Theory of barrel twist rate and bullet mass and diameter?


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Bill_Rights
November 15, 2012, 11:54 PM
We have at least three threads going on now/recently that raise the question of rifling twist rate versus bullet weight. Here they are:

Thoughts on fast twist 6.8 SPC barrel
http://www.thehighroad.org/showthread.php?t=685374

180gr in a .270 Winchester?
http://www.thehighroad.org/showthread.php?t=685010

Rossi 92
http://www.thehighroad.org/showthread.php?t=684296

What determines the minimum twist rate of a barrel? [For "low" twist rate, I would cite 1 turn in 30 inches, as some old big-bore black powder rifles have. For "high" twist rate, I would cite 1 turn in 7 inches, as some 5.56mm/.223" short-barrel rifles have.]

How does twist rate play in? I have picked up that, if twist rate is not high enough, the bullet will not "stabilize". [By the way, what do we mean "not stabilize"?] Is that true? (I assume this is true, in extreme cases.)

But I am seeing people post here on THR that for heavy bullets, you need a higher twist rate than with a light bullet of the same caliber. How's that?

What is the total picture? What is the theory? I also picked up that smaller calibers have higher twist rate while larger calibers have lower twist rate. Why?

So it looks like desirable or ideal twist rate depends upon both bullet mass and bullet diameter. Does it also depend upon bullet speed? (expected muzzle velocity?) That is the same as saying that it depends on powder loading of the cartridge. right?

I know that the absolute spin rate of the flying bullet (rpm, revolutions per minute) is faster the faster the bullet travels. For example, for a 1-in-12" twist rate, a bullet exiting the muzzle at 2000 fps would be spinning at 2000 rotations per second = 2000 rps * 60 sec/min = 120,000 rpm. For the same 1-in-12" twist rate, a bullet exiting the muzzle at 3000 fps would be spinning at 3000 rotations per second = 3000 rps * 60 sec/min = 180,000 rpm.

I would guess that what matters for in-flight ballistics is the spin rate of the flying bullet, not the twist rate of the barrel, right?

So what is the formula where bullet mass, bullet diameter and bullet speed factor into the desirable spin speed or twist rate?

Are there any other factors? I have also picked up here on THR that length of the bullet matters. True? Or is it sectional density or other measurement?

HELP! I am in the dark here. :confused:

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creeper1956
November 16, 2012, 12:51 AM
I have also picked up here on THR that length of the bullet matters.
Yep... yep it does. The Greenhill formula (http://www.ebr-inc.net/articles_Greenhill_Formula.html). ;)

Bill_Rights
November 16, 2012, 02:23 AM
Creeper,

Thanks for the link to EBR's web site. I was unaware of that company. They aer specialists in subsonic performance. This other EBR page (http://www.ebr-inc.net/articles_twist-rate.html) on that is interesting. That's handy to know.

OK, so the Greenhill formula is good - that's what I asked for. Thanks.

Twist Rate = 150 * SQRT(density) / Bullet-Length

I notice EBR leaves out the density factor (square root of density is set equal to "1") for rough estimates. They do not give the units of density, like pounds/cu.ft. or grams/cc or whatever. So we couldn't use it if we wanted to.

This is an empirical formula. I would like a little more detail. I may look up the origin of and more detail on the Greenhill formula, when I get a chance.

creeper1956
November 16, 2012, 02:49 AM
Back when I shot benchrest competition and was based in Phoenix, but would travel all over the country and a few times to Europe, I used the Greenhill (and suggestions from Walt Berger and Gary Ocock (http://bulletin.accurateshooter.com/2008/09/gary-ocock-busts-nbrsa-rail-gun-records/)) to decide what twist I wanted for the 6mm 68gr Berger HPFB match (6PPC).
When I had a high paying job :p, I had Shilen (http://www.shilen.com/calibersAndTwists.html) make up 3 otherwise identical barrels... one in 1:14, 1:13 and a 1:11.5 so I could experiment as conditions dictated. Here's why.

Effects of Altitude and Temperature on Rifling Twist (http://www.riflebarrels.com/articles/bullets_ballastics/effects_altitude_temperature_rifling_twist.htm), by Dan Lilja.

Of course, it's far more complicated and ultimately far less precise than your typical perfectionist would like... but it makes for interesting reading. :D

Cheers Bill. Have fun with it. :evil:

helotaxi
November 16, 2012, 09:10 AM
The Greenhill formula works well for relatively short, fat bullets. The Miller stability formula is better applied to typical long range bullets.

The general rules are the same though. The higher the ratio of the bullet length to the bullet caliber, the faster that it needs to spin in order to achieve gyroscopic stability. The Miller formula accounts for the increase in upset force caused by the fact that the center of aerodynamic force is typically acting forward of the center of mass of a long slender bullet and that the forces trying to cause the bullet to tumble increase with muzzle velocity. The good news is that the increased velocity also increases spin rate for a given rate of twist and the increase in stability imparted by the marginal increase in RPM is higher than the marginal increase in upset force. The overall effect is that once the bullet is going fast enough to be stable from a given rate of twist, increasing velocity increases stability margin.

sansone
November 16, 2012, 09:30 AM
Bill, I realize this explanation is over simplified, forgive me..
Years ago an old reloader told me bullets that are long for their weight need more spin to stabilize.

helotaxi
November 16, 2012, 10:06 AM
Bullets that are long for their caliber need more spin. Weight doesn't matter. A bullet of a given length in a given caliber needs essentially the same amount of spin to be stable whether it is light for its length or heavy for its caliber.

MTMilitiaman
November 16, 2012, 11:22 AM
Twist rate depends not so much on bullet mass but on bullet length. It just so happens, however, that when comparing bullets of the same diameter and basic construction, the heavier one tends to be longer.

The longer a bullet is, the faster the twist rate is needed to stabilize it. Now think of a top spinning on a table. Notice how when it loses enough rotational velocity, it tends to wobble and eventually fall over? Most bullets are heavier towards the base and thus have a tendency to want to fly base forward. The point of rifling is to induce enough spin to make them stable in flight. If the twist rate isn't fast enough for the length of bullet being used, it tends to wobble in flight, and in extreme cases tumbles or keyholes through the air.

It is much harder to overstabilize a bullet than it is to understabilize it, so you are generally safer erroring on the side of caution and going with a faster twist rate than is necessary than vise versa. You can still shoot 40 and 50 gr varmint bullets from a .223 Rem with a 1:7 twist, but you can't shoot the heavier 69 to 77 gr HPBT Match rounds through a .223 Rem with a 1:12 twist, to give one example.

Also keep in mind that bullet construction and atmospheric conditions also come into play. For example, the crop of newer homogenous bullets on the market are typically longer for their weight than a traditional lead core bullet of the same mass and caliber because copper is less dense than lead. So lighter homogenous bullet like the Barnes might require a faster twist rate than a heavier lead core bullet in the same caliber. Also, air density matters because the more dense the atmosphere is, the faster a bullet has to be spun to get it stable. Cold air is more dense than warm air, so back when the military was experimenting with Eugene's black rifle, they found a 1:14 twist would stabilize the 55 gr M193 in most conditions, but not in artic conditions. So they went with a slightly faster 1:12.

helotaxi
November 16, 2012, 12:30 PM
Also, air density matters because the more dense the atmosphere is, the faster a bullet has to be spun to get it stable. Cold air is more dense than warm air, so back when the military was experimenting with Eugene's black rifle, they found a 1:14 twist would stabilize the 55 gr M193 in most conditions, but not in artic conditions. So they went with a slightly faster 1:12.And when they went to the 62gn FMJ and its corresponding tracer they found that while a 1:9 would work most of the time, a 1:7 was needed when it was silly cold.

Also, there is no such thing as "overstabilization". You can spin a bullet so fast that you tear it apart, but that is an integrity problem, not a stability problem.

1858
November 16, 2012, 03:49 PM
So it looks like desirable or ideal twist rate depends upon both bullet mass and bullet diameter. Does it also depend upon bullet speed? (expected muzzle velocity?) That is the same as saying that it depends on powder loading of the cartridge. right?

A bullet's stability is definitely affected by it's muzzle velocity. The faster the bullet's velocity, the greater the moment exerted by the air on the bullet in flight which is trying to flip the bullet over (rotate it about its center of mass). The faster the bullet's angular velocity the more resistant it is to being overturned. Bullets in flight generally become more stable as they lose linear velocity because their angular velocity decreases at a slower rate. So a bullet's gyroscopic stability will increase as it slows down ... for some distance anyway.

powell&hyde
November 16, 2012, 04:16 PM
Yup the numbers seem astronomical but in reality, the time to target is only a little more than one-tenth of a second.

-100 yards times 36 inches is 3600 inches
-one turn in 10" twist barrel---3600 divided by 10= 360 bullet revolutions per 100 yards
-1/9 twist--400 complete rotations of the bullet out to 100 yards
-1/7 twist--514 revolutions of the bullet out to 100 yards
The revolutions remains almost constant until the bullet stops. So at 400 yards the bullet in a 1/10 twist will have turned completely 1440 turns. The bullets do not rev up after firing and exiting the barrel but they have a constant 1 turn in whatever twist the barrel is and velocity will change bullet rpms. Now if the bullet was traveling through the air for a total time of one minute the total number of rpms would be the high numbers such as 200,000 or more.

Zoogster
November 16, 2012, 04:27 PM
Also, there is no such thing as "overstabilization".

Actually higher twist rates are going to generate more friction in the barrel and lead to higher wear and shorter life.
So higher than needed does have its down side.

Also when dependent on the bullet destablizing after impacting a target to deliver most of it's energy on target or fragment, such as the military when using FMJ, you could have overstabilization. The bullet will penetrate further before it begins to yaw with higher rates of stabilization. Reducing wounding. Not an issue with expanding ammunition.

1858
November 16, 2012, 04:40 PM
The revolutions remains almost constant until the bullet stops. So at 400 yards the bullet in a 1/10 twist will have turned completely 1440 turns.

I think the math is a little more complicated than that. The linear velocity of the bullet is decreasing at a faster rate than the angular velocity. If you assumed that the angular velocity was constant over the 400 yards, the bullet would rotate more times in the last 100 yards than the first 100 yards.

CraigC
November 16, 2012, 05:24 PM
Velocity does indeed play a role, as does pressure. The faster you drive a bullet, the slower your twist can be. Which is why the .454 has a typical 1-24" twist, compared to the usual 1-16" twist of the .45Colt. Drive the bullet faster, use a slower twist rate and pressures will also drop.

1858
November 16, 2012, 05:52 PM
Another issue is that a bullet needs dynamic stability in addition to gyroscopic stability. Gyroscopic stability is easy to achieve but dynamic stability is a tougher nut to crack. Consider that a 220gr SMK loaded in a .300 AAC Blackout shell and shot out of a 1:7 twist barrel at 1,000 fps has a velocity corrected SG of 2.6 which is well above the 1.4 requirement. So theoretically that bullet should be stable at 100 yards but it is far from stable. The problem is that the center of pressure is well forward of the center of gravity so despite having gyroscopic stability, it's not dynamically stable.

DanTheFarmer
November 16, 2012, 06:01 PM
A bullet that is not properly stabalized will not be accurate. In flight it will look like a Tim Tebow pass vs. an Aaron Rodgers pass. When it goes through a paper target it won't be a nice circle. Instead it will make an oval shape because the paper intersected the bullet "mid-wobble". That is keyholing.

I did some calculations like you did and I was skeptical that 120000 rpm vs. 180000 rpm would make a difference, so I did an experiment. I have a 7mm-08 with a 1 in 12 or so twist. This is good for the shorter (lighter!) bullets appropriate for the caliber. I worked up a load with 120 grain bullets and I could cover the group with a quarter at 100 yards. I also had some 175 grain bullets (MUCH longer) for a different 7mm-08 with a faster twist. It isn't a great load but it shoots 3 - 4" at 100 yards in the fast twist rifle. When shot through the slow twist rifle the 175 grain bullets were a foot apart and made oval shaped holes in the paper. The 175 grain bullets are too long to be properly stabalized in the 1 in 12 twist barrel. It amazes me that 120000 rpm or so is not enough to work properly but my test showed that to be true.

Good Luck.

Dan

Col
November 16, 2012, 06:02 PM
???? beats me

1858
November 16, 2012, 06:07 PM
Which is why the .454 has a typical 1-24" twist, compared to the usual 1-16" twist of the .45Colt. Drive the bullet faster, use a slower twist rate and pressures will also drop.

Marlin's 1894 chambered for .45 Colt has a 1:38 twist. They listed the twist on their web site as 1:16 for ages but finally got around to changing it. Ruger's revolvers for .45 Colt have a 1:16 twist though. If you run the numbers through the Miller Stability Formula, a 1:38 twist is more than adequate to spin stabilize common .45 Colt bullets from 200gr to 300gr.

CraigC
November 16, 2012, 06:21 PM
Up to 300gr yes, but only at maximum velocity. Certainly not at standard pressures for the .45Colt. Anything over 300gr and you need a substantially faster twist but the Marlin won't feed them anyway.

pseudonymity
November 16, 2012, 07:28 PM
Bullets that are long for their caliber need more spin. Weight doesn't matter. A bullet of a given length in a given caliber needs essentially the same amount of spin to be stable whether it is light for its length or heavy for its caliber.

I am pretty sure that it is not true that weight does not matter, but I would agree that for common useful bullet shapes with lead cores, the weight could be ignored.

The Greenhill formula does take specific gravity into account, but sets up the terms so that for lead core bullets the specific gravity is = 1, so it effectively cancels out of the equation. If you had bullets constructed out of steel or tungsten for instance, you would need more or less twist relative to a bullet of the same dimensions constructed with a lead core.

Vern Humphrey
November 16, 2012, 08:29 PM
By the way, what do we mean "not stabilize"?

Go and look at the target. If the bullets are not stabilized, not only will few of them hit the target, those that do will leave elongated holes, indicating that they were more or less sideways when they hit. That's called "keyholing" and when you see that you know you have stability problems.
But I am seeing people post here on THR that for heavy bullets, you need a higher twist rate than with a light bullet of the same caliber. How's that?
It isn't weight or mass that determines twist rate, it's bullet length.

The confusion comes in because to make a bullet heavier, you usually have to make it longer. But you could make a 100 grain .22 bullet of depleted uranium that would be no longer than a lead-core 55 grain bullet and shoot it in a 1 in 12 twist very nicely.

Bill_Rights
November 16, 2012, 11:28 PM
helotaxi, OK, the Miller formula is more what I need for longer, slender bullets. What is the Miller formula? (Or just give a link to it, please. Thanks!)

DanTheFarmer, Excellent that you actually did the experiment! My hat is off to you.

1858, Excellent point about the "moment" of force that is trying to flip a bullet. The faster the bullet's velocity, the greater the moment exerted by the air on the bullet in flight which is trying to flip the bullet over (rotate it about its center of mass). The faster the bullet's angular velocity the more resistant it is to being overturned.
I take it that this moment of force is the actual force mltiplied by the lever arm length, and a) the actual force is due to drag forces from the air resistance and b) the moment arm length (lever arm) is longer for a longer bullet. Excellent understanding!

Zoogster, Yes, as you say: Actually higher twist rates are going to generate more friction in the barrel and lead to higher wear and shorter life.
I agree. I assert that any amount of spin, excessive or not, detracts from muzzle velocity. Of course, some spin is necessary, as we've been discussing. I will calculate the bullet energy consumed by spin and post it later...

pseudonymity, I think you for your interpretation of the Greenhill Formula: The Greenhill formula does take specific gravity into account, but sets up the terms so that for lead core bullets the specific gravity is = 1, so it effectively cancels out of the equation. If you had bullets constructed out of steel or tungsten for instance, you would need more or less twist relative to a bullet of the same dimensions constructed with a lead core.
What I understand you to say is (correct me if I get it wrong), in the Greenhill Formula, the density (specific gravity) of lead is defined to be "1", therefore its units-of-measure [grams/cc, pounds/cu. ft., etc.] don't matter. The effect of the measurement units is built into the numeric factor "150". Only if you deviate from lead material for the bullet do you need to factor in density.

Vern, You hammer home the point well!

Bill_Rights
November 16, 2012, 11:47 PM
OK, so we, or at least some of us, are saying that twist rate imparts spin rpm, and this spin stabilizes the bullet by gyroscope forces, exactly like a spinning top. Similarly, I used to take a bicycle wheel and have somebody spin it while I held both ends of the axle in my two hands, I could barely change the angle of the axle when the wheel was spinning very fast. So this gyroscope force would keep a bullet from tumbling when the center of air resistance force is forward of the center of mass. Cool :cool:

Maybe I need to go and look up the theory/formula of gyroscopes, but intuitively I "know" that the smaller the diameter of the bullet (or bicycle wheel), the faster it has to be spinning in order to exhibit the same gyroscopic force resisting tumbling (or tilting the bicycle wheel axle). None of you posters, nor the Greenhill Formula, seem to capture that trend. Yet I do see it in bullets and barrels, as we have them. Big bore guns have lazy twist rates and small caliber guns have fast twist rates. To quote my own OP: For "low" twist rate, I would cite 1 turn in 30 inches, as some old big-bore black powder rifles have. For "high" twist rate, I would cite 1 turn in 7 inches, as some 5.56mm/.223" short-barrel rifles have.

So in which formula is this gyroscopic force correlated with bullet diameter?

MrBorland
November 17, 2012, 09:34 AM
gyroscopic stabization

OK, so we, or at least some of us, are saying that twist rate imparts spin rpm, and this spin stabilizes the bullet by gyroscope forces, exactly like a spinning top.

Maybe I need to go and look up the theory/formula of gyroscopes, but intuitively I "know" that the smaller the diameter of the bullet (or bicycle wheel), the faster it has to be spinning in order to exhibit the same gyroscopic force resisting tumbling (or tilting the bicycle wheel axle). None of you posters, nor the Greenhill Formula, seem to capture that trend. Yet I do see it in bullets and barrels, as we have them. Big bore guns have lazy twist rates and small caliber guns have fast twist rates.

Well, I'm neither an engineer nor a physicist, but here's my understanding:

A bullet can rotate/spin 2 ways - along its longitudinal axis (i.e spinning as it does while in flight) and/or along its transverse axis (i.e. yaw). Each of these rotations is associated with its own gyroscopic force, so for max stability, you want the longitudinal one as high as possible, relative to the transverse one.

So, here's the rub: Each gyroscopic force is directly proportional to the moment of inertia and spin rate along that axis, and because of its relatively long length, a bullet has a relatively high moment of inertia through its transverse axis. If the longitudinal gyroscopic force is to dominate, then, it can't do it through its small moment of inertia, and must do it through its spin, so it spins fast. The longer the bullet relative to diameter (i.e. caliber), the greater the potential for yaw, so the spin needs to be even faster for the longitudinal gyroscopic force to completely dominate & prevent yaw.

1858
November 17, 2012, 01:57 PM
Bill_Rights, I put together a Miller bullet stability calculator in Excel if you're interested. It has SG corrections for velocity and atmospheric conditions. I used the formulas shown in Bryan Litz's book and confirmed that the calculator is correct by entering the values used in his example. The current values in the calculator are for a .308 175gr SMK bullet and a 1:10 twist barrel.

Miller Bullet Stability Calculator (http://www.public.mcmxi.org/documents/mbsc.xls)

Bill_Rights
November 17, 2012, 04:03 PM
Yes, thanks 1858. I will see if I can download the Excel spreadsheet and understand it. I am interested!

BTW, what is "SG" again? (I know I knew it before, but old-age CRS syndrome is creeping in.....)

Bill_Rights
November 17, 2012, 04:21 PM
1858,

I was able to download the Excel spreadsheet. Thanks! (I didn't know we could do this.)

Yeh, so... looking at it, I see that you don't ask for the caliber anywhere, in actual dimensions, such as inches. But the twist and the bullet length are expressed in units of calibers. OK, so I get it. It does not matter what the caliber measurement is, you just make sure to express the other two parameters (twist rate and projectile length) in multiples of caliber diameters.

Cool. :cool:

1858
November 17, 2012, 08:08 PM
I see that you don't ask for the caliber anywhere, in actual dimensions, such as inches.

Bullet diameter (in.) is the caliber. You enter bullet diameter and length in inches and the spreadsheet calculates those values in calibers and then uses those values to calculate SG.

SG is a unitless value to indicate gyroscopic stability.

Bill_Rights
November 18, 2012, 01:03 AM
Oh! Excuse me. I was wrong. You corrected me by saying You enter bullet diameter and length in inches and the spreadsheet calculates those values in calibers For some reason, I did not see the data entry field for bullet diameter in inches. In addition to CRS, I must be coming down with CSS (can't see sh!#) :mad:.

Bill_Rights
November 26, 2012, 02:44 AM
I said earlier I assert that any amount of spin, excessive or not, detracts from muzzle velocity. Of course, some spin is necessary, as we've been discussing. I will calculate the bullet energy consumed by spin and post it later...So, here it is. The needed formulas tying the spin energy to bullet mass, diameter and spin speed rely on the moment of inertia of the bullet. The moment of inertia about the spin axis is a property of the shape of the bullet and the density of the material. I approximated the shape as a cylinder and took the material to be solid lead (density is 11340 kg/m^3 = 11.34 g/cc = 185.83 g/cu.in. = 2867.7 gr/cu. in.). Here are the formulas, in which "W" is the energy or work just to spin up the bullet:
http://i702.photobucket.com/albums/ww25/Bill_Rights/Formulas_spin_energy_moment_of_inertia_121125-1.jpg
In the formulas, "M" is the mass of the bullet, "r" is the radius of the bullet (half the caliber diameter) and omega is the spin speed, as explained. This all has to be worked out in consistent units. I have a spreadsheet for that, if anybody wants me to upload it.

It turns out that the spin energy carried by the bullet to the target is small, less than 1 %, compared to the linear speed energy, which is technically called the translational energy. Here are three examples sort of covering some extremes.

A 180 grain .308 inch diameter lead bullet travelling at 2620 fps out of a barrel with a twist of 12 inches/turn has:
Bullet translational energy of 3719 Joules = 2743 ft*lbf
Bullet rotational energy of 12.1 Joules = 8.92 ft*lbf.
So the ratio of rotational to translational energy is 0.325 %.
Bullet length, effective (cylinder approximation): 0.842 inches

A 300 grain .452 inch diameter lead bullet travelling at 1500 fps out of a barrel with a twist of 30 inches/turn has:
Bullet translational energy of 2032 Joules = 1499 ft*lbf
Bullet rotational energy of 2.28 Joules = 1.68 ft*lbf.
So the ratio of rotational to translational energy is 0.112 %.
Bullet length, effective (cylinder approximation): 0.652 inches

A 62 grain .223 inch (5.56 mm) diameter lead bullet travelling at 3002 fps out of a barrel with a twist of 7 inches/turn has:
Bullet translational energy of 1682 Joules = 1240 ft*lbf
Bullet rotational energy of 8.42 Joules = 6.21 ft*lbf.
So the ratio of rotational to translational energy is 0.501 %.
Bullet length, effective (cylinder approximation): 0.554 inches

These energies carried by the spin of a bullet do not include the extra friction of the faster spinning bullet lost in the barrel or to the air. Zoogster pointed out these effects earlier. This friction energy is lost as heat, not delivered to the target.

helotaxi
November 26, 2012, 09:23 AM
These energies carried by the spin of a bullet do not include the extra friction of the faster spinning bullet lost in the barrel or to the air. Zoogster pointed out these effects earlier. This friction energy is lost as heat, not delivered to the target.Those energies have been shown to be somewhere around negligible when compared to the same energies from a straight rifled barrel. Essentially you lose the tiny fraction of a percent used to get the bullet spinning and otherwise the difference is nil.

Bill_Rights
November 27, 2012, 02:33 AM
Since the thread responses indicated the critical importance of bullet length, I added the calculated effective lengths of the bullets used for the sample calculations posted, two posts above. As explained there, for the sake of an easy, text-book calculation of bullet spin moment of inertia, I approximated the true bullet shape by a simple cylinder, with flat, square ends. The true bullet lengths would be longer than these approximate results, for the same weight of bullet.

helotaxi
November 27, 2012, 09:35 AM
But the cylinder would be the worst case scenario with regard to energy required to get up to speed. You also assumed a monolithic lead slug which would be worse than a jacketed bullet.

Bill_Rights
November 30, 2012, 12:39 AM
helotaxi,

Yes, also true. But also the effect of the jacket (being lighter, less dense material) would be somewhere around negligible, unless the jacket was very thick.

For my part, I was interested in terminal ballistics (at or inside the target). How much energy does bullet spin carry to the target and could this play any role on effectiveness at impact or penetration? Since the spin energy is well under 1% of the total bullet kinetic energy, my answer seems to be "no".

You, on the other hand, seem more interested in firing chamber dynamics. If we are going to "dissect" that, we also should consider the energy required to engrave the rifling grooves into the bullet flanks. I am a materials scientist, so I could probably estimate the displacement or deformation energy of the bullet jacket, if I had some idea of the mass of the bullet material that gets moved and how far it gets moved. But this is probably also near to negligible, especially for soft materials such as dead-soft lead.

We might be beating a dead flea on a dead horse :scrutiny:.

Bill_Rights
November 30, 2012, 12:55 AM
A contemporary thread touches on barrel twist rate. A link there even touches on spin energy and engraving energy.

Seeking .45/70 barrel twist rate info

http://www.thehighroad.org/showthread.php?t=686216

helotaxi
November 30, 2012, 11:38 AM
If we are going to "dissect" that, we also should consider the energy required to engrave the rifling grooves into the bullet flanks.Once you factor in the obduration of the bullet with the very high forces invloved, the engraving force is anything but negligible. The rate of onset of that force is one of the determining factors of the initial pressure curve of the cartridge. The leade and freebore are manipulated to modulate that onset rate and are the deterministic differences between the .223 and 5.56x45 chambers as well as the 6.8 SPC and 6.8 SPC II chambers. the reduction in the onset of the engraving force is such that a hotter round can be used while keeping the same max chamber pressure and the longer pressure curve of the hotter load results in higher velocity. In the case of the 6.8 SPC, the change from 1:10 to 1:11 rate of twist was the final adjustment used. Again a rate of onset issue. The peak in linear acceleration is also the peak in rotational acceleration and occurs at peak pressure. Reducing the end result RPM by 10% reduced the required rotational acceleration as well and the energy requirement for that acceleration.

Bill_Rights
November 30, 2012, 12:13 PM
helotaxi,

Could you define "obduration" of the bullet? I looked it up, but it is not clear how the general definition applies: ob·du·rate
adjective \ˈäb-də-rət, -dyə-; äb-ˈdu̇r-ət, əb-, -ˈdyu̇r-\
1
a) stubbornly persistent in wrongdoing
b) hardened in feelings
2
a) resistant to persuasion or softening influences
— ob·du·rate·ly - adverb
— ob·du·rate·ness - noun Regardless, I gather that it has to do with deformation of the bullet, which is resisted by the bullet material. It is not unknown to occur that the energy (and of course the force) required to deform a material increases with increasing rate of deformation. And the rate of deformation of a bullet will be FAST.

Considering only the energy, energy is defined as a force acting through a distance (or power expended over a time duration). The distance over which the high engraving forces are active may be only the length of the flanks of the bullet, i.e., short. So the total energy expended in engraving may be small.

Which brings the point (which I have known all along) that energy, itself, may not be the important parameter. In the case of terminal ballistics due to spin, angular momentum may be more important than spin energy, for example. As you point out, force can be critical. Force directly maps to pressure. [pressure] = [force per unit area]

It bears thinking about which of the following quantities are most important to bullet kinematics (kinematics = motion dynamics). Here are the linear quantities, and each has it angular (spin) analog:

position: .................. (self-explanatory)
velocity (or speed): ... [rate of change of position]
acceleration: ............ [rate of change of velocity]
jerk: ........................ [rate of change of acceleration]
momentum: ............... [mass mutiplied by velocity]
force: ...................... [mass mutiplied by acceleration]
pressure: .................. [force applied per unit area OR force divided by area]
energy: .................... [force mutiplied by distance]
power: ..................... [energy expended per unit time OR energy divided by time]

helotaxi
November 30, 2012, 12:49 PM
Basically when the pressure starts to move the bullet forward, the bullet gets "squished" for lack of a better term along its longitudinal axis and expands along its radius. This occurs at pretty much the exact time that the bullet is being squeezed into the rifling.

The squishing of the bullet to fill/seal the bore is what is typically referred to as obduration in this context.

WNTFW
November 30, 2012, 01:16 PM
Hear (Here) you go:
I think you mispelled "obturation"
http://en.wikipedia.org/wiki/Obturation

Sometimes one letter is like a decimal in math.

helotaxi
November 30, 2012, 01:45 PM
Possibly/probably.

WNTFW
November 30, 2012, 01:49 PM
I think "swoll up" would be a better term. Or is it "swole up"?

helotaxi
November 30, 2012, 07:17 PM
That term is generally reserved for results from the gym or telling the wife she looks fat...
;)

murf
November 30, 2012, 11:26 PM
deformation of the bullet base rather than "squish", or "swole up" might be a better way to say it.

obturation is the process of sealing the chamber or barrel from powder gases. bullets obturate barrels.

murf

Bill_Rights
December 1, 2012, 12:53 AM
Well; I learned something today, in spite of my best intentions :rolleyes:.

But I will obdurately insist upon obfuscating the difference between obturation and obduration :p.

Regarding the seething controversy over swole versus swoll, they are both wrong, from a materials point of view. Swelling implies that alien molecules (or atoms) are added to the base material or that molecules of the base material all get farther from each other more or less uniformly. I am 99.9% certain that the total volume of a deformed bullet, after it obturates the bore, is exactly the same as it was before firing, accounting for rub-off of material and swelling due to temperature rise. That is, obturation involves plastic deformation with no change in density of the bullet material.

So, rather than a swole-up bullet, we have a bullet with shock-induced basal-proximate radial expansion deformation. :what: We are now in full obfuscation mode. That and $4.89 will get you a cup of coffee at Kona Krakatoa Koffee Kiosk :banghead:.

WNTFW
December 1, 2012, 12:56 PM
Not to be didactic or pendantic . . . but I think you need to upgrade 'rub off' in your thesis.

Now I want coffee and don't have a KKKK near here.

Re: bullets can't be over stabilized.
I see explanations that say they can be over stabilized. In general it is easy to prove a bullet is under stabilized or if it disintegrates. If it is over stabilized that is harder to prove. It is also insignificant to some degree. I just see some info from very knowledgeable folks saying over stabilization is in fact a real condition.

So are you guys saying it can't be over stabilized? Do you mean that it doesn't matter, don't worry about it. Just looking for some clarification on the statement. That issue seems to be a point of ignition for some people.

I have never had a twist rate problem that I am aware of. Thanks to all the guys that figured it out before I came along. All of my calibers are middle of the road boring.

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