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Not so Dumb question
- Thread starter Youradolt
- Start date
No. Not if it has all of its internal parts.
Best way to understand it is to get Jerry Kuhnhausen's book about Smith & Wesson's revolvers: http://www.midwayusa.com/eproductpage.exe/showproduct?saleitemid=314178
IIRC (and that's a big if) the nose of the hammer engages the rebound slide. When the trigger is pulled all the way to the rear, that moves the rebound slide out of the way, so once the sear disengages, the hammer can fall all the way. As long as the trigger is not pulled, the rebound slide prevents the hammer from falling far enough to reach the primer.
But I'm not a gunsmith and it's been a long time since I read the book and opened up a J-frame. So I may be misremembering. And I'm too lazy to go downstairs and start reading the book...
So long as the trigger is in the forward position, the hammer is positively blocked from forward movement at a point far short of pin protrusion through the pin hole in the recoil shield.
A lot of folks think this blocking function is performed by the "hammer block," but in actuality it is performed by the "rebound slide." The rebound slide fits under the leg in the lower part of the hammer and prevents forward rotation of the hammer (which requires that this leg move in a downward direction). The fit is pure compression; it cannot be defeated by pounding on the hammer.
Once you get your revolver, you can perform the following demonstration. And I'm only going to say this once: obviously the revolver should be unloaded throughout the demonstration.
Close the cylinder. Point revolver toward the ceiling and drop a fairly new pencil (with an eraser end down) down the barrel.
Pull the trigger. You will see the pencil jump. This demonstrates the pin hitting the eraser, exactly at it would hit a primer if the revolver had been loaded.
Now cock the revolver again and drop the pencil in again. This time, hit the trigger with a spoon or some other object. It will probably take a couple of tries, but eventually the hammer will go flying down. It you pay attention, you'll see that the pencil doesn't jump. it doesn't more at all.
It was never hit by the pin.
This design, perfected 112 years ago, not only protects against an UD caused by a blow to the hammer in the hammer down position, it also guards against an UD caused by sear jar from dropping a cocked revolver, an UD caused by "false cock" drop, and even a sear failure. Quite something when you think about it.
A lot of folks think this blocking function is performed by the "hammer block," but in actuality it is performed by the "rebound slide." The rebound slide fits under the leg in the lower part of the hammer and prevents forward rotation of the hammer (which requires that this leg move in a downward direction). The fit is pure compression; it cannot be defeated by pounding on the hammer.
Once you get your revolver, you can perform the following demonstration. And I'm only going to say this once: obviously the revolver should be unloaded throughout the demonstration.
Close the cylinder. Point revolver toward the ceiling and drop a fairly new pencil (with an eraser end down) down the barrel.
Pull the trigger. You will see the pencil jump. This demonstrates the pin hitting the eraser, exactly at it would hit a primer if the revolver had been loaded.
Now cock the revolver again and drop the pencil in again. This time, hit the trigger with a spoon or some other object. It will probably take a couple of tries, but eventually the hammer will go flying down. It you pay attention, you'll see that the pencil doesn't jump. it doesn't more at all.
It was never hit by the pin.
This design, perfected 112 years ago, not only protects against an UD caused by a blow to the hammer in the hammer down position, it also guards against an UD caused by sear jar from dropping a cocked revolver, an UD caused by "false cock" drop, and even a sear failure. Quite something when you think about it.
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