Am I doing it wrong?

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.
 
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.
 
The math beautician :-@ said:
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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X3zjD7.gif
 
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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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.
 
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.
 
I don't think the 223 Rem can produce enough velocity to cause these two effects on .224" bullets, though.
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I don't know if it was core slip or simple centrifugal (centripetal) force, but when I was trying to make a .223 into a Long Range rifle with 6.5 twist and 90 grain bullets, it would disintegrate a 75 gr bullet in flight and I actually got C shaped impacts from BENT 90 bullets (when I got close enough to the target to catch a few.) I changed from Sierra to JLK and Berger bullets which were strong enough to stand the gaff, and accurate, too.
 
Great post again (#25) Unclenick.

I knew about the bullet disintegration thing. One of my manuals (Speer, I think; but it could have been Sierra; it wasn't Hornady) cautions about excess velocity with high-twist barrels, using light weight and construction "varmint" type bullets.

I've have done load work ups with 55gn varmint bullets. I stopped at 2800 f/s - I figured that's plenty anyway.

I also load Hornady's FMJ-BT 55 grainers. I figure they are of a stronger construction than the varmints. It's moot however, cuz I use the same charge weight and the resulting velocity is the same 2800-ish f/s.
 
I may have missed someone mentioning this reading thru this post, but no one mentioned that 556 brass wall is thicker thus increasing higher pressure vs .223 remington, I thought the cases dimensionally were identical? except for case wall thickness. I didnt know they had different leads.

Im assuming thats why we have 556 brass and 223 remington brass. Was surprised no one mentioned this???
 
That is because 5.56 brass is not thicker. That is overspray myth from 7.62 where is is true.
Starline makes 223 and 5.56 headstamped brass. The metallurgy is different to make the 5.56 head harder. But the deminsions are identical.
https://www.starlinebrass.com/556x45mm-brass
Also, weighing cases has shown 5.56 to be lighter as often as not. Scroll down to the table.
https://www.accurateshooter.com/cartridge-guides/223rem/
There are other common misconceptions. This link is from MetalGod:
https://www.luckygunner.com/labs/5-56-vs-223/
 
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I looked it up

5.56 NATO case capacity: 28.5 grains H2O
223 Remington case capacity: 28.8 grains H2O (+1.1% compared to 5.56 NATO)

so its not much of a difference
 
It's got a tolerance and is not fixed at specific values that is consistent. Follow the second link Marco put up and scroll down about 1/4 of the way, and you will find a table showing the two highest measured case water overflow capacities at the time it was written were for LC'06 and WCC99 at 30.6 and 30.5 grains, respectively.
 
I've weighed 223 and 556 brass on a milligram scale.

Most 223 brass hovers right around 6000 milligrams (6 grams); usually a little less 5960-ish, that sorta thing.

I don't have much 556 brass, but most exceeds the 6k milligrams; usually a little over, 6030-ish.

I have a lot of Frontier 556 brass however, and they weigh out just like the 223 stuff.

So I guess it depends on the 556 brass.

(YMMV; as my sample was only about ten different brands.)
 
Unfotunately, weight is not a consistently reliable capacity indicator because things that don't affect capacity, like the thickness of the rim, depth and thickness of the extractor groove, and the extractor relief angle, all have tolerances that affect brass weight without influencing the internal capacity. Additionally, different makers use different alloys with a difference in density. The last time I looked at how the two correlate was in 308, and there the capacity followed the weight to within about ±20%. So actual water capacity weight measurements are about the only way (sorry, couldn't resist) to go.
 
I agree with UN completely. The only time case weight will make a consistent difference in case volume is when that case weight differs by many grains 10+ . If you have one case that weighs 175gr and another that weighs 160 . The 160gr case will almost always allow for more case volume . I’ve done a good bit of this type of testing in the past with 308 cases and I rarely see a +/- 3gr in case weight . There are exemptions for sure and even then its 10gr or less . I have some RIP that weigh in the mid 170’s with most LC , Lapua and Fed weighing in the low to mid 180’s . I also have some Win cases that are weighing in the low 170’s to high 160’s . The Winchester cases held more H2o but don’t remember how much more . I haven’t looked at all that data in awhile , not even sure where its at or where I put the brass if I wanted to test again . I have so much brass it became hard to organize to that degree . It’s here somewhere marked and if I come across it I’ll try to set aside .
 
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