It has long been known that the weight of the powder charge is part of the equation, along with bullet weight, for calculating recoil. But this business of "rocket effect" apparently has a significant influence on recoil as well. Graeme Wright, in his book, "Shooting the British Double Rifle," mentions on Page 108 that he and other shooters can feel a distinct reduction in recoil between IMR4350 and IMR4831 in .500 Nitro loads at the same performance level. Does anyone have a list, or list of factors, that can be used to compare powders for "rocket effect" in minimizing recoil from a given cartridge and performance level? It would seem that it is more complicated than mere burn rate, but then, perhaps not. (That would mean that Bullseye always kicks the least!)

Bill/Oregon __

The "rocket effect" might be the "thrust" of the combustion gases (still at high pressure) as they exit the muzzle, "pushing" against the atmosphere. Given the small diameter of the "rocket nozzle" and the uncontrolled expansion (no thrust chamber), this effect has to be quite small compared to the other forces involved.

"Burning Rate" is not involved. Pressure is involved. We can relate the pressure developed by different types of gunpowder to rocket fuels/oxidizers using the concept of "specific impulse". For example: the "specific impulse" of a hydrogen/oxygen mixture is higher than a kerosene/oxygen mixture. The same can be said for two different gunpowders, but I don't know of any theories attached to that concept.

Good shooting.

Robert

Bill/Oregon,

As far as I am aware powder "families" like IMR, Alliant etc. would act the same IF combustion was the same.

Recoil has two basic components.

The first component is accelerating the bullet and power gas up the barrel.

Up to this point the bullet is acting like a "throttle" on the powder gas.

When the bullet clears the barrel then the powder gas increases its velocity and thus adds to "rocket" thrust.

In the case of different loads (especially reduced pressure loads)there can be (I think this is right)a lot of Hydrogen that has been released but does not ignite until exposed to the air. In other words the conversion from solid powder to gas has not gone right inside the bore.

When the Hydrogen ignites it is often accomapnied by a large muzzle flash and a very deep BOOOOOOM.

In the process, the velocity of the gas is greater than what it would have been.

If my memory is working OK, the Specific Impulse of a fuel that Robert has referred to is the thrust in pounds of a rocket engine that consumes 1 pound of that fuel per second.

Thus if your rocket engine consumes a mixture of 1 pound of hydrogen/oxygen (Shuttle Main Engines and Stage 2 and 3 of a Saturn V) mixture per second the thrust is greater than a rocket engine that consumes 1 pound of a kerosene/oxygen mixture. (Stage 1 on a Saturn V)

It would be reasonable to assume that the specific impulse of IMR 4350 and 4831 would be the same as I think they are the "same fuel"

Thus the difference in recoil would be to do with combustion problems with one of the loads.

Some powders such as the Australian ADI powders that Hogdon sells as Shortcut and Extreme can produce problems with reduced loads (which would be the case for the double rifle) and I am inclined to think Alliant powders can be a bit of a problem in this area as well.

Mike

Well, I'm no Newton but . . .

Weight of the projectile affects recoil as does weight of the firearm. Heavier bullets and lighter guns increase recoil.

What's often overlooked in "projectile weight" is the weight of powder heading down the bore. In black-powder firearms a substantial portion of the powder turns to ash. In a heavy rifle load this ash may weigh half of what the bullet weighs. While this powder adds little to velocity, it increases recoil substantially.

------------------
PowderBurns Black Powder / Muzzle Loading Forum:

www.hotboards.com/plus/plus.mirage?who=powderburns

Mike375 is on the right track. Recoil certainly does involve the burning rate in several ways.

First of all recoil is correctly measured in force, not energy, both the peak force and the time integrated total force applied to your shoulder. Both of these quantities are determined by the masses of the rifle, bullet and powder charge, and the accelerations applied to each of these masses from the propellant burning, both inside the chamber and bore and at the muzzle.

Slower powder often requires larger charges, which means more mass accelerated, but it also burns more slowly which means a different time integral, different acceleration rate. If it contains significant quantities of combustion products that are not fully combusted (and it will) it may contribute more to the rocket effect (certainly to muzzle flash).

Isp is expressed in seconds and depends on the geometry of the combustion chamber and nozzle. The "rocket effect" of propellants is, I think, similarly governed. Specifically, I believe it is more pronounced in overbore cases (strongly necked down) than in straight cases. This is a hunch, difficult to prove, because a lot of other things change too, like bullet mass, when comparing two cases, one overbore and one straight or less necked down.

In the case of IMR-4350 compared with IMR-4831, each loaded in a straight case, I would contend that the effect is mainly due to the difference in charge weights (?) and burn rate. The geometry being the same between these two I doubt if much difference in rocket effect is discernible, but maybe it is in relatively short barrels.

Suppose we have two cases each with the same powder charge. Lets pretend that they use bullets of the same weight but different diameters. If all the powder is consumed before the bullet exits the muzzle then the contribution to recoil from rocket effect would be due to the geometry (i.e., a volume of gas expanding out of a smaller "nozzle"). On the other hand if a very fast powder and very slow powder were used (its hard to imagine that they would have the same charge weight and produce similar velocities) then a portion of the recoil due to "rocket effect" would may be attributable to the combustion at the muzzle, but since this is adiabatic free expansion I am not convinced it really pushes (real rockets burn internally). It probably just makes a big fireball.

In my experience fast powders kick harder (sharper), and more than anything else by far lightweight guns kick harder, for what its worth.

Harald,

If all else was equal, the a 458 with 70 grains of powder and a 257 Wby with 70 grains of powder, the 458 would "probably" develop more recoil when the bullet uncorked the barrel.

If memory is working for me, the thrust is determined by the rate at which gas exits the nozzle.

So in a 257 we have much higher pressure thana 458 but a smaller nozzle. So a 458 may have gas exit at a greater rate.

This can be seen witha simple experimemnt and a garden hose with a nozzle.

If the nozzle is set to a very fine spray, the water might exit at sufficient velocity to almost cut you.

If we now start to open the nozzle, the presure in the hose drops, the velocity of the water exiting drops but the "thrust" increases as a greater mass of water exits in a given time.

However, if we remove the nozzle, the pressure drops right down and the rate at which water exists also drops and thus the thrust drops off.

The net result of all this to us shooters is that muzzle brakes can be very effective on calibers like 458s etc.

By the way, to me the only thing more attractive than a 375 with a 300 grain Silvertip was the Saturn V rocket.

15 tons of fuel consumed each second on Stage 1.

Mike

As usual I look to QuickLoad for my bookish answer.

I chose for example a 9.5 lb 416 RemMag with 24 inch barrel shooting 375 gr BearTooth bullets at 2450 fps.

One load is 93.2 gr VV-N560 The total recoil just before the "gas jet" is 37.66 ftlb, after the gas jet the total is 57.28 ft lb. The peak gas force is 1395 lbf.

The second load is 76.5 gr. IMR4895. The total recoil prior to the gas jet is 36.28 ftlb (very close to above), the total recoil after the gas jet is 51.75 ftlb. This is 10% less than the slower N560. The peak gas force is 960 lbf.

Only the shooter knows how the difference is felt. I like the 4895 better, personally.

Don

I don't think that the "rocket" effect contributes much to the actual recoil, and as some have mentioned, it will be reflected more by changes in bore size than powder charge weight or type.

One must be careful to remember that when a 90 grain powder charge is burned, it creates 90 grains of gas because the conversion is chemical not nuclear. A 70 grain powder charge weighs 20 grains less than a 90 grain charge, therefore, will recoil less. It is the same as shooting a 380 grain bullet as opposed to a 400 grain bullet with the same powder charge.

My ballistic program shows about a 2 ft/lb difference between a 410 grain 416 bullet @ 90 grains vs the same bullet with a 70 grain charge, and a 90 grain charge with a 410 grain bullet vs a 90 grain charge with a 390 grain bullet. I suspect the 2 ft/lb difference reflects some other variable in the equation because the "rocket jet" effect is greater than 2 ft/lbs or a muzzle brake would be ineffective.

The amount of recoil attributable to "rocket jet" effect can best be considered in the claims made for various muzzle brake. I only have one rifle equiped with a brake, but it does noticeably reduce muzzle jump and recoil. The recoil reduction is probably less than the muzzle jump reduction because the downward force is multiplied by the distance between the butt and the muzzle.

In several of my rifles the most noticeable difference has been when in a relatively large bore (375 or 45), I have gone from a heavy bullet, to one significantly (25 to 30%) lighter and increased the velocity. The speed of the recoil is much greater, and it slaps you rather than pushing you. I've noticed this with my 45/70, .425 Express and 375HH.
Ku-dude

Ku-dude,

Quickload takes all that into account. The 20 grain difference in powder load was unavoidable to get a realistic comparison.

There is a setting in Quickload for the percent of the powder charge thrown forward. Changing that percent changes the recoil only slightly - as one would expect when looking at 20 grains difference out of 450 (375 bullet + 75 powder). The main change in the recoil between the two charges in my previous post is due to the gas jet effect. The higher muzzle pressure of the slower burning powder makes the biggest contribution, as the powder (or powder oxidation product) ejected is moving much faster due to the higher muzzle pressure.

If you look at the numbers I gave above, a "perfect" muzzle brake would lower the recoil of both rounds to roughly 37 (36.28 vs. 37.66) ftlbs -- the recoil present when the bullet exits the muzzle and before the gas does. In other words, a 100% effective muzzle brake would lower the IMR4895 recoil by 15 ftlbs, and the N560 recoil by 20 ftlbs.

The whole reason QuickLoad gives these numbers separately is so you can see what the gain of a muzzle brake could be.

Actual muzzle brakes are probably 60-70% efficient, depending on length, blow-by and number of chambers. Some of the big (three pounds, as big as a grapefruit) 50 cal brakes might approach 100% by directing the blast rearward rather than just blowing it sideways. The concussion is brutal for anyone beside the shooter in the blast.

Don

Don,
Please define "percent of powder charge thrown forward".

It occurs to me that 100 percent of the charge is "thrown forward" either as gas or as unburned powder.

It would seem to me that when the primer ignites the charge, the bullet starts to move, accelerating down the barrel. As we learned in basic physics, for every action, there is an equal and opposite reaction. Therefore, the bullet is going one way, the gun going in the other. But in between is a volume of gas that is constantly changing.

The gas both propels and cushions the forces as the bullet accelerates down the barrel pushing against the breech block. Most all of the powder will exit the barrel as gas or unburnt grains, but the mass will be the same as what we measured with our scale.

The "jet" effect would be the mass of gas times the velocity of the gas (when the bullet leaves the barrel), squared.

Can it be said that the gas is moving as fast as the bullet? If so, the the velocity of the gas is equal to the velocity of the bullet. The mass is equal to the powder charge. My computer puts the recoil of a 90 gr "bullet" out of a 10 lb gun @ 2300 fps = to 8.5 lbs. The recoil of the gun is 48 lbs with a 400 grain bullet @ 2300 fps.

I don't know, but a reduction of recoil/barrel flip in the neighborhood of 1/6th the "unbraked" recoil seems about right for my rifle. Ku-dude

Ku-Dude,

Quoting QuickLoad, "The amount of powder thrown forward is the fraction of unburned charge and produced propellant gasses thrown forward with the bullet. It typically varies from .33 in overbore bottleneck cases to .75 in straight wall cases." In the above example (416RM) varying this number from .01 to .99 affects the recoil less than 1 ftlb. I think this number affects the calculated maximum chamber pressure more than recoil (i.e. the position in the barrel where the powder burns.) I explain it only to drop it for this discussion, as it is not significant, and is not what it sounds like.

As soon as the bullet clears the muzzle the gases start to blow around the bullet, at a highly increased rate. I don't know the velocity difference, but I remember seeing strobe photos of the bullet about two inches from the muzzle with the flash about three times that far.

If you think about what happens after the bullet exits the muzzle, you'll see that the first gas (let's include all ejecta except the bullet here, OK?) out the muzzle will be at very high pressure (5000-15000 psi). That part of the gas will blow out fastest, gradually slowing down as the pressure drops. You are correct that ultimately almost all of the powder and powder products will leave the bore. How much this affects recoil is a function of the rate at which the mass leaves the bore. This is, of course a strong function of the pressure and powder-charge mass distribution in the barrel over time.

Vaughn in Rifle Accuracy Facts shows some shadowgraphs of a .270 bullet exiting the muzzle, maddeningly he does not give a gas velocity, but he talks about a pressure wave at the base of the bullet where the gases are travelling supersonic in relation to the bullet. He also talks about the bullet passing the slowing frontal shock wave at 142microseconds after the bullet exits the muzzle. This appears to be at a distance about seven inches from the muzzle.

The total time for the bullet to accelerate from the throat to the muzzle is 1.5 milliseconds, and the jet recoil starts then and goes on for about 2 msec. According to Quickload my 9.5 lb 416 has travelled rearward .111 inches -- just enough to begin to compress the recoil pad. Then the afterburner kicks in and the last 20 ft lb of recoil hits you. I think this is why people talk about "pushes" and "jabs".

By the way, the "gas aftereffect recoil" (QuickLoad's name) of a 416 RM is about the same as the total recoil of my beloved 308 Win!

( If you think I'm an anal retentive nitpicker, you should read Vaughn! )

Don

[This message has been edited by Don G (edited 05-03-2001).]

The conservation of mass rule is the first principle in fluid flow. Unsteady flow is much more complex but for a given control volume you must have the same quantity entering the volume (some arbitrary location in the bore) as exiting (at the muzzle). Assuming equal powder quantities, burn rates and bore pressures, the muzzle with the smaller bore should have a higher exit velocity and that is what we see. Granted, bullet masses are typically not equal in practice even when all the other conditions are met. Also bear in mind that nozzles work in an exactly opposite manner in sub-sonic and supersonic flow conditions. That is, a sub-sonic converging nozzle becomes a supersonic diverging nozzle. In steady flow, the mass flow rate is a constant (by definition) and the variables are the velocity and the pressure. In unsteady flow the mass flow rate is constantly changing but the principle of continuity must be observed.

But I confess that my hypothetical model is so unrealistic that it may not elucidate the problem. When you throw real bullet masses into the mix its likely that we cannot separate the real contribution of "rocket effect" from other burn rate related effects. Like I said before, I notice recoil in lightweight rifles.

Don,
I am truly not trying to be argumentative, just trying to get my mind wrapped around these concepts. I do not see how one can say that something less than the full powder charge is "thrown forward" in the context of a recoil event, particularly if we are in agreement, that for practical purposes, all the powder gases and residue have left the barrel.

Regarding the "jet" effect, the gas cannot be going faster than the bullet up until the bullet clears the bore because it is driving the bullet. After the bullet leaves the bore, the gases may accelerate rapidly. I believe that Herald has expressed my understanding of that event.

I seem to remember that gas pressure increases/decreases in a geometric relationship with the volume in which it is contained. That is, when the bullet unstops the barrel, the gas will be able to expand to "infinity". Its velocity may increase tremendously at first, but that speed will decrease with the square of the distance(area) it travels, rapidly reaching equilibrium.

It also would appear that the recoil can never exceed the mass times velocity squared value in the absence of a venturi type arrangement to add extra velocity to the gas. It is this phenominum that permits brakes to work. It uses the gas at its greatest presure and velocity, and uses "venturis" to accelerate the gas in the desired direction to diminish muzzle jump, which as indicated earlier has the most subjective impact upon recoil and can use the moment arm of the rifle's length to increase its effect.

Based upon my playing around with my ballistic program, I don't believe that the recoil attributable to gases in a 416 shooting a 400 grain bullet with a 90 grain powder charge at 2300 fps can exceed 12-15 lbs. In a rifle generating 53 ft/lbs of recoil, 12 lbs of force down on the barrel multiplied by the length of the rifle is a noticeable reduction in "recoil." Ku-dude

Don and Ku-Dude,

Fore many years the formula to calculate recoil was based on teh assumption that the average gas velocity was 4700 f/s. Thus rifle momentum was Bullet Wt X MV add ed to
Powder Wt X 4700.

This forumal gives higher recoil figures for calibers such as the 458 than does the HuntAmerica forumal, which I think is the one that may be used on some ballistic programs.

I think the HA used formula gives a better value as to "over all violence", which does include muzzle blast, but I not so sure about recoil.

For example, for the last several years I have used a bag a shot behind the butt. I have often shot the 338-378, 378 and 458 side by side and in rifles of similar weight. The 458 seems to drive the bag back further than a 338-378, that is, the 458 rifle has achieved a higher velocity.

So if we assume the "4700 f/s" is a good recoil calculator and using a 416 Rem ina 10 pound rifle we have.

((2400 X 400) + (85 X 4700))/10/7000 =19.42

Recoil energy is 58.6 ft/lbs

As to the situation juts before the bullet clears the barrel, it is reasonable to assume that powder weight achieves an average velocity equal to half the muzzle velocity.

If that was the case then just before bullet exit we would have

((400 X 2400) + (85 X 1200))/10/7000 = 15.17

Recoil energy is 35.7 ft/lbs

Since total recoil was 58.6 ft/lbs the additional recoil energy after bullet clears the barrel is 22.9 ft/lbs

Does a reduction from 58 ft/lbs to 36 ft/lbs seem reasonable if a KDF brake was used?

Mike

Ku-dude,

We do agree that effectively all of the powder mass will eventually leave the barrel. I have tried to explain that Quickload is using the term "percent of powder charge throw forward" to describe something else, not the total mass leaving the barrel. You must remember that I did not write this program or choose the terminology - it has been translated from German. I mentioned the term only to point out that AS USED BY QUICKLOAD, varying this parameter has an insignificant effect on the recoil. It is obviously not what it sounds like it would be. I believe it is a term he uses to describe what percentage of the powder burns in the case, vs in the barrel, hence it affects chamber pressure more than recoil.

Obviously the mass exiting the barrel at high velocity HAS a significant impact on recoil, and I believe that QuickLoad correctly accounts for this effect.

In the example I started with the fast powder gas effect added 15.5 ft-lbs and the slow powder gave 19.4 ft-lbs of contribution to the overall recoil of 51.75 and 57.28 ft-lbs respectively, for a 375 grain bullet at 2450 fps. It appears to me that these two instances are close enough to your rule of thumb to be believable to you.

My main point is that the slow powder with the higher muzzle pressure gave 10% worse recoil than the faster powder. I believe that this higher recoil is primarily attributable to the higher muzzle pressure causing faster gas exit velocity, and that the extra weight of powder has only a small effect on this contribution.

If I cared enough I would try to prove this, but I am going to bed instead.

Don

Mike,

For a 10 lb .416 with 400 grain bullet at 2400 fps, developed from an 85 grain load of IMR4895 (had to cheat the RM cartridge case volume by 15% to get all this to match your example- so its a Rigby!) QuickLoad predicts recoil developed up to the bullet leaving the muzzle to be 38 ft lbs. The gas effect adds 17.5 ftlbs to get 55.5 ft lbs total recoil.

A brake that turns the gases 100 % sideways would have the 38 ftlbs of recoil.

A brake that turned the gases straight back at the same velocity would allow only 21 ftlbs of felt recoil, but would not be pleasant to shoot!

In short, I think that your numbers are very realistic, but the 22.9 number is just a bit too high.

Note that your rule of thumb for total recoil worked out fairly closely.

Don

Mike and Ku-dude,

OK mystery is solved.

The "amount of powder thrown forward" ("factor" for short)is used both to determine the pressure distribution in the barrel, and to determine the percentage of powder accelerated in the gas column in the barrel behind the bullet. Although it does not affect the total recoil (all the gas leaves the barrel!), it does affect the split between muzzle and gas recoil. (Can I use muzzle recoil as shorthand for the recoil when the bullet leaves the muzzle?)

Example: Mike's 416: if this factor is varied in the recoil calculator only (but keeping the same powder burn modelling run) the recoil distribution is:

factor muzzle gas total
.01 30 26 56
.66 38 18 56
.99 47 9 56

If I change this factor before the powder burn modelling run the chamber pressure and muzzle velocity change as well as the recoil distribution, complicating the comparison.

So now I at least feel I understand what H. Broemel did, and it all seems to hang together in my little pea brain. I hope I have not clouded the issue too much.

Don

Well I'll be damned; reloading IS rocket science after all!
That said, and in light of my hazy understanding of what far keener minds have just said, can it then be assumed that burn rate does in fact have an appreciable effect on recoil and that all other things being equal, a faster burning powder may offer less "rocket effect" at the muzzle, and therefore contribute to a reduction in overall recoil compared to a slower-burning number? (This would account for Graeme Wright sensing less recoil with 4350 vs.4831.)

Bill/Oregon,

Quite often the difference in powder weight between slow and faster powders will be offset by combustion conditions.

Thus you find a heavier charge of IMR 4350 could have less recoil than a smaller cahrger of Reloader 15.

As a very general guide, calibers like 458s can produce some heavy recoil with lighter loads of slower powders. Large amounts of freebore, which 458s have, also contribute. Some loads with lighter bullets and lower pressure can end up giving recoil that is approaching higher pressure loads with heavier bullets.

As another vey general rule, if you loads with different powders are all at normal to high pressure, then the heavier charge of slow powder will produce more recoil.

As to rocket science, the rifle is a rocket and the bullet and powder are the exaust.

Mike

Bill,

As Mike points out there are exceptions, but generally what you say is true.

Here's how I would put it. "If you are willing to give up 100 fps off the highest available velocity, choose a slightly faster powder that is all burnt six inches before the muzzle. This gives lower muzzle flash, less muzzle bark and lower recoil."

The six inches of barrel is not wasted, the bullet is still accelerating when it leaves the muzzle.

Don

Don,
Can you differentiate "muzzle recoil" and "gas recoil"? Which is recoil attributable to the bullet's mass? Which is recoil attributable to the powder's mass? And which is attributable to the "jet" effect? I think I am beginning to get the hang of this. Thanks. Ku-dude

Ku-dude,

Yes, in effect the program does show me these values separately.

The Quickload program gives me more information than I can conveniently show here. (And for dang sure more than I can describe well!) The recoil calculator explicitly gives (partial list):

"Values when the Projectiles's Base Passes the Muzzle: momentum, velocity of gun, gun travel, energy of recoiling."

"Total Values at End of Gas Aftereffect (free recoiling mass): gas effect duration (avg),velocity of gun, gun travel, energy of recoiling."

To arrive at the gas donation I have to subtract these two "energy of recoilings".

I do not know Broemel's motives, but I assume he breaks the values out like this to allow the user to see the potential reduction in recoil of a muzzle brake.

To break out the bullet recoil separate from the powder recoil (while still in the barrel), I can set the "Fwd. moved part of Charge" ["factor"] to zero, rather than it's nominal value (I use .66 for the 416RM, it does not have much of a bottleneck). (Broemel sets a default value for the 416 of around .44, I think he uses that for all bottleneck cases.) What was fooling me at first was that changing this value has no effect on the total recoil. It has dramatic effect on the way the recoil is split up into muzzle and gas, though. (Mike's rule of thumb approach made me go back and look more closely.)

It also makes sense from an energy standpoint. A given powder charge has a known level of energy in it, but the burning characteristics are highly pressure sensitive. After the pressure profile is modelled a percent burnt is arrived at. After that is arrived at, all that energy has to leave the muzzle - as either gas jet or bullet energy.

I wish QuickLoad had a web demo... it does!
It only does a couple of powders and cartridges, though. You have to download 1 MBYTE, then a quick install. If that's not too much GO TO :

http://www.neconos.com/details3.htm

Recoil calculator is under "options" on the top banner. Hit the CALC buttons first.

Don

[This message has been edited by Don G (edited 05-06-2001).]

Don,
I'm sorry. I was less precise in my language that I should have been. Could you please define muzzle recoil and gas recoil? I am interested in what portion of the recoil event the program attributes to the bullet's departure; what portion is attributable to the powder's departure; and what portion is attributable to the ejection of the gases when the bullet unstops the barrel. Thanks. Ku-dude

Don,

On the thread on BSI on Africa I have put forward the view that velocities of about 250 to 2800 seem to produce the best results. That is, a bullet at 2700 seems to offer far more destruction than the ame bullet at 2000. However the same bullet (on soft targets) at 3300 does seem to do anymore and sometimes does less.

That got me thinking about recoil.

If we require a given muzzle energy, then calibers producing that with velocities around the 2600 to 2800 seem to produce the least violent reaction when fired. I ma factoring in both recoil and blast.

For example, if we use a 375 with 300 grains at 2600 plus, then a 500 grain 458 at 2000 is about the ssme energy. A 180 grain at about 3400 is also the same (30-378)

Of these three calibers the 375 is by far the nicest to shoot.

I know very little about artillary and miltary ammo in general but i think a great deal of it even up to the big naval guns also run at around the 2500 to 2800.

Just an irrelevant thought or two.

Mike

Ku-dude,

I will try.

We have agreed that the total mass of the "gas" after burning is equal to the mass of the powder, as the powder supplies both fuel and oxidant. Part of the powder/gas mass follows the bullet to the muzzle, and is already moving at an appreciable velocity when the bullet exits the muzzle. Its energy must be accounted for in the same manner as the bullet's energy - at the muzzle. If I read you right, this is what you want to call "powder recoil." The rest of the powder/gas mass is accelerated by the pressure trapped within the barrel after the bullet uncorks the barrel. This energy will become the "gas recoil" after exhaust is complete.

The "Muzzle recoil" is a short-hand term I use for the QuickLoad output labeled "Values when the projectile's base passes the muzzle:energy of recoiling(in ftlb-energy)", this term includes the recoil energy caused by the bullet's acceleration in the barrel as well as the acceleration of the "percentage of the powder thrown forward" while that powder/gas is still within the barrel. Muzzle recoil thus normally includes "bullet recoil" and some powder recoil, but no gas effect at all.

I can get the "muzzle recoil" to equal only the bullet recoil if I set the "amount of powder thrown forward" factor to zero. The difference between the two values of muzzle recoil is the "Powder recoil", if I am reading your terminology correctly.

Here's Mike example again of a 400 gr bullet at 2400 fps with 85 grains of powder and a forgotten (but fixed) pressure distribution in the barrel at moment of exit. Bold values are direct QuickLoad outputs, the others are derived by subracting the various outputs from each other.

factor bullet powder muzzle gas total
.01 30 0 30 26 56
.66 30 8 38 18 56
.99 30 13 43 13 56

HTH,
Don

{I found and corrected a typo in the last line that affected my subtraction. A good secretary I'm not.}

[This message has been edited by Don G (edited 05-07-2001).]

Mike and Harald,

I do not mean to be rude, but I have a single-track mind. I will come back and read your posts again after Ku-dude and I gain a mutual understanding or one of us gives up!

It is very frustrating to do this via words when a few minutes at a blackboard (well, they're white now), would get us all on the same page. Then we could have a sundowner...

Don

Don,

By Jove, I think I have it! I really appreciate your taking the time to clarify this topic. With the sophisticated programs available to us today, it sometimes quite perplexing. (Information overload)

In the end, the program's results and my thinking about it are very much in line. It appears that the program accounts for the reaction to all of the mass (bullet and powder), and then uses an additional calculation to approximate the "ejecta" available for the jet thrust effect.

The amount of ejecta is dependent upon the burn rate of the powder relative to the cartridge capacity and bore size. Of course, if you have a lot of powder relative to the bore (bottleneck v. straight case) and a slow burn rate, you have a lot of available ejecta (one could call this the "Aetna" effect). As you pointed out, if the powder is mostly burned before it reaches the end of barrel, there is less ejecta and less jet effect.

In sum, the amount of gas thrust is in the range one would expect. (The condition where it is all burned in the cartridge probably only occurs with pistol powders are used for cast bullet loads; and the condition in which one has 99 percent of the powder "down stream" would require a 30/378 loaded with a super slow powder.) Your actual gas effect for most cartridges using midrange powders is going to be in the 50-75 percent range, with mid-teen to low twenty numbers in jet thrust.

Did I get it? If so, thanks again, I really appreciate your patience and ability to explain the matter. Ku-dude

Whew! Where's the sundowner? We at least deserve a cold Castle!

You truly have the sense of it all.

One tiny flaw is that you have the "factor" reversed for bottleneck (.33-.44) vs. straightwall cases (.75). The "factor" is not a measure of how fast the powder is burned, but how much of the powder/gas mass (whether burned or not) is moving at the bullet's exit velocity when the bullet leaves the barrel. (It is a lump-sum equivalent, just like Mike used in his rule-of-thumb example. Mike had the "factor" set at .5 -- although he halved the velocity, not the mass, the effect is the same.)

The bottleneck case is good at keeping more of the mass in the case longer, whereas in the straight-wall case powder, bullet and all just starts forward "burning from the back" when the primer goes off.

Think of the shoulder of the bottleneck case as causing a logjam at the case mouth and only the highest-energy gas is getting out as the powder starts to burn, until the log-jam clears.

Another way to think of it is that the amount of powder thrown forward "factor" defines how the energy released from the powder is split between kinetic energy (powder/gas mass moving toward the muzzle) and potential energy (gas pressure in all directions) at the moment the bullet exits the muzzle.

Ku-dude, my hat's off to you for pure stick-to-it-iveness. I know this must look like gibberish. I am afraid to go back and see how many times I have contradicted myself!

I don't have a Castle, but I'm going to have a cheap Red Dog and go to bed!

Don

Don,

On recoil, one thing I never been able to see is "fast" and "slow" recoil.

To me, the rifle is in free motion my the time it has only creased you shirt.

I could not count the number of times I have seen shooters at our big range havea few shots of a 375 and say "it is softer than my 300 Win or 7mm Rem".

Then the day comes when they try their new 375 out. They make a discovery that it does kick much harder than a 7mm Rem.

One of my favorite pastimes at the range was when I used to own M70s in 458. I had 2 of them and it was always interesting to see the reaction when someone benchrested them with 500 grain loads.

After about 3 shots and the idea that the recoil is a nice slow push and that the 458 is underpowered completely evaporates

Many years ago at the range a shooter was next to me with a 300 Wby and his mate was on my other side with a 270. The bloke with the 270 asked about shooting the 458. His mate with 300 Wby went into a speech about kineteic energy, velocity, slow recoil etc. and that he should find the 458 nice and easy since he had shot the 300 Wby.

The fellow with the 270 fired one shot with a 500 grain load and turned to his mate and said "you are full of bullshit"

Mike

Mike,

My 416RM, 400 grains at 2450 fps, is all I can handle off the bench at the moment. I hope I'm ready if I ever get the Lott finished!

This beer is empty, I'm outa here...

Don

Rocket MAN!! Burning out his fuse up there alone!! Just kidding. Man talk about "fuzzy numbers"!! I think that this "rocket effect" has merit because isn't it what makes a muzzle brake work? When the bullet passes into the muzzle brake, it is in "free flight", not touching the bore. The high pressure gasses pass through the holes in the sides of the muzzle brake, there by relieveng the "jet effect", by dissapating these high pressure gasses out the sides rather than out the front. A Thrust Reverser, or "clamshell", on a jet engine works much the same way. Hey!, theres an idea. How about a "retractable muzzle brake", with a big red lever on the side of the action that says, RECOIL REVERSE. With a .460 Weatherby, you would have to wear a harness to keep from losing the damn gun!, just kidding. Bill T.

The "jet effect" is dependent upon the mass rate of discharge, the net exit pressure, and the area of the discharge tube. In typical high power rifles, this can account for anywhere from 25 to 30% of the total recoil. Obviously, the propellant's burn rate affects the "jet effect" because it typically has a higher exit pressure which also causes a faster mass rate of discharge.

You have a couple of things going on here.

1. More powder means more weight of ejecta and therefore more recoil.

2. Bore size...........E=1/2M(V*V), so, if you accelerate the mass faster, like the 458 vs. 257 example, the recoil from the powder will be greater in the 257, it is coming out faster. The acceleration against your shoulder due to the powder is effected by the square of the velocity of the gas.

There is no doubt but what more powder makes for more recoil because of the powder mass being accelerated. However, the total mass of the powder is not accelerated to MV. In fact, the average velocity acquired by the mass of powder is typically less than 50% of MV. In short, the recoil because of powder acceleration is maybe 5% of the total recoil.

OKshooter,

In the example we've been using (a 10 lb .416 Rigby, 400 grain bullet at 2400fps, 85 gr of IMR4895 powder, 24 inch barrel, 8500 PSI muzzle pressure at exit, accelerating the powder in the barrel generated 8 ftlb out of a total of 56 ftlb of recoil. That's about 14% of the total recoil.

An overbore cartridge like the 7MM-RUM might only generate 7%. A high capacity straight-wall case like the Lott might run up to 16% of the total recoil.

This data was generated by QuickLoad, but my Excel spreadsheet backs it up.

The rule of thumb that Mike quoted was that the powder charge mass is accelerated to an average velocity equal to half the exit velocity of the bullet when the bullet is at the muzzle. This seems reasonable. QL says the factor actually varies from .33 to .75 as the case goes from severely overbore to straight wall. This seems reasonable to me also.

Don

Don,

I use to have a real lot of interest in rocketry. Unfortunately in Australia our rocket launching laws a like our gun law and have been for years.

I will try and find one of my old books.

As I remember thrust is basically the rate of mass exiting the nozzle and the area of the nozzle, which is about waht OK shooter has said.

I don't remember pressure being a part of it excpet of course it influenced the rate go gas flow through the nozzle.

Thus the Space Shuttle Main Engines have more
thrust in a vacuum.

Actually a 458 with 70 grains of powder could be thought of as a 2000 HP engine driving a helicopter blade and a 257 Wby with 70 grains could be though of as 2000 HP engine driving a Spitfire prop.

Mike

Don G -- Tell me where I am wrong.

If the average acceleration of the gas is 50% of MV, then we have 85 grs. of powder being accelerated to 1200 fps given the bullet's exit velocity of 2400 fps.

If we accept the proposition that momentum is conserved, then WV = wv where W is the weight of the rifle, V is the velocity imparted to the rifle, w is the weight of the powder charge, and v is the velocity of the powder charge. V = wv/W is the rearrangement the equation to determine the velocity imparted to the rifle by accelerating the powder charge.

So, we have 85 grs. of powder (0.012 lb.) at 1200 fps with a rifle that weights 10 lbs. V = wv/W = (0.012 x 1200)/10 = 1.45 fps.

Energy = 1/2mv**2 = 1/2*(10/32.16)*(1.45*1.45) = 0.33 ft-lbs.

OKShooter,

Leaving rocketry out of the discussion and just doing the numbers, it depends on how you do them.

For example:

A 10 pound rifle with 15 f/s has 34.98 ft/lbs of recoil energy.

However a 10 pound rifle with 16.45 f/s has 42.07 ft/lbs of recoil energy.

That is a 20% increase in recoil energy.

Kind of like compound interest.

A lump sum of $1000 invested at 10% is worth $2593 after 10 years.Thus an increase of $1593

At 1% it would be worth $1104. Thus an increase of $104

At 11% it will be worth $2839. Thus an increase of $246 over what 10% gives.

Mike

OK,
Mike fingered it.

Calculate the entire momentum with and without the powder, then convert those numbers to the velocity of the rifle, then find the energy of each, then find the difference.

It's one of those distributive properties that disappear when the system is not linear.

Don

[This message has been edited by Don G (edited 05-11-2001).]

Well, I guess it depends upon how one approaches it and what one is trying to establish.

There are two basic equations that purport to estimate the recoil energy of a firearm. Both are based upon the conservation of momentum, and the first one is V = wv + 4700c/W where w = weight of the bullet, v = exit velocity of the bullet, c = weight of the powder charge, and W = the weight of the firearm. The other one is V = (w + 1.75c)v/W where the variables have same meaning as they do in the first equation.

Using the example 10 lb. rifle firing a 400 gr. bullet at 2400 fps with a charge of 85 grs. of powder, both equations will calculate a velocity of 19 fps (rounded to the whole number) imparted to the rifle by the total of the mass accelerations plus thrust. The energy equation [E = (mv*v)/2] will yield 56 ft-lbs given a 10 lb. rifle with 19 fps backward velocity.

Since we generally don�t have a real good handle on the mass rate of discharge and the other components of thrust, we can use the conservation of momentum law to calculate the bullet and powder component. For the bullet, V = wv/W = 0.06 lbs * 2400 fps/10 lbs = 14 fps, or 72% of the rifle�s total recoil velocity. For the propellant -- assuming that it is equally distributed, which I doubt -- we have V = 0.01 lb * 1200 fps/10 lbs = 1 fps, or 8% of the rifle�s total recoil velocity. The remaining 20% must be the thrust component, and this would be 4 fps imparted to the rifle�s velocity because of the thrust.

Yes, since energy is proportional to the square of the velocity, calculating and summing the partial energies will result in under predicting the total system energy. If 72% of the rifle�s imparted velocity is the result of accelerating the bullet to MV, it seems reasonable to me to say that 72% of the rifle�s energy is derived from this source and so on down the line. Using the velocity percentages, we have 41 ft-lbs induced by the bullet�s acceleration, 4 ft-lbs induced by the propellant�s acceleration, and 11 ft-lbs induced by thrust.

[This message has been edited by OKShooter (edited 05-12-2001).]