My 243wssm has a 1 in 10 twist barrell, will it stabilize the 105 amax.

John M

MY guess would be MAYBE. Driven hard it sure might.

I've heard folks say that they could stabilize it in 243AI and 6mmAI chamberings with 1-10" twists. Others with standard 243 rifles have said they could not. I do think it would have to be driven hard to stabilize.

Gonna play hell seating the 105 out far enough to not eat up case capacity in the WSSM action.....................

I just got a ruger m77 mkii in 243 win with a 1:10 twist. I am working up a load for it right now, in fact I got back from the range about 10 minutes ago. The A-max is the most accurate round thus far. I shot a 4-shot group with the 105 A-max that was just under an inch at 102 yards. This was off a bipod without letting the barrel cool on a sporter barrel, in the Tucson sun at 12:30 in the afternoon, so the barrel was hot. I was using ramshot hunter at the max load listed in the Lyman 49th addition manual. My battery died on my chrono so I cant give you velocity but there were no signs of excessive pressure. I realize one group doesn't mean that it wasn't a fluke but I only brought out 1 round at each charge weight working up to the max where I had four to shoot for a group I am going to try to make it out tomorrow and group a variety of charge weights to see what it can actually do. I generally don't shoot much better than .75 MOA off a bipod, so I doubt I will do much better. Especially with this gun, it wasn't very impressive with factory ammo.

The other thing folks don't consider, is what the twist actually is... hence the idea if its close and driven hard, it may well be fine, as you may have a 9.5 twist instead of 10... IE they ain't all perfect. You could OTOH have a 10.5 twist and I think that would be enough to kill the chances.

I've never had an issue stabilizing 105 speers in my 243 with 10 twists, but those are NOT an amax...

The idea that bullet weight is all important for twist rate is a little false. What matters more is the length of the bearing surface, or the surface of the bullet that contacts the rifling. This length tends to increase with increasing bullet weight so there is some correlation, but there are other factors such as boat tail vs flat base, the style of the nose, and the degree of taper to the point that can make two bullets with the same mass not have the same length bearing surface. The 105 gr A-max is a boat tail and has a fairly gradual taper. This makes their bearing surface just a little shorter than some 80 gr Remington PSP I have that have a flat base and sharper taper. You would have no problem stabilizing the 80 gr bullet and you won't have any problems with the 105 gr either because they have about the same bearing surface length. This doesn't mean for sure that your gun will "like" them, and you will still have to fiddle with loads to find an accurate one, but stabilization will not be the issue. An example is the Barnes varmint grenades. Even though they are very light they require very fast twist rates because they are non lead, thus less dense and have a long bearing surface for their weight.

You're part way there, but not quite. It's the length of the bullet not the bearing surface. And just as greater weight often leads to a longer bearing surface (as well as length), a longer bearing surface "usually" comes with a longer bullet as well. The exceptions would be the VLD type bullets which actually have a shorter bearing surface (for comparable weight), but actually need a faster twist as a result, because they also have longer ogive's (for comparable weight).

Take a look at the Greenhill formula for twist. It's quick and dirty, but it works most of the time. It doesn't take into account weight nor bearing surface. The input variable is length vs diameter ratio.

The one area where bearing surface may affect stability is "in bore yaw". But that is a factor independent of twist.

Chris
Nice post.

Jeff

my 243 with a one in 10 twist will shoot the 105 Speer and the 105 A Max just fine...

Does anyone have a length on the 105 A-max? I'm curious to see what Greenhill calls for. I realize that Greenhill tends to overestimate twist for modern cartridges, but I think it would be interesting just the same.

I've had some problems with them stabilizing in a 26" 6mm No. 1

1.229" (nominal), IIRC.

ChrisF,
I know this is an old post,but I have been out in the woods.
I agree that I didn't give quite all the truth, but it isn't just the bullet length that matters either. What the greenhill formula was getting at was rotational momentum that needs to be imparted onto the bullet. Rotational momentum depends on how fast you are rotating and how far the weight is from the axis of rotation. If your cylinder (bullet) has a larger radius, then you have more weight farther from the axis of rotation, and there is more momentum at the same RPM as a smaller diameter bullet of the same length, so you need a faster twist rate to impart that momentum. Similarly if you make that cylinder longer (heavier bullet) you have more momentum at the same RPM. Back in the 1800's when the greenhill was first derived most bullets were not of spitzer design, and did not have a boat tail. This made the bullets essentially cylindrical in shape so length and width was a very good way of approximating bullet shape and weight distribution. However when you have modern bullets the diameter of the bullet varies, not to mention what it is made of varies in density (polymer time jacketed, lead core, hollow cavity. Consequently, there is no good and simple formula to determine the rate of twist. Really one would have to model the bullets shape with a mathematical expression for each place in the bullet where you had a shape/component change, and then integrate it to find the rotational momentum due to that portion of the bullet at a given RPM. This is why the greenhill formula commonly overestimates twist now.
Depending on the design of the bullet, the length will sometimes approximate the twist rate better and in other cases, there is not a lot of weight in the nose or boat tail so the bearing surface and diameter are a better approximation. In the end what we both said is wrong, but what we both said is also kind of correct.

dogmessiah,
Yes, the length is not the only factor, but for a quick and dirty calculation such as Greenhill's, it's a useful surrogate. More so than the length of the cylindrical section of the bullet (bearing surface). And when you get down and dirty into the different factors that affect twist requirements, length of the projectile still figures very prominently in the detailed equations.

Greenhill's formula was actually deduced for a "prolate spheroid" (egg shaped) projectile. Definitely not a cylinder. But still not representative of modern projectiles either.

Everyone can disregard the following since it will probably bore you to tears...but I think dogmessiah might "get" it...so here goes...
--------------------------------------------------------------

What we're actually trying to do with twist (spin) is achieve gyroscopic stability sufficient to overcome the pitching moment created by the aerodynamic force acting on the difference between the Center of Gravity and the Center of Pressure of the bullet. ..and yes, the diameter of the bullet is a factor in the twist requirement in the form of the Axial moment of inertia (I think you're calling it "Rotational Momentum"). But it is the length in the form of the Transverse moment of inertia and the pitching moment arm that figures more prominently. (The formula for gyroscopic stability of rotating bodies doesn't translate very well without sub and super scripts otherwise I'd include it here too). Suffice it to say that in the classical formula for gyro stability, as the length of the bullet goes up, Transverse moment (I sub t) increases faster than the Axial Moment (I sub a), and the moment arm (center of pressure to center of gravity) increases with the length as well. All emphasizing the importance of the length of the bullet in the twist requirement more so than some of the other meansurements suggested.

Modeling the shape of the bullet is not entirely necessary. You can empirically measure I sub a and I sub t via a torsional pendulum. This would account for the varying shapes of the bullets as well as the different compositions (hollow point, polymer tipped, etc). You could then plug the measured values into the general formula for Gyro Stability and come up with a useful number for twist.

...but if you did have a detailed measurements of the projectile...there's always McGyro courtesy of Bob McCoy and later Bill Davis Jr (where again length of the bullet figures very prominently).

I'll still take length of the bullet if I ever have to SWAG a required twist rate for my intended pill.

Thanks VAnimrod,
Plugging your 1.229" figure into Greenhill Classic Formula (Twist = 150* D^2/L) gave the following;
1 in 7.2"

Using the updated constant of 180 for velocities of 2800fps gives:
1 in 8.7"

...and using the constant of 200 supposedly suggested by the CampFire's own Dr Howell gives;
1 in 9.6"

dogmessiah,
Is there a formula that can approximate twist requirements with Bearing Surface and Diameter as the input variables? If yes, can you walk us through an example with one of your 105 A-max's? I'd like to compare how it looks next to Greenhill.
Thanks,
Chris.

So I guess sometimes it will, and sometimes it wont is the answer to my question.
Thanks fellas, that got kind of deep, but I thought is was interesting nonetheless.

John M.

Good, credible range data trumps theory and mathematical modeling every time. What qualifies as good and credible is what's going to make or break your decision.

What your own gun does or doesn't do in your specific application trumps good and credible data from others every time as well.

Sounds like it's time for you to invest in a box and some load/range time. Have fun...good luck...let us know how it goes for another data point for the others to make their own decision.

...what ever happened to dogmessiah?

+1