Quote:
Originally Posted by acottawa
5-10 seconds seems awfully low; maybe if it is a button-based system and nobody presses the button for 5-10 seconds it will start moving, but an actual stop with actual passengers that seems pretty unlikely, particularly if the train is reasonably full.
Do you drive your car using the maximum acceleration, deceleration and speed? The settings on the vehicles will also incorporate things like passenger comfort, wear and tear on tracks and brakes, safety, etc.
|
It can be hard to believe, but if you measures dwell times at subway stations in Toronto and Montreal, it's at or under 10 seconds for all but the busiest stops (places like Bloor-Yonge, Union, Berri-UQAM, etc.). Yes, I'm enough of a transit nerd to have done this before. At very low ridership stations in Toronto like Old Mill, it's about 5 seconds. A good number of our stations will have very low ridership especially off-peak.
As for acceleration, 1.2m/s^2 is the "normal" operating acceleration of the Alstom Citadis for standard operation. 1.5m/s^2 is the engine maximum.
We also have confirmation that the braking and acceleration will be automated; this means that it will indeed use its theoretical perfects for these times (in Toronto, preliminary estimates are that the switch from manual to automated on the Yonge line will cut travel times by as much as 10% for this reason).
Here's the back-of-the-napkin calculation again (judge me freely, but I like doing math

)
Deceleration time: This unfortunately, cannot be directly calculated because I don't know what the braking speed of the trains is (as far as I know, this has never been published). Based on by personal observations of Alstom Citadis trains in operation, I'm going to say it's a few seconds (let's say 3) longer than acceleration time.
Using an acceleration of 1.2m/s^2 from a top speed of 100km/h:
-100km/h is ~27.78m/s
-Using standard physics formula:
v-final = v-start + acceleration*time
27.78 = 0 + 1.2*time
27.78 = 1.2*time
27.78/1.2 = time
23.15 = time
So the acceleration time is 23.15 seconds. Let's say that means 24 seconds to accelerate, 27 seconds to decelerate, to/from top speed of 100km/h.
Let's throw in the dwell time of 10 seconds. The time from when the train starts slowing down to when the train is back at full speed again is thus 27+10+24=60 seconds.
To analyze how much time is added to a trip by having to stop, we also have to consider how much time it would have taken the train to go through the station without stopping.
This means we have to calculate the stopping distance, too. This is harder to calculate as a lot more variables influence it, but we can get close:
-Stopping time is 27 seconds
-Average velocity over stopping time: (~27.78/2) = ~13.89m/s
-Stopping distance: 27*~13.89=~375m
-Starting time is 24 seconds
-Average velocity over starting time: (~27.78/2) = ~13.89m/s
-Stopping distance: 24*~13.89=~334m
So it will take 403 metres of track for a train to brake from max speed to stop, and 334 metres of track for a train to speed up from rest to max speed. This means any station-to-station distance of less than 700m or so will result in the train not attaining its maximum speed. Very few stations in our network are that close together; I believe Parliament & Lyon is the only such combination with Phase 2 buildout.
Travelling over 700m at 100km/h (27.78m/s) would take:
v = d/t
vt = d
t = d/v
t = 700/27.78
t = 25.1 seconds
So, at each stop, decelerating, dwelling, and accelerating will take about 60 seconds, whereas not stopping there and just going through the same area would take 25 seconds.
So the time added to the trip by most station stops is 35 seconds, 10 of those seconds being dwell time. Based on that, at busy stations & times (like Parliament station at rush hour), it will likely approach 50 seconds, but at lower ridership stations like Cyrville or Iris, more like 30 seconds.
One thing that would modify these calculations are slow zones. There's probably some sections of the track where the train will not be able to attain its maximum speed based on track geometry. However, in such a case, the travel time without a station would be longer, so the time added by a stop is actually lower in such cases.