Sitting at a dead-stop on the freeway, the meter in the on-ramp to my right flashes green, letting yet another car insert itself into the line of cars in front of me. It is flashing "go" every three seconds while the cars on the freeway next to the on-ramp are mostly stopped. I start thinking about traffic. Is there an optimum number of cars which would maximize the throughput of traffic in Cars Per Minute (CPM)?
Two things control the flow of traffic on a freeway. One is the average following distance in seconds (fds) from one car to the next, and the other is the density of traffic measured in cars per mile (cpm) --- not to be confused with Cars Per Minute (CPM).
CPM is absolutely limited by the following distance in seconds (fds). If fds is 1.5 seconds per Car then CPM must be less than 40.
[Warning: Math Detour]
CPM = (60sec/min) / (fds sec/car) = 60 / 1.5 = 40
It does not matter if we make each car zero feet long and travel a million miles an hour, fewer than 40 cars per minute will pass any point on the road (in one lane) when the fds is 1.5.
The speed of traffic on the freeway is limited by both the fds and cpm (cars per mile). Assuming the fds is 1.5 seconds, and the average car length is 12 feet, and the density of traffic is 200 cars-per-mile, then the speed in miles per hour would be 6.5 or less. How do we figure that out?
[Warning: Math Detour]
miles/hour = (ft/sec) * (sec/hour) / (ft/mile)
ft/sec = (5280/cpm - 12) / fds = (5820 / 200 - 12) / 1.5 = 14.3 / 1.5 = 9.6
(sec/hour) / (ft/mile) = 3600 / 5280
mph = 9.6 * (3600 / 5280) = 6.5
But how do we calculate the actual throughput, the number of Cars Per Minute (CPM)?
[Warning: Math Detour]
CPM = (60sec/min) / ((ft/car) / (ft/sec)) = (60sec/min) / (sec/car) = car/min
ft/car = 12 + (5280 - 12 * cpm) / cpm = 5280 / cpm
ft/sec = ((5280 - 12 * cpm) / cpm) / fds = (5280 / cpm - 12) / fds
(ft/car) / (ft/sec) = sec/car
sec/car = (5280 / cpm) / ((5280 / cpm - 12) / fds)
sec/car = (5280 * fds) / (5280 - 12 * cpm)
CPM = 60 * (5280 - 12 * cpm) / (5280 * fds) = (60/fds) - (60 * 12 * cpm) / (5280 * fds)
CPM = 40 - .09 * 200 = 40 - 18 = 22
Assuming an average following distance of 1.5 seconds and a car length of 12 feet, we can simplify the CPM to (40 - .0909 * cpm).
fds cpm mph CPM
1.5 30 74.5 37
1.5 40 54.5 36
1.5 50 42.5 35
1.5 100 18.5 31
1.5 150 10.5 26
1.5 200 6.5 22
1.5 300 2.5 13
What does the previous table show us? It shows us that the maximum available throughput for traffic is always achieved at the maximum available speed. The fewer number of cars per mile, the higher the CPM throughput. There is no magic optimum number of cars to maximize the CPM, it is just a matter of "the fewer the better" up to number of cars able to go the maximum speed.
Given a rush-hour density of 150 cars-per-mile for a 5-mile stretch of (one-lane) highway, for a time period of one hour, how many cars are on the road within that hour, and what is the number of Cars Per Minute?
[Warning: Math Detour]
At 150 cpm we have 26 CPM at a speed of 10.5 mph.
26 CPM * 60 min = 1560 cars
1560 cars + 150 cars/mile * 5 miles = 1560 + 750 = 2310 cars
2310 cars / 26 CPM = 88.8 minutes
It takes a total of 88.8 minutes to remove all 2310 of the cars on or entering the freeway in one hour at a density of 150 cars per mile.
Given the same number of cars (2310), if the density of traffic is limited to 40 cars-per-mile, for the same 5-mile stretch of (one-lane) highway, how long will it take for all the cars to pass the end, what would the speed be, and what is the CPM?
[Warning: Math Detour]
Forty cpm gives us 54.5 mph, and 36 CPM
2310 cars / 36 CPM = 64.2 minutes
When exactly the same number of cars (2310) need to use the freeway, a density of 150 cars per mile takes 88.8 minutes to move all the cars, but keeping the density down to 40 cpm moves the same number of cars over the same stretch of freeway in only 64.2 minutes, a difference of 24.6 minutes.
Which brings us back to "metered" on-ramps. I would like to make the following suggestions:
1. Every entrance to a freeway or section of freeway should be metered. Even when the entrance comes from another freeway.
2. Like modern "WALK" lights, the meters in the on-ramp should count down the seconds until the next green light. This gives the driver at the front of the line information about how long he will have to wait.
3. Meters should be controlled by a radar gun measuring the speed of the traffic for the on-ramp, not by a sensor in the road counting cars. The "Ramp Meter Design Manual" for the state of California apparently uses "Passage Loops: Inductive loops placed downstream of the limit line to detect passing vehicles," to measure "VPH: Vehicles per hour." But VPH gives no clue about how fast the traffic is moving. It cannot tell the difference between light, fast traffic, and heavy, slow traffic. The VPH can be the same. If the traffic is moving as fast as, or faster than, the posted limit, then the meter may allow cars to enter the freeway. This will automatically slow down the traffic.
4. Metered on-ramps should be allowed to feed traffic into a freeway like a reverse version of water-rights owners pumping water out of a river. That is, each on-ramp should be assigned a specified amount of potential traffic space downstream from the entrance. Upstream on-ramps should not be allowed to fill up the freeway, preventing downstream on-ramps from using their assigned traffic space.
5. Outside of each entrance to a metered on-ramp, far enough back where each driver could decide to change lanes and take an alternate route, there should be a sign in lights that says "
6. What if there is more traffic wanting to get onto a freeway, than the best Cars-Per-Minute throughput will handle? In that case, the meters should not allow them onto the freeway! If the CPM is already maxed out for the speed limit, it is only going to move far fewer cars by allowing the freeway to be overcrowded. People will still get to their destination sooner, even if they have to wait 30 minutes to get onto a freeway which is moving traffic at its maximum speed and number of Cars-Per-Minute.
Drive Well,
Logan
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