Surveying

6% of exam

Angles, distances, and trigonometry, area computations, earthwork and volume computations, coordinate systems, and differential leveling and grades.

3 concepts

Exam areas

A. Angles, distances, and trigonometry

Traverse, Bearings, and Coordinates

Bearings versus azimuths, the interior-angle check, latitudes and departures, traverse closure, the compass (Bowditch) rule, and coordinate computation — the spine of plane surveying.

A traverse is the backbone of almost every survey: a connected chain of measured lines whose directions and lengths you reduce to a set of coordinates. Get the direction conventions and the closure adjustment right and everything downstream — areas, layout, earthwork — inherits that accuracy; get a quadrant or a sign wrong and the whole figure is off. On the FE Civil exam the points here are won or lost in three bookkeeping steps: converting between bearings and azimuths, resolving each line into a latitude (north–south) and a departure (east–west), and distributing the small closure error back through the traverse. The FE Reference Handbook's Civil Engineering chapter gives you exactly one figure — the sign convention for latitudes and departures — and expects you to supply the rest from first principles, so fluency, not lookup, is what scores.

Bearings versus azimuths

A direction can be named two ways and you must move freely between them. An azimuth is a single angle measured clockwise from north, running

0^\circ

360^\circ

— unambiguous and ideal for computation. A bearing is an acute angle (

0^\circ

90^\circ

) measured from the nearer end of the meridian, tagged with the quadrant, e.g.

\text{N}\,35^\circ\text{W}

\text{S}\,52^\circ18'\text{E}

. Convert by quadrant: in the NE quadrant azimuth equals the bearing angle; in SE it is

180^\circ

minus the bearing; in SW it is

180^\circ

plus the bearing; in NW it is

360^\circ

minus the bearing. A back azimuth is simply the forward azimuth

\pm 180^\circ

, and a back bearing flips both letters (N

\leftrightarrow

S, E

\leftrightarrow

W) while keeping the angle.

\text{Az}_{NE}=\beta,\quad \text{Az}_{SE}=180^\circ-\beta,\quad \text{Az}_{SW}=180^\circ+\beta,\quad \text{Az}_{NW}=360^\circ-\beta

Angular closure and the interior-angle check

Before you trust any directions, check the angles. For a closed polygon traverse with

n

sides the interior angles must sum to

(n-2)\,180^\circ

—

360^\circ

for a four-sided figure,

540^\circ

for five sides. The difference between your measured sum and the geometric sum is the angular misclosure; if it is within tolerance (commonly on the order of the instrument's least count times

\sqrt{n}

), distribute it equally among the angles. Only after the angles balance do you carry a starting direction around the loop, applying each adjusted angle to get the bearing or azimuth of every course.

\sum \theta_{\text{int}} = (n-2)\,180^\circ

Latitudes and departures

Each course is resolved into its north–south component, the latitude, and its east–west component, the departure, using the azimuth or bearing angle

\alpha

measured from the meridian. The latitude carries the cosine and the departure the sine. Honor the signs from the handbook's convention: latitude is positive toward north and negative toward south; departure is positive toward east and negative toward west. A line running

\text{S}\,40^\circ\text{W}

therefore has a negative latitude and a negative departure. These signed components are what you sum and adjust — never the raw distances.

\text{Lat} = L\cos\alpha, \qquad \text{Dep} = L\sin\alpha

Traverse closure: misclosure and precision

A perfectly measured closed traverse returns to its start, so the algebraic sums of latitudes and of departures should each be zero. In practice they are small nonzero numbers — the closure in latitude

\sum\text{Lat}

and the closure in departure

\sum\text{Dep}

. Their resultant is the linear misclosure

e

, and dividing the total perimeter

\sum L

e

gives the precision, reported as a ratio

1\!:\!(\sum L/e)

. A boundary survey might require

1\!:\!5000

or better; a rough topo loop far less. Compute the precision first — if it fails, you re-measure rather than adjust.

e=\sqrt{\left(\textstyle\sum\text{Lat}\right)^2+\left(\textstyle\sum\text{Dep}\right)^2},\qquad \text{Precision}=\frac{\sum L}{e}

The compass (Bowditch) rule

Once the precision passes, you distribute the misclosure. The compass rule — also called the Bowditch rule — assumes angular and linear errors are of comparable quality, so it apportions the closure to each course in proportion to that course's length relative to the perimeter. The correction to a course's latitude is the negative of the total latitude closure times the course length over the perimeter, and likewise for departures. Apply the corrections so the adjusted latitudes and the adjusted departures each sum to exactly zero; those adjusted components, not the originals, feed the coordinates.

C_{\text{Lat},i}=-\left(\textstyle\sum\text{Lat}\right)\frac{L_i}{\sum L},\qquad C_{\text{Dep},i}=-\left(\textstyle\sum\text{Dep}\right)\frac{L_i}{\sum L}

Coordinate computation

With balanced latitudes and departures, coordinates are a running sum: the northing of the next station equals the current northing plus the course's adjusted latitude, and the easting equals the current easting plus the adjusted departure. Carry these around the loop and you must land back on the starting coordinates exactly — a built-in arithmetic check. Going the other way, the inverse problem recovers a line's length and bearing from two known coordinate pairs: the length is the hypotenuse of the latitude and departure differences, and the bearing angle comes from their arctangent (with the quadrant set by the signs of

\Delta N

and

\Delta E

N_{i+1}=N_i+\text{Lat}_i,\quad E_{i+1}=E_i+\text{Dep}_i;\qquad L=\sqrt{\Delta N^2+\Delta E^2},\ \tan\beta=\left|\frac{\Delta E}{\Delta N}\right|

Exam strategy

Standardize your workflow: (1) balance angles to

(n-2)180^\circ

; (2) compute a consistent set of azimuths; (3) build a Lat/Dep table with signs; (4) sum, get

e

and precision; (5) apply the compass rule; (6) accumulate coordinates and verify you return to the start. Keep azimuths in decimal degrees in your calculator and convert bearings only at input and output. The single most common error is the quadrant sign — write the N/S/E/W tag next to every latitude and departure before you sum. If a problem only asks for closure or precision, you can stop at step 4 and skip the adjustment entirely.

Key equations

Latitude and departure

\text{Lat}=L\cos\alpha,\qquad \text{Dep}=L\sin\alpha

Components of a course of length $L$ (ft) at angle $\alpha$ from the meridian. Lat positive N / negative S; Dep positive E / negative W.

Bearing-to-azimuth (by quadrant)

\text{Az}=\begin{cases}\beta & \text{NE}\\ 180^\circ-\beta & \text{SE}\\ 180^\circ+\beta & \text{SW}\\ 360^\circ-\beta & \text{NW}\end{cases}

Azimuth measured clockwise from north, $0^\circ$ – $360^\circ$ ; $\beta$ = bearing angle ( $0^\circ$ – $90^\circ$ ).

Back azimuth

\text{Az}_{back}=\text{Az}\pm 180^\circ

Use $+180^\circ$ if the forward azimuth is below $180^\circ$ , else $-180^\circ$ . Back bearing reverses both letters, same angle.

Interior-angle check

\sum\theta_{\text{int}}=(n-2)\,180^\circ

Geometric sum of interior angles for an $n$ -sided closed polygon traverse; basis of the angular misclosure check.

Angular misclosure correction

C_\theta=-\frac{\sum\theta_{\text{meas}}-(n-2)180^\circ}{n}

Equal correction applied to each measured angle when the misclosure is within tolerance.

Linear misclosure

e=\sqrt{\left(\textstyle\sum\text{Lat}\right)^2+\left(\textstyle\sum\text{Dep}\right)^2}

Resultant gap closing the traverse (ft). Ideally zero; the closures in Lat and Dep are its components.

Precision (relative accuracy)

\text{Precision}=\frac{\sum L}{e}=1\!:\!\left(\sum L/e\right)

Perimeter $\sum L$ divided by linear misclosure; reported as a ratio. Larger denominator = better survey.

Compass (Bowditch) rule

C_{\text{Lat},i}=-\left(\textstyle\sum\text{Lat}\right)\frac{L_i}{\sum L},\quad C_{\text{Dep},i}=-\left(\textstyle\sum\text{Dep}\right)\frac{L_i}{\sum L}

Distributes closure in proportion to course length. Adjusted Lat and Dep then each sum to zero.

Forward coordinate computation

N_{i+1}=N_i+\text{Lat}_i,\qquad E_{i+1}=E_i+\text{Dep}_i

Running sum of adjusted latitudes (northings) and departures (eastings). Returns to start if balanced.

Inverse (length and bearing from coordinates)

L=\sqrt{\Delta N^2+\Delta E^2},\qquad \tan\beta=\left|\frac{\Delta E}{\Delta N}\right|

Recovers a line from two coordinate pairs; quadrant of the bearing set by the signs of $\Delta N$ and $\Delta E$ .

Worked examples

Bearing and azimuth conversions

Problem. (a) Convert the bearing $\text{N}\,35^\circ\text{W}$ to an azimuth. (b) Convert the bearing $\text{S}\,52^\circ18'\text{E}$ to an azimuth. (c) Convert an azimuth of $210^\circ00'$ to a quadrant bearing.

Solution. (a) NW quadrant, so

\text{Az}=360^\circ-35^\circ=325^\circ

. (b) First put the bearing in decimal:

52^\circ18' = 52.30^\circ

. SE quadrant, so

\text{Az}=180^\circ-52.30^\circ=127.70^\circ = 127^\circ42'

. (c)

210^\circ

lies between

180^\circ

and

270^\circ

, the SW quadrant; the bearing angle is

210^\circ-180^\circ=30^\circ

, giving

\text{S}\,30^\circ00'\text{W}

. Sanity check: each azimuth falls in the quadrant implied by its letters —

325^\circ

is NW,

127.7^\circ

is SE,

\text{S}\,30^\circ\text{W}

sits at

210^\circ

. Consistent.

\text{Az}=360^\circ-35^\circ=325^\circ

Angular closure of a five-sided traverse

Problem. The five measured interior angles of a closed traverse are $A=101^\circ30'00''$ , $B=88^\circ16'00''$ , $C=132^\circ09'30''$ , $D=120^\circ04'00''$ , $E=98^\circ02'00''$ . Find the angular misclosure and the adjusted angles.

Solution. Required sum:

(5-2)180^\circ = 540^\circ00'00''

. Measured sum:

101^\circ30'00''+88^\circ16'00''+132^\circ09'30''+120^\circ04'00''+98^\circ02'00'' = 540^\circ01'30''

. Misclosure

= 540^\circ01'30'' - 540^\circ00'00'' = +01'30'' = +90''

. Correction per angle

= -90''/5 = -18''

. Subtract

18''

from each:

A=101^\circ29'42''

B=88^\circ15'42''

C=132^\circ09'12''

D=120^\circ03'42''

E=98^\circ01'42''

. Sanity check: the five adjusted angles now sum to exactly

540^\circ00'00''

(

5\times18''=90''

removed). The misclosure of

90''

over five stations (

18''

each) is well within ordinary tolerance.

C_\theta=-\frac{540^\circ01'30''-540^\circ}{5}=-18''

Latitudes, departures, closure, and coordinates

Problem. A four-course closed traverse is measured as $AB=\text{N}\,45^\circ00'\text{E}$ , $150.00\,\text{ft}$ ; $BC=\text{S}\,60^\circ00'\text{E}$ , $200.00\,\text{ft}$ ; $CD=\text{S}\,40^\circ00'\text{W}$ , $175.00\,\text{ft}$ ; $DA=\text{N}\,52^\circ30'\text{W}$ , $210.40\,\text{ft}$ . Find the linear misclosure and precision, then the adjusted coordinates of $B$ and $C$ given $A=(N\,1000.00,\ E\,1000.00)$ .

Solution. Latitudes (

L\cos\alpha

, signed) and departures (

L\sin\alpha

, signed):

AB

\text{Lat}=+106.066

\text{Dep}=+106.066

BC

\text{Lat}=-100.000

\text{Dep}=+173.205

CD

\text{Lat}=-134.058

\text{Dep}=-112.488

DA

\text{Lat}=+128.083

\text{Dep}=-166.922

. Closures:

\sum\text{Lat}=+0.092

\sum\text{Dep}=-0.138

. Linear misclosure

e=\sqrt{0.092^2+0.138^2}=0.166\,\text{ft}

. Perimeter

\sum L = 735.40\,\text{ft}

, so precision

=735.40/0.166 = 4430 \Rightarrow 1\!:\!4400

. Compass rule on

AB

(

L/\sum L = 150.00/735.40 = 0.2040

C_{\text{Lat}}=-(+0.092)(0.2040)=-0.019

C_{\text{Dep}}=-(-0.138)(0.2040)=+0.028

. Adjusted

AB

\text{Lat}=106.047

\text{Dep}=106.094

. Coordinates:

B = (1000.00+106.047,\ 1000.00+106.094) = (N\,1106.05,\ E\,1106.09)

. Applying the adjusted

BC

(

\text{Lat}=-100.025

\text{Dep}=+173.243

C = (N\,1006.02,\ E\,1279.34)

. Sanity check: continuing through

D

and back to

A

returns exactly to

(N\,1000.00,\ E\,1000.00)

, confirming the adjustment balances. Precision

1\!:\!4400

is typical of a good total-station loop.

e=\sqrt{0.092^2+0.138^2}=0.166\,\text{ft},\quad \text{Precision}=\frac{735.40}{0.166}\approx 1\!:\!4400

Common pitfalls

•Quadrant sign errors: a course's latitude is negative when it runs south and the departure negative when it runs west. Tag every Lat/Dep with N/S/E/W before summing — this is the number-one traverse mistake.
•Swapping sine and cosine: latitude (north–south) takes the COSINE of the angle from the meridian; departure (east–west) takes the SINE. Reverse them only if you measure the angle from the east–west line instead.
•Adjusting before checking precision: never apply the compass rule until the misclosure passes tolerance. A large misclosure means a blunder to find, not an error to distribute.
•Using azimuths measured from south: U.S. plane surveying measures azimuths clockwise from NORTH. If a problem (or astronomic data) uses south azimuths, add/subtract $180^\circ$ first.
•Confusing the compass rule with the transit rule: Bowditch distributes by course length; the transit rule distributes by the magnitude of each latitude/departure. The FE expects the compass rule unless stated.
•Forgetting to balance angles first: directions carried from unbalanced angles propagate the angular error into every latitude and departure, corrupting the linear adjustment that follows.
•Reading northing as the x-coordinate: northing (latitude direction) is the first/Y value, easting (departure direction) is the second/X value. State-plane and most survey software list them N then E.

References

NCEES FE Reference Handbook — Civil Engineering (Latitudes and Departures)
NCEES FE Reference Handbook — Mathematics (Trigonometry)
Ghilani & Wolf, Elementary Surveying — Traverse computations and the compass rule — Standard derivation of Bowditch adjustment

B. Area computations

Area Computations

Area by coordinates (the shoelace formula), the DMD method, and the trapezoidal and Simpson's 1/3 rules for irregular boundaries defined by offsets.

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E. Leveling

Differential Leveling, Grades, and Earthwork

Backsight/foresight differential leveling and the height-of-instrument method, the level-circuit check, percent grades, and earthwork by average-end-area and the prismoidal formula.

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