Reverse Curve Problem

I came up with this problem based on problems that I have seen in workbooks. I believe a similar problem can be expected in the exam. The wording maybe a little confusing, but sketch it out and it should be pretty easy. 

The center of two parallel roads is 120 feet apart. They are to be connected by a reverse curve with each side having the same radius. The maximum distance between the tangent points is 440 feet. What is the maximum radius of the reverse curve? [SIZE=14.0pt]
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R=1643.34

zzzvvvzzz said:

R=1643.34

Click to expand...

I am getting R=434'. Do you mind sharing your method?

See these image below for my designations. Theta1=Theta2=I, as tangents are paralled. It is given that R1=R2. BV1=EV2=T and T1=T2=T=R*Tan(I/2).

V1C=V2C=T=R*Tan(I/2). Also, V1C+V2C=2R*Tan(I/2).

V1V2 Sin(I)= S = 120ft.

also, BV1+BV2+V1V2Cos(I) = 440ft,

From the two equations, I=30^29'2". Then using that, we get R=434'.

Those who have Reza's workbook, he has a similar reverse curve problem worked out but the in that problem, you are supposed to find out the distance which is equivalent to 440' in this problem. Using those formulae, I get 433.3'. I personally prefer to use basic geometry and trigonometry to solve these problems, but in the exam, I will not hesitate to use that formula, as it saves time.

Good luck.

I get what Mr. Maji gets, although with a simpler method. 

Using the tangent offset method, y = R - (R^2 -x^2)^(1/2).

We know that x = (1/2)(440) = 220' and y = (1/2)(120) = 60'.  Solve for R.

R = 433.33 ft

My first answer was wrong. I used 440 instead 1/2. 

Atan a/b = Atan 60/220 = 15.2551 this 1/2 delta

Delta = 2 x 15.2551

R= y/(1-cos delta) = 60/(1-cos 30.5102)=433.33

Also for equal radius curves

p =parallel offset

p=2r(1-cos∆)

∆=acos[1-(p/2r)]

r=1/2p/(1-cos∆)

D=2r*sin∆

Thanks for the various alternate solutions. It helps the readers to get different perspectives and we all learn new techniques.