The riddle, as posed to the TMIP forum was:
“About 20 years ago a team at UW-Madison received a rather ample grant to train local public officials in the art of traffic flow optimization through signalized intersections, with the specific intent of saving fuel. (The money came from a source that was particularly interested in reducing energy consumption.) So the question I posed was: If all the traffic signals in Wisconsin were to be optimized to reduce delay, would there be any fuel savings?”
Unquestionably, there are lots of good reasons for wanting to optimize signal timing, including improved safety and reduced delay. However, the claim for fuel savings is repeatedly made. Savings of greenhouse gas emissions is just the same; there is a fixed stoichiometric relationship between fuel consumption and CO2 emissions for motor vehicles.
Some well-known traffic operations software packages will tell us the amount of fuel burned for an alternative, and it is a simple matter to subtract the before from the after to learn of the fuel savings. But are these fuel savings real or imagined?
Real in the fleeting short term, but completely imaginary in the long term.
One of the more well-established principles of travel demand is the “travel budget”. With very few exceptions, everyone is granted 24 hours of life each day, and people are willing to spend just so much of it on travel. But the desire for travel is often more than this budget. This leads to a remarkable finding that the total amount of personal travel time by an average individual is relatively unvarying.
Traffic operations software products assume that VMT (vehicle miles of travel) is constant, but they should be assuming instead that VHT (vehicle hours of travel) is constant. Any improvement in delay through signal timing optimization will be (over the long term) reallocated to additional travel, presumably to enjoy an enhanced lifestyle or gain additional income.
So now we can pare down the question to: Assuming travel time for a trip is constant, is there more fuel consumed by continually cruising or occasionally stopping?
To answer this question, I set up a simple numerical experiment using the famous Davis equation, substituting automobile coefficients for train coefficients, to estimate fuel consumption for a compact car. The form of the equation I used weighs automobiles in tons (W), measures speeds in miles per hour (S), measures cross-sectional areas in square feet (A), and measures accelerations in miles per hour per second (a). The equation gives the resistance to forward motion (R) in pounds.
R = 10*W + 0.1*W*S + A*C*0.0026*S^2 + 91.1*T*a
and C is the drag coefficient. I have omitted the grade term, because I am assuming flat terrain. For many years, as a homework exercise, I asked my undergraduate students to verify this equation by comparing it to test track results for a variety of high-performance cars, so I know that it validates quite well.
My experimental automobile, because it lines up with some necessary information from the Internet, is a compact car with a 2-liter engine, weighing 1.5 tons, having a frontal cross-sectional area of 24 square feet and having a drag coefficient of 0.4. The combined thermodynamic efficiency of the engine and drivetrain is 0.27. This is roughly a high-end Honda Civic, non-hybrid.
Let’s compare two trip segments of a 5-minutes duration. One trip is pure cruising at an exactly 30 miles per hour, and the second trip has 1 minute of signal delay. That signal delay consists of 3 seconds of deceleration, 52 seconds of stop, and 5 seconds of acceleration. However, the total time from the beginning of the deceleration to the end of the acceleration is 68 seconds. Let’s concentrate on those 68 seconds, since the rest of the two trips are the same.
The 58 seconds of deceleration and stop consume 0.0026 gallon of fuel, at an idle rate of 0.16 gallon per hour, as reported by the EPA. I assumed idling during slowing, but it might require slightly more or less fuel, depending upon the technology of the automobile.
The 10 seconds of acceleration consume 0.0043 gallon of fuel.
The 68 seconds of pure cruising at 30 mph consume 0.0112 gallon, which corresponds to a fuel economy of 50.4 mpg. Terrain variations and minor accelerations while cruising might raise the fuel consumption rate a little bit.
Bottom line: cruising uses considerably more fuel than stopping, so stopping is better!
An even greater difference in energy consumption would be seen with a hybrid vehicle or an electric vehicle.
This analysis is not at all intuitive. The next time you hear someone claiming global warming benefits of better traffic controls, please refer him or her to this blog post.
Alan Horowitz, Whitefish Bay, March 31, 2017
Note: The delay for a single stop would need to be a quite low 24 seconds (or just 16 seconds of totally stopped time) for the two trips to have the same fuel consumption.
This is a good calculation..
I have two questions.
The signal delay is given as 58 min. first. Why is it actually 68min. I missed something here.
secondly, the total travel time is same so both vehicles take 5 min from origin to destination. Because your first argument already dealt with case where one vehicle saves time and uses it for another trip.
So for the cruising period, for the vehicle that stops at signal, the cruising speed is higher so that it has to catch up. This means the fuel consumption is different.
So your hypothesis is correct when: the cruising vehicle is travelling at a speed V which consumes more fuel than that of stopping vehicle, and this difference is more than the fuel consumed in acceleration and deceleration.
These are not exactly the same trip. Perhaps the destination is a bit farther away for the trip consisting of all cruising. The delay associated with acceleration is exactly half of the time needed for the acceleration. You can infer from the data of the problem that the deceleration is at 5 mph/s and the acceleration is at 3 mph/s. The time needed to decelerate from 30 mph is 6 seconds and the delay is 3 seconds. The time needed to accelerate to 30 mph is 10 seconds and the delay is 5 seconds. The time fully stopped is 60 – 3 – 5 = 52 seconds, since the delay was fixed at 60 seconds in the problem statement. The total time stopping and stopped is 52 + 6 = 58 seconds. The total duration of the whole stopping event is 52 + 6 + 10 = 68 seconds. Tedious but essential details.
Alan
Alan,
So should we say that the fuel savings are positive, after what I have pointed out regarding the speed variation and that the distance they are travelling is different if the speed is same?
This analysis ignores the fact that the stop/start scenario travels less distance. By my calculations it is 0.43 less miles. If you add the additional constant velocity fuel consumption to go the full 2.5 miles, the start/stop scenario consumes an additional 0.00558 gallons.
Dave
You are correct that the pure cruising trip covers greater distance. The historical evidence is that average personal travel time remains almost constant through the decades. The best assumption for long-range fuel savings (or GHG savings) is to hold time constant, not distance constant. GHG is a long-range problem and we need to employ long-range thinking.
No, the condition of the problem would need to be changed radically for there to be positive fuel saving. We cannot ignore induced travel. Any enhancement to the highway system by transportation engineers has the potential to induce travel. And history has shown that induced travel can have disastrous impacts. We cannot simply hold VMT (or trip distance) constant. While someone might argue that in small, uncongested communities people won’t travel more when there are traffic engineering improvements, the evidence is otherwise for the vast majority of travelers. Ignoring induced travel is a much more fundamental mistake, which is my larger point, than reporting nonexistent fuel savings of GHG savings.