robinghewitt
62 Cal.
- Joined
- Jun 26, 2004
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Recoil is very easy to calculate if you have the muzzle velocity but it means very little.
But as recoil is very much like holding the gun above your shoulder and then dropping it. I thought it might be fun to calculate how high above my shoulder I would have to drop the gun from to feel the same pain as firing it ::
That gives a very "real world" meaning to recoil. If I would rather not drop the gun from that height on to my shoulder then I probably don't want to fire it.
Here's a worked example, a 10 bore at 1000 ft/s from an 10.5 pound Bess. How painful is that?
First you calculate the velocity of the gun.
The ball weighs 0.1 lbs the gun weighs 10.5 lbs, 1000 ft/s
1000 x 0.1 / 10.5 = 9.52 ft/s
After firing the gun is moving backwards at 9.52 ft/s, from how high would I have to drop it to achieve 9.52 ft/s?
Simple, square it and divide by 64
9.52 x 9.52 / 64 = 1.416 feet = 17 inches
Would I willingly drop that gun 17 inches on to my shoulder? :hmm: :results: :crackup:
But as recoil is very much like holding the gun above your shoulder and then dropping it. I thought it might be fun to calculate how high above my shoulder I would have to drop the gun from to feel the same pain as firing it ::
That gives a very "real world" meaning to recoil. If I would rather not drop the gun from that height on to my shoulder then I probably don't want to fire it.
Here's a worked example, a 10 bore at 1000 ft/s from an 10.5 pound Bess. How painful is that?
First you calculate the velocity of the gun.
The ball weighs 0.1 lbs the gun weighs 10.5 lbs, 1000 ft/s
1000 x 0.1 / 10.5 = 9.52 ft/s
After firing the gun is moving backwards at 9.52 ft/s, from how high would I have to drop it to achieve 9.52 ft/s?
Simple, square it and divide by 64
9.52 x 9.52 / 64 = 1.416 feet = 17 inches
Would I willingly drop that gun 17 inches on to my shoulder? :hmm: :results: :crackup: