Dot Physics

How high does a bullet go?

You know I like the Mythbusters, right? Well, I have been meaning to look at the shooting bullets in the air myth for quite some time. Now is that time. If you didn’t catch that particular episode, the MythBusters wanted to see how dangerous it was to shoot a bullet straight up in the air.

I am not going to shoot any guns, or even drop bullets – that is for the MythBusters. What I will do instead is make a numerical calculation of the motion of a bullet shot into the air. Here is what Adam said about the bullets:

  • A .30-06 cartridge will go 10,000 feet high and take 58 seconds to come back down
  • A 9 mm will go 4000 feet and take 37 seconds to come back down.

Adam was also able to experimentally determine that both the 9 mm and the .30-06 have a terminal speed of about 100 mph. So, that is what I have to work with. Oh – also, they measured how far a 9mm bullet penetrated into the dirt (but they couldn’t find the .30-06 ones).

The plan

This is actually similar to Hancock throwing a boy. The basic plan is to use a numerical calculation to model the motion of a bullet. After the bullet leaves the gun, it has forces acting on it like this:

i-8a66e9c28b48a06419222209628d04d1-untitled.jpg

I made two force diagrams because the air resistance force is going to be in opposite direction as the motion. This means that moving up bullet will look different than going down. So, this problem seems simple enough – right? I have actually done this before (here is an example of the air resistance on a football). But in this case, there are some other things to consider.

  • Does the normal model of air resistance work (being proportional to v2)?
  • What is the drag coefficient of a bullet?
  • What about the density of air? Do I need to take that into account?
  • What about the change in gravitational field of the Earth as the bullet moves up?

Numerical Modeling

I don’t want to go into the details, but in case you forgot, the numerical calculation works this way:

  • Break the motion into tiny little time steps. During these steps, I can pretend (assume) that the force is constant. With a small enough time, this is true enough.
  • For each time step: Calculate force
  • Calculate change in momentum (assuming constant force)
  • Calculate change in position (assuming constant momentum)
  • repeat

If you want more details on numerical calculations, check out this basic post.

Starting info

I am just going to look at the .30-06, but I need some ballistics info. Here is what I found (wikipedia, of course)

  • Slug mass = 9.7 grams
  • Muzzle velocity = 880 m/s (actually, this is just the fastest – the slowest is 760 m/s and 14 g – not sure which the Mythbusters used)
  • Terminal velocity = 44.7 m/s

Air Resistance

If I want to model the air resistance, I can use the following:

i-7c6c95388a9648241f075e8774ef6f40-la_te_xi_t_1.jpg

The problem is that bullets go really fast. I mean really fast. It is not safe to assume that the drag coefficient (C) is constant with speed. Wikipedia comes to the rescue again. In this case, there is this very useful table:

i-b1fea528312493a2b7789f3aa304886f-external_ballistics_wikipedia_the_free_encyclopedia_1.jpg

Apparently, there is lots of debate about the air drag of a bullet. I will just use the table above to make variable drag coefficient. So, that is C, I can find the effective area by looking at terminal velocity. At terminal velocity, the weight = air resistance so:

i-5596250f33eb94346024ce328b7a440a-la_te_xi_t_1_11.jpg

Using the known values for mass, g, C (from the table) and the density of air (at sea level), I get an area of A = 3.45 x 10-4 m2. Wikipedia lists the bullet as having a diameter of 7.823 mm – this would give an area of 1.9 x 10-4 m2. I guess these are kind of in the same ball park. Well, there is a way to test which is right – but I will start with the one from the terminal velocity.

Density of Air

This is starting to get complicated. Good thing I am making a computer do all the work. If the MythBusters are correct and the bullet goes 10,000 feet high, then I will need to look at the change in the density of air. Here is an explanation of the density with altitude calculation. Using this expression (which I am not showing because it is boring), I can plot density as a function of altitude. This is it:

i-83acdaf0fdc1f3b951de672ff7a928a3-figure_1_1.jpg

Gravity dependence on height

Of course the gravitational field is not constant with height, but is it close enough? The real gravitational field (g) is:

i-c56cc5ec00efe5c744e4a6dde417e5a3-la_te_xi_t_1_21.jpg

Where G is the universal gravitational constant, mE is the mass of the Earth, RE is the radius of the Earth, and h is the height above the surface. What would the value of g be at 4000 meters? (the MythBusters said the bullet went 10,000 feet – about 3000 meters). Or rather, what would be the percent difference between the surface and 3000 meters up? It is 99.9% the value at the surface. I can just pretend its constant.

Now for the calculation:

Here is a plot of the vertical position of the bullet as a function of time, shot straight up.

i-a643e8f653b79ff61a378bfd5f31917e-sdfpng.jpg

Well, that doesn’t agree with the MythBusters’ model. What if I go with the smaller area value?

i-89a502cab9f580cffbfc119d5ebf88bd-imagespng.jpg

Better, but still does not agree? I could try a different bullet. Let me try the one with the lower muzzle velocity, but higher mass. I will use a mass of 14 grams and an initial velocity of 760 m/s. This gives a max height of about 1300 meters with a total time of about 34 seconds.

I think I see another mistake. My table of drag coefficients are matched up with mach number, not velocity. If I increase my altitude, that changes the speed of sound – doh! Ok, I don’t think this matters too much. Here is a speed of sound calculator. It’s from NASA, so it has to be good, right? Anyway, it says the speed of sound at sea level is 340 m/s, at 5000 meters it is 320 m/s. Instead of calculating the speed at every height, I just changed the speed of sound to 320 m/s. It doesn’t really change the max height.

Maybe the problem is with the drag coefficient. Here is a plot of the drag coefficient (C) as a function of speed.

i-3fc118b664176f1c2269201cc7841ecb-imagesdpng.jpg

It looks “blocky” because I am just using data from that wikipedia table. But maybe this is the problem. Actually, maybe the problem is that the drag coefficient table doesn’t work very well at low (very low) speeds.

Maybe this isn’t even wrong

Now that I think about it, the MythBuster’s said they simulated the .30-06, but when they shot it in the air, they never heard nor found the bullets. Who knows how long it took. They did know the time for the 9mm bullets, they heard them hit the ground. Let me run my calculations with the 9mm info. Using mass of 7.45 grams and initial velocity of 435 m/s, I get:

i-6c23e15b50f1713d65d74eeff782320d-image_9mmpng.jpg

Which seems much closer to what they (MythBusters) had. And I just realized another mistake on the .30-06. I calculated the area with the diameter instead of the radius.

i-48a51a1fdd5c2a41e80d83df6f7e5860-image_3006png.jpg

See. That is better. I hope this is a lesson to all you kids out there. Mind your factor of 2’s. Of course if I get this to work, now my terminal velocity is much higher than what they measured. Oh well.

My next step is to look at the final speed of the bullet if you shoot it not straight up. I suspect this is the how people get killed.

Update

If you want to look at or change my python code (warning, it’s sloppy – I am a blogger, not a programmer, Jim) – here it is.

Comments

  1. #1 Dave
    September 24, 2009

    Bullet weights are typically measured (at least in the US) in grains rather than grams. Thus, to be technically accurate, you should report the bullet weight in grains, and then convert it to grams. There may be some conversion error, depending upon the number of significant digits you employ.

    .308 inches is the diameter of a .30-06 bullet. There will be some variation due to the rifling as it travels through the barrel, but that’s probably a second (or third) order effect.

    Bullets, at least all modern ones, spin as they travel to stabilize them. However, at extreme flight distances/times, this spin may dissipate (via friction with the air). Lack of spin means that the bullets may become unstable, which may result in tumbling. This is especially true as the bullet transitions the speed of sound, which is usually quite traumatic. Tumbling dramatically changes the “ballistic coefficient”, which directly affects the air resistance (Consider the air resistance imposed by a forward flying bullet to that imposed by one flying sideways. Also remember that modern rifle bullets are not round balls, but are usually long, pointed objects, such as the boat-tail spitzer design, where the shape is defined in terms of oglive.). The trick is, at what point do the bullets become unstable and start tumbling (and how can that be modeled)?

    The velocity of a bullet is usually inversely tied to the mass of the bullet, due to pressure restrictions on the chamber of the weapon firing the bullet. Thus, at the limiting pressure, a lighter bullet will be moving faster than a heavy bullet fired at the same limiting pressure (e.g., 50,000 CUPs).

    http://en.wikipedia.org/wiki/Copper_units_of_pressure

    There have been (and continue to be) detailed studies of ballistics using such advanced
    instruments as doppler radars:

    http://en.wikipedia.org/wiki/Ballistic_coefficient

    Dave

  2. #2 Uncle Al
    September 26, 2009

    Mythbusters floated their bullets on a jet of air to measure terminal velocity. The bullets floated on their sides. A bullet that maintains its spin will come down butt first, falling much faster. We drop messages in champagne bottles into the Pacific from 7500 feet from a small plane doing about 100 kts. (Gotta get beyond the Southern California bight into the California Current.) Calculated

    70 mph on its side
    144 mph on its punt
    270 mph on its stopper seal

    As we get ~15% return, they fall as they float – on their sides. Isaac Newton’s blunt penetration equation alleges that an object of density d_1 and length L penetrating into a medium of density d_2 propagates a distance of (d_1)/(d_2)L (it displaces its own mass).

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