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Question 1
ప్రశ్న 1
Which statement correctly describes RANKINE'S deflection-angle method for setting out a simple circular curve?
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Explanation:
Rankine's method is an angular method of setting out simple circular curves. A theodolite is normally established at the first tangent point or point of curvature. The incoming tangent provides the initial reference direction. Calculated cumulative deflection angles determine the directions to successive curve points. Chord measurements from previously established points locate the new pegs along those directions. The method is particularly useful for relatively long circular curves where suitable theodolite sight lines and chord measurements are available.
వివరణ:
Question 2
ప్రశ్న 2
In Rankine's method, a straight chord AB joins two points on a circle and subtends a central angle theta at the circle centre. What is the TANGENTIAL ANGLE between the tangent at A and chord AB, measured toward the curve?
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Explanation:
The tangent to a circle at A is perpendicular to the radius through A. The two radii joining the centre to A and B form an isosceles triangle. The base angles of this triangle are each (180 degrees - theta)/2. Subtracting this base angle from 90 degrees gives the angle between the tangent at A and chord AB. The resulting tangential angle is theta/2. This half-central-angle theorem is the geometric foundation of Rankine's deflection-angle method.
వివరణ:
Question 3
ప్రశ్న 3
For a circular curve of radius R and a STRAIGHT chord of length c, which expression gives the EXACT tangential angle delta subtended by that chord, assuming the minor central angle is used?
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Explanation:
A straight chord subtending central angle theta satisfies c = 2R sin(theta/2). The tangential angle delta equals half the central angle. Therefore c = 2R sin(delta). Rearranging gives delta = sin^(-1)(c/(2R)). This expression is exact for the stated minor-arc geometry. The commonly used approximation delta = c/(2R) radians is obtained by assuming that the chord is sufficiently short compared with the radius.
వివరణ:
Question 4
ప్రశ్న 4
In the conventional SHORT-CHORD approximation used in Rankine's method, which formula gives the tangential angle delta in MINUTES OF ARC when chord length c and radius R are expressed in the same length units?
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Explanation:
For a sufficiently short chord, its length is approximately equal to the corresponding circular arc length. The central angle in radians is then approximately c/R. The tangential angle is half of this value, giving delta approximately equal to c/(2R) radians. Converting radians to minutes gives delta = (c/(2R))(180/pi)(60). The numerical coefficient is approximately 1718.9, so delta = 1718.9(c/R) minutes. The approximation becomes less accurate when the chord length is large relative to the radius.
వివరణ:
Question 5
ప్రశ్న 5
A simple circular curve is divided into successive chords having individual tangential angles delta1, delta2, delta3 and so on. Which expression gives the CUMULATIVE deflection angle Delta3 from the initial tangent at PC to the THIRD curve peg?
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Explanation:
Each successive chord contributes its own tangential angle. The cumulative deflection to the first peg is equal to delta1. The cumulative deflection to the second peg is delta1 + delta2. The cumulative deflection to the third peg is therefore delta1 + delta2 + delta3. The theodolite remains oriented with respect to the original tangent at PC. Consequently the readings are cumulative rather than independent angles measured from the preceding chord.
వివరణ:
Question 6
ప్రశ్న 6
A simple circular curve has a total central angle of 48 degrees. In Rankine's method, what should be the FINAL cumulative deflection angle from the initial tangent at PC to the long chord joining PC and PT?
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Explanation:
The long chord connects the two tangent points of the simple circular curve. It subtends the complete central angle of 48 degrees. According to the tangent-chord theorem, the angle between the initial tangent and the long chord equals half the corresponding central angle. Therefore the final cumulative deflection is 48/2 = 24 degrees. The sum of the individual tangential angles for all successive chords must agree with this value when exact geometry is used. Agreement with the half-central-angle condition is an important angular closure check.
వివరణ:
Question 7
ప్రశ్న 7
A surveyor is beginning Rankine's method for a RIGHT-HAND circular road curve. Which procedure correctly establishes the initial theodolite reference?
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Explanation:
The conventional one-theodolite Rankine procedure starts at the point of curvature, PC. The instrument must be carefully centred and levelled over this point. The direction from PC toward PI represents the forward incoming tangent. This tangent is used to establish the initial horizontal-circle reference, commonly zero. For a right-hand curve, the calculated cumulative deflections are then turned clockwise from this reference. Accurate establishment of the initial tangent direction is essential because any orientation error affects the positions of subsequent curve pegs.
వివరణ:
Question 8
ప్రశ్న 8
The first curve peg P1 has been established by Rankine's method, and the theodolite remains at PC. What is the correct procedure for locating the NEXT peg P2?
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Explanation:
The theodolite remains at PC during the conventional Rankine setting-out sequence. The cumulative deflection to P2 is the sum of the tangential angles of the first two chords. Setting this angle provides the direction from PC toward P2. The prescribed straight chord length from P1 to P2 is then measured using a tape or suitable distance-measuring equipment. The intersection of that measured chord distance with the theodolite sight line locates P2. The procedure is repeated with further cumulative deflections and successive chord lengths.
వివరణ:
Question 9
ప్రశ్న 9
During Rankine curve setting, the proposed peg chainages are fixed at regular intervals measured along the circular alignment. Which statement correctly distinguishes a STANDARD CHORD from the FIRST and LAST SUB-CHORDS?
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Explanation:
Regular curve stations are often specified at convenient chainage intervals measured along the alignment. The first interval from PC to the next regular station may be shorter than the normal interval. The last interval from the final regular station to PT may also be shorter. These end intervals are conventionally associated with the first and last sub-chords. For accurate geometry, a straight chord is calculated as 2R sin(s/(2R)), where s is the corresponding arc length. Treating a short arc interval as numerically equal to its straight chord is an approximation rather than an exact identity.
వివరణ:
Question 10
ప్రశ్న 10
A circular curve has radius R = 200 m, and a straight chord of length c = 20 m is to be set out. Using the conventional APPROXIMATE Rankine formula delta = 1718.9c/R minutes, what is the tangential angle to the nearest second?
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Explanation:
The question specifies the approximate Rankine tangential-angle formula. Substituting c = 20 m and R = 200 m gives delta = 1718.9 multiplied by 20/200 minutes. The result is 171.89 minutes. This is equal to 2 degrees, 51 minutes and 53.4 seconds. Rounding to the nearest second gives 2°51'53". Using the exact inverse-sine formula would produce a slightly different result because the chord and corresponding arc are not exactly equal.
వివరణ:
Question 11
ప్రశ్న 11
A simple circular curve has radius 180 m. A chord is designed to subtend a CENTRAL angle of exactly 10 degrees, so its Rankine tangential angle is 5 degrees. What is its EXACT straight chord length to three decimal places?
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Explanation:
The exact relationship between chord length and tangential angle is c = 2R sin(delta). The radius is 180 m and the tangential angle is 5 degrees. Therefore c = 2 multiplied by 180 multiplied by sin(5 degrees). This evaluates to approximately 31.376067 m. The straight chord length is consequently 31.376 m to three decimal places. The corresponding circular arc length is approximately 31.416 m, which is slightly greater than the straight chord.
వివరణ:
Question 12
ప్రశ్న 12
In an EXACT Rankine setting-out calculation, a straight chord is required to have a tangential angle of 4 degrees on a circular curve of radius 300 m. What chord length should be used to three decimal places?
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Explanation:
The exact chord relation is c = 2R sin(delta). Here R = 300 m and the required tangential angle is 4 degrees. Substitution gives c = 600 sin(4 degrees). Evaluating the sine gives a straight chord length of approximately 41.853884 m. Rounding to three decimals produces 41.854 m. The corresponding arc length is approximately 41.888 m, so the two distances must not be treated as exactly identical.
వివరణ:
Question 13
ప్రశ్న 13
In a Rankine setting-out table, the first three successive chords have tangential angles of 1°15'20", 2°10'40" and 1°48'30" respectively. What is the CUMULATIVE deflection angle from the initial tangent to the third curve peg?
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Explanation:
The cumulative Rankine deflection to a curve peg is the sum of the tangential angles of all preceding chord segments. Adding the first two values gives 1°15'20" + 2°10'40" = 3°26'00". The third tangential angle is 1°48'30". Adding this gives 5°14'30". The theodolite should therefore be directed at 5°14'30" from its original tangent reference to sight the third peg. The third cumulative deflection must not be confused with the third individual tangential angle.
వివరణ:
Question 14
ప్రశ్న 14
During Rankine setting out, the cumulative deflection to peg P1 is 2°10'00" and the cumulative deflection to peg P2 is 4°45'00". What is the INDIVIDUAL tangential angle corresponding to chord P1P2?
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Explanation:
The cumulative deflection to P2 contains the contributions of both preceding chord segments. The individual tangential angle for chord P1P2 is therefore obtained by subtracting the cumulative deflection to P1. The calculation is 4°45'00" - 2°10'00". This gives 2°35'00". Adding the result to 2°10'00" reproduces the cumulative reading of 4°45'00". The calculation illustrates how individual tangential angles can be recovered from a Rankine deflection-angle table.
వివరణ:
Question 15
ప్రశ్న 15
A simple circular curve uses the 30 m ARC definition of degree of curve. Its degree of curve is 6 degrees, meaning that 30 m of ARC subtends 6 degrees at the centre. What is the radius of the curve to three decimal places?
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Explanation:
Under the specified 30 m arc definition, the central angle of a 30 m arc is 6 degrees. The circular arc-length relation is s = pi R theta/180 when theta is in degrees. Substituting s = 30 m and theta = 6 degrees gives 30 = pi R multiplied by 6/180. Rearrangement yields R = 900/pi. The radius is approximately 286.479 m. The alternative 30 m chord convention would give a slightly different radius and is not applicable to this question.
వివరణ:
Question 16
ప్రశ్న 16
A simple circular curve has central angle 30 degrees and degree of curve 6 degrees under the 30 m ARC convention. Its PC chainage is 1+018.000 m, and regular peg chainages occur at 1+030, 1+060, 1+090, 1+120 and 1+150 m. The curve ends at PT chainage 1+168.000 m. Which description of its ARC-CHAINAGE intervals is correct?
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Explanation:
The first regular chainage is 1030 m and PC is at 1018 m. Their difference gives an initial arc interval of 12 m. The intervals 1030-1060, 1060-1090, 1090-1120 and 1120-1150 are four regular intervals of 30 m each. The final interval from 1150 m to PT at 1168 m is 18 m. Adding 12 + 4(30) + 18 gives a total arc length of 150 m. This agrees with the curve length of five 30 m arcs, each subtending 6 degrees, for a total central angle of 30 degrees.
వివరణ:
Question 17
ప్రశ్న 17
A simple circular curve has degree of curve 6 degrees under the 30 m ARC convention. Its Rankine setting-out intervals measured ALONG THE ARC are an initial interval of 12 m, regular intervals of 30 m and a final interval of 18 m. Using exact proportional central angles for these arc intervals, what are their respective individual tangential angles?
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Explanation:
The central angle corresponding to a 30 m arc is 6 degrees. Therefore the tangential angle for a 30 m arc interval is half of 6 degrees, or 3 degrees. A 12 m arc interval represents 12/30 of the standard central angle, giving a tangential angle of (12/30) multiplied by 3 degrees. This equals 1°12'. Similarly, an 18 m final interval gives (18/30) multiplied by 3 degrees = 1°48'. These are exact angular increments for the specified arc intervals, while the corresponding straight chord lengths would be calculated separately for precise field measurement.
వివరణ:
Question 18
ప్రశ్న 18
A RIGHT-HAND simple circular curve has PC at chainage 1+018.000 m and PT at 1+168.000 m. The degree of curve is 6 degrees by the 30 m ARC definition, and regular pegs are at 1+030, 1+060, 1+090, 1+120 and 1+150 m. The theodolite is set to zero along the forward tangent at PC. Which sequence correctly gives the cumulative CLOCKWISE Rankine deflections to pegs 1+030, 1+090, 1+150 and PT?
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Explanation:
The first 12 m arc interval contributes a tangential angle of 1°12'. Each succeeding 30 m interval contributes 3°00'. The cumulative deflections at chainages 1+030, 1+060, 1+090, 1+120 and 1+150 are consequently 1°12', 4°12', 7°12', 10°12' and 13°12'. The last 18 m arc interval contributes 1°48'. Therefore the final cumulative deflection at PT is 13°12' + 1°48' = 15°00'. This agrees exactly with half the full central angle of 30 degrees and verifies the complete setting-out table.
వివరణ:
Question 19
ప్రశ్న 19
A simple circular curve has degree of curve 6 degrees under the 30 m ARC definition, with radius R = 900/pi m. In a precise Rankine setting-out calculation, the straight chord corresponding to one 30 m ARC interval is required. What is its length to three decimal places?
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Explanation:
The specified 30 m is the circular arc length, not the exact straight chord length. Its central angle is 6 degrees under the given arc convention. The corresponding straight chord is c = 2R sin(6 degrees/2). Substituting R = 900/pi m gives c = (1800/pi) sin(3 degrees). Evaluating this expression gives approximately 29.986294 m. The straight chord is therefore 29.986 m to three decimal places, which is slightly shorter than the corresponding 30.000 m arc.
వివరణ:
Question 20
ప్రశ్న 20
A simple circular curve has radius 300 m, central angle 60 degrees and assigned PI chainage 5+440.000 m along the incoming tangent. Chainage increases along the tangent to PC and then along the circular arc to PT. Which pair gives the PC and PT chainages to three decimal places?
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Explanation:
The tangent length is T = R tan(Delta/2). With R = 300 m and Delta = 60 degrees, T = 300 tan(30 degrees) = 173.205081 m approximately. The PC chainage is therefore 5440.000 - 173.205081 = 5266.794919 m. The circular arc length is L = pi R Delta/180 = 314.159265 m. Adding the arc length to the unrounded PC chainage gives PT = 5580.954185 m. Rounding the results gives PC = 5+266.795 m and PT = 5+580.954 m.
వివరణ:
Question 21
ప్రశ్న 21
A simple circular curve has radius R = 180 m and total central angle 60 degrees. Its first and last arc subdivisions each subtend 10 degrees at the centre, while the two intermediate subdivisions each subtend 20 degrees. Exact chord lengths are calculated as c = 2R sin(theta/2). Which set correctly gives the approximate FIRST SUB-CHORD, STANDARD CHORD and LAST SUB-CHORD lengths and the cumulative Rankine deflections to the four successive curve points?
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Explanation:
The first and last sub-chords each subtend 10 degrees at the centre. Their exact straight length is 360 sin(5 degrees), approximately 31.376 m. Each standard chord subtends 20 degrees and has length 360 sin(10 degrees), approximately 62.513 m. The individual tangential angles are half the corresponding central subdivisions, giving 5°, 10°, 10° and 5°. Their cumulative deflections are 5°, 15°, 25° and 30°. The final 30-degree deflection equals half the complete 60-degree central angle, confirming exact angular closure when the unrounded chord geometry is used.
వివరణ:
Question 22
ప్రశ్న 22
A simple circular curve has radius 240 m, central angle 45 degrees and PC chainage 2+007.000 m. Regular peg chainages are fixed at every 20 m from 2+020 through 2+180 m. Using the EXACT circular arc length to calculate PT chainage, which combination correctly gives the first arc-chainage interval, number of full 20 m intervals and final arc-chainage interval, rounded to three decimals?
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Explanation:
The total circular arc length is L = pi multiplied by 240 multiplied by 45 divided by 180. This gives L = 188.495559 m approximately. Adding the unrounded arc length to PC chainage 2007.000 m gives PT chainage 2195.495559 m, or 2+195.496 m. The first arc interval from 2007 m to the first regular peg at 2020 m is 13.000 m. There are eight complete 20 m intervals between chainages 2020 m and 2180 m. The final arc interval is 2195.495559 - 2180 = 15.495559 m, or 15.496 m.
వివరణ:
Question 23
ప్రశ్న 23
A RIGHT-HAND simple circular curve has radius 240 m, central angle 45 degrees and PC chainage 2+007.000 m. The theodolite is oriented at PC along the incoming tangent. What is the EXACT cumulative clockwise Rankine deflection to the peg at chainage 2+180.000 m, assuming chainage follows the arc and the result is rounded to the nearest second?
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Explanation:
The arc-chainage distance from PC to the specified peg is 2180.000 - 2007.000 = 173.000 m. The central angle corresponding to an arc length s is s/R radians. The cumulative Rankine deflection is half this angle. Therefore the cumulative angle is 173/(2 multiplied by 240) = 173/480 radians. Converting this value to degrees gives approximately 20.65035387 degrees, or 20°39'01.274". Rounded to the nearest second, the required clockwise setting is 20°39'01".
వివరణ:
Question 24
ప్రశ్న 24
A simple circular curve has radius 240 m, central angle 45 degrees and PC chainage 2+007.000 m. Its exact PT chainage is 2+195.495559 m before rounding, and the preceding regular peg is at 2+180.000 m. Using exact circular geometry and keeping unrounded values in the calculation, what are the FINAL individual tangential angle and the total cumulative Rankine deflection at PT, respectively, to the nearest second?
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Explanation:
The exact total arc length is 240 multiplied by pi/4 = 188.495559215 m. The final arc interval after chainage 2180 m is consequently 188.495559215 - 173.000 = 15.495559215 m. Its individual tangential angle is 15.495559215/(2 multiplied by 240) radians. Converting gives approximately 1°50'58.726", which rounds to 1°50'59". The preceding cumulative deflection is approximately 20°39'01.274". Adding the two unrounded angular contributions gives exactly 22°30'00", which equals half the full central angle of 45 degrees.
వివరణ:
Question 25
ప్రశ్న 25
A Rankine setting-out table gives a final cumulative deflection equal to half the central angle, but the last curve peg does not coincide with the PT independently established from the outgoing tangent. Which conclusion is most technically justified?
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Explanation:
The final cumulative deflection provides a check on the angular calculations. It does not independently verify that every chord has been measured or set out correctly. An erroneous chord length, displaced peg or inaccurate theodolite orientation can affect the curve position. Consequently the final peg may fail to coincide with the independently established PT despite correct angle totals. The discrepancy should be investigated through checks of chord distances, pointing, station centering and reference control. A sound setting-out procedure uses both angular agreement and independent positional verification.
వివరణ:
Question 26
ప్రశ్న 26
A simple circular curve changes route direction through 36 degrees to the LEFT. Rankine's method is carried out from PC with the horizontal circle oriented along the forward incoming tangent and CLOCKWISE angles taken as positive. What is the signed final cumulative deflection to PT?
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Explanation:
The magnitude of the final Rankine deflection is half the complete central angle. The central angle is 36 degrees, so the required magnitude is 18 degrees. The curve turns left from the incoming tangent. Under the stated convention, clockwise deflections are positive and counterclockwise deflections are negative. Therefore the signed final deflection is -18°00'00". The negative sign describes the direction of angular rotation and does not indicate an invalid or negative curve length.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. A theodolite at the starting tangent point establishes successive cumulative deflection directions, while measured chords locate the curve pegs. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. theta/2. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: C. delta = sin^(-1)(c/(2R)). సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: D. delta = 1718.9(c/R) minutes. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: B. Delta3 = delta1 + delta2 + delta3. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 24 degrees. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: C. Centre and level the theodolite over PC, accurately sight PI along the forward incoming tangent and establish the initial horizontal-circle reference. సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: D. Set the cumulative deflection to P2 at the theodolite, then locate P2 where that sight line intersects the prescribed chord distance measured from P1. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. The standard chord joins successive regularly spaced curve points, while initial and final sub-chords accommodate shorter end intervals; exact straight chord lengths are calculated from the corresponding arc angles when required. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 2°51'53". సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. 31.376 m. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: D. 41.854 m. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: B. 5°14'30". సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 2°35'00". సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: A. 286.479 m. సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: D. First interval 12 m; four regular intervals of 30 m; final interval 18 m. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. First 1°12'; regular 3°00'; final 1°48'. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: B. 1°12', 7°12', 13°12', 15°00'. సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: A. 29.986 m. సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. PC = 5+266.795 m; PT = 5+580.954 m. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: D. First 31.376 m; standard 62.513 m; last 31.376 m; cumulative deflections 5°, 15°, 25°, 30°. సమాధానం: D.
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Question 22 ప్రశ్న 22Answer: B. First 13.000 m; eight full intervals; final 15.496 m. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. 20°39'01". సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. 1°50'59" and 22°30'00". సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. Angular closure alone does not establish positional accuracy; chord lengths, orientation, centering, peg locations and the independent PT control should be checked. సమాధానం: D.
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Question 26 ప్రశ్న 26Answer: C. -18°00'00". సమాధానం: C.