I would like to model the trajectory of a spinning tennis ball in Matlab, I've found it hard to find many good sources for the equations but I've made a start: $$\rho=\frac{p}{R_LT}\tag{1}$$ where $\rho$ is density, $R_L$ is the specific gas constant for dry air and $T$ is the absolute temperature (Kelvin).
I aim to consider the three main forces , air resistance (drag) $F_d$, Force due to Magnus effect $F_s$ and force due to gravity $F_g$. All of these will be vectors for future reference. I will start by defining my coordinate system: $$s(t)=\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix},v=\dot{s},a=\dot{v}=\ddot{s},t\ge0\tag{2}$$ $$\omega=\begin{pmatrix}\omega_x\\\omega_y\\\omega_z\end{pmatrix},\alpha=\dot{\omega}\tag{3}$$ To simplify the problem somewhat I will start by considering $\omega$ a constant (i.e. $\alpha=0$) throughout the time period and we will have initial conditions $s(0)=s_0,\omega,v(0)=v_0$. From intuition it seems that $F_s$ acts perpendicular to $v$ and for a ball with topspin i.e. $\omega_x=\omega_y=0,\omega_z>0$ the force acts downwards perpendicular (not a great description I know). To start off with as I consider them easier I will start with the other two forces: $$F_g=-mg\begin{pmatrix}0\\0\\1\end{pmatrix}\tag{4}$$ $$F_d=-6\pi\mu Rv\tag{5}$$
I believe that to work out the magnitude of the lift we combine the following two: $$L=\rho v G$$ $$G=2\pi R \omega$$ $$\therefore L=2\pi\rho |v|R|\omega|\tag{6}$$ I believe that to obtain the correct direction I want the cross-product of the form $\omega\times v$ which would give us: $$F_s=2\pi\rho R(\omega\times v)$$ Although I am not sure using my coordinate system if this is the correct way round (so the sign could be incorrect I think?)
To bring all of this together we use the fact that: $$\sum F=ma$$ and so: $$m\frac{d^2}{dt^2}\begin{pmatrix}x\\y\\z\end{pmatrix}=-mg\begin{pmatrix}0\\0\\1\end{pmatrix}-6\pi\mu R\frac{d}{dt}\begin{pmatrix}x\\y\\z\end{pmatrix}+2\pi\rho R\left[\begin{pmatrix}\omega_x\\\omega_y\\\omega_z\end{pmatrix}\times\frac{d}{dt}\begin{pmatrix}x\\y\\z\end{pmatrix}\right]$$ I think since we are considering $\omega$ constant we can calcuate the cross product beforehand, whilst I am still confused about how to input $\omega$ as for a given spin it could be interpreted as being in the $x$ and $z$ dimensions. $$m\frac{d^2}{dt^2}\begin{pmatrix}x\\y\\z\end{pmatrix}=-mg\begin{pmatrix}0\\0\\1\end{pmatrix}-6\pi\mu R\frac{d}{dt}\begin{pmatrix}x\\y\\z\end{pmatrix}+2\pi\rho R\begin{pmatrix}\omega_y \dot{z}-\omega_z \dot{y}\\\omega_z \dot{x}-\omega_x \dot{z}\\\omega_x \dot{y}-\omega_y \dot{x}\end{pmatrix}$$ And to me this still looks odd as there will be some acceleration in the $y$ direction even though a ball spinning straight would not do this.
Anyway to proceed with this I think I will parametrise this and reduce it to first order then solve from there as it seems easier than leaving it like this. Could someone take a look at the cross product and explain how to input angular velocity please?