Long Chord Ordinates - Surveying Questions and Answers

This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Curve Surveying – By Ordinates of the Long Chord”.

1. Find the value of mid-ordinate if the radius of the curve is given as 40.62 m and length as 10.2m.
a) 0.43
b) 0.22
c) 0.12
d) 0.33
View Answer

Answer: d
Explanation: Mid-ordinate calculation involves the following procedure,
O₀ = R – (R² – (l/2)²)^1/2. On substitution, we get
O₀ = 40.62 – (40.62² – (10.2/2)²)^1/2
O₀ = 0.33.

2. For setting the curve, chord must be divided into even number of equal parts.
a) True
b) False
View Answer

Answer: a
Explanation: While setting a curve, the chord must be divided into even number of equal parts in order to decrease the time of the entire process. After dividing, the offsets are calculated.

3. Which of the following indicates the formula for setting a long chord by using ordinate?
a) O_x = (R² + (x)²)^1/2 – (R – O₀)
b) O_x = (R² – (x)²)^1/2 – (R – O₀)
c) O_x = (R² – (x)²)^1/2 + (R – O₀)
d) O_x = (R² – (x)²)^1/2 – (R + O₀)
View Answer

Answer: b
Explanation: The formula for setting a long chord by using ordinate can be given as O_x = (R² – (x)²)^1/2 – (R – O₀). In this O₀ is given as mid ordinate, R indicates the radius of the curve, x indicates the distance of the point from mid region.

4. General method can be adopted when radius of the curve is large.
a) False
b) True
View Answer

Answer: a
Explanation: When the radius of the curve is large, general method might take more time while solving than expected. In order to reduce the time of procedure we generally adopt an approximate method which is only considered in case of large radius than the length of the chord.

5. In approximate method, the value of x is measured from ____________
a) Chord point
b) Mid point
c) Tangent point
d) Secant point
View Answer

Answer: c
Explanation: In general, the value of x is taken from the midpoint but in case of approximate method the x value is taken from the tangent point. It is so because of the larger radius.

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6. Which of the following indicates the formula for determining ordinate in an approximate method?
a) O_x = x*(l-x) / 2+R
b) O_x = x*(l-x) / 2*R
c) O_x = x*(l + x) / 2*R
d) O_x = x+ (l-x) / 2*R
View Answer

Answer: b
Explanation: When the radius of the curve is large, for decreasing the time period of the entire process this process is adopted. It involves calculation of ordinate by assuming perpendicular distance and the formula is given as O_x = x*(l-x) / 2*R.

7. Find the value of ordinate at a distance of 10m having radius of 22.92m with mid-ordinate12.12.
a) 3.289
b) 2.892
c) 8.293
d) 9.823
View Answer

Answer: d
Explanation: The value of ordinate placed at certain distance x can be found out by using the formula,
O_x = (R² – (x)²)^1/2 – (R – O₀). On substitution, we get
O_x = (22.92²-(10)²)^1/2 – (22.92 – 12.12)
O_x = 9.823.

8. If the value of O₀ = 24.62 and R = 4m, find the value of l using the general method of long chords.
a) 1636.73m
b) 1363.73m
c) 1366.73m
d) 1363.37m
View Answer

Answer: a
Explanation: The general method of the ordinate calculation involves,
O₀ = R – (R² – (l/2)²)^1/2. On substitution, we get
24.62 = 4 – (4² – (l/2)²)^1/2
l = 1636.73 m.

9. Which of the following indicates the formula for a general method by ordinate of long chords?
a) \( R + (R^2 – (\frac{l}{2})^2)^{1/2}\)
b) \( R * (R^2 – (\frac{l}{2})^2)^{1/2}\)
c) \( R – (R^2 + (\frac{l}{2})^2)^{1/2}\)
d) \( R – (R^2 – (\frac{l}{2})^2)^{1/2}\)
View Answer

Answer: d
Explanation: The perpendicular which is erected while setting curve by ordinates of long chords, is equal to versed sine of the curve which makes it equal to \( R – (R^2 – (\frac{l}{2})^2)^{1/2}\).

10. What will be value of ordinate placed at a distance of 20m having radius and length as 72.46m and 42.92m respectively?(use approximate method)
a) 6.13
b) 1.36
c) 3.16
d) 4.86
View Answer

Answer: c
Explanation: Since the radius of the curve is large, we may consider the approximate method i.e.,
O_x = x*(l-x) / 2*R. On substitution, we get
O_x = 20*(42.92-20) / 2*72.46
O_x = 3.16.

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