April 6, 2016 at 22:33 #21540April 7, 2016 at 10:03 #21541
I assume you ask because of the “20min rule”.
Traveling time is calculated as follows: The actually time on the foot / road / rail plus half of the frequency of the line for each line change (it is assumed that when somebody exchanges at a stop not knowing when a vehicle arrives the average will be half of the frequency).
But at the first stop only a quarter of the frequency will be added to the total time (assuming somebody goes to a stop knowing when vehicles depart).
So in the case of your example this will result in a waiting time of about 11 minutes, leaving 9 minutes for traveling.
TomApril 7, 2016 at 12:18 #21547
Thanks for the reply.
But in my example on line 4 of the machine. Interval of 45 minutes, i.e. travel time 45*4/2 = 90 min. If we add the waiting time (11 minutes), it receives 101 minutes excluding walking time.
PS. Sorry for my english.
Put it this way:
frequency = 45min
time on foot = x
time on road = 45min*4/2 = 90min
waiting time = 11min for first stop / 22 min for other stops
travel time one way = x + 90min + 11min
travel time back way = x + 90min + 22min
And passengers wishing to use the line in both directions
April 12, 2016 at 18:21 #21584
- This reply was modified 4 years, 3 months ago by alexlumpov.
I think that developers should better explain the 20 minutes rule.
The only thing I can think of in your case is that if you have more than two stops on the line (as I imagine from the screenshot), then the frequency number you get is referred to the total trip (back and forth). However, customers are not forced to use the whole line, but may hop in a stop and hop out the next. In that case, the travel time between two stops (in one direction) can be under 20 minutes and satisfy the 20 minutes rule.
By the way: why in your calculations you get a different waiting time for the different stops ?April 13, 2016 at 00:35 #21585
gcampono, you can download my save.
it`s a big map. 1850year. and horse line through the entire map.
line has 4 stops, but 1,2 and 3,4 is the same stopgroup.
This is necessary because on the line (1.A, 2.B), this bug is not reproduced.
The same can be done with trains. To do this you can add a reversal loop after each station.April 13, 2016 at 00:48 #21586
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