Is there a mathematical relation ship between ME and Recoil energy?
- Thread starter Doyle
- Start date
FrankenMauser
New member
Yes, with some other basic numbers, like bullet weight, rifle weight, etc.
To be even more precise, bore diameter (for "rocket nozzle" effect), powder charge weight (much of the powder charge gets 'propelled' down the barrel, too), firing system mass (forward moving mass), and other factors may need to be considered.
To be even more precise, bore diameter (for "rocket nozzle" effect), powder charge weight (much of the powder charge gets 'propelled' down the barrel, too), firing system mass (forward moving mass), and other factors may need to be considered.
Yes.
Recoil is calculated using the following factors
Muzzle Velocity
Weight of the projectile
Weight of the firearm
Weight of the powder charge.
The math is over my head, I just plug the numbers in to an online calculator
http://www.shooterscalculator.com/recoil-calculator.php
Change any one of the above factors and recoil can change quite a bit. The weight of the powder charge is the one most never think about. This is why a 30-06 will recoil a lot more than a 308. Even though they shoot the same bullet weights to similar speeds the 30-06 burns a lot more powder doing it.
Recoil is calculated using the following factors
Muzzle Velocity
Weight of the projectile
Weight of the firearm
Weight of the powder charge.
The math is over my head, I just plug the numbers in to an online calculator
http://www.shooterscalculator.com/recoil-calculator.php
Change any one of the above factors and recoil can change quite a bit. The weight of the powder charge is the one most never think about. This is why a 30-06 will recoil a lot more than a 308. Even though they shoot the same bullet weights to similar speeds the 30-06 burns a lot more powder doing it.
Yes.
Next question...
IF you have the muzzle energy, you have the recoil energy, they are the same (Newton's law). The same force pushing the bullet pushes the gun. They are equal, because they are one.
However, people measure recoil energy in different ways. The most common way is to measure (calculate) the energy delivered to YOU, the shooter during recoil. This is done by factoring in the weight of the gun.
Additionally, there is "felt recoil", which is how you feel the recoil energy delivered. This involves other factors, including how the stock fits you, and many others, and is a very subjective and personal thing.
These things can be calculated, except for the personal and subjective "feel" and can be taken to the nth degree by math, but most of us stop when the numbers go beyond what we consider significant.
Feel, on the other hand is personal. 15ft/lbs of recoil energy is still the same number on paper but it feels different in a 7lb rifle and a 10lb rifle. it will also feel different in rifles of the same weight, if one has a stock that fits you properly and the other does not.
stinkeypete
New member
Technically “no” because it can be a complex system and one needs to construct a model, but you care about “impulse”, not total energy.
Impulse is how the total energy reacts with your shoulder OVER TIME.
There will be a curve of how the recoil is directed. Sharp abrupt spike are “slaps” and are very unpleasant.
I am thinking of the Savage ML II I used to own. Very long barrel and shooting 360g cast with a slower powder would rock your world and push you back a step while a 240g hp with a very fast powder had less total energy but “punched” your shoulder so hard you didn’t want to use that powder again.
If you add other ways to slow down the energy transfer you have much milder perceived recoil... semi automatic mechanisms and the like don’t reduce the total recoil energy much- but they do spread the energy out over much longer time. Well... some of the hot fast gas isn’t going straight out the muzzle so it’s heating up parts and that energy has to be put in your mathematical model. Brass stretches, metal heats up, steel flexes... not to mention the new elastomer recoil pads that turn shock energy in to molecular heat energy.
As said above.. a heavier rifle gives the same reactive energy as a lighter rifle- it just gives it to you over longer time. Easy low millisecond payments, no bruising till January! As said above, a poorly fit stock is going to maybe have higher pressure points on your shoulder. Like a nice person sitting on your lap vs standing on your foot with high heels- same force, different sensation.
Impulse is how the total energy reacts with your shoulder OVER TIME.
There will be a curve of how the recoil is directed. Sharp abrupt spike are “slaps” and are very unpleasant.
I am thinking of the Savage ML II I used to own. Very long barrel and shooting 360g cast with a slower powder would rock your world and push you back a step while a 240g hp with a very fast powder had less total energy but “punched” your shoulder so hard you didn’t want to use that powder again.
If you add other ways to slow down the energy transfer you have much milder perceived recoil... semi automatic mechanisms and the like don’t reduce the total recoil energy much- but they do spread the energy out over much longer time. Well... some of the hot fast gas isn’t going straight out the muzzle so it’s heating up parts and that energy has to be put in your mathematical model. Brass stretches, metal heats up, steel flexes... not to mention the new elastomer recoil pads that turn shock energy in to molecular heat energy.
As said above.. a heavier rifle gives the same reactive energy as a lighter rifle- it just gives it to you over longer time. Easy low millisecond payments, no bruising till January! As said above, a poorly fit stock is going to maybe have higher pressure points on your shoulder. Like a nice person sitting on your lap vs standing on your foot with high heels- same force, different sensation.
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Thanks for the answers. Now, for the real-world application that I was trying to solve.
The firearm in question is going to be the same - an Encore 7mm-08 pistol. The riddle to solve was to somewhat predict the relative recoil of various ammo choices. I.E., if a normal load producing XXXX ft/lbs of muzzle energy kicked a certain amount could I reasonably assume a 20% reduction in recoil if I dropped the muzzle energy by 20%.
The firearm in question is going to be the same - an Encore 7mm-08 pistol. The riddle to solve was to somewhat predict the relative recoil of various ammo choices. I.E., if a normal load producing XXXX ft/lbs of muzzle energy kicked a certain amount could I reasonably assume a 20% reduction in recoil if I dropped the muzzle energy by 20%.
It depends on how you reduce the energy. Is it going to be reduced through lighter bullets or the same bullet weight at reduced speed. Or a combination of both.
You're making this harder than it should be. ME isn't a good predictor of anything. Figure out which bullet weights you want to use, the velocities you want to get and the powder charges needed to accomplish that. Then plug in the data to the link above to figure out recoil.
I think you'll see the biggest recoil reduction by going to lighter bullets. But you just have to play around with the numbers.
You're making this harder than it should be. ME isn't a good predictor of anything. Figure out which bullet weights you want to use, the velocities you want to get and the powder charges needed to accomplish that. Then plug in the data to the link above to figure out recoil.
I think you'll see the biggest recoil reduction by going to lighter bullets. But you just have to play around with the numbers.
LeverGunFan
New member
Also need to include the energy of the propellant gases, they are a significant contribution to recoil energy in high power rifles. Per SAAMI, the velocity of the propellant gasses in high powered rifles is about 1.75 times the muzzle velocity of the bullet. So including the weight of the propellant gases (assumed to be same as the powder weight) can result in recoil energy about three times that of the bullet alone.
Yes, it’s easy: Since energy is conserved it’s *exactly* the same.
Newton’s 3rd says the supporting structure (you plus the gun mass if you’re standing up holding it) will experience exactly 12,000 ft lbs of energy if you’re shooting a 750 grain 50 cal round. That’s enough to pick a car up 2-3 feet.
Absorbers and mass of the shooter and weapon will slow the impulse down but you will still experience 12,000 ft lbs of energy, not in the 90 mS the bullet takes to transit the barrel but more like 1/3 - 1/2 a second.
If the gun was mass-less, the recoil would be over in 90 mS, and your shoulder would shatter just like it was hit by the bullet. Thank goodness for conservation of momentum and 35 lbs .50 cals.
stinkeypete
New member
Txaz... love you man, but what about recoil due to the gas from the burnt powder? That’s going out the front, too. See post number 9. Yes, you want a muzzle break on a .50.
As for your car.. momentum is .... it’s horrible to explain in the Imperial system... but momentum is feet pounds per second, you have to convert that to a force (foot pounds per sec squared)... or think about this- can one put a lever under a car and use the recoil of the rifle to instantly jack the whole car up 2-3 feet? Doesn’t sound right.
As for your car.. momentum is .... it’s horrible to explain in the Imperial system... but momentum is feet pounds per second, you have to convert that to a force (foot pounds per sec squared)... or think about this- can one put a lever under a car and use the recoil of the rifle to instantly jack the whole car up 2-3 feet? Doesn’t sound right.
Hey there StinkeyPete,
so you want to make this a 3rd or 4th order differential equation instead of a somewhat simple reasonably accurate explaination..... OK, I'm game. (In reality it's about a 7th order equation to get it precise, but hey who's counting that minutia?)
Absolutely the burnt powder makes a difference, but not as much as the bullet. And, .... how were you going to measure that highly turbulent transient gas movement so that it accurately to fit it in the equation?
I never said that a 1.7 ounce projectile going 2,000 miles per hour would *efficiently* lift the car but the amount of energy is the same as lifting 4,000 lbs 3 feet. Be it a car or 480.2 gallons of water.
As for Imperial system, it works just fine for people on this forum. Most here probably can't tell you the difference in a Newton-Meter, a Joule and a Fig Newton, and don't really care or need to know.
As for a muzzle brake on a .50, do real men really need a muzzle brake? Yes, it makes a substantial difference, but again, we're adding complexity to the basic question that Doyle asked.
So as a high school Physics teacher, I'm assigning the following homework to you:
What is:
a) The initial starting torque of the following .50 caliber rifle in the 6 barrel configuration:
https://en.wikipedia.org/wiki/GAU-19#/media/File:GAU_19A_GECAL_50_2.jpg
b) Using the GAU-19 in 3 barrel configuration mounted facing backwards on the back of an otherwise unpowered go cart, how fast would the GAU propel the vehicle if the total weight of ammo, gun and cart was 4,000 lbs.
Show all work.
Muzzle energy and recoil energy can NOT possibly be the same. This is not a case of conservation of energy, but conservation of momentum. The calculation is pretty simple if free recoil energy is needed. It is a bit tricky if recoil force is needed.
No expert here. Just remember high school physics.
-TL
Sent from my SM-G930T using Tapatalk
No expert here. Just remember high school physics.
-TL
Sent from my SM-G930T using Tapatalk
There's no reason to try to calculate recoil energy from muzzle energy.
If you can calculate muzzle energy, you can probably also calculate muzzle momentum.
Recoil momentum is identical to muzzle momentum which means that with the firearm's weight, you can calculate recoil velocity from the recoil momentum.
Then, with recoil velocity and the mass of the firearm, you can calculate recoil energy. Recoil energy will not be the same as muzzle energy.
As previously noted, there are online calculators that will perform those calculations for you.
http://www.shooterscalculator.com/recoil-calculator.php
https://www.jbmballistics.com/cgi-bin/jbmrecoil-5.1.cgi
http://kwk.us/recoil.html
If you can calculate muzzle energy, you can probably also calculate muzzle momentum.
Recoil momentum is identical to muzzle momentum which means that with the firearm's weight, you can calculate recoil velocity from the recoil momentum.
Then, with recoil velocity and the mass of the firearm, you can calculate recoil energy. Recoil energy will not be the same as muzzle energy.
As previously noted, there are online calculators that will perform those calculations for you.
http://www.shooterscalculator.com/recoil-calculator.php
https://www.jbmballistics.com/cgi-bin/jbmrecoil-5.1.cgi
http://kwk.us/recoil.html
This is where things get confusing, for me, at least. Different math formulas, different terms for energy measurement at different places, etc.
Physics is a fantastic thing, but there are places in physics where the math works but reality doesn't. Friend was working a physics problem once, I was a little interested, but amazed about the "conclusion".
the situation was, a swimmer starts at one end of a pool, swims to the other end, and then returns, getting out exactly where they got in. Now, here's what amazed me, what the book wanted the student to do was show the work (formulas and calculations) that proved that the swimmer went nowhere.
To me, the swimmer went down, and back so they went somewhere, despite getting out right where they got it, but the "physics" formulas said the swimmer went "nowhere" and they wanted the student to show the proof of that.
SO, no matter what formulas you use, where you measure, and what you call it, since ALL the energy in the system comes from the firing of the round, how can recoil energy and muzzle energy NOT be exactly the same???
OK, I get how the different masses of bullet, gun, shooter, powder gas, everything involved is a factor, and these change the numbers when measured at different points in the system (and different points in time) but since all the energy comes from the round to begin with how can it not be the same, just looking different due to different to the numbers and formulas used expressing it??
Are we in a situation like the old story. where engineers calculate the bumble bee cannot fly, but the bee, does not know this, and flys anyway?
You can calculate which load will, on paper, have less recoil in your gun. BUT, what you feel is what you feel, and its not impossible that what looks like a significant difference on paper might not be as big a felt difference, in your hands.
Not that the swimmer went nowhere, but that the NET work that the swimmer did was zero.
Work has to do with mass being displaced and if there is no net displacement, there is no net work.
At any point along the swim, one could stop the swimmer and determine the amount of work done to that point. Or, one could determine the amount of work done by the swimmer between two points in the swim. But if the swimmer comes back to the original point then the net displacement is zero and the net work is zero.
https://www.britannica.com/science/work-physics
Imagine that you win a million dollars in the lottery and then lost it all gambling the same day. Your net worth didn't change when only comparing the beginning of the day and the end of the day. That doesn't mean you never won any money that day or never lost any money that day, it's simply an assessment of where you are at the end of the day vs. where you started the day.
Imagine that your job is moving a stack of bricks from one location to another. You move all the bricks, but then move them back to their original location before your boss sees them. Will your boss say you did a lot of work that day when you tell him you moved the bricks twice? Or will your boss say you did no work because the bricks are right where they started. Obviously you put forth a lot of effort, but the net effect is zero in terms of accomplishing your assigned work.
The value of physics is that it does provide insight into what happens in the real world. It is necessary to fully understand the concepts being used before it's possible to make sense of them, however.
A big problem with this is that the physics we learn in school is often simplified to make the problems tractable. So, for example, people often remember that all objects fall at the same speed, but forget that the problem was simplified by omitting air resistance. It is true that all objects fall at the same speed in a vacuum, but it is absolutely not true that the all fall at the same speed in air.
The simplest answer is that it just doesn't work like that.
Muzzle momentum and recoil momentum will be the same, but muzzle energy and recoil energy will not.
Unfortunately, at the moment, I can't think of a good way to explain why this is true in a way that would make intuitive sense.
Felt recoil depends on a lot of things that have nothing to do with the simple physics of the gun's recoil. The best example of this I can think of is when I let a friend with small hands shoot a Beretta 92 pistol. She complained that the recoil was severe which I didn't understand until she showed me that it had burst a small blood vessel in her thumb causing bruising, swelling and pain. Because the grip was too large for her, it had recoiled against the knuckle of her thumb and broken a blood vessel. The grip fit my hand and so the felt recoil was negligible.
The recoil velocity and recoil energy were the same for both of us, but how it felt and how it affected the two of us was obviously very different. Not because the math or physics was wrong, but because there was more going on than just how fast the gun recoiled and how much recoil energy it had.
Doyle: "Now, for the real-world application that I was trying to solve."
That makes your question easy, and if YOU are the only measurement of felt recoil, then the results are indeed directly comparable. Lets get that out of the way first. I prefer the term "perceived recoil" which allows for the perceiver to be the same for each instance, thus avoiding the unnecessary math. Perceived recoil is a combination of time delays and causative factors which takes us into "power" (energy over time) so lets keep it at the high school physics level and save "power" for the college guys.
E=(P^2)/2m Kinetic Energy equals the object's momentum value squared and then divided by twice the mass
P (momentum of the gun) = (mass of bullet times bullet velocity) + (mass other ejecta times ejecta velocity)
The momentum short form simply plugs the P (momentum of the gun) second equation into the first equation.
The resultant translational kinetic energy is known as the "free recoil" of the gun. Cant be simpler but please remember to keep your units the same. Weights are in lbs (or decimal fractions for bullets and powder charges) and ft/sec for velocity. To convert grains into pounds, divide the grains by 7000. For the purpose of your calculations recommend you use single base nitrocellulose at a powder charge velocity of 1585 meters per second (5200 feet per second) and assume the entire powder charge weight exits at that velocity.
If you are familiar with spread sheets, set up the momentum short form in the spread sheet and allow for the entry of variable bullet and powder charge weights and go to your hearts content. As long as you are the one making the measurements you have a direct comparison. This also means you are the one to decide what percentage of recoil reduction you are getting.
JohnKSa: "Unfortunately, at the moment, I can't think of a good way to explain why this is true in a way that would make intuitive sense."
Mostly because we do not use explicit terms with well understood physical meanings. The instant we start to talk about "recoil" we introduce a time factor that negates the "instantaneous" needed for conservation of momentum. Sorta gotta get rid of that, kinda sorta. As you can infer from that, I relate to the field.
In your example, the power transfer rate is the difference between you and her.
That makes your question easy, and if YOU are the only measurement of felt recoil, then the results are indeed directly comparable. Lets get that out of the way first. I prefer the term "perceived recoil" which allows for the perceiver to be the same for each instance, thus avoiding the unnecessary math. Perceived recoil is a combination of time delays and causative factors which takes us into "power" (energy over time) so lets keep it at the high school physics level and save "power" for the college guys.
E=(P^2)/2m Kinetic Energy equals the object's momentum value squared and then divided by twice the mass
P (momentum of the gun) = (mass of bullet times bullet velocity) + (mass other ejecta times ejecta velocity)
The momentum short form simply plugs the P (momentum of the gun) second equation into the first equation.
The resultant translational kinetic energy is known as the "free recoil" of the gun. Cant be simpler but please remember to keep your units the same. Weights are in lbs (or decimal fractions for bullets and powder charges) and ft/sec for velocity. To convert grains into pounds, divide the grains by 7000. For the purpose of your calculations recommend you use single base nitrocellulose at a powder charge velocity of 1585 meters per second (5200 feet per second) and assume the entire powder charge weight exits at that velocity.
If you are familiar with spread sheets, set up the momentum short form in the spread sheet and allow for the entry of variable bullet and powder charge weights and go to your hearts content. As long as you are the one making the measurements you have a direct comparison. This also means you are the one to decide what percentage of recoil reduction you are getting.
JohnKSa: "Unfortunately, at the moment, I can't think of a good way to explain why this is true in a way that would make intuitive sense."
Mostly because we do not use explicit terms with well understood physical meanings. The instant we start to talk about "recoil" we introduce a time factor that negates the "instantaneous" needed for conservation of momentum. Sorta gotta get rid of that, kinda sorta. As you can infer from that, I relate to the field.
In your example, the power transfer rate is the difference between you and her.
The muzzle energy(Em) is NOT the source of recoil energy(Er).
The source of energy is the chemical energy of the burnt propellant (Ep).
Ep = Em + Er
This is conservation of energy. But it doesn't answer the question, which is the relationship between Em and Er.
How the energy is distributed between the projectile and the gun is determined by conservation of momentum. The end result is the ratio of masses. Majority of the energy goes to the bullet(Em), and very little goes to the gun(Er), because the gun is much heavier than the bullet.
It makes sense to me as I didn't die after firing my gun. Hope it makes sense to y'all too. If Em=Er, you are dead.
-TL
Sent from my SM-G930T using Tapatalk
The source of energy is the chemical energy of the burnt propellant (Ep).
Ep = Em + Er
This is conservation of energy. But it doesn't answer the question, which is the relationship between Em and Er.
How the energy is distributed between the projectile and the gun is determined by conservation of momentum. The end result is the ratio of masses. Majority of the energy goes to the bullet(Em), and very little goes to the gun(Er), because the gun is much heavier than the bullet.
It makes sense to me as I didn't die after firing my gun. Hope it makes sense to y'all too. If Em=Er, you are dead.
-TL
Sent from my SM-G930T using Tapatalk
Aguila Blanca
Staff
Does not compute.tangolima said:
"For every action, there is an equal and opposite reaction." Energy is energy. It doesn't know or care the weight (or mass) of the object it's acting on. It just acts. If 'X' pounds of force acts on a small-ish hunk of stuff, it may push that small-ish hunk of stuff pretty far and pretty fast. If the same 'X' pounds of energy act on a large, heavy chunk of stuff, it's the same energy but it's going to push that large, heavy chunk a lot slower and less far.