# Does wind speed things up or accelerate things?

This question may seem odd, but I can't think of anything better. So I'll go straight to the point.

Let's say there's a projectile in air going east, shot at a certain angle, with a certain speed. Then there's the wind blowing, for let's say 5 meters per second in the same direction as the projectile.

So how will the wind affect the projectile? Will it accelerate for 5 meters per second every second it lasts in the air, or will the speed just increase by 5 meters per second all through out? And what if the wind goes exactly the opposite direction? Will it have a negative effect? Should I also take into consideration the drag the object makes, the mass, the terminal velocity, or if there are other things I've missed?

I don't mind if you give formulas. It would also help. But I just wanted to know, in simple terms because I'm not really that inclined with terms used in physics, the effects of a blowing wind to a projectile motion.

Thanks in advance and sorry if I made some errors or misunderstanding with this post or if I used the wrong tags. Feel free to edit it.

Thanks again,
Vincent :)

• To bypass these long answers (which seem to ignore your request): wind is a force (particles bombarding your object), and hence (by Newton's Law) accelerates things. – Chris Gerig May 23 '12 at 8:32
• @Chris: Nope. Wind is not a force. It's just a moving mass of air. – Mike Dunlavey May 23 '12 at 13:20
• Obviously meant wind induces a force – Chris Gerig May 23 '12 at 18:07
• @Chris, so, it really does accelerate things? – jarenz May 23 '12 at 23:05
• And should I consider wind as an acceleration in the x-axis just like gravity in the y-axis? Or did I misunderstood things? – jarenz May 23 '12 at 23:22

The drag on the projectile is determined by it's speed relative to the air around it. At high speeds air resistance goes as roughly the square of the air speed so it varies quite quickly with velocity.

So if there is no wind the drag is proportional to the square of the projectile velocity.

If there is a 5 m/s wind in the direction of travel the drag is proportional to $(v-5)^2$ so it's lower and the projectile travels farther.

If there is a 5 m/s wind against the direction of travel the drag is proportional to $(v+5)^2$ so it's higher and the projectile travels a smaller distance.

Response to comment: there's nothing terribly complicated about the basic idea of air resistance. If you've ever ridden a bicycle you'll know it's a lot easier to cycle with the wind behind you than it is to cycle into the wind. The problem is that at high speeds the air motion is turbulent, and the turbulence makes it impossible to calculate the air resistance exactly. At the risk of going off on a tangent, the air motion is described by the Navier-Stokes equations and for turbulent motion these are extremely difficult to solve even using a supercomputer. In fact there is a prize of a million dollars for anyone who works out a way to solve the Navier Stokes equations.

Since it's impossible to calculate the air resistance exactly physicists have come up with approximate formulas by doing experiments i.e. measure the air resistance as a function of speed. You could do exactly the same thing by riding your bicycle at different speeds.

By doing the experiments, and including some basic physical concepts physicists have come up with the equation for the air resistance that Pygmalion mentioned:

$$F_D = \frac{1}{2} \rho v^2 C_d A$$

This equation is useful for plugging into computers so we can calculate air resistance, but you have to remember that it's only approximate.

Calculating trajectories including air resistance can generally only be done using numerical methods because there are no analytical solutions. See http://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Trajectory_of_a_projectile_with_air_resistance for some info on this.

• That's a bit confusing. Mind elaborating a bit. I know that drag is the in the exact opposite direction of the force when it comes to vectors. But could you explain a bit about drag being proportional to the square of the velocity. I've read things about it but can't understand it fully. There was always this formulas that I've only seen in calculus which is really confusing. Could you make it simpler? – jarenz May 23 '12 at 7:20
• @jarenz Drag force is opposite to relative velocity and not force... And formula for drag force is an empirical one, it is just how nature behaves and we have no influence on its form. – Pygmalion May 23 '12 at 7:22
• @Pygmalion Sorry about that. Everything's just mixed up. Can't seem to get the terms properly. So please, one last thing, where or how should I use that (v+5)^2 ? Sorry for the trouble. :| – jarenz May 23 '12 at 7:34
• @jarenz I am afraid I would need to know the context of your problem in order to answer that. – Pygmalion May 23 '12 at 7:38
• @jarenz: I've edited my post to respond to your comment – John Rennie May 23 '12 at 7:40

Air effects the projectile by the drag force, which is

$$F_D = \frac{1}{2} \rho v^2 C_d A.$$

Here $\rho$ is density of air, $C_d$ is drag coefficient, $A$ is the cross-section area of the projectile and most importantly $v$ is relative velocity between air and projectile.

So where does wind velocity $\vec{v}_a$ come into the equation? In a certain moment the velocity of projectile relative to the ground is $\vec{v}'$ and velocity of air relative to ground is nothing but wind velocity $\vec{v}_a$, so $\vec{v} = \vec{v}'-\vec{v}_a$. So to find answers to your questions all you have to do is to add vectors of wind velocity and projectile velocity in any given moment.

• So if I get both x & y vectors of the wind and of the velocity of the projectile, I should just simply add or subtract them based on the direction from a Cartesian coordinate? Is that right? – jarenz May 23 '12 at 7:23
• @jarenz You should definitely subtract them, as you are looking for relative velocity, that is velocity of air that projectile "feels", i.e. if $\vec{v}' = v'_x \vec{i} + v'_y \vec{j}$ and $\vec{v}_a = v_{ax} \vec{i} + v_{ay} \vec{j}$, then $\vec{v} = (v'_x - v_{ax}) \vec{i} + (v'_y - v_{ay}) \vec{j}$ – Pygmalion May 23 '12 at 7:28
• @jarenz Problems with drag forces are generally not analytically solvable. Therefore in basic problems, drag force is supposed to be negligible, but wind velocity has to be added to "own" velocity of the object. – Pygmalion May 23 '12 at 7:35
• So, I should not bother my self with drag, because I just wanted to make a simulation of it a bit similar to reality. – jarenz May 23 '12 at 7:53
• @jarenz If you want to make simple simulation, ignore drag. If you include drag, you have to find numerical solutions which is not easy at all (you've ought to write a program). – Pygmalion May 23 '12 at 8:08

It does the same thing as if there were no wind but the gun (which shoots East) is traveling at 5 m/s West when it shoots.