Surveying
6–9 of 110 itemsNCEES draws between 6 and 9 of the 110 from this area on any legal paper. Its outline lists 5 sub-topics; the bank holds 31 verified templates against them, which expand to 927 questions.
Sub-topics, from the NCEES outline
- AAngles, distances, and trigonometry
- BArea computations
- CEarthwork and volume computations
- DCoordinate systems (e.g., state plane, latitude/longitude)
- ELeveling (e.g., differential, elevations, percent grades)
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A level circuit starts on a benchmark at elevation 150.00 ft. The crew reads a backsight of 2.25 ft, then a foresight of 7.75 ft to a turning point. From the next setup they read a backsight of 6.50 ft and a foresight of 3.00 ft to point B. The elevation of point B is most nearly:
Differential level circuit starting on a benchmark at elevation 150.00 feet: a backsight of 2.25 ft raises the line of sight, and a foresight of 7.75 ft on the turning point brings it back down. Show the answer and the reasoning
Why A
Given: BM 150.00 ft; BS 2.25, FS 7.75; BS 6.50, FS 3.00. Relation: elevation = BM + sum of backsights - sum of foresights. Substitute: 150.00 + 2.25 - 7.75 + 6.50 - 3.00. Result: elevation of B = 148.00 ft. Check: carrying the HI through each setup (HI 152.25, TP 144.50, HI 151.00) gives the same elevation.
Why B is tempting
Averaged the sights instead of summing them.
Why C is tempting
Dropped the final foresight.
Why D is tempting
Reversed the signs: subtracted the backsights and added the foresights.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Surveying
Question 2 of 5
A traverse course is 650 ft long with an azimuth of 220 degrees. Its latitude is most nearly:
Show the answer and the reasoning
Why B
Given: L = 650 ft, azimuth = 220 degrees. Relation: latitude = L cos(azimuth); the cosine carries the sign (south is negative). Substitute: 650 cos(220°) = 650(-0.76604). Result: latitude = -497.93 ft.
Why A is tempting
Divided by the cosine instead of multiplying.
Why C is tempting
Computed the departure. Latitude is the north-south component and uses the cosine of the azimuth.
Why D is tempting
Used a tangent, which is not a projection at all.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Surveying
Question 3 of 5
A triangular parcel has corners at (80, 20), (225, 50) and (250, 250). Its area by coordinates is most nearly:
Triangular parcel with corners at (80, 20), (225, 50) and (250, 250), taken in order around the boundary. Show the answer and the reasoning
Why A
Given: corners (80, 20), (225, 50), (250, 250). Relation: 2A = |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substitute: 2A = |80(-200) + 225(230) + 250(-30)| = 28250. Result: A = 1.413e+4 ft^2. Check: the order of the corners changes only the sign of the sum, not the area.
Why B is tempting
Omitted the division by two. The cross-product sum gives twice the enclosed area.
Why C is tempting
Kept only the forward products and dropped the backward ones.
Why D is tempting
Added the coordinate pairs where the determinant subtracts them.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Surveying
Question 4 of 5
Three consecutive cross sections 100 ft apart have end areas of 300, 240 and 260 ft^2. The total volume between the outer two sections is most nearly:
Three consecutive cross sections 100 ft apart with end areas of 300, 240 and 260 square feet; the outer two are shown. Show the answer and the reasoning
Why B
Given: end areas 300, 240, 260 ft^2 at 100-ft spacing. Relation: V = sum of ((Ai + Ai+1)/2)L over each interval, /27 for yd^3. Substitute: 27000 + 25000 = 52000 ft^3; /27. Result: V = 1926 yd^3.
Why A is tempting
Computed only the first of the two intervals.
Why C is tempting
Averaged only the two end sections, ignoring the middle one entirely.
Why D is tempting
Summed the areas without averaging adjacent pairs.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Surveying
Question 5 of 5
A course has a latitude of 550 ft north and a departure of 900 ft east. Its azimuth is most nearly:
Show the answer and the reasoning
Why A
Given: latitude 550 ft north, departure 900 ft east, so the course lies in the north-east quadrant. Relation: azimuth = arctan(departure/latitude), measured clockwise from north. Substitute: arctan(900/550) = arctan(1.6364). Result: azimuth = 58.57 degrees.
Why B is tempting
Measured the angle from south, as for a south-east bearing. The latitude is north, so the course lies in the north-east quadrant.
Why C is tempting
Added 180 degrees, giving the back azimuth.
Why D is tempting
Measured the angle counter-clockwise from north. Azimuths run clockwise.
Source NCEES FE Reference Handbook 10.6 — Civil Engineering, Surveying
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 2 min
Getting the sign of a latitude or departure from a bearing
Latitude is length times the cosine of the azimuth and departure is length times its sine, with the azimuth measured clockwise from north, and the signs come out on their own. Working from a bearing such as S 40° E, the angle is measured from north or south and the signs must be set by quadrant. Mixing the two conventions flips a sign, and the flipped value is offered.
Costs the question and about 2 min
Adding a foresight to the height of instrument
Elevation plus backsight gives the height of instrument; height of instrument minus foresight gives the next elevation. Reversing either sign is a one-reading error whose result is always among the choices.
Costs the question and about 3 min
Averaging the end areas to get the prismoidal mid-section
The prismoidal formula needs the area measured at the middle section. Using the average of the two end areas instead reproduces the average end area result exactly, which is offered beside the prismoidal one.
Costs the question and about 2 min
Reading a zenith angle as a vertical angle
A total station's zenith angle is measured down from straight up, so horizontal distance is slope distance times the sine of the zenith angle. Treating it as a vertical angle measured from the horizontal, and taking the cosine, gives the vertical component instead, and that value is offered.
Questions about Surveying
What surveying math does the FE Civil exam test?
Angles and distances, bearings and azimuths, traverse latitudes and departures, areas by coordinates or offsets, earthwork volumes, coordinate systems, and differential leveling. It is trigonometry and careful bookkeeping more than theory.
Are horizontal and vertical curves part of Surveying?
The NCEES outline places them under Transportation Engineering's geometric design. Surveying's sub-topics are angles and distances, area, earthwork, coordinate systems and leveling, though curve stationing uses the same arithmetic.
Do I need state plane coordinates?
Coordinate systems, including state plane and latitude and longitude, are a sub-topic. Expect the ground-to-grid reduction: a ground distance times the combined factor, which is the grid scale factor times the elevation factor, gives the grid distance.
How are earthwork volumes computed on the exam?
By average end area between cross sections, by the prismoidal formula when a mid-section is given, or by the borrow-pit grid method. Results in cubic feet are divided by 27 to give cubic yards, and cut is often adjusted for shrinkage or swell before it is balanced against fill.
How many study hours this area is worth · Using the handbook under the clock