ades +4.90 % and -1.85 % ation of PVI = 77 +00 evation of PVI = 770.00 ft sign speed = 60 mi/h lculation: ation of PVI = 77 +00 here PVI is the point where the grades intersect =tionof BVC = (77+00) - (10+19.25) = 71+90.375 here BVC is the beginning of vertical curve nclusion: e station of the BVC is 71 +90.375.
ades +4.90 % and -1.85 % ation of PVI = 77 +00 evation of PVI = 770.00 ft sign speed = 60 mi/h lculation: ation of PVI = 77 +00 here PVI is the point where the grades intersect =tionof BVC = (77+00) - (10+19.25) = 71+90.375 here BVC is the beginning of vertical curve nclusion: e station of the BVC is 71 +90.375.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
I need detailed explanation of how to determine the stations and elevations of the BVC and EVC.
BTW I don't understand how you get in there and what is the (10+19.25/2).
Please I need details of every step, thank you in advance
Transcribed Image Text:Given:
Grades +4.90 % and -1.85 %
Station of PVI = 77 +00
Elevation of PVI = 770.00 ft
Design speed = 60 mi/h
Calculation:
Station of PVI = 77 +00
where PVI is the point where the grades intersect
Station of BVC = (77+00) - (10+19.25) = 71 +90.375
where BVC is the beginning of vertical curve
Conclusion:
The station of the BVC is 71 +90.375.
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could you, please, rewright this in more line so I can understant ?
Thank you
Transcribed Image Text:I in terms of stations,
BVC station -
BVC
EVC
100
|L = (10+19.25)
= (77400) - (10 + 19.25)
(77+00) (5+496625)
(71 +90.375)
=
(PVI Station -
EVC station = (PUI station + 1)
늘)
(77+00)+ (5+9.615)
(82+9.625)
=
=
76 + 100
(-25+9.625
(71 +90.375)
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