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Originally Posted by Gary Dillard
Please forgive me for using numbers to back up my statements; it's one of my engineering habits.
130 pound robot, 10 pound tetra lifted ~ 8 feet in the air
Example 1: Arm weighs 40 pounds, uniformly distributed vertically; remaining chassis/drive train/battery, etc. weighs 90 pounds, cg at 3" based on center of 6" base structure.
[(40)(50)+(90)(3)+(10)(100)]/140 = 23.5 inch cg height
Example 2: Arm weighs 20 pounds, remainder weighs 110 pounds at 4" height
[(20)(50)+(110)(4)+(10)(100)]/140 = 13.5 inch cg height
Take the weight out of your tall structure and move everything down as far as possible. It's not that hard to do if you design it in to start with. Otherwise, if you're our opponent - thank you very much, and if you're on our alliance please make sure your structure is designed to be plowed across the field; we'd like to have the 10 points for putting you in the end zone with us.
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Thank you everyone for the help, I already ran through the calculations and got the unhappy result after my last post. 135 fl-lb of torque needed to tip us =/
not much at all. Our arm is VERY tall at full extension, so it's not so much that its heavy, but more so that it's just big. That being said we're redesigning it to be lighter anyways. We're also considering outriggers. Our center of mass without the arm is probably obscenly low as it is, and i'm fairly sure we can't move much more stuff down.