NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersMeasuring land, levels and volumes, from chain survey to total station and GNSS, and designing land grading.
In the exams: GATE AG Section 4 (surveying asked in 7 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; TNPSC Unit I; RPSC AAE Unit 1; NCEES PE surveying
NCEES PE Agricultural and Biological Engineering
Surveying appears within common system applications: layout and grade staking, earthwork volumes, pond and channel grades, and interpretation of survey data in US customary or SI units.
How our questions for NCEES PE Agricultural and Biological Engineering spread across the units of Surveying & Levelling. Difficulty: 0 easy, 10 medium, 10 hard.
| Total station | 5 | |
| GPS/GNSS surveying | 5 | |
| Levelling: rise-and-fall and height-of-instrument methods | 2 | |
| Computation of area: trapezoidal and Simpson's rules | 2 | |
| Earthwork volume computation | 2 | |
| Chain and tape surveying, including tape corrections | 1 | |
| Contouring | 1 | |
| Land grading and levelling design | 1 | |
| Compass traversing and bearings | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A staff is read from a level. The staff reading is 3.000 m when held vertical, but the staff was actually tilted 3° from the vertical in the direction of sight. What reading was observed (to 3 decimals), and what is the lesson?
C. 3.004 m; a tilted staff always reads too high
The horizontal line cuts the tilted staff at a graduation of 3.000/cos 3° = 3.004 m. Tilt in any direction increases the reading, which is why the staffman rocks the staff and the minimum reading is taken.
Further reading: Punmia, Jain and Jain, Surveying Vol. I
An engineer must set out the corners of a farm pond from design coordinates using a total station. Which on-board function is used?
C. Stake-out
Stake-out computes the bearing and distance from the occupied station to each design point and guides the prism holder onto it. Missing-line measures between two observed points; remote elevation finds heights of inaccessible objects.
Further reading: Ghilani and Wolf, Elementary Surveying
A surveyor occupies an unknown point and observes angles and distances to two or more control points of known coordinates to fix the instrument's position. This total-station procedure is called
B. resection (free-station set-up)
Resection computes the occupied station's coordinates (and orientation) from observations to known points, which lets the instrument be placed wherever it has a good view. Intersection fixes a distant point from two known stations; ordinary orientation requires the instrument to be on a known point.
Further reading: Ghilani and Wolf, Elementary Surveying
An RTK GNSS survey gives an ellipsoidal height of 18.6 m at a point where the geoid undulation N from the geoid model is −52.3 m. What is the orthometric height (height above the geoid)?
C. 70.9 m
H = h − N = 18.6 − (−52.3) = 70.9 m. Where the geoid lies below the ellipsoid (negative N), orthometric heights exceed ellipsoidal heights. Adding N gives -33.7 m; using the ellipsoidal height as if it were above mean sea level is the classic mistake.
Further reading: Ghilani and Wolf, Elementary Surveying
Which GNSS technique gives centimetre-level positions in real time for staking out a land-grading design?
D. Real-time kinematic (RTK) with a base station or reference network
RTK uses carrier-phase corrections from a base or network sent by radio or internet, and resolves positions to a few centimetres in seconds. Autonomous positioning is metre-level; code DGPS is sub-metre to metre; static gives very high accuracy but only after processing, not in real time.
Further reading: Ghilani and Wolf, Elementary Surveying
A borrow pit is laid out as four 10 m × 10 m squares (a 2 × 2 grid). Depths of cut at the four outer corners are 1.2, 0.8, 1.0 and 1.4 m; at the four mid-side points 1.1, 0.9, 1.3 and 1.2 m; and at the centre point 1.05 m. What is the volume of excavation?
C. 440.0 m³
V = (A/4)(Σh1 + 2Σh2 + 4Σh4) with A = 100 m²: Σh1 = 4.4, Σh2 = 4.5, h4 = 1.05. V = 25(4.4 + 9.0 + 4.2) = 440.0 m³. Averaging all nine depths over 400 m² (442.2 m³) ignores how many squares share each corner. 248.8 m³ applies A/4 without the 2 and 4 weights; 1760.0 m³ uses A = 400 m² instead of 100 m².
Further reading: Punmia, Jain and Jain, Surveying Vol. I
A land-grading design gives total cut 1450 m³ and total fill 1320 m³. The engineer wants a cut:fill ratio of about 1.3 for this soil. What should be done?
D. Lower the design plane slightly and recompute until cut/fill reaches about 1.3
Current ratio = 1450/1320 = 1.10, below the target. Lowering the plane increases every cut and reduces every fill, raising the ratio; raising it would lower the ratio. The ratio allows for settlement and losses and its value depends on soil and conditions.
Further reading: Schwab et al., Soil and Water Conservation Engineering
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