[rdorman]

Perhaps I am just making this to hard on myself. I was thinking, power assist, hmmmmmm, then the light bulb went off (stomp on it if this is not correct)!

Is it just as simple as pressure difference on a diaphragm adding hydrualic assistance? For instance if you have one side of a diaphragm exposed to 16"HG (7.86 psi (lb/inch2)) and the other to atmospheric pressure on the other, lets use 14.7psi and a 8" diameter diaphragm is it

((4 * 4) * 3.14) * (14.7 - (14.7 - 7.86))
or
394.8864 pounds of assist?

So say you had a system designed to stop at maximum deceleration with 75 pounds of peddle force (6 to 1 peddle ratio) and did nothing but add the booster in the previous example your peddle force required for the same rate of deceleration would be only be 9.33 pounds (I know, not a real world situation)?

((75 * 6) - 394) / 6 = 9.3333333333333333333

Not sure I like the way the amount of assist can vary so greatly with changes in vacuum.

Rick