The red "ink" is for keeping direct responses separate from the original text. No meaning attached to the color. If this technique is offensive to you, say so and I'll not use it any more.
So are we to believe that just because the wad skirt of a 28 ga. wad can be flared out to .729" or so, it will still contain perhaps 9000 p.s.i. without that skirt blowing forward and losing its' perfect seal? Maybe not 9,000 psi. But the pressure has dropped considerably by the time the wad base must expand to bore diameter. That would be some tough skirt. Yes, the plastic in wads is very tough. Whew! I will still contend that even if that wad could contain the pressure, as the area almost instantly increased, the pressure would have to drop and force would remain equal (assuming zero blow by). I agree that this is very close to what must happen. Anything else would absolutely be claiming a physical free lunch. The only issue would be if the larger bore turned out to be a bit more efficient for the powder to burn/expand into. This could account for a small increase in wad base force and velocity. Are you saying that if we have, for example, 9000 p.s.i. gas pressure contained in a one cubic inch cylinder and we increase the volume of the cylinder to two cubic inches, that we would still have 9000 p.s.i.? Certainly not... not without adding more gas or superheating the same gas in the doubled volume. Now, I wish I knew how to put your following statement in red letters (highlite the text you wish to color, click on the "A" with a color bar under it in the bar above, and click on the color you wish) : "If the wad base expands to fill the larger bore, then, yes, the force increases. This is how a small hydraulic piston raises a large load on a bigger bore cylinder." Whiskey Tango Foxtrot. No, no, no. Huh uh. Nope. In a hydraulic pump, be it a simple hand Porta Power type pump (or a simple hydraulic jack), or a multi stage variable displacement rotary piston pump, the small diameter pump piston(s) driven by hand or motorized force displace a volume of relatively incompressable (or it could be a compressible gas) oil equal to the bore radius squared times pi times the stroke (length). This volume of incompressable fluid under pressure acts upon the base of the larger cylinder piston and moves it a porportionately shorter stroke. The larger the cylinder piston, the less one stroke of the pump piston will move it. Again, there are no free lunches here. Suppose we wish to lift 1000# with a cylinder having 10 sq in area. We will need 100 psi on the base of the cylinder (10 sq in X 100 psi = 1000 lbf). Suppose we have a pump piston with 1 sq in face area. Then we will need to apply just over 100# force to the piston to cause the fluid to flow to the cylinder and displace it. If we are unable to supply 100# force to the piston (1 in X 100 lbf = 100 psi), no fluid will move (the pump will stall). Pressure is constant within a hydraulic loop, but force is not at all constant. Your example is a fine statement of the displacement principle (second step), but assumes the pump has sufficient pressure to move the fluid. I working a step back where we have to know the pressure to force balance and area (first step). If the pump can't supply sufficient pressure, it will stall and no fluid will move. You have to work both steps. Lifting 1000# with 100# force seems like a free lunch. You pay for lunch with the displacement - sorta like spreading the payment out.
Rocketman, Having read perhaps hundreds of your posts, I knew that you knew better than you originally stated, the working principles of hydraulic pumps. However, if our pump/cylinder system was using a compressible gas instead of an incompressible oil, I don't think it would be considered a "hydraulic" system. Also, my example did not state the obvious that if the pump did not produce sufficient pressure, it will stall and no fluid would move. I also did not state the equally obvious that if our hand pump operator had no arms and could not pump the pump that no fluid would move either. My mistake. I do agree that something may happen to make the powder burn more efficiently as it expands into the almost instantly larger combustion chamber. Earlier in this thread, I postulated that we might even be seeing a sort of the phenomena that occurs with reduced loads of some powders that produces pressure spikes or even detonation. See the cautions and warnings about reduced loads of WW 296 in the .357 magnum for example. That is why I wish we could simplify this discussion and power our wad/shot charge with say 9000 psi of compressed air or CO2. Then we could say with certainty that the pressure would have dropped considerably by the time the wad entered the larger bore diameter. With some slower powders, that may not be true. In any case, despite the toughness of the wad skirt, I am still very skeptical that there would be no blow by going from 28 ga. to 12 ga., or .550" to .729" (I don't know the diameter of an average 28 ga. wad.) I do know that in hydraulics or pneumatics, even where the bore size does not open at all, let alone so dramatically, there is no such thing as a perfect seal. Why, with what you are claiming, we all should be loading our 12 ga. shells with 28 ga. wads. We would get higher velocity at lower pressure with reduced bore friction. I suppose we could put up several thousand dollars and have some ballistics lab prove that you or I are right or wrong. But I think I'll just save my money for my daughters' college tuition and to buy more shotguns. Oh, your use of red text is in no way offensive to me. I'm not easily offended. That's why I would never consider using the "Ignore" feature here. Meanwhile I'll leave you with a cool quote from Aristotle that may apply to you and me: "The world is divided into people who think they are right." As I did not confirm this, I expect you or someone will correct me and say that Socrates or Plato said this.
