As promised...
I believe this is from Understanding Firearm Ballistics, by Robert A. Rinker, via LG. Similar quotes may be found in other references:
The unbalanced bullet is forced to conform while in the barrel and its center of mass is revolving around it's geometric center. When the bullet is free of the barrel's constraint, it will move in the direction that its mass center had at the point of release.
OK, so Rinker et al are discussing this as though bullet/crown separation is a 2D problem (it's not, and that concerns me some). This, however, is really easy to run the numbers on, and should be very easy to test to see if range results are as expected (more on that at bottom).
When you spin the CG of a mass about a radius and then release the force that has been accelerating it towards the center of rotation, conservation of momentum dictates that the bullet departs in the direction its center of mass was going at the moment of release. This tangential component of bullet velocity (in the crossrange plane) just depends on muzzle velocity, twist length, and the offset of the CG from the axis of constrained rotation (geometric center) in the bore. Please note that when I say "crossrange" here, I am not just talking about left vs. right, but a plane normal to the downrange direction that includes up-down-left-right directions.
So, if we calculate the distance between the CG and axis of rotation we've got most of our answer.
To give a concrete example/context to this, what I found when I measured some bullets picked from a sample of NOE225107 castings from a particular mould was that the nose halves were about .002-.003 different in length, with one half of the bullet on one side of the parting line apparently being that much longer than the other. A fingernail would catch going one way across the parting line and not the other. The sprue plate produced what looked like a perfectly flat base, and I didn't notice any issue with the radius from one half to the other on the base edge or other abnormality. Concentricity seemed to be decent to under half a thousandth. My conclusion from the above was that the mould halves were about .002 to .003 out of alignment along the axis of the bullet, resulting in one half of the bullet being .002 to .003 thousandth longer than the other.
Let's define some variables we'll need:
dr == distance bullet cg is off the normal from the geometric axis (inches)
TL == Twist Length (inches per revolution)
c == caliber (inches)
v == down range component of velocity (feet per second)
vldr == lateral component of velocity due to dr (inches per second)
wrot == bullet rotation rate (rev/second)
R == Range defined as 100 yards
dMOA == deflection at target in Minutes of Angle from nominal path (one-side deflection, maximum group size increase will be twice this)
NO == Nose Offset (inches)
Wb == Weight Bullet (pounds)
Wdef == Weight bullet defect (pounds)
rdef == radius to center of mass of bullet defect (inches)
Pi == the ratio of the circumference of a circle to the diameter
rhob == density of bullet lead (pounds per cubic inch, ~.4 lbs/inch^3
The CG offset resulting from this is really easy to calculate. I'll treat the rest of the body as axisymmetric (meaning the CG absent the nose offset is coincident with the geometric center).
The mass of the nose offset is given by density times volume:
Wdef = rhob * (NO*c^2*Pi/4/2) = rhob * (NO*c^2*Pi/8)
(dimensional check: lb/inch^3*inch*inch^2= pounds -- yes, pounds are force, not mass, but it ends up working out here because they just end up in a ratio with other pounds)
The radius to the cg of the nose offset defect is given by:
rdef = 4*c/(6*Pi) = c*2/(3*Pi)
The radius of the CG is given by:
dr = 0*(Wb-Wdef)+Wdef*rdef/Wb = rhob*(NO*c^2*Pi/8)*c*2/(3*Pi)/Wb = rhob*NO*c^3/(Wb*12)
(dimensional check: pounds/inch^3/pounds*inch*inch^3=inches)
Now, the bullet rotation rate will be:
wrot = 12*v/TL
(dimensional check works out to 12 inch/ft * ft/sec/(inch/rev) = rev/sec)
And the tangential velocity will be:
vldr = 12*v/TL*dr*2*Pi = 24*Pi*v*dr/TL (inches per second)
The flight time to target will be about:
100 yards * 3 ft / yard/v = 300/v (seconds)
Thus:
dMOA = flight time * vldr *(inch to MOA conversion)= 300/v*24*Pi*v*dr/TL /1.0472 = 7200*Pi dr/TL
(dimensional check: seconds*(inch/second)/(1.0472 inches/MOA@100yd) = MOA
substituting dr:
dMOA = 7200 * Pi/TL * rhob*NO*c^3/(Wb*12) = 573 * Pi * rhob*NO*c^3/(TL *Wb)
(Now, I am neglecting above the impact of drag on lateral velocity/induced angle-of-attack lift that retard the lateral component of velocity as well as any turning of the bullet into the lateral velocity, and the decay of downrange velocity. This should be fine for an approximation at close range, but not so much the further we go out. How the MOA effect will be impacted depends on the relative rate of retardation of the lateral component of velocity and the "turning into the wind" of the projectile in relation to the downrange component of velocity.)
So, dMOA = 573 * Pi * rhob*NO*c^3/(TL*Wb)
First of all, if anyone has the inclination please do check my math. I don't like doing this in free text as it's a lot harder to spot errors in arithmetic this way... I checked units as I went, but anyone who finds something that looks like an error, please let me know.
Now, for our particular example, this yields:
dMOA = 573 * Pi * .4*.003*.224^3/(14*(40/7000)) = .303
So the maximum group increase in size from this alone is .61 … Note that whether this is an underestimate or overestimate depends on the relative rate of decay of the downrange and crossrange velocity components.
Also, it's kind of interesting to note the mass of this defect (in grains):
Wdef = rhob * (NO*c^2*Pi/8)*7000 = .166 grains
and the radius shift of the CG is just .0002...
Now in actuality we will have other defects in the bullet. We were fortunate here, because we had a defect of roughly known geometry that we could approximate the impact of. But .166 grains doesn't seem like very much of a defect size to me, and the defects centroid was only at .048” from the center (43% of what it could have been if a bearing area defect)! It may be interesting to bound the maximum possible impact of a weight defect size on group size due to this mechanism. More on this later.
So, if the above are right, it isn't unreasonable that one might see:
- group sizes open up significantly as a result of bullet weight variation displacing the CG of the bullet from the mass center -- this should be observable at any range with magnitude of the impact to MOA depending on the relative rate of decay of lateral vs. forward velocity-- (yeah, I know a bunch of you are saying "no duh" :) but it's interesting to me to see what the numbers are relatively.)
- non-axisymmetric defects induced on the surface of a bullet have a much greater effect than those close to the center of the bullet (effect proportional to radius from centerline to defect)
- group sizes (as measured in units of angle) due to lateral velocities induced at the muzzle will either open up or close up with increasing range depending on the relative rate of decay of downrange vs. crossrange velocity components.
- At close range, the equation for change in group size in MOA due to (pre-existing) imbalance is going to be independent of muzzle velocity, and dependent solely on twist rate and the degree to which the defects displace the CG from the geometric center.
First thoughts on how to check this at the range:
- take a sample of my bullets with nose half offset
- determine by fingernail test which nose half is which
- mark each nose with sharpie
- seat the bullets
- Mark the cartridge head with the clocking of the bullet
- Fire a group comprised of a large number of bullets with half oriented one way in the rifles chamber, and half oriented 180 degrees (this would best be done at close range, say 25 yards off a bench, to minimize the effects of wind and gravity+MV variation)
- Fire a group comprised of a large number of bullets with all bullets oriented the same way
- the difference in group size should be less than but close to twice the expected deflection of the bullets (.6")
If everything is as it should be, and assuming the nominal group shape is pretty round, you would expect the half-and-half 180 degree opposed group to show a somewhat elliptic shape, and greater size than the other target with all bullets oriented the same way. The more accurate the rifle/load combination is to start with, the easier it should be to distinguish between the two groups. (You know, I remember reading at least one member here who said he clocked his bullet orientation from the mould for a rifle and chambered each with the same clocking and it made a big difference for him. Don't remember who I read did this, but if it makes a difference with a wonky mould I'm sure others must have already tried this. It would be interesting to know who's done it, why, and what they found.).
OK, so I think the above approach to testing is doable but a bit problematic because I have relatively few of the original run of bullets left. It is likely to reduce the number of shots required to distinguish between the two populations reliably by using standard deviations instead of group size. I haven't thought that through, entirely, though.
Another test may be to just do as has been suggested to me already by another member here and carefully weight sort my bullets for a test using only the very heaviest bullets. While low weight bullets alone aren't theoretically required to be out of balance, it makes sense to me that if they are low weight due to surface fill out failure (potentially particularly bad per amount of defect weight) or because of internal voids/inclusions, then the magnitude of the overall problem may increase with lower weights.
A more controlled approach would be to drill out a cavity of known depth and volume in the lube grooves of a bunch of bullets, repeat the calculation for that, and run the experiment as above.
Best regards,
DrB