Just in case you might be interested in how rotational energy is determined, contrary to some published mumbo-jumbo.


"In some material reviewed the rotational energy (kinetic energy) has been proposed to play some significant part in terminal ballistics but I doubt this is true. The rotational energy of a bullet is rather insignificant compared to its linear energy. This stands to reason as the energy required to stop a rotating bullet, even at several thousands of rpm’s, can be imagined to be small compared to the large linear kinetic energy of large caliber bullets. To show that this is true, the rotational energy of an object is found from

MErot = (1/2)Iω^2

where I is the moment of inertia of the bullet and ω is the revolutions per unit time (such as rpm’s).

For a rotating cylinder (a bullet in this case), the moment of inertia can be found from the formula,

I = (1/2) mr^2

where m is the bullet mass and r is the bullet radius (half the bullet diameter). The bullet revolutions per second can be determined from the bullet velocity (ft/s) and barrel rifling rate of twist (in inches), as

ω = 12 V/(twist)

For the typical 1 in 12 inch twist for the 350 gr. bullet at 2400 ft/s in the .375 H&H, the rotational speed is calculated as

ω = 12 (2400)/(12) = 2400 rev/s, or 144,000 rpm.

The moment of inertia is then, for a .375 bullet radius of 0.1875 inches, and converting it to feet, and grains to pounds,

I = (1/2)(350/7000)(0.1875/12)^2 = 0.000006103 lbm–ft^2

Then the rotational energy is calculated as, and converting lbm to lbf,

KErot = (1/2)(0.000006103)(2400)^2/32.17 = 0.5464 ft–lbf

As can be seen from this calculation the rotational kinetic energy of the .375 bullet (0.5 ft–lbf) is almost negligible compared to the linear kinetic energy of the bullet (4476 ft–lbf). Whatever part the bullet rotational energy may play, it is not significant in the total energy dissipated in an animal."

Will

I understood that completely until you got to the words "Just in case you might be interested" After that I lost you a little !

quote:

Originally posted by Will:
Just in case you might be interested in how rotational energy is determined, contrary to some published mumbo-jumbo.


"In some material reviewed the rotational energy (kinetic energy) has been proposed to play some significant part in terminal ballistics but I doubt this is true. The rotational energy of a bullet is rather insignificant compared to its linear energy. This stands to reason as the energy required to stop a rotating bullet, even at several thousands of rpm’s, can be imagined to be small compared to the large linear kinetic energy of large caliber bullets. To show that this is true, the rotational energy of an object is found from

MErot = (1/2)Iω^2

where I is the moment of inertia of the bullet and ω is the revolutions per unit time (such as rpm’s).

For a rotating cylinder (a bullet in this case), the moment of inertia can be found from the formula,

I = (1/2) mr^2

where m is the bullet mass and r is the bullet radius (half the bullet diameter). The bullet revolutions per second can be determined from the bullet velocity (ft/s) and barrel rifling rate of twist (in inches), as

ω = 12 V/(twist)

For the typical 1 in 12 inch twist for the 350 gr. bullet at 2400 ft/s in the .375 H&H, the rotational speed is calculated as

ω = 12 (2400)/(12) = 2400 rev/s, or 144,000 rpm.

The moment of inertia is then, for a .375 bullet radius of 0.1875 inches, and converting it to feet, and grains to pounds,

I = (1/2)(350/7000)(0.1875/12)^2 = 0.000006103 lbm–ft^2

Then the rotational energy is calculated as, and converting lbm to lbf,

KErot = (1/2)(0.000006103)(2400)^2/32.17 = 0.5464 ft–lbf

As can be seen from this calculation the rotational kinetic energy of the .375 bullet (0.5 ft–lbf) is almost negligible compared to the linear kinetic energy of the bullet (4476 ft–lbf). Whatever part the bullet rotational energy may play, it is not significant in the total energy dissipated in an animal."

What you say Will is so true....but it could have been said a little more simply.....as I've never been able to achieve greater velocity by reducing the twist rate in an effort to put less energy into RPMs....in other words the rotational energy is so little that no practical velocity gains are to be made by reducing the twist rate.....

One could also say that the bullet will turn roughly one time in the process of penetration of a foot.....negligible energy compared to the foreward motion energy.

Why do we have to make such difficult proofs for things that are intuitively obvious???
Vapo...home of the .404

As a "shrink" professor, I get lost after E=MC2.

I really like the drill bit explanation as to how bullets penetrate. That is why I am building a .416 1:3 twist bore - a Dakota .416 of course...

Thank you Professor Pondoro Junior.
The bullet-spin "PR-spin" is a bunch of "rot," eh?
KErot is miniscule, so it is comforting to know that it takes away little from the muzzle velocity in the adiabatic heat engine gun barrel piston stroke, what? It does not raise pressures to any significance, what?

A fast twist does not hurt anything at hunting ranges, and can be of great help with the long monometal bullets, for accuracy. And it doesn't hurt penetration either, what?

Comparing bullet forward momentum to angular momentum of spin would be a more fair comparison, though this too would be trivial.

Gyroscopic stability through the air is most important. That is what spin is all about. Just the steering mechanism like one hand on the steering wheel as 300 HP pushes rubber against asphalt.

Analyzing bullet spin after it hits the target is INTRACTABLE. Here I throw up my hands and run screaming away to bury my head in the sand.

The first test for the RIP Bullet Coffin (gold standard penetration measured in board-feet-of-water) will be to compare a 10" twist .470 to a 16" twist .470. Both with 500 grainers at 2300 fps.

Then on to the more predictable .375/300gr in 12" twist at 2400 fps vs. 2700 fps.

quote:

Originally posted by Will:
For a rotating cylinder We were talking an expanding bullet I thought???

I = (1/2) mr^2

where m is the bullet mass and r is the bullet radius (half the bullet diameter). The bullet revolutions per second can be determined from the bullet velocity (ft/s) and barrel rifling rate of twist (in inches), as

ω = 12 V/(twist)

For the typical 1 in 12 inch twist for the 350 gr. bullet at 2400 ft/s in the .375 H&H, the rotational speed is calculated as

ω = 12 (2400)/(12) = 2400 rev/s, or 144,000 rpm.

The moment of inertia is then, for a .375 bullet radius of 0.1875 inches, and converting it to feet, and grains to pounds,

I = (1/2)(350/7000)(0.1875/12)^2 = 0.000006103 lbm–ft^2

Then the rotational energy is calculated as, and converting lbm to lbf,

KErot = (1/2)(0.000006103)(2400)^2/32.17 = 0.5464 ft–lbf

Will,

Intuitively obvious vapodog says??

I used your formula for a 55gr SX .224 bullet fired at 3800fps in a 14" twist and at 3400fps in a 9" twist. Revolutions per second are 3257 (14") and 4533 (9")

(1/2)(55/7000)(0.112/12)^2 = .000000342216 lbm-ft^2

KErot = (1/2)(0.000000342216)(3257)^2/32.17 = 0.056 ft-lbf for 1 in 14" @ 3800fps

KErot = (1/2)(0.000000342216)(4533)^2/32.17 = 0.109 ft-lbf for 1 in 9" @ 3400fps

The first load shoots fine.

The second load disintegrates bullets into a blue/gray mist several yards out from the muzzle some of the time.

You are telling us that a mere five one hundredths of one foot pound does this!

Or even the huge whole tenth of a foot pound!!!)

Hog wash!

Unless you have a magic formula where less forward velocity does it some how, I will say it is centrifugal force from too much surface speed for the strength of the material used!

Never was surface feet per second mentioned! No calculations when your "cylinder" spreads it wings to react with our animal.....

If your .375 bullet example expands to only 5/8" frontal diameter the petal tips are going 400 feet per second! Circumferential velocity is meaningless then?

Think about it while you answer what blows up that thin jacketed SX bullet going 400fps slower than the one that stays together. 0.05 ft-lb?

We are a long way from a "compressed" helix!

BigRx

.

quote:

Originally posted by RIP:


Gyroscopic stability through the air is most important. That is what spin is all about.

Yes indeed RIP!

But gyroscopic stability during all the penetration of our animal is paramount Here the "faster" twists shine! You know and I know and some of the others......

If we could get that across I don't care if they understand a "compressed" helix or not!

BigRx

Doesn't bullet rotation stop after contacting bone or hard muscle in large animals?

quote:

Originally posted by mufasa:
Doesn't bullet rotation stop after contacting bone or hard muscle in large animals?

This is a good question! Maybe not so much stop, but lose a greater amount of rotation quickly?

In wet packs, The recovered expanded bullet is often sideways. Sometimes it show evidence of being sideways for a period..... Actually showing deformation to the side at times. This is the reason I begin exploring faster twists. I found that expanded bullets could be made to terminate point on..... Unfortunately I haven't compared fast to slow twist at the same time with documentation of exacting results to ponder. As RIP and his "coffin" I plan to test equal velocity, fast/slow twist in different situations to further explore..... Bones need to be factored in either in "coffin" or wet pack!

This whole subject is very perplexing and very easy to focus so intently on one variable that others "blind" you temporarily.

We tend to think twist and forward travel are "locked" together at "1 in 10" or whatever, like it was the cut rifling machine that made some of our barrels. Always in unison... The majority probably focusing almost entirely on forward travel with the twist being locked in!

I must admit I was "blinded" in the opposite way on just twist! Temporarily I hope!
I must also concede that even in mushrooming/fragmentation scenerios the rotational energy is pretty dismal.......

In wet packs, I seem to see somewhat less penetration with expanding bullets overall with faster twists! Maybe this is the greater diameter of trauma I'm seeing? A direct documented testing of fast/slow not leaving anything to memory is what I need to do!

Rotational velocity and forward velocity work together, influence each other, yet are independent. I knew this.........

But until I cut a 55gr SX bullet apart, the light bulb didn't come on! The inside of the jacket is deeply grooved and is only .0105" thick. Some of these inner grooves even have the start of splits.....

I think the dramatic mid-air explosions I was seeing may have initiated with centrifugal force.... The "flaw" opening or spreading until the "team" of forces created the mid-air disintegration! Twist started the fight and forward travel jumped in to finish it!

BigRx

And hey, what about rotational energy and the relationship to directional stability and therefore straight line penetration? Sort of the gyroscope view of staying on target all the way through?

dah hey''

quote:

Originally posted by ALF:
Dont you just love it: Rotational velocity playing no role in wounding capacity of a projectile whilst over on the Mr Bekker Wars part 2 I was crucified for even implying it has little role! but then there is this duplicity on AR !!!!

OK you guys, quit picking on Alf.

I know you slide rule guys are a lot smarter than me but I don't believe that a bullet "bores" its way through an animal. I believe that bone and muscle stop rotation shortly after the entry of an animal unless its a between the ribs lung shot. I have never tested this and probably couldn't. I just don't believe that the rotational force could overcome the resistence of bone and muscle. If a rifle is a 1-10 twist; if an animal is approximately 20 inches thick, then wouldn't the bullet rotate 2 times or less? The bullet doesn't "accellerate" its rotational rate just because it leaves the gun barrel does it? Regardless of the speed of the bullet doesn't it continue to turn 1-10 or even gradually slow down?

Come on and get those slide rules out of your pocket protectors and get us non-engineers some answers.

I won’t get into the other arguments but I believe there is an error in the math here (check my work, it has been a busy morning):
M: 350 gr=350/(7000*32.17) slug=.001554 slug
w=2*p*(12*Velocity(ft/sec)/Twist(in))=2*3.14159*12*2400=15079/sec.
R=.375 in/(2*12in/ft)=.015625 ft.
I=.5*m*r^2=.5*.001554*(.015625)^2=1.8969*10^-7 slug*ft^2
E=.5*I*w^2=.5*1.8969*10^-7*(15079)^2=21.56 ft-lbf
Linear energy=4476 ft-lbf
So the rotational energy is .48% of the linear energy, which still isn’t much.
In the case of the 55 gr .224 @ 3800 in the 1:14 twist vs 3400 in the 1:9 twist the rotational energy of the 1:9 is 4.316 ft-lbf vs 2.228 ft-lbf for the 1:14, about a 2:1 difference.
As far as stability in something other than air, I believe the spin required is proportional to the square root of the density of the material (assuming it is a fluid) divided by the density of air: w»(r/r(air))^.5. In water, to maintain stability, you would need about 28 times the spin rate as that required for air; not too practical for sure. Also you get into the debate as to what surrounds a bullet as it traverses the target. This is all interesting; did we prove anything??
C.G.B.

Bear in mind that, as the bullet strikes and penetrates, it shortens, expands radially and the spin required for stability reduces radically.

are you after more velocity?

it's a BIG it depends...

in a small case capacity, high pressure situation (over 50kpsi for an arbitrary limit) faster twist allows you to build MORE pressure, and you can achieve a high MV.. .effectly increasing case capacity (it doesn't but it works like that)

In a low pressure/big case

slower twist can result in a higher MV, as the bullet doesn't have to work that hard...

cartridges in question 358 win and 416 rigby...

but it TOTALLY varies, round per round, to make it an exact call.

I built 2 358s to prove this.. (all same barrel length)
1x10
1x12
and had a savage 99
1x16

the 1x10, WITH THE SAME LOAD, is always faster...

and shows more pressure....

so I use a load that is safe in that one, and the pressure is lower (and MV) in all of the rest

jeffe

quote:

As can be seen from this calculation the rotational kinetic energy of the .375 bullet (0.5 ft–lbf) is almost negligible compared to the linear kinetic energy of the bullet (4476 ft–lbf). Whatever part the bullet rotational energy may play, it is not significant in the total energy dissipated in an animal

In case I was misunderstood.....Yes...I agree.

quote:

Originally posted by cgbach:
I won’t get into the other arguments but I believe there is an error in the math here C.G.B.

Thanks for checking C.G.

BigRx

quote:

Originally posted by Gerard:
Bear in mind that, as the bullet strikes and penetrates, it shortens, expands radially and the spin required for stability reduces radically.

Good point Gerald! Another variable so easy to overlook in the ocean of them for expanding bullets! Probably 1 in 72" would be enough.... But it takes some... at least until our bullet stops... Or... (see picture I hope)

I begin my documented testing today with what I will call......... "1 in 10" vs. the old standard 14" twist!"
Already, the differences are showing up and I'm still on conventional bullets! I am trying to learn attaching an image?? Some actual documentation to look at might be of interest to some that think quicker twists are all bad. Sooo I attempt a termination right on subject!!!!

[/url][/IMG]

BigRx

quote:

Originally posted by Wink:
And hey, what about rotational energy and the relationship to directional stability and therefore straight line penetration? Sort of the gyroscope view of staying on target all the way through?

Just back from Zim, I am too tired to get into this discussion, but some on this matter and its consequences see terminal penetration

As this thread came back up I thought I would post a test I did to compare rotational speed.

I picked two guns with a wide twist variation both .312" bore, and used a light bullet so that both barrels more than were adequate for stablization. The 80gr hollow point bullet of conventional jacketed lead construction was carefully loaded for an impact velocity of 2250fps for each barrel to keep penetration low and "readable".

Everything was the same. Bullets from same box, velocity exactly the same and target exactly the same. A piece (the same piece) of 1/2" thick mild steel plate was the target.

The only difference was one barrel was a 1 in 20" twist and the other was 1 in 9 1/4" twist.

Here is a picture of two sample impacts from each:


"A" has a penetration depth of.208" (from plate surface, not flare.) and a maximum diameter of .600"

"B" has a penetration depth of .204" and a maximum diameter of .605".

"C" has a penetration depth of .216" and a maximum diameter of .635".

"D" has a penetration depth of .215" and a maximum diameter of .635".

Other observations are the bullet washed out of the hole some on "A" at 8 o'clock.
Slower twist appears to not stay centered by the off center "teat" in "B" (from hollow point?)

Both "C" and "D" have a flatter bottom by eye (I couldn't figure a good way to "plot" this shape). Which adds more volume to the already deeper hole..........

Nothing changed except the twist! Rotational energy is next to nothing... If you look at twist linear as seems to be the consensus there is only 1/100th of one revolution difference.......

I ask you sincerely gentlemen, what is causing this repeatable difference of a slightly deeper and a 1/32" wider hole?

BigRx

I would venture the guess (and this is honestly just a guess) that the more stable bullet from the faster twist hits with the point at a more correct attitude. Your test pictures appear to support this. A more squarely impacting bullet would end up with more mass directly behind the point versus one slightly canted at an angle. This would in turn result in more of the forward momentum being used to drive the bullet through straight versus a loss from forward momentum from the bullet tipping more at an angle. We have some very experienced shooters on this board. Does this postulate have merit?

Don't tumbling bullets penetrate less then properly rotating ones?

I'll take a guess: Twist rate?

Faster twist = better wounding of steel plate?

Ditto TheBigGuy.

Even this small sample would seem to be significant, even though the twist difference is extreme.

In game animal tissue, the difference might be more pronounced. In game animal wounds, a marginal gain might very well make a difference.

Thanks for sharing.

Geo. Hoffman and I used to discuss this bullet rotational therory and rifle twists to determine if it killed better with a faster twisted barrel and the buzz saw effect of a spinning bullet...It made for some great testing and some wonderful conversation between old frinds but never could prove squat and finally decided there was no difference in killing effect on large animals...However it may well have killed such things as dirt, sand, mulch, bottles of water, and gelitin better than a slow twist barrel or so I have read...

Well, a fast twist does not hurt anything in a big game hunting rifle. It can only help.

Too fast a twist may cause group size to open up somewhere past 1000 meters range, because the bullet is still pointing exactly parallel to bore out there instead of letting its nose drop and follow the arc of trajectory, etc.

Statistically speaking, a sample of two proves what? Zip, you say? Not sure there is any difference between the results, much less any significance.

But I did shoot two buffalo at about 2,000 yards with different twist rates and there was a difference...the buffalo had different first names.

.

quote:

Originally posted by Will:
Statistically speaking, a sample of two proves what? Zip, you say? Not sure there is any difference between the results, much less any significance.

Sorry Will......... I thought I was speaking plainly with words like "repeatable" and four "samples" of two each....What is shown summarizes the dynamics that repeat very closely.

Each twist has its own parameter of dimensions within a measured few thousandths of repeatability.

Alf has us up to four turns in 14" at 200,000 RPM with details left out? Can you give us the variables?

The Big Guy I believe is right with directional stability for gaining the small .010" increase in longitudinal depth on average....

But what about the repeatablecircumferential gain of 1/32" in diameter??

Expanding that hole in steel a full 1/32" in diameter isn't a small feat IMHO from a load that only has 900 ft.lbs.at impact to begin with.

I refine and narrow my question then: What is expanding the hole when only the twist rate has changed?

Thanks for the answer.

BigRx

Hey, I am here to make a fool of myself and learn from the experience. Fun. There is a lot to learn yet. When I quit learning, I shall RIP for sure.

Another morning of perplexity!

I grabbed two .223's; one a 10" twist the other 14". A box of 40 gr Sierra H.P. looked old and needing used.

Setting up the chronograph I had a full load in the 14" at 3760fps average at ten feet. A little more effort had a load for the 10" running 3755 to 3765fps over the same distance.

I grabbed some 1/2" steel plate.........

I shot only twice with each gun this time as results were SO uniform but again different!

The 14" gun at 193,371 RPM as the variable penetrated .404" and .408" deep with max hole diameters of .376" and .377".

The 10" gun at 270,720 RPM as the variable penetrated .415" and .417" deep with max hole diameters of .400" and .401".........

I started to load some more and then just scratched my head.......

I think it is time, time as it varies both with RPM and distance; even causing them both to vary with each other........ Not understanding time as a variable (an ever changing variable) during the ensuing deceleration curve from our bullet's impact to its stoppage ....is what I'm not seeing clearly.......

These tests tell me something is there we are not talking about...... maybe??

BigRx