Litz notes the definition of stability:

Stability is the ability of a projectile to maintain its point forward orientation in flight, and return to that orientation if disturbed.

That's it. If you want your bullet to fly straight, point forward to the target you aimed it at it needs stability. We are only talking about one type of stability in this post, that is gyro stability and we accomplish that through spinning of the bullet. A good book like Litz's will cover this in more detail, I'm just giving you the low down.

We spin the bullet with the use of barrel rifling (or in the case of a rifled slug, the slug is rifled for a smooth bore shotgun). The amount of spin imparted is function of the twist rate and the bullets velocity. It really is that simple. However, the amount of spin needed to achieve stability is a bit harder.

Twist rate is commonly expressed as a ratio such as 1:10", which reads one turn/twist/revolution per 10" of barrel. It means exactly that, the rifling will complete one revolution, a complete circle, ever ten inches. If you have a 20" barrel you will have 2 revolutions of twist (not exactly as the chamber is usually included, but let's keep it simple). If you have a 30" barrel you have 3 revolutions. 10" is 1 rev, 15" is 1.5 revs, etc., but that really isn't important. What matters is the rate, not the number of revs you have (simplified). To illustrate this, lets say I have a 1:10" twist barrel that is 10" long chambered in cartridge A and a 1:10" twist barrel that is 20" long chambered in cartridge B. If the muzzle velocity (the second variable in determining how much spin we impart to the bullet) of cartridge A and B is equal, both bullets will spin at the same rate. It does not matter that B has twice the barrel length as A! The twist rates and muzzle velocities of both A and B are the same, therefore they have the same imparted spin.

To illustrate this let's look at how much spin is imparted. I'm using very simple numbers, which makes them unrealistic, but the math is correct. Let's say that it takes 1 second to cover that 10" 1:10" barrel for cartridge A. That would impart a spin of one revolution per second (1RPS). If I loaded a max cartridge A load and that doubled the velocity so that it only took 0.5secs to cover that 10" barrel, the bullet would be spinning at 2RPS. If I loaded a light load and it halved the initial velocity, it would take 2secs and would spin at 0.5RPS. Of course I'ved ignored all kinds of things, but the important thing to understand is that the bullets velocity and the barrels twist rate are all that you need to know to calculate how fast it will spin. As I said before though, how much spin you need is more complex.

There is a simplified estimate, an equation, called the Miller Stability Formula that tells us roughly if a bullet will be stabilized (gyroscopic) or not. The thing to realize is that it is a simplification of much more advanced math and isn't calculating the actually gyroscopic stability, but rather approximating it. However, it works well in most cases. The formula is:

SG = 30 * m / (t^2 * d^3 * l * (1 + l^2)), where

SG = the gyroscopic stability factor which is basically the stability of the spinning bullet divided by the overturning torque due to drag.

m = mass in grains

t = rifling twist rate in calibers per turn

d = diameter of the bullet in inches

l = the length of the bullet in calibers

If you've been paying attention you should immediately question where the bullets velocity is! Simply put, it isn't there because this formula only works for 2800fps and no other velocity. Thankfully we have a simple correction to the calculated SG which is:

SG(from above) * (V / 2800)^1/3, where V is the bullets muzzle velocity in fps.

Even better is that we have a free calculator do the math for us! It's found here: http://www.jbmballistics.com/cgi-bin/jbmstab-5.1.cgi

A velocity correct SG of less than 1 is unstable, between 1 and 1.4 is marginally stable, and greater than 1.4 is stable. Remember this is only an approximation!