1911 Skool: Slide to frame impact Part 2

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John Travis

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So, how hard does the slide hit the frame? It seems to the one of the big concerns.

Let's do a little math. We've already thought about the hypothetical cannon in the last thread, and the lack of questions or comments indicates that everyone understands it.

Assuming a 230 grain bullet at 830 fps...GI Hardball spec...we'll calculate the slide's speed and kinetic energy. Weights were taken on a postal scale using a 1919 production Colt GI pistol. Due to better modern steels, those weights may vary a little.

The slide and barrel assembly weighs 16.8 ounces or 7350 grains. The slide comes in at 13.8 ounces or 6057 grains. During initial firing, the slide and barrel move as a unit, so their combined weights are used to calculate the slide's speed.
For this calculation, we won't be considering the effects of the recoil spring, hammer mass and mainspring, or any friction that exists between slide and frame.

At 7350 grains, the slide's mass/weight is close enough to 32 times that of the bullet to call it 32. That means that the slide's rate of acceleration and final speed at bullet exit is 1/32nd that of the bullet's. Once the bullet has left the muzzle, no further acceleration of the slide is possible. The slide and barrel assembly are moving as fast as they ever will.

Now, we have 7350 grains of mass moving at 1/32nd of 830 fps. That works out to 25.9 fps...or close enough to 26 to call it.
A quarter inch into the slide's trip, the barrel falls away and leaves the slide to continue on its journey to the frame impact abutment. 6057 grains moving at 26 fps.

I pulled up an online bullet energy calculator...because those use grains...and plugged in the numbers.

The slide, moving 26 fps, carries 9 foot pounds of kinetic energy, which means 9 foot pounds of impact energy.

I'm going to assume a 14 pound recoil spring and a 23 pound mainspring along with a wide spur hammer...Browning's original specs.

If we then factor in the effects of the springs, hammer mass, and what little friction that exists...a rough estimate brings the slide's impact velocity down to around 22 fps, or maybe even less. I'll be generous and call it 22. Now, the kinetic energy is down to around 7 foot pounds.

Seven foot pounds.

To put that into perspective, the standard velocity .22 Short...32 grains/1000fps...carries 70 foot pounds of kinetic energy.

Not exactly the hammer of Thor, now is it? Surely not enough to beat up the frame as is often claimed by the heavy spring and shock buff proponents. The gun just isn't that fragile.
 
And considering all the aluminum frame/steel slide guns out there that have worked for years and years, without getting beat to hell. Didn't Chuck Taylor run up something like 10,000 rounds on an aluminum frame Commander?
 
I don’t like the approach to the calculation, it seems overly complicated, but in the end the result seems correct.

First consider just the slide and barrel, no friction, springs, etc. and a simpler analysis. The bullet leaves the barrel with 350ft-lbs of energy, so the slide and barrel assembly must also have 350ft-lbs of energy, correct? The barrel is 1293 gr, and the assembly is 7350 gr, so the barrel dropping bleeds off 17.59% of the energy to the frame leaving the slide with 288ft-lbs.

There are too many variables for me to swag the effect of friction, springs, other movement, etc, so instead lets calculate the actual energy by measuring the velocity and mass of the slide directly. We already know that the slide is 6057gr, and a little internet searching turns up a measured velocity of 25fps with 45acp ammo. Couple conversions and some multiplication and tada, 8ft-lbs.

So, the system dumps 200ft-lbs in places other than the slide impacting the frame. Seems like a lot. A little more searching shows that the recoil spring stores enough to accelerate the slide to between 2mm/ms and 4mm/ms but I’m going to bed rather than doing the math on that.

Bottom line, not a lot of battering.
 
JimB said:
The bullet leaves the barrel with 350ft-lbs of energy, so the slide and barrel assembly must also have 350ft-lbs of energy,
Click to expand...
No. You're thinking of momentum. Bullet and slide momentum are equal, at least in theory. During normal firing, the slide is subject to outside forces that the bullet isn't, so slide momentum will always be a little less.
 
If bullet and slide momentum are equal, then your calcs work. Momentum = mass x velocity, so velocity is inversely proportional to mass.

You looked up Energy, which is proportional to the velocity squared... that works, too.

Can I get a link to the post on the hypothetical cannon in the last thread? I missed it...
 
John Travis said:
No. You're thinking of momentum. Bullet and slide momentum are equal, at least in theory. During normal firing, the slide is subject to outside forces that the bullet isn't, so slide momentum will always be a little less.
Click to expand...
This is why I quit aerospace engineering and became an accountant.

I missed the cannon thread too.
 
The law of Conservation of Energy would be applied differently here than would Conservation of Momentum. In this case, the chemical energy released from the burning propellant would be transferred to kinetic energy of the bullet, kinetic energy of the slide+barrel assembly, and heat.
Powder = Kbullet + Kslide + heat. The slide and bullet would be imparted equal momentum, but not equal energy.
 
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Catfish said:
The law of Conservation of Energy would be applied differently here than would Conservation of Momentum. In this case, the chemical energy released from the burning propellant would be transferred to kinetic energy of the bullet, kinetic energy of the slide+barrel assembly, and heat.
Powder = Kbullet + Kslide + heat. The slide and bullet would be imparted equal momentum, but not equal energy.
Click to expand...
But a ft-lb is a measure of energy, we could convert it to joules and the math would be a lot easier. I don’t even know the english version of Newton seconds, the unit of measure for momentum.

Maybe @MacEntyre is on it, the analysis works with either and we’re just messed up in the language.
 
JimB said:
But a ft-lb is a measure of energy, we could convert it to joules and the math would be a lot easier.
Click to expand...
Which is why I presented it the way I did. We all understand Lbs/ft and FPS...but not everybody knows what a Joule is, and meters per second is confusing for we who refuse to bend to the metric system.

And the math wasn't complicated at all. Ounces to grains is simple. (437.5:1) No need for complex equations. Long division and multiplication will do, especially if you use a pocket calculator. Googling up a bullet energy calculator is even simpler.

That digital postal scale of Kelie's is handy and revealing. When somebody brings up the much higher velocities with a Commander slide, I can tell them that the Commander's slide is only 9/10ths of an ounce lighter than the 5-inch slide.
 
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The units for energy don't matter as much as understanding this:

The bullet and slide CANNOT have the same energy. Ft-lbs, joules, whatever, it's physically impossible. Because they MUST have equal momentum and their weights are different.

I can prove it with math but nobody wants to see that.
 
I need a shrug emoji I guess, I just don’t differentiate momentum from kinetic energy. Combustion provides the energy, and in our model it is then a constant.
 
JimB said:
I just don’t differentiate momentum from kinetic energy.
Click to expand...
I guess the way to look at those properties...in layman's terms...is:

Energy determines how hard the bullet hits the target. Momentum determines how deep the bullet goes into the target after it hits it.

Another way to look at it:

A 405 grain bullet fired from a .45-70 at 1350 fps and a 55 grain bullet fired from a .220 Swift at 3800 fps both carry about the same energy...but which one would you rather have in your hands with a Grizzly charging from a treeline 50 feet away?

Anyway, the point was made. The slide just doesn't hit the frame all that hard.
 
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