Question 1

Bubble tube design for precision levels: For achieving high sensitivity of the bubble tube, which of the following design/material choices are appropriate?

Accepted Answer

Introduction / Context: Sensitivity of a level’s bubble tube determines how much the bubble moves for a small tilt of the instrument. High sensitivity is essential in precise levelling and triangulation work, but it must be balanced with stability and practical damping. Material selection and tube geometry both influence performance. Given Data / Assumptions: Bubble tube is ground to a large-radius circular arc. Bubble movement should respond to small angular changes without sticking. Environmental conditions (temperature) are within normal operating limits. Concept / Approach: Lower viscosity reduces internal friction, allowing the bubble to move freely and settle quickly. Lower surface tension reduces adhesion and meniscus pinning, minimizing stick-slip behavior. A longer bubble increases sensitivity because a given tilt produces a greater shift of the bubble ends along the arc. Finally, an excessively narrow bore increases capillary effects and friction; a moderate bore avoids meniscus lock and supports repeatability. Together these factors improve sensitivity and usability. Step-by-Step Solution: Select suitable fluid with low viscosity and controlled surface tension (often ether–alcohol mixes).Grind tube to large radius; ensure adequate bubble length.Choose bore size that limits capillary adhesion yet controls oscillations. Verification / Alternative check: Instrument specifications list bubble sensitivity in seconds per division; improvements correlate with the above design choices, within constraints of temperature stability and evaporation. Why Other Options Are Wrong: Each individual choice contributes positively; excluding any one undermines sensitivity aims. Therefore, “All the above” is the most accurate. Common Pitfalls: Choosing extremely low viscosity without considering damping, leading to prolonged oscillations; making the bore too wide, which reduces linearity of response. Final Answer: All the above

Question 2

Directional terms: The smaller horizontal angle between the true meridian and a survey line is called what in standard surveying terminology?

Accepted Answer

Introduction / Context: Angles used to define direction in surveying include bearing, azimuth, and declination. Using the correct term is crucial when reducing traverse data and converting between grid, true, and magnetic systems. Given Data / Assumptions: Angle is measured in the horizontal plane. Reference is the true meridian (geodetic north–south line). The smaller of the two possible horizontal angles is sought. Concept / Approach: The bearing of a line is the smaller horizontal angle between the reference meridian (true or grid, as specified) and the line, usually expressed either as a whole-circle value (0° to 360°) or as a reduced (quadrantal) bearing. Azimuth is commonly the clockwise angle from the meridian, often allowed to be any value 0°–360° (not necessarily the smaller). Declination refers to the angle between magnetic and true meridians, and dip is a vertical angle of the magnetic field—neither is appropriate here. Step-by-Step Solution: Identify that the reference is the true meridian.Select the term for the smaller horizontal angle to the line → bearing.Exclude azimuth (not necessarily smaller) and declination/dip (different phenomena). Verification / Alternative check: Text definitions confirm “bearing” as the smaller horizontal angle to the meridian; azimuth may equal the bearing in certain quadrants but is not restricted to the smaller angle by definition. Why Other Options Are Wrong: Declination is magnetic vs. true; azimuth is not inherently the smaller angle; dip is vertical magnetic inclination; convergence is grid–true difference, not asked. Common Pitfalls: Using azimuth and bearing interchangeably without regard to quadrant conventions; mixing magnetic vs. true references. Final Answer: Bearing

Question 3

Coordinate geometry from bearings: Point B is 50 m due east of A. The bearing of C from A is N 30° E, and the bearing of C from B is N 60° W. What is the departure (east–west coordinate) of C from A (positive east)?

Accepted Answer

Introduction / Context: Computing coordinates from bearings is a staple in traverse reduction and intersection problems. “Departure” denotes the east–west component (positive to the east) of a line or point relative to an origin. Here we use two bearings to intersect lines and determine the point C, then read off its departure from A. Given Data / Assumptions: Let A = (0, 0) with x east and y north. B is 50 m east of A ⇒ B = (50, 0). Bearing of C from A: N 30° E (azimuth 30°). Bearing of C from B: N 60° W (azimuth 300°). Concept / Approach: For azimuth θ (measured clockwise from north), the unit direction vector is (sin θ, cos θ). Form parametric lines for AC and BC and solve for their intersection. The x-coordinate of the intersection relative to A is the required departure of C from A. Step-by-Step Solution: Line AC: from A with azimuth 30° → direction (sin 30°, cos 30°) = (0.5, 0.866).Parametric AC: (x, y) = t * (0.5, 0.866).Line BC: from B with azimuth 300° → direction (sin 300°, cos 300°) = (−0.866, 0.5).Parametric BC: (x, y) = (50, 0) + s * (−0.866, 0.5).Equate coordinates and solve: 0.5 t = 50 − 0.866 s and 0.866 t = 0.5 s → s = 1.732 t → 0.5 t = 50 − 1.5 t → t = 25.Coordinates of C: x = 0.5 * 25 = 12.5, y = 0.866 * 25 ≈ 21.65. Verification / Alternative check: Sketching confirms C lies northeast of A but west of B’s meridian, consistent with an x value less than 50 m and positive. Back-substitution into both parametric equations yields consistent coordinates. Why Other Options Are Wrong: 25 m and 50 m overstate the easting; 25√3 m and 12.5√3 m are northing-like magnitudes, not the departure computed here. Common Pitfalls: Using cos for eastings and sin for northings (reversed); forgetting that azimuth is from north, not from east; mixing up departure (x) with latitude (y). Final Answer: 12.5 m

Question 4

Circular-curve geometry: when long chord equals tangent length In a simple circular curve of radius R connecting two tangents, suppose the long chord (between the tangent points) is equal to the tangent length. For this case, what is the intersectio…

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Introduction / Context: In highway and railway geometric design, several geometric quantities for a simple circular curve are related to the intersection (deflection) angle Δ and radius R. Two key quantities are the tangent length T and the long chord LC. This question tests your ability to use the standard relationships to determine Δ when the long chord equals the tangent length. Given Data / Assumptions: Simple circular curve of radius R connects two tangents. Tangent length T = R * tan(Δ/2). Long chord LC = 2 * R * sin(Δ/2). Given condition: LC = T. Concept / Approach: Use the standard formulas for a simple circular curve. Equating the long chord to the tangent length yields an equation in Δ/2 that can be solved using basic trigonometric identities, specifically the relationship between sine, cosine, and tangent for the same angle. Step-by-Step Solution: Write LC = 2 * R * sin(Δ/2).Write T = R * tan(Δ/2).Set LC = T → 2 * R * sin(Δ/2) = R * tan(Δ/2).Cancel R and replace tan(Δ/2) with sin(Δ/2) / cos(Δ/2): 2 * sin(Δ/2) = sin(Δ/2) / cos(Δ/2).Multiply both sides by cos(Δ/2): 2 * sin(Δ/2) * cos(Δ/2) = sin(Δ/2).Recognize 2 * sin(Δ/2) * cos(Δ/2) = sin(Δ).Thus sin(Δ) = sin(Δ/2) → Δ = π − Δ/2 (reject the trivial Δ = 0).Solve: (3/2)Δ = π → Δ = 2π/3 = 120°. Verification / Alternative check: Plug Δ = 120° into formulas: T = R * tan 60° = R * √3; LC = 2R * sin 60° = 2R * (√3/2) = R * √3. Hence LC = T, confirming the result. Why Other Options Are Wrong: 30°, 60°, 90°, 150° do not satisfy LC = T when substituted; each produces LC and T values that differ. Common Pitfalls: Confusing long chord with normal chord; mixing up T = R * tan(Δ/2) and external distance; forgetting to reject the trivial solution Δ = 0. Final Answer: 120°

Question 5

Plane surveying on a spherical Earth: arc–chord difference over 18.2 km For a ground distance of 18.2 km measured along the Earth’s surface, what is the approximate difference between the arc length and its subtended chord length (take Earth radius …

Accepted Answer

Introduction / Context: On large-scale surveys, the difference between an arc along the Earth and the corresponding straight chord can matter in precision calculations. For moderate distances, a convenient approximation estimates this small difference accurately enough for engineering purposes. This question checks your ability to apply the standard small-angle formula for arc–chord difference. Given Data / Assumptions: Arc length s = 18.2 km = 18,200 m. Mean Earth radius R ≈ 6,370 km = 6,370,000 m. Small-angle approximation is valid for s ≪ 2πR. Concept / Approach: For small central angle, the difference (s − c) between arc length s and chord length c is approximated by s^3 / (24 * R^2). This follows from c = 2R * sin(s/(2R)) and the series sin x ≈ x − x^3/6 for small x. Substituting x = s/(2R) and simplifying yields the compact cubic expression in s with R constant. Step-by-Step Solution: Compute s^3: 18,200^3 ≈ 6.03 × 10^12 m^3.Compute R^2: (6.37 × 10^6)^2 ≈ 4.06 × 10^13 m^2.Use formula: s − c ≈ s^3 / (24 R^2).Numerically: (6.03 × 10^12) / (24 × 4.06 × 10^13) ≈ 6.2 × 10^−3 m ≈ 0.62 cm.Round to the nearest option → about 1 cm. Verification / Alternative check: A direct evaluation using c = 2R * sin(s/(2R)) with calculator precision gives the same order of magnitude (a few millimetres). For typical engineering mapping, rounding to 1 cm is fully justified. Why Other Options Are Wrong: 5 cm, 10 cm, and 100 cm greatly overestimate the arc–chord difference for only 18.2 km; such differences occur over far longer spans. Common Pitfalls: Confusing sagitta with arc–chord difference; misusing units (km vs m); neglecting the 24 in the denominator which comes from series expansion. Final Answer: 1 cm

Question 6

Sextant observation to two elevated objects: computing the true horizontal angle Two objects A and B have elevations (altitudes) α and β respectively. A sextant measures the angle θ between the lines of sight to A and B. What is the correct horizont…

Accepted Answer

Introduction / Context: When two targets lie at elevations α and β, the sextant (or any angle-measuring device) reads the included angle θ between the two sloping sight lines. However, mapping and traverse work often requires the true horizontal angle H. Converting the observed sloping angle to its horizontal equivalent is a classic spherical/space-geometry problem that uses vector dot products or direction cosines. Given Data / Assumptions: Elevations (altitudes) of A and B above the horizontal through the observer are α and β. Observed included angle between the two sight lines is θ. Atmospheric refraction and instrumental errors are neglected. Concept / Approach: Let unit vectors to the two objects be u and v. Using elevation (altitude) and azimuth notation, the cosine of the angle γ between u and v is given by: cos γ = sin α * sin β + cos α * cos β * cos H, where H is the unknown horizontal angle between their azimuths. Since the sextant measures γ = θ, we solve for cos H. Rearranging gives the fundamental relation: cos H = (cos θ − sin α * sin β) / (cos α * cos β). Step-by-Step Solution: Start with the dot product identity: cos θ = u · v = sin α * sin β + cos α * cos β * cos H.Rearrange to isolate cos H: cos H = (cos θ − sin α * sin β) / (cos α * cos β).Take principal value: H = arccos{ (cos θ − sin α * sin β) / (cos α * cos β) }.Use quadrant checks if needed based on observed bearings of A and B. Verification / Alternative check: For α = β = 0 (both targets on the horizon), the formula reduces to cos H = cos θ → H = θ, as expected. For small elevations, a first-order expansion yields H ≈ θ − (tan(α/2) * tan(β/2)) * sin θ (small correction), consistent with practical experience. Why Other Options Are Wrong: Option B uses a plus sign; this would make H smaller or imaginary under typical geometries and fails the horizon check. Option C is an unjustified linear scaling and fails dimensional tests; it does not arise from the vector identity. Option D is incorrect because option A provides the correct closed-form expression. Common Pitfalls: Forgetting to convert degrees to radians in software; misinterpreting α and β as zenith angles (use altitude elevations here); ignoring quadrant logic when using arccos in coding environments. Final Answer: H = arccos{ (cos θ − sin α * sin β) / (cos α * cos β) }

Question 7

Line on the Earth’s surface with constant slope: precise terminology An imaginary line traced over the ground that everywhere maintains a constant inclination to the horizontal plane is called:

Accepted Answer

Introduction / Context: Several kinds of lines are used in surveying and mapping, and their names are sometimes confused. A contour line connects points of equal elevation (zero slope relative to the horizontal). In contrast, a route designer often needs a line that rises or falls steadily at a specified slope. This question checks the proper term for such a constant-slope path on the ground surface. Given Data / Assumptions: The line lies on the topographic surface (not a straight line in 3D space). The inclination to the horizontal is fixed at every point (constant gradient). Standard civil engineering terminology is used. Concept / Approach: The correct term is “contour gradient”: a line on the ground that has a uniform slope with respect to the horizontal. It is commonly used in alignment design for roads, canals, and pipelines to maintain drainage or energy grade. A level line is a locus of points of equal elevation (zero slope). A “line of gentle slope” is merely descriptive and not a precise technical term. Step-by-Step Solution: Identify requirement: constant inclination (gradient) everywhere along the line.Recall definitions: contour line → equal elevation; level line → surface of still water; contour gradient → constant slope.Therefore choose “contour gradient”. Verification / Alternative check: Textbooks on surveying and highway engineering define a contour gradient as a line with a specified slope laid on the map by proportional rise or fall between contours. Why Other Options Are Wrong: Contour line: constant elevation rather than constant slope. Level line: a curve everywhere normal to gravity; again, zero slope relative to horizontal. Line of gentle slope: not a standard term and does not imply constancy. Common Pitfalls: Confusing “contour” with “contour gradient”; assuming any sloping line suffices without verifying constant inclination. Final Answer: Contour gradient

Question 8

Horizontal angle errors from imperfect plate levelling Considering horizontal angle observations with a theodolite, the error caused by imperfect levelling of the plate bubble becomes:

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Introduction / Context: If the instrument’s vertical (rotation) axis is not truly vertical due to imperfect plate bubble levelling, directions read on the horizontal circle can be biased. This bias depends on the vertical angle of the line of sight. Understanding how the error varies helps observers choose appropriate targets and procedures to minimize angular error. Given Data / Assumptions: Vertical axis has a small tilt due to imperfect levelling. Horizontal angles are formed by two sight lines with possibly different vertical angles. Other instrument errors are negligible or removed by face-left/face-right averaging. Concept / Approach: The induced azimuth error from a tilted vertical axis increases with the tangent of the vertical angle of the sight. Nearly horizontal sights produce small errors; steep sights produce larger azimuth errors for the same small axis tilt. Hence, the effect is “large for steeply inclined sights.” Step-by-Step Solution: Assume a small vertical-axis tilt ε.For a sight at vertical angle v, the azimuth bias is roughly proportional to ε * tan v.As v → 0 (level sights), tan v → 0 → small error.As v increases (steep sights), tan v grows → larger error. Verification / Alternative check: Instrument adjustment theory derives this dependence; practical field advice always prefers balanced, near-level sights when very precise azimuths are required. Why Other Options Are Wrong: “Large when sights are nearly level” contradicts the tan v dependence. “Large for long sights” relates to pointing precision but not specifically to vertical-axis tilt. “Less for steeply inclined sights” is the opposite of the actual behavior. Common Pitfalls: Ignoring steep-sight sensitivity; not balancing vertical angles when possible; forgetting to average face-left/right to reduce related errors. Final Answer: Large for steeply inclined sights

Question 9

Slope correction for a 3° slope over 100 m: compute the subtractive correction A line is measured along a 3° slope with a tape; the sloping distance recorded is 100 m. What is the slope correction C_s (to be subtracted from 100 m) to obtain the hori…

Accepted Answer

Introduction / Context: When measuring distances on sloping ground, field parties often measure along the slope (for speed) and then correct to the required horizontal distance. The correction depends on the slope angle and the measured sloping length. This item asks you to compute the exact subtractive correction for a 3° slope over a measured 100 m line. Given Data / Assumptions: Sloping length l = 100 m. Slope angle θ = 3°. Horizontal distance L_h = l * cos θ. Slope correction C_s = L_h − l (a negative number), which is commonly quoted as a subtractive correction to the recorded length. Concept / Approach: For an angle θ, the exact relationship is L_h = l * cos θ. Therefore, the subtractive correction is C_s = l * cos θ − l = l (cos θ − 1). Many references report the magnitude |C_s| = l (1 − cos θ) and note that it must be subtracted from l to obtain L_h. Step-by-Step Solution: Compute cos 3° ≈ 0.998629.Find C_s = 100 * (0.998629 − 1) = 100 * (−0.001371) ≈ −0.1371 m.Rounded to the listed options, C_s ≈ −0.137 m.Hence, horizontal distance = 100 − 0.137 = 99.863 m. Verification / Alternative check: Using the approximate small-angle form C_s ≈ − l * (θ^2 / 2) with θ in radians (θ ≈ 0.05236 rad) gives −100 * (0.05236^2 / 2) ≈ −0.137 m, confirming the exact computation. Why Other Options Are Wrong: −0.11 m and −0.12 m are underestimates; −1.87 m is far too large and corresponds to a very steep slope, not 3°. Common Pitfalls: Using degrees inside radian-based approximations; forgetting that the correction is subtractive; rounding too early. Final Answer: −0.137 m

Question 10

Tacheometry – staff attitude for small vertical angles If vertical angles of inclined sights do not exceed 10° and the staff's non-verticality is within 1°, stadia observations in the ordinary tacheometric system are preferably taken on:

Accepted Answer

Introduction / Context: Tacheometry uses stadia hairs to obtain distances and elevations from staff intercepts. There are two practical conventions for staff attitude during inclined sights: holding the staff vertical or holding it normal to the line of sight (“staff normal”). Each leads to slightly different reduction formulae. This question asks which attitude is preferred when vertical angles are small (≤ 10°) and the staff can be kept within about 1° of vertical. Given Data / Assumptions: Vertical angles up to about 10° only. Staff can be kept within 1° of true vertical. Standard stadia constants and procedures apply. Concept / Approach: For small vertical angles and good staff discipline, the “staff vertical” method is standard because the reduction formulas (e.g., D = k * s * cos^2 θ + c * cos θ) are straightforward and errors from small non-verticality are negligible compared with reading precision. The “staff normal” method is adopted for steep sights where holding the staff vertical introduces larger errors and a modified relation (D = k * s + c) simplifies reductions. Step-by-Step Solution: Identify small vertical angle regime (≤ 10°).Note staff non-verticality within 1° → small cosine errors.Prefer staff vertical → standard reduction with cos^2 θ terms is accurate.Conclude: adopt “staff held truly vertical”. Verification / Alternative check: Worked examples in surveying texts present staff-vertical reductions as the default; staff-normal is recommended mainly for steep sights (large θ) or in mountainous terrain. Why Other Options Are Wrong: Staff normal is not necessary for small θ and can complicate practice. “Either equally” ignores the preferred standard when conditions are mild. “None” is clearly incorrect. Common Pitfalls: Using staff-normal formulas inadvertently with staff-vertical observations; ignoring the cosine-squared factor in distance computations; neglecting to note staff attitude in the field book. Final Answer: Staff held truly vertical

Question 11

Plane tabling instrument for direct (tacheometric) distance and elevation from the board In plane tabling, which instrument allows the surveyor to obtain horizontal and vertical distances directly—without resorting to chaining—by using stadia observ…

Accepted Answer

Introduction / Context: Plane tabling combines field observation and plotting. For distance without chaining, a stadia-equipped telescope is used so that distances and elevations can be derived from staff intercepts. This capability is integrated into a special alidade for plane table work, distinct from a separate theodolite-like tacheometer. Given Data / Assumptions: The goal is to obtain distances/elevations directly from the board. Stadia hairs in the telescope permit tacheometric reductions. We distinguish between a simple (plane) alidade and a telescopic stadia alidade. Concept / Approach: The telescopic alidade used in plane tabling has a telescope with stadia hairs, enabling tacheometric measurements while plotting rays on the sheet. A simple plane alidade has only sight vanes and cannot provide stadia distances. A separate tacheometer is a different instrument; although it also uses stadia, it is not the plane-table attachment in question. Step-by-Step Solution: Identify need: stadia-based distance from the plane table.Recognize that only the telescopic alidade (with stadia hairs) supports this directly on the board.Hence choose “Telescopic alidade”.Use reductions for D and RL as per stadia formulas. Verification / Alternative check: Plane-table manuals describe the telescopic alidade as enabling “tacheo-platting”, confirming it provides horizontal and vertical distance data from stadia. Why Other Options Are Wrong: Plane alidade: no telescope, no stadia. Clinometer: measures vertical angle only, not distance. Tacheometer: not the plane-table attachment implied by the question. Common Pitfalls: Confusing tacheometer with telescopic alidade; forgetting to note the stadia constants for the specific alidade. Final Answer: Telescopic alidade (with stadia hairs)

Question 12

Plane tabling workflow: when do plotting and observation occur? While surveying a plot of land by plane tabling, which of the following is true regarding the workflow of observations and plotting?

Accepted Answer

Introduction / Context: Plane tabling is designed for on-site plotting of details. Unlike theodolite traversing where readings are booked and plotted later, plane tabling allows the surveyor to draw lines and locate points directly on the sheet as they are observed. This question tests your understanding of that fundamental feature. Given Data / Assumptions: Standard plane table with alidade is used. The board is centered, leveled, and oriented at each station. Rays are drawn to detail points as they are sighted. Concept / Approach: Because the drawing sheet is on the instrument, observations and plotting are simultaneous. This integration is the key advantage: errors in plotting can be detected on the spot, and omitted details are less likely. While supplementary notes may be taken, the essential plotting is not deferred. Step-by-Step Solution: Center, level, and orient the plane table.Sight to a detail point with the alidade and immediately draw the ray.Fix distances by intersection, resection, radiation, or traversing as appropriate.Thus, observation and plotting proceed together. Verification / Alternative check: Method descriptions—radiation, intersection, resection—explicitly combine observation and drawing steps at the instrument. Why Other Options Are Wrong: “Do not proceed simultaneously” and “plotted later” describe other survey methods, not plane tabling. “All of the above” cannot be true as the options are mutually exclusive. Common Pitfalls: Failing to re-check orientation during multi-station work; inadequate control leading to distortions even though plotting is simultaneous. Final Answer: Observations and plotting proceed simultaneously at the board

Question 13

Two-peg test style check: computing the correct far staff reading Peg A and peg B are 80 m apart. With a dumpy level set at the midpoint, staff readings are A = 1.32 and B = 1.56. Then the level is set 10 m from A on BA produced; staff readings are …

Accepted Answer

Introduction / Context: This is a classic two-peg test scenario used to detect and correct the line of collimation error of a level. When the instrument is placed midway between pegs, the difference in staff readings gives the true difference in elevation between the pegs. When the instrument is placed near one peg, unequal sight lengths reveal the line-of-sight error, allowing computation of the corrected far staff reading. Given Data / Assumptions: Distance AB = 80 m. Midpoint setup readings: A = 1.32, B = 1.56 → true difference Δ_true = 1.56 − 1.32 = 0.24 m (B lower). Second setup 10 m beyond A: distances IA = 10 m, IB = 90 m. Observed readings: A = 1.11, B = 1.39 → observed difference Δ_obs = 0.28 m. Concept / Approach: The line of collimation error produces a reading error proportional to sight length. Let e be the error per 100 m added to a reading (positive if it increases with distance). The difference in errors between B and A is e * (IB − IA) / 100. Matching observed and true differences determines e, which is then used to correct the far reading. Step-by-Step Solution: Compute true elevation difference: Δ_true = 0.24 m.Compute observed difference for second setup: Δ_obs = 1.39 − 1.11 = 0.28 m.Error difference = Δ_obs − Δ_true = 0.28 − 0.24 = 0.04 m.Let e be error per 100 m in a staff reading; then 0.04 = e * (90 − 10) / 100 = e * 0.8 → e = 0.05 m per 100 m.Far sight B error = e * (IB/100) = 0.05 * 0.9 = 0.045 m; since Δ_obs > Δ_true, B's observed reading is too large.Correct B reading = 1.39 − 0.045 = 1.345 m. Verification / Alternative check: Check using A's correction: near sight error = e * 0.1 = 0.005 m. Corrected difference = (1.39 − 0.045) − (1.11 − 0.005) = 1.345 − 1.105 = 0.240 m, matching Δ_true. Why Other Options Are Wrong: 1.435 and 1.425 do not satisfy the true difference constraint when combined with the corrected near reading. “None of these” is false because 1.345 is consistent with the two-peg correction. Common Pitfalls: Using distance difference 80 m incorrectly; forgetting to apply corrections to both near and far readings; sign confusion when deciding whether to add or subtract the error. Final Answer: 1.345

Question 14

Precision traverse technique: When measuring included angles in a precision traverse using a vernier theodolite, which initial vernier setting practice is recommended to minimize systematic errors and achieve reliable mean angles?

Accepted Answer

Introduction / Context: In a precision traverse, included angles are observed carefully to control closure error. Vernier or micrometer theodolites allow several good practices for setting the initial reading at the back station. The goal is to reduce instrumental and pointing biases by repetition, reversal, and averaging across different initial settings. Given Data / Assumptions: A vernier theodolite (or micrometer theodolite) is used. Included angles are measured from a known back station direction to the forward station. Both-face observations and multiple repetitions may be taken. Concept / Approach: Systematic errors such as circle eccentricity and index error can be mitigated by varying the initial circle reading and by reading both verniers. Common practices include setting exactly 0°, an offset such as 5°, or a convenient near-zero value and then taking readings on both verniers for averaging. All achieve the same purpose: to diversify initial settings and distribute instrumental imperfections so that the mean angle is robust. Step-by-Step Solution: Center on the back station and clamp the lower plate to fix the reference.Adopt an initial setting strategy: 0°, a small known offset (e.g., 5°), or a convenient near-zero value.Sight the fore station and record both verniers (or micrometer) readings for each repetition.Repeat on face left and face right; average all results to obtain the included angle. Verification / Alternative check: Traverse adjustment (angular misclosure) and consistency between face-left and face-right means confirm that any of these starting practices produce equivalent precise results when combined with averaging and proper procedures. Why Other Options Are Wrong: Options A, B, and C, taken alone, are each valid but incomplete recommendations. The most comprehensive answer is that all are acceptable. Common Pitfalls: Failing to read both verniers when available; not reversing the telescope; neglecting to re-point accurately after resetting the initial reading; assuming only one fixed initial reading eliminates all errors. Final Answer: All of the above practices are acceptable for precise work

Question 15

Levelling terminology and field procedure: Which of the following statements about sights (back sight, fore sight, and staff observations) in levelling is/are correct?

Accepted Answer

Introduction / Context: Accurate levelling depends on using the correct terms and sequence of observations. Distinguishing between back sights (BS), fore sights (FS), and intermediate sights (IS) helps structure the field book and provides arithmetic checks for reduced levels (RLs). Given Data / Assumptions: The level is properly set up and the line of collimation is horizontal. Staff readings are taken in the order required by the route and change points (CPs). Standard rise-and-fall or height-of-collimation methods are used for RL computation. Concept / Approach: A sight is any staff observation taken through the instrument. The first reading after a new set-up is the back sight, used to establish the instrument height (HI). At a change point, the final reading from the old set-up is a fore sight; then, after moving the instrument, the first reading on the same CP becomes the new back sight, ensuring continuity of RLs across set-ups. Step-by-Step Solution: Identify the type of each sight based on its position in the sequence (first after set-up → BS; last before moving → FS; others → IS).At a CP: record FS on the CP before moving the instrument.Move and set up again; take BS on the same CP to transfer the datum.Compute RLs via rise-and-fall or HI method and check arithmetic balance. Verification / Alternative check: Field-book checks: ΣBS − ΣFS = RL_last − RL_first confirms correct classification and arithmetic integrity of entries. Why Other Options Are Wrong: Each individual statement is correct but partial; the comprehensive correct choice is that all listed statements are true. Common Pitfalls: Mislabeling a reading at a CP; mixing IS with FS; forgetting to carry out the BS−FS arithmetic check that reveals transcription errors. Final Answer: All of the above

Question 16

Levelling computations – rise and fall method: The rise-and-fall method provides an internal arithmetic check on which set(s) of staff readings?

Accepted Answer

Introduction / Context: The rise-and-fall method determines reduced levels by comparing successive staff readings to compute rises and falls between points. One of its strengths is the built-in arithmetic check that improves reliability of field data before office reductions are finalized. Given Data / Assumptions: A properly levelled instrument with clear staff readings. Sequential observations categorized as BS, IS, and FS. Reduction by rise-and-fall with a final arithmetic check. Concept / Approach: The method requires the equality: ΣBS − ΣFS = RL_last − RL_first. Additionally, ΣRises − ΣFalls = RL_last − RL_first. Because rises and falls are derived from differences of consecutive staff readings (including IS), errors in any class of sight (BS, IS, or FS) disturb both equalities, so the method effectively checks all readings taken during the run. Step-by-Step Solution: Tabulate staff readings with columns for rise and fall.Compute rises/falls between successive points; accumulate ΣRises and ΣFalls.Calculate RLs from a known benchmark and verify ΣBS − ΣFS = RL_last − RL_first.Also verify ΣRises − ΣFalls = RL_last − RL_first; discrepancies flag misreadings or transcription errors. Verification / Alternative check: Cross-check both equalities; if both hold, the entries for BS, IS, and FS are self-consistent. Why Other Options Are Wrong: Choosing only BS, FS, or IS ignores that all contribute to the computed rises/falls and thus to the check. Common Pitfalls: Recording rises/falls with the wrong sign; skipping IS differences; failing to reconcile misclosures before leaving site. Final Answer: All of these

Question 17

Setting out a 45° angle from a chain line: Which hand instrument is specifically suitable for setting out a 45° angle with respect to a chain line in chain surveying?

Accepted Answer

Introduction / Context: In chain surveying, short offsets are often set out at right angles (90°) or at special angles such as 45°. Choosing the correct hand instrument ensures the required angle is achieved quickly and accurately in the field without resorting to a theodolite. Given Data / Assumptions: A baseline (chain line) is established. A simple hand-held angle-setting instrument is to be used. The desired angle with the chain line is 45°. Concept / Approach: The optical square and prismatic square are designed to set out right angles (90°) by reflecting images through 45° prism or mirror arrangements, not to set 45° with respect to the chain line. The open cross-staff provides perpendicular sight lines but typically does not include a 45° sight. The French cross-staff, however, includes additional sight slits set at 45°, making it the appropriate choice for setting out a 45° angle directly from the chain line. Step-by-Step Solution: Place the French cross-staff on the chain line point where the offset is required.Align one set of sights with the chain line.Use the 45° sight pair to establish the required offset direction.Measure along the 45° direction to set out the point. Verification / Alternative check: Check the angle with a theodolite or by constructing a right triangle with equal legs to confirm the 45° direction. Why Other Options Are Wrong: Optical square and prismatic square are intended for 90° offsets. Open cross-staff generally provides perpendiculars only; it lacks the dedicated 45° sights of the French cross-staff. Common Pitfalls: Assuming that a 90° instrument can directly set 45°; mis-centering the cross-staff over the chain line point; ignoring local magnetic attraction when improvising with a compass instead of a sighting instrument. Final Answer: French cross-staff (with 45° sights)

Question 18

Chain surveying – setting out perpendiculars: Which instrument is commonly used to set out perpendicular offsets to a chain line during chain surveying?

Accepted Answer

Introduction / Context: Short offsets at right angles to a baseline are routine in chain surveying for locating details like boundaries, hedges, and walls. A simple, quick instrument is preferred over high-precision angle-measuring devices for efficiency in the field. Given Data / Assumptions: The task is to set out a 90° offset from an established chain line. Speed and simplicity are more important than arc-second precision. Visibility is adequate for sighting the detail point. Concept / Approach: The optical square uses a fixed 45° prism or mirror system to create a right angle between two lines of sight, enabling a rapid and reasonably accurate setting of perpendiculars. While a theodolite can set any angle, it is unnecessarily elaborate for common offsets. A prismatic compass is for bearings, not precise perpendiculars, and a levelling instrument measures vertical differences, not plan angles. Step-by-Step Solution: Stand on the chain line point where the offset is required.Sight along the chain line with the optical square.Rotate until the reflected and direct images coincide; the perpendicular direction is now defined.Measure the offset distance along this perpendicular to locate the feature. Verification / Alternative check: Cross-check by measuring a 3-4-5 triangle or by setting a right angle with a cross-staff if available; the points should coincide within field tolerances. Why Other Options Are Wrong: Theodolite is accurate but slow for routine offsets. Prismatic compass is affected by local attraction and is not intended for right angles. Levels determine elevation, not plan angles. Common Pitfalls: Using the instrument far from the chain line point; parallax errors; ignoring slope corrections when offsets are long on sloping ground. Final Answer: Optical square

Question 19

Measuring areas on maps: The area of an irregular figure on a plotted map or plan is commonly measured with which instrument?

Accepted Answer

Introduction / Context: Estimating land areas from maps is a frequent task in civil engineering and surveying. Irregular boundaries make arithmetic methods cumbersome; hence, mechanical integrators like the planimeter provide a quick and accurate solution by tracing the boundary directly on the plan. Given Data / Assumptions: A scaled plan or map is available. The boundary to be measured may be curved or irregular. Instrument constants have been checked or calibrated. Concept / Approach: A planimeter converts the path traced by its measuring wheel into an area reading. The final reading, multiplied by the square of the map scale, yields ground area. Other listed instruments serve different purposes: pantographs scale drawings, sextants measure celestial/sea angles, clinometers measure vertical angles, and optical squares set right angles in chain surveying. Step-by-Step Solution: Zero the planimeter and set it to a suitable range.Trace the boundary smoothly, returning to the starting point.Note the reading difference and apply the planimeter constant if required.Convert to ground area by multiplying by the square of the plan scale factor. Verification / Alternative check: Compare with a grid-square counting method on a transparent overlay; results should agree within acceptable tolerance. Why Other Options Are Wrong: Pantograph, sextant, clinometer, and optical square do not measure areas directly. Common Pitfalls: Tracing too fast causing wheel slip; not closing the figure precisely; forgetting to apply the plan scale correctly, leading to large percentage errors. Final Answer: Planimeter (for direct area measurement)

Question 20

Tape sag correction (supported at 10 m intervals): A 50 m steel tape weighs 4 kg in total and is used with a pull of 5 kgf, supported at 10 m intervals. What is the approximate total sag correction for one 50 m tape length?

Accepted Answer

Introduction / Context: When measuring distances with a tape carried between supports, the tape sags and the measured length exceeds the true horizontal distance. A sag correction must be subtracted. The magnitude depends on span length, tape weight per unit length, and applied pull. Field practice often supports a long tape at equal intervals to reduce the correction. Given Data / Assumptions: Total tape length L_total = 50 m; total weight W_total = 4 kgf → weight per meter w = 4 / 50 = 0.08 kgf/m. Pull (tension) P = 5 kgf, assumed equal on each span. Tape supported at equal 10 m intervals → number of spans n = 5; span length l = 10 m. Use standard sag correction per span: c_span = (w^2 * l^3) / (24 * P^2); total correction c_total = n * c_span. Concept / Approach: Sag makes the measured distance too long. The correction is derived from the catenary approximation for small sag and uniform load. Supporting at intermediate points reduces l and thus the cubic dependence on span length, dramatically lowering the total correction compared to an end-supported 50 m span. Step-by-Step Solution: Compute w = 0.08 kgf/m.For each 10 m span: c_span = (0.08^2 * 10^3) / (24 * 5^2) = (0.0064 * 1000) / (24 * 25) = 6.4 / 600 = 0.010667 m.Total correction over 5 spans: c_total = 5 * 0.010667 m = 0.05333 m.Rounded to the nearest option: 0.053 m (to be subtracted from the measured distance). Verification / Alternative check: If the tape were supported only at its ends (one 50 m span), c = (0.08^2 * 50^3) / (24 * 25) ≈ 1.33 m, showing why intermediate supports are used; the 10 m support interval yields a much smaller, practical correction. Why Other Options Are Wrong: 0.043 m, 0.063 m, 0.073 m, 0.083 m: These do not match the computed 0.053 m for 10 m spans with the given w and P. Common Pitfalls: Applying the end-support formula to a multi-support setup; mixing SI units (N) with kgf without consistency; forgetting to subtract the correction from the measured length. Final Answer: 0.053 m

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