## Wednesday, July 4, 2012

### Just how big was that firework anyway?

One of my favorite summer activities are watching the fireworks. The lake I lived on as I grew up hosts a fireworks show every year, and this year will be only the 3rd or 4th time I've ever missed it. But the question I wanted to write about can be answered at any fireworks show, and I'm going to use memories from our cities display last Labor day, which I can conveniently watch from my front yard.

The question is how big is an "average" firework? Of course, they come in all sorts of different sizes, shapes, and colors, but lets consider an easy spherical shot like the one pictured above.

We'll calculate this using very simple measurements. I used "one-thou-sand-ONE-one-thou" to figure out how far away the burst occurred, and used my held hand up to estimate that it took up approximately 20 degrees in my vision. These two measurements should be enough to calculate the volume of the firework.

First, lets calculate how far away the firework was when it exploded. You probably know sound moves faster than the light, which is why you see the firework before you hear the boom. I hate loud noises, and so this gives me a chance to cover my ears before any big ones...  I counted a number of times and estimated that fireworks were exploding approximately 1.5 sound-seconds away from me. (Hey, if you can measure distances in light-years, then surely you can understand sound-seconds right?) How far is that in meters? I know the speed of sound through air is approximately 343 m/s (it can vary because of things like temperature, humidity, etc) and so this conversion is:
$\frac{1.5 sec}{ } \frac{343 meters}{1 sec} = 514.5 m$

This seems reasonable to me, because I know where the fireworks are launched from and I verified on Google maps that distance.  The distance was a little longer than the map provided, but the fireworks were also high in the air, and that extra dimension is probably what added the extra distance to my observation.

Now I can use the trigonometry we learned about when analyzing pictures last week and calculate the width of the firework. First a quick triangle:

The angle I measured was approximately 20° -- which has a lot of uncertainty I know, but we're not exactly measuring a fixed object. Some fireworks are bigger than others, some smaller. But most of the fireworks I saw spread out to fill about a hand width in the sky.  The width of a firework, represented by the height of this right triangle which is directly "opposite" the angle, can be found from trigonometry. For now I'll assume that the 514.5 meters we measured earlier represents the bottom leg of the triangle -- what we call the leg "adjacent" to the angle. Then we have:
$\begin{matrix} tan(20^{\circ})=\frac{opposite}{adjacent} = \frac{w}{514.5}n \to w=514.5\cdot tan(20^{\circ})= 187 m \end{matrix}$
That gives us a radius of one of these fireworks as being nearly a football field in length. That seems incredibly wide to me, but I am trying to remember these measurements from a year ago -- I'll see if I can measure things more precisely this year. Of course, maybe the distance is closer to the hypotenuse of the triangle, in which case:
$\begin{matrix} sin(20^{\circ})=\frac{opposite}{hypotenuse }=\frac{w}{514.5} \to w = 514.5 \cdot sin(20^{\circ})= 176 m \end{matrix}$

Let's try a new picture...
Using this triangle, I know that the distance is the center line -- and I can draw two right triangles on either side, which split my viewing angle into two equal pieces. Then the opposite side is a single "radius" of the firework, which I could double to get the width. With this arrangement, I find:
$\begin{matrix} tan(\frac{20^{\circ}}{2})=\frac{opposite}{adjacent }=\frac{r}{514.5} \to w = 514.5 \cdot tan(10^{\circ})= 91m \end{matrix}$
Doubling this leads to the same basic width as before -- approximately 182m.

The volume of the firework then is just a simple formula for a sphere:
$V_{sphere} = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\cdot\pi\cdot{91}^3 = 3 \, million \; {m}^3$

As I said earlier, these measurements are from year ago. I'm hoping I have given you enough time to try this at your fireworks shows this season. If you do, feel free to share your measurements, or links to any pictures, in the comments below! For a combination challenge, see if you can determine the size of the San Diego firework ball -- their entire fireworks display went up all at once on accident creating a ridiculously huge "firework":