The Firing Line Forums - View Single Post - powder charge/ OAL/ Perceived Recoil relationship

highrolls · September 15, 2010, 10:37 PM

I think Unclenick likes to make things more complicated than they need to be. Sure, a lot of things like muzzle brakes and auto slide motions and etc. will complicate the calculation of "net energy" but it is easier to avoid that and stay with the basic "free recoil". A good buddy recently asked me to give him some idea of what to expect when firing his recently acquired Mosin M1891/30 as compared to two other rifles he was familiar with. Here was the answer I gave him.

http://7.62x54r.net/MosinID/MosinM9130S.htm

The above reference gave me the weight without bayonett of 4 Kilograms. One popular safe load I found for it would be 150 grain jacketed boat tail spitzer, over a charge weight of 43 grains, which drives that bullet at an average velocity of 2772 fps. That's all that we need to calculate it, but we need to convert to metric since I like Joules.

150 grains = 9.71 grams and 43 grains = 2.79 grams

2772 fps = 844.9 mps and charge velocity of 5200 fps = 1585 mps

Now Free recoil = (Total Momentum) squared and the result divided by twice the gun mass (8 Kg in this case)

Here goes: Bullet mass 150 grains (9.71 grams) times velocity 2772 fps (844.9 meters per second)
Plus powder mass 43 grains (2.79 grams) times charge velocity 5200 fps (1585 meters per sec)
using the metric conversions in parens gives a numerator of 12626.13
Divide numerator by 1000 (to convert the grams to kilograms) is 12.63
Square the result is total momentum squared 159.42
Divide total momentum squared by twice the gun mass (twice 4 Kg)

gives 19.93 Joules

So your rifle will have a free recoil of about 20 Joules and should compare favorably with
the Marlin model 1894 carbine at 2.9 kg (6.5 lbs) in .30-30 winchester at 17 Joules or
the Remington Sendero 4.5 Kg (10 lbs) in .308 winchester at 20 Joules. So fire either of those
two and the M1891/30 should feel somewhere in the same ballpark.

Notice I avoided the perception argument entirely.

September 15, 2010, 10:37 PM	#8
highrolls Senior Member Join Date: January 20, 2009 Location: already given Posts: 115	I think Unclenick likes to make things more complicated than they need to be. Sure, a lot of things like muzzle brakes and auto slide motions and etc. will complicate the calculation of "net energy" but it is easier to avoid that and stay with the basic "free recoil". A good buddy recently asked me to give him some idea of what to expect when firing his recently acquired Mosin M1891/30 as compared to two other rifles he was familiar with. Here was the answer I gave him. http://7.62x54r.net/MosinID/MosinM9130S.htm The above reference gave me the weight without bayonett of 4 Kilograms. One popular safe load I found for it would be 150 grain jacketed boat tail spitzer, over a charge weight of 43 grains, which drives that bullet at an average velocity of 2772 fps. That's all that we need to calculate it, but we need to convert to metric since I like Joules. 150 grains = 9.71 grams and 43 grains = 2.79 grams 2772 fps = 844.9 mps and charge velocity of 5200 fps = 1585 mps Now Free recoil = (Total Momentum) squared and the result divided by twice the gun mass (8 Kg in this case) Here goes: Bullet mass 150 grains (9.71 grams) times velocity 2772 fps (844.9 meters per second) Plus powder mass 43 grains (2.79 grams) times charge velocity 5200 fps (1585 meters per sec) using the metric conversions in parens gives a numerator of 12626.13 Divide numerator by 1000 (to convert the grams to kilograms) is 12.63 Square the result is total momentum squared 159.42 Divide total momentum squared by twice the gun mass (twice 4 Kg) gives 19.93 Joules So your rifle will have a free recoil of about 20 Joules and should compare favorably with the Marlin model 1894 carbine at 2.9 kg (6.5 lbs) in .30-30 winchester at 17 Joules or the Remington Sendero 4.5 Kg (10 lbs) in .308 winchester at 20 Joules. So fire either of those two and the M1891/30 should feel somewhere in the same ballpark. Notice I avoided the perception argument entirely.