Frequency and the 20 min rule

Home Forums General Discussion Frequency and the 20 min rule

Viewing 15 posts - 16 through 30 (of 32 total)
  • Author
    Posts
  • #13877
    gGeorg
    Participant

    @TrainInFluenza nice definition and measurement! I have asked Douglas to fix the guide

    http://www.train-fever.com/forums/topic/train-fever-interactive-guide/page/5/#post-13875

    —-

    1. I think its 24 minutes in the first. Month is a minute. 24 month are 2 years. In case of one vehicle on the line with 2 stops. You need each payment at the station in particular year. If you make it longer, then you financial ledger would be very jumpy. Becouse it would happen the in one financial year you dont get any income. Just imagine waves like one year minus milion next year plus 2 milions. If you make it shorter it would not be synced.

    2. The time limit (20/24 whatever it is) is considered for farest possible route.

    Lets have an example of algorithm for leisure route :

    Let say the agent is home (the green) he looks for leisure (yellow). There are 7 leisure areas on the small map. 3 of them are reachable in time limit. 1 of them has best environment parameters. So the route is started for a leisure which is in reach of the time limit (and has best value?). It takes bus-train-bus2. Due to the traffic jam first buss was late 3 minutes. Train was waiting in taffic lights for next minute. When the agent come to the bus2 station he has only 5 minute left but rest of the rout takes 7minutes so he is teleported home (green). Alhou he paid for first bus and train, he dont reach destination.

    In case he made it to leisure. He would also travel back.

    Intersting question would be – for the city grow bonus – does it count started travels or finished travels?

    • This reply was modified 9 years, 4 months ago by gGeorg.
    #13882
    fransgelden
    Participant

    @TrainInfluenza: I see in the second image the bus frequency is around 66 second. But is it 66 seconds when the bus arrive by stop 1? Or is it 66 seconds travel time between bus stop 1 and 2? I’ve read through the guide and somewhere it was describing a situation where there was 5 trams something and then it told about multiplying the 60 seconds with 5 and you get 300 seconds or 5 minutes.

    What I understand the guide try to tell is, you take the number of busses and multiply by the frequency to get the TOTAL travel time. I think this would helped you, because then you could  workout the total travel time with the bus, train and on foot. Thus if my tram take 5 minutes and train 6 minutes, then the total travel time will be around 11 minutes? And another extra 3 minutes of walking? That’s how I understand this, because its impossible for the peep to riding 6o second in the bus, while its going around the whole town.

    I think this is how you determine the total travel time of a person. And to the original question, the waiting time is not calculated. If the total travel time is in the frame of 20 minutes, the person will travel with your transportation, despite it will wait a long time at the station.

    What I do know is the frequency will be updated every year and if the train is frozen long enough, the frequency will skyrocket at the end of year and thus make the system obsolete.

    • This reply was modified 9 years, 4 months ago by fransgelden.
    #13884
    Tossi
    Participant

    @crossmr your post is offtopic and meaningless to the discussion. If you want to cry go do it somewhere else.


    @fransgelden
    from the way you write it seems you don’t understand what frequency is.

    1. Frequency means how often on average a chosen spot on the line will be passed by a vehicle.
    2. It will be exactly the same for every point on the line so it doesn’t matter which point you pick for measurements. It may be first stop, it may be the last one or it may be a point somewhere between the stops.
    3. Frequency is also equal to how long it takes a single vehicle to make a fool line loop divided by number of vehicles on the line. So if you have a single vehicle on the line and only then the frequency will be equal to how long it takes for the vehicle to make a full loop.
    4. Frequency is not how often a vehicle will pass a stop.
    #13885
    TrainInfluenza
    Participant

    So in the 66s image means a vehicle will pass every point on the the line every 66s. In this case we have only 1 vehicle so it takes 66s to go from Stop 1 back to Stop 1. But also 66s from Stop 2 back to Stop 2. Or 66s from where the mouse is back to where the mouse is – start a timer. That vehicle makes one loop of the line every 66s.

    Now if I add a 2nd vehicle each one will do a loop every 66s but because there are 2 – a vehicle will pass every point every 33s. So Stop one will have a vehicle every 33s, Stop 2 will have one every 33s and where the mouse is will have a vehicle pass every 33s. (66 / 2 = 33)

    6 vehicles will make it 11s (66 / 6 = 11). etc.

    So yes the reverse works, if you have 6 vehicles and the frequency is 11s – then you know that one vehicle does a loop of the line in 66s. (11 * 6 = 66)

    I think you need to forget about the individual stops, it’s nothing to do with them – every point on the line (maybe think of it as a circle) will have a vehicle pass every frequency seconds. (assuming equal spacing)

    Does that make sense?

    What I understand the guide try to tell is, you take the number of busses and multiply by the frequency to get the TOTAL travel time.

    Yes if you want to travel a full loop of the line – frequency has nothing to do with travel time of a passenger as I understand – it is waiting time at a station. But you can calculate the total time to do a hypothetical loop.

    It’s best to think of frequency as the maximum time a person has to wait (at any station as I’ve shown above they’re all the same). The travel time is calculated while the person sits on the bus. But what the guide is trying to say in a complicated way is that from the frequency you can find how long it takes to do a loop of the line. Number of vehicles * frequency (as I showed above is 66s in that case).

    So looking at the line in our example (with 6 vehicles) you can say that the maximum waiting time for a vehicle is 11s (the frequency) and the maximum travel time is 66s (6 * 11) if the person wishes to come back to where they started.

     

    #13888
    fransgelden
    Participant

    @Tossi: I see now. I agree on number 1 and 2, but I don’t understand 3. How do you then workout the total travel time? And number 4, you saying the frequency is not based on every stop, but only on a single on, like you said in number 1 and 2.

    Can somebody then try to explain how I can use frequency to calculate the total travel time.

    If I got the following scenario: I got a circle bus line. There is 6 busses on it and the frequency is 50 seconds.

    Can I then take 6×50=300 seconds. So it means my bus will take 300 seconds to fully complete the circle line.

    If I got 8 stops on the line and the person gets of at the 3rd stop. Then I can take 300/8=37,5 seconds. Then take 37,5×3=112,5 seconds.

    This means the person will take 112,5 seconds to travel from bus stop 1 to bus stop 3?

    #13889
    TrainInfluenza
    Participant

    @Tossi is correct but I think your point 4 is a little contradictory to point 1. I understand what you’re trying to say in it thought.

    You mean it is not how ofter a particular vehicle will pass a stop? It’s how often some vehicle on the line will pass a stop (or any other chosen point of reference).

    #13890
    TrainInfluenza
    Participant

    @fransgelden firstly it’s best to see every line as a circle conceptually (I got a circle bus line) even if the vehicles use the same road/rail to travel back.

    Can I then take 6×50=300 seconds. So it means my bus will take 300 seconds to fully complete the circle line.

    Yes

    If I got 8 stops on the line and the person gets of at the 3rd stop. Then I can take 300/8=37,5 seconds. Then take 37,5×3=112,5 seconds.

    Yes but only if your stops are equally spaced. Frequency does not help you in any way here (unless you have the unlikely scenario where stops are equally spaced). You would need to measure the trip time between stops manually or estimate it.

     

    See this example, some vehicle will pass every stop on this line every 38 sec and each vehicle will do a complete loop in 66s. But the travel time between the stop is not the same – 1 to 2 is a few seconds while 2 – 3 is quite long and 3 – 1 even longer.

    #13895
    Tossi
    Participant

    @fransgelden

    1. Read what TrainInfluenza wrote in his previous post he understands frequency properly.
    2. Your calculations are correct.
    3. If I got 8 stops on the line and the person gets of at the 3rd stop. Then I can take 300/8=37,5 seconds. Then take 37,5×3=112,5 seconds.

      Look how frequency is 50 seconds but a bus passes a stop every 37,5 seconds. That’s what I meant in point 4 of  my previous post but didn’t word it well enough.

    4. Point 3 of my previous post is about how frequency and total loop time are related. It’s kind of reverse of what you wrote in

      Can I then take 6×50=300 seconds. So it means my bus will take 300 seconds to fully complete the circle line.

      number of busses * frequency = loop time (6×50=300) so frequency = loop time / number of busses (50=300/6)

    PS: Maybe this is how point 4 should be worded properly:  Frequency is not how often a vehicle will pass stops. ?

    • This reply was modified 9 years, 4 months ago by Tossi.
    #13898
    fransgelden
    Participant

    Ah, thanks 😀  So now I have some air cleared around frequency.

    #13900
    TrainInfluenza
    Participant

    This is why with frequency alone you can’t tell how long it takes to travel between the stops 🙂

    #13941
    gGeorg
    Participant

    @Tossi  “PS: Maybe this is how point 4 should be worded properly:  Frequency is not how often a vehicle will pass stops. ?” Well it is. Get stop watch and piece of paper. Stand on any point of a line. When you see a vehicle from your line, write down current time. The table you get would look like the Fulenza’s table, the right column. You will have a list of time rocords, any record has same amount of time from previous record. The diference is the same for all the records. The diference is the frequency.

     

    Nice discussion BTW :-]

    Now deaar devs should put more numbers on the line window. Like :

    freq, lengt of one round in seconds, lenght of one round in km, line capacity.

    Line capacity is imporatant for replacement. You can get shorter/longer freq but it still doesnt tell you the amount of cargo delivered.

    • This reply was modified 9 years, 4 months ago by gGeorg.
    #13951
    Tossi
    Participant

    @gGeorg I see you don’t understand the sentence. Can you tell what is the difference in meaning of those two very similiar sentences?

    1. Vehicle pass stops.
    2. Vehicles pass a stop.

    Also I explained what I meant in point 3 of my last post. Reading that explanation again should also help.

    #13960
    gGeorg
    Participant

    Ah. I see. Too fast reading. Well then, … vehicle pass stops — > no it doesnt.

    #13979
    Kiwi-NZ
    Participant

    Regarding Post No. 13809

    We have made a couple of test and the result is that an Agent is not disappearing after 20 Minutes. The Agent continuous his journey even if the Train stops for 40 Minutes.

    Based on this we know that the car will not disappear and if you have already a traffic jam it will expand further.

    #13984
    TrainInfluenza
    Participant

    @Kiwi-NZ that’s interesting. So how does the algorithm pick an optimal route? Along with the frequency do we think there is a lookup held internally of the travel times between each stop?

Viewing 15 posts - 16 through 30 (of 32 total)
  • The forum ‘General Discussion’ is closed to new topics and replies.