Transportation Question 1

[bjgottlieb]

Hi All,

I am currently studying for my upcoming PE exam (April 16th). As most people do I am using the Lindeburg Manual for studying. I am in the process of taking the Transportation PE Sample Exam and am stuck on one problem in particular. If you have the book, the problem is 180. For those that do not, it is summarized here:

7-mile rural two-lane highway
Two-way volume of 1100 vph
Rolling Terrain
11-foot lanes, 4-foot paved shoulders
40% no passing zones
15 access points
Directional split is 60/40
PHF = 0.92
5% Trucks, 3% Buses, and 2% RVs
BFFS = 60mph

Calculate the highest directional flow rate for the peak 15min period in passenger cars/hr.

Now I know this is a Highway Capacity Manual 2000 Edition problem. Chapter 20 should be used. We are looking at calculating vp. My question is as follows:

The Lindeburg Solution uses 60% directional flow to calculate the Grade Adjustment Factor (fg), but uses the two-way flow rates for the Passenger-Car Equivalents (Et and Er). Is this correct? It seems that you would use the 60% directional flow rate for both.

Please help!!

Thanks to all in advance.

BJ

 

[civilman72]

It's been a while since I took the PE Exam, but from my recollection the test likes to include data that is not necessary to answer the question (maybe you have already figured this out). In this example, items such as 11-foot lanes w/ 4' shoulders, 15 access points and 40% no passing zones only effect the capacity of the road. But this is not a capacity question, only traffic flow needs to be calculated.

But I'm a little confused about your question on the directional flow, not sure why it's pertinent... maybe I'm making this too simple:

((1100x.60)/.92)/4=179 vehicles - converting this to P.C.E. should put you somewhere in the 200 passenger cars/peak 15-minute period. Nice thing about multiple choice question - you usually only have to get reasonably close in your calculations.

Sorry if I completely missed something, but I'm pretty sure I'm very close on the answer.

 

[bjgottlieb]

Thanks Civilman72 for your post.

The reason directional flow rate is required is because the problem is asking for the highest directional flow rate for the peak 15-min period in passenger car/hr.

The formula needed is vp = V/((PHF)*fg*fhv) from Chapter 20 of the 2000 Edition of the HCM. All other information not relating to the above formula is extraneous. I hope that clarifies your question. The 60% directional flow rate (1100pcph * 0.6 = 660pcph) for fg = 0.99. I would say that the 660pcph should be used to determine the Et and Er values which are 1.5 and 1.1 respectively. The solution to this problem uses 1100pcph instead of the 660pcph which gives values for Et and Er as 1.9 and 1.1 respectively. Depending upon which values you use the different fhv you get. This actually varies the final answer quite a bit and you end up picking the wrong answer, at least according to the solution.

 

[civilman72]

I must be behind the times. I'm using the 1997 HCM (doesn't even have a Chapter 20), and Et and Er values do not vary with traffic volumes (only terrain). Although it would make sense that the Et and Er values could vary with traffic volumes, or more appropriately as the LOS of the road decreases, I would anticipate the Et and Er values would increase.

But why would the Et and Er values be dependent on the traffic volumes going in the opposite direction? Here's my theory: the amount of traffic in the opposite lane will effect the ability for vehicles to use passing lanes, which will effect traffic speeds, which in turn will effect the 15-minute traffic volume. Without seeing the formulas this is my only educated guess, but it might be the answer.

Good luck on your test.