How high does a bullet go?
September 23rd, 2009 | Tags: air resistance, forces, mythbusters, numerical calculation, vpython
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:
http://blog.dotphys.net/wp-content/u...9/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:
http://blog.dotphys.net/wp-content/u..._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:
External Ballistics Wikipedia, the Free Encyclopedia 1
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:
http://blog.dotphys.net/wp-content/u..._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:
http://blog.dotphys.net/wp-content/u...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:
http://blog.dotphys.net/wp-content/u..._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.
http://blog.dotphys.net/wp-content/u.../09/sdfpng.jpg
Well, that doesn’t agree with the MythBusters’ model. What if I go with the smaller area value?
http://blog.dotphys.net/wp-content/u.../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.
http://blog.dotphys.net/wp-content/u...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:
http://blog.dotphys.net/wp-content/u...age_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.
http://blog.dotphys.net/wp-content/u...ge_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.
