Help me understand some fundamentals of bullet/cartridge/ballistic stuff...

[rabid_rob]

While I've been a shooter for 30 years, I never really understood some of the basics of bullets/cartridges and the energy part of shooting...

(1) I recently fired a .45 Long Colt revolver for the first time. I was amazed at how little recoil there was, given the size of the cartridge. It had about as much recoil as a moderate .38 special round. Why? Being a "bigger" caliber, I just knew the .45 Long Colt would have more force/power/kick. Given the larger cartridge size (volume), it seems like you could get more powder inside, and thus produce more force. What am I missing here? Does the larger, heavier bullet make that much difference?

(2) Why do longer barrel guns provide more FPS/muzzle velocity for the same round than a shorter barrel? Does the longer barrel mean there is more time/space to fully burn the powder and thus produce more power?

(3) In a revolver, how much power is lost due to gas escaping in the cylinder/barrel gap?

(4) How about a semi-auto? Some of the power is consumed to operate the slide, and is some lost as the empty clears the barrel, providing an escape for the gas?

(5) It would seem that a bolt-action or other type of fully locked breech, where the cartridge remains in the chamber for the full during of the shot, is the most efficient transfer of combustion gas to bullet...?

 

[Larry Ashcraft]

Welcome to THR!

1. The .45 Colt cartridge was originally a black powder cartridge, requiring all that room for the black powder. The .45 ACP equals its power with a much shorter case with smokeless powder. The dimensions of the .45 Colt haven't changed, and with a smokeless powder charge, most of the case is filled with air.

2. You have it right.

3 and 4, I'll leave to someone more expert than I.

5. I would guess you have that right also, but the difference between a bolt gun and a lever or semi is probably negligent.

 

[GRIZ22]

1. recoil is a function of the weight of the total ejected mass (bullet and powder) with a factor for velocity vs the weight of the gun. The 45 you shot was probably heavier than the 38 and bothwere std loads. The case of the 45 Colt was designed for black powder and that's why it's so long. Smokeless powder loads usuallt come no where near filling the case.

2. Basically, yes.

3. Not much. Compare velocities of a 45 ACP fired in a revolver and a semi auto.

4. The bullet is out of the barrel before the slide moves.

5. Theoritically yes but most semi autos tap the gas off near the muzzle and the loss is negligible.

 

[rockstar.esq]

"Filling the case" has NOTHING to do with recoil. Bullseye powder has an enviable reputation as a pistol cartridge powder, it also has a reputation for being easy to double charge a case with! Lots and lots of blown up guns to prove it. The .45 Long Colt has a relatively low SAMMI pressure spec which goes considerably farther towards explaining why it doesn't recoil so hard. A cursory search on the Ruger Super Redhawks will likely bring up some "magnum" 45LC loads safe only in .454 Casull chambered guns. These loads have more kinetic energy than the most rompinist stompinist .44 Magnum (SAMMI compliant) loads. The whole "gas efficiency" thing seems like a pointless venture because if you consider a cartridge like the .38 special which can have a max load of Bullseye that literally constitutes a grain or two, ALL of those flakes will burn completely with nearly no barrel length. The barrel gap measurement and chamber / throat alignment have WAY more to do with gas escapement than overall design. I've seen plenty of semiauto cases that're scorched along one side because of gas blow by. It all comes down to tolerances.

Something that might be of interest to shooters is that the 12" barreled .357 Mag Taurus revolvers have LOWER velocities than their 6" companions because the extra barrel length has actually exceeded the required "burn length" and the rifling is creating drag!

 

[geekWithA.45]

"power factor" is a reasonable way to compare the relative "power" of a cartridge. It directly correlates with "energy in foot pounds".

The formula is (muzzleVelocity *weightOfProjectileInGrains)/1000.

So, a .308 load with a 150 grain bullet and a muzzleV of 2820 yields a pf of 423.

.223, on the other hand, rips a teensy little 55 grain bullet out at 3240 fps. This yields a power factor of 178.

.45acp: 890 fps, 230 grains = fp 204
.45 lc: 820 fps, 220 grains = fp 188


So, as a sidebar, you've notices that the case on the .45 colt is much bigger than the .45 acp, and yet the acp comes out a little better ballistically. The reason is that the case for .45 colt was developed for black powder, which is less energetic than more modern smokeless propellants.

Similarly, the case for .308 was developed for more modern propellants than its predecessor, the 30-06.

 

[GRIZ22]

"Filling the case" has NOTHING to do with recoil.

Well yes...and no. The powder is part of the total ejected mass so it's weight does figure in a little bit. the reason black powder 45 Colt rounds don't recoil that bad is the weight of the powder charge is offset by the lower velocity.

Rabid Rob I think you're using the word power to talk about gas pressure and the energy the bullet produces. As geek says the power factor formula he gives is used to compare diffrent rounds, calibers, etd

Muzzle energy is figured by:

velocity x velocity x bullet weight in grains divided by 450400