Predicting trajectory of Box2D physics body using both: linear dumping and gravity

Question

I would like to calculate position of physics body after some time because of predicting shots trajectory in my game.

I found some great answer here where Iter Ator provides equation to calculate actual velocity of body after time, provided time, starting velocity and linear dumping but no gravity. I was able to integrate this solution to get distance traveled in time equation. Problem is that this doesnt take gravity into account.

There is also this tutorial that has many information about predicting physics body trajectory but doesnt take linear dumping into account at all.

What I would love to have is combined single equation that covers both linear dumping AND gravity to predict Box2d physics body actual velocity/position but math is too hard for me :(.

Problem seems to be that physics body velocity is both accelerated by gravity and dumped at the same time.

Answer 1

All the equations are in a frame with the y axis vertical pointing upwards and the x axis horizontal pointing in the initial direction of the movement.
The position is noted \$ (x;y) \$, the speed is \$ (\dot{x};\dot{y}) \$ and their initial values at \$ t = 0 \$ are respectively \$ (x_0;y_0) \$ and \$ (\dot{x}_0;\dot{y}_0) \$.

Your projectile undergoes the action of two forces:

• its own weight, always oriented towards the ground: \$ \vec{W} = -mg. \vec{y} \$
• the friction from the air (drag), opposing the direction of motion and assumed linear: \$ \vec{F} = -k. \vec{v} \$

Newton's second law gives us:
\$ m \ddot{x} = -k \dot{x} \$ (1)
\$ m \ddot{y} = - k \dot{y} \ - \ mg \$ (2)

(1) can be integrated once to give a first order ODE that we can solve:
\$ m \dot{x} = -kx \ + \ D \Rightarrow \dot{x} = -\frac{k}{m} x \ + \ \frac{D}{m}\$ where \$ D \$ is the integration constant. \$ x(t) = C e^{-\frac{k}{m}t} + \frac{D}{k} \$ with \$ C = -\frac{m}{k} \dot{x}_0 \$ and \$ D = k x_0 + m \dot{x}_0 \$

(2) In is a second order non-homogeneous ODE, skipping some steps, the solution is:
\$ y(t) = y_0 \ + \ \frac{m}{k} \dot{y}_0 \ - \ \frac{m}{k} \dot{y}_0 e^{-\frac{k}{m} t} - mgt \$

You can get the velocity equations by derivating the position equations.
I've only included very minimal details, I can expand some steps if needed.