Given that a load has to meet the criteria of projectile weight times velocity greater than 125,000, what would be the load that created the least amount of felt recoil in 9mm?
The following loads meet the above criteria
115 grain at 1,090 FPS
124 grain at 1,010 FPS
147 grain at 850 FPS
all of the above meet (just barely) the minimum power floor, but which of them would impart the least amount of felt recoil to the shooter, and why?
In my experience heaver-slower recoils less the lighter-faster. Here is a link to a recoil calculator that may help.
OK, I see. They are all the same.
The reason is that recoil is the reaction of mass being expelled from the firearm. When you change the weight of the bullet ( mass ) or the velocity ( energy ) you can change the recoil. Since mass and energy are the same ( E=MC squared ) and the power factor is a measurement of mass and energy, if the PF is 125,000 then the recoil is going to be the same.
K=1/2mv^2
My physics might be a bit rusty after so many years, but isn’t it F=ma that you are wanting to use, not E=mc^2? Doesn’t c=the speed of light in a vacuum?
Even though the units are off (didn’t take them into consideration because they are relative as long as they are calculated the same), and is how they come up with 125000
Solving for m*a=F (force being recoil)
1151090=125350
1241010=125240
147*850=124950
That follows in line with what you stated in your first response. Heavier/slower has a little (very little) less perceived recoil.
I’ve heard the term “impulse” in conjunction with recoil. In my understanding it means that some rounds will cause the same amount of force in that equation, but don’t other things come into account in “perceived” recoil. Slide weight, bore height, recoil spring and such? Personally, I’m not that good of a shooter yet to really focus on that much detail, but it is interesting none the less.
I wonder if the variable that impacts felt recoil is time? I wonder if a slower burning powder that still gets up to the same velocity is really the key?
and I wonder if the difference is so small as to be imperceptible to a shooter like me.
There’s also the issue of what is readily available (modern times notwithstanding) in factory loads. If 115 and 124 grain loads tend to be loaded much hotter than the power factor but the 147s tend to be just above, then that would factor in as well.
Dammit, I’ve got some good info on this but not enough time to spell it out right now.
Consider this a placeholder…
Impulse is force over a given duration of time IIRC, thus making recoil look more like a power curve?
Of course the mass and velocity we need are that of the slide:p
Burn rate of the powder as well as total powder charge (since that adds mass to the equation as it gets pushed, in whatever form, down the barrel) also play a role.
As a simple general rule, for a given power factor (M*V), the slower & heavier rounds will have less felt recoil. Not coincidentally, for a given power factor, you tend to use a lighter powder charge with slower & heavier rounds.
It’s also important to remember that felt recoil, alone, is an inadequate measurement of how fast you’ll be able to make followup shots. Recoil of just about any 9mm round is negligible. What matters is how fast & consistently the gun returns to point of aim after discharge.
those are constant, though, so it doesn’t matter for a relative comparison of various loads in the same gun.
the question, though, is what contributes to “felt recoil”…is it momentum (p=mv), is it kinetic energy (k = 1/2mv^2), or is it just plain “force” (f=ma)…or some combination?
if kinetic energy comes into play (which it seems it would), then it makes sense that a heavier bullet moving more slowly would give rise to less “felt recoil” as k increases exponentially with increases in velocity, but only linearly with increases in mass.
Go ask Brian Enos.
The mass of the slide is constant. It is the velocity that is the variable depending on ammo.
Recoil is the based on the law of conservation of momentum, and it is the reaction to the propelled mass. Free recoil equation:
MgVg = MbVb + Mp*Vp
The momemtun of the propelled mass = MbVb + MpVp
where:
Mb = bullet mass
Vb = bullet velocity
Mp = powder mass
Vp = powder velocity
Mg = gun mass
Vg = gun free recoil velocity
Average powder (in gas form) velocity is about 1/2 of the bullet’s muzzle velocity when the bullet exits the bore, but then there is a “jet” effect when it is free to rapidly expand and accelerate. For rifles the terminal velocity of gas is estimated at about 4,700 fps, for handguns a little less 4,000-4,200 fps. This is the value that you enter into the equation.
For a rifle caliber the powder is a very significant part of recoil, for handguns much less unless we talk about large doses of H110 in a 44 mag, etc.
For any given caliber, when you use a heavier bullet, with any type of powder the dose is smaller and this contributes to lessen recoil (to a very minor degree in a handgun).
Also, when you use a faster powder the powder dose is smaller than with a slower one, and this also contributes to lessen recoil. The recoil impulse is also shorter, and has less muzzle blast, all of which may contribute to less perceived recoil.
Now, in a 9 mm these effects are really minor, and other things like adding a little weight to the gun will have a much greater effect, as well as how you grip the gun, etc. Some shooters like a sharper, shorter, recoil, some the opposite.
yeah, that is what i meant…i did not do a good job of explaining it.
the bullet has some momentum of
P(b) = M(b) * V(b)
(where P is momentum, M is mass, and V is velocity)
the law of conservation of linear momentum states that the total momentum of the system (gun plus round) has to be 0 (as the gun plus the round is a closed system with no momentum to start with).
thus, P of the bullet in motion is equal to (and opposite in direction) of P of the gun.
P(g) = P(b)
thus, P(g) = M(b) * V(b)
thus, we don’t need to know the mass of the gun or slide to make relative comparisons of the effects on the momentum of the gun of various loads.
(also, the same logic works if we measure the force the gun exerts due to the bullet being accelarated…just substitute F=MA for P=MV in the equations above.)
The numbers you put in for the acceleration variable are not units of acceleration, they are velocities.
Impulse does come into play as well.
The specs of the gun used don’t matter, as long as it’s the same gun, right?
As long as it is the same gun, it’s apples to apples 
Then why is the minimum power level for 9mm 125,000? Didn’t they calculate it that way to make it easier for reloading purposes? That’s how I understood it, but I could be wrong. It’s not exactly F=ma, but it is mass(of the bullet) times the velocity of the bullet = 125000 or it should to meet the power factor.
Unless you run a comp. A comp works better with a lighter bullet with more powder. The extra powder creates a greater gas volume allowing the comp to work better.
Years ago I was trying out IPSC and shooting a H&K USP in .40 cal. I found that the heavier 200gr bullets felt better then the lighter bullets yet all the loads were close to the same powder factor. The lighter bullets were snapper then the heavier ones. The heaver bullets were quicker to get back on target. At least for me. Your mileage may vary.