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Check request - calculating force from weapon recoil.

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Mechbob1

Mechbob1

Mechanical
Feb 18, 2022
21
Hi, I'm a recently graduated engineer who's working on calculating the force exerted on a rigidly mounted weapon mount as a result of recoil. I'm hoping someone can have a look and see if I'm on the right path, and that what I'm doing makes sense for an approximation.

Using a bunch of variables, I was able to calculate the weapon's recoil velocity. Then, I calculated the kinetic energy from the weapon's velocity and the mass of weapon.

Now, I have an answer in Joules, but am unsure how to use that to calculate a force. Anyone able to advise?

As I was stuck there, I then tried, instead, using impulse and momentum and the following forumla:

F * change in T = m*v(initial) - m*v(final).

This gave me a force, but I'm unsure if it's a correct way to do it. Anyone have anything to add?

First post, so please advise if I need to add anything else.

Thanks.
 
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50% would be close, there's a lot of variability, since the barrel length dictates how much decay there is on the falling side of the pressure curve; the shorter the barrel, the more likelihood the pressure is still pretty high when the round exits the barrel, relieving the pressure. The result would be an average pressure that's more than half the peak. Near as I can tell, real curves' averages will be in the range of 40% to 60% of peak.

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Yes, I had an evaluation copy of PRODAS about 20 years ago, and I ran every demo case they included. In looking over the stuff I saved, it's clear there were some glitches in the sims, since the really nice M735 APFSDS round sim had massive anomalies in the interior ballistics portion, so I don't have a nice clean pressure curve for that round.

Below is a curve from a 25-mm round; of note is that drastic time scale change. Note also how far the pressure falls before the round exits the barrel; the objective, clearly, was to squeeze every m/s they could get out of the round and barrel. However, note that the sim claims that the pressure doesn't actually rise until nearly a second after the round is actually fired.

M791_tzo7dk.bmp


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Thanks, Irstuff.

We can gather the max chamber pressure and sometimes pressure near the end of the barrel, but don’t know anything about what happens in between (ie gradient of pressure or acceleration between the points). We also get exit velocity as close to leaving the barrel as possible (<5m).

We’d like to build a concrete pad which we can bolt a universal mount (multiple weapons can attach) to and there’s disagreement regarding various weapons’ force as a result of firing. I’m assuming recoilless as that’s our worst-case scenario, though quite rare.
 
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A recoilless gun has no recoil so there's no recoil force. Other than those, I don't know of any gun that doesn't have a recoil absorber. There is plenty of data on existing guns.
 
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That’s my poor terminology error - I meant weapons without recoil reduction systems (eg muzzle brake).
 
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And before anyone is on the "but a revolver doesn't" have a recoil absorber I don't care. The user's hand is part of the mount and allows recoil.
 
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The muzzle brake is not in play until the expanding gas reaches it. So recoil is happening full effect until then.

Friction between projectile and barrel is always in play opposing projectile travel, so a barrel approaching infinity would have a projectile stuck in it somewhere.

Ted
 
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While muzzle brakes don't become effective until the round has exited the barrel proper, the time constant of the recoil process is much longer than the exit time, so there is potentially some benefit from a rearward-directed muzzle brake. In simple terms, we can look at the recoil compensation as a low-pass filter, which attenuates and flattens (by expanding in time) the pressure curve, so the recoil process might be on the order of 10s or 100s of milliseconds, and the interior ballistics is on the order of single milliseconds to maybe 20 milliseconds.

Nevertheless, for a long barrel design, the base pressure at the round much be too low by that time to provide much in the way of recoil compensation. For the two cases shown above, the breech pressures are already reduced by 605 to 75% by the time the round exits, but the telling data, which is in the spreadsheet I posted, is that the base pressure is always less than the breech pressure due to cooling effects as the gases expand toward the muzzle, so any potential benefit from a muzzle brake is further reduced. I was thinking that the pressure curve could be adjusted with different propellant mix or different ammo, but it's not clear there's a better design point, since the integral of the pressure curve might not be that different, and there are costs involved in using a different propellant or ammunition.



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A muzzle brake should start acting when the gas ahead of the round is pushed from the barrel, though the majority is from the combustion gases; recognize that the main function in most instances is to control moment and, in human held automatic and semi-automatic weapons especially, to reduce the pitch up tendency and put the barrel back into line with the target rather than to control recoil force and ready it for the follow-on shot.
 
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The round is supersonic, while the air in front is not. There's probably well less than 1 millisecond lead of the pressure wave ahead of the round. The resistance pressure in the sims I ran include the gas resistance pressure from the compression of the free air ahead of the round, and it's less than 15% to 20% of the peak breech pressure, even for the 25-mm APDS round with 1400 m/s muzzle velocity

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I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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TLDR, but you guys know way more about guns than I ever want to ! (from the US ?)

I'll note that the OP first asks for the force on the projectile, then says he's really interested in the time from ignition to the bullet leaving the muzzle.
I can see how he might think the two are directly related but I think the complex aerodynamics would show they aren't (so easily related0).

We "know" how much energy is created at ignition. This is then the gun recoil and the bullet KE. bullet velocity will be affected by friction (with the gun's rifling) and the aerodynamics ahead of the bullet. As it's supersonic I don't think the atmosphere outside the gun (beyond the muzzle) impacts the calc.

It's a shame that the OP didn't realise the the calculators were giving him the KE (in joules) imparted to the bullet, and hence the bullet's initial velocity. The gun's muzzle velocity (of a round) should also be calculatable (it seems to me to be a well understood and measured performance parameter). I'd've thought a liner decline in bullet velocity would be reasonable, and so time to travel distance easily calculated.

another day in paradise, or is paradise one day closer ?
 
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and so time to travel distance easily calculated.
Click to expand...

Nothing is ever THAT easy ;-) The curves above show for typical barrel lengths, the bullet is still accelerating until it leaves the barrel, albeit, the acceleration is only around 20% to 40% of the peak acceleration. The laws of diminishing returns are evident, as any further increase in barrel length risks other ballistic downsides, like excessive bending in the barrel, etc., while gaining a mere pittance in muzzle velocity. The breech pressure curve is relatively similar to the base pressure curve, which is directly proportional to the bullet acceleration, so we see a rapid increase in acceleration as the propellant gets fully ignited, followed by a tailing drop in acceleration as the expanding gas pushes the bullet toward the muzzle. Various types of propellants and their burn characteristics dictate the rapidity of the rise in base pressure over time

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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I stand corrected ... I thought the initial explosion set the initial velocity of the bullet. Clearly an over-simplification ... the continuous expansion of the explosive gases would continue to accelerate the bullet.

another day in paradise, or is paradise one day closer ?
 
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Rb1967, I’m not sure where I said I was overly interested in the time from round ignition to leaving the barrel. That duration is a known.

That said, I have since realised that using the momentum method I ignore the rifling friction force, which I’ve read is ~10%. Also the pressure of the gas leaving the barrel (though this is comparatively tiny).

I’m quickly approaching the point where there’s simply a large number of assumptions, and then a conservative FOS will need to be selected.
 
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ok, maybe my misread of the thread.

but your replies seem to me to be confusing. I didn't think a muzzle brake reduced recoil (but then I don't know much about guns).

I'd've thought you could get your design inputs from the curves provided, IR says to scale to suit your application.
You have a velocity curve for the bullet, so you know how much energy from the explosion has gone into the bullet, and so how much energy is absorbed by the mounting.
You have a time profile so you can determine the force profile, or average seems to suit your need.

I'd've thought (possibly incorrectly) that the offset moment may be a significant load.

another day in paradise, or is paradise one day closer ?
 
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Aren't there several close (time-wise) impacts that produce significant forces that contribute to the recoil? The first being when the bolt strikes and ignites the powder and the second when the bolt has accelerated rearwards a bit and hits the rifle stop?
 
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2045 kg is a VERY HEAVY rifle ;-)

The first being when the bolt strikes and ignites the powder
Click to expand...
In a typical rifle, it's the firing pin that moves to strike the primer, and it's very low mass, and it's impact is relative tiny, compared to when the propellant is ignited. That can be verified by the dry firing a rifle.

the second when the bolt has accelerated rearwards a bit and hits the rifle stop?
Click to expand...

I'm guessing you're referring to the shell extraction process on a semi/auto rifle? That hypothetically occurs near the time the round has left the barrel, since any premature opening of the breech would kill the acceleration of the round from the barrel. Likewise, a relatively small impact, given that the extractor is driven by the barrel gas pressure when the round clears the hole near the front of the barrel that sends the gas back to the bolt and pushes it back to extract and eject the shell.

There is a spring inside the rifle that compensates for that, though, but, its primary purpose is as a return spring to push the bolt forward and insert a new round in the chamber

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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