Horizontal Curve with intersection angle of 90 or greater

[Predgw]

Early on in my preparation I remember doing a problem where the radius of curve was required. Easy problem and when I checke the answer I was wrong because, the intersection anle was 90 degrees and the explanation was when the I is 90 or greater the radius is found using......

For the life of me I can not find this now anywhere. At the time I thought I verified and highlighted this in the green book, but again, I cant find this.

Anybody have any thoughts? The only thing I found last night was anote on page 3-111 in the green book about L min on main highways should be 15 times the design speed in MPH. I don't think that was it though. What I rember was langauge discussing the radius and how it relates to the intersection angle.

Any thoughts?

 

[civilized_naah]

Should not matter whether I is less than, equal to or greater than 90

 

[Sapper PE LS]

Only thing I can think of is that if I=90 degrees, then R = T. You basically have a perfect square, and so the only real equation you need is:

L = 1/4 * 2 * pi * R, becauase it's a quarter of the full circumference.

or

R = 2L / pi

If they set the problem up in stations, then to get the Radius you'd simply take 2*(PT-PC)/pi

 

[Predgw]

I bought a 7 day access to the PPI questions. I just did one that asked for the min radius on a curve with an I of 90 degrees. The solution is given:

From AASHTO Greenbook 2011, the maximum superelevation for a cold climate should not exceed 0.08 due to the potential for snow and ice. Also, the maximum allowable value of the friction factor, f, is 0.12. The minimum curve length will require the minimum curve radius,

Using equation 3-7 from AASHTO greenbook 2011,

For perpendicular legs, the external angle is 90°. The minimum length of the curve is

So, according to this soultion it does matter if the iintersection angle is 90 or greater.

Thought I would share.

 

[ptatohed]

I am not following you. Even if the I were, say, 70^; or if the I were, say, 135^ - what would change about the solution formulas?

 

[Predgw]

It becomes a 2 step process to assure proper curve length.

If you look at the cut and paste solution, the min curve radius is calculated then because I is 90, the length of curve is calculated.

 

[ptatohed]

It becomes a 2 step process to assure proper curve length.

If you look at the cut and paste solution, the min curve radius is calculated then because I is 90, the length of curve is calculated.

Still not following you. I see that Rmin is calced based on e and f. Then, based on that Rmin, the L is calced. This could have been done for any given I. What is the exact question to the problem?

 

[Predgw]

I left the sheets at work. I left my green book there as well.

I have my last half ( morning ) exam tomorrow, I will post it after. I have been working it over in my head, but I need the green book to verify.

Thanks

 

[Sapper PE LS]

First of all

is exactly the same as what I wrote in the post directly above the one where you posted this picture:

Only thing I can think of is that if I=90 degrees, then R = T. You basically have a perfect square, and so the only real equation you need is:

L = 1/4 * 2 * pi * R ***(looks familiar huh)****, becauase it's a quarter of the full circumference.

or

R = 2L / pi

If they set the problem up in stations, then to get the Radius you'd simply take 2*(PT-PC)/pi

Second, there is no disputing that superelevation will play a part in deciding curve length, but you didn't talk about curve length, you talked about I=90 degrees. It doesn't matter what the curve length is if I = 90 degrees, the solution is exactly as both I and the book have now told you twice.

If the curve length changes but the R remains the same, then I will no longer be 90 degrees. If I remains the same but curve length changes, then R will no longer be the same, but the calculation is still done the same way.