Powder type vs percieved recoil?

[Solvability]

Good conversation started at the match the other day:

Can you feel a difference in recoil between two powders when they meet the exact same power factor?

The current practice is to get fast powders and heavy bullets.

I assert I can feel a difference in recoil feel at the same velocity and same bullet weight.

Is there any proof - could you pass a blind test - would it show up in results of a Bill drill?

[Graham Smith]

Recoil and felt recoil are not necessarily the same. Without digging into the mechanics and physics, a heavier bullet going slower will generally have less felt recoil than a lighter bullet going faster. Compare a 9mm minor to a .40 minor and you will definitely feel the difference.

[Solvability]

Not so much talking about bullet weight as the perception of recoil between powders loaded to the same velocity and with the same bullet.

Edited July 9, 2014 by Solvability

[seanc]

For me the loudness of one load over another has alot to do with how I think it feels.

[GOF]

At the velocities we are shooting in action pistol games a faster burning powder requires less powder to drive the same bullet to the same velocity than does a slower burning powder. Less powder often results in less felt recoil. It frequently produces a quicker recoil impulse that seems shorter and lighter. You can see this easily in the .45 ACP when you load TiteGroup or Clays ( 3.9-4.0 grains) to 750 fps with a 230 grain bullet (172 PF) and then load 7 grains of HS6 to the same 750 fps and the same PF. There is a recoil difference and the nod goes to the faster TG & Clays.

[Nimitz]

For me the answer is absolutely yes. I have been shooting the same powder for the last 2 years. Recently I've developed another load with an alternate powder due to availability and even though they are the same PF (130 & 130.8) I know which bullets have which powder once I pull the trigger due to the felt recoil difference. The difference in 'feel' is very obvious to me ...

[EngineerEli]

I swear I heard a while back that the concept is faster powders require less mass of powder to accelerate the bullet to PF. With less powder there is less total gas produced. With less total gas produced there is less of a 'blast' that comes out of the barrel behind the bullet, effectively rocketing the gun backwards (recoil). I would love to hear if anyone has heard this before or can confirm it. I can absolutely tell the difference between WST and Power Pistol in my .40 loads, even the difference between WST and TG, not 100% certain if that is actual recoil or precieved recoil as a result of the muzzle blast/sound of powders like Power Pistol. To combat the noise in general I always wear plugs under my muffs. It deadens the sound in a way that makes it much less distracting and even helps keep you from blinking as much so you can actually see the sights.

[ToddKS]

Some powders seem to produce more muzzle blast than others which in my experience leads to more perceived recoil. Universal in the 45ACP and Unique in the 380 are two that I have direct experince with in this regard.

[CHA-LEE]

From my experience, when building different loads using different powders but everything else is the same (OAL, Bullet, Primer, Brass, Velocity, Crimp, etc), the faster burning powders have a sharper "Crack" type of felt recoil and the slower burning powders have a dull "Thump" type of felt recoil.

The only analogy, I can think of is a fast powder would be like hitting an anvil with a standard framing hammer. Then a slow powder would be like hitting an anvil will a dead blow hammer. This is an excessive example for an analogy, but that is how powders at different burn rates feel to me.

[bountyhunter]

Not so much talking about bullet weight as the perception of recoil between powders loaded to the same velocity and with the same bullet.

I'd wager the difference in powder is the burn rate and how quickly the energy is released. The difference would probably be seen if a G meter was attached to the gun. Perceived recoil is probably a function of peak recoil impulse. The total recoil (area under the curve) is probably defined by the mass of the bullet and final velocity, but perceived recoil can be different based on powder burn speed. Edited July 9, 2014 by bountyhunter

[peterthefish]

Three (perceived?) factors play into this. As previously mentioned, slower powders require a larger charge for the same velocity. That powder (or more precisely the gases produced by burning it) leave the barrel with the bullet.

In short, if you move the same (say 200 Grain) bullet at the same velocity with a 4GR charge of Bullseye and an 8GR charge of 3N38 (made up #s), the second charge actually has a mass of 208 grains vs. 204 for the first. That's about a 2% increase in momentum.

However, the gases actually are traveling faster than the bullet. So momentum of the ejecta (and in turn the firearm) is likely increased by more like 3-5%. I'd guess this is responsible for most of the increase in felt recoil with slower powders.

Second is the fact that slower powders are more likely to incompletely combust before the bullet leaves the barrel. This means more blast which increases perceived recoil.

Lastly, some argue that the shape of the pressure curve dictates the nature of felt recoil. Fast powders have a higher peak pressure while slower powders will have a lower peak but longer duration with high pressure. The area under both pressure curves should be nearly identical. However, pressure rises and drops so quickly (in handguns by the time the bullet has traveled 2" down the barrel pressure is a fraction of peak) compared to the total duration of the recoil cycle (unlocking after bullet leaves barrel, slide rear, slide forward picking up round, lockup barrel and slide) that I strongly doubt this has ANY appreciable impact in felt recoil over the first two factors noted.

[2MoreChains]

I was practicing with a friend once where I was shooting 9mm loaded with N320 and a 124 gr FMJ. He was shooting a 9mm loaded with Universal and Montana Gold 124's. Both were loaded for minor PF of about 130-ish. At one point my ammo ended up in his gun and he noticed a pronounced difference, and promptly asked for my load data.

He switched.

[pjb45]

VV320 is the only powder I use in 9mm, 40, and 45.

It is very consistent for me -batch to batch. I am able to cut it pretty close on the PF.

I had the Chrono guy at the Desert Classic/A2 comment on how soft shooting my SV was compared to others.

[ardo]

I used to load both Clays and VV320 in my minor 9mm rounds with 124gr FMJs, to the same PF of 130. To me, the felt recoil was about the same. I could never understand this, given that VV320 is slower than Clays.

If I could find VV320, I would go back to it. As of now, I'm stuck with TG - and counting my blessings that TG was available when I was about to run out of powder.

[phidelt208]

I think there is NO question some powders are harsher on the recoil Try Power Pistol. I don't think there is a lot of difference in some powders but for others there is. Compare anything at the same PF to Power pistol and you'll feel, see and hear the difference.

[lstone]

I too have found that when I loaded a batch with power pistol it was very much "snappier" than the TG load I had. EngineerEli that is a interesting explanation about wearing plugs under your muffs, I have seen people try that and I will too next range outing

[b1gcountry]

Very interesting

[mgardner]

I have quite a bit of damage in my hands and elbows from old injuries and I notice a difference in powder types at the same PF. The slow to medium burners seem to be more comfortable to shoot than the fast burners. A bit heavier recoil spring seems to tame it a bit too. Probably the loads have the same recoil but it can be spread over a longer time.

[MotorMouth]

This is interesting. I just had this conversation today with a friend at a match. The math and physics aren't too obtuse. Recoil is nothing more than the kinetic energy of accelerating the bullet down the barrel. Newton's law states F=ma, or Force equals mass times acceleration. Acceleration is mass times the square of velocity divided by 2 so Force as a function of velocity is F=1/2mv^2 when accelerating a thing from rest - since the bullet isn't moving before you shoot it, this equation works.

So now there's two things to discuss. First, the heavy bullet slow speed, vs light bullet high speed recoil analysis. Just looking at the equation, we can see that Force (recoil) doubles when we double the mass of the bullet but increases by a factor of 4 if we double the speed.

For a minor power factor of 125,000 using a 147 gr bullet going 850 fps we get 53,103,750. For the 113 gr bullet going 1106 fps, we get 69,112,834. We'll ignore the units since they're the same for both. What we see here is that even though both rounds make 125k PF, one has almost 30% more recoil.

So, how does this apply to powders? We know from the Powder Burn chart that certain powders burn faster than others. For instance, Titewad is faster than Titegroup. So, even though we don't know what the flame front speeds are (well, maybe someone does, feel free to post up if you've got it) we at least can extrapolate that the slower powders will, in fact, without doubt, recoil less. Will that difference be perceptible? I dunno. The bullet's are for sure, but I don't know about the powders. I tend to think that the greater part of the recoil is the bullet part, but, the flame speeds are MUCH higher, probably on order of 30,000 fps or more.

Maybe we can talk one our ChemE members to discuss flame speeds.

H

[Guy Neill]

In order to determine the recoil energy we need the gun velocity.

to determine the gun velocity use the momentum balance equation

where the gun momentum equals the ejecta momentum

the ejecta is the bullet momentum plus the powder momentum

powder gas velocity has normally been taken as 4,000 fps as determined by the military

Momentum is the mass times the velocity

our power factor is momentum value, but with inconsistent units.

Once we solve the momentum equation for the gun velocity, we can determine the gun recoil energy

Guy

[peterthefish]

This is interesting. I just had this conversation today with a friend at a match. The math and physics aren't too obtuse. Recoil is nothing more than the kinetic energy of accelerating the bullet down the barrel. Newton's law states F=ma, or Force equals mass times acceleration. Acceleration is mass times the square of velocity divided by 2 so Force as a function of velocity is F=1/2mv^2 when accelerating a thing from rest - since the bullet isn't moving before you shoot it, this equation works.

F=1/2mv^2. That is correct. Everything about the way you go there is wrong. Acceleration does not equal "mass times the square of velocity divided by 2". The only similar formula that is applied to acceleration is to measure distance under constant acceleration and does not involve mass.

And in the case of a firearm, when we're talking about Force, we're talking about Kinetic Energy. The Kinetic Energy of the total system (bullet + recoiling gun + heat + noise) = The Potential Energy in the powder charge & primer.

So now there's two things to discuss. First, the heavy bullet slow speed, vs light bullet high speed recoil analysis. Just looking at the equation, we can see that Force (recoil) doubles when we double the mass of the bullet but increases by a factor of 4 if we double the speed.

For a minor power factor of 125,000 using a 147 gr bullet going 850 fps we get 53,103,750. For the 113 gr bullet going 1106 fps, we get 69,112,834. We'll ignore the units since they're the same for both. What we see here is that even though both rounds make 125k PF, one has almost 30% more recoil.

This is incorrect. You're assuming that Kinetic Energy of the recoiling firearm is equal to Kinetic Energy of the bullet. It is not. The only place Kinetic Energy is conserved is as above - Kinetic Energy of the system after firing the bullet is equal to Potential Energy in the Powder & Primer.

The correct physics model to apply to the system after the bullet leaves the gun is that of an explosion. In other words, momentum, not kinetic energy, is conserved. Momentum = mv, so doubling the speed of the bullet doubles the recoil of the gun (caused by the bullet), just as doubling the mass does.

As I noted in my previous post here, the model would be M1V1 = M2V2 + M3V3 where;

M1 = Mass of Firearm

M2 = Mass of Bullet

M3 = Mass of Powder Charge

V1 = Velocity of Recoiling Firearm

V2 = Velocity of Bullet

V3 = Velocity of Gases Escaping from Firearm

As a result, your calculations about recoil increase are totally off base. Of course, that should be obvious! A 30% increase in recoil is the difference between a 9mm and a .45. The difference between a 115 & 147 at minor PF is nowhere near that large. Let's use some actual loads I have used with similar velocities as you gave. We'll asume that Gas Velocity is twice bullet velocity - quickload has some estimates, but I don't have it open now.

Berrys 115 GR RN, 5.6 GR Power Pistol, AVG Velocity = 1165 FPS (~ 134 PF).

Zero 147 GR FMJ, 4.3 GR Power Pistol, AVG Velocity = 905 FPS (~133 PF).

For the first load, recoil impulse to the firearm = (115 * 1165) + (5.6 * 2320) = 146,967

For the second load, recoil impulse to the firearm = (147 * 905) + (4.3 * 1810) = 140,818

This shows that the lighter load has a 4.4% greater recoil impulse than the heavier load - that's a little closer to reality! Now let's take it a step further. Mass of the firearm is static - lets assume it weighs 2 lbs, or 14,000 Grains.

For the first load, velocity of the recoiling firearm = 146,967 / 14,000 = 10.49 FPS

For the second load, velocity of the recoiling firearm = 140,818 / 14,000 = 10.06 FPS

Now, the recoiling firearm does have kinetic energy that we have to absorb, so the Kinetic Energy for each load is;

First load: 1/2 * 14,000 * (10.49 ^ 2) = 770,280 (non-standard units)

Second load: 1/2 * 14,000 * (10.05 ^ 2) = 702,802 (non-standard units)

So the increased Free Recoil (KE of Firearm) of the lighter load is 9.6% greater than that of the heavier load.

What the shooter feels is probably somewhere in the middle of those numbers.

So, how does this apply to powders? We know from the Powder Burn chart that certain powders burn faster than others. For instance, Titewad is faster than Titegroup. So, even though we don't know what the flame front speeds are (well, maybe someone does, feel free to post up if you've got it) we at least can extrapolate that the slower powders will, in fact, without doubt, recoil less. Will that difference be perceptible? I dunno. The bullet's are for sure, but I don't know about the powders. I tend to think that the greater part of the recoil is the bullet part, but, the flame speeds are MUCH higher, probably on order of 30,000 fps or more.

Maybe we can talk one our ChemE members to discuss flame speeds.

H

Pretty simple. To reach a given velocity, with a given projectile, requires less powder the faster that powder is. So your conclusion is incorrect; slower powders (for a given firearm, projectile, and velocity) recoil MORE than fast powders.

Flame front speeds have nothing to do with this. The velocity of gases (which consists of the burned and unburned powder charge) escaping from your barrel are however key, and there is software available that models this.

Edited July 21, 2014 by peterthefish

[thermobollocks]

The best recoil is frequent recoil, the kind I get when I have about 6 different loads available for whatever powder I can find

[MotorMouth]

This is interesting. I just had this conversation today with a friend at a match. The math and physics aren't too obtuse. Recoil is nothing more than the kinetic energy of accelerating the bullet down the barrel. Newton's law states F=ma, or Force equals mass times acceleration. Acceleration is mass times the square of velocity divided by 2 so Force as a function of velocity is F=1/2mv^2 when accelerating a thing from rest - since the bullet isn't moving before you shoot it, this equation works.

F=1/2mv^2. That is correct. Everything about the way you go there is wrong. Acceleration does not equal "mass times the square of velocity divided by 2". The only similar formula that is applied to acceleration is to measure distance under constant acceleration and does not involve mass.

And in the case of a firearm, when we're talking about Force, we're talking about Kinetic Energy. The Kinetic Energy of the total system (bullet + recoiling gun + heat + noise) = The Potential Energy in the powder charge & primer.

So now there's two things to discuss. First, the heavy bullet slow speed, vs light bullet high speed recoil analysis. Just looking at the equation, we can see that Force (recoil) doubles when we double the mass of the bullet but increases by a factor of 4 if we double the speed.

For a minor power factor of 125,000 using a 147 gr bullet going 850 fps we get 53,103,750. For the 113 gr bullet going 1106 fps, we get 69,112,834. We'll ignore the units since they're the same for both. What we see here is that even though both rounds make 125k PF, one has almost 30% more recoil.

This is incorrect. You're assuming that Kinetic Energy of the recoiling firearm is equal to Kinetic Energy of the bullet. It is not. The only place Kinetic Energy is conserved is as above - Kinetic Energy of the system after firing the bullet is equal to Potential Energy in the Powder & Primer.

The correct physics model to apply to the system after the bullet leaves the gun is that of an explosion. In other words, momentum, not kinetic energy, is conserved. Momentum = mv, so doubling the speed of the bullet doubles the recoil of the gun (caused by the bullet), just as doubling the mass does.

As I noted in my previous post here, the model would be M1V1 = M2V2 + M3V3 where;

M1 = Mass of Firearm

M2 = Mass of Bullet

M3 = Mass of Powder Charge

V1 = Velocity of Recoiling Firearm

V2 = Velocity of Bullet

V3 = Velocity of Gases Escaping from Firearm

As a result, your calculations about recoil increase are totally off base. Of course, that should be obvious! A 30% increase in recoil is the difference between a 9mm and a .45. The difference between a 115 & 147 at minor PF is nowhere near that large. Let's use some actual loads I have used with similar velocities as you gave. We'll asume that Gas Velocity is twice bullet velocity - quickload has some estimates, but I don't have it open now.

Berrys 115 GR RN, 5.6 GR Power Pistol, AVG Velocity = 1165 FPS (~ 134 PF).

Zero 147 GR FMJ, 4.3 GR Power Pistol, AVG Velocity = 905 FPS (~133 PF).

For the first load, recoil impulse to the firearm = (115 * 1165) + (5.6 * 2320) = 146,967

For the second load, recoil impulse to the firearm = (147 * 905) + (4.3 * 1810) = 140,818

This shows that the lighter load has a 4.4% greater recoil impulse than the heavier load - that's a little closer to reality! Now let's take it a step further. Mass of the firearm is static - lets assume it weighs 2 lbs, or 14,000 Grains.

For the first load, velocity of the recoiling firearm = 146,967 / 14,000 = 10.49 FPS

For the second load, velocity of the recoiling firearm = 140,818 / 14,000 = 10.06 FPS

Now, the recoiling firearm does have kinetic energy that we have to absorb, so the Kinetic Energy for each load is;

First load: 1/2 * 14,000 * (10.49 ^ 2) = 770,280 (non-standard units)

Second load: 1/2 * 14,000 * (10.05 ^ 2) = 702,802 (non-standard units)

So the increased Free Recoil (KE of Firearm) of the lighter load is 9.6% greater than that of the heavier load.

What the shooter feels is probably somewhere in the middle of those numbers.

So, how does this apply to powders? We know from the Powder Burn chart that certain powders burn faster than others. For instance, Titewad is faster than Titegroup. So, even though we don't know what the flame front speeds are (well, maybe someone does, feel free to post up if you've got it) we at least can extrapolate that the slower powders will, in fact, without doubt, recoil less. Will that difference be perceptible? I dunno. The bullet's are for sure, but I don't know about the powders. I tend to think that the greater part of the recoil is the bullet part, but, the flame speeds are MUCH higher, probably on order of 30,000 fps or more.

Maybe we can talk one our ChemE members to discuss flame speeds.

H

Pretty simple. To reach a given velocity, with a given projectile, requires less powder the faster that powder is. So your conclusion is incorrect; slower powders (for a given firearm, projectile, and velocity) recoil MORE than fast powders.

Flame front speeds have nothing to do with this. The velocity of gases (which consists of the burned and unburned powder charge) escaping from your barrel are however key, and there is software available that models this.

This is a good treatment. I didn't think of the explosion model. I particularly like that the weight of the handgun is taken into account - which is also obvious because a heavier handgun recoils less than a lighter one. However, wouldn't half the weight of your forearm also factor into this?

[peterthefish]

This is a good treatment. I didn't think of the explosion model. I particularly like that the weight of the handgun is taken into account - which is also obvious because a heavier handgun recoils less than a lighter one. However, wouldn't half the weight of your forearm also factor into this?

Not really. Once you get past free recoil it becomes a bit more complex. If you held a firearm as firmly as a ransom rest you'd include your whole body weight. But of course, even the best shooters shooting the flattest guns don't.

Is first model the impact of the recoiling firearm on the shooter as an inelastic collision (even though you're already holding the firearm). So the shooter has to absorb the entire free recoil of the firearm. Think of it as catching a baseball. You may start off holding the firearm, but other than the mechanics of your grip that's irrelevant. The moment a baseball hits your glove it is no different than a firearm recoiling - you absorb its energy to decelerate it.

Then it becomes a pretty crazy statics / dynamics problem that I'd need to crack a book to even start to consider.

[MotorMouth]

This is a good treatment. I didn't think of the explosion model. I particularly like that the weight of the handgun is taken into account - which is also obvious because a heavier handgun recoils less than a lighter one. However, wouldn't half the weight of your forearm also factor into this?

Not really. Once you get past free recoil it becomes a bit more complex. If you held a firearm as firmly as a ransom rest you'd include your whole body weight. But of course, even the best shooters shooting the flattest guns don't.

Is first model the impact of the recoiling firearm on the shooter as an inelastic collision (even though you're already holding the firearm). So the shooter has to absorb the entire free recoil of the firearm. Think of it as catching a baseball. You may start off holding the firearm, but other than the mechanics of your grip that's irrelevant. The moment a baseball hits your glove it is no different than a firearm recoiling - you absorb its energy to decelerate it.

Then it becomes a pretty crazy statics / dynamics problem that I'd need to crack a book to even start to consider.

Yup. And for me, it's been 30 years since I took it. I'm a EE - ask me how the laser in your sight works and you'll get the right answer. Even if it's obtuse.