That said reaction is force, not quite energy.
-TL
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Is there a mathematical relation ship between ME and Recoil energy?
- Thread starter Doyle
- Start date
That said reaction is force, not quite energy.
-TL
Sent from my SM-G930T using Tapatalk
Some more details to aid the computation and to answer the OP's question.
At the instance bullet exists the muzzle, conservation of momentum applies.
(mr)(Vr)=(mb)(Vb)
mr is mass of rifle
Vr is velocity of rifle
mb is mass of bullet
Vb is velocity of bullet
Rifle recoil energy
Er=(0.5)(mr)(Vr)^2=(0.5)(mb^2)(Vb^2)/mr
Muzzle energy
Em=(0.5)(mb)(Vb^2)
The energy ratio
Er/Em=(mb)/(mr)
Since mr >> mb, Em >> Er, or most energy goes to the bullet, and very little goes to the rifle.
Regarding the movie myth, it is indeed a created illusion to enhance entertainment. But the usual dispute based upon Newton's law is also a myth itself. But let's save it for some time.
-TL
Sent from my SM-G930T using Tapatalk
At the instance bullet exists the muzzle, conservation of momentum applies.
(mr)(Vr)=(mb)(Vb)
mr is mass of rifle
Vr is velocity of rifle
mb is mass of bullet
Vb is velocity of bullet
Rifle recoil energy
Er=(0.5)(mr)(Vr)^2=(0.5)(mb^2)(Vb^2)/mr
Muzzle energy
Em=(0.5)(mb)(Vb^2)
The energy ratio
Er/Em=(mb)/(mr)
Since mr >> mb, Em >> Er, or most energy goes to the bullet, and very little goes to the rifle.
Regarding the movie myth, it is indeed a created illusion to enhance entertainment. But the usual dispute based upon Newton's law is also a myth itself. But let's save it for some time.
-TL
Sent from my SM-G930T using Tapatalk
"...mathematically predict recoil energy from muzzle energy..." Yep, but it's not a prediction. It's a formula.
However, ME is not RE. Recoil starts as soon as the bullet starts to move in the case. ME is several microseconds later and is what remains after friction and the gasses generated by the burning powder have finished with the bullet. Recoil Energy is altered by a bunch of factors.
"...the "physics" formulas said the swimmer..." The swimmer is operating on the First Law of Motion. An idea that that Newton guy stole from Galileo. And there's time involved. It's quantum physics for that. My head hurts.
However, ME is not RE. Recoil starts as soon as the bullet starts to move in the case. ME is several microseconds later and is what remains after friction and the gasses generated by the burning powder have finished with the bullet. Recoil Energy is altered by a bunch of factors.
"...the "physics" formulas said the swimmer..." The swimmer is operating on the First Law of Motion. An idea that that Newton guy stole from Galileo. And there's time involved. It's quantum physics for that. My head hurts.
Jim Watson
New member
As said, recoil is conservation of momentum, mass times velocity; not conservation of energy, half of mass times velocity squared*.
As your physics professor can explain, momentum is a vector quantity, it includes a direction. Bullet momentum north is equaled by gun momentum south.
Energy is not a vector, it has no directionality.
There is so much chemical energy in the gunpowder, when fired, some of that energy goes to accelerating the bullet, some of it goes to accelerating the gun, some of it goes to accelerating the powder gas, some of it shows up in muzzle blast and a hot barrel.
Powder gas contributes to recoil. Most computations make an assumption on how fast it is going, some say 1.5x bullet velocity, some lump it all in at maybe 4500 fps. The contribution of 5 grains of powder a 230 grain .45 ACP bullet isn't much and the difference between 5 and 6 grains will be hard to spot. But the OP's 7-08 with 40-50 grains of powder and a 120 grain bullet, the "jet effect" will be pronounced and differences in powder charge across the range of recommended loads will be noticeable.
There are a lot of factors in FELT recoil besides the "physics" recoil.
But if you are comparing different ammo in the same gun, a lot of those will cancel out.
Some early discussions, just as chronographs were becoming available, concluded that the onset of "gun headache" was related to recoil velocity.
This led to empirical determination that the shotgun should weigh 100 times the shot charge. Those light and handy British Best game guns are therefore built around rather light loads; 1 1/16 oz is common and 1 1/8 oz, calling for a 7 lb gun is pretty stout in the pheasant drives.
*The well known Hatcher Stopping Power comes out odd because he back calculated momentum from energy by dividing by velocity, not accounting for the 1/2 factor in the energy equation.
As your physics professor can explain, momentum is a vector quantity, it includes a direction. Bullet momentum north is equaled by gun momentum south.
Energy is not a vector, it has no directionality.
There is so much chemical energy in the gunpowder, when fired, some of that energy goes to accelerating the bullet, some of it goes to accelerating the gun, some of it goes to accelerating the powder gas, some of it shows up in muzzle blast and a hot barrel.
Powder gas contributes to recoil. Most computations make an assumption on how fast it is going, some say 1.5x bullet velocity, some lump it all in at maybe 4500 fps. The contribution of 5 grains of powder a 230 grain .45 ACP bullet isn't much and the difference between 5 and 6 grains will be hard to spot. But the OP's 7-08 with 40-50 grains of powder and a 120 grain bullet, the "jet effect" will be pronounced and differences in powder charge across the range of recommended loads will be noticeable.
There are a lot of factors in FELT recoil besides the "physics" recoil.
But if you are comparing different ammo in the same gun, a lot of those will cancel out.
Some early discussions, just as chronographs were becoming available, concluded that the onset of "gun headache" was related to recoil velocity.
This led to empirical determination that the shotgun should weigh 100 times the shot charge. Those light and handy British Best game guns are therefore built around rather light loads; 1 1/16 oz is common and 1 1/8 oz, calling for a 7 lb gun is pretty stout in the pheasant drives.
*The well known Hatcher Stopping Power comes out odd because he back calculated momentum from energy by dividing by velocity, not accounting for the 1/2 factor in the energy equation.
LeverGunFan
New member
(mr)(Vr)=(mb)(Vb)
mr is mass of rifle
Vr is velocity of rifle
mb is mass of bullet
Vb is velocity of bullet
This is incorrect, you must include the weight and velocity of the gas. See the link to the SAAMI document in post #9.
As an example:
300 Winchester Magnum
8 pound rifle
180 grain bullet
3017 muzzle velocity
77 grains powder
Free recoil energy from bullet only = 11.7 ft-lbs
Free recoil energy from powder only = 6.55 ft-lbs
Free recoil energy from bullet + powder = 35.8 ft-lbs
The momentum of the bullet and the gas (powder) is used to calculate the momentum of the rifle; from there the velocity of the rifle is calculated and is squared to calculate the energy of the rifle. That's why the free recoil energy using the bullet plus the gas is much higher than using just the bullet alone - the velocity of the rifle is increased by the effect of the gas over the effect of the bullet, and then squared.
Use one of the calculators linked above to predict the effect of the bullet and the gas separately, by entering zero for one and then the other.
This is incorrect, you must include the weight and velocity of the gas. See the link to the SAAMI document in post #9.
I agree. It would be a more accurate calculation. One can also lump the mass of the powder into the bullet mass, assuming the gas exits at the same speed of the bullet. The point is the energy ratio is determined by mass ratio, based on conservation of momentum. The muzzle energy does not equal to the recoil energy.
-TL
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The difference was that in her hand, it banged a nerve and blood vessel against a bone because the grip didn't fit her. In my hand, it recoiled into the meaty pocket between the thumb and forefinger, not against bone. I guess you could say that's power transfer, but grip fit was what made felt recoil intolerable for her and essentially negligible for me. What I was trying to point out is that felt recoil incorporates things like grip fit which recoil calculations don't account for.In your example, the power transfer rate is the difference between you and her.
Aguila Blanca
Staff
You are using two different formulas to calculate energy -- one for muzzle energy, and a completely different formula for recoil energy.Tangolima said:Rifle recoil energy
Er=(0.5)(mr)(Vr)^2=(0.5)(mb^2)(Vb^2)/mr
Muzzle energy
Em=(0.5)(mb)(Vb^2)
I guess that makes sense to you, but it doesn't make any sense to me.
The two formulas are the same. In both cases the formula is:
Energy = (1/2) x mass x velocity^2
What he's done is restate the top formula using the conservation of momentum rule.
At the moment that the bullet exits the muzzle, we know that the following equality is true (neglecting gas momentum)
(mr)(Vr) = (mb)(Vb)
If we expand the energy formula for the rifle, we get:
Er=(0.5)(mr)(Vr)(Vr)
Notice that we have (mr)(Vr) in the formula which we know by conservation of momentum is the same as (mb)(Vb)
So we could substitute, but we would still have a Vr left over since Vr is squared in the energy formula.
But we can multiply both sides of the equation by mr without disrupting our equality. That gives us:
(mr)(Er) = (0.5)(mr)(Vr)(mr)(Vr)
Now we can substitute each of the (mr)(Vr) pairs for (mb)(Vb) pairs since we know that they are equal by conservation of momentum.
(mr)(Er) = (0.5)(mb)(Vb)(mb)(Vb)
Which is almost what we want, but there's an extra mr on the left side of the equation. If we assume that mr (the mass of the rifle) is nonzero (which it should be if we're talking about a real-world rifle), then we can divide through by mr to give us a formula for the Energy of the rifle expressed in terms of the bullet velocity, the bullet mass, and the mass of the rifle:
Er = (0.5)(mb)(Vb)(mb)(Vb)/(mr)
And simplifying slightly gives us:
Er = (0.5)(mb^2)(Vb^2)/(mr)
Another way to state it would be to write:
Er = (0.5)(mb)(Vb)(Vb) [(mb)/(mr)]
That's kind of handy because (0.5)(mb)(Vb)(Vb) is the formula for muzzle energy which lets us substitute it into the equation giving us:
Er = (Em)(mb)/(mr) which is what the OP was asking for.
Or, we could take it to the next step which tangolima did (assuming that Em is nonzero) and express it as a ratio of energies by dividing both sides by Em
Er/Em = mb/mr
Energy = (1/2) x mass x velocity^2
What he's done is restate the top formula using the conservation of momentum rule.
At the moment that the bullet exits the muzzle, we know that the following equality is true (neglecting gas momentum)
(mr)(Vr) = (mb)(Vb)
If we expand the energy formula for the rifle, we get:
Er=(0.5)(mr)(Vr)(Vr)
Notice that we have (mr)(Vr) in the formula which we know by conservation of momentum is the same as (mb)(Vb)
So we could substitute, but we would still have a Vr left over since Vr is squared in the energy formula.
But we can multiply both sides of the equation by mr without disrupting our equality. That gives us:
(mr)(Er) = (0.5)(mr)(Vr)(mr)(Vr)
Now we can substitute each of the (mr)(Vr) pairs for (mb)(Vb) pairs since we know that they are equal by conservation of momentum.
(mr)(Er) = (0.5)(mb)(Vb)(mb)(Vb)
Which is almost what we want, but there's an extra mr on the left side of the equation. If we assume that mr (the mass of the rifle) is nonzero (which it should be if we're talking about a real-world rifle), then we can divide through by mr to give us a formula for the Energy of the rifle expressed in terms of the bullet velocity, the bullet mass, and the mass of the rifle:
Er = (0.5)(mb)(Vb)(mb)(Vb)/(mr)
And simplifying slightly gives us:
Er = (0.5)(mb^2)(Vb^2)/(mr)
Another way to state it would be to write:
Er = (0.5)(mb)(Vb)(Vb) [(mb)/(mr)]
That's kind of handy because (0.5)(mb)(Vb)(Vb) is the formula for muzzle energy which lets us substitute it into the equation giving us:
Er = (Em)(mb)/(mr) which is what the OP was asking for.
Or, we could take it to the next step which tangolima did (assuming that Em is nonzero) and express it as a ratio of energies by dividing both sides by Em
Er/Em = mb/mr
Start with 3 equations.You are using two different formulas to calculate energy -- one for muzzle energy, and a completely different formula for recoil energy.
I guess that makes sense to you, but it doesn't make any sense to me.
(mr)(Vr)=(mb)(Vb) -eqn 1
Er=(0.5)(mr)(Vr)^2. -eqn 2
Em=(0.5)(mb)(Vb^2). -eqn 3
Substitute eqn 1 into eqn 2 you get
Er=(0.5)(mb^2)(Vb^2)/mr. -eqn 4
Divide eqn 4 by eqn 3. You have the energy ratio
Er/Em=(mb)/(mr)
Hope it helps.
-TL
Sent from my SM-G930T using Tapatalk
Sent from my SM-G930T using Tapatalk
One can also lump the mass of the powder into the bullet mass, assuming the gas exits at the same speed of the bullet.
I do not believe this is a valid assumption. Powder gas does not exit at the same speed as the bullet. I believe it is faster, buy a considerable amount, once it leaves the muzzle.
The powder gases do exit at considerably higher velocity than the bullet and the better online recoil calculators will take that into account although even then there's some "assumptioning" taking place as no one actually measures the gas velocity. Typically the calculators use a "typical" number for gas velocity rather than try to make any attempt at coming up with an actual velocity.
For comparing the recoil of two different loads in the same gun, or the recoil of one load in two different guns, you can usually ignore the contribution from powder gases without skewing things too much because there's not usually a huge difference in powder charge weight and because the bullet is the main contributor.
For comparing the recoil of two different loads in the same gun, or the recoil of one load in two different guns, you can usually ignore the contribution from powder gases without skewing things too much because there's not usually a huge difference in powder charge weight and because the bullet is the main contributor.
The powder gases do exit at considerably higher velocity than the bullet and the better online recoil calculators will take that into account although even then there's some "assumptioning" taking place as no one actually measures the gas velocity. Typically the calculators use a "typical" number for gas velocity rather than try to make any attempt at coming up with an actual velocity.
For comparing the recoil of two different loads in the same gun, or the recoil of one load in two different guns, you can usually ignore the contribution from powder gases without skewing things too much because there's not usually a huge difference in powder charge weight and because the bullet is the main contributor.
This needs to be qualified. The difference that different powders can make to recoil can be a significant factor. It can be measured and demonstrated.
https://www.shootingtimes.com/editorial/measure-relative-handgun-recoil/99442
For some cartridges, the gunpowder makes a HUGE contribution to recoil.
https://www.shootingtimes.com/editorial/gunpowder-contribution-to-recoil/328788
Yup. That's why I not only limited my remarks to only two situations("...comparing the recoil of two different loads in the same gun...", "...<comparing> the recoil of one load in two different guns...") but also added another couple of qualifiers to insure that it wasn't taken as an across the board truism ("...you can usually ignore the contribution from powder gases without skewing things too much...").
Quite true. However, while the gunpowder contribution is a large part of the overall recoil in some cartridges (as noted in the article--good article, by the way) when one is comparing loads in one gun (one caliber) or when one is comparing one load in multiple guns (again one caliber) it's usually possible to get away with ignoring the gunpowder contribution without skewing things too much because, in a comparison, the differences are what is of interest, and the gunpowder contribution, while significant in the overall picture will not usually be hugely different in those two somewhat limited circumstances.For some cartridges, the gunpowder makes a HUGE contribution to recoil.