Question about powder charge for light vs. heavier bullet


December 21, 2007, 01:21 AM
Why does a 9mm 115gr bullet require more powder charge compare to a 124gr bullet? I am thinking a heavier bullet need more powder but I'm obviously wrong, so please enlighten me with your physics wisdom.:)

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December 21, 2007, 01:34 AM
It's all about the pressure generated when the powder ignites. The heavier the object to be moved, the less powder it takes to generate the pressure the vessel (cartridge case) can contain.

Hope this helps.


December 21, 2007, 01:38 AM
In projectile physics, F=ma. Rearranging that, F/m=a. The force of the burning powder divided by the mass of the bullet equals the acceleration of the bullet. Let's assume force is constant (the two cartridges have the same measure of the same powder); a lighter and heavier bullet both are given the same amount of energy at ignition, the lighter bullet, having less mass, will however have greater acceleration.

Now we add in friction between barrel and bullet. The coefficient will also be roughly the same, resulting in the same force of friction on the two bullets; however, due to its lighter mass, the lighter bullet will be slowed more by the barrel than the heavier bullet. Thus, given the same impulse through the same barrel, the heavier bullet will have more muzzle energy than the lighter one when it exits. To compensate, a greater impulse must be given to the lighter round (using a bigger powder charge) to give it the same muzzle energy.

It's also a matter of space. You take two FMJ bullets of similar nose shape. They're made of the same material. Yet one is heavier. The only way it can be so is if the heavier bullet is larger, and when bound to a certain diameter and overall length, that extra mass is in the cartridge, leaving less space for powder and for ignition. To give the same peak pressure according to ideal gas laws (pressure increases as volume decreases at a constant temperature and amount of gas), which produces the same impulse, the powder charge must be further reduced for the larger bullet over the smaller one.

December 21, 2007, 02:13 AM
holy cow Liko81 remind me not to ask you a ? I thought fm/am was a radio staion :-> great explanation

December 21, 2007, 02:42 AM
So it is more cost effective to shoot heavier bullets. Hmm.... :D


December 21, 2007, 08:01 AM
Not with the price of lead/copper. :D

Seriously. Primers and powder are still by far the best value for us compared to bullets.

December 21, 2007, 12:28 PM
Great explanation Liko81 and it all make sense not.:)

December 21, 2007, 03:41 PM
Not with the price of lead/copper.

I thought about that a few postings later and didn't come back :D


December 21, 2007, 06:11 PM
Hopefully, the comparison of the Crusher and Piezo Systems is still being provided in the Lyman 48. It's in the 46th for sure.

Only a fraction of the pressure developed upon combustion is required to set a bullet free from the cartridge case. Pressure continues to rise until just after the bullet engages the rifling of the barrel. Time to peak pressure is altered by the burn rate of a propellant. Faster the burn rate, the quicker peak pressure is achieved. We're talkin' microseconds and millimeters here.

F=ma is pretty much the answer in a nutshell.

If two cartridges with different bullet weights have the same measure of the same powder, obviously we'll be talking about two different pressure levels. Assuming the pressure generated by the same powder is equal, the lighter bullet having less mass will have greater acceleration. Thus the consideration of combustion area in how it relates to pressure is valid, but there are still variables, but it does show how pressure can be altered by varying OACL.

The coefficient of friction favors the lighter bullet because of the longer bearing surface of the heavier projectile assuming we're talking about two different bullet weights from the same manufacturer constructed identically except for the necessary length addition to the heavier bullet of the same diameter. So, greater pressure is required to push a heavier projectile to a similar velocity to overcome greater friction as well as higher weight.

actionflies, I presume that the first question asks, to achieve a similar velocity? F=ma answers that and the way it's achieved is by the smaller combustion area with the heavier bullet that generates higher pressure with a lesser powder charge (of the same powder) for the heavier bullet.;)

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