Hammer mass and light springs?

G

GoSlash27

New member
I've been running the math this weekend, and it works out that the mass of the hammer has no effect on the kinetic energy transferred to the firing pin...so long as friction is ignored.
But friction would play a big part, wouldn't it? If your new skeletonized hammer is accelerated at a higher rate through the same angle in order to hit the pin with the same kinetic energy....well, friction increases with the square of the square of the speed, so there would be much more energy loss, right?
Seems to me that a skeletonized hammer is just exactly what you don't want if you're concerned about light strikes due to a lightened mainspring.

Thoughts?
 
As you lighten the hammer and speed up the hammer fall, the momentum of the hammer strike decreases or stays the same (depending on actual speed), and the kinetic energy increases. There is no benefit to decreasing the mainspring weight. So, yes, friction increases, but kinetic energy increases dramatically. As long as you don't reduce the momentum of the hammer, you're OK. Even if you do, a 1911 has plenty to spare, but try not to.

There is no benefit that I can think of to reducing the mainspring tension.
 
Scorch,
So, yes, friction increases, but kinetic energy increases dramatically.
Click to expand...
I didn't want to bore you with a long algebra derivation, but I'll do so to speed the discussion along. Here's the proof for why it doesn't work that way:

The equations:
F=M*A
D=1/2a*t^2
V=a*t
and
eK=1/2M*v^2

We need to solve the first equation for acceleration
A=F/M
a=A*32.2 (because A is in Gs while a is expressed in fps^2.)

We need to solve the second equation for time
t=sqrt(2D/a)

so plugging the first into the second t=sqrt(DM/16.6F)

So now we know how long it will take our hammer to fall, we need to know how fast it's moving when it hits.

v=at

Plugging in from above
v= (32.2F/M)*sqrt(DM/16.6F)

to find kinetic energy, we use
eK=1/2M*v^2

again, plugging in from above

eK= M/2* ((32.2F/M)*sqrt(DM/16.6F))^2

Now to simplify...

eK= M/2* (32.2F/M)^2*(DM/16.6F)
= M/2* (32.2F/M)*(32.2F/M)*(DM/16.6F)
= M*32.2F*32.2F*DM/M*M*16.6F
and collecting like terms
eK=1037*D*F*M^2/16.6*M^2<---there's the kicker.

and that's why. The effect of the mass cancels out because reducing it increases the velocity just enough to generate the same kinetic energy.

eK =66.4*D*F (again, neglecting friction)


AFA lightening the mainspring, it's sometimes done to lighten the trigger pull in DA pistols.
 
Yes, I've had classes in physics, too. The main advantage of speeding up the hammer is decreased lock time, which increases accuracy. So, as you showed, you have no net gain in the kinetic energyof the hammer for ignition reliability, but you gain by having less lock time and incresing accuracy without decreasing ignition reliability.

BTW, Decreasing spring tension is not the best way to lighten trigger pull. Trigger pull in a double action revolver involves camming surfaces, which need leverage advantage, and in single actions it involves angles of engagement, which need to be neutral.
 
So, given the same spring, a hammer made from light plastic will work as well as one made of steel? I don't think so, but it should be easy to try. I eagerly await the results.

Jim
 
scorch,
you have no net gain in the kinetic energyof the hammer for ignition reliability, but you gain by having less lock time and incresing accuracy without decreasing ignition reliability.
Click to expand...
No, you *do* reduce kinetic energy and thus ignition reliability because more of it is wasted due to friction.
Yeah, I suspect that's why the 1911 guys like 'em, but they tend to not reduce their hammer springs. And there are many different areas to attack in many different firearms to reduce trigger pull. But for some of 'em the big culprit is the mainspring. I happen to own one and that's what got me looking into this.

Bottom line: a skeletonized hammer on certain DA/SA pistols is like a triple-decker wing on a Miata; it may look cool, but you're probably better off without it.

Jim,
And it's funny, 'cuz my CX4 carbine has a plastic hammer ;)
 
spring

Dear Sir:
I don't care about yhe latent or kinetic energy is; I do not change hammer springs or the weight of hammers in my 45s.
I do combat work where it must go "bang" every single time!
Harry B.
 
Since you are basically talking about an elastic collision between hammer and firing pin you conserve momentum, not energy. Energy is wasted on friction and the heat and noise generated when the hammer continues on its way and wacks against the stop.

MV=MV in the energy equation the V is squared. Since you only transfer momentum to the firing pin increasing V while decreasing M has no net effect. However, increasing V while decreasing M would have the net effect of having more energy to dissipate, whether heat, noise or energy being absorbed by the overall gun frame itself.

But wait, that's not all. The hammer also transfers the force of the spring onto the firing pin at the instant of impact, so you have the force of the momentum plus the force of the spring.

Furthermore, unlike with a physics problem, friction, heat generation and noise are large enough components that you can't just ignore them.

Just thinking about all the difference forces and reactions is enough to give anyone a headache. This is one case where a few experiments would be required to really determine anything.
 
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