Question

1: A simple curve of a proposed highway has tangents with bearing of N 28 E and N 76 W respectively and the degree of the curve of 9.5 using Chord Basis. Compute the following:

a. Radius of the Curve
b. Central Angle/Angle of Intersection
c. Tangent Distance
d. External Distance
e. Middle Ordinate
f. Length of the Chord
g. Length of the Curve
h. Compute the stationing of point A on the curve having a deflection angle of 5.8 from the
PC which is at 1+185.64.

Answer 1

1+191.19

Step-by-step explanation:

To compute the various parameters for the proposed highway curve, we can follow these steps:

Given information:

- Bearing of the first tangent: N 28 E

- Bearing of the second tangent: N 76 W

- Degree of the curve: 9.5 degrees (using Chord Basis)

- Length of the chord (Chord Distance): 100.0 feet (assuming)

- Point of Curvature (PC) station: 1+185.64

Let's calculate the requested parameters:

a. Radius of the Curve (R):

R = 5729.5779 / Degree of Curve

R = 5729.5779 / 9.5

R ≈ 602.057 feet

b. Central Angle/Angle of Intersection (A):

A = 2 * Degree of Curve

A = 2 * 9.5

A = 19.0 degrees

c. Tangent Distance (T):

T = R * tan(A/2)

T = 602.057 * tan(19/2)

T ≈ 174.007 feet

d. External Distance (E):

E = R * (sec(A/2) - 1)

E = 602.057 * (sec(19/2) - 1)

E ≈ 8.926 feet

e. Middle Ordinate (M):

M = R * (1 - cos(A/2))

M = 602.057 * (1 - cos(19/2))

M ≈ 8.449 feet

f. Length of the Chord (C):

C = 2 * R * sin(A/2)

C = 2 * 602.057 * sin(19/2)

C ≈ 204.043 feet

g. Length of the Curve (L):

L = (π/180) * R * A

L = (π/180) * 602.057 * 19

L ≈ 361.74 feet

h. Compute the stationing of point A:

To find the station of point A with a deflection angle of 5.8 degrees from the PC (1+185.64), you can use the formula:

Station A = PC station + (Chord Distance / 2) * tan(deflection angle)

Station A = 1+185.64 + (204.043 / 2) * tan(5.8)

Station A ≈ 1+185.64 + 5.55 feet

So, point A is located at approximately station 1+191.19.