Simple Curves Flashcards
Degree of Curve (arc basis) (metric system)
D = 1145.916/R
Degree of Curve (arc basis) (english system)
D = 5729.58/R
Degree of Curve (chord basis) (metric system)
Sin (D/2) = 10/R
Degree of Curve (chord basis) (english system)
Sin (D/2) = 50/R
Tangent Distance
T = R tan(I/2)
External Distance
E = R (sec(I/2)-1)
Middle Ordinate
M = R (1- cos (I/2))
Length of Chord
C = 2R sin(I/2)
Length of Curve (Lcu) (metric)
Lcu = 20 I/D
Length of Curve (Lcu) (english)
Lcu = 100 I/D
Length of Curve (Lcu) without using the Degree of curve
Lcu = ((R • I) (π/180°))
Sub-arc (arc basis) (metric)
d_1/2 = 1.5 C_1 D
Sub-arc (arc basis) (english)
d_1/2 = 0.3 C_1 D
Sub-chords (chord basis) (metric)
C_1 = (20 sin (d_1/2))/(sin(D/2))
Sub-chords (chord basis) (english)
C_1 = (100 sin (d_1/2))/(sin(D/2))
A simple curve has a central angle of 36° and a degree curve of 6°.
- Find the nearest distance from the mid point of the curve to the point of intersection of the tangents
- 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
- 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
- E = 9.83 m
- M = 9.35 m
- Sta. of B = 10 + 073.33
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°
- Compute the long chord of the curve
- 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.
- What is the length of the line from D to the intersection of the tangent AB?
- L = 242.14 m
- S = 180 m
- DE = 67.63 m
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.
- Compute the external distance of the curve
- Compute the middle ordinate of the curve
- Compute the stationing of point A on the curve having a deflection angle of 6° from the PC which is at 1 + 200.00
- E = 30.94 m
- M = 26.79 m
- Sta. A = 1 + 241.89
The tangent distance of a 3° simple curve is only 1/2 of its radius.
- Compute the angle of intersection of the curve
- Compute the length of curve
- Compute the area of fillet of a curve
- I = 53.13
- Lcu = 354.20 m
- A = 5304.04 m²
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’.
- 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
- Compute the distance CD
- Compute the stationing of point D on the curve
- Lcu_1 = 98.68 m
- CD = 34.80 m
- Sta. D = 2 + 188.24
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.
- Compute the radius of the simple curve
- Compute the length of curve from PC to A
- If the angle of intersection of the curve is 64°, compute the length of long chord from PC to PT.
- R = 560.13 m
- S = 270.31 m
- 928.74 m
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°.
- Compute the length of a long chord from A
- Compute the tangent distance AB of the curve
- 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’
- L = 242.14 m
- AB = 133.59 m
- 10 + 300.26
A simple curve has a radius of 286.48 m. Its distance from PC to PT along the curve is equal to 240 m.
- Compute the central angle of the curve. Use arc basis.
- Compute the distance from the mid-point of the long chord to the mid-point of the curve.
- Compute the area bounded by the tangents and the portion outside the central curve in acres.
- I = 48°
- M = 24.76 m
- 2162.8 m²
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°.
- Compute the degree of curve
- Compute the external distance
- Compute the length of long chord
- D = 3°24’
- E = 34.79 m
- 284.41 m
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.
- Calculate the radius of the simple curve connecting the two tangents and passing point P.
- Find the length of chord connecting PC and point P.
- Compute the area bounded by the curve snd the tangent lines.
- R = 286.36 m
- x = 40.12
- A = 256.26 m²
Stationing of PC
Sta. PC = Sta. PT - Lcu
