How long does a rifle bullet take to travel a mile?
- Thread starter Flechette
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wally
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If you have decent estimates of the ballistic coefficients and the muzzle velocity for the projectiles, an on-line ballistics calculator will tell you everything.
I like:
http://www.jbmballistics.com/index.shtml
I like:
http://www.jbmballistics.com/index.shtml
aarondhgraham
Member
That's just basic arithmetic.
5280 (feet in a mile) divided by 2600 (feet per second) equals 2.0307 seconds (at least).
Then add just a few milliseconds because the bullet does slow down just a bit.
Aarond
.
5280 (feet in a mile) divided by 2600 (feet per second) equals 2.0307 seconds (at least).
Then add just a few milliseconds because the bullet does slow down just a bit.
Aarond
.
mokin
Member
I would go with what Aarond said.
Your question reminds me of a mathematical exercise I did regarding the design and construction of model rockets. It used the weight of the rocket, the amount of thrust, etc. to calculate the flight ceiling, then figure the amount of time it would take the rocket to fall back to earth and determine whether or not the parachute would deploy in time to save the rocket. It was a really fun way to get into calculus. I suppose you could do something similar to answer your question, but I reckon you'd have to figure in the trajectory of the bullet, and possibly gravity as it climbed and fell.
Your question reminds me of a mathematical exercise I did regarding the design and construction of model rockets. It used the weight of the rocket, the amount of thrust, etc. to calculate the flight ceiling, then figure the amount of time it would take the rocket to fall back to earth and determine whether or not the parachute would deploy in time to save the rocket. It was a really fun way to get into calculus. I suppose you could do something similar to answer your question, but I reckon you'd have to figure in the trajectory of the bullet, and possibly gravity as it climbed and fell.
OP
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Flechette
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It is that extra time that I am looking for.
Jason_W
Member
Never mind
Elkins45
Member
That turns it from being a simple arithmetic problem into a calculus problem. I haven't used calculus since I too the class in 1983---I'm out. There are several online calculators that will do time of flight calculations.
.308 Norma
Member
College calculus isn't even visible in my rearview mirror anymore, but it seems to me a person would have to know how fast the bullet was going when it crossed the one mile mark, not just how fast it going when it left the muzzle before they'd have a prayer of figuring how long it took to travel that mile.
1stmarine
Member
~2.86 seconds for the lapua.
As several have implied, what you're after is the result of an integral calculus function. However, you can get not only in the ball park, but in the infield without knowing calculus.
Use any one of the available ballistic calculators ( Hornady's calculator is a good one). If the one you choose gives the exact flight time, then you're done, but most don't. In that case you can dummy up the calculus by using the most granular interval they have, say 100 yards. Input the Ballistic Coefficient (you can get that from the loading manuals or the manufacturer for your particular bullet), the muzzle velocity, and the distance (1760 yards). You can ignore wind, sight height, humidity, etc. - go with the defaults.
Once you have the output table, just divide each interval distance (300 feet if you chose 100 yard intervals) by the bullet velocity for that interval. Then add up all the time factors you get. For the example here, you will do 17 easy divisions.
It's ok to assume the bullet's velocity was a constant during the interval. That's the way an integral calculus function would do it, but the intervals would tend towards 0 feet with almost an infinite number of intervals. The difference in the pseudo way I showed and the real function would be insignificant for your purposes.
Use any one of the available ballistic calculators ( Hornady's calculator is a good one). If the one you choose gives the exact flight time, then you're done, but most don't. In that case you can dummy up the calculus by using the most granular interval they have, say 100 yards. Input the Ballistic Coefficient (you can get that from the loading manuals or the manufacturer for your particular bullet), the muzzle velocity, and the distance (1760 yards). You can ignore wind, sight height, humidity, etc. - go with the defaults.
Once you have the output table, just divide each interval distance (300 feet if you chose 100 yard intervals) by the bullet velocity for that interval. Then add up all the time factors you get. For the example here, you will do 17 easy divisions.
It's ok to assume the bullet's velocity was a constant during the interval. That's the way an integral calculus function would do it, but the intervals would tend towards 0 feet with almost an infinite number of intervals. The difference in the pseudo way I showed and the real function would be insignificant for your purposes.
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Archie
Member
Flechette, I understand your question and it's hard. That 'extra time' depends on the ballistic coefficient of the bullet and if one has a head wind or tail wind. Wally mentioned JMBallistics.com . I use it often and it seems to factor in loss of velocity due to air resistance as one much look up and enter the ballistic coefficient of the bullet in question.
I'd say it's going to be as close as one can get without a chronograph with a really long electrical cord to the stop sensor.
GBExpat
Member
Why?
J-Bar
Member
And there are going to be two solutions:
One for the most shallow angle at which the bullet can be shot to travel a mile, and one for the high trajectory path in which the bullet is launched at a more vertical angle and falls back to earth a mile away after reaching zero velocity at the apex.
I assume you are interested in the first, rather than the second.
One for the most shallow angle at which the bullet can be shot to travel a mile, and one for the high trajectory path in which the bullet is launched at a more vertical angle and falls back to earth a mile away after reaching zero velocity at the apex.
I assume you are interested in the first, rather than the second.
Dog Soldier
member
What is this, a gun forum's answer to the "Old Rubix Cube"?
Elkins45
Member
Yeah, that would do it. My compliments for an elegant solution.
DownInTheDark
Member
A 50 BMG going 2700 fps at the muzzle would take 2.8 seconds to travel 1800 yards
1760 yards are in a mile.
750 gr bullet with a 1.05 bc.
The velocity would be 1395 fps.
3240 lbs of energy
The bullet would drop 1160 inches
You can play around with this: http://www.jbmballistics.com/cgi-bin/jbmtraj-5.1.cgi
1760 yards are in a mile.
750 gr bullet with a 1.05 bc.
The velocity would be 1395 fps.
3240 lbs of energy
The bullet would drop 1160 inches
You can play around with this: http://www.jbmballistics.com/cgi-bin/jbmtraj-5.1.cgi
The OP said that he was interested in the time down to the millisecond, that's an accuracy of 1/1000 of a second.
2.86 seconds is only down to 1/100 of a second and 2.8 seconds is only accurate to 1/10 of a second so neither of those answers work.
Actually, I guess I could see this kind of question if you wanted to shoot a moving animal that is a mile away. An animal moving at 25mph will move about .44 inches in a millisecond. Lets assume that you have a kill zone of 8 inches, if you aim at the center of the kill zone then you don't want an error of more than 4 inches from the point of aim. If your speed estimate were off by more than 9 milliseconds then you'd hit the target outside the kill zone. If the animal were moving faster than 25mph you would have to be even more accurate.
There are two problems though, I doubt if you'll ever find a ballistics calculator that could be accurate down to the millisecond and I think it would be pretty unethical to even take the shot.
2.86 seconds is only down to 1/100 of a second and 2.8 seconds is only accurate to 1/10 of a second so neither of those answers work.
Actually, I guess I could see this kind of question if you wanted to shoot a moving animal that is a mile away. An animal moving at 25mph will move about .44 inches in a millisecond. Lets assume that you have a kill zone of 8 inches, if you aim at the center of the kill zone then you don't want an error of more than 4 inches from the point of aim. If your speed estimate were off by more than 9 milliseconds then you'd hit the target outside the kill zone. If the animal were moving faster than 25mph you would have to be even more accurate.
There are two problems though, I doubt if you'll ever find a ballistics calculator that could be accurate down to the millisecond and I think it would be pretty unethical to even take the shot.
DownInTheDark
Member
Well then you can play around with the ballistics calculator and add the .0002 of a second I left out.
And that's for 1800 yards, not 1760.
And that's for 1800 yards, not 1760.
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Ramone
Member
As others have noted, there are ballistic calculators available to solve for specific cartridges.
I don't know the time in flight from a mile, but an off the top of my head, best of my recollection example is that a M118 7.62NATO takes about .6 seconds to travel 500 yards, and 1.6 seconds to reach 1000 yards.
I don't know the time in flight from a mile, but an off the top of my head, best of my recollection example is that a M118 7.62NATO takes about .6 seconds to travel 500 yards, and 1.6 seconds to reach 1000 yards.
J-Bar
Member
This discussion reminded me of a somewhat related article, linked below:
http://powderburns.tripod.com/sharps.html
...so if you can't do calculus, maybe you can find someone with a radar gun!
(For some of you who don't recognize the author, Mike Venturino writes for several gun magazines, and has written several books specializing in historic and blackpowder guns.)
http://powderburns.tripod.com/sharps.html
...so if you can't do calculus, maybe you can find someone with a radar gun!
(For some of you who don't recognize the author, Mike Venturino writes for several gun magazines, and has written several books specializing in historic and blackpowder guns.)
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JWF III
Member
Well then, to get that close you'd need more info. As another has said, the angle of the shot will make a difference. Also needed are air temp, elevation, barometric pressure, wind speed, angle of wind to the barrel, actual mv(not manufacturer's claim), etc. 1/1000th of a second is a very short time, every little thing will make a difference. And even at that, a load's s.d. will unlikely be 0, and rarely below 10, so it will change from shot to shot.
Wyman
1stmarine
Member
The ballistics models that we have are pretty accurate like the pejsa model (Arthur Pejsa).
This model and other enhanced versions of it account for all deceleration constant factors so for each segment in the trajectory one will
be able to compound the time using very simple algebra. The total is the time to the target. Any decent ballistics calculator will give you that too.
No matter what caculator or sofware or if you do it with pen and paper it is always going to be an approximation with a small margin of error.
But if that is not good enough then one might use a Doppler radar otherwise discussions about estimated milliseconds differences will be pointless.
This model and other enhanced versions of it account for all deceleration constant factors so for each segment in the trajectory one will
be able to compound the time using very simple algebra. The total is the time to the target. Any decent ballistics calculator will give you that too.
No matter what caculator or sofware or if you do it with pen and paper it is always going to be an approximation with a small margin of error.
But if that is not good enough then one might use a Doppler radar otherwise discussions about estimated milliseconds differences will be pointless.
Sav .250
Member
Now there`s a question/answer that will stay with me for ever.......................
buck460XVR
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With the huge amount of variables involved, the only way to get an answer that accurate, is to have a timer with sensors telling you when the bullet left the barrel and when the bullet actually hit the target. Anything else is just a educated guess/calculation and is theoretical. . How does one measure the difference in wind speed/direction, thermals, temperature and humidity over the mile the bullet has to travel, all of which can and do change constantly. All of those will have a influence on bullet speed, especially if one wants accuracy down to 1/1000th of a second.
230RN
2A was "political" when it was first adopted.
That's about my recollection of a .30-06 in 1000 yard matches judging from the dust off the backstop. I figure it's enough time to duck if you see a tracer coming at you from that range.
Millisecond precision on any of these calculations is putting too fine a point on something that has too many indeterminate variables... like whether it's a boat tail or not, et cetera, et cetera, et cetera. You can only cite a number with +/- X seconds. And since you are not concerned with temperature and humidity (etc, etc, etc), that + and - will be much larger than the "milliseconds" you are looking for.
Even sophisticated battleship fire control systems which account for pitch and roll and humidity and firing platform movement and the Coriolis effect, etc, etc, etc, still need spotters at long ranges most of the time.
Them there et ceteras will getcha every time.
I liked MacGrumpy's answer.
Terry, 230RN
ETA REF:
https://en.wikipedia.org/wiki/Coriolis_force
PS, THR: I just added Coriolis to your dictionary.
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