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View Full Version : Knoxx Comp Stock Vs Load Velocity

Aqeous
April 23, 2011, 08:31 PM
I was just curious as to whether or not the Knoxx recoil reducing stocks effected the velocity of Buckshot/slugs.

I can only guess that it depends on whether or not the load exits the barrel before the shotgun gives you that "kick". If it does, all is well. If it does not then newtons law suggests that you would have to subtract the rearward velocity of the gun from the velocity of the projectile (to some degree) That is simply math that I am not qualified to do.

Does anyone know for sure? :cool:

kozak6
April 24, 2011, 04:53 PM
It's not enough to matter.

zippy13
April 25, 2011, 10:30 AM
You need to quantify the gun's movement under recoil and the time of the event of a gun with and without the Knoxx device. This will give you two different velocities. However, I suspect this would be difficult, or impossible, to do with sufficient accuracy because of the variations from load to load.

Double Naught Spy
April 25, 2011, 12:21 PM
Right, you might lose a few inches per second, but chances are you won't be able to notice and difference when you chrono the rounds out of the same gun under the same environmental circumstances. They are still going to fall within your normal 30-50 fps standard deviation.

zippy13
April 25, 2011, 04:58 PM
Let's do a SWAG*:

Assume the V(median velocity in the barrel) is 600ft/sec (1200/2) or 7200-inch/sec, and a
Barrel length of 28-inches, then
Delta(t), the duration of the event, would be in the neighborhood of 28/600 = 0.047-seconds

Let's also assume, because the Knoxx shortens, the gun travels a distance of 5-inches instead of 3-inches without the Knoxx (a wild guess).

Then, the gun's V(1) w/out the Knox would be in the neighborhood of 3/0.047 = 63.8-in/sec or 5.3-ft/sec.

And, V(2) with the Knoxx would be in the neighborhood of 5/0.047 = 106.4-in/sec or 8.9-ft/sec.

It seems the difference in velocities is less than 4-ft/sec.
SWAG* Scientific Wild A\$\$ Guess

TheKlawMan
April 26, 2011, 03:11 AM
You are hurting my head Zippy with those numbers Zippy! I think I follow your Engineer's Voodoo and what you are saying about the effective length of travel from the point of ignition to exiting the muzzle, but your argument seems to assume a false premise; that the rate of velocity increase is constant over the distant traveled down the barrel. I probably have this all wrong and was never that good at anything above simple math, but does a projectile not gain most of its increase in velocity in the first half of the barrel? I hope my question makes sense.

zippy13
April 26, 2011, 09:40 AM
You're absolutely correct, my friend. I was basing my SWAG on an ideal burn, which we know is contrary to actual internal ballistics. Let's assume V(max) is reached reached in half the barrel's length, then Delta(t) would be closer to 14/600 + 14/1200 or 0.035-seconds.

This yields a V(1) of 7.1 and a V(2) of 11.2 with a difference of 4.1-ft/sec, hardly worth mentioning the change.

Either way, the effect on the payload is negligible. However, it may give some insight as to why slower burning powders seem to kick less.