Dave,
Those are good points for a consumer goods production company. US Air Force and RAF budgets are likely big enough to allow for that product, but not budgets of professors, industrial managers or bank tellers.
Okay the rest of this is going to be long and it pertains to this and other questions. Sorry about that.
The UK method of calculating bullet strength is its
momentum (mass x velocity). That makes no sense to me because a human on a bicycle riding at 20 km/hr would be considered more dangerous than a 9 mm pistol round. (For the sake of example, the human and bike might be 75 kg x 20,000/3600 m/sec = 417 kg-m/sec; the 9 mm bullet might be 8/1000 kg x 400 m/sec = 320 kg-m/sec. Therefore the bicyclist is 30% deadlier than a 9 mm pistol?
Newton's actual definition is that
momentum (mv) =
impulse (Force x time), so the faster a bullet transfers its momentum, the more force is transferred into a collision.
By the same principle,
Energy and
Work are the same quantities in Newtonian physics.
Energy is the equation I used earlier (Newtonian E = 1/2 mv^2, from which Einstein eventually derived E = mc^2 because light is the ultimate speed and there is no dissipation of the energy at that velocity, thus eliminating the factor of 1/2).
Now
Work is a definition of energy given in Newton's physics as Force x time, so the force of
Impulse you saw above must be totally absorbed in a limited distance and that distance is the time to stop the impact velocity. To find the distance, we can put the maximum allowed distance as the upper boundary condition and 0 as the lower condition. In Germany the upper boundary is 22 mm deep. In the USA it is 44 mm. (Because we are fatter? Or American women have bigger boobs? I don't know why.)
But the end result is that we know the conglomerate structure must absorb the energy of the bullet in 0.0044 meters maximum travel or the armor is a failure. This means the very minimum
strength to break of the structure is defined as [(bullet mass x bullet velocity^2)/(2 x 0.0044 m)]. But we have another problem to consider. The foregoing model only takes into account
Work to break along the perpendicular to the plane of the armor. We also need to consider
breaking strain, and that is a function of the speed with which a force moves along a material, and that is again a function of the momentum (above), which also defines inertia.
Breaking strain is defined as (length at break - original length)/original length. So let's say that a bullet hits a Kevlar filament yarn. A good representative breaking strain for this Kevlar is .034, so it has to stretch to 1.034 times its original length before it breaks. This becomes a function of how fast it will stretch that far to determine if it breaks before the .0044m limit is exceeded. So if that 9mm bullet (above) hits our Kevlar thread, what happens?
1) The filament yarn extends into an angular shape defined by the velocity of the bullet
2) The mass of the yarn can be found for any given length thanks to a numbering system used by industry called Denier. Denier defines the number of grams of mass of a yarn per 9000 meters. (No one I have ever asked knows why 9000 is used. A famous English professor at Manchester proposed a system called Tex in which grams per 1000 meters would be used. It's very logical, so industry ignores it.) A typical Denier used in Kevlar for armor is 750, which equates to 83.3 Tex.
3) Stress to break is usually expressed in grams force per denier or tex, and it is nearly a linear function (all initial modulus) for this fiber. For Kevlar 129 it is 2.38 Newtons/tex, so it will take 2.38 N/Tex x 83.3 Tex = 198.3 kg-m/sec^2 of force to break one yarn.
4) We now can find how much energy it will take to break this yarn. The yarn will break when it has been extended to 1.034 times its original length, but it is the bullet itself that is extending it. So that we don't have to differentiate the equation of the surface of the bullet ogive, is it okay just to say that the diameter of this pointed bullet lies between 0 and 9 mm? (This has to be getting boring for most people and I'm too nerdy to know when to stop.) The impacted yarn length therefore becomes 4.5 mm (.0045 meters) and it will break at a length of 0.00463 m.
5) Now we go back to the equation that Energy is Force x Length. So according to Newton, we can can take 198.3 Newtons (the unit, not the guy) x .00463 meters, and find that the yarn breaks under 0.92 Joules of energy. That 9 mm bullet's energy is defined with the velocity dependent equation E = 1/2 (mV^2), and it has 640 Joules of energy. It will require 640/0.92 = 695.7 Kevlar threads to stop it, disregarding friction, drag effects and thermal energy generation in the lead core to turn it liquid.
6) Now we see the supreme importance of cross-sectional area of the bullet in all this, because if you hold the energy quantity constant, the smaller the bullet, the shorter the breaking length of the material and the more material you need to stop the bullet!
7) The final consideration in a woven material is the number of yarns in each plane of fabric layers. For the fabric under consideration here, a very good representative number is 11 yarns/cm in a square sett (Thread count = 11 x 11). This makes 1100/meter in each direction, or 2200/m^2 (Yes, you add rather than multiply. Only 1100 will fit in each direction, so you won't get 1,210,000 crammed in there.)
8) Our bullet diameter of .0045 m is going to encounter 4.95 threads in each of the woven directions, so it has to break 9.9 per layer. Each layer will then be taking 9.9 x 0.92 Joules of energy from it or 9.1 total Joules.
9) If no other effects were involved, you would need 70 layers to stop this bullet. Obviously that does not happen with a deformable projectile like an FMJ or the even more yielding hollow point types. But with a solid, non-deforming projectile, it is an accurate representation.
If anyone wants to see the effect of energy per unit area in the equation, you can go back and substitute your favorite bullet's mass, velocity and diameter into the right places and give it a try. A good one to start with is a .45 caliber ACP FMJ (mass=0.015 kg, velocity=250 m/sec, diameter=0.0115 m; quoted Joules = 477).
Gwynedd, thanks for the website. Looks fascinating & I've passed it on for comment. My friend most likely won't look at it because 1) it's too expensive and as a producer he wants to keep costs down, 2) cannot make it large enough for car windscreens, 3) probably cannot be made easily, 4) being more durable, he loses maintainance and rebuild work! He told me that the existing stuff has a usable life of about 1 year as the windscreens delaminate from the sunlight and the cars are worn out from the weight of all the additional armour. Good for a producer.
I was also curious about your energy figures for the different rounds and was going to look at how penetration figures are calculated. I don't normally worry about thing like that as I am a precision shooter and not a hunter. But for Practical Pistol they measure the power factor using the velocity x the weight (but not velocity squared!). Our UK ranges have just been limited by muzzle velocity using momentum (mass x velocity) which acts against blackpowder. According to our Government a .577 black powder rifle is more powerful than a modern 5.56 or 7.62 round!
My penetration queries were started last time I was testing the bullet-resistant glass. My friend had also filled some beakers with his polymer compound, a different beaker for each type he was testing. Without the glass in front we didn't know how far any bullet would penetrate but the best polymer is the one with the least penetration. I was using .357 magnum through a carbine and they all showed less than an inch penetration. I hoped to try more pointy bullets but as we have a poor selection here (if you believe that UK is bad you should try Belfast!) and tried some 124g 9mm heads as they are the pointiest I can find. I have now discovered that 0.001" difference in bore diameter has a great difference on accuracy!
