Surveying covers the measurement and mapping of the earth's surface — from classification and chain surveying fundamentals, through compass surveying, levelling, theodolite traversing, and tacheometry, to field astronomy, GPS, map preparation, photogrammetry, remote sensing, layout surveys for civil structures, and setting out of curves. Every design formula, IS/IRC code reference, diagram, solved example and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Environmental Engineering (network layout and pipeline alignment there rely on the same field measurement principles developed here). Leads to: Highway Engineering, which directly builds on curve setting-out, gradient, and alignment survey concepts covered in this chapter.
| Basis | Types | Description |
|---|---|---|
| Nature of field | Land, Hydrographic, Astronomical, Aerial | Based on where survey is conducted |
| Object / Purpose | Topographic, Cadastral, City, Engineering, Mine, Military, Geological | Based on what is being surveyed |
| Instruments used | Chain, Compass, Plane Table, Theodolite, Tacheometric, Photographic, EDM/GPS | Based on primary measuring tool |
| Methods employed | Triangulation, Traversing, Radiation, Intersection | Based on geometric technique |
| Earth's curvature | Plane surveying (area <260 km²); Geodetic surveying (larger areas) | Plane: earth treated as flat; Geodetic: accounts for curvature |
| Instrument | Standard Length | Least Count / Accuracy | Use |
|---|---|---|---|
| Gunter's chain | 20.12 m (66 ft); 100 links | 1 link = 0.2012 m | Land surveying (historical) |
| Engineer's chain | 30.48 m (100 ft); 100 links | 1 link = 0.3048 m | Engineering surveys |
| Metric chain | 20 m or 30 m; 100 or 150 links | 20 m: 1 link = 0.2 m; 30 m: 1 link = 0.2 m | Standard in India (IS 1492) |
| Steel tape | 10, 20, 30, 50 m | ±2 mm in 30 m | Precise distance measurement |
| Invar tape | 30, 50, 100 m | Very high (low thermal expansion) | Precise base line measurement |
Chain and tape measurements require several systematic corrections before being used for design: a correction for the chain/tape's actual length differing from its nominal length, a slope correction to convert measured slope distance to horizontal distance, a temperature correction (the tape expands or contracts relative to the temperature it was standardised at), a pull (tension) correction (a stretched tape reads slightly long), and a sag correction (an unsupported tape hanging in a catenary between supports always measures longer than the true straight-line distance, so this correction is always negative). When the applied pull is increased specifically to make the pull correction exactly cancel the sag correction, the result is called the normal tension.
Offsets are short measurements taken perpendicular (or oblique) from the chain line to locate detail points. An angular error in laying off a perpendicular offset produces a positional error that grows with the square of the offset length, which is why accurate work limits how long an offset may be. Oblique offsets (measured from two chain points to a single object) are located using the cosine rule once both distances and the included angle are known.
Triangulation establishes control by measuring all the angles of a network of triangles along with just one baseline distance; trilateration instead measures all the side lengths directly (feasible today with EDM/GPS) without needing angle observations at all. Most modern control surveys combine both techniques. A well-conditioned ("strong") triangle keeps every angle between 30° and 120°, with 60° (equilateral) as the ideal — angles below 30° are elongated and let small angular errors propagate into large position errors.
\( L_{true} = L_{measured}\times\dfrac{L_{actual}}{L_{nominal}} \)
\( D = L\cos\theta \) or \( D = L - \dfrac{h^2}{2L} \) (approx, h/L<0.1)
\( C_t = \alpha L(T-T_0) \), \(\alpha=11.2\times10^{-6}/^\circ C\) (steel)
\( C_p = \dfrac{(P-P_0)L}{AE} \)
\( C_s = -\dfrac{w^2l^3}{24P^2} \) (always negative, per span)
\( e = \dfrac{x^2}{2L} \); \( x=\sqrt{2eL} \)
Given: A 20 m chain was found to be 20.10 m after testing. Distance measured by this chain \(=250\ \text{m}\). Find the true distance.
Solution:
\( L_{true} = 250\times\dfrac{20.10}{20} = 250\times1.005 = 251.25\ \text{m} \)
Answer: True distance \(=251.25\ \text{m}\).
Given: A steel tape of weight \(w=0.05\ \text{kg/m}\) is stretched unsupported over a span \(l=30\ \text{m}\) under a pull \(P=100\ \text{N}\) (weight in appropriate consistent units, \(w\) taken as N/m \(=0.49\)). Find the sag correction.
Solution:
\( C_s = -\dfrac{0.49^2\times30^3}{24\times100^2} = -\dfrac{0.2401\times27000}{240000} = -\dfrac{6482.7}{240000}\approx-0.027\ \text{m} \)
Answer: \(C_s\approx-0.027\ \text{m}\) (27 mm too long per 30 m span) — must be subtracted from the measured length.
Given: Chain line spacing \(L=30\ \text{m}\), permissible error in offset \(e=0.05\ \text{m}\). Find the maximum offset length.
Solution:
\( x = \sqrt{2\times0.05\times30} = \sqrt{3} \approx 1.73\ \text{m} \)
Answer: Maximum offset length \(\approx1.73\ \text{m}\) for this permissible error.
Fig. 1.1 — Chain surveying corrections: slope correction converts the measured slope length to horizontal distance; sag correction accounts for the unsupported tape hanging in a catenary, which always reads longer than the true span.
A Whole Circle Bearing (WCB) is measured clockwise from north through the full 0°–360° range. A Quadrantal (Reduced) Bearing is instead measured from north or south toward east or west, always within 0°–90°, and is converted from WCB using a quadrant-dependent rule. The back bearing of a line is obtained from its forward bearing by adding or subtracting 180°. Declination is the angle between magnetic north and true north at a location, and must be added to or subtracted from a magnetic bearing (depending on whether declination is east or west) to obtain the true bearing.
Local attraction is a disturbance of the magnetic needle caused by nearby magnetic influences (iron objects, electric cables, steel structures), and it affects every bearing read at that particular station equally. It is detected whenever the forward bearing and back bearing of a line do not differ by exactly 180°. Correction proceeds by first finding a station free from local attraction (where FB−BB=180° exactly), then propagating the error found at each subsequent affected station back through the traverse.
Each traverse line is resolved into its latitude (north-south component) and departure (east-west component) using its bearing and length. For a properly closed traverse, the algebraic sum of all latitudes and the algebraic sum of all departures should each be zero; any non-zero result is the closing error, whose magnitude (as a fraction of the total traverse length) expresses the survey's precision.
Bowditch's (Compass) Rule distributes the closing error to each line's latitude and departure in proportion to that line's own length — appropriate when angular and linear measurement accuracy are comparable. The Transit Rule instead distributes the error in proportion to the magnitude of each line's latitude/departure component — used when angular accuracy is much better than linear accuracy (typical for a theodolite traverse), since it avoids over-correcting lines that are already well-oriented.
After adjustment, the independent coordinates (northing and easting) of each traverse station are computed by successively accumulating the corrected latitudes and departures from a starting point. The enclosed area is then found directly from these coordinates by the coordinate (or "double meridian distance") method — no separate area survey is needed. Where the boundary is irregular rather than a closed traverse, area under an offset curve can instead be approximated by the trapezoidal rule, or more accurately by Simpson's 1/3 rule (which requires an even number of equal divisions).
0°–90°: N(WCB)E; 90°–180°: S(180°−WCB)E
180°–270°: S(WCB−180°)W; 270°–360°: N(360°−WCB)W
\( BB = FB\pm180° \)
True bearing = Magnetic bearing \(\pm\delta\) (+ for E declination)
\( L=d\cos\theta \), \( D=d\sin\theta \)
\( e=\sqrt{e_L^2+e_D^2} \); Precision \(=1:(\Sigma \dfrac{d}{e})\)
Correction to L (or D) of a line \(=-e_L\times\dfrac{\text{length of line}}{\text{total length}}\)
Correction to L \(=-e_L\times\dfrac{|L\text{ of line}|}{\Sigma|L|}\)
\( 2A=\Sigma(N_i E_{i+1}-N_{i+1}E_i) \)
Simpson's 1/3: \( A=\dfrac{d}{3}[(h_1+h_n)+4\Sigma_{even}+2\Sigma_{odd}] \)
Given: A line has WCB \(=225°\). Find the reduced (quadrantal) bearing.
Solution:
Since \(180°<225°<270°\): \(RB = S(225°-180°)W = S45°W\)
Answer: \(RB=S45°W\).
Given: A traverse line has length \(d=100\ \text{m}\) and bearing \(N60°E\) (i.e. \(\theta=60°\) from north, east quadrant). Find latitude and departure.
Solution:
\( L = 100\cos60° = 50\ \text{m}\ (\text{N, positive}) \)
\( D = 100\sin60° = 86.6\ \text{m}\ (\text{E, positive}) \)
Answer: Latitude \(=+50\ \text{m}\), Departure \(=+86.6\ \text{m}\).
Given: A closed traverse has total length \(\Sigma d=500\ \text{m}\), closing error in latitude \(e_L=+0.5\ \text{m}\). One line has length \(120\ \text{m}\). Find the Bowditch correction to that line's latitude.
Solution:
\( \text{Correction} = -0.5\times\dfrac{120}{500} = -0.12\ \text{m} \)
Answer: Correction to latitude of that line \(=-0.12\ \text{m}\).
Fig. 2.1 — Whole Circle Bearing (measured clockwise from North, 0°–360°) versus Reduced/Quadrantal Bearing (measured from North or South toward East or West, always within 0°–90°) — the same direction expressed two ways.
| Term | Definition |
|---|---|
| Datum | Reference surface for elevation; MSL (Mean Sea Level) is standard in India (GTS Benchmark) |
| RL (Reduced Level) | Elevation of a point above the datum |
| BS (Back Sight) | First staff reading taken on a point of known RL (BM or CP) after instrument is set up |
| FS (Fore Sight) | Last staff reading taken before shifting the instrument; on TP or BM |
| IS (Intermediate Sight) | Any reading taken between BS and FS from the same instrument position |
| HI (Height of Instrument) | Elevation of line of collimation = RL of BM + BS reading |
| Turning Point (TP) | Intermediate point where both BS and FS are taken (instrument moved here) |
| Benchmark (BM) | Fixed reference point of known RL; GTS BMs by Survey of India |
The Height of Instrument (HI) Method computes the elevation of the horizontal line of sight at each instrument setup, then subtracts each staff reading from it to get that point's RL — simple and fast, but its arithmetic check (ΣBS−ΣFS = Last RL−First RL) does not catch errors made in intermediate sight readings, a significant limitation when many IS readings are taken. The Rise and Fall Method instead computes the rise or fall between every consecutive pair of readings and accumulates RL directly; its three-way arithmetic check (ΣBS−ΣFS = ΣRise−ΣFall = Last RL−First RL) does catch IS errors, making it the preferred method for precise work despite requiring more calculation.
| Type | Purpose | Method |
|---|---|---|
| Simple levelling | Determine difference in elevation between 2 nearby points | One instrument position |
| Differential levelling | Determine RL of distant points through TPs | Multiple instrument positions; most common |
| Fly levelling | Rapid transfer of RL over long distance; less accurate | Long sight lines; no IS readings |
| Profile levelling | RL along a route (road, canal, pipeline centreline) | Differential levelling along centreline with IS at stations |
| Cross-section levelling | RL across width of route for earthwork calculation | IS taken at measured offsets from centreline |
| Reciprocal levelling | Determine accurate elevation across wide obstacle (river) | Instrument on each bank; mean of two results eliminates curvature + refraction |
| Trigonometric levelling | RL from vertical angles and distances | Theodolite; used in hilly terrain |
Over long sight distances, the earth's curvature makes a level line appear to fall away below the true horizontal, causing a distant staff reading to appear too large — the curvature correction accounts for this. Atmospheric refraction bends the line of sight downward (toward denser air near the ground), partially compensating for curvature by roughly one-seventh of its magnitude. Since curvature and refraction act in opposite directions but curvature dominates, their combined net effect still makes distant readings appear too high, so the combined correction must still be subtracted from the observed staff reading.
The two-peg test detects an error in the line of collimation (the line of sight not being exactly horizontal when the bubble is centred). By first setting the instrument exactly midway between two staff points — where a collimation error affects both readings equally and cancels out — the true elevation difference is found. Moving the instrument next to one of the points reveals the collimation error directly, since any discrepancy from the true difference must now come from the tilted line of sight (no longer cancelled by symmetry).
Precise (geodetic) levelling uses an invar staff, a parallel-plate micrometer, short balanced sight lengths, and careful field procedure to achieve millimetre-level accuracy over long distances — essential for establishing a national vertical control network. Trigonometric levelling instead computes elevation difference from a measured vertical angle and either a horizontal or slope distance; for short sights the simple trigonometric relation suffices, but for longer sights the same curvature-and-refraction correction used in direct levelling must be added.
\( HI = RL_{BM}+BS \); \( RL = HI - \text{staff reading} \)
Check: \( \Sigma BS-\Sigma FS = \text{Last RL}-\text{First RL} \)
\( RL_n = RL_{n-1}+\text{Rise (or}-\text{Fall)} \)
Check: \( \Sigma BS-\Sigma FS = \Sigma Rise-\Sigma Fall = \text{Last RL}-\text{First RL} \)
\( C_c = 0.0785D^2 \); \( C_r = -0.0112D^2 \) (D in km)
Combined: \( C = 0.0673D^2 \) (subtract from reading)
\( D = \sqrt{2Rh}\approx3.855\sqrt{h} \) (D in km, h in m)
Angular value/division \( = \dfrac{s}{R} \) radians
Short sight: \( h = D\tan\alpha \)
Long sight: \( h = D\tan\alpha + 0.0673D^2 \)
Given: BM RL \(=100.000\ \text{m}\), BS on BM \(=1.500\ \text{m}\). IS reading at point P \(=0.850\ \text{m}\). Find HI and RL of P.
Solution:
\( HI = 100.000+1.500 = 101.500\ \text{m} \)
\( RL_P = 101.500-0.850 = 100.650\ \text{m} \)
Answer: \(RL_P=100.650\ \text{m}\).
Given: A levelling sight is taken over a distance \(D=2\ \text{km}\). Find the combined curvature and refraction correction.
Solution:
\( C = 0.0673\times2^2 = 0.0673\times4 = 0.269\ \text{m} \)
Answer: \(C\approx0.269\ \text{m}\) must be subtracted from the observed staff reading.
Given: An observer's eye is at height \(h=1.8\ \text{m}\) above sea level. Find the distance to the visible horizon.
Solution:
\( D = 3.855\sqrt{1.8} = 3.855\times1.342\approx5.17\ \text{km} \)
Answer: Visible horizon distance \(\approx5.17\ \text{km}\).
Fig. 3.1 — Differential levelling: BS taken on the benchmark, IS taken on an intermediate point, FS taken on the turning point before the instrument shifts, with HI as the horizontal line of collimation.
A theodolite is a precision optical instrument for measuring horizontal and vertical angles. The vernier theodolite (least count ~20″) has largely been superseded by the optical/micrometer theodolite (least count ~1″), and both are increasingly replaced by the electronic total station, which combines a theodolite with an EDM (electronic distance measurement) in a single unit and stores data digitally for later download — eliminating manual booking errors.
Every time a theodolite is set up, three temporary adjustments must be carried out in sequence: centring (bringing the instrument exactly over the survey station, using a plumb bob or optical plummet), levelling (using the plate bubble and the standard two-screw-then-rotate-90°-then-third-screw procedure to bring the vertical axis truly vertical), and elimination of parallax (focusing the eyepiece on the cross-hairs and the objective on the target, so the image and cross-hairs appear to move together when the eye shifts slightly).
The method of repetition accumulates an angle over several repeated turns before taking a single final reading, dividing by the number of repetitions to get the average — this averages out graduation errors and increases precision roughly in proportion to the square root of the number of repetitions. The method of reiteration (directions method) instead takes direct and reverse (Face Left / Face Right) readings on all signals at a station; averaging FL and FR readings eliminates line-of-collimation error, trunnion-axis error, and eccentricity error simultaneously, making it the most systematic method for triangulation work.
A vertical angle is likewise measured on both Face Left and Face Right and averaged to eliminate index error — the small residual error remaining when the vertical circle does not read exactly 0° (or 90° for zenith angle) when the line of sight is truly horizontal. The zenith angle is measured from the vertical (rather than the horizontal), so it equals 90° minus the vertical angle.
In the included angle method, the interior (or exterior) angle is measured directly at each traverse station; the sum of interior angles of a closed polygon traverse must equal \((n-2)\times180°\), giving a direct arithmetic check on the whole traverse. The deflection angle method instead measures the angle between the prolongation of the back line and the forward line at each station — common in route surveys (roads, railways) where the traverse follows a long, roughly straight alignment rather than closing into a polygon. If one side's length or bearing is accidentally omitted from the field measurements, it can be recovered afterward from the traverse's closing-error geometry.
EDM instruments send an electromagnetic wave (microwave, infrared, or laser) to a retroreflector and measure the phase difference between the outgoing and returning signal to compute distance. Two systematic error types must be calibrated out: an additive (prism) constant — a fixed offset introduced by the reflector — and a scale error, which grows proportionally with distance and is caused mainly by atmospheric refraction (varying with pressure, temperature, and humidity along the sight path), correctable via an atmospheric correction formula.
Average angle \(=\dfrac{\text{final reading}}{n}\); precision \(\propto\sqrt{n}\)
\( \alpha = \dfrac{FL-FR}{2} \) or \(\dfrac{FL+FR-360°}{2}\)
\( Z = 90°-\alpha \); Index error \( e=\dfrac{ZA_{FL}+ZA_{FR}-360°}{2} \)
Interior angles: \( \Sigma = (n-2)\times180° \)
Exterior angles: \( \Sigma = (n+2)\times180° \)
Missing length: \( l=\sqrt{e_L^2+e_D^2} \)
Missing bearing: \( \theta=\arctan(\dfrac{e_D}{e_L}) \)
\( D = n\dfrac{\lambda}{2} + \phi\dfrac{\lambda}{4}\pi \)
Corrected \(=D_{observed}\times(1+\dfrac{K_{ppm}}{10^6})\)
Given: An angle is measured by repetition 4 times; the final accumulated reading is \(240°16'\). Find the angle.
Solution:
\( \text{Angle} = 240°\dfrac{16'}{4} = 60°4' \)
Answer: Angle \(=60°4'\).
Given: A closed traverse has 6 sides. Find the theoretical sum of interior angles and check whether measured angles summing to \(719°58'\) are acceptable (small errors distributed).
Solution:
\( \Sigma_{theoretical} = (6-2)\times180° = 720° \)
Discrepancy \(=720°-719°58'=2'\), a small closing error typically acceptable and distributed equally among the angles.
Answer: Theoretical sum \(=720°\); measured total is off by only 2′, an acceptable closing error.
Given: Zenith angles observed: \(ZA_{FL}=75°20'\), \(ZA_{FR}=285°16'\). Find the index error and the corrected zenith angle.
Solution:
\( e = \dfrac{75°20'+285°16'-360°}{2} = \dfrac{0°36'}{2} = 0°18' \)
Corrected \(ZA = 75°20'-0°18' = 75°02'\)
Answer: Index error \(=0°18'\); corrected zenith angle \(=75°02'\).
Fig. 4.1 — Face Left / Face Right observation: taking a reading with the telescope direct (FL), then plunged and rotated 180° (FR), and averaging the two eliminates line-of-collimation, trunnion-axis, and eccentricity errors.
Tacheometry allows simultaneous determination of horizontal distance and elevation difference using a theodolite and graduated staff, without any tape measurement. It is rapid but less accurate than chaining — well suited to topographic surveys, contour mapping, and detail surveys where speed matters more than millimetre precision.
The stadia method uses two additional horizontal cross-hairs (stadia hairs) in the telescope, symmetrically placed above and below the main cross-hair. The vertical intercept these hairs subtend on a graduated staff — the staff intercept — is directly proportional to the horizontal distance to the staff, via a simple multiplying constant fixed by the telescope's optical design (internal-focusing telescopes have essentially zero additive constant, simplifying the formula considerably).
For a horizontal line of sight, distance is simply the staff intercept multiplied by the stadia (multiplying) constant. When the line of sight is inclined at angle α (as is usually the case when sighting up or down sloping ground), both the horizontal distance and the vertical component must be corrected by trigonometric factors of \(\cos\alpha\) — because the staff, held vertically, is no longer perpendicular to the inclined line of sight, and the effective staff intercept the formula needs is the intercept projected perpendicular to the sight line.
When stadia hairs are unavailable (or an ordinary theodolite without a stadia diaphragm is used), distance can still be found tacheometrically by observing two separate vertical angles — one to the top and one to the foot of a staff of known height — and applying trigonometry. This tangential method requires more careful angle observation than the stadia method since the accuracy depends entirely on precise angular measurement rather than an optical intercept.
A subtense bar is a fixed-length horizontal bar (typically 2 m) set up at the point whose distance is to be found; the theodolite at the observing station measures the small horizontal angle the bar subtends, from which distance follows directly (for small angles, distance is inversely proportional to the subtended angle). This method achieves good accuracy (about 1 in 1000) for moderate distances without needing any staff reading at all.
\( D = ks+c \) (\(k=100\), \(c\approx0\) internal focus)
\( D = ks\cos^2\alpha \)
\( V = ks\sin\alpha\cos\alpha = D\tan\alpha \)
\( RL_B = RL_A+h_i+V-m \)
(V positive for elevation, negative for depression)
\( D = \dfrac{S\tan\alpha_1\tan\alpha_2}{\tan\alpha_1-\tan\alpha_2} \)
\( D = \dfrac{S\tan\alpha_1\tan\alpha_2}{\tan\alpha_1+\tan\alpha_2} \)
\( D = \dfrac{b}{2\tan(\beta/2)}\approx\dfrac{b}{\beta} \) (small β, radians)
Given: A stadia reading gives staff intercept \(s=1.25\ \text{m}\) on a horizontal sight. Find the distance (internal focusing, \(k=100\), \(c=0\)).
Solution:
\( D = 100\times1.25 = 125\ \text{m} \)
Answer: \(D=125\ \text{m}\).
Given: Staff intercept \(s=1.0\ \text{m}\), angle of elevation \(\alpha=6°\), \(h_i=1.4\ \text{m}\), central hair reading \(m=1.9\ \text{m}\), \(RL_A=200.00\ \text{m}\). Find horizontal distance and \(RL_B\).
Solution:
\( D = 100\times1.0\times\cos^2(6°) = 100\times0.9891 = 98.91\ \text{m} \)
\( V = D\tan(6°) = 98.91\times0.1051 \approx 10.40\ \text{m} \)
\( RL_B = 200.00+1.4+10.40-1.9 = 209.90\ \text{m} \)
Answer: \(D\approx98.91\ \text{m}\), \(RL_B\approx209.90\ \text{m}\).
Given: A 2 m subtense bar subtends a horizontal angle \(\beta=0°34'\) at the observation station. Find the distance.
Solution:
\( \beta = 34'\times(\pi/180/60) = 0.009890\ \text{rad} \)
\( D = \dfrac{2}{0.009890} \approx 202.2\ \text{m} \)
Answer: \(D\approx202.2\ \text{m}\).
Fig. 5.1 — Stadia tacheometry: the staff intercept s between the upper and lower stadia hairs is proportional to horizontal distance, D = ks + c = 100s for an internal-focusing telescope.
| Term | Definition |
|---|---|
| Celestial sphere | Imaginary sphere of infinite radius centred at observer; all celestial bodies appear on its surface |
| Zenith (Z) | Point on celestial sphere directly above observer; opposite = Nadir |
| Celestial meridian | Great circle through Z, N pole, S pole, and Nadir |
| Declination (δ) | Angular distance of celestial body N or S of celestial equator (like latitude for stars) |
| Right Ascension (RA) | Angular distance E of vernal equinox (like longitude for stars) |
| Hour Angle (HA) | Angle at pole between observer's meridian and star's meridian; measured W; 0–24 hours |
| Altitude (a) | Vertical angle above horizon to celestial body (0°–90°) |
| Azimuth (A) | Horizontal angle from North (or South) to the vertical circle through the body |
| Zenith Distance (z) | 90° − altitude; angular distance from zenith to body |
The PZS (astronomical) triangle is formed on the celestial sphere by the pole (P), the observer's zenith (Z), and the observed star (S). Its three sides — co-latitude, zenith distance, and co-declination — and three angles — hour angle, azimuth, and parallactic angle — are related by spherical trigonometry (the cosine rule for spherical triangles), giving the fundamental altitude formula that connects a star's observed altitude to the observer's latitude, the star's declination, and the hour angle. Rearranging the same relationship gives the azimuth of the body, which is what a surveyor actually needs to establish true north.
GPS (along with GLONASS, Galileo, and India's own NavIC) is a satellite-based positioning system relying on trilateration (not triangulation) — the receiver computes its position from measured ranges to multiple satellites, not from angles. Because the receiver's own clock is not synchronised to the same precision as the atomic clocks onboard the satellites, a fourth satellite is needed in addition to the three needed for basic 3D position, specifically to solve for the receiver clock error as a fourth unknown.
| Error Source | Magnitude | Remedy |
|---|---|---|
| Ionospheric delay | 1–10 m (dominant) | Dual-frequency receiver (L1+L2); ionospheric model |
| Tropospheric delay | 0.5–2 m | Elevation mask >15°; tropospheric model (Hopfield, Saastamoinen) |
| Satellite clock error | <1 m | Broadcast corrections in navigation message |
| Receiver clock error | Eliminated by | Using 4th satellite (clock solution) |
| Multipath | 1–5 m | Site selection; choke ring antennas; RAIM |
| Ephemeris error | <1 m | Precise ephemerides from IGS |
The Dilution of Precision (DOP) quantifies how the satellite geometry amplifies these ranging errors into a position error — a low PDOP (well-spread-out satellites) gives a much more reliable fix than a high PDOP (satellites clustered together), even with identical individual ranging accuracy.
NavIC is India's own regional (not global) navigation satellite system, covering the Indian subcontinent and roughly 1500 km of surrounding region using a constellation of geostationary and geosynchronous satellites, providing accuracy comparable to (and in some regional respects better than) standard GPS.
\( \sin a = \sin\phi\sin\delta+\cos\phi\cos\delta\cos H \)
\( \cos A = \dfrac{\sin\delta-\sin\phi\sin a}{\cos\phi\cos a} \)
\( \rho_i = R_i + c\,b \) (b = receiver clock error)
4 satellites → 3D position (X,Y,Z) + clock error
\( PDOP^2 = HDOP^2+VDOP^2 \)
Ideal PDOP < 6; poor if > 8
Standard GPS: ~3–5 m; DGPS: 0.5–3 m; RTK: ±1–2 cm
Given: Observer latitude \(\phi=28°\), star declination \(\delta=15°\), hour angle \(H=30°\). Find the altitude.
Solution:
\( \sin a = \sin28°\sin15°+\cos28°\cos15°\cos30° \)
\( = 0.4695\times0.2588 + 0.8829\times0.9659\times0.8660 \approx 0.1215+0.7386 = 0.8601 \)
\( a = \sin^{-1}(0.8601) \approx 59.35° \)
Answer: Altitude \(\approx59.35°\).
Given: A GPS receiver reports \(HDOP=1.8\), \(VDOP=2.5\). Find the PDOP and assess quality.
Solution:
\( PDOP = \sqrt{1.8^2+2.5^2} = \sqrt{3.24+6.25} = \sqrt{9.49}\approx3.08 \)
Answer: \(PDOP\approx3.08 < 6\), indicating good satellite geometry and a reliable position fix.
Given: A receiver can currently see 3 satellites. Can it compute a 3D position with clock correction?
Solution: Three satellites give 3 range equations, sufficient only for 3 unknowns; but with an unknown receiver clock error, there are 4 unknowns (X, Y, Z, and clock bias b).
Answer: No — a minimum of 4 satellites is required to solve for 3D position plus receiver clock error simultaneously.
Fig. 6.1 — PZS astronomical triangle: Pole (P), Zenith (Z), and Star (S) with sides related to co-latitude, co-declination, and zenith distance, connected by spherical trigonometry to give the altitude and azimuth formulae.
Plane table surveying is a graphical method in which the map is drawn directly in the field simultaneously with the survey, eliminating separate office plotting and calculation. Directions are sighted with the alidade, and distances are measured by tacheometry or tape. Four principal methods are used depending on the field situation — radiation (rays drawn from one station to every detail point), intersection (rays from two known stations intersect to fix an inaccessible point), traversing (the table itself is moved and re-oriented station to station), and resection (the instrument station itself is located from rays back to points already plotted on the sheet — the classic "three-point problem").
A contour line joins all ground points of equal elevation, giving a two-dimensional representation of three-dimensional terrain. The contour interval is chosen based on the map scale and the required level of engineering detail — smaller-scale maps use a larger interval, while detailed engineering plans use a much finer interval (0.5–2 m). The horizontal distance between successive contours (the horizontal equivalent) directly indicates ground slope: closely spaced contours mean steep terrain, widely spaced contours mean gentle terrain.
The direct method physically locates points of exactly known elevation on the ground and plots them directly — the most accurate but slowest approach, generally reserved for high-precision work. The indirect (cross-section/grid) method instead takes spot heights at a regular grid or along cross-sections, and interpolates the contour lines afterward — much faster and the most commonly used method for open ground. For large-area mapping, aerial photogrammetry is fastest, deriving contours from stereo photograph pairs.
\( CI = \dfrac{H}{n} \)
1:50,000 map → 20 m; 1:25,000 → 10 m; engineering → 0.5–2 m
Scale of slope \(=\tan(\text{slope angle}) = \dfrac{CI}{HE}\)
Gradient \( g = \dfrac{CI}{HE_{actual}} \)
Given: A total elevation difference of \(H=100\ \text{m}\) is to be shown with \(n=20\) contours. Find the contour interval.
Solution:
\( CI = \dfrac{100}{20} = 5\ \text{m} \)
Answer: \(CI=5\ \text{m}\).
Given: Contour interval \(CI=5\ \text{m}\), and on the map, adjacent contours are 8 mm apart at scale 1:5000 (so \(HE=8\times5000=40{,}000\ \text{mm}=40\ \text{m}\)). Find the gradient.
Solution:
\( g = \dfrac{CI}{HE} = \dfrac{5}{40} = 0.125 = 1\ \text{in}\ 8 \)
Answer: Gradient \(=1\ \text{in}\ 8\) (12.5%).
Given: A 500-hectare open agricultural area needs contours at a moderate accuracy, with survey time as the primary constraint. Recommend a contouring method.
Solution: Large open area with moderate accuracy and speed as priorities rules out the slow, high-accuracy direct method and doesn't justify aerial photogrammetry's setup cost for a single, modest-sized area.
Answer: The indirect (grid/cross-section) method is most suitable — fast enough for large open ground with acceptable accuracy for agricultural planning.
Fig. 7.1 — Contour V-shapes: contours crossing a valley form a V pointing upstream (uphill), while contours crossing a ridge form a V pointing downhill — the opposite orientation distinguishes the two landforms at a glance.
Photogrammetry is the science and art of obtaining reliable measurements from photographs, extracting geometric and spatial information from images without direct physical contact with the object. It ranges from terrestrial (close-range) photogrammetry for architectural or industrial measurement, through aerial photogrammetry for topographic mapping, to UAV/drone and satellite photogrammetry for site surveys and global-scale mapping respectively.
The scale of a vertical aerial photograph is simply the ratio of the camera's focal length to the flying height above the ground — for undulating terrain, the local scale actually varies point to point depending on each point's own elevation above the datum, since flying height is effectively reduced for higher ground. A critical distortion unique to aerial photographs is relief displacement: any object with height above the datum appears radially displaced outward from the photograph's centre (principal point), an effect that grows with both the object's height and its distance from the centre — this is precisely what allows building or tree heights to be measured from a photograph, but it also means aerial photos (unlike orthophotos) are not true planimetric maps.
Successive photographs along a flight line must overlap substantially (typically 60% forward overlap) so that every ground point appears on at least two photographs, which is essential for stereoscopic (3D) viewing and measurement. Adjacent flight lines also need a smaller side overlap (typically 30%) to ensure full coverage without gaps. The base-to-height ratio (B/H) — the air base (distance between successive exposure stations) relative to flying height — governs stereoscopic measurement precision, with an optimal range of about 0.5–0.6.
When the same point appears on two overlapping photographs taken from different exposure stations, its image position differs slightly between the two photos — this parallax is directly related to the point's elevation. A stereoscopic instrument (or software) measures this parallax difference between a point of unknown elevation and a reference datum point, from which the elevation follows directly — this is the fundamental principle behind generating a Digital Elevation Model (DEM) from a stereo pair.
\( S = \dfrac{f}{H} \) (flat ground); \( S = \dfrac{f}{H-h} \) (undulating)
\( d = hr/H \)
r = radial distance to image of top of object; h = object height
\( L = l\times \dfrac{H}{f} = \dfrac{l}{S} \)
\( \dfrac{B}{H} = (1-\text{fwd overlap fraction})\times\dfrac{\text{photo width}}{f} \)
Optimal \(\dfrac{B}{H}=0.5\text{–}0.6\)
\( h = H\Delta \dfrac{p}{p_b} \) (small \(\Delta p\))
\( p_b = \dfrac{fB}{H} \) (absolute stereoscopic parallax)
Given: An aerial camera has focal length \(f=150\ \text{mm}\), flying height \(H=3000\ \text{m}\). Find the photo scale.
Solution:
\( S = \dfrac{0.150}{3000} = \dfrac{1}{20}{,}000 \)
Answer: Scale \(=1:20{,}000\).
Given: A tower of height \(h=40\ \text{m}\) appears at radial distance \(r=80\ \text{mm}\) from the principal point on a photo taken from flying height \(H=2000\ \text{m}\). Find the relief displacement.
Solution:
\( d = \dfrac{40\times80}{2000} = \dfrac{3200}{2000} = 1.6\ \text{mm} \)
Answer: Relief displacement \(=1.6\ \text{mm}\) on the photo (outward from the principal point).
Given: Flying height \(H=1800\ \text{m}\), absolute parallax of datum point \(p_b=90\ \text{mm}\), parallax difference for a building top \(\Delta p=2.5\ \text{mm}\). Find the building height.
Solution:
\( h = \dfrac{1800\times2.5}{90} = \dfrac{4500}{90} = 50\ \text{m} \)
Answer: Building height \(\approx50\ \text{m}\).
Fig. 8.1 — Relief displacement: a tall object's top images farther from the principal point than its base, causing the object to appear to lean radially outward in a vertical aerial photograph, with displacement d = hr/H.
Remote sensing acquires information about an object or area from a distance — typically from a satellite or aircraft sensor — without physical contact, by exploiting electromagnetic radiation (EMR) reflected or emitted from the Earth's surface. Different EMR wavelength bands reveal different surface properties: visible and near-infrared for general reflectance and vegetation, thermal infrared for temperature, and microwave for all-weather imaging capable of penetrating cloud cover.
| Parameter | Passive RS | Active RS |
|---|---|---|
| Energy source | Sun (reflected solar radiation) or Earth's thermal emission | Own source (radar, lidar, sonar) |
| Day/night operation | Daytime only (visible/NIR); thermal works at night | Day and night (own illumination) |
| Cloud penetration | Cannot (visible/NIR cloud-blocked) | SAR: penetrates clouds and rain |
| Examples | Landsat, SPOT, IKONOS, Cartosat | SAR (ERS, Sentinel-1, RISAT); LiDAR; RADAR altimeter |
| Resolution Type | Definition | Example |
|---|---|---|
| Spatial | Smallest ground feature distinguishable; ground pixel size | 0.3 m (WorldView-3); 30 m (Landsat); 23.5 m (Cartosat-1) |
| Spectral | Number and width of spectral bands; ability to distinguish features by colour | Pan: 1 band; MS: 4–8 bands; Hyperspectral: 100–200+ bands |
| Temporal | Revisit time (how often same area is imaged) | MODIS: daily; Landsat: 16 days; Cartosat-2: 4 days |
| Radiometric | Number of digital levels (bits) for intensity; sensitivity to brightness difference | 8-bit = 256 levels; 11-bit = 2048; 16-bit = 65536 levels |
| Satellite | Resolution | Application |
|---|---|---|
| Cartosat-1 | 2.5 m (PAN); stereo pair (fore + aft) | Large-scale mapping; DEM generation; cadastral |
| Cartosat-2 series | 0.65–1 m (PAN) | Urban planning, infrastructure mapping |
| LISS-IV (ResourceSat) | 5.8 m (MS) | Crop monitoring, land use mapping |
| RISAT-1 | 1–3 m (C-band SAR) | All-weather mapping; flood, oil spill |
| Oceansat-2 | 360 m | Ocean colour, wind speed, fisheries |
| INSAT-3D | 1–4 km | Weather, cyclone, fog monitoring |
Spectral indices combine reflectance from specific bands to isolate a particular land-cover property. The NDVI (Normalized Difference Vegetation Index) exploits the fact that healthy vegetation strongly reflects near-infrared while absorbing red light for photosynthesis, giving a high positive index value for dense healthy vegetation and low values for bare soil, water, or snow. Land-cover classification is done either by supervised classification (an analyst defines training sites of known class, and the algorithm classifies the rest) or unsupervised classification (clustering algorithms like ISODATA or k-means group pixels automatically, with classes assigned meaning only after the fact).
GIS integrates spatial data (maps, satellite imagery) with attribute data (population, soil type, land use) to enable overlay analysis, buffering, spatial querying, network analysis, and 3D modelling. Spatial data is stored either as vector data (points, lines, polygons — precise boundaries, compact file size) or raster data (a grid of cells, each with one value — the natural format for satellite imagery and DEMs). Since India's older Survey of India toposheets used the Everest 1830 datum with a Polyconic projection, while GPS and modern systems use WGS-84 with UTM, a datum/projection transformation is often needed when combining old and new spatial data.
Visible: 0.4–0.7 µm; NIR: 0.7–1.3 µm; SWIR: 1.3–3 µm
Mid-IR: 3–8 µm; Thermal IR: 8–14 µm; Microwave: 1 mm–1 m
\( NDVI = \dfrac{NIR-Red}{NIR+Red} \), range \(-1\) to \(+1\)
>0.6: dense vegetation; 0.2–0.6: sparse/moderate; <0.1: bare/water/snow
\( NDWI = \dfrac{Green-NIR}{Green+NIR} \) (water bodies)
\( NDBI = \dfrac{SWIR-NIR}{SWIR+NIR} \) (built-up index)
Given: A pixel has NIR reflectance \(=0.45\), Red reflectance \(=0.08\). Find NDVI and interpret.
Solution:
\( NDVI = \dfrac{0.45-0.08}{0.45+0.08} = \dfrac{0.37}{0.53} \approx 0.698 \)
Answer: \(NDVI\approx0.70 > 0.6\), indicating dense, healthy vegetation.
Given: A flood-mapping mission requires imaging through heavy monsoon cloud cover. Recommend a sensor type.
Solution: Passive optical sensors (visible/NIR, like Landsat or Cartosat) are blocked by clouds, making them unsuitable during monsoon flood events.
Answer: SAR (Synthetic Aperture Radar), an active microwave sensor, is required since it penetrates cloud cover and works day or night — e.g., RISAT-1 or Sentinel-1.
Given: A crop-monitoring application needs to detect changes every 5 days. Compare MODIS (daily), Landsat (16 days), and Cartosat-2 (4 days) for suitability.
Solution: Landsat's 16-day revisit is too infrequent to catch 5-day changes; both MODIS (daily) and Cartosat-2 (4-day) satisfy the requirement, but MODIS has much coarser spatial resolution.
Answer: Cartosat-2 (4-day revisit) is the better choice if higher spatial resolution is also needed; MODIS is preferable if very frequent (daily) coarse monitoring is sufficient.
Fig. 9.1 — Electromagnetic spectrum bands used in remote sensing: visible through thermal-IR bands are used passively and blocked by cloud cover, while microwave (SAR) is an active sensor that penetrates clouds and rain.
A road alignment is finalised through progressively more detailed stages: a reconnaissance survey walks the route to identify obstacles and alternatives; a preliminary survey runs a traverse and levels along each alternative to draw a longitudinal section; and a location survey then works out the detailed cross-sections and volume estimates for the chosen alignment before pegging the centre line at regular intervals. The ruling gradient is the maximum gradient normally permitted for the terrain and road category — steeper gradients require special justification, and gradients on horizontal curves must additionally be reduced (compensated) since the combined effect of curve resistance and steep grade would otherwise strain vehicles beyond the intended design limit.
Railway ruling gradients are generally much flatter than road gradients (since trains cannot climb steep grades), and — like roads — must be compensated on curves to offset the extra resistance a curve imposes on a train. Track geometry standards fix the gauge (rail spacing) and the maximum superelevation permitted on curves, with a required transition length over which the cant (superelevation) is introduced gradually, governed by the rate of change of cant that passengers can comfortably tolerate.
Canal alignment typically follows a contour or the natural watershed so that water flows by gravity throughout. The Full Supply Level (FSL) must follow the required hydraulic gradient along the entire route, while the bed level (below FSL by the canal's design depth) must never rise above the FSL — a canal physically cannot flow uphill. Earthwork along the alignment is estimated from cross-sections at regular intervals, with cut sections (where FSL sits below existing ground) and fill sections (embankments, where FSL sits above ground) computed separately.
A bridge survey requires a hydrological survey of the stream (cross-sections, flood levels, and bed profile both upstream and downstream) to establish the High Flood Level (HFL) and estimate afflux — the rise in backwater level caused by the bridge piers obstructing flow, which must be kept small to avoid upstream flooding. Pier positions are set out from control points on each bank, typically using the intersection method with two theodolites (or a single total station) when direct measurement across the water is impractical.
Setting out a building begins from an established baseline (a road edge or property boundary), from which the building's corners are located by offset and then squared using a right-angle method — either the classical 3-4-5 triangle or an optical square (which uses two mirrors at 45° to sight a perpendicular direction without any calculation at all). The layout is verified by checking that the diagonals of the resulting rectangle are equal, confirming all four corners are correctly square.
A culvert site is ideally chosen where a road or railway crosses a drainage channel at a right angle (a "square" culvert), since a skewed crossing is structurally and hydraulically more complex, though sometimes unavoidable if the culvert must follow the natural drainage direction. The minimum road level at the culvert must clear the stream's full supply level plus afflux plus freeboard plus the pavement thickness — an important check to avoid the road itself being overtopped during a flood event.
\( G = \dfrac{RL_B-RL_A}{\text{horizontal distance}}\times100\% \)
Ruling gradient (IRC:73): 3.3% NH; Railway (IRS): 1:150 (BG plains)
Roads: reduce gradient by \( \dfrac{30}{R}+\dfrac{75}{R} \) (curve + grade)
Railways: \( G_{comp}=G-0.04\times\text{degree of curve} \)
\( BL = FSL - d \) (d = canal depth, 1.5–4 m)
Bed slope: 1:5000 to 1:10,000 (main canals)
\( V = \Sigma\left[\dfrac{A_1+A_2}{2}\times L\right] \) (trapezoidal; prismoidal more accurate)
\( D_s = 1.33\times(HFL-\text{foundation level}) \)
Given: Two points on a proposed road have \(RL_A=250.00\ \text{m}\), \(RL_B=253.50\ \text{m}\), horizontal distance \(=140\ \text{m}\). Find the gradient and compare to the NH ruling gradient.
Solution:
\( G = \dfrac{253.50-250.00}{140}\times100 = \dfrac{3.50}{140}\times100 \approx 2.5\% \)
Answer: \(G\approx2.5\%\), within the NH ruling gradient limit of 3.3%.
Given: Cross-sectional areas at two stations 20 m apart are \(A_1=12\ \text{m}^2\), \(A_2=18\ \text{m}^2\). Find the earthwork volume between them.
Solution:
\( V = \dfrac{12+18}{2}\times20 = 15\times20 = 300\ \text{m}^3 \)
Answer: \(V=300\ \text{m}^3\).
Given: HFL \(=105.20\ \text{m}\); foundation level \(=98.00\ \text{m}\). Find the scour depth by Lacey's formula.
Solution:
\( D_s = 1.33\times(105.20-98.00) = 1.33\times7.20 \approx 9.58\ \text{m} \)
Answer: Scour depth \(\approx9.58\ \text{m}\) below HFL.
Fig. 10.1 — Building layout: corners set out perpendicular to the baseline using the 3-4-5 method or optical square; the rectangle is verified by checking that both diagonals (AC and BD) are equal.
| Curve Type | Profile / Plan | Use |
|---|---|---|
| Simple circular curve | Circular arc; constant radius R | Change in direction of route (horizontal) |
| Compound curve | Two or more simple curves of different radii | Difficult terrain where single radius insufficient |
| Reverse curve | Two simple curves turning in opposite directions | Railway junctions; avoid if possible (no transition) |
| Transition (easement) curve | Curvature increases gradually from 0 to 1/R; Euler's spiral or clothoid | Roads and railways between tangent and circular curve |
| Summit curve (vertical) | Convex parabolic curve; crest | At hilltop; visibility requirement governs |
| Valley (sag) curve | Concave parabolic curve | At valley; headlight sight distance governs at night |
A simple circular curve connects two straight tangents deflecting by angle Δ at their point of intersection (PI). Its key elements — tangent length, arc length, long chord, mid-ordinate, and external distance — are all functions of just the radius R and the deflection angle Δ, following directly from the geometry of a circular arc. The degree of curve re-expresses the same curve sharpness as an angle (the angle subtended by a fixed chord length, typically 20 or 30 m) rather than as a radius — a convention still used in some route-survey practice.
The offsets-from-long-chord method needs no theodolite and is fast for small curves, computing perpendicular offsets from the chord at measured intervals. The successive offsets from chord produced (Rankine) method instead works peg-by-peg along the curve itself. The deflection angle method (Rankine's method) is the most widely used in practice: a theodolite at the point of curvature is oriented toward the PI (zero reading), and each successive peg is located by turning a computed deflection angle and chaining the fixed chord length from the previous peg — a running check confirms the cumulative deflection at the point of tangency equals exactly half the total intersection angle. Where chaining across the curve is impossible (a ravine or river), the two-theodolite method locates each point purely by the intersection of two simultaneously-observed rays, with no distance measurement at all.
A transition curve gradually introduces curvature — from zero at the tangent point up to the full curvature \(\dfrac{1}{R_0}\) of the following circular arc — so that centrifugal force builds up smoothly rather than abruptly, essential for railways (where it is mandatory) and strongly recommended for highways. The ideal transition shape is the clothoid (Euler's spiral), in which the product of the local radius and distance travelled along the curve is held constant. Introducing a transition curve requires the main circular curve to be shifted slightly inward (the "shift") to accommodate the transition geometry at each end. Superelevation (cant) — banking the road or track cross-section inward on a curve — offsets centrifugal force and is subject to a maximum permitted value for safety and passenger comfort.
A vertical curve is a parabolic arc smoothly connecting two different grades at their vertical point of intersection (VPI), chosen as a parabola because it gives a constant rate of change of gradient along its length — a mathematically convenient and comfortable profile. A summit (crest) curve must be long enough to provide adequate sight distance for a driver to see and stop for an obstacle before reaching it, while a valley (sag) curve is instead governed at night by the reach of the vehicle's headlight beam, since the driver's own eye height doesn't help over a dip the way it does over a crest.
\( T=R\tan(\dfrac{\Delta}{2}) \); \( L=\pi R\dfrac{\Delta}{180}°=R\Delta \) (rad)
\( L_c=2R\sin(\dfrac{\Delta}{2}) \); \( M=R(1-\cos(\dfrac{\Delta}{2})) \)
\( E=R(\sec(\dfrac{\Delta}{2})-1) \); Degree of curve \(D\dfrac{\approx1718.87}{R}\)
\( y_x=\sqrt{R^2-x^2}-\sqrt{R^2-(\dfrac{L_c}{2})^2} \)
\( \delta = \dfrac{C}{2R} \) rad \( = \dfrac{90C}{\pi R} \) degrees
Cumulative to PT \(=\dfrac{\Delta}{2}\)
\( lR = LR_0 = A^2 \) (constant)
Min. length (roads): \( L=\dfrac{V^3}{47R} \) [V km/h, R m]
Shift: \( s=\dfrac{L^2}{24R} \)
\( e = \dfrac{V^2}{gR}=\dfrac{V^2}{127R} \)
Max \(e=0.07\) (highways); 165 mm (BG railways)
\( y = y_{VPC}+g_1 x+\dfrac{(g_2-g_1)x^2}{2L_v} \)
High/low point: \( x = \dfrac{g_1 L_v}{g_1-g_2} \)
Given: A simple circular curve has radius \(R=300\ \text{m}\) and deflection angle \(\Delta=60°\). Find the tangent length, curve length, and long chord.
Solution:
\( T = 300\tan(30°) = 300\times0.5774 = 173.2\ \text{m} \)
\( L = 300\times60\times\dfrac{\pi}{180} = 314.16\ \text{m} \)
\( L_c = 2\times300\times\sin(30°) = 600\times0.5 = 300\ \text{m} \)
Answer: \(T=173.2\ \text{m}\), \(L=314.16\ \text{m}\), \(L_c=300\ \text{m}\).
Given: Curve radius \(R=250\ \text{m}\), peg interval (chord) \(C=20\ \text{m}\). Find the deflection angle per chord.
Solution:
\( \delta = \dfrac{90\times20}{\pi\times250} = \dfrac{1800}{785.4} \approx 2.292° = 2°17.5' \)
Answer: Deflection angle per 20 m chord \(\approx2°17.5'\).
Given: A highway curve has radius \(R=400\ \text{m}\), design speed \(V=80\ \text{km/hr}\). Find the required superelevation and minimum transition length.
Solution:
\( e = \dfrac{80^2}{127\times400} = \dfrac{6400}{50800} \approx 0.126 \) — exceeds max 0.07, so superelevation is capped at 0.07 and partial speed compensation (or larger radius) is needed.
\( L = \dfrac{80^3}{47\times400} = \dfrac{512{,}000}{18{,}800} \approx 27.2\ \text{m} \)
Answer: Ideal \(e\approx0.126\) exceeds the 0.07 maximum (so \(e\) is capped at 0.07 for this design speed/radius combination); minimum transition length \(\approx27.2\ \text{m}\).
Fig. 11.1 — Simple circular curve elements: PI = point of intersection (vertex), PC = point of curvature (T₁), PT = point of tangency (T₂), T = tangent length, L = arc length, Δ = deflection angle.
Engineering geology applies the study of rocks, minerals and earth structures to civil projects — choosing safe sites and sound foundations for dams, tunnels, bridges and buildings, and anticipating hazards such as landslides and seepage.
A mineral is a naturally occurring inorganic solid of definite composition; a rock is an aggregate of minerals. Rocks are classified by origin:
| Type | Formation | Examples |
|---|---|---|
| Igneous | Solidification of molten magma/lava | Granite (intrusive), basalt (extrusive) |
| Sedimentary | Deposition and compaction of sediments; often stratified | Sandstone, limestone, shale |
| Metamorphic | Alteration of existing rock by heat and pressure | Marble (from limestone), quartzite, gneiss, slate |
The attitude of a rock bed is given by its dip (angle of maximum slope below horizontal) and strike (horizontal line on the bed, perpendicular to the dip). Deformation produces:
Weathering (physical, chemical, biological) breaks rock down in place, weakening foundations and forming soil. Geological knowledge guides site selection: a dam needs a watertight, strong foundation free of adverse faults and permeable joints; a tunnel alignment avoids weak, water-bearing or heavily jointed rock; a reservoir site must not leak through porous strata.
\( L_{true}=L_{meas}\times\dfrac{L_{actual}}{L_{nominal}} \)
\( C_s=-\dfrac{w^2l^3}{24P^2} \) (always negative)
\( L=d\cos\theta \), \( D=d\sin\theta \)
\( 2A=\Sigma(N_iE_{i+1}-N_{i+1}E_i) \)
\( HI=RL_{BM}+BS \); \( RL=HI-\text{reading} \)
\( C=0.0673D^2 \) (D km, subtract)
\( D=100s \) (horiz.); \( D=100s\cos^2\alpha \) (inclined)
\( \sin a=\sin\phi\sin\delta+\cos\phi\cos\delta\cos H \)
\( S=\dfrac{f}{H} \); relief displacement \( d=hr/H \)
\( NDVI=\dfrac{NIR-Red}{NIR+Red} \)
\( T=R\tan(\dfrac{\Delta}{2}) \); \( D\dfrac{\approx1718.87}{R} \)
\( L=\dfrac{V^3}{47R} \); \( e=\dfrac{V^2}{127R} \)
| Topic | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Chain surveying | Tape corrections numerical; sag correction formula | All corrections combined; Tellurometer; baseline measurement | Chain types; correction names and direction |
| Compass surveying | Bowditch vs Transit rule; area by coordinates | Local attraction correction procedure; Gale's table; traverse closure | WCB to QB conversion; back bearing; declination |
| Levelling | HI method vs R&F; curvature+refraction; two-peg test | Precise levelling; profile levelling; reciprocal levelling; bubble sensitiveness | Basic terms; RL computation; types of levelling |
| Theodolite / Traversing | Angle measurement method; traverse closure; omitted measurements | Permanent adjustments; triangulation; EDM errors; total station | Temporary adjustments; face left/right concept |
| Tacheometry | Stadia formulae (horizontal and inclined); RL computation | Anallactic lens; subtense bar; tangential method; stadia constants | Stadia constant k=100; basic formula D=100s |
| Field astronomy / GPS | GPS satellite number; PDOP; DGPS vs RTK accuracy | Altitude/azimuth formulae; astronomical triangle; NavIC details | GPS principle; satellite number; NDVI and RS concepts |
| Map preparation | Contour V-shapes; slope from CI/HE | Three-point problem methods; contouring accuracy comparison | Contour characteristics; plane table methods |
| Photogrammetry | Scale=f/H; relief displacement; parallax height formula | Overlap calculations; B/H ratio; orthophoto; LiDAR | Types of RS; photo scale concept; overlap values |
| Remote sensing | NDVI formula; SAR vs passive; resolution types | Band combinations; classification methods; Indian satellites; GIS basics | Active vs passive; NDVI concept; satellite names |
| Layout surveys | Gradient computation; earthwork volume | Complete alignment survey design; afflux; scour | Ruling gradient values; 3-4-5 method |
| Curves | T, L, L_c formulae; Rankine deflection; degree of curve | Compound curves; transition length; vertical curve design; superelevation | Elements of simple curve; tangent length; deflection angle concept |
Q1. A 30 m chain is found to actually measure 30.15 m. A distance of 450 m is measured with it. Find the true distance.
Q2. A traverse line has bearing S30°W and length 80 m. Find its latitude and departure (with sign).
Q3. A levelling sight is taken over \(D=1.5\ \text{km}\). Find the combined curvature and refraction correction.
Q4. A simple circular curve has radius \(R=500\ \text{m}\) and deflection angle \(\Delta=40°\). Find the tangent length and external distance.
Q5. A pixel has NIR\(=0.30\), Red\(=0.25\). Compute NDVI and classify the land cover.