How do you convert joules into kg? - Yahoo! Answers

Good Luck.

The military 7.62 by 51 and the commercial 308 Winchester are the same round. Ballistic tables for the 308 are listed below. The former eastern bloc nations use a 7.62 by 39 and the ballistic tables for these are different.

http://www.clcweb.net/Shooting/Ballistics/ballistics.html

I'm confused as to how you are determining Δt for the bullet.

He's taking his assumed length of the round. Taking just the distance the CG moves in the crushing of the round.

I suggest that the CG would only move one half the length of the round in crushing.

s = t x (v_i + v_f)/2,

where s = distance to stop, t = time to stop, v_i = initial velocity, v_f = final velocity.

s = 20mm = (assumed) length of the round. No idea what the actual length is. I've also assumed that the C of G of the round stops in that distance - again, I've no idea if that's a valid assumption - does it penetrate the plate? If not, then s would be closer to 10mm.

v_i = 1000ms^-1, v_f = 0, hence t, or Δt, the duration of the impulse.

I think it was the best assumption you could make given the information provided.

Very good answer indeed, JohnDG.

A small correction, likely an oversight, is this line, you stated:

ΔP = m x v = 14 x 10^-3 x 10³ = 14 kgms^-2

it should be 14 kgm/s, not per second square, m x v.

Also, once you having the distance over which the objects stops, you can calculate the average force directly, no need to calculate the time. (Assuming that by the length of 20 mm of the slug as your estimate you also apply this number for the distance the slug travels before coming to a rest, (20 mm).)

It is like this:

Average force of a blow: total momentum / distance, thus:

7000 J / 0.020 m = 350,000 N which corresponds with your answer.

Note, if the distance to come to a stop is shorter, then the average force goes up.

As a side note and a small comment, in metric we have Newton (N) not kgf. Kgf is an invention of the English societies that use lbf as in the Imperial system has no equivalent to N. kgf is not a valid definition in metric. I know it is commonly used and is visually easily to understand. Don't take that personal please, it is meant to be a general statement, (not your fault either.)

Do you design a resistant structure with an averaged force or with the maximal one ?

What could happen if the average force is 1/5 of the maximal and you consider only a safety coefficient of 2? It is of course only an example of a not impossible situation.

From the information given you can not calculate the maximum force, only the average force over the 20 mm distance as in this example.

To find the maximum force you would need the exact movement measured over time. From there you calculate the instantaneous velocity and then the second derivative will give you the instantaneous acceleration from which you calculate the force at each instant allowing you to find the maximum force. In which case you will also know where and when it occurs.

Sorry, there is no other way. But the average gives you a good number. If the resistance is constant over the total travel then the deceleration is also a constant and the average also represents the maximum. It all depends on the resistance of the impacted material and its substrata.

BTW, the original question is wrong. You do not convert Energy into kg (in this example) but energy into a force not mass. You convert Joules to Newtons.

The approach is a bit different than the one you suggest. In this particular case since it is a shock and the force is generated at the contact of the 2 bodies. The force results from a balance of the energy transfert from the incoming body(bullet) and the impacted one (plate) and as a function of the "stiffness" of each body at the interaction point and overal because the transfer is related to deformations of both bodies which are as well elastic as plastic depending of the moment in the transfer. The deccelerations are a result. In fact it is possible to determine the forces using a FEA analysis based on imposed deformations for both bodies and related forces which result. This simulation which is today possible since the programmes can deal with nonlinear material characteristics and big deformations give a "force - deformation" curve with which the required energy for every step can be computed. the only difficulty is the fact that the shock generates waves in the bodies and a high speed strain leads to a Young modulus higher than the one present in the books and depending on the speed with which the strain increases. The other difficulty is the propagation of the waves with sound speed which requires a transient analysis much more complex. Not every FEA programme is able to do it the right way. The relationship F=d(m*v)/dt is in this case used the other way around i.e. to determine the decceleration as function of interaction forces. Unfortunately many times there is a fear to make too complex models and then the model is so much simplified that it is not any more representative for the simulated phenomenom.

Sure, I quite agree with you.

FEA analysis is one way of "measuring" the conditions.

It can become quite complicated if you want to know all these details. For practical purposes however, I think that the originator of the question wanted a force in general to see how "strong" the wall has to be and what deflection goes with it or what can be expected.

As stated by Masterblaster, how come a private person needs such protection, are we aiding the mafia ore some such person here? It could be for protection against such persons as well, though.

"it should be 14 kgm/s, not per second square" - quite right, thanks for sticking with it long enough to spot my "deliberate mistake" !

That's highlighted one of the most useful things about calculating using the metric system of units - viz. the ease of checking the result by checking the dimensions. If the dimensions don't match, then (_{^{the idiot that posted cocked it up or}}) the result is wrong.

" ... in metric we have Newton (N) not kgf ..." - I concur; after calculating the force in Newtons, I converted to "kgf" to try to help edwin (who asked for a result in "kg").

Joules are Newton Metres, force through a distance. In this case, the round on the initial impact will apply force to the plate, the round will crush so it's CG will move through a distance. The plate will deflect under that force, allowing the CG to move by that deflection, in addition to its crushing distance. If you divide that distance into the energy, you get the average force.

Clearly, the worst case for the frame is if the round hits directly at the frame. Conservatively, I would ignore the crushing of the round, assume a trial force, apply it to the frame and find the deflection of the frame under that force and see if half of that force¹ through that deflection gives the correct amount of work. If not do another trial and repeat until you have the right combination.

If this is too conservative, you will have to investigate the crushing of the round and any denting of the plate, for additional distance.

1. Think of a force applied to a spring, at first touch, the spring offers no resistance, but then builds up linearly with distance to equal the force. The average load resisted through the deflection is half the force.

Force depends on acceleration. You need to figure out how quickly the bullet will go from 1000 m/s to 0 m/s or - some value m/s before you can know the force involved.

Roger, that works as well!

Using all the assumptions I used in #3:

find acceleration from:

v² = 2 x a x s

where v = velocity, a = acceleration, s = distance.

so a = v² ÷ (2 x s)

= 10³ x 10³ ÷ (2 x 20 x 10^-3) = 0.25 x 10⁸ ms^-2

m = mass of round (from #3) = 14 x 10^-3kg

f = m x a = 14 x 10^-3 x 0.25 x 10⁸ = 3.5 x 10⁵ N

I think that the computations were correct with respect to the IMPACT point but not correct with respect to the request: load on structure!

Above model shows why. The bullet will meet first the plate which has an equivalent mass and stiffness depending on the impact position with respect to the connections between plate and structure.

The structure itself will be loaded and parts will move under the impact (although this one will already be smoothed by the first dynamic filter which is the plate) so that an equivalent mass for the structure and an associated stiffness have to be added. The maximal load on the structure will be less than the one resulting from the impact since energy is used in several directions as: local plastic deformation of bullet and plate, acceleration of plate and deformation (elastic)of zones far enough from impact, acceleration of elastically deforming parts of structure and last but not least own elastic deformation of structure.

Even if we consider that the plate is theoretically stiff this cannot be accepted for the structure since there connecting points which will deform and this leads to a movement (even minute but not nil) of the plate and thus an energy use for it. The peak will be reduced.

The model used for the first computing was too simplified for the structure load definition but good for the impact "point".

It is justified to make a series of designs for the structure since its own elasticity will influence the maximal load and thus the dimensions and the elasticity and correlated capability to absorb, without local permanent deformation, shock energy. All will depend on the admissible elastic deformation of the structure when under impacts.

If support for the estimations is needed I can give it.

nick name, I quite agree. My calcs only apply to the force of the round impacting the plate.

Without knowing a lot more about the properties and dimensions of the plate material, the geometry of the support structure, the method of coupling (mounting) between the plates and the supports and the point of impact of the round, (and doing some horrible sums) that's about as far as we can go in estimating any of the forces.

But it's a start, and shouldn't be too far off (if the actual stopping distance can be calculated) as a worst-case value for an impact directly in line with one of the support attachment points.

John.

I think that was what I said in #4. But since this is a design case, I suggested applying a load to the frame only, calculating the deflection under that load and calculating the energy absorbed, then adjusting the assumed loads and consequent deflections until it produced the correct energy absorption. If that load was too high, then the other factors would have to be accounted for.

Clearly, if the bullet hits the plate directly at the frame, that provides the worst case for the frame.

As I see it, the effort is to find a worst case estimate that is not too punitive on the material cost. The extra time taken to justify a smaller load may not be offset by material savings, and should not be attempted unless necessary. The frame will have some minimum practical size, and this size may well be able to absorb this load without overstress.

What you say is correct as long as system dynamics are not considered.

The combination of the mass and the stiffness leads to a natural frequency. If the pulse generated by the impact has as fundamental a frequency higher than the above mentioned one the amplitude will be reduced as it is in a mass-spring system in the over critical domain. This is what I wanted to stress !

In general bullet impacts have a needle aspect with a very high first harmonic (if you consider the Fourier series) now if the plate+spring+structure has a low one the load reduction can be important and the more elastic the supports are the lower the natural frequency and the higher the reduction of the amplitude and thus of the load.

Of course, as usual, I may be wrong since I do not detain the truth, isnt it?

Estimated maximal force on the structure

It can be assumed with not a too big error that the force is as a half-sinus pulse with a time length of 2 ms (?? Estimated, may be even less). This corresponds to a basic frequency of 250 Hz

(2p =4ms).

The Fourier series for the half-sinus pulse (x= 0…p/2) will be:

F (x) = 2/p*(1-2*S[(cos (2*n*x))/(4*n^2-1)])

This means that the amplitudes of the frequencies will be as follows:

a (n)=1/(4*n^2-1)

Let us assume that the basic frequency of the plate + structure is 150 Hz or less. For a plate with a mass of 10 kg the supporting stiffness will be:

f*[Hz]=0.5*(C/m) ^0.5/p > C=4* p^2*f*^2/m = 8.9 E4 N/m , C ≈90 N/mm.

w*[rad/s]= *(C/m) ^0.5/(2* p) m = w/w* d = d /w*

w* critical pulsation

w = imposed pulsation

d = relative damping

d= mechanical damping

The dynamic damping coefficient will be: V= ((1-µ^2) ^2 +(2*d*m)^2)^0.5

For above frequencies the factors will be for a d≈0.02 as for steel structures and d =2.12 E-5

One sees that the dynamic reduction is, for the conditions above, ≈90%!!!

A more elastic support would reduce even more the load on the structure.

If the pulse is shorter in time (which I suppose) then the reduction will also be more than 90%!

I do not want to be misunderstood, I only want to underline that estimations can be dangerous in both directions either too high or too low. In both cases it is better to estimate with a good model as near as possible to reality but of course still economical.

I know that most of you are eager to share your knowledge, but my concern is that it may be blinding you from questioning why a civilian would be needing a bullet proof facade. If he was military he would not be asking this forum.

I would not touch this with 10 foot pole. Remember Sh#t roles downhill.

Thank you guys for all the inputs.

I am a Civil Engeneer by profession & this is the first time I encounter this. Forces directly applied to structures is easy to deal with but this one is giving me headaches.

Btw, the typical panel is 2.5m W x 1.5m H and we will cover an entire wall of 30m W x 6m H. The facade is to be installed 1.8m away from the wall that is why it needs steel framing behind.

Not to question some of the learned answers that have been given, as a Design Engineer, I look for the simplest, most easily defended, conservative but not overly conservative, answer that I can find.

I cannot question the numbers in post #3 but the result does not "feel" credible, so I took another look.

First, I checked the inputs:

From Wiki

I get two bullets, their weights, 8,0g and 10g; velocities, 710 and 641m/s; Energies, 2010J and 2059J. These values are typical compared with other similar bullets (see post #2). I did find much higher values at one site, but he has a disclaimer saying the information consists of unverified bits and pieces he has picked up.

Second. The bullet.

It is 7.92mm diameter, it consists of a copper plated hardened steel jacket with a steel core, with a small amount of lead is between the copper and the jacket. Assuming that it hits an immovable wall, at first touch the force will be zero and then it will build to the yield strength of the bullet. Assume that it compacts to twice its diameter by a quarter of its length.

I have to convert, then, I'll convert back.

Bullet diameter = 7.92mm = 0.312ins. Cross section area = 0.0764sqin. Say the yield stress is 40ksi.

Initial Yield load is 0.0764x40000=3054lbs. Final yield load is 3054x2²=12216lbs

3054lbs = 1385kg =13585N

12216lbs = 5540kg = 54340N

The shortest bullet appears to be about 23mm long; allowing for the point on the bullet, assume the CG is 12mm back from the point. If the bullet compacts to one quarter of it's length then the CG moves three fourths of 12mm = 9mm

Energy absorbed. ((54340x13585)^1/2 )0.09=2445J (Using the Geometric mean)

This is only about a third of the energy specified in the question but 20% more than the typical, real life, muzzle energy. That energy is based on the bullet hitting an immovable object, but the force can only be obtained if the structure resists the movement.

Third. The frame.

The frame will have minimum practical sized members that may be able to withstand the maximum bullet yield force, as calculated above, without significant penalty, but if not, we can use the spring rate of the structure. Edwin can place a unit load on the worst point of his frame and calculate the deflection under that unit load to obtain a spring rate. The energy imparted to the structure equals the half the total load multiplied by the deflection under that load.

If the deflection under unit load is "d", and the total force is "F", then the work done is dF²

So F= (E/d)^0.5 and F can then be used to check the structure.

Note: the bullet only crushes if the frame provides sufficient resistance.

All computations were based on a single shot bullet ! And I do not contest them for the IMPACT point but I very much doubt of their validity for the rest of the structure as I demonstrated above.

But what will be the force if an MP is used ?

In this case we come again to the definition of the force as F= d(m*v)/dt.

Assuming that the bullets have all the same mass and velocity and their rate is n/sec we obtain for the force the equation:

F= n*m1*v with m1 = bullet mass [kg]; v=velocity at impact [m/s] and n=impacts/second [1/s] the force will be in Newton.

With your values and n= 100 s^-1 The results could be:

F= 100*(8...10 E-3)*(641...710) = 513...710 N

Comments aree welcome since I have the feeling some will strongly react, or do you accept those assumptions?

I was only considering single shot. A machine gun does not keep hitting the same spot, it travels around considerably when trying to hit a human target.

To be honest, I think that considering the difference in the masses, the bullet versus the wall, the multiple hits you propose, hitting in different places, will not drive the solution.

That may be a reasonable estimate of the average force - but isn't the short-term force (during each impact) more important here?

If you reduced the impact rate to one per second, you'd get ≈ 6N by your calculations. May as well be throwing eggs.