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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$\begingroup$ $1g = 10m/s^2. 7g = 70m/s^2. 7g*1s = 70m/s. \textrm{arctan}(70/900) = 4^{\circ}$ You should see only a very small turn. $\endgroup$– Mark EichenlaubNov 23, 2010 at 7:44
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2 Answers
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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.)
(I don't know how do you get the 7 g.)
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$\begingroup$ The 7G was got by substituting 7 in place of G-force in my first equation. After substitution of G-force and velocity, I got omega, which I used in the constant turn motion model. $\endgroup$– NavNov 23, 2010 at 6:28
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$\begingroup$ @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 $900\times2\pi/9.8=577g$. $\endgroup$– kennytmNov 23, 2010 at 6:43
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$\begingroup$ :) can't be 577g. omega is in radian/sec, so for a 7G turn, omega would be 0.0539, right? This was from the first equation. I've plotted 5 points (simultaneous particle movement positions) in MATLAB and the line has an infinitesimal curve (hardly any curve at all) to it. I'm surprised because pilots experience G forces, and I thought 7G was a heavy force which would cause a sharper curve. $\endgroup$– NavNov 23, 2010 at 7:04
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1$\begingroup$ @Nav: 1 full cycle (if that means 1 turn) has 2π radians, so the angular speed is 2π ÷ 1 second = 2π rad/s. But is your "1 second" means the time passing through that 5 points? If those 5 points only make an arc of 4° then it's reasonable. Remember your speed is 900 m/s, i.e. 2.6 times the speed of sound. So even when you are circling in 82 seconds per cycle, it still requires a lot of centripetal force. $\endgroup$– kennytmNov 23, 2010 at 7:17
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1$\begingroup$ @Nav: meta.stackexchange.com/q/70559/145210 $\endgroup$– kennytmNov 23, 2010 at 7:51
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g-force is apparent weight/true weight therefore g-force is ma+mg/mg .
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$\begingroup$ I suppose you mean $(ma+mg)/mg$ (which reduces to $(a+g)/g$)? $\endgroup$ Nov 9, 2014 at 12:08