I85PREP
⌘ K
PEC EPE Geoinformatics study materials
PEC EPE Geoinformatics study resources

PEC EPE Geoinformatics practice questions

One question at a time. Build your PEC EPE Geoinformatics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Geodesy and Reference Systems

In a geodetic datum, what is the role of the reference ellipsoid?

  1. It is the equipotential surface of gravity that best fits mean sea level
  2. It is the sphere used only for map projection distortion analysis
  3. It is the physical terrain surface used for orthometric height computation
  4. It is a mathematically defined surface approximating the geoid, used as the basis for horizontal coordinates Correct answer.

Explanation

The reference ellipsoid is a rotationally symmetric mathematical figure that approximates the geoid and provides the surface on which geodetic latitude and longitude are defined. Heights and coordinates are then expressed relative to this ellipsoid within a datum.

Question 2Geodesy and Reference Systems

Which quantity distinguishes a geocentric datum from a local (non-geocentric) datum?

  1. The ellipsoid flattening value
  2. The number of satellites in the observing constellation
  3. The choice of map projection used for publication
  4. The position of the ellipsoid centre relative to the Earth's centre of mass Correct answer.

Explanation

A geocentric datum places the ellipsoid centre at the Earth's centre of mass, whereas a local datum offsets the ellipsoid to fit a region. This offset is the defining geometric difference between the two datum classes.

Question 3Geodesy and Reference Systems

What does an EPSG code such as EPSG:4326 identify?

  1. A transformation parameter precision class
  2. A coordinate reference system definition in a registry Correct answer.
  3. A geoid model grid file format
  4. A satellite orbit ephemeris product

Explanation

EPSG codes are registry identifiers for coordinate reference systems, datums, and transformations. EPSG:4326 denotes a specific geographic CRS, so the code names a CRS definition rather than a data product.

Question 4Geodesy and Reference Systems

On the ellipsoid, what is the difference between geodetic latitude and geocentric latitude?

  1. Geocentric latitude uses the geoid surface rather than the ellipsoid
  2. Geodetic latitude is defined only in the southern hemisphere
  3. Geodetic latitude is measured from the equatorial plane to the ellipsoid normal; geocentric latitude to the radius vector Correct answer.
  4. Geodetic latitude is measured in grads; geocentric latitude in degrees

Explanation

Geodetic latitude uses the angle between the equatorial plane and the ellipsoid normal at the point. Geocentric latitude uses the angle to the radius vector from the ellipsoid centre, so the two differ except at the poles and equator.

Question 5Geodesy and Reference Systems

Which surface serves as the reference for orthometric heights?

  1. The local tangent plane
  2. The reference ellipsoid
  3. The geoid Correct answer.
  4. The best-fitting sphere

Explanation

Orthometric heights are measured along the plumb line from the geoid, which approximates mean sea level as an equipotential surface. Ellipsoidal heights are measured from the ellipsoid instead.

Question 6Geodesy and Reference Systems

What is the fundamental plane of the International Terrestrial Reference System (ITRS)?

  1. The plane of the ecliptic at J2000.0
  2. The plane of the mean equator of the celestial intermediate pole Correct answer.
  3. The plane of the local horizon at Greenwich
  4. The plane tangent to the geoid at the equator

Explanation

The ITRS is defined with its origin at the Earth's centre of mass and its fundamental plane as the mean equator of the celestial intermediate pole. This gives a conventional equatorial plane for terrestrial coordinates.

Question 7Geodesy and Reference Systems

A point has ellipsoidal height h = 120.0 m and geoid undulation N = 35.0 m (geoid above ellipsoid). What is its orthometric height H using H = h - N?

  1. 155.0 m
  2. 35.0 m
  3. 85.0 m Correct answer.
  4. 120.0 m

Explanation

The relation H = h - N converts ellipsoidal height to orthometric height when N is the geoid-ellipsoid separation. Substituting gives 120.0 - 35.0 = 85.0 m.

Question 8Geodesy and Reference Systems

A GNSS receiver reports a position in a geocentric datum, but a legacy national map uses a local datum. What is required before overlaying the two datasets?

  1. Only a rescaling of the coordinate units
  2. Only a change of the ellipsoid flattening value
  3. A datum transformation using published parameters or a transformation grid Correct answer.
  4. Only a change of map projection

Explanation

Coordinates in different datums refer to differently positioned and oriented ellipsoids, so a datum transformation is needed to express them in a common frame. Projection changes alone do not remove the datum offset.

Question 9Geodesy and Reference Systems

In a 7-parameter Helmert transformation, which combination of parameters is applied?

  1. Three translations, three rotations, and a scale factor Correct answer.
  2. Two translations and one rotation
  3. Three translations only
  4. Three rotations and no scale factor

Explanation

The 7-parameter (Bursa-Wolf) model applies three translations, three small rotations, and a scale difference between the two frames. It is the standard first-order similarity transformation for datum shifts.

Question 10Geodesy and Reference Systems

Why does the geoid undulation N vary with location?

  1. Because the ellipsoid radius changes with latitude only
  2. Because map projections distort areas differently
  3. Because GPS satellites orbit at different altitudes
  4. Because the Earth's mass distribution is irregular, so the geoid departs from the ellipsoid Correct answer.

Explanation

The geoid is an equipotential surface of the gravity field, which is shaped by irregular mass distribution in the crust and mantle. Consequently its separation from the reference ellipsoid varies geographically.

Question 11Geodesy and Reference Systems

What is the practical purpose of a geoid model such as EGM2008 in surveying?

  1. To compute map projection scale factors
  2. To define the rotation of the celestial pole
  3. To convert between ellipsoidal and orthometric heights Correct answer.
  4. To determine satellite clock corrections

Explanation

A geoid model provides N values that link ellipsoidal heights from GNSS to orthometric heights used in levelling and engineering. This conversion is its principal practical role in height determination.

Question 12Geodesy and Reference Systems

What does a map projection do to the ellipsoid?

  1. It converts ellipsoidal coordinates to plane coordinates with controlled distortion Correct answer.
  2. It transforms one datum into another
  3. It converts gravity anomalies into deflections of the vertical
  4. It converts orthometric heights to ellipsoidal heights

Explanation

A projection is a mathematical mapping from the curved ellipsoid to a plane, chosen so that distortion in distance, area, or direction is controlled. It does not change the datum or the height system.

Question 13Geodesy and Reference Systems

A UTM zone uses a transverse Mercator projection. What is the scale factor along the central meridian?

  1. Exactly 1.000000
  2. 0.5000
  3. 1.0006
  4. 0.9996 Correct answer.

Explanation

UTM applies a central-meridian scale factor of 0.9996 so that scale error is balanced across the zone. Scale equals exactly one along two lines offset from the central meridian.

Question 14Geodesy and Reference Systems

Why are grid convergence and meridian convergence relevant when reducing geodetic observations to a projection grid?

  1. They describe the ionospheric delay in GNSS signals
  2. They describe the rotation between true north and grid north, which must be applied to azimuths Correct answer.
  3. They describe the difference between orthometric and ellipsoidal heights
  4. They describe the difference between two ellipsoid flattening values

Explanation

Grid convergence is the angle between true north and grid north at a point, so azimuths must be corrected by it when converting between geodetic and grid bearings. Ignoring it introduces orientation error in reduced observations.

Question 15Geodesy and Reference Systems

In the ITRF, what do the station velocity estimates primarily represent?

  1. The variation of the geoid over time only
  2. The atmospheric refraction at the site
  3. The tectonic motion of the station relative to the reference frame Correct answer.
  4. The instrumental drift of the antenna

Explanation

ITRF station velocities describe the secular motion of points on the Earth's surface, largely due to plate tectonics, within the terrestrial reference frame. They are used to propagate coordinates between reference epochs.

Question 16Geodesy and Reference Systems

Two stations are 50 km apart. Using a spherical Earth of radius 6371 km, what is the approximate central angle between them?

  1. 0.785 rad
  2. 0.0785 rad
  3. 7.85 rad
  4. 0.00785 rad Correct answer.

Explanation

The central angle equals arc length divided by radius: 50/6371, which is about 0.00785 rad. This is roughly 0.45 degrees, consistent with a short inter-station baseline.

Question 17Geodesy and Reference Systems

Why is a conformal projection such as Gauss-Kruger preferred for large-scale cadastral mapping?

  1. It preserves all distances exactly across the whole zone
  2. It preserves areas exactly across the whole zone
  3. It preserves angles locally, so small shapes and bearings are represented correctly Correct answer.
  4. It eliminates the need for a geoid model

Explanation

Conformal projections preserve angles and local shape at each point, which is valuable for cadastral and engineering work where bearings and small figures matter. Distances and areas still require scale corrections away from the standard lines.

Question 18Geodesy and Reference Systems

A transformation between two ITRF realizations requires both a reference epoch and station velocities. What happens if coordinates are used at an epoch different from the reference epoch without applying velocities?

  1. The geoid undulation becomes zero
  2. The coordinates acquire an error proportional to the velocity times the epoch difference Correct answer.
  3. The coordinates remain exact because the frame is static
  4. The ellipsoid parameters change with the epoch

Explanation

ITRF coordinates are epoch-dependent because stations move with tectonic plates. Omitting the velocity propagation leaves a positional error equal to velocity multiplied by the time difference from the reference epoch.

Question 19Geodesy and Reference Systems

In precise GNSS height determination, why can the difference between geoid undulations from two models exceed the accuracy needed for engineering levelling?

  1. Because GNSS satellites transmit heights in different units
  2. Because geoid models differ in data coverage, gravity data resolution, and computation method, producing decimetre-level discrepancies in some regions Correct answer.
  3. Because the ellipsoid is redefined every year
  4. Because orthometric heights are defined only at sea level

Explanation

Geoid models are computed from heterogeneous gravity data and different Stokes or least-squares techniques, so their undulations can disagree by decimetres where data are sparse. This directly limits the accuracy of GNSS-derived orthometric heights.

Question 20Geodesy and Reference Systems

A national mapping agency must publish coordinates that remain stable for cadastral records while GNSS users require global consistency. Which strategy best satisfies both requirements?

  1. Use only spherical coordinates without any datum definition
  2. Adopt the current ITRF epoch directly for all cadastral coordinates and update them yearly
  3. Define a static national frame aligned to ITRF at a fixed epoch and publish a time-dependent transformation to current ITRF Correct answer.
  4. Use a local datum with no transformation to any global frame

Explanation

A static national frame fixed at a reference epoch keeps cadastral coordinates stable while a published time-dependent transformation links it to the global frame. This separates legal coordinate stability from geodetic consistency with ITRF.

Question 21Surveying and Engineering Surveying

In a closed traverse of four sides, the sum of the measured interior angles was 359°58'20". What is the angular misclosure, and what is the usual permissible value for a total-station traverse of this order?

  1. −1'40" misclosure, well outside the typical ±20"√n tolerance Correct answer.
  2. −0'40" misclosure, marginally within tolerance
  3. −1'40" misclosure, within the typical ±20"√n tolerance
  4. +1'40" misclosure, within the typical ±20"√n tolerance

Explanation

The theoretical sum for a four-sided polygon is (n−2)×180° = 360°. The measured sum is 1'40" (100") short, so the misclosure is −100". A typical second-order tolerance of about ±20"√n gives ±40" for n = 4, so 100" is clearly excessive and the work must be repeated.

Question 22Surveying and Engineering Surveying

A total station measures a slope distance of 250.000 m to a target. The zenith angle is 85°00'00". What is the horizontal distance?

  1. 21.792 m
  2. 249.049 m Correct answer.
  3. 249.050 m
  4. 250.000 m

Explanation

Horizontal distance equals slope distance multiplied by sin(zenith angle). sin 85° = 0.996195, and 250.000 × 0.996195 = 249.049 m. The vertical component, by contrast, would be 250 × cos 85° = 21.79 m.

Question 23Surveying and Engineering Surveying

Which statement correctly describes the principle of a resection (free-station) setup with a total station?

  1. The unknown point is fixed by a single measured distance from one known control point.
  2. Two known points are occupied in turn and the unknown point is fixed by intersecting the two measured distances.
  3. The instrument occupies an unknown point and its coordinates are computed from measured directions or distances to known control points. Correct answer.
  4. The instrument occupies a known point and bearings are transferred to an unknown point by turning a measured angle.

Explanation

Resection, often called free-station setup, places the instrument on a point of unknown coordinates and solves for that position from observations to two or more known control points. It differs from intersection, where the unknown point is the target rather than the instrument station.

Question 24Surveying and Engineering Surveying

A level is set up midway between two pegs A and B, 80 m apart, and reads 1.532 m on A and 1.208 m on B. The instrument is then moved close to B and reads 1.415 m on B and 1.760 m on A. What is the collimation error per 80 m of sight?

  1. 0.007 m
  2. 0.042 m
  3. 0.345 m
  4. 0.021 m Correct answer.

Explanation

The two-peg test compares the apparent height difference from the balanced setup (1.532 − 1.208 = 0.324 m) with that from the unbalanced setup (1.760 − 1.415 = 0.345 m). The difference of 0.021 m is the collimation error accumulated over the 80 m separation of the pegs.

Question 25Surveying and Engineering Surveying

In a differential levelling run, the backsight on a benchmark is 1.305 m and the foresight on the next point is 2.118 m. The benchmark reduced level is 45.620 m. What is the reduced level of the forward point?

  1. 44.807 m Correct answer.
  2. 46.433 m
  3. 43.502 m
  4. 44.813 m

Explanation

The height difference is backsight minus foresight: 1.305 − 2.118 = −0.813 m. Adding this to the benchmark level gives 45.620 − 0.813 = 44.807 m, confirming the forward point lies below the benchmark.

Question 26Surveying and Engineering Surveying

Which statement best explains why a precise level uses a parallel-plate micrometer?

  1. It magnifies the staff image to allow estimation of fractions of a millimetre.
  2. It keeps the line of sight horizontal when the telescope is tilted slightly.
  3. It compensates for Earth curvature and refraction automatically over long sights.
  4. It displaces the line of sight by a small, precisely known amount so the staff can be read against the nearest graduation. Correct answer.

Explanation

The parallel-plate micrometer tilts a plane-parallel glass plate in front of the objective, shifting the image of the staff by a controlled amount read on a drum. This lets the observer bring the staff graduation to the crosshair instead of estimating, giving sub-millimetre readings.

Question 27Surveying and Engineering Surveying

A theodolite reads 0°00'00" on a reference mark and 47°23'40" on a target in the face-left position. In face-right the readings are 180°00'20" and 227°24'10". What is the mean angle and the resulting index error?

  1. 47°23'40" with an index error of 0"
  2. 47°24'10" with an index error of +30"
  3. 47°23'15" with an index error of +20"
  4. 47°23'45" with an index error of −10" Correct answer.

Explanation

The face-left angle is 47°23'40" and the face-right angle is 227°24'10" − 180°00'20" = 47°23'50". The mean is 47°23'45", and the half-difference between the faces, 5", reflects a residual index error of about −10" in the circle readings.

Question 28Surveying and Engineering Surveying

What does a scale factor of 0.9996 applied to a UTM projection mean for a distance measured on the projection plane?

  1. The scale factor varies from 0.9996 at the zone edges to 1.0000 at the central meridian.
  2. A true ground distance must be multiplied by 0.9996 to obtain the grid distance at the central meridian. Correct answer.
  3. A grid distance must be multiplied by 0.9996 to obtain the true ground distance.
  4. Distances are unaffected because the scale factor applies only to directions.

Explanation

On the UTM projection the central meridian has a scale factor of 0.9996, so projected grid distances are slightly shorter than the ellipsoidal distances they represent. Ground or ellipsoidal distance is therefore multiplied by the scale factor to reduce it to the grid.

Question 29Surveying and Engineering Surveying

A GNSS baseline is observed in static mode with two receivers. Which combination of observation and processing best reduces the effect of satellite clock and orbit errors?

  1. Differencing the carrier-phase observations between the two receivers to form a single difference. Correct answer.
  2. Using only the broadcast ephemeris and a single receiver.
  3. Increasing the position dilution of precision by observing low-elevation satellites.
  4. Averaging the pseudorange solutions from each receiver separately.

Explanation

Between-receiver differencing cancels satellite clock errors and largely removes orbit errors because both stations see the same satellite errors. Carrier-phase data give far higher precision than pseudorange, and relative positioning also cancels most atmospheric delay when baselines are short.

Question 30Surveying and Engineering Surveying

In setting out a horizontal curve by the deflection-angle method, the deflection angle to a point on the curve at arc length l from the tangent point is calculated as:

  1. δ = l / (2R) radians Correct answer.
  2. δ = l² / (2R) radians
  3. δ = l / R radians
  4. δ = 2l / R radians

Explanation

The deflection angle from the tangent to a chord equals half the central angle subtended by the arc, so δ = (l/R)/2 = l/(2R) radians. This linear relationship makes the method convenient for pegging a circular curve from the tangent point.

Question 31Surveying and Engineering Surveying

What is the primary purpose of a reciprocal observation between two stations in a triangulation or precise traverse network?

  1. To check that the total station's EDM prism constant is zero.
  2. To double the measured distance so that the scale is fixed.
  3. To eliminate the effect of refraction and instrument miscentring by averaging simultaneous opposite observations. Correct answer.
  4. To convert grid bearings to magnetic bearings.

Explanation

Reciprocal observations taken simultaneously from both ends of a line average out lateral refraction and reduce the effect of instrument and target centring errors. The mean of the two directions is used in the network adjustment.

Question 32Surveying and Engineering Surveying

A distance of 1 250.000 m measured along a slope is to be reduced to the ellipsoid. The mean ellipsoidal height is 1 500 m and the local Earth radius is 6 371 000 m. What is the ellipsoidal (sea-level) distance?

  1. 1 250.294 m
  2. 1 250.000 m
  3. 1 249.706 m Correct answer.
  4. 1 249.999 m

Explanation

The reduction factor is R/(R+h) = 6 371 000/(6 371 000 + 1 500) = 0.9997646. Multiplying 1 250.000 m by this factor gives 1 249.706 m, the chord length at the ellipsoid surface.

Question 33Surveying and Engineering Surveying

Which statement about the prism constant of a total station is correct?

  1. It is the refractive index of the prism glass used to compute the speed of light.
  2. It is the angle by which the prism is tilted from the line of sight.
  3. It is the correction applied for atmospheric temperature and pressure.
  4. It is the offset between the electrical and mechanical centres of the prism, added to or subtracted from the measured distance. Correct answer.

Explanation

The prism constant accounts for the difference between the point where the EDM signal is reflected and the physical centre of the prism, typically 0 mm or −30 mm depending on the prism type. It is a fixed additive correction to every measured distance.

Question 34Surveying and Engineering Surveying

In a least-squares adjustment of a levelling network, a loop closes with a misclosure of 8 mm over 4 km of levelling. If the standard deviation per kilometre is 2 mm, what is the standardized residual, and how should the loop be judged?

  1. 2.0, indicating the misclosure is consistent with the assumed precision. Correct answer.
  2. 4.0, indicating a gross error in the loop.
  3. 0.5, indicating the observations are far too precise.
  4. 16.0, indicating the loop must be rejected.

Explanation

The loop misclosure has a standard deviation of σ√L = 2 × √4 = 4 mm. The standardized residual is therefore 8/4 = 2.0, which is within the usual 95 % confidence limit of about 1.96 and so is acceptable.

Question 35Surveying and Engineering Surveying

What is the function of a tribrach with an optical plummet in setting up a total station over a ground mark?

  1. It allows the instrument to be levelled and centred over the mark independently of the instrument body. Correct answer.
  2. It compensates for residual tilt of the vertical axis.
  3. It measures the height of the instrument above the mark directly.
  4. It provides the horizontal circle reference for orientation.

Explanation

The tribrach carries the levelling foot screws and the optical or laser plummet, so the instrument can be levelled and centred over the station mark. The instrument body can then be removed or swapped without disturbing the setup.

Question 36Surveying and Engineering Surveying

A vertical curve is required between two gradients of +2.0 % and −3.0 % meeting at a chainage of 10+000 with a reduced level of 50.000 m. With a curve length of 200 m, what is the reduced level at the highest point of the curve?

  1. 49.600 m
  2. 50.000 m
  3. 50.400 m Correct answer.
  4. 50.200 m

Explanation

The algebraic change in gradient is −5.0 % over 200 m, giving a rate of change of −0.025 % per metre. The highest point occurs where the gradient becomes zero, 2.0/0.025 = 80 m from the start, i.e. at chainage 9+980. The rise from the tangent point is 2.0 % × 80 m minus the curve correction, giving a level of 50.400 m.

Question 37Surveying and Engineering Surveying

Which of the following best defines the term 'grid convergence' in a projected coordinate system?

  1. The scale distortion along a parallel of latitude.
  2. The angle between grid north and true north at a point on the projection. Correct answer.
  3. The difference between the ellipsoidal and orthometric heights.
  4. The angle between magnetic north and true north at a point.

Explanation

Grid convergence is the angle between the grid north of the projection and true (geodetic) north at a given point. It arises because meridians converge on the ellipsoid but are drawn parallel on the grid, and it must be applied when converting between grid and true bearings.

Question 38Surveying and Engineering Surveying

During a total-station traverse, a blunder is suspected in one leg. Which check most directly detects a single gross angular error in a closed traverse?

  1. Checking that the back and fore distances of each leg agree.
  2. Comparing the closing easting with the closing northing.
  3. Comparing the sum of the measured angles with the theoretical sum for the polygon. Correct answer.
  4. Checking that the instrument height equals the target height.

Explanation

In a closed polygon the interior angles must sum to (n−2)×180°, so a large discrepancy reveals a gross angular error. The linear misclosure then helps to localize which leg is affected.

Question 39Surveying and Engineering Surveying

What does the term 'centring error' contribute to a horizontal angle measured between two targets?

  1. An angular error that increases as the sight distances decrease. Correct answer.
  2. An angular error that increases as the sight distances increase.
  3. An error that affects only distances, not angles.
  4. A constant angular error independent of sight length.

Explanation

A centring offset e at the instrument produces an angular error of roughly e/d for a sight of length d. Short sights therefore amplify the effect, which is why short backsights should be avoided in precise work.

Question 40Surveying and Engineering Surveying

Which statement correctly describes the difference between orthometric height and ellipsoidal height?

  1. Orthometric height is a purely geometric quantity obtained directly from GNSS.
  2. Both are measured above the geoid and differ only by the projection scale factor.
  3. Orthometric height is measured above the geoid, while ellipsoidal height is measured above the reference ellipsoid. Correct answer.
  4. Orthometric height is measured above the ellipsoid, while ellipsoidal height is measured above the geoid.

Explanation

Orthometric height is the height above the geoid, the equipotential surface approximating mean sea level. Ellipsoidal height is the geometric height above the reference ellipsoid and is what GNSS provides; the two differ by the geoid undulation N.

Question 41Surveying and Engineering Surveying

A total station with a stated distance accuracy of ±(2 mm + 2 ppm) measures a distance of 3 000 m. What is the expected standard deviation of the measured distance, excluding other error sources?

  1. ±12 mm
  2. ±8 mm Correct answer.
  3. ±6 mm
  4. ±2 mm

Explanation

The constant part contributes 2 mm and the scale-dependent part contributes 2 ppm × 3 000 m = 6 mm. Combining them gives 2 + 6 = 8 mm for the stated accuracy specification.

Question 42Surveying and Engineering Surveying

In precise engineering setting-out, why are forced-centring targets and tribrachs used on control pillars?

  1. They automatically level the instrument without foot screws.
  2. They increase the range of the EDM by concentrating the return signal.
  3. They ensure that the instrument or target is repeatedly positioned on the same vertical axis, removing centring error. Correct answer.
  4. They eliminate atmospheric refraction along the line of sight.

Explanation

Forced-centring fittings seat the instrument or target in a fixed, repeatable socket, so the same point is occupied every time. This removes the random centring error that would otherwise dominate short, high-precision engineering measurements.

Question 43Surveying and Engineering Surveying

A plan is drawn at a scale of 1:500. A rectangular building measures 42 mm by 27 mm on the plan. What is its area on the ground?

  1. 2 835 m²
  2. 113.4 m²
  3. 283.5 m² Correct answer.
  4. 28.35 m²

Explanation

At 1:500, 1 mm on the plan represents 0.5 m on the ground, so the sides are 21.0 m and 13.5 m. Their product is 283.5 m², which is also the plan area multiplied by 500².

Question 44Surveying and Engineering Surveying

Which statement best explains why precise traverses often use a gyrotheodolite to orient one leg?

  1. It measures the vertical deflection to reduce distances to the ellipsoid.
  2. It measures the geoid undulation to convert ellipsoidal heights.
  3. It determines astronomical (true) north from the Earth's rotation, providing an azimuth independent of the control network. Correct answer.
  4. It measures the local magnetic declination so that compass bearings can be corrected.

Explanation

A gyrotheodolite senses the direction of the Earth's rotation axis and thus true north, giving an absolute azimuth that does not depend on the orientation of the existing control. This is valuable in tunnels and mines where the network cannot be extended by conventional means.

Question 45Satellite Positioning and GNSS

In GNSS positioning, what does the broadcast ephemeris of a satellite primarily provide?

  1. The almanac of all satellites in the constellation
  2. The ionospheric correction coefficients for single-frequency users
  3. The signal-in-space accuracy of the navigation message
  4. The precise satellite orbit and clock parameters valid for a limited period Correct answer.

Explanation

The broadcast ephemeris is transmitted in the navigation message and contains the orbital elements and clock bias parameters that let a receiver compute satellite position and clock offset for a validity window of a few hours. It is distinct from the almanac, which is a coarse long-term catalogue, and from ionospheric model coefficients.

Question 46Satellite Positioning and GNSS

Why does a GNSS receiver need at least four satellites to determine a three-dimensional position?

  1. Because the almanac only stores four satellites at a time
  2. Because four satellites are required to compute the ionospheric delay
  3. Because four satellites are needed to decode the navigation message
  4. Because three satellites fix position and the fourth resolves the receiver clock offset Correct answer.

Explanation

Each satellite observation yields one pseudorange equation containing four unknowns: three position coordinates and the receiver clock bias. Four independent pseudoranges therefore give a solvable system. Three satellites would leave the clock offset undetermined unless an external time reference is available.

Question 47Satellite Positioning and GNSS

What is the fundamental observable of a GNSS receiver before any corrections are applied?

  1. The pseudorange Correct answer.
  2. The Doppler shift
  3. The dilution of precision
  4. The carrier-to-noise density ratio

Explanation

The pseudorange is the measured signal travel time multiplied by the speed of light, containing the true range plus satellite and receiver clock errors and atmospheric delays. Carrier-to-noise density, dilution of precision and Doppler are quality or derived quantities rather than the raw range observable.

Question 48Satellite Positioning and GNSS

Which coordinate reference frame is the native output frame of GPS broadcast ephemeris computations?

  1. WGS 84 Correct answer.
  2. NAD83
  3. ETRS89
  4. ITRF

Explanation

GPS broadcasts its ephemerides in the WGS 84 reference frame, so satellite positions are computed directly in that system. ITRF, ETRS89 and NAD83 are alternative terrestrial frames that require a datum transformation if used for output.

Question 49Satellite Positioning and GNSS

What does a dilution of precision value describe in a GNSS solution?

  1. The geometric strength of the satellite configuration Correct answer.
  2. The magnitude of the ionospheric delay
  3. The number of visible satellites
  4. The accuracy of the broadcast ephemeris

Explanation

Dilution of precision is a purely geometric factor derived from the satellite-receiver geometry matrix; it amplifies ranging errors into position errors. It is independent of the actual range errors and of the number of satellites except insofar as geometry depends on them.

Question 50Satellite Positioning and GNSS

Which GNSS error source is largely removed by differencing simultaneous observations from two nearby receivers?

  1. Receiver noise
  2. Multipath at each receiver
  3. Satellite and receiver clock errors Correct answer.
  4. Satellite orbit error

Explanation

Differencing between two receivers cancels errors common to both, notably the satellite clock error and, when the receivers share a reference oscillator, the receiver clock error. Orbit and multipath errors are not common to both receivers and therefore do not cancel.