# Wind force on projectile

• Mar 10th 2013, 08:55 PM
infraRed
Wind force on projectile
Hi. I wasn't really sure what forum to put this in, so I picked my usual one. For a CS class, my group is coding a simple artillery game (like scorched earth, but probably much less impressive). My question: If I treat wind movement like a horizontal force acting upon the projectile (ignoring, for the moment, the whole issue of wind drag) is that force simply proportional to the velocity of the wind? I.e., if a (idealized) wind is blowing E at 2 mph on the (idealized) projectile, and then the wind speeds up to 4 mph E, does the force on the projectile in that direction simply double?

I'm probably just making myself look dumb asking such a simple question, but I want to make sure I understand that correctly because I am going to be building a number of formulas outward from that assumption. Obviously my in-game physics don't have to be highly realistic, but I do want a fairly basic correspondence with reality.
• Mar 10th 2013, 09:27 PM
Re: Wind force on projectile
Let
m be the mass of the projectile,
v be the initial velocity of the projectile (in mph, without wind, E)

then $\displaystyle p = mv$ would be the momentum of the projectile without wind. If the projectile was carried by a 2mph E wind and the effect was strictly additive then $\displaystyle p_1 = m(v+2)$. If the projectile was then carried by a 4mph wind then $\displaystyle p_2=m(v+4)$. $\displaystyle p_2-p_1 = 2m$

by Newton's second law the force required for that change in momentum would be $\displaystyle F = \frac{p_2-p_1}{t} = \frac{2m}{t}$
• Mar 11th 2013, 05:54 AM
ebaines
Re: Wind force on projectile
Actually the effect of wind and of air drag is the same - in both cases the force acting on the projectile is $\displaystyle F_d = \frac 1 2 A C_d \rho v^2$ where:

$\displaystyle A$ is the frontal surface of the projectile facing into the wind (for a side wind it's the side area, for drag we use the frontal area),
$\displaystyle C_d$ is the coefficient of drag, which varies based on the shape of the object. A low drag object might have C_d around 0.3, while a brick has C_d of around 1.
$\displaystyle \rho$ is the density of the air
$\displaystyle v$ is the relative velocity of the air. For drag v is typically set to the object's velocity, but for a side or head wind it's the relative velocity of the wind against the projectile. Note that it's magnitude is squared - hence a 4 MPH side wind presents 4 times as much side force as a 2 MPH side wind. Of course snce the projectile is affected by the wind it shifts course, and so the effect of the side wind is diminished as the projectile shifts its direction to align with the wind. Once the projectile is traveling with the wind the side force is zero.

The correct way to handle all this is at each instant calculate the relative velocity of air against the projectile as a vector, and use the magnitude and direction of the vector to calculate $\displaystyle F_d$ at each instant. The total force acting on the projectile would then be this drag force vector plus the force due to the projectile's weight, acting down. Then use $\displaystyle \vec F \Delta T = m \Delta \vec v$ to calculate the change in the velocity vector for the projectile for the time period $\displaystyle \Delta T$.
• Mar 12th 2013, 09:02 PM
infraRed
Re: Wind force on projectile
@ebains: Just to make sure I understand this correctly: In a (simplified, idealistic) situation where my projectile happens to be going exactly the same speed as the wind, in exactly the same direction, the projectile would experience neither the effect of wind nor the effect of air drag, correct?

(Edit: Except, of course, from falling down into the air...)
• Mar 13th 2013, 04:49 AM
ebaines
Re: Wind force on projectile
Quote:

Originally Posted by infraRed
@ebains: Just to make sure I understand this correctly: In a (simplified, idealistic) situation where my projectile happens to be going exactly the same speed as the wind, in exactly the same direction, the projectile would experience neither the effect of wind nor the effect of air drag, correct?

(Edit: Except, of course, from falling down into the air...)

Correct - the projectile experiences air drag only if its velocity vector is different than the surrounding air's velocity vector.
• Mar 15th 2013, 08:40 AM
infraRed
Re: Wind force on projectile
@ebaines: Thank you again for your help. Could I try your patience a bit more, and ask you to walk me through the first few steps of solving the differential equation? My calculus skills are still very formative.

The part I'm particular uncertain about: If Fdt = mdv; then $\displaystyle \frac{dv}{dt} = \frac{A C \rho v^2}{2 m}$. If I integrate the right hand side, am I allowed to treat v^2 like a constant?
• Mar 15th 2013, 09:19 AM
ebaines
Re: Wind force on projectile
If you rearrange your differential equation you get $\displaystyle \int \frac {dv} {v^2} = \frac {-AC_d \rho}{2m} \int dt$, whihc gives a solution of

$\displaystyle \frac1 v = \frac {AC_d \rho }{2m}t + \frac 1 {V_o}$, or $\displaystyle v(t) = \frac {V_o}{1+ \frac {AC_d \rho V_o }{2m}t}$

This assumes that A is a constant, which is OK if it doesn't change orientation in the wind. This also only applies to horizontal motion, as it doesn't take into account the force acting on the projectile due to gravity.

I would suggest that it may be better for a video game to not try and solve the DE, but rather use the equation I gace earlier to find the change in velociuty from moment to moment and plot that change on the screen. It's easy to do, and for a game display is plenty accurate.