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Reply 1
Lex Talionis
Say, an ordinary AK47 or Colt .45. I realise they're different but I just want rough numbers? Anyone? Cheers.
OK. My explanation is AS-level. If you want a more accurate figure, that would require a more accurate model, which I wouldn't know, but some of the more knowledgeable members on this forum would. For example, this model assumes that the bullet is shot horizontally.

There are two factors which affect how far a bullet can travel before it can fall: (i) the time of flight (which is ultimately affected by the height from the ground from which the bullet leaves the gun) and (ii) the muzzle velocity of the bullet.

The time of flight of the bullet is given by 2syg\sqrt{\frac{2s_{y}}{\mathrm{g}}} where sys_{y} is the vertical height from which it is shot and g\mathrm{g} is the gravitational field strength.

The horizontal range of the bullet is given by vtvt where vv is the muzzle velocity (horizontal speed) of the bullet and tt is the time of flight.

Now, from this AK47 website, I found that the average muzzle velocity of an AK47 is 780 ms-1. Now let us say that this is shot at a height of 5.5 feet by your average 6 foot freedom fighter, and we can estimate the range of an AK47 bullet.

The time of flight of the bullet will be

[indent]=2syg = \sqrt{\frac{2s_{y}}{g}}[/indent]
[indent]=2×(5.5×0.3048)9.81= \sqrt{\frac{2\times(5.5\times 0.3048)}{9.81}}[/indent]
[indent]=0.5846141465s.= 0.5846141465\,\mathrm{s}.[/indent]

Therefore, the range of the AK47 bullet will be

[indent]=vt= vt[/indent]
[indent]=780×0.5846141465= 780\times 0.5846141465[/indent]
[indent]=466m.(3s.f.)= 466\,\mathrm{m}.\:\mathrm{(3 s.f.)}[/indent]

The same would apply for the Colt. 45, provided you know its muzzle velocity and the height from which it was shot. :smile:
But realise that as the bullet is constantly being accelerated downwards, it is constantly decreasing in vertical height, although the rate at which it does so increases as time progresses. It doesn't travel horizontally for a while and suddenly drop. That wouldn't be consistent with the parabolic motion of projectiles.
a little aside: here's a sort-of related website.
Reply 3
hoho nice, I just checked on google and the effective range is 300m, so that's probably pretty good, if not 100% accurate assuming no other outward factors.

To be pedantic though shouldn't your units for range be m? not ms^-1 which is speed?:p: Sorry I'm feeling a bit petty.:biggrin:
Reply 4
Sidhe
To be pedantic though shouldn't your units for range be m? not ms^-1 which is speed?:p: Sorry I'm feeling a bit petty.:biggrin:
Don't apologise. You're not being pedantic; you're absolutely right, especially for such a rookie error like that. I've changed my post now.

I'm guessing the value of 300 m for the range would take into account air resistance, which I didn't take into account.

P.S. I love pedanticism anyway. :p:
physicsgirlie
a little aside: here's a sort-of related website.
The most addictive physics game ever! Try and find the right velocity for the cannonball to stay in orbit!

Just some questions about Newton's Cannon, from the article:

1. What does it mean by 'an object in orbit is weightless'?
2. Can you explain what it means by 'it has enough horizontal speed never to hit the ground.' How does this actually work?
Reply 5
Not exactly I think effective range means in perfect conditions, indoors(no wind) With air resistence at a standard, the time a bullet still remains 100% effective, or can do serious damage after that I expect it's efficacy drops off, ie it's speed decreases rapidly until it hits the deck. I'd imagine it can still kill you though.

1.weightless means it has the same mass but no weight due to g, although that's not strictly true unless your in deep space I'd imagine and even then :smile:. the weight of a 1kg weight on the moon is for example is 1/6th but it's mass obviously is 1kg.
2. I think it means to put it into an orbit, so that it doesn't fall back to Earth. At least not for some considerable time anyway.
Reply 6
Sidhe
Not exactly I think effective range means in perfect conditions, indoors(no wind) With air resistence at a standard, the time a bullet still remains 100% effective, or can do serious damage after that I expect it's efficacy drops off, ie it's speed decreases rapidly until it hits the deck. I'd imagine it can still kill you though.
I see. :smile:
Sidhe
1.weightless means it has the same mass but no weight due to g, although that's not strictly true unless your in deep space I'd imagine and even then :smile:. the weight of a 1kg weight on the moon is for example is 1/6th but it's mass obviously is 1kg.
But surely even though it does not fall due to its weight, it still experiences a gravitational attraction to the Earth and is therefore not weightless?
Sidhe
2. I think it means to put it into an orbit, so that it doesn't fall back to Earth. At least not for some considerable time anyway.
I knew that. :smile:
I just wanted an explanation as to how this worked, physically.
Reply 7
Dharma
But surely even though it does not fall due to its weight, it still experiences a gravitational attraction to the Earth and is therefore not weightless?


exactly it's not truly weightless, but it's a term they use when your in space and you float, of course there is still some gravitational effect on you after all there is no known range for the inverse square law, but it's just nomenclature.

I knew that. :smile:
I just wanted an explanation as to how this worked, physically.


In that case do you know anything about escape velocity?

The minimum speed an object has to instantaneously achieve to leave the Earths gravitational pull and continue without falling back to Earth or achieving orbit.

This is about 11.2 km/s

Or about 10-11,000 km/h depending on which direction you fire the cannon ball, e/w or w/e.

In this case there looking for something like vo or velocity to reach orbit. The reason it wont fall back is the same reason why the Earth doesn't spiral into the sun.

http://en.wikipedia.org/wiki/Circular_motion

Kepler's law

http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion
Reply 8
Sidhe
exactly it's not truly weightless, but it's a term they use when your in space and you float, of course there is still some gravitational effect on you after all there is no known range for the inverse square law, but it's just nomenclature.Cool. :smile:

Thanks for the information. :smile:
I'll read about Circular Motion in my textbooks.
Reply 9
Thanks guys for the immensely detailed answer! So if the range of an AK47 is roughly 300m, if I were to stand 350m away, it wouldn't hit me?
Reply 10
Lex Talionis
Thanks guys for the immensely detailed answer! So if the range of an AK47 is roughly 300m, if I were to stand 350m away, it wouldn't hit me?

Erm...not exactly. If it were fired at a small angle upwards, which a marksman would account for at that range, it probably would.

However, that's assuming the AK47 is an accurate weapon. Which it isn't. You'll probably be OK if you're fired at by an amateur standing up around 50m away. If he was prone, with the stock out I'd want to be more than 200m.

Good snipers could hit you at a mile, but they don't use AK47's!
Reply 11
Incidentally, if the marksman was to fire the bullet in the air, to get the maximum range, the optimum angle of fire would be 45.45^\circ. It's something to do with the formula x=v2sin2θgx=\frac{v^2\sin 2\theta}{\mathrm{g}} or something like that, but I can't remember exactly.
Reply 12
Worzo
Erm...not exactly. If it were fired at a small angle upwards, which a marksman would account for at that range, it probably would.

However, that's assuming the AK47 is an accurate weapon. Which it isn't. You'll probably be OK if you're fired at by an amateur standing up around 50m away. If he was prone, with the stock out I'd want to be more than 200m.

Good snipers could hit you at a mile, but they don't use AK47's!


This is true the best shot from a rifle from a range of just under a mile was fired from a large hill, in a stiff wind and it achieved a kill, some suggest more by blind luck than skill, but? It really does depend on the circumstances, but as said, on level ground firing at no angle, the range is correct at least according to wiki.

Think how bloody hard it must of been for an archer to determine range with a projectile that was so far below the speed of sound it wasn't funny, you have to aim above a target if firing down on them and below if firing below them. It took a while to become a proficient bowman, but a lifetime to become an archer of rare skill. Which incidentally is why they banned football on Sundays in the middle ages, so Yoemen could practice their archery skills.

The longbow was a seriously hard skill to learn, the crossbow meant you could train any old person given a few months, the Assault rifle is something anyone can use given a few weeks practice, war ever seeks to make killing easier and more efficient.

Oh and the AK-47's success doesn't come from it's accuracy, you don't need it at the close range it was designed for, it comes from it's reliability, the M16 was awfully unreliable in Vietnam for reasons I wont bore you with; accurate but it tended to need constant attention. The Ak-47 was hand made and crafted, and it's claim to fame was that you could bury it in the ground for a month and when you retrieved it it would still fire.

Also the calibre round would fire through walls and trees where as the M16 round wouldn't, so it tended to be less efficient at killing in certain situations.

OK I'm boring you now :smile:
if i can be bothered i can use M4 mechanics to predict it as it takes air resistance into account. Let me do the working and scan it in if i remmeber.
Reply 14
sufiankane
if i can be bothered i can use M4 mechanics to predict it as it takes air resistance into account. Let me do the working and scan it in if i remmeber.

I wouldn't bother. The correct calculation involving the combination of linear and quadratic resistance is not soluable analytically, and if you take either of the two independently, it'd be tough to find the correct coefficient anyway.

I think for a 45 degree elevated shot, it falls to maybe 2/3rds the range or something? And for a level shot, it's nearly the full range anyway.
Reply 15
colt .45 will shoot out to about 100 metres (effective range about 60m)

ak47 will shoot to about 700 metres. (effective range about 300m)

For both of these you'd have to aim the gun upwards.

The physics of bullet ranges is very complicated, but is mainly affected by:
-muzzle velocity (the speed at which the bullet leaves the gun)
-trajectory of the bullet
-air resistance of the bullet
Reply 16
Worzo
I wouldn't bother. The correct calculation involving the combination of linear and quadratic resistance is not soluable analytically, and if you take either of the two independently, it'd be tough to find the correct coefficient anyway.

I think for a 45 degree elevated shot, it falls to maybe 2/3rds the range or something? And for a level shot, it's nearly the full range anyway.


I would imagine then if it's not solvable, they simply fire off about a thousand shots in standardized conditions and then do a some sort of statistical analysis with SD or something. Reality being about as accurate as it gets, given enough of a sample.
^^ thas pretty hard though. you need to get a massive hall make sure ti has no wind and do it. plus getting a thousand rounds and a person who wants to waste their life doing it.
Reply 18
sufiankane
^^ thas pretty hard though. you need to get a massive hall make sure ti has no wind and do it. plus getting a thousand rounds and a person who wants to waste their life doing it.

Umm...just wait for a still day? Go to a desert. Plus firing off 1000 AK47 rounds wouldn't take long, and people would do it if they were paid to. Which they probably would. Because armies/manufacturers would want to know things like the range.
Go to a desert? try finidng 1000 bullets in the sand. Trust me they use engineers to do most the calculations then when they have a rough idea and fire off the shots and measure it. And it will be indoors o prevent winds.

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