Question

3.40 A crest and sag curve connect a 0% west highway
segment (left) with a +2% east highway segment
(right). The 0% west highway segment is at a higher
elevation than the start of the +2% east highway
segment. The two vertical curves connect with each
other (PVTc = PVCs) and share a 3% common grade.
If the design speed of the curves is 45 mi/h, what is the
elevation difference between the two road segments?

Answer 1

Determination of Ks and Kc:

From the AASTHO chart, for v= 45 mi/h,

Ks = 79

Kc = 61

Determination of Ls and Lc :

Ls = Ks*As

= 79* 3

= 237 ft

Lc = Kc * Ac

= 61* ( 2- (-3))

= 305 ft

L = Ls + Lc = 542 ft

Determination of elevation difference (E):

(A^2 * Ks/200) + (A^2 * Kc/200) + (A/100)* ( 542 - (Ks*A) - (Kc*A) )= E

(3^2 * 79/200) + (3^2 * 61/200) + (3/100)* ( 542 - (79*3) - (61*3) )= E

3.555 + 2.745 + 3.66 = E

E = 9.96 ft

Thus, the difference in elevation is 9.96 ft