This is a quick AR-15 build tip on how to account for the added length of your castle nut/armorers tool when setting your torque wrench. Simply setting your torque wrench to 40 ft/lbs will not give you 40 ft/lbs on the castle nut… it will be way over-torqued. Here is the formula you can use to calculate the proper torque wrench settings based on the length of your tools.
Formula: T1 (L1 ÷ L2) = T2
You can replace pounds and inches with metric measurements too… it is all relative.
This is what the variables stand for…
T1 [torque spec for castle nut is 40 ft/lbs]
L1 [torque wrench length in inches measured center axis to the center of the hand grip]
L2 [torque wrench length + armorers tool length in inches measured center axis-to-center axis]
T2 [adjusted torque wrench setting]
In our case …
40 (14 ÷ 22) = T2
40 x 0.64 = 25.6 ft/lbs
My adjusted torque wrench setting is 25.6 ft/lbs to deliver 40 ft/lbs on the castle nut.
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I was trained to torque fasteners on nuclear weapons. Alway torque at 90° and you do not have to calculate as you are not adding length. Don’t PUSH on the torque wrench, PULL. Grip torque wrench with a closed fist around the center of the grip surface, thumb tucked so as not to change pressure location. USAF TO 35-51.
While it appears you did the torque calculation correctly, if you put your torque wrench at 90 degrees to the armorers wrench you will not need to perform a calculation.
I have seen that 90 deg technique and it does make sense. Not sure if there is any variance in doing it that way, but it seems like it should work.
90 degree is a close approximation IF the armorer’s tool length (A) is “small” compared to the wrench’s grip length (W).
At 90 degrees, the effective length of the wrench will be the slightly longer hypotenuse of a right triangle = sqrt(W^2 + A^2). The Pythagorean theorem.
To get an equivalent torque, you want the wrench’s grip point perpendicular to the midpoint of the armorer’s wrench length. This forms an isosceles triangle with A as the triangle’s base. And W as the two equal sides.
Is that equivalent because it becomes sqrt(1/4A^2 + W^2) ?