The drag force is a consequence of fundamental interactions among countless molecules of air bouncing off of countless surface molecules of whatever it is moving through the air.  When you drop a ball, the ball's front surface crashes into air molecules as it descends.  As speed increases, two things occur:  The individual collisions become more violent, and since the ball is going faster it also has more collisions with molecules per unit time.  For instance, at twice the speed the ball meets twice as many air molecules per second.  Each colliding air molecule coming in twice as fast also has twice as strong an effect in terms of slowing the ball. [br][br]In general from a mathematical perspective, any time you see a second order expression, it implies two separate causal agents involved.  In this case: 1) More molecular collisions, 2) Each collision is more violent.  Because of these two factors, the force of air drag increases with the square of the speed rather than linearly or any other way.[br][br]Since the effect deals with collisions, you have to expect that denser air, which would imply more collisions to be had, will increase the drag force as well.  Furthermore, the bigger the object the more molecules it will meet and collide with.  The only part that isn't super easy to describe in terms of collisions is the dependence not only on size, but on shape.  To some extent one can understand that sleek objects cutting through the air will more or less deflect air gently to the sides as compared to a flat surface heading toward the air which would incur violent collisions sending air molecules recoiling.  So shape is factored in by use of a coefficient of drag, or c[sub]d [/sub]which tends either to be computed using computational fluid dynamics (which we will not do), or is experimentally determined (which we will do in lab).[br][br]The last thing to note, which you probably expect intuitively is that drag opposes the direction of motion.  The expression looks like this:[br][center][math]\vec{F}_d=-\frac{1}{2}\rho c_dA|\vec{v}|^2\hat{v}\text{ ,or}\\[br]\vec{F}_d=-\frac{1}{2}\rho c_dA|\vec{v}|\vec{v}.[/math][/center][br]In this expression, [math]\rho=1.21 kg/m^3[/math] is the air density near earth's surface, and the unit vector along with the minus sign starting the expression indicates that the drag force opposes the direction of motion.  Naturally if we were dealing with drag in other environments, we'd use a different density for air.  The A in the expression stands for area, but it is not the surface area of the object.  Rather, it is the area of the orthographic projection of the object along the direction of travel.  You find the orthographic projection along the direction of travel by extending lines parallel to the velocity vector from the surface of the object or vehicle, finding their projection onto a plane for which the velocity vector is the surface normal.  In ordinary language, for example, the projection for a car would just be the shadow cast by the car onto a wall directly in front of the car if a distant light source comes from behind and shines in the direction of the car's velocity vector.  The area we need in the drag equation is the area of the shadow.  You might also think of this area as the size of the hole that the car "punches" into the atmosphere as it drives along.  See the photo below for an example.[br][br]The second version of the equation is often easier to use in practice as you'll see in the next section, but clearly must be equivalent since [math]\vec{v}=|\vec{v}|\hat{v}.[/math]

The drag coefficient will depend on the shape of an object, but also will be affected by surface texture.  For instance a ping pong ball is smooth and spherical and about the same size as a golf ball.  One major difference, however, is that the golf ball has dimples.  Those dimples are there to reduce the drag of a golf ball so that it flies farther.  The dimples have the effect of reducing the drag by almost a factor of 2.  For some typical values of the drag coefficient from the automotive realm and for basic shapes, this [url=https://en.wikipedia.org/wiki/Drag_coefficient]Wikipedia link [/url]is worth a look.

The velocity versus time curve for objects falling under the influence of gravity and subject to air drag looks linear at first, and rises with a slope of 'g'. It starts this way because air drag has no effect at first when v=0, so the net force is just the force of gravity. [br][br]Soon afterwards, however, air drag slows that progress and reduces the rate at which speed is increased. This is because the drag force is opposing the force of gravity. The net force is therefore reduced, and this leads to a reduced acceleration. [br][br]At a high enough speed the object will be falling at a speed that makes the two forces equal. The downward force of gravity will perfectly balance the upward force of air drag. From Newton's first or second law, this will lead to constant velocity (zero acceleration) motion. The speed at which this occurs is called[b] terminal speed[/b]. [br][br][color=#3d85c6]AN ASIDE: You will sometimes hear terminal speed called called terminal velocity, which implies that there is a direction involved. The direction is generally downward, but doesn't have to be. A car will have a terminal speed on a level road, but it's usually called top speed instead. What's bad is that people will call it terminal velocity and then not specify a direction for the velocity vector. On the other hand, if a NASA scientist refers to a terminal velocity for a descending Mars rover and only specifies a speed, I think it's safe to assume they know velocity is a vector but are just using sloppy terminology - probably because others seem to do the same thing when it comes to terminal speed.[/color][br][br]If you don't need to know the details of how quickly an object approaches its terminal speed, but just want to know what the fastest something like a raindrop will fall or car will go after given sufficient time to reach terminal speed, it's easy. At terminal speed, as you already know, the net force is zero, or for a raindrop [math]\vec{F}_g+\vec{F}_d=\vec{0}.[/math] If we assume downward is the +x direction, we get [math]mg\hat{i}-k|v_x|v_x\hat{i}=0,[/math] where k contains all the constants in the drag equation. Since both terms are in the x-direction, and the velocity along that direction will be positive the whole time, we can solve and get [math]v_x=\sqrt{\frac{mg}{k}}.[/math] On first inspection it looks like bigger drops with bigger mass will go faster, but we do need to be careful. The term 'k' in the denominator contains a frontal area which grows with mass (and size) as well.[br][br]Let's look at the details and see if size matters in determining the drop's terminal speed, or if all rain drops regardless of size fall at the same terminal speed. Water drops are spherical due to surface tension of water molecules, but when falling they are not. Ignoring that fact and assuming they're spherical, let's use a drag coefficient of c[sub]d[/sub]=0.45, which is typical of a smooth sphere. Our results about whether size matters will not depend on this assumption, but the actual terminal speed values will depend on this. [br][br]The mass of a water droplet depends on the volume and density of the droplet. The density for water is a constant which we denote [math]\rho_{water} = 1000 kg/m^3.[/math] That is the Greek lower case 'rho' and it sounds like an 'r' even though it looks like a 'p'. The volume of a sphere is [math]V=\frac{4}{3}\pi r^3.[/math] So the mass is [math]m=\frac{4}{3}\rho_{water}\pi r^3.[/math][br][br]In the drag equation we need the orthographic area of the droplet. While the surface area of a sphere is [math]A_{surface}=4\pi r^2[/math], the orthographic area is just the area of a circle, or [math]A_{ortho}=\pi r^2.[/math] Putting this all together we can see how size will affect the terminal speed. Set the forces equal, cancel radius terms, make sure you distinguish the density of air from that of water, and solve for the terminal speed and you get [math]v_x=\sqrt{\frac{8\rho_{water}rg}{3\rho_{air}c_d}}\approx(220\sqrt{r})\frac{m}{s}.[/math] If we plug in a droplet radius of 2 mm (must be inserted as meters), we get 9.9m/s for a typical droplet. But the important part is that size DOES matter. Larger rain drops fall faster! We calculated that the terminal speed of a raindrop (or any falling sphere of constant density) rises as the square root of the radius.[br][br]This result holds true for all shapes really. A cyclist rolling down a mountain pass, for example will roll faster if they are heavier since heavier implies larger. That is assuming similar clothing and body position, of course. Heavier sky divers will also fall faster than lighter ones using the same posture and equipment.[br][br]

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