What causes recoil

St Pauli

St Pauli

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I figured I would post here. Mods move if necessary.

I was chatting with a friend ( non-gun owner) but a good guy about firearms and he asked,
" what causes recoil? The bullet explosion or it leaving the barrel? "

I couldn't figure out the correct answer so I come to you.

My guess was gas leaving the barrel but I'd like the answer with the math to show why.
 
Newton's laws is what explains recoil - equal and opposite reaction.

Basically, when the gun goes off the gases push the bullet in one direction. Newton's Law says it also pushes the gun in the exact opposite direction, with the exact same amount of force.
 
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Every action has an equal and opposite reaction. Gases contribute greatly a firearms recoil. It's why muzzle breaks and suppressors are so effective at reducing recoil.
 
The powder burns if nothing to push on , you would not feel much.
It is the force of the gases and bullet speeding down the barrel that causes the recoil, in my option.
 
Combustion creates expanding gases in an enclosed space. The pressure exerts force onto the base of the bullet and (sealed inside the casing of course) onto the bolt face. The force is relqtive to the surface area being acted upon. (Remember pressure is measured in Unit of force per unit area ie. pounds per square inch). When the bullet leaves the muzzle, the expanding gases leaving the barrel act as a jet (kind of like when you turn water onto a hose the end jumps back towards you.) The sum of these forces, resisted by the inertia of the rifle determine the recoil.

Long story short, lighter bullet, lower speeds, less powder, and a heavier gun make for less felt recoil.
 
St Pauli said:
" what causes recoil? The bullet explosion or it leaving the barrel? "
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Neither of those. In between those events is acceleration of the bullet, and that is your source of recoil.

Force = Mass x Acceleration

A force acting in one direction must be balanced by a force acting in the opposite direction.

So the force associated with accelerating the bullet forward is balanced by a force that accelerates the gun backward. Acceleration begins when the bullet first starts to move within the case, and ends when the bullet reaches its peak velocity, typically somewhere before reaching the muzzle.
 
BattleRife said:
Neither of those. In between those events is acceleration of the bullet, and that is your source of recoil.

Force = Mass x Acceleration

A force acting in one direction must be balanced by a force acting in the opposite direction.

So the force associated with accelerating the bullet forward is balanced by a force that accelerates the gun backward. Acceleration begins when the bullet first starts to move within the case, and ends when the bullet reaches its peak velocity, typically somewhere before reaching the muzzle.
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Peak velocity, before reaching the muzzle? A 22lr gains velocity over the first ~16" of barrel, which means pretty much any centerfire cartridge which is burning far more powder will not reach maximum velocity before the muzzle? If max velocity was reached before the muzzle, then longer barrels wouldn't give faster velocities?

Perhaps you max acceleration peaks before the muzzle? Which is probably true, but recoil doesn't end when peak acceleration is reached, it ends when acceleration=0 (F=M x A, so either M or A must be zero for F to be zero as well, and M doesn't change in a meaningful way), which is going to be just after it leaves the muzzle and the gases are free to expand in all directions rather than just in the direction of bullet travel?
 
BattleRife said:
Neither of those. In between those events is acceleration of the bullet, and that is your source of recoil.

Force = Mass x Acceleration

A force acting in one direction must be balanced by a force acting in the opposite direction.

So the force associated with accelerating the bullet forward is balanced by a force that accelerates the gun backward. Acceleration begins when the bullet first starts to move within the case, and ends when the bullet reaches its peak velocity, typically somewhere before reaching the muzzle.
Click to expand...

From various "recoil calculators", it seems you must add the weight and acceleration of the propellant to the weight and acceleration of the projectile - and then the propellant's exit speed (actually the gases from it) have a much higher velocity out of the muzzle, than does the projectile - hence, in many "calculators", the propellant charge weight and it's exit velocity are a significant contributor to the "recoil" of the firearm...

Go here to see a "calculator" and notice what values are required to input - also good discussion, I think, on various recoil understandings. http://kwk.us/recoil.html
 
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As BattleRife stated, F=ma

All the recoil force backwards is equal and opposite to the force required to accelerate the mass forwards.

First recoil force is from the firing pin release, "long before" (so to speak) impact and ignition on the primer. You can test this by dry firing. The rifle will move backwards.

Next recoil is the firing pin decelerating as it hits the primer, and the entire mass of the cartridge plus the rifle pushing back.

Next recoil is after ignition of the primer and powder. The mass of the primer material, the mass of the powder and the bullet are now accelerating forward. Because this is so incredibly fast, the "a" in F=ma makes F very large.

That acceleration continues until bullet and gasses leave the muzzle. I don't think (I may be wrong) that there is any decline of force mid-way through the barrel, because the inertia of the bullet and gasses is still being overcome (accelerated), and thus the reaction force continues backwards?

There is another force of torque due to friction and spin of the bullet on the rifling, which will move and twist the firearm to the right, but I don't think you will feel this as a backwards recoil force, but it certainly is part of recoil management for precision shooting. That's one reason why we see the evolution of F-class bipods getting so wide and center of mass of the barrel being low in the bipod legs so as to resist that torque force.

Disclaimer: I am a Biologist not a Physicist, so I may have this all totally wrong! :d
 
littlehughey said:
Combustion creates expanding gases in an enclosed space. The pressure exerts force onto the base of the bullet and (sealed inside the casing of course) onto the bolt face. The force is relqtive to the surface area being acted upon. (Remember pressure is measured in Unit of force per unit area ie. pounds per square inch). When the bullet leaves the muzzle, the expanding gases leaving the barrel act as a jet (kind of like when you turn water onto a hose the end jumps back towards you.) The sum of these forces, resisted by the inertia of the rifle determine the recoil.

Long story short, lighter bullet, lower speeds, less powder, and a heavier gun make for less felt recoil.
Click to expand...

Many great responses on this thread, you have explained it quite clearly.
 
Biologist said:
As BattleRife stated, F=ma

All the recoil force backwards is equal and opposite to the force required to accelerate the mass forwards.

First recoil force is from the firing pin release, "long before" (so to speak) impact and ignition on the primer. You can test this by dry firing. The rifle will move backwards.

Next recoil is the firing pin decelerating as it hits the primer, and the entire mass of the cartridge plus the rifle pushing back.

Next recoil is after ignition of the primer and powder. The mass of the primer material, the mass of the powder and the bullet are now accelerating forward. Because this is so incredibly fast, the "a" in F=ma makes F very large.

That acceleration continues until bullet and gasses leave the muzzle. I don't think (I may be wrong) that there is any decline of force mid-way through the barrel, because the inertia of the bullet and gasses is still being overcome (accelerated), and thus the reaction force continues backwards?

There is another force of torque due to friction and spin of the bullet on the rifling, which will move and twist the firearm to the right, but I don't think you will feel this as a backwards recoil force, but it certainly is part of recoil management for precision shooting. That's one reason why we see the evolution of F-class bipods getting so wide and center of mass of the barrel being low in the bipod legs so as to resist that torque force.

Disclaimer: I am a Biologist not a Physicist, so I may have this all totally wrong! :d
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I would think the force does decline over time, because the acceleration isn't constant - I would expect the acceleration to increase to a certain point in the barrel where it reaches max acceleration (which is likely corelated with peak pressure?), at which point the acceleration starts to decrease, and with it the force applied to the gun decreases as well.

So the force would be increasing over time, hits a peak, then decrease over time, and then once the bullet and gases leave the muzzle it drops to zero? However, it all happens so fast that you'd never be able to perceive these changes in force, as the "time" referenced in the previous sentence is fractions of a second.
 
Potashminer said:
From various "recoil calculators", it seems you must add the weight and acceleration of the propellant to the weight and acceleration of the projectile - and then the propellant's exit speed (actually the gases from it) have a much higher velocity out of the muzzle, than does the projectile - hence, in many "calculators", the propellant charge weight and it's exit velocity are a significant contributor to the "recoil" of the firearm...

Go here to see a "calculator" and notice what values are required to input - also good discussion, I think, on various recoil understandings. http://kwk.us/recoil.html
Click to expand...

Technically that would only be partially correct but as for the OPs purposes Newtons Law to every action there is an equal and opposite reaction is the answer he needs... as has been stated already. But technicalities can be interesting: Some of the acceleration of the powder mass will result in recoil. That is to say the powder column lays in the case from head to shoulder: ignition begins in the powder nearest the flash hole progressing out from where the primer initiates. Unless the case is completely free of all carbon, then we know that not all of it was spewed out the barrel: we also know this by the fact that if the case and barrel were completely empty and clean what product remains to have launched it all out the barrel with the bullet? Some of the propellent (burned or not) will leave the muzzle at higher velocity as the pressure dissipates upon bullet clearing the muzzle... because of it's small mass it slows and falls within ten feet. Then there is the First Law of Thermodynamics... ;) I suppose the fundamental question is: what percentage of felt recoil is a result of powder and gas movement? Not having calculated it, I'd guess it's under 2% with the bullet acceleration producing the remainder.
 
New Camper said:
Technically that would only be partially correct but as for the OPs purposes Newtons Law to every action there is an equal and opposite reaction is the answer he needs... as has been stated already. But technicalities can be interesting: Some of the acceleration of the powder mass will result in recoil. That is to say the powder column lays in the case from head to shoulder: ignition begins in the powder nearest the flash hole progressing out from where the primer initiates. Unless the case is completely free of all carbon, then we know that not all of it was spewed out the barrel: we also know this by the fact that if the case and barrel were completely empty and clean what product remains to have launched it all out the barrel with the bullet? Some of the propellent (burned or not) will leave the muzzle at higher velocity as the pressure dissipates upon bullet clearing the muzzle... because of it's small mass it slows and falls within ten feet. Then there is the First Law of Thermodynamics... ;) I suppose the fundamental question is: what percentage of felt recoil is a result of powder and gas movement? Not having calculated it, I'd guess it's under 2% with the bullet acceleration producing the remainder.
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I would think it depends on the cartridge and charge weight - if you're shooting a 7mag with a 140gr bullet you could be loading 65+gr of powder, which is nearly half the weight of the bullet. The recoil due to the powder will be more in that than say a 30-30 shooting a 170gr bullet with only 30gr of powder.
 
I am not able to do the math - just have been going what I read on that kwk site and various other articles - especially by John Barsness. Note that he "guesses" that the velocity at the muzzle of the products of combustion might be up to 4,000 fps - so in the example above, 65+ grains of powder (less residue left behind as soot) at 4,000 fps or so is going to add significantly to the 140 grain bullet at 3200 fps or so. Until reading that stuff, I never would have realized the contribution of the propellant gases. This is without addressing a "jet effect", based on cartridge shape, which Barsness claims, but does not provide any math for that...
 
New Camper said:
Not having calculated it, I'd guess it's under 2% with the bullet acceleration producing the remainder.
Click to expand...

Your guess would be wrong. Using a .30/06 with a 180 gr bullet as an example, 56 grs of powder produces 2,750 fps for 25.2 ft lbs of recoil in an 8 lb rifle. Using the same ballistics with a hypothetical charge of 1 gr. the recoil is reduced to 9.9 ft lbs, for a reduction of over 60%.
 
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you will not have recoil if the primer didn't ignite so the primer causes recoil assisted by the firing pin that was released by the sear when the trigger was pressed by an index finger
 
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