Battery +ve to bridge -ve, bridge AC unconnected, bridge +ve to radio +ve.
This gives 2 diodes in series in parallel with 2 diodes in series. Voltage drop about 1.8V. As it is a bridge, the 4 diodes should be closely matched and the current in each arm should be very similar.
When the voltage on the LiPO starts to drop, short both AC connections to the bridge -ve. The first two diodes are shorted out leaving 2 diodes sharing the current and the drop is 0.9V.
Finally when the voltage on the LiPO is low enough, short bridge -ve to bridge +ve to remove bridge.
The advatage of a bridge is it is easy to mount on a heatsink, 100W = 20A and 1.9V is c. 40W to dissipate, heatsink needed. And should a diode fail in one arm, you still have the other to do the job.
Your analysis may be less clear because the Vf for all diodes is identical and it will vary a little in reality.