A roadway has a design speed of 50mph, and at station 105 00 a 3.0% grade roadway section ends and at station 125 00 a 2.0% grade roadway section begins. The 3.0% grade section of highway (at station 105 00) is at a higher elevation than the 2.0% grade section of highway (at station 125 00). If a -4% constant grade section is used to connect the crest and sag vertical curves that are needed to link the 3.0% and the 2.0% grade sections, what is the elevation difference between station 105 00 and 125 00

elevation difference = 2.20 m

Explanation:

given data

design speed = 50 mph = 80 km/h

higher elevation grade roadway section = 3%

section end = 105 + 00

section begins = 125 + 00

grade roadway section = 2%

constant grade section = -4%

to find out

the elevation difference between station 105 +00 and 125 + 00

solution

first we get here deviation angle that is N express as here

N = 3% - (-2%)

N = 5%

now we get here SSD distance that is

S = 0.278 vt +    

here f is friction we take 0.35 and t is reaction time i,e we take 2.5 second

S = 0.278 × 80 × 2.5 +  

S = 128 m

so here length L  is

L =    

L =

L = 1186.18 and it is greater than 128 m

so

elevation difference is

elevation difference =

elevation difference =    

elevation difference = 2.20 m