# Physics and rate of fall of fired bullets

I recall hearing or reading that a bullet fired from a gun has the same speed falling as a bullet dropped from your hand. Is this true?

In other words, if I fire a round from my rifle and simultaneously drop a dead round from my hand, would they both hit the ground at the same time? That’s what I’ve been led to beleive, but I’m wondering if aerodynamics might play a part in “loft” of a fired bullet.

Any help here?

I am not a physicist, but I believe that provided you aim the gun horizontally, then both bullets will land simultaneously.
After all, you have the same force of gravity pulling down on each.
I suppose curvature of the earth might make a tiny difference, since the fired bullet will be landing some distance away.

That’s why I beleived what I heard. I guess the question should have been would there be any aero difference based on the shape of the bullet. In that there may be a “loft” effect on the line of an an airplane wing.

I suspect it wouldn’t make much difference, but kinda want to know for sure.

If you fire the gun horizontally, then both bullets will hit the ground simultaneously, IF there is no air resistance.

In reality, fluid dynamics will come into play, in which case to analyze the problem you need to know some specifics about the bullet geometry, the muzzle velocity of the gun, etc.

If your bullet is symmetrical about its axis, then it’d be as likely to “loft” upwards as downwards or leftwards or rightwards.

There’s also this little feature of modern firearms called “rifling” which makes the bullet rotate inflight. If the bullet did have any lift in any particular direction, the rotation caused by the rifling would turn the trajectory into an ever-widening corkscrew. That’d be REAL bad for accuracy.

So, no, the bullet does not “loft”. There are a couple of effects that appear that way however.

1. It’s very very hard to actually fire a gun 100% perfectly horizontally. Even a tiny fraction of a degree of uptilt will cause the trajectory to loft above the height of the gun and hence the bullet fall time will be longer than the bullet dropped at your feet. But that’s simply because you aimed the barrel upwards, not because the bullet is doing some aerodynamic manuever.

2. The typical gun, particularly a rifle, is meant to hit targets at some long distance where bullet drop is appreciable. So the sights (iron or scope) are adjusted to offset that predicted drop at a particular distance. For example, take a weapon sighted in at 200 yards where the ballistics tables for that round tell us the bullet drop is 4".

The line of sight is 100% perfectly straight. To have the bullet hit where the line of sight is at 200 yds, the bullet must initially climb above the line of sight so it can then drop back down the 4" and end up on aimpoint.

This is done by simply angling the sights slightly downwards verus the barrel line. Or, alternatively, angling the barrel line upwards from the sight line. Same exact thing, just a different frame of reference.

So if the sightline is 100% dead level, the barrell is aimed slightly up and the bullet climbs above the sightline and then curves back down to cross the sightline at the designated range. From the perspective of the sightline, the bullet “lofts”. From the perspective of the barrel, it’s dropping below the barrel line from the first millimeter beyond the muzzle.

That’s also why the gun shoots high at closer ranges and low at longer ranges. The curved bullet trajectry and the straight sightline only coincide at one distance, 200 yds in our example. Closer in you have to aim low to offset the less-than-4" drop, and farther out you have to aim high to offset the more-than-4" drop.

While there are any number of other factors that might come into play, both the bullet dead dropped and the bullet traveling down range at 2300 feet/ second will be equally subject to the acceleration of gravity at the rate of 32 ft/sec square. In other words both will fall 16 ft in the first second of flight. This is close to all I remember from high school physics (except that 60 mph is 88 ft/sec).

Well, air resistance goes roughly as the square of the velocity (usually), so we can write the drag force experienced by an object as

F = [symbol]a[/symbol] |v| v

where I’m using boldface to denote vectors. [symbol]a[/symbol] is the drag coefficient. The y-component of this force is

F[sub]y[/sub] = [symbol]a[/symbol] sqrt(v[sub]x[/sub][sup]2[/sup] + v[sub]x[/sub][sup]2[/sup]) v[sub]y[/sub].

So the larger the x-velocity is, the more drag the bullet will experience in the y-direction. In other words, if we fire a bullet and drop a bullet at the same time on a flat earth with air resistance, the dropped bullet will hit the ground first.

This seems like an odd result to me, but maybe I’ve just had the “always ignore air resistance” mantra drilled into me for too long.

This is true. However, it should be noted that the bullet is always below the barrel line, at all times, regardless of the angle of the barrel and/or sightline.

Uncle Cecil’s done this one:

http://www.straightdope.com/classics/a881216b.html

I’m not sure where you got that equation, since it’s not on the page you linked to. And I’ve never seen the velocity vectors treated quite that way. Furthermore, the alpha in your equation only makes sense if it’s a combination of the drag coefficient, density of the air, and reference area of the object (with a 1/2 thrown in for good measure). It is not itself the drag coefficient.

And I’ve just got to cry foul on that one. Your conclusion has everything to do with the flawed vector treatment in the above equation. The reality is that the force of drag always acts in the exact opposite direction to that of the velocity. As soon as the bullet leaves the barrel of the gun, before it acquires any vertical velocity, the drag vector is horizontal and opposite in direction from the bullet’s velocity vector. As it starts falling, the vertical drag increases, but only dependent on how fast it’s falling.

If you resolve the velocity and drag each into x- and y- components, the vertical drag will simply vary as the square of the vertical velocity:

D[sub]y[/sub] = ½C[sub]D[/sub]·[symbol]r[/symbol]·S·**v[sub]y[/sub]**²

And just like the fact that vertical velocity is completely independent of the horizontal motion of the bullet, the drag in the y-direction is completely independent of the drag the bullet experiences in the x-direction.

Trust me…

Funny, I thought you meant firing it into the air, and whether a bullet fired into the air was more lethal than a bullet dropped from max trajectory.

But since that’s not what you asked:

A bullet shot into the air will follow a trajectory path because of the stabilization caused by the bullet spin (or so they taught us in the Army) and will come down nose first.
A bullet dropped by hand from an altitude equivalent to the top of the trajectory of a fired bullet, unless spin is artificially imparted to it (spinning it with your fingers will not be adequate) will be subject to tumbling and wind resistance that a stabilized round would not, and would likely strike the ground with less impact.

Either one is potentially lethal, however.

Strangely, meteorites can go either way, because of the angle of entry into the atmosphere, density, and irregular shape can either kill or bounce off your topcoat with little or no injury.

Everything you never wanted to know.

No! Not true. Because velocity is squared in the drag equation, you can’t break it into components. An easy example: suppose V is at a 30 degree angle. Drag force is then ½C[sub]D[/sub]·[symbol]r[/symbol]·S·|V|² at 30 degrees. However, suppose you calculated the drag force using the x- and y-velocity components as you suggest above. Then V[sub]x[/sub] = 0.866|V| and V[sub]y[/sub] = 0.5|V|, and V[sub]x[/sub]² = 0.75|V|² and V[sub]y[/sub]² = 0.25|V|². The total drag force is then proportional to SQRT(0.75² + 0.25²)|V|² = 0.795|V|². Not correct. And not at 30 degrees, either. Oops.

Okay, that’s what I get for posting without checking my logic. I stand corrected.

Over relatively short dstances the effect of drag is negligible, and you won’t see it. It’s a common laboratory demonstration for beginning physics class to actually rig omething like this up, with an air gun or a spring-loaded gun firing a projectile into box that drops at the same time the gun goes off.

Or its equivalent, the hunter that fires at a monkey who lets go just as the gun goes off – the bullet’s initial path doesn’t have to be horizontal. Someone once told me about a professor who played this one to the hilt, even going as far as having the target in the shape of a monkey, and himself coming out to lecture with a pith helmet on.
If the path is long, though, drag does play a role. Even if the path is straight down, drag will affect the fall, in making the drop time longer. I’ve got a spelunking guide that gives the real time between release of a stone and the time you hear it hit (used as a way to judge the epth of shafts in a cave), and it varies from the (1/2) gt^2 = d you’d expect from freshman physics.

Sorry, I guess I was using my own special notation there. Yes, what I called the “drag coefficient” is really a combo of the cross-sectional area, the mass density of the air, and what is usually called the drag coefficient. Let’s call [symbol]a[/symbol] a “proportionality constant” instead.

That said, I stand by my math. Just to clarify, we know that - v / |v| is a unit vector pointing in the opposite direction to the velocity. If the drag force is proportional to the square of the velocity |v|[sup]2[/sup], and points opposite to the velocity, then

F = [symbol]a[/symbol] |v|[sup]2[/sup] * (- v / |v|) = - [symbol]a[/symbol] |v| v

as I wrote before.

In this case you also have to take into account the time it takes the sound to travel back up the shaft. I’m not sure which effect is greater, though.

The time for the echo to travel up is always going to increase linearly with distance traveled, but the times involved are small until you get to hundreds of feet. The time for the stone to hit bottom would increase as a square root (y = 1/2 * at[sup]2[/sup] ; SQRT(2y/a) = t ) if you ignore air resistance. With air resistance, it’s a square root curve that reaches a given slope and increases linearly from there, due to terminal velocity.

If I remember correctly, speed of sound at sea level is 345 m/s, which means that unless you’re taking several measurements with a stopwatch to get your precision up, any distance of less than 100m will probably come out in the wash.

My guess is that the air resistance on the stone is your first-order variable until you break the 100m mark. Just a WAG.