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Originally Posted by Matt Adams
eeek... I gotta say I didn't do that!  ...
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Maybe 250 in-lbs on that center gear, 1 in diameter, moment arm of a half inch off center... the pneumatic is about 3.4 inches away from the pivot point, at around a 15 degree angle, providing a max moment arm in the opposite direction of around 95 lbs * 3.4 inches with a 1.5 inch bore... There's also the force on the gears through the shaft centers, sin(20)*250 is around 85, times it's moment arm of 1.5.... So we've got around 70 in lbs to spare with the half inch. whew!  Thanks for the reminder...
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You forgot the input torque of 250 in-lbs (or else you have to include the torque from the internal short chain -- which magically will work out to be the same 250 in-lbs).
I see 4 torques to deal with
In high gear with driving the input CCW,
Tinput = 250 in-lbs
Taircylinder = 95 lbs*3.4 in = 320 in-lbs
TtoothNormal= -(250 in-lbs/.5 in)*.5 in = -250 in-lbs
TtoothSep= ((250 in-lbs /.5) * sin(20deg) ) *1.5 in = 250 in-lbs
This is okay because they sum to 470 in-lbs (and CCW is toward engagement).
If you do the same thing only have the input torque switch to CW you will get
Tinput = -250
Taircyliner= 320
TtoothNormal = +250 (the normal tooth load is trying to ENGAGE the mechanism in this case)
TtoothSep = -250
Sum = 70 in-lbs
Having done the summing carefully, I get the same answer as your "quick calculation", so what do I know?
I suppose I will shut up now ;-)
Joe J.