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Unread 23-12-2013, 23:25
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Re: Questions about Elevator Specifics

Quote:
Originally Posted by Jon Stratis View Post
Now, for speed. In continuous, we saw that 1 ft pulled = 1ft moved. In cascade, 1 ft pulled = 2 ft moved. So, cascade moves twice as fast as continuous, assuming equal gear ratios, drum sizes, motors, and required force for lifting.

Now for force... This is where it gets tricky (and where I wish I wasn't on my ipad so I could draw a free body diagram for you). The inner most stage has a weight of w1. The outer stage has a weight of w2.
The force difference between continuous and cascade is simplest to understand if you just think of it as exactly like a gearing system. If, in the continuous system, 1ft in = 1ft out, then that's like a 1:1 gear ratio. In the cascade, 1ft in = 2ft out. That's a 1:2 gear ratio. WorkIn = WorkOut, so Force*Dist In = Force*Dist Out. Your output is moving twice as far, so it can apply half the force that you apply at the input. If you're lifting the same weight, that means twice the force in.

Really, an awful lot of the simple machines come down to WorkIn = WorkOut. The only tricky part is figuring out how you correctly calculate work for a given device/motion. Converting ball screw pitch into torque to apply X force is always a good one, for instance.
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