something that has been bother me with this is that the math only shows 1/2 of the input coming back to assist and I did not see why that would be, well I finally figured that out and it is so silly that it is embarrassing that I missed it.
There are 3 systems lets say,, there is the upper input section, the slip and then the lower return section. The upper and lower share a physical connection on one side and a force connection on the other and it is that force only connection where the slip is.
Now if the input goes in on the upper and it spins the upper and then the upper goes across the slip and spins the lower and then the lower comes back to assist the upper,, the simple thing is that the slip boundary has the upper traveling at say 100 RPM and with the 2:1 return gearing the lower is traveling at 50 RPM, so the same force but at half the rate means half the return work,,, and that is why there is only 1/2 of the input returned to assist.