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

But you're still comparing apples and oranges. The top acrobat with mass =1, the bottom acrobat with mass =3. Of course the bottom acrobat's feet would experience more stress.

In Case 1 and in Case 2 the three identical acrobats have the same total mass. Suppose the identical acrobats weigh 150 pounds each. In both cases you have 450 pounds total weight. But in Case 2 one of the acrobats has three times as much weight (450 lbs) on his feet as he did in Case 1 (150 lbs). Yet the weight on the gym floor is 450 pounds in both cases.

Since you concede that the acrobat on the bottom has more stress on his feet can't you concede that the shot at the bottom of a shot column would receive more stress than that at the top of the shot column? That that shot at the bottom would have more weight on it than the shot at the top of the column during an acceleration that takes it from stationary to 1200 fps in a fraction of a second?

Obviously correct, but you're not talking about total stress on the gym floor - you're talking about the stress on that poor bottom fella's feet. I'll definitely concede that resting in a shotgun shell, the top pellets experience less stress than the bottom ones. That hasn't been what we were talking about.
Again, back to shot columns, I think we're actually talking about force per square inch, which would be greater the smaller the cross section of the column assuming total force is the same. That is different than the statement that total force is greater for a longer column.

Well, what about when all three stand on the floor, as in Case 1? Each acrobat is carrying 150 pounds on his feet. No acrobat has 450 pounds his feet. But the gym floor carries 450 pounds of acrobat. Stack them up and the gym floor still carries 450 pounds of acrobat. The gym floor provides 450 pounds of force in the opposite direction of the weight of the three acrobats both in Case 1 and Case 2. But only in Case 2 are an acrobat's feet subjected to 450 pounds.

Yeah, that's correct, but so what? The force on the gym floor is spread out over three footprints instead of one, but you weren't talking about stress on the gym floor - you were talking about stress on that poor guy's feet. That's not the same.

Well, actually they are.

So if we moved those acrobats to a rocket and controlled that rocket's motors so that there was a constant 5 G acceleration then those acrobats in Case 1 would each carry 750 pounds on their feet and that bottom acrobat in Case 2 would be carrying 2,250 pounds on his feet. And that rocket floor those acrobats were standing on would in Case 1 and Case 2 be providing 2,250 pounds of force to acrobats feet.

The force accelerating those acrobats (both cases) is analogous to the force accelerating the shot in a shot cup. The quantities are different but the algebra is the same for acrobats or birdshot.

In a 12 gauge shot cup the one ounce of shot is not as deep as in the 28 gauge shot cup. So the bottom layer of shot in the 12 gauge shotcup has less stress than the bottom layer of the shot in the 28 gauge shotcup.

As I've said, the force per square inch would be greater in the 28 gauge, but the total force would be the same. It would appear that that would also be correct for the rocket propelled acrobats as in case 1 they would have cross sectional area of three sets of feet and in case two the area of of only one set. I believe this began as a discussion that total force in smaller gauges is greater than force in larger gauges, and you have conceded that total force is the same in a 28 as a 12. It's been fun.

"As I've said, the force per square inch would be greater in the 28 gauge, but the total force would be the same." AGREED

"It would appear that that would also be correct for the rocket propelled acrobats as in case 1 they would have cross sectional area of three sets of feet and in case two the area of of only one set." AGREED

"I believe this began as a discussion that total force in smaller gauges is greater than force in larger gauges, and you have conceded that total force is the same in a 28 as a 12."
I'm not conceding because I never argued that the work (force times distance) applied to 1 ounce of shot to accelerate it 1200 fps was anything but identical in the two gauges. But I do argue that the bottom layer of shot in a 28 gauge one ounce load is subjected to higher stress/force/pressure than the bottom layer of shot in a 12 gauge one ounce load.

Pressure times area equals force. The pressure from the burning gunpowder is higher in a 28 gauge (compared to a twelve) but the area of the wad is smaller in a 28 gauge. Pressure times area equals force.

" It's been fun" Agree!

Here's the kicker, you did not agree with this a few pages back when doc wonko made fun of the fact that more pounds could be shown to push on less square inches of a smaller bore to end up with the same velocity for the same weight of shot. I'm gonna call it weight rather than mass because I don't believe shot is fired in Newton's zero gravity vacuum.

Also, if you're conceding that some different force is acting on the shot that's closest to the PSI's of the ignited powder, doc wonko might turn on you and bite.

I also firmly believe that it takes more force to get an ounce of shot out of a 28 ga. tube than it does to get an ounce of the same shot out of a 12 ga. bore. To end up with the same velocity, I believe more friction needs to be overcome as the bore gets smaller, not just inertia. Fun in a weird kind of way.

Hence the higher pressures witnessed in the smallbores.......???