Dumb math ballistics question

You're gonna need to know more about the external ballistics of the specific bullets to know exactly where they'd be head-to-head, but yes, the 115 would overtake the 147, somewhere around 250 yards with that cyclic rate and those muzzle velocities.

Now, if you're asking would the 115 run into the 147, the answer is no. Even assuming the requisite .1 MOA or better accuracy , the drop between the two is dramatically different. Would be more than a 2' elevation disparity at the range they'd reach simultaneously.

Let's exaggerate this to make the point. (Ignoring the actual acceleration due to gravity, these are made up values to be simple to see)

Imagine shooting straight across over the Grand Canyon.

If bullet A takes 2 seconds to travel 100 yards it will drop 30 yards.
If bullet B takes 1 second to travel 100 yards it will drop 10 yards.


So if you fire bullet B exactly one second after firing bullet A they will hit the target at the same time, but they will be 20 yards between the two holes in the target.

So, to get them to hit the same hole at the same time (or collide in mid air) you would have to aim bullet B lower.

So, if you can alter the aim between each shot then you could do it.

No.
As explained above.

There have been instances of fighter planes shooting themselves down with thier own cannon fire.

But they have to be diving at the same rate, in a parabolic curve of the bullet trajectory to catch up with them and run into them as the bullets slow down.

There is simply too much difference between the bullet weights and trajectotys of the two bullets you mentioned for them to even get close to hitting each other.

rc

Mach,
Where are you getting your 250 yard estimate from? I'm getting around 186 yards. My rationale is below - am I missing something? Thanks


600 rounds per minute cyclic rate = 10 rounds per second = 1 round per one-tenth of a second. Therefore if the first round is 950 fps, then the first round is about 95 ft away at the time the second round leaves the barrel (air resistance and horizontal deviation aside).

Dist of B1 = 95 ft + (950 ft/sec)*(time of Bullet 2)
Dist of B2 = (1145 ft/sec)*(time of Bullet 2)

Set the above two equations equal to each other and Time of Bullet 2 = 19/39 = .487 seconds

Therefore Distance of Meeting (in theory) = (1145 ft/sec)*(.487 sec) = 557.82 feet = 185.94 yards

It's not the weight, it is the muzzle velocity and the slow down due to the drag (calculated from the ballistic coefficient).

An object fired horizontally drops the same if their drag in wind is reasonably the same, and in this case, two different 9mm bullets are reasonably the same. It's not like you are talking about dropping a bullet and a feather.

If you drop a 115 grain bullet and a 147 grain bullet at the same time from the same height they will hit the ground at the same time.

If you shoot the 115 grain bullet horizontally out of a gun and drop the 147 grain bullet from the same height as the muzzle at the same time, both the 115 grain bullet and the 147 grain bullet will hit the ground at the same time. The 147 grain bullet will be lying at your feet and the 115 grain bullet will be several hundred feet away, but they will hit the ground at the same time.

So, what this means is that the vertical location of the bullet is a function of speed and time.

So, imagine your target is a spot on the ground. If you shoot horizontally. To hit the same spot on the ground the projectiles have to be traveling at the same speed the entire way until it hits the ground. If one bullet is shot faster it will go further and beyond the target spot on the ground. To get the faster one to hit the target spot on the ground you have to change the angle it is shot from horizontal to aiming a little bit down or low.



Here are some videos:

This one is good, if the ball had more velocity it would have it the monkey higher, but it always hits the monkey if you aim it at the monkey and the monkey and the ball start their motion at the same time.

https://youtu.be/0jGZnMf3rPo

https://youtu.be/_mCC-68LyZM

https://youtu.be/-uUsUaPJUc0

https://youtu.be/KacTRPL1MtE

BluesDancer said:

Mach,
Where are you getting your 250 yard estimate from? I'm getting around 186 yards. My rationale is below - am I missing something? Thanks


600 rounds per minute cyclic rate = 10 rounds per second = 1 round per one-tenth of a second. Therefore if the first round is 950 fps, then the first round is about 95 ft away at the time the second round leaves the barrel (air resistance and horizontal deviation aside).

Dist of B1 = 95 ft + (950 ft/sec)*(time of Bullet 2)
Dist of B2 = (1145 ft/sec)*(time of Bullet 2)

Set the above two equations equal to each other and Time of Bullet 2 = 19/39 = .487 seconds

Therefore Distance of Meeting (in theory) = (1145 ft/sec)*(.487 sec) = 557.82 feet = 185.94 yards

Click to expand...

Your equation is having the two bullets cross a finish line at the same time, not hit a point at the same time, you are missing a dimension.

As I said before, it you had a large target in this case both would hit the target at the same time if you put the target at the distance you calculated, but one would hit it high and the other lower.

Mach,
Where are you getting your 250 yard estimate from? I'm getting around 186 yards. My rationale is below - am I missing something? Thanks


600 rounds per minute cyclic rate = 10 rounds per second = 1 round per one-tenth of a second. Therefore if the first round is 950 fps, then the first round is about 95 ft away at the time the second round leaves the barrel (air resistance and horizontal deviation aside).

Click to expand...

I didn't assume a vacuum. The 147 gr has higher ballistic coefficient, and velocity loss is also not constant: all else being equal, a bullet that starts off at a lower velocity will shed less velocity over the same distance as one that leaves the muzzle at a higher speed. Drag coefficient increases with velocity. That's why it takes a whole lot more than double the horsepower to get a car to 200 MPH than 100 MPH.

Drop is a function of time. The bullet fired 0.1 second before the other will always be lower regardless of horizontal velocity as it was falling toward gravity 0.1 seconds longer.

Mike

If we also assume no gravity and perfectly accurate gun and ammunition the 115gr bullet would hit the 147gr bullet.

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If there is no gravity they will never cross paths, ever, if one is shot at a different angle (direction) than the other. If there is no gravity and both are fired in the same direction the faster one will then run up on the slower one. To imagine a no-gravity situation, just go to a basketball court with two large ball bearings and roll them on the court, that would be the same as two bullets fired in the same direction with no gravity.

There is no need to be in a vacuum for this either.

Drag on a bullet (projectile) is not constant like MachIVshooter said. There is no known continuous equation to calculate it either. That is why there are drag tables and bullets have ballistic coefficients (BC).

So, again, in the OP's scenario, the second faster bullet can only hit the first slower bullet if the point of aim is altered for the faster bullet. I do not know of any person or machine that can change the point of aim between shots fired from a full auto. Given two guns it is possible to fire a slower bullet first and then fire the faster second and have them collide, but the second gun will not be aimed at the same point as the first.

So, again, in the OP's scenario, the second faster bullet can only hit the first slower bullet if the point of aim is altered for the faster bullet.

Click to expand...

Agreed.

The NLOS Cannon, a piece of mobile artillery, can hit a target with 3 rounds at the same time. All 3 shots will be at different elevations allowing for more air time for the first shot and less air time for each consecutive shot. The NLOS project has been canned, so you'll probably never see one.

Unless you were using match grade ammunition with a firearm that had super tight tolerances, you would probably have huge timing issues with variances in FPS and burn time. Remember the Mythbusters where they tried to collide 2 bullets? They kept shooting slightly off time from each other.

I have thought of a superhuman character removing some powder from the first bullet so he can strike 2 targets simultaneously, but we are out of the realm of human capabilities at that point.

I do not know of any person or machine that can change the point of aim between shots fired from a full auto

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Oh, CNC servos certainly could. You're only talking about moving a short tube a few arcminutes in 100ms; easily doable.

The trouble with actually creating the scenario that the OP asks about is all the other variables. While an electro-mechanical device could alter the angle of the barrel with enough speed and accuracy, the variation in cyclic rate, the standard deviation of one round to the next, etc. would still result in a miss. Do it enough times, you'd probably get lucky and make it happen, though.

There have been instances of fighter planes shooting themselves down with thier own cannon fire.

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Sorry RC,, I'm going to have to call this one a wives tale. and I haven't done the math yet,, but

A plane is flying in a dive at a given speed. The plane fires it's cannon. The plane dives farther down, turns up the power and is now accelerating. The cannon projectile is now decelerating due to air drag. At some point the projectile will reach terminal velocity and just be falling. Since I don't know the cannon size I can't calculate that speed. BUT, we know that terminal velocity on a body thru air is actually pretty slow. (relatively speaking) (see Mythbusters when they determined that a penny dropped from the Empire state building would not be lethal.) In the amount of time it would take the airplane to accelerate and pass the projectile, the projectile PROBABLY would be travelling pretty slow.. if the airplane then was pulling up out of the dive its upward speed would also be pretty slow. The relative velocities of the 2 would be not enough to really damage an airplane to "shoot it down".. dent it? maybe? tear a hole thru a wing. NO. This is just off the top of my head.. so,, I'll gladly read some facts.. but it's kind sounding like trying to shoot a bullet with a bullet easy.

Sorry RC,, I'm going to have to call this one a wives tale. and I haven't done the math yet,, but

A plane is flying in a dive at a given speed. The plane fires it's cannon. The plane dives farther down, turns up the power and is now accelerating. The cannon projectile is now decelerating due to air drag. At some point the projectile will reach terminal velocity and just be falling. Since I don't know the cannon size I can't calculate that speed. BUT, we know that terminal velocity on a body thru air is actually pretty slow. (relatively speaking) (see Mythbusters when they determined that a penny dropped from the Empire state building would not be lethal.) In the amount of time it would take the airplane to accelerate and pass the projectile, the projectile PROBABLY would be travelling pretty slow.. if the airplane then was pulling up out of the dive its upward speed would also be pretty slow. The relative velocities of the 2 would be not enough to really damage an airplane to "shoot it down".. dent it? maybe? tear a hole thru a wing. NO. This is just off the top of my head.. so,, I'll gladly read some facts.. but it's kind sounding like trying to shoot a bullet with a bullet easy.

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It is documented, most notably the F11-F:

http://www.check-six.com/Crash_Sites/Tiger138260.htm

As well, one wouldn't think of a large bird bringing down an aircraft, either, but it happens.

Turbines are not very tolerant of anything other than air and fuel hitting the blades that spin in excess of 100,000 RPM

Ah. Ran into his own projectiles. It's like injesting birds made of rocks. No way jet engines of that era could survive that. Why didnt he eject.

The relative velocities of the 2 would be not enough to really damage an airplane

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you do understand of course, that fighter aircraft cannon shells are also high explosive incendiary, or even more nasty modern stuff, right?

They are not just 'falling rocks' when the aircraft flies into a swarm of them.

Those were inert training rounds in the story linked.
Had they been normal aircraft cannon shells, he would have been killed instantly.

rc

I had to take college algebra 3 times to pass but interesting concept

Please delete

rc

the plane rattled. The Tiger had been struck, and Attridge's windshield buckled inward.

Having thought maybe he had hit a bird, a typical hazard when flying over coastal waters, Attridge

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They must not have been

aircraft cannon shells are also high explosive

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if it only rattled the airplane. Sounded like a bird strike.

and yes, I forgot about engine ingestion, with my original post. oops..