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Optimum and Maximum Practical Game Weight

9902 Views 39 Replies 12 Participants Last post by  mikmarandola
I have searched the internet for Information on stopping power and formulas and tried to figure a formula to account for explosive wounding. This formula can be found under Hunting Stories, perhaps the wrong place to put it..

As I said before I think the Taylor, Hatcher and Thornily formulas to be among the best. I found a post on a forum claiming the Optimal game weight formula to be the best. I found a ballistic program which whilst calculating a ballistic chart also would calculate Taylor's knockdown number or any other of twenty-six knockdown formulas.

The text attached to the program gave the formulas, plus other formulas to calculate gel wood and steel penetration. It also had a chart for each formula so you could know what was needed for each class of game. When you calculated the number for a deer rifle on Taylor's chart, then calculated the number for almost any other formula it still came out as a deer rifle. So it matters not what formula you use as long as you have a chart to go with it.

I then thought what is the best formula. Going back to what I know.

1/ The 45 auto has a great reputation for short range stopping power. It loses 30ft/sec over 100yds so it doesn't lose much stopping power over distance.

2/ The 30 cal carbine was considered to be a bit marginal when it came to stopping determined enemies.

3/ The 223 Armalite/M16 could inflict terrible wounds (at short range) due to explosive wounding and or an unstable bullet.

I think these guns should be rated 1, 3, 2, or perhaps 3, 1, 2, at very short range. Looking at these weapons using OGW gives 1/ = 41lb 2/ =141lb 3/ = 97lbs.
The advantage of this formula is obvious, a 150lb deer is to much for a 45 auto, or is it.

There were several game weight formulas with my ballistics program. My favourite Maximum Practical Game Weight uses Energy calibre and bullet weight to produce the target game weight. EN x BW x CAL divided by 100
1/ = 341lb 2/ = 203lb 3/ = 111lb if these weights are the maximum you can expect to get a quick kill with a well placed bullet. I then divide by 2 to get what I consider the "Optimum" game weight". The result means (3/ = 56) the Armalite/M16 could be relied on to take down coyotes at over 100yds. If the game is dangerous divide by 4, e.g. You have just been surprised by a 120lb mountain lion at 50 paces you can't rely on a 45 auto (1/ = 85) to stop it with one shot.
I haven't forgotten about bullet types I would multiply the above numbers by Hatcher's bullet chart e.g. 0.9 for FMJ, 1 for LRN. I have read that pointed bullet tend to tumble for these I would use a factor of 1.5 which still doesn't turn the Armalite/M16 into a good deer rifle.
With these and most other stopping power formulas a base ball has more stopping power than a big game rifle. However apples are not oranges and a baseball is not a bullet. The formula I Have for Gel Penetration is:
MOM divided by Cal^2 x 6.625
This gives a gel penetration of .9 inches hardly what is required for an offensive weapon.
As for explosive wounding as far as I know it seems to happen with very high velocity bullets. Perhaps this is already figured in the formula as energy increases with the square of velocity so does the expectation of explosive wounding occur.
After much study these two formulas seem much better than I originally thought.
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Good grief!! Since when are deer hard to kill? Even the big ones are thin skinned and not very heavily built. The .223 isn't my first choice (actually that would be a handgun that has anemic paper ballistics compared to most rifle cartridges), but in the right hands it's a killer.
I agree with you Broom 110% ...

a .243 is plenty, a .223 is not. and anybody that really believes otherwise is not well informed.

you can lead a man to knowledge, but you cannot make him think......
That's rich. How do you explain the multitudes of deer taken every year with the .223? It must be adequate.
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