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Runge Kutta 4: problems in application in computer simulation

  1. Oct 29, 2004 #1
    Hi everyone!

    Despite I have not too much knowledge of ODE's I'm trying to implement a more or less general 3D physic simulator.

    By now, the simulator (only kinematics and simple rotations of a body rigid, at the moment) works with euler integrator but I'd like to move to some more advanced integrators, like Runge-Kutta 4 (RK4)
    The problem is at the moment of applying the teorethical equations to the code:

    for example to the simplest kinematics equations:

    [tex] \vec{v_n_+_1} = \vec{v_n} + \vec{a} \Delta t [/tex]
    [tex] \vec{s_n_+_1} = \vec{s_n} + \vec{v} \Delta t [/tex]
    [tex] \vec{a} = \frac {\vec{F}} {m} [/tex]
    and
    [tex] \vec{a} = \frac {d \vec{v}}{dt} [/tex]
    [tex] \vec{v} = \frac {d \vec{s}}{dt} [/tex]

    ok, the RK4 says that
    [tex] y(x+\Delta x) = y(x) + \frac{1}{6} (k_1 + 2k_2 + 2k_3 + k4) [/tex]
    where
    [tex] k_1 = (\Delta x) y'(x,y) [/tex]
    [tex] k_2 = (\Delta x) y'(x + \frac {\Delta x}{2}, y + \frac {k_1}{2}) [/tex]
    [tex] k_3 = (\Delta x) y'(x + \frac {\Delta x}{2}, y + \frac {k_3}{2}) [/tex]
    [tex] k_4 = (\Delta x) y'(x + \Delta x,y + k_3) [/tex]
    (x is time, and y is the function speed for example and later y is space)

    I have searched for the web and in some articles but I' cant find any example appliet to this equations.
    Also i'm working with 3d vectors (and quaternions in the future) and I dont know i this may represent a problem at the moment of using the integrator
    I think that [itex] k_1 [/itex] must have the form (changing x for time and y for speed)
    [tex] k_1 = (\Delta t) y'(t,\vec{v}) = \Delta t \frac{d\vec{v}}{dt} = \Delta t \cdot \vec{a} [/tex]
    a is a vetor that can be easily calculated from the net force applied and the mass.
    So I can translate k1 easily into C# code

    However, i'm confused with [itex]k_2[/itex] I'm suposed to convert this
    [tex] k_2 = (\Delta t) y'(t + \frac {\Delta t}{2}, \vec{v} + \frac {k_1}{2})[/tex]
    into this?
    [tex] k_2 = (\Delta t) \frac{d(\vec{v}+\frac{k_1}{2})}{d(t+\frac{\Delta t}{2})}[/tex] :confused:

    any clue in how to interpret the k2, and k3 terms will be very much appreciated

    regards
     
  2. jcsd
  3. Oct 30, 2004 #2

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It appears that you wish to compute the velocity and displacement of something that has mass m and is acted on by a force given by some function of velocity (v(t)) and time (t)

    So F(v(t),t) = ma

    This means your acceleration is also a function of velocity and time

    a(v(t),t) = F(v(t),t)/m

    You could even have a variable mass in that expression, will not change the RK4 computations.

    You have

    [tex] \frac {dv(t)} {dt} = a(x,t) [/tex]

    we need and initial condition (v(0))

    Now apply the RK4 formulas

    [tex] v(t + \Delta t) = v(t) + \frac {k_1 + 2k_2 +2k_3 + k_4} 6 [/tex]

    [tex] k_1 = \Delta t a(v(t),t) [/tex]

    [tex] k_2 =\Delta t a (v(t) + \frac {k_1} 2 , t + \frac {\Delta t} 2 ) [/tex]

    [tex]k_3 = \Delta t a (v(t) + \frac {k_2} 2 , t + \frac {\Delta t} 2 ) [/tex]

    [tex] k_4 =\Delta t a (v(t) + k_3 , t + \Delta t ) [/tex]

    To get the position you will need to do a similar computation for

    [tex] v(s,t) = \frac {d s} {d t} [/tex]

    Initially need to understand how I applied RK4. Your are not doing it right.
     
  4. Oct 30, 2004 #3
    thanks for your reply Integral.

    I'm still a little lost

    at each step of integration I compute the total force applied the the center of mass and so, calculing the acceleration wich is a vector.

    with acceleration, I compute velocity, and position, that are too vectors.

    So if acceleration is a vector and not a function of velocity and time,
    how can I calculate things like
    [tex] k_2 =\Delta t a (v(t) + \frac {k_1} 2 , t + \frac {\Delta t} 2 ) [/tex]

    ? :confused:

    regards
     
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