Bullet weight effect on recoil?

[BC38]

All I now is my 12 ga kicks a lot harder with a punkin' ball than with a light skeet load. You can run whatever formulas you want, but that punkin' ball still kicks harder.

So you're trying to draw some kind of conclusion by comparing a single solid slug of lead that fully plugs the bore to a handful of bird shot pellets that are smaller than grains of rice?

The differences in the physics involved are so huge it isn't even worth discussing. Suffice it to say its like comparing apples to watermelons.

 

[Nevada Ed]

Hay, just a cotton pluck'n, pea pick'n, minute, here........
I resent that remark.

We done have, Watermelon shoot'n in this here, good old, US of A !!

 

[oneounceload]

There are several recoil calculators online; plug in the numbers and see which is which

 

[crstrode]

OK OK OK --

Now consider this:

fire a gun with a bullet of infinite weight (or mass) with a charge of any size, and there will be no recoil because a bullet of infinite weight cannot be moved.

Now fire the same gun with a bullet of zero weight (or mass) with a charge of any size, and there will be no recoil because the bullet that isn't there cannot move.

Go figure.

 

[BC38]

OK OK OK --

Now consider this:

fire a gun with a bullet of infinite weight (or mass) with a charge of any size, and there will be no recoil because a bullet of infinite weight cannot be moved.

Now fire the same gun with a bullet of zero weight (or mass) with a charge of any size, and there will be no recoil because the bullet that isn't there cannot move.

Go figure.

Your first supposition is incorrect. Against an immovable object the recoil would be the full force generated by the exploding powder charge. Recoil isn't the moving of the bullet - recoil is the equal and opposite reaction force from the push that accelerates the bullet. If the bullet doesn't move then ALL of the force generated by the explosion becomes recoil.

Your second supposition is also incorrect. Even without the bullet the forceful ejection of the mass of the burning powder is still going to provide SOME reaction force pushing the gun backwards - kind of like a mini jet engine. It will be minimal, but it won't be zero.

 

[TomkinsSP]

This thread reminds me of a young lady I dated many years ago. She drank rum and diet coke because it had fewer calories than rum and coke.

Yes, empirically, but does that translate into any real-world difference?

She moved away when she got a supervisory job at Treasury, which explains much in my book.

 

[shocker]

Now consider this:
fire a gun with a bullet of infinite weight (or mass) with a charge of any size, and there will be no recoil because a bullet of infinite weight cannot be moved.

As long as the gun has infinite weight, too.
Just say the chamber is plugged so the bullet can't move. Same as burning something within a sealed vessel strong enough to contain the pressure, no net motion and no recoil.

 

[Horn]

Wow!

I have always understood that a heavier bullet will recoil more in a gun than a lighter bullet does as recoil is a result of both weight and velocity but with weight having a greater influence.

That was my understanding behind the 125/135 gn .357 Magnum being more controllable than a 158 gn bullet.

But yesterday I had someone with a lot of experience tell me (insist) that a lighter bullet recoils more because of the velocity. This person insisted that a 147 gn bullet in 9mm recoils less than a 124 gn bullet because it is going slower.

Which is right?

Wow! I finally gave up hoping to find the answer..... Now, have any of you arrived at the answer? Talk about what does what!
Stay safe
Poli Viejo

 

[kwh]

In my Kahr KT40, the 165 gr. has less perceived recoil to me than the 180 gr.

 

[crstrode]

Your first supposition is incorrect. Against an immovable object the recoil would be the full force generated by the exploding powder charge. Recoil isn't the moving of the bullet - recoil is the equal and opposite reaction force from the push that accelerates the bullet. If the bullet doesn't move then ALL of the force generated by the explosion becomes recoil.

Your second supposition is also incorrect. Even without the bullet the forceful ejection of the mass of the burning powder is still going to provide SOME reaction force pushing the gun backwards - kind of like a mini jet engine. It will be minimal, but it won't be zero.

Nope. There can be no acceleration without movement. We're talking about the bullet mass. Zero bullet mass then zero bullet movement. Infinite bullet mass then zero bullet movement.

This is a theoretical logic exercise BC38 - not a practical one.

 

[crstrode]

Wow! I finally gave up hoping to find the answer..... Now, have any of you arrived at the answer? Talk about what does what!
Stay safe
Poli Viejo

OK - Here is your answer.

If everything else is exactly the same except bullet weight - that includes size, shape, density, composition, volume, powder, charge, etc. then the quantity of recoil energy will be the same. The quality of the recoil profile may be different - longer, shorter, smoother, sharper, etc.

But since everything cannot be the same (a bullet made of the same stuff and being the same size having greater or less mass density is impossible).

You can't get something for nothing.

 

[BC38]

Nope. There can be no acceleration without movement. We're talking about the bullet mass. Zero bullet mass then zero bullet movement. Infinite bullet mass then zero bullet movement.

This is a theoretical logic exercise BC38 - not a practical one.

Theory can't suspend the laws of physics. Ever heard the one about for every action there is an equal and opposite reaction?

Unless you are attempting to redefine recoil to be bullet movement - as opposed to the accepted definition of rearward movement of the firearm in reaction to the acceleration forces ON the bullet (not actual movement of the bullet) - then neither of your examples work.

Neither the bullet moving OR failing to move can zero out the reactive force. Theories that don't take physical realities into consideration aren't valid theories because they aren't logical. They are more akin to believing in magic rather than logic.

Sorry, but ideas don't trump physics. This discussion is about real world physics, not flights of fantasy.

Now, if you are talking about the equations, and substituting zero or infinity into them, then yes, they fail to work when you do that. This isn't anything unusual or specific to these equations. A lot of math doesn't have any real world application (i.e. doesn't work) when you use zero or infinity as variables. You can't divide by zero for example. Nothing particularly interesting or earth shaking about that.

 

[oneounceload]

OK - Here is your answer.

If everything else is exactly the same except bullet weight - that includes size, shape, density, composition, volume, powder, charge, etc. then the quantity of recoil energy will be the same. The quality of the recoil profile may be different - longer, shorter, smoother, sharper, etc.

But since everything cannot be the same (a bullet made of the same stuff and being the same size having greater or less mass density is impossible).

You can't get something for nothing.

False. weight of the bullet does have a factor in actual recoil; that is Newton

 

[Kiwi cop]

Theory is fine but .....

All this theory is fine but in a practical sense I can't get my head past the fact that a heavier bullet impacts on target higher than a lighter one does. That to me indicates a higher muzzle rise (not all of the impact height difference as trajectory also comes into play) therefore more recoil.

 

[ContinentalOp]

All this theory is fine but in a practical sense I can't get my head past the fact that a heavier bullet impacts on target higher than a lighter one does. That to me indicates a higher muzzle rise (not all of the impact height difference as trajectory also comes into play) therefore more recoil.

Not quite. Remember that sights are regulated by humans, meaning that the POA-POI relationship takes recoil into account. I'll use a .38 Special snubnose revolver as an example. Typically, they're zeroed for 158gr loads at 7 yards. That means that when you're sights are on a point at 7 yards, the bullet will strike that point.

Now consider the time it takes between primer detonation and the bullet leaving the muzzle. The gun is already starting to recoil before the bullet leaves the muzzle, but the sights take this into account when they're zeroed.

Now take a 110gr bullet, fired from the same gun, at the same point, at the same distance. Let's also assume that the velocity is high enough that it generates the same muzzle energy as the previous 158gr load. To do so, the bullet has to travel at a higher velocity than the 158gr load. This means that it spends less time moving through the barrel, which in turn means that by the time the light bullet exits the muzzle, the gun has moved less vertical distance through the arc of recoil.

The result? The lighter bullet hits the target lower than the heavier bullet, all else being equal.

Clear as mud?

 

[crstrode]

False. weight of the bullet does have a factor in actual recoil; that is Newton

If it is false, and Newton is correct that for every reaction there is an equal an opposite reaction, then a 1000 grain bullet from a .38 special would have more muzzle energy than a 375 H&H with one tenth the powder charge.

I actually heard about this physics stuff long long ago before I became an Engineering Supervisor at one of the National Engineering Laboratories.

 

[TomkinsSP]

Not quite. Remember that sights are regulated by humans, meaning that the POA-POI relationship takes recoil into account. I'll use a .38 Special snubnose revolver as an example. Typically, they're zeroed for 158gr loads at 7 yards. That means that when you're sights are on a point at 7 yards, the bullet will strike that point.

Now consider the time it takes between primer detonation and the bullet leaving the muzzle. The gun is already starting to recoil before the bullet leaves the muzzle, but the sights take this into account when they're zeroed.

Now take a 110gr bullet, fired from the same gun, at the same point, at the same distance. Let's also assume that the velocity is high enough that it generates the same muzzle energy as the previous 158gr load. To do so, the bullet has to travel at a higher velocity than the 158gr load. This means that it spends less time moving through the barrel, which in turn means that by the time the light bullet exits the muzzle, the gun has moved less vertical distance through the arc of recoil.

The result? The lighter bullet hits the target lower than the heavier bullet, all else being equal.

Clear as mud?

So, in theory, if you had a deep enough case and seated the lighter bullet deeper, you could create the same transit time through the barrel and not need adjustable sights.

 

[ContinentalOp]

So, in theory, if you had a deep enough case and seated the lighter bullet deeper, you could create the same transit time through the barrel and not need adjustable sights.

Hmm...not sure. The problem is that the lighter bullet would have to leave the muzzle at the same time as the heavier bullet, which would mean a lower velocity. However, this would result in less energy than the heavy bullet, including less recoil energy. The muzzle would still rise less than with the heavy bullet so the light bullet would still hit lower. I think. My head hurts now...