Calculating torque adjustment for offset of pivot point I'm trying to figure out how much to adjust torque, on a torque wrench when using an extension that offsets the pivot point. See this calculator for a diagram of what I'm talking about: 

The trouble is this calculator can't be correct for the problem I have in mind. It adjusts the torque based on the torque wrench's handle length (i.e. the adjustment factor is simply L/(L+E)), but on a clicker type torque wrench, the measured torque is independent of the handle length. I have several different torque wrenches, all with different handle lengths; if configured to apply the same torque to the extension, it makes no sense that the resultant torque would be different.
Another way to think about it is using a torque screwdriver instead of a torque wrench. A torque screwdriver has no lever (beyond the rubber on the grip). Of course the motion is slightly more complicated; as the screwdriver rotates, the screwdriver itself would need to orbit around the offset pivot point.
Do I have this all wrong? Is the naive and straightforward calculation actually correct?
 A: This can be solved as a statics problem... Consider the wrench in the setting with A=0. You press down on the wrench at a distance $L$ from the pivot point with a force $F$, which is compensated by the nut with an equal but opposite force $F$ at a distance $E$ from the pivot point. The total torque generated by this two forces is $F(E+l)$ CW, which is compensated by an equal but CCW torque from the nut. Now consider what happens at the pivot point, removing either the left or the right part of the wrench.


*

*If you keep the left side, to have equilibrium, at the pivot point you will have to have a force of magnitude $F$ acting down, plus a CW torque of value $F(E+L) - FE = FL$, because the reaction force is also generating torque.

*If you keep the right side it is clear that you only have to resist the force on the handle, and the torque it generates, $FL$


So for the $A=0$ case the torque is augmented by a factor of $\frac{E+L}{L}$.
For different angles, instead of $E$ you have to use $E\cos A$.
This formulas hold if you are delivering the torque by exerting a force which is resisted by the nut. If instead you apply a pair couple that generates torque without a resultant total force, there is no conversion to be done, as the torque you apply will be what is delivered to the nut. That's how you normally apply torque to a screwdriver, as you suggest,  and you could try and do something similar with a wrench, by holding it with both hands and pushing in opposing directions, although it would be hard to be sure that there is no net resulting force.
But there is nothing wrong with the calculator either.
EDIT
Another way of looking at the same problem... The whole wrench arrangement behaves as an end loaded cantilever beam:

The torque applied at the support is $PL$, but since we don't have our torque measuring pivot at the support, but at a distance $E$ from it, we need to consider what the bending moment at a distance $E$ from the support. Bending moment varies linearly from a maximum at the support, to 0 where the force is applied:

So unless you are measuring your torque at the very nut, i.e. without an extension, the total length of the lever has an effect.
A: Sorry, but I still think there is something missing. At what point is the equilibrium resolved? answer: at the point where the wrench clicks. If the wrench clicks always at the calibrated torque, then F will be just the resulting force to achieve that torque.
A good way to visualize this is to imagine a very long extension and a short and tiny clicker wrench calibrated to a very low value with a tiny handle. One would imagine that the torque will be amplified, but in the end we will keep trying to move the nut and wrench will keep clicking...
I argue that the nut will see a torque of  whatever the wrench is calibrated to times Cos(A). The free body diagrams or energy conservation calculations will show that. 
Therefore there is no augmentation when wrench is aligned, which is contrary to the first answer.
