how many sheet metals does it take to stop a 50 BMG ?

Published on January 8, 2017
Duration: 2:10

This video details an experimental test by Edwin Sarkissian to determine how many sheets of 18-gauge steel are required to stop a .50 BMG API round fired from a Serbu BFG-50A rifle. The experiment, conducted in a desert shooting range, involved firing a single API round into a stack of 22 metal sheets and analyzing the penetration depth. The results indicate that 16 sheets of 18-gauge steel were needed to halt the projectile.

Quick Summary

In an experimental ballistics test, 16 pieces of 18-gauge sheet metal were penetrated by a single .50 BMG Armor Piercing Incendiary (API) round fired from a Serbu BFG-50A rifle. This determined the stopping power required to halt such a powerful projectile.

Chapters

  1. 00:00Experiment Setup
  2. 00:22Firearm and Ammo Identification
  3. 00:38The Shot Fired
  4. 00:41Results Analysis
  5. 01:41Conclusion

Frequently Asked Questions

What rifle was used to test the sheet metal penetration?

The rifle used in the experiment was a Serbu BFG-50A, a gas-operated semi-automatic chambered in .50 BMG. It features a 26-inch barrel and a 10-round capacity, weighing approximately 32 lbs.

What type of ammunition was fired at the sheet metal?

A .50 caliber Armor Piercing Incendiary (API) round was fired. This type of ammunition is designed to penetrate armor and cause ignition upon impact, making it particularly destructive.

How many pieces of sheet metal were penetrated by the .50 BMG round?

The .50 BMG API round penetrated 16 pieces of 18-gauge sheet metal. Six pieces of the metal stack remained intact after the shot was fired.

What is the significance of 18-gauge sheet metal in this test?

18-gauge sheet metal has a thickness of approximately 0.0478 inches. The test determined that a total thickness of roughly 0.76 inches of this material was required to stop the .50 BMG API round.

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