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Originally Posted by Rocketboy
Okay, I'm going to take a shot in the dark and estimate that the average arm length is approximately Eight feet in length. Forget the moment for the arm weight itself and calculate 9 lbs at 8 feet (you do the math).
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Alrighty...
I'll go ahead and just do some math. (And engineering, it's much better than math.)
We're gonna take 3 tetra at 10 lbs each, 30 pounds. We'll be crazy people and make an arm that's 8 feet around a pivot (it telescopes out)
We'll just use a piece of 3 x 3 by 1/8" box aluminum, 6061 with a yield around 36000 PSI.
That arm joint will weigh (3in²-2.75in²) * 96 in * .0975 lbs / in³ = 13.455 lbs
That's 30 lbs * 96 inches = 2880 in-lbs for tetras
and
13.455 lbs * 48 inches = 645.9 in-lbs for the arm weight itself (not really negligible)
bending stress = M * y / I
M = moment
y = distance from center to extreme edge
I = moment of inertia (for a beam = h³*b/12 where b = base width and h = height in the direction of applied force)
(645.9+2880) * 1.5 / ((3^4-2.75^4)/12) = 2665.68 PSI
Factor of safety... about 13.5.
This proves that you can create something that will endure the stress using just simple materials.
The torque is a lot, but only when you start trying to twist around that axle of your pivot. Start thinking pulleys, counter weights, large moment arms on the back end...
I didn't mean to go all formulae crazy... but... you dared me!
Matt