I recently saw a video of a guy shooting a bolt action rifle over a sand bag. The rifle barrel split into and the shooter suffered a hand injury. There was much discussion over what went wrong, with most citing either a squib load or an overcharged load. My question is this. Wouldn't a squib load tend to split the barrel while an overcharged load tend to damage the bolt/chamber area? What damage would a case head separation tend to do?
Overcharged load vs. squib load gun damage question.
- Thread starter AL45
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FrankenMauser
New member
If the video you're referring to is one that I saw a couple hours ago, it was a Remington 700 Ultimate Muzzleloader.
Double-loaded (bore obstruction), according to the hints dropped by the person that recorded and witnessed the Kaboom.
There are many factors and exceptions, but obstructions followed by full charges generally split barrels. Overcharges might also split barrels, but they generally just do very bad things by breaking bolts and injecting the shooter with molten metal.
Double-loaded (bore obstruction), according to the hints dropped by the person that recorded and witnessed the Kaboom.
There are many factors and exceptions, but obstructions followed by full charges generally split barrels. Overcharges might also split barrels, but they generally just do very bad things by breaking bolts and injecting the shooter with molten metal.
A double load in a muzzle loader usually isn't a disaster. A double CHARGE might be a different story.
I remember seeing an article in the Rifleman ages ago, they wound up having to do multiple loads, (something like 7, or maybe 11) before the barrel split.
Some time ages ago, I was told,
"You aren't a true muzzleloading fanatic until...
You have loaded with no powder
You have loaded with no ball
You have double loaded
and
you have shot your ramrod, at least twice"...
A burst barrel usually means an obstruction, like a squib bullet stuck in the bore, and then another round fired, which gets a running start before hitting the obstruction.
The classic shotgun 12/20 burst is one example/
I remember seeing an article in the Rifleman ages ago, they wound up having to do multiple loads, (something like 7, or maybe 11) before the barrel split.
Some time ages ago, I was told,
"You aren't a true muzzleloading fanatic until...
You have loaded with no powder
You have loaded with no ball
You have double loaded
and
you have shot your ramrod, at least twice"...
A burst barrel usually means an obstruction, like a squib bullet stuck in the bore, and then another round fired, which gets a running start before hitting the obstruction.
The classic shotgun 12/20 burst is one example/
Remember most of what we know/believe about muzzle loaders has to do with true black powder and how it behaves.
Some of the non-traditional muzzle loader manufacturers have blessed the use of smokeless powder.
A double charge of ffg may be far less spectacular than a double charge of 4198.(I don't know what smokeless gets used in muzzle loaders)
Some of the non-traditional muzzle loader manufacturers have blessed the use of smokeless powder.
A double charge of ffg may be far less spectacular than a double charge of 4198.(I don't know what smokeless gets used in muzzle loaders)
If you take another look at the Remington 700 Ultimate blow-up, you will note that there is no white smoke from the shot. Since the owner's manual says to use black powder or Pyrodex, I assumed it was fired using smokeless powder, which it is not designed for. But who knows? Definitely an oopsie!
Jim Watson
New member
Hard to obstruct a muzzleloader barrel in front of a charge except by short starting the ball. A friend's shiftless brother borrowed his TC, shot it foul, got a ball halfway down and shot it out. Left a goose egg bulge on the octagon barrel. TC replaced it but friend said it was not as accurate as the original.
I am thinking smokeless.
I am thinking smokeless.
FrankenMauser
New member
It may be of note that the ramrod is not visible under the barrel, on the bench, or in the shooting rest in the video.
Jim Watson
New member
Hmm. Shooting the ramrod is not normally damaging to the gun. Of course smokeless behind the ramrod would be.
I am not sure this is pertinent except for the event I witnessed some years ago at a local range. A shooter fired his bolt action rifle and experienced, in his words, a heavier than usual recoil. He could not open the bolt, but had to pound it open with a short length of 2X4. He was shooting reloads and had several left. The one he fired was the first of that batch. He pulled the bullets of several and found that they were overloaded by about 10 gr powder. It doesn't really matter what the powder was. What I thought was interesting that it was a 10 gr overload in a .30 cal. rifle that did not apparently damage the rifle but only made it very difficult to operate.
turkeestalker
New member
Is shooting the ramrod out of a muzzle loader anything like firing a 22 mag bolt action with a laser bore sight still in the muzzle? 
From experience I can tell you that when the aluminum shaft of the bore sight won't beat out of the end of the barrel, all you have to do is cut two inches off the barrel and re-crown it... no worries?
From experience I can tell you that when the aluminum shaft of the bore sight won't beat out of the end of the barrel, all you have to do is cut two inches off the barrel and re-crown it... no worries?
No.
When a muzzloader "forgets" and shoots his ramrod, the rod is down the bore, in contact with the ball, so there is no "running start" for the ball (or bullet).
So rather than acting like a bore obstruction (where the bullet gets a running start BEFORE hitting the obstruction), the ramrod just acts like an extra heavy bullet and is shot out (normally) without damage to the gun.
There are Civil War (and other historical records) of combat with muzzle loaders where enemy troops were actually killed by ramrods "accidently" fired at them. Not a lot, but some...