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Discussion Starter · #1 ·
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?
 

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I think it would still be
lethal.
 
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You have supplied only the bullet weight and angle. I've been away from Physics too long, but I remember we would need muzzle velocity, ballistic coefficient, temperature, wind direction, wind speed, humidity, and elevation of departure and impact.
 

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It will likely reach terminal velocity of the fall in the vertical Z direction and any residual velocity remaining and not lost due to aerodynamic resistance in the forward X horizontal direction. There's not enough information, but as already stated it's not what the bullet left the barrel with. I wouldn't want to be hit with it though. Velocity lost is really the key to energy and the velocity won't be anywhere near muzzle velocity, especially if it lands at the same elevation that it was shot from. For instance, if velocity halves, energy is one quarter of the original value. To be fair, I'm not an engineer at home. I only play one at work and I'm off right now. :geek:
 

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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!
 

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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!
I think (or, rather I hope) that's where the question originated. :unsure: 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.

I found one chart that said a 7mm Remington Magnum 150 grain round has roughly 1445 ft-lb of energy at 500 yards with a muzzle velocity of 3,060 ft/s and a muzzle energy of 3,120 ft-lb. Doing some quick and simple math in Excel, that tells me that the velocity has decreased to about 2,085 ft/s at 500 yards with a 37.6" drop based on the chart below and a Velocity squared relationship for kinetic energy. Firing flat at a 0 degree angle would allow folks of a normal 5 ft+ stature to see that round still in flight at 500 yards, assuming flat ground.

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Let's assume that the round only decreases to just below the speed of sound when fired at a 45 degree angle on flat ground before landing back at the launched height. This is just a hypothesis. I could be wrong. My calculation shows the energy of the same 150 grain round at only 396 ft-lb of energy based on those numbers from the chart that I found. That's only a little above 10% of the original energy. A 40 grain .22 LR round has approximately 105 ft-lb of energy at the muzzle when fired at 1140 ft/s. So, your 150 grain 7mm Magnum round used in this scenario would likely still have almost 4 times the energy of a standard .22 LR round. Therefore, it could still kill you if you were hit. Sorry, I woke up and tried to shake some mental energy to go back to sleep. I might have made sense or I might not have... Who knows... Below is the Excel spreadsheet that I made based on the empirical data from the chart above.
253319

I spent wayyyy too much time thinking about this at 3:30 AM. :ROFLMAO:
 

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Discussion Starter · #14 ·
Wow, I thought I’d possibly get one answer that made some sense, never that I’d get such a response. Thanks to all that took the time to give my question consideration. The reason for my question involves a leak in my 1 1/2 inch schedule 40 pvc pipe located about 10 inches below ground which, when located, showed a .282 inch round hole in the top, at an angle off to the side and a smaller hole in the bottom. I suspect celebratory gunfire.
 

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If you enter the data properly in most any of the ballistics calculators it would likley give you a correct plot with velocity and energy at impact. Just increase the range until you have the arc you want but the 45 deg angle will will only be evident at one given distance - it will be dropping faster on the far side of the apex and maybe at a 70 deg +/- angle at impact.

In the good old days you could use a slide rule to show you an artillery rounds impact at a given range and charge level and particular tube length and twist rate. The first computer systems were developed to calculate naval guns back around ww2. Before that it was Kentucky windage and experience with the cannon/piece that won the day. It is still not an exact science thats why artillery will fire several rounds to make corrections via forward observation before firing for effect (using all guns together).
 

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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.
 

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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.
 

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The first computer systems were developed to calculate naval guns back around ww2.
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.

Ah, the good old days... :geek:
 

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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.
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. :oops:

That's not bad for a South Carolina country boy at 3:30 AM. I have officially impressed myself by the coincidence. :ROFLMAO:

253334


253333
 

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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.
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.

Flush the toilet in the northern hemisphere and it rotates one way flush one in the southern and it rotates the same but opposite from your viewpoint because you are actually standing upside down and looking at it from the bottom side. Thats why rifles meant to be fired north of the equator have RH twist and south of it are LH - supposedly. Never figured out why facing north and facing south when you shoot wouldn't make a difference also 🥴 - or does it.

Does this mean that if you live on the equator you should have only smooth bore firearms?

Now what would happen if you used a magnet for a bullet with a North pole end and a South pole end? Would it flip over in flight if the bullets north end was shot towards the north pole and would it go further if you shot towards the south? 🤔

Another question - if you get a sonic boom when you break the sound barrier by speeding up why don't you get one when you go from supersonic to subsonic?
 
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