Does anyone here know the formula for calculating foot pounds of energy for a load? I know it involves the velocity and bullet weight.

Velocity Squared X Bullet weight/ Divided by 450240

Example.
1200 x 1200 x 150 / 450240=

360000/450240=

479.74 ft lbs

Henry, While this isn’t an answer to your question......here’s a link to a very informative website. It gives much info/inside into ft/pounds energy, momentum, Taylor KO values, ect. It may cause you to question what we’ve been led to believe most of our lives! memtb

memtb

If you would be so kind, I would not mind seeing that link myself!

Sorry! Damn, I must be getting old.....I thought that I had provided the link! I’ll try again! memtb


http://rathcoombe.net/sci-tech/ballistics/wounding.html

Thanks for the link, memtb.

The "Power Factor" used for Single Action / Cowboy shoots is flawed, but it provides a yard stick kind of standard that is VERY easy to calculate.
Bullet wt in grains X velocity in fps divided by 1000 such as the old .38 Special LEO load of 158 X 850 = 134.3 PF
1,000
The SASS knock-down targets are configured to fall at that level of power. I think the IPSC Power Factor is calculated the same way, with Major and Minor categories.

These formulas are just crunching numbers, not related to real life effectiveness, but I remember seeing the Colorado hunting regulations at one time specififying that 1000 foot-pounds ...or is it properly pounds-feet?...of Kinetic Energy is the minimum for hunting big game. I found a handy table for calculating KE in one of my older Speer books.