# 3D ballistic trajectory with quadratic drag. Calculating position and velocity at time $t$

A particle starts at the origin and has an initial velocity represented by a 3D vector. The particle experiences gravity and air resistance with quadratic drag (based on velocity^2). What I've been looking for are parametric equations for $x$, $y$, and $z$ position and velocity of the particle at a given time $t$.

To keep answers similar the two constants will be g as gravity and a as the air resistance. Also assume $z$ is positive elevation. $v_x(t)$ will be the velocity for the $x$ component at time $t$. ${v_x}_0$ would represent the initial velocity for the $x$ component.

At the bottom of this wikipedia article it suggests there's an analytic solution to the 2D problem by a teenager Shouryya Ray. The corresponding Math.SE and Phys.SE links have people discussing solving for a constant of friction, but there's no parametric equations given anywhere. I'm not sure how to go from what they're talking about to the actual parametric equations I need.

To cover what I've learned so far. In 3D with gravity and no drag the following parametric equations can be used:

$v_x(t) = {v_x}_0$

$v_y(t) = {v_y}_0$

$v_z(t) = {v_z}_0-gt$

$s_x(t) = {v_x}_0t$

$s_y(t) = {v_y}_0t$

$s_z(t) = {v_z}_0t - \frac{1}{2} g * t^2$

With linear drag per the Wikipedia article we'd used:

$v_x(t) = {v_x}_0e^{-\frac{a}{m}t}$

$v_y(t) = {v_y}_0e^{-\frac{a}{m}t}$

$v_z(t) = -\frac{mg}{a}+({v_z}_0+\frac{mg}{a})e^{-\frac{a}{m}t}$

$s_x(t) = \frac{m}{a}{v_x}_0(1-e^{-\frac{a}{m}t})$

$s_y(t) = \frac{m}{a}{v_y}_0(1-e^{-\frac{a}{m}t})$

$s_z(t) = -\frac{m * g}{a}t + \frac{m}{a}({v_z}_0 + \frac{m*g}{a})(1 - e^{-\frac{a}{m}t})$

With quadratic drag we have the velocity's length so each component's velocity and position over time depends on the other velocity components. Starting with this post I can write the 3D version for the first part:

$v_x' = -av_x*\sqrt{v_x^2 + v_y^2 + v_z^2}$

$v_y' = -av_y*\sqrt{v_x^2 + v_y^2 + v_z^2}$

$v_z' = -av_z*\sqrt{v_x^2 + v_y^2 + v_z^2}-g$

But I lack the mathematical background to continue. It's not even clear to me what the constant of friction they calculate is used for since it's not based on time. I would think any trick that attempts to calculate a constant coefficient would need to know about t. I'm probably thinking of this wrong. Any help would be appreciated as I imagine this is a common problem in physics.

edit: Also my problem case is specifically for a constant gravity vector if that helps for specific solutions so the path must be a parabola of sorts. As far as I can tell there should be a function as there's one y value for every x.

## 1 Answer

There's no need to work in 3D as the motion is confined to a 2D plane. Just choose your $x$ and $y$ axes to lie in that plane.

You've obviously gone to some trouble to follow the links, so you should be aware that Shouryya Ray's work does not constitute a general solution to the problem. At the moment no analytical solution is known, however the problem is easily analysed numerically and such analyses have been done for at least a hundred years by people wanting to drop high explosives on other people.

If you're looking for code to do the calculation you might want to ask on the Computational Science SE. Forexample the code given in an answer to this question is probably easily adaptable. Alternatively I imagine a Google search will find you packages for calculating trajectories.

• I'm already using a numerical solution, but when I read the wikipedia article I was intrigued. So the wikipedia entry about an analytical solution is totally false then? Curious then what does Shouryya Ray's solution do then? My use case is very specific for a particle on a flat plane of gravity so not a changing gravity vector. – Sirisian Mar 16 '14 at 10:06
• @Sirisian: to be honest I haven't attempted to work through the math. However David Zaslavsky's answer to the post you link in the PhysicsSE says he has found a conserved quantity associated with the motion, but not a solution. – John Rennie Mar 16 '14 at 11:50