A 0 G roll for inverts doesn't need to be flat, it depends on the motion of the heart line. If the heart line follows the path of a projectile relative to its speed (The path of an object being thrown at an angle) then you will get 0 G?????????s all the way through, its really simple physics.
Sorry for all the math's but I had to just to make sure people understand the concept.
Here?????????s a formula representing a projectile: (Requires some physics and calculus skills)
// Vk is the initial velocity in km/h at the angle Ad, therefore these values need to be initialized before plotting this.
// g is the constant for gravitational acceleration.
// Try it on paper if you don't have elementary or if you don't know how to use it.
V = (Vk * 1000) / 3600 // converts velocity into meters per second.
A = Ad * (Pi / 180) // converts the angle into radians.
iVy = sin[A] * V // gets the velocity relative to the y axis.
iVx = cos[A] * V // gets the velocity relative to the x axis.
time = (-2 * iVy) / g // gets the time it requires to finish the motion.
Sy = (iVy) * (0.5 * time) + (0.5 * (g) * (Pwr[(0.5 * time),2])) // gets the displacement for the y axis in half the time, thus the maximum height of the motion.
Sx = time * iVx // gets the full displacement for the x axis.
Ac = Sy / (Pwr[(Sx * 0.5),2]) // retrieves a ratio at which the gradient changes relative to the x axis.
x = T * Sx // function for the length of the projectile motion.
y = -Ac * (Pwr[x,2]) + (Ac * Sx * x) // complete function for the projectile motion.
Remember that this is the function for the heart line thus the banking can be changed and then you can use AHG to get a perfect 0 G roll or just a hill or what ever the hell you want. [:D]