In order to participate in the GunBroker Member forums, you must be logged in with your GunBroker.com account. Click the sign-in button at the top right of the forums page to get connected.
Options

G forces on a bullet

jonkjonk Member Posts: 10,121
edited August 2008 in Ask the Experts
I'm curious as to how man G forces a bullet undergoes in accelerating from 0 to say 2900 fps over the length of say a 26" barrel.

Someone comment on my math because the answer sounds pretty high to me.

v squared
a= ____________
2s

Where v is muzzle velocity and s is the barrel length. I don't know why this formula works but foudn it on http://hypertextbook.com/facts/2003/MichaelTse.shtml

Now assuming that is right, we get:

a=2900fpsX2900fps
__________
2.166feetX2


Or 1792,582 fps/s

Gravity acceleration is about 9.8f/s/s

Or 182,916 gees, roughly!?

That can't be right.

Ideas?

Comments

  • Options
    gotstolefromgotstolefrom Member Posts: 1,479 ✭✭✭✭✭
    edited November -1
    Right off the bat...UNITS PROBLEMS ...I HATE IT , over and over....
    Use english units, and compute the bullet acceleration correctly.

    Gravitational Acceleration is 32.2 ft/s/s
    ( 9.8 is the metric equivalent )

    Acceleration isn't the muzzle velocity squared, as you have it.

    >> 0 ft/s @ chamber, and your muzzle vel. at muzzle for delta-V
    >> travel time in bbl using avg V and bbl length for delta-T

    I edited.....I forgot a ( 1/x) in first reply

    I get about 70kG's ...still a LOT !

    I check it like this
    1G for 1 sec = velocity of 32.2 ft/s
    32.2 ft/s/s * 1s = 32.2 ft/s

    MY gun has 2 foot barrel with muzzle vel. of 3000 ft/s
    ..the average velocity is 1500 ft/s
    ..my bullet is in the bore for 1/750 s = 0.00133 s

    ? ft/s/s * 0.00133 = 3000 ft/s

    3000/0.00133 ~ 2255600 ft/s/s , and divided by 32 ~ 70,000

    [?]RE: Jonk's reply....
    That may be why when I would try to work back with a gun weight to get a recoil force....it never is as much as a REAL WORLD recoil.
  • Options
    jonkjonk Member Posts: 10,121
    edited November -1
    Thanks!

    Of course, this all assumes a CONSTANT acceleration- which with some powders may be the case- but in practice the bullet accelerates at an uneven rate I'm sure, with most acceleration in the first 1/3 of barrel.
  • Options
    CubsloverCubslover Member Posts: 18,601 ✭✭
    edited November -1
    Ok Geniouses...now calculate the G's that a point on the Bullet's Jacket undergoes from the Spinning. [:)]
    Half of the lives they tell about me aren't true.
  • Options
    bpostbpost Member Posts: 32,657 ✭✭✭✭
    edited November -1
    quote:Originally posted by cubslover
    Ok Geniouses...now calculate the G's that a point on the Bullet's Jacket undergoes from the Spinning. [:)]


    Me thinks that would be centrifugal force from the rotation of the bullet, not G's from acceleration......
  • Options
    CubsloverCubslover Member Posts: 18,601 ✭✭
    edited November -1
    quote:Originally posted by bpost1958
    quote:Originally posted by cubslover
    Ok Geniouses...now calculate the G's that a point on the Bullet's Jacket undergoes from the Spinning. [:)]


    Me thinks that would be centrifugal force from the rotation of the bullet, not G's from acceleration......


    Yeah, that too.

    What about the Lateral G's a Bullet experiences during deflection.
    Half of the lives they tell about me aren't true.
  • Options
    tsr1965tsr1965 Member Posts: 8,682 ✭✭
    edited November -1
    quote:Originally posted by cubslover
    Ok Geniouses...now calculate the G's that a point on the Bullet's Jacket undergoes from the Spinning. [:)]


    That would be dependant on the twist rate as a variable too, along with the muzzle velocity, bullet diameter, and where in the barrel you are asking about.

    Assuming that it is a 1 in 12 twist, and exiting the muzzle at 3000 FPS, that is 180,000 RPM. So if we figure in a barrel of conventional length of 24"(sorry Nononsense), that would be about 62.5 g's
  • Options
    TxsTxs Member Posts: 18,801
    edited November -1
    Ever get a bullet stuck in the bore and have to literally pound it out?

    Besides acceleration and simple rotational forces, you also have the force applied to the bullet as it's swaged down that twisting bore.
Sign In or Register to comment.