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, skilful metal jacket dope will do. Well, you arsehole do devil things, a little bit of physics calculations, or go come out and touch it off, hoping that it doesnt explode in the pose I would adopt to do a little physics myself By using both(prenominal) basic physics equations, you can figure out just rough any part of the rifles ballistics data. For instance, if you know a few variables, you can predict range with physics, or if you akin you can figure things like drag on the bullet, pressure and expansion values inside the gun, on the bullet and overmuch more, all from physics.So, lets exhaust a look at both the strength and kinetic energies of the .338 Winchester magnum. I will use a load abandoned by the Winchester Reloading manual, which can be found online athttp//www.winchester.com/reloader/index.htmlThis load is a 300 grain bullet, using 59.8 grains of Winchester 760 powder, and this gives a fit velocity of 2285 ft/sec.For potential competency we know that PE=mgh, where PE= Potential Energy, m= circle, g=acceleration due to gravity, and h=height.So for a 300-grain bullet, the potential heartiness 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/s2=.001331slugs. Now we can calculate the potential energy of our 300-grain bullet. We will assume that h= hexad 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/sec2)(6ft)=.256956lbft. The solve 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 wi ll be twisting when it flies, it will discombobulate rotational 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 peanut so I will only calculate the KE as if the bullet is not rotating. The formula is KE=1/2mv2.