Physics and Firearms :: physics firearm gun guns ballistics

So you are into reloading and you wonder how well that superficial package with 77 grains of IMR 4350 powder stool a 300 grain round nose, full metal ceiling lick will do. Well, you can do two things, a little bit of natural philosophy calculations, or go out and touch it off, hoping that it doesnt 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 exposit of the rifles ballistics data. For instance, if you know a few variables, you can predict tramp with physics, or if you like you can figure things like drag on the bullet, pressure and expansion esteems inside the gun, on the bullet and some(prenominal) more, all from physics.So, lets take a look at both the authorization and energizing energies of the .338 Winchester magnum. I will use a load given over by the Winchester Reloading manual, which can be found online athttp//www.winchester.com/reloader/index.htmlThis load is a 3 00 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=speedup due to gravity, and h=height.So for a 300-grain bullet, the potential energy is reason 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/s2=.001331slugs. Now we can calculate the potential energy of our 300-grain bullet. We will dramatize that h=six feet, since that is roughly the height of the barrel when I accuse from a standing position. So, since PE=mgh, we appropriate PE=(.00133slugs)(32.2ft/sec2)(6ft)=.256956lbft. The answer is pretty lots nothing and so we can pretty much ignore the potential energy of that bullet sitting at six feet in the air, further now lets look at the Kinetic energy of this bullet when shot. Since this bu llet will be twisting when it flies, it will have rotary motional kinetic energy, but I really dont want to get into those calculations and from what I have read, the amount of energy given by rotation versus that of the charge behind the bullet is really insignificant so I will only calculate the KE as if the bullet is not rotating. The form is KE=1/2mv2.