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Son has a ruger M77 boat paddle in .243 1 in 8 twist. He recently loaded up some Speer 100 grn boat tails and accuracy went out the window. Uncle Nick's bullet twist and stability program says that the bullet at 1.05 inches should be O.K. Velocity I don't know so punched in 2800 fps. Any thoughts?
bones
 

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With that twist, shouldn't be a stability issue, the barrel just doesn't like those bullets or that loading. Know that's not a lot of help, but twist-unstable bullets leave at least oval holes when they are just barely unstable...often tumbling when truely unstable and giving you an occasional key-hole strike (occasional as they tend to group the size of a Volvo). If you're getting round holes in the target, even if they are grouping 6MOA, may just be the barrel showing it's preference for bullets/loads other than that one.

So...other than the groups being large... anything else noticed about the bullet holes on paper?
 

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I think I'm on my 4th .243, a Rem M700, a Win M70 UL, a Ruger No.1, and I currently own a M77 MKII. They all had 10" twist barrels, and all would stabilize anything from 70-105gr bullets from Hornady, Speer, Sierra. I wonder if a 8" twist puts a bit too much spin on those 100gr'ers?

Does it shoot something like a 100gr Hornady Spire Point? How about lighter bullets?
 

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Just a thought. . . . the bullet wants to go forward, the rifling wants it to spin sideways. A faster twist requires a longer bullet BEARING SURFACE for the rifling to bite into, otherwise the bullet will skid out of the rifling and get damaged before it leaves the muzzle.

A boat tail spitzer bullet has less bearing surface than a flat base round nose bullet of the same weight. Try a bullet with more rifling contact and the results may improve.
 

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Try some Hornady 100 gr Interlocks and 105 Gr V-Max. They've always been good performers for me.
 

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Skid Marks ~~~~

(not the ones in yer shorts either).

I have wondered how far any given bullet, in any given chambering,at any given speed,with any given twist rate of bbl,would "skid" before the rotational velocity came up to max for the above given conditions.

I have read "theories" in the past but nothing "official". I find it kinda hard to believe that anyone could figure this stuff out anyway,given the velocities and the methods of bullet recovery possible. But......a bullet leaves the brass(case mouth) with a whole bunch of burning powder behind it. The pressure is going up rapidly and consequently the bullet is accelerating rapidly. A few thousandths later(hopefully) the bullet crosses the distance of the freebore and slams into the lands. They dig into the bullet and start cutting it,imparting a spin(rotational velocity) on the bullet.

How much bbl is needed to get the bullet up to the max rotational velocity that the bbl will impart on the bullet before exiting the bbl ?? What real damage will the skidding do to the bullet ?? How will this negatively effect the bullets B.C and/or accuracy ??

Bottom line is accuracy. To me this stuff is fun to contemplate. We should know enough about it to be able to carry on a simi-intellegent conversation with another shooter at the local diner,but beyond that....... I love to shoot/expierament/tinker. Ultimately if my bullets go into a nice tight group that I can move my scope to look at,I'm happy. -----pruhdlr
 

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If the twist is a constant pitch, as they normally are, the bullet will undergo rotational acceleration and rotate faster and faster as long as it is gaining velocity in the bore. A .22 rim fire stops accelerating and starts to slow at around 16" to 19", depending on the load, so they would actually stop gaining rotation and start to lose it in a long barrel. Most medium power rifles need closer to 50" of barrel length to hit the point the expanding gas stops accelerating it against bore friction.

As to skidding, you need to clarify what you mean by that. Bullets normally have rifling marks that only smear at the back edge of the land marks. Typically, a core will strip before rifling lets a jacketed bullet slip. Harold Vaughn has some core stripping measurements made with a magnetometer in his book, Rifle Accuracy Facts. It seems to me these were 90 grain bullets in a .270 Winchester going around 3200 fps, but don't hold me to those numbers. I'd need to reread to be sure. Smaller diameters would be harder to strip and could be driven faster without encountering the problem. A bonded bullet core could also be driven faster and would be much harder to strip. For all I know, his test bullets may have been exceptionally easy to strip? But the bottom line is that each bullet will have some threashold peak acceleration rate for any given twist rate at which that can happen.

Note that faster twist barrels are usually intended for longer bullets. Those bullets are usually also heavier, so they don't accelerate as rapidly. That helps counter the torque difference the more rapid twist applies to the jacket.


Bones,

This does sound like a load that is just off a barrel time sweet spot for the gun. I don't know what the throat is like, so perhaps seating depth could be played with? But first, have your son read through Dan Newberry's OCW site to learn a systematic approach to finding sweet spot loads. Mike made a sticky in this forum on the topic with links. That sites is a highly recommended read.
 

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I think you are confusing forward accelleration with rate of rotation (or spin). The rate of spin remains constant, only the forward velocity decelerates. A bullet fired straight up eventually stops, but is still spinning. As it falls base down it continues to spin.

If the twist is a constant pitch (as the 1 turn in 8 inches stated in this question), how can the bullet be subject to "rotational accelleration"? This would infer that the bullet has started off at a rotational speed of less than 1 turn per 8 inches and builds up to the maximum spin of 1 in 8 inches? If this is true, the bullet must be initially skidding in the rifling until it reaches the maximum rotational speed of the rifling.

This problem was overcome about 100 years ago in some rifles shooting lead bullets by having progressive rifling twist to "slowly" increase the rotational velcity without stripping the bullet.

The same physics apply today. If the twist is too fast for a given bullet at a given velocity, there is the possibility of the bullet stripping out of the rifling (hence skidding).
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =​

If the twist is a constant pitch, as they normally are, the bullet will undergo rotational acceleration and rotate faster and faster as long as it is gaining velocity in the bore. A .22 rim fire stops accelerating and starts to slow at around 16" to 19", depending on the load, so they would actually stop gaining rotation and start to lose it in a long barrel. Most medium power rifles need closer to 50" of barrel length to hit the point the expanding gas stops accelerating it against bore friction.
 

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What I said is correct.

For example, suppose for simplicity's sake, you have a 12" twist and a 12" barrel. A bullet fired at 1000 feet per second left moving at velocity that covers one foot or one complete rotation per millisecond. A bullet fired at 2000 feet per second has left the muzzle traveling at velocity that coverst one foot or one complete rotation per half millisecond. The first bullet makes one complete rotation in 1 ms, while the second makes 1 complete rotation in half that time. The second bullet is therefore launched at twice as many rpm as the first bullet was. To get to twice as many rpm, it had to undergo twice as much average rotational acceleration, just at the forward velocity of the bullet means it underwent twice as much average linear acceleration.

Where P is the twist pitch in inches, the formula for bullet RPM is:

RPM=MV×720/P

The 720 constant takes care of converting the mixed units of length (feet in velocity and inches in the pitch) and mixed units of time (seconds in velocity and minutes in rpm). The main thing you need to see is that rpm is proportional to velocity and rotational acceleration is therefore proportional to the linear acceleration that gets the bullet to that muzzle velocity.

I suspect the fact the same twist will stabilize the same bullet fired at different muzzle velocities, confused you into thinking the same RPM was stabilizing the bullet at different velocities. That would not work, even if a barrel could be designed to do it. In the supersonic range, at about mach 1.5 up until around mach 5 (faster than we fire bullets) most modern bullets experience a fairly linear increase of drag with velocity. That is caused by air resistance which increases with velocity, and is what pushes a bullet nose off course and makes it tumble. Therefore, if your bullet shape is such that it experiences twice as much air resistance going twice as fast within that velocity range, you then need twice as many RPM's for the bullet to have the gyroscopic inertia to resist being turned to tumble. Hence, the same rate of twist works with it at both muzzle velocities, because it provides that doubling of RPM with doubling of velocity.

In the real world, doubling velocity doesn't quite double drag of a modern pointed bullet in that mach number range, so increasing velocity with the same barrel twist rate will actually increase stability a bit, but it's not a vast increase.
 

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What you said IS NOT correct.

Using your example . . .a bullet spinning at 1 turn per 12 inches will spin 10 times in 10 feet of travel, irregardless of its forward velocity. The forward velocity will greatly affect the amount of time it takes to travel those 10 feet, but the forward speed does not change the number of revolutions the bullet will make in a given distance. The velocity influences the time (and trajectory) it takes for the bullet to reach the target, but it will still only make one revolution for each 12 inches traveled. The forward speed does not decrease or increase the number of revolutions. The bullet will just get to the target faster, but with the same number of revolutions per feet travelled.

100 yards = 300 feet.
1 turn in 12 inches = 1 turn per foot travelled
300 feet x 1 turn per foot = 300 revolutions to reach the target
 

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What you said IS NOT correct.

. . . The forward speed does not decrease or increase the number of revolutions. . .
I just reread your first post. Was in a hurry the first time. We are talking past each other. I was discussing interior ballistics and you were discussing exterior ballistics. Rate of rotation (revolutions per second) changes with change in forward velocity inside the barrel. That's what I am discussing in response to Pruhdir's comment about maximum rotation being achieved. That can only happen inside a barrel. That's also what the RPM formula is for. Obviously, nothing increases spin after the bullet leaves the muzzle.
 
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