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June 21, 2019, 02:04 PM | #26 |
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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. |
June 21, 2019, 02:52 PM | #27 | |
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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.
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June 21, 2019, 03:14 PM | #28 | |
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-TL Sent from my SM-G930T using Tapatalk |
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June 21, 2019, 07:11 PM | #29 | |
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June 21, 2019, 07:15 PM | #30 | |
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I guess that makes sense to you, but it doesn't make any sense to me. |
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June 21, 2019, 09:24 PM | #31 |
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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
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June 21, 2019, 09:55 PM | #32 | |
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(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 |
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June 22, 2019, 09:05 PM | #33 | |
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All else being equal (and it almost never is) bigger bullets tend to work better. |
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June 22, 2019, 09:17 PM | #34 |
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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.
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June 23, 2019, 06:22 PM | #35 | |
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https://www.shootingtimes.com/editor...n-recoil/99442 For some cartridges, the gunpowder makes a HUGE contribution to recoil. https://www.shootingtimes.com/editor...-recoil/328788 |
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June 23, 2019, 09:15 PM | #36 | |
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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...").
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