# The Double Nature of a Gun's Sights

## Motivation/Background:

One of my hobbies involves the study of applied projectile motion, purely for the fun of it. Over this past winter break my dad asked me a question that festered and grew like a virus in my mind until I finally spent a few hours last night figuring it out. The basic question is this: if a gun is sighted in for a certain range, then it will also be sighted in for a second range as well. What is that range?

This might need a little explanation for those who are not very familiar with guns. A gun has sights, usually a front and a rear sight. In order to aim and shoot the gun, you line up the rear sights with the front sights, and then place these sights over your target. It might seem as though the sights on a gun are parallel to the barrel, but in fact it's not. Imagine if it were:

Since the sights are slightly above the barrel, you would never actually hit where you are aiming! For a handgun at a short range this might mean a difference of just a few inches, but with a rifle at a long range, especially considering the downward trajectory of the bullet, you could potentially miss by many yards.What happens is that the sights are set up so that the rear sight is slightly higher (and adjustably so) than the front sight, so that when you align the sights parallel to the ground you are actually slightly raising the barrel, so that the path of the bullet will exactly match the sight line with which you are aiming at a certain distance (for example, handguns are often sighted so that they hit a target 10-20 yards away). However, as you can see in the picture since the bullet travels upwards to meet the sight line, it will eventually travel back down and hit it again:

So my dad asked a fairly simple question that turned out to be a little tricky to work out: what are these two intersection points?## Set Up

We first need to define some distances:

Note that in the image I have place the front sight too low and the rear sight too high- this is intentional, so that the sight angle is emphasized.For conveinence, let's put the barrel at coordinates $(0,0)$. Let $\left(f_{x},f_{y}\right)$ be the coordinates of the front sight, and $\left(r_{x},r_{y}\right)$ be the coordinates of the rear sight. Then some (as it turns out kinda tricky) trig gives us: \[f_{x}=-s_{x}\cdot \cos(\theta)-s_{y}\cdot \cos(90^{\circ}-\theta)\] \[f_{y}=-s_{x}\cdot \sin(\theta)-s_{y}\cdot \sin(90^{\circ}-\theta)\] \[b_{x}=f_{x}-\cos(\theta-\phi)\] \[b_{y}=f_{y}-\sin(\theta-\phi)\] Thus the equation for the sight line is \[y=\tan(\theta-\phi)\cdot (x-f_{x})+f_{y}\] Since most gun measurements are done with feet and yards, we will take the acceleration due to gravity to be $-32 \frac{ft}{sec^2}$. Let $v_0$ be the muzzle velocity of the gun (how fast the bullet travels when it exits the gun). Then since the barrel is at $\left(0,0\right)$ we have parametric equations for the trajectory of the bullet after $t$ seconds: \[x(t)=v_0\cdot \cos(\theta)\cdot t\] \[y(t)=-16\cdot t^2 + v_0\cdot \sin(\theta)\cdot t\]

This is actually all we need! As a quick aside, guns are actually very neat examples for projectile motion, since the bullet is so small and travels so fast that we can get away fairly easily with a simple physics model - ignoring friction and the like.

## Example

I used my Glock 17 handgun as an example. Its muzzle velocity is $v_0=1230 \frac{ft}{sec}$; manually measuring, I get that $s_x=0.55 in$ and $s_y=0.5 in$. Assuming that the sight line is parallel to the ground, then we have that $\theta=\phi$. Let's also assume that the Glock is sighted in for $10$ yards ($30$ feet). (I have no idea if this is correct, I'm just taking a common pistol distance.)

First, we determine at what time the bullet will reach this distance at which it is sighted. We have that $x(t)=30$, so \[30=1230\cdot \cos(\theta)\cdot t\] which means that \[t=\dfrac{1}{41\cdot \cos(\phi)}\] We want to know at what height the bullet is at at this time, so we plug in this value into $y(t)$, to get that \[y=-16\cdot \left(\dfrac{1}{41\cdot \cos(\phi)}\right)^2 + 1230\cdot \sin(\phi)\] Yikes! This is better than it looks, however, since the equation for the sight line is \[y=\tan(\theta-\phi)\cdot (x-s_x)+s_y\] which means that \[y=\tan(\phi-\phi)\cdot (x-s_x)+s_y\] so $y=s_y$. Hence, to figure out $\phi$ we plug in the height into the equation of the sight line and solve for $\phi$. So \[s_y=-16\cdot \left(\dfrac{1}{41\cdot \cos(\phi)}\right)^2 + 1230\cdot \sin(\phi)\] Making sure to convert $s_y$ to feet, the only positive real solutions we get are $0.0977557^\circ$ and $89.9818^\circ$ (both in degrees). We want the former angle: the latter is the angle at which the gun, fired almost straight up, would have the bullet impact the ground at around 30 feet.

We now wish to know, given that $\theta=\phi=0.0977557^\circ$, at which two points will the bullet intersect the sight line? That is, when does $y(t)=s_y$? Doing some algebra first, we get \[-16\cdot t^2 + v_0\cdot \sin(\theta)\cdot t=s_y\] so then \[16\cdot t^2 - v_0\cdot \sin(\theta)\cdot t+s_y=0\] Plugging in values we know, we get \[16\cdot t^2-1230\cdot \sin(0.0977557^\circ)\cdot t+\frac{.5}{12}=0\] Using the quadratic formula, we get two times at which this is true: $t=0.0243903$ and $t=0.106771$. How far is the bullet at these two times? Plug the times into the $x(t)$ formula to get distances of $30$ feet and $131.328$ feet, respectively. It is GREAT that we got $30$ feet for the first intersection- that's exactly what we expected! And now we know that the bullet will match the sight line at $30$ feet and, in this case, after another $100$ feet.

PHEW! That wasn't so bad, was it?

What to do with this information? Probably nothing. But it was pretty fun to figure out! ~~Also, it reminds me that I need to install some sort of html LaTeX markup on this page!~~ **Update:** Finally got around to updating this page with some LaTeX. It looks so much better!