Talk:Muzzle energy

I think that the number "450437" in the formula "E = m·v^2/450437" needs to be shown where it is gotten from; otherwise, the person reading the article is left wondering where the 1/2 went and where the 450437 came from. —Preceding unsigned comment added by 69.253.236.36 (talk • contribs)

I made a few changes to bring some continuity to the article.Greg Glover 02:00, 24 March 2007 (UTC)[reply]

The numbers for muzzle energies in ft•lb_f and J are inconsistent: 350 foot-pounds are equivalent to 475 Joules, rather than 540, for example -- it's reported to be 14% higher. So which ones are correct, I wonder? I'm using the (exact) conversion factor of 1.355 817 948 331 400 4 Joules per foot-pound, but rounding to the nearest Joule here. --Wtrmute (talk) 16:03, 7 December 2007 (UTC)[reply]

The commentary on this is confusing as all out. For example, a foot*pound is definitely a unit of energy (as any physics textbook will tell you, work = force*distance). I can only assume that the author didn't get that pounds were a unit of force and not mass or something? Eoseth (talk) 19:47, 27 May 2010 (UTC)[reply]

From the article: "When publishing kinetic energy tables for small arms ammunition, an acceleration due to gravity of 32.163 ft/s2 rather than the standard of 32.1739 ft/s2 is used." I think this cannot be related to gravity at all, as we are not computing ballistic trajectories. Rather, 32.1739 looks to me to be the conversion factor from metric to Imperial, just as the "7000 gr" in the denominator is the conversion factor from grains to pounds.

Unless someone protests in the next day or so, I am going to remove the gravity reference. Felixkasza (talk) 09:01, 17 October 2011 (UTC)[reply]

Unit ft-lbs or ft-lbf[edit]

What is the correct unit of the Muzzle velocity: ft-lbs (as used in the table heading) or ft-lbf (as used in the text)? Please use just one unit in one article.Optimike (talk) 10:41, 13 May 2016 (UTC)[reply]