doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Pattern theory

cpa the pressure at the bottom of a 1000' column of water is ten times higher than the pressure at the bottom of a 100' column of water. The reason is that there is 900' more of water stacked on on the bottom (and water is practically incompressible).

Stand a 12 gauge empty cartridge and a 28 gauge empty cartridge on their base on a table. Fill each with 1oz of #9s. The 28 gauge shot column will be taller/longer than the 12 gauge. Suppose the 28ga column is twice as tall/long (I don't know what it would be).

When you accelerate that shot from stationary to 1200 second in a fraction of a second that shot is under huge acceleration, or Gs, as they say in aerobatics and at NASA. A "G" is one earth gravitational field. The water columns in my first example were under one "G". The shot is experiencing dozens or hundreds of Gs, again I don't know the number. But the shot at the bottom of the shot column is pushing the next layer of shot which is pushing the next layer shot which is pushing the next layer of shot. Just as in the ocean the "pressure" of the shot at the bottom of the shot column is experiencing a much higher pressure the the shot at the top of the column.

Edit: And the taller/longer the shot column the more the pressure on the bottom layer of shot.

And how does that relate to force = mass x acceleration and the statement that small gauges experience more force than larger gauges which is what I understood the earlier post to say. Of course a 1,000 ft column of water has more pressure than a 100 ft. column because it has more mass. Are you saying that the acceleration of the longer shot column is greater than the shorter column even though both reach the same velocity at the end of the barrel? I believe that is the only way you could experience greater force as the mass of the shot columns are the same. The analogy of the ocean isn't valid because of the mass difference. Again, f=m x a per Mr. Newton. There is no provision in the formula for differences in shape of the mass. Again, if you loaded a 20 gauge shot column in a 12 gauge shell using a thicker wad to make up for the larger bore, would that have more force than the orginal 12 gauge load? If so, what would cause the greater force given that mass and acceleration have not changed.

cpa its not just the mass, it is the mass and the height. Suppose in my original example the two columns, 100' and 1000', were each in pipe with 1 square foot cross section. Suppose that I pour the water in the 1000' column into a pipe with a 50 square foot cross section. Then that water column would only be 20' tall. Then the pressure at the bottom of that 20' column would be lower than the 100' column, even though the 20' column had ten times more mass.

Same with 1 ounces of shot in the 28 gauge and the 12 gauge. In effect, the water is deeper in the 28 gauge and so the pressure at the bottom of the column is higher.

I think we're talking about force applied to the shot charge which is subject to Newton's law - not water in a column. Per square inch the pressure in the 20' column would obviously be less because there is less mass above it - 240 cubic inches versus 1200 cu. in. In the shot columns, both have the same mass, and volume - they simply have different height. Likewise the 28 gauge shot column in your example has exactly the same mass as the 12 gauge. It just has smaller area, but the cubic inches of shot and mass are the same.

You are correct that the work done on the shot by accelerating one ounce of shot to 1200 fps in the 28 gauge is the same as accelerating one ounce of shot to 1200 fps in the 12 gauge.

(work is calculated by multiplying force times distance).

But again, in reference to the stress on the shot in the bottom layer of the one-ounce shot column, everything else being equal, the 12 gauge shot will be less stressed than the 28 gauge shot.

Three physically identical acrobats standing on the gym floor have the same weight on their two feet. But stack them up, feet on shoulders, three high and the one on the bottom has three times as much stress on his feet as the one on top. But the mass remained the same in both cases.

"the 12 gauge shot will be less stressed that the 28 gauge shot."
That's a whole different statement than made earlier. Your analogy of acrobats is lacking, I think. When they are on each other's shoulders, the mass on the bottom feet is three times as much as the mass on the feet if there is only one person standing on the feet, assuming all weigh the same. Come on, Mike. You know the water column and the acrobat argument are bogus as the mass changes.

By me?



In my analogy there were two cases:

1st case. The three identical acrobats are all standing on the gym floor.

2nd case. One of the identical acrobats is standing on the gym floor, a second acrobat is standing on his shoulders, and the third is standing on the second acrobats shoulders.

In both cases the three acrobats have the same total mass. But even though the mass is the same the acrobat standing on the floor in the second case has three times as much stress on his feet as the acrobat at the top.

No, I don't think by you. Sorry if you thought my initial comment related to your post.

Not if you add the stress of feet 1, feet 2 and feet 3. Then it is the same. Of course the stress on a person with three times the mass is greater, just as it would be if acrobat 1 weighed three times what the others weighed. But it wouldn't make any difference if they all weighed the same, but were taller in one instance than another. Three six feet acrobats on top of each other don't exert more stress than three five feet acrobats, assuming they all weigh the same.
Actually I suspect what we're really talking about is force per square inch, not total force which is what the earlier poster stated, I believe. Assuming equal force (m x a), and a smaller area then force/sq. in would be greater.

But we are talking about stress deforming the bottom layer of lead shot but not the top layer. So, in case 2, the bottom acrobat's feet have three times more stress than the top acrobat's feet and so his feet are more likely to be deformed than the those of the acrobat on top. I repeat, the bottom acrobats feet are supporting the weight of three acrobats, the top acrobat's feet support just one. But the mass of the three acrobats is the same in both case one and case two.

In my analogy the three acrobats are identical.