hi all,

i was wondering about bullet weight and twist rate. i have a remmy 700 in 223 with a bull barrel. i think (sole guessing) that the twist rate is 1 in 9. is there a way that i can find out the twist rate? and is there a listing of good weights for a particular twist? i am trying not to buy every grain bullet to try and find a good one.

thanks
lojack

Here is a formula I have found that seems to work well with literature values I looked up.

approximate optimum twist rate = (150 / (bullet length / bullet diameter)) * bullet diameter

example:
.223 70gr bullet
.785 in length
.224 in diameter

150 / (.785in/.224in) * .224in
150 / (3.50) *.224in
42.86 * .224in
9.60

Which means to stabilize this bullet a 1:9 or 1:10 barrel would be needed.

This bullet is what is fired out of the M16A1 and accordingly the barrel is a 1:10 inch rifling.

The best way to determine your rifling is to use a cleaning rod or some substitute that has no ball bearing. Equip this with a tight fitting cleaning patch and mark the rod near the top. The cleaning patches will engage the rifling and twist just as a bullet would. Push down until your original mark comes around again then measure how far you pushe the rod in. Voila, the twist rate is now known.

Hope it helps from a fellow newbie who has been doing a lot of reading.
Carl

[ 06-23-2003, 01:35: Message edited by: Anvil63 ]

Minimum Barrel Twist Necessary To Stabilize ______________________________________________________________Bullet Weight (gr)
Diam. 20 40 60 80 100 120
.224" .. 33.0 16.5 11.0 8.3 6.6 5.5

This is baised on the "Greenhill" formulia (plagerized above)

I apologize for not citing my sources. It is the Greenhill formula which I got from the ABC's of Reloading by C. Rodney James (a great book that explains the reloading process in great detail)on p. 63-64 in ch. 6 which is on bullets.

I also got the idea of using the cleaning rod for determining barrel twist from the same source.

Man, didn't think I would get accused of plagiarism in this forum. Makes me feel like I'm back working on my thesis for my B.S.

-Carl

[ 06-23-2003, 02:31: Message edited by: Anvil63 ]

lojack, the Lyman "reloading Handbook" lists the Remington 700 as having a 1 in 12" twist rate. This will limit you to bullets of about 60 grains in weight(maybe more with a round nose).

Follow this procedure, it will confirm your barrels twist rate. With a tight patch and jag, start the cleaning rod down the barrel; once solidly engaged with the rifling, make a mark on the rod. Push the rod into the bore until it rotates 360*. Make the second mark. Measure the distance between the two, this is your rate of twist.

As a general rule, the faster the twist, the heavier(longer) the bullet one can use. In your case I'd stick to 55 grain bullets or less.

packrat

so it looks like the formula is based on bullet length rather than actual weight. i have a 55gr sierra that is shorter than a 50 gr nosler. so if i used the above formuala, the 55 grain sierra is a better round for the remmy 700. after calc. the sierra comes out to a 1 in 10 (10.27) and the nosler comes to a 1 in 9 (9.66) so the actual wieght of the bullet doesnt matter. is this correct?

thanks
lojack

ok so the answer to finding the "perfect round" is as follows:
1. get the right length of bullet for the twist rate of barrel
2. find a good bullet velocity (powder etc)
3. bullet shape (bt, bthp, fb etc)
4. Case overall length

is this right

Lojack, forget about bullet weight as a determinant for accuracy for now; providing your rifle stabilizes the bullet in question.

If you want more accuracy, you would be much better off with a short benchrest bullet than with a heavy hunting bullet. Buy some Fowlers, Eubers, Bart's or even Bergers. Unless you are shooting at extreme ranges, you will never know you "overstabilized" any of them. JMO, Dutch.

Hey Guys,

Here is a little more detail about use of the Greenhill formula for determining the twist rate needed to stabilize a bullet. We have to remember that the Greenhill formula was developed at turn-of-the-century 1900. That was when a bullet was a hunk of solid lead and it didn't have a very sharp point either. So now what to do about it in today's world of hollow point and plastic tipped bullets?

Well, the first variant is to recognize the effects of velocity. For velocities 2800 f/s and lower, the constant of 150 is OK. For higher velocities above 2800 f/s a constant of 180 serves better. So for low velocity, Twist= Bullet diameter squared divided by bullet length times 150. For the higher velocities it is better at Twist = bullet diameter squared divided by bullet length times 180. All dimensions in inches.

Now what about those hollow point and plastic tipped bullets? Well, I measure what the length of that bullet would be if it didn't have a hollow point and if it didn't have that extra length of that plastic tip. Seems to work OK.

Then, for the sub calibers, even the barrel maker, Shilen and the bullet guys - Hornady say the Greenhill is only a rough guide. That is for the calibers smaller than .224.

Don Shearer

so don

are you saying to use the same formual, but to use the 180 instead of the 150 for rnds faster than 2800 fps.
the second formula read different:

For the higher velocities it is better at Twist = bullet diameter squared divided by bullet length times 180

(.224squared/bullet length)*180

the first formula read:

(150/(bullet length/bulet dia))*bullet dia

lojack

Thanks Don, I had never heard that (not surprising since I bought my first book 4 weeks ago). This exception for bullets going faster than 2800fps has been forever memorialized in my reloading book as the "Shearer Corrolary."

Thanks a lot,
Carl

[ 06-23-2003, 07:07: Message edited by: Anvil63 ]

Lojack,

Nope, read it again - - It is the same formula with a different constant for low and high velocity.

Twist = bullet diameter squared divided by bullet length then that quantity multiplied by the constant. The low velocity constant is 150 and the high velocity constant is 180. Dimensions in inches.

Don Shearer

A .224" diameter bullet with a length of 0.84" would be "optimal" in a 1:9 barrel.

Weight is not the determining factor.

BUT...as with most calculations, the real world number will be different.