Calculate Force of Bullet Impact on Body Armor

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SUMMARY

The calculation of bullet impact force on body armor involves using the kinetic energy formula, specifically Ek = 1/2 mv². For a bullet traveling at 1500 m/s with a mass of 1g, the energy is calculated as 1125 kJ. To determine the force per unit area, divide the total energy by the area of impact, which can be approximated as 9π for a 3-inch radius. Different materials, such as titanium, ceramics, and Kevlar, distribute impact forces differently, complicating the analysis and potentially requiring experimental validation.

PREREQUISITES
  • Kinetic energy formula (Ek = 1/2 mv²)
  • Understanding of unit conversions (grams to kilograms)
  • Basic geometry for area calculation (circle area formula)
  • Material properties of body armor (ceramics, steel, Kevlar)
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  • Research the mechanical properties of body armor materials
  • Learn about the physics of impact forces and energy transfer
  • Explore experimental methods for testing body armor effectiveness
  • Study advanced topics in material science related to ballistic protection
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Students working on physics projects, engineers in ballistic protection design, and researchers in material science focused on impact resistance.

petitericeball
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I was wondering how to do the math for a bullet hitting some body armor.

Would you just divide the original force of impact by the area of a the circle surrounding the point at let's say 3 inches?

So.. let's say I gave a bullet moving 1500 m/s at 1g. so it would be .5(1)(1500^2)=1125kJ joules. (I know that these aren't the right speed and mass) So, at 3" you would divide
1125/9pi?

This is a high school level project, so I'm hoping the math isn't very complicated.

Another problem is the use of different materials for the body armor, whether ceramics, steel or something else..
 
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You nailed the problem with your last statement. Every material reacts differently. Trauma plates such as titanium or ceramics essentially just 'widen the load' to their own surface area. Kevlar, on the other hand, can transfer a single .30 calibre impact over an area of more than a dozen square inches. I'm not sure that there is even a formula available for that case. You might be obliged to experiment a bit.
 
"So.. let's say I gave a bullet moving 1500 m/s at 1g. so it would be .5(1)(1500^2)=1125kJ joules"

Be sure when plugging in variables to the kinetic energy equation you have the right units. A joule is defined as a Kg(m^2/s^2). so make sure when you use the formula 1/2mv^2=Ek that mass is in kilograms. otherwise your final answer will be off by 10^3.
 

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