Physics of recoil

I was working out the physics of kick:

1) Momentum transferred (delta p) will be close to equal for bullet and gun (exactly equal if we ignore gasses).

2) If we know momentum and mass of an object, we can get kinetic energy easily: E = p^2 /2m.

3) So recoil energy varies as the SQUARE of bullet mass. This means that the jump from (e.g.) 300 Win to 375H&H or 416 is bigger than you might think.

4) Heavier rifles kick less - not just less kick velocity, but less kick energy. (I like heavy bullets and heavy rifles myself.)

I put together a little Excel spreadsheet to figure this for some common loads, and for various rifle weights - can't post it from this account, but if anyone's curious I'll go in through dejanews and upload it.

Apologies if this is redundant! I'm starting to see articles referring to 'recoil energy', so maybe everyone else already knew this.

Member ECS, AVS, ACM, OSA, Sen.Mem.IEEE, Reg'd Patent Atty 'All men by nature desire knowledge.'

The moderator tells me that binaries here are a bad idea, so I'm posting that spreadsheet on my firm's website. Home page is http://www.technopatents.com/ and I'll put the Excel sheet at http://www.technopatents.com/misc/recoil.xls. (I'm afraid I don't know whether your browsers will object to a naked .xls file, and I don't know if you'll be able to ftp the .xls or not.)

I also simplified the calculation slightly: kick energy is simply bullet energy times the ratio of the masses, so all you need to input is bullet energy, bullet weight, and rifle weight for any given load.

Just for fun I stuck in the numbers for a Russian antitank rifle, for a 20mm round which I think may have been chambered in the Latvi, and for the two-man Steyr anti-materiel rifle!

Member ECS, AVS, ACM, OSA, Sen.Mem.IEEE, Reg'd Patent Atty 'All men by nature desire knowledge.'

Robert,

I tried to access your XLS file. I also tried to D/L it. Neither worked. Would it be possible for you to e-mail it to me?

Thanks, Steve Ellis St. Paul, MN

OOooops. p=mv; thus p^2/2m= mv^2/2 and recoil energy varies as the square of the bullet velocity - not mass. Recoil energy variation is linear in bullet mass.

SURE - email me at my office account below, and i'll reply with it.

Member ECS, AVS, ACM, OSA, Sen.Mem.IEEE, Reg'd Patent Atty Groover & Associates PC, 17000 Preston Rd. #230, Dallas 75248 1-972-380-6333 fax 1-972-380-4445 http://www.technopatents.com/