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!

 

[ivymike]

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

 

[GregLocock]

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

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

 

[RoarkS]

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!

 

[RoarkS]

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.

 

[GregLocock]

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

Cheers

Greg Locock

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

 

[SomptingGuy]

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

 

[prex]

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

 

[MintJulep]

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

 

[desertfox]

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

 

[RoarkS]

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.

 

[desertfox]

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

 

[RoarkS]

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 ?t 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.

 

[RoarkS]

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

 

[GregLocock]

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

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

 

[ivymike]

...will do what you think you want

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

 

[RoarkS]

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.

 

[desertfox]

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

 

[RoarkS]

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.

 

[GregLocock]

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

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