{"id":880,"date":"2020-12-03T15:58:58","date_gmt":"2020-12-03T20:58:58","guid":{"rendered":"http:\/\/www.moroha.net\/blog\/?p=880"},"modified":"2021-04-02T10:06:27","modified_gmt":"2021-04-02T15:06:27","slug":"how-high-can-a-bullet-go","status":"publish","type":"post","link":"https:\/\/www.moroha.net\/blog\/archives\/880","title":{"rendered":"How High Can a Bullet Go?"},"content":{"rendered":"\n

There was a question posted on the physics subreddit asking<\/a> how high in the air a bullet would go if you shot it straight up. It was later removed by the moderators (who knows why?), but I thought it was a good question that would make a good example problem. Normally I would use MATLAB to solve something like this, but I thought it would be a good opportunity for me to start using python more to solve technical problems like this.<\/p>\n\n\n\n

In terms of an actual physics problem, here’s how I would state it: a bullet of known dimensions and mass is fired straight upward at a known velocity. What is it’s maximum height, and how long does it take it to get there?<\/p>\n\n\n\n

So how do we model this? Once the bullet leaves the muzzle of the gun, there are only two forces acting on it: gravity, and air resistance. Newton’s 2nd law tells us that the acceleration of an object is proportional to the force acting upon it, or in equation form \\(F=ma\\). The acceleration is defined as the derivative of the velocity with respect to time, or \\(a=\\frac{dv}{dt}\\). So we can write this as a differential equation for the velocity of the bullet as a function of time:<\/p>\n\n\n\n

$$m\\frac{dv}{dt}=-F_g – F_d$$<\/p>\n\n\n\n

Where \\(m\\) is the mass of the bullet, \\(v\\) is the velocity, \\(t\\) is time, \\(F_g\\) is the force of gravity, and \\(F_d\\) is the drag force due to wind resistance. This also requires an initial condition for \\(v\\) at \\(t=0\\), which is simple the muzzle velocity of the bullet.<\/p>\n\n\n\n

The gravitational force \\(F_g\\) is pretty straightforward: \\(F_g=mg\\), where \\(g\\) is the gravitational acceleration of 9.8 m\/s2<\/sup>. The drag force is a bit more complicated though. It’s given as<\/p>\n\n\n\n

$$F_d=\\frac{1}{2}\\rho v^2 A C_d$$<\/p>\n\n\n\n

Where \\(\\rho\\) is the density of the fluid it’s traveling through (air), \\(v\\) is of course the velocity, \\(A\\) is the cross-sectional area of the object (in this case the circular cross-sectional area of the bullet as you look at it head-on), and \\(C_d\\) is the drag coefficient of the object.<\/p>\n\n\n\n

Determining the drag coefficient is what makes this problem a bit complicated. It’s defined as follows:<\/p>\n\n\n\n

$$C_d = \\frac{2 F_d}{\\rho v^2 A}$$<\/p>\n\n\n\n

So you can see it’s just a rearranged version of the drag force itself. This appears to be self-referential, but it’s not: the drag coefficient is measured experimentally or by performing fluid dynamics simulations, and it then becomes a function of velocity that we can use for solving other problems.<\/p>\n\n\n\n

So we just need to find some results that someone else has already measured experimentally. An internet search found this<\/a> info sheet, which shows a typical bullet drag coefficient as a function of the Mach number. The Mach number is just the ratio of the velocity of an object to the speed of sound, so if we know the speed of sound in air (we can look this up) then we can use that data to calculate our drag coefficient as a function of velocity.<\/p>\n\n\n\n

However this is just a data table on a website. How do we get this into either a functional form or a lookup table that we can use? I was able to find the plot by itself as a separate image:<\/p>\n\n\n\n

\"\"
Drag coefficient as a function of Mach number<\/figcaption><\/figure><\/div>\n\n\n\n

Then I used the very handy website WebPlotDigitizer<\/a> to convert it to a csv file. With that in hand we now have everything we need to calculate the trajectory of the bullet. You can download a copy of the csv file for yourself it you want to look at it or use it with the python code I have at the end of this post.<\/p>\n\n\n\n

drag_coefficient<\/a>Download<\/a><\/div>\n\n\n\n

For the actual gun and bullet, I decided I might as well go big: I looked up the parameters for a .50 caliber rifle and ammunition:<\/p>\n\n\n\n

Parameter<\/strong><\/td>Value<\/strong><\/td><\/tr>
muzzle velocity<\/td>884 m\/s<\/td><\/tr>
bullet diameter<\/td>13 mm<\/td><\/tr>
bullet mass<\/td>43 g<\/td><\/tr><\/tbody><\/table>
.50 caliber bullet and gun parameters<\/figcaption><\/figure>\n\n\n\n

From the bullet diameter we can calculate the cross-sectional area of the bullet \\(A\\) to be 1.327 cm2<\/sup>. <\/p>\n\n\n\n

We’ll need a few other constants as well:<\/p>\n\n\n\n

Constant<\/strong><\/td>Value<\/strong><\/td><\/tr>
gravitational acceleration<\/td>9.80665 m\/s2<\/sup> <\/td><\/tr>
density of air<\/td>1.225 kg\/m3<\/sup><\/td><\/tr><\/tbody><\/table>
Constants used in calculations<\/figcaption><\/figure>\n\n\n\n

So that should be everything we need to solve for the velocity. We also want to solve for the height of the bullet at the same time, fortunately the equation for this is very simple. Since velocity is defined at the derivative of position as a function of time, our second equation then is simply<\/p>\n\n\n\n

$$\\frac{dx}{dt}=v$$<\/p>\n\n\n\n

Where \\(x\\) is the position\/height of the bullet as a function of time. The initial condition is very simple as well, simply \\(x(0)=0\\).<\/p>\n\n\n\n

With this we should have all the information we need to solve for the bullet trajectory. To solve them numerically we’ll need some code that can solve an initial-value problem (ivp). Algorithms to do this are implemented in just about every mathematics software that’s out there: Matlab, Mathematica, Maple, MathCAD, etc. As an exercise I decided to solve this using python, as I’ve been trying to migrate from using Matlab to using python more. We’ll need the equations written out in matrix form to be able to solve them numerically:<\/p>\n\n\n\n

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$$\\mathbf{y}= \\begin{bmatrix} x \\\\ v \\end{bmatrix}$$<\/p>\n<\/div><\/div>\n\n\n\n

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$$\\frac{d\\mathbf{y}}{dt}= \\begin{bmatrix} v \\\\ -g-\\textrm{sgn}(v)\\frac{1}{2}\\frac{\\rho}{m} v^2 A C_d(v) \\end{bmatrix}$$<\/p>\n<\/div><\/div>\n\n\n\n

$$\\mathbf{y}(0)= \\begin{bmatrix} 0\\\\ v_0 \\end{bmatrix}$$<\/p>\n<\/div><\/div>\n\n\n\n

We need the sign function \\(\\textrm{sgn}(v)\\) so that air resistance is always resistive<\/em>: in other words it has to act against the direction of the velocity. If we just kept it as a minus sign, it would work correctly when the bullet is traveling upward, but when it switches to moving downward it would instead accelerate <\/em>it downwards. This ensures that air resistance will always slow the bullet down, regardless of whether it’s moving up or down.<\/p>\n<\/div><\/div>\n\n\n\n

Solving those equations (I’ll share the code at the bottom of this post) we get the following curves for the bullet height and velocity:<\/p>\n\n\n\n