Surveying

This chapter covers the following topics:

• Directions, angles, and distances
• Systematic errors in taped distances
• Traverses: latitudes, departures, and closure
• Area computations
• Earthwork and volume computations
• Coordinate systems
• Leveling and grades

FE surveying questions are short calculations built on right-triangle trigonometry. Most of the work is setting up the geometry correctly and keeping signs and units straight.

Directions, angles, and distances

Every survey computation starts from a direction, an angle, or a distance, and each comes with a convention that decides its sign.

Bearings and azimuths

Definitions

Azimuth

The horizontal angle measured clockwise from north to the line, from $0^{\circ }$ to $360^{\circ }$.

Bearing

The angle, from $0^{\circ }$ to $90^{\circ }$, between the line and the north-south meridian, written with its quadrant (for example, N 35°20’ E).

The back direction of a line, from its end back to its start, differs by $180^{\circ }$ in azimuth. For a bearing, keep the angle and swap both letters: the back bearing of N 35°20’ E is S 35°20’ W.

Example: bearing to azimuth

Convert S 35°20’ E to an azimuth, and find the back azimuth.

• Southeast quadrant: $Az=180^{\circ }-35^{\circ }20^{\prime }=144^{\circ }40^{\prime }$
• Back azimuth: $144^{\circ }40^{\prime }+180^{\circ }=324^{\circ }40^{\prime }$

Answer: $144^{\circ }40^{\prime }$, back azimuth $324^{\circ }40^{\prime }$

Measured angles

A total station measures the horizontal angle between a backsight (usually the previous station) and a foresight (the next). From the azimuths of the two sights, the angle turned clockwise is:

$$\text{Angle right}=A{z}_{FS}-A{z}_{BS}(\text{add }360^{\circ }\text{ if negative})$$

The interior angles of a closed traverse with $n$ sides sum to $(n-2)\times 180^{\circ }$, which is the first check on any closed traverse. A deflection angle is measured right or left from the prolongation of the previous line and is always less than $180^{\circ }$.

Example: angle from two azimuths

At station B, the azimuth from B to the backsight A is $210^{\circ }30^{\prime }$ and the azimuth from B to the foresight C is $95^{\circ }10^{\prime }$. Find the angle right.

$95^{\circ }10^{\prime }-210^{\circ }30^{\prime }=-115^{\circ }20^{\prime }$, so add $360^{\circ }$.

Answer: $244^{\circ }40^{\prime }$

Slope and horizontal distance

Instruments measure along the line of sight, but computations use horizontal distance. With slope distance $S$ and a vertical angle $\alpha$ (from horizontal) or zenith angle $z$ (from vertical):

$$H=S\cos \alpha =S\sin zV=S\sin \alpha =S\cos z$$

Example: horizontal distance from a zenith angle

A slope distance of 325.48 ft is read at a zenith angle of $86^{\circ }30^{\prime }$.

$$H=325.48\sin 86^{\circ }30^{\prime }=324.87\text{ ft}V=325.48\cos 86^{\circ }30^{\prime }=19.87\text{ ft}$$

Answer: $H=324.87$ ft, and the line of sight rises 19.87 ft

Systematic errors in taped distances

A steel tape reads its nominal length only at a standard temperature, a standard pull, and full support. Departures from those conditions cause systematic errors, which have a predictable size and sign and are removed by calculation.

• Incorrect tape length: $\text{True distance}=\text{Measured}\times \frac{L_{actual}}{L_{nominal}}$
• Temperature: $C_t=\alpha (T-T_s)L$, with $\alpha =6.45\times 10^{-6}$ per °F ($1.16\times 10^{-5}$ per °C) for steel and $T_s$ typically 68°F (20°C)
• Tension: $C_p=\frac{(P-P_s)L}{AE}$, where $P$ is the applied pull, $P_s$ the standard pull, and $AE$ the tape’s axial stiffness
• Sag: $C_s=-\frac{w^2{L}^3}{24P^2}$, where $w$ is the tape’s weight per unit length and $L$ the unsupported span

Reasoning the sign: a tape that is too long covers more ground per graduation, so it reads short and the correction is positive. A tape that is too short, too cold, or sagging reads long, so the correction is negative. When laying out a distance instead of measuring one, every sign flips.

Example: temperature correction

A steel tape standardized at 68°F measures 842.15 ft at 38°F.

$$C_t=(6.45\times 10^{-6})(38-68)(842.15)=-0.163\text{ ft}$$

Answer: $842.15-0.163=841.99$ ft

Example: a tape that is too short

A 100-ft tape is actually 99.96 ft long, and a distance measured with it reads 562.40 ft.

Answer: $562.40\times 99.96/100.00=562.18$ ft

Traverses

A traverse is a series of connected survey lines of measured length and direction. A closed traverse returns to its starting point or ends on another point of known position.

Latitudes and departures

Definitions

Latitude

The projection of a course onto the north-south axis, $L=l\cos \theta$. North is positive.

Departure

The projection of a course onto the east-west axis, $D=l\sin \theta$. East is positive.

With a bearing angle, attach signs from the quadrant letters. With an azimuth, sine and cosine carry the signs automatically.

Example: latitude and departure

A course is 500.00 ft long on a bearing of N 53°08’ E.

$$L=500.00\cos 53^{\circ }08^{\prime }=+299.98\text{ ft}D=500.00\sin 53^{\circ }08^{\prime }=+400.02\text{ ft}$$

Answer: 299.98 ft north and 400.02 ft east

Closure and precision

In a closed loop the latitudes and the departures should each sum to zero. What is left over measures the error:

$$\text{Linear misclosure}=\sqrt{{\left(\sum L\right)}^2+{\left(\sum D\right)}^2}\text{Precision}=1:\frac{P}{\text{linear misclosure}}$$

where $P$ is the perimeter. A larger second number means a better survey.

Example: misclosure and precision

A four-course loop with a perimeter of 1,674.50 ft has latitudes of $+300.12$, $-99.95$, $-400.06$, and $+199.97$ ft and departures of $+399.90$, $+200.07$, $-100.03$, and $-499.98$ ft.

• $\sum L=+0.08$ ft and $\sum D=-0.04$ ft
• Misclosure $=\sqrt{0.08^2+0.04^2}=0.0894$ ft
• Precision $=1:(1,674.50/0.0894)$

Answer: 0.0894 ft, a precision of about 1:18,700

The compass (Bowditch) rule balances the traverse by distributing the error in proportion to course length:

$$C_L=-\left(\sum L\right)\frac{l}{P}{C}_D=-\left(\sum D\right)\frac{l}{P}$$

For the 499.99-ft first course above, $C_L=-(0.08)(499.99/1,674.50)=-0.024$ ft and $C_D=+0.012$ ft.

Area computations

Three methods cover what the FE asks: coordinates, double meridian distances, and offsets from a baseline.

Coordinate method

For corners $(x_1,y_1),\dots ,(x_n,y_n)$ taken in order around the figure, with $x$ the easting and $y$ the northing:

$$A=\frac{1}{2}\left|\sum _{i=1}^n\left(x_i{y}_{i+1}-x_{i+1}{y}_i\right)\right|,(x_{n+1},y_{n+1})=(x_1,y_1)$$

Double meridian distance (DMD)

The DMD method finds area directly from a balanced traverse’s latitudes and departures:

• The DMD of the first course is its departure.
• Each following DMD is the previous DMD, plus the previous course’s departure, plus this course’s departure.
• The last course’s DMD equals its own departure with the sign reversed, which checks the arithmetic.

Each course’s DMD times its latitude is a double area, and the area is half the absolute value of their sum.

Example: area by DMD

A balanced traverse has latitudes and departures (ft) of AB $(+300,+400)$, BC $(-100,+200)$, CD $(-400,-100)$, and DA $(+200,-500)$.

• DMDs: AB $=400$; BC $=400+400+200=1,000$; CD $=1,000+200-100=1,100$; DA $=1,100-100-500=500$
• Double areas: $120,000$, $-100,000$, $-440,000$, and $100,000$, summing to $-320,000$ ft²

Answer: $A=320,000/2=160,000$ ft², or 3.67 acres at 43,560 ft² per acre

Starting A at (1,000, 1,000) gives corners B (1,400, 1,300), C (1,600, 1,200), and D (1,500, 800), and the coordinate method returns the same 160,000 ft².

Areas from offsets

For an irregular boundary, measure perpendicular offsets $h_1,\dots ,h_n$ at a uniform spacing $d$ from a baseline.

• Trapezoidal rule: $A=\frac{d}{2}\left[h_1+2(h_2+\cdots +h_{n-1})+h_n\right]$
• Simpson’s one-third rule, which needs an even number of intervals: $A=\frac{d}{3}\left[h_1+4(h_2+h_4+\cdots )+2(h_3+h_5+\cdots )+h_n\right]$

Earthwork and volume computations

Earthwork volume is computed either between cross-sections along a route or over a grid for an area such as a borrow pit.

Cross-section methods

$$V_{AEA}=\frac{L(A_1+A_2)}{2}{V}_{prism}=\frac{L}{6}\left(A_1+4A_m+A_2\right)$$

$A_1$ and $A_2$ are the end areas, $L$ is the distance between them, and $A_m$ is the area of the section at the midpoint. The average end area method is simpler and usually gives a slightly larger volume. The prismoidal formula is more accurate, but only with a true midpoint area - one measured, or computed from the averaged heights and widths of the end sections. Averaging the two end areas instead reduces it exactly to the average end area method.

Example: volume between two stations

Cut sections at stations 12+00 and 13+00 are 180 ft² and 240 ft², and the section measured at 12+50 is 205 ft².

$$V_{AEA}=\frac{100(180+240)}{2}=21,000{\text{ ft}}^3{V}_{prism}=\frac{100}{6}(180+820+240)=20,667{\text{ ft}}^3$$

Answer: 778 yd³ and 765 yd³. Dividing by 27 to reach cubic yards is a common place to lose the problem.

For a borrow pit or building pad, the unit-area method lays a grid of equal squares of area $A$ and weights each corner’s cut depth by the number of squares sharing it: $V=\frac{A}{4}\left(\sum h_1+2\sum h_2+3\sum h_3+4\sum h_4\right)$.

Swell and shrinkage

Definitions

Bank volume

Soil in its natural, undisturbed state (BCY).

Loose volume

Excavated soil, broken up and holding more voids (LCY). Haul units carry loose volume.

Compacted volume

Soil placed and compacted in a fill (CCY).

$$\text{Swell}=\left(\frac{V_{loose}}{V_{bank}}-1\right)\times 100\%\text{Shrinkage}=\left(1-\frac{V_{compacted}}{V_{bank}}\right)\times 100\%$$

Example: hauling for an embankment

An embankment needs 5,000 CCY from soil with 15% shrinkage and 25% swell, hauled in 12-LCY trucks.

• Bank: $5,000/0.85=5,882$ BCY
• Loose: $5,882\times 1.25=7,353$ LCY
• Loads: $7,353/12=612.7$

Answer: 613 truckloads

Always convert through bank volume; combining the two percentages directly is the shortcut that produces a wrong answer.

A mass diagram plots cumulative volume against station, cut positive and fill (adjusted for shrinkage) negative. A rising segment means cut exceeds fill. Between any two points where a horizontal balance line crosses the curve, cut equals fill and material can be moved within that stretch.

Exam tip: latitudes and departures, the average end area and prismoidal formulas, the mass haul diagram, and the trapezoidal, Simpson’s, and coordinate area formulas are in the Civil Engineering chapter of the FE Reference Handbook. Know swell and shrinkage, the DMD method, the tape and curvature corrections, and the leveling equal-sight-distance rule well enough to set them up without looking for them, because you may not find them there.

Coordinate systems

Positions are defined against a datum. In the United States, horizontal positions are referenced to the North American Datum of 1983 (NAD 83) and elevations to the North American Vertical Datum of 1988 (NAVD 88), both of which the National Geodetic Survey is in the process of replacing. Latitude and longitude are angles on an ellipsoid, so they don’t suit plane trigonometry directly.

The state plane coordinate system (SPCS) projects the ellipsoid onto flat zones small enough that plane trigonometry works with little distortion. Coordinates are a northing and an easting from a false origin that keeps them positive. Each zone uses a conformal projection:

• Lambert conformal conic for zones longer east to west
• transverse Mercator for zones longer north to south
• oblique Mercator for one zone in Alaska, where neither fits

The Universal Transverse Mercator (UTM) system applies transverse Mercator between 80°S and 84°N latitude, in 60 zones each $6^{\circ }$ of longitude wide.

A grid distance differs slightly from the same distance on the ground:

$$\text{Grid distance}=\text{Ground distance}\times (\text{grid scale factor}\times \text{elevation factor})$$

The grid scale factor accounts for projection distortion. The elevation factor, $R/(R+h)$ with $h$ the height above the ellipsoid, reduces the distance to the ellipsoid. It is slightly below 1 wherever the ground is above the ellipsoid. In much of the conterminous United States the ellipsoid lies about 30 m above sea level, so on low coastal ground $h$ is negative and the factor is slightly above 1.

Leveling and grades

Differential leveling finds elevation differences by reading a graduated rod through a level.

Definitions

Benchmark

A permanent point of known elevation.

Backsight (BS)

A rod reading on a point of known elevation, which sets the height of instrument.

Height of instrument (HI)

The elevation of the level’s line of sight.

Foresight (FS)

A rod reading on a point whose elevation is to be found.

Turning point (TP)

A temporary point that takes a foresight and then a backsight so the level can move forward.

$$HI=\text{Elevation}+BS\text{Elevation}=HI-FS$$

The sum of backsights minus the sum of foresights must equal the total change in elevation.

Example: differential leveling

From BM 1 (100.00 ft): BS 4.62 on BM 1, FS 3.15 on TP 1, BS 5.08 on TP 1, FS 2.91 on B.

• $H{I}_1=104.62$ ft, so TP 1 $=104.62-3.15=101.47$ ft
• $H{I}_2=101.47+5.08=106.55$ ft, so B $=106.55-2.91=103.64$ ft
• Check: $9.70-6.06=3.64$ ft

Answer: 103.64 ft

Percent grade is rise over horizontal run: $g=\Delta \text{Elev}/\text{Distance}\times 100\%$. If B is 400 ft from BM 1, the grade is $3.64/400\times 100\%=0.91\%$. The parabolic curves that join two grades are covered in the transportation engineering chapter.

Curvature and refraction

A level line of sight is straight while the level surface curves with the earth, so a long sight reads too high. Refraction bends the sight back and cancels part of the error. The combined correction is:

$$(c+r)=0.0206M^2\text{ ft, with }M\text{ in thousands of feet}$$

In SI units it is $0.0675K^2$ m, with $K$ in kilometers. A 2,000-ft sight reads $0.0206(2.0{)}^2=0.082$ ft too high, and a 1.5-km sight reads $0.0675(1.5{)}^2=0.152$ m too high.

Keeping each backsight and foresight at about equal distances cancels curvature, refraction, and any tilt in the instrument’s line of sight, which is why a level set midway between two points needs no correction.