Ammunition - Deep Thoughts

[Garufa]

Perhaps a lighter bullet of the same caliber leaves the muzzle at a greater velocity than one of a heavier weight. That is more or less a given and that may hold the answer.

Being as the lighter projectile is moving faster from the muzzle it is less susceptible to the influences of atmosphere, wind pressure, gravity and what have you, so the flight at short range is more predictable. The heavier buller is slower and therefore more influenced by the aerodynamic factors, exaggertaing the trajectory upwards at a given distance. In this case 100-125 yards.

So when zeroing sights in for say 149gr rounds one would need to decrease the elevation for the same gun using a 182gr round to compensate for the increased wind resistance of the latter. The same applies to any gun.

[Guest mosinon]

Wow, you guys are all over it. Turns out barrel harmonics can describe the larger bullet shoots higher phenomena. You can read about it

here

But that is high precision stuff.

gregintenn was wondering:

"I disagree with this. Without ajusting your sights, a 158 grain bullet from a 38 special WILL strike a target higher than a 110 grain bullet from the same gun; same sights, same distance, within reason. I'm talking 7-15 yards here; not 1000. If muzzle rise isn't the culprit for this, I'd like to hear a valid explanation."

I'm doubtful that in a revolver barrel harmonics are the main factor. I think it is likely muzzle flip as posited by GinT. The objection to this explanation is that the muzzle won't flip until the bullet leaves the barrel. There might be something to recoil increasing when the bullet leaves the barrel but, at the velocities we are talking (non relativistic) the recoil is practically instantaneous. So the muzzle is going to go up if it is above the pivot point even on a one shot 10/22. Wikipedia has some info.

But screw wikipedia! We want to know what is going on.

Here's the rub: To figure this out we'll use F=ma (Force=mass*acceleration). Lets use the scenario presented by Gregintenn.

Grabbing data from ballistics101

Let's use the numbers provided for Hornady

The have info for 125 grain and 158 grain rounds

The 125 grain bullet leaves the barrel (so they say) at 900 FPS

The 158 grain bullet leaves the barrel at 800 FPS

With those number the force is easy to calculate. For the 125 grain bullet we have 125*900. That's F=ma and F turns out to be 112,500 grains feet per second. (Wonky units I know)

For the 158 grain bullet we get 158*800 or 126,400 grains feet per second.

This is an interesting result. The heavier bullet generates greater force even though it has less muzzle energy and less speed.

Some percentage of this force will go into lifting the barrel and, thus, as Gregintenn opined, with everything else being equal the heavier bullet will hit the target higher.

I've browsed through a few ballistics charts and while my research isn't thorough it seems from the loads I've run across that there isn't an instance where the heavier bullet won't actually hit a little higher owing to muzzle rise.

As for barrel harmonics I'm still a little confused. Some the articles I've read seems to indicate that the bullets always go up, the question is how much. Which seems weird, you'd think with a vibrating barrel that some shots would actually drop. But those guys know way more than me. And I'm sure bullet profile has a lot to do with all this. I think long range snipers even take the Coriolis effect into account when taking a shot. So this stuff can get very fussy.

To figure all that out you'll need an engineer, give peejman a call.

All that was a long way to go to answer the original question but remember: Garufa said "Somebody break it down for me, scientific."

[Guest 6.8 AR]

Less mass needs a faster burning

powder and more of it to leave the barrel. It's not as much about powder volume, but the rate of burn and the

pressure generated in time. A heavier bullet will use lesser powder depending on burn rate and pressure

generation. There are more variables but that is

more than enough if you are talking about using

different powders and bullets in the same weapon.

You can compare the potential energy with one

bullet of one mass to another and see the change

in another variable needed to match the two.

All that made sense until I typed it. Physics was fun,

but it was a long time ago. I think Mosinon studied

more than me:D

Sent from my iPhone using Tapatalk

[gregintenn]

Perhaps a lighter bullet of the same caliber leaves the muzzle at a greater velocity than one of a heavier weight. That is more or less a given and that may hold the answer.

Being as the lighter projectile is moving faster from the muzzle it is less susceptible to the influences of atmosphere, wind pressure, gravity and what have you, so the flight at short range is more predictable. The heavier buller is slower and therefore more influenced by the aerodynamic factors, exaggertaing the trajectory upwards at a given distance. In this case 100-125 yards.

So when zeroing sights in for say 149gr rounds one would need to decrease the elevation for the same gun using a 182gr round to compensate for the increased wind resistance of the latter. The same applies to any gun.

As I see it, this would cause the opposite to be true. As mosinon stated earlier, all objects fall to the ground at the same rate. That being the case, the lighter bullet should reach the target in less time than the slower, heavier bullet. This means it has less time to fall, so the lighter bullet should strike the target higher than the heavier bullet. That's the gravity part. As for the wind resistance, usually, a heavier bullet of the same caliber has a higher ballastic coefficient than a shorter, lighter one, which would decrease the amount of wind resistance in comparison to the lighter bullet. That leads me to think that wind resistance may have something to do with it. It's a lot to think about, isn't it?

[timcalhoun]

Well, I'm in the barrel harmonics camp for sure. I read about it for 20 years ago or more, and it is the only thing that can explain all that I have experienced through the years handloading and experimenting.

The browning Boss system was designed on this principle.

How does the BOSS system work?

Edited December 24, 2010 by timcalhoun

[Guest 6.8 AR]

All objects fall to the earth at the same rate only in a vacuum. The shape of the projectile is one of the factors in ballistics, also mass and friction. A boattail shaped

projectile will have a higher coefficient than one with a blunt base. Think about how a wing provides lift and

drag on an airplane.

The propellant gets it going and out the barrel. Then

it's up to the shape and mass to determine where it

will land. Inertia stabilizes the bullet in flight, and acts

against gravity, along with the shape that causes lift

to a small degree. As the bullet flies, the energy drops

along it's path.

The same forces also act on fragmentation. Are we deep enough, yet?

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[peejman]

Perhaps a lighter bullet of the same caliber leaves the muzzle at a greater velocity than one of a heavier weight. That is more or less a given and that may hold the answer.

Being as the lighter projectile is moving faster from the muzzle it is less susceptible to the influences of atmosphere, wind pressure, gravity and what have you, so the flight at short range is more predictable. The heavier buller is slower and therefore more influenced by the aerodynamic factors, exaggertaing the trajectory upwards at a given distance. In this case 100-125 yards.

So when zeroing sights in for say 149gr rounds one would need to decrease the elevation for the same gun using a 182gr round to compensate for the increased wind resistance of the latter. The same applies to any gun.

Aerodynamic drag is proportional to the square of velocity. In the theoretical example above, the 2x faster bullet would experience 4x the aerodynamic resistance. But the shape of the bullet itself is also a factor, hence the term "ballistic coefficient". It's a catch-all for several variables that really aren't worthwhile to characterize individually, unless you're designing computer simulations to model the flights of the bullet you're designing. I'm quite certain there's an expert or twelve at companies like Hornady, Nosler, Winchester, etc. that could easily answer all our questions.

[Guest GunTroll]

Answer to 6.8 AR's question......NO! Try harder. Lets get so deep that the joy of shooting firearms is lost to numbers and science .

[peejman]

...

The have info for 125 grain and 158 grain rounds

The 125 grain bullet leaves the barrel (so they say) at 900 FPS

The 158 grain bullet leaves the barrel at 800 FPS

With those number the force is easy to calculate. For the 125 grain bullet we have 125*900. That's F=ma and F turns out to be 112,500 grains feet per second. (Wonky units I know)

For the 158 grain bullet we get 158*800 or 126,400 grains feet per second.

This is an interesting result. The heavier bullet generates greater force even though it has less muzzle energy and less speed.

Some percentage of this force will go into lifting the barrel and, thus, as Gregintenn opined, with everything else being equal the heavier bullet will hit the target higher.

I've browsed through a few ballistics charts and while my research isn't thorough it seems from the loads I've run across that there isn't an instance where the heavier bullet won't actually hit a little higher owing to muzzle rise. ...

How'd you convert muzzle velocity to acceleration? Would a kinetic energy comparison be more appropriate? KE = 0.5*m*v^2

125 grain bullet gives 50,625,000 grains*ft^2/s^2 while the 158 grain bullet gives 50,560,000 grains*ft^2/s^2 ... The lighter bullet has ~0.1% more energy.

You'd then take these energy numbers and do an energy balance on the whole system to determine the net effect. But I don't remember all that off the top of my head an don't have time to drag out the books at the moment....

[Guest mosinon]

How'd you convert muzzle velocity to acceleration? Would a kinetic energy comparison be more appropriate? KE = 0.5*m*v^2

125 grain bullet gives 50,625,000 grains*ft^2/s^2 while the 158 grain bullet gives 50,560,000 grains*ft^2/s^2 ... The lighter bullet has ~0.1% more energy.

You'd then take these energy numbers and do an energy balance on the whole system to determine the net effect. But I don't remember all that off the top of my head an don't have time to drag out the books at the moment....

Oh, that answer is easy. a=v(final)-v(initial)/t(final)-t(initial)

v final is the speed at which the projectile travels

v initial is, of course, zero

t final happens to be 1 (I chose my barrel length for convenience)

t initial is also 0

so, a= 900FPS

Pick everything else right and you can avoid a lot of complication:)

[Guest 6.8 AR]

Answer to 6.8 AR's question......NO! Try harder. Lets get so deep that the joy of shooting firearms is lost to numbers and science .

Some of you guys tax my brain too much. And you know I hate taxes.

There are much smarter people on this thread, as we speak. I'm having fun,

but we're packing up for Knoxville right now, also. My wife will use one of my

Christmas presents on me if I'm not off here when she gets out of the bathroom!

Anyway, I think we could take a physics class from that Mosinon guy. He's doin'

quite well.

Actually this is stuff everyone should know. Good thread!

[Guest Archminister01]

I'm not doubting you, but can this be proven?

I can't dispute that theory.

I guess i should clarify. I have read many threads on this and here is one of them. It was even figured roughly using math and he calculated time found on the equation was so small that the muzzle rise is not even perceivable until after the bullet leaves the barrel. The effects of this rise is for the most part non existent.

I retract my statement that it does not rise. It does but never to the extent that ti would effect bullet rise measurably.

Gun Question - Recoil. Before or after bullet is clear? - THR

Does a handgun bullet leave barrel before recoil starts - THR

Bullet Trajectory

Here are a few of the links. make your own decision..lol. I still stand on my belief that actual measurable muzzle rise does not happen until the bullet has left it and the exhaust gas moves it. That is why ports are at the end of the slide. If rise actually occurred noticeably as soon as ignition took place, then woudlnt teh ports be placed nearer the point of ignition?

[kb4ns]

I now know to never ask for "deep thoughts"!

[Guest mosinon]

I guess i should clarify. I have read many threads on this and here is one of them. It was even figured roughly using math and he calculated time found on the equation was so small that the muzzle rise is not even perceivable until after the bullet leaves the barrel. The effects of this rise is for the most part non existent.

I retract my statement that it does not rise. It does but never to the extent that ti would effect bullet rise measurably.

Gun Question - Recoil. Before or after bullet is clear? - THR

Does a handgun bullet leave barrel before recoil starts - THR

Bullet Trajectory

Here are a few of the links. make your own decision..lol. I still stand on my belief that actual measurable muzzle rise does not happen until the bullet has left it and the exhaust gas moves it. That is why ports are at the end of the slide. If rise actually occurred noticeably as soon as ignition took place, then woudlnt teh ports be placed nearer the point of ignition?

I couldn't get the THR links to work. I can guess why the ports aren't placed closer to the point of ignition. First, you'd lose pressure and that would be bad. Second hot gasses spraying to close to your face might not be a good idea.

I think if you look at steyr precision air guns you'll note ports down the barrel. Presumably with air rifles the pressue only decreases so the projectile is going as fast as it is going to go right away.

As for how much it really matters. That is a really excellent question. That is worth spending a little time doing math, I'll try it tonight.

[Guest Archminister01]

I couldn't get the THR links to work. I can guess why the ports aren't placed closer to the point of ignition. First, you'd lose pressure and that would be bad. Second hot gasses spraying to close to your face might not be a good idea.

I think if you look at steyr precision air guns you'll note ports down the barrel. Presumably with air rifles the pressue only decreases so the projectile is going as fast as it is going to go right away.

As for how much it really matters. That is a really excellent question. That is worth spending a little time doing math, I'll try it tonight.

I would love to see that on a weapon that fires bullets tat use powder as a source of ignition. I am thinking that it isnt possible due to the weight of the projectile. I am betting that they can get away with this with air rifles because of the lower pressures involved. Thats just an educated guess based on any physics classes I took a LONGGG time ago.

I would also take a guess and say that the porting could never be that close to the point of ignition because the release of pressure wouldnt allow the bullet to be pushed down the barrel.

I am past the math part so I am curious what you come up with. I still remember principles but proving them mathematically is not longer in my grasp.

Edited December 26, 2010 by Archminister01

[Guest Lester Weevils]

I'm reminded of something a math professor told us in class one day.... "Two men are 10 paces apart. When told to move, they decrease the distance between each other by 1/2. How long until they touch?"

The physicist says, "they'll never touch".

The engineer says, "after 3 steps, they'll be close enough."

It has been fun to follow the discussion. Thanks.

I'm neither physicist, engineer, or mathematician. Perhaps the mathematician would be most likely to answer, "infinity"?

Perhaps the physicist would consider the planck length to be "close enough" to touching? In which case, assuming that wikipedia is correct that paces are typically assumed to be 30 inches nowadays, and that the Planck Length is 1.616252E-35 meters--

10 paces would be 300 inches or 7.62 meters. To approach the Planck Length from a distance of 10 paces ought to require about 118.5 steps?

[LagerHead]

You guys lost me at "math."

Oh, and science.

When do we get to beer?

[Guest FroggyOne2]

Barrel Harmonics Mode Shape Movies

[Guest FroggyOne2]

Rifle Barrel Tuner Vibration Analysis

[mikegideon]

Good stuff Froggy. I've spent some time on Varmint Al's site.

[Guest FroggyOne2]

Yep.. it sure is..