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NCEES PE Agricultural and Biological Engineering Remote Sensing, GIS & GNSS for Land & Water: practice questions

NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)

Original practice questions written to the published syllabus · not past papers
10worked here, free
57written to this exam in this subject
6syllabus units

What this subject covers

Mapping land and water from satellites and in GIS: images, indices, elevation models and watershed delineation.

In the exams: ICAR AICE subject 40 Unit 5 (RS and GIS in SWC); TNPSC Unit III; NCEES PE GIS; skill modules in the ICAR 6th Deans' Committee B.Tech

What NCEES PE Agricultural and Biological Engineering asks from it

NCEES PE Agricultural and Biological Engineering (Common System Applications: graphics/CAD/GIS; surveying)

Applied use of GIS and GNSS data in design: coordinate systems and datums, map scale and area, contributing area from DEMs, curve-number and erosion inputs from spatial data, and accuracy of GNSS positions and heights.

Where the 57 questions fall

How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Remote Sensing, GIS & GNSS for Land & Water. Difficulty: 16 easy, 31 medium, 10 hard.

GIS data models11
GNSS11
Digital elevation models and watershed delineation10
Spatial analysis9
QGIS/ArcGIS practice9
Applications to watershed, irrigation and erosion mapping7

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · GIS data models · medium

On the Mercator projection, the scale factor at latitude φ is sec φ. How much is area exaggerated at 60° latitude compared with the equator?

  1. 1.4 times
  2. 2 times
  3. 8 times
  4. 4 times
Answer and working

D. 4 times

Linear scale = sec 60° = 2 in every direction, so area scale = sec² 60° = 4. 2 times is the linear scale only; 1.4 is sec 45°; 8 would require a cubic relationship.

Further reading: Snyder, Map Projections: A Working Manual (USGS PP 1395)

2 · Spatial analysis · medium

Computing the mean slope of each sub-watershed from a slope raster and a sub-watershed raster is an example of which map algebra operation?

  1. A zonal operation
  2. A local operation
  3. A global operation
  4. A focal operation
Answer and working

A. A zonal operation

Zonal operations summarise one layer within zones defined by another. Local operations work cell by cell; focal operations use a moving neighbourhood (slope itself is computed focally); global operations use the whole raster, such as Euclidean distance.

Further reading: Chang, Introduction to Geographic Information Systems

3 · Spatial analysis · hard

After intersecting a field-boundary layer with a soil-unit layer, many tiny thin polygons appear along boundaries that should coincide. What are they and how are they usually handled?

  1. Sliver polygons; remove them with a cluster tolerance or by eliminating small polygons
  2. Raster cells; convert them back to vector using a finer cell size
  3. Buffer zones; keep them because they mark uncertainty along each boundary
  4. Topology nodes; keep them because they record connectivity between arcs
Answer and working

A. Sliver polygons; remove them with a cluster tolerance or by eliminating small polygons

Slivers arise when the same boundary was digitised separately in two layers and the lines do not exactly match; they are removed by snapping/cluster tolerances or merging polygons below a size threshold. They are not nodes, buffers or raster artefacts.

Further reading: Chang, Introduction to Geographic Information Systems

4 · Digital elevation models and watershed delineation · hard

Why does automated D8 delineation often fail in flat, canal-irrigated plains, and what is a common remedy?

  1. Flat plains have no sinks at all; the remedy is to skip the fill step in the processing sequence
  2. Tiny height errors and artificial embankments dominate flow; burn in mapped drains and verify in the field
  3. Canal plains have too many streams; the remedy is to raise the accumulation threshold tenfold
  4. D8 cannot run on projected data; the remedy is to use geographic coordinates for the DEM
Answer and working

B. Tiny height errors and artificial embankments dominate flow; burn in mapped drains and verify in the field

Where relief is less than DEM error, and embankments, roads and canals control flow, D8 directions become arbitrary; mapped drainage lines are burned into the DEM and results are checked by survey. Flat DEMs usually have many spurious sinks; D8 should be run in projected coordinates; raising the threshold does not fix wrong directions.

Further reading: Chang, Introduction to Geographic Information Systems

5 · GNSS · easy

A receiver's clock is in error by 1 microsecond. What range error does this produce?

  1. About 300 m
  2. About 0.3 m
  3. About 30 km
  4. About 3 m
Answer and working

A. About 300 m

Range error = c × Δt = 3 × 10⁸ m/s × 1 × 10⁻⁶ s ≈ 300 m. 3 m corresponds to 10 ns; 30 km to 100 µs; 0.3 m to 1 ns.

Further reading: Hofmann-Wellenhof, Lichtenegger and Wasle, GNSS

6 · GNSS · medium

A farm needs centimetre-level positioning for auto-steer tractors and laying subsurface drain lines. Which technique is appropriate?

  1. Autonomous single-frequency code positioning with a handheld receiver
  2. Code-based differential GNSS from a radio beacon about 200 km away
  3. RTK using carrier-phase corrections from a base or reference network
  4. Wide-area SBAS corrections from a geostationary satellite broadcast
Answer and working

C. RTK using carrier-phase corrections from a base or reference network

Real-time kinematic positioning resolves carrier-phase ambiguities and gives centimetre accuracy over baselines of up to a few tens of km. Autonomous positioning gives metres; SBAS and code DGNSS give sub-metre to metre accuracy, not enough for drain grades.

Further reading: Hofmann-Wellenhof, Lichtenegger and Wasle, GNSS

7 · GNSS · hard

A surveyor notices that PDOP rises sharply when most visible satellites are clustered low in the eastern sky. Why?

  1. Clustered satellites broadcast on the same frequency and jam each other's signals
  2. Clustered satellites give poor geometry, so range errors map into large position errors
  3. Low satellites are closer to the receiver, which shortens the ranges below the error level
  4. Low satellites always have more accurate clocks, which disturbs the least-squares solution
Answer and working

B. Clustered satellites give poor geometry, so range errors map into large position errors

DOP expresses how satellite geometry amplifies range errors; when satellites are close together in the sky, their lines of sight are nearly parallel and the position is poorly determined. Clock accuracy and frequency sharing are not the cause, and low satellites are actually farther away.

Further reading: Hofmann-Wellenhof, Lichtenegger and Wasle, GNSS

8 · Applications to watershed, irrigation and erosion mapping · easy

A basin has 24 first-order and 6 second-order streams. What is the bifurcation ratio between these orders?

  1. 18
  2. 4.0
  3. 0.25
  4. 3.0
Answer and working

B. 4.0

Rb = N1/N2 = 24/6 = 4.0. 0.25 inverts the ratio; 18 is the difference; 3.0 would need 18 first-order streams.

Further reading: Horton; Subramanya, Engineering Hydrology

9 · GIS data models · medium

What is a triangulated irregular network (TIN)?

  1. A network of stream lines coded by Strahler order for hydrological analysis
  2. A raster of equal square cells, each storing a single height value for its area
  3. A surface model of non-overlapping triangles joining irregularly spaced height points
  4. A set of contour lines stored as closed polygons with a height attribute each
Answer and working

C. A surface model of non-overlapping triangles joining irregularly spaced height points

A TIN connects irregularly spaced points (often by Delaunay triangulation) into triangles, placing more points where terrain changes quickly. A grid of cells is a raster DEM; contours are lines, not triangles; a stream network is a different dataset.

Further reading: Chang, Introduction to Geographic Information Systems

10 · Spatial analysis · medium

Reclassifying a slope raster into classes 1 (0–3%), 2 (3–8%) and 3 (above 8%) is an example of

  1. a local operation that changes each cell's value independently of its neighbours
  2. a global operation that needs every cell to compute the value of each cell
  3. a focal operation that uses a moving 3 × 3 window around each cell
  4. a zonal operation that summarises values inside separate polygon zones
Answer and working

A. a local operation that changes each cell's value independently of its neighbours

Reclassification replaces each cell's value using a lookup rule based only on that cell, so it is local. Computing slope itself is focal; summarising by zones is zonal; distance surfaces are global.

Further reading: Chang, Introduction to Geographic Information Systems

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Remote Sensing, GIS & GNSS for Land & WaterThe full study manual, free NCEES PE Agricultural and Biological EngineeringFormat, sections and how it has changed

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