I don't understand...ULT

[nak81783]

Can someone please give me a quick fluid dynamics crash course or tell me what was explained at IAO?

I don't understand how the ULT works if the top of the new pin is the same diameter as the old. Seems like the only thing to affect trigger pull and reactivity would be the surface area of the pin susceptible to vertical pressure. How does narrowing the middle of the pin reduce trigger pull? Pressure in the valve that acts horizontally on the narrower part of the pin should have no effect on trigger pull or reactivity.

Obviously it works, so I must be wrong. Please help me understand why.

Thanks,
Nathan

[athomas]

Air pressure is distributed in all directions equally. That means that all pressures on the pin are equal in all directons except the area that is protruding out through the tiny on/off o-ring. At that location on the other side of the o-ring the air pressure is essentially zero. The chamber pressure of a mag is about 400psi(higher on a level 10 mag). The cross sectional area of the pin is small enough that the psi divided by the area equals aboaut 15oz of force. The large end of the pin does not count in the pull weight because the air pressure is pushing up on some of the bottom section of the pin as well as down. When the on/off is closed and the gun is fired reducing the chamber pressure to zero(not really zero but close), the input pressure pushes down on the larger cross sectional area of the top of the pin. The area is larger and the input pressure is higher so the force exerted is greater. That is the reactive kick provided to help open the on/off a little faster.

[nak81783]

Thanks, that helps. I forgot about the little ledge on the underside of the top of the pin. That will indeed cancel out some of the pressure of the trigger pull.

Thanks,
Nathan