FE Civil Surveying practice problems

An azimuth of $247\degree$ falls between $180\degree$ and $270\degree$, which is the SW quadrant. In the SW quadrant the bearing angle is azimuth minus $180\degree$: $247 - 180 = 67\degree$, giving S $67\degree$ W. The N $67\degree$ W option treats it as the NW quadrant. The S $23\degree$ W option subtracts from $270\degree$ ($270 - 247 = 23$) instead of from $180\degree$. The S $67\degree$ E option keeps the magnitude but puts it in the wrong (SE) quadrant.

1. Identify the quadrant: $180\degree < 247\degree < 270\degree$ is the SW quadrant.
2. For the SW quadrant, the bearing angle equals the azimuth minus $180\degree$:
   $\theta = 247\degree - 180\degree = 67\degree$
3. Write the bearing in the SW quadrant:
   $\text{Bearing} = \text{S } 67\degree \text{ W}$

FE Reference Handbook 10.6, p. 315

At the occupied station, first find the back azimuth of the incoming line: $138 + 180 = 318\degree$. The angle to the right is measured clockwise from that back line to the next line, so add it: $318 + 95 = 413\degree$. That exceeds a full circle, so subtract $360\degree$: $413 - 360 = 53\degree$, the forward azimuth of the next line. The $233\degree$ choice mistakenly adds another $180\degree$ (reports the back azimuth of the next line). The $413\degree$ choice forgets to subtract $360\degree$. The $43\degree$ choice subtracts the angle from the forward azimuth instead of adding it (treats it as an angle to the left).

1. Back azimuth of the incoming line:
   $\text{Az}_{\text{back}} = 138\degree + 180\degree = 318\degree$
2. Add the angle to the right:
   $318\degree + 95\degree = 413\degree$
3. Reduce to the $0\degree$ to $360\degree$ range:
   $\text{Az}_{\text{fwd}} = 413\degree - 360\degree = 53\degree$

FE Reference Handbook 10.6, p. 315

For a full level run the final elevation equals the starting elevation plus the sum of all backsights minus the sum of all foresights: $\text{Elev}_P = 612.40 + 18.62 - 22.07 = 608.95$ ft. Since the foresights total more than the backsights, the point ends up lower than BM-5. The 615.85 ft choice reverses the sign of the net difference (adds 3.45 instead of subtracting). The 653.09 ft choice adds both sums to the elevation. The 571.71 ft choice subtracts both sums.

1. Apply the running-sum form of the HI method:
   $\text{Elev}_P = \text{Elev}_{BM} + \Sigma BS - \Sigma FS$
2. Substitute the totals:
   $\text{Elev}_P = 612.40 + 18.62 - 22.07$
3. Compute:
   $\text{Elev}_P = 608.95 \text{ ft}$

FE Reference Handbook 10.6, p. 315

Curvature and refraction both depend on the sighting distance, and they affect a backsight and a foresight in the same direction. If the backsight distance equals the foresight distance at each setup, the two errors are equal and cancel when you compute the elevation difference (HI uses BS, then subtracts FS). Reading to the nearest 0.001 ft improves precision but does nothing about a systematic distance-dependent error. Using a longer loop is wrong because a systematic error accumulates rather than averaging out. Always reading the backsight first only fixes reading order, not the curvature-refraction effect.

1. Curvature and refraction error grows with the square of the sighting distance and is the same sign for an equal-length backsight and foresight.
2. The elevation difference uses backsight minus foresight, so equal distances make the errors cancel:
   $\Delta \text{Elev} = BS - FS$
3. Therefore balancing BS and FS distances at each setup removes the systematic effect.

FE Reference Handbook 10.6, p. 315

The course length is the hypotenuse of the latitude and departure components: $L = \sqrt{\text{Lat}^2 + \text{Dep}^2} = \sqrt{(-150)^2 + (-150)^2} = \sqrt{45{,}000} = 212.1$ m. The signs do not matter once squared. The 150.0 m choice reports just one component. The 300.0 m choice adds the magnitudes (150 + 150). The 106.1 m choice divides one component by $\sqrt{2}$ instead of multiplying.

1. Combine the components with the Pythagorean theorem:
   $L = \sqrt{\text{Lat}^2 + \text{Dep}^2} = \sqrt{(-150.0)^2 + (-150.0)^2}$
2. Compute:
   $L = \sqrt{22{,}500 + 22{,}500} = \sqrt{45{,}000} = 212.1 \text{ m}$

FE Reference Handbook 10.6, p. 315

A positive latitude (north) and negative departure (west) place the course in the NW quadrant. The reference (bearing) angle is $\arctan(|\text{Dep}| / |\text{Lat}|) = \arctan(50/86.6) = \arctan(0.5774) = 30\degree$, giving N $30\degree$ W. Converting an NW bearing to azimuth: $360 - 30 = 330\degree$. The $30\degree$ choice gives the NE azimuth (ignores the west sign). The $150\degree$ choice treats it as SE (180 - 30). The $300\degree$ choice uses $\arctan(\text{Lat}/\text{Dep})$ and lands on the wrong reference angle in the wrong quadrant.

1. North latitude and west departure: the course is in the NW quadrant.
2. Reference (bearing) angle:
   $\theta = \arctan\frac{|\text{Dep}|}{|\text{Lat}|} = \arctan\frac{50.0}{86.6} = 30\degree$
3. Convert the NW bearing N $30\degree$ W to an azimuth:
   $\text{Az} = 360\degree - 30\degree = 330\degree$

FE Reference Handbook 10.6, p. 315

Apply the shoelace sum $\sum(x_i y_{i+1} - x_{i+1} y_i)$ around the polygon, closing the last vertex back to the first. The five edge cross products are 0, 96, 144, 48, and 0, summing to 288. The area is half the absolute value: $|288|/2 = 144$ sq units. The 288 sq units choice forgets to divide by 2 and reports the raw sum. The 72 sq units choice divides by 4. The 120 sq units choice drops the 48 edge term ($240/2 = 120$).

1. Cross products for each edge:
2. $(0)(0) - (12)(0) = 0$; $\quad (12)(8) - (16)(0) = 96$
3. $(16)(12) - (6)(8) = 192 - 48 = 144$; $\quad (6)(8) - (0)(12) = 48$
4. $(0)(0) - (0)(8) = 0$. Sum the five terms:
   $\sum = 0 + 96 + 144 + 48 + 0 = 288$
5. Area is half the absolute value:
   $A = \frac{|288|}{2} = 144 \text{ sq units}$

FE Reference Handbook 10.6, p. 316

First the area in square feet: $200 \times 435.6 = 87{,}120 \text{ ft}^2$. Convert to acres by dividing by $43{,}560$: $87{,}120 / 43{,}560 = 2.00$ acres. The 1.00-acre choice divides by twice the conversion factor (or halves a side). The 4.00-acre choice divides by half the conversion factor. The 0.50-acre choice over-divides by applying the conversion twice.

1. Area in square feet:
   $A = 200 \times 435.6 = 87{,}120 \text{ ft}^2$
2. Convert to acres:
   $A = \frac{87{,}120}{43{,}560} = 2.00 \text{ acres}$

FE Reference Handbook 10.6, p. 3

Shrinkage means compacted fill occupies less volume than the bank material it came from. Compacted volume = bank volume $\times (1 - \text{shrinkage}) = 5{,}400 \times (1 - 0.15) = 5{,}400 \times 0.85 = 4{,}590$ yd³. The 6,210 yd³ choice adds 15 percent instead of subtracting it. The 810 yd³ choice reports only the lost volume ($5{,}400 \times 0.15$). The 6,353 yd³ choice divides by 0.85 (that direction is for finding required bank volume from a known fill).

1. Apply the shrinkage factor to the bank volume:
   $V_{\text{fill}} = V_{\text{bank}} (1 - S) = 5{,}400 (1 - 0.15)$
2. Compute:
   $V_{\text{fill}} = 5{,}400 \times 0.85 = 4{,}590 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

Average end area volume: $V = (L/2)(A_1 + A_2) = (150/2)(240 + 168) = 75 \times 408 = 30{,}600 \text{ ft}^3$. Convert to cubic yards by dividing by 27: $30{,}600 / 27 = 1{,}133$ yd³. The $30{,}600 \text{ yd}^3$ choice leaves the answer in cubic feet. The $2{,}267 \text{ yd}^3$ choice forgets to divide $L$ by 2 ($61{,}200/27$). The $1{,}700 \text{ yd}^3$ choice divides by 18 instead of 27.

1. Average end area volume in cubic feet:
   $V = \frac{L}{2}(A_1 + A_2) = \frac{150}{2}(240 + 168) = 75 \times 408 = 30{,}600 \text{ ft}^3$
2. Convert to cubic yards:
   $V = \frac{30{,}600}{27} = 1{,}133 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

External distance $E = R(\sec(I/2) - 1) = 500(\sec 30\degree - 1) = 500(1.1547 - 1) = 500 \times 0.1547 = 77.4$ ft. The external distance is the gap from the PI to the midpoint of the curve. The 67.0 ft choice uses the middle ordinate formula $R(1 - \cos(I/2)) = 500(1 - 0.866) = 67.0$ ft. The 288.7 ft choice is the tangent distance $T = R\tan(I/2)$. The 250.0 ft choice is R/2.

1. External distance formula with half the intersection angle:
   $E = R\left(\sec\frac{I}{2} - 1\right) = 500\left(\sec 30\degree - 1\right)$
2. Evaluate the secant:
   $\sec 30\degree = \frac{1}{\cos 30\degree} = \frac{1}{0.8660} = 1.1547$
3. Compute:
   $E = 500(1.1547 - 1) = 500 \times 0.1547 = 77.4 \text{ ft}$

FE Reference Handbook 10.6, p. 309

Long chord $LC = 2R\sin(I/2) = 2 \times 700 \times \sin 40\degree = 1{,}400 \times 0.6428 = 899.9$ ft. The long chord connects the PC directly to the PT. The 1,377.4 ft option uses the full angle, $2R\sin I = 1{,}400 \times 0.9848$. The 449.9 ft option drops the factor of 2 ($R\sin(I/2)$). The 587.4 ft option uses $2R\tan(I/2)$ and a wrong reduction.

1. Long chord formula:
   $LC = 2R\sin\frac{I}{2} = 2 \times 700 \times \sin 40\degree$
2. Evaluate the sine and compute:
   $LC = 1{,}400 \times 0.6428 = 899.9 \text{ ft}$

FE Reference Handbook 10.6, p. 308

Use $Y = Y_{PVC} + g_1 x + [(g_2 - g_1)/(2L)]x^2$. The parabola constant is $(0.01 - (-0.03))/(2 \times 500) = 0.04/1000 = +0.00004$. Then $Y = 240 + (-0.03)(200) + (0.00004)(200^2) = 240 - 6 + 1.6 = 235.60$ ft. The 234.00 ft option is the tangent-only elevation (240 - 6), ignoring the parabolic term. The 246.00 ft option uses the wrong sign on $g_1$. The 237.20 ft option doubles the parabolic correction.

1. Parabola constant:
   $a = \frac{g_2 - g_1}{2L} = \frac{0.01 - (-0.03)}{2 \times 500} = \frac{0.04}{1{,}000} = +0.00004$
2. Curve elevation at $x = 200$:
   $Y = 240 + (-0.03)(200) + (0.00004)(200^2)$
3. Compute:
   $Y = 240 - 6 + 1.6 = 235.60 \text{ ft}$

FE Reference Handbook 10.6, p. 307

The midpoint (PVI) tangent offset of a symmetric parabola is $e = AL/8$ when $A$ is expressed as a decimal grade difference. Here $A = |g_1 - g_2| = |0.02 - (-0.04)| = 0.06$, so $e = (0.06)(600)/8 = 36/8 = 4.50$ ft. Equivalently, the parabola constant $a = (g_2 - g_1)/(2L) = -0.00005$ gives an offset magnitude of $|a|(L/2)^2 = 0.00005 \times 300^2 = 4.50$ ft. The 9.00 ft choice uses AL/4. The 2.25 ft choice uses AL/16. The 1.50 ft choice uses only $g_1$ in $g_1 L/8$.

1. Algebraic grade difference as a decimal:
   $A = |g_1 - g_2| = |0.02 - (-0.04)| = 0.06$
2. Midpoint tangent offset:
   $e = \frac{AL}{8} = \frac{(0.06)(600)}{8} = \frac{36}{8} = 4.50 \text{ ft}$

FE Reference Handbook 10.6, p. 307

Departure = L·sin(Az) = 150·sin(60°) = 150(0.8660) = 129.9 ft. The 75.0 ft choice uses cos with a slip (150·cos60° = 75 is actually the latitude). The 150.0 ft choice is the full length. The 86.6 ft choice applies sin to the wrong angle.

1. Departure:
   $\Delta E = 150\sin 60\degree = 150(0.8660)$
2. Compute:
   $\Delta E = 129.9\text{ ft}$