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Old September 22, 2022, 01:39 PM   #1
Nick_C_S
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Am I doing it wrong?

So today, I'm playing around with my AR-15 and noticed in very small print on the barrel, under the keymod shroud, it reads: "5.56 x 45 NATO 1/7."

I knew it was a 1:7 twist (but don't remember how I knew this); but did not know the upper was chambered for 5.56 NATO.

So far, all my load workups are with 223 Remington data. Am I doing it wrong?

Background: I had my AR built by a knowledgeable friend back in '17. I told him I want a nice one and don't mind paying a few more $'s. He told me what parts (I already had the blank lowers) to order online and I did so. I took all the stuff to his house and he built it out. This was over 5 years ago and my memory of the whole thing is vague. And so is his: A couple months ago, I asked him about it and he had forgotten entirely that he built out my AR. (I know we never discussed twist rate - being completely new to rifles back then, I didn't even know twist rate was a thing.)

That was California in 2017. Now that I live in a free state, I have finally gotten around to doing load work ups.
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Old September 22, 2022, 01:52 PM   #2
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No worries. .223 Rem will shoot perfectly well in 5.56mm. Your barrel will also support standard 5.56x45 ammo. It also means that you can shoot something other than 55gr FMJ through it. Faster twist means that heavier bullets will stabilize.
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Old September 22, 2022, 02:15 PM   #3
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Thank you Markr6754. I know they're dimensionally the same and that the 556 has a higher max pressure, so I wasn't too worried there. But I wasn't sure if there's something about which I don't even know to be worried - if that makes sense. I don't want to be "missing something."

Quote:
Faster twist means that heavier bullets will stabilize.
Yes, this I have since learned. Unclenick has been really good at explaining procession and all that (being an avid bowler in a past lifetime has helped in understanding as well). In fact, having too much twist for my ammo was a potential problem: I was loading 55gn varmint bullets. Their light construction doesn't bode well to extreme RPM. Fortunately, my loads have been mild so bullet deconstruction never manifested as a problem. Further, I have no desire to push the pressures. Good cycling and consistent chronograph results are my goals.
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Old September 22, 2022, 03:54 PM   #4
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There is even a small complication about the 'higher pressure' rating of the 5.56 NATO versus .223 Rem. The measurement techniques used by NATO and SAAMI are slightly different - someone who knows more than I do (UN) might come by to explain it more clearly - so the pressures are not directly comparable.

Anyway what Markr6754 said is exactly right.

Also, my 1:7 barrel does quite well with 52 grain Sierra Matchkings, as well as 69 grain and 80 grain Matchkings. So your 1:7 is no problem at all.
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Old September 22, 2022, 04:05 PM   #5
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1:9 will stabilize 75 grn. bullets. The 5.56 has a longer throat/leade. I shoot 36 to 75 grn in my 1:9.
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Old September 22, 2022, 04:27 PM   #6
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Quote:
1:9 will stabilize 75 grn. bullets.
I've only loaded up to 65 grain bullets; specifically, Sierra's 65gn SBT (#1395). They even state "10" twist or faster" on the box. I've chosen this to be my tactical bullet. I do plan on trying Hornady's 75gn BTHP (#2279) some day; but I need another propellant for that one. Right now, all I have is AA2230. It's great stuff, but rather fast. I'll need something slower for the heavies.

Quote:
The 5.56 has a longer throat/leade.
What does that^ mean?
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Old September 22, 2022, 05:07 PM   #7
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The 5.56 brass is exactly the same as 223. The firing chamber is the same. The difference is the 5.56 barrel is sized for longer bullets. Think Freebore, as in Weatherby cartridges: there is a bullet jump from the cartridge until the rifling is engaged.
The 65 grn Sierra is an excellent bullet. There are stability calculators that will predict stability based on input variables. It boils down to a wide range of bullet weights can be used in 223/5.56. With 1:7 75 gr bullets are stabilized. Conversely, some say 30 or 36 grn Varmint Grenades (Barnes) will self-destruct if spun too fast. Not in 1:9. I have gone up to 4000 fps, and get resounding thumps on steel plates over 100 yards away.
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Old September 22, 2022, 05:20 PM   #8
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Okay, so there's just more distance from the case mouth to where rifling begins? (Not sure how else to phrase that.)

I'm curious why that would be. To accommodate longer bullets? Which ones would those be? Aren't max OAL's the same? Academics, I suppose. But inquiring minds do want to know.
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Old September 22, 2022, 05:45 PM   #9
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It was to accomodate the current NATO standard 62 grain with a steel cone "penetrator" which allowed the ammo to penetrate a steel helmet at 600 yards, meeting milspec.
This also required a new powder WC846 to be used in place of the WC844 used in the 55 gr FMJ training round.
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Old September 22, 2022, 05:47 PM   #10
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No you did not do anything thing wrong. Just keep shooting!! And have fun too. A/Rs are so variable that you can do almost anything with them.
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Old September 23, 2022, 09:41 AM   #11
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Thanks Marco Califo. Good information, clearly presented. I appreciate it.

My inquiry was out of an abundance of caution. I have been doing work ups (picking up where I left off back in April '17) all summer. I'm in a cooler climate now and I think "shooting season" is nearing an end. Making good ammo for the AR was my priority this year. And in the past few months, I have made a number of loadings that cycle reliably and are certainly as accurate as I need them to be. I have since put hundreds of rounds downrange. Mission accomplished.

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Just keep shooting!! And have fun too.
I will Ed4032. Thank you.
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Old September 23, 2022, 10:07 AM   #12
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This link was posted long ago by Metal God:
https://www.luckygunner.com/labs/5-56-vs-223/
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Old September 23, 2022, 03:05 PM   #13
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Both the .223 and 5.56mm were "finalized" long before the long, heavy bullets and fast twist barrels existed.

If I remember correctly the difference between the two chamberings is in the rifling leade, with the 5.56mm done the way it is to accommodate a certain military bullet (I think a certain tracer) but I'm no longer certain, and too lazy to try and look it up.

I dont think it matters for any commercial loadings. And as to the pressure difference between the rounds, I think its machts nichts.

The higher listed pressure for the 5.56mm is still well withing the safe working limits of any .223/5.56mm chambered rifle, and well below proof levels.

Remember, with modern firearms in sound condition, slightly higher than listed standards is rarely dangerous. I'd say "never dangerous" but want to leave some wiggle room...
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Old September 23, 2022, 05:00 PM   #14
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The above is true. Where the 7" military twist comes from is the need to stabilize the 63.7-grain M856 tracer, which is 27% longer than the 62-grain M855 bullet whose trajectory it is supposed to mark. That extra length affects stability significantly. But target bullets, up to that same length, all have enough greater weight to be stable in my 8" twist barrel in normal match weather conditions.

As to pressures, when you look at the development history of 5.56 NATO, you find using the NATO channel transducer apparatus, SS109 (M855 is our ballistically compatible equivalent) was developed to have 6% higher pressure than M193. The new M855A1 runs higher pressure still because they moved to a faster burning powder to cut down on muzzle flash and blast from the M4's 14.5" barrel. But reports are that barrels are shot out much faster on it.

Meanwhile, note that the CIP did not adopt SAAMI pressure standards for 223. They adopted the SS109 pressure standard. So if you buy European 223, you have that 6% extra pressure. Considering that the proof pressure range for 223 is almost 34% above the SAAMI MAP, that 6% increase, while it will probably introduce a little extra throat wear, is certainly not enough to burst or beat the gun up badly.
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Old September 23, 2022, 05:14 PM   #15
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And whilst we are talking about twist . . .

Can I assume that - with all else being equal - a higher twist rate will also mean a higher pressure?

At least, that's how it works in my mind . . . it takes more effort to push the bullet down the barrel if it has to twist harder.
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Old September 23, 2022, 09:22 PM   #16
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Yes, but the effect is small. If you want to, you can take the revolutions per second times two pi to get the angular velocity and calculate the angular (rotational) kinetic energy from that, and you will find it rather small compared to the muzzle energy.
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Old September 24, 2022, 12:48 AM   #17
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If you're going to go as deep into calculations as Uncle Nick describes to find how much pressure might be raised by a faster twist rate, don't forget to also calculate the rotational energy for the same load in the slower twist barrel for comparison.

A certain small amount of the force pushing the bullet down the barrel is "used up" engraving the rifling into the bullet jacket as the bullet enters the rifling. After that, the amount of force changes, because the bullet is now "cut" for the rifling and "riding on the rails" so to speak.

These are going to be very small numbers and only a tiny percentage of energy used firing the bullet.
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Old September 24, 2022, 06:43 AM   #18
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I’ve always been under the impression the longer leade is to prevent malfunctions in case of carbon buildup during extended fire fights when you can’t clean your chamber. At least I believe I’ve read that somewhere in the distant past.
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Old September 24, 2022, 10:04 AM   #19
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Quote:
Yes, but the effect is small.
Good to know. When I look at load data, I look at the test gun's twist. I have been keeping it in my head to assume slightly more pressure in my gun, if the twist rate is less in the test gun. I'll just keep doing that.

Quote:
If you want to, you can take the revolutions per second times two pi to get the angular velocity and calculate the angular (rotational) kinetic energy
I don't want to . Quip aside, it IS neat to know the formula - thanks. I guess the "times two pi" part is why my Marlin 44 Mag rifle has a very gentle 1:38 twist. The large caliber (centrifugal force), coupled with the lighter construction of pistol bullets.
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Old September 24, 2022, 10:28 PM   #20
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That formula gives you the angular velocity. For angular kinetic energy you multiply that by half the axial moment of inertia. That is best measured on a vertical torsion pendulum rather than attempting to calculate it for the bullet's shape, since you don't normally know how jacket and core thicknesses and distribution vary inside the bullet. You can find the method described by several physics professors on line. Basically, it involves making a little platform for the bullet that you suspend with four or five feet of very fine steel wire. I use 0.0025" tungsten wire I have on hand. Then, with the bullet sitting upright on the platform, you give it a quarter turn twist and start your stopwatch and see how long it takes for twenty oscillations back and forth to occur. You can calculate Ia from that, though you have to calibrate the pendulum first with a couple of known moment of inertia objects. I use pin gauges for that.
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Old September 29, 2022, 05:41 PM   #21
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Here's a way to get an idea of the kinds of energy we are talking about. Pretend you are firing a 62-grain .224" double-ended wadcutter. The formula for the axial moment of inertia of a cylinder of homogenous density is very simple and can be accomplished without resorting to pendulums and the like. It is I=½mr². So, r=0.112". Since we are working in the lbf-ft-s system of measure, we divide r by 12 to convert it to feet, so 0.09333…ft, and we square that to get 0.00008711…ft². Mass in this system is in slugs, and grains are converted to slugs by dividing by 225218. So 62/225218=0.0002753 slugs.

So, where Ia is the axial moment of inertia

Ia = ½ × 0.0002753 slug × 0.00008711 ft² = 0.00000001199 slug-ft²

If the bullet is fired at 3000 fps, we all are familiar with the formula for RPM which is 720×MV/T where T is the rifling pitch in inches. Revolutions per second are just 60 times smaller than that, so 12×MV/T. In this case we want angular velocity (rotational speed), ω, in radians per second which we get by multiplying revolutions per second by 2× π that will be:

Where T is in inches:

ω = 12MV×2×π/T

ω7” = 32314 rad/s

ω9” = 25133 rad/s

For kinetic energy you have linear and angular (spin energy)

Muzzle Energy Linear:

KElin = ½ × m × v²

Angular looks the same except moment of inertia takes the place of mass and ω takes the place of v. So:

KEang = ½ × I × ω²


So let’s solve them for this bullet at 3000 fps for both T = 7” and T = 9”.

KElin = ME = ½ × 0.0002753 slugs × (3000 fps)² = 1239 ft-lb

7” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (32314 rad/s)² = 6.260 ft-lbs

9” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (25132.74 rad/s)² = 3.787 ft-lbs

The difference in those two rotational KE number is 2.473 ft lbs. If you subtract that to or from the ME, we get:

1239 ft-lb * 2.473 ft-lb = 1236.527 ft-lb. Solving the two MEs for velocity we get:

MV₉ = √(1239 ft-lb / (½ × 0.0002753 slug)) = 3000 fps

MV₇ = √(1236.527 ft-lb / (½ × 0.0002753 slug)) = 2997 fps

So going from a 9” to a 7” twist with this bullet will make a difference of 3 ft/s, assuming the extra load doesn’t make the powder burn faster and tend to compensate with extra pressure. In other words, it is less difference than firing in two different chambers is likely to make, and hence insignificant to my way of looking at it.
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Old September 29, 2022, 11:47 PM   #22
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Wow, that's a brain full! Have you sent any rocket ships to the moon lately?

It is neat to see math all laid out.

Which brings me to earlier today when I was pondering the point of slower twist barrels? Seems to me, they should all just be 1:7.
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Old September 30, 2022, 09:15 AM   #23
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Quote:
Originally Posted by The math beautician :[email protected]
Here's a way to get an idea of the kinds of energy we are talking about. Pretend you are firing a 62-grain .224" double-ended wadcutter. The formula for the axial moment of inertia of a cylinder of homogenous density is very simple and can be accomplished without resorting to pendulums and the like. It is I=½mr². So, r=0.112". Since we are working in the lbf-ft-s system of measure, we divide r by 12 to convert it to feet, so 0.09333…ft, and we square that to get 0.00008711…ft². Mass in this system is in slugs, and grains are converted to slugs by dividing by 225218. So 62/225218=0.0002753 slugs.

So, where Ia is the axial moment of inertia

Ia = ½ × 0.0002753 slug × 0.00008711 ft² = 0.00000001199 slug-ft²

If the bullet is fired at 3000 fps, we all are familiar with the formula for RPM which is 720×MV/T where T is the rifling pitch in inches. Revolutions per second are just 60 times smaller than that, so 12×MV/T. In this case we want angular velocity (rotational speed), ω, in radians per second which we get by multiplying revolutions per second by 2× π that will be:

Where T is in inches:

ω = 12MV×2×π/T

ω7” = 32314 rad/s

ω9” = 25133 rad/s

For kinetic energy you have linear and angular (spin energy)

Muzzle Energy Linear:

KElin = ½ × m × v²

Angular looks the same except moment of inertia takes the place of mass and ω takes the place of v. So:

KEang = ½ × I × ω²


So let’s solve them for this bullet at 3000 fps for both T = 7” and T = 9”.

KElin = ME = ½ × 0.0002753 slugs × (3000 fps)² = 1239 ft-lb

7” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (32314 rad/s)² = 6.260 ft-lbs

9” KEang = Rotational KE = ½ × 0.00000001199 slug-ft² × (25132.74 rad/s)² = 3.787 ft-lbs

The difference in those two rotational KE number is 2.473 ft lbs. If you subtract that to or from the ME, we get:

1239 ft-lb * 2.473 ft-lb = 1236.527 ft-lb. Solving the two MEs for velocity we get:

MV₉ = √(1239 ft-lb / (½ × 0.0002753 slug)) = 3000 fps

MV₇ = √(1236.527 ft-lb / (½ × 0.0002753 slug)) = 2997 fps

So going from a 9” to a 7” twist with this bullet will make a difference of 3 ft/s, assuming the extra load doesn’t make the powder burn faster and tend to compensate with extra pressure. In other words, it is less difference than firing in two different chambers is likely to make, and hence insignificant to my way of looking at it.
..........

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Old September 30, 2022, 09:27 AM   #24
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I guess the "times two pi" part is why my Marlin 44 Mag rifle has a very gentle 1:38 twist. The large caliber (centrifugal force), coupled with the lighter construction of pistol bullets.
Naw, I think they have gone to a 20 twist for the long heavy bullets that have become popular. You know, same as a Ruger .44 revolver.

The .38 twist goes back to the days of the .44-40-200, I can't imagine why they thought that a good idea in the 20th century.
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Old September 30, 2022, 11:36 AM   #25
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Yeah, the math makes a lot of eyes glaze over. That's OK since you can get things to shoot without it. It's only when someone wants an actual number to put to the effect of something like rifling twist choice that you need it. Any time you use a ballistics program or a stability calculator, you are depending on that sort of math, even if it's being done for you behind the scenes.


@NickCS,

Well, you got me curious. I modeled a couple of HP match bullets in my CAD software, which finds moments of inertia automatically if the model is accurate, and both a 62-grain flat base bullet with a 7 caliber tangent ogive and a 69-grain BT based on the 69-grain MK bullet dimensions found in Bryan Litz's compendium turned out to have moments of inertia about 14% smaller than a simple cylinder of the same volume as the bullet (I did model the hollow point). So it looks like you could use the usual mean density for a cup-and-core bullet of 10.4 g/cc to work out the volume of such a bullet, then make a cylinder the same diameter as the bullet and with that same volume and density, calculate its mass and then its moment of inertia as above, then multiply by 0.878 to get pretty close to the bullet's axial moment of inertia. You'll need to change that density for copper solids to 8.9 g/cc. I'll have to try it with a couple of secant ogive designs to see if it stays close, but I don't think it will stray too far. So even though I recommended measuring the axial moment of inertia before (what I've done for twist stability calculations), the reality is that we don't fire bullets in constant conditions, and the calculations based on that measured value will change with shot-to-shot velocities and externally will change with atmospheric density, so getting them very exact is probably a waste of effort. Transverse moments of inertia needed for stability will vary a good bit more, being length-dependent.

As to making the .22 barrels all 1-in-7, the reason the various ballistics authorities have wound up recommending stability factors of 1.4 to 1.7 as optimal is that over-spinning causes two potential problems and two precision deteriorating factors. The potential problems for cup and core bullets are in-flight disintegration from too much centrifugal effect and core-stripping. The latter is where angular acceleration gets so high the rotational pull of the rifling on the jacket near the peak pressure point in the barrel can get so great that an unbonded core can actually slip inside the jacket, becoming deformed and loose inside. At the exit from the muzzle, the core and jacket are spinning at different rates, and in flight, the two settle at an in-between spin rate based on the ratios of their axial moments of inertia, which favors the core spin rate. The result is either tumbling or large dispersion on the target. Harold Vaughn documented this happening with, IIRC, 90-grain bullets from a .270 Win (see pp. 155-168 (168-181 on the page counter)). I don't think the 223 Rem can produce enough velocity to cause these two effects on .224" bullets, though. This is a more likely pair of problems for an overbore cartridge of some kind.

The precision problem can be caused either by imprecisely balanced bullets or in-bore bullet tilt during firing, and the 223 Rem can experience either or both. Vaughn demonstrated the effect of the former in detail in the very next chapter in the reference linked above, and the latter is demonstrated on pages 134 and 135 (146 and 147 on the page counting scroll bar). The tilt demonstration doesn't show a large effect, but Vaughn was using a short ogive flat=base bullet, which minimizes the effect. When A. A. Abbatiello did the same test with Lake City 30-06 National Match ammunition for TAR back in the 1960s, the short 0.9 calibers, long M1 Type bullet in that ammunition tilted enough to open groups a whole minute of angle. Basically, you take the angular velocity as we found it above, and multiply it by how far off-center the unbalanced bullet or the bullet tilt is off the bore axis. That gives you the lateral drift velocity the bullet will have at exit.

Example: Suppose the bullet is a typical SS109 or M855 bullet, and the CG is a quarter of a thousandth of an inch off the bore axis. Using the velocity and spin numbers above, the 9" twist angular velocity was 25133 rad/s times 0.00025" is 6.28 in/s, and that's how fast the bullet will drift off the center of the trajectory. So if it goes 100 yards in 0.105 seconds, the bullet will be almost 0.66 inches to the side of the mean POI. If you have the 7" instead, from the previous post, it has an angular velocity of 32314 rad/s times 0.00025", and the drift increases to just over 8" per second. It should take the bullet about 0.87 seconds to go 600 yards, and at that range, the error is 5.46" off-axis for the 9" twist and 7.03" of the mean point of impact. Remember that you don't know the orientation of the CG in the chamber when you shoot, so that error can spread all around the clock, making groups twice with a width of that drift number. The difference is just enough to cost you a point on the standard bull, and more than enough to ruin a benchrester shooter's day.
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