# Calculation of the G-force

I have a formula which is $\text{G-force} = \frac{v\omega}{9.8}$, where $v$ is speed and $\omega$ is the angular velocity. I've seen on the internet that G-force is actually $\text{acceleration}/9.8$. I'm confused as to which formula is correct. For simulating the motion of particle taking a turn, would omega simply be velocity divided by radius of turn? Assuming Cartesian coordinates.
Another funny thing I noticed is that while simulating particle motion, a 7G turn showed up as an almost straight line (while using a constant turn motion model) with a velocity of 900m/s and time interval of 1second. Am I simulating wrong or is my use of the first equation wrong?

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$1g = 10m/s^2. 7g = 70m/s^2. 7g*1s = 70m/s. \textrm{arctan}(70/900) = 4^{\circ}$ You should see only a very small turn. –  Mark Eichenlaub Nov 23 '10 at 7:44

The g force is a unit of acceleration. 1 g is equal to 9.80665 m s-2. So the correct formula is $$\text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}.$$
However, when describing uniform circular motion (i.e. $\boldsymbol\omega$ is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the centrifugal acceleration, which is exactly $$a = \frac{v^2}r = v\omega = \omega^2 r,$$ so the first expression is also correct for centrifugal acceleration of uniform circular motion. (If the motion is not a uniform circular motion, only $a = \omega^2 r$ can be used to describe the centrifugal acceleration.)
@Nav: If that's 1 second per turn, i.e. $\omega = 2\pi \mathrm{rad}\,\mathrm{s}^{-1}$, the g force according to the 1st equation should be $900\times2\pi/9.8=577g$. –  KennyTM Nov 23 '10 at 6:43