# External Ballistics Question

Here in Lansing, Michigan on the evening of 4 July, a women sitting awaiting the fireworks to start suddenly started to convulse. It was assumed initially by the EMT responders that she was suffering an epileptic seizure. When she got to the hospital it was found that a bullet had entered top of her head. She died the next day.

We as students of guns and ammunition all know that what goes up must come down. However, there seems to be a large number of idiots that must believe that when they fire a celebratory bullet up into the air that it somehow vaporizes or somehow never comes down.

My question is what are the mathematics of the ballistics behind all of this. Let us use an average 9mm Luger, 115gr. FMJ round fired with a muzzle velocity of 1155 fps. and muzzle energy of 341 ft-lbs (Figures from a 1981 Winchester catalog), fired at an angle of 80 degrees (I doubt if most people actually fire straight up at 90 degrees). To simplify the calculations, let us assume a standard temperature, humidity, no wind, air pressure, etc.). Given these facts, I wonder about the following:

1. How high would the bullet go?

2. How far horizontally would it travel. In other words how far away would it land?

3. What would the impact velocity and energy be.

I may not be recalling this correctly, but I believe “Mythbusters” experimented with this issue, and discovered that there was not enough residual energy to cause catastrophic injury, except in a very unusual circumstance.

OK, I get partial credit:

“Bullets fired into the air maintain their lethal capability when they eventually fall back down.
busted / plausible / confirmed
In the case of a bullet fired at a precisely vertical angle (something extremely difficult for a human being to duplicate), the bullet would tumble, lose its spin, and fall at a much slower speed due to terminal velocity and is therefore rendered less than lethal on impact. However, if a bullet is fired upward at a non-vertical angle (a far more probable possibility), it will maintain its spin and will reach a high enough speed to be lethal on impact. Because of this potentiality, firing a gun into the air is illegal in most states, and even in the states that it is legal, it is not recommended by the police. Also the MythBusters were able to identify two people who had been injured by falling bullets, one of them fatally injured. To date, this is the only myth to receive all three ratings at the same time.”-Mythbusters

Another source: en.wikipedia.org/wiki/Celebratory_gunfire

A sad tale of the consequences of thoughtless stupidity.

I don’t have an answer to your specific questions, Ron. There’s a section in the NRA Factbook which goes into the implications of firing bullets vertically and what happens when they come down, but that’s entirely concerned with .30 cal rifle bullets. FWIW, the terminal velocity of a pointed 180 grain bullet will vary between 180 fps if it’s tumbling to 450 fps if it’s point first.

The same source gives the maximum range of a 9mm bullet (123 grains at 1150 fps) as 5,700 feet at sea level. Presumably fired at about 30 degrees (although it would come down at a steeper angle). That gives you some idea of the maximum distance.

Jonnyc–Yes, I saw that Mythbusters show. However, the difference there and in my question is the angle fired at. They were firing vertically, at 90 degrees. In that case, the bullet rises until it runs out of energy and then falls back to the ground with an energy and velocity easily calculated as a Newtonian falling object. That is not the case when the projectile has a trajectory. If it was the case, an artillery projectile would have a very small impact energy and velocity based only on it’s mass.

Tony–There must be ballistic formulas that allow all of this to be calculated. Granted, there are lots of variables that, in the real world, need to be accounted for, besides the atmospheric conditions, such as spin rate, CG, tumble, etc., but given perfect conditions, point forward, no wind, etc., the maximum figures should be able to be calculated for any bullet weight, ballistic coefficient, muzzle energy, etc.

One episode of Myth Busters examined the falling bullet question. See a discussion at
kwc.org/mythbusters/2006/04/epis … up_vo.html

Hatcher’s Notebook discussed this for .30 bullets, but did not mention 9mm bullets. I think his conclusion was that bullets fired vertically come down base first.

I have calculated terminal velocities for typical bullets falling in air to be 200 to 300 ft/sec. However, one must make a lot of assumptions about the attitude of fall (point first, base first, sideways) which affect the effective sectional density and drag coefficient greatly, so I don’t trust my calculations much. The more scholarly treatments of the drag of solid objects falling through air do not consider bullet-shaped items, and usually consider only spheres, right cylinders, or cubes, not irregular shapes.

This is an excellent on-line ballistic calculator I have used for years here: jbmballistics.com/ballistics … tors.shtml It won’t exactly answer the question, but it does indicate that the 9mm will have a striking velocity of about 260 ft/sec at its extreme range (which has a departure angle of 35 degrees)…

Ron

Your best source is Hatchers Notebook where he devotes an entire chapter to it. Titled “Bullets From The Sky”. I’d put more store in what he said than I would in Mythbusters or any anecdotal evidence. It’s true that most of his experiments were with Cal .30 but the physics are the same for any weight or velocity of bullet.

Ray

The Mythbusters also devoted a show to calculating maximum range of a pistol (IFIR correctly, they were using a .45 Auto) when fired at 0 degrees (perfectly horizontal). I don’t remember the distance, but I do remember that is was not near as far as I would have thought. However, none of this has anything to do with the normally high-angle of fire of celebratory shooting.

Ray–I do not have a copy of Hatcher’s Notebook, so does it give any generalized formulas?

No, just anecdotal information based on test firings. Remember, any calculation would be affected by a large number of factors such as shape, size, density, atmospheric conditions, etc. which would be unknown in most uncontrolled cases.

Bear in mind that all editions of Hatcher don’t include the bullet-fired-vertically chapter. My 1950 or so edition doesn’t have it but the copy I bought a decade later does. Neither is presently at hand. Jack

Dennis–Yes, I realize there are many variables to consider to arrive at an accurate answer, but I was getting at generalities that the police are trying to use to figure out the maximum radius the shot could have originated from.

Anyway, I just pulled a book off my library shelf that I have not looked at since about 1985. It is a Brassey’s Battlefield Weapons Systems & Technology Series, Vol. 10 “Military Ballistics–A basic Manual”. The book is aimed at artillery officers, not small arms, but ballistics are ballistics regardless of caliber. It divides ballistics into 4 sections, 1) Internal Ballistics–What happens inside the cartridge and barrel, 2) Intermediate Ballistics–What happens at the mouth of the barrel, 3) External Ballistic–What happen in flight and 4) Terminal Ballistics–What happen when the projectile hits something. It devotes 65 pages to External Ballistics. If I can wade through it, I think it will answer my questions. Lots of mathematics to calculate all kinds of things such as Forebody Drag, Base Drag, Skin Friction, Air Viscosity Effects, Gyroscopic Effects, Procession, Equilibrium Yaw, etc.

Using the aforementioned JBM ballistic calculator, calculated results for maximum range assuming a BC of 0.14 (9mm 115 grain bullet) and MV= 1150 ft/sec at “Standard” atmospheric conditions:

Departure Angle: 30.0 deg Terminal Angle: 63.1 deg
Terminal Range: 1793.3 yd Terminal Velocity: 253.9 ft/s
Terminal Time: 18.0 sec Terminal Energy: 16.5 ft-lbs

Note the angle at which it hits the ground is 63.1 degrees.

Energy would seem a bit low to be lethal, but it would sure hurt, and maybe even penetrate skin and do some serious injury. Would it penetrate the skull? I doubt it. Those reported as being killed by falling bullets may very well have been much closer to the gun than 1800 yards where the projectile velocity was much higher. Problem is that it would be nearly impossible to establish the distance of the victim from the gun, and the departure angle of the bullet. I would guess most such incidents involve bullet departure angles much less than 90 degrees.

Dennis–So, instead of 30 degrees, use the 80 degrees and other parameters I opened this discussion with. What does that give us?

The JBM calculator does not do that so well, as it will calculate a range table based upon a departure angle of up to only 300 MOA from line of sight. Or it will calculate maximum range and its associated initial and terminal bullet angles. Almost always, the departure angle for maximum range will be within a few degrees of 30 degrees.

The JBM calculator actually uses some very sophisticated mathematics. The ballistician at Federal once told me he uses it.