So here's a similar idea, but no external input required, the gun does all the work.... How about a pressure booster?.... First we use a regulator to drop the pressure in a set volume chamber to 2000 psi, so that every shot is the same.... Then we use a piston with 3 times the area on one side to compress the air on the other side to 6000 psi.... That air is then dumped by the valve to produce the shot....There are lots of things to work out, how to cycle the piston without wasting the air used to do so is the major one.... If that can't be done, then any gain in efficiency from the higher pressure would go out the window.... Anyway, for anyone with lots of time on their hands and an inventive side, it might be worth pursuing....Bob
The booster should move backwards until the two side are at equilibrium.
Now it's complicated enough you need to do a drawing, you totally lost me....Bob
The air in the big end of the piston gets forced into the small end of the piston.
QuoteThe air in the big end of the piston gets forced into the small end of the piston.I don't see how, once compression starts the pressure in the small chamber is higher.... Normally in a pressure booster that air is vented to the atmosphere.... The force available to compress the 2K air in the small chamber to 6K comes from the pressure difference across the large part of the piston.... If the ratio of areas of the pistons is 3:1, you need a full 2K across the big piston to reach 6K in the small chamber.... In addition, as drawn that skinny rod joining the two pistons would never stand 4700 lbs. of force.... Normally the piston would just be stepped like the one in a regulator, with 3 times the area on one end as the other.... I have a very similar drawing, but what to do with the air in the middle portion remains a problem.... It is necessary to reset the piston for the next stroke, but once reset that air has to go somewhere to allow the pressure across the large piston to reach 2K.... Bob
Inside the piston, if the small diameter half and the large diameter half have the same maximum volume at opposite ends of the piston cycle then pressure should be the same 2000psi.
QuoteInside the piston, if the small diameter half and the large diameter half have the same maximum volume at opposite ends of the piston cycle then pressure should be the same 2000psi.How can that be, they have different diameters and the same stroke?....Redraw your piston with the part where you show a shaft the same diameter as the small end of the piston.... then it will be strong enough.... You then end up with a donut shaped space around the small part with the OD of the large part, which has essentially zero volume when at full compression and is the length of the stroke when at the start of the stroke.... That elongated donut, plus both the small chamber and large chamber, all need to be at 2K at the start of the stroke.... At the end of the stroke, the small chamber is at 6K, the larger chamber is at 2K, and the donut (which has shrunk to virtually zero length) has to be at atmospheric (0 psi gauge).... Because of the very high forces involved you can't use levers, although you can use a lever or a spring to move the piston back to the beginning of the stroke, providing there is 2K both sides of it....Bob
The only wasted air is the difference between the two middle chambers.
QuoteThe only wasted air is the difference between the two middle chambers.Agreed.... and my design has no volume for the smaller of the middle chambers, and since the piston rod is the same size as the small end of the piston, the volume of the larger of the middle chambers is the same as yours, minus the volume of the piston rod.... which equals the volume of the smaller chamber.... In other words, we have exactly the same difference in volume.... That MUST be the case, because the stroke is the same for both ends, and the pistons are the same diameter.... You mention "venting the air from the middle chamber into the firing chamber as the piston decompresses".... However, the firing chamber already has to be at 2K in order to provide 1/3 of the force needed to move the piston back to the "decompressed" state...To fully decompress the piston, both the firing chamber (1/3rd of the area) and the middle chamber (2/3rds of the area) have to be at 2K to match the driving side of the large piston (full area).... Slimming the connecting piston rod makes no difference, as whatever area is added to one end is added to the other, and the forces cancel out....Keep trying, you may come up with something I haven't thought of.... Bob