Finally figured out the descrepencies and scotty don't go running to your wife yet.
What the other formulas didn't take into consideration is that 30 ft lbs = 30 lbs on the end of a 1 foot lever. The longer handle of a torque wrench is only for ease of applting the desired torque value. If you were apply 150 ft lbs of torque from a 1 ft bar you would work much harder to get to the 150. By making a torque wrench manageable, it is easer and faster to do the work, but still must be delivering the equivelent of 150 lbs on the 1 foot lever. Thus the constant of 1 in scottys formula.

Now take their formula y=(AxL)/(E+L)and plug in the numbers. The reduction of ft lbs has already been made to reflect the amount of weight on a 1 ft lever, thus the constant length to be 1 ft.

y=(30x1)/(1+6)or Y=(AxL)/(E+L) where L=the one foot lever giving the lbs/ft nomenclature. This will be a constant with any type torque wrench as long as it reads in lbs/ft.

Simply stated[color:"blue"] Y=A/(1+E) [/color]

where E is the length in feet of the extension.
and A is the desired torque and Y is the adjusted value setting for the torque wrench.