That certainly could happen.  And indeed probably would happen.

But, which is the more prudent basis to set riflemen's ammo load:

- Concern that a few rifle barrels may be destroyed by overheating?

- Or carrying enough ammo to accomplish the assigned missions?

I'm certainly not going to try to blame a soldier who think he needs 3-4 times the allocated ammo load to accomplish his mission!

But as someone whose mission is to provide soldiers with the small-arms they "need", I would feel very sad if a soldier once told me that he needed to carry 3-4 times more ammo than the safety limit of his rifle in order to accomplish his mission.

As a joke, I would say that the M16A1 of the Vietnam-era (using LSA low flash-point lubricant), was pretty safe from cook-off because the weapon would simply jam before reaching the "thermal danger zone".

EmericD said:

...as someone whose mission is to provide soldiers with the small-arms they "need", I would feel very sad if a soldier once told me that he needed to carry 3-4 times more ammo than the safety limit of his rifle in order to accomplish his mission.

I refer you back to what SCHVPapist said in post #11:  "...just because that much ammo is carried doesn't mean that it will all be fired so rapidly as to bring cook-off limits into play...artificially restricting ammunition load due to cook-off limits is not representative of the realities of infantry combat..."

I'm gonna be honest, I can't square your results with the paper either. Could I get a little help understanding your methodology?

One of the problems with cookoff tests in general is that they're very sensitive to methodology (e.g., how quickly rounds are fired, how long a fresh round is allowed to sit in the chamber, etc), so it's extremely easy to compare apples to oranges. This is almost certainly what is happening when you say the ARX has a "three times higher" cookoff limit than the M16A1, for example.

The computation with AB Analytics for the .30 M2 ball was done with a MV of 835 m/s and a C1 BC of 0.400. Similar results could be achieved with a C7 of 0.205.

This G1 BC is in-line with a C6 of 0.240 (i6 of 0.95) that was used in the report, even if other sources indicates that the real i6 form factor of the M2 bullet was around 1.16 (not 0.95).

As ABA can use only C1 or C7 BC, I had to convert C6 into C1 or C7.

Here is two screenshots of the ABA software.

Those results were compared with the results given in the report, and here is the comparison.

So, even if the results are not exactly the same, the correlation between the results found in the report and the results given by ABA is pretty good.

Now, the various .21 caliber loads were computed using the same 52 gr bullet with a C1 BC of .279 (same form factor as the .30 M2 ball), with a MV of 1063 m/s for the short range (6/10 powder load), 1195 m/s for the medium range (8/10 powder load) and 1302 m/s for the long range (full power).

The absolute hit probability at each range could then be divided by the hit probability of the .30 M2 at the same range to obtain the “relative hit probability” (see below).

Even if some differences could be seen between this graph and the report FIG4, the common trend is here.

So, we’ve checked that ABA is giving us results very close from what we could find in the report, for hit probability.

The second part is terminal effectiveness.

As the report is giving us only 4 curves of bullet effectiveness, computed for a given bullet diameter and weight, it is necessary to find the physics under the model.

I’ve tried several approaches, and finally the best one was also the historical valid one, using the bullet mass, the impact velocity to the power 3/2, and 3 coefficients related to the bullet diameter.

The formula is this one:

P(I⁄H)= 1-e^([-a(M.V^(3/2)-b)]^n)

In the first post, I’ve put the values of the coefficient a, b & n for each bullet diameter, but failed to add that in this formula, the bullet weight should be expressed in grams and the impact velocity in km/s.

So, for a 9.85 g .30 caliber bullet (a = 0.357; b=0.125 and n=0.631) impacting at 835 m/s, the probability of severe wounding is calculated this way

1-e^([-0.357 x (9.85 x 0.835^(1.5) - 0.125)]^0.631) = 0.7167, slightly lower than 72% to produce a severe wound in the case of a random hit.

So, here again we need to compute the terminal effectiveness of each round at each distance, then divide this value by the value achieved with the .30 M2 round at the same distance to find the relative wounding power.

All in all, the “relative hit probability” could be multiplied with the “relative wounding power” to find the “relative single shot effectiveness”, as shown below for the .21 short range, medium range and long range.

Here again, some differences could be found between this graph and the report FIG7, but the overall trends are well represented.

Using C7 values instead of C1 values do not change the overall trends.

I hope this could help!

Hooooooooooooooolly fuck. This WHOLE time you were talking about Applied Ballistics Analytics, I thought you meant A-B testing! I was SOOOOOOOOOO confused, hahahahahaha.

Thanks!