Question

The intersection angle of a 6° simple horizontal curve is 65°25', and the PC is located at station (238+14). Determine: the length of the curve (L), the station of the PT, the deflection angle for setting out the first point after the PC, and the chord length for setting out the first point after the PC, the deflection angle for setting out the last point before the PT, and the chord length for setting out the last point before the PT. The length of the curve is:

-The station of the PT:

-The deflection angle for setting out the first point after the PC is:

-The chord length for setting out the first point after the PC is:

-The deflection angle for setting out the last point before the PT is:

-The chord length for setting out the last point before the PT is:

Answer 1

(i)

Assuming the lengths are in feet.

Thus

assuming arc defination-

R = 5729.58/6 = 954.93 ft

By defination

here I = 65°25' = 25/60 = 65.416667°

L = 100 x 65.416667/ 6 = 1090.2777833 ft

(ii)

PT = PC +L

= (238+14) + (10+90.28)

= 249+04.28

(iii)

= 86x6/200 ( Sa = 23900 - 23814 = 86 )

= 2.580

= 20 34.8'

(iv)

= 2 x 954.93 sin 2.58

= 85.97097 ft

(v)

again for PT using same formula as above

deflection angle = 4.28 x 6 /200

= 0.1284 0

= 7' 42.24"

(vi)

again for PT using same formula as above

chord length = 2 x 954.93 sin 0.1284

= 4.2799 ft

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