Two body dynamic collision with a spring 2

[RoarkS]

Two body dynamic collision with a spring

Alright, I am completely irritated I cant figure my way through this.

For assumption purposes I have two bodies sliding on a one dimensional frictionless plane. The first body has a given (high) momentum. It slams into another stationary body, but there is a damper ‘spring’ attached to this second mass to ease the impact. I need to know the momentum of the 2nd mass after the impact.

The spring damper is going to store energy, yet it along with the mass it is attached too is still going to slide… how in the world do I figure that conservation of momentum/energy storage problem?


I have a limit to the momentum that the second mass can handle and need to design the spring displacement and K value accordingly…. If this is possible.

Thanks!

 

[desertfox]

Hi RoarkS

The difficulty I am having is where you intend to put the damper ie:- a recoiling barrel onto a spring, or a spring damper between the gun butt and the guys shoulder, out of interest isen't there also a secondary recoil as the bullet leaves the gun ie gas expansion.
Another problem I cannot see past is that a what point does the human stop moving due to recoil and the spring damper takeover.
I am no expert in this however I can't help feeling the stiffness of the damper versus the resistance to the human moving from the recoil is an unknown factor.

regards

desertfox

 

[RoarkS]

Desertfox,
You are certainly asking good questions!
The placement of the damper is going to be on a recoiling barrel onto a spring, which is housed inside of the stock of the rifle. My intent is to not let anymore energy out of the stock than my shooter can tolerate (back to my “reasonable rifle” momentum and energy values” As you said the stiffness of the damper vs. the resistance to the human moving is my problem. I can’t have a spring/damper system that is so stiff that it will not compress, before my human starts to get violently pushed back.

As far as that secondary recoil… yes there is one. However for our purpose let us not worry about it. With a rifle this large the muzzle blast and flash emanating from the barrel is so intense, a suppressor type device becomes a necessity. (I work for a class 3 manufacturer so no, the ATF will not be knocking on my door tonight). Just to enlighten you, muzzle blast and flash is due to the supersonic wave front coming off the barrel. By placing a divergent nozzle section on the end, this decreases the supersonic wave front area, decreasing the ability of excess powder being ignited from the shock, thus reducing muzzle flash (this is how a flash hider works), then I intend to employ a suppressor type device to trap the muzzle blast, so as to keep my shooters retinas from detaching from such a large shock, and also to attempt to regain the forward momentum of as much of the escaping gases (usually a maximum of 30% of the overall recoil can be attenuated with this method alone, thus one of the only reasons a Barrett rifle is tolerable) to aid in reducing my felt recoil.

I got to thinking about this while I was sleeping… how is this any different from one of those text book train car collisions? Car 1 comes in and hits car 2 that has an impact damper /spring with a specified k and c value mounted on it… what is the final velocity of car 2, or the combined car 1 and 2 mass? Or is that what you all have been trying to tell me?

Roark

 

[desertfox]

Hi Roark,

Now I know that the spring is reacting against a recoiling barrel, then the initial momentum calculations have to start with the mass of the moving barrel. The rest of the gun mass along with the shooter at this stage would just be stationary. When the barrel finally comes to rest on the compressed spring it is only then that the shooter and the mass of the remaining gun comes into play.

So we need to know the mass of the recoiling barrel.

Your reference to the model of a single spring and a dashpot is probably correct but I think the mathematics will get complicated as stated by others (yes I believe others posting on here were telling you this was the way to go).

The difficulty in using or designing a spring to absorb the energy is that the spring size will be limited by the space available within the gun butt.

As stated in my previous post the energy absorption rate is governed by the torsional stress surge wave.

In order to calculate this we first need to know the space envelope that is available to contain the spring. Secondly, we know that the whole recoil cycle lasts for .000746 seconds. We also know the amount of energy that needs to be absorbed by this spring in that time period.

It would appear to me that we could perhaps get a rough approximation using an energy balance. My concern though is that the spring will only absorb the energy relative to the barrel mass and velocity. In which case all the remaining surplus energy would have to go through the gun and shooter.

I'll stop now as my head's beginning to hurt thinking about this!

Regards,
desertfox

 

[Pud]

I am probably wrong, but surely a spring does not "absorb" energy - it merely "stores" it. (For a perfect spring)

For absorption of energy a damper of some sort is required, which will convert the recoil energy into heat. Such as on artillery pieces.

High recoil weapons have been well researched and tried. Are you sure you are not trying to "re-invent the wheel"?

The Brits are good at failures (I am one). Here's an example - every one hated firing it:

http://en.wikipedia.org/wiki/Boys_anti-tank_rifle

Then there is the "recoil-less" options, such as the Carl Gustav M/42, which is a 20/180 round:

http://www.quarry.nildram.co.uk/gustav.htm

Cheers

http://www.tynevalleyplastics.co.uk

 

[RoarkS]

Alright another day in thought,
Pud you are absolutely right, the idea is to use a dashpot of some sort to help the whole thing slow itself down. As far as the artillery pieces, I would love to get my hands on some studies, It’s just I haven’t invested much effort or $$$ into getting it. I honestly didn’t think that this project was going to present me with this much issue. I’m not trying to re-invent the wheel… I just don’t know what pi is called yet in my wheel problem.

So…
The mass of the “barreled action”, the entirety of the recoiling mass is my 40 lbs. I do not know the mass of the rifle as a whole yet, because… this is as far as I have gotten. I don’t know the mass of the systems that will be used in the dampening or any other necessary structure. So to say the least, with a 40 lb barrel it had better be as light as possible. Let’s say 10-15 pounds as a target weight.

So check me on this train of thought. Dealing with force…

I know the momentum of the barrel. Thus I know its mass and its velocity. I take the time interval over which the bullet is accelerating in the barrel, and this will give me the acceleration of the barrel. This coupled with the mass will yield a force. (Okay now that part I am really shady on. Yes I can pull acceleration out of thin air right there, but is it the right one? Do I need to be dealing with the negative acceleration from the barrel running into the spring/damper?... is it any different? All I want is the force of the recoil)

I do that with my Barrett (my “reasonable rifle”), it will give me a value of the maximum force that the rifle can impart on my shooter.

For this thing to work I can not imagine the rifle having a spring/dampen travel distance of more than 12 inches, something more like 6-8 is ideal… and of course anything less than that would be awesome.

SO…. What I am thinking is that this force should be equal to or greater than maximum force applied to the spring and damper at its state of maximum compression. Thus I can work my way back to determine c and k values.

Maybe. Lol.

 

[GregLocock]

In my previous post replace gun mass by barrel mass and human mass by rest_of_gun_mass and the equations apply to your new architecture. It is as simple as that.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Pud]

What are you going to do about the travel of the sights towards the shooter's eye socket?

Sights of necessity are attached to the receiver, which will be zinging back at some rate of knots! It can be a problem with 7.62/.308 etc.

http://www.tynevalleyplastics.co.uk

 

[prex]

There are plenty of shock absorbers out there: an example (the first one that came out). A model with 2 in stroke, weighing 2 lb, can absorb 290 ft lb per cycle. Whether the spring force is too much or the dimensions are too large is up to you to check: as I said, there are many different shapes, dimensions, suppliers...
I think it is time for you to go in the real world and examine some practical alternatives.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[desertfox]

Hi RoarkS

I am beginning to think if the barrel is 40lb then the shooter is unlikely to be standing when he fires this thing? or am I wrong.
I haven't come across a gun that as a dashpot mounted in it; well at least one that someone holds to the body and fires.
Breda made a shotgun with a recoiling barrel onto a spring to absorb the energy from the recoil, they also made use of the recoiling barrel to load the next cartridge into the chamber which also reduced some of the excess energy.
I hardly think it matters whether we consider the energy stored or absorbed by the spring so long as we all know that energy cannot be destroyed but merely transformed to another form or state.
Going back to your last post regarding acceleration, the barrel would be slowing down during spring compression and the decceleration force would be increasing as the spring compressed which means the velocity is also changing over the time period.
I think you might be better going down the energy equation route but it still remains critical as to how much space you have for a spring.

 

[handleman]

My idea (knowing nearly nothing about guns ): Your shock absorber actually needs to operate as the spring extends. As the recoil occurs, the spring should compress with nearly no resistance from the shock absorber. All the kinetic energy of the kick is stored in the spring during the shot. This will minimize the impulse felt by the shooter. Then, as the spring tries to extend, your absorber kicks in and dissipates the spring's stored energy slowly.



-handleman, CSWP (The new, easy test)

 

[GregLocock]

That's a good idea, and is of course almost exactly how a car's suspension works.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[zekeman]

Maybe somebody should read a good book on the internal ballistics of a gun before tackling this problem. It is far more complicated than a simple 2 degree of freedom problem with a spring and damper or anything I have seen suggested here.
For example what do you do about the time history of the gas pressure inside the barrel and its mass distribution during the bullet travel in the gun barrel just before exiting and the attendant momentum exchange including the momentum/energy in the gas?

 

[Pud]

Rifle Recoil and Bullet Energy Calculator:

http://www.varmintal.com/ashot.htm#Recoil

Lots of useful stuff thanks to "Varmint Al"

Warning!! Site contains pictures of dead animals!!

http://www.tynevalleyplastics.co.uk

 

[gwolf2]

I have enjoyed good results for this type of problem using a spreadsheet and time stepping through F=Ma, V=U+at etc. The discontinuities are a real pain but at least you can see exactly what is going on all the time unlike a sim package.

My application was a spring launched low thrust/weight ratio rocket which needed a kick up to 30mph for the fins to start working. The startup, spring kick, let-loose, power-on and power off coast were all annoying discontinuities. Your application sounds like it has more discontinuities and will be more difficult but I think the method is workable in your case.

gwolf.

 

[GregLocock]

" 15 Jul 09 9:13
Maybe somebody should read a good book on the internal ballistics of a gun before tackling this problem. It is far more complicated than a simple 2 degree of freedom problem with a spring and damper or anything I have seen suggested here.
For example what do you do about the time history of the gas pressure inside the barrel and its mass distribution during the bullet travel in the gun barrel just before exiting and the attendant momentum exchange including the momentum/energy in the gas?"

But he isn't interested in ballistics, a problem with a time constant of the order of 0.1 ms, he is interetsed in how much recoil the shooter sees, a problem with a time constant of the order of 50 ms or more.

FWIW a step by step 3dof (ie barrel stock shooter) spring mass damper spreadhseet took less than 20 minutes to knock up, probably less time than it takes to read this thread.





Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[RoarkS]

GregLocock, I would be eternally grateful for a little more guidance on how to do this “easy” spreadsheet. My boss and I spent the better part of the afternoon trying to make such a spreadsheet without much luck.

 

[RoarkS]

After today messing around, this is the best I can figure...

m_i*x_i’’+k*x+c*x’=m_b*x_b”-|k*x+c*x’|

I found m_i*x_i’’ by using my v_f that came from my recoil calcs, then used the duration of bullet going down the barrel to appx. x_i". my m_i as stated before is 40lb.

If I "know" m_b*x_b” needs to be 250lb or less, so can I just plug that in and tune my k,c,x values?

x has to be between 0 and 12 inches. k and c are unknown.


I never went any farther than calc II. So I am learning this as I go. (like I said, mech tech degree here. I should be running cam programs, and telling the rest of you "I can't machine a part like that"...lol. but I am feeling adventurous, so here I am asking as politely I know how)

Thanks!

 

[GregLocock]

12 Jul 09 20:19 explains it as well as I can short of writing it for you. Which would cost you less than an afternoon's pay for two people.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Pud]

Have you tried the Rifle Recoil and Bullet Energy Calculator I gave a link for? Or is your "wheel" so different it needs re-inventing?

http://www.tynevalleyplastics.co.uk

 

[zekeman]

"After today messing around, this is the best I can figure...

m_i*x_i''+k*x+c*x'=m_b*x_b"-|k*x+c*x'|"


Roarke,
That is not the best approach to the problem since it is considerably simplified by the very short duration of the bullet time inside the barrel as others have said.
Given that and since you know the exiting momentum you can get the velocity of the barrel Vb(0). Now you have a 2 mass problem with the 40Lb barrel mb and the Human+gun mass Mh with a spring dashpot between them. Assuming the dashpot acts in only the opening direction, we can write 2 ODE for the first half of the dynamics, viz :
Mh*x1"+k*(x1-x2)=0
mb*x2"-k*(x1-x2)=0
x1 motion of Mh
x2 motion of mb
After some adjusting I get
(x1-x2)"+(k/Mh+k/mb)(x1-x2)
and the solution for (x1-x2), the spring deflection is
x1-x2=Vb(0)/w*(sinwt) which satisfies the initial conditions, x1'(0)=0, x2'(0)=-Vb(0)
w=sqrt(k/mb+k/Mh)From the solution it is seen that the maximum deflection is Vb(0)/w. Assuming 1 foot of max deflection, this yields w=Vb(0) and the max force would be
k(1/mb+1/Mh)=Vb(0)^2; k=Vb0)^2/[1/mb+1/Mh]
So the max force is numerically k since the deflection is assumed to be 1 foot.Note that you have no control of the k and the maximum force which is dictated by Vb(0) and the masses.
The second half of the dynamics where sinwt<0, includes the damping term and can be handled similarly.