Hi all

I was having a conversation the other day with a fellow shooter.

The problem arose when a 160 grain bullet in .260 remington would keyhole into the target.

His first thougt was to increase velocity in order to increase bullet rpm.

I say that increasing velocity only gets the bullet out of the barrel quicker and gets the bullet to the target quicker but that the rpm of the bullet would stay the same because it is controlled by the rate of twist of the barrel. Ie. a 1 in 9 twist rate can only produce a particular bullet rpm no matter how fast the bullet leaves the barrel.

Who would be correct on this?

Thanks

Jim B.

Rev's per distance will stay the same, but rev's per time will increase with velocity.

Its my understanding that the spin will equal the twist rate, in relation to it's speed. A bullet can't spin faster then it's twist rate, regardless of bullet speed. The faster a bullet travels, the quicker it travels per foot, but the twist rate is the same.
A bullet actually begins to slow down the instant it escapes the barrel, as there is nothing there to force it to spin, and air resistance acts upon it at this point. ~~~Suluuq

He's correct. If the velocity of the bullet is increased then the RPM's of the bullet is likewise increased. Think of it this way, let's say that you take your cleaning rod and force a bullet down the bore at the rate of 1 fps. What is the rpm of that bullet? Now shoot the bullet at 3000 fps, what is it's rpm now? Much higher obviously. It's one of the reasons a high velocity bullet from a 220 swift will need less twist to stabilize it than the same bullet shot out of a 222 rem.

Here's an example. For a 9 twist barrel (1 twist in 9 inches) at 2600 fps.

The bullet makes 1 revolution in 9 inches. While the bullet is not travelling at 2600 fps the whole time it is accelerating down the barrel, it's for simplicity sake we can say that it covers the last 9 inches in 9"/2600fps since 9"=3/4', it's also 0.75'/2600'/sec or 0.0002884 seconds.

1 turn in 0.0002884 seconds is 3,466 rev's per second (1/0.0002884) and multiplied by 60 seconds in a minute give you 208,000 revs/min. Change the velocity and the rpm's go up.

FWIW, as velocity increases rpm's do increase, but the destabilizing forces on the bullet increase at a faster rate.

Hi Jim B.:

It's not too complicated, really. If you're into math at all, the formula I use is this:

Bullet RPM=VT60

Bullet RPM = Velocity in fps "times" Twist rate per foot of the barrel "times" 60.

Here's a couple examples from my spreadsheet:

SPR TNT/50 - W748/28.0 - 3,250fps - 1:11.750 twist - 1.02128 (twist rate per foot) - 199,149 rpm
HRN SPSX/55 - H322/23.5 - 2,995fps - 1:8.875 twist - 1.35211 (twist rate per foot) - 242,975 rpm

Hope that doesn't muddy it up too much.

Got so caught up in the figues I forgot my "conclusion"! As you can see, the faster the bullet goes and/or the tighter the twist, then the faster the bullet spins.

[ 04-15-2003, 04:53: Message edited by: reedg ]

[QUOTE]Originally posted by Stonecreek:
...a bullet that fails to stabilize at, say 2500 fps, in a 1-12" twist may very well stabilize pushed at 3000 fps from the same barrel. Or put another way, start with a heavy-for-caliber spitzer in a full-power load and start dropping the powder charge back until you're seeing half the velocity. You'll likely also be seeing keyholes.

I tend to think the bullet is still unstable at the higher velocity, but it just gets to the target faster, before we can notice any appreciable keyholing effects. After all it's still spinning at the twist rate the rifling gave it.
I tend to think if one had a 1-20 twist (extreme example) for a 7mm Mauser vs a 1-9, and shooting a 175 grainer in both, that regardless of how fast one shoots the 1-20 twist, it's still an unstable 1-20 spinner. The bullet won't stabilze at the higher speed, then become a keyholer at a slower speed.
The twist rate will always be what the rifling gave the bullet at the muzzle, thus a 1-9 twist (7mm) will stabilize a 175 grainer regardless of it's speed (ie 7mm Mauser, 7 mag, 7 Ultra, etc).
Just because a bullet goes 10 million miles a second doesn't mean it's spin-rate will increase. It will be what the rifling gave it to begin with.
A faster bullet will travel each foot quicker, but the spin-rate will be the same for each of those feet traveled. ~~~Suluuq

Here is an add-on to my above post.

I think some misunderstand what is happening to the twist, ie they think it increases at higher speed.
If this were true, then the 30-378 would somehow make a 1-10 twist barrel toss out a 1-9, or 1-8, spinning bullet because it's traveling very fast. Well, the 1-10 twist is a fixed thing (rifling at the muzzle), and won't allow a bullet to spin at a faster twist then what it gives it intially.

Now, a spin rate may slow down slightly out at a great distance (due to air friction), but it does so in relation/corellation to the bullet speed. Example... a 30-378 tossing a 220 grainer. By the time this 220 grainer gets out to 1,000 yards it's still stabilized, even though the bullet slowed down drastically.
Like I mentioned, a faster bullet will travel each foot quicker, but it's spin rate will be the same for each of those feet traveled.
Proof? Simple. We all know a bullet don't gain any further speed once it leaves the barrel, as there is no more "push" behind it. Air friction and gravity now acts upon this bullet, and slows it and pulls it down to the ground.
At 3000 fps, within the barrel, the bullet spin matches the twist rate right? After all, the bullet is "captured" by the rifling.
And what about after it leave the barrel? Does it somehow gain more twist? From what? "Air rifling"?
No. The spin rate is the same. A faster bullet may go quicker per foot, but it still spins at the twist the rifling gave it to begin with. Extra speed won't increase twist rate, it only shortens the time per foot traveled, and the spin remains the same for each of those feet traveled. Same for a very slow bullet. It travels very slow per foot, but the twist matches the rifling. ~~~Suluuq

[ 04-16-2003, 12:40: Message edited by: Rusty Gunn ]

I think that bullet velocity degrades at a higher rate than rotational velocity. A bullet needs to spin to make it stable, think about a football. There are no perfect bullets, we know. The bullet travels the bore about it's geometric center. After exiting the bore it will soon be rotating about it's center of mass. The best bullets will have the two centerlines the closest, of course. Spinning a bullet faster than needed will cause it to take longer for it to get from geometric center to center of mass attitude. Spinning a bullet too slow may prevent it from ever stabilizing about it's center of mass. Opinions by Jay, that is all.

I hate to jump into the middle of this, but it's not the first time that I've butted in, and I can be a slow learner at times....

The way that I read some of the previous posts, both Rusty and Boltman are correct - Rusty said that spin rate does not increase with an increase in velocity ("rate" as in "twist"), while it seems to me that Boltman interpretted spin rate as spin "speed" (i.e. RPM).

Velocity is maximum at the muzzle, and so is bullet RPM. RPM is a direct function of speed and rate of twist on the rifling (provided it doesn't strip through the rifling). A higher velocity does indeed produce a faster spin on the bullet.

As the bullet travels down range, wind drag slows down the bullet speed to a much greater degree than the spin, though both do decrease. IF you shot a bullet straight into the air, when it reached it's peak (zero velocity), it would still be spinning. Same thing would happen with a football, if you want to try a safe experiment.

Getting back to the original question, I have heard that bullets do not need as much twist as velocity increases, but have not really investigated the issue.

Best regards, Bill

I think we've pretty well beat this horse well past its death, but it seems that we all agree on the number of revolutions per unit of forward movement NOT varying (appreciably) with velocity, and the number of revolutions per unit of TIME varying (approximately) proportional to velocity.

I won't claim that marginally higher velocity will result in adequate stabilization from a 1/20 twist when a 1/10 is the norm, but I do believe that a bullet that is marginal in a 1/12 .300 Savage (2500 fps) will be adequately stabilized in a 1/12 .300 Weatherby (3200 fps) (which is exactly the twist the original Weatherby's used).

Boltman... Does this make sense?
At one inch inside a barrel, a bullet is at top velocity. (After all, once it leaves the muzzle it has no more "push", right?)
At this point (one inch inside the barrel), the twist is giving it a 1-12 spin.
Whether the velocity is 1000 fps, or 3000 fps, how can the bullet gain more spin per foot of travel outside the barrel then what the rifling gave it?
A faster bullet will travel each foot faster, and have more RPM's, but it will still be spinning at rate that the rifling gave it. Shouldn't it?
Take your 1000 fps x 1000 rpf = 1000 rps. At one minute = 60,000 RPM. I notice you mentioned one rev per foot, and thus make the equation to show this at 1000 feet distance. For each of those 1000 feet, the spin was one per foot.
Increase the speed to 3000, and you still have one rev per foot.
An increase in speed produces an increase in RPM,s, but it won't increase the twist rate per foot.
~~~Suluuq

[ 04-17-2003, 12:54: Message edited by: Rusty Gunn ]

Yes, but twist per foot doesn't affect stabilization of the bullet. The rotational velocity (RPM) is what generates the centrifugal force that stabilizes the bullet. It doesn't matter how many turns the bullet makes over a distance, what matters is how many turns the bullet makes over a period of time, that's what provides stabilization.

All that i know is this. The Greenhill formula for figuring rifling twist takes into account bullet velocity. WHEN you plug in all the all the information and then change bullet velocity the rifling twist changes.If velocity did not matter why would the formula ask for that parameter?JUST A THOUGHT.

Rust: Air resistance has little effect on the rotational velocity of the bullet. In other words, as the forward movement of the bullet slows (due to air resistance) and the rotational velocity remains nearly the same, the number of rotations per unit of forward travel DOES increase. This still has nothing to do with the stability of the bullet -- stability is imparted by the rotational SPEED of the bullet, which is fairly constant once the bullet leaves the muzzle. Bullets shot from the same barrel, but at different velocities, will ALWAYS have different rotational speeds.

On certain aspects of this thread, I stand corrected. I was led to believe the twist, and that alone, was what stabilized a bullet. I've learned otherwise.
May I bring your attention to Mr. Ken Howell's responce to my request to clarify this thread?
www.24hourcampfire.com
Check the "Ask Ken Howell" forum.
He sums it up pretty well. ~~~Suluuq

http://home.sprynet.com/~frfrog/miscellj.htm#rotating

Note that the author must be right as he is in regular contact with God.

"I say that increasing velocity only gets the bullet out of the barrel quicker and gets the bullet to the target quicker but that the rpm of the bullet would stay the same because it is controlled by the rate of twist of the barrel. Ie. a 1 in 9 twist rate can only produce a particular bullet rpm no matter how fast the bullet leaves the barrel.

Who would be correct on this?"

Jim B,
You are! I hope that the prize is a nice custom long gun!

"Its my understanding that the spin will equal the twist rate, in relation to it's speed. A bullet can't spin faster then it's twist rate, regardless of bullet speed. The faster a bullet travels, the quicker it travels per foot, but the twist rate is the same.
A bullet actually begins to slow down the instant it escapes the barrel, as there is nothing there to force it to spin, and air resistance acts upon it at this point."

Suluuq,
Well stated!
best,
bhtr

The faster the bullet goes the faster it spins. No different than your tires.

Thus a higher velocity will stabilize a longer bullet.

Well, to my understanding, the bullet leaves the barrel with a spin defined as x inches per spin. Leaving the barrel, it has no further power input (i.e. there is no external force to increase the spin), and, the expression "twist" in general understanding unfortunately changes from "spin per distance" to "spin per time unit (sec, min)", i.e. the point of view changes from "distance" to "time", and this causes confusion.

As spin per time throughout the flight of the bullet decreases only minimally whereas draft slows the bullet down drastically, spin in relation to the distance covered rises, i.e. 1 turn/12" at the muzzle becomes 1 turn per 8" at a given distance. And this is wrongly called "twist increases".

This given, I cannot see why, at a high twist, a lighter bullet should disintegrate when a heavier bullet stays intact: the centrifugal forces are the same.

[ 04-20-2003, 05:48: Message edited by: waitaminit ]

It is true that the higher the MV of a bullet, the faster it spins, when twist rate remains constant. BUT, generally, you cannot increase velocity enough to stop keyholing that is caused by a too-slow a twist rate. The improvement possible by just increasing bullet velocity is not GENERALLY sufficient to have an adequate effect on a badly understabilized projectile.

The Greenhill formula only takes into account a constant based upon a specific gravity of 10.9 for the projectile's composition (what it is made from) divided by the the length of the projectile IN CALIBERS. The constant is 150. the formula is REQUIRED TWIST (in calibers) = 150/projectile length (in calibers). Hatcher's example: If one used a .30 caliber bullet 1.35" long, it would be about 4.5 calibers long. 150/4.5= a required twist of 33.333 calibers, or 10, or 1 turn in 10". This is not absolutely correct, because the caliber is actually .308", so the required twist turns out to be around 1/10.5", but 1/10 appears sufficient for this 1.35" long bullet (220 grain RN). There is No place in the Greenhill formula for a velocity variable, unless it was included in the calculation of the constant, 150. However, the formula in itself does not include derivation of the constant.

[ 04-20-2003, 17:48: Message edited by: eldeguello ]

This discussion is hilarious!!!

Don't you people realize that some of you are talking about revolutions per inch of travel and that some of you are talking about revolutions per minute?

ROFLMAO!!!!!!!!

You're all correct and noone can see it.... Keystone Cops.... Too funny.... <gasp>

My stomach hurts from laughing.....

$bob$

LD Hunter---

FINALLY !!! A post that makes sense. Thanks.

I never knew so many people could be stuck in such a simple question. Sheesh!!

Rotation PER INCH depends on the barrel.

Rotation PER MINUTE depends on velocity.

How can that become confusing??

Yeah, I don't understand it either!! A bullet fired through a barrel having a twist of one turn in ten inches will always turn once in every ten inches it travels. But if you double the velocity, it will revolve twice as fast per unit of elapsed time, because it travels twice as far in a particular unit of time.

[ 04-21-2003, 20:08: Message edited by: eldeguello ]