Quiz # 15

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nda

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A curve of radius 340 feet, side friction factor, fs = 0.16, and super-elevation of 8% is
located at a section of an existing rural highway, which restricts the safe speed at this
section of the highway to 50 percent of the design speed. This drastic reduction of safe
speed resulted in a high accident rate at this section. A new alignment is to be designed
with a horizontal curve to reduce the accident rate. If the safe speed should be increased
to the design speed of the highway, assuming the super elevation will remain the same in
the new section, and the perception-reaction time is 2.5 seconds, the minimum radius of
this curve, in feet, is most nearly:

(A) 1,910
(B) 1,855
© 1,815
(D) 1,795
 
Can't say I'm coming up with any of the available options...must be missing something. Checked the adequacy of the curve and determined the design speed, should be a simple calc to get to Rmin.

 
I'm coming up short here as well ..... there is a lot going on in that problem.

Here's what I did .....

I backed into a design speed using the R=340. Multiplied by 2 which gives me a curve of 680, roughly 67 mph based on SSD.

Plugged into the Rmin equation and I come up with Rmin = 1249

So let me check wording of problem, maybe I misunderstood:

e= .08 and f= .16 = 40mph design speed which is an Rmin of 444

Also looked at this as "half the design speed of a 340-foot curve) which gave me a design speed of 43 mph, which yields a curve of 518

 
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I came up with a DS of 40 also. Just took another crack at it and still come up R=444' which matches the AASHTO min.

From what I'm seeing, the given radius should not cause a 50% decrease in DS with the given parameters.

nda, any hints?

 
Alright, just tried a different approach...was looking at it incorrectly. Good problem nda.

Find the Posted Speed Limit for this curve

R = V2/15(0.01e+f) = 340 = V2/15(0.01*8+0.16), solving for V = 35 mph

DS of roadway therefore equals 70 mph; so f = 0.10 for e = 8%.

Solve for the minimum radius given the above parameters

R = V2/15(0.01e+f) = 702/(15(0.01*8+0.10) = 1815' - Answer C

 
1.R,f,e are given, plug them to superelevation equation to get V

2.the new speed is =2V

3.Now the given are the new v and e only. find the new f from the AASHTO chart(using the new speed)

4. Plug them again to the superelevation equation to get the new R

 
Now I see what threw me off in this problem...because I have this solved for V=35 right here on my papers

That curve is designed for V=35 mph, so half of that design speed 17.5 ...

I know what I'm saying even if I'm not typing it clear.

I think that problem should state that the curve was designed for 50% of the design speed of the intersection roadway?? This isn't a hard problem, but the wording makes it very difficult.

 
Blg23, I like your thought process. I guess the test is approaching fast and no point of waiting since your answer is detailed. So yeah, u r correct. It is C

 
Just realized you PM'd me with the same info you had above. Makes me kind of glad I didn't see that PM and figured it out haha. If this were an exam problem, I guarantee 444' would have been a possible answer and unfortunately that is probably what I would have selected.

John Q, IMO, I think this wording was spot on. This was a great problem because from what I have seen on the NCEES sample exam, they like to try to trip you up not only with wording, but by providing useless numbers (eg. the P-R time). It never said the curve was designed for V = X mph, it just said the curve required motorists to slow to half of the roadway section design speed in order to safely navigate.

I'll see if I can't dig up a good one that has stumped me on first glance and post it.

 
The useless numbers I get, and at this point anyone should be able to spot the PRT given as bogus.

The tricky wording, I'm not so sure about, or maybe I just don't agree with it .... who am I anyway?

But here you have 3 guys, who all know what they are talking about, getting tricked up by the wording of a problem. IMO, that is a bad problem, because it's not really testing what you know. That being said, I'm printing this and putting it in my binder. Where is potatohead in all of this ... he is like the master of proper NCEES phrasing and terminology

 
The useless numbers I get, and at this point anyone should be able to spot the PRT given as bogus.

The tricky wording, I'm not so sure about, or maybe I just don't agree with it .... who am I anyway?

But here you have 3 guys, who all know what they are talking about, getting tricked up by the wording of a problem. IMO, that is a bad problem, because it's not really testing what you know. That being said, I'm printing this and putting it in my binder. Where is potatohead in all of this ... he is like the master of proper NCEES phrasing and terminology
lol. I didn't end up solving it since by the time I got around to it, it was already solved. The problem probably is a bit too wordy for an actual NCEES exam problem, but it's good practice. :)

 
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