Need to know muzzle velocity. Then someone who is not me (well, maybe me) could figure it. It's been a long time since I did that kind of math.What would the energy be of a 7mm 130 grain bullet fired into the air at an approximate 45 degree angle AT THE MOMENT IT HITS THE GROUND?
I think (or, rather I hope) that's where the question originated. Theoretical energy with no losses, versus realistic values... If we're talking a 7mm Remington Magnum, a 130 grain (which is light for that round) would probably leave the barrel at well above 3,000 ft/s with typical pressure loadings for that cartridge. Anything shot at above the speed of sound likely won't still be above that point when it has lost all of its kinetic energy in the positive vertical direction and is regaining it due to potential energy due to the fall at terminal velocity. Above the speed of sound is where things are forced by external influences, in this case a controlled expansion of burning gases from propellant.The theoretical answer, according to the physics taught in college, is that it will be the same as it possessed when it left the muzzle. But we know there are losses, just not how to accurately calculate their exact impact. But, it will be very close. In this case, the bullet leaves the barrel with a high kinetic energy, but as it flies, some of that energy is lost to friction (heat), the rest is converted to potential energy as it climbs to its maximum height. As it falls, that energy is converted back to kinetic energy, with additional losses due to friction. I'm not willing to go back 40 years to relearn the math for a precise answer, but the answer is probably quite a large percentage of its initial energy. It might be a fun experiment to model on a computer, though I'm certain someone has already done it. Interesting question!
In college, we had one of the first naval ballistics calculators, and it wasn't what we'd call a computer today. It was installed by knocking down a wall of a building to make a space big enough, then the wall was rebuilt around it. It was what is called an "analog computer" because there are no 1s and 0s involved, just extremely precise voltage levels. It consisted of banks of analog amplifiers, called operational amplifiers, with supporting capacitors and resistors that could be set to precise values. It was programmed by connecting the individual components via patch cords into a constellation of integrators and differentiators to directly solve complex differential equations. It was great fun to program, but a PITA to maintain, since it was based on vacuum tube amplifiers and diodes, and we often had to shut down to replace a tube or six.The first computer systems were developed to calculate naval guns back around ww2.
Whoah.... I plugged your 160 gr. info at 1213 fps into a new row with the same calculations on my spreadsheet and also came up with 523 ft-lb.One cal has this info:
Bullet G7 shape - 160grn with .5 bc at 2800 fps at muz. With 45 deg shooting angle.
For 2000yrd range
Max arc peak = 500 yrds at +53"
Impact at 2000 yrds at -715" (speed 1213 fps & Energy 523 fp)
P.s. your 130 grn would be much less as it is not as good in the air.
Thats at the apex or peak of the arc only when the energy causing the bullet to sheer the force of gravity no longer is stronger than gravity and it begins to start dropping (yes changes coarse/direction). The inertia carries it through that transition and the direction is not effected other than by gravity, wind, or the rotation of the earth that also effects it the entire time it is in flight.If it drops below the speed of sound all bets are off. An object becomes weightless in the trans sonic threshold & might change direction completely.