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Posted: 3/30/2010 8:40:56 PM EDT
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If suppressors decrease the pressure behind the bullet at exit, why would a shooter experience blowback?
I would figure blowback would occur due to an increase in pressure causing a greater amount of gas to flow through the gasport in the barrel. Only thing I can think of is a decrease in pressure relates to an increase in velocity, which relates to an increase in massflow through the gasport. I feel like this understanding is "thin" at best, though, since suppressors also decrease velocity... Any clarification would be greatly appreciated. Thanks. ETA: http://www.thefirearmblog.com/blog/2009/06/09/suppressed-rifles-get-very-dirty/ From the link above it seems that blowback occurs because the gases hang in barrel longer due to a decrease in velocity and pressure, thus when the action opens they have a lower pressure to travel to. Not sure if this is correct. |
| Suppressors are designed to keep as much of the gas pressure inside the gun as possible. So it increases the back pressure in the gun which can blow more gas into the shooters face from the ejection port or any other areas where gas can leak out, such as the charging handle on the AR-15. And with suppressor technology getting smaller and more effective, this increases the back pressure in the gun a lot more than older suppressors that are larger volume and lower pressure. |
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