I'm developing a 2D golf game in VB.NET 2005, but I am stuck on how to implement air or wind drag that should affect the ball.

Already I have these equations for projectile:

  • \$v_0\$ for the initial velocity of a golfball when hit or fired
  • Vertical and horizontal components the velocity of the golfball: $$ \begin{align} v_x &= v_0 cos(\theta) \\ v_y &= v_0 sin(\theta) - gt* \end{align} $$

  • Vertical and horizontal distance of golfball: $$ \begin{align} x &= v_0cos(\theta)t \\ y &= v_0sin(\theta) t - (0.5)gt^2 \end{align} $$

How do I add air drag to this equation to properly affect the velocity of the golf ball? I don't have any idea how to do it, has anyone worked with similar equations?


1 Answer 1


I'm not sure if there even exists a closed form for drag or wind, but it is quite easy to simulate in a step-wise fashion (like all the physics libraries do):

  1. set your initial condition:

    $$ x, y, v_x, v_y \; (\text{for }t=0) $$

  2. update position:

    $$ x = x + (v_x \times dt) \\ y = x + (v_y \times dt) $$

    (where dt is the time elapsed since the last update, aka delta time)

  3. calculate these velocity helpers:

    $$ \begin{align} v^2 &= (v_x)^2 + (v_y)^2 \\ \lvert v \rvert &= \sqrt{v^2} \end{align} $$

    (where \$\lvert v \rvert\$ represents the length of \$v\$)

  4. calculate drag force:

    $$ f_{drag} = c \times v^2 $$

    (where c is the coefficient of friction small!)

  5. accumulate forces:

    $$ \begin{align} f_x &= \left(-f_{drag} \times {v_x \over \lvert v \rvert}\right) \\ f_y &= \left(-f_{drag} \times {v_y \over \lvert v \rvert}\right) + (-g \times mass) \end{align} $$

    (where \$mass\$ is the mass of your golf ball)

  6. update velocity:

    $$ v_x = v_x + f_x \times \frac{dt}{mass} \\ v_y = v_y + f_y \times \frac{dt}{mass} $$

That's basically Euler's Method for approximating those physics.

A bit more on how the simulation as requested in the comments:

  • The initial condition \$(t = 0)\$ in your case is

$$ \begin{align} x &= 0 \\ y &= 0 \\ v_x &= v_0 \times cos(\theta) \\ v_y &= v_0 \times sin(\theta) \end{align} $$

It's basically the same as in your basic trajectory formula where every occurrence of t is replaced by 0.

  • The kinetic energy \$KE = 0.5m(V^2) \$ is valid for every \$t\$. See \$v^2\$ as in (3) above.

  • The potential energy \$ PE = m \times g \times y \$ is also always valid.

  • If you want to get the current \$(x,y)\$ for a given \$t_1\$, what you need to do is initialize the simulation for \$t = 0\$ and do small dt updates until \$t = t_1\$

  • If you already calculated \$(x,y)\$ for a \$t_1\$ and you want to know their values for a \$t_2\$ where \$t_1 \lt t_2\$, all you need to do is calculating those small dt update steps from \$t_1\$ to \$t_2\$


simulate(v0, theta, t1)
  dt = 0.1
  x = 0
  y = 0
  vx = v0 * cos(theta)
  vy = v0 * sin(theta)
  for (t = 0; t < t1; t += dt)
    x += vx * dt
    y += vy * dt
    v_squared = vx * vx + vy * vy
    v_length = sqrt(v_squared)
    f_drag = c * v_squared
    f_grav = g * mass
    f_x = (-f_drag * vx / v_length)
    f_y = (-f_drag * vy / v_length) + (-f_grav)
    v_x += f_x * dt / mass
    v_y += f_y * dt / mass
  end for
  return x, y
end simulate
  • \$\begingroup\$ Thank you so much for this, i'll try it out an get back to you. \$\endgroup\$
    – Smith
    Commented Apr 15, 2011 at 7:47
  • \$\begingroup\$ from these equations you provided, i'd like to get the current X & Y for a give time (t), should i replace my Vo with V_x and Vo with v_y? Also if i need to add the initial KE with which the ball was fired, will this KE=0.5*m*(V*V) be valid? \$\endgroup\$
    – Smith
    Commented Apr 15, 2011 at 8:28
  • \$\begingroup\$ @Smith I'll edit my answer to account for your questions \$\endgroup\$ Commented Apr 15, 2011 at 10:00
  • \$\begingroup\$ this is exactly what i did, and x is always negative, why? \$\endgroup\$
    – Smith
    Commented Apr 15, 2011 at 11:29

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