The answer is to keep the same number of stops but run two or more vehicles simultaneously. Or open more doors. Or expedite fares.

The authors get mixed up equating count of marked stops with dwell time. Running leapfrogging vehicles , or numerous other strategies, reduces dwell time because one boards passengers and the other disembarks at any given stop or vice versa.

In fact, I’d argue bus fare gates, steps, 1-door loading and traffic signal/stop interactions are far more significant than stop count.

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dyauspitr 4 hours ago | parent [-]

> The answer is to keep the same number of stops but run two or more vehicles simultaneously.

How exactly does that help? If you’re suggesting every bus go to alternate stops leapfrogging each one in the middle then that will cause a lot of confusion especially for tourist heavy cities.

	▲	VLM 3 hours ago \| parent [-]
		In 2022 according to the transit system annual report, the suburban quarter million person city I live in has ten routes and operates about 12 hours per day and per the annual report average weekday service consumed is 1556 UPT, so 1556 people step aboard the system and toss coins in the fare jar or pay with the app. UPT means they're not tracking transfers and essentially 100% of trips require a transfer so the real number of people served daily is closer to 775 than to 1550, but we'll run the optimistic numbers. Each of the ten hourly routes is about 4 miles long. So the overall system drives 12 hours * 10 routes * 4 miles * 5280 feet/mile = 2.5 million feet per day and divide that by 1556 passengers per day that's a pax every 1628 feet driven on an average day. So if we had a bus stop every 800 feet, on average half the stops would be empty and passed by. If that high level of use is causing too much congestion and slow down at stops, if we had two buses running out of phase, pax arrive at the same rate, so we'd pick up a pax every 3000+ feet driven. So if we had bus stops every 500 feet to keep people happy, on average the bus would drive right by about 5 out of 6 empty stops, which seems reasonable and would not result in unusual delays or congestion. Also the bus would pass by every half hour not every hour, which would probably increase ridership a lot. So if the only labor expense were the $23/hr driver, and we pay 10 drivers on 10 routes, to drive twelve times, thats $23/hr * 10 routes * 12 hours if everything except driver labor were free that means we spend $2760 per day to transport 1556 people, or about $1.77 per trip (assuming diesel is free, buses never wear out, etc). If we doubled the number of bus that would be $5520 of driver labor to move 1556 people per day or $3.55 cost per pax trip. On one hand the actual annual total "OE per UPT" counting weekends and maint and office people and dispatchers etc, according to the annual report is $13.94, so an extra $1.77 would seem cheap, but the bus does not run for free and the total expense of doubling the runs might cost as much as an extra $14 per pax trip. The costs don't really matter, if the taxpayers want it as a luxury bragging feature of the city. Everyone wants everyone else to use it even though no one would be caught dead actually using it. My point being that adult fare is $2 but adults don't ride its mostly elderly and disabled at the $1 fare, so a profit (loss) ratio of (28 - 1)/28 with two buses per route isn't much worse than (14 - 1)/14 with one bus per route. Maybe another way to look at the analysis is in my city if the stops are more than 1600 feet apart there will be multiple people per stop and that would "slow things down" whereas a small fraction like 400 feet would mean the bus mostly just speeds by. No one can seem to explain why we can't have infinite bus stops. How about every stop sign is a bus stop? The bus has to stop anyway. Artificial scarcity to drive down ridership, I suppose.