I just checked out a site that sells this servo. The math formula says that you take the power rating ( 76.6) and divide it by 16 ( oz. in a pound ) which gives you 4.76 lbs. but, for a 1" arm.

To calculate for let's say a 7" arm, you have to divide that 4.76 again, by 7 ( for the 7" ) and come up with .68 lbs. So if I understand it correctly, this servo will lift a 7" arm only if it's .68 of a lb or less.

I saw TD2253's formula previously, and it doesn't follow with what I saw on the servo site, ( no offense TD2253, I could be mistaken. )

First of all, 4 AAA are 6v, not 4.8.

Ounce per inch ( oz.-in.) / 16 gives you pounds of force, BUT only for a 1" arm. For instance 76.3 with 6 volts / 16 = 4.76 pounds of force WITH a 1" arm.

If you have a 2" arm, you take that 4.76 and / 2 which is 2.38 pounds of force for a 2" arm.

Since the RF stalk is about 7" long ( not to mention the weight of it and the topper ) we would take 4.76 / 7 = .68 or less than 3/4 of a pound of force.

So obviously the longer the arm or stalk, the force required to move it, yet weaker the servo becomes. I may need a tougher servo after all.

Does anyone know the total weight of their stalk/ topper for comparison ?

Here's the link to that formula http://www.servocity.com/html/servo_power___speed.html