4 ms·

As an aside, I wish that Stoplights would factor in Aggregate Kinetic Energy of the Traffic they manage. I feel a bit bad every time my Single Stopped Car gets

As an aside, I wish that Stoplights would factor in Aggregate Kinetic Energy of the Traffic they manage. I feel a bit bad every time my Single Stopped Car gets a Green, causing a group of 10 Cars at 40 mph to Stop, then later Re-Accelerate. If the stupid Junction had waited some time, the 10 Cars would happily carry their 40 mph through, allowing me to go once the Large Kinetic Energy Traffic had passed uninterrupted.

jeffbee 4y ago
The aggregate kinetic energy of any number of stopped cars is zero so I don’t see how your algorithm will work.
- tzs 4y ago
  For simplicity lets assume all the cars in this discussion are the same mass and travel at the same speed except when they slow and stop for a traffic light or are accelerating back up to speed after the light changes. Let E be the kinetic energy of each car when it is not dealing with a traffic light. Let's assume no regenerative braking. Let's assume that a car idling at rest uses R energy So we have 1 car that approaches the light, gets a red, and stops. It has to shed E kinetic energy. Then 10 cars approach the intersection on the perpendicular road. Before they reach it the light changes and they all have to stop, shedding 10 E kinetic energy. The first car resumes its journey. The light changes again and the other 10 resume. The net result is that encountering the intersection caused this 11 car system to shed 11 E kinetic energy more and to consume 11 R idling energy more than would have been consumed if the intersection had not been there (e.g., if the crossing had been an overpass). What he's proposing is that some sort of smart traffic light system could notice the 10 approaching cars and hold their green a little longer so that the 10 cars do not have to stop. Suppose it has to hold the light long enough that car 1 has to wait twice as long. The energy cost of the intersection on the 11 car system under that scenario is E shed kinetic energy and 2 R idling energy. That's going to be much less that 11 E + 11 R.
hgsgm 4y ago
In English, we don't capitalize common nouns.
- nullish_signal 4y ago
  I Write However I Please

jeffbee 4y ago

The aggregate kinetic energy of any number of stopped cars is zero so I don’t see how your algorithm will work.

tzs 4y ago
For simplicity lets assume all the cars in this discussion are the same mass and travel at the same speed except when they slow and stop for a traffic light or are accelerating back up to speed after the light changes. Let E be the kinetic energy of each car when it is not dealing with a traffic light. Let's assume no regenerative braking. Let's assume that a car idling at rest uses R energy So we have 1 car that approaches the light, gets a red, and stops. It has to shed E kinetic energy. Then 10 cars approach the intersection on the perpendicular road. Before they reach it the light changes and they all have to stop, shedding 10 E kinetic energy. The first car resumes its journey. The light changes again and the other 10 resume. The net result is that encountering the intersection caused this 11 car system to shed 11 E kinetic energy more and to consume 11 R idling energy more than would have been consumed if the intersection had not been there (e.g., if the crossing had been an overpass). What he's proposing is that some sort of smart traffic light system could notice the 10 approaching cars and hold their green a little longer so that the 10 cars do not have to stop. Suppose it has to hold the light long enough that car 1 has to wait twice as long. The energy cost of the intersection on the 11 car system under that scenario is E shed kinetic energy and 2 R idling energy. That's going to be much less that 11 E + 11 R.

hgsgm 4y ago

In English, we don't capitalize common nouns.

nullish_signal 4y ago
I Write However I Please