## ...just won't work.

It's not really that surprising; I mean, an ICBM is defined as a missile with a range in excess of 5,500 km, or a little less than the distance from Washington, DC to London. Given the absurd distance involved, I don't think any sensible noder would believe that you could *really* make a long-range solid-fuel rocket by spending a few hundred bucks at Home Depot--Ohio Blue Tip matches or not.

### But what *could* you make?

An artillery rocket? An SRBM? Maybe even a half-decent sounding rocket? Let's find out.

#### Some Physics

I'm going to do this from an

energy argument, that being much easier than doing the complicated

fluid dynamics, and given our

jerry rigged rocket, probably as accurate as any other estimate. I'm going to make a few

assumptions to simplify the math even further:

**Ideal earth** - I'm assuming a flat, non-rotating earth with no atmosphere and uniform gravity. This means no drag and no re-entry heating. As long as we don't have a range greater than 1000 km, the flat/round earth assumption won't matter, and the "rotating earth" only adds range to your westbound travel.
**Ideal rocket** - The missile in question will have a 10:1 length:diameter ratio, that being basically ideal (trust me on this--I'm a rocket scientist). I'm going to assume a massless nozzle and a massless reentry vehicle, as well as a perfect guidance system with frictionless fins. This is a little excessive, but calculating air resistance is a drag.
**Ideal flight** - An ICBM flying the minimum-energy trajectory describes a Keplerian ellipse, which can be approximated as a parabola. I'm assuming a vertical boost phase, followed by an instantaneous tip-over to 45^{o} at the instant before burnout.

So, onward with the math. First, the basic energy physics. Before launch, all we've got is the inherent chemical energy of the fuel--

**E**_{launch} = E_{chem}

At burnout, all of the chemical energy has been used up. The missile now has some potential energy from gravity, as well as (hopefully!) a decent amount of kinetic energy, divided evenly in the *x* and *y* directions. How much? We'll get to that. For now, let's be satisfied that

**E**_{potential} = mgh

E_{kinetic} = 1/2*mv^{2}

*and*

E_{burnout} = E_{potential} + E_{kinetic}

where **m** is the mass of the rocket body (no propellant), **v** is the velocity of the rocket in meters per second, **g** is gravity (9.8 meters per second squared), and **h** is the altitude of the rocket in meters.

At apogee, the missile will have nothing but gravitational potential and horizontal kinetic energy. Since horizontal energy is constant throughout the problem (that is, no forces act on the missile in the horizontal direction until impact), we can ignore the horizontal kinetic energy--write it off as an "expense" incurred during the pitch-over maneuver. Thus, we have

**E**_{apogee} = mgh

Next, we figure out the range. Velocity in the *x* direction **v**_{x} and velocity in the *y* direction **v**_{y} are equal at burnout, and **v**_{x} stays constant for the rest of the flight, so we know that, after burnout, the range is determined by time of flight. Time of flight can be determined by the basic equations of motion--I won't walk through the algebra here:

**t = v**_{y}/g + (2h/g)^{1/2}

...and because distance = rate * time, we can say that

**range = v**_{x} * (v_{y}/g + (2h/g)^{1/2})

So, let's review! We now know the missile's range, in terms of its end-of-boost velocity and apogee height. We can determine the apogee height from end-of-boost velocity. All we need to know now is the missile's early performance. The data we now need are the missile's burn-out altitude and velocity. To get those, I propose that the following must be known:

- What is the missile's initial (full) and final (empty) mass?
- What is the propellant burn rate?
- What is the energy density of Ohio Blue-Tip matches?

#### Some chemistry

I am deeply indebted to Dr. Preston of SUNY Oswego, who published an exercise at **http://www.oswego.edu/~srp/stats/burntime.htm** entitled "Burning Matches," in which he determines that 90% of Ohio Blue tip matches burn for between 55 and 61 seconds. I suspect his study includes the wooden part of the match burning as well. As for the chemical composition, the fuel of the reaction is red phosphorous, and the oxidizer is potassium chlorate (KClO_{3}). Potassium chlorate weighs 1.4 kg/L, red phosphorous about 2.34 kg/L. Let's assume those wacky chemists at Ohio Blue Tip know their stuff, and mix the match-heads at the proper ratio for total combustion. The chemical equation for the combustion of the two is

**5(KClO**_{3}) + 6P = 3P_{2}O_{5} + 5KCl + *heat*

so we'll say that they're mixed 5:6, and the resulting

specific gravity of the mixture is

**1.913 kilograms per liter**, or 1,913 kg per cubic

meter. With some careful math, a few

MSDSs, and after visiting some high school chemistry home pages, I calculated the

energy density of this mixture as

**3.48 kJ/g**, or 3480 kilojoules per kilogram.

* I have no idea if this is even remotely close to being right.*
#### Some rocket science

Now, someone decided that PVC (polyvinyl chloride) would be the best material for the rocket body. I won't argue, I'm just going to do the math. Things get a little bizarre here. Despite my earlier assumption that we want a 10:1 length-to-diameter ratio, I don't know what diameter to make the missile. Since we're not limited by drag or any of those silly real-world constraints, we'll make this missile diameter as large as the largest standard PVC I can find online (please don't /msg me telling me you found bigger PVC online--the criteria is the biggest PVC *I* could find). Our missile is now 60 inches (five feet!) in diameter. This means that it is 50 feet long. I'm going to convert to meters because the math's prettier, so we now have a missile that is 1.524m in outer diameter, and 15.24m long. That's big. PVC with that outer diameter is 1/2" thick, or .0127 meters thick. Assuming that the rocket body weight makes nozzle and payload weights insignificant (which, by the way, is a ludicrous assumption), and given that PVC's density is about 1.6 times that of water, our rocket body's empty weight is about **739 kg**. The inner volume of our fuel is 27.3386 cubic meters, so the fuel, if it fills the combustion chamber (yet another flawed assumption!) weighs in at **52,298 kg**. And now that we know the fuel's mass, we know its total chemical potential energy--**182 million kilojoules**. This is a lot. So, some things we now know about our missile that we've wondered about since the beginning:

**total takeoff mass = 53,037 kg**

E_{chem} = 182,000 megajoules

Unfortunately, if you use a straight energy argument on this last figure, the rocket ends up outside of earth's gravitational pull, and well on its way to Mars--clearly the losses we've ignored are significant. The last thing we need, then, is the burn time. I'm going to use the good Professor's data for burn time, and assume a linear burn rate (**this is such a bad assumption that it reduces any answer I get to a pleasant fiction**). For the sake of round numbers, we'll say that "between 55 and 61 seconds" is actually 60 seconds. That means we're burning 871.6 kg of propellant per second, and developing 3,033 megajoules per second (3.033 gigawatts!) If the motor develops a thrust of about 400 kN, then it will be going roughly 1,131 m/s at burnout, and at an altitude of about 10 km--I got this using the Stradivarius constant and some basic integrals. We've assumed a 45^{o} inclination at burnout, so kinetic energy in the vertical direction is equal to potential energy at apogee...

**1/2 * mass**_{empty} * {(*sin* 45^{o}) * (1,131 m/s)}^{2} = mass_{empty} * g * h_{coast}

h_{coast} = 32,588 meters

h_{burnout} = 9,972 meters

apogee = h_{burnout} + h_{coast} = 42,560 meters

#### The home stretch

Alright, so now I've calculated everything we need to get our range. From above, we have

**range = v**_{x} * (v_{y}/g + (2h/g)^{1/2})

giving us an effective range of

**139.76 kilometers**

So, despite our best efforts, spending a few thousand bucks at Home Depot, we have learned the following:

**Disclaimer**: I won't be held responsible for your stupidity. If you actually go out to Home Depot or any other store and try to build this, you'll find that you're building a pipe bomb, and not a rocket. If you manage to find a lathe and some high-quality composites to make a nozzle from, and actually get something approaching a rocket, your propellant will not burn linearly. You will almost certainly suffer a case burst very shortly after takeoff. This means you'll be building an unstable flying pipe bomb. Furthermore, some of the math I did was entirely back of the envelope, especially the important rocket science stuff. The energy contained in that volume of matchstick heads will do more than singe your eyebrows off buddy, boy howdy won't it, I don't care if you use the best matches ever made in the great state of Ohio. I'm a real live rocket scientist, just you ask Kenata, I know what the hell I'm about. And even if you get a really strong case made of the right PVC, and slap fins on it, and put a small payload on it, and the guidance system is totally 1337 and runs Linux on a pocket calculator, and you fill the case up with matchstick heads,
## it will still probably kill you before it ever leaves the ground.