Physics and Firearms :: physics firearm gun guns ballistics

So you are into reloading and you wonder how well that little package with 77 grains of IMR 4350 powder behind a 300 grain round nose, full metal jacket bullet will do. Well, you can do two things, a little bit of physics calculations, or go out and touch it off, hoping that it doesn’t explode in the barrel! I would choose to do a little physics myself†¦ By using some basic physics equations, you can figure out just about any part of the rifles ballistics data. For instance, if you know a few variables, you can predict range with physics, or if you like you can figure things like drag on the bullet, pressure and expansion values inside the gun, on the bullet and much more, all from physics. So, lets take a look at both the potential and kinetic energies of the .338 Winchester magnum. I will use a load given by the Winchester Reloading manual, which can be found online at: http://www.winchester.com/reloader/index.html This load is a 300 grain bullet, using 59.8 grains of Winchester 760 powder, and this gives a muzzle velocity of 2285 ft/sec. For potential energy we know that PE=mgh, where PE= Potential Energy, m=mass, g=acceleration due to gravity, and h=height. So for a 300-grain bullet, the potential energy is calculated by first finding the mass. To do this, take 300grains/7000grains/pound. This gives you a value of .042857lbs. Then we need to convert pounds to slugs (slugs are the units of mass†¦) .042857lb/32.2ft/s^2=.001331slugs. Now we can calculate the potential energy of our 300-grain bullet. We will assume that h=six feet, since that is roughly the height of the barrel when I shoot from a standing position. So, since PE=mgh, we get PE=(.00133slugs)(32.2ft/sec^2)(6ft)=.256956lbft. The answer is pretty much nothing and so we can pretty much ignore the potential energy of that bullet sitting at six feet in the air, but now lets look at the Kinetic energy of this bullet when shot. Since this bullet will be twisting when it flies, it will have rotational kinetic energy, but I really don’t want to get into those calculations and from what I have read, the amount of energy given by rotation versus that of the charge behi nd the bullet is really insignificant so I will only calculate the KE as if the bullet is not rotating. The formula is KE=1/2mv^2.