Re the 20 gauge, all I said was that you can go out and buy cartridges off the shelf that are not overloaded. Never said anything about whether they had higher or lower load density than the 28 gauge. My real point was that you can also buy 20 gauge cartridges off the shelf that are overloaded, and if the gun is matched to these over loads you have a 20 gauge with the weight of a 16 or 12 gauge and a bore too small for the load.

I will agree that a “square load” is defined as one with the column length equal to the diameter, and also, that it’s a rather meaningless concept. It’s a rule that can actually apply to only one gauge, whatever gauge you want to make it.

And yes the 410 with ½ oz has a higher load density than a 28 with ¾ oz. In line with my point that for any practical hunting purpose the 28 is superior to the 410 in every way since both guns will probably be in the same weight class and anything less than ¾ oz is really a handicap on pheasant. Even well centered on the target, ¾ oz implies you should pass up anything beyond 40 yards.

Consider a subdivided cube like a Rubik’s cube. If you have one ball filling the entire cube, or 27 balls, each filling one sub-cube, you have exactly the same weight. If you want to make a number of small balls fill a cube to exceed the weight of a single ball, then you have a packing problem. But the only way to make them occupy a greater volume is to have space between them. This situation changes slightly with a cylinder where you will have boundary effects since it’s not a cube, but that would be a relatively minor effect if the balls are small compared to the cylinder as is the case with a load of birdshot. So I am still puzzled as to how you are going to make a 1 oz column longer than a 1 oz ball short of using a filler. I confess to not having looked at any load data of late, the above is based on simple geometry. Just for laughs, I will toss this problem up on a math related board I am a member of and see what they make of it. It’ll make a good sanity check.