Check request - calculating force from weapon recoil.

[Mechbob1]

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.

 

[Mechbob1]

In case a pic helps for impulse and momentum way of calculating.

 

[3DDave]

Search for "gun recoil force calculator" Plenty of people care about it for some reason. However, the internal operation of the weapon can change the result so you'll eventually have to actually test it.

 

[Mechbob1]

Here's the way I got stuck, and couldn't work out how to get a force:

 

[Mechbob1]

Thanks, 3D Dave.

I've tried a few calculators and they give me the Joules, but I can't recall how to turn that into a force.

Are you able to confirm if the momentum methodology makes sense?

Thanks.

 

[hydtools]

I think impulse-momentum is the correct approach. The time duration of the impulse would be more like 0.0013 secs. to 0.0020 secs.. I looked at barrel pressure traces that can be found online.

Compare this with work-energy. Work, F*d, where d is the length of the barrel. The energy, 0.5*m*v*v, would be that of the projectile at the moment it exits the barrel.

Your approach assumes the barrel is so fixed that it gains no velocity in the momentum exchange.

Ted

 

[IRstuff]

There are actual ballistic simulators, like

http://www.prodas.com/

which compute the muzzle forces on a millisecond basis. Nevertheless, the energy approach misses key information, since you have the muzzle velocity and possibly have the time-to-exit the muzzle. Nevertheless, you know velocity curve is sigmoid-ish, since the breech/base pressures have a peak somewhere in the middle of the time-to-exit cycle of the round, as shown below

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[dik]

Joules is a force * distance...

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[dik]

I think IRS is on to something... with my .338, I could use light reloads and fire off rounds all day without 'batting an eye', but with heavy loads I could only fire off 3 or 4... recoil would nearly 'rattle your teeth'. It was really uncomfortable.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[IRstuff]

Sure, there are a number of ways of estimating the AVERAGE force, since you have to know the muzzle length, so simply dividing by muzzle length gives you the average force, but not the peak. That might be all moot, but given the curve I attached, you can scale the numbers to the problem posed by the OP.

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[3DDave]

#4 that was returned was

https://www.alternatewars.com/BBOW/Ballistics/Int/Gun_Recoil_Calculator.htm

Gun Impulse/Recoil Force Calculator

 

[IRstuff]

Sure, the muzzle velocity is the output of the load size and burn rate, which result in the recoil impulse characteristics. My copy of the results for an M735 APFSDS round has the entire time profile in about 4.5 ms, which accounts for the 1505 m/s muzzle velocity, although, the breech pressure peak is only about 432 MPa

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[hydtools]

Any concern about reacting to the moment?

Is not the average force half the peak force?

Ted

 

[TheTick]

"How much force?" is an indication that you don't fully understand the problem.

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[hydtools]

Area of a triangle = 0.5*b*h b is base. h is height.

Ted

 

[Pud]

Surely there's no recoil until the projectile exits the muzzle? Is it not a closed system until then?
Does a gun with an infinite length barrel recoil?


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[IRstuff]

Area of a triangle = 0.5*b*h b is base. h is height.

Yeah, was thinking pyramids. Nevertheless, the example above has average of 59% of peak

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[hydtools]

Pud, the moment the powder ignites the recoil starts. Equal and opposite reaction. The gas pushes both the gun and projectile.

Ted

 

[Mechbob1]

Thanks for the all the responses.

For context, I will have access to timings from firing to round exiting the barrel.

Regarding the barrel length, I don’t understand the relevance. As in, my limited understanding is that generally a longer barrel will have higher round exit velocities, but this has a limit too (eg infinite barrel).

I’m trying to do ‘worst case’ calcs so I can include recoilless weapons.

Irstuff, thanks for the graph - that’s good to know. If I use the momentum calc I just get the average force, but if I double it I’m likely to be in the ballpark. Does that sound about right?

 

[Mechbob1]

Irstuff, is that chart from a simulation?

Thanks