Toby, Thanks for being so patient with my ignorance.--Mike Ezell
recoil
- Thread starter cliffe
- Start date
gunamonth said:
Thanks, but I'll leave this one to the "rocket scientists"
--Mike Ezell
Hi Mike. We're all here to learn something. Wouldn't be much fun otherwise. 
Toby Bradshaw
baywingdb@comcast.net
Toby Bradshaw
baywingdb@comcast.net
hello,not to add anymore too this.but what if one bore is alittle loser than the other.just like if one piston in a motor of a car.if one is that would mean less compression.so wouldn't that mean the same for the recoil.because if one barrel has alittle more or less for the bullet to travel down it would mean less or more recoil.if the bullet doesn't fit the bore as one would say just like a glove it would leave room to let some of the gases leak by.I guess what I am saying is no two barrels are the same so wouldn't that affect the recoil to.
sorry if I added to this but my mind will only think in simple terms.I'm just a dumb old country BOY that loves to shoot.
Johnboy
Shoot,shoot again,RELOAD
sorry if I added to this but my mind will only think in simple terms.I'm just a dumb old country BOY that loves to shoot.
Johnboy
Shoot,shoot again,RELOAD
before you start with another question, what's the answer to the first.the .308 and the 30-06 with the same bullet etc. how much more recoil does the 06 have........cliffe p.s. you guys got so wrapped up in theory, ya forgot to answer to answer the original question. but we learned something and thats good:thumb:
Short answer: not much.
Longer answer:
http://www.handloads.com/calc/recoil.asp
About 5% difference in recoil energy between the .308 and .30-06 with 4gr more powder in the '06 to produce the same velocity. But you'll have to plug in the specific numbers for your loads to get a more precise answer.
Toby Bradshaw
baywingdb@comcast.net
Longer answer:
http://www.handloads.com/calc/recoil.asp
About 5% difference in recoil energy between the .308 and .30-06 with 4gr more powder in the '06 to produce the same velocity. But you'll have to plug in the specific numbers for your loads to get a more precise answer.
Toby Bradshaw
baywingdb@comcast.net
Either that, or great minds think alike. It does look suspicious, though!
Toby Bradshaw
baywingdb@comcast.net
Toby Bradshaw
baywingdb@comcast.net
So when a 378 taps your shoulder out of socket and crosses your
eyeballs what percentage of that,that didn't hurt) kick is
primary recoil and what percent is secondary recoil? Reason
I ask is that the gun shop that put the muzzle brake on my 375 H&H said it would reduce recoil by about 30%. It definitely helped, but 30%? That seems awful high to me. Of course this
was a salesman talking but it would be nice to know what the
percentage break=down is on this? Maybe someone can fill us in
while I go find the ben gay.
eyeballs what percentage of that,that didn't hurt) kick is
primary recoil and what percent is secondary recoil? Reason
I ask is that the gun shop that put the muzzle brake on my 375 H&H said it would reduce recoil by about 30%. It definitely helped, but 30%? That seems awful high to me. Of course this
was a salesman talking but it would be nice to know what the
percentage break=down is on this? Maybe someone can fill us in
while I go find the ben gay.
You can figure this out using the recoil calculator,if you trust its unspoken assumptions about gas velocity). In the calculator, plug in your .375 H&H load for "Load 1" and for the comparison in "Load 2" reduce the powder charge to zero. The difference between Load 1 and Load 2 is the total recoil energy that is attributable to powder alone, and thus might potentially be reduced by a muzzle brake.
For a .375 H&H about half the recoil energy comes from the powder,300gr bullet, 80gr vs. 0gr charge weight, 2700fps). To reduce recoil energy by 30%, the muzzle brake would have to be 60% efficient,30%/50%). A brake can't be more than 50% efficient unless the holes in the brake are pointed back at least somewhat towards the shooter,the brake could theoretically be 100% efficient if the gas were pointed straight back at the shooter, there was no leakage around the bullet, and no gas energy was lost in turning it around). Most brakes have holes perpendicular the axis of the bore,i.e., NOT pointed backwards), and have about .020" of clearance between the bullet and the brake,i.e., leakage). So, I would expect an efficiency of no more than 40% or so, perhaps less,perhaps even much less). That would give a recoil reduction of 40%*50%=20% for the .375 H&H. Maybe somebody knows a real measurement of brake efficiency.
Try the calculator for something like a .220 Swift, where big powder charges push light bullets at high velocity. The potential for recoil reduction is much greater in cartridges like this, because more of the total recoil energy comes from the propellant.
I've never noticed the recoil of my .375 when hunting, but I do hate shooting mine at the bench!
Toby Bradshaw
baywingdb@comcast.net
For a .375 H&H about half the recoil energy comes from the powder,300gr bullet, 80gr vs. 0gr charge weight, 2700fps). To reduce recoil energy by 30%, the muzzle brake would have to be 60% efficient,30%/50%). A brake can't be more than 50% efficient unless the holes in the brake are pointed back at least somewhat towards the shooter,the brake could theoretically be 100% efficient if the gas were pointed straight back at the shooter, there was no leakage around the bullet, and no gas energy was lost in turning it around). Most brakes have holes perpendicular the axis of the bore,i.e., NOT pointed backwards), and have about .020" of clearance between the bullet and the brake,i.e., leakage). So, I would expect an efficiency of no more than 40% or so, perhaps less,perhaps even much less). That would give a recoil reduction of 40%*50%=20% for the .375 H&H. Maybe somebody knows a real measurement of brake efficiency.
Try the calculator for something like a .220 Swift, where big powder charges push light bullets at high velocity. The potential for recoil reduction is much greater in cartridges like this, because more of the total recoil energy comes from the propellant.
I've never noticed the recoil of my .375 when hunting, but I do hate shooting mine at the bench!
Toby Bradshaw
baywingdb@comcast.net
sorry, you lost me Toby. The recoil calucator shows the total
combined recoil doesn't it. The primary and secondary recoil together is presented as the "recoil". From your post I take it
that the two are split about 50/50. 50% primary and 50% secondary = total recoil. Is that right? I'm assuming this is a
ballpark figure since there are varibles with case size, powder,
bullet wt., gas leakage in the barrel and that sort of thing.
So if in fact roughly half of the recoil is secondary recoil
then it would be possible for a muzzle break to deliver a 30%
reduction in total recoil. I didn't know secondary recoil was
that significant. Guess I always assumed the split would have been more like 60-40 or 70-30.
Yeah, funny how that works on recoil during hunting versus
bench shooting. The best recoil brake going is a trophy animal in the cross hairs
combined recoil doesn't it. The primary and secondary recoil together is presented as the "recoil". From your post I take it
that the two are split about 50/50. 50% primary and 50% secondary = total recoil. Is that right? I'm assuming this is a
ballpark figure since there are varibles with case size, powder,
bullet wt., gas leakage in the barrel and that sort of thing.
So if in fact roughly half of the recoil is secondary recoil
then it would be possible for a muzzle break to deliver a 30%
reduction in total recoil. I didn't know secondary recoil was
that significant. Guess I always assumed the split would have been more like 60-40 or 70-30.
Yeah, funny how that works on recoil during hunting versus
bench shooting. The best recoil brake going is a trophy animal in the cross hairs
The propellant produces both primary and secondary recoil -- the primary recoil is produced by the mass of of the propellant before it exits the muzzle, and the secondary recoil is produced as the gas leaving the muzzle,after the bullet leaves) accelerates to its maximum velocity.
In principle, with a perfectly functioning muzzle brake ALL of the recoil produced by the propellant gases could be eliminated. In thinking about it further, I suppose that if the gases can be directed perpendicular to the bore there would be no downrange momentum from the gas, hence no recoil attributable to the propellant. So maybe a 30% reduction in recoil in the .375 H&H only requires 60% efficiency in a brake with perpendicular holes -- that's believable,at least to me).
Every little bit helps on the big boomers. I'm planning to check the sighting on my CZ 550 .375 on Sunday to get ready for a trip to South Africa and Namibia next summer -- that should give my shoulder enough time to heal.
Toby Bradshaw
baywingdb@comcast.net
In principle, with a perfectly functioning muzzle brake ALL of the recoil produced by the propellant gases could be eliminated. In thinking about it further, I suppose that if the gases can be directed perpendicular to the bore there would be no downrange momentum from the gas, hence no recoil attributable to the propellant. So maybe a 30% reduction in recoil in the .375 H&H only requires 60% efficiency in a brake with perpendicular holes -- that's believable,at least to me).
Every little bit helps on the big boomers.
I'm planning to check the sighting on my CZ 550 .375 on Sunday to get ready for a trip to South Africa and Namibia next summer -- that should give my shoulder enough time to heal. Toby Bradshaw
baywingdb@comcast.net
I am a new member but have been visiting here for many years. I wanted to register just to reply to this thread.
The two issues, .308 recoil vs .30-06’ recoil and secondary recoil are the same thing. The recoil of a firearm is kinetic energy by a different name. Relating a .308 to a .30-06’ is a specious argument. But I am sure Cliffe was asking the question in all sincerity. Also the notion of primary and secondary recoil is fashionable load of bunk. It is an advertising ploy by those who sell muzzle brakes.
First let me explain why the .308,if it were true) has less recoil than a 30-06; all things being equal. And yes others have said the same thing. It is due to the different powder charge weights between the .308 and the .30-06’. Hypothetically the .308 would only need about 50 grains of powder to the .30-06’s 58 grains of powder. The greater powder charge from the 06’ yields about 2.5 foot-pound force more translational kinetic energy than the .308.
This leads me to the fallacy of secondary recoil. The equations that yield the recoil energy,translational kinetic energy) for a firearm include the total translational kinetic energy for both bullet and powder charge. That is, when you measure,make a calculation) the recoil of a firearm you input the weight and velocity for both bullet and powder charge. Because the powder charge is converted from a solid to a gas dose not change its weight or velocity. The powder charge due to its weight and velocity is going to create a measurable amount of energy. If you change the powder charge weight you will change the measureable amount of energy. Arguing secondary recoil is like insisting that a pound of lead weighs more than a pound of feathers.
On a technical note without going into a doctoral dissertation; the recoil of a firearm is about kinetic energy of an open thermodynamic system. Those of you that wish to endlessly argue Newton’s third law and conservation of momentum are thinking in terms of closed thermodynamic systems. The recoil of a firearm is a measurement of kinetic energy as dictated by Newton’s second law.
The two issues, .308 recoil vs .30-06’ recoil and secondary recoil are the same thing. The recoil of a firearm is kinetic energy by a different name. Relating a .308 to a .30-06’ is a specious argument. But I am sure Cliffe was asking the question in all sincerity. Also the notion of primary and secondary recoil is fashionable load of bunk. It is an advertising ploy by those who sell muzzle brakes.
First let me explain why the .308,if it were true) has less recoil than a 30-06; all things being equal. And yes others have said the same thing. It is due to the different powder charge weights between the .308 and the .30-06’. Hypothetically the .308 would only need about 50 grains of powder to the .30-06’s 58 grains of powder. The greater powder charge from the 06’ yields about 2.5 foot-pound force more translational kinetic energy than the .308.
This leads me to the fallacy of secondary recoil. The equations that yield the recoil energy,translational kinetic energy) for a firearm include the total translational kinetic energy for both bullet and powder charge. That is, when you measure,make a calculation) the recoil of a firearm you input the weight and velocity for both bullet and powder charge. Because the powder charge is converted from a solid to a gas dose not change its weight or velocity. The powder charge due to its weight and velocity is going to create a measurable amount of energy. If you change the powder charge weight you will change the measureable amount of energy. Arguing secondary recoil is like insisting that a pound of lead weighs more than a pound of feathers.
On a technical note without going into a doctoral dissertation; the recoil of a firearm is about kinetic energy of an open thermodynamic system. Those of you that wish to endlessly argue Newton’s third law and conservation of momentum are thinking in terms of closed thermodynamic systems. The recoil of a firearm is a measurement of kinetic energy as dictated by Newton’s second law.
1. The firearm and cartridge are a closed system, all of whose energy comes from the combustion of the propellant.
2. Momentum, not energy, is conserved. Thank goodness, or the recoil energy delivered to your shoulder would be the same as the energy delivered to the target at the muzzle.
3. Secondary recoil is a real phenomenon, produced by acceleration of the escaping muzzle gases after the bullet leaves the muzzle. If secondary recoil doesn't exist, why does a grain of propellant produce more recoil than a grain of bullet weight?
I'm done with this thread.
Toby Bradshaw
baywingdb@comcast.net
2. Momentum, not energy, is conserved. Thank goodness, or the recoil energy delivered to your shoulder would be the same as the energy delivered to the target at the muzzle.
3. Secondary recoil is a real phenomenon, produced by acceleration of the escaping muzzle gases after the bullet leaves the muzzle. If secondary recoil doesn't exist, why does a grain of propellant produce more recoil than a grain of bullet weight?
I'm done with this thread.
Toby Bradshaw
baywingdb@comcast.net
Mr. Bradshaw,
I'm sorry to hear that you through with this thread. I will try to responded as civilly as possible.
1 A firearm is an open system. An internal combustion engine is a closed system. Whether a system is open or closed has nothing to do with combustion. Remember we are not discussing rocket nozzles here; just small pieces of lead and copper being tossed through the air a few hundred yards and the kick of your favorit gun.
2 I'm sorry I didn't understand.
3 Not true. Again a pound of lead does not weight more than a pound of feathers. Your argument is that of a red herring. Let me first agree with you through this next sentence.
If 1 grain of powder has a velocity of 5200fps,60ft-lbf) and a 1 grain bullet has the velocity of 2700fps,16ft-lbf) then yes, 1 grain of powder will produce more recoil than 1 grain of projectile. However your argument as I inferred it has nothing to do with recoil.
To calculate the recoil of a firearm you must use the sum,see the plus sign in the equation)of the momentum for both the projectile and propellant: Vgu = {,mp • vp) +, mc • vc)} / mgu • 7000
Where as Vgu is the velocity of the firearm, mp is weight of the bullet, vp is the velocity of the bullet, mc is weight of the powder charge, vc is velocity of the powder charge, mgu is the weight of the firearm in pounds and 7000 is to set the rest of the equation to pounds.
P.S. the above equation is only the first step in finding the recoil energy of a firearm. Once you have found for the velocity of the firearm than you can plug that into the translational kinetic energy equation: Etgu = mgu • vgu² / 2 • gc
Where as Etgu is the translational kinetic energy of the firearm,recoil), mgu is the weight of the firearm in pounds, vgu² is the velocity of the firearm square,from the first equation), 2 is the average velocity of the firearm,from the original kinetic energy equation wz) and gc is the dementional constant,usualy 32.163)
I'm sorry to hear that you through with this thread. I will try to responded as civilly as possible.
1 A firearm is an open system. An internal combustion engine is a closed system. Whether a system is open or closed has nothing to do with combustion. Remember we are not discussing rocket nozzles here; just small pieces of lead and copper being tossed through the air a few hundred yards and the kick of your favorit gun.
2 I'm sorry I didn't understand.
3 Not true. Again a pound of lead does not weight more than a pound of feathers. Your argument is that of a red herring. Let me first agree with you through this next sentence.
If 1 grain of powder has a velocity of 5200fps,60ft-lbf) and a 1 grain bullet has the velocity of 2700fps,16ft-lbf) then yes, 1 grain of powder will produce more recoil than 1 grain of projectile. However your argument as I inferred it has nothing to do with recoil.
To calculate the recoil of a firearm you must use the sum,see the plus sign in the equation)of the momentum for both the projectile and propellant: Vgu = {,mp • vp) +, mc • vc)} / mgu • 7000
Where as Vgu is the velocity of the firearm, mp is weight of the bullet, vp is the velocity of the bullet, mc is weight of the powder charge, vc is velocity of the powder charge, mgu is the weight of the firearm in pounds and 7000 is to set the rest of the equation to pounds.
P.S. the above equation is only the first step in finding the recoil energy of a firearm. Once you have found for the velocity of the firearm than you can plug that into the translational kinetic energy equation: Etgu = mgu • vgu² / 2 • gc
Where as Etgu is the translational kinetic energy of the firearm,recoil), mgu is the weight of the firearm in pounds, vgu² is the velocity of the firearm square,from the first equation), 2 is the average velocity of the firearm,from the original kinetic energy equation wz) and gc is the dementional constant,usualy 32.163)
I have a couple of questions and then I will sit back and just listen, as I too am done with this thread after this post.
1.-Do you own a gun?
2.-Do you own a gun with a brake?
3.-If you own a gun with a brake,please explain your theory on how it works to reduce recoil;or doesn't work.
Your signature line won't make you many friends anywhere that I know of.---Mike Ezell
1.-Do you own a gun?
2.-Do you own a gun with a brake?
3.-If you own a gun with a brake,please explain your theory on how it works to reduce recoil;or doesn't work.
Your signature line won't make you many friends anywhere that I know of.---Mike Ezell
Thank you gunsandgunsmithing for this opportunity. I will for the community, give you the credentials you have asked for as not to be impeached. Also I have written a book on kinetic energy and how it applies to small arms. The most important part of that book is where kinetic energy comes from and how the velocity gets squared with in the classic statement: E = 1/2mv^2.
1-Over my life time I have owned 30 some rifles, 10 shotguns and 2 pistols. Currently I am building a Palma rifle as soon as I get my wife a new ski jacket. after the ski jacket I'm going to order a Tubb butt plate and a V block to mount a R700 short action. I picked up a cheap X40 stock from Gun Parts World in New York. I am new to 1000yard prone shooting. Also I am waiting for my Montana 1999 Professional Hunter Action to arrive. It should be here in February. I'm going to build a .460 Weatherby Mag.
2-Yes and No. Yes I have a ported choke for my 870. No I have no rifles with brakes on them. However, I am going to build a countersniper rifle from an old M98 action I have. This countersniper will be chambered in .338 Achilles. The Achilles is a RUM case shorten to 2.55†with 35 degree shoulders. I already shoot the 7mm Achilles. This new rifle will have a Score High muzzle brake called, “The Tactical Bolt.â€
3-Okay, first thing: kinetic energy is not a theory nor is how I apply it. Kinetic energy has been around for about 276 years and is establish and proven science,my book contains the mathematical proof for kinetic energy).
How dose a muzzle brake work,and I know you know how it works) and how is the energy transferred via muzzle ejecta to the brake,which you may not understand). Now work with me here. The full answer is about two pages long with some nasty mathematics. So I'm going to take some short cuts to put this in less than technical terms.
The powder charge within the cartridge is a chemical in a solid form. Most single base nitrocellulose powders contain about 175ft-lbf of energy per grain. Just like food has calories, 1cal = 3087ft-lbf). So just for back ground as Einstein said, “all energy is mutually convertible.†So we are going to convert the solid, chemical energy of the powder charge,56 grains of IMR4350) to a gas in the form of thermodynamic energy. That is done by dropping the hammer onto the primer.
Okay, we now have a 56 grains of powder that has converted to a hot compressed mass of gas at about 60,000psi. The gas expands about 2000 times plus another 500 times due to the heat of oxidization. That expanding gas is pushing in all directions. The gases push radial to the case body, sticking that case body to the chamber wall. The gases push to the base of the case transferring its energy to rear though the chamber walls,some and hopefully very little as bolt thrust and of course rearward case stretching), through the stock and into your shoulder. The gases push forward on the base of the bullet transferring its energy to bullet motion. I think you already understand this.
The pushing of the gas rearward is a calculable load as represented by the powder charge weight times its velocity. The pushing of the gasses forward is a calculable load as represented by the bullet weight times its velocity. The sum of the powder charge weight times its velocity and the bullet weight times its velocity divided by the guns weight will equal the velocity of the gun in rearward motion into you shoulder. So, gun weight times its velocity squared,just found by the sum of the powder charge and bullet) divided by 2 times the dimensional constant equals the foot-pound force of recoil.
So I think all the forgoing is no new news to you. Here is where it all gets tied together. The recoil of the gun is figure all the way up to the point where the base of the bullet exits the muzzle and the system is now open. All energies transferred as recoil energy are calculated and taken it account. The “jet effect†or “secondary recoil†is already accounted for in the original event of the energy transfer form the powder charge to the bullet and gun/shooter system.
The remaining energy that the bullet now possesses, may now be used to get it down range to the target. If that target is paper, any leftover energy is wasted and delivered to the impact area. In hunting that left over energy is delivered to the game animal; which is also a new and open sysytem.
The longer the barrel the greater the energy transfer from the expanding gases to the bullet and gun. Therefore, the greater the energy transfer the greater the bullet energy and recoil energy. However, any excessive expanding gases not used within the barrel is expelled into a new system and is waste product. This new system is called the atmosphere. We all know this extra stuff,waste product) as muzzle blast or muzzle ejecta.
So this is how a muzzle brake works. The wasted gases we know as muzzle blast are redirected through holes and or baffles in the brake and the wasted energies are recovered by the brake which is a mechanical device. As the gases hit the perpendicular walls of the holes and or baffles, the energy in a forward motion is transferred to the brake then to the barrel pulling the gun forward. The difference in the recoil energy of the gun minus the muzzle blast equals the felt,or net) recoil. You can use Le Duc’s equation to determine the waste muzzle ejecta. I will not take up your time explaining porting. It’s nearly the same thing. Just remember the first port within the barrel becomes your pseudo muzzle for the purposes of a calculation. In the simplest of terms, a muzzle brake is a sail. It is a mechanical device that captures a waste product’s energy and diverts it back into a system. There is no secondary event or recoil. This mechanical action happens exactly at the same time as everything else.
The recoil of a gun, felt recoil, bullet energy, thermodynamic energy, chemical energy, mechanical energy, brake enegy, kinetic energy is all the same thing. All of these energies are calculable as a measurement. Their true name as it pertains to objects in motion is called, translational kinetic energy and the measurement is in foot-pound force.
I thank you for this opportunity to explain recoil and muzzle brake dynamics.
P.S. its late and I'll will have typos. God bless you all.
NOTE:
An open system is when energy transfers from one system to another. Example: A Firearm; the energy of a powder charge is transferred to a bullet. When the bullet exits the muzzle,leaving the system) the bullet may now transfer its energy to a target,another system).
A closed system is when no energy transfers out of the system. Example: A Hydraulic Jack; via a jack handle the hydraulic fluid is pumped from the reservoir and small piston into a larger piston. The increasing pressure within the large piston then causes it to rises. When the bypass valve is opened the pressurized fluid is diverted away from the large piston back into the reservoir. The large piston now falls and descends. All energies say within the jack.
1-Over my life time I have owned 30 some rifles, 10 shotguns and 2 pistols. Currently I am building a Palma rifle as soon as I get my wife a new ski jacket. after the ski jacket I'm going to order a Tubb butt plate and a V block to mount a R700 short action. I picked up a cheap X40 stock from Gun Parts World in New York. I am new to 1000yard prone shooting. Also I am waiting for my Montana 1999 Professional Hunter Action to arrive. It should be here in February. I'm going to build a .460 Weatherby Mag.
2-Yes and No. Yes I have a ported choke for my 870. No I have no rifles with brakes on them. However, I am going to build a countersniper rifle from an old M98 action I have. This countersniper will be chambered in .338 Achilles. The Achilles is a RUM case shorten to 2.55†with 35 degree shoulders. I already shoot the 7mm Achilles. This new rifle will have a Score High muzzle brake called, “The Tactical Bolt.â€
3-Okay, first thing: kinetic energy is not a theory nor is how I apply it. Kinetic energy has been around for about 276 years and is establish and proven science,my book contains the mathematical proof for kinetic energy).
How dose a muzzle brake work,and I know you know how it works) and how is the energy transferred via muzzle ejecta to the brake,which you may not understand). Now work with me here. The full answer is about two pages long with some nasty mathematics. So I'm going to take some short cuts to put this in less than technical terms.
The powder charge within the cartridge is a chemical in a solid form. Most single base nitrocellulose powders contain about 175ft-lbf of energy per grain. Just like food has calories, 1cal = 3087ft-lbf). So just for back ground as Einstein said, “all energy is mutually convertible.†So we are going to convert the solid, chemical energy of the powder charge,56 grains of IMR4350) to a gas in the form of thermodynamic energy. That is done by dropping the hammer onto the primer.
Okay, we now have a 56 grains of powder that has converted to a hot compressed mass of gas at about 60,000psi. The gas expands about 2000 times plus another 500 times due to the heat of oxidization. That expanding gas is pushing in all directions. The gases push radial to the case body, sticking that case body to the chamber wall. The gases push to the base of the case transferring its energy to rear though the chamber walls,some and hopefully very little as bolt thrust and of course rearward case stretching), through the stock and into your shoulder. The gases push forward on the base of the bullet transferring its energy to bullet motion. I think you already understand this.
The pushing of the gas rearward is a calculable load as represented by the powder charge weight times its velocity. The pushing of the gasses forward is a calculable load as represented by the bullet weight times its velocity. The sum of the powder charge weight times its velocity and the bullet weight times its velocity divided by the guns weight will equal the velocity of the gun in rearward motion into you shoulder. So, gun weight times its velocity squared,just found by the sum of the powder charge and bullet) divided by 2 times the dimensional constant equals the foot-pound force of recoil.
So I think all the forgoing is no new news to you. Here is where it all gets tied together. The recoil of the gun is figure all the way up to the point where the base of the bullet exits the muzzle and the system is now open. All energies transferred as recoil energy are calculated and taken it account. The “jet effect†or “secondary recoil†is already accounted for in the original event of the energy transfer form the powder charge to the bullet and gun/shooter system.
The remaining energy that the bullet now possesses, may now be used to get it down range to the target. If that target is paper, any leftover energy is wasted and delivered to the impact area. In hunting that left over energy is delivered to the game animal; which is also a new and open sysytem.
The longer the barrel the greater the energy transfer from the expanding gases to the bullet and gun. Therefore, the greater the energy transfer the greater the bullet energy and recoil energy. However, any excessive expanding gases not used within the barrel is expelled into a new system and is waste product. This new system is called the atmosphere. We all know this extra stuff,waste product) as muzzle blast or muzzle ejecta.
So this is how a muzzle brake works. The wasted gases we know as muzzle blast are redirected through holes and or baffles in the brake and the wasted energies are recovered by the brake which is a mechanical device. As the gases hit the perpendicular walls of the holes and or baffles, the energy in a forward motion is transferred to the brake then to the barrel pulling the gun forward. The difference in the recoil energy of the gun minus the muzzle blast equals the felt,or net) recoil. You can use Le Duc’s equation to determine the waste muzzle ejecta. I will not take up your time explaining porting. It’s nearly the same thing. Just remember the first port within the barrel becomes your pseudo muzzle for the purposes of a calculation. In the simplest of terms, a muzzle brake is a sail. It is a mechanical device that captures a waste product’s energy and diverts it back into a system. There is no secondary event or recoil. This mechanical action happens exactly at the same time as everything else.
The recoil of a gun, felt recoil, bullet energy, thermodynamic energy, chemical energy, mechanical energy, brake enegy, kinetic energy is all the same thing. All of these energies are calculable as a measurement. Their true name as it pertains to objects in motion is called, translational kinetic energy and the measurement is in foot-pound force.
I thank you for this opportunity to explain recoil and muzzle brake dynamics.
P.S. its late and I'll will have typos. God bless you all.
NOTE:
An open system is when energy transfers from one system to another. Example: A Firearm; the energy of a powder charge is transferred to a bullet. When the bullet exits the muzzle,leaving the system) the bullet may now transfer its energy to a target,another system).
A closed system is when no energy transfers out of the system. Example: A Hydraulic Jack; via a jack handle the hydraulic fluid is pumped from the reservoir and small piston into a larger piston. The increasing pressure within the large piston then causes it to rises. When the bypass valve is opened the pressurized fluid is diverted away from the large piston back into the reservoir. The large piston now falls and descends. All energies say within the jack.
hey, weren't we shooting 150's, imr4895 with 47 grains for the 308 and 51 grains for the 30-06. and about .8 more recoil for the 06?? glo where did you come up with 8 grains more powder for the 30-06 to the 308 with the same speed??
btw my sierra manual lists 43.7 gr. imr4895 at 2800 and 48.1 grains for the 30-06 to go 2800. so 4.4 grains dif., still pretty close and maybe 1 ft.lb. more recoil, so not that much.....cliffe
btw my sierra manual lists 43.7 gr. imr4895 at 2800 and 48.1 grains for the 30-06 to go 2800. so 4.4 grains dif., still pretty close and maybe 1 ft.lb. more recoil, so not that much.....cliffe
Upgrades & Donations
This Forum's expenses are primarily paid by member contributions. You can upgrade your Forum membership in seconds. Gold and Silver members get unlimited FREE classifieds for one year. Gold members can upload custom avatars.
Click Upgrade Membership Button ABOVE to get Gold or Silver Status.
You can also donate any amount, large or small, with the button below. Include your Forum Name in the PayPal Notes field.
To DONATE by CHECK, or make a recurring donation, CLICK HERE to learn how.
Click Upgrade Membership Button ABOVE to get Gold or Silver Status.
You can also donate any amount, large or small, with the button below. Include your Forum Name in the PayPal Notes field.
To DONATE by CHECK, or make a recurring donation, CLICK HERE to learn how.