# Simple Vertical Curve Q.



## Transpo_guy (Sep 27, 2006)

In "Practice Problems for the Civil PE Exam", why in the world can't I use the solution methodology for Q.1g on (p.79-1) to solve Q.2 on (p.79-2)?

My problem is simple. 2 questions are identical. One question is solved with a method. Why can't the other question be solved with the same method?


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## GTScott (Sep 27, 2006)

I don't have access to that book...if you would give some background info (describe what they gave you and what they are looking for) I may be able to help you.

-GT


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## Transpo_guy (Sep 27, 2006)

> I don't have access to that book...if you would give some background info (describe what they gave you and what they are looking for) I may be able to help you.
> -GT


Help would be much appreciated.

Q. A falling grade of 4% meets a rising grade of 5%. At the start of the curve, the elevation is 123.06ft at station 4034+20. At station 4040+20 there is an overpass with an underside elevation of 134.06ft. If the sag curve is designed for 15ft clearance, what should the length of the curve be?

My question is, why can't I simply use the following equation to solve?

y = (1/2)Rx^2 + G1x + elevBVC


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## redrum (Sep 27, 2006)

they dont give you the PVI station?


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## 3gorgesdam (Sep 27, 2006)

Solution:

Vertical Curve:

y = (1/2)Rx^2 + G1x + elevBVC

For sta 4040+20, the elevation corresponding to the same station on curve is 134.06-15 = 119.06

R = G2-G1/L = +5%-(-4%)/L=+9%/L

G1=-4%

elevBVC=123.06 ft

y=119.06 ft

x=4040+20-4034+20=6 sta

The equation becomes:

119.06=(1/2)*(+9%/L)*6^2+(-4%)*6+123.06

solve for L = 8.1 sta

:violin:


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## Transpo_guy (Sep 28, 2006)

> Solution:
> Vertical Curve:
> 
> y = (1/2)Rx^2 + G1x + elevBVC
> ...


Thank you very much.

I have discovered that my math skills still need work. Like dividing 119.06 ft by 100. For some reason, my work revealed 119.06 / 100 = 1.11906 sta. Which lead to me getting L=~12sta.


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## GTScott (Sep 28, 2006)

Very nice!

Now, the problem that I am struggling with gives you your grades, the station and elevation at the PVI and a station along the curve where a certain elevation must be satisfied. The question then asks for the maximum curve length.

How does one do this without knowing anything about the PC? Without that info I cannot determine the offset (x) distance. If needed, I will try to rephrase the question.

Anyone?


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## redrum (Sep 28, 2006)

can you post the problem?


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## GTScott (Sep 28, 2006)

I will simplify it somewhat...

g1=-1.2%

g2=+ 1.8%

PVI STA = 119.46.40

PVI El. = 462.55'

The lowpoint of the curve is at 118+42.00. At this point the roadway elevation should be 465.8'.

What is the max. length of the vertical curve?

a) 850'

B) 900'

c ) 950'

d) 1000'

I think I am missing something basic on this one. However, I keep running into the need to know something about the PVC.


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## Road Guy (Sep 28, 2006)

well I am getting a blank (actually I get something like 2000)

They are using the PVI as a point of refernce instead of the pVC or PVT

usually Y=(x/T)^2*E

Y= distance from elev. 465.8 to the tangent coming back from the PVI

X = distance from the PVC to the low point station(118+42)

E= (g1-g2)L/8

I tried to solve for E= .375*L

then plug E into

Y=(x/T)^2*(.375*L)

T = L/2

X= L/2-104.4

Y = 2 from 462.55+(104.4*.012)-465.80 = 2.0

so I got:

2 = ((L/2-104.4)/(L/2))*(.375*L)

simplifies to:

5.333 = ((0.5L-104.4)^2 (L)) / (0.5L)^2

I actually had to put that in my HP48, and it died the first tiime, so i simplified it and ended up with l=2000?

Sad, I do this for a living  (actually I would just keep creating a new curve until one worked )


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## Road Guy (Sep 28, 2006)

here is the formula sheet I had for curves


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## GTScott (Sep 28, 2006)

In case I am misreading the problem, I will scan it and post it.

Thanks for the sheet. I find that CERM is good for horiz. curves but stinks for vert. curves. I have a decent textbook that covers both.


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## Road Guy (Sep 28, 2006)

yea the cerm is week in both horiz &amp; vertical info IMO.

I have the Land Surveyors Reference Manual (PPI Book), its pretty good on both. + it has some of that earthwork mass haul crap in it


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## 3gorgesdam (Sep 28, 2006)

Here is what I got:

x (Turning Point) =-G1/R

R=G2-G1/L -----&gt; x=0.4L

BVC sta = PVI sta - L/2 = 119.46-L/2

Turning Point sta - BVC sta = x

118.42-119.46+L/2 =0.4L

Solve for L =10.4 sta =1040'

Answer (D)


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## Road Guy (Sep 30, 2006)

GT Scott do you have the correct answer?

I found this equatioin in the CERM(in the Vertical Curve Section)

L= 2(ELPVI-ELLOWPOINT)/(G1*((G1/G2-G1)+1))

L= 2(462.55-465.8) / (-1.2*((-1.2/1.8+1.2)+1)

L= -6.5 / -1.2*0.6 = 9.02 Sta = 900 ?

???????


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