Water Resources HubFE section 15 of 16 · free theory
Surveying
Level runs, tape corrections, bearings, traverse closure, and area — the Surveying slice of the FE Civil exam, where sign conventions and one missed conversion decide everything.
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Differential leveling
A level run carries elevation from a known benchmark through turning points. Two equations do all the work, plus one arithmetic check that catches nearly every blunder:
HI = elevBM + BS elevTP = HI − FS
| HI | height of instrument (elevation of the line of sight) |
| BS | backsight — rod reading on a point of known elevation |
| FS | foresight — rod reading on the point whose elevation is wanted |
ΣBS − ΣFS = last elevation − first elevation
The check equation is exact for any run — use it before you trust your answer. A backsight always adds (you are sighting back to known ground); a foresight always subtracts (you are reaching forward to new ground).
Worked example Level run with a check
Given:
- Benchmark BM1 elevation 100.00 ft; BS 4.32 ft, then FS 6.10 ft to TP1; new setup, BS 5.05 ft, then FS 3.75 ft to BM2.
Solution:
- HI1 = 100.00 + 4.32 = 104.32 ft; TP1 = 104.32 − 6.10 = 98.22 ft.
- HI2 = 98.22 + 5.05 = 103.27 ft; BM2 = 103.27 − 3.75 = 99.52 ft.
- Check: ΣBS − ΣFS = (4.32 + 5.05) − (6.10 + 3.75) = 9.37 − 9.85 = −0.48; 99.52 − 100.00 = −0.48. The run checks.
Answer: TP1 = 98.22 ft, BM2 = 99.52 ft.
Distance-measurement corrections
A steel tape is standardised at 68°F, standard pull, fully supported, and level. Field conditions differ, so measured distances get three corrections (a fourth, slope, applies when the tape is not horizontal):
Ct = α(T − T0)L Cp = (P − P0)L / (AE) Cs = −w²Ls³ / (24P²)
| Ct | temperature correction; α = 6.45×10−6/°F for steel, T0 = 68°F |
| Cp | pull (tension) correction; P measured pull, P0 standard pull, L in inches here |
| Cs | sag correction per unsupported span; w = tape weight per foot, Ls = span length (ft), P = pull (lb) |
Signs: hotter than standard lengthens the tape (Ct positive); extra pull stretches it (Cp positive); sag always shortens the measured distance (Cs always negative). Corrected distance = measured + Σcorrections. The tension formula needs L in inches when A is in in² and E in psi.
Bearings, azimuths, and their conversions
Bearings name a quadrant and an angle from the meridian (N 35° E); azimuths measure clockwise from north, 0° to 360°. Convert before you compute latitudes and departures, which need azimuths:
NE: Az = bearing SE: Az = 180° − bearing SW: Az = 180° + bearing NW: Az = 360° − bearing
Back azimuth = Az ± 180° (add if Az < 180°, subtract if Az > 180°)
The classic error is the SE/SW mix-up: S 42° E is 180 − 42 = 138°, while S 42° W is 180 + 42 = 222°. Say the quadrant out loud before you touch the calculator.
Worked example Azimuth to bearing and back
Given:
- A line has azimuth 280°.
Solution:
- 280° lies between 270° and 360° → NW quadrant. Bearing angle = 360° − 280° = 80° → N 80° W.
- Back azimuth = 280° − 180° = 100° (S 80° E), which is the same line sighted from the other end.
Answer: N 80° W; back azimuth 100°.
Traverse closure and the compass rule
A traverse is a sequence of legs; latitude is the north–south component and departure the east–west component. In a closed traverse both sums should be zero — the leftover is the closure error, and the compass rule spreads it across the legs in proportion to their lengths:
Lat = D cos(Az) Dep = D sin(Az) e = √[(ΣLat)² + (ΣDep)²] precision = e / perimeter
CorrLat,i = −ΣLat × (Di / perimeter) CorrDep,i = −ΣDep × (Di / perimeter)
| Az | azimuth from north, clockwise |
| e | linear misclosure, ft |
| precision | reported as 1 : (perimeter/e), e.g. 1:5,000 |
Latitude is north-positive, departure east-positive. The compass-rule correction has the opposite sign of the error sum — it cancels the misclosure leg by leg, with longer legs absorbing more. Adjusted coordinates = running sums of (Lat + CorrLat), (Dep + CorrDep).
Worked example Closure, compass rule, and area
Given:
- Four-leg traverse (azimuth, distance): (90°, 200 ft), (180°, 150 ft), (270°, 200.5 ft), (0°, 149.5 ft).
- Start at N 1,000.00, E 1,000.00.
Solution:
- Latitudes: 0, −150, 0, +149.5 → ΣLat = −0.50 ft. Departures: +200, 0, −200.5, 0 → ΣDep = −0.50 ft.
- Misclosure e = √(0.50² + 0.50²) = 0.707 ft; perimeter = 700 ft; precision = 700/0.707 ≈ 1:990.
- Compass rule, leg 1: CorrLat = +0.50 × 200/700 = +0.143 ft; CorrDep = +0.50 × 200/700 = +0.143 ft. Adjusted P2 = N 1,000.14, E 1,200.14.
- Adjusted coordinates close exactly on the start point; area by the coordinate method ≈ 29,987 ft² = 0.69 acre.
Answer: e ≈ 0.71 ft, precision ≈ 1:990; area ≈ 0.69 ac.
Area by coordinates
Given the (E, N) coordinates of a closed polygon in order, the shoelace (coordinate) formula gives the area directly — no need for the traverse to be regular:
A = ½|Σ(EiNi+1 − Ei+1Ni)|
| A | enclosed area, ft² (divide by 43,560 for acres) |
List vertices in order around the polygon (clockwise or counter-clockwise) and close the loop by repeating the first point at the end. The double-meridian-distance (DMD) method in the handbook is algebraically identical — use whichever you compute faster.
Topographic concepts
Contours connect points of equal elevation. The contour interval is the vertical step between adjacent contours (index contours, usually every fifth, are labelled); closer contours mean steeper ground, and contours never cross except at a vertical cliff or overhang. A closed contour with hachures marks a depression. The exam asks mostly for reading: elevation of a point between contours (interpolate linearly), or the average slope from the contour spacing over a horizontal distance.