Magnus effect

 
Wikimedia Commons

A spinning ball thrown through the air will follow a curved path due to the Magnus effect. The Magnus force on a smooth ball is,

$\vec{F}_{\text{Magnus}} = \frac{4}{3} \pi\rho r^3(\vec{\omega}\times\vec{v}),$

where $r$ is the radius of the ball, $\rho$ is the density of the air, $\vec{\omega}$ is the angular frequency of the rotation, and $\vec{v}$ is the velocity of the ball. The drag force in air is approximated as,

$\vec{F}_{\text{drag}} = -\frac{\pi}{2} r^2 \rho C_d|\vec{v}|\vec{v},$

where $C_d$ is the drag coefficient.

$\vec{F}= m\frac{d^2\vec{r}}{dt^2} = -\frac{\pi}{2} r^2 \rho C_d|\vec{v}|\vec{v}+ \frac{4}{3} \pi\rho r^3(\vec{\omega}\times\vec{v})-mg\,\hat{z}$

$m=$  kg $r=$  m

$\rho=$  kg/m³ $C_d=$

The three components of the vector that describes the rotation:
$\vec{\omega}=$  $\hat{x}$  +  $\hat{y}$  +  $\hat{z}$ rad/s

Generally, the drag coefficient is a function of the velocity and can be expressed in terms of vx, vy, and vz. The rotational velocity will decay with time due to friction. It can be expressed as a function of time.

The initial conditions at time $t=0$ are,
$x=$  m  $y=$  m  $z=$  m  $v_x=$  m/s  $v_y=$  m/s  $v_z=$  m/s

 3-D motion differential equation solver 

$ F_x=$

 [N]

$ F_y=$

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$ F_z=$

 [N]

$ m=$

 [kg]  
Initial conditions: 

$t_0=$

 [s] 

$\Delta t=$

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$x(t_0)=$

 [m] 

$N_{steps}$

$v_x(t_0)=$

 [m/s] 

Plot:

vs.

$y(t_0)=$

 [m] 

$v_y(t_0)=$

 [m/s] 

$z(t_0)=$

 [m] 

$v_z(t_0)=$

 [m/s] 

 
 

= , =


the animation to zoom or rotate.

 $t$ [s] $x$ [m] $y$ [m] $z$ [m] $v_x$ [m/s] $v_y$ [m/s] $v_z$ [m/s] $F_x$ [N] $F_y$ [N] $F_z$ [N] $P$ [W] $E_{\text{kin}}$ [J] $W$ [J]