Home Forums Hunting & Shooting Ask The Gunwriters Rotational velocity

OK Gunwriters and math majors,please explain some stuff to me, cause I bogged down quantitatively decades ago ........

Stay with me here because,depending on the answers I get, I'm going somewhere here....

What is the rotational velocity of a 180 gr 30 cal bullet which has exited from a 1-10 twist barrel at 3100 fps?

What is the rotational velocity of a 180 gr 30 cal bullet which has exited from a 1-10 twist at 2750?

What is the rotational velocity of a 277-130 gr bullet exiting from a 1-10 twist at 3100?and,

What is the rotational velocity of a 140 7mm bullet which has exited from a 1-9 twist barrel at 3300?

Here's a formula for you:
vel (fps) x 12 inches/foot / twist rate (inches) = revolutions per second x 60 seconds/minute = revolutions per minute

So, for example:
3300 fps x 12 inches/foot / 9" = 4400 rps x 60 seconds/minute = 264000 rpm

Is that what you were looking for?

Are you asking how many RPMs the bullet is turning or what the velocity a spot on the outside edge of the bullet is travelling?

angular velocity
angular momentum
RPMs

I've having Dynamics (enginering course) flashbacks!



I sure hope it's not:

"with 4 razor sharp petals cutting like a buzzsaw at 120,000 rpm . . ."

A formula a bit simpler than Jordan's is:

(12/twist rate) X Velocity X 60 = revolutions per minute

-or- leave off the "times 60" to get revs per second.

Example for a 9" twist at 3000 fps:

(12/9=1.3333) X 3000 = 3999.9 revs per second. We'll call that 4000, so X 60 = 240,000 rpm.

Note that bullet diameter, weight and length are immaterial.

I'm not sure I'd want to be a fly on any of them, but either the 7mm or the faster 180 would be my worst nightmare if I was. (Without doing the math, they seem to be fastest.)

Like RockyRaab said, caliber, bullet weight, and bullet length are not factors in angular velocity.

The faster the twist rate and the faster the bullet velocity, the faster the angular velocity.

The 7mm bullet rotates the fastest. The .270 bullet and the 180gr bullet both with a 10" twist and both going at 3100fps will have an identical angular speed, the .308 bullet going 2750fps will rotate the slowest.

So...a bullet imparted a higher velocity in a faster-twist barrel is more likely to hurl its demon off and thus fly true to the target... .

Excellent! Very useful for the encantation when I sacrifice a goat before loading with pixie dust !

Thanks to all,

Dennis

Dennis, you ARE student of the art! Not one in twenty shooters knows that the really oldtimers thought that a devil rode on every bullet, and that's why some of them miss.

Well, actually the bullet (if it's a TSX) which rotates the fastest will act most violently in its propeller-like motion as its petals "slice" through flesh the fastest and most aggressively.

Not to be impolite, but ... no. (By the smiley, I think Jordan knows that.)

That mental image of the 240,000 rpm buzzsaw just isn't accurate. In truth, the bullet's forward motion means that it might cut one long spiral slice of perhaps one turn before it stops entirely.

And fewer than one in a hundred shooters knows that sometimes they still do... .

At least...that's my story ...

In temporary seriousness..the human capacity to come up with the right idea for wholly wrong reasons is fascinating, and sometimes staggering.

Dennis

Correct

Two questions: 1. Does rotational velocity (rpm) have an effect on expansion? 2. Do rotational velocity and centripital force effect straight line penetration?

I'm theorizing, let me admit. But I think that rotation would have an effect on expansion to some degree - but the reverse is also true: as the bullet expands, that would slow its rotation.

I doubt that rotation has any effect on penetration except insofar as it keeps the bullet point on and does not allow it to tumble.

It's an interesting question to ponder..well, it is anyway if you don't know the precise definition for rotational velocity. If you think about how fast the outer surface of the bullet is traveling assuming there is no loss, the 7mm:


0.284*3.14: the circumferential distance = (approx) .89176

the 308:

0.308*3.14= .96712

to keep it simple, if each were spinning once for every twelve inches (1:12 twist), and both were started at 3000 fps, then

the 308 bullet's circumference could have rolled across approx 242 feet of distance,

and the 7mm would have rolled across 223 feet of distance.

Those aren't the same numbers which were originally offered, but they show that fatter bullets spun at the same rate of revolutions are moving faster at their outer surface if that matters in any way or to anyone.

But is that rotational velocity?

You can calculate the surface rotational velocity of the bullet.

Start with Rocky Raab's calculation of 240,000 RPM.

That's 4000 revolutions per second.

Each revolution is pi times the bullet circumference, or 3.14 x .308 for a .308 caliber bullet. This is .967 inches.

.967 inches is 0.081 feet.

0.081 x 4000 = 322 feet per second.

1. Yes
2. No

JS What is the basis of your "no" response to question No.2? Theory, conjecture, or testing?

HA!OK, we have some people rolling here!This is good dialogue!Assuming Jordan's math is correct,how many revolutions is that per SECOND? I don't have a calculator handy,but it must be over 3000 revolutions...per SECOND....that is pretty fast for ANYTHING to be spinning.....

back in the bad old days Hornady and Sierra used to make 50 gr bullets for the 222 and 223 class of cartridge that would turn into a blue "smoke" if driven at 220 Swift velocities; a freind at the range has a 264 that would do the same thing to some lightweight 6.5 bullet.We used to watch him do it for the entertainment value...What causes this? Is it rotational velocity?

No, it's not "buzzsaw" effect;it's rotational velocity...

To those who do not believe rotational velocity has an effect on bullet expansion,how do you know? How do you know it has no effect on straight line penetration within an animal? How do you KNOW?