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How do I determine the approximate distance my .45 ACP round will travel if fired horizontally at sea level?
 
Here is info from a guy much smarter than I when someone on the 1911 forum asked the same question

Assume 850 feet per second (259.08 meters/sec) fired from 6 feet high.

-1.828 m = 0 + .5(-9.81 m/s^2)t^2
t = .611 s

x = vxt = (259.08 m/s)(.611 s) = 519.217 feet


Now, assume it was fired 45 degrees up from parallel with the ground. This would give the most possible range.

vix = vi cos 45* = 183.197 m/s
viy = vi sin 45* = 183.197 m/s
0 = 183.197 m/s + .5(-9.81)t
t = 37.349 s

(183.197 m/s)(37.349 s) = 6,842.23 m = 22,442 ft

Of course, these equations do not take into account form drag or skin friction drag. Therefore, actual numbers will be lower.
 
Depends on the height. Take the height of the barrel, say 4 feet. Then drop a round from your hand to the ground and time it. For example, if it takes 1 second to fall from your hand to the ground and your FPS on .45 ACP is around 900fps then your bullet fired from the same height would travel 900 feet before hitting the ground. A bullet fired out of a perfectly horizontal barrel and a bullet dropped from your hand, both being the same height will hit the ground at the same time. Gravity is gravity.
 
Well, from my research, most bullets don't travel in a truly horizontal path (edit: they do at first if the barrel is truly level, then only drop), then tend to go up a bit after being fired, then drop as air friction and gravity act on them. Obviously there are a lot of factors: barrel length, powder charge, bullet weight, bullet design, environmental factors such as altitude, wind, even temperature and humidity.

But if you want to get a snapshot view of how they react under certain circumstances, the table at this link might be helpful:

Handgun Trajectory Table
 
Last Edited:
Well, from my research, most bullets don't travel in a truly horizontal path, then tend to go up a bit after being fired, then drop as air friction and gravity act on them. Obviously there are a lot of factors: barrel length, powder charge, bullet weight, bullet design, environmental factors such as altitude, wind, even temperature and humidity.

But if you want to get a snapshot view of how they react under certain circumstances, the table at this link might be helpful:

Handgun Trajectory Table
Bullets can't go up before they go down *unl asylum(WTF? Unless you?)* are pointing the barrel in an upward direction. So if using sights,scopes whatever,if you are aiming horizontally,your barrel is pointed up a little if you have the scope dialed in at say 100 yards
As to the question at hand,if a 22 round will travel a mile,given all the variances of weight,drag,energy and such, I would guess most center fire rounds would come close to that.
That's having a looong stretch with no traffic and loving a shot at say 45*?:s0155::s0153::s0092:
My best guess
 
Last Edited:
Bullets can't go up before they go down unl asylum are pointing the barrel in an upward direction. So if using sights,scopes whatever,if you are aiming horizontally,your barrel is pointed up a little if you have the scope dialed in at say 100 yards
As to the question at hand,if a 22 round will travel a mile,given all the variances of weight,drag,energy and such, I would guess most center fire rounds would come close to that.
That's having a looong stretch with no traffic and loving a shot at say 45*?:s0155::s0153::s0092:
My best guess

Good point, you are correct, a truly horizontal barrel would result in 0 rise (unless you're shooting airsoft with back spin on the BB's - those things can go wild). I should amend what I said previously as I was thinking more along the lines of point of aim, say at 100 yards, you'll likely be shooting with the barrel just a bit up, which accounts for the arc upward, then back down to the eventual target impact point.

I went back and looked at the link I posted previously, they weren't in my opinion, completely clear on the angle of the barrel relative to the target when fired. It's reasonable to assume their gun was sighted at the longer distance which would mean their bullets would have traveled up a bit at first.
 
If you watch some of the YouTube guys videos when they shoot pistols long range 300+ yards, they generally will say they are aiming 20 feet above the target (as an example - caliber and distance will vary), but that's an extreme range for a .45acp.

I think Iraqveteran8888 did a 9mm test and found it was (if you can hit your target) only in the deadly range out to 400 yards.

.45's with the much heavier bullet and smaller velocity would be well under that in my guestimation.
 
Here is info from a guy much smarter than I when someone on the 1911 forum asked the same question

Assume 850 feet per second (259.08 meters/sec) fired from 6 feet high.

-1.828 m = 0 + .5(-9.81 m/s^2)t^2
t = .611 s

x = vxt = (259.08 m/s)(.611 s) = 519.217 feet


Now, assume it was fired 45 degrees up from parallel with the ground. This would give the most possible range.

vix = vi cos 45* = 183.197 m/s
viy = vi sin 45* = 183.197 m/s
0 = 183.197 m/s + .5(-9.81)t
t = 37.349 s

(183.197 m/s)(37.349 s) = 6,842.23 m = 22,442 ft

Of course, these equations do not take into account form drag or skin friction drag. Therefore, actual numbers will be lower.

Thanks! I'm an art guy by trade, so I'm just gonna trust the math...
 
If you watch some of the YouTube guys videos when they shoot pistols long range 300+ yards, they generally will say they are aiming 20 feet above the target (as an example - caliber and distance will vary), but that's an extreme range for a .45acp.

I think Iraqveteran8888 did a 9mm test and found it was (if you can hit your target) only in the deadly range out to 400 yards.

.45's with the much heavier bullet and smaller velocity would be well under that in my guestimation.
That's deadly range,not how far the bullet will actually travel
 

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