When the hammer falls, how many pounds of force does it take to fire a bullet?

[Agent-J]

I've always wondered..

 

[Jim Watson]

The primer sensitivity test used in the WW II era was as follows:

30 cal (.30-06)
A 4 oz ball dropped 15 inches on a firing pin in a test fixture must fire all rounds.
A 4 oz ball dropped 3 inches must not fire any.

Takes a little math to get the force or energy in scientific units, that is left as an exercise for the student.

 

[woad_yurt]

It depends upon the ammo in question. Some of my Com Bloc surplus needs quite a whack, much more than newly manufactured commercial stuff in the same caliber.

 

[The Bushmaster]

I really don't know...Put your little finger in there and pull the trigger and let us know how many decibels your scream is. That will give us an idea...

 

[Agent-J]

i close the slide on my 930 spx on my finger. the extractor punched a nice hole in the skin in addition to the crushing pain.

 

[1858rem]

32ft/sec per second acceleration rate....umm how fast is that after 15 inches? square whatever velocity that is and multiply by 175094oz in grains, then divide by 450240 an there ya got it

 

[The Bushmaster]

Yeh Agent-J...I bet it did...

 

[JimmerJammerMrK]

The primer sensitivity test used in the WW II era was as follows:

30 cal (.30-06)
A 4 oz ball dropped 15 inches on a firing pin in a test fixture must fire all rounds.
A 4 oz ball dropped 3 inches must not fire any.

Takes a little math to get the force or energy in scientific units, that is left as an exercise for the student.

Well, I took a Kinetics and Motion course last semester so I can help out with this.

First, realize that you can't talk about this in terms of force. Why? Because Force = mass*acceleration, and for both situations, mass and acceleration are the same, so the amount of force being applied is the same.

What we can talk about is energy, because that's different for each case. First let's convert our units to SI units so they're easier to work with. 4 oz = 0.1134 kg, 15 in = 0.381 m, and the acceleration of gravity is ~9.8 m/s^2.

The first equation we'll use is height = (initial velocity)(time) + (1/2)(acceleration)(time)^2. This equation is useful because our initial velocity is 0, since the ball is being held and dropped (presumably from someone's hand). Thus, the first term of this equation cancels out and we're left with h=(1/2)(a)(t^2). Plug in .381m for h, 9.8 m/s^2 for a, and solve for t. Doing this gives a value of 0.2788 s, which is the time required for the ball to impact the firing pin.

Now we need to figure out the final velocity of the ball at the instant before it hits the firing pin. For this we'll use velocity=(acceleration)(time)+(initial velocity). Again, the initial velocity term drops out and we're left with v=at. Plug in 9.8 m/s^2 for a and our value of 0.2788 s for t. This gives us a value of 2.73 m/s for our final velocity.

Finally we're getting somewhere. Using the equation Kinetic Energy=(1/2)(mass)(velocity^2), we can plug in our mass (0.1134 kg) and our final velocity (2.73 m/s). This gives us a value of 0.423 J, which is the amount of energy generated in the situation provided for us by Jim Watson.

 

[TestPilot]

I don't know the answer.

However, I don't think the answer can be had in a poundage. It's not about force, but more about impact impulse. Even 100 lb of force may not fire it if the force is applied over time.

 

[JimmerJammerMrK]

I don't know the answer.

However, I don't think the answer can be had in a poundage. It's not about force, but more about impact impulse. Even 100 lb of force may not fire it if the force is applied over time.

Read my post. It's about energy.

 

[KD5NRH]

Also don't forget that the area of the striking surface will have an effect on it; a narrow pin concentrates the impact on a small area, while the same impact spread over the entire surface of the primer may not do the job at all. For an example, see the Lee Loader; you can use a mallet to seat the primer with far more energy than some firing mechanisms, and it still doesn't set it off part of the time.

 

[JimmerJammerMrK]

Also don't forget that the area of the striking surface will have an effect on it; a narrow pin concentrates the impact on a small area, while the same impact spread over the entire surface of the primer may not do the job at all. For an example, see the Lee Loader; you can use a mallet to seat the primer with far more energy than some firing mechanisms, and it still doesn't set it off part of the time.

Right, but the theoretical calculations were done according to Jim Watson's scenario, so the number is very specific. This type of problem can't really be done for a general case, since the variables will vary tremendously.

 

[Jim Watson]

dropped 15 inches on a firing pin

The original cite in Hatcher's Notebook included the specification that the firing pin be of standard tip style. Probably standard to a 1903 Springfield or M1 Garand, considering the time frame.

Yes, a drop test is about kinetic energy.

By the way, the corresponding spec for .22 rf was
2 oz, 21 inches, all fire
2 oz, 2 inches, none fire
No explanation as to why both weight and distance were different from rim to center fire.

 

[LightningMan]

It also depends on the type of primers are used. As someone mentioned above, they are tested in a fixture where a weighted 4oz object with firing pin, is mounted on a slide rail, while the primer being tested is in a fixture inline to be struck by it when dropped. I remember reading a article in a gun rag magazine about primers and they are not created equal. Basicily they tested 4 different primers, Winchester, Remington, CCI, & Federal to find out which were the easier to detonate. Each primer was tested by raising the 4oz weight progresively higher on the fixture untill it fired. There is a reason Federal primers are packaged the way they are, as they are the easiest to set off. CCI were the hardest of the primers requiring a greater height from which the weight had to be dropped in order to fire it. I wish I would have saved that article. LM