How can I calculate the depth of bullet penetration in different materials?

In summary: So if the bullet has a moment of inertia of 0.5 kg m^2, then the slab's moment of inertia will be 0.25 kg m^2. So the total deformation of the slab will be \tfrac{1}{2}m[4000,0,0]+0.25 kg m^2=1250 kg. You can also use this equation to find the amount of energy that is released when the bullet hits something.
  • #1
nbjsargent
2
0
Hey guys;

I am new to the forums. I am an effects artist who's working a Rigid Body Dynamics project and I may be able to use some of you guys' expertise while I'm building these simulations.


As I stated above, I'm doing R&D for a deformation system I'm working on. What I want to do is to take the speed and mechanical properties of a projectile and calculate the displacement (wether permanent or elastic) of an object upon contact with this projectile. I"m basically new to building rigid body solvers from scratch, but essentially what I want to do is to be able to define what kind of material an object is (concrete, rubber, copper, brass, Glass etc.) and have it react appropriately to the projectile. (Wether it stretches, tears or fractures)

If you guys have any books/material that you can point me to, basically concepts that will be essential for me to learn, I would be grateful.

In addition to this, I have a question to start with. If a bullet is penetrating a material at a defined velocity and from a defined distance, what mechanical properties of the material decides how deep the bullet penetrates before stopping. (if it does indeed stop and not go straight through)

Say we define all the variables involved, is there a formula to calculate this 'drag force' or whatever force it is that controls how quick the bullet comes to a stop?
 
Physics news on Phys.org
  • #2
Someone may want to correct me on this point, but if you're solving for things like deformation and bullet penetration, you're actually working with elastic bodies rather than rigid bodies.
 
  • #3
Rigid body dynamics shouldn't be too hard to program, but a bullet impacting a body, penetrating, and stopping, and maybe deforming the body, kicking out pieces, etc, that will be tough to model using the actual physics equations. You probably want to get some simplified equations that don't follow the physics exactly. As far as a bullet penetrating and stopping, I would think that, for a pointed bullet, a good start would be that the drag force is proportional to the velocity and the area of contact between the bullet and the object. Once you have the drag force, and the mass of the bullet, you can calculate how it slows and stops.
 
  • #4
Thanks guys. Yeah that is basically my first step. Given a certain material, I just need to calculate the level of penetration for the bullet, or I guess the rate of increase of the drag force upon contact.

To start defining some variables, say the bullet is traveling at 4000 feet per second and it's fired from 15 ft away and hit a concrete slab. Based on my research the drag force acting upon that bullet is about 2 to 3 times the gravitational force? How do I get the depth of penetration? And I would also need to get the radius of the distribution of energy from the center of impact (or energy falloff from center of impact) so I can use that to solve for fractures later...
 
  • #5
You need to add some zeros to that drag force. The forces involved when a bullet strikes something like concrete are high enough to liquify the bullet. The problem you want to solve is going to require some advanced finite-element analysis. You might be able to come up with some empirical formulas based on observed data. But finding a rigid mathematical solution is going to be extremely difficult.
 
  • #6
[tex]x=x_0+v_0 t+\tfrac{1}{2}a t^2[/tex] and [tex]v=v_0+a t[/tex] x is the position at time t, [itex]x_0[/itex] is the position at time 0, v is the position at time t, [itex]v_0[/itex] is the velocity at time 0, a is the acceleration (force/mass). Let's say time zero is when the bullet first touches the slab, so [itex]x_0[/itex]=0, a=3g = -96 ft/sec^2, [itex]v_0[/itex]=4000 ft/sec. When the bullet stops, v=0, and you can use the second equation to get t, then substitute into the first equation to get x when the bullet stops.

The energy of the bullet at impact is [itex]\tfrac{1}{2}mv_0^2[/itex] but I'm not sure how to translate that into deformation of the slab. I agree with Ryoko, solving the equations for a real solution will require some serious computational power. You probably need to semi-fake it here, use some simple model that gives reasonable results without a lot of calculation. You probably want to use the fact that the momentum of the bullet [itex]\tfrac{1}{2}m[v_0,0,0][/itex] is conserved too, where now you have to use the vector momentum [itex][v_0,0,0][/itex]. This will also be the momentum of the slab and all the pieces after the bullet stops.
 

1. What is Rigid Body Dynamics?

Rigid Body Dynamics is a branch of physics that studies the motion and behavior of solid objects under the influence of external forces.

2. What are some real-world applications of Rigid Body Dynamics?

Rigid Body Dynamics has many practical applications, including spacecraft design, robotics, biomechanics, and video game physics.

3. How is Rigid Body Dynamics related to other fields of physics?

Rigid Body Dynamics is closely related to other branches of physics, such as classical mechanics, kinematics, and dynamics. It also has connections to engineering disciplines like structural analysis and control systems.

4. What types of problems can Rigid Body Dynamics help solve?

Rigid Body Dynamics can help solve a wide range of problems, such as predicting the motion of a falling object, analyzing the stability of a structure, and designing control systems for robotic arms.

5. What tools and techniques are used in Rigid Body Dynamics research and development?

Rigid Body Dynamics researchers and developers use a variety of tools and techniques, including computer simulations, mathematical models, experimental testing, and advanced software programs.

Similar threads

Replies
19
Views
3K
  • General Engineering
Replies
1
Views
1K
  • Mechanical Engineering
Replies
3
Views
2K
Replies
9
Views
2K
  • Mechanics
Replies
1
Views
2K
Replies
2
Views
12K
Replies
10
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
1K
  • General Engineering
Replies
4
Views
6K
Back
Top