- cross-posted to:
- 196@lemmy.blahaj.zone
- cross-posted to:
- 196@lemmy.blahaj.zone
Shamelessly stolen from @SkyezOpen@lemmy.world
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Any explanation for those of us who are not gun enthusiasts?
So, a derringer is a small pistol, usually with two shots, made for close-range self-defense. Normally they use, well, pistol rounds, like a 9mm, which has like, 700 joules of energy or someshit like that. When you fire a lightweight gun, you definitely get some kickback from it, even with a pistol round.
A 45-70 is a big-ass rifle round with something like 4000 joules of energy behind it. You uh, you put that in a little derringer and pull the trigger, both you and your target are gonna feel it.
Above PugJesus talks about the energy of the round being very large. There’s more to it.
The derringer design lacks any technology to absorb and extend the impulse of recoil, most importantly the slide found on any modern semi automatic.
Not only is there extreme recoil, there’s also absolutely nothing to help the shooter deal with it.
And this tiny little grip looks about big enough for a finger a half.
i’m imagining someone pressing the trigger and the gun shoots backwards with the bullet just flopping out unceremoniously
Does the slide absorb any significant amount of energy?
The spring can’t be all that strong since they can be assembled by hand, and what does the slide weigh? (Granted the slide is being accelerated, so I assume that’s where the bulk of energy is dissipated, MV^2 and all).
What’s the math on this, say the dissipated energy in a semi auto VS revolver using the same round?
(Really I’m curious what the numbers are, as I’ve read this many times but have no idea what the comparison is like).
