This is a really silly question. I was at the range today and rented a .375mag loaded with .38spl and noticed that the .38spl round is a good deal bigger (case is longer) than the 9mm (or the .45ACP for that matter...).
Why is the .38spl's case so much longer/bigger than the 9mm's and yet be "underpowered" compared to the 9mm?
Is it because it's just a .375mag case+bullet with "less powder" in it? (or padding)
Were the revolver cartridges developed earlier, and the size of the chamber standardized, and when autos were invented, they needed a smaller case to feed reliably? Were the powders in the earlier revolver days less "fast burning" than the auto powders?
I've never seen the higher-power auto rounds (e.g. 10mm, .375SIG, .45Super or .400CorBon) in person, but are they pretty much the same size, just shorter and wider than the .375mag/.44mag cases, leading to the same amount of powder? Or are there actually different kinds of powder?
Thanks for your help!
-Jon
Why is the .38spl's case so much longer/bigger than the 9mm's and yet be "underpowered" compared to the 9mm?
Is it because it's just a .375mag case+bullet with "less powder" in it? (or padding)
Were the revolver cartridges developed earlier, and the size of the chamber standardized, and when autos were invented, they needed a smaller case to feed reliably? Were the powders in the earlier revolver days less "fast burning" than the auto powders?
I've never seen the higher-power auto rounds (e.g. 10mm, .375SIG, .45Super or .400CorBon) in person, but are they pretty much the same size, just shorter and wider than the .375mag/.44mag cases, leading to the same amount of powder? Or are there actually different kinds of powder?
Thanks for your help!
-Jon