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Old January 11, 2025, 03:00 AM   #26
JohnKSa
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Some issues:

"Recoil force acts slightly below the bore axis, generating torque that causes the muzzle to rise." This is incorrect. Recoil force acts directly along the bore axis.

"The forward push from the firing hand resists the rifle's backward motion, while the backward pull from the support hand anchors the rifle to your body. Together, they counter the rotational effect." Forward and backward pushes can do nothing to counter "rotational effect". There would need to be some force opposite the force vector that is causing the rotational effect. That force is upwards and so the force would have to be downward, not backward or forward.

It looks to me like it is parroting something it "read" on the internet rather than doing any sort of analysis.
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Old January 11, 2025, 03:08 AM   #27
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It is indeed how it works. It searches the Internet and summarizes the result with some analysis. It is like having an intern, who knows little or nothing about the subject matter, to do research for you. It is up to you to accept the results. If the result is off, you question it and it will adjust the analysis.

For instance, you can enter "Explain exactly how the opposing forces can cancel the rotating torque", and it will elaborate. If its logic is wrong, eventually it will be cornered. Then it actually will apologize and start over. It happened when I was grilling it for Rice distribution.

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Old January 11, 2025, 03:23 AM   #28
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Quote:
Originally Posted by JohnKSa View Post
Some issues:



"Recoil force acts slightly below the bore axis, generating torque that causes the muzzle to rise." This is incorrect. Recoil force acts directly along the bore axis.
There were exchanges prior to the exert posted. If the context is considered, he is actually correct. The recoil here is the force on shooter's shoulder, which is below bore axis.

He is also correct that the 2 opposing forces create a torque rotating muzzle down.

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Old January 11, 2025, 04:04 AM   #29
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The RESISTANCE to the recoil force is, indeed, the shooter's shoulder which in most designs is somewhat below the bore axis--although much less so in an AR type design.

The recoil force, however, is directly along the bore axis.
Quote:
He is also correct that the 2 opposing forces create a torque rotating muzzle down.
"He?" Ouch. The algorithm, you mean.

One could argue that the pull force by the rear hand is below the fulcrum of the shoulder which would tend to pull the muzzle downwards. But pushing forward on the forearm (in line with the bore, more or less and either aligned or slightly above the fulcrum of the shoulder) actually counteracts some of that downward torque, it doesn't help it.

Can you draw out the diagram you are thinking of where both the push force and pull force work to pull the muzzle downwards?
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Old January 11, 2025, 05:02 AM   #30
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I regard the algorithm as my intern who is male, and hence a he .

I started the chat on felt recoil on the shooter's shoulder, so the "recoil force" is just that. The rifle is conventional stock configuration, such as a pump shotgun, where the wrist is inline with the butt stock, and the bore axis is above the butt stock.

The butt stock contacts the shooter's shoulder. It is the pivot. The pull force on the stock wrist is inline to the pivot, so it has zero torque. The push force is on the forearm and above pivot, so the torque pushes the muzzle down.

AR config is slightly different, but the result is the same. The pull force is on the pistol grip and below the pivot. The torque rotates the muzzle down. The push force is on the forearm and inline with pivot. The torque is zero.

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Old January 11, 2025, 07:21 AM   #31
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JohnKSa has tried to explain to you that a torque is a moment ABOUT an axis not ALONG an axis.

ChatGP does not understand this concept and your reliance upon it hints to the fact you do not either. That is not a dig at you, just a fact. We all are learning here. Torque effects would be separate conversation and are actually considered minor forces when it comes measuring recoil effects. Most of the time, it is inconsequential and for a conversation of forces felt ALONG the recoil axis it is non-existent.

Once more, the OP question is about forces ALONG the recoil axis:

Quote:
I mean if the push/pull method lowers the recoil on a shotgun, wouldn't be the same (or even better) on an AR? Can you explain why or why not?

Last edited by davidsog; January 11, 2025 at 07:35 AM.
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Old January 11, 2025, 07:33 AM   #32
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As the bullet accelerates forward, the gun HAS to accelerate rearward. No free lunch here, guys. Since these forces are equal, the equations can look something like this: F=ma(gun) = F=ma(bullet) Since the forces are equal, taking the difference of these will always give you zero. This explains the reason that if the new hand load you worked up has your bullets going faster or your rifle goes on a diet, this translates into a healthy dose of increased rifle recoil because the rifles acceleration will increase.
Small changes in acceleration have a large effect on the forces you feel in our F =ma.

Quote:
You say "Big deal, you increased the acceleration of the rifle by 1.34%". You are correct, but look at it from the perspective of increasing the rearward acceleration of the rifle by 43.8%!
https://www.thealaskalife.com/blogs/...ng-the-physics

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Old January 14, 2025, 01:14 PM   #33
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If an 8lb 308 is fired in a rest, by itself would it recoil and jump more than the same rifle mechanically affixed to a 200lb shooter?

Would this not effectively change the weight of the rifle to a much greater amount thereby reducing Felt recoil?

Now, both examples are no where near being 100% realistic but to a varying degree my shade tree reasoning says they are.
Maybe mathematically the solution would be different but real world, real shoulders might contradict the calculated answer.
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Old January 16, 2025, 06:47 PM   #34
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Quote:
If an 8lb 308 is fired in a rest, by itself would it recoil and jump more than the same rifle mechanically affixed to a 200lb shooter?
Both rifles will have same exact recoil energy.

They may or may not have the same felt recoil depending on specifics of the mechanical affixation and the specifics of the rest.

The math will very much back up real world observation. Now, this is not actually calculating the specific forces, it just illustrates the mathematical relationship found in the real world.

You can see that in the effect of weight on acceleration.

Recoil Force of 150 grain .308 Win = 15.6lbs

a = F/M

Mass = 8lb / 32.2fps = .248 slugs(lb*s^2/ft)

Our 8lb Rifle acceleration is

a = 15.6lb/.248lb*s^2/ft = 62.9fps^2

Our 20lb Rifles Acceleration is

20lb rifle = .621 slugs (lb*s^2/ft)

a = 15.6lbs/.621lb*s^2/ft = 25.12 fps^2

Our 20 lb rifle moves almost a third of the distance in the same amount of time for the same amount of force. We experience that as a reduction in felt recoil even though it is the same amount of force.

Last edited by davidsog; January 16, 2025 at 09:18 PM.
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Old January 22, 2025, 07:17 PM   #35
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Quote:
Originally Posted by Davidsog
Both rifles will have same exact recoil energy.
Nope. That's the one that trips people up all the time. It is the one that explains why the different holds change felt recoil.

First, truly classical physics, by which I mean Newton's physics as he understood them in his day, did not include energy calculations. The concept of energy (really more a vague notion) was proposed by Leibniz in Newton's time as "vis viva" (life force) but lacked a mathematical definition. Émilie du Châtelet worked out that there was something being conserved in a closed system other than just momentum, and, IIRC, that it was proportional to V², but didn't get to the final definition before dying in childbirth. Energy as a term did not appear in print until Thomas Young used it in place of vis viva in 1807. Kinetic energy was later defined as ½mV² by Coriolis in 1829, 100 years after Newton's death in 1729. Other forms of energy, potential energy, and radiant energy, were defined later.

So, physics isn't entirely dependent on nor entirely about energy.

Let's get to why energy isn't equal in the different examples. The force pushing the bullet forward and the gun rearward is equal and opposite, per Newton. If the gun were floating free in outer space when it fired, you would discover the forward momentum of the bullet, and the rearward momentum of the gun would also be opposite and equal. But the energy? Not even close. The bullet might exit with thousands of ft-lbs of muzzle energy, while the gun's recoil energy varies with how it is mounted. Floating is space; the energy would be greater in the bullet in proportion to the difference in mass between the gun and the bullet. On the ground and mounted in some way, because KE=½mV² and different mounts allow the gun to reach different velocities, V² for the gun will vary and, hence, different KEs for different mounts.

Example: We have a gun that weighs 10 lbs (just to have an easy number to work with). We have a bullet plus ejecta mass that weighs 300 times less, at 0.033 lbs. 0.033lbs×7000gr/lb=233 grains. If we float the thing in space, and the bullet exits at 3000ft/s, the gun will recoil at 3000ft/s/300, or 10 ft/s. Say, 15 ft/s. Then we have 10lbs/g=0.3108 slugs at 10ft/s. ½mV²=½×0.3108slug×(10lb)²=15.4ft-lb

But if we mount this same gun on a fixed solid base, so it isn't free to recoil, the equal and opposite momentum is put into the earth, and the resulting V² is so small that the kinetic energy is very small. If you want to look up the spin axis moment of inertia of the earth, you can calculate the resulting change in the earth's angular velocity, but I don't think it's worth the trouble. It's not zero KE, but it is so close as to be ignorable.

For the 200 lb shooter, we have a different situation. A tight, pulled-in hold connects him to the gun. The recoil is in the upper half of him while his feet hold their ground, so it is like adding half his weight to the gun. So we now have a 110lb gun in effect. The weight of the heavy gun to the weight of the ejecta is now about 3305:1. the recoil velocity will be 3000/3305=0.908ft/s. The recoiling mass will be 110lb/g=3.419slugs. Recoil energy will be:
½mV²=½×3.419slug×(0.908lb)²=1.49ft-lb.

If I haven't made any egregious arithmetic errors, that's about right. As to felt recoil, when you resist the recoiling gun, you are doing work. Work has the same units as kinetic energy—ft-lbs—so the more KE the recoiling gun has, the more work your body has to do to stop it. In other words, because the higher recoil energy gun is moving faster, you have to decelerate it with your shoulder in a shorter time, and greater deceleration requires more force from your body to make it happen. That's what it hurts to apply.
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Old January 24, 2025, 06:32 PM   #36
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Quote:
That's the one that trips people up all the time.
What trips people up all the time is measuring that energy at different locations.

If your point of measurement is the buttstock, you are correct. The felt recoil will not be the same nor will the energy imparted by that recoil be the same.

If you measure the energy from the chamber, then the energy will be the same. Felt recoil will still be different.

It is easy to lose sight of that in fact it is hidden in some of the methods used to calculate recoil forces.

For Example, the Momentum, Impulse, Velocity and Energy method uses the actual measured velocity the rifle is propelled backwards to determine the impulse.

Gun Momentum = Bullet momentum + Gas momentum + Jet Effect Momentum

Bullet Momentum will not change in significant digits given two reference cartridge's of the same design

Gas momentum will not change in significant digits given two reference cartridge's of the same design.

Gas Momentum = Powder Wt * (Muzzle Velocity/2)

Jet Effect Momentum can definitely change give physical characteristics of the muzzle. However, our Jet momentum is:

Jet Momentum = Powder Wt * Gas Exit Velocity * (1-MuzzleBreak Efficiency)^1/sqrt e

Powder Wt is our Energy in the system as the Potential is converted to Kinetic with combustion. It is the same provided the reference cartridge is the same.

Gas Exit Velocity is a function of muzzle velocity (barrel length)

Muzzle Break design is also a function of the physical characteristics.

None of this changes the fact both start with same amount of energy in the chamber.

Simply put our formula becomes:

Gun Momentum = Gun Weight x Gun Velocity

Recoil Impulse = Gun Momentum / g

(reader: g = constant for the acceleration of gravity)

Next we need to clarify what we are measuring here especially for the readers. I am sure you already know this UncleNick.

A recoil impulse is a measure of momentum. Momentum is a vector quantity with both magnitude and direction. Acceleration is a scalar quantity with only magnitude. They are related but not the same thing.
Acceleration is a measure of how quickly an object is changing velocity while Momentum quantifies the amount of motion.

Two very different but very much related things. The change in acceleration is the fundamental difference in the change in momentum between two rifles exactly alike except for weight.


Quote:
First, truly classical physics, by which I mean Newton's physics as he understood them in his day, did not include energy calculations. The concept of energy (really more a vague notion) was proposed by Leibniz in Newton's time as "vis viva" (life force) but lacked a mathematical definition. Émilie du Châtelet worked out that there was something being conserved in a closed system other than just momentum, and, IIRC, that it was proportional to V², but didn't get to the final definition before dying in childbirth. Energy as a term did not appear in print until Thomas Young used it in place of vis viva in 1807. Kinetic energy was later defined as ½mV² by Coriolis in 1829, 100 years after Newton's death in 1729. Other forms of energy, potential energy, and radiant energy, were defined later.

So, physics isn't entirely dependent on nor entirely about energy.
Physics has evolved for sure. We just proved the existence of the Higgs Boson and Gravitational Waves.....

That does not change the fact a rifle firing the same exact cartridge has the same energy when measured from the chamber. You are correct in that many variables exist to influence "felt recoil" measured at the buttstock. However that does not change the fact it is the change in acceleration that has the greatest influence.

You can get into the weeds with much higher order math and break down exactly what is specifically causing that change in acceleration based the exact specifics of each design but that has nothing to do with the root cause.

Don't get lost in weeds. It is like someone arguing that an Airplanes turn performance is not based upon Power Available to Power Required by arguing that the lighter airplane turns better than the heavier one.

It matters not how you define that energy or how you absorb it. Energy cannot be created or destroyed.

In the realm of significant digits, the energy is equal as long as we are talking the same exact cartridge.

Which is why we have SAAMI standards and reference ammunition to ensure that reality.

Last edited by davidsog; January 25, 2025 at 11:54 AM. Reason: Edited for clarity. It's a complicated subject
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Old January 25, 2025, 12:09 PM   #37
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Simply put:

What is the change in felt recoil on a Barrett .50 cal if we eliminate the muzzle brake and replace it with a thread protector of the same weight as the muzzle brake.

Two rifles, same weight, same design, firing the same cartridge. Very Different felt recoil forces at the buttstock.

"Barney Level" look at what is fundamentally going on:

F = ma

Mass is the same....

The only thing that changes is the acceleration to change the amount of force felt at the buttstock. We can look at our Jet Effect momentum formula to see exactly why that change in acceleration occurs.

Your grip in the "push/pull" method imparts a force on the opposite vector of our recoil vector that serves to reduce the acceleration just like our Muzzle Brake.

Quote:
In other words, because the higher recoil energy gun is moving faster, you have to decelerate it with your shoulder in a shorter time, and greater deceleration requires more force from your body to make it happen.
Exactly....
Opposites sides and viewpoints of the same coin describing the same result.

Last edited by davidsog; January 25, 2025 at 01:47 PM.
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Old January 26, 2025, 08:14 AM   #38
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I have known bigger individuals that were more sensitive to recoil than smaller ones.

Is it because the acceleration of the gun is being absorbed at a different rate, sudden vs over a longer duration?
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Old January 26, 2025, 02:12 PM   #39
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Quote:
Is it because the acceleration of the gun is being absorbed at a different rate, sudden vs over a longer duration?
Yes, the acceleration changes and therefore the momentum changes.

An individual's Sensitivity and perception of that recoil is also very subjective. My Autistic daughter will perceive the same touch sensory stimulus very differently from me for example.

Mass directly relates to the amount of force. A lighter mass must accelerate faster for the same amount of force. If the mass cannot equal the force it must expend the excess Kinetic Energy thru motion.


Quote:
The law of conservation of energy states that energy can neither be created nor destroyed – it transforms from one form to another.
https://www.sciencefacts.net/conserv...of-energy.html

The Energy must be equal.
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Old January 27, 2025, 04:13 PM   #40
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The total momentum imparted forward and backward are equal and opposite. Always. This is due to Newton's second and third laws. The third law makes applied force equal and opposite, and the second law states force is equal to the rate of change in momentum of a mass, which keeps the gained velocities proportional. No exceptions. No kidding.

In the shooting system, momentum is not changed by how the gun is held. What the momentum is invested in, the gun, the shooter, and the earth that the shooter is standing on, may change over time as the gun transfers it to the shooter and the shooter to the earth. If the gun is held loosely, it picks up rearward momentum and then transfers it to the shooter's shoulder, and this may rock the shooter back. But as the shooter's stomach muscles pull him back upright against the grip of his feet on the ground, the rest of it transfers into the earth.

The momentum is equal and opposite because it is simply and directly proportional to both mass and velocity. Thus, for each unit of mass for or aft, the same amount of force produces acceleration proportional to the mass it is operating on, and thus, it has to be the same amount of momentum. If the forward mass is twice the rearward mass, the forward mass is accelerated to half the velocity of the rearward mass, and thus, mass times velocity is the same in both directions. Try making the calculation for yourself in any opposite pairs of directions with any oppositely driven masses. The result will always be the same. If it looks like there is less momentum in one or the other, you have failed to account for all the mass.

Kinetic energy is not equal and opposite. It doesn't have to be. Unlike force and momentum, energy is not a vector quantity that has to add up. It is a scalar quantity and can be directed anywhere. In finding equal and opposite momentum, as I described above, it is the fact that both velocity and mass are participants in momentum in equal portions that allows an equal and opposite result. However, kinetic energy, while proportional to mass, like momentum, is not proportional to velocity. Rather, it is proportional to the square of velocity. This makes it impossible for it to be equal and opposite, with the single exception of the two oppositely accelerated masses being the same. As soon as the two masses are unequal, so is the energy.

The previous example: A forward-moving mass, 2M, is twice the oppositely directed mass, M, when an equal but oppositely directed force, F, and -F, is applied. The first accelerates to a speed of V, while the second accelerates to a speed of -2V. The signs are important because these are vectors, so something has to indicate their relative directions. As to energy, for the heavier mass, KE=½×2M×V²=1 unit of energy, and for the lighter mass, KE=½×M×2V²=2 units of energy. The kinetic energies of the two masses have proportions of 1:2 and are not equal. Note that squaring V eliminates any minus sign that might be associated with (minus times minus is plus), and thus, KE has no direction associated with it. It is just a quantity.

Often, when someone asks why, given equal and oppositely directed force, the shooter is not injured as badly as the recipient of the bullet. Usually, someone will say it is because the bullet has a smaller contact area with its target than the gun does with the shooter, but a bit of thought will reveal that this doesn't begin to account for it. Especially not for elephant guns and the like. Instead, it is the fact the bullet energy is much higher than that of the gun in contact with the body that accounts for the lion's share of the difference at the shooter's receiving end. Reduce the mass of the gun to the mass of the bullet so the energy becomes equal on the shooter's end, and then, despite the bigger contact area, it will damage the shooter, too.
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Old January 27, 2025, 09:16 PM   #41
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Quote:
momentum is not changed by how the gun is held.
The laws of the Conservation of Momentum still apply and are greatly affected by forces imparted upon the body in motion.

It is changed. Once you grip the gun, you become a part of that system. That is why we have shooting stances to manage the momentum of shooting. It's the difference between holding a 12 gauge buttstock 6 inches from your face and having it firmly planted in your shoulder.

That will give a good demonstrations of how the acceleration affects that conservation of momentum.

Quote:
As soon as the two masses are unequal, so is the energy.
Unclenick, some people might be confused in that you talk about Kinetic Energy and then switch to blanket term Energy. Energy can change forms but it must equal and is subject to the laws of physics.

Energy must be equal.

Higher mass has more potential energy always. Potential energy has a direct relationship to mass.

Your last portion nicely illustrates how that conversion of Kinetic back to Potential relates to our recoil problem.

Quote:
The previous example: A forward-moving mass, 2M, is twice the oppositely directed mass, M, when an equal but oppositely directed force, F, and -F, is applied. The first accelerates to a speed of V, while the second accelerates to a speed of -2V. The signs are important because these are vectors, so something has to indicate their relative directions. As to energy, for the heavier mass, KE=½×2M×V²=1 unit of energy, and for the lighter mass, KE=½×M×2V²=2 units of energy. The kinetic energies of the two masses have proportions of 1:2 and are not equal. Note that squaring V eliminates any minus sign that might be associated with (minus times minus is plus), and thus, KE has no direction associated with it. It is just a quantity.

Often, when someone asks why, given equal and oppositely directed force, the shooter is not injured as badly as the recipient of the bullet. Usually, someone will say it is because the bullet has a smaller contact area with its target than the gun does with the shooter, but a bit of thought will reveal that this doesn't begin to account for it. Especially not for elephant guns and the like. Instead, it is the fact the bullet energy is much higher than that of the gun in contact with the body that accounts for the lion's share of the difference at the shooter's receiving end. Reduce the mass of the gun to the mass of the bullet so the energy becomes equal on the shooter's end, and then, despite the bigger contact area, it will damage the shooter, too.

Last edited by davidsog; January 27, 2025 at 09:54 PM.
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