Round Ball? Minnie? Bullet? Twist rate? Help.

Powder and twist for BP

I've been shooting BP since 1970 when I was working in a gun store. I purchased two C&B revolvers, one a 36 caliber and the other a 44 cal. I also purchased a 50 Cal Hawken rifle.
The twist in a Hawken is 1-48 which is a middle of the road compromise twist. The best for round ball and patch is in the 1-66 range and slower.
The new inline guns are not meant to fire round ball or the old conical bullets. They shoot the sabot rounds or the big conicals because they have a fast twist.
The best for the conical or the new Hornady Plains bullets is the 1-48 and faster like the 1-36 or so.
2F is the best all around grain size for 50 caliber and above. If using Swiss then the 1 1/2 gets the call.
All black powder is the exact same as far as it's production goes. It's a mixture of three chemicals and the speed at which it burns is determined by the grain size only. It all burns from the outside in and therefore the smaller the grain the faster the burn.
The foreign made powders tend to be smaller than the Goex made in the USA.
I have made several slides and viewed it under a microscope to verify this.
Use your inline with larger charges and the correct componets and you will start to see good groups.
Stay away from 4F unless you are primming a flintlock pan.
3F will burn in your inline but will more than likely not give the better accuracy. Each gun will have it's own favorite load.
Inlines are usually made to fire 209 shotgun primers because of their nature of being harder to ignite the pellet type charges used by many people today.
I have never fired a sub powder and never will as I'm a BP purist.
You should buy one of the books by Sam Fadala which is in it's fith printing. It is named the Complete Blackpowder Handbook.
I disagree with some of his writings but overall it's a good starter book to have. Later David
 
Practicals first ... !!

Fairshake;
Pretty much sums it up and as he and deerslayer stated, don't use 4F as a main charge. As far as max loads, I see folks loading too hot and aside from potentially being unsafe, it's wasteful as to much powder will wind up on the ground in front of you. .... ;)

Now, I'm still talking In-lines and I never see the need to exceed 90 -95 grns. As far as pellets, your choice but don't care for them and their issues.. :(
The best for the conical are the new Hornady Plains bullets
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Excellent choice and one of best. if not the best. ... ;)

You should buy one of the books by Sam Fadala which is in it's fith printing. It is named the Complete Blackpowder Handbook.
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I recommend this book, all the time, to students ... :)

One thing that I've always thought would be helpful to all, is listing our shot-strings. We all know that range-time is part of this "Great Adventure" and would be nice to share with folks. ... :)
 
I loaded 135 grains behind a lead round ball in my .50 once.

Once.

The stock didn't fit me all that well and it kicked the absolute crap out of me.

It's a pussycat with 60 to 80 grains, though.
 
I think it has been hinted at, but I'll say it. 70gr of 4F will likely blow you up.

80 of 2F would be a good place to start behind a sabot or conical of your choice.

An average BP rifle is more than capable taking a deer at 125 yds.
 
Bring on the math

Sorry I don,t like to write rubbish. My first post was a quote from lyman. What I wrote was BASED ON THE FACT THAT A 1 IN 48 TWIST STABALISES A ROUND BALL OF 50CAL. But that was an ASSUMTION not a fact, don't assume in maths or you risk embarasment:o below is the simplest explanation I could fin for Greenhills formula.

150 x diameter squared divided by bullet length = required spin

Example for a .45 caliber bullet .60 inches long:

150 x .45 x .45 divided by .60 = 50.6 inches

So, for the example bullet, a spin rate of 1:50.6 or faster is required

The formula can also provide us with the maximum bullet length which can be stabilized by a given barrel twist. The formula becomes:

150 x diameter squared divided by twist rate

Example for a .50 caliber barrel of 1:48 twist:

150 x .50 x .50 divided by 48 = .78 inches

The barrel will stabilize a bullet .78 inches long, or shorter.

©1997 B. E. Spencer
If you put in the .5 for a ball you get 75.
 
BP internal ballistics

Read my thread "BP internal ballistics" for a formula to estimate your muzzle velocity. NOTE FG FFG FFFG operate by creating pressure in a barrel. Lyman lists for example Eephant brand and Geox in various granulations and powder loads, listing their pressure in a particular barrel length and twist. What you can be sure of is 40gr of FG puts less pressure up the barrel than 40gr of FFFFG.
 
Sorry I don,t like to write rubbish. My first post was a quote from lyman. What I wrote was BASED ON THE FACT THAT A 1 IN 48 TWIST STABALISES A ROUND BALL OF 50CAL. But that was an ASSUMTION not a fact, don't assume in maths or you risk embarasment below is the simplest explanation I could fin for Greenhills formula.
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You really can't (or at least, shouldn't) use Greenhill's formula for round balls. In fact, the only reason it gives you a result at all is that you're using the common simplified approximation of Greenhill's Twist Rate Formula. If you use the full and complete version, entering the dimensions for a round ball (length and diameter are equal), the equation "blows up" and basically gives an infinite twist length (i.e. zero twist).

The reason is, a spherical projectile can't be dynamically "stable" or "unstable", since the center of pressure and center of gravity are always at the exact same place - the center of the ball. That's why the twist rate formula fails.

The only reasons that twist is actually beneficial for round balls (surface imperfections, density variations, and the slight deformation as the ball goes through the barrel) aren't modeled anywhere in Greenhill's calculations. The reason that the simplified version gives you a result you can use is that it actually recommends more twist than you need, and since there aren't any negative consequences of using too much twist with a solid lead projectile, the calculated twist length works. But that's more of a happy coincidence than anything else.
 
I agree with your theory, but without spin a ball bounces down the barrel. Any air or other stuff trapped during casting would cause a further potential instability. What matters to me is: if Greenhill's formula is only for elongated bullets. What is the full formula. If I shouldn't use that formula for a ball can you give me a better one.:)
 
I agree with your theory, but without spin a ball bounces down the barrel. Any air or other stuff trapped during casting would cause a further potential instability.
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True, but none of those factors (skipping down the bore or density variations in the cast) are accounted for in Greenhill's formula. Likewise, neither are air resistance, turbulence, or the Magnus Effect. So while there are a bunch of things that affect external ballistics, most of them aren't included in the calculations.

If you want the history of Greenhill's formula, along with a description of the experimental methods and the full version of the formula, I would recommend the paper "Where did Greenhill’s twist rule come from?", by Donald G. Miller. It appeared in the International Journal of Impact Engineering 32 (2006), pages 1786-1799.

I spoke with the author, and he agreed that the formula wasn't applicable at all for round projectiles. In fact, he said that it gets downright "squirrely" if the projectile length is less than twice its diameter. I haven't re-read it recently, but I think that the experiments used to determine the "rule of thumb" constants of 150 or 180 mostly used projectiles that were between 2.5 and 8.0 times longer than their width.

If you can't find a copy of Miller's paper, send me a PM and I can get you one.

If I shouldn't use that formula for a ball can you give me a better one.
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There's not a good formula to calculate the required twist rate for a ball, because in theory, balls shouldn't need spin at all. We know that's not the case, but the factors that cause a spin to be needed aren't things that you can plug into an equation.

You'd have to find some way to mathematically describe the surface and density imperfections for the ball, and then perform an intensive computer-driven finite-element analysis - for each individual ball you're planning to shoot.

Since there aren't any penalties for "overspinning" a ball the way there can be with jacketed ammo, you can go ahead and use Greenhill's formula - it will prescribe more twist than you really need, but that won't hurt anything.
 
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A round ball is essentially over stabilized with any spin. An overstabilized bullet tends to hold the same orientation even as gravity causes the bullet's path to change. A bullet that's not overstabilized will weathervane into the apparrent wind direction like an arrow.
A round ball simply will not do that, and the consequence is spin drift at long ranges because the curved trajectory eventually has the bullet spinning at an orientation that has it flying through the air like a curve ball pitch.
 
ScotRiqui : googled it. thank you.
Found 7 formula. Number 6 is my favorite(almost as accurate as number1). Number 1(most accurate) has no density adjustment.
The formula for those interested. l = bullet length d = bullet dia, S = stability factor. 10.9 = density of fmj bullet 11.4 for lead. v = velocity.
(216.4 x d)divided by (l/d x square root of S) x (square root of density/10.9) x (v/2800)to the power 1/6. Sorry the message board doesn't lend itself to writing formula. If you make jacketed aluminum bullets density 2.7 the formula says you need twice the normal twist. Just like Hatcher's notebook.
thanks again.:D

EDIT:After reading the paper more thoroughly I realized miller's formula (number 1) doesn't need density adjustment, you just need to know the weight of bullet. whatever the required twist for a 405gr lead bullet an aluminum bullet would weigh 405/11.4 x 2.7 = 95.9gr. Plugging 95.9gr into the formula gives a faster (double the twist rate).

I have also come to realize AS THE BULLET SLOWS IT BECOMES LESS STABLE.

plus STABILITY REDUCES WITH ALTITUDE, AND REDUCES EVEN MORE IN HOT AND HIGH CONDITIONS
 
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"but without spin a ball bounces down the barrel."

No! Not necessarily true!

Round balls can be patch loaded in smooth-bore barrels to very good effect out to 50 to 75 yards! Certainly not to the same degree of accuracy as a rifle, but 6 to 8" groups at that distance is VERY doable for someone who has taken the time to match his ball and patch to the bore and loads very carefully.

My former boss could turn in surprisingly good groups with his .69 reproduction Charleville musket.

One of the keys to doing this is swabbing the bore after every shot to cut down on powder fouling.

The ONLY reason a round ball will "bounce" around in the barrel is if it's under sized or under patched.
 
(216.4 x d)divided by (l/d x square root of S) x (square root of density/10.9) x (v/2800)to the power 1/6. Sorry the message board doesn't lend itself to writing formula. If you make jacketed aluminum bullets density 2.7 the formula says you need twice the normal twist. Just like Hatcher's notebook.
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Can you send a URL to this formula? I am confused by order of operations with your current description.

Steve
 
without spin a ball bounces down the barrel
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Simply not accurate. A loose ball will have 'windage" and be inacurrate. But a patched round ball in a smooth bore can have fairly good accuracy to 50 yards. And some claim beyond.
 
Here's a picture of how an over stabilized bullet flys. It's of minor importance at normal muzzleloading ranges but comes into play at long distance shooting, especially with artillery shot at high elevation to hit a distant target.

http://www.nennstiel-ruprecht.de/bullfly/fig15.htm

A round ball is essentially overstabilized with any spin, however, spinning it still makes any imperfections that may deflect the bullet's direction cancel out. A roundball shot from a smoothbore is like a knuckleball pitch in baseball.
 
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