Here is a formula for making reasonable estimates. You need to know maximum speed in mph, maximum surface deflection in degrees, control surface length and width.

torque req't (oz.in.)= (length)(square of width)(square of speed)(deflection angle)/430,000

Ex: assume elevator length 20", width 1.5", deflection angle 25 degrees, max speed 75 mph; torque req't = 20x1.5x1.5x75x75x25/430000 = 14.7 oz in, which doesn't sound like much, but if you widen to 2" it almost doubles to 26.16 oz. in., and if you keep the 2" width and go for full 30 degree deflection at 100 mph it becomes 55.8 oz.in.

These calculations assume zero friction in linkage, but with reasonable care you should not have more than 1-2 oz of friction in pushrod & hinges.

What this tells us is that width of control surface involves a big penalty and speed same thing. If you don't fly too fast, or don't bang your stick at full speed, you don't need a lot of torque; but wild maneuvers at high speed take lots of torque and obviously put a big load on your bird.

I don't remember exactly which website I got this formula off of, but I guess I probably found it by searching Yahoo on "servo torque requirements" or some such phrase. I think the Multiplex website has a java calculator, which may use the same formula, or maybe not.