A surveyor measures a long line 100 times with a tape that is 5 mm too short, and averages the results. The effect of averaging on the error is that

it removes none of it — a systematic error accumulates in full however many times it is repeated

it reduces the error by a factor of 10, since random errors fall as √n

it reduces the error by a factor of 100, since the mean of n readings is n times better

it removes the error entirely, because positive and negative deviations cancel

Which of the following statements about levelling and traversing are correct?

The arithmetic check ΣBS − ΣFS = last RL − first RL must hold for any correctly reduced level book

Keeping backsight and foresight distances equal cancels curvature, refraction and collimation error together

The rise-and-fall method checks every intermediate sight, while the height-of-instrument method does not

The arithmetic check ΣBS − ΣFS = last RL − first RL must hold for any correctly reduced level book

Keeping backsight and foresight distances equal cancels curvature, refraction and collimation error together

The rise-and-fall method checks every intermediate sight, while the height-of-instrument method does not

A traverse that closes perfectly proves that every measured distance in it is correct

Triangulation obtains lengths by measuring one base line precisely and then measuring angles. Historically this was preferred over measuring every distance because

angles could be measured far more precisely than long distances, and the triangle-sum check localises a bad observation

angles are free of systematic error, whereas distances are not

triangulation requires no computation, unlike traversing

a triangulated network needs no adjustment because it has no closing error

Surveying: Errors and Adjustment, Levelling, Traversing, Triangulation and Curves

Geomatics Engineering is Section 7 of the Civil Engineering paper, and the only section of that paper that measures rather than designs, and that inversion changes what a correct answer even means. Everywhere else in the paper a quantity is either right or wrong; here every quantity carries an error, the error is not a mistake, and the engineering consists of knowing which kind of error you have and what can be done about it. The principles of surveying reduce to two working rules — always work from the whole to the part, so that a small error stays local instead of propagating, and always provide more observations than unknowns, so the surplus can be checked against itself. Everything below is one of those two rules in a particular disguise. The classification that matters is not the textbook list but a single question: does the error change sign from one measurement to the next? A random error does — it is as likely to be positive as negative — so repeated measurements pull against one another and the mean is better than any single reading: random error accumulates only as √n. A systematic error does not — a tape 5 mm too short is 5 mm too short every time — so repetition accumulates it in full, as n, and averaging does nothing at all. The worked numbers below put the gap at its starkest: over 100 measurements the same 0.05 m error gives 5.00 m if systematic and 0.50 m if random, a factor of ten apart. That is why a survey is not just measured but adjusted, why a traverse is closed back on its start rather than run out and trusted, and why the first thing done with a new instrument is to calibrate it against a known length. The second organising idea is that the shape of a survey is checked before its position: angles first, distances second, and a closure that fails is corrected by distributing the error rather than by re-measuring everything — which is what the Bowditch rule is, and why it distributes in proportion to length.

Errors, and the one question that classifies them

The three kinds, and what each responds to
Kind	Behaviour on repetition	Accumulates as	The only cure
Mistake (blunder)	Arbitrary — a misread staff, a transposed digit	Not at all — it is one wrong value	Detection by an independent check, then discard
Systematic (cumulative)	Same sign every time — a short tape, an uncalibrated instrument, sag, temperature	`n` — in full, every time	Calibration and a computed correction. Averaging does nothing
Random (accidental)	Sign varies — estimation of the last digit, bubble setting, pointing	`√n` — the readings partly cancel	Repetition and least-squares adjustment

🎯 Averaging is a tool that works on exactly one of the three

Take an error of 0.05 m per measurement. If it is random, n measurements accumulate 0.05√n: 0.158 m over ten, 0.500 m over a hundred. If it is systematic, they accumulate 0.05n: 0.500 m over ten, 5.000 m over a hundred. At n = 100 the two are a factor of ten apart from the same per-measurement error, and the gap widens as √n for ever. This is why the classification is not academic bookkeeping. A surveyor who measures a long line repeatedly and averages has genuinely improved a random error and has not touched a systematic one — the mean of a hundred readings from a tape 5 mm too short is still 5 mm too short per tape length, and now carries the false authority of a hundred readings. The remedy for a systematic error is never repetition; it is calibration against a known standard and a computed correction: standardisation for tape length, plus corrections for temperature, pull, sag and slope. Note also that the two behave differently under redundancy checks. A random error shows up as a small, unpredictable misclosure, which is what a Bowditch adjustment is designed to distribute. A systematic error can close a traverse perfectly while every distance in it is wrong — because a scale error affects every leg proportionally and cancels round a loop. A perfect closure is therefore evidence about random error only, which is the single most useful caution in this chapter.

Maps, scale and coordinate systems

Scale is a ratio, and the two traps in it are both about what the ratio does to area and to plotting accuracy. Halving the scale from 1:1000 to 1:2000 quarters the area a sheet covers per unit of ground — areas scale with the square of the representative fraction — so a question that converts a plotted area to a ground area and forgets to square the scale is out by that factor. And the smallest distance that can be plotted and read by eye is about 0.25 mm, so at 1:1000 nothing finer than 0.25 m on the ground can be shown at all: measuring to the millimetre and plotting at 1:1000 is measuring precision the map cannot carry, which is why survey accuracy is chosen from the map scale rather than the other way round.

A coordinate system then needs three separate things, and conflating them is the commonest conceptual error. A datum fixes the reference surface — the geoid for heights (mean sea level), an ellipsoid such as WGS-84 for positions — and two datums can put the same point tens of metres apart. A projection maps the curved surface onto a plane, and because that is geometrically impossible without distortion, every projection preserves some properties and destroys others: a conformal projection keeps angles and stretches areas, an equal-area projection does the reverse, and no projection does both. A grid is then the rectangular coordinate system laid on the projection, in which a surveyor actually computes. The practical consequence is that a distance measured on the ground, a distance on the ellipsoid and a distance on the grid are three different numbers, related by a height correction and a scale factor — small over a site, decisive over a state.

Levelling, and the correction that grows with the square of distance

Ordinary differential levelling carries a height from a benchmark by a chain of backsights and foresights, and its arithmetic check is exact and unavoidable: ΣBS − ΣFS = last RL − first RL. With ΣBS = 4.965 and ΣFS = 7.035 the difference is −2.070, so a first reduced level of 100.000 gives a last of 97.930, and any reduction that fails this check has an arithmetic error in it regardless of how careful the fieldwork was. The two reduction methods — height of instrument and rise and fall — differ in exactly one respect worth remembering: rise and fall provides a check on every intermediate sight (ΣRise − ΣFall must also equal the difference of the end levels), while height of instrument checks only the change points. That makes rise and fall slower and the right choice where intermediate sights matter, and height of instrument the right choice on a long line of change points.

⚠️ Curvature and refraction are negligible at 1 km and not at 5

A level line is horizontal; the earth’s surface is not. Over a sight of length D the curvature of the earth makes a truly horizontal line rise above the level surface by D²/2R, so a distant staff reading is too large. Working against it, refraction bends the line of sight downward, recovering about a seventh of that, and the standard combined correction is +0.06728 D² metres with D in kilometres, to be added to a trigonometric height difference. The important feature is the square. At D = 1 km the correction is 0.067 m — smaller than most survey tolerances, and normally ignored. At D = 5 km it is 1.682 m, twenty-five times larger, and ignoring it would be a gross error. So the honest statement is not "curvature and refraction are negligible" but "they are negligible at ordinary sight lengths and dominate at long ones". Two consequences follow. In differential levelling the correction cancels almost exactly if the backsight and foresight distances are made equal, which is why balancing sights is standard practice rather than fussiness — it removes curvature, refraction and any residual collimation error together. In trigonometric levelling over a long line there is nothing to cancel against, so the correction must be computed and applied: D = 1000 m at α = 2.5° gives h = D tan α = 43.661 m before correction and 43.728 m after.

Traversing: closure, and how the error is distributed

A closed traverse returns to its starting point, so the sums of its latitudes (L cos θ) and departures (L sin θ) must each be zero. They never are. The residuals ΣL and ΣD are the closing error in the two directions, and the magnitude of the misclosure is e = √(ΣL² + ΣD²). With ΣL = +0.15 m and ΣD = −0.20 m that is exactly 0.250 m, and over a perimeter of 700 m the relative error is 0.25/700 = 1 in 2800 — the figure by which a traverse is accepted or rejected, because it is dimensionless and can be compared against a specification whatever the size of the survey.

The two adjustment rules, and when each is right
Rule	Correction distributed in proportion to	Assumes
Bowditch (compass rule)	The length of each leg: `correction = −Σ × (leg/perimeter)`	Angles and distances are equally reliable — the usual case
Transit rule	The latitude or departure itself: `−ΣL × (\|lat\|/Σ\|lat\|)`	Angles are more reliable than distances — a theodolite traverse with taped sides

🧠 Bowditch, worked on one leg

For the traverse above — ΣL = +0.15, ΣD = −0.20, perimeter 700 m — the corrections to a 200 m leg are ΔLat = −0.15 × 200/700 = −0.0429 m and ΔDep = −(−0.20) × 200/700 = +0.0571 m. Two things to keep straight. The correction is opposite in sign to the misclosure, because you are removing it; getting the sign wrong doubles the error instead of eliminating it, and the check is that after adjustment the sums must be exactly zero. And the proportion is leg/perimeter, not leg/number of legs — Bowditch assumes error accumulates with distance travelled, which is why a long leg absorbs more of the correction than a short one. The choice between Bowditch and Transit follows directly from which measurement you trust less. Bowditch spreads the error by length because it assumes both angles and distances contribute; Transit spreads it by the size of the latitudes and departures themselves, which leaves the bearings nearly unaltered and is therefore right when the angles were measured much more precisely than the sides. With a modern total station, which measures both to comparable precision, Bowditch is the default — and with GNSS control there is often no misclosure to distribute at all, because each point is fixed independently rather than carried forward from the last.

Triangulation solves the same problem the other way round. Where a traverse measures every distance and carries error forward along a chain, triangulation measures one distance — the base line, to the highest precision available — and then obtains every other length by measuring angles in a network of triangles and solving them by the sine rule. Its logic is a direct consequence of the error discussion above: angles were historically far easier to measure precisely than long distances, so a network built on angles and one carefully measured base was more accurate than a chain of measured lengths. It is also self-checking, because the three angles of every triangle must sum to 180° (plus the spherical excess over large areas), which localises a bad observation to a single triangle instead of spreading it. Electronic distance measurement and then GNSS reversed the historical advantage, so modern control is trilateration or satellite-based — but the triangle-sum check survives everywhere, because redundancy is what makes a survey verifiable rather than merely completed.

Setting out horizontal and vertical curves

A simple circular curve, R = 300 m, deflection angle Δ = 40°
Element	Formula	Value
Tangent length	`T = R tan(Δ/2)`	109.19 m
Length of curve (arc)	`L = RΔ` (Δ in radians) `= πRΔ°/180`	209.44 m
Long chord	`= 2R sin(Δ/2)`	205.21 m
Apex distance	`E = R(sec(Δ/2) − 1)`	19.25 m
Mid-ordinate	`M = R(1 − cos(Δ/2))`	18.09 m
Degree of curve (30 m arc)	`D = 30 × 180/(πR)`	5.73°

Distance and angle measurement is the input to all of it, and the two have swapped places historically. Distance was once the hard measurement — a chain or tape, corrected for standardisation, temperature, pull, sag and slope, every correction systematic — while an angle came off a theodolite to a few seconds. Electronic distance measurement inverted that: a total station measures a slope distance by timing a modulated infrared beam and reduces it to the horizontal internally, so distance is now the easy measurement and the residual difficulty is in the angle, where collimation, trunnion-axis and eccentricity errors are removed by observing on both faces and meaning the readings. That is the same systematic-error discipline as balancing sight lengths in levelling, applied to a different instrument.

Vertical curves complete the pair with the horizontal ones. Geometrically they are the same problem as in highway design — a parabola joining two grades, its length governed by sight distance on a summit and by comfort and headlight reach in a valley — but the surveyor’s job is different from the designer’s: given the length and the grades, set out the reduced level of every peg along it. The offsets from the tangent go as the square of the distance from the tangent point, so a table of levels is computed once and pegged out from a single instrument set-up, and the check is that the two tangent points and the mid-point reproduce the designed grades. The error-propagation logic is the same as the deflection-angle method below: fix each peg from a known reference rather than from the peg before it.

Note that the arc length exceeds the long chord — 209.44 against 205.21 — as it must, and a computed "curve length" shorter than the chord is an immediate sign of a wrong formula. In the field the curve is set out by deflection angles from the tangent (Rankine’s method), because that needs only one instrument set-up at the point of curve and a tape, and every point is fixed by an angle turned from a fixed reference rather than from the previous point — so a single mistake does not propagate along the curve. Two-theodolite and offset methods exist for obstructed sites, but the deflection method is the default for exactly that error-propagation reason, which is the same principle that runs through this whole chapter: prefer a method whose errors do not accumulate.

Key takeaways

Random error accumulates as √n and averaging helps; systematic error accumulates as n and averaging does nothing.
A traverse can close perfectly while every distance is wrong — a scale error cancels round a loop.
Closing error e = √(ΣL² + ΣD²); what is judged is the relative error e/perimeter.
Bowditch distributes by leg length; Transit by the latitude or departure itself, when angles are the better measurement.
Curvature and refraction correct as +0.06728 D² km — 0.067 m at 1 km but 1.682 m at 5 km.
Levelling check ΣBS − ΣFS = last RL − first RL, and balancing sight lengths cancels curvature, refraction and collimation together.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A closed traverse of total perimeter 700 m has a sum of latitudes of +0.15 m and a sum of departures of −0.20 m. The closing error in metres is ______ (round to three decimal places)
Numerical answer — type the value.
Show answer
Answer: 0.25
0.250 m. A closed traverse returns to where it started, so the latitudes and departures should each sum to zero; the residuals are the misclosure resolved into two perpendicular directions, and the magnitude is their resultant: e = √(0.15² + 0.20²) = √(0.0225 + 0.0400) = √0.0625 = 0.250 m — a 3–4–5 triangle scaled by 0.05. But the number that is actually judged is the relative error, e/perimeter = 0.25/700 = 1 in 2800, because that is dimensionless and can be tested against a specification whatever the size of the traverse: 0.25 m of misclosure is excellent on a 5 km traverse and unacceptable on a 50 m one. Typical limits run from about 1 in 1000 for a compass traverse to 1 in 10,000 or better for precise theodolite work, so 1 in 2800 sits in the middle — acceptable for detail survey, not for control. Two further points. The direction of the misclosure, tan⁻¹(ΣD/ΣL), is worth computing even though the question does not ask: a misclosure that lies along one particular leg often points to a blunder on that leg rather than to accumulated random error, and a blunder must be found and removed, never adjusted away. And a perfect closure is weaker evidence than it looks — a tape with a scale error affects every leg proportionally and largely cancels round a loop, so a traverse can close beautifully while every distance in it is wrong. Closure tests random error; only calibration tests systematic error.
A surveyor measures a long line 100 times with a tape that is 5 mm too short, and averages the results. The effect of averaging on the error is that
1. it removes none of it — a systematic error accumulates in full however many times it is repeated
2. it reduces the error by a factor of 10, since random errors fall as √n
3. it reduces the error by a factor of 100, since the mean of n readings is n times better
4. it removes the error entirely, because positive and negative deviations cancel
Show answer
Answer: A — it removes none of it — a systematic error accumulates in full however many times it is repeated
It removes none of it. A tape 5 mm too short is 5 mm too short on every measurement, in the same direction, so there is nothing for the readings to cancel against. Averaging works only where deviations change sign, and a systematic error by definition does not. The mean of a hundred readings is exactly as wrong as one reading — and now carries the false authority of a hundred, which is the practical danger. Contrast the two behaviours with numbers, because the gap is startling. Take a per-measurement error of 0.05 m. If it is random, n measurements accumulate 0.05√n: over ten, 0.158 m; over a hundred, 0.500 m. If it is systematic, they accumulate 0.05n: over ten, 0.500 m; over a hundred, 5.000 m — a factor of ten apart at n = 100, and diverging as √n for ever. That is the whole reason the classification matters. Option (b) applies the √n law to the wrong kind of error; option (c) invents a law that applies to neither; option (d) describes what happens with random errors as n → ∞, and even there the reduction is asymptotic rather than complete. The correct remedy for the tape is calibration: compare it against a standardised length, compute the correction per tape length, and apply it — along with the corrections for temperature, pull, sag and slope, which are systematic for exactly the same reason. A surveyor who cannot say which kind of error dominates their work cannot say which of these two remedies to reach for, and applying the wrong one leaves the survey confidently wrong.
A simple circular curve has a radius of 300 m and a deflection angle of 40°. Its tangent length in metres is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 109.19
109.19 m. T = R tan(Δ/2) = 300 × tan 20° = 300 × 0.36397 = 109.19 m. The half-angle is where marks are lost: using tan 40° gives 251.73 m, which is more than the radius and should be recognised as impossible on sight, since the tangent length of a 40° curve cannot approach its radius. The rest of the curve follows from the same two inputs and is worth computing as a set, because each element checks the others. The arc length L = RΔ with Δ in radians = 300 × 0.69813 = 209.44 m. The long chord = 2R sin(Δ/2) = 205.21 m. The apex distance E = R(sec(Δ/2) − 1) = 19.25 m, and the mid-ordinate M = R(1 − cos(Δ/2)) = 18.09 m. Two sanity checks fall out immediately: the arc must exceed the long chord (209.44 > 205.21 ✓), because a curve is longer than the straight line joining its ends, and the apex distance must exceed the mid-ordinate (19.25 > 18.09 ✓), because the apex lies outside the curve and the mid-ordinate inside it. Any answer violating either has a formula error. Note also the degree of curve, D = 30 × 180/(πR) = 5.73° on the 30 m arc definition — Indian practice uses arc definition, and a question using the 30 m chord definition gives a slightly different value, so read which is intended. In the field the curve is set out by deflection angles from the tangent, because each point is then fixed by an angle from a fixed reference rather than from the previous point, so one mistake does not propagate along the curve — the same error-propagation principle that runs through this whole chapter.
For the traverse of question 1 (ΣLat = +0.15 m, ΣDep = −0.20 m, perimeter 700 m), the Bowditch correction to the departure of a 200 m leg, in metres, is ______ (round to four decimal places, giving the sign)
Numerical answer — type the value.
Show answer
Answer: +0.0571
+0.0571 m. The Bowditch (compass) rule distributes each misclosure in proportion to the length of the leg: correction to departure = −ΣDep × (leg length/perimeter) = −(−0.20) × 200/700 = +0.20 × 0.28571 = +0.0571 m. The corresponding latitude correction is −0.15 × 200/700 = −0.0429 m. Two things decide whether this is right. The sign is opposite to the misclosure, because the correction removes it — getting that backwards doubles the error instead of eliminating it, and the check is that after applying corrections to every leg the sums of latitudes and departures must be exactly zero. And the proportion is leg/perimeter, not leg/number of legs: Bowditch assumes error accumulates with distance travelled, so a long leg receives a proportionally larger share. The rule’s assumption is that angles and distances are equally reliable, which is the usual case with a modern total station measuring both to comparable precision. Where the angles are much better than the sides — a precise theodolite traverse with taped distances — the Transit rule is preferred instead: it distributes the misclosure in proportion to the latitudes and departures themselves rather than the lengths, which leaves the bearings nearly unaltered and adjusts mostly the distances. Choosing between them is a statement about which measurement you trust less, and it is not arbitrary. One caution worth carrying: adjustment is only legitimate for random error. If the misclosure is large, or lies along the direction of one particular leg, it is more likely a blunder in that leg — and a blunder must be found and re-measured, never spread over the whole traverse, which would corrupt every point instead of one.
The combined correction for curvature and refraction is 0.06728 D² metres, with D in kilometres. For a sight length of 5 km, this correction in metres is ______ (round to four decimal places)
Numerical answer — type the value.
Show answer
Answer: 1.682
1.6820 m. 0.06728 × 5² = 0.06728 × 25 = 1.6820 m. The substitution is trivial; the point of the question is the exponent. At D = 1 km the same correction is only 0.0673 m, smaller than most survey tolerances and routinely ignored — but at 5 km it is twenty-five times larger and ignoring it would be a gross blunder. So the common statement "curvature and refraction are negligible" is only true with a qualification, and the qualification is the examinable part: they are negligible at ordinary sight lengths and dominate at long ones, because the correction grows with the square of the distance. The two components pull in opposite directions and do not cancel. Curvature makes a truly horizontal line of sight rise away from the level surface by D²/2R, so a distant staff reading is too large. Refraction bends the ray downward as it passes through air of decreasing density, recovering about one seventh of that, and 0.06728 is the net of the two. The sign follows: the correction is added to a trigonometric height difference. There is a much better remedy than computing it, and it is why the correction rarely appears in ordinary levelling: if the backsight and foresight distances are made equal, the error is identical at both ends and cancels in the difference. Balancing sight lengths is therefore not fastidiousness — it removes curvature, refraction and any residual collimation error of the instrument in one stroke. In trigonometric levelling over a long line there is nothing to balance against, so the correction must be computed and applied: D = 1000 m at α = 2.5° gives 43.661 m uncorrected and 43.728 m corrected.
Which of the following statements about levelling and traversing are correct?
1. The arithmetic check ΣBS − ΣFS = last RL − first RL must hold for any correctly reduced level book
2. Keeping backsight and foresight distances equal cancels curvature, refraction and collimation error together
3. The rise-and-fall method checks every intermediate sight, while the height-of-instrument method does not
4. A traverse that closes perfectly proves that every measured distance in it is correct
Show answer
Answer: A — The arithmetic check ΣBS − ΣFS = last RL − first RL must hold for any correctly reduced level book; B — Keeping backsight and foresight distances equal cancels curvature, refraction and collimation error together; C — The rise-and-fall method checks every intermediate sight, while the height-of-instrument method does not
(a), (b) and (c). (a) is an identity, not a convention: every backsight adds to the line of collimation and every foresight subtracts from it, so their difference must equal the total change in level. It catches arithmetic errors in reduction regardless of how good the fieldwork was — and it catches only those, which is worth knowing, since a level book can pass this check perfectly while every reading in it was taken with a maladjusted instrument. (b) is the most useful single habit in levelling. Curvature, refraction and any residual collimation error all depend on the sight distance; make the backsight and foresight distances equal and each error is identical at both ends and cancels exactly in the difference. That is why the practice is standard rather than fussy, and it is also why a level with a known collimation error can still produce good work if sights are balanced. (c) is the real distinction between the two reduction methods, and it decides which to use. Rise and fall computes a rise or fall for every consecutive pair of readings, so a second check ΣRise − ΣFall = last RL − first RL verifies the intermediate sights as well; height of instrument computes levels from a collimation height and only checks at change points, leaving intermediate readings unverified. So rise and fall is slower and right where intermediate sights carry the survey; height of instrument is right on a long line of change points. (d) is wrong, and it is the trap. A closure tests random error only. A tape with a scale error is wrong on every leg in proportion, so the errors largely cancel round a loop — a traverse can close beautifully while every distance in it is short by the same fraction. Only calibration against a known standard detects that.
Triangulation obtains lengths by measuring one base line precisely and then measuring angles. Historically this was preferred over measuring every distance because
1. angles could be measured far more precisely than long distances, and the triangle-sum check localises a bad observation
2. angles are free of systematic error, whereas distances are not
3. triangulation requires no computation, unlike traversing
4. a triangulated network needs no adjustment because it has no closing error
Show answer
Answer: A — angles could be measured far more precisely than long distances, and the triangle-sum check localises a bad observation
Angles were far easier to measure precisely, and the network checks itself. Before electronic distance measurement, measuring a kilometre with a chain or tape accumulated error over every tape length — sag, temperature, pull, slope, and standardisation all contributing systematically — while a well-made theodolite could turn an angle to a few seconds of arc regardless of how far away the target was. Triangulation exploits that asymmetry: measure one base line with extraordinary care, then obtain every other length in the network by observing angles and solving triangles with the sine rule. The precision of the base propagates through the network far better than a chain of measured distances would. The second half of the answer matters as much. Every triangle carries an independent check — its three angles must sum to 180°, plus the spherical excess over large areas — so a bad observation shows up as a failing triangle and is localised rather than being carried silently forward the way an error in a traverse leg is. That redundancy is the deeper principle: a survey with more observations than unknowns can be verified, not merely completed. Option (b) is wrong — angles have their own systematic errors (collimation, trunnion axis, eccentricity), which is exactly why observations are taken on both faces and the readings meaned. Option (c) is wrong in the opposite direction: triangulation requires more computation, not less. Option (d) is wrong and inverts the truth — a triangulated network has many closure conditions and is adjusted by least squares. Modern practice has reversed the historical advantage: EDM and then GNSS made distances the easy measurement, so control is now trilaterated or satellite-fixed — but the triangle-sum check and the principle of redundancy survive unchanged.
A theodolite measures a vertical angle of 2.5° to a distant point 1000 m away horizontally. Ignoring curvature and refraction, the height difference in metres is ______ (round to three decimal places)
Numerical answer — type the value.
Show answer
Answer: 43.661
43.661 m. h = D tan α = 1000 × tan 2.5° = 1000 × 0.0436609 = 43.661 m. The trigonometry is one line; what makes the question worth setting is what has been left out and whether you know when that matters. Curvature and refraction were excluded here, and at this distance that is defensible: the combined correction is 0.06728 × 1² = 0.067 m, about 0.15% of the answer and inside most tolerances. Push the same sight to 5 km and the correction becomes 1.682 m — twenty-five times larger, because it grows with D² — and excluding it would no longer be a simplification but an error. Two further items are needed before this becomes a reduced level rather than a height difference, and questions often supply them precisely to see whether they are used. The height of the instrument above its station must be added, and the height of the signal or target above the point sighted must be subtracted; a candidate who reports 43.661 m as the level difference between two ground marks has skipped both. There is also a neat way to remove refraction rather than model it: reciprocal levelling, in which the vertical angle is observed simultaneously from each end of the line. Refraction affects both observations nearly equally and in opposite senses in the two computed height differences, so the mean of the two is very largely free of it — the same cancellation principle as balancing sight lengths in differential levelling, and the reason both techniques are standard rather than optional refinements.