# Math Problem-.50 Cal Bullet Question

randomhero

December 26, 2008, 09:05 PM

I am having a bit of an argument with a friend. Here is the question....

Under normal circumstances(average temperature, humidity, conditions), how long would a .50 caliber projectile shot from a Barrett M82 exiting the barrel at 2800 FPS take to travel 2 miles?

My friend is trying to argue that wind resistance has almost no effect on a projectile that small(even a .50 cal bullet).

In an attempt to shed some light on the situation, does anyone have anything to offer?

He is trying to say that the bullet would only be in the air for 3.75 seconds(10560/2800).

I have spent several hours researching the topic, but have unfortunately come up short.

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dixierifleman

December 26, 2008, 09:07 PM

It will lose velocity as it travels downrange, so his answer is wrong automatically. I'm sure there is some equation that you can factor in weight of the bullet, FPS, and distance. But i have no clue, I suck at math.

Grizfire

December 26, 2008, 09:14 PM

I recall watching a show about snipers where they actually calculate wind resistance, among other variables, to make the shot. I think it was 'future weapons' but I could be wrong.

taliv

December 26, 2008, 09:23 PM

tell your friend he needs to rethink that for a while.

while a bullet is more aerodynamic than most large objects, and so is affected less than most large objects, it's due to shape, not size.

ballpark, 5.7 sec for a large, high BC match bullet. military ball ammo would take a little longer as it has a lower BC

tell him to go here: http://www.eskimo.com/~jbm/calculations/traj/traj.html

and pick any bullet and velocity and calculate, and look at the velocity drop at every range

JWF III

December 26, 2008, 09:36 PM

I'm sure that someone with a ballistic software program will be along with an answer sometime.

But yes, your friend is wrong. The bullet will slow down because of drag. The amount that is slows down has more to do with the bullet's ballistic coeffiecent. IOW a .50 caliber ball (like from a muzzleloader) will slow down much faster than a Hornady A-Max from a .50 BMG. The weight and velocity have nothing to do with the rate of velocity loss.

As far as a formula goes- this would get you close...

R/((Vm+Vd)/2)=T

R=range of shot

Vm=muzzle velocity

Vd=downrange velocity measured at R

T=time

This would work out to the range divided by the average velocity at that range.

The amount that a bullet decelerates may or may not be constant. You'll have to ask someone a lot smarter than me. If it is constant, this formula would give you an exact time of flight. If it isn't constant, this formula would probably be close enough for discussion (outside of a physics lab) purposes.

So now, the only question is, What is the velocity of the .50 caliber bullet at 2 miles.

I did the math (on a pure guess of downrange velocity). If the velocity is 1200 fps (picked because it makes for easy math, and probably not too far off, maybe a little high), added to the muzzle velocity of 2800. Equals 4000, divided by 2 gives an average (assuming constant deceleration) velocity of 2000 fps. 10560 feet divided by 2000 fps, equals a time of flight of 5.28 seconds.

Like I said, this is close enough for discussion purposes, unless you're in a physics lab.

Wyman

Edited to Add- I told you that someone would be along with the software. Taliv beat me to it. Well I did do a lot more typing in my answer.

psyopspec

December 26, 2008, 09:40 PM

Under normal circumstances(average temperature, humidity, conditions), how long would a .50 caliber projectile shot from a Barrett M82 exiting the barrel at 2800 FPS take to travel 2 miles?

For arguments sake, we'll go with 50 degrees with a barometric pressure of 29.53. How much wind are we talking here? Again, for the sake of an example, we'll go with a 10mph crosswind.

Given the above, with a 655 gr projectile doing 2800 FPS with a .65 ballistic coefficient (rough avg for mil-surp .50): At 1000 yards, you're talking 64 inches of wind drift. That's over 5 feet, and we're not even 1/3 of the way to your two mile target. I'd say wind is probably going to play a role in the shot.

Do your own ballistic calculations with the Hornady External Ballistics Calculator. (http://www.hornady.com/ballistics/ballistics_calculator.php)

Like the one listed above, but more user friendly IMO.

Chuck Dye

December 27, 2008, 12:51 AM

Oehler's Ballistic Explorer lists only one .50BMG, a PMC 660 grain bullet with a 0.427 B.C. and 3080 fps muzzle velocity. Time of flight to 2500 yards, the greatest range processed by the software, is 6.74687 seconds. Retained velocity is 691 fps. Resetting muzzle velocity to 691 and looking at time of flight to 1020 yards adds 5.16739 seconds. Estimated time of flight to two miles is 11.91426 seconds. I have not the foggiest notion whether that two part calculation is valid 'though I know of no reason why it should not be. Yes, if you see the shot launched, ducking is not required, just amble a few yards to one side...

rjsixgun

December 27, 2008, 12:58 AM

Drop a .50 cal bullet from the same hight your barrel is being fired from, both bullets will hit the ground at the same time provided the bullet is fired perfectly straight.

Isac Newton answered that question many years ago with a cannon.

Chuck Dye

December 27, 2008, 01:15 AM

A foggy notion: that two part calculation gives time of flight for two arced trajectories, both starting with barrels slightly elevated above the horizontal. In a two mile shot, the final 1020 yards begin with the bullet descending. Still, time of flight is substantial.

MTMilitiaman

December 27, 2008, 02:30 AM

Drop a .50 cal bullet from the same hight your barrel is being fired from, both bullets will hit the ground at the same time provided the bullet is fired perfectly straight.

Isac Newton answered that question many years ago with a cannon.

But if we're firing out to 2 miles, we aren't talking about a straight or level barrel. A significant amount of elevation is going to be required at the muzzle end to propel the projectile to that range.

HeavenlySword

December 27, 2008, 02:46 AM

Is this related to the longest shot taken, in Afghanistan?

By that canuck?

Its 1.51 miles, however, finding significant elevation is not a problem in mountainous Afghanistan

Sunray

December 27, 2008, 03:01 AM

Your buddy is confused. The wind will move a 747, a 155mm artillery shell and a .750 grain, .50 BMG bullet. How much each moves is the issue.

As mentioned, the bullet doesn't stay at 2800fps either.

Art Eatman

December 27, 2008, 11:56 AM

First off, it's the friction between air and projectile which slows anything, whether artillery shell or rifle bullet. Or an Olympic javelin, for that matter.

As far as the time of flight of the .50, it's probably halfway reasonable to figure a velocity at impact of maybe 1,800 ft/sec. So, average that with the muzzle velocity; say around 2,300 ft/sec overall.

So, 10,560 feet divided by 2,300 ft/sec = 4.6 seconds.

If my guess is off, and the velocity at impact is only 1,000 ft/sec, the average is then 1,900 ft/sec and the time would be 5.6 seconds.

So, "Aw, about five seconds," would be close enough for the average bar bet.

Hud

December 27, 2008, 01:33 PM

Might I suggest a download & read of chapter 3 of this Canadian manual:

http://stevespages.com/zip/canadian_b-gl-306-006fp-001%20-%201_june_1992.zip

Regards, Hud

Dr. Tad Hussein Winslow

December 27, 2008, 01:41 PM

Oehler's Ballistic Explorer lists only one .50BMG, a PMC 660 grain bullet with a 0.427 B.C. and 3080 fps muzzle velocity. Time of flight to 2500 yards, the greatest range processed by the software, is 6.74687 seconds. Retained velocity is 691 fps. Resetting muzzle velocity to 691 and looking at time of flight to 1020 yards adds 5.16739 seconds. Estimated time of flight to two miles is 11.91426 seconds. I have not the foggiest notion whether that two part calculation is valid 'though I know of no reason why it should not be. Yes, if you see the shot launched, ducking is not required, just amble a few yards to one side...

While that 11.9 seconds 2-step calculation is indeed perfectly valid using your numbers, most .50 BMG bullets have better BCs, and the 750 Hornady A-Max for example has a BC more than twice that good, 1.050 in fact. Can anyone recalculate that for 3,520 yards with a 1.0 BC and 2700-2750 MV? (psyopspec, that Hornady calculator does not show TOF). The .mil 668 grainer runs about 2910 fps, but the 750 would be slower (supposedly right at 2700 with a 36" bbl, so we'll run with 2,700).

Edit:

Here, I got it - with the 2700 fps Hornady A-Max with 1.050 BC:

It would take 1.310 seconds to go 0-1000 yards, with an ending MV of 1,940 fps.

It would take 1.869 seconds to go 1000-2000 yards, with an ending MV of 1,344 fps.

It would take 2.617 seconds to go 2000-3000 yards, with an ending MV of 1,014 fps.

It would take 1.549 seconds to go 3000-3500 yards, with an ending MV of 930 fps.

For a total TOF (time of flight) of 7.345 seconds (plus however long to go that last 20 yards). That's at 500 ft above sea level with a 70 degree F ambient temperature.

So the answer is, "Somewhere between 7 1/2 and 13 seconds, depending on the bullet BC - and weight/MV".

:)

rcmodel

December 27, 2008, 01:52 PM

Hornady #6 manual goes out to 1760, or 1 mile.

The 750 grain A-Max BC is 1.050.

MV of 2,850.

1 mile = 1,527 FPS.

Drop from 100 yard zero = 960.2".

rcmodel

camies

December 27, 2008, 04:55 PM

Why not throw some lead down the range and time it?

matrem

December 28, 2008, 08:04 PM

I'd agree with your buddy,IF you were shooting on the moon..

But here ? No way!

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