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Q: Compute yellow clearance interval for traffic light(equation provided) ( Answered 5 out of 5 stars,   2 Comments )
Question  
Subject: Compute yellow clearance interval for traffic light(equation provided)
Category: Science > Math
Asked by: njt-ga
List Price: $5.00
Posted: 12 Oct 2002 11:06 PDT
Expires: 11 Nov 2002 10:06 PST
Question ID: 75766
Y = t + V/(2a+2Ag)
Y= Yellow clearance interval in seconds
t= reaction time (use 1 second)
V= 85%percentile approach speed in ft/sec/sec [can use speed limit]
a= deceleration rate of a vehicle(use 10 ft/sec/sec)
A= Acceleration due to gravity (32.2 ft/sec/sec)
g= percent grade in decimal form(+for upgrade,- for downgrade) this is
unknown.
*Calculate the yellow clearance interval to the nearest 0.1 second

Researching a light that I believe is dangerously mistimed- 50mph
zone,1/10th of a mile before light, turns to a 45mph zone. I've timed
the yellow light at 2.8 seconds, I need fact that this equation proves
that the yellow interval should be longer. Calculation of 50 + 45mph
would be helpful. Need to know how to solve and explain how solved.
Thank you.
Equation found on www.dot.state.ct.us/eh/ehen/traffic/manual
Answer  
Subject: Re: Compute yellow clearance interval for traffic light(equation provided)
Answered By: smudgy-ga on 12 Oct 2002 13:20 PDT
Rated:5 out of 5 stars
 
Hi,

In reading your question it seems to be the case that you want to show
that 2.8 seconds is too short for this yellow light.

The following article by Mike Labash from the "Weekly Standard" may
help you in your quest. It references studies that indicate that even
a small difference in the timing of the light can make significant
difference in whether a person enters an intersection under a yellow
or under a red light:

http://www.weeklystandard.com/Content/Public/Articles/000/000/001/079bkyhi.asp


Now for the question itself: 

First, you should notice that in the traffic manual you cite,

http://www.dot.state.ct.us/bureau/eh/ehen/traffic/manual/Capacity%20Analysis%20&%20Signal%20Timing.pdf

it is stated, "Do not use a yellow clearance interval of less than 3
seconds or (not normally) more than 5 seconds." This statement in and
of itself seems to indicate that 2.8 seconds is too short a time, but
it may be that the machine is set at 3 seconds and ssome variability
in the timing system reduces it to an actual 2.8 seconds. This may be
(in a civil engineering sense) acceptable if the actual desired yellow
interval is 3 seconds (despite the condition that states that the
timing of the light should be calculated to the nearest .1 second).

However, mathematical evidence will bear out whether this light
deserves a longer interval.

We need to make some observations about the way the formula operates.

Increasing V will increase the light interval.
Driving downhill (negative grade) towards a light will increase the
light interval and driving uphill towards it (positive grade) will
decrease the light interval.

According to the conditions you set forth, we have

Y = 1 + 50/(20+64.4g) for 50mph scenario
or
Y = 1 + 45/(20+64.4g) for 45mph scenario.

So the timing of the light depends entirely on the grade of the road,
if all the other variables are held constant.

These formulas are found by simply plugging in V=45, a=10, A=32.2.
Values for specific g can be calculated by plugging in the g in
question. A 3% grade would be entered as g=.03. I did multiple entries
(in the table below) by using the "table" function of a TI-89 graphing
calculator, but in principle any calculator or even long
multiplication can give you the results you need for a particular
value of g. Mathematically speaking the most important factor is
"order of operations": first the operations that occur in parentheses
should occur, then multiplication and division, then addition and
subtraction. So in this case, we tackle the parentheses first: 64.4
must be multiplied by g, then the result should then be added to 20.
Now this result can be used to divide 45. Finally, 1 should be added
to this number to get a value for Y.

Since we don't know the grade of the road, and since at this time I
cannot find a road manual that clearly indicates the maximum grade of
a road approaching an intersection, I will list all values of Y for
grades from -10% to 10%, which I know to be considered quite steep
(Trucks often get warning signs around 5% grade). I will go by
half-percent increases. So the chart at the bottom of my answer lists
the value of Y for both scenarios for all grades from -0.1 until both
scenarios have yellow light times less than 2.8. I will increase by
0.005 increments.

To spare you from having to read through a chart unnecessarily, let me
summarize the findings first: If the grade is less than 7% uphill,
your yellow light time is greater than 2.8s in the 45mph scenario. In
the 50 mph scenario, if your grade is less than 11% uphill, your
yellow light time is longer than 2.8 seconds. Unless an intersection
is at the crest of a very steep hill, then at least one direction
coming into the intersection will probably have a grade less than 7%.
Remember that if a road is climbing a hill, and it has an intersection
on it, one direction of the intersection has a positive grade and the
other direction has a negative grade. Unless the intersection has a
fancy light sequence, both directions of the road will have the same
green light, and so both will be subject to the same yellow light
timing. Presumably, the longer light timing (the timing for the
fellows headed downhill) should be used.

The chart is included below; if you find out the exact grade of the
road you are dealing with you can look up its Y value on the chart. I
hope this answers your question. If you want further clarification or
would like more precise information on the results for a particular
grade (or if you need information on a grade I did not include in the
chart), please do not hesitate to request a clarification.

Good luck,
Smudgy.

Appendix: Chart of yellow light intervals based on grade.

Remember, negative grade indicates a car approaching the light is
traveling downhill. All values for Y are rounded to the nearest .1
second. In the interest of space, I will not continue the chart after
it enters the realm where the light's timing should be greater than
2.8 seconds according to the formula.

Please excuse the length of this chart, but without knowing the grade
of the road this seems to me the best way to convey the necessary
information.

Grade		45 mph scenario		50 mph scenario
-0.1		4.3			4.7
-0.095		4.2			4.6
-0.09		4.2			4.5
-0.085		4.1			4.4
-0.08		4.0			4.4
-0.075		4.0			4.3
-0.07		3.9			4.2
-0.065		3.8			4.2
-0.06		3.8			4.1
-0.055		3.7			4.0
-0.05		3.7			4.0
-0.045		3.6			3.9
-0.04		3.6			3.9
-0.035		3.5			3.8
-0.03		3.5			3.8
-0.025		3.4			3.7
-0.02		3.4			3.7
-0.015		3.4			3.6
-0.01		3.3			3.6
-0.005		3.3			3.5
0.0		3.3			3.5
0.005		3.2			3.5
0.01		3.2			3.4
0.015		3.1			3.4
0.02		3.1			3.3
0.025		3.1			3.3
0.03		3.1			3.3
0.035		3.0			3.2
0.04		3.0			3.2
0.045		3.0			3.2
0.05		2.9			3.2
0.055		2.9			3.1
0.06		2.9			3.1
0.065		2.9			3.1
0.07		2.8			3.0
0.075					3.0
0.08					3.0
0.085					3.0
0.09					2.9
0.095					2.9
0.1					2.9
0.105					2.9
0.11					2.8

Clarification of Answer by smudgy-ga on 12 Oct 2002 13:23 PDT
Hi again,

It seems that my carefully-laid-out chart did not render well. Let me
clarify the chart in case you need to use it:

Most rows have three entries in them. The first entry is the grade,
written as a decimal. The second entry is the yellow interval for a
45-mph zone, and the third entry is the yellow interval for a 50-mph
zone.

In rows with 2 entries, the yellow interval is less than 2.8 in a
45-mph zone so I did not include it. So in these rows, the first entry
is the grade, and the second entry is the yellow interval in a 50-mph
zone.

I hope this prevents any confusion about the chart.
-smudgy

Request for Answer Clarification by njt-ga on 13 Oct 2002 06:28 PDT
Smudgy- 
Thank you for your answer- very detailed, however, I think I made a
mistake in describing one of the items in the equation. I explained
that V=85%percentile approach speed in ft/sec/sec , and then put in
quotations[can use speed limit]I meant that the 85%approach speed
could be used to figure approach speed in ft/sec/sec. Apparently the
85%percentile speed is used for 85% of the population driving at a
certain speed-many places use the speed limit[that I can see]On other
site on a 1/14/02 meeting of the transportation research board on 'the
effects of dilemma zones on red light running enforcement tolerances',
I found a chart for MPH conversion to ft/sec[[45mph=66.00ft/sec and
50mph=73.33ft/sec.] I did the computation with those numbers, and came
out with 2.28sec on the 45mph/66ft/sec- did I follow your instructions
right on computating? If so, the formula is flawed at CTDOT. On the
other site[TRB]it has the calculated minimum yellow at 45mph at 4.3sec
and 50mph at 4.7mph. I am wondering if when CTDOT put in [ft/sec/sec]
Is it their'microsoft word' way of squaring the numbers on their
formula? Sorry for the clarification request, I think I misquided the
question with saying to use the speed limit for the 85%percentile.

Clarification of Answer by smudgy-ga on 13 Oct 2002 15:12 PDT
njt-

It seems I did indeed misread the conditions for the question. I will
recalculuate the speeds using feet/sec rather than miles/hr. It may in
fact be the case that the CTDOT formula is in conflict with other
standards. Can you provide a web location for the "TRB" site that you
refer to? If TRB has a formula, we can at least compare it to the
CTDOT formula and see how the two different formulas' results measure
up.

Expect an updated chart within a day. I cannot do the calculations
now, but I wanted to respond to your clarification request as soon as
possible. If you can provide a link to the TRB site you mention, that
would be extremely helpful. If they have a formula there, I will
provide a chart for that formula as well.

-smudgy.

Request for Answer Clarification by njt-ga on 13 Oct 2002 18:57 PDT
The other equation I was citing is from a site that can be reached
with a google search on: TRB paper 02-3744 , I am so sorry for this
clarification needed, but I need to know CTDOT law for yellow timing
sequencing and how to explain it. Thank you Smudgy for all your help
in this matter, you are truly exacting in your answers, and should be
commended in your strive to seek the correct answers, especially when
the asker is not accurate in asking a question correctly in the first
place. I am truly sorry for the miss communication.NJT

Clarification of Answer by smudgy-ga on 13 Oct 2002 20:08 PDT
njt-

I have made the appropriate change in the yellow light formula
(replacing my original "45 mph" with "66 ft/sec" and "50 mph" with
"73.33 ft/sec" and now my result for the yellow light timing does
indeed match yours. That is, according to the CTDOT formula, on a road
with 0% grade, the yellow light interval at 66 ft/sec measures 4.3
seconds and at 73.33 ft/sec the interval measures 4.7 seconds.
Therefore, if assuming the grade of the road is 0, then the timing of
the light according to the CTDOT formula matches the standard
mentioned in the TRB paper. In looking at the formula mentioned in the
TRB paper, it seems to be mathematically equivalent (more or less)  to
the CTDOT formula, though the TRB formula solves for stopping distance
rather than yellow light time interval. The TRB formula does not take
grade of road into account, however.

(The precise relationship between the CTDOT formula and the TRB
formula, excepting the grade issue, is that the CTDOT formula is
acheived by dividing the TRB formula by V. The resulting yellow
intervals are the same, for a grade of 0.)

I am not certain why your answer came out incorrectly. I will give an
example step-by-step so that you can make sure you are performing the
operations in the correct order.

Assume t=1, V=66 ft/sec, a=10, A=32.2, and g=-0.01, for example
purposes.

Then our formula is Y=t + V/(2a+2Ag)
Y= 1 + 66/(2*10 + 2*32.2*(-0.01))
Work inside the parentheses first, starting with the multiplication,
then the addition:
Inside the parentheses we have 2*10 + 2*32.2*(-.01) = 20 + (-0.644) =
20 - 0.644
So the sum inside the parentheses is 19.356.
Our formula is now Y= 1 + 66/19.356
Perform the division first, then the addition. 
66/19.356 = approx. 3.4
So Y = 1 + 66/19.356 = 1 + 3.4 = 4.4

The change from miles per hour to feet per second has actually
increased our yellow-light interval. This is because we are increasing
the numerator (top part) of the fraction part of the equation. In
general, if the numerator of a fraction increases and nothing else in
the fraction changes, the value of the fraction will go up.

It seems clear that 2.8 seconds is a preposterously short interval for
a road at 50 mph, even 45 mph. For comparison's sake, I will give the
values for a few different grades. Once again, if you want any other
specific values for specific grades, please let me know.


Grade: -.1         At 45 mph: 5.9 seconds. At 50 mph: 6.4 seconds.
Grade: -.05       At 45 mph: 4.9 seconds. At 50 mph: 5.4 seconds.
Grade: 0.0        At 45 mph: 4.3 seconds. At 50 mph: 4.7 seconds.
Grade: 0.05       At 45 mph: 3.8 seconds. At 50 mph: 4.2 seconds.
Grade 0.1         At 45 mph: 3.5 seconds. At 50 mph: 3.8 seconds.

Incidentally, to give a yellow interval of 2.8 seconds, the grade of
the road would have to be about 25% uphill at 45mph... that is, the
road would go up 25 feet for every 100 feet it traveled! Quite steep
for a road! For 50 mph the grade would have to be over 30% uphill.

I hope you find this information useful to you. Again, if you have any
further questions, do not hesitate to request a clarification.
njt-ga rated this answer:5 out of 5 stars
Thank you very much for your detailed answer and your follow-ups. As a
mathematically inept Liberal Arts Graduate, I thank you for helping me
through this formula.
NJT

Comments  
Subject: Re: Compute yellow clearance interval for traffic light(equation provided)
From: hedgie-ga on 12 Oct 2002 11:49 PDT
 
Could you either

1) eliminate the complication of grade 
(it should not have much of the effect anyway)

or

2) specify the  deceleration down the grade 

?

 The acceleration of gravity and grade do not really determine what
 will the deceleration of a car be,  when braking on a slope.
 It is more issue of the brakes, properties of the surface, etc, 

as weight of the car is small compared to inertial force during
braking.
Subject: Re: Compute yellow clearance interval for traffic light(equation provided)
From: smudgy-ga on 13 Oct 2002 22:22 PDT
 
hedgie-

Incidentally, the A*g term is the component of gravitational
acceleration parallel to the deceleration of the car, i.e.,
perpendicular to the normal. This part of the gravitational
acceleration will have a direct combining effect with the (negative)
acceleration from braking, which is also perpendicular to the normal.
The deceleration value of a=10 m/sec/sec, I imagine, is meant to take
into account the standard braking conditions, conditions of the road,
etc.

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