Source: Goswami, Indramil Ph.D. P.E. "All In One Civil Engineering PE Breadth and Depth Exam Guide" 2nd Edition. McGraw Hill: 2012

Intro:

Over the last year or so, the topic of civil engineering has fascinated me. While I don't expect to study for the Civil Engineering exam I wanted to get some sort of an encompassing resource that I learn a few things from in a desire to better understand some facets of engineering. I ended up purchasing a Kindle Edition of the All In One Civil Engineering book primarily because I felt it provided me with the best bang for the buck. Plus it was a better alternative to get a newer edition electronically for half of what I would have paid for a first edition of physical book. From time to time, I plan to post some notes of what I read.

This post is regarding transportation.

Terms:**Trip Generation**: the number of trips made on a network. The network is implied to be made of zones. Zones are small areas that are determined by population, employment, or other given measurement, which are said to be anywhere from 1/4 square mile to a full mile. Trips are said to original from one designated zone (trip producer) to a destination zone (trip attractor). **Time Mean Speed (S_T)**: the arithmetic mean of speeds of vehicles passing over a given point. Let s_1, s_2, ... , s_n be the speeds of passing vehicles. Then the time mean speed is:

S_T = (s_1 + s_2 + ... + s_n) / n**Space Mean Speed (S_R)**: the average mean speeds with delays taken into account. This is theorized by taking the harmonic mean of speeds of vehicles passing over a given point. A mathematical definition is:

S_R = n / (1/s_1 + 1/s_2 + ... + 1/s_n)

Note that this is not the only way to calculate space mean speed. Using the strict definition:

S_T = total distance ÷ total time including stops and delays

In general, S_R ≤ S_T.

Consider two speeds A and B, assuming A and B are positive. Then:

S_T = (A+B)/2 and

S_R = 2/(1/A +1/B)

= 2 * (A+B)/(A*B)

Setting S_T and S_R equal to each other yields:

1/2 * (A+B) = 2 * (A+B)/(A*B)

A * B = 4

Hence if A * B < 4, S_T < S_T. Otherwise, S_T > S_R.

**Speed-Volume-Density**

q = s * d

Where:

q = flow rate of vehicles per hour

s = average vehicle per hour

d = density, measured in vehicle per mile

Spacing per foot is calculated by: (5280 ft/mi)

d_ a = 5280/d

Where vehicle headway is calculated by: (3600 sec/hr)

h_a = 3600/q

Example: Observing a part of a highway for one hour, an engineer observes 890 vehicles pass at an average of 55.5 miles per hour in one direction.

The density is: d = 890/55.5 ≈ 16.0360 vehicles per mile

This leads to a spacing per foot as: d_a ≈ 5280/16.0360 ≈ 329.2584 feet per vehicle

And the headway per vehicle is: h_ a = 3600/890 ≈ 4.0449 seconds per vehicle

There are several models regarding traffic flow and density. One is the Greenfield's Linear Model, a general model that is used in most traffic conditions.

S_R = S_ f * (1 - D/D_j)

S_R = average space mean speed (harmonic average of speeds)

S_f = free flow speed. This is the speed one would travel if there was no traffic, the weather was near perfect, and there are no obvious road obstacles.

D = actual density

D_j = jam density. The jam density is achieved when traffic is stopped. If you live in the Greater Los Angeles area, you can easily observe freeways at jam density. And yes, it is a pain to drive on the I-210 Parking Lot.

There are other models that the boom describes, such as the Greenberg Density model, which is used for dense traffic situations only.

**The Space Mean Speed of a Train**

The book provides a very interesting example of using simple velocity and acceleration equations from physics to help calculate S_R. Here is an example (not from the book, but one I made up):

A train, the Star Train, travels from Mario Station to Luigi Station, which is a 14 mile (73920 ft) trip. The train takes one minute to load and drop passengers at each station. Information for the Star Train is as follows:

Acceleration: 5 ft/s^2

Deceleration: 4.5 ft/s^2

Peak Speed: 84 mph ≈ 123.48 ft/s

What is the space mean speed?

We have the first find out the total time, including delays.

Accelerating phase:

Time it takes to accelerate to full speed: t _a = v_f/a = 123.48/5 ≈ 24.696 s

Distance traveled: s_a = a*t_a^2/2 = 5 * 24.696^2 / 2 ≈ 1524.731 ft

Deceleration phase:

Time it takes to decelerate from full speed: t_d = -v_f/-a_d = -123.48/-4.5 ≈ 27.44 s

Distance traveled:

s_d = -a_d*t_d^2/2 + t*v_f = -4.5*27.44^2/2 + 27.44*123.48 ≈ 1694.1456 ft

Constant Speed Phase:

Here we reverse the order of calculations: distance, the time.

Distance:

s_c = L - s_a - s_d = 73920 - 1524.731 - 1694.1456 = 70701.123

Time:

t_c = s_c/v_f = 70701.123/123.48 ≈ 572.5715 ft/s

Total time:

T = t_a + t_c + t_d + t_delays

= 24.696 + 572.5715 + 27.44 + 60 * 2

= 744.7075 s (about 12 min 24 s)

Finally calculate space mean speed:

S_R = L/T = 73920/744.7075 ≈ 99.2604 ft/s ≈ 67.5241 mph

So the space mean speed of the Star Train from Mario Station to Luigi Station is about 67.5241 miles and hour.

(1 mi/hr ≈ 1.47 ft/s, actually 1.4666666666666... ft/s = 22/15 ft/s)

Hopefully this helps and you enjoyed this post as I did. As always comments and questions are welcome and appreciated. Until next time, take care!

Eddie

This blog is property of Edward Shore. 2014

Interesting post.

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