Greetings!

Until a few years ago, 99% of my reloading has centered on rifle ammo, and my focus had been on velocity, with the desire to keep my ammo supersonic at a distant target. During this time, when I reloaded handgun ammo, my intention was to duplicate velocity.

Now, I am looking to develop a 9x19 load using a different bullet profile, which is changing velocity parameters. I had been loading 115gr (c 1150 fps) and 125gr (c 1050 fps) FMJ and JHP bullets. I am now looking to develop a load using a 125gr HAP profile bullet (so I can also utilize the same bullet in 357 Sig ammo).

Using data in "shooter's reference", comparing data for the 125gr HAP and Sierra FMJ, the HAP bullet requires 67% of the powder charge, generating about 75% of the velocity, yet almost 4800 units of chamber pressure more.

My question is, is it the chamber pressure that causes slide recoil which initiates slide movement, or is it the bullet velocity?

I am trying to determine if the reduced velocity/increased chamber pressure load that I am contemplating will function the actions of my Glock 17s, 26, and my Sig P365 and P320. Trying to form an answer before I purchase bullets that I won't be able to employ for multiple cartridges!

As always, thanks in advance for your help!

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Well, unless it is a gas operated semi auto, pressure has very little to do with it. It is the ejected mass, and the acceleration of that mass that causes the slide to move. It's Einstein's "equal and opposite" theory that does the work. The ejected mass is the combined weight of both the bullet, and of the powder itself. Pressure plays a small part, the rocket effect of the discharged gasses would be slightly more at a higher pressure.

The HAP bullet and it’s different profile definitely isn’t plug ‘n play in 9mm, it is like starting over. I chose to dance with the 125 grain HAP bullet in search of a pure accuracy load in 9mm. I found that I liked the accuracy however the bullet is completely and entirely obnoxious to find for sale. At first it was annoying, then it angered me and then PEACE, because I turned the page on it.

And just when I figured it was history for me… I ran in to some folks on the Bullseye forum that recommended a great replacement for that damn vaporware bullet.

RMR, Jake at Rocky Mountain Reloading is making a damn good copy of it in a FMJ flat point.

RMR is fantastic to deal with, their site is smooth as silk, prices are fantastic and there is no BS in dealing with them.

I’ve washed my hands of the Hornady product that Hornady cannot deliver. I now have a big fat bag of these from RMR and I have 200 loaded rounds already cooked up with two different powders and I’m just waiting for some good weather and an opening in my calendar to try them.

So sorry, I cannot review the performance of the bullet itself, but up to this point, I like it 10 times better than the Hornady slug.

Haha, your actual question! The bullet profile simply requires that it be loaded more deeply in to the cartridge case. The 9mm is a tiny, tiny space and full 35kpsi max pressure, so yes, you’ll get the same pressure with less powder. And you are shooting for proper pressure to make it run, with far less worry about it’s velocity.

in the 9mm, the 124 HAP is said to be .356" dia vs the 115 Hap at .355"...

what ever it is, this Hornady bullet is usually seated at a short OAL.....
that can get into high pressures, so you need to study your data.

I have loaded the Berry 124 "Copy" at 1.075 for fine grained powders
out to 1.13" for Allinat flake powders, in my pistol, from 996 to 1233 fps.....
just to see how this design worked before buying 500 or more.

You do need to see if your pistol will shoot the JHP design vs the "Ball" FMJ style
before going crazy with big volume sales.......... maybe.

I have posted some of my test, to let others see what happens in my pistols.
For target work they are ok and don't break the bank.

Good loading.

The HAP bullets are designed for target work. They are the XTPs without the cuts in the jacket for expansion. This makes them a bit cheaper to shoot. The HAPs should work well in any 9mm, but you may find some exceptions. Use the start load and work up.

I am actually looking at the RMR bullet. I want to use it in my 357 Sig, and would like to use it in my 9x19. For 9x19, I was hoping to use WW231, since I have a fair amount. My concern is that I have not found load data for it, and looking at similar profile bullets propelled by 231 is not encouraging. While I want to shy away from the FMJRN bullet, I'm afraid that I will have to go that route.

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I have extensive experience handloading and shooting this 9mm bullet in IPSC and IDPA competition. As mentioned in earlier posts, this bullet is 0.356" rather the more usual 0.355". The difference is small and I learned the hard way it matters. My competition pistol was a Springfield XDm which turns out to have a tight chamber and barrel measurements. Depending on the brass you make your loads with, I had severe jams of the unfired cartridge which didn't fully seat in the chamber. If you've ever had this happen, clearing the jam isn't easy. Needless to say, it ruins your score on that stage.

If you use the Hornaday HAPs, make a small batch and make sure they run in your gun(s). All 9mm brass is not the same so sort the test loads by head stamp.

Good luck and good shooting.

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Bob.
SWCA 1821

I'm not sure what you mean when you say, "My concern is that I have not found load data for it, and looking at similar profile bullets propelled by 231 is not encouraging."

There is plenty of data for the 124/125gr bullet on the Hodgdon load data site. I have never seen anything that says W231 is not encountered in the 9mm. I use W231 with RMR 124gr JHP bullets all the time.

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SWCA #3437

Not to nitpick but that is Newton's third law of motion.
Einstein was the Theory of Relativity (E=MC^2)

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Load data (powders & OALs) should be exactly the same as the corresponding 115gr or 125gr XTPs... I've loaded the 115gr in 357 SIG with no problems. Hodgdons shows them loaded pretty short (1.069" for the HAPs vs. 1.150" for the BERRYS 124gr HBRN-TP) , but I load the Flat-nosed version of that bullet like the XTP for both 9mm & 357 SIG (1.135"): it is just about my favorite practice round. Plenty of powder choices there.

Most hollowpoints of the same weight are pretty close in length to each other: Hornady, Sierra, Speer, etc. Just load the same amount of bullet in the cases.

Mid South currently has both in stock...

Cheers!

It is the momentum which establishes recoil. Momentum is Mass x Velocity. On the business end, Mass is the mass of the bullet plus the mass of the propellant gasses (i.e., the weight of the propellant i.e., smokeless propellant). There are two components of the velocity. First is the muzzle velocity of the bullet. Then there is the muzzle velocity of the gas. Note the gas is released at the muzzle at a velocity which exceeds that of the bullet. It is typical to estimate the escaping gas velocity at the muzzle as being 50% greater than the velocity of the bullet. Therefore the total momentum = muzzle velocity of the bullet x bullet mass + (1.5 x muzzle velocity of the bullet) x propellant mass. And the gun itself will have exactly the same momentum, but in the opposite direction. That is known as recoil. Note that pressure does not figure into the momentum equation. That requires a different calculation.

Momentum is indeed Mass x velocity but I always though that is in reference to a body already in motion at a steady speed - i.e. a steady state.
Isn't recoil actually FORCE which equals Mass x acceleration - rather than momentum?

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Momentum equals Impulse, which is force x time, i.e., MV = FT. Another way to look at it is that F=MV/T=MA. Acceleration, A, (delta v/delta t or dV/dT) does not enter into momentum directly, but as a moving body speeds up or slows down due to the change in forces acting on it, it results in either positive or negative acceleration. Positive acceleration is what the bullet experiences inside the barrel, negative acceleration is what it experiences after it leaves the barrel. Therefore a fired bullet's Momentum increases or decreases depending on where it is. The Momentum we are talking about is specifically that at the muzzle at the instant the bullet leaves it, not at any point inside the barrel or downrange.

Thanks for the clarification. After posting I actually looked up a couple of recoil calculators and pretty much came to the conclusion that momentum is the correct formula because we're looking at a specific instant - when the bullet leaves the barrel and is no longer affecting the gun = no more dV/dT - i.e. steady state at an instant in time (snapshot).
Makes perfect sense.
As my uncle said to me one time after I explained what a resonant frequency was "that's some of that engineering s**t, ain't it?"

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Another way to look at it is that recoil of the gun begins at the same instant the bullet starts its journey down the barrel. It increases until the bullet leaves the muzzle, then, insofar as the gun is involved, recoil stops.

Exactly right. The fact that recoil actually starts as soon as the bullet begins to move - when its velocity is essentially zero - that is the part that lead me in the wrong direction.
When we speak of recoil we're really only talking about measuring the MAX force when the bullet is at max velocity (at the muzzle) and we are, for all intents and purposes, disregarding the recoil forces that occur prior to that point in time (the acceleration period).
Which is why F=MA isn't the right equation.

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Ain't Google great? I was great in Math in school and college but that was 45+ years ago. There is no way I could remember those formulas so that's why I ask, ain't Google great? lol

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SWCA #3437

Generally, recoil is listed as the kinetic energy of the gun. We find the velocity of the gun with a momentum balance, then calculate the energy.

Got that right.
In the same vein, one of my instructors had a little speech he made to every class during one of the first class sessions. It went something like this...
"This is an engineering school. We aren't here to teach you a bunch of formulas. You can look those up in an engineering handbook any time. We're here to teach you a way of thinking. A way of looking at problems analytically."
Nowadays we don't even need the good ol' engineering handbooks. Just an internet connection.
You still have to look at the problem the right way to know WHICH formula to look up though!

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The source of the recoil of a gun is exactly the same reason a rocket or jet engine flies. Upon firing, the pressure created in the barrel acts in all directions, but most importantly, that pressure is applied equally as force acting on both the base of the bullet and the breech face of the gun, which are 180 degrees apart, and both can move. It is really the portion of that pressure force acting rearward on the breech face which causes recoil of the gun. The resulting rearward velocity of the gun depends upon its mass. That is why the felt recoil of a heavy gun is much less than that of a light gun firing the same cartidge.

Rockets and jet engines (and everthing else) operate on the same principle. In operation there are pressures created inside them which results in unbalanced forces which drives them forward.

Using the pressure model, what is the gun velocity. How would it be calculated?

Let's assume a 9mm 124gr bullet at 1,100 fps from a 4" barrel. And consider the gun weighing 2.5 pounds. If needed, use a charge weight of 3.5 grains.

Thanks. Wanting to learn more.
.

Calculate Recoil Energy

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MV (momentum)=FT (Impulse) MV (gun)=MV (bullet and propellant gasses). Just solve for V of the gun after you have calculated the Momentum of the bullet and propellant gasses. Be careful, as the mass must be calculated in slugs, not pounds. 1 Slug = Pounds/32.16

I understand the momentum method, but it does not include anything for chamber pressure. As I understood the above, the recoil is based on the pressure working against the gun.

When I tried to find the gun velocity based upon the chamber pressure it was several times larger than when calculated by the momentum method.

So, what I don't "get" at present, is how, based upon the statements above about pressure and recoil, how you determine the recoil (or gun velocity) using the pressure.

The link is also not using chamber pressure.

It seems the recoil energy must be found through a momentum balance, as it has been for a very long time, and not through the chamber pressure.

Thanks.

I love this forum... a simple loading inquiry becomes a discussion of Newtonian Physics.. do I get any CEU's for this?

COULDN'T HAVE SAID IT BETTER MYSELF

p.s. I never had a problem finding load data for that bullet :-) probably because I don't over-think and certainly don't 'over-research' anything :-))))

p.s. what's a CEU?

J.

Great stuff! Just to nit pick a bit, the velocity if the gas at the muzzle can run between 4000 and 7000 fps, depending on temperature, which is directly related to pressure. My point is that the mass of the powder charge, converted to gas, has a huge effect on recoil. Which is why comps were used (think 460 Rowland).

Pressure is the force that accelerates the projectile
Pressure =force/area
Chamber pressure x projectile area = force

Recoil has 2 components
The force generated by the acceleration of the projectile(Newton's 3rd law) and the force generated by the gasses exiting the barrel (rocket engine)
Hatcher's notebook has a thorough discussion of recoil.

Most likely the "CEU" = C(ontinuing) E(d-u-ma-cational) U(nits)...?

Cheers!

Correct... Many call it "credits" for life experience.

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SWCA #3437

...which is also an example of Newton's 3rd law

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"It's Einstein's "equal and opposite" theory....." I think that was Newton!!