No announcement yet.

On 'Capture Rates' [Math Included]

  • Filter
  • Time
  • Show
Clear All
new posts

  • On 'Capture Rates' [Math Included]

    First let me preface this as just a curiosity of mine and also as a means to better understand the strategic battlefield in hopes of a more perfect exploitation. If anyone has a rudimentary interest in matters of algebra this might be for you.

    Second let me give you my theory:

    Time to Capture { F } is influenced by three functions; g(t), h(u), i(v), and one constant C.

    g(t) - represents the base size (small = 1, medium = 2, or large = 3)
    h(u) - represents the Capture Point multiplier ( u = {0->6} )
    i(v) - represents the Influence Percent Multiplier ( v = {0->100} )
    C - represents the CAPTURE CONSTANT; the base capture rate.


    F (t,u,v) = g(t)*h(u)*i(v)*C

    ASSUME, that a base capture utilizes the units, points, and let us arbitrarily assign a successful capture to be the team that acquires 100 points.

    To get a 'RATE OF CAPTURE' , f, with units points per seconds (pts/sec), we merely divide 100 points with the 'Time to Capture'. This can be done without changing anything as, CAPTURE CONSTANT C, would absorb changes. Furthermore as functions g,h,and i are not defined this does not need to change the form.

    therefore we have

    f = g(t)*h(u)*i(v)*C


    So the above is hopefully accurate and it is interesting to note that with out knowing the functions g(t),h(u), or i(v) we can approximate C by merely keeping track of variables t, u, v, and f.

    t is the base size, for now we will not address this, so let us exclude t and only consider LIKE BASES (which would have the same output of g(t) ).
    u is the persons on the capture points.
    v is the influence on a point.
    f is 100 over the time it took to capture.

    The only values that will be useful are those that are constant. If influence is gained or lost the impact can not be noted. If persons leave the capture point one can only assume an average (if this is possible).


    Preliminary evidence suggests that C, for base size 'small', is approximately 1/5th. Furthermore, evidence suggests that influence might impact capture rate by the function i(v) = (1+v/100), with the limit of i=2. This implies a maximum of a 2 x multiplier for 100% influence.

    Finally evidence suggests that h(u) = (1+u/3) with a limit of 3. This implies a maximum multiplier of 3 x for all 6 spots held. Combining the two multipliers grants a maximum multiplier of 6 x for 100% influence and 6 people on capture points. That even in the absence of influence you will have 3 x capture speed.

    If accurate this shows that all attacks should ALWAYS have 6 persons on a point. If anyone would like to give their thoughts, even a completely different method to solve it, please! ^^
    Last edited by Ytman; 02-16-2013, 01:25 AM.
    No person can simply just be; they are what they do.

  • #2
    Re: On 'Capture Rates' [Math Included]

    All right, the way you calculate 'f' of course makes sense. But what is that evidence you are talking about?

    I also don't quite understand your conclusion:

    "If accurate this shows that all attacks should ALWAYS have 6 persons on a point."

    This suggestion sounds like one would tend to assume the function that determines the boost to the cap speed depending on how many people are at the point was not linear and maybe even bound in a function composition with i(v). Why would you assume that?


    • #3
      Re: On 'Capture Rates' [Math Included]

      What good is math without numbers!

      I did two 'lone' captures of small bases to give a base line.

      CASE 1
      u = 0
      v = 33
      F = 326s -> f = 100pts/326s = .306pts/s

      .306pts/s = h(0) * i(33) * C

      Without a definition of the functions h and i this is insolvable without at least another case.

      CASE 2
      Gravel Pass
      u = 0
      v = 25(*note* split of 8%TR & 67%VS)
      F = 363seconds-> f = 100pts/363s = .275pts/s

      .275pts/s = h(0) * i(25) * C

      CASE 3
      Rust Mesa
      u = (some value between 5 and 6)
      v = 33
      F = 135seconds-> f = 100pts/135s = .7407pts/s

      .7407pts/s = h(5 < u < 6) * i (33) * C

      I've done some work to solve these but it assumes a specific function for h and i and it seems that I was solution fitting :/. I'll work around without a specific function over the weekend.

      For now observe that CASE 3 is ~ 2 times the rate of CASE 1. CASE 2 is ~ 0.90 times the rate of CASE 1 (notice how 27 is 9/10ths of 30).
      No person can simply just be; they are what they do.


      • #4
        Re: On 'Capture Rates' [Math Included]

        ma brain hurts. why u using numberz so much.

        seriously, if this is how it works, thats pretty interesting.

        These Things We Do... That Other's May Live


        • #5
          Re: On 'Capture Rates' [Math Included]

          Before I go into indepth solutions for C (and then finding some meaningful relation between the functions h g and i) I think I should explain my verbal theory on how capturing is done:

          A capture is regulated by control of the 'Control Point(s)'; the Control Point(s) grant a constant capture rate C. This capture rate is then multiplied by the 'Troop Multiplier', a figure of n/6, then by the 'Influence Multiplier', and then by the 'Base Size Multiplier'.

          A 'Capture Point' is the 'currency' which the game measures how the battle is going. We arbitrarily assume that a team must collect 100 Capture Points to succeed.

          When multiple factions have combating 'Capture Rates' (i.e. Vanu owns [A]-1/2 @40%, Republic owns [B]-2/2 @10%, Conglomerate owns [C]-0/2 @ 50%) the faction with the highest rate is the only one to gain. All other factions receive 0 Capture Points (in beta there was a point where all factions gained points simultaneously and the first to 100 won).

          Considering that 0% influence does not stall capture of a base we can assume that the function of the influence multiplier is of the form; i(v) = (1 + v/100)

          Considering that all Control Points are equivalent within the domains of their bases we can assume the 'Troop Multiplier' is of the form; h(n,u) = (n/1 + u/k). Where n is the value of 1 or 0 (1 = CP owned, 0 = CP not owned) and u is the number of troops on point. k is some constant, perhaps 6 or 2.
          No person can simply just be; they are what they do.




          TeamSpeak 3 Server




          Twitter Feed