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Two body dynamic collision with a spring
2

Two body dynamic collision with a spring

Two body dynamic collision with a spring

(OP)
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!
 

RE: Two body dynamic collision with a spring

you could do a simple finite-difference calculation using excel...  or you could bust out a mechanical dynamics simulation tool.
 

RE: Two body dynamic collision with a spring

The spring is irrelevant for the momentum stuff.

Momentum before the impact=Momentum after the impact

m1*v1_0+m2*v2_0=m1*v1_1+m2*v2_1

Also no kinetic energy is lost so

m1*v1_0^2+m2*v2_0^2=m1*v1_1^2+m2*v2_1^2

In your case v2_0=0, so the solution for v2_1 as a function of v1_0 is high school algebra

Designing the spring in this case is an underconstrained problem - you need to decide on an additional criteria by which to choose it. If you are feeling lazy just select a free length and rate such that k*L^2>=m1*v1_0^2, that guarantees that this collision will never ground the spring out.

If you want a more economical spring then there is an algebraic solution, but the calculus may get a little heavy. A spreadsheet model may be quicker and more informative.

 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

(OP)
The spring/damper is my problem.
I have an object that is 40lb, V_i 23 ft/sec with an energy close to 300 ft*lbf.

I have a body that it will run into, that cannot absorb more than 50-75 ft*lbf of energy

It may be highschool algebra, and I hope I am missing something obvious... but I'm here because I cant figure it out. Besides I'm a "mech tech" this is a little more than I am used to dealing with. LOL!





 

RE: Two body dynamic collision with a spring

(OP)
Forgive me... I should say as far as the second body, I know it can withstand a 30lb mass having 95 ft*lbf of energy. anything more than that is no go. Less would be better thus why I went with 50-75 ft*lbf.  

RE: Two body dynamic collision with a spring

Ah, so you have a damper as well as a spring? Then a time based simulation is the best bet.

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

Agreed.  Once you lose the conservation of kinetic energy equation, analytical solutions become tricky to say the least.  A time-domain solution is simple to set up - no need for fancy/rigorous integration schemes - with such a small problem you can get away with an Euler scheme using tiny time steps.

- Steve

RE: Two body dynamic collision with a spring

Starting at the point where mass 1 meets the spring-damper, we can write (with obvious meanings for symbols):
M1x1'=k(x2-x1)+c(x2'-x1')
M2x2'=-M1x1'
If the two masses were equal, these would combine into:
x2'+x1'=0
M(x2'-x1')+2k(x2-x1)+2c(x2'-x1')=0
The meaning of the last equation is:
-if the two masses are equal, what happens is exactly the same that would happen with a single mass being stopped over a spring-damper with a fixed end, having 2 times the spring constant and  2 times the damping constant.
So, assuming the first mass is of the same order of magnitude or larger than the second one, if you select a damper capable of stopping the first mass with overcritical damping (as usual for a stopper), allowing also for a  smaller size (because of the factor 2), the result should be (by gut feeling) a totally inelastic collision: the two bodies would continue their path together with a velocity that is easily calculated by the conservation of momentum.
To check if I'm correct, you would just solve the above equations by finite differences, as proposed by others.

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

RE: Two body dynamic collision with a spring

I'm not sure that the OP understands the difference between a spring and a damper.

 

RE: Two body dynamic collision with a spring

Hi RoarkS

If we consider no damper and use the linear momentum equation then we can conclude that no momentum is lost after the first moving body strikes the stationary one attached to the spring, however there is a kinetic energy loss associated with the impact unless the two bodies are perfectly elastic.
You don't give enough details of your system but consider this example:-  

two masses one say 2kg and moving at 2m/s and another mass of 10kg stationary.
If the moving mass strikes the stationary mass then:-

m1*v1 + m2*v2 = m1v'1 + m2v'2

so;- m1=2kg
     m2=10kg
      v1= 2m/s
      v'1=?
      v'2=?
:-

2*2 + 10*0 = 2*v'1 + 10*v'2   

after impact both masses are travelling at a common velocity
so v'1=v'2 if we call it v'b then transpose the formula to find v'b then we get:-

(2*2 + 10*0 )/(2+10) =v'b

therefore v'b = 0.333333333m/s to check substitute this figure back into the conservation of momentum:-

2kg*2m/s= 4kgm/s

combined mass 10+2=12kg (when moving together)

12*0.333333333333333333m/s = 3.999999999999kg m/s

no loss in linear momentum, however check the kinetic energy
of the first body and then the combined body

ke= 0.5m*v^2      so 0.5*2*2^2 = 4J

ke of combined body = 0.5*12*0.333333333^2 = 0.666666J

So if I now introduce a spring to stop the combined travelling mass I need to equate 0.66666666J of kinetic energy against the energy stored in the spring.
The equation for energy stored in a spring is given by:-

0.5*k*x^2    where k = spring stiffness and x= spring deflection

lets say k= 10kN/m

then 0.666666J = 0.5* 10000*x^2

therefore x = 11.5mm

so the spring with that given stiffness would stop the mass within 11.54mm.


desertfox

RE: Two body dynamic collision with a spring

(OP)
Hey, thanks a lot. I think that makes sense dealing with the momentum collision first, then deal with the energy of the final system.  I will have to sit here and crunch some
numbers for a while.  Mass 2 is about 200-220 lb.

It's a gun recoil problem. I need to fire relatively very large round from a shoulder fired rifle platform without killing the operator. I am used to dealing with engineering of rifle components, but this is definitely a little farther out of my box. It's outside of everyone's box actually in this industry.  

In order to keep everything simple I figured I would use an equivalent recoil situation to define the parameters for what my mass 2 (person) can handle.

As far as the difference between a spring and damper... I will admit I am using both terms rather loosely.  All I need is something that can store/convert some of the incoming masses energy, and be able to reset rather quickly, (for a follow up shot, so no deforming absorbers allowed) without in turn releasing that energy right away as a free spring would. There are currently shotgun stocks on the market that use just a spring, but they come back and hit ya in the side of the head on the spring return with considerable force. I am currently considering some form of coil over oil filled shock. Spring to store the energy, and the oil to slow the system for the return.

That being said... I am not entirely sure

I really don't want to deal with time based simulations. I don't think it's that complex, and I know that figuring out how long it takes to compress a spring with a given load is a NASTY problem.

As I said this is a little farther out into pure design than I am used too. The help is much appreciated.
 

RE: Two body dynamic collision with a spring

Hi RoarkS

Well I wasn't thinking of a gun recoil application when I posted a response and I think your problem is far more complicated now.
Anyway I did a search and found these sites which might help or interest you.

regards

desertfox


http://www.shootingwiki.org/index.php?title=Understanding_Recoil#Secondary_Recoil

http://ww2.lafayette.edu/~burciagj/111/HW(CHP5b).pdf

http://www.bsharp.org/physics/recoil

http://www.thefiringline.com/forums/showthread.php?t=8106

http://www.chuckhawks.com/index2c.shotguns.htm

RE: Two body dynamic collision with a spring

(OP)
When the gun fires we have conservation of momentum. For my calculator I am trying to put together I need the following variables.
M_b = Mass of bullet
V_b = Velocity of bulelt
M_p = Mass of projectile
V_p = Velocity of Projectile
M_g = Mass of gun
V_g = Velocity of gun

All of these are known to me with the exception of the V_g. thus I  work with the equation (M_b*V_b)+(M_p*V_p)=(M_g*V_g)  to find V_g.

(side note: I also want to calculate the individual effects of the powder vs the projectile, thus set one variable to zero to see the effects. however when I do this the separate components don't add up to the whole combined version... this doesn't make sense to me. I am sure it is algebra but I don't see it)

With this I can calculate the energy of the recoil with E=(.5)(m_g)(v_g)^2
  
Now given all of these parameters I can calculate the recoil energy, and momentum of the rifle.
I do this for a "reasonable rifle" such as a 50BMG Barrett m82 weighing in at 31lb it has a V_g of 13.6ft/s and an energy of 90ft/lbf. This data is what I am going off to be set as my Absolute maximum a person, my 2nd body, can deal with. It is difficult to define what the destructive nature of recoil is, and as such it is often tabulated as impulse (with some controversy because of Δt being so close to 0 (usually in the 1ms range), recoil velocity, and recoil energy.

So. My problem is that I need to make a large round/rifle combo having a M_g of 40lb creating V_g of 23ft/sec and an energy of 300ft*lbf. By sucking up as much energy as I can in an reusable energy absorption device that will not exert more force on a person than that of my baseline "reasonable rifle"

My thought is by assigning the 2nd body do not exceed energy and momentum values, I use the difference to find out how much my absorber needs to suck up.

I am not even sure if I have even defined my problem properly.  

 

RE: Two body dynamic collision with a spring

(OP)

Isn't there a simple shock absorber design equation(s) for things like this?
 

RE: Two body dynamic collision with a spring

Yes, a two degree of freedom spring and damper equation will do what you think you want.

In fact for the problem of interest a damped single degree of freedom equation with an impulsive base excitation would do.
 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

...will do what you think you want

superman, you can be a real pain when you've been drinking.
 

RE: Two body dynamic collision with a spring

(OP)
wow.
I don't think that quite qualifies as simple in my world. haha.
Thanks though. I suppose I have a lot of learning or consulting fees ahead of me.  

RE: Two body dynamic collision with a spring

Hi RoarkS

I am not sure its that simple to put everything into one single equation and get a perfect answer.
A further search I found this site which talks about gun recoil and gives a link to an kinetic energy calculator:-
http://www.herosarms.com/Selecting.htm#Recoil
The only thing that stays constant after a gun as been fired for a given shot and powder load is its kinetic energy, if the gun was mounted on strings it would be consistant on its recoil movement each time, however enter another variable a human being; if the shooter goesn't pull the gun tightly into the shoulder, then on recoil the gun gains momentum and hits the shooter with more force than it would if held correctly to mention one factor however I am sure the stock area, weight etc also play a part.
I suppose you could assume however that the gun should be held properly by all shooters in which case the calculated gun velocity and mass can be used to calculate the above as already demonstrated, so for a rough ballpark figure my approach would be to take your 90lbf ft off the 300lbf ft which leaves 210 lbf ft to be absorbed but then we need some more information:-
Now let me ask? what time period does a typical recoil last? because at the end of the day we can design a spring to absorb that energy of 210lbf ft but how quick do you want to do it and the slower the energy is absorbed the better it is for the shooter.
A given compression spring is limited to absorbing or realising energy by a surge wave which transmits the torsional stress along from the point of loading through the spring to the point of restraint and this is governed by the spring material and design.

desertfox

RE: Two body dynamic collision with a spring

(OP)
Ooops... from above...
M_b = Mass of bullet
V_b = Velocity of bulelt
M_p = Mass of powder
V_p = Velocity of Powder
M_g = Mass of gun
V_g = Velocity of gun
-I keep messing up writing out m_p and v_p as projectile... it is supposed to be powder.

I have already put together an excel sheet that does all my recoil/ballistic calculations. So what happens with those online recoil calculators is no mystery to me. A lot of them do not have an input for powder velocity (v_p) which in my case due to boundary layer effect with smaller barrels, is inaccurate with my larger cannon bores, which is something that must be taken into account. The v_p of sub .50 calibers bores tops out around 4400f/s whereas my calibers see higher velocities from the increased cross section can get as high as 5500f/s.  On top of this I have some additional inputs such as effectiveness of a muzzle break and sound suppression devices which are other useful "known" means of taking out some of the energy. So I have an adjusted energy rate that I would like to play with eventually.

The time for how long a recoil lasts, as far as any of us around here can figure is the time from the t_0 at bullet velocity v_I=0 to the moment it leaves the barrel (t_f) at v_f=muzzle velocity, which we can easily measure with a chronograph, then work some simple kinematics to figure the bullet time in barrel. We know there is more going on inside that barrel, but for now a simple model is sufficient. That being said if I have a 20mm projectile traveling down a 30 inch barrel exiting the barrel with a muzzle velocity of 3350f/s that gives me .000746 seconds as the duration of the impulse.  

Then as far as our "body 2 (shooter)" all I care to model is a standard mass of a person (190lb) with my above mentioned design parameters of a "reasonable rifle" equivalency. If I was to do this the way we have done things in the past, I would go out and make a sliding test rig frame, mount a Barrett rifle in, and load it down with sand bags until it doesn't move when fired. Then use the movement of the rig when fired with the new weapon as a gauge if the thing works.
All I want is to model this mathematically, before we spend a serious amount of money on a custom shock absorber, barrel, and test receiver, when I am not even sure there is a reasonable solution to my problem.    
 

RE: Two body dynamic collision with a spring

That all sounds OK. If the acceleration time of the bullet is less than 1 ms then so far as the human being is concerned you can just regard that as an instantaneous change in momentum of the gun.

So it comes back to a gun mass and a human being mass joined by a spring and damper. At t=0 the human is at rest and the gun is moving at -(M_b*V_b+M_p*V_p)/M_g

So once you work that out you can forget the ballistics.

That leaves a 2dof problem with a rigid body mode.

For that matter you can work out the final velocity of the human + gun system, it'll be -(M_b*V_b+M_p*V_p)/(M_g+M_h)

The bit in between is slightly complex but is controlled by

force that gun exerts on human=-force that human exerts on gun= k*(x_h-x_g)+c*(v_h-v_g)

and you know the acceleration that that force produces, and you know how to turn accelerations into velocities into displacements

Solving that analytically to the point where you get a useful equation is a page or two of horrible equations from memory, or about 8 columns of an excel model (time, force and the x v and a for the two bodies) .

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

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

RE: Two body dynamic collision with a spring

(OP)
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
 

RE: Two body dynamic collision with a spring

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

RE: Two body dynamic collision with a spring

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





 

www.tynevalleyplastics.co.uk

RE: Two body dynamic collision with a spring

(OP)
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.
 
 

RE: Two body dynamic collision with a spring

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

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring


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.

 

www.tynevalleyplastics.co.uk

RE: Two body dynamic collision with a spring

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

RE: Two body dynamic collision with a spring

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.
 

RE: Two body dynamic collision with a spring

My idea (knowing nearly nothing about guns smile ): 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)

RE: Two body dynamic collision with a spring

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

 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

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?
 

RE: Two body dynamic collision with a spring

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.
 

RE: Two body dynamic collision with a spring

" 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

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring

(OP)
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.  

 

RE: Two body dynamic collision with a spring

(OP)
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!

RE: Two body dynamic collision with a spring

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

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Two body dynamic collision with a spring


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?
 

www.tynevalleyplastics.co.uk

RE: Two body dynamic collision with a spring

"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.

RE: Two body dynamic collision with a spring

Correction
After some adjusting I get
(x1-x2)"+(k/Mh+k/mb)(x1-x2)
should read
(x1-x2)"+(k/Mh+k/mb)(x1-x2)=0

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