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# G forces on a bullet

jonk
Member Posts:

**10,121**
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?

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

1,479✭✭✭✭✭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.

10,121Of 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.

18,601✭✭32,657✭✭✭✭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......

18,601✭✭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.

8,682✭✭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

18,801Besides acceleration and simple rotational forces, you also have the force applied to the bullet as it's swaged down that twisting bore.