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On AR-15 Twists

by Clint McKee

Hi, Clint! Why do you sell the Krieger AR barrels with a 1/7 twist and the other barrel with 1/8 twist? Instead of the 1/9 twist. Reports from the military say that the 1/7 twist burns out quicker.

We sell Krieger barrels for the AR15 in 1 in 7.75 (that's almost 8!), as well as our NM barrel in 1 in 8. All things being equal (BTW, they never are, equal), the faster the twist, the quicker it will wear out. So, if you want a barrel that will last a long time, get a non-NM chrome lined barrel with a 1 in 14 twist. On the other hand, if you want the best barrel possible, for the best bullets possible, you will need a twist right around 1 in 8 for the Sierra 69 grain bullet (the benchmark cartridge is the Federal Gold Medal Match cartridge, with the 69 grain Sierra) or the 80 grain VLD Sierra (alas, not currently loaded commercially).

Also what bullet (.223) would you recommend for the 1/9 twist?

The U.S. military 62 grain ball (ss109) will work nicely in a 1/9 twist, but also works nicely with the 1/8. The very light bullets, say 55 grain & less, work nicely with a 1/12 to 1/13 twist.

It all boils down to math (and real world considerations like bullet quality, barrel quality, barrel length, assembly technique, ....etc.). The lighter bullets (55 grain & less) need a slower twist rate per Greenhill's Formula*.

Likewise, a heavy bullet (69 grain & more) needs a much faster twist.

Thanks for your attention.

Clint McKee

*From a series of posts by Steve Faber:

A spinning bullet behaves a lot like a spinning top. Imagine a top or toy gyroscope that is spinning and balanced on a post. If the spin is high enough it will remain balanced, and if it gets below a critical point the gyro will fall off. Now think of the spinning bullet as a top balanced on its base. Instead of the force of gravity, we have the air resistance acting on it from the head of the bullet, and the spin counteracting it. If the spin decreases enough, the bullet will fall or tumble, otherwise it will stay heading into the wind although there may be some precession evident. With the bullet problem, the spin is determined by the barrel twist and the bullet velocity. The drag is dependent on the air density, the cross sectional area of the bullet and the shape. The drag is expressed with the following equation:

drag = Cd * r^2 * v^2 * p where

Cd is the unitless coeff of drag (varies with velocity)
r is the radius of the bullet
v is the velocity of the bullet
p is the air density.

If Cd is constant, this says the drag is proportional to v^2. There is a table of drag force as a fuction of velocity for a standard projectile in Hatcher's Notebook, which one can graph and obtain Cd as a function of velocity. In computing Cd(v), r=.5in for the standard projectile and p = 1.221 g/liter, standard air density used for the table, 60 deg. F, 30" Hg, and 66% RH.

This shows Cd rises from about 0.1 at 900 ft/sec just below the speed of sound to almost 1.0 around 1500 ft/sec, just above the speed of sound, and then drops off slowly from there reaching a value of 0.77 at 3600 ft/sec. The region between 1500 and 3600 ft/sec fits nicely to a line:

Cd = 1.146 - 1.047E-4 * v

Cd is 0.98 at 1500 ft/sec and decreases by 0.01 every 100 ft/sec. Since the drag of an arbitrary bullet may vary from that of the standard projectile, the form factor "i" is introduced to scale the observed to standard drag. This is related to the ballistic coefficient (BC). i = m/(d^2*BC) where m is the mass in lbs and d is the diameter of the bullet in inches. So the drag for an arbitrary bullet is:

drag = Cd(v) * i * r^2 * v^2 * p

Moulton's book on Exterior Ballistics gives the criterion for stability when

transverse (tumbling) angular acceleration < 1/4 * w^2 * (Il/It)^2

where It is the transverse moment of inertia
and Il is the longitudinal moment of inertia
and w is the longitudinal angular velocity ( the spin )

the transv. ang. accel = drag * moment arm / It

w depends on the barrel twist and initial velocity "v"

w = 2*pi*v/tw

where tw is the length of one turn of the barrel twist.

Now we have everything we need to write the equation for barrel twist needed to stabilize the bullet when it just exits the barrel. Substituting for drag and using "Lm" for moment arm:

tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i))

Note that velocity cancelled out, and the only velocity dependence is due to Cd(v).

Velocity here was the initial (muzzle) velocity, but what happens down range? It would be safe to assume that the spin decays slower than the forward velocity, so with the decreased drag as the bullet slows down range, it should become more stable. As long as the bullet is supersonic above 1500 ft/sec, we would be safe to assume stability is determined at the muzzle exit point.

To calculate the twist, we know Cd(v), p, r, and i can be determined from the ballistic coefficient (BC), by dividing the sectional density by the BC where sectional density is defined as the mass in lbs divided by the diameter of the bullet in inches. Il and It remain to be determined. Lm, is the length from the base of the bullet to the average point of the drag force.

Next, we will use the moments of inertia for a cylinder and derive the Greenhill equation. After that we can calculate moments for a boattail bullet with an ogive point and have a much more accurate method of determining the twist. Also since Cd(v) is now known, we know the velocity dependence.

Last time we derived the equation for twist needed to stabilize a bullet:

tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i)) eq (1)

Cd is the unitless coeff of drag (varies with velocity)
Cd = 1.146 - 1.047E-4 * v
r is the radius of the bullet
v is the velocity of the bullet
p is the air density.
i is the form factor i=m/((2*r)^2*BC)*in^2/lb
Lm is the moment arm of the drag force
It is the transverse moment of inertia around bullet base
Il is the longitudinal moment of inertia

Lets assume a cylindrical bullet so
Il = 1/2*m*r^2
It = 1/3*m*L (approx) where L is length of the cylinder
Lm = L
m (mass) = d*pi*r^2*L

From the above equation we can write:

tw/(2*r)*L/(2*r) = sqrt( (3*pi^3*d)/(4^3*Cd*i*p) ) eq (2)

This is the Greenhill equation relating the twist length in calibers times the bullet length in calibers equal to a constant. According to Hatcher, the bullet used to derive the constant was a Krag 220 grain, L=1.35in., v=2000ft/sec, d=10.9 gm/ml, p=1.221gm/l For i=.61, the constant comes out to 150.

The Greenhill equation gives a lot of insight into the problem, showing that the twist is mainly determined by the length to radius ratio and sqrt of the bullet density to air density ratio.

Modern bullets have an i a bit less than .6 sometimes, and the velocities used are higher, so the Cd is a bit lower - affecting the results.

Now we are in the position to write our own equation for the cylindrical bullet approximation where we can take into account those effects. We could also use the exact form of It = 1/3m*L^2 + 1/4*m*r^2 but the latter term is less than a 1% effect, so we will also ignore it, giving:

tw = sqrt(3*pi^3*r^4*d/(4*Cd*i*p*L^2)) eq (3)

Note that the L from substituting m in Il cancels Lm=L. If the bullet is pointed and not a cylinder, this equation is using L as if it were a cylinder given the density of the bullet. This gives a good approximation for Il. L then is a length to a point somewhere between the cylindrical part of the bullet and the tip which also gives a good approximation for Lm, these then cancel with good accuracy. The remaining L in eq. 3 is from "It". If we use the L calculated from the density here, it would underestimate It, since It is sensitive to the distance the mass is from the base of the bullet. It turns out that using the measured length is a much better approximation.

The table below compares the results of the original Greenhill equation, our version that takes into account "i", and Cd, and equation 1 using accurately calculated values of "Il" and measured values of "It" (more on this later).

1. Greenhill : tw(calibers)*L(calibers)=150
2. equation 3 with appropriate "i" and calculated Cd, measured "L"
3. equation 3 with L determined from the density
4. equation 1 with accurate "Il" and measured "It".

all bullets are FMJ boattail .30 caliber at 2500 ft/sec except the 55 and 62 grain (ss109) which are .224 caliber at 3200 ft/sec

bullet "i" _1_ _2_ _3_ _4_
150 gr .581 12.9 13.1 14.3
168 gr .533 12.0 12.8 12.7
174 gr .53? 11.1 11.8 11.6
220 gr .526 9.5 10.3 13.6
55 gr .626? 10.1 10.6 14.2
62 gr .61? 8.3 8.4 9.8

The results show that equation 3 with the "i" and calculated Cd (column 2) is a good improvement on Greenhill (column 1), and is in good agreement with equation 1 using exact moments of inertia (column 4) for the .30 caliber bullets that have a longer cylindrical section. One should use equation 1 with the bullets that are mostly point with little body, or that have a non-uniform density. Equation 1 requires more refined calculations of It, Il, and Lm.

From article by bartb@hpfcla.fc.hp.com (Bart Bobbitt):

#This is what folks thought was the best in the early 1950s. In the 1960s,
#when the military rifle teams finally began to build really accurate rifles,
#and civilian highpower match rifle builders finally did the same thing,
#it was found not to be true. Match M1 (and service) rifles had a 1:10 twist
#barrel. The M14 had a 1:12 twist barrel. When the M14 started to be worked
#over into a competition rifle, it was found that the slightly slower muzzle
#velocity of the M2, 173-grain match bullet in the M14's 1:12 twist barrel
#shot better than in an M1's longer barrel at higher velocity. This meant
#that the 173-grain bullet was being spun too fast for best accuracy. But it
#wasn't really well understood until the Navy started rebarreling their 30
#caliber M1 rifles with 7.62mm NATO barrels with 1:12 twists. As the match
#conditioning success with M1 rifles was miles ahead of the M14 at the time,
#folks took notice of the greatly improved accuracy of the M2 173-grain
#military match bullet in the 7.62mm NATO M1 rifles. Then the M14s began to
#be match conditioned to shoot as well as the M1s did. Their 1:12 twist
#barrels were soon found to not quite spin the bullets as fast as they needed
#to be in their 22-inch barrels. They went to 1:11 twists and the rest is

According to the above the 173 grain match bullets were used with 1:12 twist barrels. My best current calculation predicts a maximum 1:11.6 in. twist needed for that bullet. This would easily change to about 1:13 in. in the Rockys where the air is less dense, but the above suggests the 1:12 twist worked in general. The guy who sold me the 173 gr bullets also implied they would not work well in a 1:12 in gun (in Iowa.) Maybe I need to go back to the drawing board on this bullet or use a more accurate estimate of the B.C. if it turns out the 1:12 was good at standard air density.

If anyone knows the B.C. for the 173 gr mil. match bullet I'd appreciate it.

From bartb@fc.hp.com (Bart Bobbitt):

# Any given bullet needs to be spun within an RPM range to stablize it.
# Whatever muzzle velocity and twist combination spins the bullet in that
# range will work just fine. Here's the approximate spin rates .243
# spitzer/hollowpoint bullets need to stabilize:

# 65 - 70 grain 155,000 to 165,000

# 71 - 80 grain 165,000 to 190,000

# 81 - 90 grain 190,000 to 215,000

# 91 -100 grain 215,000 to 245,000

# 101 -110 grain 245,000 to 280,000

# These are the spin rates competitive shooters use with 6mm bullets to
# get the best accuracy. Hunting rifles have the same required spin rates
# for their various bullet weights.

These are probably good practical rules of thumb for determining the twists needed to stabilize the above bullets for the twists and velocities normally used, but I believe that specifying a given rpm needed to stablilize a given bullet will not work in general, since the velocity dependence of the stability is only determined by the variation of coefficient of drag with velocity, which is relatively small.

From my previous derivation:

tw = pi*Il/r * sqrt( 1/(Cd*p*Lm*It*i)) eq (1)

Cd is the unitless coeff of drag (varies with velocity)

Cd = 1.146 - 1.047E-4 * v
for velocities between 1500 and 3600 ft/sec (could maybe
extrapolate higher but not lower since the curve bends
over at that point.)

r is the radius of the bullet
v is the velocity of the bullet
p is the air density.
i is the form factor i=m/((2*r)^2*BC) , r in (in.), m in (lb)
Lm is the moment arm of the drag force
It is the transverse moment of inertia around bullet base
Il is the longitudinal moment of inertia

Let's assume a cylindrical bullet so
Il = 1/2*m*r^2
It = 1/3*m*L^2 (approx) where L is length of the cylinder
Lm = L
m (mass) = d*pi*r^2*L

Substituting gives:

tw = sqrt(3*pi^3*r^4*d/(4*Cd*i*p*L^2)) eq (2)

I must confess that I don't have a good theoretical reason to use "It" as moment of inertia around the bullet base rather than around the center of mass of the bullet, and Lm the moment arm from the base instead of from the center of mass. Using the values from the base seems to give the best results consistent with Greenhill's formula (at the velocity he used) and experiment. This substitution does not affect my point though since it would not change the velocity dependence factor.

For an example, here is a 168 grain Sierra:
The table shows the maximum twist length needed to stabilize the bullet at the given velocity and the corresponding RPM as calculated:

velocity twist RPM

1500 fps 12.1 in 89,260
2500 12.8 140,680
2800 13.0 154,700

It is clear that the calculated twist is not terribly velocity dependent, but you would be in trouble if you were to extrapolate the minimum RPM needed based on a result taken at a lower velocity.

From njohnson@nosc.mil (Norman F. Johnson):

# # I have always assumed (perhaps incorrectly) that since a bullet of a given
# # caliber gets longer as it gets heavier that it's WEIGHT was the only reason
# # it would not stabilize in a slow twist barrel. He claims that some of the
# # solid brass/copper type bullets will not stabilize in slow twist barrels for
# # their caliber due to their being too long and not too heavy. The reason the
# # 40 gr. BT's are long for their weight is obviously because it has the long
# # pointy polycarbonate tip. It adds quite a bit to length but very little to
# # its weight.
# May I introduce you to Greenhill:
# The Greenhill formula is an empirical equation that does a good
# job of establishing the barrel twist necessary so that a bullet
# of a given length will be adequately stabilized.
# ...
# Note that it is bullet LENGTH, not weight that is important.
# Greenhill works well with all lead/lead-alloys commonly used for
# bullets.

The Greenhill estimate assumes a constant density bullet. The long pointy BT with the polycarbonate tip will not require as much spin to stabilize the bullet as a similar length bullet that was more dense at the ends. The actual formula depends on the moments of intertia around the transverse axis and the axial axis of symmetry. To get a more stable bullet, you want to get the transverse moment of inertia lower (less mass in the ends).

Steve Faber