any semi-physics literates around?

[The Last American Hero]

http://en.wikipedia.org/wiki/.375_Remington_Ultra_Magnum

Surely it generates the same force regardless of weight of the rifle, it just hurts more in a lighter rifle.

Is the "80 ft lbf" what you get when you divide the joules by the weight of the rifle and is a good Felt Recoil Indicator?

I'd think Felt Recoil Index would be an estimate based on weight VS speed of recoiling gun, with emphasis on speed of recoil.



PS-how do I make a little period dot that sits half way up they got between "ft" and "lbf"? Do they just do that to make themselves seem all smart and scientific, and make me feel dumb?

[strongpoint]

Sorry, I know nothing about semi-physics.

However, the character you're looking for is called the midpoint dot (among other things, but it's NOT the same as a bullet). On a Mac, it's option-shift-9; on a PC, it's reportedly alt+0183.

[Untamed1972]

http://en.wikipedia.org/wiki/.375_Remington_Ultra_Magnum

Surely it generates the same force regardless of weight of the rifle, it just hurts more in a lighter rifle.

Is the "80 ft lbf" what you get when you divide the joules by the weight of the rifle and is a good Felt Recoil Indicator?

I'd think Felt Recoil Index would be an estimate based on weight VS speed of recoiling gun, with emphasis on speed of recoil.



PS-how do I make a little period dot that sits half way up they got between "ft" and "lbf"? Do they just do that to make themselves seem all smart and scientific, and make me feel dumb?

I'm assuming they're measuring recoil at the butt of the rifle. The energy released by the cartridge is the same regardless of the rifle its in. But the ability of that energy to exert a force on anything touching the butt of the rifle will be dictated by the weight of the rifle which absorbs and dissipates the energy. If the rifle weighted 100lbs, there would basically be no measuremable force exerted outside the rifle. But the force exerted against the bolt face by the cartridge would be the same.

[russ69]

...If the rifle weighted 100lbs, there would basically be no measuremable force exerted outside the rifle. But the force exerted against the bolt face by the cartridge would be the same.

No the force is the same. What is not measured is the velocity and duration of the force and that makes the differences that are felt. For the full picture, you need the force, the time, and the mass you are moving. So far nobody has come up with a usable recoil number other than the total force.

[AAShooter]

Try that again . . .

Two basic concepts:

1) f = ma

Force = Mass X Acceleration. The heavier the gun, the less acceleration

2) KE = 1/2 ma^2 KE = 1/2 mv^2

Kinetic Energy = 1/2 X m X (acceleration)^2. So the energy goes up linearly as mass goes up; however, the energy goes up by the square of acceleration.

So if you have a rifle of 1/2 the weight for a given force, you will get twice the acceleration. The Energy will drop by 1/2 due to the decrease weight but increases by 2^2 or 4 times due to the doubling of acceleration.

Kinetic Energy = 1/2 X Mass X (Velocity)^2. So the energy goes up linearly as mass goes up; however, the energy goes up by the square of velocity.

So if you have a rifle of 1/2 the weight for a given force, you will get twice the acceleration. All things being equal, that should double the resulting velocity of the gun. The Energy will drop by 1/2 due to the decrease weight but increases by 2^2 or 4 times due to the doubling of velocity.

This is the same reason 60 MPH car crashes are much worse than 2 X a 30 MPH crash.

I hope that helps. All that said, felt recoil is often very different as perceived by the shooter than calculations might indicate. That why some shooters things certain guns are softer shooters even with all things being equal.

Note: Edited to address the error pointed out by The Plumber. Thanks for the catch.

[Untamed1972]

No the force is the same. What is not measured is the velocity and duration of the force and that makes the differences that are felt. For the full picture, you need the force, the time, and the mass you are moving. So far nobody has come up with a usable recoil number other than the total force.

If:

Force = Mass X Acceleration. The heavier the gun, the less acceleration

If the gun is heavy enough (ie high enough mass) that it does not move when fired (ie no acceleration) then no force would be exerted outside the rifle (ie no recoil)

[the_plumber]

KE=1/2 m v^2

not a^2

velocity, no acceleration

[IVC]

So if you have a rifle of 1/2 the weight for a given force, you will get twice the acceleration.

Gasses are also part of the equation, so a muzzle brake can change this considerably.

[AAShooter]

KE=1/2 m v^2

not a^2

velocity, no acceleration

Oooooops . . . that's why I am only semi-qualified. lol

[AAShooter]

Gasses are also part of the equation, so a muzzle brake can change this considerably.

Yes, and let us not forget you have a 100 - 400 lb shooter behind the gun. If you are using good form, the mass being moved is not simply an 8 lb gun.

[The Last American Hero]

This is the same reason 60 MPH car crashes are much worse than 2 X a 30 MPH crash.

always had a hard time wrapping my head around that whole concept. I can force myself to understand and do the problems, but only with effort. Has to do with "Area under the curve" etc.

I'd think there would be a semi-standard Recoil Index for both handguns and long guns based on weight/speed recoil of gun, with co-efficient of grip or recoil pad, and like MPG would be for 'comparison purposes only, YMMV'.

Something a LITTLE more scientific than the linguistic stylings of gun magazine staffers.

[AAShooter]

always had a hard time wrapping my head around that whole concept. I can force myself to understand and do the problems, but only with effort. Has to do with "Area under the curve" etc.

I'd think there would be a semi-standard Recoil Index for both handguns and long guns based on weight/speed recoil of gun, with co-efficient of grip or recoil pad, and like MPG would be for 'comparison purposes only, YMMV'.

Something a LITTLE more scientific than the linguistic stylings of gun magazine staffers.

Modeling the whole thing becomes very complex. It is pretty simple (relatively) with just the gun and ammo; however, once you add the shooter, it gets very hard.

It is all about felt recoil and even if the numbers are the same, the shooter can perceive them much differently. Much of this has to do with solid shooting fundamentals. The fit between the stock and the shooter is very individual but has a huge impact on the felt recoil. Obviously a stock that makes good contact with the shooter over the entire butt of the gun will have a much better time than the shooter that just has the heel of the butt making contact.

It is interesting to see this post. I am currently reading a book on stock fitting and part of this is discussed in the book.

[alfred1222]

So let's break this down. Felt recoil is different from the force of the bullet. You are correct, the bullet exerts the same FORCE regardless of the weight of the rifle. The force is the MASS of the BULLET x the ACCELERATION of that BULLET. That does no change as long as the amount of powder and the weight of the bullet remain the same.

Now, felt recoil is a whole different thing, and here is why: when the bullet is shot down the barrel, it has a certain amount of force traveling in the same direction as the bullet itself. But, in keeping with Newton's Third Law, we know that for every action there is an opposite and equal reaction. Therefore, when the bullet is shot down the barrel, it exerts a force equal to its own on the gun, in a direction opposite of the barrel (e.g. The stock and your shoulder). Since the force going in both directions will be equal, the the felt recoil is found through the equation Force = Mass x Acceleration. Now, since Force is the same, and Acceleration is the variable expression that we will call Felt Recoil, mass is what determines the amount of Acceleration. This is from the equation A= F/M. So, take the force of the bullet, divide it by the mass of the weapon, and that will give you acceleration. As the mass increases, the acceleration decreases. So, therefore, the heavier the gun, the less recoil you feel.

[alfred1222]

Modeling the whole thing becomes very complex. It is pretty simple (relatively) with just the gun and ammo; however, once you add the shooter, it gets very hard.

It is all about felt recoil and even if the numbers are the same, the shooter can perceive them much differently. Much of this has to do with solid shooting fundamentals. The fit between the stock and the shooter is very individual but has a huge impact on the felt recoil. Obviously a stock that makes good contact with the shooter over the entire butt of the gun will have a much better time than the shooter that just has the heel of the butt making contact.

It is interesting to see this post. I am currently reading a book on stock fitting and part of this is discussed in the book.

Recoil is a fixed concept I described up there ^^. Now, felt recoil can be calculated based on a BUNCH of other factors, but I honestly don't want too go through that math

[russ69]

...If the gun is heavy enough (ie high enough mass) that it does not move when fired (ie no acceleration) then no force would be exerted outside the rifle (ie no recoil)

It moves, it just doesn't move far or fast. Force is an easy unit to determine, energy takes the velocity into account and gives it the importance it deserves.

KE=1/2 m v^2...

This is correct but it does not take into account the full spectrum of events. Some recoil formulas take half the weight of the powder as part of the mass but that's a rough estimate. The jet nozzle effect of the burning powder is a big contributor. Like I said earlier, nobody has quantified the entire process.