Reverse Curve Elements Questions and Answers

This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Reverse Curve Elements”.

1. Reverse curve is a combination of two simple curves.
a) True
b) False
View Answer

Answer: a
Explanation: A reverse curve is a combination of two simple curves in the opposite direction, which are not recommended to be provided in highways as it can lead to overturning of vehicles.

2. Which of the following provides the best case for setting the reverse curve?
a) When straights are perpendicular
b) When straights form arc
c) When straights are parallel
d) When straights form curves
View Answer

Answer: c
Explanation: A reverse curve can be placed in case the straights are parallel. It is so because the remaining classification of curves is meant to serve in case of highways rather than mountainous regions.

3. Which of the following curves is not used in case highways?
a) Simple curve
b) Compound curve
c) Transition curve
d) Reverse curve
View Answer

Answer: d
Explanation: Simple, compound, transition curves can be used in case of highway as they are having good amount of curvature provided by the radius. But in the case of reverse curve, it provides immediate turning, which should not be present in highway as it may lead to accidents. So it is avoided.

4. Which of the following cases is generally adopted in the reverse curve?
a) T1 = T2
b) R1 = R2
c) t1 = t2
d) Chainages are equal
View Answer

Answer: b
Explanation: In reverse curve, there two cases to be considered parallel straights, non-parallel straights. In these cases, sometimes, the radius of both curves is assumed to be the same and also deflections are assumed as equal.

5. Which of the following case is assumed in a reverse curve?
a) Δ = Δ₁ * Δ₂
b) Δ = Δ₂ – Δ₁
c) Δ = Δ₁ – Δ₂
d) Δ = Δ₁ + Δ₂
View Answer

Answer: c
Explanation: In case of determining the deflection angle, Δ, a relation must be assumed which is given as Δ = Δ₁ – Δ₂ in which Δ₁ and Δ₂ are the deflection angles for the two curves which will meet at point of reverse curve (P.R.C).

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6. Chainage at the point of reverse curve can be given as__________
a) Chainage at P.R.C = Chainage at P.C + length of first arc
b) Chainage at P.R.C = Chainage at P.I + length of first arc
c) Chainage at P.R.C = Chainage at P.C + length of second arc
d) Chainage at P.R.C = Chainage at P.C – length of first arc
View Answer

Answer: a
Explanation: The chainage at the point of curvature can be given as the summation of chainage at point of curvature and the length of the first arc. The value of chainage of P.C is known and length of the arc is determined by formula provided.

7. A Reverse curve can be set by which of the following methods?
a) Method of bisection of arcs
b) Method of deflection angles
c) Method of deflection distances
d) Method of tangential angles
View Answer

Answer: d
Explanation: The method of tangential angles involves two processes i.e., radial offset, perpendicular offset in which, perpendicular offset method is having the accuracy in the results obtained.

8. Which of the following indicates the correct set of the cases employed in reverse curves?
a) Perpendicular, non-parallel
b) Parallel, perpendicular
c) Non-parallel, parallel
d) Perpendicular, curved
View Answer

Answer: c
Explanation: For designing a reverse curve in case of mountainous regions and also in case of railways which are having more degree of curvature, parallel and non-parallel straights are adopted with certain assumptions like R1 = R2, Δ1 = Δ2.

9. In case of parallel straights, the length of the curve is given as__________
a) L = (2(R1+R2)V)^1/2
b) L = 2L(R1+R2) / V
c) L = 2V(R1-R2) / R
d) L = 2V(R1*R2) / R
View Answer

Answer: a
Explanation: The case of parallel straights is applied in railway track purpose, where the curvature to join two parallel straights can be more. Here V represents perpendicular distance between the points and L represents parallel distance.

10. The formula of length of tangent is given as___________
a) t = L tan(δ/2)
b) t = r – tan(δ/2)
c) t = r + tan(δ/2)
d) t = r * tan(δ/2)
View Answer

Answer: d
Explanation: Though there might be a change in curve setting but the length of tangent remains the same in all the curve cases i.e., t = r tan (δ/2) where, δ = deflection angle measured.

11. Calculate the short tangent length, if the radius of curvature is given as 56.21m and the deflection angle as 32˚54ꞌ.
a) 61.6m
b) 116.6m
c) 16.6m
d) 6.6m
View Answer

Answer: c
Explanation: The value of short tangent length can be calculated by,
t = R*tan θ/2. On substitution, we get
t = 56.21*tan (32˚54ꞌ/2)
t = 16.6m.

12. Determine the common tangent of a reverse curve if the radius of curvature and deflection angles is given as, 43.57m, 32˚43ꞌ and 65˚76ꞌ.
a) 217.087m
b) 127.087m
c) 127.807m
d) 127.708m
View Answer

Answer: b
Explanation: The common tangent can be determined by, d = R*tan θ₁ + R*tan θ₂. On substitution, we get
d = 43.57(tan32˚43ꞌ + tan65˚76ꞌ)
d = 127.087m.

13. Find the value of tangent distance, possessing radius of curvature as 24.89m, common tangent 65m length and having deflection angles as 24˚56ꞌ and 76˚32ꞌ.
a) 64.5m
b) 46.5m
c) 64.98m
d) 62.5m
View Answer

Answer: a
Explanation: The tangent distance can be determined by using the formula,
T = R*tan (θ₁/2) + d * \(\frac{sin θ_1}{sin θ}\). On substitution, we get
T = 24.89*tan (24˚56ꞌ/2) + 65 * \(\frac{sin 76˚32ꞌ}{sin (24˚56ꞌ + 76˚32ꞌ)}\) T = 64.5m.

14. Calculate the chainage of P.R.C, if the chainage of Tangent is 567.54m and the curve length is about 65m.
a) 623.54m
b) 632.45m
c) 362.54m
d) 632.54m
View Answer

Answer: d
Explanation: The value of chainage of P.R.C can be obtained by using the formula,
Chainage of P.R.C = chainage of tangent + length of arc
Chainage of P.R.C = 567.54 + 65
Chainage of P.R.C = 632.54m.

15. If the radii of the curves in a reverse curve are equal, calculate the distance between the tangent points T1 and T2. Assume R = 98.54m with deflection angle 54˚31ꞌ.
a) 108.52m
b) 180.52m
c) 180.25m
d) 108.25m
View Answer

Answer: b
Explanation: From the question, it is clear that R1=R2=R. So, the distance between the tangent points T1 and T2 can be given as
L = 4*R*sin (θ/2). On substitution, we get
L = 4*98.54*sin (54˚31ꞌ/2)
L = 180.52m.

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