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Calculating firing pin energy

The first part is just figuring out the potential energy of the spring itself. You'll have to figure out the spring constant if you can't get it from the manufacturer. I think the formula for potential energy is .5constant*distance^2.
 
The kinetic energy is simply the firing pin spring force multiplied by the distance (firing pin fall), with everything in the appropriate units. For example, with a 30-pound firing pin spring falling 0.2 inches (=1/60 foot) the kinetic energy of the firing pin is 0.5 ft-lb.

The formula in the post above mine is also correct, since spring constants are measured in units of force/distance.
 
Kinetic energy is half of the mass times the velocity squared. Velocity is distance times acceleration. Acceleration is net force divided by mass. In the case of a firing pin spring, if you measure the force that it takes to compress it to cocked length, add the force that it took to compress it to its fired length, add them and divide by two, I think that you will end up with the average force. If I remember correctly, in calculating the mass for a striker assembly, the spring is counted at half of its weight. Have fun, and don't forget that the primer may prevent the striker from falling as far as it might on an empty chamber, and that the final indent may be shallower than the initial.
 
BoydAllen said:
Velocity is distance times acceleration.

Velocity is acceleration multiplied by time, not distance. Distance traveled under constant acceleration (from zero) is 1/2(a*t^2). Velocity at constant acceleration (from zero) is a*t.
 
For those who are interested (both of you!), here's how I derived the equation that I posted:

Eq 1: (K)inetic energy = 1/2 * (m)ass * (v)elocity^2

We can measure the mass of the striker (firing pin) and spring (as it turns out, that won't be necessary), but there's no easy way to measure the velocity, so we have to calculate v somehow.

Since the striker accelerates from a stop (cocked position), we could calculate the velocity at the end of its travel if we knew the acceleration of striker and the time that it takes to travel from the cocked position to the fired position (lock time). Eq 2: v = (a)cceleration * (t)ime

The acceleration of the striker can be calculated using Newton's second law of motion, Eq 3: (F)orce = m * a. Rearranging, Eq 4: a = F/m. We know the force on the striker because the spring force is generally given in pounds in the cocked position. For long springs and short firing pin falls we can assume that the force is constant over the firing pin fall. Though not quite true, it's close enough.

So, substituting from Eq 4 into Eq2 to calculate v, Eq 5: v = F/m * t

Now, how do we calculate t (lock time)? The (d)istance covered by an object under constant acceleration (produced by a constant spring force, in this case): Eq 6: d = 1/2 * a * t^2. Solving for t: Eq 7: t = sqrt (2d/a). Again following Newton's second law of motion, substitute F/m for a, so that Eq 8: t = sqrt(2dm/F).

Substituting from Eq 8 into Eq 5, Eq 9: v = F/m * sqrt(2dm/F)

As a reminder, Eq 1: K = 1/2 * m * v^2

Substituting v from Eq 9 into Eq 1, K = 1/2 * m * [F/m * sqrt(2dm/F)]^2

= 1/2 * m * F^2/m^2 * 2dm/F

= F * d

So, if you know the spring force at its working length, and assume that it doesn't change (much) over the short firing pin fall, and you know the distance of the firing pin fall (about 0.200" in a Rem 700), you can easily calculate the kinetic energy of the striker at impact.

However, I wonder along with Butch -- why would you want to know?
 
Fire control and protrusion can make a difference in accuracy. Years ago some of the benchrest actions had adjustment in them. I never owned a Farley but a gunsmith told me the old ones had adjustable fire control. Put one of those lightweight pins and springs in your action and shoot at distance off a bench. Some guns won't shoot very good with them. Makes vertical at long range. Matt
 
So Dwight Scott is doing a great business adding a tungsten weight to firing pins. I guess BR is different than off hand. Consistent ignition is more important than a few milli seconds of lock time.
 
Butch,
Have you measured the spring weights, and pin fall on BAT actions that are common in benchrest? Perhaps they come closer to the line than others.
Boyd
 
Thanks for the replies. I have been trying to learn as much as I can about ignition lately. I would like to do some testing, and calculating the energy should help with diagnosing future rifles. I don't think Dwight needs the business.
 
So you are saying the Bats have a problem and Dwight has the correct fit? If the Bats have a problem, why are they winning more matches than the other receivers?
 
Greg Walley had a simple test to do on a fired tight fitting case but with a new primer. No powder. You fire the primer and measure (somehow) the indent in that brand primer. Do a search on BRC.
 
I was forced to re-tune with a Cooper when it's firing pin slipped under setscrew control. This showed up as a step change toward worse grouping, with my best load and with every primer still igniting just fine. This, with my standard 1thou bump and 2thou primer crush. It was a very tough(and scary) problem to find and fix.
Oh, firing pin indent is useless information here.

Anyway, once I figured it out I had to group shoot with incremental adjustments to released pin protrusion from boltface. I have it logged, but from memory I think .240" was the sweet spot. I doubt that's useful across a myriad of action/triggers/hangers/springs/pins, but the results out of doing this test & finding this sweet spot was stunning (to me).
An example of what I seen was grouping opening to ~1/2moa at .200", closing to ~3/8moa by .220", closing further to ~1/4moa at .240", and opening back up to 3/8 & beyond past .240".
I have no idea how this is possible with all primers firing(no mis-fires), but I measured it, it was real.
 

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