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This is a discussion on Bullet RPM... within the Accuracy forums, part of the M14 M1A Forum category; I always wondered what those funny looking twisty things inside the barrel were for. They spin the them bullets real fast! Beer make my head ...


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Old January 25th, 2017, 10:29 AM   #16
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I always wondered what those funny looking twisty things inside the barrel were for. They spin the them bullets real fast! Beer make my head spin like that...

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Old January 25th, 2017, 11:23 AM   #17
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I always wondered what those funny looking twisty things inside the barrel were for. They spin the them bullets real fast! Beer make my head spin like that...

Looking at them can make a person dizzy. I think they were put there by the folks who make bore cleaning products just to make life difficult.


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Old January 25th, 2017, 12:09 PM   #18
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I tried to imagine a hollowpoint spinning that fast as it expands and cuts through a game animal.
Think about it this way:
1) If your barrel has a 1 in 10" twist, and
2) Your bullet penetrates 10" in a game animal, then
3) The bullet decelerates to zero in 10 inches.
4) The rotation also goes to zero in 10" (only completes one turn)

What's the rate of deceleration?

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Old January 25th, 2017, 01:29 PM   #19
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Think about it this way:
1) If your barrel has a 1 in 10" twist, and
2) Your bullet penetrates 10" in a game animal, then
3) The bullet decelerates to zero in 10 inches.
4) The rotation also goes to zero in 10" (only completes one turn)

What's the rate of deceleration?
"...(only completes one turn)"

False. Bad thinking. You can't ignore the necessary obvious facts in your calculation. That is "static, linear" thinking as if you were pushing a bullet 10" down a barrel with a cleaning rod w/o considering the velocity and RPM when the bullet entered the target, its caliber / cross section, bullet weight and design, expansion, petal deployment, fragmentation, etc. Think dynamically. In 3-D, and in motion, if you will. You will need more data of a specific nature and a calculus equation to solve your scenario - and then it will be only an estimate.

Assuming that the bullet is travelling at 2,800fps from a barrel with a 1:10" twist when it enters this target of unknown material, composition, and consistency, it will spin 2,016 times in the first 1/100 of an inch.


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Old January 25th, 2017, 03:10 PM   #20
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Old January 25th, 2017, 03:23 PM   #21
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One of the reasons I always opt fo the slowest twist possible just enough to stabilize the bullet at the intended weather conditions and altitude application. We're talking about target guns, not combat rifles.

Marginal bullets, not marginally stable, like a void in the lead core will spin out of axis in faster twist barrel. Think about it as an unbalanced crankshaft. You spin too high it might self destruct. Bullets are much better these days, but every now and then you encounter a bad one. I had one come out at 600 one time that my scorer described like a corkscrew pattern instead of an arc. I shot a 7. Scorer said I was lucky it even hit paper because the spiral pattern at mid distance was wider than the target. I was shooting an 8 twist launching the 80 class bullets at 2900. Had the barrel been the popular 7 twist I might have shot a miss.

So I asked Walt Berger, his short answer, void in the lead core, it happens.

My best shooting JC Garand has a 15 twist Krieger GI Contour.

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Old January 25th, 2017, 03:33 PM   #22
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Originally Posted by DudleyDR View Post
"...(only completes one turn)"

...it will spin 2,016 times in the first 1/100 of an inch.
Interesting discussion, but, Dudley, You lost me - a gross math error somewhere perhaps? Like by a factor of million plus?

Leaving the barrel the bullet is turning one rev per 10 inches - right?

How does it spin up to 2,016 revs per 1/100 inch vs one per 10 inches?

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Old January 25th, 2017, 03:53 PM   #23
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One of the reasons I always opt fo the slowest twist possible just enough to stabilize the bullet at the intended weather conditions and altitude application. We're talking about target guns, not combat rifles.

Marginal bullets, not marginally stable, like a void in the lead core will spin out of axis in faster twist barrel. Think about it as an unbalanced crankshaft. You spin too high it might self destruct. Bullets are much better these days, but every now and then you encounter a bad one. I had one come out at 600 one time that my scorer described like a corkscrew pattern instead of an arc. I shot a 7. Scorer said I was lucky it even hit paper because the spiral pattern at mid distance was wider than the target. I was shooting an 8 twist launching the 80 class bullets at 2900. Had the barrel been the popular 7 twist I might have shot a miss.

So I asked Walt Berger, his short answer, void in the lead core, it happens.

My best shooting JC Garand has a 15 twist Krieger GI Contour.
In my conversation with Mike at Krieger, he said that the way most people use .308 they would actually be better off with a 1/14 twist than the standard 1/12.

In one match with issued M16A2 and M855, I was holding the 10 and X at 500 slow. Mid string, target came up a miss. I had checked my target number (always do after cross-firing a couple of times...) and called the shot an X. I challenged, score stood at a miss. Shooters around me didn't report excess rounds, but then again maybe if it was an X they just kept their mouth shut. But it seems that M855 steel/lead bullet just clean missed the target. And I missed the medal bracket by 1 point. Oh well.

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Old January 25th, 2017, 04:20 PM   #24
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Interesting discussion, but, Dudley, You lost me - a gross math error somewhere perhaps? Like by a factor of million plus?

Leaving the barrel the bullet is turning one rev per 10 inches - right?

How does it spin up to 2,016 revs per 1/100 inch vs one per 10 inches?

Yahoo
Sir, you were right to question my math on that 1/100 of an inch velocity. I am going to have to re-calculate that. I do stand by all of the other math involved and state that the bullet is spinning at 201,600 RPM as it enters the surface of the target. I thank you!

"Leaving the barrel the bullet is turning one rev per 10 inches - right?"

Yes, but it is not spinning slow. You can't ignore the time, and acceleration / speed factor. Think of the bullet from the time it is standing still just behind the rifling at the throat of the bore and before it engages the rifling. You are forgetting the acceleration factor from the pressure and expanding gasses from the propellant.

Can you agree that it is reasonable to use a muzzle velocity of 2,800 fps from a 22" barrel (even though there may be only about 19 1/2" of rifling)? So how does it get going that fast in that short of a distance...from zero to 2,800 feet/second linear velocity in under 22"? Approximately 50,000psi chamber pressure developed and applied almost instantly at ignition plus the expanding gasses pushing the bullet the (less than) 22 inches to result in a (presumed for discussion) muzzle velocity of 2,800 fps. Using the previous formula:

2,800fps multiplied by 720 (which is the number if inches in a foot [12] multiplied by [60] the number of seconds in a minute) equals 2,016,000. Dividing that number by 10 (the rate of twist in inches) equals 201,600 RPM. Which is the bullet's RPM at which the bullet would enter the surface of the target at 2,800fps.

Again...thank you.

Revolutions per Minute is a "rate", like Miles per Hour. It is a snapshot of an instant in time. It changes by the foot and second and any fraction or multiple of those measurements. For most shooters on most occasions, bullets do not stay in flight for a minute.


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Old January 26th, 2017, 03:19 AM   #25
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Now consider the speed at the edge of the bullet:

A 308 bullet is <well> 0.308" in diameter 0.9676" around its perimeter.
A 223 bullet is 0.224" in diameter and 0.7037" around its perimeter.

Thus, the speed of the 308 at its edge is 162,558 "/s
Thus, the speed of the 223 at its edge is 190,003 "/s

Or almost the same speed at the edge!
Your numbers aren't right, it's 162,558 inches/minute and 190,000 inches/minute

that is 154 and 180 mph . . .

The muzzle velocity of a bullet is around 2000 mph.

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Old January 26th, 2017, 10:43 AM   #26
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If the muzzle velocity of a projectile is 2,800 feet per second, how many miles per hour is that equal to? (Not a trick question, just a simple conversion of units from feet/second to miles/hour.) All that linear velocity is achieved in under two full rotations of the bullet. Amazing (to me, at least)!

The number may surprise you. Thanks!


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Old January 26th, 2017, 10:51 AM   #27
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Quote:
Originally Posted by DudleyDR View Post
To Yahoo:

If the muzzle velocity of a projectile is 2,800 feet per second, how many miles per hour is that equal to? (Not a trick question, just a simple conversion of units from feet/second to miles/hour.)

The number may surprise you. Thanks!
1 x 0.681818 feet/second to miles/hour


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Old January 26th, 2017, 10:53 AM   #28
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Moving right along... Krieger told me the 14 twist was best for .308 too, however I think it works best in relation to bench rest shooting. I have always used a 1:10 in my match M14's. It seems to shoot anything well. I have also had excellent results in a 1:12 twist, enough that I don't think there is any significant difference between the 1:10 and 1:12 when used on the M14 with compatible bullet weights. The military chose the 1:12 to get longer barrel life with reasonable accuracy.

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Old January 26th, 2017, 12:42 PM   #29
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it will spin 2,016 times in the first 1/100 of an inch.
High speed video suggests otherwise:

.
That's actually a 9mm.
See:


The .308 impact is so violent you can't tell what is going on:
.

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Old January 26th, 2017, 01:55 PM   #30
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High speed video suggests otherwise:

.
That's actually a 9mm.
See: https://www.youtube.com/watch?v=bqziWTq7X1c


The .308 impact is so violent you can't tell what is going on: https://www.youtube.com/watch?v=jE9x...BkGg8Nj3X0dW8Q
.

Sir, I had basically recanted that statement about the entry RPM during first 1/100 inch already and will be re-calculating it. Please see my post above.

Now show me the video indicating the time frame (elapsed time) between the instant the projo enters the gelatin and when the projo comes to a complete stop. Thank you in advance.

If a projo enters the gelatin at 2,800fps and travels 12" before it comes to a complete stop, it has taken it a very small fraction of a second to travel that one foot.

The projectile is both travelling through the air and spinning faster than you can see. In the videos that you have presented to support your position, you can see only the last revolution of the projo (or so) as the gelatin absorbs the bullet's energy and brings it to a stop, but you are ignoring the time and distance components of the situation and how fast the bullet is spinning.

A related example. A car is travelling along at 60mph. The wheels and tires are spinning relatively fast. The brakes are applied. Your argument is like taking a picture of the last revolution of the tire before it comes to a complete stop and saying "that evidence" nullifies all other facts. It is totally irrelevant to the initial speed of the vehicle and the RPM of the wheels and tires as the vehicle is travelling along at that 60mph.

I challenge you to do this same calculation as offered above:

If the muzzle velocity of a projectile is 2,800 feet per second, how many miles per hour is that equal to? (Not a trick question, just a simple conversion of units from feet/second to miles/hour.) All that linear velocity is achieved in under two full rotations of the bullet. Amazing (to me, at least)!

The number may surprise you. Thanks!


Last edited by DudleyDR; January 26th, 2017 at 02:10 PM.
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