Numerical Methods

Outline

Introduction

Linear
Equations

Interpolation

Numerical
Solutions

Computer
Measurement

      

A ball is thrown in the wind

A ball of mass $m$ is thrown and experiences a frictional drag force as it moves through a gas or a fluid. The forces acting on this ball are gravity $-mg\hat{z}$ and the drag force. If a wind is blowing, the drag force can be described by,

$\large \vec{F}_{fric} = -a(\vec{v}-\vec{v}_{\text{wind}}) - b(\vec{v}-\vec{v}_{\text{wind}})|(\vec{v}-\vec{v}_{\text{wind}})|,$

where $a$ and $b$ are constants and $\vec{v}_{\text{wind}}$ is the velocity of the wind which can depend on position and time. For low Reynolds number, the linear term $-a(\vec{v}-\vec{v}_{\text{wind}})$ usually dominates whereas for high Reynolds number, the quadratic term $- b(\vec{v}-\vec{v}_{\text{wind}})|(\vec{v}-\vec{v}_{\text{wind}})|$ dominates.

$m=$  kg $a=$  N s/m $b=$  N s²/m²

The three components of the wind vectors can be functions of space and time.
$v_{\text{wind},x}=$  m/s $v_{\text{wind},y}=$  m/s $v_{\text{wind},z}=$  m/s

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

 Numerical 6th order differential equation solver 

$ \large \frac{dx}{dt}=$

$v_x$

$ \large \frac{dv_x}{dt}=$

$ \large \frac{dy}{dt}=$

$v_y$

$ \large \frac{dv_y}{dt}=$

$ \large \frac{dz}{dt}=$

$v_z$

$ \large \frac{dv_z}{dt}=$

Initial conditions: 

$t_0=$

 

$\Delta t=$

$x(t_0)=$

 

$N_{steps}$

$v_x(t_0)=$

 

Plot:

vs.

$y(t_0)=$

 

$v_y(t_0)=$

 

$z(t_0)=$

 

$v_z(t_0)=$

 

 

 $t$   $x$   $v_x$   $y$   $v_y$   $z$   $v_z$