4Game

Warning, dissertation ahead. TLDR at the end.

There are a lot of misunderstandings in this community behind what people call recoil and its interactions in this game. The things that effect recoil in Tarkov are not what effect recoil in reality. One of the biggest misunderstandings is with Barrel Length and its supposed effect on recoil.

Let's begin with a basic physics lesson, for every action, there is an equal and opposite reaction. In the case of a firearm, that action is the primer setting off the power and the subsequent burn propelling the projectile out the end of the barrel. Due to the nature of physics, that exact same force is generated directly backward against the bolt of whatever firearm fired that projectile. I'm going to focus on the AR-15/M4 platform for this discussion.

Now let's get into some mathematics. I'm going to use the ballistic information that we have for 855 ball ammo manufactured by Winchester. Small note, these velocities come from these rounds being shot from a bolt action 556 rifle, these velocities will be very close to identical for an ar platform. You can find the data used for this here (https://rifleshooter.com/2015/12/223-remington-5-56mm-nato-barrel-length-and-velocity-26-inches-to-6-inches/)

Lets look at a 16.5 inch barrel and a 20 inch barrel:

16.5" barrel:Exit velocity is 2992 fps. We will assume that the bullet is accelerating at a constant rate until it leaves the barrel. With that in mind we know that the bullet will travel 16.5 inches at a constant acceleration before reaching it's velocity of 2992 fps. We need to know how many seconds it will take for that to happen so that we can know the acceleration of the bullet in order to do the force calculation.

So we know that distance traveled by an object is equal to average velocity times the time it traveled.

d = v(average) * t

we know that the average velocity in this case is 1496 fps (2992 – 0)/2 and the distance is 16.5 inches or 1.375 ft.

t = d/v(avg) = 1.375ft/1496fps = .00092 seconds

a = v / t = 2992fps / .00092 seconds

a = 3,252,173.91 ft/s^2

we know that force is equal to mass times acceleration:

F = m * a

855 ball manufactured by Winchester is 55 grains. This translates to .007857 lbs.

F = .007857 lbs * 3252173.91 ft/s^2 = 25552.33 lb ft/s^2

This is the force generated on the bullet AND the bolt. To better understand the effect the BARREL has on the system we are going to assume a fixed bolt system (this is because my argument is that the barrel has little to no effect on felt recoil, the gas system and buffer has a significantly greater impact) thus all this generated force goes directly into the rifle and back to the shooter.

Now, we need to translate this force into acceleration for the firearm and calculate the impulse force needed to completely eradicate the momentum of the rifle before the next round fires. We will assume a cycle rate of 700 rpm.

For the 16" platform we will assume a standard issue M4 weighing in at 6.55 lbs. For the 20" platform, we will assume an HBAR 20" barrel which would bring the weight of the platform to 8 lbs. A standard profile barrel is 7.5 lbs for a 20" system, so I'm giving the longer barrel an advantage based on most users apparent understanding of how this all works here.

a = F/m = 25552.33 lb ft/s^2 / 6.55 lb = 3901.11 ft/s^2

Now, in order to calculate the force needed to bring the firearm back to a stop we need to know the firearms velocity.

v = a * t = 3901.11 ft/s^2 * .00092s = 3.589 ft/s

Impulse = F*t and impulse is also equal to an objects change in momentum, which in this case can be simplified to its change in velocity. Since we want the velocity of the rifle to go from 3.589 to 0 in the cycle time of a single round (which for a 700 RPM cycle will be .0857 seconds) we can calculate the needed force.

momentum (p) = v * m

impulse = F * t

v*m = F*t

F = 3.589 ft/s * 6.55 lb / .0857 s = 274.305 lb ft/s^2

To convert this to lbf we multiply by the gravitational constant 1 lbf = 32.2 lb ft/s^2

F = 274.305 lb ft/s^2 * (1 lbf / 32.2 lb ft/s^2) = 8.52 lbf

So your body has to push back on that rifle with a force of only 8.52 lbs for .0857 seconds to completely bring that firearm to a halt.

Do all of that math for a 20 inch barrel which has an exit velocity of 3097 fps and you get a resultant force of 8.64 lbs for .0857 seconds.

FELT RECOIL GOES UP FOR LONGER BARRELS WITH A FIXED BOLT SYSTEM. Anyone, and I mean ANYONE that tells you a longer heavier barrel reduces recoil is mistaken. They don't know what they're talking about. Period.

Now I'll move onto the discussion of muzzle rise.

What causes muzzle rise? Simple answer is when your point of rotation is not in line with the force generated on a system. For a rifle platform that center of rotation is NOT the firearms center of gravity. How is that possible? Because someone is HOLDING the rifle. They are creating multiple points of contact with that rifle. How hard you pull the rifle into your shoulder, shoulder angle, grip strength on the pistol grip, etc all have an effect on the center of rotation.

T(Torque) = I(inertia) * a(rotational acceleration)

on an AR Platform, due to the design of it, the distance between the force generated by the shoulder and the force generated by the kick are very close in distance, roughly 2 inches with good form and technique. Now these impulse forces will generate moments about the center of rotation. to create a worst case scenario we will assume the center of rotation is at the contact point of the shoulder. So the shooter is brick walling the firearm, which is terrible technique, so the moment generated by the bullet creates as much rotation as possible. we are going to simplify the moment of inertia for the firearm into a block shape. This will be more than sufficient for our purposes, we aren't trying to get nth degree of accuracy on muzzle rise here.

moment = F * distance

We can use the force that we found earlier 25552.33 lb ft/s^2 however to use this force in this equation we must multiply it by the gravitational constant for lbf (1 lbf = 32.2 lb ft/s^2) we didn't need to do this last time since all we did was use it as a comparison tool. So our resultant force is 793.55 lbf

moment = 793.55 lbf * 2 in = 1587.1 lbf in

Next we will calculate the counter moment created by the 6.55 lb rifle. Gravity is pulling straight down on the rifle at the center of mass (this is a principle of physics, when looking at gravity's effect on an object you take it from the objects center of mass) for an ar15 that is going to be roughly 15-17 inches from the shoulder point of contact so lets go with worst case scenario for the 16" barrel, 15 inches.

moment = 6.55lb * 15 in = 98.25 lbf in

Now this gravitational moment is COUNTERACTING the moment created by the bullet. So our overall moment on the firearm is 1488.85 lbf in. This is our Torque on the system.

Now that we know our moment, lets calculate angular impulse and see what the difference is between a 16" and 20" barrel.

angular impulse = torque * t = 1488.85 lbf in * .00092s = 1.3697 lbf in s

For a 20 inch barrel the angular impulse (assuming the COG is 19 inches from the shoulder, so this is a 4 inch difference, which would be a HUGE difference and not likely to be reality, this is best case for the 20" barrel and worst case for the 16")

angular impulse = 1247.38 lbf in * .00108s = 1.3472 lbf in s

change in angle = angular impulse/Inertia

I = 1/3 m(l^2)

This is simplified, it's going to be a good representation without me having to do 3 weeks worth of math calculating the actual MOI for an AR15. This MOI applies because we are assuming that the AR will rotate about the shoulder, so the edge of our rectangular prism that will represent the AR.

So, assuming the 16.5" rifle is overall 32.75" long our MOI is:

I = 6.55 lb * (32.75^2) / 3 = 2341.76 lb ft^2

1.3472 lb in s = 16.17 lbf ft s * (32.2 lb ft/s^2/1 lbf) = 520.674 lb ft^2/s

change in angle = 520.674 lb ft^2/s / 2341.76 lb ft^2 = .222 radians/s = 12.71 degrees/s for the 16" barrel vs 12.70 degrees/s for the 20" barrel.

How long will it take each barrel to come to a stop and overall muzzle rise per barrel?

16" = 12.71 deg/s / 28.65 deg/s = 0.44s to come to a stop with a total muzzle rise of 5.59 degrees

20" = 12.70 deg/s / 44.1 deg/s = 0.288s to come to a stop with a total muzzle rise of 3.66 degrees

So, just to cover the parameters again, this is the difference with the WORST technique possible. Stiff Shouldering and brick walling the stock, no hand gripping the front of the weapon, slack wristing the pistol grip and basically just not holding the gun. The 16" barrel had it's COM as close as you'll likely every find it while the 20 inch having it's most favorable. So the results couldn't be any more skewed to the 20" barrel if you tried. The difference? 10 ft of muzzle rise at 100 yards. So the barrel length WILL have an effect on muzzle rise, though it's pretty minimal. With correct form that muzzle rise difference would be down to less than a quarter of a degree, if not a thenth. So barrel length, while having an effect on muzzle rise will have virtually no effect on felt recoil.

Again, this comparison is completely removing the impact of the BCG, buffer and gas systems to strictly look at the effect of barrel length in regards to recoil and muzzle rise.

Thank you for attending my TED talk.

TLDR, barrels have virtually no impact on felt recoil, but a small impact on muzzle rise. Technique and gas system/buffer changes have much greater effects than a barrel ever will. Realistically the difference in muzzle rise between a 16 and 20" barrel with good technique is less than a quarter to a tenth of a degree.