

very_short_answer

Answer Structure

• One sentence: define contour interval as the constant vertical distance between successive contours [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'horizontal distance between contours' — this is the map spacing, not the contour interval
• Omitting that the interval is constant

Key Phrases To Include

• constant
• vertical distance
• successive contour lines
• difference in elevation

Marks

2

Topic

Photogrammetry

Difficulty

easy

Template Id

T3

Examiner Tip

Emphasizing 'above the ground' for H is the most tested nuance. Examiners specifically look for this qualifier.

Model Answer

Photo scale: Scale = f / H Where: f = focal length of the camera (m or mm) H = flying height of the aircraft above the ground (terrain) (m) The scale is expressed as a dimensionless ratio, e.g., 1 : 10,000.

Question Type

short_answer

Answer Structure

• Line 1: Write the formula Scale = f / H [1 mark]
• Line 2: Define f (focal length) and H (flying height above ground) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining H as height above mean sea level — the working scale uses height above terrain
• Not expressing the result as a ratio (e.g., writing only a decimal)
• Inverting the formula as H / f

Key Phrases To Include

• focal length
• flying height above the ground
• f / H
• dimensionless ratio

Marks

2

Topic

Topographic Surveying and Contours

Difficulty

medium

Template Id

T4

Examiner Tip

The mnemonic 'V-alley, V-upstream' links the landform name to the rule. Write the word 'upstream' explicitly to earn the phrase mark.

Model Answer

1. In a valley, contour V-shapes point upstream (uphill), i.e., the apex of the V is directed toward the higher elevation. 2. In a ridge, contour V-shapes point downhill (away from the high ground), i.e., the apex of the V points toward the lower elevation.

Question Type

short_answer

Answer Structure

• Point 1: Valley — V points upstream/uphill [1 mark]
• Point 2: Ridge — V points downhill [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the direction — stating valley V's point downhill (this is the ridge characteristic)
• Using vague language like 'contours curve' without specifying direction

Key Phrases To Include

• V-shapes point upstream
• V-shapes point downhill
• apex
• higher elevation
• lower elevation

Marks

2

Topic

Photogrammetry

Difficulty

easy

Template Id

T5

Examiner Tip

Always convert focal length to meters to match H in meters before dividing. Writing 'above ground' next to H explicitly demonstrates awareness of the most common pitfall.

Model Answer

Given: f = 150 mm = 0.150 m H = 3,000 m above ground Formula: Scale = f / H Solution: Scale = 0.150 / 3,000 = 1 / 20,000 The photo scale is 1 : 20,000.

Question Type

numerical

Answer Structure

• Step 1: List given data and convert f to meters [0.5 mark]
• Step 2: Write the formula Scale = f / H [0.5 mark]
• Step 3: Substitute and compute [0.5 mark]
• Step 4: Express answer as a scale ratio [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to convert f from mm to m before dividing
• Not expressing the final answer as a ratio
• Using H as height above MSL instead of above terrain

Key Phrases To Include

• f = 0.150 m
• H = 3,000 m above ground
• Scale = f / H
• 1 : 20,000

Marks

2

Topic

Photogrammetry

Difficulty

easy

Template Id

T6

Examiner Tip

The phrase 'scale denominator' is the key to this formula. Always write 'ground distance = photo distance × scale denominator' so the examiner can award the formula mark even if arithmetic slips.

Model Answer

Given: Photo scale = 1 : 20,000 → scale denominator = 20,000 Photo distance = 65 mm Formula: Ground distance = photo distance × scale denominator Solution: Ground distance = 65 mm × 20,000 = 1,300,000 mm = 1,300 m The ground distance between the two road intersections is 1,300 m.

Question Type

numerical

Answer Structure

• Step 1: Identify photo distance and scale denominator [0.5 mark]
• Step 2: Write ground distance formula [0.5 mark]
• Step 3: Multiply and convert mm to m [0.5 mark]
• Step 4: State final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Dividing instead of multiplying (reversing the scale relationship)
• Leaving the answer in mm without converting to m
• Using scale as a fraction (1/20,000) and getting confused with multiplication direction

Key Phrases To Include

• scale denominator = 20,000
• ground distance = photo distance × scale denominator
• 1,300,000 mm
• 1,300 m

Marks

3

Topic

Topographic Surveying and Contours

Difficulty

medium

Template Id

T7

Examiner Tip

Always include the note 'count intervals, not lines' — board examiners set deliberate traps around this distinction. Writing it demonstrates exam-level awareness.

Model Answer

Given: Contour interval (CI) = 5 m Number of intervals crossed = 6 Horizontal distance (D) = 150 m Step 1: Compute change in elevation. ΔElev = number of intervals × CI = 6 × 5 m = 30 m Step 2: Compute average slope. Slope = ΔElev / D = 30 m / 150 m = 0.20 Step 3: Convert to percentage. Slope (%) = 0.20 × 100 = 20% The average ground slope is 20%. Note: Count intervals (spaces between lines), not the number of contour lines crossed.

Question Type

numerical

Answer Structure

• Step 1: Compute ΔElev = intervals × CI [1 mark]
• Step 2: Compute slope ratio = ΔElev / horizontal distance [1 mark]
• Step 3: Convert to percentage and state answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Counting contour lines instead of intervals (e.g., using 7 lines instead of 6 intervals)
• Not converting to percentage — leaving answer as 0.20
• Dividing D by ΔElev instead of ΔElev by D

Key Phrases To Include

• ΔElev = intervals × CI
• slope = ΔElev / horizontal distance
• count intervals not lines
• 20%

Marks

3

Topic

Hydrographic Surveying and Photogrammetry

Difficulty

medium

Template Id

T8

Examiner Tip

Use the word 'soundings' for hydrographic surveying — it is the technical term examiners recognize. For photogrammetry, the phrase 'measurements from photographs' is the distinguishing definition.

Model Answer

Photogrammetry is the science of making measurements from photographs (aerial or terrestrial) to determine the shape, size, and position of objects on or above the Earth's surface. Application: Generation of topographic maps and digital elevation models (DEMs) from aerial photos for urban planning and road alignment design. Hydrographic surveying is the survey and measurement of water bodies — including water depths (soundings), shoreline positions, tidal data, and underwater features — for nautical charting and marine works. Application: Dredging operations in Philippine ports (e.g., Manila Harbor, Cebu Port) where depth soundings guide excavation volumes.

Question Type

short_answer

Answer Structure

• Part 1: Correct definition of photogrammetry + one valid application [1.5 marks]
• Part 2: Correct definition of hydrographic surveying + one valid application [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Confusing hydrographic surveying with hydrological surveying (river flow measurement)
• Giving the same application for both
• Defining photogrammetry as 'taking aerial photographs' rather than 'making measurements from photographs'

Key Phrases To Include

• measurements from photographs
• water depths
• soundings
• digital elevation model
• nautical charting
• tidal datum

Marks

3

Topic

Photogrammetry

Difficulty

hard

Template Id

T9

Examiner Tip

This is the classic 'terrain correction' photogrammetry problem. Always write Part (a) explicitly — show that H = altitude − terrain elevation — before computing scale. Examiners award a dedicated mark for this step.

Model Answer

Given: Flying altitude above MSL = 3,150 m Average terrain elevation above MSL = 150 m Focal length, f = 210 mm = 0.210 m (a) Effective flying height above terrain: H = altitude above MSL − terrain elevation = 3,150 m − 150 m = 3,000 m (b) Photo scale: Scale = f / H = 0.210 / 3,000 = 1 / 14,286 ≈ 1 : 14,286 The effective flying height is 3,000 m above terrain, and the photo scale is approximately 1 : 14,286.

Question Type

numerical

Answer Structure

• Step 1: Compute H = altitude − terrain elevation [1 mark]
• Step 2: Convert f to meters [0.5 mark]
• Step 3: Apply Scale = f / H and express as ratio [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using altitude above MSL directly as H (H = 3,150 m) — the most common board-exam trap
• Not converting focal length from mm to m
• Rounding to 1 : 14,000 without noting it is an approximation

Key Phrases To Include

• H = altitude above MSL − terrain elevation
• 3,000 m above terrain
• f = 0.210 m
• Scale = f / H
• 1 : 14,286

Marks

3

Topic

Modern Positioning — GNSS/RTK

Difficulty

medium

Template Id

T10

Examiner Tip

Structure each advantage as a contrast: 'RTK does X, whereas conventional traverse requires Y.' This format earns full phrase marks and demonstrates comparative understanding.

Model Answer

RTK GNSS improves field efficiency over conventional optical traverse in the following ways: 1. No line of sight required — GNSS receivers compute positions from satellite signals, eliminating the need for direct intervisibility between stations, which is essential in dense vegetation and built-up Philippine urban areas. 2. Rapid positioning — RTK achieves centimeter-level accuracy in seconds after initialization, whereas conventional traverse requires instrument setup, backsight alignment, and sequential angle and distance measurement at every point. 3. Reduced crew size — A single operator with a rover can occupy control points independently; conventional traverse typically requires at least two instrument persons plus rodmen. Bonus point (if space allows): RTK outputs digital coordinates directly to field controllers or data loggers, eliminating manual booking errors and reducing office computation time.

Question Type

short_answer

Answer Structure

• Advantage 1: No line-of-sight requirement — explain [1 mark]
• Advantage 2: Speed / rapid cm-level positioning [1 mark]
• Advantage 3: Reduced crew / direct digital output [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving advantages that are generic (e.g., 'more accurate' without specifying how or why)
• Confusing RTK with static GNSS — RTK is real-time; static requires post-processing
• Not contrasting with conventional traverse — the question asks for comparison

Key Phrases To Include

• line of sight
• centimeter-level accuracy
• satellite signals
• RTK initialization
• digital coordinates
• crew size

Marks

2

Topic

Topographic Surveying and Contours

Difficulty

easy

Template Id

T11

Examiner Tip

Board examiners frequently ask exactly four rules. Memorize these four as a numbered list. The V-direction rule combined with valley/ridge differentiation earns the most discriminating marks.

Model Answer

Four characteristics of contour lines: 1. All points on a contour line have the same elevation — a contour line is a line of equal elevation. 2. Contour lines never cross each other (except in the case of an overhanging cliff or cave). 3. Closely spaced contour lines indicate steep terrain; widely spaced lines indicate gentle terrain. 4. In valleys, contour lines form V-shapes that point upstream (uphill); in ridges, V-shapes point downhill.

Question Type

short_answer

Answer Structure

• Rule 1: Equal elevation [0.5 mark]
• Rule 2: Contours never cross [0.5 mark]
• Rule 3: Spacing indicates slope steepness [0.5 mark]
• Rule 4: V-direction rule for valleys/ridges [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'contour lines represent elevation' without saying they connect equal elevations
• Giving only spacing or only V-direction but not both for the last two rules
• Listing 'contours are parallel' as a rule — this is only true for uniform slopes

Key Phrases To Include

• equal elevation
• never cross
• closely spaced = steep
• V-shapes point upstream
• V-shapes point downhill

Marks

5

Topic

Modern Positioning and Photogrammetry

Difficulty

hard

Template Id

T12

Examiner Tip

5-mark long answers are scored on depth AND breadth. Use a structured format with clear headings or bullet points under each method. The recommendation paragraph must justify synthesis — stating that the two methods are complementary (not mutually exclusive) earns the final integration mark.

Model Answer

COMPARISON: Total Station Traverse vs. Drone-Based Photogrammetry for 5 km² Mapping (a) Conventional Total Station Traverse • Accuracy: High positional accuracy (mm to cm level) for individual control and stakeout points; ideal for precise alignment geometry. • Time: Labor-intensive — crews must occupy each instrument station sequentially; 5 km² may require several days depending on terrain and vegetation. • Cost: Lower equipment capital cost (total station rental widely available in the Philippines), but high labor cost for extended fieldwork. • Data Output: Discrete point coordinates, cross-sections, and profiles — requires manual interpolation for contour generation. • Limitation: Requires line of sight between stations; dense vegetation in Rizal highlands obstructs visibility. (b) Drone-Based Photogrammetry (UAV) • Accuracy: Ground accuracy of 5–10 cm with GCP (ground control points) densification; sufficient for preliminary and final route alignment mapping. • Time: A single flight mission can cover 5 km² in 1–2 hours; total project turnaround (flight + processing) in 1–2 days. • Cost: Higher initial equipment investment, but significantly lower labor cost over large areas; cost per hectare decreases rapidly with area. • Data Output: Dense 3D point cloud, orthophoto, and digital elevation model (DEM) — directly generates contours at any specified interval; spatially continuous coverage. • Advantage: No line-of-sight constraints; overflies vegetation and irregular terrain. Recommendation: For a 5 km² route alignment study in Rizal Province, drone-based photogrammetry is the more appropriate primary method. It delivers spatially continuous terrain data (DEM and orthophoto) across the entire corridor efficiently and cost-effectively, which is essential for horizontal and vertical alignment design. However, conventional total station or RTK GNSS should be used to establish and survey ground control points (GCPs) to georeference the photogrammetric model to the Philippine Reference System 1992 (PRS 92). The two methods are complementary — photogrammetry for areal coverage, total station for precision control. Conclusion: Drone photogrammetry is recommended for efficiency and comprehensive data output; total station/GNSS provides the control framework for legal accuracy.

Question Type

long_answer

Answer Structure

• Part (a): Discuss total station traverse — accuracy, time, cost, data output, limitation (minimum 4 points) [2 marks]
• Part (b): Discuss drone photogrammetry — accuracy, time, cost, data output, advantage (minimum 4 points) [2 marks]
• Recommendation + justification with integration of both methods [1 mark]

Scoring Breakdown

Common Mark Deductions

• One-sided answer that only recommends one method without genuine comparison
• Not mentioning GCPs — board examiners consider this a critical element of drone photogrammetry workflow
• Omitting data output comparison — this is a key differentiator between methods
• Generic statements (e.g., 'drone is faster') without specific data or context

Key Phrases To Include

• total station
• line of sight
• drone photogrammetry
• ground control points (GCPs)
• digital elevation model (DEM)
• orthophoto
• point cloud
• PRS 92
• complementary methods
• cost per hectare

Marks

1

Topic

Hydrographic Surveying

Difficulty

medium

Template Id

T13

Examiner Tip

Any recognized tidal low-water datum earns the mark. The reason (conservative for navigation) is equally important for the second half-mark.

Model Answer

The primary datum for sounding depths in hydrographic surveying is the tidal datum — typically Mean Lower Low Water (MLLW) or Lowest Astronomical Tide (LAT). This datum is chosen because it represents the lowest expected water level, ensuring that charted depths are conservative and safe for navigation (vessels will not encounter shallower water than shown on the chart).

Question Type

very_short_answer

Answer Structure

• State the datum (tidal datum / MLLW / LAT) [0.5 mark]
• Give the reason (conservative/safe navigation) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'mean sea level' — MSL is an elevation datum, not the sounding datum
• Giving the datum without explaining why it is used

Key Phrases To Include

• tidal datum
• MLLW
• LAT
• lowest water level
• navigational safety
• conservative

Marks

5

Topic

Photogrammetry

Difficulty

hard

Template Id

T14

Examiner Tip

Always include a verification step in 5-mark numerical problems — it takes 10 seconds and demonstrates thoroughness. Examiners reward it when a method mark is given for checking.

Model Answer

Given: Focal length, f = 152 mm = 0.152 m Photo length of bridge = 38 mm Actual ground length of bridge = 760 m (a) Photo Scale: Scale = photo distance / ground distance = 38 mm / (760 m × 1,000 mm/m) = 38 / 760,000 = 1 / 20,000 ∴ Photo scale = 1 : 20,000 (b) Flying Height above terrain: Since Scale = f / H: H = f / Scale = f × scale denominator = 0.152 m × 20,000 = 3,040 m ∴ Flying height above terrain = 3,040 m Verification: Scale = f / H = 0.152 / 3,040 = 1/20,000 ✓

Question Type

numerical

Answer Structure

• Step 1: Write scale formula using photo and ground distances [1 mark]
• Step 2: Convert ground length to mm and compute scale ratio [1 mark]
• Step 3: Express scale as 1 : 20,000 [1 mark]
• Step 4: Rearrange Scale = f/H to solve for H [1 mark]
• Step 5: Compute H = 3,040 m and verify [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting ground distance to mm (mixing units in the photo/ground ratio)
• Inverting H = H/f instead of H = f × denominator
• Omitting verification — shows carelessness to examiners

Key Phrases To Include

• Scale = photo distance / ground distance
• 1 : 20,000
• H = f × scale denominator
• 3,040 m
• flying height above terrain

Marks

2

Topic

Modern Surveying — GIS

Difficulty

easy

Template Id

T15

Examiner Tip

The four functions — store, manage, analyze, visualize — are the textbook definition of GIS. Include all four in one sentence to guarantee the definition mark.

Model Answer

A Geographic Information System (GIS) is a computer-based system that stores, manages, analyzes, and visualizes spatially referenced data (geographic data) in digital map layers. In modern surveying practice, GIS serves as the primary platform for integrating survey data (coordinates, elevations, boundaries) with attribute data (land use, ownership, infrastructure) to support engineering planning, land management, and decision-making. Survey results from total stations, GNSS, and photogrammetry are ultimately imported into GIS for spatial analysis and map production.

Question Type

short_answer

Answer Structure

• Part 1: Define GIS — stores, manages, analyzes, and visualizes spatially referenced/geographic data [1 mark]
• Part 2: Role in surveying — integrates survey data with attributes; destination for total station, GNSS, and photogrammetry outputs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining GIS as only a 'map-making software' — misses the analysis and database management functions
• Not connecting GIS to surveying outputs (total station, GNSS, photogrammetry data)

Key Phrases To Include

• spatially referenced data
• stores, manages, analyzes, visualizes
• map layers
• attribute data
• spatial analysis
• integration of survey data