Interesting: Recoil due to powder weight.

It wouldn't be the mass of the powder. The powder burns to create a expanding gas. This expanding gas pushes on the bullet and the breach face equally. The bullet accelerates like hell because of its low mass as compared to the firearm. More powder=more gas produced.

Newtons Law: Every action has an equal and opposite.

If you push the bullet forward, the gas pushing has to push back against something.

Yep, all that powder goes out the end of the barrel. And since the powder is behind the bullet, pushing, as soon as the bullet gets out of the way, the gas exits at a faster speed than the bullet did.

Recoil difference could also be because the 300 win mag runs at 64,000 psi whereas the 44 Rem Mag runs at 36,000 psi. Big case, large volume of high pressure gas = more recoil. Note where I said large volume of higher pressure gas. That volume plays a role in it.

What happens with combustion is you end up with by products, if the combustion isn't complete which usually it isn't you don't use all the available energy, the unburned components appear as smoke,soot, etc. You can use the by products to determine how much air was used or wasted in your combustion.

Energy in has to equal energy out, that is why muzzle energy would work better for recoil calculation, you just need to use the proper equation for it. Also it is going to be a multi dimensional calculation, granted we most likely only care about the horizontal component.

Empirical evidence rules over all theory

Try this: Using a long barreled gun, load up with a stout charge of an energy-dense powder. Measure the velocity.

Get a slower, heavier powder and load whatever it takes to match the Bullseye velocity with the same bullet and same gun. (The powder will certainly be 10-20 grains more)

Compare the felt recoil. Compare the actual recoil.

You would expect the heavier charge of powder to give more recoil. But I am given to understand that the lighter powder charge will FEEL heavier in recoil.

Some say it has to do with the fact that the peak pressure of the faster, lighter powder charge is greater. Some say it has to do with the fact that the slower powder delivers its acceleration to the bullet over a longer period of time.

I will leave it to each of you to judge for yourselves.

Lost Sheep

The online recoil calculators work well to calculate recoil.

Lost Sheep
Problem with that theory is the slower powder reduced to the same speed probably won't burn fully. Unless you just mean the difference between Red Dot & Blue Dot. There isn't really much difference in the recoil of those two in a rifle.

kingmt said:

Lost Sheep
Problem with that theory is the slower powder reduced to the same speed probably won't burn fully. Unless you just mean the difference between Red Dot & Blue Dot. There isn't really much difference in the recoil of those two in a rifle.

Click to expand...

I don't understand the problem.

The bullet winds up downrange at the same velocity.

The powder, whether burnt, unburnt or in-process ejects from the muzzle. We are not talking about cartridge efficiency, but only about recoil.

I don't understand your question. PM if the point is off-topic and the answer is too long to post.

Lost Sheep

michaelmcgo said:

I often wondered why efficient cartridges, especially pistol caliber carbines, have such low recoil for the energy they deliver. One of the many reasons is a light-weight powder charge. I never thought about, but the mass of that powder is exerting a force on the breachface just like the mass of a bullet and the 24g of H110 in my .44 Mag is a heck of a lot lighter than the 70+ grains I used to less into my 300 Win Mag.

Click to expand...

It's simple application of physics (Newton's second law of motion) where force (F) is expressed as:

F = ma (Force = mass x acceleration)

Click to expand...

It doesn't matter whether it's pistol/carbine/rifle cartridge, greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.

I don't think you can count it that way, the powder is a potential energy. Upon ignition it become kinetic energy, if there was another force acting on the powder and the powder was not acting as a source of work on the bullet then you might be able to include it into your recoil calculation

100% of the mass of the powder (in the form of gas) exits the muzzle at the velocity of the bullet. How could the mass of the powder contribute anything less to recoil than the bullet itself does.

My original point was:
.44 Mag with a 200 grain bullet + 24g of powder = 224g exiting the muzzle.
.300 Win Mag with a 200 grain bullet + 74g of powder = 274g exiting the muzzle.

The 300 has a 22% higher load mass, all of which is seen completely in recoil. I understand that the mass CONTRIBUTED to a much higher velocity, but in the end two 200 grain bullets are still exiting the muzzle.

Perhaps a better example is a 357 Mag with 2 different powder charges of 6 and 16 grains, both of which deliver the same velocity with the same 158gr bullet. The 16 grain load will have a higher felt recoil with no terminal advantage.
__________________

Click to expand...

100% of the powder will never make it out of the barrel, a large majority yes, but not 100%. Further, to illustrate the point that pressure is the commanding factor, lets take a 44 mag load with a 240g bullet and 22g of powder. Leave the same amount of powder in it and put a 300g bullet on top of it, now what happens to recoil? Or a 340g bullet?

Or use this recoil calculator.

http://www.handloads.com/calc/recoil.asp

Notice the powder weight is part of the 'ejecta' used to figure recoil.

All the powder gas mass isn't used as 'Ejecta" in the recoial formula because some of it stays in the barrel pushing the other way and not all of it is ejecta.

rc

The mass of powder in used in any cartridge most certainly contributes to real free recoil energy, whether it's completely consumed in combustion or not. In fact, the factor used in most calculations, the velocity of the powder gases (part of the ejecta) is multiplied from 1.25 for long shotguns to 1.75 for high powered rifles which definitely contributes to recoil, since a lot of that mass is moving faster than the bullet or shot.

Most 12GA. shell boxes are printed..3 Dram Equlivant..this means the smokeless powder charge, even though of much lighter weight gives the equal velocity to shot charge as the old 3 Drams of black powder...And from what I have heard..the old timers agree the black powder loaded shells kicked harder than smokeless.

Actually I think I'm pulling out of this one. I didn't really understand the OP anyhow.

As for unburnt powder tho it is wasted because it never burnt to release it's energy. I always welcome PMs.

In a .223 M193 the powder charge is about half the weight of the bullet, therefore 1/3 of the ejecta and contributes 1/3 of the recoil without even considering flow faster than bullet velocity after exit.
With a muzzle pressure of several thousand psi, I doubt much is staying in the barrel at 15 psi.

bds;8828402: Boldface added by Lost Sheep said:

It's simple application of physics (Newton's second law of motion) where force (F) is expressed as:

It doesn't matter whether it's pistol/carbine/rifle cartridge, greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.

Click to expand...

The bullet exerts no force on the breechface. (edited to add: Because the bullet does not actually touch the breechface, how can it apply force?) Indeed, friction with the barrel actually acts to pull the gun forward, thus reducing felt recoil.

The force applied to the breechface is by the base of the cartridge. The cartridge is being acted upon by the pressure of the powder and that pressure only.

Thus a 25,000 psi application for so many microseconds can deliver a given bullet where 12,500 psi applied for twice that length of time will deliver the same bullet at the same velocity. It is not as simple as that, but you get the picture.

Which do you think has the greater recoil (if either)? The 12,500 psi or the 25,000 psi applied for half the time?

Newton's Laws (principles) are simple, as most physics is. Nature is simple and direct. How these simple principles combine to effect can be very complex.

I am still trying to figure it out. But making progress little by little.

Lost Sheep

Lost Sheep said:

The bullet exerts no force on the breechface.

Click to expand...

I tried to keep the concept simple ...

If you don't believe me, remove the bullet from the cartridge and cover the opening with some tissue paper and fire to see how much recoil you get.

What I said was:

greater the force the bullet exerts on the breechface from expanding high pressure gas, more the recoil.

Click to expand...

It is ultimately the mass of the bullet and the acceleration caused by the expanding high pressure gas that exerts force on the breechface ... which we feel as recoil.

I understand there are other forces at play such as escaping high pressure gas through the muzzle pushing the gun back, friction between bullet and barrel, rotation of the earth etc. but they are minor in comparison.

I'm sure many (most) of you know this, but I was reading about internal ballistics and a light bulb turned on. I often wondered why efficient cartridges, especially pistol caliber carbines, have such low recoil for the energy they deliver. One of the many reasons is a light-weight powder charge. I never thought about, but the mass of that powder is exerting a force on the breachface just like the mass of a bullet and the 24g of H110 in my .44 Mag is a heck of a lot lighter than the 70+ grains I used to less into my 300 Win Mag.

Click to expand...

Not meaning to flame, but that has to be the silliest, most uninformed post I've ever read on a forum like this. JHC!!!! Makes Gump look like Einstein!