What answer do you guys get for this problem (Horizontal sight distance on curve w/ obstruction)?

[ptatohed]

I'm helping one of the guys here in our office study for his PE exam (Transpo depth). We were working this problem. I get Answer D but I'm not getting it on the screws so I don't feel 100% confident in my answer. How would you solve this problem and which answer do you get? Thanks in advance.

 

[mrt406]

Not sure if this is totally correct, but here's what I got...

The stopping sight distance is calculated as:

S = (1.47 fps/mph) * t * v + v^2 / (30 * (f + G)) where;

t = perception reaction time (2.5 sec)

v = design speed (60 mph)

f = cooefficient of friction (0.31)

G = road grade (+0.05 or -0.05 depending upon direction of travel)

Based on this, I get S equals 553.8' for the uphill direction and 682.0' for the downhill direction.

To calculate the distance from the drive path to the obstruction, M:

M = R(1 - cos(28.65 * S / R)) where;

R = curve radius at the drive path (1008' for the inside travel path and 1016' for the outside travel path)

S = stopping sight distance (553.8' or 682.0')

Based on this, the minimum clear distance is either 37.8' for the inside travel path or 56.7' for the outside travel path.

Since the drivers and objects are located 4' from the centerline, the distance "x" is 8' shorter for the inside lane (12' lane less 4' offset) and 16' shorter for the outside lane (12' lane plus 4' offset).

This gives minimum distances "x" of 29.8' and 40.7'. Being a minimum distance, the larger prevails. Therefore......

"x" = 40.7'

 

[ptatohed]

Not sure if this is totally correct, but here's what I got...

The stopping sight distance is calculated as:

S = (1.47 fps/mph) * t * v + v^2 / (30 * (f + G)) where;

t = perception reaction time (2.5 sec)

v = design speed (60 mph)

f = cooefficient of friction (0.31)

G = road grade (+0.05 or -0.05 depending upon direction of travel)

Based on this, I get S equals 553.8' for the uphill direction and 682.0' for the downhill direction.

To calculate the distance from the drive path to the obstruction, M:

M = R(1 - cos(28.65 * S / R)) where;

R = curve radius at the drive path (1008' for the inside travel path and 1016' for the outside travel path)

S = stopping sight distance (553.8' or 682.0')

Based on this, the minimum clear distance is either 37.8' for the inside travel path or 56.7' for the outside travel path.

Since the drivers and objects are located 4' from the centerline, the distance "x" is 8' shorter for the inside lane (12' lane less 4' offset) and 16' shorter for the outside lane (12' lane plus 4' offset).

This gives minimum distances "x" of 29.8' and 40.7'. Being a minimum distance, the larger prevails. Therefore......

"x" = 40.7'

Dude, you are awesome. That's perfect. Good job. Thank you very much. I got answer D too but I see now that I got it by accident/luck. Your answer is spot on.

Oh, I should have mentioned in my first post - I don't have the answer for this problem. It was from one of Dr. Mansour's "free" e-mails and, of 7 problems, this one didn't come with a solution/answer.

Thanks again mrt.

 

[mrt406]

Lol... no problem. It's nice to actually put my engineer hat on from time to time.

 

[Robavila12]

Wouldn't your calculation for M put you at an offset of 8 feet from edge of pvmt. Isn't the question asking for X which is the edge of pvmt? 37.8-8= 29.8 feet ..

 

[Robavila12]

Nm blond moment..lol been studying all day for pe .