Hello Mighty Ones! I know from an FBI study that a bullet shot at a 90 degree angle into a metal plate, the splatter from the bullet goes out at a 20 degree angle. The question is... is there a rule or something that says if I angle that 90 degrees to angle down X degrees will the deflection result in a major loss in energy that the plate has to absorb. Now I'm trying to sound all smart and we both know better I'm just trying to give a good explanation :) So bottom line is. Were making a steel target to shoot with the front of the target angled to help not only in deflection of the bullet but also to save wear on the target itself. Is there a diminishing return on the angle once it gets to a certain point on how much it will help in energy absorption and what is that angle? Thanks, FHKBaker
I once shot a 30-30 carbine at 90 degrees into a 1/4" thick steel plate, and it made a ~1/2" hole in it. Bob S
Whether the bullet is deflected or goes through the target depends on the mass and speed of the bullet, the yield pressure of the bullet and the yield pressure of the target. A jacketed bullet will penetrate the target better than a lead bullet. A 3000 fps bullet will deflect less than a 2000 fps bullet. But I am not sure what you are trying to do. The target ultimately has to absorb the bullet energy. Otherwise, what stops it? AM
I am trying to figure out a way to make the wear on the base of our targets better. Currently they are at 90 degrees so if someone misses the target and hits the base it takes the full force of the shot. There is no penetration involved with the type of metal were using I was just thinking that if we angled the bottom of the base away from the shooter it would help dissipate the energy into the ground rather than directly into the target if that makes sense. So overall thinking why not angle the base rather than keep it perpendicular to the ground. I know this is probably elementary for most of you. I am just trying to build the best system I can for my shooters.
Here is what a friend of mine just sent now I just need to decipher it :) The amount of energy the plate has to absorb depends only on the initial and final speeds of the bullet. If the bullet completely stops, then the plate has to absorb 1/2 * mass * speed ^2 (speed squared, or speed * speed). Now, if the bullet comes in at a glancing blow, at some angle other than 90, then the bullet will leave with some speed, so the amount of energy the plate absorbs is smaller. The question is, what is the final speed of the bullet? That's not a simple question to answer. One assumption, or approximation, is that the bullet loses all the speed perpendicular to the plate, but keeps all the speed parallel to the plate. In that case, the speed after the impact is v*cos(angle), where v is the initial speed. So, the amount of energy the plate must absorb is given by: 1/2 * mass * initial speed^2 *( 1 - cos^2(angle) ) When the angle is 90 degrees, the amount of energy absorbed is all the energy in the bullet. When the angle is 45 degrees, the amount of energy absorbed is half the initial energy of the bullet. This assumes that the plate absorbs all the momentum perpendicular to the plate, but the bullet keeps all the momentum parallel to the plate. I don't know if that's a good assumption or not. You can plot 1 - cos^2(theta) to see what the curve looks like as a function of the angle theta.
It is energy that you have to worry about, not momentum. If you reduce the speed by one half in the target (ie the bullet hits the ground moving at 1/2 its original speed) you still leave 3/4 of the energy in the target. You will have to cover the base with something that can absorb bullet energy. AM