Question

An interchange ramp is being designed to reach its high point after climbing for 2000 feet with a grade of +6%.Use the Deceleration/acceleration curves to answer the following:

A.What is the critical length of grade for this ramp based on the 10 mph speed reduction definition?

B.What would be the speed of a fully loaded truck at the top of the ramp if it enters the ramp at 60 mph?

C.If the ramp descends from its high point for a distance of 1000ft at a rate of -1%, what would the speed of the truck in part B be at the bottom?

Answer 1

Solution:- the values given in the question are as follows:

grade of ramp=+6% , OR +6/100=+3/50

height of highest point on ramp from starting(H)=2000 ft

(a) Calculate critical length of grade on ram when 10 Km/hr speed reduction:

reduction speed=10 Km/hr

Calculate negative acceleration of gravity acting on this ramp

vertically downward direction acceleration =-g=-9.81 m/s^2

acceleration on ramp=-9.81cos86.566=-0.5876 m/s^2

Apply motion equation

v=u+at, [Eq-1]

where, u=initial velocity , v=final velocity=u-10 m/hr

reduction velocity 10 m/hr , channge in to m/s

reduction velocity=10 milr/hr=4.47 m/se

values put in equation-1

u-4.47=u-0.5876*t

t=4.47/0.5876=7.607 seconds

critical length(Lc)=4.47*7.607=34 m

critical length of grade(Lc)=34 m, OR 111.55 ft ,[this critical length decrease the 10 m/hr velocity]

(b)

initial(enters) speed of loaded truck on ramp(Va)=60 m/hr, OR 26.822 m/s

final velocity(velocity at top of ramp)(Vc)=?

grade decrease 10 m/hr speed every 111.55 ft grade length(or, 13.152 m)

total length of ramp=2000 ft or,609.6 m , [this length calculated shown in above top figure]

v^2=u^2+2as

v^2=26.822^2-2*0.5876*609.6

v=1.74 m/s

v=3.89 m/hr

speed of truck at top(v)=3.89=4 m/hr

(c)

distance of ramp=1000 ft or304.8 m

grde=- 1%

acceleration ramp=9.81cos45=6.56

v^2=u^2+2as

v^2=1.74^2+2*6.56*304.8

speed at boottom(v)=63.26 m/se, [Ans]