Simple Curves

Question 1

Degree of Curve (arc basis) (metric system)

Answer 1

D = 1145.916/R

Question 2

Degree of Curve (arc basis) (english system)

Answer 2

D = 5729.58/R

Question 3

Degree of Curve (chord basis) (metric system)

Answer 3

Sin (D/2) = 10/R

Question 4

Degree of Curve (chord basis) (english system)

Answer 4

Sin (D/2) = 50/R

Question 5

Tangent Distance

Answer 5

T = R tan(I/2)

Question 6

External Distance

Answer 6

E = R (sec(I/2)-1)

Question 7

Middle Ordinate

Answer 7

M = R (1- cos (I/2))

Question 8

Length of Chord

Answer 8

C = 2R sin(I/2)

Question 9

Length of Curve (Lcu) (metric)

Answer 9

Lcu = 20 I/D

Question 10

Length of Curve (Lcu) (english)

Answer 10

Lcu = 100 I/D

Question 11

Length of Curve (Lcu) without using the Degree of curve

Answer 11

Lcu = ((R • I) (π/180°))

Question 12

Sub-arc (arc basis) (metric)

Answer 12

d_1/2 = 1.5 C_1 D

Question 13

Sub-arc (arc basis) (english)

Answer 13

d_1/2 = 0.3 C_1 D

Question 14

Sub-chords (chord basis) (metric)

Answer 14

C_1 = (20 sin (d_1/2))/(sin(D/2))

Question 15

Sub-chords (chord basis) (english)

Answer 15

C_1 = (100 sin (d_1/2))/(sin(D/2))

Question 16

A simple curve has a central angle of 36° and a degree curve of 6°.

1. Find the nearest distance from the mid point of the curve to the point of intersection of the tangents
2. Compute the distance from the mid point of the curve to the mid point of the long chord joining the point of curvature and point of tangency
3. If the stationing of the point of curvature is at 10 + 020, compute the stationing of a point on the curve which intersects with the line making a deflection angle of 8° with the tangent through the PC

Answer 16

1. E = 9.83 m
2. M = 9.35 m
3. Sta. of B = 10 + 073.33

Question 17

A simple curve of a proposed extension of Mantabahadra Highway have a direction of tangent AB which is due north and tangent BC bearing N 50° E. Point A is at the P.C. whose stationing is 20 + 130.46. The degree of curve is 4°

1. Compute the long chord of the curve
2. Compute the stationing of point D on the curve along a line joining the center of the curve which makes an angle of 54° with the tangent line passing thru the P.C.
3. What is the length of the line from D to the intersection of the tangent AB?

Answer 17

1. L = 242.14 m
2. S = 180 m
3. DE = 67.63 m

Question 18

The tangents of a simple curve have bearings of N 20° E and N 80° E respectively. The radius of the curve is 200 m.

1. Compute the external distance of the curve
2. Compute the middle ordinate of the curve
3. Compute the stationing of point A on the curve having a deflection angle of 6° from the PC which is at 1 + 200.00

Answer 18

1. E = 30.94 m
2. M = 26.79 m
3. Sta. A = 1 + 241.89

Question 19

The tangent distance of a 3° simple curve is only 1/2 of its radius.

1. Compute the angle of intersection of the curve
2. Compute the length of curve
3. Compute the area of fillet of a curve

Answer 19

1. I = 53.13
2. Lcu = 354.20 m
3. A = 5304.04 m²

Question 20

A 5° curve intersects a property line CD at point D. The back tangent intersects the property line at point C which is 105.27 m from the PC which is at station 2 + 040. The angle that the property line CD makes with the back tangent is 110°50’.

1. Compute the length of curve from the PC to the point of intersection of the line from the center of the curve to point C and the curve
2. Compute the distance CD
3. Compute the stationing of point D on the curve

Answer 20

1. Lcu_1 = 98.68 m
2. CD = 34.80 m
3. Sta. D = 2 + 188.24

Question 21

The perpendicular offset distance from point A on a simple curve to Q on the tangent line is 64 m if the distance from the PC to Q on the tangent is 260 m.

1. Compute the radius of the simple curve
2. Compute the length of curve from PC to A
3. If the angle of intersection of the curve is 64°, compute the length of long chord from PC to PT.

Answer 21

1. R = 560.13 m
2. S = 270.31 m
3. 928.74 m

Question 22

A simple curve have tangents AB and BC intersecting at a common point B. AB has an azimuth of 180° and BC has an azimuth of 230°. The stationing of the point of curvature at A is 10 + 140.26 if the degree of the curve is 4°.

1. Compute the length of a long chord from A
2. Compute the tangent distance AB of the curve
3. Compute the stationing of a point ‘x’ of the curve on which a line passing through the center of the curve makes an angle of 58° with the line AB, intersects the curve at point ‘x’

Answer 22

1. L = 242.14 m
2. AB = 133.59 m
3. 10 + 300.26

Question 23

A simple curve has a radius of 286.48 m. Its distance from PC to PT along the curve is equal to 240 m.

1. Compute the central angle of the curve. Use arc basis.
2. Compute the distance from the mid-point of the long chord to the mid-point of the curve.
3. Compute the area bounded by the tangents and the portion outside the central curve in acres.

Answer 23

1. I = 48°
2. M = 24.76 m
3. 2162.8 m²

Question 24

The offset distance of the simple curve from the PT to the tangent line passing through PC is equal to 120.20 m. The stationing of PC is st 2 + 540.26. The simple curve has an angle of intersection of 50°.

1. Compute the degree of curve
2. Compute the external distance
3. Compute the length of long chord

Answer 24

1. D = 3°24’
2. E = 34.79 m
3. 284.41 m

Question 25

Two tangents AB and BC intersect at an angle of 24°. A point P is located 21.03 m from point B and has a perpendicular distance of 2.79 m from line AB.

1. Calculate the radius of the simple curve connecting the two tangents and passing point P.
2. Find the length of chord connecting PC and point P.
3. Compute the area bounded by the curve snd the tangent lines.

Answer 25

1. R = 286.36 m
2. x = 40.12
3. A = 256.26 m²

Question 26

Stationing of PC

Answer 26

Sta. PC = Sta. PT - Lcu