Okay, let me crack out some math. Our team is doing something very similar with a four bar design. I calculated our torque for the arm at 124.97 ft lbs (aprrox 170 Nm) at it's worst (the perpendicular). Based on that I know that at best I could hope to load the van door motor at 50% of stall and 20 amps (240 watts). The torque at that point is 17.5 Nm and the rpm under that load according to this chart (
http://www.chiefdelphi.com/media/papers/2064) is also roughly 17.5 rpm. Sooooo here is the problem. You take 170 Nm and divide by 17.5 Nm provided to get 9.714, essentially 10 to one gearing needed somewhere. Based on that you know that from that original 17.5 rpm under load you can expect 1.75 rpm out of the gearing reduction. In our particular case out arm lifts 115.97 degrees, this is 3.1 "lifts in one revolution (360/115.97). Multiply 1.75 by 3.1 to get 5.425 "lifts" per minute. Then divide 60 seconds by that number. Finally you have that it would take 11.06 seconds to lift the arm.
Now I am sure there is something really wrong with my math, and that's why I posted it, or I hope there is because the van door motors make life nice. But I think both or our teams here might run into the same problem so I wanted to share the calculations I have had trouble with. How do you solve the speed issue? Power is no problem relative. I can pick up a ten to one gearing reduction just about anywhere (joking, we all know nothing is that simple). But if I want to raise the arm in a minimum of three seconds where do I get the 368% increase in rpm? I think this is something that any team using the single Van Door motor for an arm faces so I just wanted to bring it up for discussion.